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4900 | https://www.anirdesh.com/math/trig/constant-ratio-curve.php | Trigonometry > Constant Ratio Curve
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= Online Updated: Apr 29, 2016
Circle - The Constant Ratio Curve
There are many definitions of a circle. There may also be many ways of generating a circle. A circle is simply the locus of all points equidistant from a point, which is the center. However, there is also another way to define a circle, and it is simply remarkable. After investigating the constant product curve, the lemniscate, I decided to see what the curve for a constant ratio would look like. The constant ratio curve is the result of this.
The locus of all points such that the ratio of the distances between the point in the locus and two fixed points is a circle, with the condition that the ratio does not equal 1.
In other words, let the two points be A1(–a,0) and A2(a,0), and let the point P(x, y) be in the locus. Now, let the distance between A1 and P be d1 and let distance between A2 and P be d2. If d1/d2 = k for some constant k, then the points P lie on a circle. The image below shows our setup.
Derivation of Circle Equation
Deriving the equation of this curve is simple enough. From the image above, we can derive the distance d1 and d2 as follows:
(i)
(ii)
Since the distances are in a ratio k, then we have:
(iii)
We can manipulate equation (iii) to give us the equation of a circle:
(iv)
Thus, the equation of the circle has a radius of , for k ≠ 1.
When k = 1, the equation of the locus P is x = 0. This is the equation of a line that passes through the midpoint of a segment that connects A1 and A2 and is perpendicular to the segment. This can be visualized intuitively. The image below shows our curve - actually two curves that give the ratio of k.
These two circles are congruent. These circles are positioned symmetric to the midpoint of A1 and A2. Note that the order in which we take the ratio of the distances matters. If we take the ratio d1/d2 = k, then this represents one circle. The other circle with ratio of –k represents the ratio d2/d1 = 1/k.
Also, notice that the center of the circle lies on the segment or a line continuous with the segment passing through A1 and A2. As k approaches 1, the radius of the two circles approaches infinity—or that the size of the two circles increases. The two circles converge to form the line x = 0. This is because the ratio is approaching 1.
As k approaches 0, then the radius of the two circles approaches 0 and their centers get closer and closer to the two points A1 and A2. When k = 0, the locus P represents the two points A1 and A2.
You can interact with the Geogebra activity below to play around with how a and k affect the circles.
Anirdesh.com Last Updated: May 09 2025, 21:53:43. |
4901 | https://tasks.illustrativemathematics.org/content-standards/tasks/890 | Illustrative Mathematics
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Rolling Twice
Tags:AMC
Alignments to Content Standards:7.SP.C.8
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Task
A fair six-sided die is rolled twice. What is the theoretical probability that the first number that comes up is greater than or equal to the second number?
IM Commentary
The purpose of this task is for students to compute the theoretical probability of a compound event. Teachers may wish to emphasize the distinction between theoretical and experimental probabilities for this problem. For students learning to distinguish between theoretical and experimental probability, it would be good to find an experimental probability either before or after students have calculated the theoretical probability.
If this is the first time students have worked to compute the probability of a compound event, it is a good idea to run the experiment first to get a ball-park sense of the magnitude of the theoretical probability. A quick way to generate an experimental probability is to have students work in pairs to quickly roll the dice ten times and record the outcomes. Each pair can report on the number of times the event in question occurred and the teacher can compile the results quickly on the board or overhead. Then students should be challenged to determine the theoretical probability and comment on how close the experimental probability matched it.
If students are familiar with compound events, this task could be used to deepen their understanding of the relationship between theoretical and experimental probability. In this case, once the students have calculated the theoretical probability, they can perform the experiment and see how well the experimental probability agrees (or disagrees) with the theoretical probability. Because a common incorrect solution is 1 2, the teacher can lead a discussion about the difficulty in using an experiment to distinguish between two probabilities that are close (like 7 12 and 1 2=6 12). This is a good opportunity to discuss the fact that the more times an experiment is performed, the closer the experimental probability and the theoretical probability are likely to be.
Students who are still operating on a more concrete as opposed to abstract level of cognitive development may choose to "draw" the outcomes as listed in the table. Teachers may want to make dice available to students who are struggling to visualize the possible outcomes. Braille dice are available for those with visual impairments.
This task was adapted from problem #18 on the 2011 American Mathematics Competition (AMC) 8 Test. The responses to the multiple choice answers for the problem had the following distribution:
Choice Answer Percentage of Answers
(A)1 6 23.56
(B)5 12 17.61
(C)1 2 16.98
(D)7 12 23.38
(E)5 6 11.20
Omit-7.02
Of the 153,485 students who participated, 72,648 or 47% were in 8th grade, 50,433 or 33% were in 7th grade, and the remainder were less than 7th grade.
Solutions
Solution: 1 Plotting outcomes in a table
We can plot the different possible outcomes as the six sided die is rolled. One way of doing this is with a table as shown below.
1,11,2 1,3 1,4 1,5 1,6
2,12,22,3 2,4 2,5 2,6
3,13,23,33,4 3,5 3,6
4,14,24,34,44,5 4,6
5,15,25,35,45,55,6
6,16,26,36,46,56,6
In the table the entry marked 1,3 means that the first throw was a 1 and the second throw a 3. An asterisk next to the entry means that the first number that came up is greater than or equal to the second number. There are 36 possibilities listed in the table, each equally likely. For the 21 starred cases, the first number was greater than or equal to the second number. So the probability that the first number is greater than or equal to the second number is 21 36=7 12.
Solution: 2 Plotting outcomes in a list
If the first number is a 1, this can only be bigger than or equal to the second number if the second number is a 1. If the first number is a 2, then this is bigger than or equal to 1 or 2. We can make a table for all possibilities that the first number is greater than or equal to the second:
Number on first throw Possible numbers on second throw
1 1
2 1,2
3 1,2,3
4 1,2,3,4
5 1,2,3,4,5
6 1,2,3,4,5,6
Adding up the possibilities in the table, there are 21 ways that the number on the second roll can be greater than or equal to the number on the first roll. Since there are 6×6=36 total possible outcomes, the probability that the first number is greater than or equal to the second is 21 36=7 12.
Solution: 3 Abstract analysis of outcomes
There are 6 outcomes for the first roll and 6 for the second so the total number of possible outcomes is 6×6=36. There are 6 of these in which the two numbers are equal: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). When the two numbers are not equal, half of these cases have a greater first number and half have a greater second number: this can be seen by reversing the order to the two throws, which exchanges these two cases. So of the 30 cases where the two numbers are different, 15 of them have a greater first number. This means the first number is greater than or equal to the second in 6+15=21 of the 36 different possible outcomes. So the probability that the first throw is at least as big as the second is 21 36=7 12.
Rolling Twice
A fair six-sided die is rolled twice. What is the theoretical probability that the first number that comes up is greater than or equal to the second number?
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4902 | https://bsac.org.uk/wp-content/uploads/2012/02/BSAC-Susceptibility-testing-version-14.pdf | British Society for Antimicrobial Chemotherapy BSAC to actively support the EUCAST Disc Diffusion Method for Antimicrobial Susceptibility Testing in preference to the current BSAC Disc Diffusion Method From January 2016, the BSAC Standing Committee for Antimicrobial Susceptibility Testing, with the support of Council, will: Cease active support, maintainance and development of the BSAC disc diffusion method (queries from laboratories that continue to use the BSAC disc diffusion method will be supported during the transition period). Support UK laboratories in changing to the EUCAST (European Committee on Antimicrobial Susceptibility Testing) disc diffusion method should they wish to do this, through increased educational activities. Re-fashion the Residential Workshops to support a wider range of susceptibility testing and resistance detection methods and particularly support those using EUCAST methods. Re-fashion the current “User Days” to cover a wider range of issues in susceptibility testing. Support EUCAST in the further development and maintenance of the EUCAST susceptibility testing methods. Support UK laboratories implementing EUCAST methods and having queries about the method Background Since it was first developed and published in 2001, the BSAC standardized disc diffusion method of antimicrobial susceptibility testing has been adopted by more than 175 laboratories across the UK. Annual updates have been published since the initial launch and Version 14 of the method was published on the BSAC website in January 2015.
However, over the last five years there have been a number of developments in the field of antimicrobial susceptibility testing which have rightly led to a re-evaluation of the position of the BSAC method.
The BSAC Standing Committee has been instrumental in supporting the development of EUCAST. It signed-up to the EUCAST process for harmonised MIC breakpoint setting and EUCAST breakpoints have been incorporated into the BSAC guidelines. Although it was not part of the original EUCAST project, a standardised disc diffusion method (based on the Kirby-Bauer method using Mueller-Hinton agar) has been developed, resulting in a choice of two similar standardised disc diffusion methods (BSAC and EUCAST) that are calibrated against EUCAST breakpoints. The decision to support the EUCAST disc diffusion method in preference to the BSAC disc diffusion method has been taken for a number of reasons: The EUCAST method is a robust and standardised method. It is correlated to MICs performed according to the international standard method for testing antimicrobial susceptibility (ISO20776-1:2006). Many laboratories in the UK have already changed to using the EUCAST disc diffusion method. This leads to confusion between laboratories, particularly when reviewing NEQAS performance as the BSAC and EUCAST methods may perform differently for some challenging organisms. The EUCAST disc diffusion method has been developed to cover more antimicrobial agent/organism combinations than the BSAC disc diffusion method. A few gaps remain (e.g. Neisseria gonorrhoeae testing), but these are being actively developed. The fact that both BSAC and EUCAST methods are now used across the UK raises issues for the Standing Committee in delivery of relevant day-to-day support and also educational meetings and workshops. The EUCAST disc diffusion method is now the standard method used in most European countries and increasingly outside Europe. This means that EUCAST can draw on a wider international pool of experts and laboratories (including those in the UK) for development and support Use of the EUCAST disc diffusion method would improve international standardisation and comparability and support resistance surveillance. EUCAST is recognised by the EMA for the setting of MIC breakpoints for new agents and is increasingly seen by drug developers as the standard-setting organisation for MIC breakpoints and disc diffusion testing.
For further information please contact: Mandy Wootton (Secretary to the Standing Committee) Mandy.wootton@wales.nhs.uk Or Robin Howe (Chair of the Standing Committee) Robin.howe@wales.nhs.uk Content Page Additional information Changes in version 13.1 2 Suggestions for appropriate agents to test 3 Susceptibility testing tables: Enterobacteriaceae 6 Acinetobacter spp.
9 Pseudomonas spp.
10 Stenotrophomonas maltophilia 12 Link to guidance document on Stenotrophomonas maltophilia Link to guidance document on vancomycin susceptibility testing in S. aureus Link to guidance document on dissociated resistance in staphylococci Streptococcus pneumoniae 16 Enterococcus spp.
18 Alpha haemolytic streptococci 20 Beta haemolytic streptococci 21 Moraxella catarrhalis 22 Neisseria gonorrhoeae 24 Link to guidance document on N. gonorrhoeae Neisseria meningitidis 25 Haemophilus influenzae 26 Pasteurella multocida 28 Campylobacter spp.
29 Corynebacterium spp.
30 Gram-negative anaerobes 31 Clostridium difficile 33 Gram-positive anaerobes 34 Urinary Tract Infection comments 36 Principals of reporting 37 Further testing guidance documents Burkholderia cepacia Link to guidance document on Burkholderia cepacia group Helicobacter pylori Link to guidance document on Helicobacter pylori Listeria species Link to guidance document on Listeria species Brucella species Link to guidance document on Brucella species Pink indicates breakpoints have restricted use.
Staphylococcus spp.
13 British Society for Antimicrobial Chemotherapy Standing Committee on Susceptibility Testing Version 14.0, 05-01-2015 1 Version 14, January 2015 Changes Changes (cells containing a change, a deletion or an addition) from version 12 are marked yellow Addition of MIC breakpoints: Ceftaroline Addition of diameter breakpoints: Ceftaroline Addition of comments • Any ceftaroline resistant isolates should be confirmed using an MIC method. • Methicillin susceptible isolates can be reported susceptible to ceftaroline without further testing. British Society for Antimicrobial Chemotherapy Standing Committee on Susceptibility Testing Enterobacteriaceae 2 Ciprofloxacin Cefpodoxime (for ESBL screening) Ertapenem Nitrofurantoin Gentamicin Ciprofloxacin or norfloxacin Imipenem or meropenem Cephalexin Suggestions for appropriate agents to include in routine antimicrobial susceptibility testing These suggestions are intended to indicate minimum sets of agents to test routinely in a diagnostic laboratory in order to give an For each organism group, suggestions are given of agents to test in systemic infection, or uncomplicated Urinary Tract Infection. In a few Organisms Systemic infections Uncomplicated UTI Enterobactericeae Ampicillin or Amoxicillin Ampicillin or Amoxicillin Ceftazidime plus cefotaxime or Amoxicillin-clavulanate [Cefpodoxime] (for ESBL screening) It is recommended that an MIC is performed for invasive Salmonella isolates Organisms Systemic infections Uncomplicated UTI Piperacillin-tazobactam Trimethoprim [Cefuroxime] Acinetobacter Ciprofloxacin Treat as systemic as likely not uncomplicated Gentamicin Imipenem or meropenem Colistin Amikacin [Piperacillin-tazobactam] MIC testing is required to establish colistin susceptibility EUCAST rule 12.7 "If intermediate or resistant to tobramycin and susceptible to gentamicin and amikacin, report amikacin as Organisms Systemic infections Uncomplicated UTI Treat as systemic as likely not uncomplicated Ceftazidime Ciprofloxacin Gentamicin Imipenem or meropenem Piperacillin-tazobactam Colistin [Tobramycin] [Amikacin] MIC testing is required to establish colistin susceptibility May be appropriate according to local use Pseudomonas spp Amikacin 3 Suggestions for appropriate agents to include in routine antimicrobial susceptibility testing Trimethoprim Organisms Systemic infections Uncomplicated UTI Staphylococci Oxacillin or cefoxitin S. saprophyticus Erythromycin Ciprofloxacin or norfloxacin Fusidic acid or rifampicin Gentamicin Gentamicin Oxacillin or cefoxitin Tetracycline Vancomycin Vancomycin Nitrofurantoin Mupirocin Trimethoprim [Linezolid] [Daptomycin] Treat as other species as systemic as likely not.
[Penicillin] [Teicoplanin] MIC testing is required to establish vancomycin susceptibility Recommended for testing in severe infection Organisms Systemic infections Tetracycline Levofloxacin or moxifloxacin [Vancomycin] Organisms Systemic infections Uncomplicated UTI S. pneumoniae Penicillin (oxacillin screen) Erythromycin Linezolid Trimethoprim Teicoplanin [additional not alternative to vancomycin] Ciprofloxacin or norfloxacin Teicoplanin [additional not alternative to vancomycin] Enterococcus spp Ampicillin or amoxicillin Ampicillin or amoxicillin Gentamicin (high level screen) Vancomycin Vancomycin Nitrofurantoin Uncomplicated UTI Beta-haemolytic streptococci Erythromycin (Group B) Penicillin Penicillin Tetracycline Nitrofurantoin Organisms Systemic infections 4 Suggestions for appropriate agents to include in routine antimicrobial susceptibility testing Tetracycline Tetracycline Trimethoprim Organisms Systemic infections M. catarrhalis Ampicillin or amoxicillin Co-amoxiclav Erythromycin Ciprofloxacin [nalidixic acid to detect any quinolone resistance] [Chloramphenicol] [Cefotaxime] Resistance to ampicillin by production of β-lactamase (BRO-1/2 β-lactamase) may be misidentified by disk diffusion technique and, because production is slow, may give weak results with in-vitro tests. Since >90% of M. catarrhalis strains produce β-lactamase, testing of penicillinase production is discouraged and isolates reported resistant to ampicillin and amoxicillin.
Organisms Systemic infections N. gonorrhoeae Penicillin Ceftriaxone Cefixime Spectinomycin Ciprofloxacin [nalidixic acid to detect any quinolone resistance] Beta-lactamase [Cefuroxime as indicator of caphalosporin resistance] Cefuroxime Tetracycline Ciprofloxacin [nalidixic acid to detect any quinolone resistance] Beta-lactamase Co-amoxiclav [Chloramphenicol] [Cefotaxime] Organisms Systemic infections H. influenzae Ampicillin or amoxicillin 5 S ≤ I R > S ≥ I R ≤ Amikacin 8 16 16 30 19 16-18 15 Gentamicin 2 4 4 10 20 17-19 16 Gentamicin (Topical only) 2 -2 10 20 -19 Tobramycin 2 4 4 10 21 18-20 17 General notes: S ≤ I R > S ≥ I R ≤ Amoxicillin 8 -8 10 15 -14 Ampicillin 8 -8 10 15 -14 Co-amoxiclav (Systemic) 8 -8 20/10 21 -20 Co-amoxiclav (see UTI comments) 32 -32 20/10 15 -14 Mecillinam (see UTI comments) 8 -8 10 14 -13 These interpretative criteria are for P. mirabilis &E. coli only. Isolates of E. coli and Klebsiella spp. that produce ESBLs often appear susceptible to mecillinam in vitro but clinical efficacy against these organisms is unproven.
Piperacillin 8 16 16 75 23 21-22 20 Piperacillin-tazobactam 8 16 16 75/10 23 21-22 20 No EUCAST BP available, based on BSAC data.
The distribution of zone diameters for ESBL and AmpC producers straddles the breakpoint. Organisms that appear resistant by disc diffusion should have resistance confirmed by MIC determination. Temocillin (see UTI comments) 32 -32 30 12 -11 No EUCAST BP available, based on BSAC data.
Ticarcillin-clavulanate 8 16 16 75/10 23 -22 The zone diameter breakpoint relates to an MIC of 8mg/L as no data for the intermediate category are currently available.
Enterobacteriaceae (including Salmonella, Shigella spp. And Yersinia enterocolitica) BSAC, Version 14, January 2015 Species that have chromosomal penicillinases (Klebsiella spp.) or inducible AmpC enzymes (e.g. Enterobacter spp., Citrobacter spp. and Serratia spp.) are intrinsically resistant to ampicillin/amoxicillin.
Salmonella spp should be reported resistant to these agents, irrespective of susceptibility testing result. Aminoglycosides are considered inactive against Salmonella spp in-vivo.
Agent specific notes: Agent specific notes: The identification of Enterobacteriaceae to species level is essential before applying Expert Rules for the interpretation of susceptibility. MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Disk content (µg) Zone diameter breakpoint (mm) Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar Inoculum: McFarland 0.5, dilute 1:100 Incubation: Air, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the back of the plate against a dark background illuminated with reflected light.
Quality control: Escherichia coli NCTC 10418 or ATCC 25922 Penicillins MIC breakpoint (mg/L) MIC breakpoints are correlated to MICs performed using fixed concentration of 2mg/L clavulanate.
Species inducible AmpC enzymes (e.g. Enterobacter spp., Citrobacter spp. and Serratia spp.) are intrinsically resistant to Co-amoxiclav. Reporting guidance Individual aminoglycoside agents must be tested; susceptibility to other aminoglycosides cannot be inferred from the gentamicin result and vice versa .
Reporting guidance -8 Temocillin 19 -20 30 8 6 Enterobacteriaceae (including Salmonella, Shigella spp. And Yersinia enterocolitica) BSAC, Version 14, January 2015 S ≤ I R > S ≥ I R ≤ Cefalexin (see UTI comments) 16 -16 30 16 -15 These interpretative critera are for E. coli and Klebsiella spp. only. Cefalexin (see UTI comments) 16 -16 30 18 -17 These interpretative critera are for P. mirabilis only. Cefepime 1 2-4 4 30 32 27-31 26 Cefixime 1 -1 5 20 -19 MIC breakpoint for UTI only Cefotaxime 1 2 2 30 30 24-29 23 For Enterobacter spp,. Citrobacter freundii, Serratia spp. & Morganella morganii: if susceptible in-vitro either supress result or add comment discouraging use of cefotaxime as monotherapy due to selection of resistance. ( Cefoxitin (AmpC screen) ---30 23 --This is an epidemiological "cut off" for AmpC detection which has high sensitivity but low specificity as susceptibility is also affected by permeability.
Cefpodoxime (ESBL screen) 1 -1 10 20 -19 If screening for ESBLs is required for infection control or epidemiological purposes, Enterobacteriaceae isolates should be screened with cefpodoxime or both cefotaxime and ceftazidime. The presence of ESBLs should be confirmed with a specific test.
Ceftaroline 0.5 -0.5 5 23 -22 Any resistant isolates should be confirmed using an MIC method Ceftazidime 1 2-4 4 30 27 23-26 22 Ceftriaxone 1 2 2 30 28 24-27 23 Cefuroxime (atexil) (see UTI comments) 8 -8 30 20 -19 Cefuroxime (parenteral) 8 -8 30 20 -19 Breakpoint relates to a dosage of 1.5g three time a day and to E. coli , Klebsiella spp. and P. mirabilis only.
S ≤ I R > S ≥ I R ≤ The doripenem MIC breakpoint has changed but a review of the data indicates that no adjustment of the zone diameter breakpoints is necessary.
Ertapenem 0.5 1 1 10 28 16-27 15 Imipenem 2 4-8 8 10 21 17-20 16 Proteus spp. and Morganella morganii are considered poor targets for imipenem Meropenem 2 4-8 8 10 27 20-26 19 General notes: Detection of carbapenem resistance is difficult. For epidemiological or cross infection purposes consideration should be given to testing isolates resistant to ceftazidime and a carbapenem for the presence of carbapenemases. Guidance on detection is given Doripenem 1 Salmonella spp. should be reported resistant to these agents, irrespective of susceptibility testing result.
2 10 19-23 18 24 Zone diameter breakpoint (mm) For Enterobacter spp,. Citrobacter freundii, Serratia spp. & Morganella morganii: if susceptible in-vitro either supress result or add comment discouraging use of cefotaxime as monotherapy due to selection of resistance. ( -MIC breakpoint (mg/L) Disk content (µg) Cefalexin results may be used to report susceptibility to cefadroxil and cefradine. Agent specific notes: Reporting guidance Cephalosporins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Screening for carbapenem producing Enterobactericeae can be performed using a cut-off of 32mm with meropenem 10ug disc. Carbapenems 7 Enterobacteriaceae (including Salmonella, Shigella spp. And Yersinia enterocolitica) BSAC, Version 14, January 2015 S ≤ I R > S ≥ I R ≤ Aztreonam 1 2-4 4 30 28 23-27 22 S ≤ I R > S ≥ I R ≤ Ciprofloxacin 0.5 1 1 1 20 17-19 16 For ciprofloxacin there is clinical evidence to indicate a poor response in systemic infections caused by Salmonella spp., with reduced susceptibility to fluoroquinolones. Isolates with MICs greater than 0.06mg/L should be reported as resistant. It is recommended that the ciprofloxacin MIC be determined for all invasive salmonellae infections.
Levofloxacin 1 2 2 1 17 14-16 13 Moxifloxacin 0.5 1 1 1 20 17-19 16 Nalidixic acid (see UTI comments) 16 -16 30 18 -17 Norfloxacin (systemic) 0.5 1 1 2 26 19-25 18 Norfloxacin (see UTI comments) 4 -4 2 16 -15 No EUCAST breakpoint. BSAC data used.
Ofloxacin 0.5 1 1 5 29 26-28 25 S ≤ I R > S ≥ I R ≤ These interpretative criteria are for S. typhi only.
Azithromycin has been used in the treatment of infections with S. typhi (MIC ≤16mg/L for wild type isolates) and some enteric infections.
S ≤ I R > S ≥ I R ≤ Tetracycline 4 -4 10 24 -23 These interpretative criteria are for Y. enterocolitica only.
Tigecycline 1 2 2 15 24 20-23 19 Disc diffusion for Enterobacteriaceae other than E. coli may not give reliable results; an MIC method is prefered if tigecycline is considered as therapy. Susceptibility of E. coli isolates appearing intermediate or resistant should be confirmed with an MIC method.
Morganell morganii, Providencia spp. & Proteus spp. are considered inherently non-susceptible to tigecycline.
S ≤ I R > S ≥ I R ≤ Chloramphenicol 8 -8 30 21 -20 Colistin 2 -2 ----Disc diffusion susceptibility testing is unreliable. Colistin susceptibility should be determined with an MIC method.
Co-trimoxazole 2 4 4 1.25/23.75 16 -15 The MIC breakpoint is based on the trimethorpim concentration in a 1:19 combination with suphamethoxazole. Link to co-trimoxazole guidance 25 -24 These interpretative criteria are for E. coli only.
37 -36 These interpretative criteria are for P. mirabilis only.
Reporting guidance Agent specific notes: Quinolones MIC breakpoint (mg/L) Disk content (µg) Disk content (µg) Zone diameter breakpoint (mm) MIC breakpoint (mg/L) Disk content (µg) -Zone diameter breakpoint (mm) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Other β-lactams MIC breakpoint (mg/L) Fosfomycin (see UTI comments) 32 -32 200/50 MIC breakpoints refer to i.v. treatment for system infections and oral treatment for uncomplicated UTI therapy.
Agent specific notes: Reporting guidance Tetracyclines MIC breakpoint (mg/L) Disk content (µg) Miscellaneous antimicrobials Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Azithromycin 18 -19 15 --Macrolides, lincosamides & streptogramins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance 8 Enterobacteriaceae (including Salmonella, Shigella spp. And Yersinia enterocolitica) BSAC, Version 14, January 2015 Nitrofurantoin (see UTI comments) 64 -64 200 17 -16 These interpretative criteria are for E. coli only.
Trimethoprim (see UTI comments) 2 4 4 2.5 17 14-16 13 9 S ≤ I R > S ≥ I R ≤ Amikacin 8 16 16 30 21 19-20 18 Gentamicin 4 -4 10 20 -19 S ≤ I R > S ≥ I R ≤ Piperacillin-tazobactam 8 16 16 75/10 22 20-21 19 No EUCAST MIC BP available due to insufficient evidence. BSAC data used.
S ≤ I R > S ≥ I R ≤ Doripenem 1 -2 10 22 15-21 14 The doripenem MIC breakpoint has changed but a review of the data indicates that no adjustment of the zone diameter breakpoints is necessary.
Imipenem 2 4-8 8 10 25 14-24 13 Meropenem 2 4-8 8 10 20 13-19 12 S ≤ I R > S ≥ I R ≤ Ciprofloxacin 1 -1 1 21 -20 S ≤ I R > S ≥ I R ≤ Tigecycline -------No EUCAST MIC BP due to insufficient clinical evidence. For determining susceptibility an MIC method should be used and the EUCAST PKPD BPs of S=0.25mg/L, R=0.5mg/L applied to interpret.
S ≤ I R > S ≥ I R ≤ Colistin 2 -2 ----Disc diffusion susceptibility testing is unreliable. Colistin susceptibility should be determined with an MIC method.
Acinetobacter spp. BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar Inoculum: McFarland 0.5, dilute 1:100 Incubation: Air, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the back of the plate against a dark background illuminated with reflected light.
Quality control: Escherichia coli NCTC 10418 or ATCC 25922 Aminoglycosides MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Reporting guidance Penicillins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Carbapenems MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Quinolones MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Tetracyclines MIC breakpoint (mg/L) Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance 9 S ≤ I R > S ≥ I R ≤ Amikacin 8 16 16 30 22 16-21 15 Gentamicin 4 -4 10 18 -17 Netilmicin 4 -4 10 14 -13 Tobramycin 4 -4 10 20 -19 General notes: S ≤ I R > S ≥ I R ≤ Piperacillin 16 -16 75 25 -24 Piperacillin-tazobactam 16 -16 75/10 25 -24 Ticarcillin 16 -16 75 20 -19 Ticarcillin-clavulanate 16 -16 75/10 20 -19 S ≤ I R > S ≥ I R ≤ Ceftazidime 8 -8 30 24 -23 S ≤ I R > S ≥ I R ≤ The doripenem MIC breakpoint has changed but a review of the data indicates that no adjustment of the zone diameter breakpoints is necessary.
Imipenem 4 8 8 10 23 17-22 16 Meropenem 2 4-8 8 10 20 16-19 15 S ≤ I R > S ≥ I R ≤ Aztreonam 1 2-16 16 30 36 20-35 19 Relates only to isolates from patients with cystic fibrosis given high dosage therapy to treat P. aeruginosa .
Aminoglycosides MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Pseudomonas BSAC, Version 14, January 2015 These interpretative criteria are not for use with other non-fermenting organisms(inclding Burkholderia spp.) Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar Inoculum: McFarland 0.5, dilute 1:100 Incubation: Air, 36±1ºC, 19±2h Reading: Read zone edges as the point showing no growth viewed from the back of the plate against a dark background illuminated with reflected light.
Quality control: Pseudomonas aeruginosa ATCC 27853 or NCTC 10662 Reporting guidance Individual aminoglycoside agents must be tested; susceptibility to other aminoglycosides cannot be inferred from the gentamicin result and vice versa .
Penicillins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Cephalosporins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Carbapenems MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Doripenem Other β-lactams MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Detection of carbapenem resistance is difficult. Guidance on detection is given at: b_C/1317138520481. 24 25-31 32 10 2 -1 10 Pseudomonas BSAC, Version 14, January 2015 S ≤ I R > S ≥ I R ≤ Ciprofloxacin 0.5 1 1 1 23 13-22 12 Ciprofloxacin 0.5 1 1 5 30 20-29 19 Levofloxacin 1 2 2 5 22 17-21 16 No EUCAST MIC breakpoint due to insufficient clinical evidence. EUCAST PKPD breakpoint and BSAC data used.
S ≤ I R > S ≥ I R ≤ Colistin 4 -4 ----Disc diffusion susceptibility testing is unreliable. Colistin susceptibility should be determined with an MIC method.
Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Quinolones MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance 11 S ≤ I R > S ≥ I R ≤ Co-trimoxazole 4 -4 1.25/23.75 20 -19 For Stenotrophomonas maltophilia, susceptibility testing is not recommended except for co-timoxazole. See
Stenotrophomonas BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar Inoculum: McFarland 0.5, dilute 1:100 Incubation: Air, 30ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the back of the plate against a dark background illuminated with reflected light.
Quality control: Pseudomonas aeruginosa ATCC 27853 or NCTC 10662 Agent specific notes: Reporting guidance Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) 12 S ≤ I R > S ≥ I R ≤ 19 16-18 15 These interpretative criteria are for S. aureus only.
25 22-24 21 These interpretative criteria are for coagulase negative staphylococci only.
Gentamicin 1 -1 10 20 -19 21 -20 These interpretative criteria are for S. aureus only.
30 -29 These interpretative criteria are for coagulase negative staphylococci only.
Neomycin ---10 17 -16 For topical use only. The zone diameter breakpoint distinguishes the "wild type" susceptible population from isolates with reduced susceptibility.
General notes: S ≤ I R > S ≥ I R ≤ Ampicillin (UTI 1,2,4) ---25 26 -25 These interpretative criteria are for S. saprophyticus only.
Cefoxitinsee incubation temp above 4 --10 22 -21 These interpretative criteria are for S. aureus only.
Cefoxitinsee incubation temp above ---10 20 -19 These interpretative criteria are for S. saprophyticus only.
These interpretative criteria are for coagulase negative staphylococci only.
For coagulase negative staphylococci (except S. saprophyticus ) with cefoxitin zone diameters 22-26mm PCR for mecA is required to determine susceptibility for treatment of deep seated infection with any β-lactam.
Ceftaroline 1 -1 5 20 -19 Any resistant isolates should be confirmed using an MIC method Oxacillin 2 --1 15 -14 For oxacillin tests on Mueller-Hinton or Columbia agar with 2% NaCl: some hyperproducers of β-lactamase give zones within range of 7-14mm and if possible, should be checked by a PCR method for mecA or a latex agglutination test for PBP2a. Increase in zone sioze in the presence of clavunanic acid is not a reliable test for hyper-producers of β-lactamase as zones of inhibition with some MRSA also increase in the presence of clavulanic acid. Rarely, hyper-producers of β-lactamase give no zone in this test and would therefore not be distinguishable from MRSA.
Penicillin 0.12 -0.12 1 unit 25 -24 These interpretative criteria are for S. aureus and S. lugdunensis only. With penicillin, check for heaped zone edge which indicates β-lactamase mediated resistance.
General notes: Tobramycin Cefoxitinsee incubation temp above Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Amikacin 30 16 16 8 Staphylococci exhibiting resistance to oxacillin/cefoxitin should be regarded as resistant to other penicillins, cephalosporins, carbapenems and combinations of β-lactam and β-lactamase inhibitors.
Most staphylococci are penicillinase producers. The benzylpenicillin will mostly, but not unequivocally, separate β-lactamase producers. Isolates positive for β-lactamase are resistant to benzylpenicillin, phenoxymethylpenicillin, amino- and ureido-penicillins. Isolates negative for β-lactamase and susceptible to cefoxitin can be reported susceptible to these drugs. Isolates positive for β-lactamase and susceptible to cefoxitin are susceptible to penicillin-β-lactamase inhibitor combinations and penicillinase-resistant penicillins (oxacillinloxacillin, dicloxacillin and flucloxacillin). Isolates resistant to cefoxitin are methicillin resistant and resistant to β-lactam agents, including β-lactamase inhibitor combinations, except for cephalosporins with approved anti-MRSA activity and clinical breakpoints.
10 --4 BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar Inoculum: McFarland 0.5, dilute 1:10 Incubation: Air, 36±1ºC, 19±1h ( cefoxitin should be tested at 35±1°C) Reading: Read zone edges as the point showing no growth viewed from the back of the plate against a dark background illuminated with reflected light.
Quality control: Staphylococcus aureus NCTC 6571 or ATCC 25923 Reporting guidance Aminoglycosides MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Staphylococci Individual aminoglycoside agents must be tested; susceptibility to other aminoglycosides cannot be inferred from the gentamicin result and vice versa .
10 1 -1 21 22-26 27 β-lactams MIC breakpoint (mg/L) 13 BSAC, Version 14, January 2015 Staphylococci S ≤ I R > S ≥ I R ≤ Ciprofloxacin 1 -1 1 14 -13 MIC breakpoints relate to high-dose therapy (750mg BD).
Ciprofloxacin (UTI 1,2,4) 1 -1 1 18 -17 These interpretative criteria are for S. saprophyticus only.
Levofloxacin 1 2 2 5 23 --Moxifloxacin 0.5 1 1 1 20 16-19 15 Ofloxacin 1 -1 5 28 -27 S ≤ I R > S ≥ I R ≤ Teicoplanin 2 -2 ----These interpretative criteria are for S. aureus only.
Teicoplanin 4 -4 ----These interpretative criteria are for coagulase negative staphylococci only.
Vancomycin 2 -2 ----These interpretative criteria are for S. aureus only.
Vancomycin 4 -4 ----These interpretative criteria are for coagulase negative staphylococci only.
General notes: S ≤ I R > S ≥ I R ≤ Azithromycin 1 2 2 15 20 -19 The zone diameter breakpoint relates to an MIC of 1mg/L as no data for the intermediate category are currently available.
Clarithromycin 1 2 2 2 18 15-17 14 Clindamycin 0.25 0.5 0.5 2 26 23-25 22 Organisms that appear resistant to erythromycin, but susceptible to clindamycin should be checked for the presence of inducible resistance (see in index tab). Place the erythromycin and clindamycin discs 12-20 mm apart (edge to edge) and look for antagonism (the D phenomenon).
Inducible resistance can only be detected in the presence of a macrolide antibiotic. If positive, report as resistant to clindamycin or report as susceptible with a warning that clinical failure during treatment with clindamycin may occur by selection of constitutively resistant mutants and the use of clindamycin best avoided in severe infection.
Erythromycin 1 2 2 5 20 17-19 16 See clindamycin note above. Erythromycin can be used to determine susceptibility to azithromycin, clarithromycin and roxithromycin.
Quinupristin-dalfopristin 1 2 2 15 22 19-21 18 The presence of blood has a marked effect on the activitiy of quinupristin-dalfopristin. On the rare ocassions when blood needs to be added to enhance the groath of staphylococci, susceptible ≥15mm, resistant ≤14mm.
S ≤ I R > S ≥ I R ≤ Doxycycline 1 2 2 30 31 -30 The zone diameter breakpoint realtes to an MIC of 1mg/L as no data for the intermediate category are currently available.
Minocyline 0.5 1 1 30 28 -27 The zone diameter breakpoint realtes to an MIC of 0.5mg/L as no data for the intermediate category are currently available.
Tetracycline 1 2 2 10 20 -19 The zone diameter breakpoint realtes to an MIC of 1mg/L as no data for the intermediate category are currently available.
Isolates susceptible to tetracycline are also susceptible to doxycycline and minocycline. Some isolates resistant to tetracyline may be susceptible to minocycline and/or doxycycline. Tigecycline 0.5 -0.5 15 26 -25 Strains with MIC values above the susceptible breakpoint are not yet reported. The identification and susceptibility tests on any such isolate must be repeated and if the result is confirmed the isolate must be sent to a reference laboratory.
Until there is further evidence regarding clinical laboratory response for confirmed isolates with MIC above the current breakpoint they should be reported as resistant.
Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Reporting guidance Tetracyclines Quinolones MIC breakpoint (mg/L) Disk content (µg) Disc diffusion for staphylococci does not give reliable results. An MIC method should be used to determine susceptibility, positive results requiring confirmation. Population analysis is the most reliable method for confirmating resistance and for distinguishing susceptible, hetero-GISA and GISA isolates. If, on clinical grounds, resistance to vancomycin is suspected, it is recommended that the organism be sent to a specialist laboratory, such as Dept. of Microbiology. Lime Walk Building, Southmead Hospital, Westbuty on Trym, Bristol BS10 5NB or the Specialist Antimicrobial Chemotherapy Unit, Public Health Wales, University Hospital of Wales, Heath Park, Cardiff, CF14 4XW. Reporting guidance Macrolides, lincosamides & streptogramins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Glycopeptides MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: 14 BSAC, Version 14, January 2015 Staphylococci S ≤ I R > S ≥ I R ≤ Strains with MIC values above the susceptible breakpoint are very rare or not yet reported. The identification and susceptibility tests on any such isolate must be repeated and if the result is confirmed the isolate must be sent to a reference laboratory. Susceptibility testing by disc diffusion is not reliable. Susceptibility should be determined using a broth dilution method with Mueller-Hinton broth or by an MIC method on Mueller-Hinton agar. The test conditions must provide 50mg Ca++ to avoid false resistance being reported.
Chloramphenicol 8 -8 10 15 -14 Co-trimoxazole 2 4 4 1.25/23.75 17 14-16 13 LINK to guidance Trimethoprim 1 -1 5 20 -19 Breakpoints are epidemiological "cut offs" based on distributions for the "wild type" population. However there is no clear evidence correlating these breakpoints with clinical efficacy.
Trimethoprim (UTI 1,2,4) 2 4 4 2.5 15 13-14 12 These interpretative criteria are for S. saprophyticus only.
Fosfomycin (IV) 32 -32 200/50 34 -33 Fusidic acid 1 -1 10 30 -29 Linezolid 4 -4 10 20 -19 Mupirocin 1 2-256 256 20 27 7-26 6 In nasal decontamination, isolates with low-level resistance to mupirocin (MICs 2-256mg/L) may be initially cleared, but early recolonisation is common.
Nitrofurantoin (UTI 1,2,4) 64 -64 200 20 -19 These interpretative criteria are for S. saprophyticus only.
Rifampicin 0.06 0.12-0.5 0.5 2 30 24-29 23 MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance -Daptomycin 1 -1 Miscellaneous antimicrobials ---Until there is evidence regarding clinical laboratory response for confirmed isolates with MIC above the current breakpoint they should be reported as resistant.
15 S ≤ I R > S ≥ I R ≤ Reduced susceptibility to penicillin in Streptococcus pneumoniae is most reliably detected with an oxacillin 1ug disc; confirm resistance with a penicillin MIC determination. Most MIC values for penicillin, ampicillin, amoxicillin and piperacillin (with or without a β-lactamase inhibitor) differ by no more than one dilution step and isolates fully susceptible to benzylpenicillin (MIC≤0.06mg/L; suceptible by oxacillin disc screen) can be reported susceptible to β-lactam agents that have been given breakpoints.
Infections with organisms with a penicillin MIC ≤2mg/L may be effectively treated if adequate doses are used except in infections of the central nervous system. In addition, cefotaxime or ceftriaxone MIC determination is advised for isolates from meningitis or other invasive infections.
S ≤ I R > S ≥ I R ≤ Cefaclor 0.03 0.06-0.5 0.5 ----Cefotaxime 0.5 1-2 2 ----Cefpodoxime 0.25 0.5 0.5 ----Ceftriaxone 0.5 1-2 2 ----Cefuroxime 0.5 1 1 ----General notes: S ≤ I R > S ≥ I R ≤ Ertapenem 0.5 -0.5 ----Imipenem 2 -2 ----Meropenem (infections other than meningitis) 2 -2 ----Meropenem is the only carbapenem used for meningitis; for use determine MIC value.
General notes: 20 -19 Penicillin 0.06 Oxacillin 1 0.12-2 2 Streptococcus pneumoniae BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood (ISA + %5 horse blood + 20mg/L NAD may also be used) Inoculum: McFarland 0.5, dilute 1:10 Incubation: 4-6% CO2, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the front of the plate, being careful not to read haemolysis.
Quality control: Streptococcus pneumoniae ATCC 49619 OR Staphylococcus aureusNCTC 6571 Reporting guidance Penicillins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Isolates with MIC values above the S/I breakpoint are very rare. The identification and susceptibility tests on any such isolate must be repeated and if the result is confirmed the isolate must be sent to a reference laboratory. Until there is evidence regarding clinical response for confirmed isolates with MIC values above the current resistant breakpoint they should be reported resistant.
Carbapenems MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Screen for β-lactam resistance with the oxacillin 1ug disc.
Zone diameter breakpoint (mm) MIC breakpoint (mg/L) Reporting guidance Cephalosporins Disk content (µg) Agent specific notes: Reporting guidance Isolates categorised as suceptible with the oxacillin 1ug disc can be reported susceptible to cefepime, cefotaxime, cefpodoxime, ceftriaxone, cefuroxime ± axetil and cefaclor.
Isolates categorised as suceptible with the oxacillin 1ug disc can be reported susceptible to imipenem, ertapenem, and meropenem.
Screen for β-lactam resistance with the oxacillin 1ug disc.
---Isolates with MIC values above the S/I breakpoint are very rare. The identification and susceptibility tests on any such isolate must be repeated and if the result is confirmed the isolate must be sent to a reference laboratory. Until there is evidence regarding clinical response for confirmed isolates with MIC values above the current resistant breakpoint they should be reported resistant.
-0.25 0.5-1 1 Meropenem (meningitis) 16 Streptococcus pneumoniae BSAC, Version 14, January 2015 S ≤ I R > S ≥ I R ≤ Ciprofloxacin 0.12 0.25-2 2 1 25 10-24 9 For systemic infection the "wild type" isolates (MIC 0.25-2mg/L) are considered intermediate in susceptibility.
Ofloxacin 0.12 0.25-4 4 5 28 16-27 15 For systemic infection the "wild type" isolates (MIC 0.25-4mg/L) are considered intermediate in susceptibility.
Levofloxacin 2 -2 1 10 -9 Moxifloxacin 0.5 -0.5 1 18 -17 S ≤ I R > S ≥ I R ≤ Vancomycin 2 -2 5 13 -12 General notes: S ≤ I R > S ≥ I R ≤ Azithromycin 0.25 0.5 0.5 15 22 20-21 19 Clarithromycin 0.25 0.5 0.5 2 22 20-21 19 Clindamycin 0.5 -0.5 2 24 -23 Organisms that appear resistant to erythromycin, but susceptible to clindamycin should be checked for the presence of inducible resistance. Place the erythromycin and clindamycin discs 12-16 mm apart (edge to edge) and look for antagonism (the D phenomenon) Inducible resistance can only be detected in the presence of a macrolide antibiotic. If positive, report as resistant to clindamycin or report as susceptible with a warning that clinical failure during treatment with clindamycin may occur by selection of constitutively resistant mutants and the use of clindamycin best avoided in severe infection.
Erythromycin 0.25 0.5 0.5 5 22 20-21 19 See clindamycin note above.
Erythromycin can be used to determine susceptibility to azithromycin, clarithromycin and roxithromycin.
Telithromycin 0.25 0.5 0.5 15 29 -28 No EUCAST breakpoint, BSAC data used. Insufficient data are available to distinguish the intermediate category.
S ≤ I R > S ≥ I R ≤ Tetracycline 1 2 2 10 20 -19 The zone diameter breakpoint relates to an MIC of 1mg/L as no data for the intermediate category are currently available.
Isolates susceptible to tetracycline are also susceptible to doxycycline, and minocycline. Some isolates resistant to tetracycline may be susceptible to minocycline and/or doxycycline.
S ≤ I R > S ≥ I R ≤ Chloramphenicol 8 -8 10 18 -17 Co-trimoxazole 1 2 2 1.25/23.75 17 -16 LINK to guidance Linezolid 2 4 4 10 20 -19 The zone diameter breakpoint relates to an MIC of 2mg/L as no data for the intermediate category are currently available.
Rifampicin 0.06 0.12-0.5 0.5 5 23 21-22 20 Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Reporting guidance Reporting guidance Agent specific notes: Zone diameter breakpoint (mm) Disk content (µg) MIC breakpoint (mg/L) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Macrolides, lincosamides & streptogramins Quinolones MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Tetracyclines MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Glycopeptides MIC breakpoint (mg/L) Disk content (µg) 17 S ≤ I R > S ≥ I R ≤ High-level gentamicin resistant enterococci usually give no zone or only a trace of inibition around a 200ug disc. Occasionally, however, the plasmid carrying the resistance may be unstable and the resistance is seen as a zone of inhibition with a few small colonies within the zone. Retesting of resistant colonies results in growth to the disc or increased numbers of colonies within the zone.
Zones should be carefully examined to avoid missing such resistant organisms. If in doubt, isolates may be sent to a reference laboratory for confirmation.
Streptomycin 128 -128 300 24 -23 The EUCAST breakpoint is 512mg/L tested on Mueller-Hinton agar which correlates with the MIC breakpoint of 128mg/L on IsoSensitest agar and the zone criteria given.
S ≤ I R > S ≥ I R ≤ Amoxicillin 4 8 8 10 20 -19 Ampicillin 4 8 8 10 20 -19 Co-amoxiclav susceptibility can be inferred from the ampicillin result.
S ≤ I R > S ≥ I R ≤ Imipenem 4 8 8 10 19 17-18 16 These interpretative criteria are for E. faecalis only.
S ≤ I R > S ≥ I R ≤ Teicoplanin 2 -2 30 20 -19 Vancomycin 4 -4 5 13 -12 S ≤ I R > S ≥ I R ≤ Quinupristin-dalfopristin 1 2-4 4 15 20 12-19 11 The presence of blood has a marked effect on the activitiy of quinupristin-dalfopristin. On the rare ocassions when blood needs to be added to enhance the growth of enterococci, susceptible ≥15mm, resistant ≤14mm.
Generally, E. faecalis are intermediate or resistant and E. faecium are susceptible. BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar Inoculum: McFarland 0.5, dilute 1:100 Incubation: Air, 36±1ºC, 18±2h (glycopeptides require full 24h incubation time) Reading: Read zone edges as the point showing no growth viewed from the back of the plate against a dark background illuminated with reflected light.
Quality control: Enterococcus faecalis NCTC 12697 (ATCC 29213) Aminoglycosides MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Gentamicin 128 -128 200 Enterococci Penicillins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Macrolides, lincosamides & streptogramins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Agent specific notes: Reporting guidance Agent specific notes: Reporting guidance Carbapenems Disk content (µg) MIC breakpoint (mg/L) To ensure that microcolonies indicating reduced susceptibility to the glycopeptides are detected, it is essential that plates are incubated for at least 24h before reporting a strain as susceptible to vancomycin or teicoplanin.
For isolates from endocarditis the MIC should be determined and interpreted according to the national endocarditis guidelines (Gould FK et al Guidelines for the diagnosis and antibiotic treatment of endocarditis in adults; report of the Working Party of the British Society for Antimicrobial Chemotherapy. J. Antimicrob. Chemother. 2012;67:269-89.
14 -15 Zone diameter breakpoint (mm) Glycopeptides MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) 18 BSAC, Version 14, January 2015 Enterococci S ≤ I R > S ≥ I R ≤ Tigecycline 0.25 0.5 0.5 15 21 -20 Isolates with MIC values above the susceptible breakpoint are very rare or not yet reported, so there is no intermediate category for disc diffusion. The identification and susceptibility tests on any such isolate must be repeated and if the result is confirmed the isolate must be sent to a reference laboratory.
Until there is further evidence regarding clinical laboratory response for confirmed isolates with MIC above the current breakpoint they should be reported as resistant.
S ≤ I R > S ≥ I R ≤ Linezolid 4 -4 10 20 -19 Nitrofurantoin (UTI 1,2,4) 64 -64 200 20 -19 Trimethoprim (UTI 1,2,4) 0.03 0.06-1 1 2.5 >50 22-50 21 There is some doubt about the clinical relevance of testing the susceptibility of enterococci to trimethoprim. The breakpoints have been set to interpret all enterococci as intermediate.
Tetracyclines MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance 19 S ≤ I R > S ≥ I R ≤ Amoxicillin 1 1-2 2 2 24 15-23 14 Penicillin 0.25 0.5-2 2 1 unit 17 11-16 10 S ≤ I R > S ≥ I R ≤ Cefotaxime 0.5 -0.5 5 23 -22 S ≤ I R > S ≥ I R ≤ Teicoplanin 2 -2 30 16 -15 Vancomycin 2 -2 5 14 -13 S ≤ I R > S ≥ I R ≤ Clindamycin 0.5 -0.5 2 20 -19 Erythromycin 2 -2 5 20 -19 S ≤ I R > S ≥ I R ≤ Linezolid 2 -2 10 20 -19 No EUCAST MIC breakpoint as there is insufficient evidence. BSAC data used.
MIC breakpoint (mg/L) Zone diameter breakpoint (mm) Zone diameter breakpoint (mm) Glycopeptides Agent specific notes: Agent specific notes: Organisms that appear resistant to erythromycin, but susceptible to clindamycin should be checked for the presence of inducible resistance (see Place the erythromycin and clindamycin discs 12-16 mm apart (edge to edge) and look for antagonism (the D phenomenon).
Disk content (µg) Disk content (µg) MIC breakpoint (mg/L) Reporting guidance Alpha haemolytic streptococci BSAC, Version 14, January 2015 For isolates from endocarditis the MIC should be determined and interpreted according to the national endocarditis guidelines (Gould FK et al Guidelines for the diagnosis and antibiotic treatment of endocarditis in adults; report of the Working Party of the British Society for Antimicrobial Chemotherapy. J. Antimicrob. Chemother. 2012;67:269-89.
Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood + 20mg/L NAD Inoculum: McFarland 0.5, dilute 1:10 Incubation: 4-6% CO2, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the front of the plate.
Quality control: Streptococcus pneumoniae ATCC 49619 OR Staphylococcus aureus NCTC 6571 Reporting guidance Zone diameter breakpoint (mm) Inducible resistance can only be detected in the presence of a macrolide antibiotic. If positive, report as resistant to clindamycin or report as susceptible with a warning that clinical failure during treatment with clindamycin may occur by selection of constitutively resistant mutants and the use of clindamycin best avoided in severe infection.
Reporting guidance Miscellaneous antimicrobials Macrolides, lincosamides & streptogramins MIC breakpoint (mg/L) Reporting guidance Zone diameter breakpoint (mm) Zone diameter breakpoint (mm) Disk content (µg) Disk content (µg) MIC breakpoint (mg/L) Agent specific notes: Agent specific notes: Penicillins Agent specific notes: Reporting guidance Disk content (µg) MIC breakpoint (mg/L) Cephalosporins 20 S ≤ I R > S ≥ I R ≤ Penicillin 0.25 -0.25 1 unit 20 -19 Susceptibility to other penicillins, carbapenems and cephalosporins can be inferred from the penicillin result.
S ≤ I R > S ≥ I R ≤ Azithromycin 0.25 0.5 0.5 15 22 20-21 19 Clarithromycin 0.25 0.5 0.5 2 22 20-21 19 Clindamycin 0.5 -0.5 2 17 -16 Erythromycin 0.25 0.5 0.5 5 22 20-21 19 Telithromycin 0.25 0.5 0.5 15 26 -25 Zone diameter breakpoints relate to the "wild type" susceptible population as no data are available for the non-susceptible population.
S ≤ I R > S ≥ I R ≤ Tetracyline 1 2 2 10 20 -19 Isolates susceptible to tetracycline are also susceptible to doxycycline and minocycline Tigecycline 0.25 0.5 0.5 15 25 20-24 19 Strains with MIC values above the susceptible breakpoint are very rare or not yet reported. The identification and antimicrobial susceptibility testing of any such isolate must be repeated and if the result is confirmed the isolate must be sent to a reference laboratory. Until there is evidence regarding clinical response for confirmed isolates with MIC above the current breakpoint they should be reported resistant.
S ≤ I R > S ≥ I R ≤ Co-trimoxazole 1 2 2 1.25/23.75 20 17-19 16 Trimethoprim (UTI 1,2,4) 2 -2 2.5 16 -15 These interpretative criteria are for Group B streptococci only.
Daptomycin 1 -1 ----No zone diameter breakpoints are given because disc diffusion susceptibility testing is unreliable.
Strains with MIC values above the susceptible breakpoint are very rare or not yet reported. The identification and antimicrobial susceptibility testing of any such isolate must be repeated and if the result is confirmed the isolate must be sent to a reference laboratory. Until there is evidence regarding clinical response for confirmed isolates with MIC above the current breakpoint they should be reported resistant.
Linezolid 2 4 4 10 20 -19 Zone diameter breakpoints relate to the MIC breakpoint of 0.12mg/L as no data for the intermediate category are currently available.
Nitrofurantoin (UTI 1,2,4) 64 -64 200 19 -18 Beta haemolytic streptococci BSAC, Version 14, January 2015 For isolates from endocarditis the MIC should be determined and interpreted according to the national endocarditis guidelines (Gould FK et al Guidelines for the diagnosis and antibiotic treatment of endocarditis in adults; report of the Working Party of the British Society for Antimicrobial Chemotherapy. J. Antimicrob. Chemother. 2012;67:269-89.
Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood (ISA + %5 horse blood + 20mg/L NAD may also be used) Inoculum: McFarland 0.5, dilute 1:100 Incubation: O2, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the front of the plate.
Quality control: Streptococcus pneumoniae ATCC 49619 OR Staphylococcus aureus NCTC 6571 Agent specific notes: Reporting guidance Penicillins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Zone diameter breakpoint (mm) Reporting guidance Macrolides, lincosamides & streptogramins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Agent specific notes: Reporting guidance Organisms that appear resistant to erythromycin, but susceptible to clindamycin should be checked for the presence of inducible resistance (see Place the erythromycin and clindamycin discs 12-16 mm apart (edge to edge) and look for antagonism (the D phenomenon) Inducible resistance can only be detected in the presence of a macrolide antibiotic. If positive, report as resistant to clindamycin or report as susceptible with a warning that clinical failure during treatment with clindamycin may occur by selection of constitutively resistant mutants and the use of clindamycin best avoided in severe infection.
Tetracyclines MIC breakpoint (mg/L) Disk content (µg) 21 S ≤ I R > S ≥ I R ≤ Ampicillin -------Resistance to ampicillin by production of β-lactamase (BRO-1/2 β-lactamase) may be misidentified by disc diffusion technique and, because β-lactamase production is slow, may give weak results with in vitro tests. Since 90% of M. catarrhalis strain produce β-lactamase, testing of penicillinase production is discouraged and isolates reported resistant to ampicillin and amoxicillin.
Co-amoxiclav 1 -1 2/1 19 -18 S ≤ I R > S ≥ I R ≤ Cephaclor 0.12 -0.12 30 28 -37 MIC breakpoints render all M. catarrhalis resistant to cefaclor.
Cefuroxime 4 8 8 5 17 -16 Zone diameter breakpoints realte to the MIC breakpoint of 4mg/L as no data for the intermediate category are currently available.
Cefuroxime axetil 0.12 0.25-4 4 5 35 17-34 16 General notes: S ≤ I R > S ≥ I R ≤ Ertapenem 0.12 -0.50 10 35 -34 S ≤ I R > S ≥ I R ≤ Ciprofloxacin 0.5 -0.5 1 18 -17 Levofloxacin 1 -1 1 20 -19 Moxifloxacin 0.5 -0.5 1 18 -17 Nalidixic acid (screen) ---30 18 -17 Ofloxacin 0.5 -0.5 5 35 -34 S ≤ I R > S ≥ I R ≤ Clarithromycin 0.25 0.5 0.5 2 22 20-21 19 Erythromycin 0.25 0.5 0.5 5 28 -27 Zone diameter breakpoints relate to the MIC breakpoint of 0.25mg/L as no data for the intemediate category are available.
Erythromycin can be used to determine susceptibility to azithromycin and clarithromycin.
Telithromycin 0.25 0.5 0.5 15 30 -29 Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Cephalosporins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) MIC breakpoint (mg/L) Quinolones Zone diameter breakpoint (mm) Disk content (µg) Moraxella catarrhalis BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood (ISA + %5 horse blood + 20mg/L NAD may also be used) Inoculum: McFarland 0.5, dilute 1:10 Incubation: Air, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the front of the plate.
Quality control: Haemophilus influenzae NCTC 11931 OR Haemophilus influenzae ATCC 49247 Penicillins Reporting guidance MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Carbapenems Agent specific notes: Disk content (µg) Quinolone resistance is most reliably detected with nalidixic acid.
Disk content (µg) Agent specific notes: Reporting guidance MIC breakpoint (mg/L) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Macrolides, lincosamides & streptogramins MIC breakpoint (mg/L) 22 Moraxella catarrhalis BSAC, Version 14, January 2015 S ≤ I R > S ≥ I R ≤ Tetracycline 1 2 2 10 22 -21 No disc diffusion data to distinguish the intermediate category available at present.
Isolates susceptible to tetracycline are also susceptible to doxycycline and minocycline. Some isolates resistant to tetracyline may be susceptible to minocycline and/or doxycycline. S ≤ I R > S ≥ I R ≤ Chloramphenicol 2 -2 10 30 -29 Breakpoints relate to topical use of chloramphenicol.
Co-trimoxazole 0.5 1 1 1.25/23.75 12 -11 Tetracyclines MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance 23 S ≤ I R > S ≥ I R ≤ Penicillin 0.06 0.12-1 1 1 unit 26 18-25 17 Always test for β-lactamase. If positive for β-lactamase report resistant to penicillin.
S ≤ I R > S ≥ I R ≤ Cefixime 0.12 -0.12 ----Cefotaxime 0.12 -0.12 5 ---Ceftriaxone 0.12 -0.12 5 ---Cefuroxime (Screen) ---5 24 -23 General notes: S ≤ I R > S ≥ I R ≤ Zone diameter breakpoints relate to the MIC breakpoint of 0.03mg/L as no data for the intermediate category are currently available.
Nalidixic acid (Screen) ---30 32 10-31 9 S ≤ I R > S ≥ I R ≤ Azithromycin 0.25 0.5 0.5 15 28 -27 Zone diameter breakpoints relate to the MIC breakpoints of >0.5mg/L as disc diffusion will not reliably differentiate between the intermediate and susceptible populations.
S ≤ I R > S ≥ I R ≤ Tetracycline 0.5 1 1 10 32 27-31 26 No disc diffusion data to distinguish the intermediate category available at present.
The tetracycline result may be used to infer susceptibility to doxycycline. Isolates susceptible to tetracycline are also susceptible to doxycycline and minocycline. S ≤ I R > S ≥ I R ≤ Spectinomycin 64 -64 25 14 -13 Cephalosporins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance For general susceptibility testing in N. gonorroheae please see: Macrolides, lincosamides & streptogramins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Neisseria gonorrhoeae BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood (ISA + %5 horse blood + 20mg/L NAD may also be used) Inoculum: McFarland 0.5, no dilution Incubation: CO2, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the front of the plate.
Quality control: Neisseria gonorrhoeae ATCC 49226 OR Staphylococcus aureus NCTC 6571 Penicillins 0.03 0.06 Quinolone resistance is most reliably detected with nalidixic acid; however there are a few isolates that are resistant to ciprofloxacin yet susceptible to nalidixic acid in disc diffusion tests. The mechanism of resistance and prevalence of these isolates in the UK is still under investigation. Isolates with reduced susceptibility to fluoroquinolones normally have no zone of inhibition with a 30ug nalidixic acid disc. For organisms with nalidixic acid zone diameters 10-31mm a ciprofloxacin MIC should be determined if the patient is to be treated with this agent.
0.06 1 29 -28 Quinolones MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Ciprofloxacin Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Tetracyclines Although cefuroxime is not recommended for clinical use, it can be used as an indicator antibiotic to detect reduced susceptibility to other oxyamino cephalosporins. For organisms with reduced zones to cefuroxime an MIC determination is needed to confirm susceptibility to ceftriaxone, cefotaxime and cefixime.
Agent specific notes: Reporting guidance Reporting guidance MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance 24 S ≤ I R > S ≥ I R ≤ Ampicillin ---2 32 -31 Amoxicillin ---2 30 -29 Penicillin 0.06 0.12-0.25 0.25 1 unit 29 15-28 14 S ≤ I R > S ≥ I R ≤ Cefotaxime 0.12 -0.12 5 40 -39 Ceftriaxone 0.12 -0.12 5 40 -39 General notes: S ≤ I R > S ≥ I R ≤ Zone diameter breakpoints relate to the MIC breakpoint of 0.03mg/L as no data for the intermediate category are currently available.
Quinolone resistance is most reliably detected with nalidixic acid. Isolates with reduced susceptibility to fluoroquinolones normally have no zone of inhibition with a 30ug nalidixic acid disc. S ≤ I R > S ≥ I R ≤ Chloramphenicol 2 4 4 10 20 -19 Zone diameter breakpoints relate to the MIC breakpoint of 2mg/L as insufficient data to distinguish the intermediate category are currently available.
Rifampicin 0.25 -0.25 2 30 -29 Epidemiological breakpoint based on an MIC breakpoint of 0.25mg/L.
Reporting guidance Ampicillin and amoxicillin are used as indicator antibiotics to detect reduced susceptibility to penicillin. The recommendations given are for this purpose only; ampicillin and amoxicillin should not be used therapeutically.
EUCAST MIC breakpoints are ≤0.12mg/L, R>1mg/L. Currently there are no BSAC MIC breakpoints and zone diameter breakpoints relating to the presence of specific mutations in the penA gene.
Cephalosporins MIC breakpoint (mg/L) Neisseria meningitidis BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood (ISA + %5 horse blood + 20mg/L NAD may also be used) Inoculum: McFarland 0.5, dilute 1:10 Incubation: CO2, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the front of the plate.
Quality control: Neisseria gonorrhoeae ATCC 49226 OR Staphylococcus aureus NCTC 6571 Reporting guidance Penicillins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Agent specific notes: Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Quinolones MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Reporting guidance -31 Ciprofloxacin 0.03 0.06 0.06 1 32 25 S ≤ I R > S ≥ I R ≤ Ampicillin 1 -1 2 18 -17 Amoxicillin 2 -2 2 14 -13 Co-amoxiclav 2 -2 2/1 14 -13 Isolates susceptible to co-amoxiclav are also susceptible to piperacillin and piperacillin/tazobactam.
General notes: S ≤ I R > S ≥ I R ≤ Cefotaxime 0.12 -0.12 5 25 -24 Ceftriaxone 0.12 -0.12 30 25 -24 Cefuroxime 1 2 2 5 17 -16 Zone diameter breakpoints relate to the MIC breakpoint of 1mg/L as no data for the intermediate category are currently available.
S ≤ I R > S ≥ I R ≤ Ertapenem 0.5 -0.5 10 33 -32 Imipenem 2 -2 10 23 -22 Meropenem (infection other than meningitis) 2 -2 10 23 -22 Meropenem (Meningitis) 1 0.5-1 0.25 ----For use in meningitis detemine MIC.
S ≤ I R > S ≥ I R ≤ Ciprofloxacin 0.5 -0.5 1 28 -27 Levofloxacin 1 -1 1 20 -19 Moxifloxacin 0.5 -0.5 1 18 -17 Nalidixic acid ---30 ---Ofloxacin 0.5 -0.5 5 37 -26 Isolates susceptible to ampicillin/amoxicillin are also susceptible to piperacillin and piperacillin/tazobactam. Susceptibility to amoxicillin can be inferred from ampicillin.
Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Strains may be resistant to penicillins, aminopenicillins and/or cephalosporins due to changes in PBPs (βLNAR β-lactamase negative ampicillin resistant) and a few strains have both resistance mechanisms (βLPACR, β-lactamase positive, co-amoxiclav resistant).
Haemophilus influenzae BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood + 20mg/L NAD Inoculum: McFarland 0.5, dilute 1:100 Incubation: CO2, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the front of the plate.
Quality control: Haemophilus influenzae ATCC 49247 OR NCTC 11931 Penicillins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Reporting guidance Carbapenems MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Agent specific notes: Reporting guidance Quinolones MIC breakpoint (mg/L) Always test for β-lactamase; β-lactamase positive isolates should always be reported resistant. Breakpoints refer to β-lactamase negative isolates only.
Cephalosporins MIC breakpoint (mg/L) Disk content (µg) Disk content (µg) Zone diameter breakpoint (mm) Meropenem is the only agent used for meningitis.
Quinolone resistance is most reliably detected in tests with nalidixic acid. Strains with reduced susceptibility to fluoroquinolones give no zone of inhibition with a 30ug nalidixic acid disc.
26 Haemophilus influenzae BSAC, Version 14, January 2015 S ≤ I R > S ≥ I R ≤ Azithromycin --4 15 --19 Clarithromycin --32 5 --8 Erythromycin can be used to determine susceptibility to azithromycin and clarithromycin.
Erythromycin --16 5 --14 Telithromycin --8 15 --15 S ≤ I R > S ≥ I R ≤ Tetracycline 1 2 2 10 22 18-21 17 Isolates susceptible to tetracycline are also susceptible to doxycycline and minocycline. Some isolates resistant to tetracyline may be susceptible to minocycline and/or doxycycline. S ≤ I R > S ≥ I R ≤ Chloramphenicol 2 -2 10 25 -24 Co-trimoxazole 0.5 1 1 25 21 18-20 17 See link in index tab.
Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Agent specific notes: Reporting guidance MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Tetracyclines Correlation between macrolide MICs and clinical outcome is weak for H. influenzae . Therefore breakpoints for macrolides and related antibiotics have been set to categorise "wild type" H. influenzae as intermediate.
Macrolides, lincosamides & streptogramins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) 27 S ≤ I R > S ≥ I R ≤ Ampicillin 1 -1 10 30 -29 Co-amoxiclav 1 -1 ----Penicillin 0.5 -0.5 1 unit 22 -21 Any resistant isolate should be confirmed by MIC method.
Resistant isolates are rare.
General notes: S ≤ I R > S ≥ I R ≤ Cefotaxime 0.03 -0.03 ----S ≤ I R > S ≥ I R ≤ Ciprofloxacin 0.06 -0.06 ----Nalidixic acid ---30 28 -27 S ≤ I R > S ≥ I R ≤ Tetracycline 2 -2 ----Tetracyclines MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Quinolone resistance is most reliably detected in tests with nalidixic acid discs.
Quinolones MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Cephalosporins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Pasteurella multocida BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood + 20mg/L NAD Inoculum: McFarland 0.5, dilute 1:100 Incubation: CO2, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the front of the plate.
Quality control: Pasteurella multocida NCTC 8489 Penicillins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance 28 S ≤ I R > S ≥ I R ≤ Ciprofloxacin 0.5 -0.5 1 26 -25 Nalidixic acid ---30 20 -19 S ≤ I R > S ≥ I R ≤ Erythromycin 4 -4 5 22 -21 The susceptibility of clarithromycin can be inferred from the erythromycin result.
Quinolone resistance is most reliably detected in tests with nalidixic acid discs.
Macrolides, lincosamides & streptogramins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Reporting guidance Campylobacter species BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood (ISA + %5 horse blood + 20mg/L NAD may also be used Inoculum: McFarland 0.5, no dilution Incubation: microaerophilic conditions, 42ºC, 24h Reading: Read zone edges as the point showing no growth viewed from the front of the plate.
Quality control: Staphylococcus aureusNCTC 6571 or ATCC 25923, For target zone sizes see: Quinolones MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: 29 S ≤ I R > S ≥ I R ≤ Gentamicin 1 -1 ----S ≤ I R > S ≥ I R ≤ Penicillin 0.12 -0.12 1 unit 20 -19 S ≤ I R > S ≥ I R ≤ Ciprofloxacin 0.5 1 1 1 17 12-16 11 The zone diameter relates to an MIC breakpoint of 0.5mg/L as no data for the internediate category are currently available. Moxifloxacin 0.5 -0.5 ----S ≤ I R > S ≥ I R ≤ Vancomycin 2 -2 5 20 -19 S ≤ I R > S ≥ I R ≤ Clindamycin 0.5 -0.5 ----S ≤ I R > S ≥ I R ≤ Tetracycline 2 -2 ----S ≤ I R > S ≥ I R ≤ Linezolid 2 -2 ----Rifampicin 0.06 -0.5 ----Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Tetracyclines MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Macrolides, lincosamides & streptogramins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Glycopeptides MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Quinolones MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance β-lactams MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Corynebacterium spp (except C.diphtheriae ) BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood + 20mg/L NAD Inoculum: McFarland 0.5, dilute 1:10 Incubation: CO2, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the back of the plate against a dark background illuminated with reflected light.
Quality control: Staphylococcus aureus NCTC 6571 or ATCC 25923 Aminoglycosides MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance 30 S ≤ I R > S ≥ I R ≤ Amoxicillin 0.5 1-2 2 ----Ampicillin 0.5 1-2 2 ----Co-amoxiclav 4 8 8 30 29 21-28 20 Zone diameter breakpoints are for B. fragilis only.
Susceptibility to ampicillin, amoxicillin and piperacillin ± tazobactam can be inferred from susceptibility to penicillin.
B. fragilis is inherently resistant to penicillin.
Piperacillin 16 -16 ----Zone diameter breakpoints are for B. fragilis only.
The zone diameter breakpoint relates to an MIC of 8mg/L as no data for the intermediate category are currently available.
Ticarcillin 16 -16 ----Ticarcillin-clavulanate 8 16 16 ----S ≤ I R > S ≥ I R ≤ Doripenem 1 -1 ----Ertapenem 1 -1 ----Imipenem 2 4-8 8 ----Meropenem 2 4-8 8 10 26 19-25 18 Zone diameter breakpoints are for B. fragilis and B. thetaiotaomicron only.
S ≤ I R > S ≥ I R ≤ Clindamycin 4 -4 2 10 -9 Zone diameter breakpoints are for B. fragilis and B. thetaiotaomicron only.
The breakpoints are based on "wild type" susceptible population as there are few clinical data relating MIC to outcome. Organisms that appear resistant in disc diffusion tests should have resistance confirmed by MIC determination and resistant isolates be sent to the Anaerobe Reference Unit, Public Health Wales, Cardiff.
Macrolides, lincosamides & streptogramins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Carbapenems MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance ---26 75/10 27 -Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance The breakpoints are based on "wild type" susceptible population as there are few clinical data relating MIC to outcome. Organisms that appear resistant in disc diffusion tests should have resistance confirmed by MIC determination and resistant isolates be sent to the Anaerobe Reference Unit, Public Health Wales, Cardiff.
Piperacillin-tazobactam 8 16 16 --0.5 Penicillin 0.25 Gram-negative anaerobes (incl. Bacteroides species) BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood + 20mg/L NAD Inoculum: McFarland 0.5, dilute 1:100 Incubation: 10% CO2 /10%H2 /80% N2, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the front of the plate.
Quality control: Bacteroides fragilis NCTC 9343 Penicillins MIC breakpoint (mg/L) 31 Gram-negative anaerobes (incl. Bacteroides species) BSAC, Version 14, January 2015 S ≤ I R > S ≥ I R ≤ Chloramphenicol 8 -8 ----Metronidazole 4 -4 5 18 -17 Zone diameter breakpoints are for B. fragilis and B. thetaiotaomicron only.
Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance 32 S ≤ I R > S ≥ I R ≤ Daptomycin --4 ----MIC breakpoint based on the ECOFF for the "wild type" population.
Not used clinically. May be tested for epidemiological purposes only.
Fusidic acid --2 ----MIC breakpoint based on the ECOFF for the "wild type" population.
Not used clinically. May be tested for epidemiological purposes only.
Metronidazole 2 -2 ----Breakpoints are based on epidemiological "cut off" values (ECOFFs) which distinguish "wild type" isolates from those with reduced susceptibility.
Moxifloxacin --4 ----MIC breakpoint based on the ECOFF for the "wild type" population.
Not used clinically. May be tested for epidemiological purposes only.
Tigecycline 0.25 ------MIC breakpoint based on the ECOFF for the "wild type" population.
Not used clinically. May be tested for epidemiological purposes only.
Rifampicin 0.004 ------MIC breakpoint based on the ECOFF for the "wild type" population.
Not used clinically. May be tested for epidemiological purposes only.
Vancomycin 2 -2 ----Breakpoints are based on epidemiological "cut off" values (ECOFFs) which distinguish "wild type" isolates from those with reduced susceptibility.
Clostridium dificile BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood + 20mg/L NAD Inoculum: McFarland 0.5, dilute 1:100 Incubation: 10% CO2 /10%H2 /80% N2, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the front of the plate.
Quality control: Bacteroides fragilis NCTC 9343 Antimicrobial MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance 33 S ≤ I R > S ≥ I R ≤ Amoxicillin 4 8 8 ----Ampicillin 4 8 8 ----Co-amoxiclav 4 8 8 Zone diameter breakpoints are for C. perfringens only.
The breakpoints are based on "wild type" susceptible population as there are few clinical data relating MIC to outcome. Organisms that appear resistant in disc diffusion tests should have resistance confirmed by MIC determination and resistant isolates be sent to the Anaerobe Reference Unit, Public Health Wales, Cardiff.
The zone diameter breakpoint relates to an MIC of 0.25mg/L as no data for the intermediate category are currently available.
Susceptibility to ampicillin, amoxicillin, co-amoxiclav and piperacillin ± tazobactam can be inferred from susceptibility to penicillin.
Piperacillin 8 16 16 ----Piperacillin-tazobactam 8 16 16 The breakpoints are based on "wild type" susceptible population as there are few clinical data relating MIC to outcome. Organisms that appear resistant in disc diffusion tests should have resistance confirmed by MIC determination and resistant isolates be sent to the Anaerobe Reference Unit, Public Health Wales, Cardiff.
Ticarcillin 8 16 16 ----Ticarcillin-clavulanate 8 16 16 ----S ≤ I R > S ≥ I R ≤ Doripenem 1 -1 ----Ertapenem 1 -1 ----Imipenem 2 4-8 8 ----Meropenem 2 4-8 8 10 26 19-25 18 Zone diameter breakpoints are for C. perfringens only.
S ≤ I R > S ≥ I R ≤ Vancomycin 2 -2 ----Gram positive anaerobes except Clostridium dificile BSAC, Version 14, January 2015 Disk diffusion method for Antimicrobial Susceptibility testing Medium: Iso-Sensitest agar supplemented with 5% defibrinated horse blood + 20mg/L NAD Inoculum: McFarland 0.5, dilute 1:10 Incubation: 10% CO2 /10%H2 /80% N2, 36±1ºC, 19±1h Reading: Read zone edges as the point showing no growth viewed from the front of the plate.
Quality control: Clostridium perfringens NCTC 8359 Penicillins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance -22 Penicillin 0.25 0.5 0.5 1 unit 23 Carbapenems MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Glycopeptides MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance 34 Gram positive anaerobes except Clostridium dificile BSAC, Version 14, January 2015 S ≤ I R > S ≥ I R ≤ Clindamycin 4 -4 2 10 -9 Zone diameter breakpoints are for C. perfringens only.
The breakpoints are based on "wild type" susceptible population as there are few clinical data relating MIC to outcome. Organisms that appear resistant in disc diffusion tests should have resistance confirmed by MIC determination and resistant isolates be sent to the Anaerobe Reference Unit, Public Health Wales, Cardiff.
S ≤ I R > S ≥ I R ≤ Chloramphenicol 8 -8 ----Metronidazole 4 -4 5 18 -17 Zone diameter breakpoints are for C. perfringens only.
Macrolides, lincosamides & streptogramins MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance Miscellaneous antimicrobials MIC breakpoint (mg/L) Disk content (µg) Zone diameter breakpoint (mm) Agent specific notes: Reporting guidance 35 1 2 3 4 5 Direct susceptibility tests on urine samples may be interpreted only if the inoculum gives semi-confluent growth. In the absence of definitive organism identification, use the recommendations most appropriate for the presumptive identification, accepting that on some occasions the interpretation may be incorrect. A more cautious approach is to use the systemic recommendations.
UTI related comments UTI recommendations are for organisms associated with uncomplicated urinary infections only. For complicated UTI systemic recommendations should be used.
If an organism is isolated from multiple sites, for example from blood and urine, interpretation of susceptibility should be made with regard to the systemic site (e.g., if the blood isolate is resistant and the urine isolate susceptible, both should be reported resistant irrespective of the results obtained using interpretative criteria for urine isolates).
For agents not listed, criteria given for systemic isolates may be used for urinary tract isolates. Intermediate susceptibility infers that the infection may respond as the agent is concentrated at the site of infection.
36 1 2 3 4 5 6 7 8 9 10 Principles for Reporting Susceptibility to Antimicrobial Agents Reporting is one of the most important parts of the service, as what a laboratory releases makes a difference to the prescribing of antimicrobial agents.
Ensure reporting is in line with local guidance on the use of antimicrobial agents. Report all clinically-relevant resistances for significant pathogens.
Report results for relevant antimicrobial agent(s) that the requestor has stated are in use, unless clinically inappropriate.
Take note of restrictions for special patient groups when reporting (e.g. tetracyclines not to be used in pregnancy or for children) Reporting should aim to reduce antimicrobial resistance and C. difficile through reducing selective pressures and targeting the most appropriate treatment for each organism reported.
The order in which the laboratory reports susceptibility results is important, as prescribers will tend to choose the first listed.
Inform clinicians that susceptibility results for further antimicrobial agents may be available.
Whenever possible, always include a susceptibility result for a non-β-lactam agent, so there is always a treatment option for those with penicillin allergy.
Whenever possible and appropriate include results for antimicrobial agents that can be given orally.
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运营两年，亚投行交出的这份成绩单令国人自豪！
运营两年，亚投行交出的这份成绩单令国人自豪！
什么是众数
众数是指一组数据中出现次数最多的那个数据，一组数据可以有多个众数，也可以没有众数。众数是由英国统计学家皮尔生首先提出来的。所谓众数是指社会经济现象中最普遍出现的标志值。从分布角度看，众数是具备明显集中趋势的数值。
例如：某制鞋厂要了解消费者最需要哪种型号的男皮鞋，调查了某百货商场某季度男皮鞋的销售状况，得到资料如下表（某商场某季度男皮鞋销售状况）：
| 男皮鞋号码/厘米 | 销售量/双 |
--- |
| 24.0 | 12 |
| 24.5 | 84 |
| 25.0 | 118 |
| 25.5 | 541 |
| 26.0 | 320 |
| 26.5 | 104 |
| 27.0 | 52 |
| 合计 | 1200 |
从表中可以看到，25.5厘米的鞋号销售量最多，如果我们计算算术平均数，则平均号码为25.65厘米，而这个号码显然是没有实际意义的，而直接用25.5厘米作为顾客对男皮鞋所需尺寸的集中趋势既便捷又符合实际。
统计上把这种在一组数据中出现次数最多的变量值叫做众数。用Mo表示。它紧要用于定类（品质标志）数据的集中趋势，当然也适用于作为定序（品质标志）数据以及定距和定比（数量标志）数据集中趋势的测度值。上面的例子中，鞋号25.5厘米就是众数。
众数的特点
1、众数是以它在所有标志值中所处的位置确定的全体单位标志值的代表值，它不受分布数列的极大或极小值的影响，从而增强了众数对分布数列的代表性。
2、当分组数列没有任何一组的次数占多数，也即分布数列中没有明显的集中趋势，而是近似于均匀分布时，则该次数分配数列无众数。若将无众数的分布数列重新分组或各组频数依序合并，又会使分配数列再现出明显的集中趋势。
3、如果与众数组相比邻的上下两组的次数相等，则众数组的组中值就是众数值；如果与众数组比邻的上一组的次数较多，而下一组的次数较少，则众数在众数组内会偏向该组下限；如果与众数组比邻的上一组的次数较少，而下一组的次数较多，则众数在众数组内会偏向该组上限。
4、缺乏敏感性。这是由于众数的计算只利用了众数组的数据信息，不象数值平均数那样利用了全部数据信息。
众数的意义
第一，众数是一组数据中出现次数最多的数据，是一组数据中的原数据，而不是相应的次数。
第二，一组数据中的众数可能不存在。
第三，众数粗糙，但众数不受个别数据的影响，可在数据缺陷较大或需要快速而粗略地寻求一组数据的代表值时用。
众数的计算
由品质数列和单项式变量数列确定众数比较容易，哪个变量值出现的次数最多，它就是众数，如上面的两个例子。
若所掌握的资料是组距式数列，则只能按一定的方式来推算众数的近似值。计算公式为：
式中：
L——众数所在组下限； U——众数所在组上限； ——众数所在组次数与其下限的邻组次数之差； ——众数所在组次数与其上限的邻组次数之差； d——众数所在组组距。
例:根据下表的数据，计算50名工人日加工零件数的众数。
解：从表中的数据可以看出，最大的频数值是14，即众数组为120~125这一组，根据公式得50名工人日加工零件的众数为：
(件)
或：(件)
众数是一种位置平均数，是总体中出现次数最多的变量值，因而在实际工作中有时有它特殊的用途。诸如，要说明一个企业中工人最普遍的技术等级，说明消费者需要的内衣、鞋袜、帽子等最普遍的号码，说明农贸市场上某种农副产品最普遍的成交价格等，都需要利用众数。但是必须注意，从分布的角度看，众数是具备明显集中趋势点的数值，一组数据分布的最高峰点所对应的数值即为众数。当然，如果数据的分布没有明显的集中趋势或最高峰点，众数也可能不存在；如果有两个最高峰点，也可以有两个众数。只有在总体单位比较多，而且又明显地集中于某个变量值时，计算众数才有意义。
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4904 | https://people.maths.ox.ac.uk/greenbj/papers/bourgain-sumset.pdf | On Arithmetic Progressions in Sums of Sets of Integers Jean Bourgain An exposition by Ben Green, May 2000 Contents 1 Introduction 1 2 Dissociated Sets 3 3 Proof of Theorem 1 9 3.1 Estimation of the g1 term. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 3.2 Estimation of the g3 term. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 3.3 Estimation of the g2 term. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 3.4 Putting everything together. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 1 Introduction This is an exposition of yet another brilliant paper by Bourgain, “Arithmetic Progressions in Sums of Sets of Integers” . In this amazingly short paper Bourgain proves the following result.
Theorem 1 Let A, B ⊆{1, . . . , N} be sets with |A| = αN and |B| = βN. Then there is a constant c = c(α, β) such that the sumset A + B contains an arithmetic progression of length ec(log n)1/3.
The result itself is clearly very interesting, but more interesting still is the method of proof. Much inspiration is taken from the theory of thin sets in harmonic analysis, and indeed it is almost impossible to understand this paper without delving into some texts on that subject such as or . Even then the necessary results can only be extracted with some difficulty, and one of our main objectives in writing this exposition is to give an account accessible to those with a background in additive number theory.
We begin by giving an overview of the proof of Theorem 1. The first thing to appreciate is the fundamental difference between this result and the result of Roth on arithmetic progressions of length 3. To prove Roth’s result the basic idea is to pass to a subprogression on which our set has increased density, and then iterate the argument. Such an approach cannot work here for the very obvious reason that a subset of A + B is unlikely to be a set of form A′ + B′, and so it is hard to see what an inductive hypothesis might be. Instead we must locate a long arithmetic progression straight away.
It is possible to conceive of a variety of ways in which we might find such a progression, and Bourgain uses the following clever approach. We begin by making the problem modular, so that harmonic analysis can be done more easily. To do this regard A and B as subsets of Zp, where p > 2N is a prime number. Let f(x) = A ∗B(x), where we are using convolutions in the conventional way 1 (so that A ∗B(x) = P y A(y)B(x −y) = |(a, b) ∈A × B : a + b = x|). Observe that, because of the choice of p, the modular version of f is the same as the Z-version. Let K be a positive integer. If we could find x, y for which max 0<k<K |f(x + ky) −f(x)| < f(x), then f(x+ky) would be positive for all x = 0, 1, . . . , K −1 and we would have a modular arithmetic progression of length K lying in A + B. It is easy to see that we can find a genuine (i.e. Z-) progression of length 1 2 √ K from this.
How are we going to find such an x and y? It seems reasonable that we are going to need to look at the arithmetic structure of A+B more closely to find y, but we might well get away with averaging over x. That is to say, we shall look for y such that X x max 0<k<K |f(x + ky) −f(x)| < X x f(x).
If we can show, for K ∼ec(log n)1/3, that such a y can be found then Theorem 1 will be proved.
As everyone knows, one of the best ways of isolating arithmetic information about a function is by looking at its Fourier coefficients. This is particularly true when we are looking for some sort of bias along an arithmetic progression as is the case here. Examples of such techniques are the classical result of Roth mentioned earlier, and the improvements of that result due to Heath-Brown and Szemer´ edi . In fact, readers familiar with these works will notice a number of similarities with the current paper.
Roth, Heath-Brown and Szemer´ edi all look for large Fourier Coefficents ˆ f(r)(r ∈R) and then procced to show that there is a bias of f along a progression with common difference d, where rd (mod p) is small for all r ∈R. The main difference between the approaches is that Roth only considers the case |R| = 1. Such an approach is of course extremely natural because if f is the characteristic function of a small arithmetic progression with common difference d then ˆ f(r) is large precisely when rd (mod p) is small.
Bourgain goes a step further. Suppose that the large Fourier Coefficients of f are ˆ f(r)(r ∈R) (for some suitable definition of “large”). Suppose for a moment that f is the characteristic function of a short AP with common difference d. Then R is the set of r for which rd (mod p) is small, and is itself an arithmetic progression. One can concieve of making the jump from this observation to the following rough statement, where f is now A ∗B again.
(∗) If we are looking for a bias of f along an arithmetic progression, then the “relevant” large Fourier coefficients ˆ f(r) ought to be those with r lying in some set R which has lots of structure.
What Bourgain does, then, is take all the large Fourier coefficients of f, and then isolate a highly structured subset of them. This subset will be used to find an appropriate arithmetic progression on which f is biased. This procedure is carried out in such a way that the other large Fourier coefficients are extremely unstructured, and their contribution has the status of an error term. It is 2 in dealing with the unstructured contribution that we must appeal to the ideas of and . The exact notion of “unstructured” that we shall use is the concept of dissociativity, which we discuss at length now.
2 Dissociated Sets At this point we take a fairly lengthy detour into the world of dissociated sets. Fix an integer N.
Then a set E ⊆ZN is said to be dissociated if no x has more than one representation as P i ǫixi, where the xi are distinct elements of E and ǫi ∈{−1, 1}. Although this is a purely additive number theory definition, the concept of dissociated set arises quite naturally in the study of sparse trigonometric series. See, for example, (Chapter 5) or .
We note that has a slightly weaker notion of dissociated set (that is to say not all of Bourgain’s dissociated sets are dissociated in our sense). We have not been able to prove the results that follow for this weaker notion. However it turns out that a small modification to a later part of enables one to get away with our rather more restrictive definition.
In the following we work in ZN. In particular the hat symbol refers to Fourier transforms in that group, so that if f : Z →C is a function then we set ˆ f(r) = X x f(x)ω−rx, where ω = e2πi/N.
From now on we fix a dissociated set E ⊆ZN. Let f : Z →C be a function supported on E, and for any positive integer s write f ⊕s(x) = X (x1,...,xs)∈Es x1+x2+···+xs=x f(x1) . . . f(xs) for the s-fold convolution of f with itself. It is easy to check that the Fourier transform (f ∗g)ˆ(r) is simply ˆ f(r)ˆ g(r), and hence that f ⊕s has transform ˆ f(r)s. Now suppose it were the case that every x ∈ZN had at most one representation in the form x1 + · · · + xs (xi ∈E), at least up to reordering of the summands. Then we would have, using Parseval’s identity, ∥ˆ f∥2s 2s = X x f ⊕s(x) 2 ≤ s!
X x X (x1,...,xs)∈Es x1+x2+···+xs=x |f(x1)|2 . . . |f(xs)|2 = s!∥ˆ f∥2s 2 .
This immediately gives the rather strong-looking inequality ∥ˆ f∥2s ≤√s∥ˆ f∥2.
(1) 3 Alas that derivation was based on the (in general) false supposition that no x has two essentially distinct representations in the form x1+· · ·+xs. Such a statement does not follow from the fact that E is dissociated, because these representations are allowed to have repeated summands, a situation which is not covered by the definition of dissociativity. One still feels, however, that a statement similar to (1) ought to be true. A reason for this is that dissociativity implies that no x can have two different “good” representations as x1 + · · · + xs with the xi all distinct, but if E is at all large then most representations are good.
A possible approach, then, would be to prove that the contribution to ∥ˆ f∥2s from “bad” represen-tations is small. I cannot make such an approach give more than the weaker bound ∥ˆ f∥2s ≪s∥ˆ f∥2 and I believe that there is a good reason for this, namely that it is hard to make such an approach use any of the information about sums x1 + · · · + xr (r > s) that dissociativity gives us. A similar situation is considered in Ruzsa , pages 270–271. In that paper Ruzsa considers sets in which no integer has two representations as x1 + · · · + xs with x1, . . . , xs distinct and s fixed. There is no interesting information about sums x1 + · · · + xr with r > s, and Ruzsa gets a bound of shape ∥ˆ f∥2s ≪s∥ˆ f∥2. It is interesting to observe that such a bound still seems to give non-trivial infor-mation along the lines of Theorem 1, but with the exponent 1/3 replaced by 1/4.
We are going to prove that a considerably cleverer approach will yield an inequality comparable to (1).
Theorem 2 Let E ⊆ZN be dissociated. Let f : ZN →C be supported on E. Then we have, for all real numbers p ≥2, the inequality ∥ˆ f∥p ≤2√p∥ˆ f∥2.
In there appears a similar inequality in which the factor 2 is replaced by 10. This would suggest to me that one can work with Bourgain’s weaker definition of dissociativity, but that I have not been clever enough to do so. The first step towards a proof of Theorem 2 is the following standard Lemma.
Lemma 3 (Young’s Inequality) Let m be a positive integer, and let f, g : ZN →C be two functions. Then ∥f ∗g∥m ≤N∥f∥1∥g∥m.
4 Proof We have N∥f ∗g∥m = X x X y f(y)g(x −y) m ≤ X x X y1 · · · X ym |f(y1)| . . . |f(ym)||g(x −y1)| . . . |g(x −ym)| = X y1 · · · X ym |f(y1)| . . . |f(ym)| X x |g(x −y1)| . . . |g(x −ym)| ≤ X y1 · · · X ym |f(y1)| . . . |f(ym)| X x |g(x −y1)|m !1/m . . . X x |g(x −ym)|m !1/m = N m∥f∥m 1 · N∥g∥m m.
The result follows immediately.
□ Now we come to the key part of the proof of Theorem 2. Our strategy will be as follows. We shall consider “twists” of f by functions ǫ : E →T. Writing fǫ(x) = f(x)ǫ(x) we shall prove in Proposition 4 that, for any positive integer m, one has ∥ˆ f∥m ≤2∥ˆ fǫ∥m. This is a rather surprising result which depends heavily on the dissociativity of E. Next, in Proposition 6, we shall show that one can choose ǫ such that the function fǫ is well-behaved and, in particular, satisfies ∥ˆ fǫ∥m ≪√m∥ˆ fǫ∥2 = √m∥ˆ f∥2.
This is considerably less surprising because one feels that ǫ can be chosen so that contributions from the “bad” representations cancel out.
Proposition 4 Let m be a positive integer, and let ǫ : E →T. Define fǫ(x) = f(x)ǫ(x) as above.
Then ∥ˆ f∥m ≤2∥ˆ fǫ∥m.
Proof It is far from clear to me how one is supposed to come up with a proof like this. Define pǫ(x) = 2 Y r∈E 1 2ǫ(r)ω−rx + 1 + 1 2ǫ(r)ωrx .
This object is, as the reader may appreciate, some kind of Riesz product. The first thing to observe is that pǫ is a product of non-negative real factors, and so ∥pǫ∥1 is just 1 N P x pǫ(x). If one expands out the product for pǫ in full one gets a sum of terms of the form 21−kǫ(r1) . . . ǫ(rl)ǫ(rl+1) . . . ǫ(rk)ω(r1+···+rl−rl+1−···−rk)x, where the ri are distinct.
When one sums over x all such terms disappear except those with r1 + · · · + rl −rl+1 −· · · −rk = 0. Since E is dissociated there is only one such term, corresponding to k = l = 0. It follows immediately that ∥pǫ∥1 = 2.
Very similar considerations allow us to find ˆ pǫ(r) for certain r. Suppose that r ∈E. Then ˆ pǫ(r) is a sum of terms of the form 21−kǫ(r1) . . . ǫ(rl)ǫ(rl+1) . . . ǫ(rk)ω(r1+···+rl−rl+1−···−rk−r)x.
5 When we sum over x the only terms that remain are those with r1+· · ·+rl−rl+1−· · ·−rk = r. The dissociativity of E tells us that this can only happen if l = k = 1 and r1 = r , and it follows quickly that ˆ pǫ(r) = Nǫ(r). This information can be used to establish the following interesting result.
Claim 5 We have, for all r, ˆ fǫ ∗pǫ = N ˆ f.
Proof We check this by taking Fourier transforms of both sides. The Fourier transfrom of ˆ fǫ ∗pǫ is ˆ ˆ fǫ(x) ˆ pǫ(x), and it is easy to check that ˆ ˆ fǫ(x) = Nf(−x)ǫ(−x).
This is zero unless x ∈−E, and for such values we can say what ˆ pǫ is: indeed since pǫ is real we have, when x ∈−E, that ˆ pǫ(x) = ˆ pǫ(−x) = Nǫ(−x).
Putting these remarks together gives ( ˆ fǫ ∗pǫ)ˆ(x) = N 2f(−x) for all x. It is a simple matter to check that this equals the Fourier transform of N ˆ f, thereby establishing the claim.
□ It is now an easy matter to prove Proposition 4, for using Young’s Inequality we have N∥ˆ f∥m = ∥ˆ fǫ ∗pǫ∥m ≤N∥ˆ fǫ∥m∥pǫ∥1 = 2N∥ˆ fǫ∥m.
We turn now to the issue of finding an ǫ : E →T for which ∥ˆ fǫ∥m is not large, and we hope the reader will agree that it would be a surprise if we were not to do this by averaging over a suitable set of ǫ. We present two arguments of which the second is a little more complicated, but contains some interesting points.
Proposition 6 Let Ω= T|E| be the group of all possible maps ǫ. Let Ωhave normalised Haar measure µ, and let s be a positive integer. Then Z Ω ∥ˆ fǫ∥2s 2s dµ ≤s!∥ˆ f∥2s 2 .
Proof We have | ˆ fǫ(x)|s = X m X r1+···+rs=m f(r1) . . . f(rs)ǫ(r1) . . . ǫ(rs) !
ω−mx and so, by Parseval’s identity, ∥ˆ fǫ∥2s 2s = X m X r1+···+rs=m f(r1) . . . f(rs)ǫ(r1) . . . ǫ(rs) 2 = X m X r1+···+rs=m X t1+···+ts=m f(r1) . . . f(rs)f(t1) . . . f(ts)ǫ(r1) . . . ǫ(rs)ǫ(t1) . . . ǫ(ts).
6 Now it is quite easy to check that if r1, . . . , rs and t1, . . . , ts are fixed then Z Ω ǫ(r1) . . . ǫ(rs)ǫ(t1) . . . ǫ(ts) dµ is zero unless the ri are a rearrangement of the ti, in which case it equals 1. It follows from (2) and the Cauchy-Schwarz inequality that Z Ω ∥ˆ fǫ∥2s 2s dµ ≤ s!
X m X r1+···+rs=m |f(r1)|2 . . . |f(rs)|2 = s!∥ˆ f∥2s 2 .
This completes the proof of the proposition.
□ In a moment we shall give an alternative derivation of something like Proposition 6, in which it is only necessary to consider maps ǫ : E →{−1, 1}. First, however, we complete the proof of Theorem 2. It is immediate from Propositions 4 and 6 that, for any positive integer s, we have ∥ˆ f∥2s ≤2(s!)1/2s∥ˆ f∥2 ≤2√s∥ˆ f∥2.
Now if p ≥2 is any real number then we certainly have ∥∥p ≤∥∥2s where s = ⌈p/2⌉≤(p + 2)/2.
Hence ∥ˆ f∥p ≤2 r p + 2 2 ∥ˆ f∥2 ≤2√p∥ˆ f∥2.
This completes the proof of Theorem 2.
□ We now give an alternative proof of Proposition 6 using only maps ǫ : E →{−1, 1}. This proof comes from the same circle of ideas as Khintchin’s Inequality, as discussed in . Let Ωbe the probability space consisting of all ǫ : E →{−1, 1}, chosen with equal probability. Our aim is to show that E∥ˆ fǫ∥p is not too large. If we try to copy the above proof it becomes necessary to show that Eǫ(r1) . . . ǫ(rs)ǫ(t1) . . . ǫ(ts) = 0 unless the ri are a rearrangement of the ti. Unfortunately this is simply false, and we must proceed in a different way. Let Xr denote the random variable Xr(ǫ) = X x f(x)ǫ(x)ω−rx, so that ∥ˆ fǫ∥p = P r Xp r N 1/p .
(2) Now Xr is itself a sum of N independent random variables Y (r) i , where Y (r) i takes the values ±f(x)ω−rx with equal probability. In such a situation the following inequality of Hoeffding tells us that Xr is strongly concentrated about its mean, 0.
7 Theorem 7 (Hoeffding’s Inequality) Let Z1, Z2, . . . , Zn be independent random variables such that ai ≤Zi ≤bi for all i and some real numbers ai, bi. Write Z = (Z1 + · · · + Zn)/n and µ = EZ.
Let t > 0. Then P Z −µ ≥t ≤e−2n2t2/ P i(bi−ai)2.
Applying this in the present situation gives, using Parseval’s Identity, P (|Xr| > λ) ≤eλ2/2∥ˆ f∥2 2.
It follows immediately that P (|Xr|p > λ) ≤2e−λ2/p/∥ˆ f∥2 2, and so E|Xr|p = Z ∞ 0 P (|Xr|p > λ) dλ ≤ 2 Z ∞ 0 e−λ2/p/∥ˆ f∥2 2 dλ = 2∥ˆ f∥p 2 · p 2 Z ∞ 0 u p 2 −1e−u du = pΓ(p/2)∥ˆ f∥p 2.
From (2) we have that E∥ˆ fǫ∥p p ≤pΓ(p/2)∥ˆ f∥p 2. This is, as claimed, rather similar to Proposition 6.
In Theorem 2 is applied through the following corollary. Before stating and proving this, we observe that Theorem 2 has a rather straightforward dual formulation. This states that if f : ZN → C is a function with Supp ˆ f ⊆E, where E is dissociated, then ∥f∥p ≤2√p∥f∥2 for all real numbers p ≥2. We shall call functions of this form “Fourier-dissociated”.
Corollary 8 Let f : ZN →C be Fourier-dissociated, and let I be a subset of ZN. For i ∈I denote by fi the translate of f by i, so that fi(x) = f(i + x). Then max i∈I |fi| 2 ≤5 p log |I|∥f∥2.
8 Proof This is best done without words. We have, for any p ≥2, max i∈I |fi| 2 = 1 N X x max i∈I |f(x + i)|2 ≤ 1 N X x X i∈I |f(x + i)|p !2/p ≤ N −2/p X x X i∈I |f(x + i)|p !2/p = |I|2/p∥f∥2 p ≤ 4p|I|2/p∥f∥2 2.
Take p = 2 log |I| and we get that this is at most 8e log |I|∥f∥2 2, which concludes the proof.
□ If the reader desires a few words on this then we might add that considering Lp norms for large p is an obvious tactic given that we are interested in an L∞object (i.e. maxi∈I |fi|). Of more relevance, perhaps, is some explanation of what this corollary is saying.
In some vague sense, Fourier-dissociated functions have very little additive structure.
Thus it is not unreasonable to expect that their translates should be rather regularly behaved (this would certainly not be the case if f were supported on some arithmetic progression, for example).
3 Proof of Theorem 1 The reader who has survived that exposition of dissociativity may wish, at this juncture, to read the introductory outline once more. Suppose that A, B ⊆ZN, where N is prime (N has taken the place of the prime p that we used earlier, because I like my Z’s to be subscripted with N’s, and also so that we can keep the notation of §2). Suppose that |A| = αN and |B| = βN, and let f = A ∗B.
Lemma 9 We have P x f(x) = N 2αβ and X r | ˆ f(r)| ≤N 2α1/2β1/2.
Proof The first statement is obvious. The second is an easy consequence of Cauchy-Schwarz and Parseval’s identity: X r | ˆ f(r)| = X r | ˆ A(r)|| ˆ B(r)| ≤ X r | ˆ A(r)|2 !1/2 X r | ˆ B(r)|2 !1/2 = N 2α1/2β1/2.
That concludes the proof of the lemma.
□ 9 Although rather trivial, this lemma is the reason the argument works. Indeed the reader may care to check that we use nothing more about the sumset A + B than that P r | ˆ f(r)| is small compared to ∥f∥1.
Recall from the introduction that we are trying to find K ∼ec(log n)1/3 and y ∈ZN such that X x max 0<k<K |f(x + ky) −f(x)| < X x f(x).
(3) Since this inequality is homogeneous in f, we can renormalise f and it turns out to be convenient to do so. We shall abandon the notion that f = A ∗B, and work only with the facts that X r | ˆ f(r)| ≤1 (4) and X x f(x) = p αβ.
(5) Recall that our plan was to identify some structure in the set of large Fourier coefficients ˆ f(r)(r ∈R).
As a first step, let ν be an integer to be chosen later and set Es = n r : 2−s ≤| ˆ f(r)| ≤2−s+1o for s = 1, . . . , ν and Esmall = n r : | ˆ f(r)| ≤2−vo .
We have divided up into powers of two because it gives us increased control which turns out to be crucial (and we can be fairly safe in the knowledge that the eventual bound will not be affected too greatly). The next lemma describes our method for dividing an arbitrary set into a highly-structured set and some totally unstructed sets.
Lemma 10 Let X ⊆ZN and let l be a positive integer. Then we can decompose X as X = m [ i=1 X(i) ∪X(st), where each X(i) is a dissociated set of size l and X(st) is highly structured. By this we mean that there is a set H = {h1, . . . , hk}, k ≤l, such that all elements of X(st) can be written in the form x = k X i=1 ηihi with ηi ∈{−2, −1, −1 2, 0, 1 2, 1, 2}.
10 Proof We proceed by induction on l, the result being trivial for l = 0, 1. Suppose that we have achieved a decomposition of the required type for some l ≥1. Let X(1) = {x(1) 1 , . . . , x(1) l }. We try to find an element of X \X(1) that can be added to X(1) so that the resulting set is still dissociated.
The only way in which we could fail to locate such an element would be if everything in X \ X(1) could be written in the form l X i=1 ηix(1) i (6) with ηi ∈{−2, −1, −1 2, 0, 1 2, 1, 2}, in which case the result is proved.
Indeed suppose that, for every y ∈X \ X(1), there exists a t with two different representations as P i ǫix(1) i + ǫy with ǫi, ǫ ∈{−1, 1}. The ǫ’s in these two representations must be different or else dissociativity of X(1) would be contradicted, and this allows us, by eliminating t, to express y in the form (6).
If we succeed in adding an element to X(1) then we try our luck with X(2), and a similar analysis applies. If we succeed in adding an element to each of X(1), . . . , X(m) then we are particularly happy, because then we may take the new X(st) to be just what is left of the old one.
□ Now we are ready for our grand “cutting” of the Fourier transform. We apply the decomposition of Lemma 10 to each of the sets Es defined above, getting Es = [ i E(i) s ∪E(st) s for s = 1, . . . , ν.
We use this to decompose f as f = ν X s=1 X i f (i) s + ν X s=1 f (st) s + fsmall (7) = g1 + g2 + g3.
The notation here will be clear if we say that f (i) s has Fourier coeffiecents ˆ f(r) for r ∈E(i) s , and 0 for all other r. One can check that f (i) s is given by the formula f (i) s (x) = 1 N X r∈E(i) s ˆ f(r)ωrx, and that similar results hold for the other functions featuring in (7). One has X x max 0<k<K |f(x + ky) −f(x)| ≤ 3 X i=1 X x max 0<k<K |gi(x + ky) −gi(x)|.
(8) Our aim is, the reader will recall, to show that y can be chosen so that this is less than P x f(x) = √αβ. We begin by bounding the two “error terms”, by which we mean the terms involving g1 and g3. These estimates will hold regardless of the value of y; we will then show that y can be chosen to make the term with g2 small as well.
11 3.1 Estimation of the g1 term.
In this subsection we shall prove the following lemma.
Lemma 11 X x max 0<k<K |g1(x + ky) −g1(x)| ≤20 r log K l .
Proof This is the only place in which the results of §2 will be used, and even here we only require Corollary 8. We start by observing that max 0<k<K |g1(x + ky) −g1(x)| ≤2 max 0≤k<K |g1(x + ky)|.
However we have max 0≤k<K |g1(x + ky)| 1 ≤ ν X s=1 X i max 0≤k<K |f (i) s (x + ky)| 1 ≤ ν X s=1 X i max 0≤k<K |f (i) s (x + ky)| 2 ≤ 5 p log K ν X s=1 X i ∥f (i) s ∥2 = 5√log K N ν X s=1 X i X r∈E(i) s | ˆ f(r)|2 1/2 , where we have applied Corollary 8 in deriving the third line from the second. However X r∈E(i) s | ˆ f(r)|2 1/2 ≤ 2−s+1√ l ≤ 2 √ l X r∈E(i) s | ˆ f(r)|, since |E(i) s | = l and 2−s ≤| ˆ f(r)| ≤2−s+1 for r ∈E(i) s . It follows that max 0<k<K |g1(x + ky) −g1(x)| ≤ 2N max 0≤k<K |g1(x + ky)| 1 ≤ 20 r log K l ν X s=1 X i X r∈E(i) s | ˆ f(r)| ≤ 20 r log K l .
This proves the lemma.
□ 12 3.2 Estimation of the g3 term.
Lemma 12 X x max 0<k<K |g3(x + ky) −g3(x)| ≤2 √ 2−νK.
Proof No words are required. We have X x max 0<k<K |g3(x + ky) −g3(x)| ≤ 2 X x max 0≤k<K |g3(x + ky)| = 2 N X x max 0≤k<K X r∈Esmall ˆ f(r)ωr(x+ky) ≤ 2 N 1/2 X x max 0≤k<K X r∈Esmall ˆ f(r)ωr(x+ky) 2 1/2 ≤ 2 N 1/2 X x X 0≤k<K X r∈Esmall ˆ f(r)ωr(x+ky) 2 1/2 = 2 √ K X r∈Esmall | ˆ f(r)|2 1/2 ≤ 2 √ 2−νK, which proves the lemma.
□ 3.3 Estimation of the g2 term.
From now on we write B = Sν s=1 E(st) s , the union of all our highly-structured sets of large Fourier coefficients. We have arrived at what is in some sense the heart of the proof of Theorem 1. The previous two sections have consisted in estimating what are, morally, error terms and our inequalities were valid for all y. We now come to actually picking a value of y which makes X x max 0<k<K |g2(x + ky) −g2(x)| small. Our main tool is an argument of Dirichlet on simultaneous approximation, and in this respect Bourgain’s paper bears a substantial resemblance to the papers and on progressions of length 3. We start with a chain of straightforward inequalities.
X x max 0<k<K |g2(x + ky) −g2(x)| = 1 N X x max 0<k≤K X n∈B ˆ f(n) ωnky −1 ωnx ≤ max n∈B 0 δ−q then there is some y for which the fractional parts {uiy/N}, i = 1, . . . , q, are all at most δ in magnitude. For such a y one then has, by an easy computation, that max n∈U |1 −ωny| ≤2πδ.
Taking δ = N −1/3lν and recalling (9), we have X x max 0<k<K |g2(x + ky) −g2(x)| ≤7KlN −1/3lν for a suitable y.
3.4 Putting everything together.
From (3), (8) and the last three subsections it suffices to show that, for K ∼ec(log N)1/3, there are choices of l and ν for which 20 r log K l + 2 √ 2−νK + 7KlN −1/3lv < p αβ.
One can check that K = e 1 100 (αβ log N)1/3 works for N > N(α, β). A more precise result can be obtained if desired (which might well be the case, if one is considering sets whose density tends to 0 slowly with N), but we leave this to the reader. One gets a constant c for which one can take K = ec((αβ log N)1/3−log log N).
We remark that the fact that we are considering modular arithmetic progressions rather than genuine ones makes almost no difference to the bound we have obtained. This follows from the observation, made at the start of our discussion, that a modular progression of length L contains a genuine progression of length at least 1 2 √ L. Of even less consequence is the fact that the N we have been using is actually the least prime greater than twice the original N, as we stated at the very beginning of the proof.
14 References Bourgain, J. On arithmetic progressions in sums of sets of integers, in A Tribute to Paul Erd¨ os, CUP 1990.
Heath-Brown, D.R. Integer Sets Containing No Arithmetic Progressions, J. London Math. Soc (2) 35 (1987) 385 – 394.
L´ opez, J.M and Ross, K.A. Sidon Sets, Lecture Notes in Pure and Applied Mathematics 13, Marcel Dekker (New York) 1975.
Rudin, W. Fourier Analysis on Groups, Wiley 1990 (Reprint of the original 1962 edition).
Ruzsa, I.Z. Solving a linear equation in a set of integers I, Acta Arith. 65 (1993) 259–282.
Szemer´ edi, E. Integer Sets Containing No Arithmetic Progressions, Acta. Math. Hung. 56 (1990) 155 – 158.
Tao, T. Lecture Notes from a course on the Restriction and Kakeya Conjectures, available at tao.
15 |
4905 | https://blog.hayespump.com/blog/wire-to-water-efficiency | Wire to Water Efficiency: Why Is It Important? - Hayes Pump
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Home » Blog » Wire to Water Efficiency: Why Is It Important?
Wire to Water Efficiency: Why Is It Important?
Apr 22, 2022 7:30:00 AM
Efficiency is an essential factor that influences pump selection. Wire-to-water efficiency is often a critical parameter when selecting a pump, especially when the pump needs to operate in an industrial environment. In large-scale industrial applications, even a difference of 10% in efficiency can significantly impact overall operating costs in a year.
Let’s find out what wire to water efficiency means and how you can calculate this performance metric to ensure that your pump is functioning at optimum efficiency levels.
What is Wire-to-Water Efficiency?
Wire-to-water efficiency measures how efficient a pump is in converting electricity for electro-mechanical work. In other words, it is a measure of how well the pump performs against the input electrical power that it receives.
Wire to water efficiency helps you calculate the power required to pump water at a specific flow rate and pressure head. By varying the different parameters here, you can get the maximum efficiency setting from your pump.
While you can use wire-to-water efficiency charts that are already available for standard pump types, the type of pump system used also plays a critical role in getting the desired efficiency. Therefore, for larger pumps or a series of large pumps, fitting losses and the efficiency of variable speed drive combinations are also considered.
Mathematically,
Wire to water efficiency = (water horsepower/motor horsepower) x 100
In the case of variable speed motors, the denominator is the kilowatt energy supplied to the variable frequency drive instead of the motor.
The Importance of Wire-to-Water Efficiency
Wire to water efficiency is critical because the real-life performance of any pump will always differ from the theoretical values. Knowing the wire to water efficiency helps you set a pump to deliver desired flow rate and make the most out of the input electrical energy fed into the system.
A fluid pumping system consists of two major parts: a motor and a pump. A general belief is that the pump’s overall efficiency is the average of these two numbers. That means, if the pump efficiency is 85% and the motor efficiency is 90%, the average would be 87.5%. However, this calculation is incorrect. The actual efficiency is the product of both these numbers, which comes out to be 78%, and is much lesser than the presumed value.
For example, running a single pump at its maximum capacity results in a 20% efficiency loss compared to a system with two pumps maintaining the same flow. This is where a wire-to-water efficiency chart helps in determining the best system settings to maintain high-efficiency flow.
For industrial applications, the pump flow rate and input current will profoundly impact the operating costs. In such cases, it is possible to select and run pumps at the correct settings to get the best results with a wire-to-water efficiency chart.
How to Achieve Better Pump Efficiency?
Better pump efficiency translates to an efficient pump and less wastage of power, leading to less CO2 emission. However, achieving better efficiency does not always mean picking the most efficient motors and pumps and connecting them. As discussed earlier, it can even lead to poor performance.
For better efficiency, particularly in industrial applications, it is recommended to perform wire-to-water calculations that consider both the pump and motor parameters. With wire-to-water efficiency, you can design a system so that the pump curve intersects the system curve in the preferred operating region.
Get Assistance from Experts at Hayes Pump
Designing an efficient pump system can get overwhelming and complex if you ignore some of the key performance metrics that make an efficient pump system. At Hayes Pump, our highly skilled Field Sales Engineers can offer training and education for your personnel in pump and system operation, troubleshooting, equipment repair, reassembly, and more.
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4906 | https://math.stackexchange.com/questions/1016052/number-of-length-8-binary-strings-with-no-consecutive-0s | combinatorics - Number of length $8$ binary strings with no consecutive $0$'s - Mathematics Stack Exchange
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Number of length 8 8 binary strings with no consecutive 0 0's
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How many 8 8 bit strings are there with no consecutive 0 0's?
I just sat an exam, and the only question I think I got wrong was the above(The decider for a high-distinction or a distinction :SSS)
I took Number with consecutive 0 0's
1 zero 0 0
2 zeros 7!6!7!6!
3 zeros above + 6!5!6!5!
4 zeroes above + 5!4!5!4!
5 zeros above + 4!3!4!3!
6 zeroes above + 3!2!3!2!
7 zeroes above + 2!1!2!1!
8 zeroes above + 1 1
(7∗7)+(6∗6)+(5∗5)+(4∗4)+(3∗3)+(2∗2)+1=140(7∗7)+(6∗6)+(5∗5)+(4∗4)+(3∗3)+(2∗2)+1=140
and we have 2 8−140=116 2 8−140=116
Is this correct?
combinatorics
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edited Nov 11, 2014 at 1:29
CombinatoricsCombinatorics
asked Nov 11, 2014 at 0:59
CombinatoricsCombinatorics
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@BrianM.Scott Yes sorry Combinatorics –Combinatorics 2014-11-11 01:02:13 +00:00 Commented Nov 11, 2014 at 1:02
1 Since you're no longer in a timed exam, can you rewrite your solution with sentences to make it easier to read? I can't understand what you're getting at.NoName –NoName 2014-11-11 01:09:08 +00:00 Commented Nov 11, 2014 at 1:09
Also, 2 8−77=256−77=179 2 8−77=256−77=179.Brian M. Scott –Brian M. Scott 2014-11-11 01:24:24 +00:00 Commented Nov 11, 2014 at 1:24
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(Edited): The number of n n-bit strings with no consecutive zeros is the Fibonacci number F n+2 F n+2. There are F n+1 F n+1 of these strings ending in 1 1 and F n F n ending in 0 0 (prove by induction).
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edited Nov 11, 2014 at 3:04
answered Nov 11, 2014 at 1:16
Robert IsraelRobert Israel
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1 Form the sequence for n n by adding to the sequences for n−1 n−1. The number of sequences equals the number of sequences in the n−1 n−1 set, to which we can add 1, plus the number of sequences ending in 1 in the n−1 n−1 set, to which we can add 0. The number of sequences ending in 1 in the n−1 n−1 set is just the number of sequences in the n−2 n−2 set, since these all join the n−1 n−1 set by adding 1. Thus |X n|=|X n−1|+|X n−2||X n|=|X n−1|+|X n−2|.Suzu Hirose –Suzu Hirose 2014-11-11 02:56:19 +00:00 Commented Nov 11, 2014 at 2:56
@BrianM.Scott Thanks for catching that. I edited.Robert Israel –Robert Israel 2014-11-11 03:04:34 +00:00 Commented Nov 11, 2014 at 3:04
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I can’t follow all of your work, but your final answer is off by quite a bit. Here’s a straightforward analysis by cases that’s probably about as quick as any if you’re not aware of the connection with Fibonacci numbers that Robert Israel gives in his answer.
Call a string good if it does not have consecutive 0 0 s. Clearly there is one good string with no 0 0 s, and there are 8 8 good strings with one 0 0. There are (8 2)=28(8 2)=28 strings with two 0 0 s, but 7 7 of them are bad, so 21 21 of them are good.
With three 0 0 s you have to have something like 01_01_0 01_01_0 , where each of the underscores may but need not represent one or more 1 1 s. This leaves three 1 1 s to be distributed into those underscores. You can put all of them into the same blank in 4 4 ways. You can put two of them into one blank and one into another, and there are 4⋅3=12 4⋅3=12 ways to choose the blanks. Finally, you can put them into 3 3 different blanks, and there are (4 3)=4(4 3)=4 ways to do this. This yields a grand total of 4+12+4=20 4+12+4=20 good strings with three 0 0 s.
With four 0 0 s you have to have 01_01_01_0 01_01_01_0 , and the remaining 1 1 can go into any of the five blanks.
A little thought will show that if you have more than four 0 0 s, two of them must be adjacent, so we’re done: there are
1+8+21+20+5=55 1+8+21+20+5=55
good strings. (In passing I note that 55 55 is indeed F 10 F 10.)
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answered Nov 11, 2014 at 1:27
Brian M. ScottBrian M. Scott
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Ahhh, my friend got 55 55 so I suppose that is the silver-lining, thanks! Your answer is very intuitive, thank you very much!Combinatorics –Combinatorics 2014-11-11 01:36:53 +00:00 Commented Nov 11, 2014 at 1:36
@Combinatorics: You’ve very welcome.Brian M. Scott –Brian M. Scott 2014-11-11 01:38:23 +00:00 Commented Nov 11, 2014 at 1:38
"In passing I note that 55 is indeed F10." - some interesting thing here?Suzu Hirose –Suzu Hirose 2014-11-11 02:41:20 +00:00 Commented Nov 11, 2014 at 2:41
@SuzuHirose: See Robert Israel’s answer to the question.Brian M. Scott –Brian M. Scott 2014-11-11 02:42:55 +00:00 Commented Nov 11, 2014 at 2:42
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I can't say what the cleanest answer is, but here's how I was able to work it. In such a sequence, there are at most 4 zeros. So we have 5 cases.
If there are no zeros, there is 1 solution.
If there is 1 zero, there are 8 solutions.
If there are 2 zeros, we can count the solutions as follows: Put the 1st 0 0 in the 1st place. Then there are 6 places that the 2nd 0 0 can go. If the 1st 0 0 is in the 2nd place, then there are 5 places for the 2nd 0 0, and so on. So we get 6+5+4+3+2+1=21 6+5+4+3+2+1=21 solutions.
If there are 3 zeros... This case is trickier but the argument is similar. Put the zeros as close to the beginning as possible, eg. 01010111 01010111. The last 0 0 can be where it is, or it can move to occupy the place of the last three 1 1 s, so we count 4 4 solutions from this configuration. Next, we move the second 0 0 to get the sequence 01101011 01101011. Here the last 0 0 has 3 3 places to choose from.
By counting this way, we get (4+3+2+1)+(3+2+1)+(2+1)+(1)=20(4+3+2+1)+(3+2+1)+(2+1)+(1)=20 solutions.
Finally, if there are 4 4 zeros, you can use the same technique to show there are 5 5 solutions.
So the final answer is 1+8+21+20+5=55 1+8+21+20+5=55.
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answered Nov 11, 2014 at 1:34
NoNameNoName
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A bit string with no consecutive zeros is composed of 1 and 10 units, but may also begin with one 0 unit. (That is, every zero is either preceeded by a one or is at the start of the string).
Let N x,y N x,y count the permutations of a bit string composed of x x1 and y y10 units.
N x,y=(x+y)!x!y!N x,y=(x+y)!x!y!
So N 6,1 N 6,1 counts permutations of an eight bit string of 6 61 and 1 110.
N 5,1 N 5,1 counts permutations of an eight bit string beginning with 0, and the remainder composed of 5 51 and 1 110. (The starting 0 is fixed and cannot move.)
And so forth ...
You want to calculate: N 8,0+N 7,0+N 6,1+N 5,1+N 4,2+N 3,2+N 2,3+N 1,3+N 0,4=8!8!0!+7!7!0!+7!6!1!+6!5!1!+6!4!2!+5!3!2!+5!2!3!+4!1!3!+4!0!4!=1+1+7+6+15+10+10+4+1=55 N 8,0+N 7,0+N 6,1+N 5,1+N 4,2+N 3,2+N 2,3+N 1,3+N 0,4=8!8!0!+7!7!0!+7!6!1!+6!5!1!+6!4!2!+5!3!2!+5!2!3!+4!1!3!+4!0!4!=1+1+7+6+15+10+10+4+1=55
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edited Nov 11, 2014 at 1:38
answered Nov 11, 2014 at 1:23
Graham KempGraham Kemp
133k 7 7 gold badges 55 55 silver badges 128 128 bronze badges
2
Is there anything wrong with taking 1, and 01 as the building block? I was just trying to formalize this approach.Nishant –Nishant 2018-10-18 08:10:51 +00:00 Commented Oct 18, 2018 at 8:10
1 Nope, @Nishant . Symmetry. That's just building from the other end.Graham Kemp –Graham Kemp 2018-10-18 08:16:06 +00:00 Commented Oct 18, 2018 at 8:16
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4907 | https://artofproblemsolving.com/wiki/index.php/Trigonometry?srsltid=AfmBOoqSUJI3G-_Jda6T2XAaCgbbRf1yBuZm-kbzUNSFum6rtyB06r-Q | Art of Problem Solving
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Trigonometry
Trigonometry is the study of relations between the side lengths and angles of triangles through the trigonometric functions. It is a fundamental branch of mathematics, and its discovery paved the way towards countless famous results.
In contest math, trigonometry is an integral subfield of both geometry and algebra. Many essential results in geometry are written in terms of the trigonometric functions, such as the Law of Sines and the Law of Cosines; many more, such as Stewart's Theorem, are most easily proven using trigonometry. In algebra, expressions involving the trigonometric functions appear frequently on contests. These are solved by clever usage of the trigonometric functions' countless identities, which can simplify otherwise unwieldy equations.
Outside of competition math, trigonometry is the backbone of much of analysis. In particular, Fourier Analysis is written almost entirely in the language of the trigonometric functions.
Contents
1 Definitions
1.1 Right triangle definition
1.2 Unit circle definition
1.3 Taylor series definition
2 Applications in Geometry
2.1 Law of Sines
2.2 Law of cosines
3 Trigonometric identities
4 See also
Definitions
The trigonometric functions can be defined in several equivalent ways. The definition usually taught first is the right triangle definition, for its ease of access. An intermediate to olympiad geometry course usually uses the unit circle definition of trigonometry. Beyond the scope of contest math, the Taylor series definition of trigonometry is preferred in order to extend trigonometry to a complex domain.
Right triangle definition
The right triangle definition of trigonometry involves the ratios between edges of a right triangle, with respect to a given angle. The definitions below will be referring to angle , with side lengths specified in the diagram. Because angle must be less than for the triangle to stay right, these definitions only work for acute angles.
Sine: The sine of angle , denoted , is defined as the ratio of the side opposite to the hypotenuse.
Cosine: The cosine of angle , denoted , is defined as the ratio of the side adjacent to the hypotenuse.
Tangent: The tangent of angle , denoted , is defined as the ratio of the side opposite to the side adjacent to .
A common mnemonic to remember this is SOH-CAH-TOA, where Sine = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, and Tangent = Opposite / Adjacent
More uncommon are the reciprocals of the trigonometric functions, listed below.
Cosecant: The cosecant of angle , denoted , is defined as the reciprocal of the sine of .
Secant: The secant of angle , denoted , is defined as the reciprocal of the cosine of .
Cotangent: The cotangent of angle , denoted , is defined as the reciprocal of the tangent of .
The right triangle definition is most commonly taught in introductory geometry classes for its simplicity. However, it has its limitations. It only works if is right, which means that the trigonometric functions are only defined when angle is acute.
Even though it is defined using right triangles, trigonometry is just as useful when used on acute and obtuse triangles. The Law of Sines and Law of Cosines mentioned below generalize the right triangle definition to include all triangles.
Unit circle definition
Consider the unit circle, the circle with radius one centered at the origin. Starting at , walk a distance counterclockwise around the unit circle, as shown in the diagram. The coordinates of this point are defined to be .
As for the other trigonometric functions, is defined to be the ratio of to , and cosecant, secant, and cotangent are defined to be the reciprocals of sine, cosine, and tangent, respectively.
The benefit of this definition is that it matches the right triangle definition for acute angles, but extends their domain from acute angles to all real-valued angles. As such, this definition is usually preferred in intermediate to olympiad geometry settings.
Taylor series definition
The Taylor series for sine and cosine are used as their definitions in all higher mathematics. This meets the rigorous standards of real analysis, and gives a concrete way to extend the definitions of trigonometric functions from the real numbers to the full complex plane. The Taylor series for sine and cosine are shown below: These formulas are not used in high school math competitions. However, they do appear on the Putnam and other undergraduate competitions.
Applications in Geometry
While trigonometry is useful at any level, intermediate competitions are particularly fond of geometry problems demanding trigonometry. In addition to those mentioned, here are some highlights of the applications of trigonometry to geometry:
Law of Sines
The Law of Sines states that in any , where is the side opposite to , opposite to , opposite to , and is the circumradius of . The law of sines is particularly handy in problems involving the circumradius, seeing extremely wide usage in intermediate geometry.
Law of cosines
The Law of Cosines states that in any , where is the side opposite to , opposite to , and opposite to . It is a generalization of the Pythagorean Theorem and is used to prove several famous results, such as Heron's Formula and Stewart's Theorem. However, it sees limited applicability compared to the Law of Sines, as usage of the Law of Cosines can get algebra-heavy. It is helpful to memorize common, "nicer" values of sine and cosine as it can come in handy in contests, especially if you wish to apply either this or the Law of Sines to problems.
Trigonometric identities
Trigonometric identities are expressions true for all inputs involving the trigonometric functions. Due to the natural relationship between their definitions, these identities run numerous. In contest math, the most useful of these are:
Pythagorean identities
Angle addition identities
Double angle identities
Half angle identities
Sum-to-product identities
Product-to-sum identities
See also
Trigonometric identities
Law of Sines
Law of Cosines
Stewart's Theorem
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4908 | https://artofproblemsolving.com/articles/files/MildorfInequalities.pdf?srsltid=AfmBOopiSmlE14gAz2V8rX7kUOnXDQY2gRpH2zBPiGy_tFXuWk9Xfx2X | Olympiad Inequalities
Thomas J. Mildorf December 22, 2005
It is the purpose of this document to familiarize the reader with a wide range of theorems and techniques that can be used to solve inequalities of the variety typically appearing on mathematical olympiads or other elementary proof contests. The Standard Dozen is an exhibition of twelve famous inequalities which can be cited and applied without proof in a solution. It is expected that most problems will fall entirely within the span of these inequalities. The Examples section provides numerous complete solutions as well as remarks on inequality-solving intuition, all intended to increase the reader’s aptitude for the material covered here. It is organized in rough order of difficulty. Finally, the Problems section contains exercises without solutions, ranging from straightforward to quite difficult, for the purpose of practicing techniques contained in this document. I have compiled much of this from posts by my peers in a number of mathematical communities, particularly the Mathlinks-Art of Problem Solving forums, 1 as well as from various MOP lectures, 2 Kiran Kedlaya’s inequalities packet, 3 and John Scholes’ site. 4 I have tried to take note of original sources where possible. This work in progress is distributed for personal educational use only. In particular, any publication of all or part of this manuscript without prior consent of the author, as well as any original sources noted herein, is strictly prohibited. Please send comments - suggestions, corrections, missing information, 5 or other interesting problems - to the author at tmildorf@mit.edu .Without further delay...
1
and respectively, though they have merged into a single, very large and robust group. The forums there are also host to a considerable wealth of additional material outside of inequalities.
2
Math Olympiad Program. Although some people would try to convince me it is the Math Olympiad Summer Program and therefore is due the acronym MOSP, those who know acknowledge that the traditional “MOP” is the preferred appellation.
3
The particularly diligent student of inequalities would be interested in this document, which is available online at Further ma-terial is also available in the books Andreescu-Cartoaje-Dospinescu-Lascu, Old and New Inequalities , GIL Publishing House, and Hardy-Littlewood-P´ olya, Inequalities , Cambridge University Press. (The former is elementary and geared towards contests, the latter is more technical.)
4
where a seemingly inexhaustible supply of Olympiads is available.
5
Such as the source of the last problem in this document.
11 The Standard Dozen
Throughout this lecture, we refer to convex and concave functions. Write I and I′ for the intervals [ a, b ] and ( a, b ) respectively. A function f is said to be convex on I if and only if
λf (x) + (1 − λ)f (y) ≥ f (λx + (1 − λ)y) for all x, y ∈ I and 0 ≤ λ ≤ 1. Conversely, if the inequality always holds in the opposite direction, the function is said to be concave on the interval. A function f that is continuous on I and twice differentiable on I′ is convex on I
if and only if f ′′ (x) ≥ 0 for all x ∈ I (Concave if the inequality is flipped.) Let x1 ≥ x2 ≥ · · · ≥ xn; y1 ≥ y2 ≥ · · · ≥ yn be two sequences of real numbers. If
x1 + · · · + xk ≥ y1 + · · · + yk for k = 1 , 2, . . . , n with equality where k = n, then the sequence
{xi} is said to majorize the sequence {yi}. An equivalent criterion is that for all real numbers
t,
|t − x1| + |t − x2| + · · · + |t − xn| ≥ | t − y1| + |t − y2| + · · · + |t − yn|
We use these definitions to introduce some famous inequalities.
Theorem 1 (Jensen) Let f : I → R be a convex function. Then for any x1, . . . , x n ∈ I
and any nonnegative reals ω1, . . . , ω n,
ω1f (x1) + · · · + ωnf (xn) ≥ (ω1 + · · · + ωn) f
(ω1x1 + · · · + ωnxn
ω1 + · · · + ωn
)
If f is concave, then the inequality is flipped.
Theorem 2 (Weighted Power Mean) If x1, . . . , x n are nonnegative reals and ω1, . . . , ω n
are nonnegative reals with a postive sum, then
f (r) :=
(ω1xr
1
· · · + ωnxrn
ω1 + · · · + ωn
)1
r
is a non-decreasing function of r, with the convention that r = 0 is the weighted geometric mean. f is strictly increasing unless all the xi are equal except possibly for r ∈ (−∞ , 0] ,where if some xi is zero f is identically 0. In particular, f (1) ≥ f (0) ≥ f (−1) gives the AM-GM-HM inequality.
Theorem 3 (H¨ older) Let a1, . . . , a n; b1, . . . , b n; · · · ; z1, . . . , z n be sequences of nonnegative real numbers, and let λa, λ b, . . . , λ z positive reals which sum to 1. Then
(a1 + · · · + an)λa (b1 + · · · + bn)λb · · · (z1 + · · · + zn)λz ≥ aλa
1
bλb
1
· · · zλz
1
· · · + aλz
n
bλb
n
· · · zλz
n
This theorem is customarily identified as Cauchy when there are just two sequences.
Theorem 4 (Rearrangement) Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be two nondecreasing sequences of real numbers. Then, for any permutation π of {1, 2, . . . , n }, we have
a1b1 + a2b2 + · · · + anbn ≥ a1bπ(1) + a2bπ(2) + · · · + anbπ(n) ≥ a1bn + a2bn−1 + · · · + anb1
with equality on the left and right holding if and only if the sequence π(1) , . . . , π (n) is de-creasing and increasing respectively.
2Theorem 5 (Chebyshev) Let a1 ≤ a2 ≤ · · · ≤ an; b1 ≤ b2 ≤ · · · ≤ bn be two nondecreas-ing sequences of real numbers. Then
a1b1 + a2b2 + · · · + anbn
n ≥ a1 + a2 + · · · + an
n ·b1 + b2 + · · · + bn
n ≥ a1bn + a2bn−1 + · · · + anb1
n
Theorem 6 (Schur) Let a, b, c be nonnegative reals and r > 0. Then
ar(a − b)( a − c) + br(b − c)( b − a) + cr(c − a)( c − b) ≥ 0
with equality if and only if a = b = c or some two of a, b, c are equal and the other is 0.
Remark - This can be improved considerably. (See the problems section.) However, they are not as well known (as of now) as this form of Schur, and so should be proven whenever used on a contest.
Theorem 7 (Newton) Let x1, . . . , x n be nonnegative real numbers. Define the symmetric polynomials s0, s 1, . . . , s n by (x + x1)( x + x2) · · · (x + xn) = snxn + · · · + s1x + s0, and define the symmetric averages by di = si/(ni
). Then
d2
i
≥ di+1 di−1
Theorem 8 (Maclaurin) Let di be defined as above. Then
d1 ≥ √d2 ≥ 3
√d3 ≥ · · · ≥ n
√dn
Theorem 9 (Majorization) Let f : I → R be a convex on I and suppose that the sequence
x1, . . . , x n majorizes the sequence y1, . . . , y n, where xi, y i ∈ I. Then
f (x1) + · · · + f (xn) ≥ f (y1) + · · · + f (yn)
Theorem 10 (Popoviciu) Let f : I → R be convex on I, and let x, y, z ∈ I. Then for any positive reals p, q, r ,
pf (x) + qf (y) + rf (z) + (p + q + r)f
(px + qy + rz p + q + r
)
≥ (p + q)f
(px + qy p + q
)
( q + r)f
(qy + rz q + r
)
( r + p)f
(rz + px r + p
)
Theorem 11 (Bernoulli) For all r ≥ 1 and x ≥ − 1,
(1 + x)r ≥ 1 + xr
3Theorem 12 (Muirhead) Suppose the sequence a1, . . . , a n majorizes the sequence b1, . . . , b n.Then for any positive reals x1, . . . , x n,
∑
sym
xa1
1
xa2
2
· · · xan
n
≥ ∑
sym
xb1
1
xb2
2
· · · xbn
n
where the sums are taken over all permutations of n variables.
Remark - Although Muirhead’s theorem is a named theorem, it is generally not favor-ably regarded as part of a formal olympiad solution. Essentially, the majorization criterion guarantees that Muirhead’s inequality can be deduced from a suitable application of AM-GM. Hence, whenever possible, you should use Muirhead’s inequality only to deduce the correct relationship and then explicitly write all of the necessary applications of AM-GM. For a particular case this is a simple matter.
We now present an array of problems and solutions based primarily on these inequalities and ideas.
2 Examples
When solving any kind of problem, we should always look for a comparatively easy solu-tion first, and only later try medium or hard approaches. Although what constitutes this notoriously indeterminate “difficulty” varies widely from person to person, I usually con-sider “Dumbassing,” AM-GM (Power Mean), Cauchy, Chebyshev (Rearrangement), Jensen, H¨ older, in that order before moving to more clever techniques. (The first technique is de-scribed in remarks after example 1.) Weak inequalities will fall to AM-GM, which blatantly pins a sum to its smallest term. Weighted Jensen and H¨ older are “smarter” in that the effect of widely unequal terms does not cost a large degree of sharpness 6 - observe what happens when a weight of 0 appears. Especially sharp inequalities may be assailable only through clever algebra. Anyway, I have arranged the following with that in mind. 1. Show that for positive reals a, b, c
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) ≥ 9a2b2c2
Solution 1. Simply use AM-GM on the terms within each factor, obtaining
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) ≥
(
3 3
√a3b3c3
) (
3 3
√a3b3c3
)
= 9 a2b2c2
6The sharpness of an inequality generally refers to the extent to which the two sides mimic each other, particularly near equality cases.
4Solution 2. Rearrange the terms of each factor and apply Cauchy,
(a2b + b2c + c2a) ( bc 2 + ca 2 + ab 2) ≥
(√a3b3c3 + √a3b3c3 + √a3b3c3
)2
= 9 a2b2c2
Solution 3. Expand the left hand side, then apply AM-GM, obtaining
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) = a3b3 + a2b2c2 + a4bc
ab 4c + b3c3 + a2b2c2
a2b2c2 + abc 4 + a3c3
≥ 9 9
√a18 b18 c18 = 9 a2b2c2
We knew this solution existed by Muirhead, since (4 , 1, 1) , (3 , 3, 0), and (2 , 2, 2) all majorize (2 , 2, 2). The strategy of multiplying out all polynomial expressions and ap-plying AM-GM in conjunction with Schur is generally knowing as dumbassing because it requires only the calculational fortitude to compute polynomial products and no real ingenuity. As we shall see, dumbassing is a valuable technique. We also remark that the AM-GM combining all of the terms together was a particularly weak inequality, but the desired was a multiple of a2b2c2’s, the smallest 6th degree symmetric polynomial of three variables; such a singular AM-GM may not always suffice. 2. Let a, b, c be positive reals such that abc = 1. Prove that
a + b + c ≤ a2 + b2 + c2
Solution. First, we homogenize the inequality. that is, apply the constraint so as to make all terms of the same degree. Once an inequality is homogenous in degree
d, we may scale all of the variables by an arbitrary factor of k, causing both sides of the inequality to scale by the factor kd. This is valid in that it does not change the correctness of an inequality for any positive k, and if d is even, for any nonzero
k. Hence, we need consider a nonhomogenous constraint no futher. In this case, we multiply the left hand side by 3
√abc , obtaining
a43 b13 c13 + a13 b43 c13 + a13 b13 c43 ≤ a2 + b2 + c2
As abc = 1 is not homogenous, the above inequality must be true for all nonnegative
a, b, c . As (2 , 0, 0) majorizes (4 /3, 1/3, 1/3), we know it is true, and the necessary AM-GM is 2a2
3 + b2
6 + c2
6 = a2 + a2 + a2 + a2 + b2 + c2
6 ≥ 6
√a8b2c2 = a43 b13 c13
Let P (x) be a polynomial with positive coefficients. Prove that if
P
( 1
x
)
≥ 1
P (x)5holds for x = 1, then it holds for all x > 0.
Solution. Let P (x) = anxn + an−1xn−1 + · · · + a1x + a0. The first thing we notice is that the given is P (1) ≥ 1. Hence, the natural strategy is to combine P (x) and P ( 1
x
)
into P (1) in some fashion. The best way to accomplish this is Cauchy, which gives
P (x)P
( 1
x
)
= (anxn + · · · + a1x + a0)
(
an
1
xn + · · · + a1
1
x + a0
)
≥ (an + · · · + a1 + a0)2 = P (1) 2 ≥ 1as desired. This illustrates a useful means of eliminating denominators - by introducing similar factors weighted by reciprocals and applying Cauchy / H¨ older. 4. (USAMO 78/1) a, b, c, d, e are real numbers such that
a + b + c + d + e = 8
a2 + b2 + c2 + d2 + e2 = 16 What is the largest possible value of e?
Solution. Observe that the givens can be effectively combined by considering squares: (a − r)2 + ( b − r)2 + ( c − r)2 + ( d − r)2 + ( e − r)2 = (a2 + b2 + c2 + d2 + e2)
− 2r(a + b + c + d + e) + 5 r2
= 16 − 16 r + 5 r2
Since these squares are nonnegative, e ≤ √5r2 − 16 r + 16 + r = f (r) for all r. Since equality e = f (r) can be achieved when a = b = c = d = r, we need only compute the smallest value f (r). Since f grows large at either infinity, the minimum occurs when
f ′(r) = 1 + 10 r−16 2√5r2−16 r+16 = 0. The resultant quadratic is easily solved for r = 65 and
r = 2, with the latter being an extraneous root introduced by squaring. The largest possible e and greatest lower bound of f (r) is then f (6 /5) = 16 /5, which occurs when
a = b = c = d = 6 /5 and e = 16 /5. Alternatively, proceed as before except write
a = b = c = d = 8−e
4
since the maximum e must occur when the other four variables are equal. The second condition becomes a quadratic, and the largest solution is seen to be e = 16 5 .The notion of equating a, b, c, d is closely related to the idea of smoothing and Jensen’s inequality. If we are working with S1 = f (x1) + · · · + f (xn) under the constraint of a fixed sum x1 + · · · + xn, we can decrease S1 by moving several xi in the same interval
I together (that is, replacing xi1 < x i2 with x′
i1
= xi1 + ≤ < x i2 − ≤ = x′
i2
for any sufficiently small ≤) for any I where f is convex. S1 can also be decreased by spreading
xi in the same interval where f is concave. When seeking the maximum of S1, we proceed in the opposite fashion, pushing xi on the concave intervals of f together and moving xi on the convex intervals apart. 65. Show that for all positive reals a, b, c, d ,1
a + 1
b + 4
c + 16
d ≥ 64
a + b + c + d
Solution. Upon noticing that the numerators are all squares with √1 + √1 + √4 + √16 = √64, Cauchy should seem a natural choice. Indeed, multiplying through by
a + b + c + d and applying Cauchy, we have (a + b + c + d)
(12
a + 12
b + 22
c + 42
d
)
≥ (1 + 1 + 2 + 4) 2 = 64 as desired. 6. (USAMO 80/5) Show that for all non-negative reals a, b, c ≤ 1,
ab + c + 1 + bc + a + 1 + ca + b + 1 + (1 − a)(1 − b)(1 − c) ≤ 1
Solution. Let f (a, b, c ) denote the left hand side of the inequality. Since ∂2
∂a 2
f =
2b
(c+a+1) 3
2c
(a+b+1) 3
≥ 0, we have that f is convex in each of the three variables; hence, the maximum must occur where a, b, c ∈ { 0, 1}. Since f is 1 at each of these 8 points, the inequality follows. Second derivative testing for convexity/concavity is one of the few places where the use of Calculus is not seriously loathed by olympiad graders. It is one of the standard techniques in inequalities and deserves to be in any mental checklist of inequality solving. In this instance, it led to an easy solution. 7. (USAMO 77/5) If a, b, c, d, e are positive reals bounded by p and q with 0 < p ≤ q,prove that (a + b + c + d + e)
( 1
a + 1
b + 1
c + 1
d + 1
e
)
≤ 25 + 6
(√ pq −
√ qp
)2
and determine when equality holds.
Solution. As a function f of five variables, the left hand side is convex in each of
a, b, c, d, e ; hence, its maximum must occur when a, b, c, d, e ∈ { p, q }. When all five variables are p or all five are q, f is 25. If one is p and the other four are q, or vice versa, f becomes 17 + 4( pq + qp ), and when three are of one value and two of the other,
f = 13 + 6( pq + qp ). pq + qp ≥ 2, with equality if and only if p = q. Clearly, equality holds where p = q. Otherwise, the largest value assumed by f is 13 + 6( pq + qp ), which is obtained only when two of a, b, c, d, e are p and the other three are q, or vice versa. In such instances, f is identically the right hand side. This is a particular case of the Schweitzer inequality, which, in its weighted form, is sometimes known as the Kantorovich inequality. 78. a, b, c, are non-negative reals such that a + b + c = 1. Prove that
a3 + b3 + c3 + 6 abc ≥ 14
Solution. Multiplying by 4 and homogenizing, we seek 4a3 + 4 b3 + 4 c3 + 24 abc ≥ (a + b + c)3
= a3 + b3 + c3 + 3 (a2(b + c) + b2(c + a) + c2(a + b)) + 6 abc
⇐⇒ a3 + b3 + c3 + 6 abc ≥ a2(b + c) + b2(c + a) + c2(a + b)Recalling that Schur’s inequality gives a3 +b3 +c3 +3 abc ≥ a2(b+c)+ b2(c+a)+ c2(a+b), the inequality follows. In particular, equality necessitates that the extra 3 abc on the left is 0. Combined with the equality condition of Schur, we have equality where two of a, b, c are 12 and the third is 0. This is a typical dumbass solution.
Solution 2. Without loss of generality, take a ≥ b ≥ c. As a+b+c = 1, we have c ≤ 13
or 1 −3c ≥ 0. Write the left hand side as ( a+b)3 −3ab (a+b−2c) = ( a+b)3 −3ab (1 −3c). This is minimized for a fixed sum a + b where ab is made as large as possible. As by AM-GM ( a + b)2 ≥ 4ab , this minimum occurs if and only if a = b. Hence, we need only consider the one variable inequality 2 (1−c
2
)3 + c3 + 6 (1−c
2
)2 c = 14 · (9 c3 − 9c2 + 3 c + 1). Since c ≤ 13 , 3 c ≥ 9c2. Dropping this term and 9 c3, the inequality follows. Particularly, 9c3 = 0 if and only if c = 0, and the equality cases are when two variables are 12 and the third is 0. 9. (IMO 74/5) If a, b, c, d are positive reals, then determine the possible values of
aa + b + d + bb + c + a + cb + c + d + da + c + d
Solution. We can obtain any real value in (1 , 2). The lower bound is approached by
a → ∞ , b = d = √a, and c = 1. The upper bound is approached by a = c → ∞ ,
b = d = 1. As the expression is a continuous function of the variables, we can obtain all of the values in between these bounds. Finally, these bounds are strict because
aa + b + d + bb + c + a + cb + c + d + da + c + d >aa + b + c + d + ba + b + c + d + ca + b + c + d + da + b + c + d = 1 and
aa + b + d + bb + c + a + cb + c + d + da + c + d <aa + b + ba + b + cc + d + dc + d = 2 Whenever extrema occur for unusual parameterizations, we should expect the need for non-classical inequalities such as those of this problem where terms were completely dropped. 810. (IMO 95/2) a, b, c are positive reals with abc = 1. Prove that 1
a3(b + c) + 1
b3(c + a) + 1
c3(a + b) ≥ 32
Solution 1. Let x = 1
a
, y = 1
b
, and z = 1
c
. We perform this substitution to move terms out of the denominator. Since abc = xyz = 1, we have 1
a3(b + c) + 1
b3(c + a) + 1
c3(a + b) = x2
y + z + y2
x + z + z2
x + y
Now, multiplying through by ( x + y)( y + z)( z + x), we seek
x4 + y4 + z4 + x3y + x3z + y3z + xy 3 + xz 3 + yz 3 + x2yz + xy 2z + xyz 2 ≥
3
√xyz ·
(
3xyz + 32 · (x2y + x2z + y2x + xy 2 + xz 2 + yz 2))
which follows immediately by AM-GM, since x2yz +xy 2z+xyz 2 ≥ 3 3
√x4y4z4, x3y+xy 3+x3z
3
≥
3
√x7y4z and 7x4+4 y4+z4
12
≥ 3
√x7y4z - as guaranteed by Muirhead’s inequality.
Solution 2. Substitute x, y, z as before. Now, consider the convex function f (x) = x−1
for x > 0. ( f (x) = xc is convex for c < 0 and c ≥ 1, and concave for 0 < c ≤ 1, verify this with the second derivative test.) Now, by Jensen,
x2
y + z + y2
z + x + z2
x + y = xf
(y + zx
)
yf
(z + xy
)
zf
(x + yz
)
≥ (x + y + z)f
((y + z) + ( z + x) + ( x + y)
x + y + z
)
= x + y + z
2But x + y + z ≥ 3 3
√xyz = 3, as desired.
Solution 3. Perform the same substitution. Now, multiplying by ( x + y + z) and applying Cauchy, we have 12 (( y + z) + ( z + x) + ( x + y))
( x2
y + z + y2
z + x + z2
x + y
)
≥ 12(x + y + z)2
Upon recalling that x+y +z ≥ 3 we are done. Incidentally, the progress of this solution with Cauchy is very similar to the weighted Jensen solution shown above. This is no coincidence, it happens for many convex f (x) = xc.
Solution 4. Apply the same substitution, and put x ≥ y ≥ z. Simultaneously,
xy+z
≥ yz+x ≥ zx+y . Hence, by Chebyshev,
x ·
( xy + z
)
y ·
( yz + x
)
z ·
( zx + y
)
≥ x + y + z
3
( xy + z + yx + z + zx + y
)
Again, x + y + z ≥ 3. But now we have Nesbitt’s inequality, xy+z + yx+z + zx+y ≥ 32 . This follows immediately from AM-HM upon adding 1 to each term. 911. Let a, b, c be positive reals such that abc = 1. Show that 2(a + 1) 2 + b2 + 1 + 2(b + 1) 2 + c2 + 1 + 2(c + 1) 2 + a2 + 1 ≤ 1
Solution. The previous problem showed the substitution offers a way to rewrite an inequality in a more convenient form. Substitution can also be used to implicity use a given. First, expand the denominators and apply AM-GM, obtaining 2(a + 1) 2 + b2 + 1 = 2
a2 + b2 + 2 a + 2 ≤ 1
ab + a + 1 Now, write a = xy , b = yz , c = zx . We have 1
ab +a+1
= 1
xz+xy+1
= yz xy +yz +zx . It is now evident that the sum of the new fractions is 1. 12. (USAMO 98/3) Let a0, . . . , a n real numbers in the interval (0 , π
2
) such that tan
(
a0 − π
4
)
tan
(
a1 − π
4
)
· · · + tan
(
an − π
4
)
≥ n − 1Prove that tan( a0) tan( a1) · · · tan( an) ≥ nn+1
Solution 1. Let yi = tan (x − π
4
). We have tan( xi) = tan ((xi − π
4
) + π
4
) = yi+1 1−yi .Hence, given s = y0 + · · · + yn ≥ n − 1 we seek to prove ∏ni=0 1+ yi
1−yi
≥ nn+1 . Observe that for any a > b and fixed sum a + b, the expression
( 1 + a
1 − a
)
·
( 1 + b
1 − b
)
= 1 + 2( a + b)(1 − a)(1 − b)can be decreased by moving a and b any amount closer together. Hence, for any sequence y0, . . . , y n, we can replace any yi > sn+1 and yj < sn+1 with y′
i
= sn+1 and
y′
j
= yi + yj − sn+1 , decreasing the product. Now we have
n
∏
i=0
1 + yi
1 − yi
≥
( 1 + sn+1
1 − sn+1
)n+1
≥
( 2nn+1 2
n+1
)n+1
= nn+1
Where the last inequality follows from the fact that 1+ x
1−x
is an increasing function of x.
Solution 2. Perform the same substitution. The given can be written as 1 + yi ≥
∑
j6=i
(1 − yj ), which by AM-GM gives 1+ yn
n
≥ ∏
j6=i
(1 − yj ) 1
n
. Now we have
n
∏
i=0
1 + yi
n ≥
n
∏
i=0
∏
j6=i
(1 − yj ) 1
n
=
n
∏
i=0
(1 − yi)as desired. 10 13. Let a, b, c be positive reals. Prove that 1
a(1 + b) + 1
b(1 + c) + 1
c(1 + a) ≥ 31 + abc
with equality if and only if a = b = c = 1.
Solution. Multiply through by 1+ abc and add three to each side, on the left obtaining 1 + a + ab + abc a(1 + b) + 1 + b + bc + abc b(1 + c) + 1 + c + ac + abc c(1 + a)= (1 + a) + ab (1 + c)
a(1 + b) + (1 + b) + bc (1 + a)
b(1 + c) + (1 + c) + ac (1 + b)
c(1 + a)which is at least 6 by AM-GM, as desired. In particular, this AM-GM asserts the equivalence of (1+ a)
a(1+ b)
and a(1+ b)1+ a , or that they are both one. Likewise, all of the other terms must be 1. Now, (1 + a)2 = a2(1 + b)2 = a2b2(1 + c)2 = a2b2c2(1 + a)2, so the product abc = 1. Hence, 1+ aa(1+ b) = bc (1+ a)1+ b = bc (1+ a)
b(1+ c)
so that 1 + b = b + bc = b + 1
a
. It is now easy to see that equality holds if and only if a = b = c = 1. 14. (Romanian TST) Let a, b, x, y, z be positive reals. Show that
xay + bz + yaz + bx + zax + by ≥ 3
a + b
Solution. Note that ( a + b)( xy + yz + xz ) = ( x(ay + bz ) + y(az + bx ) + z(ax + by )). We introduce this factor in the inequality, obtaining (x(ay + bz ) + y(az + bx ) + z(ax + by ))
( xay + bz + yaz + bx + zax + by
)
≥
(x + y + z)2 ≥ 3( xy + yz + xz )Where the last inequality is simple AM-GM. The desired follows by simple algebra. Again we have used the idea of introducing a convenient factor to clear denominators with Cauchy. 15. The numbers x1, x 2, . . . , x n obey −1 ≤ x1, x 2, . . . , x n ≤ 1 and x 31 + x 32 + · · · + x 3
n
= 0. Prove that
x1 + x2 + · · · + xn ≤ n
3
Solution 1. Substitute yi = x3
i
so that y1 + · · · + yn = 0. In maximizing 3
√y1 +
· · · + 3
√yn, we note that f (y) = y 13 is concave over [0 , 1] and convex over [ −1, 0], with
|f ′(y1)| ≥ | f ′(y2)| ⇐⇒ 0 < |y1| ≤ | y2|. Hence, we may put y1 = · · · = yk = −1;
−1 ≤ yk+1 < 0, and yk+2 = · · · = yn = k−yk+1
n−k−1
. We first show that yk+1 leads to a maximal sum of 3
√yi if it is -1 or can be made positive. If |yk+1 | < |yk+2 |, we set 11 y′
k+1
= y′
k+2
= yk+1 +yk+2
2
, increasing the sum while making yk+1 positive. Otherwise, set y′
k+1
= −1 and y′
k+2
= 1 − yk+1 − yk+2 , again increasing the sum of the 3
√yi. Now we may apply Jensen to equate all positive variables, so that we need only show
k 3
√−1 + ( n − k) 3
√ kn − k ≤ n
3But we have (n + 3 k)3 − 27( n − k)2k = n3 − 18 n2k + 81 nk 2 = n(n − 9k)2 ≥ 0as desired. Particularly, as k is an integer, equality can hold only if 9 |n and then if and only if one ninth of the variables yi are -1 and the rest are 1/8.
Solution 2. Let xi = sin( αi), and write 0 = x31 + · · · + x3
n
= sin 3(α1) + · · · + sin 3(αn) =
14
((3 sin( α1) − sin(3 α1)) + · · · + (3 sin( αn) − sin(3 αn))). It follows that x1 + · · · + xn =sin( α1) + · · · + sin( αn) = sin(3 α1)+ ··· +sin(3 αn)3 ≤ n
3
. The only values of sin( α) which lead to sin(3 α) = 1 are 12 and -1. The condition for equality follows. 16. (Turkey) Let n ≥ 2 be an integer, and x1, x 2, . . . , x n positive reals such that x21 + x22 +
· · · + x2
n
= 1. Determine the smallest possible value of
x51
x2 + x3 + · · · + xn
x52
x3 + · · · + xn + x1
· · · + x5
n
x1 + · · · + xn−1
Solution. Observe that ∑ni=1 xi
∑
j6=i
xj ≤ n − 1, so that
( n∑
i=1
xi
(∑
j6=i
xj
)) ( n∑
i=1
x5
i
∑
j6=i
xi
)
≥ (x31 + · · · + x3
n
)2
= n2
(x31 + · · · + x3
n
n
)2
≥ n2
(x21 + · · · + x2
n
n
)3
= 1
n
Leads to n∑
i=1
x5
i
∑
j6=i
xi
≥ 1
n(n − 1) with equality if and only if x1 = · · · = xn = 1√n .17. (Poland 95) Let n be a positive integer. Compute the minimum value of the sum
x1 + x22
2 + x33
3 + · · · + xnn
n
12 where x1, x 2, . . . , x n are positive reals such that 1
x1
1
x2
· · · + 1
xn
= n
Solution. The given is that the harmonic mean of x1, . . . , x n is 1, which implies that the product x1x2 · · · xn is at least 1. Now, we apply weighted AM-GM
x1 + x22
2 + x33
3 + · · · + xnn
n ≥
(
1 + 12 + 13 + · · · + 1
n
)
1+ 12 +··· + 1
n
√x1x2 · · · xn
= 1 + 12 + 13 + · · · + 1
n
Prove that for all positive reals a, b, c, d ,
a4b + b4c + c4d + d4a ≥ abcd (a + b + c + d)
Solution. By AM-GM, 23 a4b + 7 b4c + 11 c4d + 10 ad 4
51 ≥ 51
√a102 b51 c51 d51 = a2bcd
from which the desired follows easily. Indeed, the most difficult part of this problem is determining suitable weights for the AM-GM. One way is to suppose arbitrary weights
x1, x 2, x 3, x 4 for a4b, b 4c, c 4d, ad 4 respectively, and solve the system
x1 + x2 + x3 + x4 = 14x1 + x2 = 24x2 + x3 = 14x3 + x4 = 119. (USAMO 01/3) Let a, b, c be nonnegative reals such that
a2 + b2 + c2 + abc = 4 Prove that 0 ≤ ab + bc + ca − abc ≤ 2
Solution [by Tony Zhang.] For the left hand side, note that we cannot have a, b, c >
Suppose WLOG that c ≤ 1. Then ab +bc +ca −abc ≥ ab +bc +ca −ab = c(a+b) ≥ 0. For the right, 4 = a2 + b2 + c2 + abc ≥ 4( abc )43 =⇒ abc ≤ 1. Since by the pigeon hole principle, among three numbers either two exceed 1 or two are at most 1. Hence, we assume WLOG that ( a − 1)( b − 1) ≥ 0, which gives ab + 1 ≥ a + b ⇐⇒ abc + c ≥ ac +
bc ⇐⇒ c ≥ ac + bc − abc . Now, we have ab + bc + ca − abc ≤ ab + c. Either we are done or ab +c > 2. But in the latter case, 4 = ( a2 +b2)+ c(c+2 ab ) > 2ab +2 c = 2( ab +c) > 4, a contradiction. 13 20. (Vietnam 98) Let x1, . . . , x n be positive reals such that 1
x1 + 1998 + 1
x2 + 1998 + · · · + 1
xn + 1998 = 11998 Prove that n
√x1x2 · · · xn
n − 1 ≥ 1998
Solution. Let yi = 1
xi+1998
so that y1 + · · · + yn = 11998 and xi = 1
yi
− 1998. Now
n
∏
i=1
xi =
n
∏
i=1
( 1
yi
− 1998
)
= e
Pni=1 ln
“1
yi−1998
”
Hence, to minimize the product of the xi, we equivalently minimize the sum of ln
( 1
yi
− 1998
)
.In particular,
ddy
(
ln
( 1
y − 1998
))
= 1
( 1
y
− 1998
)2 · −1
y2
= −1
y − 1998 y2
d2
dy 2
(
ln
( 1
y − 1998
))
= 1 − 3996 y
(y − 1998 y2)2
So ln
( 1
y
− 1998
)
is convex on [0 , 1/3996]. If we had 0 < y i ≤ 1/3996 for all i we could apply Jensen. Since yi + yj ≤ 1/1998 for all i, j , we consider
( 1
a − 1998
) ( 1
b − 1998
)
≥
( 2
a + b − 1998
)2
⇐⇒ 1
ab − 1998
( 1
a + 1
b
)
≥ 4(a + b)2 − 4 · 1998
a + b
⇐⇒ (a + b)2 − 1998( a + b)3 ≥ 4ab − 4ab (a + b) · 1998
⇐⇒ (a − b)2 ≥ 1998( a + b)( a − b)2
which incidentally holds for any a + b ≤ 11998 . Hence, any two yi and yj may be set to their average while decreasing the sum in question; hence, we may assume yi ∈ (0 , 13996 ]. Now Jensen’s inequality shows that the minimum occurs when yi = 11998 n for all i, or when xi = 1998( n − 1) for all i. It is easy to see that this yields equality. 21. (Romania 99) Show that for all positive reals x1, . . . , x n with x1x2 · · · xn = 1, we have 1
n − 1 + x1
· · · + 1
n − 1 + xn
≤ 114 Solution. First, we prove a lemma: the maximum of the sum occurs when n − 1 of the xi are equal. Consider f (y) = 1
k+ey
for an arbitrary nonnegative constant k. We have f ′(y) = −ey
(k+ey)2
and f ′′ (y) = ey (ey −k)(k+ey )3 . Evidently f ′′ (y) ≥ 0 ⇐⇒ ey ≥ k. Hence,
f (y) has a single inflexion point where y = ln( k), where f (y) is convex over the interval ((ln( k), ∞). Now, we employ the substitution yi = ln( xi) so that y1 + · · · + yn = 0 and
n
∑
i=1
1
n − 1 + xi
=
n
∑
i=1
f (yi)We take k = n − 1 and write k0 = ln( n − 1). Suppose that y1 ≥ · · · ≥ ym ≥ k0 ≥
ym+1 ≥ · · · xn for some positive m. Then by, Majorization,
f (y1) + · · · + f (ym) ≤ (m − 1) f (k0) + f (y1 + · · · + ym − (m − 1) k0)But then, also by Majorization, (m − 1) f (k0) + f (ym+1 ) + · · · + f (yn) ≤ (n − 1) f
((m − 1) k0 + ym+1 + · · · + yn
n − 1
)
Otherwise, all of the yi are less than k0. In that case we may directly apply Majorization to equate n − 1 of the yi whilst increasing the sum in question. Hence, the lemma is valid. 7 N
Applying the lemma, it would suffice to show
kk + x + 1
k + 1
xk
≤ 1Clearing the denominators,
(
k2 + kxk
)
( k + x) ≤ k2 + k
(
x + 1
xk
)
x1−k
−xk + x + k ≤ x1−k
But now this is evident. We have Bernoulli’s inequality, since x1−k = (1 + ( x − 1)) 1−k ≥
1 + ( x − 1)(1 − k) = x + k − xk . Equality holds only where x = 1 or n = 2. 22. (Darij Grinberg) Show that for all positive reals a, b, c ,
√b + ca +
√c + ab +
√a + bc ≥ 4( a + b + c)
√(a + b)( b + c)( c + a)
7This n−1 equal value principle is particularly useful. If a differentiable function has a single inflexion point and is evaluated at narbitrary reals with a fixed sum, any minimum or maximum must occur where some n−1 variables are equal.
15 Solution 1. By Cauchy, we have √(a + b)( a + c) ≥ a + √bc . Now,
∑
cyc
√b + ca ≥ 4( a + b + c)
√(a + b)( b + c)( c + a)
⇐⇒ ∑
cyc
b + ca
√(a + b)( a + c) ≥ 4( a + b + c)Substituting our result from Cauchy, it would suffice to show
∑
cyc
(b + c)
√bc a ≥ 2( a + b + c)WLOG a ≥ b ≥ c, implying b + c ≤ c + a ≤ a + b and √bc a ≤ √ca b ≤ √ab c . Hence, by Chebyshev and AM-GM,
∑
cyc
(b + c)
√bc a ≥
(2( a + b + c))
(√bc a + √ca b + √ab c
)
3
≥ 2( a + b + c)as desired.
Solution 2. Let x = √b + c, y = √c + a, z = √a + b. Then x, y, z are the sides of acute triangle XY Z (in the typical manner), since x2 + y2 = a + b + 2 c > a + b = z2.The inequality is equivalent to
∑
cyc
xy2 + z2 − x2 ≥ x2 + y2 + z2
xyz
Recalling that y2 + z2 − x2 = 2 yz cos( X), we reduce this to the equivalent
∑
cyc
x2
cos( X) ≥ 2( x2 + y2 + z2)WLOG, we have x ≥ y ≥ z, implying 1cos( X) ≥ 1cos( Y ) ≥ 1cos( Z) , so that applying Chebyshev to the left reduces the desired to proving that the sum of the reciprocals of the cosines is at least 6. By AM-HM, 1cos( X) + 1cos( Y ) + 1cos( Z) ≥ 9cos( X) + cos( Y ) + cos( Z)But recall from triangle geometry that cos( X) + cos( Y ) + cos( Z) = 1 + rR and R ≥ 2r.The desired is now evident. 16 23. Show that for all positive numbers x1, . . . , x n,
x31
x21 + x1x2 + x22
x32
x22 + x2x3 + x23
· · · + x3
n
x2
n
xnx1 + x21
≥ x1 + · · · + xn
3
Solution. Observe that 0 = ( x1 −x2)+( x2 −x3)+ · · · +( xn −x1) = ∑ni=1
x3
i−x3
i+1
x2
i+xixi+1 +x2
i+1
.Hence, (where xn+1 = x1)
n
∑
i=1
x3
i
x2
i
xixi+1 x2
i+1
= 12
n
∑
i=1
x3
i
x3
i+1
x2
i
xixi+1 + x2
i+1
But now a3 + b3 ≥ 13 a3 + 23 a2b + 23 ab 2 + 13 b3 = 13 (a + b)( a2 + ab + b2). Hence, 12
n
∑
i=1
x3
i
x3
i+1
x2
i
xixi+1 + x2
i+1
≥ 12
n
∑
i=1
xi + xi+3
3 = 13
n
∑
i=1
xi
as desired. 24. Let a, b, c be positive reals such that a + b ≥ c; b + c ≥ a; and c + a ≥ b, we have 2a2(b + c) + 2 b2(c + a) + 2 c2(a + b) ≥ a3 + b3 + c3 + 9 abc
Solution. After checking that equality holds for ( a, b, c ) = ( t, t, t ) and (2 t, t, t ), it is apparent that more than straight AM-GM will be required. To handle the condition, put a = y + z, b = z + x, c = x + y with x, y, z ≥ 0. Now, the left hand side becomes 4x3 + 4 y3 + 4 z3 + 10 x2(y + z) + 10 y2(z + x) + 10 z2(x + y) + 24 xyz while the right hand side becomes 2 x3 + 2 y3 + 2 z3 + 12 x2(y + z) + 12 y2(z + x) + 12 z2(x + y) + 18 xyz .The desired is seen to be equivalent to x3 + y3 + z3 + 3 xyz ≥ x2(y + z) + y2(z +
x) + z2(x + y), which is Schur’s inequality. Equality holds where x = y = z, which gives ( a, b, c ) = ( t, t, t ), or when two of x, y, z are equal and the third is 0, which gives (a, b, c ) ∈ { (2 t, t, t ), (t, 2t, t ), (t, t, 2t)}.25. Let a, b, c be the lengths of the sides of a triangle. Prove that
a
√2b2 + 2 c2 − a2 + b
√2c2 + 2 a2 − b2 + c
√2a2 + 2 b2 − c2 ≥ √3
Solution 1. Again write a = y + z, b = z + x, and c = x + y, noting that x, y, z are positive. (Triangles are generally taken to be non-degenerate when used in inequalities.) We have
∑
cyc
a
√2b2 + 2 c2 − a2 = ∑
cyc
y + z
√4x2 + 4 xy + 4 xz + y2 + z2 − 2yz
17 Consider the convex function f (x) = 1√x . (As we shall see, Jensen almost always provides a tractable means of eliminating radicals from inequalities.) Put x+y +z = 1. We have
∑
cyc
(y + z)f (4x2 + 4 xy + 4 xz + y2 + z2 − 2yz ) ≥
(( y + z) + ( z + x) + ( x + y)) f
(∑
cyc
(y + z) (4 x2 + 4 xy + 4 xz + y2 + z2 − 2yz )(y + z) + ( z + x) + ( x + y)
)
= 2√2
√∑
cyc
4x2(y + z) + (4 xy 2 + 4 xyz ) + (4 xyz + 4 xz 2) + y3 + z3 − y2z − yz 2
Noting that ∑
cyc
4x2(y + z) + (4 xy 2 + 4 xyz ) + (4 xyz + 4 xz 2) + y3 + z3 − y2z − yz 2 =
∑
cyc
2x3 + 7 x2(y + z) + 8 xyz ,8( x + y + z)3 ≥ 3 ∑
cyc
2x3 + 7 x2(y + z) + 8 xyz
⇐⇒ ∑
sym
4x3 + 24 x2y + 8 xyz ≥ ∑
sym
3x3 + 21 x2y + 12 xyz
⇐⇒ 2x3 + 2 y3 + 2 z3 + 3 (x2(y + z) + y2(z + x) + z2(x + y)) ≥ 24 xyz
which follows by AM-GM. As a follow up on an earlier mentioned connection, oberserve the similarity of the above application of Jensen and the following inequality (which follows by H¨ older’s inequality)
(∑
i
αiβi
) ( ∑
i
αi
1
√βi
)2
≥
(∑
i
αi
)3
Solution 2 [by Darij Grinberg.] Let ABC be a triangle of side lengths a, b, c
in the usual order. Denote by ma, m b, m c the lengths of the medians from A, B, C
respectively. Recall from triangle goemetry that ma = 12
√2b2 + 2 c2 − a2, so that we need only show ama + bmb + cmc ≥ 2√3. But a triangle with side lengths ma, m b, m c, in turn, has medians of length 3a
4
, 3b
4
, and 3c
4
. The desired inequality is therefore equivalent to 43 ma
a
43 mb
b
43 mc
c
≥ 2√3 where we refer to the new triangle ABC . Recalling that
23
ma = AG , where G is the centroid, the desired is seen to be equivalent to the geometric inequality AG a + BG b + CG c ≥ √3. But we are done as we recall from triangle geometry that AM a + BM b + CM c ≥ √3 holds for any point inside triangle ABC .8
8For a complete proof of this last inequality, see post #14.
18 26. (IMO 99/2) For n ≥ 2 a fixed positive integer, find the smallest constant C such that for all nonnegative reals x1, . . . , x n,
∑
1≤i<j ≤n
xixj (x2
i
x2
j
) ≤ C
( n∑
i=1
xi
)4
Solution. The answer is C = 18 , which is obtained when any two xi are non-zero and equal and the rest are 0. Observe that by AM-GM, (x1 + · · · + xn)4 =
( n∑
i=1
x2
i
2 ∑
1≤i<j ≤n
xixj
)2
≥ 4
( n∑
i=1
x2
i
) (
2 ∑
1≤i<j ≤n
xixj
)
= 8 ∑
1≤i<j ≤n
xixjn∑
k=1
x2
k
But x21 + · · · + x2
n
≥ x2
i
x2
j
with equality iff xk = 0 for all k 6 = i, j . It follows that (x1 + · · · + xn)4 ≥ 8 ∑
1≤i<j ≤n
xixj
(x2
i
x2
j
)
as desired. 27. Show that for nonnegative reals a, b, c ,2a6 + 2 b6 + 2 c6 + 16 a3b3 + 16 b3c3 + 16 c3a3 ≥ 9a4(b2 + c2) + 9 b4(c2 + a2) + 9 c4(a2 + b2)
Solution 1. Consider
∑
cyc
(a − b)6 = ∑
cyc
a6 − 6a5b + 15 a4b2 − 20 a3b3 + 15 a2b4 − 6ab 5 + b6 ≥ 0and ∑
cyc
ab (a − b)4 = ∑
cyc
a5b − 4a4b2 + 6 a3b3 − 4a2b4 + ab 5 ≥ 0Adding six times the latter to the former yields the desired result.
Solution 2. We shall prove a6 − 9a4b2 + 16 a3b3 − 9a2b4 + b6 ≥ 0. We have
a6 − 2a3b3 + b6 = (a3 − b3)2
= ((a − b)( a2 + ab + b2))2
≥ (a − b)2(3 ab )2 = 9 a4b2 − 18 a3b3 + 9 a2b4
19 As desired. The result now follows from adding this lemma cyclicly. The main difficulty with this problem is the absence of a5b terms on the right and also the presence of
a4b2 terms on the right - contrary to where Schur’s inequality would generate them. Evidently AM-GM is too weak to be applied directly, since a6 + 2 a3b3 ≥ 3a4b2 cannot be added symmetrically to deduce the problem. By introducing the factor ( a − b)2,however, we weight the AM-GM by a factor which we “know” will be zero at equality, thereby increasing its sharpness. 28. Let 0 ≤ a, b, c ≤ 12 be real numbers with a + b + c = 1. Show that
a3 + b3 + c3 + 4 abc ≤ 932
Solution. Let f (a, b, c ) = a3 + b3 + c3 + 4 abc and g(a, b, c ) = a + b + c = 1. Because
f and g are polynomials, they have continuous first partial derivatives. Moreover, the gradient of g is never zero. Hence, by the theorem of Lagrange Multipliers ,any extrema occur on the boundary or where ∇f = λ∇g for suitable scalars λ. As
∇f =< 3a2 + 4 bc, 3b2 + 4 ca, 3c2 + 4 ab > and ∇g =< 1, 1, 1 >, we have
λ = 3a2 + 4 bc
= 3b2 + 4 ca
= 3c2 + 4 ab g(a, b, c ) = a + b + c = 1 We have 3 a2 + 4 bc = 3 b2 + 4 ca or ( a − b)(3 a + 3 b − 4c) = ( a − b)(3 − 7c) = 0 for any permutation of a, b, c . Hence, without loss of generality, a = b. Now, 3 a2 + 4 ac =3c2 + 4 a2 and a2 − 4ac + 3 c2 = ( a − c)( a − 3c) = 0. The interior local extrema therefore occur when a = b = c or when two of {a, b, c } are three times as large as the third. Checking, we have f (13 , 13 , 13 ) = 7 /27 < 13 /49 = f (17 , 37 , 37 ). Recalling that f (a, b, c ) is symmetric in a, b, c , the only boundary check we need is f (12 , t, 12 −t) ≤ 932 for 0 ≤ t ≤ 12 .We solve
h(t) = f
(12, t, 12 − t
)
= 18 + t3 +
(12 − t
)3
2 t
(12 − t
)
= 14 + t
4 − t2
2
h(t) is 14 at either endpoint. Its derivative h′(t) = 14 − t is zero only at t = 14 . Checking,
h(14 ) = f (12 , 14 , 14 ) = 932 . Since h(t) has a continuous derivative, we are done. (As a further check, we could observe that h′′ (t) = −1 < 0, which guarantees that h(14 ) is a local minimum.) 20 Usage Note. The use of Lagrange Multipliers in any solution will almost certainly draw hostile review, in the sense that the tiniest of errors will be grounds for null marks. If you consider multipliers on Olympiads, be diligent and provide explicit, kosher remarks about the continuous first partial derivatives of both f (x1, . . . , x n) and the constraint g(x1, . . . , x n) = k, as well as ∇g 6 = 0, before proceeding to solve the system ∇f = λ∇g. The main reason this approach is so severely detested is that, given sufficient computational fortitude (if you are able to sort through the relevant algebra and Calculus), it can and will produce a complete solution. The example provided here is included for completeness of instruction; typical multipliers solutions will not be as clean or painless. 9
(Vascile Cartoaje) Let p ≥ 2 be a real number. Show that for all nonnegative reals
a, b, c ,
3
√
a3 + pabc
1 + p + 3
√
b3 + pabc
1 + p + 3
√
c3 + pabc
1 + p ≤ a + b + c
Solution. By H¨ older,
(∑
cyc
3
√
a3 + pabc
1 + p
)3
≤
(∑
cyc
11 + p
) ( ∑
cyc
a
) ( ∑
cyc
a2 + pbc
)
But a2 + b2 + c2 ≥ ab + bc + ca (proven by AM-GM, factoring, or a number of other methods) implies that
∑
cyc
a2 + pbc ≤ (p + 1) ∑
cyc
a2 + 2 bc
3 = p + 1 3 (a + b + c)2
From which we conclude
(∑
cyc
3
√
a3 + pabc
1 + p
)3
≤ (a + b + c)3
as desired. 30. Let a, b, c be real numbers such that abc = −1. Show that
a4 + b4 + c4 + 3( a + b + c) ≥ a2
b + a2
c + b2
c + b2
a + c2
a + c2
b
Solution. First we homogenize, obtaining a4 + b4 + c4 + a3(b + c) + b3(c + a) + c3(a +
b) − 3abc (a + b + c) ≥ 0. As this is homogenous in the fourth degree, we can scale a, b, c
9Just how painful can the calculations get? Most multipliers solutions will tend to look more like than this solution.
21 by any real k and hence may now ignore abc = −1. Equality holds at a = b = c = 1, but also at a = b = 1 , c = −2, a = 1 , b = 0 , c = −1, and a number of unusual locations with the commonality that a + b + c = 0. Indeed, c = −a − b is a parametric solution, and we discover the factorization ( a + b + c)2(a2 + b2 + c2 − ab − bc − ca ) ≥ 0. (We are motivated to work with factorizations because there are essentially no other inequalities with a + b + c = 0 as an equality condition.) 31. (MOP 2003) Show that for all nonnegative reals a, b, c ,
a4(b2 + c2) + b4(c2 + a2) + c4(a2 + b2) +2abc (a2b + a2c + b2c + b2a + c2a + c2b − a3 − b3 − c3 − 3abc )
≥ 2a3b3 + 2 b3c3 + 2 c3a3
Solution. As was suggested by the previous problem, checking for equality cases is important when deciding how to solve a problem. We see that setting a = b produces equality. As the expression is symmetric, this certainly implies that b = c and c = a are equality cases. Hence, if P (a, b, c ) is the difference LHS - RHS, then ( a − b)( b − c)( c −
a)|P (a, b, c ). Obviously, if the problem is going to be true, ( a−b) must be a double root of P , and accordingly we discover the factorization P (a, b, c ) = ( a − b)2(b − c)2(c − a)2.The result illustrated above was no accident. If ( x−y) divides a symmetric polynomial
P (x, y, z ), then ( x − y)2 divides the same polynomial. If we write P (x, y, z ) = ( x −
y)Q(x, y, z ), then ( x − y)Q(x, y, z ) = P (x, y, z ) = P (y, x, z ) = ( y − x)Q(y, x, z ), which gives Q(x, y, z ) = −Q(y, x, z ). Hence Q(x, x, z ) = 0, and ( x − y) also divides Q(x, y, z ). 32. (Cezar Lupu) Let a, b, c be positive reals such that a + b + c + abc = 4. Prove that
a
√b + c + b
√c + a + c
√a + b ≥√22 · (a + b + c)
Solution. By Cauchy
(∑
cyc
a√b + c
) ( ∑
cyc
a
√b + c
)
≥ (a + b + c)2
But, also by Cauchy,
√(a + b + c) ( a(b + c) + b(c + a) + c(a + b)) ≥ ∑
cyc
a√b + c
Hence, ∑
cyc
a
√b + c ≥√22 · (a + b + c) ·
√ a + b + cab + bc + ca
22 And we need only show a + b + c ≥ ab + bc + ca . Schur’s inequality for r = 1 can be expressed as 9abc a+b+c ≥ 4( ab + bc + ca ) − (a + b + c)2. Now, we suppose that
ab + bc + ca > a + b + c, and have 9abc a + b + c ≥ 4( ab + bc + ca ) − (a + b + c)2
(a + b + c) (4 − (a + b + c)) = abc (a + b + c)Hence, a + b + c < 3. But then abc < 1, which implies 4 = a + b + c + abc < 4. Contradiction, as desired. 33. (Iran 1996) Show that for all positive real numbers a, b, c ,(ab + bc + ca )
( 1(a + b)2 + 1(b + c)2 + 1(c + a)2
)
≥ 94
Solution. Fearless courage is the foundation of all success. 10 When everything else fails, return to the sure-fire strategy of clearing all denominators. In this case, we obtain 4( a + b)2(b + c)2(c + a)2(ab + bc + ca )
( 1(a + b)2 + 1(b + c)2 + 1(c + a)2
)
=
∑
sym
4a5b + 8 a4b2 + 10 a4bc + 6 a3b3 + 52 a3b2c + 16 a2b2c2
on the left, and on the right, 9( a + b)2(b + c)2(c + a)2 =
∑
sym
9a4b2 + 9 a4bc + 9 a3b3 + 54 a3b2c + 15 a2b2c2
Canceling like terms, we seek
∑
sym
4a5b − a4b2 + a4bc − 3a3b3 − 2a3b2c + a2b2c2
Sure enough, this is true, since 3a5b+ab 5
4
≥ a4b2 and a4b2+a2b4
2
≥ a3b3 by AM-GM, and
abc (a3 + b3 + c3 − a2(b + c) + b2(c + a) + c2(a + b) + 3 abc ) ≥ 0 by Schur. 34. (Japan 1997) Show that for all positive reals a, b, c ,(a + b − c)2
(a + b)2 + c2 + (b + c − a)2
(b + c)2 + a2 + (c + a − b)2
(c + a)2 + b2 ≥ 35
10 Found on a fortune cookie by Po-Ru Loh while grading an inequality on 2005 Mock IMO Day 2 that was solved by brutal force.
23 Solution. Put a + b + c = 3 so that equality will hold at a = b = c = 1 and suppose that there exists some k for which (b + c − a)2
(b + c)2 + a2 = (3 − 2a)2
(3 − a)2 + a2 ≥ 15 + ka − k
for all positive a, b, c ; such an inequality would allow us to add cyclicly to deduce the desired inequality. As the inequality is parametrically contrived to yield equality where
a = 1, we need to find k such that a = 1 is a double root. At a = 1, the derivative on the left is (2(3 −2a)·− 2)((3 −a)2+a2)−((3 −2a)2)(2(3 −a)·− 1+2 a)((3 −a)2+a2)2 = −18 25 . The derivative on the right is k, so we set k = −18 25 . But for this k we find (3 − 2a)2 −
(15 + ka − k
) ((3 − a)2 + a2) = 18 25 − 54 a2
25 + 36 a3
25 = 18 25 (a − 1) 2(2 a + 1) ≥ 0as desired. Alternatively, we could have used AM-GM to show a3 + a3 + 1 ≥ 3a2. As hinted at by a previous problem, inequalities are closely linked to polynomials with roots of even multiplicity. The isolated manipulation idea used in this solution offers a completely different approach to the inequalities which work with every term. 35. (MOP 02) Let a, b, c be positive reals. Prove that
( 2ab + c
)23
+
( 2bc + a
)23
+
( 2ca + b
)23
≥ 3
Solution. Suppose that there exists some r such that
( 2ab + c
)23
≥ 3ar
ar + br + cr
We could sum the inequality cyclicly to deduce what we want. Since equality holds at
a = b = c = 1, we use derivatives to find a suitable r. At the said equality case, on the left, the partial derivative with respect to a is 23 , while the same derivative on the right is 23 r. Equating the two we have r = 1. (This is necessary since otherwise the inequality will not hold for either a = 1 + ≤ or a = 1 − ≤.) 11 Now, 3aa + b + c ≤ 3a
3 3
√
a · (b+c
2
)2
11 Actually, even this is a special case of the general sense that the convexity of one side must exceed the convexity of the other. More precisely, we have the following result: Let fand gfunctions over the domain Dwith continuous partial derivatives. If f(ν)≥g(ν) for all ν∈D, then at every equality case ν0,
∇(f−g)( ν0) = 0and every component of ∇2(f−g) ( ν0) is nonnegative.
24 = a23
(b+c
2
)23
=
( 2ab + c
)23
by AM-GM, as desired. 36. (Mildorf) Let n ≥ 2 be an integer. Prove that for all reals a1, a 2, . . . , a n > 0 and reals
p, k ≥ 1, ( a1 + a2 + · · · + an
ap
1
ap
2
· · · + apn
)k
≥ ak
1
ak
2
· · · + akn
apk
1
apk
2
· · · + apk n
where inequality holds iff p = 1 or k = 1 or a1 = a2 = · · · = an, flips if instead 0 < p < 1, and flips (possibly again) if instead 0 < k < 1.
Solution. Taking the kth root of both sides, we see that the inequality is equivalent to n∑
i=1
k
√
aki
ak
1
ak
2
· · · + akn
≥
n
∑
i=1
k
√
apk i
apk
1
apk
2
· · · apk n
WLOG, suppose that a1 ≥ a2 ≥ · · · ≥ an. We prove a lemma. Let Si = api
ap
1+··· +apn
and
Ti = aqi
aq
1+··· +aqn
for i = 1 , 2, . . . , n where 0 < q < p . Then the sequence S1, S 2, . . . , S n
majorizes the sequence T1, T 2, . . . , T n.To prove the claim, we note that S1 ≥ · · · ≥ Sn and T1 ≥ · · · ≥ Tn and have, for
m ≤ n,
m
∑
i=1
Si ≥
m
∑
i=1
Ti
⇐⇒ (ap
1
· · · + apm) ( aq
1
· · · + aqn) ≥ (aq
1
· · · + aqm) ( ap
1
· · · + apn)
⇐⇒ (ap
1
· · · + apm) (aqm+1 + · · · + aqn
) ≥ (aq
1
· · · + aqm) (apm+1 + · · · + apn
)
⇐⇒ ∑
(i,j )| { 1≤i≤m<j ≤n}
api aqj − aqi apj ≥ 0Which is obvious. In particular, m = n is the equality case, and the claim is established. But now the desired is a direct consequence of the Majorization inequality applied to the sequences in question and the function f (x) = k
√x.37. (Vascile Cartoaje) Show that for all real numbers a, b, c,(a2 + b2 + c2)2 ≥ 3 (a3b + b3c + c3a)
25 Solution. We will be content to give the identity (a2 + b2 + c2)2 − 3( a3b + b3c + c3a) = 12
∑
cyc
(a2 − 2ab + bc − c2 + ca )2
Any Olympiad partipant should be comfortable constructing various inequalities through well-chosen squares. Here, we could certainly have figured we were summing the square of a quadratic that is 0 when a = b = c such that no term a2bc is left uncancelled. A good exercise is to show that equality actually holds iff a = b = c or, for some cyclic permutation, a : b : c ≡ sin 2 (4π
7
) : sin 2 (2π
7
) : sin 2 (π
7
).38. (Anh-Cuong) Show that for all nonnegative reals a, b, c ,
a3 + b3 + c3 + 3 abc ≥ ab √2a2 + 2 b2 + bc √2b2 + 2 c2 + ca √2c2 + 2 a2
Solution. Upon observing that this inequality is stronger than Schur’s inequality for
r = 1, we are inspired to prove a sharp lemma to eliminate the radical. Knowing that √2x2 + 2 y2 ≥ x + y ≥ 2xy x+y , we seek a combination of the latter two that exceeds the former. We find 3x2 + 2 xy + 3 y2
2( x + y) ≥ √2x2 + 2 y2
This follows from algebra, since (3 x2 + 2 xy + 3 y2)2 = 9 x4 + 12 x3y + 22 x2y2 + 12 xy 3 +9y4 ≥ 8x4 + 16 x3y + 16 x2y2 + 16 xy 3 + 8 y4 = 4( x + y)2(2 x2 + 2 y2), so that (3 x2 + 2 xy +3y2)2 − 4( x + y)2(2 x2 + 2 y2) = x4 − 4x3y + 6 x2y2 − 4xy 3 + y4 = ( x − y)4 ≥ 0. Now,
∑
cyc
ab √2a2 + 2 b2 ≤ ∑
cyc
(3 a2 + 2 ab + 3 b2)ab
2( a + b)So it would suffice to show
∑
cyc
a(a − b)( a − c) = ∑
cyc
(a3 + abc − ab (a + b))
≥ ∑
cyc
(3 a2 + 2 ab + 3 b2)ab
2( a + b) − ab (a + b)= ∑
cyc
3a3b + 2 a2b2 + 3 ab 3 − 2a3b − 4a2c2 − 2ab 3
2( a + b)= ∑
cyc
ab (a − b)2
2( a + b)But ∑
cyc
(b + c − a)( b − c)2 = 2 ∑
cyc
a(a − b)( a − c)26 so that the desired is
∑
cyc
(
b + c − a − bc b + c
)
(b − c)2 ≥ 0which is evident, since without loss of generality we may assume a ≥ b ≥ c and find
(
a + b − c − ab a + b
)
(a − b)2 ≥ 0
(
c + a − b − ac a + c
) ((a − c)2 − (b − c)2) ≥ 0
(
b + c − a − bc b + c
)
(b − c)2 +
(
c + a − b − ac a + c
)
(b − c)2 ≥ 0The key to this solution was the sharp upper bound on the root-mean-square. At first glance our lemma seems rather arbitrary and contrived. Actually, it is a special case of a very sharp bound on the two variable power mean that I have conjectured and proved.
Mildorf’s Lemma 1 Let k ≥ − 1 be an integer. Then for all positive reals a and b,
(1 + k)( a − b)2 + 8 ab
4( a + b) ≥ k
√ak + bk
2
with equality if and only if a = b or k = ±1, where the power mean k = 0 is interpreted to be the geometric mean √ab . Moreover, if k < −1, then the inequality holds in the reverse direction, with equality if and only if a = b.
Usage Note. As of early November 2005, I have proven an extension of this lemma to additional values of k.12 Thus, you may rest assured that the result stated above is true. I was unable to get this result published, so I have instead posted the proof here as “ASharpBound.pdf.” However, the proof is rather difficult (or at least so I think, being as though it took me nearly half a year) and the lemma is far from mainstream. Thus, should you require it on an Olympiad, you should prove it for whatever particular value of k you are invoking. This is not terribly difficult if k is a small integer. One simply takes the kth power of both sides and factors the difference of the two sides as
(a − b)4 · P (a, b ), etc.
For x ≥ y ≥ 1, prove that
x
√x + y + y
√y + 1 + 1
√x + 1 ≥ y
√x + y + x
√x + 1 + 1
√y + 1
12 In particular, the inequality holds for all kin ( −∞ ,−1) ,{− 1,0,1},(1 ,3/2] ,[2 ,∞) with the signs ≤,≥,≤
,≥respectively, with equality iff a=bor k=±1.
27 Solution. By observation, equality holds when y = 1 and when x = y. Combining this with the restriction, it makes sense to write x = y + a and y = 1 + b where a, b ≥ 0. Now we can write
x − y
√x + y + y − 1
√y + 1 + 1 − x
√1 + x ≥ 0
⇐⇒ a
√2 + a + 2 b + b
√2 + b ≥ a + b
√2 + a + b
But this is evident by Jensen’s inequality applied to the convex function f (x) = 1√x ,since
af (2 + a + 2 b) + bf (2 + b) ≥ (a + b)f
(a(2 + a + 2 b) + b(2 + b)
a + b
)
= (a + b)f
((a + b)2 + 2( a + b)
a + b
)
= a + b
√2 + a + b
as desired. 40. (MOP) For n ≥ 2 a fixed positive integer, let x1, . . . , x n be positive reals such that
x1 + x2 + · · · + xn = 1
x1
1
x2
· · · + 1
xn
Prove that 1
n − 1 + x1
1
n − 1 + x2
· · · + 1
n − 1 + xn
≤ 1
Solution. We will prove the contrapositive. (We are motivated to do this for two good reasons: 1) it is usually difficult the show that the sum of some reciprocals is bounded above, and 2) the given relation in its current form is an abomination.) Take
yi = 1
n−1+ xi
, and for the sake of contradiction assume y1 + · · · + yn > 1. Since the yi
are too large, the xi are too small and we shall prove 1
x1
· · · + 1
xn
x 1 + · · · + xn.Since xiyi = 1 − (n − 1) yi, we have (n − 1) yi > (n − 1)
(
yi + 1 −
n
∑
j=1
yj
)
= (n − 1) yi − 1 +
n
∑
j=1
(1 − (n − 1) yj )= −xiyi +
n
∑
j=1
xj yj (∗)=⇒ n − 1
xi
−1 +
n
∑
j=1
xj yj
xiyi
(∗∗ )28 Summing () over i,(n − 1)
( 1
x1
· · · + 1
xn
)
n
∑
i=1
xiyi
(( n∑
j=1
1
xj yj
)
− 1
xiyi
)
But by Cauchy and (), we have
( n∑
j=1
1
xj yj
)
− 1
xiyi
≥ (n − 1) 2
(∑nj=1 xj yj
)
− xiyi
(n − 1) 2
(n − 1) yi
= n − 1
yi
Hence, (n − 1)
( 1
x1
· · · + 1
xn
)
n
∑
i=1
xiyi
(n − 1
yi
)
= ( n − 1)( x1 + · · · + xn)as desired. 41. (Vascile Cartoaje) Show that for positive reals a, b, c ,14a2 − ab + 4 b2 + 14b2 − bc + 4 c2 + 14c2 − ca + 4 a2 ≥ 97( a2 + b2 + c2)
Solution. Upon expansion, we see that it is equivalent to
∑
sym
56 a6 − 28 a5b + 128 a4b2 + 44 a3b3 + 95 2 a4bc + 31 a3b2c − 45 2 a2b2c2 ≥ 0We conjure up the following inequalities:
∑
sym
a6 − 2a5b + a4bc ≥ 0 (1)
∑
sym
a5b − 4a4b2 + 3 a3b3 ≥ 0 (2)
∑
sym
a4b2 − a4bc − a3b3 + 2 a3b2c − a2b2c2 ≥ 0 (3)
∑
sym
a4bc − 2a3b2c + a2b2c2 ≥ 0 (4) (1) and (4) follow from Schur’s inequality for r = 4 and r = 1 (multiplied by abc )respectively. (2) is the result of expanding ∑
cyc
ab (a − b)4 ≥ 0, and (3) is the expanded form of the famous ( a − b)2(b − c)2(c − a)2 ≥ 0. The desired now follows by subtracting 56 times (1), 84 times (2), 208 times (3), 399 2 times (4), and then simple AM-GM to clear the remaining a2b2c2.29 This is about as difficult as a dumbass solution can get. A good general strategy is to work with the sharpest inequalities you can find until you reduce a problem to something obvious, starting with the most powerful (most bunched, in this case ∑
sym
a6) term and work your way down to the weak terms while keeping the most powerful term’s coefficient positive. My solution to this problem starts with (1), Schur with r = 4 (Schur is stronger for larger r), which is almost certainly sharper than the inequality in question. Next, inequality (2) is a sharp cyclic sum to use the a5b terms. In particular, it relates terms involving only two of the three variables. Most of the time, the only inequality that can “pull up” symmetric sums involving three variables to stronger ones involving just two is Schur, although it does so at the expense of a very strong term with only one variable. Hence, we made a logical choice. Inequality (3) is extremely sharp, and allowed us to obtain more a4bc and a3b3 terms simultaneously. In particular, it was necessary to cancel the a3b3 terms. I’ll note that this inequality is peculiar to sixth degree symmetry in three variables - it does not belong to a family of similar, nice inequalities. Finally, inequality (4), which is a handy corollary to (3), is another Schur. Every inequality we have used so far is quite sharp, and so it is no surprise that the leftovers are the comparatively weak AM-GM. 42. (Reid Barton, IMO Shortlist 03/A6.) Let n ≥ 2 be a positive integer and x1, x 2, . . . , x n,y1, y 2, . . . , y n a sequence of 2 n positive reals. Suppose z2, z 3, . . . , z 2n is such that z2
i+j
≥
xiyj for all i, j ∈ { 1, . . . , n }. Let M = max {z2, z 3, . . . , z 2n}. Prove that
(M + z2 + z3 + · · · + z2n
2n
)2
≥
(x1 + · · · + xn
n
) ( y1 + · · · + yn
n
)
Reid’s official solution. Let max( x1, . . . , x n) = max( y1, . . . , y n) = 1. (We can do this by factoring X from every xi, Y from every yj , and √XY from every zi+j
without changing the sign of the inequality.) We will prove M + z2 + · · · + z2n ≥
x1 + x2 + · · · + xn + y1 + y2 + · · · + yn, after which the desired follows by AM-GM. We will show that the number of terms on the left which are greater than r is at least as large as the number of terms on the right which are greater than r, for all r ≥ 0. For r ≥ 1, the claim is obvious, since all terms on the right are at most 1. Now take
r < 1. Let A and B denote the set of i for which xi > r and the set of j for which
yj > r respectively, and write a = |A|, b = |B|. Evidently, from our scaling, a, b ≥ 1. Now, xi > r and yj > r implies zi+j ≥ √xiyj ≥ r. Hence, if C is the set of k for which zk > r , we have |C| ≥ | A + B|, where the set addition is defined by the set of possible values if we take an element of A and add it to an element of B. How-ever, |A + B| ≥ | A| + |B| − 1, since if A and B consist of the values p1 < · · · < p a and
q1 < · · · < q b respectively we have all of the values p1 +q1 < . . . < p a +q1 < · · · < p a +qb
in A + B. Hence, |C| ≥ a + b − 1. Since |C| ≥ 1, there is some zk > r , and hence,
M > r . Therefore, the left side of the inequality in question has at least a + b terms which exceed r, as desired. •
30 The preponderance of difficulty here stemmed from dealing with the superabundance of givens, especially the mysterious M . Scaling allowed us to introduce some degree of control and, with marked audacity, a profoundly clever idea. As it turned out, the in-equality was no sharper than simple AM-GM! It is my opinion that it is highly unlikely that a problem as staggeringly pernicious as this one will appear on an Olympiad - at least in the foreseeable future. Nevertheless, I have included it here for the purpose of illustrating just how unusual and creative a solution can be.
3 Problems
(MOP 04) Show that for all positive reals a, b, c ,
( a + 2 ba + 2 c
)3
+
( b + 2 cb + 2 a
)3
+
(c + 2 ac + 2 b
)3
≥ 32. (MOP) Show that if k is a positive integer and a1, a 2, . . . , a n are positive reals which sum to 1, then n∏
i=1
1 − aki
aki
≥ (nk − 1)n
Let a1, a 2, . . . , a n be nonnegative reals with a sum of 1. Prove that
a1a2 + a2a3 + · · · + an−1an ≤ 144. (Ukraine 01) Let a, b, c, x, y, z be nonnegative reals such that x + y + z = 1. Show that
ax + by + cz + 2 √(ab + bc + ca )( xy + yz + zx ) ≤ a + b + c
Let n > 1 be a positive integer and a1, a 2, . . . , a n positive reals such that a1a2 . . . a n = 1. Show that 11 + a1
· · · + 11 + an
≤ a1 + · · · + an + n
(Aaron Pixton) Let a, b, c be positive reals with product 1. Show that 5 + ab + bc + ca ≥ (1 + a)(1 + b)(1 + c)7. (Valentin Vornicu 13 ) Let a, b, c, x, y, z be arbitrary reals such that a ≥ b ≥ c and either
x ≥ y ≥ z or x ≤ y ≤ z. Let f : R → R+0 be either monotonic or convex, and let k be a positive integer. Prove that
f (x)( a − b)k(a − c)k + f (y)( b − c)k(b − a)k + f (z)( c − a)k(c − b)k ≥ 0
13 This improvement is more widely known than the other one in this packet, and is published in his book,
Olimpiada de Matematica... de la provocare la experienta , GIL Publishing House, Zalau, Romania. (In English, “The Math Olympiad... from challenge to experience.”)
31 8. (IMO 01/2) Let a, b, c be positive reals. Prove that
a
√a2 + 8 bc + b
√b2 + 8 ca + c
√c2 + 8 ab ≥ 19. (USAMO 04/5) Let a, b, c be positive reals. Prove that
(a5 − a2 + 3 ) ( b5 − b2 + 3 ) ( c5 − c2 + 3 ) ≥ (a + b + c)3
(Titu Andreescu) Show that for all nonzero reals a, b, c ,
a2
b2 + b2
c2 + c2
a2 ≥ ac + cb + ba
(IMO 96 Shortlist) Let a, b, c be positive reals with abc = 1. Show that
ab a5 + b5 + ab + bc b5 + c5 + bc + ca c5 + a5 + ca ≤ 112. Let a, b, c be positive reals such that a + b + c = 1. Prove that
√ab + c + √bc + a + √ca + b ≥ 1 + √ab + √bc + √ca
(APMO 2005/2) Let a, b, c be positive reals with abc = 8. Prove that
a2
√(a3 + 1) ( b3 + 1) + b2
√(b3 + 1) ( c3 + 1) + c2
√(c3 + 1) ( a3 + 1) ≥ 4314. Show that for all positive reals a, b, c ,
a3
b2 − bc + c2 + b3
c2 − ca + a2 + c3
a2 − ab + b2 ≥ a + b + c
(USAMO 97/5) Prove that for all positive reals a, b, c ,1
a3 + b3 + abc + 1
b3 + c3 + abc + 1
c3 + a3 + abc ≤ 1
abc
(Mathlinks Lore) Show that for all positive reals a, b, c, d with abcd = 1, and k ≥ 2, 1(1 + a)k + 1(1 + b)k + 1(1 + c)k + 1(1 + d)k ≥ 22−k
(IMO 05/3) Prove that for all positive a, b, c with product at least 1,
a5 − a2
a5 + b2 + c2 + b5 − b2
b5 + c2 + a2 + c5 − c2
c5 + a2 + b2 ≥ 032 18. (Mildorf) Let a, b, c, k be positive reals. Determine a simple, necessary and sufficient condition for the following inequality to hold: (a + b + c)k (akbk + bkck + ckak) ≤ (ab + bc + ca )k(ak + bk + ck)19. Let a, b, c be reals with a + b + c = 1 and a, b, c ≥ − 34 . Prove that
aa2 + 1 + bb2 + 1 + cc2 + 1 ≤ 910 20. (Mildorf) Show that for all positive reals a, b, c ,
3
√4a3 + 4 b3 + 3
√4b3 + 4 c3 + 3
√4c3 + 4 a3 ≤ 4a2
a + b + 4b2
b + c + 4c2
c + a
Let a, b, c, x, y, z be real numbers such that (a + b + c)( x + y + z) = 3 , (a2 + b2 + c2)( x2 + y2 + z2) = 4 Prove that
ax + by + cz ≥ 022. (Po-Ru Loh) Let a, b, c be reals with a, b, c > 1 such that 1
a2 − 1 + 1
b2 − 1 + 1
c2 − 1 = 1 Prove that 1
a + 1 + 1
b + 1 + 1
c + 1 ≤ 123. (Weighao Wu) Prove that (sin x)sin x < (cos x)cos x
for all real numbers 0 < x < π
4
.24. (Mock IMO 05/2) Let a, b, c be positive reals. Show that 1 < a
√a2 + b2 + b
√b2 + c2 + c
√c2 + a2 ≤ 3√2225. (Gabriel Dospinescu) Let n ≥ 2 be a positive integer. Show that for all positive reals
a1, a 2, . . . , a n with a1a2 . . . a n = 1,
√a21 + 1 2 + · · · +
√a2
n
1 2 ≤ a1 + · · · + an
33 26. Let n ≥ 2 be a positive integer, and let k ≥ n−1
n
be a real number. Show that for all positive reals a1, a 2, . . . , a n,
( (n − 1) a1
a2 + · · · + an
)k
+
( (n − 1) a2
a3 + · · · + an + a1
)k
· · · +
( (n − 1) an
a1 + · · · + an−1
)k
≥ n
(Mildorf) Let a, b, c be arbitrary reals such that a ≥ b ≥ c, and let x, y, z be nonnegative reals with x + z ≥ y. Prove that
x2(a − b)( a − c) + y2(b − c)( b − a) + z2(c − a)( c − b) ≥ 0and determine where equality holds. 28. (USAMO 00/6) Let n ≥ 2 be an integer and S = {1, 2, . . . , n }. Show that for all nonnegative reals a1, a 2, . . . , a n, b 1, b 2, . . . , b n,
∑
i,j ∈S
min {aiaj , b ibj } ≤ ∑
i,j ∈S
min {aibj , a j bi}
(Kiran Kedlaya) Show that for all nonnegative a1, a 2, . . . , a n,
a1 + √a1a2 + · · · + n
√a1 · · · an
n ≤ n
√
a1 · a1 + a2
2 · · · a1 + · · · + an
n
(Vascile Cartoaje) Prove that for all positive reals a, b, c such that a + b + c = 3,
aab + 1 + bbc + 1 + cca + 1 ≥ 3231. (Gabriel Dospinescu) Prove that ∀a, b, c, x, y, z ∈ R+| xy + yz + zx = 3,
a(y + z)
b + c + b(z + x)
c + a + c(x + y)
a + b ≥ 332. (Mildorf) Let a, b, c be non-negative reals. Show that for all real k,
∑
cyc
max( ak, b k)( a − b)2
2 ≥ ∑
cyc
ak(a − b)( a − c) ≥ ∑
cyc
min( ak, b k)( a − b)2
2(where a, b, c 6 = 0 if k ≤ 0) and determine where equality holds for k > 0, k = 0, and
k < 0 respectively. 33. (Vascile Cartoaje) Let a, b, c, k be positive reals. Prove that
ab + ( k − 3) bc + ca
(b − c)2 + kbc + bc + ( k − 3) ca + ab
(c − a)2 + kca + ca + ( k − 3) ab + bc
(a − b)2 + kab ≥ 3( k − 1)
k
(Taiwan? 02) Show that for all positive a, b, c, d ≤ k, we have
abcd
(2 k − a)(2 k − b)(2 k − c)(2 k − d) ≤ a4 + b4 + c4 + d4
(2 k − a)4 + (2 k − b)4 + (2 k − c)4 + (2 k − d)4
34 |
4909 | https://courses.lumenlearning.com/cuny-kbcc-microeconomics/chapter/interpreting-slope/ | Module 1: Economic Thinking
Interpreting Slope
Learning Objectives
Differentiate between a positive relationship and a negative relationship
Figure 1. This skier speeds down the slope in an Olympic race. What’s your guess as to the steepness, or slope, of this ski hill?
What the Slope Means
The concept of slope is very useful in economics, because it measures the relationship between two variables. A positive slope means that two variables are positively related—that is, when x increases, so does y, and when x decreases, y also decreases. Graphically, a positive slope means that as a line on the line graph moves from left to right, the line rises. We will learn in other sections that “price” and “quantity supplied” have a positive relationship; that is, firms will supply more when the price is higher.
Figure 1. Positive Slope.
A negative slope means that two variables are negatively related; that is, when x increases, y decreases, and when x decreases, y increases. Graphically, a negative slope means that as the line on the line graph moves from left to right, the line falls. We will learn that “price” and “quantity demanded” have a negative relationship; that is, consumers will purchase less when the price is higher.
Figure 2. Negative slope.
A slope of zero means that y is constant no matter the value of x. Graphically, the line is flat; the rise over run is zero.
Figure 3. Slope of Zero
The unemployment-rate graph in Figure 4, below, illustrates a common pattern of many line graphs: some segments where the slope is positive, other segments where the slope is negative, and still other segments where the slope is close to zero.
Figure 4. U.S. Unemployment Rate, 1975–2014.
Try It
Calculating Slope
The slope of a straight line between two points can be calculated in numerical terms. To calculate slope, begin by designating one point as the “starting point” and the other point as the “end point” and then calculating the rise over run between these two points.
Try It
Use the graph to find the slope of the line.
Figure 5.
Show Answer
Start from a point on the line, such as latex[/latex] and move vertically until in line with another point on the line, such as latex[/latex]. The rise is 2 units. It is positive as you moved up.
Next, move horizontally to the point latex[/latex]. Count the number of units. The run is 4 units. It is positive as you moved to the right.
Then solve using the formula:
[latex]\displaystyle \text{Slope }=\frac{\text{rise}}{\text{run}}[/latex]
so
[latex]\displaystyle \text{Slope}=\frac{2}{4}=\frac{1}{2}[/latex]
Try It
These next questions allow you to get as much practice as you need, as you can click the link at the top of the first question (“Try another version of these questions”) to get a new set of questions. Practice until you feel comfortable doing the questions and then move on.
Graphs of economic relationships are not always straight lines. In this course, you will often see nonlinear (curved) lines, like Figure 6, which shows the relationship between quantity of output being produced and the cost of producing that output. As the quantity of output increases, the total cost increases at a faster rate. Table 1 shows the data behind this graph.
| |
| Table 1: Total Cost Curve |
| Quantity of Output (Q) | Total Cost (TC) |
| 1 | $1 |
| 2 | $4 |
| 3 | $9 |
| “Point A” | 4 | $16 |
| “Point B” | 5 | $25 |
| 6 | $36 |
| 7 | $49 |
| 8 | $64 |
| 9 | $81 |
| 10 | $100 |
Figure 6. In this example, the total cost of production increase at a faster rate when the quantity of output increases.
We can interpret nonlinear relationships similarly to the way we interpret linear relationships. Their slopes can be positive or negative. We can calculate the slopes similarly also, looking at the rise over the run of a segment of a curve.
As an example, consider the slope of the total cost curve, above, between points A & B. Going from point A to point B, the rise is the change in total cost (i.e. the variable on the vertical axis):
$25 – $16 = $9
Similarly, the run is the change in quantity (i.e. the variable on the horizontal axis):
5 – 4 = 1
Thus, the slope of a straight line between these two points would be 9/1 = 9. In other words, as we increase the quantity of output produced by one unit, the total cost of production increases by $9.
Try It
Suppose the slope of a line were to increase. Graphically, that means it would get steeper. Suppose the slope of a line were to decrease. Then it would get flatter. These conditions are true whether or not the slope was positive or negative to begin with. A lower positive slope means a flatter upward tilt to the curve, which you can see in Figure 6 at low levels of output. A higher positive slope means a steeper upward tilt to the curve, which you can see at higher output levels.
A negative slope that is larger in absolute value (that is, more negative) means a steeper downward tilt to the line. A slope of zero is a horizontal line. A vertical line has an infinite slope.
Suppose a line has a larger intercept. Graphically, that means it would shift out (or up) from the old origin, parallel to the old line. This is shown in Figure 7, below, as the shift from the line labeled Y to the line labeled Y1. If a line has a smaller intercept, it would shift in (or down), parallel to the old line.
Figure 7. A larger y-intercept shifts the entire graph to cross the y-axis at a higher point.
Glossary
negative slope:
: indicates that two variables are negatively related; when one variable increases, the other decreases, and when one variable decreases, the other increases
positive slope:
: indicates that two variables are positively related; when one variable increases, so does the other, and when one variable decreases, the other also decreases
slope:
: the change in the vertical axis divided by the change in the horizontal axis
slope of zero:
: indicates that there is a constant relationship between two variables: when one variable changes, the other does not change
Candela Citations
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Principles of Microeconomics Appendix. Authored by: OpenStax College. Provided by: Rice University. Located at: License: CC BY: Attribution. License Terms: Download for free at
Go Time. Authored by: Jon Wick. Located at: License: CC BY: Attribution |
4910 | https://www.bgcs.org.uk/wp-content/uploads/2020/05/FINAL-Cx-Ca-Version-for-submission.pdf | 1 British Gynaecological Cancer Society (BGCS) Cervical Cancer Guidelines: Recommendations for Practice Authors: Nick Reed, Janos Balega, Tara Barwick, Lynn Buckley, Kevin Burton, Gemma Eminowicz, Jenny Forrest, Raji Ganesan, Cathrine Holland, Tamara Howe, Thomas Ind, Rema Iyer, Sonali Kaushik, Robert Music, Azmat Sadozye, Smruta Shanbhag, Nadeem Siddiqui, Victoria Stewart, Sheeba Syed, Natalie Percival, Natasha Lauren Whitham, Andy Nordin, Christina Fotopoulou, International Reviewers: Gemma Kenter, Frederic Amant, David Cibula Editorial Assistant: Natasha Lauren Whitham The remit of this guideline is to collate and propose evidence-based guidelines for the diagnosis and management of adult patients with cervical cancer being treated in the United Kingdom. Grades of recommendations Recommendations are graded as per the Royal College of Obstetricians and Gynaecologists document. Clinical Governance Advice No. 1: Guidance for the Development of RCOG Green-top Guidelines, available on the RCOG website at: Evidence was searched in the Cochrane Central Register of Controlled Trials (CENTRAL, The Cochrane Library 2014, Issue 12), MEDLINE and EMBASE up to April 2018, registers of clinical trials, abstracts of scientific meetings, reference lists of included studies and contacted experts in the field. Guidelines development process 1) These guidelines are the property of the BGCS, and the Society reserves the right to amend/withdraw the guidelines. 2) The guideline development process is detailed below: a) Chair, officers, council and guidelines committee (GC) nominated a lead for each guideline topic; b) Lead then identified a team called the guideline team (GT) to develop the 1st draft; c) 1st draft was submitted to the GC; d) GC approved draft and recommended changes; e) Changes were accepted by the GT who produced the guidelines; f) 2nd draft was then submitted to council members and officers; g) Council and officers approved 2nd draft and recommended changes; h) Changes were then accepted by GC and GT; 2 i) 3rd draft was sent to national and international peer review; j) GC and GT then made changes based on peer review comments; k) 4th draft was sent back to council for approval; l) 4th draft was sent to BGCS members for feedback; m) GC and GT then made changes based on members’ feedback; n) 5th draft was sent to public consultation including patient support groups; o) GC and GT then made changes based on non-members’ feedback; p) Final draft approved by council and officers. Tables of Contents 3 1. Introduction page 6 2. Presentation and referral: Presenting symptoms and diagnostic methods page 6 2.1 Presenting symptoms page 6 2.2 History and examination page 7 3. Pathways for Management of cervical cancer page 7 3.1 Patient care pathways associated with a local MDT page 7 3.2 Governance and failsafe page 8 4 . Preoperative assessment and imaging page 9 5 . Sentinel Lymph node biopsy page 10 6. Pathology of cervix cancer page 12 6.1 Clinical information required on the specimen request form page 12 6.2 Reporting of small biopsy specimens page 13 6.3 Reporting lymphovascular space invasion page 13 6.4 Reporting of frozen sections page 13 6.5 Testing for HPV/p 16 page 14 6.5.1 HPV testing page 14 6.5.2 P16 staining page 14 7. Surgical management of cervical cancer per stage (the new FIGO classification version staging 2018 applies) page 15 7.1 Stage IA1 cervical cancer page 15 7.2 Stage IA2 – IB2 cervical cancer page 15 7.3 Ovarian conservation page 17 7.4 Laparoscopy versus laparotomy page 17 7.5 Adjuvant treatment after surgery page 18 8 Management of locally advanced cervix cancer stage IB3-IVA (FIGO 2018) page 19 8.1 Stage IB3/IIA2 cervical cancer page 19 8.2 Locally advanced disease and involved lymph nodes on radiological staging page 19 8.3 Stage IIB, IIIA/IIIB, IVA cervical cancer page 19 8.4 Cervical stump cancer page 20 8.5 Principles of radiotherapy page 20 8.5.1 Definitive chemoradiotherapy and brachytherapy: general aspects page 20 8.6 Definitive brachytherapy page 21 9. Adjuvant treatment post hysterectomy: Use of external beam radiotherapy, vagina vault brachytherapy, chemotherapy and combinations page 22 9.1 External Beam radiotherapy page 22 9.2 Concurrent chemotherapy with external beam radiotherapy page 23 9.3 Radiotherapy technique page 23 4 9.4 Vaginal vault brachytherapy page 24 9.5 Adjuvant chemotherapy page 24 10 Neoadjuvant chemotherapy in cervix cancer page 25 10.1 Neoadjuvant chemotherapy followed by surgery page 25 11. Management of cervical cancer in pregnancy page 26 11.1 Pre-invasive disease in pregnancy page 26 11.2 Diagnosis of cervical cancer in pregnancy page 26 11.3 Staging of cervical cancer in pregnancy page 27 11.4 Treatment options page 27 12. Surgical management of relapsed cervical cancer page 28 12.1 Intra-pelvic relapse page 28 12.2 Laterally extended endopelvic resection in pelvic side wall relapse page 30 12.3 Surgical morbidity and mortality of exenteration page 30 12.4 Reconstructive procedures page 31 12.4.1 Urinary system page 31 12.4.2 Gastrointestinal system page 32 12.4.3 Neovagina page 32 12.5 Fistulation/Cloaca (urinary or colonic) page 32 13 Re-irradiation in Cervix cancer page 33 13.1 Vaginal cancer following radiotherapy for cervix cancer page 33 13.2 Isolated recurrence following previous pelvic radiotherapy page 33 14. Chemotherapy in metastatic or advanced cervical cancer page 33 15. Follow-up of Cervical Cancer page 35 15.1 Identifying relapsed disease/Imaging page 35 15.2. Imaging page 36 16. Supportive care- psychological care for patients and carers with cervix cancer page 36 16.1 Management of complications and late effect/quality of life page 39 16.2 Sexuality/Sexual morbidity page 39 16.3 Menopause/Hormone replacement therapy page 39 16.4 Lymphoedema page 40 16.5 Bowel/bladder function page 40 16.6 Psychosocial page 40 17. Rare malignancies: small cell, mucinous and clear cell carcinomas, and sarcomas of the uterine cervix page 40 5 17.1 High grade NEC of small cell type page 40 17.2. Neuroendocrine tumours page 41 17.3 Mucinous tumours page 42 17.4 Clear cell carcinomas page 43 17.5 Basaloid and adenoid cystic carcinomas page 43 17.6 Sarcomatous tumours of the cervix page 43 18 Fertility sparing surgery in cervical cancer page 44 18.1 Service delivery page 44 18.2 Treatment options page 44 18.3 Simple trachelectomy and cone biopsy for selected subsets of women with stage IB1 (FIGO 2018) cervical cancer page 45 18.4.1 radical vaginal trachelectomy (RVT) for stage IB1 (FIGO 2018) disease page 45 18.4.2. abdominal radical trachelectomy (ART) for stage IB1 disease page 46 18.6 Ovarian transposition page 48 18.7 Neo-adjuvant chemotherapy prior to fertility sparing surgery page 48 1. Introduction 6 Despite the presence of well-organized cervical screening programmes in the UK and the introduction of HPV vaccination in 2008 for schoolgirls, the incidence of cervical cancer is not expected to significantly decrease over the next few years. Over the last decade, cervical cancer incidence rates have increased by around 4% in females in the UK, although higher rates have been witnessed in Northern Ireland. Incidence rates for cervical cancer are projected to rise by 43% in the UK between 2014 and 2035, to 17 cases per 100,000 females by 2035. The present document ranges from screening and investigation through to management of early and advanced invasive cancer and on to metastatic disease. In addition, we have included survivorship and quality of life and we increasingly rely on nurse specialists to provide support to our patients throughout their cancer journey. The aspect of fertility preserving therapeutic approaches will also be addressed at the various aspects of management. 2. Presentation and referral: Presenting symptoms and diagnostic methods 2.1 Presenting symptoms The reader is directed to the National Institute for Health and Care Excellence (NICE) guidance (NG12) on referrals for suspected cancer (1), the Scottish Intercollegiate Guidelines network Guidance on management of cervical cancer (2) and the NHS Cervical Screening Programme NHSCP20 document (3). Cancer of the cervix often has no symptoms in its early stages and may be detected after an abnormal screening smear test. Symptoms can be subtle and attributed to benign gynaecological conditions or remain asymptomatic until the cancer has reached an advanced stage. When symptomatic, the most common symptoms are abnormal vaginal bleeding, intermenstrual (IMB), postcoital (PCB) or postmenopausal bleeding (PMB). Other symptoms of cervical cancer may include dyspareunia and abnormal vaginal discharge. Abnormal appearance of the cervix during examination should also raise suspicion and referral for further investigations (4). It is possible for women of all ages to develop cervical cancer, but traditionally the condition mainly affected sexually active women aged between 30 and 45 years of age, however, recent data from Cancer Research UK shows the peak age of incidence has reduced to 25-29 years of age. Cervical cancer is very rare in women under 25 years old (4) and may be more difficult to prevent in younger age women. ( ). If there is extra cervical advanced stage disease with spread into surrounding tissue and organs, it can cause other symptoms, including haematuria, urinary incontinence, bone pain, lower limb oedema, flank or loin pain (due to hydroureter or hydronephrosis), changes to bladder and bowel habits, loss of appetite, weight loss and fatigue. Many of the signs and symptoms suggestive of cervical cancer are common to genital Chlamydia trachomatis infection. Women who have symptoms of irregular or contact bleeding or have an inflamed or friable cervix should be tested for Chlamydia trachomatis and treated if appropriate (5). The point prevalence of PCB in women in the community is 0.7-9% (6), but only a small proportion of these women are seen in secondary care. The probability that a woman under the age of 25 with PCB has cervical cancer is very low (6). Two per cent of women attending secondary care with PCB are diagnosed with cervical cancer (6). The risk of having a cervical cancer is not related to the duration and extent of symptoms (7). Women referred with PCB, where cervical cancer is excluded, have no increased risk of cervical cancer in the future (8). A systematic review identified no evidence to support performing a smear when a woman presents with PCB outside of the scheduled screening intervals (6) unless it is opportunistic in someone who has failed to attend regular screening. 7 Although the risk is not entirely removed in a woman presenting with symptoms, a woman with a negative screening history has a greatly reduced risk of cervical cancer compared to a woman with positive cytology (6,9). 2.2 History and examination Current guidance in the UK, recommendations for diagnosis and referral are based on guidance from NICE, SIGN, NHSCSP and BSCCP (1-3). The clinician should obtain a detailed account of the presenting symptoms, a full history including smoking status, previous smears and any previous cervical treatment. Women presenting with abnormal vaginal bleeding should receive an abdominal, speculum and pelvic examination at their clinical assessment (8). Sample-takers must visualise the cervix when taking a sample. If they notice abnormalities suggesting possible malignancy, the patient should be urgently referred for a colposcopic examination within a two weeks referral pathway (3). Treatment for patients referred via the two week wait rule or following high grade cervical cytology abnormality should be commenced within 62 days of referral or within 28 days from the date of decision to treat as per the Cancer Waiting Times guidelines. (1) (Grade D) Screen-detected cancers which are not directly visible will be referred to colposcopy clinic. For clinically visible cervix cancers, formal colposcopy may be omitted. However, biopsies are the recommended diagnostic investigations. Biopsy should be a punch biopsy, multiple punch biopsies are usually more reliable than a single one, or even an excisional biopsy, such as a LLETZ especially if referral screening test indicates high grade lesion (3). Recommendations: • Patients with suspected cervical cancer should be seen urgently, ideally in the colposcopy clinic to obtain directed biopsies to make a diagnosis. (Grade D) 3. Pathways for management of cervical cancer Staging System This document reflects the new FIGO staging 2018. Since this is a new system, all the presented supporting evidence reflects the earlier staging system, and this will be delineated and specified throughout these guidelines. The, previous and updated, FIGO classification systems are attached in the appendix of this document. 3.1 Patient care pathways associated with a local MDT The National Cancer Intelligence Network report on routes to diagnosis revealed that in 2013 the majority of cervical cancers (61%) were diagnosed by the two week wait / direct outpatient referral from general practitioners (GPs), 17% were screen detected and 10% presented as emergency admissions. (12). All women with a confirmed or suspected diagnosis of cervix cancer should be discussed at a specialist gynaecological cancer multidisciplinary team meeting. (SMDT) (Grade D). The SMDT should comprise a minimum of two surgical gynaecological oncologists, clinical oncologist (radiotherapy specialist), medical oncologist (chemotherapy specialist), radiologist, histopathologist, cytopathologist, clinical nurse specialist and a multi-disciplinary team co-ordinator. (13) (Grade D). Table one highlights the associated pathways with a local MDT. 8 Table One- Patient care pathways associated with a local MDT Source: Manual for cancer services: Gynaecology measures 2014 3.2 Governance and failsafe Failsafe mechanisms should be in place at each stage of the patient’s journey, from referral to diagnostics and treatment. Healthcare providers should have systems in place to ensure that cervical smears suspicious of malignancy and cervical biopsies confirming malignancies are appropriately managed. Patients with recurrent ‘red flag’ symptoms such as intermenstrual and post coital bleeding should be encouraged to seek help, even if they have previously had normal findings following investigations. The cancer centres should comply with requirements of submission of data items, including Cancer Outcomes and Services Dataset (COSD), to enable meaningful performance and outcomes assessments by the National Cancer Registration and Analysis Service (NCRAS). Women <25 years of age should be managed in conjunction with the relevant teenage and young adult cancer network co-ordinating group (TYACNCG). The teenage and young adult (TYA), gynaecology cancer patient pathways provide additional guidance for initial management, treatment and follow up. (13,14). For women seeking fertility preserving options, please see section 10. Management of cancer in pregnancy will be found in section 12. Recommendations: • ‘Red flag’ symptoms include intermenstrual and post-coital bleeding should trigger referral for investigation in secondary care. • All appropriate cases should be discussed at the Multi-disciplinary team meeting. (Grade D) Patient Presents Specialist MDT Stand alone diagnostic service or Either or Local MDT Cervical Cancer Colposcopy/Biopsy ? SCC st. IA1 or > Histology review by histopathology core member of specialist MDT. Where the specialist MDT considers a microinvasive squamous cancer is very low risk and indicates that cone biopsy alone is adequate management, the patient may be managed by the local team.
No further treatment Complete excision biopsy by local team No further treatment Incomplete excision biopsy ?higher stage Management by specialist MDT SCC stage 1A2 or > Or non-squamous cancer Management by specialist MDT 9 4. Preoperative assessment and imaging All women with cervical cancer beyond stage IA1 should have a visual inspection of the cervix and vagina along with a bimanual vaginal examination to assess vaginal and parametrial extension. A rectal or rectovaginal examination is strongly recommended, especially when assessing larger tumours, or if there is uncertainty on bimanual vaginal examination or discordance between imaging and initial examination. (Grade D). However, this is usually carried out in the clinic before MRI scanning is undertaken. Examination under anaesthesia (EUA) may be required to gain tissue for diagnosis and to adequately assess for vaginal/parametrial extension in some cases, although the routine use of an EUA and cystoscopy for staging has been superseded by modern cross-sectional imaging techniques. However, the use of imaging does not negate the need for a thorough clinical examination. Discordance between initial clinical examination and cross-sectional imaging would be a further indication for EUA. Cross-sectional imaging does not have a role in staging IA1 tumours. MRI is more accurate than CT in pre-treatment local staging of cervical cancers and thus patient selection for either surgery or primary chemoradiotherapy. In particular, candidates for fertility sparing surgery can be identified more accurately with specialised MRI techniques, compared to clinical examination alone (grade B). A systematic review of 57 studies comparing CT and MRI for staging of cervical cancer reported that MRI was more accurate than CT, particularly for parametrial involvement, bladder and rectal invasion (19). A systematic review of four studies with 366 patients with cervical cancer FIGO stage IIB or below reported a high level of accuracy of MRI in detecting involvement of the uterine internal os in cervical cancer (20). MRI should be performed to set protocols and interpreted by radiologists with expertise in gynaecological cancer. {RCR guidelines cervical cancer imaging 2014}. (Grade B). Radiologists should be aware that post-biopsy changes can mimic malignancy and adversely affect the assessment of tumour volume particularly in small tumours (21). Staging MRI should be performed at least 7-10 days post biopsy to reduce post biopsy artefact, although there are no studies that evaluate the chronology of post-biopsy changes and they can be observed for longer. Pre-operative chest radiology is part of staging and should be performed in all women with cervical cancer prior to surgery (Grade D). Plain chest X-ray is more cost-effective than CT in clinical early stage tumours as the risk of lung metastases is low. Vaginal involvement can be seen on MRI by disruption of the low T2 vaginal wall by intermediate signal intensity tumour (22). Early involvement is best determined by clinical examination, as MRI may overestimate the degree of vaginal involvement especially at the fornices or when tumours are large or in cases of local inflammation (21,23). With regard to parametrial assessment, MRI has been shown to outperform clinical examination (23,24). High resolution T2 axial oblique sequences tangential to the cervix are essential to evaluate the parametrium and when performed accurately achieve specificity of up to 97% and negative predictive value of 100% (25). When seen, the presence of an intact low T2 stromal ring has a high negative predictive value (22). Endorectal surface coil assessment of cervical cancer is more sensitive in the detection of parametrial invasion than standard body coil (26), but due to its more invasive nature is not used routinely. Due to the very high negative predictive value in the assessment of bladder and rectal involvement, MRI can safely obviate the need for invasive cystoscopic or endoscopic staging in the majority of patients, if imaging does not show any bowel or bladder involvement (27). 10 In relation to nodal involvement, cross sectional imaging is essential. Studies have investigated both CT and MRI and show that either can be used, although both may present limited sensitivity and specificity, depending on the study (19, 27). Some studies have shown that diffusion weighted imaging (DWI) MRI improves detection of nodal metastases but further studies with standardized protocols are required to provide conclusive evidence (28). If nodal disease is suspected, PET/CT has shown to be the most sensitive in locally advanced cervical cancer. The MAPPING study (Clin Trials Ident: NCT01836484) will present findings around the diagnostic accuracy of MRI, DWI MRI, FDG-PET/CT and FEC PET/CT in the detection of lymph node metastases in surgically staged endometrial and cervical carcinoma. FDG PET/CT should be considered in patients staged>IB1 planned for radical chemoradiation therapy (grade B). A meta-analysis to compare the diagnostic performance of CT, MRI and PET or PET/CT for detection of metastatic nodes in cervical cancer reported PET or PET/CT had the highest pooled sensitivity (82%) and specificity (95%) whilst sensitivity and specificity for CT (50% and 92% ) and MRI (56% and 91%) were less accurate (29). A systematic review and meta-analysis of diagnostic accuracy of tests for lymph node status in primary cervical cancer reported the greatest accuracy with sentinel node biopsy and that PET/CT was superior to MRI and CT (30). Surgical staging with lymphadenectomy remains the gold standard. Recommendations: • Visual inspection of the cervix is essential with thorough clinical staging • MRI is the preferred imaging technique. DWI may increase the sensitivity • FDG-PET scanning should be carried out in cases more advanced than stage IB2 (FIGO 2018) and in all cases planned for combined chemotherapy and radiation therapy (Grade B) • For further reading and references, the reader is directed to the following (32-48). 5. Sentinel lymph node biopsy The recent consensus statement on sentinel lymph node biopsy (SLNB) from the BGCS is accessible using the link below. This also includes recommendation on the pathological processing of sentinel lymph nodes (SLNs). SLNB has a high sensitivity for detection of metastases in tumours <2 cm, but is less accurate for larger tumours, (Grade B). SLNB is a technique in which the first draining lymph nodes (known as sentinel lymph nodes) are removed and examined histologically in order to provide information about the status of the nodal basin (metastatic or non-metastatic) without the need for full lymphadenectomy. SLNB is now integrated into staging protocols for breast and vulval cancers and has been of increasing interest in the management of cervical cancer. The identification of positive nodes pre-operatively may allow the selection of women that would benefit from primary chemo-radiotherapy and can spare these women the additional morbidity associated with radical hysterectomy and pelvic lymphadenectomy. Taken overall, the results of studies to date indicate that the negative predictive value of an uninvolved sentinel node is high (95%) and is superior to all other imaging modalities, including PET/CT. In one multi-centre prospective study, when combined with pre-operative lymphoscintigraphy, 17% of involved SLNs were identified in unexpected anatomical locations. In addition, in 39.1% of cases, histological ultra-staging and immunohistochemistry detected micro-metastases that would have been undetected by conventional histopathological assessment (49). Therefore, SLNB may be more reliable in determining lymph node status than conventional full lymphadenectomy. Combined use of Technetium-99m colloid (99mTc) lymphoscintigraphy and blue dye currently appears to be the most reliable detection method for SLNB. (Grade A) 11 Of the three most commonly used detection methods for SLNB (Technetium-99m colloid, blue dye or a combination of the two), the combined method using Technetium-99m colloid (99mTc) and blue dye together is consistently shown to be the most reliable in high quality meta-analyses (50-53). The sensitivity of a combination of 99mTC and blue dye for detection of SLN metastasis in early cervix cancer is 88-92%. Bilateral detection rates are 92-97% for the combined method compared with 81-88% for blue dye alone and 88-90% for 99mTc alone. Near infrared (NIR) fluorescence imaging is a more recent technique which uses medical dyes that fluorescence in the NIR spectrum. Indocyanine green is the most commonly used dye and has gained interest with the increased use of robotic surgical platforms that are compatible with NIR detection. Because there is no requirement for the lengthy regulatory processes associated with the use of radioactive methods, NIR imaging is also seen as attractive method, due to its simplicity and ease of use. The formal evaluation of NIR fluorescence imaging for SLNB in cervix cancer compared with full pelvic lymphadenectomy is limited to a few studies with small patient numbers. These show promising results and indicate successful bilateral detection rate for SLNs in 60-89% of cases with a negative predictive value of 93-100% rate for SLNs in 60-89% of cases with a negative predictive value of 93-100% (54-56). Until these findings are confirmed in larger studies, directly comparing the use of NIR fluorescence imaging with combined radiocolloid and blue dye detection, or in high quality meta-analyses, results should be interpreted with caution. Failure to identify a SLN in one hemi-pelvis, necessitates a full lymphadenectomy on that side. (Grade D). As a midline structure, the cervix drains to nodal basins on both sides of the pelvis. Therefore, where SLNB is used, sentinel nodes must be identified in both sides of the pelvis, as the histological status of the SLN in one hemi-pelvis does not predict the histological status of the nodes in the contralateral hemi-pelvis (57-58). Lymph node metastasis in one hemi-pelvis may prevent normal drainage to that side and lead to false negatives. This phenomenon is seen in other cancers where SLNB is used in management. Therefore, failure to identify a SLN in one hemi-pelvis mandates full pelvic lymphadenectomy on that side. SLNs should be processed by serial sectioning and immunohistochemistry. (Grade B). Multiple studies have found that intra-operative frozen section is limited in its ability to identify small metastases and micro-metastatic disease with sensitivity as low as 21% reported (53-55). A high false negative is particularly associated with tumours of >20 cm3 volume (56-58). More recent reports show improved outcome with lower recurrence rate. Histopathological ultra-staging of the lymph node with serial sectioning and immunohistochemistry is required. As such, intra-operative evaluation of non-enlarged nodes is limited, in terms of clinical utility. Therefore, until more accurate methods of intra-operative histopathological assessment have been developed, a two-stage procedure is required in which the SLNB biopsy is performed initially, followed by definitive surgery on another date, in the case of a histopathologically negative SLN. Assessment of lymph node status with SLNB can be considered in centres with sufficient expertise and training in this technique (Grade B). Systematic reviews of SLNB in clinically staged, early cervix cancer show that it is a reliable method for detection of lymph node metastases in this setting, particularly for small tumours (<2cm) (59-61). In a recent randomised prospective multicentre study of 206 women with stage IA-IIA1 cervical cancer (FIGO 2009 staging system), women with a negative intra-operative frozen section of SLNs were randomised to either no further nodal dissection (SLNB alone) versus SLN surgery plus complete pelvic lymphadenectomy. No false negatives were identified in the women undergoing SLN surgery plus complete lymphadenectomy. Morbidity was significantly reduced in the SLNB only group compared to those having complete lymphadenectomy (31.4% and 51.5% respectively, P=0.0046). Quality of life scores were better in the SLNB only group and lymphoedema rates were reduced. Two isolated lymph node recurrences were reported in the SLNB-only group, compared with none in the group having SLNB plus complete lymphadenectomy. The study did not have sufficient statistical power to assess survival 12 outcomes, although recurrence-free survival does not appear significantly different in the two groups at 3 years (62). Longer term outcomes following SLNB alone are awaited. The SENTIX trial, a prospective multicentre trial of sentinel lymph node surgery alone in SLN negative patients which commenced enrolment in June 2016, has recruited 340 patients and will report on 5-year oncological outcomes at the end of 2021. The SENTICOL III study, started accrual in 2018 and randomizes women with a negative SLN on intra-operative frozen section to SLN biopsy alone versus SLN and pelvic lymphadenectomy. The end points are 3-year disease-free survival and quality of life. There is currently not enough data on oncological outcomes and centres adopting this technique should maintain prospective data on patients. Multiple protocols have been used for SLNB across different studies using different detection methods, timing of injections, numbers of injections, and volumes and sites of injection. To date there is no standard accepted protocol. There is also a learning curve associated with the technique of SLNB across all tumour sites. Improved detection rates are consistently seen with increased numbers of procedures performed. Therefore, surgeons performing SLNB are strongly advised to undertake this procedure as part of a full lymphadenectomy initially, in order to determine their false negative rate as a quality assurance process during their learning curve. Results of local practice should be audited. Women considering SLNB should be advised that the long-term survival outcomes following sentinel node surgery alone are not yet available. SLNB should be offered within the context of relevant clinical trials, such as SHAPE (63). More recent studies offer alternative opinions, the lower sensitivity of SLN for pelvic LN staging had only been shown in early reports however more recent papers did not show this demonstrating a higher sensitivity (64, 65). Recommendations: • SLNB can be considered in the management of women with tumours of <2 cm in diameter in addition to formal PLND. • Strict adherence to protocol is important and surgeons must perform side-specific nodal dissection in any cases of failed mapping. • There should be an initial training period with high exposure to ensure quality. • All suspicious or grossly enlarged nodes must be removed regardless of findings at SLN mapping. • Long-term safety of SLNB alone has yet to be established in prospective studies comparing SLNB with SLNB and full lymphadenectomy and women considering SLNB should be informed of this. (Grade B) 6. Pathology of cervix cancer Detailed guidelines are available on Royal College of Pathologists (RCPath) website ( This section addresses some areas of clinical importance in the pathology of cervical cancer. 6.1 Clinical information required on the specimen request form Given the increasing centralization of laboratories this may result in pathology departments being sited away from the clinical areas and not always sharing the same information systems. The request form has become an important medium of communication. In addition to demographic details, the request form should contain cervical screening history (at least the latest cytology result), results of previous biopsies including the dimensions of the cancer, radiology results and the details of all specimens sent. 13 6.2 Reporting of small biopsy specimens Recording of dimensions of a large loop excision of the transformation zone (LLETZ) or loop biopsy of the cervix should be in three dimensions. The diameter of the ectocervix (two dimensions) and the depth (thickness) should be recorded. This should be as accurate and reproducible as possible, as it is relevant for staging, assessing adequacy of colposcopy (66) and therapeutic decision-making. A cervical LLETZ specimen is usually trimmed by serial slicing at 2-3 mm in a sagittal/parasagittal plane. The slices are submitted in sequential, individually labelled cassettes such that the sequence of slices is unambiguous. This enables assessment of a carcinoma whose third dimension may exceed 7 mm. Core histological data items that should be present in all histological reports of cervical carcinoma include tumour type, tumour grade, tumour dimensions, lymphovascular invasion, resection margins and FIGO stage. All cervical carcinomas should be typed according to the 2014 WHO classification (67). Although a detailed description of different tumour types is beyond the scope of these guidelines, a few points should be noted. Nearly all-squamous carcinomas of the cervix are aetiologically related to HPV infections; although rare HPV-negative cancers have been described . HPV-negative cervical adenocarcinomas are increasingly recognized, and it is now well established that gastric type adenocarcinomas of the cervix are unrelated to HPV. A new classification, based on HPV association, has been proposed (69). A system of assessing cervical adenocarcinomas based on the pattern of invasion has been developed and has been shown to be reproducible amongst pathologists and to correlate with the risk of lymph node involvement and outcomes (70). Measurement of tumour dimensions in cervical carcinomas is important for accurate FIGO staging of early cervical cancers. Carcinomas must be measured in millimetres in three dimensions. Accurate measurements are paramount in distinguishing between the different early FIGO stages (71). Tumour measurements are one of the parameters, which affect decisions regarding type of surgery and need for adjuvant therapy. The measurements must be provided in three dimensions, depth of invasion can be replaced by thickness of tumour in some cases especially in ulcerated cancers and adenocarcinomas. The term ‘microinvasive carcinoma’ does not feature in the FIGO staging systems and should not be used. Detailed descriptions of preferred methods of measurement are beyond the scope of these guidelines. 6.3 Reporting lymphovascular space invasion Lymphovascular space invasion (LVSI) does not affect staging of cervical carcinoma. Correlation of LVSI with outcome has been difficult to assess due to variability in recognition (72). Variability in fixation leading to retraction around tumour groups is a well-recognized mimic. Assessment is difficult in fragmented and diathermized specimens. Features that may help in recognition of LVSI are presence of endothelial lining, adherence of the tumour cells to the side of the space, presence of adherent fibrin and matching of tumour contours to the outline of the space. LVSI has been shown as an independent predictive factor of adverse outcome although there is not yet enough evidence to support quantification of LVSI, distinguishing between blood and lymphatic channels or description of site of LVSI (73-75). 6.4 Reporting of frozen sections In most institutions in the United Kingdom, frozen sections for cervical tumours are not routinely used for the assessment of resection margins. However, in some specialist centres, frozen sections are used for the intraoperative evaluation of the upper limit of trachelectomy specimens. The resected cervix is sent for frozen section evaluation of the proximal margin with a recommended tumour (invasive and in- 14 situ) clearance of 5-10 mm . If this clearance is not achieved, additional cervical tissue is excised to obtain that clearance, or a completion hysterectomy is performed if intraoperative assessment indicates that an adequate margin clearance is not possible (77). There are limitations of frozen section on trachelectomy specimens and the discrepancies are mainly due to inherent difficulties in discrimination of subtle invasion from benign mimics. Tubo-endometrioid metaplasia, which commonly occurs- particularly after loop excisions, may simulate cervical glandular intraepithelial neoplasia on frozen section giving rise to a false positive report on frozen section. Assessment of margins in trachelectomy specimens is possible in some macroscopically visible tumours (78). In cervical cancers, lymph node involvement is a strong predictor of survival . The 5-year survival rate decreases from 92% to 64% in cases of positive pelvic lymph nodes (80-81). In the context of fertility-preserving procedures, the intra-operative assessment of clinically suspicious lymph nodes may be of value. Frozen section is also highly crucial within the context of Sentinel LN technique (83). (Grade B) 6.5 Testing for HPV/p 16 6.5.1 HPV testing The causative role of distinct types of human papilloma viruses (HPVs), referred to as high risk (HR) HPV types, in most types of cervical cancer has been unequivocally established . The oncogenic potential of these viruses is quite divergent: HPV16, 18 and 45 are clearly very potent oncogenic agents, whereas oncogenicity of most of the other viruses of this group is less potent . Molecular testing for HPV may occasionally be useful in a diagnostic scenario when, for example, in the context of a metastatic neoplasm the differential includes an HPV-related versus a non-HPV-related primary carcinoma. PCR based assays are usually sufficiently sensitive, but when the viral load is low, in-situ hybridization (ISH) techniques may be needed. HPV typing may be of value when considering patient selection for immunotherapy trials (86). 6.5.2 P16 staining In cervical pre-neoplastic and neoplastic lesions associated with high-risk human papillomavirus (HPV) infection, there is functional inactivation of Rb by HPV E7 protein. This results in an accumulation of p16 protein, because normally Rb inhibits transcription of p16 (88). As a consequence, in the cervix diffuse p16 positivity can be regarded as a surrogate marker of the presence of high-risk HPV (84). This is of value in that most histopathology laboratories do not have the facilities to undertake molecular techniques to identify HPV, whereas p16 immunohistochemistry is easy to perform. It is also clear that focal, or even diffuse, p16 expression may occur as a result of non-HPV- related mechanisms (89). Diffuse (as opposed to focal) positivity with p16 in the cervix can be regarded as surrogate marker of the presence of high-risk HPV. Most cervical carcinomas (squamous, glandular and high grade neuroendocrine) are p16 positive, as a result of their association with high risk HPV (89). Some types of endocervical adenocarcinoma have been reported to be p16-negative (90). P16 is a useful marker to help in grading intraepithelial neoplasia. (90). It can be used to ascertain the tumour origin as endocervical adenocarcinoma, as discussed, is usually associated with high-risk HPV, whereas this is rarely the case with an endometrial adenocarcinoma (91). As a consequence, most cervical adenocarcinomas exhibit diffuse positivity with p16 whereas endometrial adenocarcinoma of endometrioid type is usually negative or there is focal positivity. Recommendations: • These would adhere to the RCPath guidance. (Grade B) 15 7 Surgical management of cervical cancer per stage (the new FIGO classification version 2018 applies) 7.1. Stage IA1 Conisation for stage IA1 cervical cancer Conisation for stage IA1 cervical cancer has been recommended by numerous authors and is included in other established guidelines. The aim of conisation is to achieve negative margins to both cancer and dysplasia. Cold knife conisation used to be seen as having the advantage over loop conisation (LLETZ) with the margins being easier to assess. However, LLETZ is now the acceptable and preferred option in the UK, as long as adequate margins are achieved along with a non-fragmented specimen, correct histological orientation, and no electrosurgical artefact interfering with marginal assessment. Conisation (LLETZ, knife, & LASER) has been systematically reviewed extensively in patients with cervical intra-epithelial neoplasia (CIN). Complications include cervical incompetence and stenosis, risks of premature delivery, leading to neonatal death and complications of extreme prematurity. A detailed assessment of these risks is outside the scope of this guideline, but it should be noted that when treating stage IA1 disease, patients often receive more than one conisation or large cones greater than 15mm deep, which are associated with a greater risk of these complications. This must be explained to patients undergoing conisation for cervical cancer (92-96). Especially in view of the fact that the lateral extent of the lesion is no longer taken into consideration for the early stages IA1 and 2, lesions with wider lateral extent will need larger cones. individual discussions and decision-making processes will need to happen whether a cone excision alone would be sufficient for the larger tumours, although they will probably represent a rarity. When the patient has completed her family or does not require fertility sparing approach, a simple hysterectomy may be advised. Recommendations: • Conisation (cold knife or LLETZ) is an acceptable treatment for stage IA1 squamous cell and adenocarcinoma of the cervix (Grade B) • Re-conisation is recommended if there are positive margins to intra-epithelial neoplasia (Grade A) • Re-conisation is recommended if the specimen cannot be orientated, is fragmented, or has diathermy artefact that makes margin assessment impossible • Conisation (LLETZ, knife, & LASER) for cervical cancer is often large or multiple and carries a higher risk of obstetric complications in the future. This must be explained to women who have this treatment and alternative options discussed • Recent evidence has indicated that SLND approach may be extended to use in Stage IA1 (Grade C) • In patients with LVSI a pelvic LND and/or SLNB should be performed (Grade A) • 7.2 Stage IA2 – IB2 cervical cancer (FIGO 2018) The standard management for patients with FIGO stages IA2, IB1 and Ib2 (FIGO 2018) cervical cancer is radical hysterectomy and bilateral salpingectomy +/- bilateral oophorectomy with bilateral pelvic lymphadenectomy. For younger patients who wish to preserve their ovarian function, opportunistic bilateral salpingectomy should be discussed and offered. The extent of radicality of the local resection has been an issue of debate for decades, however two randomised controlled trials found that following a less radical approach in terms of a Piver-Rutledge type I instead of III (Piver 1 or 2 – Class A/B) has equal oncologic safety, but a lower surgical morbidity profile (97,98). Wright et al found that patients with stage IB1 cervical cancer (FIGO 2018) with no LVSI and negative pelvic nodes had only 0.4% risk for 16 parametrial invasion and, therefore, could have received simple hysterectomy (95-96). Based on this and similar retrospective studies (101-103), three prospective randomised clinical trials have been designed to study the role of less radical surgery in this patient group. Patients eligible for these trials should be offered trial participation. The largest ongoing trial is the SHAPE Trial (ClinicalTrials.gov Identifier: NCT01658930), a randomized phase III trial comparing radical hysterectomy and pelvic node dissection vs simple hysterectomy and pelvic node dissection in patients with low risk early stage cervical cancer less than 2cm, including those with IA1 disease. The study has completed recruitment with seven hundred patients in 2019 and survival results are awaited. Traditionally, the removal of lymph nodes is recommended from the following anatomical regions: external iliac lymph nodes from the bifurcation of common iliac artery to the deep circumflex iliac vein; the nodes around the internal iliac vessels; the nodes from the obturator fossa down to the level of the obturator nerve. (For SLN protocols and value please see section five.) For women with stage IA2 disease who wish to preserve their fertility, individualised management should be discussed ranging from simple cone excision , simple or radical hysterectomy depending on the patient’s previous cervical surgeries, lateral extension of the tumour, and presence of LVSI. Recommendations: • Women with stage IA2 cervical carcinoma who wish to preserve their fertility are eligible for conization/ simple or radical trachelectomy irrespective of grade and lymph-vascular space invasion, as long as clear margins can be achieved, and no pathological LN are identified in staging. (Grade B) • The standard management for stage IB1 and IB2 cervical cancer (FIGO 2018), outside of clinical trials, is radical hysterectomy and bilateral salpingectomy with bilateral pelvic lymphadenectomy +/- bilateral oophorectomy. (Grade A) • Patients eligible for clinical trials evaluating extent of surgery should be offered trial participation • SLNB is a promising way of diagnosing lymph node metastases and reducing the morbidity of systematic pelvic LND (Grade C) • Young patients with adenocarcinoma should be counselled carefully regarding bilateral salpingoophorectomy, due to the higher risk of metastases and/or relapse in the adnexa compared to squamous cell histology (Grade A) • Chemoradiotherapy is a valid alternative to radical surgery but usually offered for those unfit for surgery or where a double treatment modality (i.e., surgery and radiotherapy) is to be avoided (Grade A) • Radical trachelectomy with cerclage and bilateral pelvic lymphadenectomy and/or SLN is an option for patients with cervical cancer up to stage Ib1 (FIGO 2018) who wish to preserve their fertility. Trachelectomy may be an option also for Ib2 (FIGO 2018) tumours if patients are properly counselled by an experienced team about the higher oncologic risks. The estimated figures of global outcomes following trachelectomy can be seen in Table two below. There are protocols that apply neoadjuvant chemotherapy followed by trachelectomy, however these are not standard practice. Adjuvant chemoradiotherapy is recommended for patients with high-risk pathological prognostic factors (positive nodes, positive/close surgical margin, positive parametrium) (Grade C) Table Two, Recommended Estimated Figures of Global Outcomes Following Different Types of Trachelectomy for Stage IB1 Cervical Cancer to be used During Patient Counselling. 17 RVT ART (open) Minimally invasive ART Does not receive fertility sparing surgery Circa 11% Circa 17% Circa 8% Recurrence rate Circa 5% Circa 5% Circa 6% Cervical erosion rate Circa 3% Circa 3% Circa 5% Cervical stenosis rate Circa 8% Circa 11% Circa 5% Achieves at least one pregnancy Circa 50% Circa 50% Circa 66% Proportion of pregnancies secondary to fertility treatment About a ¼ About a ½ About a 1/3 Women who achieve a live birth Circa 50% Circa 25% Circa 57% Premature delivery rate Circa 40% Circa 50% Circa 49% Extreme premature delivery rate Circa 17% Circa 20% Circa 22% This figure is affected by the proportion of elective Caesarean Sections before 37 weeks’ gestation. 7.3. Ovarian conservation The incidence of ovarian metastasis is not significant in patients with stage IB squamous cell carcinoma (0.2%) and therefore, ovarian preservation in young patients is safe (104). Published data on rates of ovarian metastasis in cervical adenocarcinomas are conflicting. Shimada et al found that 3.72% of patients with stage IB adenocarcinoma had ovarian metastases, which suggests that clinicians should be cautious offering ovarian preservation (104). Others, however found no difference in outcomes in patients with adenocarcinoma (105-108). One recent study of 312 patients found ovarian metastases in 14 women (4.5%) (109). Nine of these had stage IIA (FIGO 2009) disease or above and would not have qualified for fertility sparing surgery. Of the five patients with stage IB (FIGO 2009) disease, all had uterine corpus involvement at final histology and would also have been unsuitable for fertility-sparing surgery. The fact that most series include patients with disease greater than 4cm makes the estimation of risk of ovarian metastasis in smaller cases, so patients should be counselled carefully. Ovarian preservation is acceptable in adenocarcinoma patients who wish to preserve their fertility as on the balance of probability, women at risk of ovarian metastases will have completion treatment including the ovaries. 7.4. Laparoscopy versus laparotomy Recent evidence from a prospective randomized trial (311) and a large SEER meta-analysis (312) showed a significant compromise of overall oncologic outcome in terms of both progression-free and overall survival of the minimal invasive compared to the open approach for patients with early stage cervical cancer up to 4cm. The surgical morbidity and mortality, as well as quality of life scores were all similar between the two modalities. These data are in contradiction with previous retrospective studies which showed equivalent oncologic outcome and even lower morbidity in the minimal invasive arm. The UK National Cancer Registration and Analysis Service in their analysis of 929 patients who underwent either open or minimally invasive radical hysterectomy replicated the impaired overall survival outcome of the minimally invasive surgery group. ( In their May 2019 statement, ESGO advised that in radical hysterectomy, open approach is the gold standard. ( Following the recently published evidence, patients should be carefully counselled about existing evidence to enable them to make an informed decision about their treatment. In light of this analysis from English data, the BGCS recommends that clinicians and patients exercise caution, taking into 18 account factors such as tumour size, when considering laparoscopic surgery for cervical cancer. We recommend that gynaecological cancer surgeons and nurse specialists discuss in detail the risks and benefits of the different surgical options with patients to enable women to make an informed choice. NCRAS analysis of the size of cervical tumour as a variable in predicting survival is ongoing and will provide very useful data in guiding future BGCS recommendations in this area. Furthermore, In an effort to further elucidate variations in practice in Europe and the impact of minimally invasive surgery (MIS) in cervical cancer in European Centres; the European Society of Gynecological Oncology (ESGO), has launched the SUCCOR Study NCT03958305 (313): “An international european cohort observational study comparing MIS versus open abdominal radical hysterectomy in patients with stage IB1 cervical cancer operated in 2013-2014”. The FIGO staging that applied was still the one from FIGO 2009. Even though this retrospective study also confirmed, that patients with IB1 (FIGO 2009) cervical cancer that underwent radical hysterectomy by MIS showed a significantly higher risk of relapse and death, patients with tumours smaller than 2 cm and those with previous cone biopsy did not show differences in disease free survival (DFS) by the surgical approach. Furthermore, the investigators showed that the use of a uterine manipulator in MIS impacted the DFS negatively in this population. Patients that underwent radical hysterectomy by MIS without the use of a manipulator showed the same outcome as those operated by open surgery and even more, protective manoeuvres to avoid tumour spillage at the time of the colpotomy in MIS improved the DFS in these patients. The currently ongoing phase III prospective randomised controlled international trial has been established (Robot-assisted Approach to Cervical Cancer “RACC” NCT03719547) to further assess the risks and compare oncologic and surgical outcomes of open radical hysterectomy versus robotic hysterectomy in centres with strictly established quality assurance criteria of surgical training and appropriate infrastructure. 7.5 Adjuvant treatment after surgery There will be some patients post operatively who, despite adequate preoperative staging, at final pathology have adverse prognostic factors that will require discussion about adjuvant treatment, such as microscopically involved margins, positive LN or larger tumour size. The following histopathological risk factors should be considered when considering postoperative adjuvant chemoradiotherapy. High-risk factors: • Positive pelvic/para-aortic lymph nodes • Parametrial spread • Positive surgical margins (microscopic) • Patients with high-risk factors present in the radical hysterectomy specimen should be offered adjuvant concurrent chemoradiotherapy. Concurrent chemoradiation therapy is superior to radiation alone (118) Intermediate-risk factors: • Presence of LVSI • Tumour maximum diameter >4cm at final pathology • Deep cervical stromal invasion (>1/3) The GOG 92 prospective trial recruited patients with two or more intermediate-risk factors (according to Sedlis-criteria, which can be seen below in Table Three) after radical hysterectomy and pelvic lymphadenectomy and compared outcomes of patient groups with or without adjuvant pelvic radiotherapy (but no chemotherapy) (118-124). Patients who received adjuvant radiotherapy had a significantly improved recurrence-free survival, but no statistically significant overall survival benefit. It is unknown if the addition of chemotherapy would increase the overall survival. 19 Table Three. Sedlis Criteria (117) LVSI Depth of invasion in thirds Tumour size in cm + Deep Any + Middle ≥2 + Superficial ≥5 - Middle or deep ≥4 Patients with intermediate-risk factors, therefore, should be considered for adjuvant chemoradiotherapy or observation and the GOG 92 results should be discussed. Patients were eligible for the trial according to the Sedlis-criteria (121). This study was carried out some time ago and more recent studies show better surgical outcome without the need for adjuvant treatment (122). 8 Management of locally advanced cervix cancer stage IB3-IVA (FIGO 2018) The standard treatment for locally advanced cervical cancer is currently chemo-radiotherapy consisting of external beam radiotherapy (EBRT), intracavitary brachytherapy (BT) and concomitant chemotherapy with Cisplatin (125-128). A randomised controlled trial by Landoni et al (126) compared radical surgery with radical radiotherapy in patients with stage IB-IIA and found no survival difference in patients with squamous cell carcinoma but higher morbidity in patients with a combined treatment modality. For that reason, combined treatment should be avoided, and surgery should not be pursued in patients who are expected to need postoperative chemoradiotherapy. There is controversy about how to manage patients with positive nodes and whether the proposed surgery should be abandoned, or completion carried out. No trial has ever been completed to address this and individual MDTs/tumour boards should determine a local policy. 8.1 Stage IB3/IIA2 cervical cancer (FIGO 2018) 1) Treatment strategy should aim to avoid the combination of radical surgery and postoperative EBRT, due to significant increase of morbidity and no impact on survival. (Grade A) 2) Definitive platinum-based chemo-radiotherapy and brachytherapy is preferred (see Principles of radiotherapy). (Grade A) 3) Pelvic exenteration is an option in selected cases with stage IVA disease. This should be especially considered when need for symptom control applies, eg for fistulae. (Grade C) 8.2 Locally advanced disease and involved lymph nodes on radiological staging 1) Definitive chemoradiotherapy and brachytherapy with an additional radiation boost to the involved lymph nodes is recommended in patients with unequivocally involved pelvic lymph nodes on imaging (see principles of radiotherapy) 2) Para-aortic (at least up to the inferior mesenteric artery) lymph node dissection maybe considered before treatment for staging purposes if no evidence of disease on imaging (PET-CT recommended for imaging nodes). 8.3 Stage IIB, IIIA/IIIB, IVA cervical cancer (FIGO 2018) 1) Definitive chemoradiotherapy and brachytherapy with an additional radiation boost to the involved lymph nodes is recommended in patients with unequivocally involved pelvic lymph nodes on imaging (see principles of radiotherapy). The Uterus-11 trial, (NCT01049100) a 20 multicentre phase III Intergroup trial of the German Radiation Oncology Group (ARO) and the Gynaecologic Cancer Group (AGO) was designed to evaluate the role of surgical staging in patients with stage IIB-IV cervical cancer before primary chemoradiation therapy within a prospective randomised design. The study has finished recruiting, results are awaited. 2) Paraaortic (at least up to the inferior mesenteric artery) lymph node dissection maybe considered before treatment for staging purposes if no evidence of disease on imaging (PET-CT recommended for imaging nodes). 3) Pelvic exenteration is an option in selected cases with stage 4A (T4M0) disease. 8.4 Cervical stump cancer Management of cervical stump cancer follows the recommendations for patients with subtotal hysterectomy. Adaptation of radiotherapy and brachytherapy may be necessary. 8.5 Principles of radiotherapy 8.5.1 Definitive chemoradiotherapy and brachytherapy: general aspects Definitive management (without tumour related surgery) consists of concomitant pelvic radiotherapy (platinum-based) and brachytherapy or pelvic EBRT alone and brachytherapy. 1) Overall treatment time for the definitive treatment should not exceed 7-8 weeks. Overall treatment time for EBRT should not exceed 5-6 weeks (Grade A). 2) Delay of treatment and/or treatment interruptions must be avoided, i.e. patients should be treated as category 1 patients (Grade A). There is evidence that overall treatment time (OTT), including brachytherapy, should be as short as possible and should not exceed 56 days for squamous carcinoma (124). Recent data from retro EMBRACE study indicates that the effect of OTT shortening by one week was equivalent to escalating CTVHR dose by 5 Gy (D90), resulting in increase of local control by 1.0% (small tumours) to 2.5% (large tumours). The EMBRACE data (125) is based on a treatment time of 7 weeks and ideally OTT should be less than 7 weeks. It is recommended to use pelvic intensity modulated radiotherapy to deliver EBRT with an appropriate IGRT (image guided radiotherapy) programme (Grade B). Pelvic EBRT is currently delivered with different techniques: 3D conformal EBRT, intensity modulated radiotherapy (IMRT), volumetric arc techniques (VMAT), and tomotherapy (131-135). Application of IMRT in cervix cancer significantly reduces the volume of tissue irradiated to intermediate doses such as 30-40 Gy for bladder, rectum, sigmoid and bowel (133). The progress from 3D conformal EBRT to IMRT has demonstrated a reduction of treatment related morbidity in mono-institutional and retrospective settings (135-137). Furthermore, EMBRACE quality of life data has shown a significantly lower incidence of bowel symptoms in patients treated with IMRT compared to 3D conformal EBRT with the four-field box technique (131). However studies of intra and inter-fraction motion show that there is significant internal motion of the cervix and uterus due to rectal and bladder filling and tumour regression as shown by Beadle (132), Van de Bunt (136), Taylor & Powell (137) and adequate margins must allow for this when using IMRT as well as a comprehensive online cone beam based imaging protocol ideally including daily imaging with a back-up plan. 1) A total dose of 45-50 Gy using 1.8-2 Gy per fraction should be used. Randomised studies of radiotherapy have utilised fractionation regimens of 40-50.4 Gy in daily 1.8-2Gy fractions (Grade B) (RCR Guidelines). An RCR audit by Forrest & Clarke in 2012 (140) showed the most common UK regimens were 45 Gy in 25 fractions or 50.4 Gy in 28 fractions. Pelvic EBRT dose should be 45-50.4 Gy in 1.8 Gy per fraction to the central disease and the elective nodes. The central CTV volume 21 should include all known disease (GTV), entire cervix, parametrium, upper half of the vagina / at least 2 cm below GTV and entire uterus. The elective nodal volume should include all involved nodes, common iliac lymph nodes, internal and external iliac lymph nodes, obturator nodes and pre-sacral lymph nodes. These should be contoured as per a contouring atlas (Taylor and Powell 2008) (134-139,135). 2) Tumour and lymph node related target volume for IMRT includes the primary cervical tumour, the parametria, uterine corpus, upper half vagina (at least 2 cm below disease) and the pelvic lymph nodes (obturator, internal, external, common iliac and pre-sacral). In case of extensive pelvic node involvement / common iliac/ para aortic lymph node involvement the nodal target should include the para-aortic lymph node at least up to the level of the renal vessels. 3) Data from the EMBRACE studies (132) showed a significant incidence of para-aortic failure in patients with known pelvic nodes (11.5%) and the role of para-aortic nodal irradiation electively is being studied in EMBRACE II. Significant risk factors for para-aortic nodal recurrence is common iliac lymph node involvement and >/=3LN involved. 4) A reduced target volume for EBRT excluding the common iliac nodes may be considered in low and intermediate risk IB1 patients with negative nodes on imaging. 5) Boost treatment for involved lymph node(s) may be applied by simultaneous integrated boost within the IMRT or as a sequential boost. The total dose including the contribution from brachytherapy should be 55-60Gy (EQD2). An alternative treatment option is debulking of enlarged nodes. The role of dose escalation to involved nodes is being explored in the DEPICT trial (137). Other studies indicate that pelvic nodes are a poor prognostic marker with increased likelihood of pelvic and para-aortic nodal recurrence. At this time nodal dose escalation can be considered but there is no good data to recommend this definitely. There is no definite survival advantage to treating the para-aortic area electively in patients at high risk of lymph node involvement. Single agent radio-sensitising chemotherapy, preferably cisplatin (weekly 40 mg/m2) should be used unless contraindicated. If cisplatin is not applicable alternative options are 5FU and carboplatin. 6) The benefit of concurrent chemo-radiotherapy is seen across all tumour stages. Standard treatments would be weekly cisplatin chemotherapy but where cisplatin is contra-indicated 5FU based chemotherapy is an alternative option (CCMAC 2008 & 2009) (141,142). The currently recruiting INTERLACE study (NCT01566240) is a phase III multicentre trial comparing weekly induction chemotherapy followed by standard chemoradiation versus standard chemoradiation alone in patients with locally advanced cervical cancer (143) and the OUTBACK study (NCT01414608) which is a Phase III Trial of Adjuvant Chemotherapy Following Chemoradiation as Primary Treatment for Locally Advanced Cervical Cancer Compared to Chemoradiation Alone (144). Results are awaited. 8.6 Definitive brachytherapy 1) Image-guided adaptive brachytherapy (IGABT) is recommended, preferably using MRI at the time of brachytherapy (Grade A). 2) The tumour-related target for brachytherapy includes the residual gross tumour volume, the adaptive high-risk clinical target volume and the intermediate risk clinical target volume (Grade B). 3) Intracavity and combine intracavity / interstitial brachytherapy should be performed under anaesthesia. 4) The brachytherapy applicator should consist of a uterine tandem and vaginal applicator (ovoid’s or ring). Combined intracavity / interstitial brachytherapy should be used where appropriate 22 (such as significant residual disease in parametria) in order to achieve a sufficiently high radiation dose to the target and / or reduce dose to organs at risk (Grade B). 5) In IGABT the aim should be to deliver a brachytherapy dose of 40-45 Gy (EQD2)(D90)to reach a total EBRT + brachytherapy dose of 85-90 Gy to the high-risk volume and 60 Gy (D98) to the intermediate risk volume. 3D dose volume constraints for rectum, vagina, sigmoid and bowel are recommended as published in the literature (Grade B). 6) Generally, point A dose normalisation should be used as the starting point for stepwise treatment plan optimisation. 7) Brachytherapy should be delivered in several fractions as high dose rate (usually 3-4) or in 1-2 fractions as pulsed dose rate (Grade A). Evidence from cohort series supports the use of image guided brachytherapy to maximise survival and local control while minimising late toxicity. Results from EMBRACE and retro EMBRACE supports the recommended dose of > /= 85 Gy EQD2 to the high-risk CTV D90 which is predicted to lead to a 3-year actuarial local control of >96% in tumours ≦30 cm3 and >91% in tumours >30 cm3. Utilization of combined intracavitary / interstitial applicators is an essential tool for dose escalation in large tumours. The EMBRACE studies provide the best data on doses to organs at risk based on organ volume rather than point doses (D2cc Bladder 80 Gy, D2cc rectum 70-75 Gy, D2cc Bowel / sigmoid 70 Gy). These should be the treatment aim where possible but should not preclude adequate tumour doses. Of note there is not a good correlation between bowel / sigmoid dose and toxicity due to the mobility of this organ. Recommendations: • External beam radiation should be planned to use newer technologies • Concurrent platinum-based chemotherapy should usually be administered • Brachytherapy remains an integral part of the treatment • Treatment times should not exceed 56 days (Grade A) 9 Adjuvant treatment post hysterectomy: Use of external beam radiotherapy, vagina vault brachytherapy, chemotherapy and combinations This section provides evidence-based information on the adjuvant treatment options post hysterectomy for cervical cancer. In this setting, it describes the role of external beam radiotherapy, vaginal vault brachytherapy, chemotherapy and combinations after a hysterectomy for early stage cervical cancer. The purpose of adjuvant treatment is to reduce recurrence risk, increase chance of cure and prolong life if possible. It is therefore important to understand the risk factors that contribute to higher recurrence and lower survival rates. 9.1 External beam radiotherapy Large tumour size, LVSI and deep stromal invasion have been shown to independently predict local recurrence after surgery in FIGO stage I cervical squamous cell cancer (123). The GOG 92 phase 3 trial therefore randomized patients with stage IB disease with at least two of these poor prognostic features to post-operative pelvic radiotherapy (46-50.4 Gy) or observation alone (121). The adjuvant radiotherapy arm showed a 46% reduction of recurrence (HR 0.54) and a reduction in risk of progression or death (HR 0.58). Both local recurrence (13.9% vs 20.7%) and distant recurrence (2.9% vs 8.6%) were reduced with the use of radiotherapy. This progression free survival benefit was balanced against a 6.6% grade 3 or 4 adverse event rate, (haematological RR 2.38; gastrointestinal RR 7.32) compared with 2.1% without radiotherapy. Post–operative radiotherapy was most beneficial in patients with adenocarcinoma or adenosquamous histology (120). A subsequent 2012 Cochrane Review which pooled this trial with a German randomized controlled trial (148) evaluated 397 patients and concluded that 23 although progression free survival was improved, overall survival was not affected by adjuvant radiotherapy in stage IB disease (149). In this cohort of patients, it remains unclear whether concurrent chemotherapy with radiation is superior. Retrospective data suggests lower recurrence rate but no definite survival improvement with the addition of chemotherapy concurrently (150). The answer may hopefully be provided by the currently recruiting GOG 263 prospective randomised study of chemo-radiation versus radiation alone. Post-operative radiotherapy in patients with LVSI, deep cervical stromal invasion and large tumour size is associated with a progression free survival benefit. It can therefore be recommended to well-informed women with these risk factors providing they fully understand the risk of toxicity. The addition of concurrent chemotherapy in this setting should be investigated within a clinical trial or on an individual patient basis. 9.2 Concurrent chemotherapy with external beam radiotherapy High risk factors for recurrence after surgery for early stage cervical cancer include pathological confirmation of lymph node positivity, parametrial involvement and positive surgical margins. The recurrence risk in this setting is in the region of 40% if treated with surgery alone. The landmark randomized phase III intergroup trial {SWOG 8797, RTOG 91-12, GOG 109} (151) evaluated concurrent chemotherapy with radiation or radiation alone in patients with any of these risk factors. The chemotherapy was 4 cycles of 3 weekly cisplatin and fluorouracil started with radiotherapy. Progression free survival (HR 2.01 4yr PFS 80% vs 63%) and overall survival (HR 1.96, 4yr OS 81% vs 71%) were both significantly improved with the addition of chemotherapy (146). Interestingly, the 5-year survival benefit was significantly higher in tumours >2cm (19% vs 5%) and if more than 1 lymph node was involved compared to only one (20% vs 4%). Grade 3 and 4 haematological and gastrointestinal toxicity was increased with chemotherapy with 17% of chemo-radiation patients experiencing grade 4 toxicity, mainly hematologic compared to 4% with radiation alone. The use of single agent cisplatin is likely to reduce this toxicity risk and, as seen in locally advanced disease, may be equivalent in efficacy. A 2016 Cochrane analysis combined this phase III intergroup trial with three other randomised controlled studies to evaluate 401 patients in total. This concluded that use of concurrent platinum-based chemotherapy seems appropriate (153). High risk patients, defined as lymph node positive, parametrial involvement and positive margins, gain a survival benefit from receiving post-operative cisplatin-based chemo-radiation. This should therefore be recommended to women with these pathological risk factors. 9.3 Radiotherapy technique Guidance regarding the delivery of postoperative radiotherapy should be followed. Consensus guidelines have been published for delineation (153). Newer radiotherapy techniques such as Intensity Modulated Radiotherapy (IMRT) can facilitate reduced toxicity with radiotherapy and achieve equivalent survival outcomes (155-157). The role of IMRT has been investigated in the phase III trial of IMRT versus standard therapy in postoperative treatment of endometrial and cervical cancer (TIME-C/RTOG 1203 CCOP), now closed to recruitment. As for locally advanced cervical cancer patients, the radiotherapy field includes the para-aortic nodes if the common iliac nodes or the para-aortics themselves are proven positive. Treatment time between surgery and start date is important. A recent National Cancer Database multivariate analysis of 3051 patients revealed a surgery to radiotherapy interval of more than 8 weeks is a poor prognostic factor (HR 1.20). Along with the overall radiotherapy treatment time of more than 7 weeks (HR 1.21). (157) 24 9.4 Vaginal vault brachytherapy Within the GOG 92 and SWOG 8797/GOG 109/RTOG 91-12 protocols no brachytherapy was permitted (120-124). However, in cases of vaginal margin involvement or suboptimal surgery (less than radical hysterectomy) it is recommended to deliver a vaginal vault brachytherapy boost of an additional 10-15 Gy in view of the risk of residual tumour cells (158,159). There is no randomised evidence proving the additional role of brachytherapy in this situation. A retrospective analysis of 142 women who underwent post-operative radiotherapy for cervical cancer found the use of a brachytherapy boost in patients with close or involved vaginal resection margins did not have inferior local control, distant control or survival. This suggests a benefit, but patient numbers were small (160). In this cohort parametrial resection margin positivity was associated with a very poor prognosis (5 yr. OS 19%) and the majority of these patients did not receive brachytherapy. A further retrospective analysis of 292 patients demonstrated that systematic application of vaginal vault brachytherapy after EBRT (10-14 Gy) in patients with at least 2 of the following, (adenocarcinoma, nodal involvement and parametrial extension) significantly reduced the poor prognostic impact of histology on local and distant recurrence. This was not associated with any increase in complication risk (161). Without a significant increase in complication risk it is therefore reasonable to consider brachytherapy boost in all high-risk patients (158). Vaginal vault brachytherapy boost (10-15 Gy) after adjuvant EBRT should be administered if less than radical hysterectomy has been performed or there are close or involved vaginal margins. It should also be considered in other high-risk patients with appropriate counselling of patients regarding the uncertain benefit and potential toxicity; adenocarcinoma, parametrial or vaginal involvement, extensive LVSI, larger tumours or deeply invasive tumours. In the unlikely scenario of vaginal margin involvement following adequate surgery with no other high-risk factors then consideration of adjuvant vaginal vault brachytherapy boost alone could be discussed (162). 9.5 Adjuvant chemotherapy The four randomised controlled trials analysed within the 2016 Cochrane review compared differing chemotherapy regimens in addition to radiotherapy post-operatively (152). Only one study compared chemoradiation with chemoradiation plus further adjuvant chemotherapy (163). Unfortunately, due to significant toxicity with the combination chemotherapy (topotecan and cisplatin) this study was terminated early with only 39 patients being recruited. The benefit of further chemotherapy after chemo-radiation in high risk patients is therefore still unknown. The currently recruiting joint GOG and RTOG 0724 trial is addressing that question by randomising patients to receive concurrent cisplatin with radiotherapy with or without 4 cycles of adjuvant carboplatin and paclitaxel. The NCT 00806117 is a similar randomised study recruiting in China. Retrospective reviews have analysed the potential role of post-operative chemotherapy in certain high-risk patients (e.g. node positive, adenocarcinoma) as an alternative to chemo-radiation (164-166). They have demonstrated feasibility of platinum-based chemotherapy, but there is no randomised evidence to support its application. Adenocarcinomas have a worse prognosis than squamous cell cancers with a retrospective review suggesting adjuvant chemotherapy improves progression free survival compared to chemo-radiation in this cohort (167). A non-statistically significant improvement in survival of 9% was reported for node positive patients treated with chemotherapy after surgery compared to a historical cohort managed surgically only (164). As chemo-radiation has now been shown to provide a survival benefit in node positive patients the relevance of this result is unclear. Adjuvant chemotherapy (not concurrent with radiation) in early stage cervical cancer should only be used in the context of a clinical trial. (Grade B) 25 10 Neoadjuvant chemotherapy in cervix cancer Recommendations: Neoadjuvant chemotherapy is not standard practise and was shown to have poorer oncologic outcome in earlier studies compared to cisplatin-based concomitant chemoradiation (Grade A). However more recent use of combinations which include a taxane has challenged this, hence current clinical studies such as INTERLACE comparing NACT followed by CCRT with CCRT alone (168-170). Neoadjuvant chemotherapy (NACT) is usually defined as the use of chemotherapy prior to the definitive treatment. Thus, it may be used as an induction or neoadjuvant prior to planned surgery or prior to planned radiation therapy. The rationale for NACT is that it may: • Reduce the size of the tumour and convert it from being surgically unresectable to resectable, • Allow a lesser surgical procedure to be carried out such as fertility-preserving surgery or • Shrink the tumour to allow definitive chemoradiation a more effective chance of achieving improved tumour control. Furthermore, it may have a systemic effect and reduce risk of metastatic spread and/or eliminate micro-metastatic disease. NACT may also be considered in some low-income nations where there are scarce radiotherapy facilities and chemotherapy is given prior to surgery, but this is beyond the scope of this chapter. As discussed below, NACT may also have a place in management of cervix cancer in pregnancy. (Grade D) 10.1 Neoadjuvant chemotherapy followed by surgery Two large randomised clinical trials have been conducted, the first by the EORTC GCG study 55994 which although this began in 1999 and results were presented at ASCO 2019. This trial compared neoadjuvant chemotherapy with cisplatin or cisplatin-based chemotherapy (minimal cisplatin dose of 225 mg/m² over 9 weeks) followed by radical hysterectomy with pelvic node dissection (RHND) and compared it with concomitant cisplatin chemoradiation. The second trial was a large single-centre, phase III, randomized controlled trial by Gupta et.al (171) comparing the efficacy and toxicity of neoadjuvant chemotherapy followed by radical surgery versus standard cisplatin-based chemoradiation in patients with locally advanced squamous cervical cancer in stage IB2, IIA, or IIB squamous cervical cancer showing the 5-year disease free survival in the neoadjuvant chemotherapy plus surgery group was 69.3% compared with 76.7% in the concomitant chemoradiation group (HR, 1.38; 95% CI, 1.02 to 1.87; P = 0.038), whereas the corresponding 5-year OS rates were 75.4% and 74.7%, respectively (HR, 1.025; 95% CI, 0.752 to 1.398; P = 0.87). The delayed toxicities at 24 months or later after treatment completion in the neoadjuvant chemotherapy plus surgery group versus the concomitant chemoradiation group were rectal (2.2% v 3.5%, respectively), bladder (1.6% v 3.5%, respectively), and vaginal (12.0% v 25.6%, respectively). This study showed that cisplatin-based concomitant chemoradiation resulted in superior progression free survival (PFS) compared with neoadjuvant chemotherapy followed by radical surgery in locally advanced cervical cancer. (165). The EORTC 55994 study (172) was presented at ASCO 2019 and shows remarkably similar results and would suggest no advantage to NACT prior to surgery for larger tumours, whereas for stage IB2 tumours (FIGO 2009) there may be a small benefit to NACT. Thus, concomitant chemoradiation should be the current standard of care until INTERLACE is concluded and reported. Recommendations: • Concomitant chemoradiation should be the current standard of care for locally advanced cervical cancers. Results of further studies are awaited. (Grade A) 26 11 Management of cervical cancer in pregnancy Cancer of the cervix is the most common gynaecological malignancy diagnosed in pregnancy with an incidence rate of 0.1 to 12 per 10,000 pregnancies. (173) At present, management is mostly based on small case series, expert opinion and anecdotal case reports. (174, 175) 11.1 Pre-invasive disease in pregnancy 11.1 a Cytology An abnormal cytology in pregnancy should trigger a colposcopy and/or gynaecological oncology referral; investigation should not be deferred until after pregnancy (176-178). (Grade C) 11.1 b Colposcopy The aim of colposcopy during pregnancy is to exclude invasion, and in appropriate patients, i.e. with pre-invasive disease, defer treatment until after delivery (176,177). If there is suspicion of invasion at colposcopy then a biopsy adequate for diagnosis should be performed (176). (Grade C) The colposcopic features of invasive cancer do not differ between pregnant and non-pregnant women although colposcopy may be more challenging during pregnancy. These features may include abnormal vessels, irregular surface contour, mosaicism and punctation. The absence of histological invasion cannot be guaranteed by a punch biopsy only demonstrating CIN (176,177). If a loop biopsy is to be performed, it should be carried out where there are facilities to manage haemorrhage (177) and a vaginal pack can be inserted post procedure, which may reduce the risk of bleeding (179). Published case series have demonstrated the risk of significant bleeding to be as high as 25% (180), this is due to the increased vascularisation of the cervix (179). 11.2 Diagnosis of cervical cancer in pregnancy Almost two-thirds of cervical cancer cases in pregnancy are diagnosed in the first two trimesters (181). The symptomatology of cervical cancer does not differ between the pregnant and the non-pregnant state (182). Cancer related symptoms, such as painless vaginal bleeding, pelvic and lower back pain and urinary frequency, may mimic pregnancy complications (183). Consequently, a delay may mean the cancer is at a higher stage when diagnosis occurs (184). Conversely, there is evidence to show that regular examinations during pregnancy can result in earlier diagnosis (185). Some studies have suggested the chances of cervical cancer being diagnosed in its early stages are three times more likely in pregnancy compared to controls; 76% of lesions diagnosed in pregnancy are a stage IB1(186). All women presenting with a suspected cervical abnormality in pregnancy require an accurate pelvic examination (speculum and bimanual examination) including colposcopic assessment, regardless of gestational age (183,186). (Grade D). ESGO has a registry for cancers in pregnancy, refer to website, www.esgo.org. 11.2 a Signs and symptoms The clinical stage and size of tumour will dictate the symptoms of cervical cancer. Early stages of disease may be an incidental finding on routine pelvic examination or cytological investigations. However, symptoms could also include painless vaginal bleeding, abnormal vaginal discharge, post-coital bleeding and dyspareunia. Patients with more advanced disease may complain of urinary dysfunction, pelvic pain, changes in bowel habit, back pain and swelling of legs (187,188). 11.2 b Examination Examination should assess bleeding, discharge and the presence of lesions in the vagina/ on the cervix. Any lesions should be assessed in terms of size, shape and consistency (187). Cervical ectopy may make the assessment of a pregnant woman’s cervix more challenging. Necrotic, friable and exophytic lesions 27 are suspicious and require further investigation. (187) A woman with suspicious invasive disease in pregnancy should undergo careful biopsy to establish diagnosis (187). 11.2 c Referral pathway Women with any suspicious cervical pathology during pregnancy should be referred urgently to an experienced colposcopist or gynaecologist under the two-week rule (176,183). (Grade D) 11.3 Staging of cervical cancer in pregnancy Magnetic resonance imaging (MRI) MRI is the first line for staging of cervical cancer in pregnancy (190). (Grade D). In experienced centres, Ultrasound may be considered as an alternative which avoids ionizing radiation. MRI is safe in pregnancy (191) and currently the mainstay in staging of cervical cancer. It is essential to consider the challenges pregnancy poses when interpreting MRI. These difficulties can include dilated pelvic veins being misinterpreted as pelvic adenopathy (184), or a reduction in image quality due to fetal movements. (192) Computed tomography (CT) CT may be considered essential to gain diagnostic information about maternal wellbeing. Where there is a high suspicion of lung or pleural spread the benefits of undertaking the investigation can outweigh the risks to the fetus. In situations such as these, all efforts should be made to limit fetal exposure to radiation and not compromise the baby’s health. Abdominal shielding and low dose radiation should be considered (193). MRI is preferred as there is no ionizing radiation. Laparoscopic lymphadenectomy and sentinel lymph node biopsy Lymph node status is an important prognostic factor. (188). (Grade D). The agents used for SLNB are blue dye and radio-colloid technetium. Certain blue dyes, such as lymphazurine, have not been tested in pregnancy and should be avoided. There is limited evidence on the use of radio-colloids in pregnancy, however, recent studies show fetal exposure to radiation is low and pregnancy should not be considered an absolute contraindication (190,191). The use of ICG has shown safety for retinal angiography and to measure hepatic blood flow in pregnant women. At present, the role of SLNB in these patients remains unclear; more trials are needed to clarify the issues of safety and for now the routine use of SLNB should not be performed, unless in the context of clinical trials (198). 11.4 Treatment options Currently, gynaecological oncologists performing vaginal or laparoscopic radical trachelectomies routinely perform cervical cerclage as part of the procedure (199). However, radical trachelectomy undertaken via laparotomy is different, as cerclage was not included in the initial technique described. (200-205). Given the blood loss and poor fetal outcome, trachelectomy during pregnancy is not advised, in contrast, NACT maybe advocated. There is limited level A evidence demonstrating prophylactic cerclage reduces the risk of fetal loss and prematurity (199). Despite this, one study demonstrated a reduced risk of fetal loss from 50% to 22% when used after vaginal radical trachelectomy (206). Patients who have undergone trachelectomy should be delivered by caesarean section. (207) Delivery by caesarean section will aim to avoid damage to the reconstructive procedures performed, hence reducing prematurity in case of subsequent pregnancy. In addition, massive vaginal bleeding has been reported after vaginal delivery (208). 28 For patients with locally advanced cervical cancer neoadjuvant chemotherapy can help treat and control disease. This should be formally discussed within the local MDT. This will allow for fetal maturation and delivery at an appropriate gestation by caesarean section (209). (Grade D). Carboplatin is the first line chemotherapeutic agent (210). (Grade C) Carboplatin has a similar efficacy but less nephrotoxicity and ototoxicity to cisplatin (66). There is evidence to show if paclitaxel is used in conjunction with a single platinum agent, the treatment has superior outcomes with a longer period of remission (211). Though studies are limited, those that do exist have demonstrated minimal trans-placental spread of paclitaxel (212). Breastfeeding during chemotherapy treatment is contraindicated (213,214). (Grade C). Chemotherapy drugs cross into breast milk and may cause neonatal leucopenia. In turn increasing the risk of infection to the baby. An interval of 14 days from the last chemotherapy session is recommended before breastfeeding, allowing time for drug clearance from the breast milk (213). There is evidence to suggest that a short period of lactation after a stressful pregnancy can be of psychological benefit to the woman. (214) Route of delivery is determined by the presence or absence of visible tumour. If tumour is still present, delivery by caesarean section is preferable to reduce risk of cancer cell implantation in episiotomy scar (212). (Grade D) Delivery options must be discussed with the patient and obstetric contraindications to vaginal delivery also considered. Recurrence of tumour in episiotomy scars, haemorrhage and infection are well-documented complications of vaginal delivery in this cohort of women. Although less frequent, abdominal wall recurrences have also been described (215). If the tumour itself is very large, consideration to a wound protective system or corporeal uterine incision may be of benefit. In this circumstance, performing a classical caesarean section is likely to reduce blood loss. Delivery should take place in a hospital with a level three neonatal unit, (173,175,185,188) and the placenta should be sent for histology to exclude metastases (216) (Grade D) 12 Surgical management of relapsed cervical cancer The management of relapse will be influenced by the choice of initial therapy. Usually salvage surgery is considered after primary chemo-radiation, where the reverse is considered if primary management was surgical management. Surgical treatment of relapsed cervical cancer is challenging due to the commonly palliative situation of the disease, the complex disease-related clinical symptoms and the high surgical complexity associated with any surgical intervention. In the majority of cases, metastatic cervical cancer is not curable, but for some patients who present with locoregional recurrence or with limited distant metastatic disease, surgical treatment may be potentially curative. 12.1. Intra-pelvic relapse Patients with intrapelvic relapse should be considered for exenterative surgery with curative intent, including laterally extended endopelvic resection (LEER), only if a complete tumour resection with microscopically clear tumour margins is anticipated (Grade B). A PET/CT prior to exenterative surgery should exclude the presence of distant metastases. (Grade B). Patients with radiologically or clinically positive pelvic lymph nodes should not be excluded as surgical candidates, if disease appears resectable, since pelvic lymphadenopathy has been shown not to affect prognosis, as opposed to paraaortic lymphadenopathy which has been demonstrated to be a negative prognostic indicator of survival (Grade C). 29 For women who have undergone primary chemoradiotherapy, subsequent radical hysterectomy or pelvic exenteration, with the aim to achieve clear surgical margins, as management of local recurrence, is the preferred approach, as it has been shown to have five-year survival rates ranging between 30 and 40 percent (217,218). Careful patient selection is required given the perioperative and postoperative morbidity associated with this extensive surgical approach and should include psycho-sexual considerations. These should be part of the specialist team doing the procedures. Intraoperative radiation therapy (IORT) is an additional potential option for women undergoing exenterative procedures for relapse, where available. However, the efficacy of IORT has not yet been shown in prospective trials and not routinely available in clinical practice (219,220). Pelvic relapse can be divided into central versus pelvic side wall relapse. Central relapse can affect the anterior, posterior or both compartments. Pelvic side wall relapse also varies in terms of the affected structures and extent of clinical symptoms. The probability of complete resection is higher in the case of a central relapse where exenterative surgery tailored to the affected compartment can achieve microscopically clear resection margins. The most common type of central relapse is a vaginal vault relapse and hence affecting both compartments and often necessitating total exenteration. Brunschwig was the first to publish in 1948 the first successful results of pelvic exenteration in patients with central cervical cancer relapse (221,222). The original procedure included the en bloc resection of the internal and external reproductive organs including the bladder, urethra, both ureters, rectosigmoid colon together with the anus and perineum. The ureters were then implanted into the colon upstream of the colostomy site in terms of a wet colostomy. (221) There are three types of exenteration: total, anterior and posterior. Total exenteration refers to removal of the uterus and parametria, bladder, rectum, vagina, urethra, and a part of the levator muscles. In an anterior exenteration, the rectum is spared, while in a posterior exenteration, the bladder and urethra are preserved. A perineal phase, resecting the anus, urethra, and portions of the vulva, may also be required in cases where the distal part of the vagina and/ or vulva and perineum are affected. The type of procedure chosen for the patient depends on the site of relapse, previous treatment, anatomy, and the patient's wish and expectations. There are no prospective randomized studies to assess the impact of exenterative surgery on patient’s survival. In a systematic review of 21 studies of pelvic exenteration for gynaecological malignancy, one-third to one-half of women were found to have unresectable or extra pelvic disease found at exploration resulting in abortion of the procedure (223). Those who had negative resection margins and no metastatic disease had an approximately 50% cure rate; the remainder died of recurrent cancer. One of the largest monocentric case series to date is by Egger et al, which evaluated the outcome of 282 patients with various types and extents of advanced or relapsed cervical cancer (222). The majority of patients (75%) had undergone exenteration for relapsed disease. Anterior or posterior exenteration alone cleared the disease in only 5% and 2% of the cases, respectively, and 93% of the patients required a total exenteration to achieve clear margins. The 5- and 10-year overall survival (OS) rates of all patients were 41% and 37%, respectively. The 5-year disease-free survival (DFS) for all patients was 61%. The 5- and 10-year OS rates of the 133 patients that were operated on with a curative intent, was 64% and 57% compared to those that were operated on with a palliative intent, who had a 5-year OS of 19% and a 10-year OS rate of 18%. No significant OS difference was found in the patients with positive pelvic lymph nodes compared to those with clinically and radiologically normal appearing pelvic lymph nodes. In contrast, para-aortic lymph node metastasis had a highly significant negative impact on the OS constituting a prognostically unfavourable group that should not therefore undergo major exenterative procedures. 30 A preoperative PET CT is an accurate imaging tool to demonstrate extra pelvic disease that should preclude patients from pelvic exenteration and is useful for preoperative assessment of disease extent (224,225). (Grade C) 12.2 Laterally extended endopelvic resection in pelvic side wall relapse Laterally extended endopelvic resection (LEER) is a potential therapeutic option for patients with limited pelvic disease fixed to the pelvic side wall. Careful patient selection is crucial to keep surgical morbidity and mortality in limits. (Grade C) In the presence of novel surgical advances, a new surgical approach has been developed in the last decade, based on developmentally derived surgical anatomy and aiming to increase the curative resection rates, even of tumours extending to and fixed to the pelvic side wall. The so called ‘laterally extended endopelvic resection’ or LEER has been developed with the potential to salvage selected patients, traditionally not considered for surgical resection (219,226). LEER as described by Höckel, is performed in combination with at least two of the following procedures: total mesorectal excision, total mesometrial resection, and total mesovesical resection. In cases of lateral tumour fixation, the inclusion of pelvic side wall and floor muscles, such as the obturator internus muscle and pubococcygeus, iliococcygeus and coccygeus muscles, and eventually of the internal iliac vessel system assures the completeness of the multi-compartmental resection (227). Resection of major vessels such as the common or external iliac arteries, requires surgical bypass grafts post-resection. Höckel described the outcome of 91 patients with locally advanced primary (n=30) and recurrent or persistent (n=61) carcinoma of the cervix and vagina who were treated with LEER. No LEER treatment was aborted and R0 resection was histopathologically confirmed in all cases, suggestive of careful patient selection. The authors concluded that LEER could definitively control the locoregional disease in 92% of the patients. Five-year OS was up to 61% despite 74% of the patients having tumours fixed to the pelvic side walls (226,227). LEER was developed on the concept of ontogenetic anatomy. 12.3. Surgical morbidity and mortality of pelvic exenteration Due to the high surgical morbidity and mortality rates of exenterative procedures, especially following previous radiotherapy, careful patient selection and team specialization is key to optimal outcomes. (Grade C). Post-operative leaks and haemorrhage should be preferably treated conservatively with endoscopic procedures and interventional radiology, rather than re-laparotomy wherever possible, to avoid additional morbidity from the re-operation (Grade C) The significant surgical morbidity of exenterative procedures in relapsed cervical cancer is very common and can be as high as 50% (223,228,229). In retrospective reviews the most common complications are infections and wound healing problems ranging from 39% to 86%, gastrointestinal and/ or urinary fistula (10%-23%) and small bowel obstruction (11%-33%) (229,230). Perioperative mortality varies between series and patient cohorts, ranging from less than 5% to 10% (231). Sepsis, adult respiratory distress syndrome, heart failure, pulmonary embolus, and multi-organ system failure are typical terminal events. Intraoperative complications are predominantly related to haemorrhage and problems associated with pelvic reconstruction requiring high rates of blood transfusion. A significant issue can be delayed haemorrhage, even weeks following the operation, in association with pelvic abscesses leading to erosion and bleeding from major pelvic vessels, especially after previous pelvic radiotherapy. Delayed haemorrhage is best managed with percutaneous embolisation via interventional radiology, as re-laparotomy is associated with significant morbidity and mortality. (232) 31 Gastrointestinal or urinary tract fistulas should be treated conservatively, with endoscopic procedures such as stents, total parenteral nutrition and decompression with for example percutaneous catheters. Low output fistulas, in the absence of distal obstruction, may spontaneously heal (233). Delayed complications include bowel obstruction (especially small bowel that is torted around the ileal conduit), gastrointestinal fistulas, ureteral obstruction with renal compromise, and stomal stenosis. Initial conservative management is recommended, where appropriate. Multidisciplinary management and counselling with dedicated psychotherapists and clinical nurse specialists are strongly recommended prior to indication for exenteration in surgical candidates. (Grade D) Body image and sexual function are significantly affected by exenterative procedures and should therefore be discussed with surgical candidates preoperatively, even though the degree of impairment varies strongly. A study of 16 women found sexual function and body image declined during the first three postoperative months before returning to baseline by 12 months in accordance with studies evaluating quality of life after ultra-radical surgery for ovarian cancer (234). Counselling of the patient regarding postoperative changes in anatomy, body function and involvement of the patient's sexual partner are important parts of exenterative surgery care. In a recent retrospective, multicentric study investigating quality of life issues and emotional distress in 91 gynaecological cancer survivors submitted to pelvic exenteration, the following parameters have been identified as independent predictors of lower body image levels (235): Older age (HR, 11.235; P = 0.003), vaginal/vulvar cancer (HR, 7.369; P = 0.013), total/posterior pelvic exenteration (HR, 7.393; P = 0.013), higher number of ostomies (HR, 7.613; P = 0.012), the creation of a non-continent bladder (HR, 8.230; P = 0.009), and of definitive colostomy (HR, 8.516; P = 0.008). 12.4. Reconstructive procedures Reconstructive procedures of the urinary, gastrointestinal tract and the vagina are as important as the procedure of exenteration itself. It should be carefully discussed preoperatively with those requiring consideration of exenteration to identify the adequate reconstructive techniques according to patient’s needs and wishes. (Grade C) 12.4.1 Urinary system Urinary conduits and reservoirs are typically constructed from an isolated segment of bowel. Any segment of bowel can be theoretically used to form a urinary diversion, however, there are metabolic consequences to using each bowel segment, determined by its absorptive function, especially in patients with chronic renal failure (235). The most commonly used bowel segments in urinary diversions include the distal ileum and/or caecum and the ascending or sigmoid colon; these sites are associated with the fewest metabolic consequences. There are two main types of urinary diversions: incontinent and continent (237). Incontinent urine diversion is the most commonly used type of diversion and includes implanting the ureters into an isolated segmented loop of bowel, so the urine is drained via a cutaneous stoma. There, urine drains continuously and is collected by an external appliance adhered to the skin surface (stoma bag). The most common and widely used type of incontinent urinary diversion is the ileal conduit. Other types and the years where they were initially described are: Ureteroproctostomy (Simon, 1851), Ureterosigmoidostomy (Smith, 1878), Rectal bladder (Gersuny, 1898), Ileal loop (Bricker, 1950s), Ileal neobladder (Camay, 1959) Koch pouch (1970), Indiana pouch (early 1980s), Orthotopic diversion (late 1980s). Diversion into a non-continent conduit is considered less technically demanding and hence is associated with fewer postoperative complications and is the most widely used. 32 By contrast, continent reservoirs collect and store urine internally which is then actively emptied by the patient via catheterization of a cutaneous stoma. The patient is thereby freed from the need for an external appliance. The most commonly used bowel segments for continent urinary diversion are either ileum or a combination of terminal ileum and ascending colon with the appendix. The process of selecting a particular urinary diversion option is multifactorial and addresses many levels: oncological surgical complexity and patient-related factors (co-morbidities, lifestyle, and quality of life expectations). In an attempt to reduce morbidity from an additional anastomosis in the area of harvesting of the ileal conduit, an option is to implant the ureters into the distal part of the sigmoid and/ or descending colon after total exenteration. In patients where surgery is with palliative intent, or in those who do not want to have two stomatas, a safe and simple alternative is the double-barrelled wet colostomy. The technique is relatively easy to learn and reduces the time for urinary and faecal diversion, length of stay, and morbidity rate (238). 12.4.2 Gastrointestinal system The indication to perform an end-sigmoid colostomy or a low rectal anastomosis depends on the level of the resection, length of the rectal stump, the extent of surgery and the radiotherapy induced bowel damage. Due to the radiotherapy-induced changes on the bowel wall and the deeper extent of dissection into the pelvis the leak and fistula rates are higher compared to those following debulking surgery for ovarian cancer. A protective temporary, double-barrelled ileostomy can be formed in an attempt to reduce postoperative morbidity from fistula formation, even though no randomised trials have proven that such an approach is protective. An omental flap can be used to cover the resulting empty pelvis and may help prevent adhesion and possibly subsequent fistulation or obstruction of the small bowel in the empty pelvis, and functions as a carpet for the raw surfaces of the exenterated pelvis. 12.4.3 Neovagina Young and/ or sexually active women should have a discussion pre-operatively about neovaginal reconstruction. Several methods for vaginal and pelvic reconstruction have been described. Myocutaneous grafts, including rectus and gracilis muscle flaps, can be brought into the pelvis and perineum to create pelvic support and a neovagina. Split-thickness skin grafts have also been used to create neovaginas. A further technique is the sigmoid neovagina, even though radiotherapy induced changes need to be taken into consideration to avoid morbidity from anastomotic leaks. An advantage of bowel neovagina is that is does not need regular dilatation to avoid stenosis (239). 12.5. Fistulation / Cloaca (urinary or colonic) Management of urinary or colonic fistulation depends on the oncological situation of the patient and her overall prognosis. In palliative settings where prognosis is likely to be shorter, the primary approach should be conservative with diverting nephrostomies or double-barrelled stomatas. (Grade C). There are two main causes of fistula formation: cancer-induced or radiotherapy-induced. Depending on the extent of fistulation, the entire clinical picture, symptoms and the overall prognosis of the patient’s treatment options range from minimally invasive nephrostomies and double-barrelled ileostomies or colostomies to total pelvic exenteration (240). Patients with limited prognosis should not undergo the high surgical risk of an extenterative procedure but are better managed with diverting nephrostomies in case of a urinary fistula and/ or a diverting bowel stoma in case of a gastrointestinal fistula, that can be also performed minimally invasive, where clinically appropriate. While quality of life comparative data is largely lacking, the patient’s prognosis and overall clinical picture should dictate radicality of any therapeutic approach. 33 13. Re-Irradiation of cervix cancer The need for pelvic re-irradiation is increasing, this is mainly due to improved cure rates with modern cancer treatments. In the context of recurrent gynaecological malignancies, we are likely to encounter two scenarios: 1. Patients cured of their cervix cancer developing a new vaginal cancer more than a decade later 2. Patients treated for cervical cancer develop a pelvic recurrence a) isolated recurrence, either side wall or central b) widespread recurrence within the pelvis. The radiotherapy technique and dose/fractionation applied depends on the individual case. For example, for patients in scenario 2b, the recurrence is likely to be within a 5-year period and therefore further EBRT is unlikely to be either feasible, or indeed advisable, and such patients are better treated with systemic therapies either standard or within clinical trials. The following articles cover more recent events and developments including intra-operative and stereotactic ablative approaches as discussed below (241-245). 13.1 Vaginal cancer following radiotherapy for cervix cancer Almost by definition, these patients will be 10-20 years post previous treatment. Disease is often localised to the vagina and, PET/CT, where available and used for staging, shows no pelvic nodal involvement. Such patients should be treated with a combination of small field radiotherapy, encompassing the whole of the vagina but not treating pelvic nodes. Brachytherapy boost is provided either with intracavity brachytherapy or interstitial implant. Selected cases may be treated with brachytherapy alone, often this would be interstitial brachytherapy. EBRT dose would be 40-45Gy in 20-25 fractions and brachytherapy dose 24Gy/4# for intracavitary and 15Gy/3# for Interstitial. Stereotactic Ablative (Body) Radiotherapy [SABR or SBRT] is not of much use in this setting. 13.2 Isolated recurrence following previous pelvic radiotherapy This is more frequently seen less than 5 years from previous radiotherapy, either a side wall recurrence (often nodal) or a central recurrence (cervix or parametrium). A central recurrence would be amenable to re-treatment with brachytherapy which would, indeed, be the optimal choice in this setting. The most appropriate modality may either be intracavitary alone or a combination of intracavitary/interstitial. For side wall recurrences, brachytherapy is not an option. This is the group of patients who would benefit most from SABR/SBRT. The literature on this is limited although expanding (243-245). Doses used range from 21Gy/3# to 42Gy/6#, recent publications point to even higher doses. There is no robust organ-at-risk (OAR) tolerance data although the COMET trial did provide some guidance and the author has utilized this in his own practice. The reported literature is very encouraging with little or no Grade 3 toxicity in any of the reports. This is therefore a very promising area of further enquiry and clinical trials are being planned, although not in the setting of re-irradiation. Recommendations: • Discussion of management should take place through the MDT/Tumour Board. In selected cases re-irradiation may be advised but with increased risk of later toxicity. • Surgical excision of isolated nodal recurrence should be discussed where appropriate. • FDG PET imaging is essential before the procedures are discussed. (Grade D) 14 Chemotherapy in metastatic or advanced cervical cancer All patients with newly diagnosed stage 4 or relapsed cervical cancer should be treated as part of a multidisciplinary team, due to the complexity of disease relapse, especially when in the pelvis, where morbidity can be severely debilitating. Along with the obvious physical impact of disease, these patients 34 are often young with small children and emotional support for patients and families is of paramount importance. Those patients with a WHO performance status (WHO PS) 0/1 should be considered for systemic treatment, whereas those with lower performance status should be carefully risk assessed as to their suitability and likely benefit from treatment, with the patient fully informed of expectations and limitations of chemotherapy. Best supportive care or palliative radiotherapy may be a more preferable option for these patients. Platinum paclitaxel doublet has been the standard of care for some time for advanced and recurrent cervical carcinoma (246) however since 2014, Bevacizumab has been FDA approved and SMC and NICE approved from 2016 to be used with either platinum/paclitaxel or platinum/topotecan. Given alongside the chemotherapy, there was an overall survival (OS) advantage of 3 months and median progression free survival (PFS) of 2 months, with a higher response rate when compared to platinum/paclitaxel or platinum/topotecan alone (247-249), as shown in the GOG 240 trial. With extended follow up, the benefit of the addition of bevacizumab was sustained, as the overall survival curves remained separated. Furthermore, after stopping bevacizumab there does not seem to be a flare in disease burden. Blood pressure and urinary monitoring is mandatory, but bevacizumab is contra-indicated in patients with fistulas or those within 4 weeks of surgery. However, the addition of bevacizumab does not appear to impact significantly of patient’s quality of life and is well tolerated (250-252). Those patients with recurrent/persistent inoperable disease have in the majority already been exposed to chemotherapy, either in the neoadjuvant setting prior to radiotherapy, or concurrent chemotherapy in the form of Cisplatin with radiotherapy. Interestingly, bevacizumab showed benefit in those with short platinum-free intervals, in contrast with the experience in earlier GOG studies where platinum free intervals less than 12-15 months were associated with poorer response possibly due to platinum resistance. Cisplatin and carboplatin have been recognised as being equivalent in efficacy in a number of tumour sites and with their differing toxicity profile can lend themselves to help tailor treatment depending on patients’ comorbidities (253). Locally recurrent/persistent disease can involve the bowel and would make bevacizumab relatively contraindicated, however in those able to receive chemotherapy/bevacizumab combination even those with prior cisplatin exposure saw a significant benefit. In those that have pelvic reoccurrence or persistent disease who are initially not deemed salvageable with exenterative surgery, the option of upfront chemotherapy with a view to reconsideration for surgery depending on response should not be overlooked. Frumovitz, in 2017 also demonstrated improved progression free survival could be seen in those patients with recurrent small cell neuroendocrine carcinoma of the cervix (254) using topotecan and bevacizumab. Second line treatment and beyond is dependent on the interval of progression since first line treatment in those patients with a good partial response with first line treatment and are more than 6 months out, rechallenging with platinum/paclitaxel could be considered. Mitomycin/5FU, vinorelbine, docetaxel, gemcitabine, weekly paclitaxel and topotecan have some activity but there is no standard of care. Response rates are universally poor and entry into clinical trials where possible to assess novel and immunotherapeutic agents should be strongly considered depending on patient’s fitness and desires. McLachlan recently published retrospective results from the Royal Marsden 2004-2014 showing 70 % of their advanced cervical cancer patients received second line treatment. Median PFS was 3.2 months and median OS of 9.3 months. Thirty nine percent went on to receive 3rd line treatment (255). 35 Of these newer agents and combinations some have shown promising results. The CIRCCa trial used standard carboplatin/paclitaxel with or without cediranib - a vascular endothelial growth factor (VEGF) inhibitor, this showed only a PFS of 6 weeks with no OS) but subgroups showing expression of VEGF seemed to show greater benefit (256). This has led on to the COMICE trial which is combining cediranib and olaparib – a PARP inhibitor vs placebo in second line patients with stage 4 or recurrent cervical cancer. Immunotherapy using PDL1 inhibitors pembrolizumab, nivolumab and ipilimumab, pathways targeted for example Trametinib and HPV related therapy using therapeutic vaccines are in early phase one and two stages of trial and development. Recent data from the Keynote studies with pembrolizumab has also shown benefit but at time of writing is not approved or licenced in Europe. Studies with Tumour infiltrating lymphocytes (TILs) are also under development and show great potential. Recommendations: • For those patients who are chemotherapy naive with stage 4 disease, first line treatment would be systemic chemotherapy with cisplatin/paclitaxel or carboplatin/paclitaxel doublets with or without bevacizumab depending on any patient risk factors. However, bevacizumab may lead to prolonged benefit and should be offered if not contra-indicated. (Grade B) 15 Follow-up of cervical cancer Follow-up may be clinical, imaging or biochemical. It may be consultant-led by a gynaecological oncologist or clinical oncologist working in a specialist Cancer Centre or maybe by a general gynaecologist in a district general hospital or some form of shared care. The purposes of follow-up are to detect recurrence and offer appropriate salvage treatment, and to monitor for toxicities, particularly when new treatments or techniques are introduced. It is also an opportunity to seek support and care from CNS to provide psycho-sexual counselling and advice. Recent advances in surgery and changes in radiotherapeutic and chemotherapy practice require careful assessment to ensure that these changes are truly beneficial. The best evidence comes from a Canadian systematic review, a Cochrane review in 2011 and a consensus ESGO State of the Art conference in 2014 but most of this is very low certainty evidence (257-259). Following surgery without any added radiation, follow up is usually in the gynaecological oncology clinic supervised by specialist teams. Clinical examination will be undertaken, and smears may be taken. If any uncertainty a biopsy will be organised. There is no proven benefit for imaging of asymptomatic patients, but imaging will be directed by symptoms. Routine use of cervical screening after radiation therapy is not recommended. Following fertility-sparing surgery, cervical screening is advised. ( Traditionally, most recurrences were thought to occur within the first 2 years of follow-up following definitive treatment, but recent evidence suggests that this may be delayed when chemotherapy is incorporated in the treatment. Patients receiving concomitant cisplatin-based chemotherapy have a better outcome than those treated by radiation alone, thus recurrences are more frequently documented after the second year. This has significant implications as historically follow-up was more closely observed in the first 2 years, but now patients may need to be seen more frequently for longer. The BGCS has issued a comprehensive paper about the value of patient initiated follow up in gynaecological cancers, including cervical cancer, that supports this approach in patients with early disease who have not experienced any relapse (Newton et.al. Int J Gynae Cancer 2020 in press). 15.1 Identifying relapsed disease 36 However, contrary to this, there is published evidence to show that the numbers of patients picked up with relapse at the time of a clinic visit is actually relatively small and that most recurrences are interval relapses. In times of limited resources, it may be better to use these services more efficiently. However, there are other reasons for follow up such as emotional and psychosocial/psycho-sexual support and management of any menopausal symptoms and management of delayed surgical or radiation-induced sequelae. Furthermore, there are opportunities to reinforce the importance of smoking cessation. Not only is this beneficial to an individual’s health but will also reduce the risk of subsequent gynaecological tract malignancies (mainly vaginal and vulval/perineal) but also other smoking related squamous cancers in lung and head and neck. This will be discussed in the following chapter. Tumour markers such as the squamous cell carcinoma antigen (SCCA) are available, but there are very few centres in the United Kingdom who use it and it is not recommended for routine use. There are no other recommended tumour markers in clinical practice. 15.2 Imaging Imaging is also controversial in the follow up of patients with cervix cancers. This would only be useful if there is a salvage treatment option. Some experts recommend re-imaging with MRI at 3 months post chemoradiation to show whether there is residual disease or not. The only randomised trials that have looked at chemoradiation followed by hysterectomy have failed to show any survival advantage and were terminated early. Therefore, the evidence base for recommending this is dubious at best and is a questionable use of this resource. Salvage hysterectomy should be considered on an individual basis at the MDT. If surgery is considered, an MRI is essential to delineate extent of disease for optimal surgical planning. Potentially more valuable is the use of follow-up FDG PET/CT imaging. Data from Grigsby and his colleagues in Washington Missouri have shown that follow-up scans can be highly predictive of the risk of recurrence (260,261). Positive uptake on the FDG PET/CT at 3 months was usually associated with premature death whereas patients with complete metabolic response were generally in remission at five years and considered to be cured. The use of FDG PET/CT scanning at 3-6 months after treatment may help to identify those at low risk of recurrence who could then be put on minimal intervention follow-up or even nurse-led clinics whereas those with positive scans would need a higher intensity follow-up as they would be at considerable risk of developing a recurrence. Recommendations: • Follow up should be undertaken by gynaecological or oncological specialist teams arranged according to local arrangements. Access to a CNS can help to provide psycho-social and sexual support. Routine imaging has no place and should be used in symptomatic patients. (Grade C) 16 Supportive care – psychological care for patients and carers with cervix cancer Over 49,000 women are living with or beyond cervical cancer in the UK today. Every day nine women are diagnosed, and it is currently the most common cancer in women under 35 years of age. Survival is high with over two thirds of women living 10 years or more however for those affected, diagnosis and treatment can have a significant impact on a quality of life starting for many when abnormal cells are detected for the first time. In addition to pain and uncertainty, many treatment types bring multiple, long term physical and psychological consequences which can be extreme in their presentation. Impact on finances, relationships and ability to carry out day to day functions such as employment, education, shopping, housework and leisure activities cannot be underestimated. Being given the ‘all clear’ is far from the end as many women may have to live with numerous side effects and emotional and physical scars for the rest of their life. 37 In a report, using the largest known dataset of cervical cancer patients, Jo’s Cervical Cancer Trust found that far higher than anticipated numbers are affected by multiple, often complex long-term consequences of their diagnosis and treatment. High numbers of women reported unmet physical and psychological needs with many simply suffering in silence. (262) Almost ninety percent (88%) of women reported experiencing at least one long term physical problem, two thirds (63%) at least three, and a quarter (24%) six or more potentially life-changing physical problems including with bowel and urinary function (both 54%), pain (52%) and negative impact on their sex life (67%). Many problems were reported as presenting for the first time long after treatment was complete with many presenting at least a year after treatment. The further women were from treatment, the greater the number of problems they reported, clearly highlighting the longevity of the impact of a diagnosis and the need for ongoing care and support. (262) In addition to physical problems, the psychological impact of a cervical cancer diagnosis is significant. 37% of the women questioned reported depression and among young women it is far more pronounced with 79% of 25-34-year olds feeling blue, sad, down or depressed compared to 57% across all ages. The impact of cervical cancer does not stop after treatment is finished or when the five-year milestone has passed but can often be long term. For many women anxiety, fear and uncertainty start well before diagnosis with referral to colposcopy or even invitation to screening being a stressful experience. Lack of information or signposting to sources of support can contribute to anxiety over waiting for results or uncertainty over potential outcomes or next steps. Ensuring women are made aware of reliable sources such as Jo’s Cervical Cancer Trust can reduce isolation and alleviate pressure on those working within primary and secondary care to be the sole source of information and support. This also ensures women feel in control of their treatment options, decision making and fully understand the potential consequences they may face as a result of those decisions. Increased awareness of long-term consequences will enable women to identify symptoms, seek early intervention and better self-manage their care. Cervical cancer remains a less common cancer therefore many women will often never have met someone in the same situation as themselves. Signposting to relevant charities, support networks and services can enable women to make contact with others with similar experiences for mutual support. This includes through forums and support groups have been shown to reduce isolation and can significantly increase psychological and emotional wellbeing. Following completion of treatment, patients will have significantly reduced contact with their care team. For some women this may be a relief, for others it is a difficult transition leaving them feeling vulnerable. Instead of speaking with their oncologist or a familiar member of their care team, patients may present to unfamiliar healthcare professionals such as A&E or their GP. Again, this is where a key worker or CNS become invaluable. Without encouragement to seek help many women simply accept their physical or emotional challenges which can lead to poor self- management of conditions, where expert intervention is required. Without appropriate information and signposting, women are more likely to feel unsure of where to seek help, feel embarrassed or lack the confidence to speak out. Fear of having further tests and treatment can also contribute to women remaining silent. A lack of awareness of the long-term consequences of cervical cancer and treatment is contributing to women experiencing typical symptoms not being diagnosed or referred and even being told nothing can be done. Minimal research and understanding in this field have contributed to gaps in expertise, literature and information available to both professionals and patients. Just 21% of women who looked for information on long-term consequences of cervical cancer and treatment report fully finding what they need. Of 67% of women who reported having negative changes in their sex life as a direct result of treatment, a shocking 68% had not spoken to a doctor and of those affected by changes to their bowel 38 or bladder (both 54%) 39% and 42% respectively had not spoken to a doctor about these issues. Only half of women who experienced bowel and urinary problems had received treatment (41% and 54% respectively) and for those who experienced negative changes in their sex life, just 10% had received treatment. Worryingly high numbers say their emotional (75%) and physical (70%) needs have not been fully met. Long term chronic conditions bring added psychological strain therefore early referral and treatment is essential to alleviating this. The emotional impact of diagnosis and treatment is also often neglected with 34% reporting their emotional needs not being addressed at all with only 25% could say they had been completely addressed. Unmet needs were reported among women of all ages, however, was greater among younger women and women further from diagnosis. Only 7% of 25-34-year olds said that the long-term emotional impacts of their treatment had been completely addressed and treated, rising to 46% for 65-84-year olds. While women were more positive about the physical impacts being addressed, only 30% said they had been completely addressed with 20% saying they had not been addressed at all. This demonstrates a clear need for support and referral for women affected by cervical cancer. The disparity in emotional and physical impacts may demonstrate a lack of awareness about the emotional consequences of cervical cancer and treatment which urgently needs to change. Ongoing provision of and access to information and support is essential to ensuring they receive the information at a time that is appropriate to the individual patient, this can be as simple as posters on walls signposting to charities including Macmillan, Maggie’s and Jo’s Cervical Cancer Trust. The role of a designated CNS or key worker cannot be understated. This vital role offers consistent support from point of diagnosis, providing a holistic approach to the care and support needs of the patient. Each patient should have a Recovery Package. This valuable tool can support health professionals in delivery of patient-centred care, helping to identify and address changing needs from diagnosis onwards. When a patient is discharged from oncology, a detailed treatment summary helps facilitate a seamless transition from secondary to primary care and a multidisciplinary approach to long term care. If every patient and health professional is encouraged to use it to its full potential, it could significantly improve quality of life, enabling the patient to feel better informed and in control in addition to reducing delays in referrals, subsequent diagnosis and treatment or management. For women who work, lack of assistance and support to return to work can have an extremely detrimental impact not only on their financial situation but also on mental health. For women who work, or worked pre diagnosis, returning to work may be an important part of rebuilding life post cancer. Impact on employment status is high with 60% of women who have had negative changes to employment status attributing it to their diagnosis and treatment. More research is needed to fully understand the issues faced by women regarding employment and returning to work to identify where interventions are needed to assist both employers and employees. It is impossible to separate the physical and psychological impacts of a diagnosis of cervical cancer as physical difficulties caused by treatment can in themselves lead to emotional and psychological conditions and they in turn can have an impact on physical wellbeing. Physical consequences can be lifelong meaning that the psychological needs of women following diagnosis may change or increase over time. For young women, where greater numbers of diagnoses are seen, coping with loss of fertility or early menopause can be devastating and exacerbated when dealing with changes in appearance and watching friends go through pregnancy. Societal pressures and expectations around women’s ability to conceive only serve to heighten the isolation and distress that women affected by cervical cancer may already be experiencing. Avoiding psychological impact is impossible, and therefore ensuring women have the best possible experience of diagnosis, treatment and care along with continued access to support long term can alleviate some of the feelings of anxiety, fear and uncertainty. Feeling informed 39 and in control through the provision of information can lead to women feeling engaged and active in their treatment decisions leading to better patient outcomes. Psycho-oncological support and CNS support should be offered to all women with the diagnosis of cervical cancer (Grade C) 16.1 Management of complications and late effect/quality of life This section provides information on prevention, identification and management of complications, late effects and quality of life issues following a cancer of the cervix diagnosis and treatment. It aims to guide/signpost the reader to agencies/services that provide appropriate intervention and support for the woman and her family if needed. Women should have the opportunity to address symptoms attributed to their cancer and its management before, during and after treatment. Predictable side-effects are dependent on treatment modality and can include but are not limited to sexuality/sexual morbidity, menopause, lymphoedema, effects on gastrointestinal and urinary systems and psychosocial concerns. Both physiological and psychosocial factors can impact on quality of life, addressing possible and actual problems as they arise may help to reduce the negative impact experienced by women. It is good practice to talk about symptoms that could be attributed to cancer and the consequence of treatment and this should also be addressed at each follow-up appointment or through holistic needs assessment. Women should receive appropriate information so they are informed of the relevant risks of short- and long-term side-effects during the consenting process, this should be recorded on the consent form. Good quality information is available from both Macmillan and Jo’s Cervical Cancer Trust charities which the patient can source themselves or be given in clinic (this is provided free of charge). • • • Areas of specific concerns are: 16.2 Sexuality/Sexual morbidity Factual information on possible anatomical changes due to surgery or radiotherapy should be given to the women prior to treatment. This will acknowledge that the subject of sexuality is open should she need to seek further information if difficulties occur. The use of vaginal dilators following radiotherapy should be recommended to reduce the risk of stenosis. If women experience sexual difficulties these should be addressed and where possible specific suggestions given e.g. use of lubrication during intercourse. Where available, women with ongoing difficulties should be referred to psychosexual services. Fertility can play an important role in a women’s sexuality. Where possible, fertility sparing treatment should be considered in women who wish to maintain their fertility. Referral to fertility services to discuss other options should also be available. 16.3 Menopause and hormone replacement therapy Hormone replacement therapy (HRT) should be considered in women who develop treatment induced premature menopause to reduce risks associated with early menopause including cardiovascular and skeletal morbidity and menopause symptoms. There is no evidence recommending specific HRT preparations however those women with an intact uterus should receive combined HRT, this includes those women who have had radiotherapy. In women who do not want HRT, alternative health and lifestyle therapies may help – signpost appropriately. Local vaginal oestrogen application can also be used to improve side-effects of vaginal stenosis/dryness; alternatively, vaginal moisturisers may be helpful. 40 16.4 Lymphoedema Risk of developing lymphoedema ranges from 21% following surgery alone to 77.8% in those who have undergone both surgery and radiotherapy. Prophylactic information on reducing the risk of lymphoedema should be available to women ( ). And those women who develop lymphoedema should be referred to specialist lymphoedema services. 16.5 Bowel/bladder function At follow up ask if any new problems relating to bowel/bladder function, if present initially manage with simple solutions such as loperamide for diarrhoea, dietary changes for constipation, anticholinergics for bladder urgency. Consider referral to other services for persistent problems e.g. gastroenterology, colorectal, urodynamic, continence or urology as appropriate. 16.6 Psychosocial The impact of cancer and treatment can affect quality of life, the psychosocial needs of women should be addressed throughout; health needs assessment should be performed at pivotal points in the cancer pathway. Women should have the opportunity to explore ways of improving their quality of life through appropriate support and signposting to survivorship/living with and beyond cancer, and psychological services where available. The following references are recommended for further reading (263-271). Recommendations: • Prevention, identification and management of complications, late effects and quality of life issues following a cancer of the cervix diagnosis and treatment are essential part of package of care – (Grade C) • Recording of late side effects should be documented (Grade C) • Written information should be provided about treatment choices and side effects including late effects. (Grade C) • Access to a CNS or equivalent and psycho-sexual counsellors should be available as part of the multi-disciplinary team. (Grade C) 17 Rare malignancies: small cell, mucinous and clear cell carcinomas, and sarcomas of the uterine cervix 17.1 High grade NEC of small cell type These are all very rare and uncommon cancers and as such there is no evidence from randomised clinical trials data to support the optimal management. Most experience and evidence will come from case reports, small series or the reports of experience from the larger cancer centres. Crucial to the management of these tumours is the availability of a dedicated expert gynaecological oncology pathologist and commonly surgical tissues may need to be looked at for second specialist opinion. Many of these rare tumours can be difficult to interpret histologically and require a panel of immunocytochemistry, and even then, may be difficult to identify the histological subtype. Molecular pathology is evolving and in the course of the next few years will probably characterise these tumours with greater precision. A recent report looked at the genomic landscape of small cell carcinomas of the cervix. Aneuploidy was common, occurring in nearly of the cases reported (265). A variety of mutations were reported including MSH2, pi3kinase and pTEN whereas TP53 and RB1 most commonly seen in small cell lung cancer were relatively uncommon. It is hoped that identification of these mutations may eventually lead to better selection of new drugs. Currently these rare tumours include neuroendocrine tumours (small cell carcinomas and carcinoids), clear cell carcinomas, mucinous carcinomas and rare sarcomas. 41 17.2 Neuroendocrine tumours (NETs) These are rare (orphan status) especially in the gynaecological tract and are classified differently. This terminology was introduced in the 1970s but WHO 2014 classifies neuroendocrine tumours as low grade NET to include typical and atypical carcinoid tumours, and high-grade neuroendocrine carcinomas (NEC) which includes small cell and mixed large cell NECs. (273-276). Primary cervical carcinoids tumours are better differentiated and exceedingly rare. Usually it is necessary to exclude a primary of the gastro-intestinal tract. For detailed advice on carcinoids, reference should be made to the standard guidelines from National and International Societies (277,278). This guideline will focus primarily on small cell carcinoma of cervix. Small cell neuroendocrine carcinomas probably account for less than 1% of all cervix cancers and are amongst the most lethal. It is important to differentiate between pure small cell cancers and poorly differentiated squamous carcinomas that have neuroendocrine elements as these are probably best managed as G3 epithelial carcinomas. Pure small cell carcinomas are generally composed of small round cells and have characteristic immunocytochemical staining. Mixed large cell and small cell tumours are also seen and are currently included within the NET family of tumours. (279-281) Clinical presentation is usually very similar to that of any other carcinoma of the cervix, but they are slightly more common in the younger age group. They are associated with HPV exposure but more commonly HPV 16 and 34 are found. Some patients may present with hypercalcaemia, which may be due to secretion of parathyroid hormone related peptide (PTHRP), without evidence of bony metastasis, and this is usually a poor prognostic factor. Occasional patients may present with hyponatraemia and SIADH. Staging and clinical investigation are the same for all types of cervix cancer with a biopsy, MRI pelvis and 18FDG PET CT scanning. Cases should be presented to the MDT for further discussion of management on an individual basis but would not be a standard of care (279-281). Early stage disease In early stage cases the standard approach is a radical hysterectomy and lymph node dissection following further MDT discussion, these patients should be offered adjuvant chemotherapy and radiotherapy in virtually all cases, given the high risk of relapse. The published data shows that virtually no patients with bulky stage II disease or higher stage will be cured and only around 1/3 of the early-stage cases will have 5-year survival. Since no clinical trials have been carried out, there is no firm evidence to support the adjuvant treatment post-operatively, but many would advocate either concomitant cisplatin-based chemoradiation or adjuvant platinum and etoposide followed by pelvic irradiation. There is no evidence to support the use of prophylactic cranial irradiation which had been advocated by some in the past. The usual schedule of chemotherapy will be a platinum and etoposide combination, with 4-6 cycles given at 3-weekly intervals. Either carboplatin or cisplatin can be given as determined by clinician’s choice. There is no evidence to support whether 4 or 6 cycles should be given. Lung oncologists who treat small cell cancers tend to use 4 cycles whereas those treating gynaecological cancers more frequently favour up to 6 cycles. Advanced stage disease For more advanced disease, bulky stage IIB (>5cm) or higher stage, chemotherapy is normally recommended initially. Many experts would recommend that the chemotherapy drugs and doses used for small cell lung cancer are prescribed but again there is no hard evidence to support this. Most centres will use either cisplatin or carboplatin combined with etoposide and administer between 4 and 6 cycles of chemotherapy at 3 weekly intervals. In patients who have had a very good response to this induction chemotherapy and where the tumour may have shrunk to less than 4 cm and furthermore where there is no evidence of distant metastatic disease, discussion should take place at the MDT to consider the potential place of radical hysterectomy and lymph node dissection. A follow up FDG PET 42 scan should be considered before embarking on a radical surgical approach. These patients will then require adjuvant radiation post-operatively. For patients who receive neoadjuvant chemotherapy but do not achieve sufficient shrinkage to permit surgery or whether there are other contraindications, high-dose (chemo-)radiotherapy should be considered with external beam radiotherapy and intracavitary brachytherapy. Very few patients however will go on to long-term remission or cure. Concomitant weekly cisplatin may be offered but be wary of risk of neuropathy. Doses of radiation will be similar to those used in squamous cancers of cervix. Studies which have deployed high dose chemotherapy with marrow transplant and maintenance chemotherapy are unproven. Equally whilst bevacizumab has been shown to be active in squamous cell carcinomas of the cervix, there is no proven data to support its use in small cell NEC of cervix. However, a recent report of topotecan, paclitaxel and bevacizumab has suggested possible benefit, but the series was small and non-randomised. Metastatic or recurrent disease In those patients where scanning shows residual, persistent or progressive extra pelvic disease, options are very limited. Second line chemotherapy may be offered but is rarely effective. Topoisomerase inhibitors potentially should be effective but the use of topotecan has been very disappointing. Others have used the CAVE regimen used in small cell lung cancer but again with limited effect. Several small series in small cell ovarian cancers have reported on dose dense and dose intense weekly carboplatin and paclitaxel but again no long-term responders were identified. To date no new targeted agents have been identified as being effective but it is hoped that with new developments in molecular pathology over the course of the next few years, new targets will be identified leading to new drugs. As mentioned above, one small series has suggested topotecan, paclitaxel and bevacizumab may show promise and needs further assessment. Fertility preserving options Again, there is no widescale experience and these patients should be managed within a multi-disciplinary team with careful counselling that the use of fertility preserving surgery is unproven in this situation. Neoadjuvant chemotherapy with carboplatin and paclitaxel may be considered and some form of conservative surgery. Given the uncertainty of risk of recurrence, early attempts to conceive and elective post-partum surgery can be discussed. Progress in the management of these tumours is difficult but collaboration by international groups such as the Gynaecological Cancer InterGroup (GCIG) may help to take forward the management of these rare and difficult tumours. Only these organisations can manage to conduct the kinds of clinical trials needed to improve care in these rare tumours. 17.3 Mucinous tumours These are very rare, and it is important to ensure it is not an endometrial tumour that has extended down to the cervix. The clinical management would normally be very similar to that for squamous or adenocarcinoma of the cervix with standard pre-treatment workup. Recent pathology review has suggested that there is an intestinal sub-type, which behaves differently and may explain the poor responsiveness to standard chemotherapy-based approaches. More recently has been the identification of gastric type adenocarcinoma, which appears to carry a worse prognosis. It is therefore important to recognise the gastric-type adenocarcinoma. They are the largest group of non-HPV associated carcinomas. They are often larger in size, of higher stage and poorer outcome than HPV associated carcinomas. They have an increased risk of disease recurrence and a decreased 5-year disease specific survival rate. They are characterised by the pressure of 43 abundant clear, foamy or pale cytoplasm, distinct cytoplasmic borders, pleomorphic nuclei and foci of minimal deviation architecture with minimal stromal response (282,283). If the tumour is confined to the cervix and less than 4 cm, radical hysterectomy and pelvic lymph node dissection would normally be the treatment of choice. There are questions about the radio sensitivity of mucinous tumours at other sites but if the tumour is not considered suitable for surgical resection then concomitant chemoradiation would be considered for localised disease. If there is evidence of distant metastatic spread, systemic treatment would be considered. Consideration should be given to the use of combination regimens as used in colorectal cancer such as oxaliplatin or irinotecan and capecitabine/5-Fluoro-uracil and folinic acid. There is of course no randomised data to support the use of any particular combination but intuitively for a mucinous tumour, regimens used for colorectal cancer are more likely to have a better outcome. 17.4 Clear cell carcinomas (CCC) These again are rare, and the same basic principles of management apply as for mucinous or other rare tumours. These should be discussed at the multidisciplinary team and subjected to the usual clinical workup and examination. Where the tumour is suitable for surgery, the patient should be offered radical hysterectomy and lymph node dissection. If adjuvant treatment is required there is limited evidence that paclitaxel added to cisplatin or carboplatin may improve the response rates but again this is based on very limited series. Some of this is also extrapolated from Japanese experience in CCC of ovary. Again, there is some suggestion that clear cell carcinomas may have reduced radio sensitivity compared to squamous cell carcinomas, but there is no evidence to support the view that higher doses of radiation may be more effective, and the patient should be treated in the same manner. For patients with advanced disease, with extra pelvic spread, chemotherapy with carboplatin and paclitaxel is probably the preferred schedule with some limited evidence suggesting that regimens including paclitaxel may confer a slightly higher benefit. 17.5 Basaloid and adenoid cystic carcinomas These are again exceedingly rare and may include pure basaloid carcinomas, adenoid-basal carcinomas and adenoid cystic carcinomas. These are usually managed surgically and are often polypoidal. There is very little published data regarding the management. The older literature suggested these were less radio responsive. 17.6 Sarcomatous tumours of the cervix These are very rare but include adenosarcoma, leiomyosarcoma, and even endometrial stromal sarcoma as well as high-grade undifferentiated sarcoma. Even rarer are rhabdomyosarcomas which may present as a polypoid mass although not like the botryoides pattern seen in childhood. Many of these start as polypoidal tumours and can be managed surgically with very limited evidence to suggest whether they need any adjuvant treatment and they should be discussed on an individual basis at the MDT. Consideration should be given to discussing with colleagues from other centres particularly in the case of these ultra-rare tumours. Carcinosarcomas are not true sarcomas and should be managed as poorly differentiated carcinomas. It is beyond the scope of this article to go into further detail about their management. Recommendations: • These are rare tumours and all cases must be discussed at MDTs. Consideration should be given to supra-regional teams providing the treatment with shared care for follow up. (Grade D) 18. Fertility sparing surgery in cervical cancer 44 18.1 Service delivery When offering fertility sparing treatments for cervical cancer it is important to consider the efficacy in treating cancer, the effectiveness in preserving fertility, and the complications of surgery. Complications may be to the patient, such as those secondary to parametrial resection, but they may also involve infertility from cervical stenosis and adverse effects on subsequent pregnancies such as premature labour and consequential harm to a neonate. It is a difficult balance for a woman who has every right to compromise on her chances of cure in order to maximise a subsequent favourable obstetric outcome if she achieves a cure. If this right is to be respected, then it is intuitive that centres offering these treatments should be able to provide adequate counselling and support services. A woman’s perception of the relative importance of different outcomes (oncological and obstetric) must be respected by professionals. Where the issues to consider are complex, such as the risk and consequences of premature labour following trachelectomy, it seems sensible to offer support from a CNS and for treatment to occur in a centre experienced in providing fertility sparing surgery. A recent survey from Public Health England estimated that only 80 cases of trachelectomy occur in England annually (284) and as case numbers for a particular centre are likely to be low, it is considered good practice for units to prospectively audit their outcomes. Full clinical history, initial colposcopic examination and possibly hysteroscopy for measuring cervical lengths prior to trachelectomy, are important tools for the decision-making process of a fertility sparing surgery and may be used in conjunction with some form of conisation to determine those patients who can have the most conservative of fertility sparing approaches. Recommendations: • Counselling for procedures should include details of the oncological and fertility-sparing efficacy as well as potential complications to the patient, risks to subsequent pregnancies, and adverse outcomes to children who might be born prematurely • A patient who prioritizes one set of outcomes above another (oncological versus obstetric) should have their views respected and the direction of treatment modified appropriately if necessary • Centres performing fertility sparing surgery should be experienced in the techniques they use and audit their outcomes regularly. (Grade C) 18.2 Treatment options Conisation for stage IA cervical cancer Conisation for stage IA cervical cancer has been recommended by numerous authors (285,286) and is included in another established guideline (287,288). The aim of conisation is to achieve negative margins to both cancer and dysplasia. A 3mm margin to cancer has been recommended in previous guidelines (280). Cold knife conisation has the advantage over loop conisation (LLETZ) in that the margins are easier to assess. However, LLETZ is acceptable as long as adequate margins are achieved along with a non-fragmented specimen, correct histological orientation, and no electrosurgical artefact interfering with marginal assessment (289-291). A detailed assessment of these risks is outside the scope of this guideline, but it should be noted that when treating stage Ia disease, patients often receive more than one conisation or large cones greater than 15mm deep which are associated with a greater risk of these complications. This must be explained to patients undergoing conisation for cervical cancer. Recommendations: 45 • Conisation (cold knife or LLETZ) is an acceptable local treatment for stage IA1 (FIGO 2018) squamous cell and adenocarcinoma of the cervix. In stage IA2 with LVSI a pelvic LND should also be performed • Re-conisation is recommended if margins to the cancer are within 3mm, if there are positive margins to intra-epithelial neoplasia, if the specimen cannot be orientated, is fragmented, or has diathermy artefact that makes margin assessment impossible • Conisation (LLETZ, knife, & LASER) for cervical cancer is often large or multiple and carries a risk of cervical incompetence, premature labour, and neonatal death. This must be explained to women who have this treatment 18.3 Simple trachelectomy and cone biopsy for selected subsets of women with stage IB1 (FIGO 2018) cervical cancer A number of authors have identified a group of patients with stage IB1 cervical cancer in whom the risk of parametrial spread is low leading to the hypothesis that a more conservative approach to selected cases may be appropriate. Kinney et al (293) found no parametrial involvement in a group of women with less than 3mm depth of tumour irrespective of the width. Covens et al (294) found the risk of parametrial involvement in 536 women to be only 0.6% when the lesion was <20mm in maximum diameter with <10mm of stromal invasion and negative lymph nodes. Wright et al (295) found the incidence of parametrial involvement to be 0.4% in a group of 270 patients with no lymph-vascular space invasion and tumours less than 20mm while Frumovitz et al (296) found the incidence to be 0% in a similar group of 125 women. Centres offering fertility sparing surgery should have a protocol in place to offer a select group of women more conservative surgery and audit their outcomes. Recommendations : • Selected women with low volume disease are suitable for treatment by conisation alone. • A centre offering fertility sparing surgery should have a protocol for selecting women with low volume disease for cone biopsy or simple trachelectomy rather than radical trachelectomy. • Centres offering cone biopsy or simple trachelectomy for stage IB1 (FIGO 2018) disease should audit their outcomes prospectively. 18.4.1 Radical vaginal trachelectomy (RVT) for stage IB1 (FIGO 2018) disease The efficacy of RVT for the treatment of cervical cancer has been systematically reviewed (297,298). Published data on RVT includes women with stage IA1, IA2, and IB1 disease (299-301). Those patients with IB1 disease have both low volume and high-volume tumours. Where it is possible to define, only 922/1277 (72%) of women reported in the literature had the procedure performed for stage IB1 disease (300-302) . When additional treatment is reported in papers, 150/1355 (11.1%) never completed or retained fertility sparing treatment either because of adverse factors subsequently diagnosed as part of work up or because of the need for adjuvant treatment (further surgery or radiotherapy). Specific complications of RVT include cerclage erosion and cervical stenosis. Cerclage erosion is not routinely reported in the literature but seems to occur in between 0 and 30% of cases (303-305). Disadvantages of trachelectomy are lower fertility rates and risk of preterm delivery. Alternative options include discussions to include the use of neoadjuvant chemotherapy followed by conisation or simple trachelectomy. The probability of a woman achieving at least one pregnancy following RVT is unclear in the literature. This is because data is often confounded by patients not attempting a pregnancy; occasions when one woman had more than one pregnancy and this not being reported clearly; and instances of short follow-up periods. One group used an actuarial analysis to calculate the percent probability of conception to be 46 53% (305). When the numbers of women trying for a pregnancy are reported in the literature, we found 299 (54.5%) women achieved a pregnancy out of 549 trying to conceive (294,302). Where it was possible to interpret data, we found 293/541 women (54.2%) achieved at least one live birth (294,302). Although these numbers are similar, many women had more than one pregnancy and there were many cases of first and second trimester miscarriages, and even cases of pregnancies being terminated. When reported, and data possible to interpret, a pregnancy following RVT was secondary to fertility treatment in 41/159 (25.8%) of cases (305,306). It should be noted that fertility in this group of women is not only compromised by the RVT and cervical stenosis, but also by the fact that on average, women are over the age of 30 years. Of babies born to women who have had RVT, where it is possible to glean the data, 113/295 (38.3%) are born prematurely (36 weeks and less) (294,304,305). In addition, where it is possible to interpret the data 44/259 (17.0%) of babies born to women who had had RVT are born very prematurely (<32 weeks) (294,307). For this reason and because delivery is likely to be by caesarean section (possibly through a classical incision), it is advised that pregnancies are supervised by an obstetric team experienced in high risk obstetrics. Furthermore, women need to be advised of the above risks and counselled as to the risks of prematurity. Recommendations: • Radical vaginal trachelectomy is an acceptable fertility sparing treatment for women with stage IB1 (FIGO 2018) disease • At histological review, those with low risk and low volume tumours should be considered for cone biopsy instead of radical trachelectomy • When patients are considered for RVT, the risk of eventually receiving treatment that is fertility sacrificing (such as radiotherapy) is in the region of 11%. This figure should be presented to patients who should be counselled appropriately taking in to account other prognostic markers such as grade, LVSI, tumour size, and audit figures from the centre providing treatment • The risk of recurrence following RVT is in the region of 5%. This figure should be presented to patients who should be counselled appropriately taking in to account other prognostic markers such as grade, LVSI, tumour size, and audit figures from the local centre providing treatment • Prior to surgery, a woman scheduled for an RVT should be informed in detail about all surgical risks and the impact of the procedure on her fertility and future obstetric outcomes • Prior to surgery, a woman scheduled for RVT should be warned that if they do conceive, and a cerclage is in situ, a caesarean section will be required • Centres offering RVT should audit their outcomes regularly. This should include recurrence rates, complications, fertility rates, and obstetric outcomes • Given the rarity of these cases most of the above is based on expert opinion 18.4.2. Abdominal radical trachelectomy (ART) for stage IB1 disease The efficacy of ART for the treatment of cervical cancer has been systematically reviewed (297-298) for women with stage IA1 – IB1 disease for both low and high volume tumours. As with RVT, it is difficult to calculate the exact figure for recurrence following ART for stage IB1 lesions. This is as the figures reported in the literature are contaminated with patients with stage IA1, IA2 & IIA lesions and in many cases it is impossible to dissect these patients out of reported results (297,298). For those patients included in systematic reviews in which a recurrence rate can be calculated, the overall recurrence rate from all people who were selected for trachelectomy (whether or not they had completion treatment) was 31/866 (3.6%) (297). If an assumption is made that all recurrences existed in women with stage IB1 or greater, than the figure for recurrence would be 31/699 (4.4%) for this group (297). As with RVT, this figure is further distorted by series containing patients with short follow-up durations and influenced by tumour grade, presence of LVSI, tumour volume and other established prognostic factors. With the data 47 available, it seems sensible that the figure presented to patients as to the risk of recurrence should be in the region of 5% but also taking into account all the other prognostic markers available for an individual patient. Cerclage erosion as a surgical complication is not routinely reported in the literature but when reported occurs in 13/433 (3.0%) of cases (303). There was a total of 50/433 (11.3%) cases of cervical stenosis in papers where it was possible to assess for this complication. As with RVT the probability of a woman achieving at least one pregnancy following ART is not clear in the literature (303,309-312). This is for similar reasons to RVT that women not attempting a pregnancy often confound the results; there are occasions when one woman had more than one pregnancy and this is not clearly reported; and there are instances of short follow-up periods. When the numbers of women trying for a pregnancy are reported in the literature, we found 136 (41.6%) women achieved a pregnancy out of 327 trying to conceive (310-312). Of women who successfully conceived and where data could be interpreted, we found that 75/134 (56.0%) of pregnancies were secondary to fertility treatment. Where it was possible to interpret data, we found 35/142 women (24.6%) achieved at least one live birth (310-312). However, these numbers are small because much data was excluded as it was not clear in the literature how many of the live births were born to how many women. In total, 119/169 (72.6%) of pregnancies resulted in some form of live birth. Of babies born to women who had an ART, where it is possible to glean the data, 59/118 (50.0%) are born prematurely (36 weeks and less) (310-312). However, as with RVT, the data is affected by women having elective caesarean sections at 36 weeks’ gestation. Recommendations: • Abdominal radical trachelectomy (ART) is an acceptable fertility sparing treatment for women with stage IB1 disease. Especially those with larger tumours • At histological review, those with low risk and low volume tumours should be considered for cone biopsy, simple trachelectomy, or radical vaginal trachelectomy • When patients are considered for ART, the risk of receiving treatment that is fertility sacrificing (such as radiotherapy) is in the region of 17%. This figure should be presented to patients who should be counselled appropriately taking in to account other prognostic markers such as grade, LVSI, tumour size, and audit figures from the centre providing treatment • The risk of recurrence following ART is in the region of 5%. This figure should be presented to patients who should be counselled appropriately taking into account other prognostic markers such as grade, LVSI, tumour size, and audit figures from the local centre providing treatment • Prior to surgery, women should be warned of the risk of cerclage erosion in the region of 3% and the risk of cervical stenosis in the region of 11% • Prior to surgery, women scheduled for an ART should be informed that the current data suggests the proportion of women who eventually achieve a pregnancy is over 42% but over 50% of these pregnancies are secondary to fertility treatment. The proportion of women trying for a baby who have a live birth is about 25% • Prior to surgery, a woman scheduled for an ART should be informed that the premature delivery rate if a baby is achieved is just about 50% and the extreme premature labour rate about 20%. Women should also be informed that prematurity is associated with learning difficulties and slow development in children • Prior to surgery, a woman scheduled for an ART should be warned if they do conceive, with a stitch is in situ they will need a caesarean section • Centres offering ART should audit their outcomes regularly. This should include recurrence rates, complications, fertility rates, and obstetric outcomes 48 18.6 Ovarian transposition Ovarian transposition has been considered to preserve ovarian function in women who require pelvic radiotherapy. The aim is not just to preserve hormonal production but also to preserve function so that egg collection can occur, and surrogacy be considered. Ovarian transposition has been systematically reviewed (313). Of the twenty-four papers included in the systematic review, much of the data is contaminated by patients who did not actually undergo subsequent radiotherapy (314-316). Of women who had external beam radiotherapy (EBRT) 65% preserved their ovarian function but 5% of women developed ovarian cysts. Ovarian cyst development is often associated with subsequent severe pain and difficult surgery to remove entrapped ovaries. The meta-analysis did not include case reports and did not report any recurrences in this specific group of patients. There are a few unanswered questions regarding ovarian transposition. It seems sensible that not only the ovary but also the vascular pedicle needs to be moved away from the radiation field. Therefore, the ovarian ligament should be cut. Transposition of the Fallopian tube would preserve blood supply from the mesosalpinx but might carry cancer cells with it. High stage and poor histological type adenocarcinomas might have already metastasised to the ovary resulting in recurrence and perhaps this is the occasion when women with adenocarcinomas need to be treated differently. It seems intuitive that women with an adenocarcinoma and uterine spread should not have their ovaries transposed. Ovarian transposition is particularly difficult after ovarian stimulation due to the size of the ovaries. Opportunistic removal of tubes as risk reduction strategy should be discussed with the patient prior to the surgery. Recommendations: • Women due to receive external beam radiotherapy should be offered ovarian transposition if the radiation field will be away from the vascular pedicle • Opportunistic removal of tubes should be discussed with the patient • Prior to surgery a woman should be warned that the failure rate is in the region of 35% • Prior to surgery a woman should be advised that the risk of ovarian cyst formation and entrapment is in the region of 5% • Prior to surgery a woman should be warned of the potential risk of metastases to the transposed ovary • A surgeon should note that transposition following ovarian stimulation is more difficult due to the size of the ovaries (Grade D) 18.7 Neo-adjuvant chemotherapy prior to fertility sparing surgery Current literature A number of studies has looked at neo-adjuvant chemotherapy (NAC) in an attempt to reduce tumour size and allow fertility sparing surgery when otherwise it might not be possible (311,317). 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Laparoscopic ovarian transposition in patients with early cervical cancer. International journal of gynecological cancer : official journal of the International Gynecological Cancer Society. 2008 May-Jun;18(3):584-9. PubMed PMID: 18476952. Epub 2008/05/15. eng. 317) Lanowska M, Mangler M, Speiser D, Bockholdt C, Schneider A, Kohler C et al. Radical vaginal trachelectomy after laparoscopic staging and neoadjuvant chemotherapy in women with early-stage cervical cancer over 2 cm: oncologic, fertility, and neonatal outcome in a series of 20 patients. International journal of gynecological cancer : official journal of the International Gynecological Cancer Society. 2014 Mar;24(3):586-93. PubMed PMID: 24469326. Epub 2014/01/29. 318) Ramirez PT, Frumovitz M, Pareja R, Lopez A, Vieira M, Ribeiro R et al. Minimally Invasive versus Abdominal Radical Hysterectomy for Cervical Cancer. N Engl J Med. 2018 Nov 15;379(20):1895-1904. doi: 10.1056/NEJMoa1806395. 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4911 | http://www.cs.columbia.edu/~feiner/courses/csw4160/slides/S99-10.pdf | Computer Graphics - Week 10 Computer Graphics - Week 10 Bengt-Olaf Schneider Bengt-Olaf Schneider IBM T.J. Watson Research Center IBM T.J. Watson Research Center © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Questions about Last Week ?
Questions about Last Week ?
© Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Overview of Week 10 Overview of Week 10 Solid Modeling Sweeps Boundary Representations Spatial Partitioning Constructive Solid Geometry (CSG) Global Illumination (Part 1) Ray-tracing Advanced lighting models Spatial data structures © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Modeling Modeling Modeling is the process of generating the description of a 2D or 3D shape Approximation of a real shape, e.g. 3D capture or visualization Creation or design of a new shape, e.g. CAD or DCC The shape description has to satisfy several, often conflicting demands Accuracy, i.e. match between model and real shape Compactness Consistency, e.g. no "impossible" shapes Robustness, e.g. limited errors due to floating-point calculations Support of editing and queries, e.g. to support interactive manipulation © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Desirable Properties of Desirable Properties of Modeling Data Structures Modeling Data Structures Expressive Power What kind of objects can be represented using the data structure ?
How accurately can objects be represented ?
Validity Are all values of the data structure representing valid objects ?
Uniqueness Does every valid object have exactly one representation ?
Conciseness How large is the representation of interesting objects ?
Closure of operations Do operations on the data structure always generate valid objects ?
Applicability and Computational Ease What kind of algorithms are supported by the data structure ?
© Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Solid Modeling Solid Modeling © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Solid Modeling Solid Modeling Representation of solid models, i.e. models filled with material Manipulation of solids, e.g. combination or interaction of solid models Used in various applications, e.g.
Manufacturing and CAD/CAM Interference detection, e.g. robot path planning Physical properties, e.g. a part's mass and center of gravity Automatic instructions for machining of parts © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Solids Solids The representations we have discussed so far, allow to model solids but are not always describing solids The vertex-edge-face descriptions can also describe non-solid objects Dangling Edge Dangling Face © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Taking apart a Solid (1) Taking apart a Solid (1) Interior Points that are entirely interior of the object Points having a non-zero neighborhood with only interior points Boundary Points with zero distance from both the object and its complement There is no non-zero neighborhood that contains only interior points Interior Boundary © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Taking apart a Solid (2) Taking apart a Solid (2) Open Set A set with only its interior point and none of its boundary points Closed Set A set with all its interior and boundary points Closure Union of a set with its boundary points Regularization Closure of the interior points © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Taking apart a Solid (3) Taking apart a Solid (3) Object Closure Closure(Interior) Boundary Interior © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Combining Solids (1) Combining Solids (1) Construction of more complicated solids can be accomplished through Boolean set operations Standard set operations can create non-solid objects !
© Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Combining Solids (2) Combining Solids (2) Instead of standard Boolean operations, we use regularized Boolean operations: Regularized Boolean operations produce the same result as standard Boolean operations if the resulting objects are solid Otherwise, they eliminate lower-dimensional features Dangling edge, faces, points A op B A op B closure interior = b g c h © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Combining Solids (3) Combining Solids (3) A A A B B B A B ∩ A B ∩ Example Standard intersection operation retains a piece of shared boundary Regularized intersection avoid the generation of a dangling edge © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Representing Solids Representing Solids Solids can be represented in several ways Sweeps Boundary representation (b-rep) Decomposition representations Constructive Solid Geometry (CSG) © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Sweep Representation (1) Sweep Representation (1) Sweeps move an object along a trajectory Simple and natural way to describe many objects For example, the path of a cutting tool or a robot arm A sweep is described by the object and a trajectory for the sweeping process Generalized sweep 2D or 3D object Arbitrary trajectory Transformation of the object along the trajectory © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Sweep Representation (2) Sweep Representation (2) Extrusion Simplest sweep The object is a 2D area The trajectory is perpendicular to the object plane Rotational sweep Rotation of a 2D area about an axis Object Sweep Direction Sweep (Extrusion) © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Sweep Representations (3) Sweep Representations (3) General sweeps can generate complex objects Transformations can change the shape of the swept object Complex trajectories may create self-intersecting object Application of Boolean set operations to general sweeps is difficult Sweeps are not closed under Boolean operations. E.g. the union of two sweep is generally not a sweep.
Sweeping a 2D object within its plane does not generate a solid.
Therefore, sweeps are converted to another representation first Sweeps are supported because they are a natural way to model objects © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Boundary Representation (B-rep) Boundary Representation (B-rep) Only describe the boundary explicitly Interior is defined implicitly via the boundary B-reps were conceived as an extension of early ways to represent objects with vectors The boundary can be described using several techniques Polygon meshes we have discussed earlier Higher-order or freeform surfaces We will first look at ways to describe polygon meshes, before we look at modeling solids using b-reps.
© Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Polygon Meshes Polygon Meshes Polygons are popular primitives to approximate shapes Linear approximation w/ easily controlled error Conceptually simple, in particular for triangles Efficient rendering with hardware support Polygon meshes are connected sets of polygons © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Creating and Editing of Models Creating and Editing of Models Moving of single vertex Make sure all connected edges and triangles follow Moving of an edge Make sure that the delimiting vertices and triangles follow Add or delete a triangle Make sure that neighboring triangles and adjacent edges/vertices are updated We need a data structure that supports such operations © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Data Structures for Polygon Meshes Data Structures for Polygon Meshes Explicit representation Indexed representation Edge-based representations © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Explicit Representation Explicit Representation The coordinates for every vertex are stored explicitely in the polygon representation P = [(x1, y1, z1), (x2, y2, z2), (x3, y3, z3), ... , (xn, yn, zn)] Problems No concept of shared vertices, i.e. two polygons with shared vertices have to replicated the vertex coordinates Difficult to maintain Inefficient memory utilization © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Indexed Representations Indexed Representations To avoid the problem of replicated data, indexed representation use pointers to the actual data For instance, indexed-face list in VRML P = [V1, V2, V3, ... Vn] V1 = (x1, y1, z1) V2 = (x2, y2, z2) V3 = (x3, y3, z3) ...
Advantages More space-efficient than explicit representation Shared vertices are stored only once Easier to edit and maintain Problems Still not easy to find polygons sharing an edge or a vertex © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Edge-based Representations Edge-based Representations Polygons are represented by their enclosing edges Edges are stored using pointers to end points and adjacent polygons In typical meshes every edge has only 1 or 2 adjacent polygons P = [E1, E2, E3, ... En] V1 = (x1, y1, z1), V2 = (x2, y2, z2), V3 = (x3, y3, z3) ...
E1 = (V1, V2, P, -), E2 = (V2, V3, P, -), E3 = (V3, V4, P, Q) ...
Still does not support easy queries for ...
... all polygons adjacent to a vertex ... all vertices shared by 2 polygons ... all edges meeting at a vertex © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Winged-edge Data Structure (1) Winged-edge Data Structure (1) Edges are (again) the central link between vertices and polygons For each edge the following is stored Pointers to the vertices V1 and V2, edge is Pointers to the vertices V1 and V2, edge is oriented oriented from V1 to V2 from V1 to V2 Pointers to the adjacent polygons (left and right defined by edge orientation) Pointers to the adjacent polygons (left and right defined by edge orientation) Pointers to 4 additional edges (next edges cw and ccw at both ends) Pointers to 4 additional edges (next edges cw and ccw at both ends) For each vertex a pointer to an edge sharing that vertex is stored For each polygon a pointer to one of its edges is stored © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Winged-Edge Data Structure (2) Winged-Edge Data Structure (2) Determine all edges incident to a vertex !
Determine the edge associated with the vertex Follow the edges around the vertex by reading the ccw next edge Stop when the first edge is encountered Determine all faces sharing an edge !
Simply retrieve the adjacent faces from the edge information Determine the adjacent faces for a vertex !
Determine all edges incident to the vertex (see above) For all edges: Report left (right) face if edge starts (ends) at the vertex © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Winged-Edge Data Structure (3) Winged-Edge Data Structure (3) Winged-edge data structures describes 2-manifolds 2-manifolds are surfaces where every point on the surface has a (arbitrarily small) neighborhood that is a topological disk Non-manifold surfaces can be described using the radial-edge data structure © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Only describe the boundary explicitly Interior is defined implicitly via the boundary Computation of Boolean set operations is somewhat complex because of the many elements involved Solids in Boundary Representation (1) Solids in Boundary Representation (1) © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Properties Generally, there is no unique b-rep for a given solid, i.e. the same solid can be described by different b-reps Validity of B-reps is difficult to establish and enforce, e.g. dangling faces, non-manifolds, "open" objects or self-intersecting objects.
Topological integrity can be assured by data structures, e.g. winged edge data Topological integrity can be assured by data structures, e.g. winged edge data structure structure However, there is still the possibility of geometric integrity, e.g. self-intersection.
However, there is still the possibility of geometric integrity, e.g. self-intersection.
B-reps typically produce only approximations and therefore suffer from accuracy problems B-reps are not as compact as other representations but are convenient for graphics and therefore efficient to display B-rep models are tedious to generate and edit. However, there are algorithms that can convert other representations into B-rep.
Solids in Boundary Representation (2) Solids in Boundary Representation (2) © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Decomposition Representations Decomposition Representations Solids are described by combining basic building blocks.
The type of building blocks leads to various representations Exhaustive Enumeration (Voxel Representations) Cell Decomposition Space Subdivision Quadtrees and Octrees Quadtrees and Octrees Binary Space-Partitioning Trees (BSP Trees) Binary Space-Partitioning Trees (BSP Trees) © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Exhaustive Enumeration Exhaustive Enumeration The solid is decomposed into identical cells on a regular grid Grid cells are called voxels or (less often) cuberille Decomposition is similar to scan-conversion of 2D primitives Voxels Binary voxels only indicate whether they are occupied (1) or not (0) Multi-valued voxels can represent several values, e.g. color, transparency, or material properties © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Exhaustive Enumeration: Example Exhaustive Enumeration: Example Viewpoint Boundary representation © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Exhaustive Enumeration: Example Exhaustive Enumeration: Example Viewpoint Voxel representation © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Exhaustive Enumeration: Properties Exhaustive Enumeration: Properties Only approximates the actual shape (sampling !) Represents always valid solids (if connectedness is not a requirement) Voxel representations are unambiguous and unique Voxel representations are close under Boolean operations Not very compact Algorithms are typically simple but slow because of the size of the data. Simplicity and regularity of algorithms allow easy parallelization.
© Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Cell Decomposition Cell Decomposition Generalization of the exhaustive enumeration Instead of identical voxels, cells can be of different shape Cells must be topological spheres, i.e. must not contain holes Cells can be "glued" together to describe a solid Cells may touch but must not have common interior points, i.e. must not intersect Topologically, cells are either disjoint or touch in exactly one corner, edge or face Various cell types are possible, e.g.
Polyhedra Polyhedra Curved polyhedra, i.e. a polyhedron bounded by bi-quadratic or bi-cubic patches Curved polyhedra, i.e. a polyhedron bounded by bi-quadratic or bi-cubic patches © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Cell Decomposition: Properties Cell Decomposition: Properties Cell decomposition can be very accurate up to the degree of the cell boundaries (typically quadratic) Hard to establish valid decompositions. Requires test for intersection between all pairs of cells !
There is not structural support as in voxel reps or octrees.
Cell decompositions are not unique.
Can be fairly compact Cell decompositions are usually not close under Boolean set operations Only few general algorithms for manipulating cell decompositions Cell decomposition is mostly used for analysis purposes, e.g. FEA Cell decompositions are usually generated from another representation of the solid © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Space-Subdivision Representations Space-Subdivision Representations Exhaustive enumeration and cell decomposition subdivide space in fairly regular fashion Requires large amount of storage Inefficient to process Adaptive subdivision overcomes these problems Only subdivide space along the solid's boundaries There are usually large areas without a boundary Exploits the fundamental property that the number of cells needed is proportional to the surface area.
Proportional to the square O( Proportional to the square O(r r2 2) of the desired resolution ) of the desired resolution r r Exhaustive enumeration requires O( Exhaustive enumeration requires O(r r3 3) cells ) cells We will look at 2 space subdivision schemes: Octrees and BSP-trees © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Octrees Octrees Recursive, binary subdivision of space Each subdivision step generates 8 cells Subdivision is controlled by the application To describe solids, cells are subdivided if they intersects with the solid's boundary Subdivision stops at a pre-determined maximum subdivision level or when the cell interior is homogeneous © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Quadtrees Quadtrees Quadtrees are the 2D equivalent of octrees Frequently used to subdivide and compress images Concepts apply similarly to octrees and quadtrees © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Quadtrees: Example Quadtrees: Example © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Describing Quadtrees and Octrees Describing Quadtrees and Octrees The binary subdivision process is described by a tree structure The tree contains three types of nodes: Partially filled nodes Empty leaves Full leaves P P P P P F F F F F F F F E E E E E E E E © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Labeling Quadtrees and Octrees Labeling Quadtrees and Octrees The cells in the tree can be labeled and numbered by their location There is no common convention for labels and numbers Quadtrees (north/south, east/west) NW(0), NE(1), SW(2), SE(3) Octrees (front/back, up/down, left/right): FUL(0), FUR(1), FDL(2), FDR(3), BUL(4), BUR(5), BDL(6), BDR(7) Number of leaves and nodes is limited L n N n n n h i h i h ≤ ≤ = − − − = − ∑ 1 0 1 1 1 © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Quadtrees & Octrees: Linear Notations (1) Quadtrees & Octrees: Linear Notations (1) Several ways to describe the by a string Only full (or empty) leaves are enumerated Linear addresses Terminator symbol for leaves not at the lowest level: X Bits per digit: 2n+1 Bits per leave: h (2n+1) Example (n=2, h=4): 11XX, 12XX, 1301, 1303, 131X, 1320, 1322, 2XXX (Bit count: 8 4 5 = 160) P P P P P F F F F F F F F E E E E E E E E 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Quadtrees & Octrees: Linear Notations (2) Quadtrees & Octrees: Linear Notations (2) Node/leave encoding Breadth-first traversal Bit 1: Internal node (0) or leave (1) Bit 2 (only for leaves): Empty (0) or full (1) 1 bit per internal node 2 bits per leave Example: 0 10 0 11 10 10 11 11 0 0 11 0 10 10 11 10 11, 11 10 11 10 (Bit count: 37) P P P P P F F F F F F F F E E E E E E E E 0 10 11 10 10 10 10 10 10 10 11 11 11 11 11 11 11 0 0 0 0 © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Octrees: Boolean Set Operations Octrees: Boolean Set Operations Combining two octrees is performed by synchronous traversal of the input trees An output octree is built by copying nodes from the input trees Nodes are combined according to simple rules If both nodes are leaves, the leaves are combined according to the Boolean set operation and appended to the output tree.
If both nodes are internal nodes, recursive traversal of the children If one node is a leaf and the other is an internal node, the operation determines, whether only the leaf node or the leaf node and the children of the internal node are copied to the output tree.
Example: Example: Leaf=1, Operation = AND: Copy leaf node and children of internal node Leaf=1, Operation = AND: Copy leaf node and children of internal node Leaf=0, Operation = AND: Only copy leaf node Leaf=0, Operation = AND: Only copy leaf node The output octree may not be compact, Internal nodes may have children that are all full or all empty Post-processing can easily compact an octree into a unique form © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Octrees: Neighbor Finding (1) Octrees: Neighbor Finding (1) Several applications / algorithms require to find the neighbor of a given octree node in a given direction Space traversal along a given path (ray) Averaging of neighboring nodes to compute normals for rendering Problem Find that octree node that borders the original in direction D Basic Algorithm Ascend from the node to a common ancestor with the target node May require ascent all the way to the root node May require ascent all the way to the root node Descend from that common ancestor to the target node © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Octrees: Neighbor Finding (2) Octrees: Neighbor Finding (2) Finding the Common Ancestor From the original node, find the first ancestor node that has not been reached from direction D Example: Find a node's Southern neighbor.
Find a node's Southern neighbor.
Climb the tree on an "upward path".
Climb the tree on an "upward path".
For every step upwards, check For every step upwards, check whether the path is in the Southern whether the path is in the Southern half of the node.
half of the node.
If not, the common ancestor is found. If not, the common ancestor is found. P P P P P F F F F F F F F E E E E E E E E © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Octrees: Neighbor Finding (3) Octrees: Neighbor Finding (3) P P P P P F F F F F F F F E E E E E E E E © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Octrees: Properties Octrees: Properties Octrees approximate the actual shape of the solid Like spatial enumeration and cell-based descriptions, octrees are always valid A compacted octree is a unique and unambiguous description of the solid The number of nodes in the octree are roughly proportional to the the solid's surface area. Although still quite large, octrees are more compact than voxel or cell representations.
Operations on octrees are closed for Boolean ops Octree algorithms are fairly simple as they rely on tree traversal.
© Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Binary-Space Partitioning Trees Binary-Space Partitioning Trees (BSP Trees) (BSP Trees) Generalization of Octrees Octrees partition space using orthogonal planes BSP trees use arbitrary planes to subdivide space Binary Space Partitioning (BSP) Oriented planes divide space into IN cells and OUT cells IN and OUT portions can be subdivided further by more planes To account for limited numerical precision, planes have "thickness" Thickness is a numerical tolerance. Points within the thickness are ON the plane.
Thickness is a numerical tolerance. Points within the thickness are ON the plane.
BSP Trees Each plane is a node in a tree IN and OUT cells are the children of a node © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 BSP Trees: Example BSP Trees: Example a b c d e f g h i j k out in out out out in in in out out in out a b j c e k d f g h i © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 BSP Trees: Point Classification BSP Trees: Point Classification Determine whether a point is inside or outside the solid The point is passed down the BSP tree, starting at the root At every node, the point is tested against the associated plane The point is then recursively descending the tree to a leave If the point is classified as ON, it is passed to both nodes a b c d e f g h i j k out in out out out in in in out out in out a b j c e k d f g h i © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 BSP Trees: Properties BSP Trees: Properties Dimension independent Concept of dividing (hyper)planes extends into higher dimensions Linear approximation of actual solid shape BSP description are not always valid Can describe objects that are not closed (open half-spaces !) Non-unique. Several BSP trees for the same solid.
BSP trees tend to be more compact than octrees.
Algorithms for closed Boolean operations on BSP trees BSP tree algorithms are more complicated than octree algorithms Often involve splitting objects on dividing planes Require attention to numerical precision © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Space Subdivision: Applications Space Subdivision: Applications Space subdivision representations and algorithms were also developed for other application domains Hidden Surface Removal Subdivide space along polygons Space subdivision allows to traverse objects front-to-back or back-to-front Collision Detection Animations and simulations require detection of collisions between objects Dynamically updated spacial partitioning allows to quickly determine candidates for potential object collisions © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Constructive Solid Geometry (CSG) Constructive Solid Geometry (CSG) © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Constructive Solid Geometry (CSG) Constructive Solid Geometry (CSG) CSG builds solid models by hierarchically combining primitive solids Most basic primitives are half-spaces, e.g.
Planes: ax + by + cd + e < 0 Cylinders: x2 + b2 - r2 < 0 Spheres: x2 + y2 + z2 - r2 < 0 Half-spaces and resulting solids are combined using Boolean set operations Union A + B Intersection A B Difference A - B = A (-B) © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 CSG Tree CSG Tree The combination of half-spaces using the binary Boolean operators creates a binary tree, the CSG tree The tree mirrors the Boolean expression describing the CSG object CSG expressions can be manipulated to create equivalent expression of the same object Boolean algebra, e.g. distribution of terms or De Morgan's law In particular, CSG expressions can be brought into conjunctive or disjunctive normal forms Then, the CSG tree has only 2 levels, e.g. sum of intersections © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 CSG: Example CSG: Example -A B C (A+B) - C + The union of the blocks forms the bracket Subtracting the cylinder creates the hole CSG Tree © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Rendering CSG Objects Rendering CSG Objects Boundary Evaluation Calculate the boundary of the CSG solid Render the boundary using standard (polygon) rendering techniques Requires intersection of faces, edges and points Intersection calculations can be complicated when allowing higher-order halfspaces like cylinders or tori Can result in hairy case analyses, in particular in the presence of numerical errors Ray-Casting © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Casting CSG Objects (1) Ray-Casting CSG Objects (1) Ray Casting Cast rays from the eye into the scene Determine first intersection of rays with objects to determine visible object Approximation as the object is sampled by the rays Ray-casting CSG objects Much simpler than boundary evaluation as it only requires intersection of lines with half-spaces Each half-space segments the line into IN and OUT segments These segments are combined using the Boolean expression describing the CSG object Requires computation of all intersections of ray and object © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Casting CSG Objects Ray-Casting CSG Objects (2) (2) B A A B A+B A-B AB A+B A A-B AB © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Casting CSG Objects (3) Ray-Casting CSG Objects (3) The IN segments can also be used to approximate volume/mass and center of gravity of the CSG object Generalized CSG primitives Ray-casting requires only a small set of operations to be supported for a primitive, namely intersection with a ray Therefore, other primitive types than only half-spaces can easily be integrated into CSG For instance, polygonal models can be easily incorporated in a CSG modeler. This bridges the gap between CSG and b-reps.
Another example is the integration of sweeping primitives with CSG (Sweeping, moves a shape along a trajectory cover, i.e. sweeping out, a part of space) © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray Tracing Ray Tracing © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray Tracing (1) Ray Tracing (1) Ray Tracing follows rays through the scene Typically, rays are traced from the eye into the scene As rays hit objects new rays are generated Shadow rays to determine whether the object is lit If exposed to a light source, the lighting model is computed If exposed to a light source, the lighting model is computed Reflected ray to create inter-object reflections Refracted ray if the object is transparent The pixel color is computed as the combination of several contributions Light directly received from the light source(s) and Light received from other objects Ambient light © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray Tracing (2) Ray Tracing (2) Screen A B C Viewpoint Light-source © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Forward vs. Backward Ray Tracing Forward vs. Backward Ray Tracing Forward Ray Tracing Rays are starting at the light source(s) and traced through the scene until they hit the eye Approximates how light propates in the physical world Chances of finding a ray from a light source that actually hits the eye are very small.
Therefore, forward ray tracing is very inefficient Backward Ray Tracing Inverts the forward ray tracing process by tracing rays from the eye back to the light source.
Only uses rays that are in fact hitting the eye ... more efficient © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Global Illumination Global Illumination Simplistic rendering algorithms, e.g. raster pipeline (OpenGL) only account for direct interactions between light and objects Secondary effects are crudely approximated using ambient light Global Illumination describes a class of methods that try to capture the overall distribution of light (energy) in a scene Models higher-order effects, e.g. self-shadowing, inter-object reflections or light attenuation For reflective surfaces, ray-tracing is adequate (ray optics) Radiosity algorithms capture the interaction of diffuse reflectors (global energy balance) © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray Tracing: Steps Ray Tracing: Steps We will now look more carefully at the steps involved in the ray-tracing process Generate rays Find ray-object intersections and choose the closest one Cast rays towards all light sources and compute lighting model if lit Generate secondary rays and recursively trace them Combine the contribution of all rays at a surface point, shadow rays and secondary rays © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray Generation Ray Generation At least one ray has to be generated for every pixel Several rays per pixel can be used to implement anti-aliasing Rays are determined by the view geometry, i.e. position of the eye and the screen with respect to the scene Ray is described as point and direction t>0 for numerical precision Z Y Eye Screen Ray Pixel ZP ZE YE YP R R R R R E R P E P E = + ⋅ = > = = − − orig dir dir orig dir t t with (Normalized) and (Semi - infinte ray) Here: and 1 0 © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Object Intersection Ray-Object Intersection Once rays enter the scene, their intersections with the objects in the scene must be computed Each primitive type needs special intersection routine (Suggests object-oriented programming approach !) Simple for polyhedra or spheres More complicated for higher-order primitives Typically, only the nearest intersection with the object is needed CSG operations require all intersections to define the IN and OUT segments © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Object Intersection: Polygon (1) Ray-Object Intersection: Polygon (1) We assume that polygons are planar First, compute ray-plane intersection Then, determine whether intersection is inside the polygon Plane: Ray-Plane Intersection: Ax By Cz D B C A B C D x y z + + + = + + = ⋅ = = F H G G G G I K J J J J = F H G G G G I K J J J J 0 1 0 1 2 2 with A with and 2 N P N P N R R N R N R ⋅ + ⋅ = ⇒ = − ⋅ ⋅ orig i dir i orig dir t t d i 0 © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Object Intersection: Polygon (2) Ray-Object Intersection: Polygon (2) Ray intersects the plane iff: Intersection point: Now the intersection point has to be tested against the polygon (point in polygon test): Simplify this test by projecting the polygon along one of the axes Pick axis by finding the largest coordinate in the normal vector N Project by "dropping" that coordinate from all polygon vertices and the intersection point. Call the remaining axes U and V.
Example: If N = (1, 4, 2) then project onto XZ-plane, i.e. drop Y.
X axis becomes U axis and Y axis becomes V axis.
X axis becomes U axis and Y axis becomes V axis.
This process does not change the polygon topology or the location of the intersection point with respect to the polygon !
ti > 0 P R R i orig i dir t = + ⋅ © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Object Intersection: Polygon (3) Ray-Object Intersection: Polygon (3) Now we have to solve the 2D point in polygon problem For general polygons, the interior can be defined in several ways.
We will use the Jordan curve theorem: A point is inside the polygon, if an infinite ray from the point intersects an odd number of polygon edges.
© Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Object Intersection: Polygon (4) Ray-Object Intersection: Polygon (4) Practical implementation: Translate the projected intersection point falls onto the origin Pick one positive axis as the ray to be tested against the polgon Count number of intersections of that axis with the polygon edges Attention: Each vertex must belong to only one edge to avoid double-counting !
Attention: Each vertex must belong to only one edge to avoid double-counting !
True intersection calculation is only necessary if the two vertices are in diagonally opposed quadrants.
Other cases can be handled trivially.
U V Ray Intersection Point © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Object Intersections: Sphere (1) Ray-Object Intersections: Sphere (1) Spheres are frequently used in ray-traced images "Spheres floating over checker boards" Ray-Sphere intersection is very simple to compute The sphere is described by center SC and radius SR Substituting the ray coordinates for X, Y and Z: This simplifies to a quadratic expression: S X X Y Y Z Z R C C C 2 2 2 2 = − + − + − b g b g b g S X t X X Y t Y Y Z t Z Z R orig dir C orig dir C orig dir C 2 2 2 2 = + ⋅ − + + ⋅ − + + ⋅ − d i d i d i 0 2 = ⋅ + ⋅+ A t B t C A X Y Z B X X X Y Y Y Z Z X C X X Y Y Z Z S dir dir dir dir orig C dir orig C dir orig C orig C orig C orig C R = + + = = ⋅ ⋅ − + ⋅ − + ⋅ − = − + − + − − 2 2 2 2 2 2 2 1 2 if ray direction is normalized b g d i d i d i e j d i d i d i © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Object Intersections: Sphere (2) Ray-Object Intersections: Sphere (2) Solving the quadratic expression for A=1: No intersection if ti1 and ti2 are complex or both ti less than zero Ray starts inside the sphere if only one solution is real and positive For two intersections: nearest intersection for smaller ti The normal at the intersection point is t B B C t B B C i i 1 2 2 2 4 2 4 2 = − − − = − + − and N X X S Y Y S Z Z S S i C R i C R i C R = − − − F H G G G I K J J J b g b g b g © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Object Intersections: Cylinder Ray-Object Intersections: Cylinder Cylinders are frequently used to describe pipes, connections etc. or to create holes in a CSG object We will consider the intersection of an arbitrary ray with an axis-aligned infinite cylinder The cylinder is described by its radius: Substituting the ray coordinates gives: This results in a quadratic equation: C X X Y Y R C C 2 2 2 = − + − b g b g C X t X X Y t Y Y R orig dir C orig dir C 2 2 2 = + ⋅ − + + ⋅ − d i d i 0 2 2 2 2 2 2 2 = ⋅ − + ⋅ − + − + − + − − t X Y t X X X Y Y Y X X Y Y C dir dir dir orig C dir orig C orig C orig C R c h d i d i d i d i © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Object Intersections: Ray-Object Intersections: Numerical Precision (1) Numerical Precision (1) Intersection calculations are done using floating point operations, giving rise to numerical problems The computed intersection may actually fall inside the object Secondary rays will then be intersecting the object again This effect is known as self-shadowing (a.k.a. surface acne) Primary ray Secondary or shadow ray Surface Computed intersection © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Object Intersections: Ray-Object Intersections: Numerical Precision (2) Numerical Precision (2) Possible solutions Exclude the object from intersection calculations for secondary rays Introduce a numerical tolerance for parameter t indicating that the ray starts on the surface Move the intersection point along the ray to be on the proper side of the surface Primary ray Secondary or shadow ray Surface Computed intersection © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Object Intersections: Coordinates Ray-Object Intersections: Coordinates All objects and rays are specfied in world coordinates Objects are transformed from local coordinates into world coordinates: modeling transformation Pixels must be transformed from screen space into world coordinates: inverse viewport mapping Intersection of ray and object is computed by transforming the ray into the local object coordinates Simply apply the inverse of the modeling transformation to the ray Intuition: Transform ray and object back into the local coordinates © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray-Object Intersection: Nearest Object Ray-Object Intersection: Nearest Object To determine the visible object from a given point, the closest object intersect by the ray must be found For all ray-object intersections, find the one that has the smallest ray parameter t For this intersection point, compute the surface color © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Light/Shadow Rays Light/Shadow Rays From every ray-object intersection point, rays are traced to the light source(s) Light rays are tested against the scene to determine whether the intersection point can see the light source, i.e. whether it is lit Other information available at the intersection point: Surface normal Ray direction Material properties This are all the parameters needed to compute a lighting model at the intersection point © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Secondary Rays Secondary Rays At every intersection point secondary rays are generated according to ray optics Reflected rays to model surface mirroring Refracted rays for translucent materials Recursive tracing of rays creates a ray tree Eye O1 O2 O3 O4 L1 L2 E R1 T1 S1 S2 R2 T2 S3 S4 S5 S6 R3 T3 O3 O2 E R1 T1 S1 S2 R2 T2 S3 S4 S5 S6 R3 T3 O1 © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Combining Ray Intensities Combining Ray Intensities The color/intensity reflected from a surface point depends on two components Light directly received from light sources (light rays) Light received indirectly from other objects (reflected/refracted rays) The components are superimposed (summed up) to compute the color of the incident ray Incident ray Light ray 1 Light ray 2 Reflected ray Refracted ray Surface I I I I k Ray Reflected Refracted Light k = + +∑ © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray Tracing Optimizations (1) Ray Tracing Optimizations (1) Many ray tracers spend 80% of the time performing intersection calculations. Optimizations try to reduce the number of intersection calculations to speed up ray tracing Hierarchical object description Rays are first tested against higher hierarchy levels Intersections with lower hierarchy levels are only computed if the ray intersects the higher hierarchy Frequently, bounding boxes (or other bounding volumes) are computed for objects. Rays are first tested against the bounding volumes.
© Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray Tracing Optimizations (2) Ray Tracing Optimizations (2) Space partitioning To avoid testing of objects entirely, space is partitioned If the ray does not enter a cell, none of the objects in this cell are tested against the ray Popular partitioning schemes include octrees and BSP trees © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Ray Tracing: Anti-aliasing Ray Tracing: Anti-aliasing Aliasing is introduced by several sources Geometry (spatial aliasing) Object motion (temporal aliasing) Material properties (light aliasing) --> radiosity Geometric aliasing can be alleviated by supersampling Not a complete cure but reduces the artifacts Implementation is simple, shoot several rays at every pixel and filter the resulting sub-pixel values Adaptive supersampling adds samples only in pixel with high color gradient Stochastic supersampling uses randomly distributed sub-pixel. Introduces noise instead of aliasing.
© Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Summary Summary Solid modeling Definition of what makes an object a solid Different techniques to describe a solid object Properties of each representation Ray Tracing Global illumination rendering technique Simulates (backwards) the propagation of light rays Works with surfaces and solids © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Further Study Further Study Solid Modeling Christoph M. Hoffmann, Geometric & Solid Modeling, Morgan Kaufmann Publishers, 1989 Martti Mäntylä, An Introduction to Solid Modeling, Computer Science Press, 1988 Ray Tracing Andrew Glassner (editor), An Introduction to Ray Tracing, Academic Press, 1989.
Andrew Glassner, Principals of Digital Image Synthesis, Morgan Kaufmann Publishers, 1995 © Bengt-Olaf Schneider, 1999 Computer Graphics – Week 10 Homework Homework Study textbook: solid modeling (chapter 12) and ray tracing (chapter 15.10 and 16.12) Prepare for next week by reading about radiosity (chapter 16.13) Start working on final assignment To be handed out on Friday Start early !!! |
4912 | https://brians.wsu.edu/2016/05/25/dozen-of/ | dozen of
Why isn’t it “a dozen of eggs” when it’s standard to say “a couple of eggs”? The answer is that “dozen” is a precise number word, like “two” or “hundred”; we say “two eggs,” “a hundred eggs,” and “a dozen eggs.”
“Couple” is often used less precisely, to mean “a few,” so it isn’t treated grammatically as an exact number. “A couple eggs” is less standard than “a couple of eggs.”
“Dozens of eggs” is standard because you’re not specifying how many dozens you’re talking about.
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Absolute Value Inequalities – Explanation & Examples
Absolute Value Inequalities – Explanation & Examples
The absolute value of inequalities follows the same rules as the absolute value of numbers. The difference is that we have a variable in the prior and a constant in the latter.
This article will show a brief overview of the absolute value inequalities, followed by the step-by-step method to solve the absolute value inequalities.
Finally, there are examples of different scenarios for better understanding.
What is Absolute Value Inequality?
Before we can learn how to solve absolute value inequalities, let’s remind ourselves about a number’s absolute value.
By definition, the absolute value of a number is the distance of a value from the origin, regardless of the direction. Absolute value is denoted by two vertical lines enclosing the number or expression.
For example, the absolute value of x is expressed as | x | = a, which implies that, x = +a and -a. Now let’s see what the absolute value inequalities entail.
An absolute value inequality is an expression with absolute functions as well as inequality signs. For example, the expression |x + 3| > 1 is an absolute value inequality containing a greater than symbol.
There are four different inequality symbols to choose from. These are less than (<), greater than (>), less than or equal (≤), and greater than or equal (≥). So, the absolute value inequalities can possess any one of these four symbols.
How to Solve Absolute Value Inequalities?
The steps for solving absolute value inequalities are much similar to solving absolute value equations. However, there is some extra information you need to keep in mind when solving absolute value inequalities.
The following are the general rules to consider when solving absolute value inequalities:
Isolate on the left the absolute value expression.
Solve the positive and negative versions of the absolute value inequality.
When the number on the other side of the inequality sign is negative, we either conclude all real numbers as the solutions, or the inequality has no solution.
When the number on the other side is positive, we proceed by setting up a compound inequality by removing the absolute value bars.
The type of inequality sign determines the format of the compound inequality to be formed. For instance, if a problem contains greater than or greater than/equals to sign, set up a compound inequality that has the following formation:
(The values within absolute value bars) < – (The number on the other side) OR (The values within absolute value bars) > (The number on the other side).
Similarly, if a problem contains a less than or less than/equals to sign, set up a 3- part compound inequality of the following form:
– (The number on the other side of inequality sign) < (quantity within the absolute value bars) < (The number on the other side of the inequality sign)
Example 1
Solve the inequality for x: | 5 + 5x| − 3 > 2.
Solution
Isolate the absolute value expression by adding 3 to both sides of the inequality;
\=> | 5 + 5x| − 3 (+ 3) > 2 (+ 3)
\=> | 5 + 5x | > 5.
Now solve both the positive and negative “versions” of the inequality as follows;
We’ll assume absolute value symbols by solving the equation the normal way.
\=> | 5 + 5x| > 5 → 5 + 5x > 5.
\=> 5 + 5_x_> 5
Subtract 5 from both sides
5 + 5x (− 5) > 5 (− 5) 5x > 0
Now, divide both sides by 5
5x/5 > 0/5
x > 0.
Thus, x > 0 is one of the possible solutions.
To solve for negative version of the absolute value inequality, multiply the number on the other side of the inequality sign by -1, and reverse the inequality sign:
| 5 + 5x | > 5 → 5 + 5x < − 5 => 5 + 5x < -5 Subtract 5 from both sides => 5 + 5x ( −5) < −5 (− 5) => 5x < −10 => 5x/5 < −10/5 => x < −2.
x > 0 or x < −2 are the two possible solutions to the inequality. Alternatively, we can solve | 5 + 5x | > 5 using the formula:
(The values within absolute value bars) < – (The number on other side) OR (The values within absolute value bars) > (The number on other side).
Illustration:
(5 + 5x) < – 5 OR (5 + 5x) > 5
Solve the expression above to get;
x < −2 or x > 0
Example 2
Solve |x + 4| – 6 < 9
Solution
Isolate the absolute value.
|x + 4| – 6 < 9 → |x + 4| < 15
Since our absolute value expression has a less than inequality sign, we set up the a 3-part compound inequality solution as:
-15 < x + 4 < 15
-19 < x < 11
Example 3
Solve |2x – 1| – 7 ≥ -3
Solution
First, isolate the variable
|2x – 1| – 7≥-3 → |2x – 1|≥4
We will set up an “or” compound inequality because of the greater than or equal to sign in our equation.
2 – 1≤ – 4 or 2x – 1 ≥ 4
Now, solve the inequalities;
2x – 1 ≤ -4 or 2x – 1 ≥ 4
2x ≤ -3 or 2x ≥ 5
x ≤ -3/2 or x ≥ 5/2
Example 4
Solve |5x + 6| + 4 < 1
Solution
Isolate the absolute value.
|5x + 6| + 4 < 1 → |5x + 6| < -3
Since the number on the other side is negative, check also the opposite to determine the solution.
|5x + 6| < -3
Positive < negative (false). Therefore, this absolute value inequality has no solution.
Example 5
Solve |3x – 4| + 9 > 5
Solution
Isolate the absolute value.
|3x – 4| + 9 > 5 → |3x – 4| > -4
|5x + 6| < -3
Since, positive < negative (true). Therefore, the solutions to this absolute value inequality are all real numbers.
Practice Questions
1. Solve the inequality for x: $| 3 + 3x| − 2 > 1$.
$x < -2$ or $x>0$
$x < -2$ or $ x>2 $
$-2 < x < 2$
$-2 \leq x \leq 2$
Loading…
2. Solve the inequality for x: $|x + 6| – 3 < 8$.
$x < -17$ or $x>5$
$x < -5$ or $x>17$
$ -17 < x < 5$
$-17 \leq x \leq 5$
Loading…
3. Solve for the absolute value inequality, $|2x – 5| – 3 ≥ 6$.
$x < -2$ or $x > 7$
$x \leq -2$ or $x \geq 7$
$ -2 < x<7$
$-2 \leq x \leq 7$
Loading…
4. Solve for the absolute value inequality, $|3x + 2| + 6 < 2$.
$x < -2$ or $x > 2$
$x \leq -2$ or $x \geq 2$
All values of $x$ satisfy the inequality.
The inequality has no solution.
Loading…
5. Solve for the absolute value inequality, $|2x – 6| + 8 > 2$.
$x < 0$
$x > 0$
All values of $x$ satisfy the inequality.
The inequality has no solution.
Loading…
Loading…
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4914 | https://www.almabetter.com/bytes/tutorials/applied-statistics/point-estimation-and-interval-estimation | Course Outline
Point Estimation and Interval Estimation
Maximum Likelihood Estimation
Hypothesis Testing and p-Values
Type I and Type II Errors: Definition, Differences, Example
Fundamentals of Confidence Intervals and Margin of Error
Power and Sample Size Estimation
Point Estimation and Interval Estimation
Last Updated: 10th October, 2023
Point estimation and interval estimation are two important concepts in statistics used to estimate population parameters from sample data. Point estimation involves estimating a population parameter by using a single value, while interval estimation involves constructing a range of values that contains the population parameter with a certain level of confidence. Point estimation is simpler and faster to calculate but provides less information than interval estimation. Interval estimation provides more information but requires more calculations and assumptions. Ultimately, the choice between point estimation and interval estimation depends on the specific research question, sample size, level of precision, and confidence level desired by the researcher.
Introduction to Estimation
Estimation is a statistical method used to estimate unknown parameters of a population based on a sample of data. There are two types of estimation: point estimation and interval estimation.
Point Estimation
Point estimation involves using a single value, called a point estimator, to estimate the unknown population parameter. For example, the sample mean can be used as a point estimator of the population mean, and the sample proportion can be used as a point estimator of the population proportion.
The formula for the sample mean is:
```
```
x̄ = Σxi / n
```
```
where x̄ is the sample mean, Σxi is the sum of the sample values, and n is the sample size.
The formula for the sample proportion is:
```
```
p̂ = x / n
```
```
where p̂ is the sample proportion, x is the number of sample values that have the characteristic of interest, and n is the sample size.
Properties of Point Estimators
A good point estimator should have three desirable properties: unbiasedness, consistency, and efficiency.
Unbiasedness means that the expected value of the point estimator is equal to the true population parameter. A point estimator that is biased will systematically overestimate or underestimate the population parameter.
Consistency means that as the sample size increases, the point estimator becomes closer and closer to the true population parameter. A consistent point estimator will converge to the true parameter as the sample size increases.
Efficiency means that the point estimator has the smallest possible variance among all unbiased point estimators. An efficient point estimator will provide the most precise estimates of the population parameter.
The mean squared error (MSE) is a measure of the performance of a point estimator. The MSE is defined as the expected value of the squared difference between the point estimator and the true parameter. A good point estimator will have a small MSE.
Interval Estimation
Interval estimation involves constructing a range of values, called a confidence interval, that is likely to contain the unknown population parameter with a certain level of confidence. For example, a 95% confidence interval for the population mean would contain the true population mean in 95% of all possible samples.
The formula for a confidence interval for the population mean is:
```
```
x̄ ± zα/2 σ / √n
```
```
where x̄ is the sample mean, zα/2 is the z-score that corresponds to the desired level of confidence, σ is the population standard deviation (if known) or the sample standard deviation (if unknown), and n is the sample size.
The formula for a confidence interval for the population proportion is:
```
```
p̂ ± zα/2 √(p̂(1 - p̂)) / n
```
```
where p̂ is the sample proportion, zα/2 is the z-score that corresponds to the desired level of confidence, and n is the sample size.
Properties of Confidence Intervals
A good confidence interval should have two desirable properties: coverage probability and margin of error.
Coverage probability means that the confidence interval will contain the true population parameter with a certain level of confidence. For example, a 95% confidence interval should contain the true population parameter in 95% of all possible samples.
Margin of error is a measure of the precision of the confidence interval. The margin of error is defined as half the width of the confidence interval. A narrower confidence interval will have a smaller margin of error and be more precise.
Confidence Interval Estimation
Interval estimation is another method of estimation that provides a range of plausible values for a population parameter. Unlike point estimation, it provides a range of values rather than a single value. The range is called the confidence interval, and it represents a level of certainty that the true population parameter falls within the interval.
The confidence interval is computed by taking the point estimate and adding and subtracting a margin of error. The margin of error is based on the level of confidence, the sample size, and the standard error of the point estimate. A common level of confidence used in interval estimation is 95%.
For example, suppose we want to estimate the mean weight of all dogs in a population. We take a sample of 100 dogs and compute the sample mean weight to be 30 pounds with a standard deviation of 5 pounds. We want to construct a 95% confidence interval for the population mean weight.
Using the formula for a confidence interval for the population mean, we can compute the margin of error as follows:
```
```
Margin of error = Z_(α/2) (s/√n) = 1.96 (5/√100) = 0.98
```
```
We then construct the confidence interval as follows:
```
```
Confidence interval = sample mean ± margin of error = 30 ± 0.98 = (29.02, 30.98)
```
```
This means we are 95% confident that the true population mean weight falls within the interval of 29.02 to 30.98 pounds.
Types of Interval Estimation
There are different types of interval estimation, depending on the parameter being estimated and the method used to compute the confidence interval. Some common types are:
Confidence interval for the population mean: This is used to estimate the population mean when the population standard deviation is unknown. It uses the t-distribution to compute the critical value for the confidence interval.
Confidence interval for the population proportion: This is used to estimate the population proportion, such as the proportion of voters who support a certain candidate. It uses the normal distribution to compute the critical value for the confidence interval.
Confidence interval for the difference between two means: This is used to estimate the difference between two population means. It uses either the t-distribution or normal distribution, depending on the sample sizes and whether the variances are assumed to be equal or not.
Confidence interval for the difference between two proportions: This is used to estimate the difference between two population proportions. It uses either the normal distribution or the chi-square distribution, depending on the sample sizes and whether the variances are assumed to be equal or not.
Limitations and Assumptions of Interval Estimation
Interval estimation, like point estimation, relies on certain assumptions and has some limitations. Some of the key assumptions and limitations are:
The sample is representative of the population: Interval estimation assumes that the sample is randomly selected and represents the population of interest. If the sample is biased or non-random, the confidence interval may not be accurate.
Normality assumption: Interval estimation assumes that the population is normally distributed or that the sample size is large enough for the central limit theorem to apply. If the data is not normally distributed and the sample size is small, the confidence interval may not be accurate.
Independence assumption: Interval estimation assumes that the observations are independent of each other. If there is correlation or dependence between the observations, the confidence interval may not be accurate.
Finite population correction: If the sample size is a significant portion of the population size, a finite population correction factor may need to be applied to adjust the confidence interval.
Conclusion
In conclusion, point estimation and interval estimation are two important concepts in statistics used to estimate population parameters based on sample data. Point estimation provides a single value estimate for the population parameter, while interval estimation provides a range of values that the population parameter is likely to fall within. Both methods have their own advantages and limitations, and the choice of method depends on the specific research question and data at hand. It is important to understand the assumptions and conditions required for both methods, and to interpret the results appropriately.
Key Takeaways
Point estimation involves using sample data to estimate a population parameter with a single value.
The most common point estimator is the sample mean, which is an unbiased and efficient estimator of the population mean.
Interval estimation provides a range of values within which the population parameter is likely to fall.
Confidence intervals are the most commonly used type of interval estimation, and are calculated based on the sample mean and standard error.
Confidence level and sample size are two important factors that affect the width of the confidence interval.
Both point estimation and interval estimation have their own advantages and limitations, and the choice of method depends on the specific research question and data at hand.
It is important to understand the assumptions and conditions required for both methods, and to interpret the results appropriately.
Quiz
1. What is the key difference between point estimation and interval estimation?
a) Point estimation provides a range of values while interval estimation provides a single value.
b) Point estimation involves using sample statistics while interval estimation uses population parameters.
c) Point estimation provides a single value while interval estimation provides a range of values.
d) Point estimation is used for discrete variables while interval estimation is used for continuous variables.
Answer: c) Point estimation provides a single value while interval estimation provides a range of values.
2. Which of the following is an example of point estimation?
a) Estimating the average height of all students in a school based on a sample of 30 students.
b) Estimating the proportion of voters who will support a particular candidate in an upcoming election.
c) Estimating the standard deviation of a population based on a sample of 50 observations.
d) Estimating the median income of a population based on a sample of 100 households.
Answer: a) Estimating the average height of all students in a school based on a sample of 30 students.
3. What is the purpose of confidence intervals in interval estimation?
a) To provide an exact estimate of the population parameter.
b) To provide a range of values that contains the population parameter with a certain level of confidence.
c) To determine the standard error of the sampling distribution.
d) To identify the margin of error in the sample statistics. Answer: b) To provide a range of values that contains the population parameter with a certain level of confidence.
4. Which of the following factors affects the width of a confidence interval?
a) Sample size and population size.
b) Significance level and sample size.
c) Sample size and variability in the sample.
d) Significance level and population size.
Answer: c) Sample size and variability in the sample.
Module 6: Statistical Inference
Lesson 1: Point Estimation and Interval Estimation
Module 6: Statistical Inference
Lesson 1: Point Estimation and Interval Estimation
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4915 | https://www.tiger-algebra.com/en/solution/scientific-notation-conversion/22000000/ | Copyright Ⓒ 2013-2025
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Solution - Scientific notation/Standard form
Other Ways to Solve
Step-by-step explanation
1. Write the number as a decimal
22000000.0
2. Make it a new number between 1 and 10
Move the decimal point to make 22000000.0 a new number between 1 and 10. Because our number is greater than 10, we move the decimal point to the left. Drop any trailing zeros and place the decimal point after the first non-zero digit. Keep track of how many times we move the decimal point.
22000000.0 -> 2.2
Our new number is 2.2. We moved the decimal point 7 times.
3. Define the power of 10
Because our original number was greater than 10, the power of 10 is positive. Remember, we moved the decimal point 7 times, so the exponent is positive 7:
107
4. Final result
2.2⋅107
How did we do?
Why learn this
Scientific notation, or standard form, makes things easier when working with very small or very big numbers, both of which come up frequently in the fields of science and engineering. It is used in science, for example, to convey the mass of the heavenly bodies: Jupiter’s mass is 1.898⋅1027kg, which is easier to comprehend than writing the number 1,898 followed by 24 zeroes. Scientific notation also makes solving problems that use such high or low numbers more straightforward.
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4916 | https://fiveable.me/key-terms/electrical-circuits-systems-ii/v-=-l-didt | 🔦electrical circuits and systems ii review
key term - V = l (di/dt)
Citation:
Definition
The equation v = l (di/dt) describes the relationship between voltage (v) across an inductor, the inductance (l), and the rate of change of current (di/dt). This formula indicates that the voltage induced across an inductor is proportional to how quickly the current flowing through it changes. The faster the current changes, the greater the induced voltage will be, highlighting the fundamental nature of inductors in electrical circuits.
5 Must Know Facts For Your Next Test
Review Questions
Related terms
A passive electrical component that stores energy in a magnetic field when electric current flows through it.
Self-induction: The phenomenon where a changing current in a coil induces a voltage within itself, opposing the change in current.
Mutual induction: The process by which a changing current in one coil induces a voltage in another nearby coil.
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4917 | https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality?srsltid=AfmBOoqVQgvNXpqV2SMxf8M4U1pZ7WC9NtZUi-co-P980hi1PTFI3QLe | Page
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AM-GM Inequality
In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same.
In symbols, the inequality states that for any real numbers , with equality if and only if .
The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests.
Contents
Proofs
All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for .
Generalizations
The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM.
Weighted AM-GM Inequality
The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article.
Mean Inequality Chain
The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain.
Power Mean Inequality
The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality.
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4918 | https://web.stanford.edu/class/archive/cs/cs103/cs103.1152/lectures/21/Small21.pdf | Unsolvable Problems Outline for Today ●Recap from Last Time ●Wow, we covered a lot. Where are we again?
●R and RE Languages ●What does it mean to solve a problem?
●Encodings ●Computing over general objects.
●The Universal Turing Machine ●One machine to run them all.
●Impossible Problems ●Discovering truly impossible problems. Recap from Last Time What problems can we solve with a computer?
What kind of computer? The Church-Turing Thesis claims that every effective method of computation is either equivalent to or weaker than a Turing machine.
This is not a mathematical fact – it's a hypothesis about the nature of computation. Regular Languages CFLs All Languages Problems Solvable by Any Feasible Computing Machine Regular Languages CFLs All Languages Languages Recognized by Turing Machines What problems can we solve with a computer?
What does it mean to solve a problem? Some Important Terminology ● Let M be a Turing machine.
● M accepts a string w if it enters the accept state when run on w.
● M rejects a string w if it enters the reject state when run on w.
● M loops infinitely (or just loops) on a string w if when run on w it enters neither the accept or reject state.
● M does not accept w if it either rejects w or loops infinitely on w.
● M does not reject w w if it either accepts w or loops on w.
● M halts on w if it accepts w or rejects w.
Accept Loop Reject does not accept does not reject halts The Language of a TM ●The language of a Turing machine M, denoted ( ℒM), is the set of all strings that M accepts: ℒ(M) = { w ∈ Σ | M accepts w } ●For any w ∈ ( ℒM), M accepts w.
●For any w ∉ ( ℒM), M does not accept w.
●It might loop forever, or it might explicitly reject.
●A language is called recognizable if it is the language of some TM.
●Notation: the class RE is the set of all recognizable languages.
L ∈ RE iff L is recognizable The Power of TMs ●Because TMs only need to accept strings in their languages, many problems can be formulated as RE languages.
●Any context-free language: simulate all possible production rules and see if the target string can be derived.
●Solving a maze – use the worklist to explore all paths of length 0, 1, 2, … until a solution is found.
●Determining whether a polynomial has an integer zeros: try 0, -1, +1, -2, +2, -3, +3, … until a result is found. Why “Recognizable?” ●Given TM M with language ℒ(M), running M on a string w will not necessarily tell you whether w ∈ ( ℒM).
●If the machine is running, as an observer, you can't tell whether ● it is eventually going to halt, but just needs more time, or ● it is never going to halt.
●However, if you know for a fact that w ∈ ( ℒM), then the machine can confirm this (it eventually accepts).
●The machine can't decide whether or not w ∈ ( ℒM), but it can recognize strings that are in the language.
●We sometimes call a TM for a language L a recognizer for L. Is this a satisfactory definition of “solving” a problem? New Stuff! Deciders ●Some Turing machines always halt; they never go into an infinite loop.
●If M is a TM and M halts on every possible input, then we say that M is a decider.
●For deciders, accepting is the same as not rejecting and rejecting is the same as not accepting.
Accept Reject halts (always) does not accept does not reject Decidable Languages ●A language L is called decidable if there is a decider M such that ( ℒM) = L.
●Given a decider M, you can learn whether or not a string w ∈ ( ℒM).
●Run M on w.
●Although it might take a staggeringly long time, M will eventually accept or reject w.
●The class R is the set of all decidable languages.
L ∈ R iff L is decidable R and RE Languages ●Intuitively, a language is in RE if there is some way that you could exhaustively search for a proof that w ∈ L.
●If you find it, accept!
●If you don't find one, keep looking!
●Intuitively, a language is in R if there is a concrete algorithm that can determine whether w ∈ L.
●It tends to be much harder to show that a language is in R than in RE. Examples of R Languages ●All regular languages are in R.
●If L is regular, we can run the DFA for L on a string w and then either accept or reject w based on what state it ends in.
●{ 0n1n | n ∈ ℕ } is in R.
●The TM we built on Wednesday is a decider.
●{ 1n | n ∈ ℕ and n is composite} is in R.
●The TM we built on Friday is a decider. CFLs and R ●With a worklist approach, we can build a recognizer that checks membership in any CFL.
●Harder result: all CFLs are in R.
●Read Sipser, Ch. 4.1 for details.
●Or come talk to me after lecture! Why R Matters ●If a language is in R, there is an algorithm that can decide membership in that language.
● Run the decider and see what it says.
●If there is an algorithm that can decide membership in a language, that language is in R.
● By the Church-Turing thesis, any effective model of computation is equivalent in power to a Turing machine.
● Therefore, if there is any algorithm for deciding membership in the language, there is a decider for it.
● Therefore, the language is in R.
●A language is in R if and only there is an algorithm for deciding membership in that language. R ≟RE ●Every decider is a Turing machine, but not every Turing machine is a decider.
●Thus R ⊆ RE.
●Hugely important theoretical question: Is R = RE? ●That is, if we can verify that a string is in a language, can we decide whether that string is in the language? Regular Languages CFLs All Languages R RE Which Picture is Correct? Regular Languages CFLs All Languages R RE Which Picture is Correct? Encodings Computing over Objects ●Turing machines always compute over strings.
●We have seen examples of automata that can essentially compute over other objects: ●Walking your Dog: Compute over paths.
●Composite numbers: Compute over numbers.
●Up to this point, we have always said how we will encode objects: ●e.g. { 1m+1n=1mn | m, n ∈ ℕ } A Multitude of Encodings ●There can be many ways of encoding the same object.
●Example: the natural number 13 can be encoded ● in unary: 1111111111111 ● in binary: 1101 ● in decimal: 13 ● in hexadecimal: D ● in Roman numerals: XIII ● … ●Claim: Turing machines are sufficiently powerful to transform any one of these representations into any other of these representations. An Abstract Idea of Encodings ●For simplicity, from this point forward we will make the following assumption: For any finite, discrete object O, it is always possible to find some way of encoding O as a string. ●Think about how an actual computer works with mixed data – everything is 1's and 0's!
●When working with Turing machines, it really doesn't matter how we do the encoding. A TM can convert any reasonable encoding scheme into any other encoding scheme. Notation for Encodings ●For any object O, we will denote a string encoding of O by writing O in angle brackets: O is encoded as ⟨O⟩.
●This makes it much easier to specify languages.
●Examples: { ⟨R⟩ | R is a regular expression that matches ε } { ⟨n⟩ | n ∈ ℕ and the hailstone sequence terminates for n.} ●The encoding scheme can make a difference when trying to determine whether a language is regular or context-free because of the relative weakness of DFAs and CFGs. Encoding Multiple Objects ●Suppose that we want to provide an encoding of multiple objects.
●Two natural numbers and their product.
●A graph and a path in the graph.
●“I just met you” and “this is crazy.” ●We can get encodings of each individual object.
●Can we make one string encoding all of these objects? One Encoding Scheme 0 1 0 0 1 0 0 1 1 0 0 0 0 1 1 0 ⟨X1⟩ ⟨X2⟩ ⟨X1, X2⟩ Encoding Multiple Objects ●Given several different objects O1, …, On, we can represent the encoding of those n objects as ⟨O1, O2, …, On⟩.
●Examples: ●{ ⟨m, n, mn⟩ | m, n ∈ ℕ } ●{ ⟨G, w⟩ | G is a context-free grammar that generates w } Encoding Turing Machines ●Critically important fact: Any Turing machine can be represented as a string.
●One way to do this: encode the TM as a transition table, then write it out one row at a time.
●Stronger claim: Any TM M can be represented as a string in M's own alphabet.
●Analogy: program source code.
●All data fed into a program is encoded using the binary alphabet Σ = {0, 1}.
●A program's source code is itself represented in binary on disk. We can now encode TMs as strings.
TMs accept strings as input.
What can we do with this knowledge? Universal Machines The Universal Turing Machine ●Theorem: There is a Turing machine UTM called the universal Turing machine that, when run on ⟨M, w⟩, where M is a Turing machine and w is a string, simulates M running on w.
●Conceptually: UTM = “On input ⟨M, w⟩, where M is a TM and w ∈ Σ: Set up the initial configuration of M running on w. while (true) { If M accepted w, then UTM accepts ⟨M, w⟩. If M rejected w, then UTM rejects ⟨M, w⟩. Otherwise, simulate one more step of M on w. }” The Universal Turing Machine ●Theorem: There is a Turing machine UTM called the universal Turing machine that, when run on ⟨M, w⟩, where M is a Turing machine and w is a string, simulates M running on w.
●The observable behavior of UTM is the following: ●If M accepts w, then UTM accepts ⟨M, w⟩.
●If M rejects w, then UTM rejects ⟨M, w⟩.
●If M loops on w, then UTM loops on ⟨M, w⟩. An Intuition for UTM ●You can think of UTM as a general-purpose, programmable computer.
●Rather than purchasing one TM for each language, just purchase UTM and program in the “software” corresponding to the TM you actually want.
●UTM is a powerful machine: it can perform any computation that could be performed by any feasible computing device! The Language of UTM ●Recall: For any TM M, the language of M, denoted ( ℒM), is the set ℒ(M) = { w ∈ Σ | M accepts w } ●What is the language of UTM?
●UTM accepts ⟨M, w⟩ iff M is a TM that accepts w.
●Therefore: ℒ(UTM) = { ⟨M, w⟩ | M is a TM and M accepts w } ℒ(UTM) = { ⟨M, w⟩ | M is a TM and w ∈ ℒ(M) } ●For simplicity, define ATM = ℒ(UTM). This is an important language and we'll see it a lot this week. Regular Languages CFLs All Languages RE ATM Time-Out for Announcements! Midterm Logistics ●Second midterm is this Thursday, November 13, from 7PM – 10PM.
●Rooms divvied up by last name: ●Abr – Sad: Go to Cemex Auditorium.
●Sal – Zie: Go to Cubberly Auditorium.
●Exam is closed-computer, closed-book, open one page of notes (double-sided, 8.5” × 11”).
●Cumulative exam, focus is PS4 – PS6.
●Alternate exams: if you've requested an alternate exam time, you should have heard back from Maesen with location information. Contact us ASAP if that isn't the case. Practice Midterm Exam ●We're holding a practice midterm exam tonight in Annenberg Auditorium from 7PM – 10PM.
●Excellent way to get practice for the exam; anecdotally, this really, really seems to help.
●SCPD students: practice exam will be posted online later tonight. Extra Practice Problems ●There are now three sets of extra practice problems up on the course website.
●Solutions to EPP1 available in the filing cabinet.
●Solutions to EPP2 will be available in the filing cabinet later tonight (and at the practice exam.) ●Solutions to EPP3 will be released on Wednesday at the start of class.
●Have questions? Feel free to stop by office hours. We can take questions on these problems!
●Hope this helps! Problem Set 7 ●Problem Set 7 goes out today and is due next Monday at 2:15PM.
●Covers material up through and including today's discussion of LD.
●Explore Turing machines and the limits of computational problem-solving! Your Questions! “Did you get my previous email?” “In class, you mentioned that there is no physical analog to NFAs, and that quantum computers "do not really match" how an NFA works. I've read a bit about QFAs - could you give us a broad idea of how they can solve problems that we otherwise can't?” “How does quantum computing relate to Turing Machines and DFAs? Do the same bounds apply?” “I feel like I am spending the majority of my time and thought on CS related classes and projects. While they are super cool, this also means that I am exploring less. Do you have any advice?” “Could you share an example of a nonregular finite language (if such a thing exists...)?” “In proofs, why do you use 'we' instead of 'I'?” Back to CS103! The Story So Far ●We can now encode arbitrary objects, including Turing machines, as strings.
●Turing machines are capable of running other Turing machines specified through TM encodings.
●This has some deep consequences for what TMs can ever hope to accomplish... Timeline of CS103 ●Lecture 00: Unsolvable problems exist.
●Lecture 11: Proof by diagonalization.
●Lecture 20: TMs formalize computation.
●Lecture 21: TMs can be encoded as strings.
We are finally ready to start answering the following question: What problems cannot be solved by a computer? Languages, TMs, and TM Encodings ●Recall: The language of a TM M is the set ℒ(M) = { w ∈ Σ | M accepts w } ●Some of the strings in this set might be descriptions of TMs.
●What happens if we just focus on the set of strings that are legal TM descriptions? M1 M2 M0 M3 M4 M5 … All Turing machines, listed in some order.
All Turing machines, listed in some order. M1 M2 M0 M3 M4 M5 w0 w1 w2 w3 w4 w5 … … All descriptions of TMs, listed in the same order.
All descriptions of TMs, listed in the same order.
⟨M0⟩⟨M1⟩⟨M2⟩⟨M3⟩⟨M4⟩⟨M5⟩… ⟨M0⟩⟨M1⟩⟨M2⟩⟨M3⟩⟨M4⟩⟨M5⟩… M1 M2 M0 M3 M4 M5 w0 w1 w2 w3 w4 w5 … … Acc No No Acc Acc No … Acc Acc Acc Acc Acc Acc … Acc Acc Acc Acc Acc Acc … No Acc Acc No Acc Acc … Acc No Acc No Acc No … No No Acc Acc No No … … … … … … … … Acc Acc Acc No Acc No … ⟨M0⟩⟨M1⟩⟨M2⟩⟨M3⟩⟨M4⟩⟨M5⟩… ⟨M0⟩⟨M1⟩⟨M2⟩⟨M3⟩⟨M4⟩⟨M5⟩… M1 M2 M0 M3 M4 M5 w0 w1 w2 w3 w4 w5 … … Acc No No Acc Acc No … Acc Acc Acc Acc Acc Acc … Acc Acc Acc Acc Acc Acc … No Acc Acc No Acc Acc … Acc No Acc No Acc No … No No Acc Acc No No … … … … … … … … Acc Acc Acc No Acc No … ⟨M0⟩⟨M1⟩⟨M2⟩⟨M3⟩⟨M4⟩⟨M5⟩… Flip all “accept” to “no” and vice-versa Flip all “accept” to “no” and vice-versa No No No Acc No Acc … ⟨M0⟩⟨M1⟩⟨M2⟩⟨M3⟩⟨M4⟩⟨M5⟩… … … No Acc … … Acc No No Acc Acc No M1 M2 M0 M3 M4 M5 w0 w1 w2 w3 w4 w5 … … Acc No No Acc Acc No … Acc Acc Acc Acc Acc Acc … Acc Acc Acc Acc Acc Acc … No Acc Acc No Acc Acc … Acc No Acc No No … No No Acc Acc No … … … … … … … No No No Acc No TM has this behavior!
No TM has this behavior!
⟨M0⟩⟨M1⟩⟨M2⟩⟨M3⟩⟨M4⟩⟨M5⟩… ⟨M0⟩⟨M1⟩⟨M2⟩⟨M3⟩⟨M4⟩⟨M5⟩… … … No Acc … … Acc No No Acc Acc No M1 M2 M0 M3 M4 M5 w0 w1 w2 w3 w4 w5 … … Acc No No Acc Acc No … Acc Acc Acc Acc Acc Acc … Acc Acc Acc Acc Acc Acc … No Acc Acc No Acc Acc … Acc No Acc No No … No No Acc Acc No … … … … … … … No No No Acc ⟨M0⟩⟨M1⟩⟨M2⟩⟨M3⟩⟨M4⟩⟨M5⟩… { ⟨M⟩ | M is a TM and ⟨M⟩ ∉ ℒ(M) } ⟨M0⟩⟨M1⟩⟨M2⟩⟨M3⟩⟨M4⟩⟨M5⟩… Diagonalization Revisited ●The diagonalization language, which we denote LD, is defined as LD = { ⟨M⟩ | M is a TM and ⟨M⟩ ∉ ℒ(M) } ●That is, LD is the set of descriptions of Turing machines that do not accept themselves. ●This is a hugely important language and we'll see it a lot this week. LD = { ⟨M⟩ | M is a TM and ⟨M⟩ ∉ ℒ(M) } Theorem: LD ∉ RE.
Proof: By contradiction; assume that LD ∈ RE. Then there must be some TM R such that ( ℒR) = LD. Since ( ℒR) = LD, we know that if M is any TM, then ⟨M⟩ ∈ LD iff ⟨M⟩ ∈ ( ℒR) (1) From the definition of LD, we see that ⟨M⟩ ∈ LD iff ⟨M⟩ ∉ ( ℒM).
Combining this with statement (1) tells us that ⟨M⟩ ∉ ( ℒM) iff ⟨M⟩ ∈ ( ℒR) (2) Statement (2) holds for any TM M, so in particular it should hold when M = R. If we pick M = R, we see that ⟨R⟩ ∉ ( ℒR) iff ⟨R⟩ ∈ ( ℒR) (3) This is clearly impossible. We have reached a contradiction, so our assumption must have been wrong. Thus LD ∉ RE. ■ Regular Languages CFLs All Languages RE LD Why is LD ∉ RE?
●Intuitively, any TM for LD would have to be wrong on itself.
●If the TM accepts itself, then it doesn't belong to LD, so it shouldn't accept itself.
●If the TM doesn't accept itself, then it belongs to LD, so it should accept itself.
●This is indirect self-reference: the language LD isn't directly self-referential, but it still causes problems due to self-reference. |
4919 | https://www.science.org/doi/10.1126/science.adp1853 | Skip to main content
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A geological timescale for bacterial evolution and oxygen adaptation
Adrián A. Davín Ben J. Woodcroft [...] , Rochelle M. Soo Benoit Morel [...] , Ranjani Murali Dominik Schrempf James W. Clark Sandra Álvarez-Carretero Bastien Boussau [...] , Edmund R. R. Moody Lénárd L. Szánthó Etienne Richy Davide Pisani James Hemp Woodward W. Fischer Philip C. J. Donoghue Anja Spang Philip Hugenholtz Tom A. Williams and Gergely J. Szöllősi authors +15 authors +10 authors fewerAuthors Info & Affiliations
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Vol 388, Issue 6742
DOI: 10.1126/science.adp1853
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Editor’s summary
When exploring deep time, the problem is that there are few, if any, good fossils of the earliest living organisms, and it is impossible to precisely date the evolution of those that do exist. One calibration point is provided by the impact event about 4.5 billion years ago that resulted in sterilization of Earth and formation of the Moon. Davín et al. used molecular clocks, machine learning, and phylogenetic reconciliation to present a reconstruction of the evolution of Earth’s bacterial biosphere over the past 4 billion years with particular emphasis on aerobic metabolisms. Their analysis showed that the last common ancestor of bacteria likely existed 4.4 to 3.9 billion years ago, and aerobic organisms likely emerged before the Great Oxidation Event (2.43 to 2.33 billion years ago). Oxygen tolerance may have been a prerequisite for, rather than a consequence of, the evolution of oxygenic photosynthesis. —Caroline Ash
Structured Abstract
INTRODUCTION
Microbial life dominates the biosphere, but a timescale of early microbial evolution has proven elusive as a result of an inadequate fossil record. The lack of maximum age calibrations—the earliest point in time at which a given group might have emerged—is particularly problematic. However, the geochemical record bears the imprint of microbial metabolism through time, providing a complementary source of information. A pivotal event in this history was the Great Oxidation Event (GOE) ~2.43 to 2.33 billion years ago (Ga), which marked a substantial increase in atmospheric oxygen. This transition, driven by the evolution of cyanobacterial oxygenic photosynthesis and carbon burial, transformed the biosphere from predominantly anoxic to oxic, causing widespread adaptation to oxygen. In this study, we used the temporal link between atmospheric oxygenation and the evolutionary spread of aerobic metabolism to calibrate the phylogeny of the bacterial domain.
RATIONALE
To date the bacterial tree, we introduced multiple new maximum age calibrations by linking the GOE to the age of aerobic lineages. We used a Bayesian approach that assumes that aerobic nodes are unlikely to be older than the GOE but can predate it given sufficient evidence from fossils or sequence divergence. To implement this approach, we integrated phylogenetic reconciliation with machine learning to map transitions from anaerobic to aerobic lifestyles onto the bacterial tree. By aggregating signals across the genome, we could robustly infer aerobic and anaerobic phenotypes from incomplete ancestral gene repertoires.
RESULTS
We identified 84 anaerobic to aerobic transitions on a species tree of 1007 bacteria. Most transitions occurred after the GOE and were driven by horizontal acquisition of respiratory and oxygen tolerance genes. However, despite the GOE calibration, at least three transitions predated this event, suggesting that aerobic respiration evolved before widespread atmospheric oxygenation and may have facilitated the evolution of oxygenic photosynthesis in cyanobacteria. Our molecular clock analyses estimated that the last bacterial common ancestor lived in the Hadean or earliest Archaean era (4.4 to 3.9 Ga), whereas bacterial phyla originated in the Archaean and Proterozoic eras (2.5 to 1.8 Ga); most bacterial families are as old as land plants and animal phyla, dating back to the late Proterozoic (0.6 to 0.75 Ga).
CONCLUSION
We infer that the earliest aerobic bacteria emerged in the Archaean, predating the GOE by 900 million years. After the GOE, aerobic lineages experienced faster diversification than their anaerobic counterparts, highlighting the impact of atmospheric oxygenation on bacterial evolution. The approach developed here provides a framework for linking microbial traits to Earth’s geochemical history, offering a pathway for exploring the evolution of other phenotypes in the context of Earth’s history.
Abstract
Microbial life has dominated Earth’s history but left a sparse fossil record, greatly hindering our understanding of evolution in deep time. However, bacterial metabolism has left signatures in the geochemical record, most conspicuously the Great Oxidation Event (GOE). We combine machine learning and phylogenetic reconciliation to infer ancestral bacterial transitions to aerobic lifestyles, linking them to the GOE to calibrate the bacterial time tree. Extant bacterial phyla trace their diversity to the Archaean and Proterozoic, and bacterial families prior to the Phanerozoic. We infer that most bacterial phyla were ancestrally anaerobic and adopted aerobic lifestyles after the GOE. However, in the cyanobacterial ancestor, aerobic metabolism likely predated the GOE, which may have facilitated the evolution of oxygenic photosynthesis.
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Volume 388 | Issue 6742
4 April 2025
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Copyright © 2025 The Authors, some rights reserved; exclusive licensee American Association for the Advancement of Science. No claim to original U.S. Government Works.
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Received: 12 March 2024
Accepted: 19 December 2024
Published in print: 4 April 2025
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Acknowledgments
We thank J. A. Palacios for providing R scripts to compute the D2 distance for ranked trees. We are grateful for the help and support provided by the Scientific Computing and Data Analysis section of Core Facilities at OIST.
Funding:
This work was funded by the following: European Research Council (ERC) grant 714774 “GENECLOCKS” (to A.A.D., L.L.S., D.S., and G.J.S.) under the European Union’s Horizon 2020 research and innovation program; ERC grant 947317” ASymbEL” (to A.S.) under the European Union’s Horizon 2020 research and innovation program; Gordon and Betty Moore Foundation grant GBMF9741 (to T.A.W., A.S., P.C.J.D., and G.J.S.); Gordon and Betty Moore Foundation’s Symbiosis in Aquatic Systems Initiative grant GBMF9346 (to A.S.); Moore–Simons Project on the Origin of the Eukaryotic Cell, Simons Foundation 735929LPI grant 735929LP (to A.S.); Royal Society University Research Fellowship (to T.A.W.); John Templeton Foundation grant 62220 (to E.R.R.M., D.P., P.C.J.D., and T.A.W.); Leverhulme Trust Research Fellowship grant RF-2022-167 (to P.C.J.D.); Biotechnology and Biological Sciences Research Council grants BB/T012773/1 and BB/Y003624/1 (to P.C.J.D.); Australian Research Council (ARC) Future Fellowship grant FT210100521 (to B.J.W.); ARC Discovery Project grant DP230101171 (to B.J.W.); ARC Discovery Early Career Researcher Award grant DE190100008 (to R.M.S.); ARC Laureate Fellowship grant FL150100038 (to A.A.D. and P.H.); ARC Discovery Project grant DP220100900 (to A.A.D. and P.H.).
Author Contributions:
Conceptualization: A.A.D., A.S., B.B., B.J.W., D.P., E.R.R.M., G.J.S., J.H., P.C.J.D., P.H., R.M., R.M.S., T.A.W., and W.W.F. Method development and coding: A.A.D., B.B., B.J.W., B.M., D.S., G.J.S., L.L.S., and T.A.W. Taxon selection and genomic data preparation: A.A.D., A.S., P.H., and T.A.W. Phylogenetic analyses: A.A.D., B.B., E.R.R.M., G.J.S., L.L.S., and T.A.W. Fossil Calibrations: D.P., P.C.J.D., and T.A.W. Molecular dating: D.S., G.J.S., L.L.S., and S.Á.-C. Machine Learning: B.B., B.J.W., and G.J.S. HCO curation and analysis: R.M., R.M.S., and J.H. Visualization: A.A.D., A.S., B.J.W., E.R., G.J.S., J.W.C., and T.A.W. Writing: A.A.D., A.S., B.B., B.J.W., B.M., D.P., D.S., E.R., E.R.R.M., G.J.S., J.W.C., J.H., L.L.S., P.C.J.D., P.H., R.M., R.M.S., S.Á.-C., T.A.W., and W.W.F.
Competing Interests:
R.M. is currently affiliated with Dept Life Sciences, University of Nevada, Las Vegas. All other authors declare no other competing interests.
Data and Materials Availability:
All data used are described in detail in the SM and available in the figshare repository associated with the submission (92). All code used is described in detail in the SM and available either openly on github or other relevant public repository, as indicated, or included in the figshare repository (92) associated with the submission in the case of more specialized scripts.
License Information:
Copyright © 2025 the authors, some rights reserved; exclusive licensee American Association for the Advancement of Science. No claim to original US government works.
Authors
Affiliations
Adrián A. Davín aaredav@gmail.com
The University of Queensland, School of Chemistry and Molecular Biosciences, Australian Centre for Ecogenomics, Brisbane, Queensland, Australia.
Department of Biological Physics, Eötvös Loránd University, Budapest, Hungary.
Department of Biological Sciences, Graduate School of Science, University of Tokyo, Tokyo, Japan.
Roles: Conceptualization, Formal analysis, Investigation, Methodology, Project administration, Software, Supervision, Validation, Visualization, Writing - original draft, and Writing - review & editing.
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Ben J. Woodcroft aaredav@gmail.com
Centre for Microbiome Research, School of Biomedical Sciences, Queensland University of Technology (QUT), Translational Research Institute, Woolloongabba, Australia.
Roles: Conceptualization, Data curation, Formal analysis, Funding acquisition, Investigation, Methodology, Resources, Software, Validation, Visualization, Writing - original draft, and Writing - review & editing.
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Rochelle M. Soo
The University of Queensland, School of Chemistry and Molecular Biosciences, Australian Centre for Ecogenomics, Brisbane, Queensland, Australia.
Roles: Conceptualization, Investigation, Writing - original draft, and Writing - review & editing.
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Benoit Morel
Computational Molecular Evolution Group, Heidelberg Institute for Theoretical Studies, Heidelberg, Germany.
Institute for Theoretical Informatics, Karlsruhe Institute of Technology, Karlsruhe, Germany.
Roles: Software and Writing - review & editing.
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Ranjani Murali
Division of Geological and Planetary Sciences, California Institute of Technology, Pasadena, CA, USA.
Roles: Formal analysis, Methodology, Resources, Validation, and Writing - review & editing.
View all articles by this author
Dominik Schrempf
Department of Biological Physics, Eötvös Loránd University, Budapest, Hungary.
MTA-ELTE “Lendület” Evolutionary Genomics Research Group, Budapest, Hungary.
Roles: Formal analysis, Methodology, Software, Validation, and Writing - review & editing.
View all articles by this author
James W. Clark
Bristol Palaeobiology Group, School of Earth Sciences, University of Bristol, Bristol, UK.
Milner Centre for Evolution, Department of Life Sciences, University of Bath, Bath, UK.
Roles: Formal analysis, Methodology, Visualization, and Writing - review & editing.
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Sandra Álvarez-Carretero
Bristol Palaeobiology Group, School of Earth Sciences, University of Bristol, Bristol, UK.
Roles: Formal analysis, Software, Validation, and Writing - review & editing.
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Bastien Boussau
Laboratoire de Biométrie et Biologie Evolutive, Univ Lyon, Univ Lyon 1, CNRS, VetAgro Sup, Villeurbanne, France.
Roles: Conceptualization, Formal analysis, Investigation, Methodology, Software, and Writing - review & editing.
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Edmund R. R. Moody
Bristol Palaeobiology Group, School of Earth Sciences, University of Bristol, Bristol, UK.
School of Biological Sciences, University of Bristol, Bristol, UK.
Roles: Formal analysis, Investigation, Writing - original draft, and Writing - review & editing.
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Lénárd L. Szánthó
Department of Biological Physics, Eötvös Loránd University, Budapest, Hungary.
Institute of Evolution, Centre for Ecological Research, Budapest, Hungary.
Model-Based Evolutionary Genomics Unit, Okinawa Institute of Science and Technology Graduate University, Okinawa, Japan.
Roles: Data curation, Investigation, Resources, Software, Validation, Writing - original draft, and Writing - review & editing.
View all articles by this author
Etienne Richy
School of Biological Sciences, University of Bristol, Bristol, UK.
Roles: Formal analysis and Writing - review & editing.
View all articles by this author
Davide Pisani
Bristol Palaeobiology Group, School of Earth Sciences, University of Bristol, Bristol, UK.
School of Biological Sciences, University of Bristol, Bristol, UK.
Roles: Conceptualization, Supervision, Validation, Writing - original draft, and Writing - review & editing.
View all articles by this author
James Hemp
Metrodora Institute, West Valley City, UT, USA.
View all articles by this author
Woodward W. Fischer
Division of Geological and Planetary Sciences, California Institute of Technology, Pasadena, CA, USA.
View all articles by this author
Philip C. J. Donoghue
Bristol Palaeobiology Group, School of Earth Sciences, University of Bristol, Bristol, UK.
View all articles by this author
Anja Spang
Department of Marine Microbiology and Biogeochemistry, NIOZ, Royal Netherlands Institute for Sea Research, Den Burg, Netherlands.
Department of Evolutionary & Population Biology, Institute for Biodiversity and Ecosystem Dynamics (IBED), University of Amsterdam, Amsterdam, Netherlands.
View all articles by this author
Philip Hugenholtz aaredav@gmail.com
The University of Queensland, School of Chemistry and Molecular Biosciences, Australian Centre for Ecogenomics, Brisbane, Queensland, Australia.
View all articles by this author
Tom A. Williams,† aaredav@gmail.com
School of Biological Sciences, University of Bristol, Bristol, UK.
View all articles by this author
Gergely J. Szöllősi,† aaredav@gmail.com
Department of Biological Physics, Eötvös Loránd University, Budapest, Hungary.
MTA-ELTE “Lendület” Evolutionary Genomics Research Group, Budapest, Hungary.
Institute of Evolution, Centre for Ecological Research, Budapest, Hungary.
Model-Based Evolutionary Genomics Unit, Okinawa Institute of Science and Technology Graduate University, Okinawa, Japan.
View all articles by this author
Funding Information
John Templeton Foundation: 62220
Gordon and Betty Moore Foundation: GBMF9741
Gordon and Betty Moore Foundation: GBMF9346
Horizon 2020 Framework Programme: 714774
Horizon 2020 Framework Programme: 947317
Biotechnology and Biological Sciences Research Council: BB/T012773/1
Australian Resuscitation Council: FT210100521
Australian Resuscitation Council: DE190100008
Australian Resuscitation Council: DP220100900
Biotechnology and Biological Sciences Research Council: BB/Y003624/1
Agricultural Research Center: FT210100521
Agricultural Research Center: DE190100008
Agricultural Research Center: DP220100900
Leverhulme Trust Research Fellowship: RF-2022-167
Australian Research Council Future Fellowship: FT210100521
Notes
Corresponding author. Email: aaredav@gmail.com (A.A.D.); b.woodcroft@qut.edu.au (B.J.W.); p.hugenholtz@uq.edu.au (P.H.); thwillia@gmail.com (T.A.W.); gergely.szollosi@oist.jp (G.J.S.)
These authors contributed equally to this work.
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A geological timescale for bacterial evolution and oxygen adaptation.Science388,eadp1853(2025).DOI:10.1126/science.adp1853
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Figures
An integrated approach to date bacterial evolution and reconstruct the history of oxygen adaptation.
We inferred a bacterial timetree by integrating genomic, fossil, and geochemical data and linking oxygen tolerance and aerobic metabolism to the GOE. Colors denote anaerobic (blue) and aerobic (red) states, whereas shades of purple show the fraction of aerobic lineages within extant bacterial phyla. Mitochondria and plastids were included to leverage the more extensive eukaryotic fossils. Land plants and animals are indicated for temporal comparison.
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4920 | https://books.google.com/books/about/Koss_Diagnostic_Cytology_and_Its_Histopa.html?id=yUvhjjShhpEC | Koss' Diagnostic Cytology and Its Histopathologic Bases - Google Books
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Koss' Diagnostic Cytology and Its Histopathologic Bases, Volume 1 ================================================================= Leopold G. Koss, Myron R. Melamed Lippincott Williams & Wilkins, 2006 - Medical - 1752 pages The most influential and frequently cited pathology classic is now in its Fifth Edition, with thoroughly revised chapters and over 3,000 brand-new full-color illustrations. This two-volume work provides comprehensive, current information on the principles and techniques of cytopathology and the cytologic evaluation of benign and malignant disorders at every anatomic site. This edition provides greatly expanded coverage of the interpretation of aspirated cell samples. Innovations in the practice of cytopathology and data on molecular biology and cytogenetics have been incorporated into the organ system chapters. This edition also has a greater focus on avoiding diagnostic errors. A bound-in image bank DVD is included in this edition. More » Preview this book »
Selected pages
Title Page
Table of Contents
Index
References
Contents
Cerebrospinal and Miscellaneous Fluids 1023
Techniques of FineNeedle Aspiration Smear 1056
The Breast 1081
The Thyroid Parathyroid and Neck Masses Other 1148
Lymph Nodes 1186
Salivary Glands 1229
The Prostate and the Testis 1262
The Skin 1286
The Eyelids Orbit and Eye 1508
The Central Nervous System 1523
Circulating Cancer Cells 1544
Laboratory Techniques 1569
The Basic Structure of the Mammalian Cell 21 1635
Principles of Cytogenetics 79 1671
Digital Analysis of Cells and Tissues 1681
Recognizing and Classifying Cells 119 1709
More
Soft Tissue Lesions 1302
Bone Tumors 1340
The Mediastinum 1375
The Liver and Spleen 1389
Diseases of the Vagina Vulva Perineum 1417
The Kidneys Adrenals 1457
Tumors of the Ovary and Fallopian Tube 491 1495
Its Origins and Principles 3 1504
DIAGNOSTIC CYTOLOGY 1734
Index1-1
Benign Disorders of the Uterine Cervix1-3
Proliferative Disorders and Carcinoma of1-13
VOLUME II1-36
Effusions in the Absence of Cancer 9191-51
Copyright
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Koss' Diagnostic Cytology and Its Histopathologic Bases, Volume 1
Leopold G. Koss,Myron R. Melamed
No preview available - 2006
Koss' Diagnostic Cytology and Its Histopathologic Bases
Leopold G. Koss
No preview available - 2015
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Urinzytologie: Praxis und Atlas
Peter Rathert,Stephan Roth
No preview available - 2007
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Title Koss' Diagnostic Cytology and Its Histopathologic Bases, Volume 1
Koss' Diagnostic Cytology and Its Histopathologic Bases
EditorsLeopold G. Koss, Myron R. Melamed
Edition illustrated
Publisher Lippincott Williams & Wilkins, 2006
ISBN 0781719283, 9780781719285
Length 1752 pages
SubjectsMedical
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Medical / Diagnosis
Medical / Histology
Medical / Pathology
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4921 | https://math.stackexchange.com/questions/2118581/lhopitals-rule-and-frac-sin-xx | Stack Exchange Network
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L'Hopital's rule and $\frac{\sin x}x$
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$\begingroup$
I have heard people say that you can't (or shouldn't) use the L'Hopital's rule to calculate $\lim\limits_{x\to 0}\frac{\sin x}x=\lim\limits_{x\to 0}\cos x=1$, because the result $\frac d{dx}\sin x=\cos x$ is derived by using that limit.
But is that opinion justified? Why should I be vary of applying L'Hopital's rule to that limit?
I don't see any problem with it. The sine function fulfills the conditions of the L'Hopital's rule.
Also, it is a fact that the derivative of sine is cosine, no matter how we proved it. Certainly there is a way to prove $\frac d{dx}\sin x=\cos x$ without using the said limit (if someone knows how, they can post it) so we don't even have any circular logic. Even if there isn't, $\frac d{dx}\sin x=\cos x$ was proven sometime without referencing the L'Hopital's rule so we know it is true. Why wouldn't we then freely apply the L'Hopital's rule to $\frac {\sin x}x$?
PS I'm not saying that this is the best method to derive the limit or anything, but that I don't understand why it is so frowned upon and often considered invalid.
calculus
real-analysis
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edited Jan 29, 2017 at 0:52
BlazaBlaza
asked Jan 28, 2017 at 23:55
BlazaBlaza
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$\begingroup$ Once you've historically shown the limit / derivative without l'Hopital, you are principally allowed to use it here as well. Just don't do it before you ever have established what the derivative of $\sin x $ is. Then again, $\lim_{x\to0}\frac {\sin x}x=\cos 0 =1$ seems to use once limit rule less. $\endgroup$
Hagen von Eitzen
– Hagen von Eitzen
2017-01-28 23:59:59 +00:00
Commented Jan 28, 2017 at 23:59
4
$\begingroup$ It's OK to do it, as long as you don't mind looking silly when you do. $\endgroup$
zhw.
– zhw.
2017-01-29 00:04:40 +00:00
Commented Jan 29, 2017 at 0:04
$\begingroup$ Why would someone look silly when they do it? $\endgroup$
Blaza
– Blaza
2017-01-29 00:22:23 +00:00
Commented Jan 29, 2017 at 0:22
$\begingroup$ It's like Tex Avery's kangaroo, hiding in its own pouch when it's frightened¦ $\endgroup$
Bernard
– Bernard
2017-01-29 00:53:08 +00:00
Commented Jan 29, 2017 at 0:53
2
$\begingroup$ Because it's then clear the someone doesn't really know why that limit is $1.$ Instead, someone appeals to a result that depends on the limit being $1$ to show the limit is $1.$ Our someone is starting to sound like the butt of a Monty-Python joke. Why are such someones silly? Ask John Cleese. I had a litte rant on this topic here: math.stackexchange.com/questions/1286699/… $\endgroup$
zhw.
– zhw.
2017-01-29 00:59:38 +00:00
Commented Jan 29, 2017 at 0:59
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5 Answers 5
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17
$\begingroup$
It is a correct application of l'Hôpital's rule, but using it to prove that the derivative of $\sin x$ is $\cos x$ is possibly circular logic, depending on your definition of $\sin x$.
If we did not know what the derivative of $\sin x$ was, but we did know the angle sum formula, here are some steps we could take to compute that derivative, starting from first principles, i.e. the definition of the derivative.
$$ (\sin x)'= \lim_{h\to 0}\dfrac{\sin(x+h)-\sin x}{h} = \lim_{h\to 0}\dfrac{\sin x\cos h+\cos x\sin h-\sin x}{h} \ = \cos x\cdot\left(\lim_{h\to 0}\dfrac{\sin h}{h}\right)+ \sin x\cdot\left(\lim_{h\to 0}\dfrac{\cos h -1}{h}\right) $$
To complete the computation of the derivative of $\sin x$, we must first know the limit of $\sin h/h$ and $(\cos h-1)/h$. We can't use l'Hôpital at this point because to use l'Hôpital you need to know the derivative of $\sin x$ at $0$, which is the very thing we are trying to compute. Assuming what you're trying to prove is a logical error known as "begging the question". Mathematics is axiomatic, so that every result builds on previous results, and no arguments are circular.
But if we can confirm the derivative of $\sin x$ by some other means, for example by defining it in terms of its Taylor series or a differential equation, then we may use it in a l'Hôpital computation to derive $\lim \sin x/x$ without fear of being circular. While not logically incorrect, this would still be fairly redundant: if you know the derivative of $\sin x$ just observe that, by definition, $\lim_{h\to 0}\sin x/x$ is just the derivative at 0. Using the machinery of l'Hôpital is overkill to get an answer you already know.
However if you're in a situation where you don't care about logical foundations and rigor, you're not proving theorems from first principles about calculus of trigonometric functions, you just need to compute $\lim\sin x/x$ and want to allow all known formulas and techniques, feel free to use l'Hôpital. It is correct.
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edited Feb 5, 2021 at 9:55
answered Jan 29, 2017 at 0:10
ziggurismziggurism
17.5k22 gold badges5656 silver badges116116 bronze badges
$\endgroup$
9
$\begingroup$ If using l'Hopital to prove that the derivative of $\sin$ is $\cos$ is circular logic, how is using knowledge of the derivative of $\sin$ in order to use l'Hopital not circular? $\endgroup$
Arthur
– Arthur
2017-01-29 00:14:47 +00:00
Commented Jan 29, 2017 at 0:14
$\begingroup$ It isn't because you don't need l'Hopital to prove that the derivative is $\cos x$ $\endgroup$
Blaza
– Blaza
2017-01-29 00:17:55 +00:00
Commented Jan 29, 2017 at 0:17
$\begingroup$ @Blaza hi I added some details to the answer. Is it clearer why it is circular? You do need to know limit of $\sin h/h$, to compute derivative of $\sin x$, and using lhopital to compute it is circular. $\endgroup$
ziggurism
– ziggurism
2017-01-29 00:46:19 +00:00
Commented Jan 29, 2017 at 0:46
7
$\begingroup$ Supposing that you know the derivative of $\sin x$ is $\cos x$ by some means. Then it is a very roundabout way (but correct) to use L'Hospital's Rule to find the limit under question because by definition that limit is the derivative of $\sin x$ at $x=0$ and that is $\cos 0=1$. Remember that L'Hospital's Rule should not be used to calculate the limits of type $$\lim_{x\to a} \frac{f(x) - f(a)} {x - a} $$ because in the act of finding derivative of numerator you have already computed this particular limit and hence why go further. $\endgroup$
Paramanand Singh
– Paramanand Singh ♦
2017-01-29 04:04:13 +00:00
Commented Jan 29, 2017 at 4:04
$\begingroup$ @ParamanandSingh yes, that's a good point. $\endgroup$
ziggurism
– ziggurism
2017-01-29 04:23:03 +00:00
Commented Jan 29, 2017 at 4:23
| Show 4 more comments
8
$\begingroup$
I expand my comment to ziggurism's answer here.
Suppose that by a lot of hardwork, patience and ... (add more nice words if you like) I have obtained these three facts:
$\cos x$ is continuous.
$\dfrac{d}{dx}(\sin x) = \cos x$
L'Hospital's Rule
and my next goal is establish the following fact: $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{}$$ Because of all that hardwork, patience and ... I know that my goal $()$ is an immediate consequence of just the fact $(2)$ alone stated above. Then why would I combine all the three facts mentioned above to achieve my goal? To borrow an idea from user "zhw.", wouldn't a person doing this would be considered silly?
More often than not, many students don't really understand what's going behind the scenes when we use the mantra of "differentiate and plug" championed by L'Hospital's Rule. The act of differentiation itself entails that we know certain limits (and rules of differentiation) and further most of the derivatives are continuous (so that plugging works after differentiation step).
If one is so fond of L'Hospital's Rule why not put that to a better use to solve complex problems (like this and this) instead of using it to obtain limits which are immediate consequences of differentiation formulas.
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edited Apr 13, 2017 at 12:21
CommunityBot
1
answered Jan 29, 2017 at 4:35
Paramanand Singh♦Paramanand Singh
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$\endgroup$
11
$\begingroup$ Yeah, I know, it's like killing a fly with a canon. It's silly in that regard, but not necessarily incorrect, that was my point. Thanks for the answer. $\endgroup$
Blaza
– Blaza
2017-01-29 10:01:44 +00:00
Commented Jan 29, 2017 at 10:01
5
$\begingroup$ @Blaza: the situation is not like using canon to kill a fly, but rather it's like having killed the fly (by squeezing it into your hands) and then using canon on the fly's corpse. $\endgroup$
Paramanand Singh
– Paramanand Singh ♦
2017-01-29 10:48:15 +00:00
Commented Jan 29, 2017 at 10:48
$\begingroup$ That's quite a picturesque analogy. I agree, thanks $\endgroup$
Blaza
– Blaza
2017-01-29 11:11:23 +00:00
Commented Jan 29, 2017 at 11:11
$\begingroup$ @Blaza: actually I am not that an expert in creating analogies but I am happy that I was able to get my point across via some sort of analogy. And I am glad that you found it picturesque. $\endgroup$
Paramanand Singh
– Paramanand Singh ♦
2017-01-29 15:32:16 +00:00
Commented Jan 29, 2017 at 15:32
2
$\begingroup$ Nice post. +1. "Then why would I combine all the three facts mentioned above to achieve my goal?" In order to get ($$) as an immediate consequence of (2) one needs to recognize that (i) $\frac{\sin x}{x}=\frac{\sin x- \sin 0}{x-0}$ (ii) by definition, $\sin'(0)=\lim_{x\to 0}\frac{\sin x- \sin 0}{x-0}$; inexperienced students may find applying L'Hopital more "straightforward" than identifying (i) and (ii). $\endgroup$
user587192
– user587192
2018-12-22 15:45:12 +00:00
Commented Dec 22, 2018 at 15:45
| Show 6 more comments
6
$\begingroup$
My personal favorite definition of sine and cosine come from (I think) Apostol's book, where he says
$\sin, \cos : \Bbb R \to [-1, 1]$
$\sin, \cos$ are continuous
$\cos(\pi/2) = 0$
$\cos(a-b) = \cos(a) \cos (b) + \sin(a) \sin(b)$
...and perhaps there's a fourth one. From these, it's easy to show (pick $a = b = 0$) that $$ \cos(0) = 1 \ $$ and then picking $a = b$, to get that $$ \sin^2(t) + \cos^2(t) = 1 $$ for all $t$,
and picking $a = -b$, to derive the double-angle formula for cosine, and from that, the half-angle formula, and similar ones for sine.
From these, you can establish that if there are functions that satisfy these rules, then we can compute their values on $S = \Bbb Q\pi$, the set of all rational multiples of $\pi$. It's not too hard to do that, either. And then, with somewhat greater effort, you can show that these restrictions to $S$ are continuous.
Now you have continuous functions on $S$, which is evidently dense in $\Bbb R$, and hence there are unique continuous extensions of these functions to $\Bbb R$.
Finally, the cosine and sine addition and subtraction rules can be used to show that on $S$ (and hence on $\Bbb R$ as well), $\sin x \approx x$ for $x$ small, whence the limit we all love turns out to be one.
I'm pretty sure the details are all in Apostol's calculus.
An alternative approach is to define arcsine and arccos via integrals, observe that they're increasing with nonzero derivatives, and therefore have differentiable inverses which we define to be sine and cosine.
Any one of these avoids using L'h's rule to evaluate the limit. Hence using it to evaluate the limit later is just fine.
Of course, both these approaches require that you have some topology under your belt before you start calculus, and/or that you're willing to postpone the definition of trig functions until after you've defined integration, so they're not typical. That doesn't, however, make them wrong.
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answered Jan 29, 2017 at 0:38
John HughesJohn Hughes
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How do you prove that $\frac{d}{dx}[\sin(x)]=\cos(x)$? If you use the limit $$\lim_{x\to 0}\frac{\sin(x)}{x}=1 \qquad\qquad(1.1)$$ than you can't use the L'Hopital rule, because it is an example of circular logic.
You can avoid it, definying $$\sin(x)=\sum_{n=0}^{+\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\ \ \ \ \forall x\in\mathbb{R}$$ and $$\cos(x)=\sum_{n=0}^{+\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}\ \ \ \forall x\in\mathbb{R}$$.
Once you proved that $\frac{d}{dx}[\sin(x)]=\cos(x)$ and $\cos(0)=1$ than you can use the l'Hopital rule to calculate the limit (1.1).
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edited Jan 29, 2017 at 1:05
Harry
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answered Jan 29, 2017 at 0:29
IxionIxion
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Proving $\mathrm D \sin=\cos$ does not necessarily involve the use of l'Hopital if you define sine as a power series like Rudin.
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answered Jan 29, 2017 at 2:02
Henricus V.Henricus V.
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4922 | https://www.youtube.com/watch?v=vtzMopWOGKg | Chords on a Parabola (2 of 2: Exploring the focal chord visually)
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now on the reference sheet and you if you want you can pull it up right now there are a bunch of equations and i've deliberately not shown you them yet um related to all of this stuff this equation here it's not one of them okay here's why what was it well knowledge was required to prove this result right it's not a rhetorical question what do we need to know you need to find the equation of a line well two-point formula that's that's part of the course you need to know that okay what else do you need to know we need to know where the points are but that's part of the definition of what this is something you expected to know so therefore this process you actually have to be able to recreate okay and people don't necessarily like that welcome to the extension one course okay now that's nice but so what well let me show you some nice a really really nice piece of geometry here okay um this is not just any parabola this is the parabola in a locus form right with like focal length and directrix and all that kind of thing okay speaking of focal length where is the focus on this parabola where is it what are the coordinates of the focus if the focus is like say here what were those coordinates being what's the x coordinate zero very good and the y coordinate would be if it's the focal length you go up from the vertex and you go a units right so this should be zero comma a yeah now watch what happens some chords go through the focus some of them unoriginally these chords are called vocal chords a focal chord is a chord that goes through the focus if that thing goes through there then this is really badly messily drawn coordinates sorry that's a little more legible those coordinates ought to satisfy this equation do you agree okay so therefore let's that into the equation of chord and just see what happens right into equation of chord okay 0 comma a it's really really simple because 0 is just going to make this term disappear what comes up on the left hand side have a look it's just going to be a what's on the right hand side this becomes 0 take away what just apq okay what would i'm going to do i think we should divide through by a we can do one more thing what were p and q again they're they're the parameters at these particular spots aren't they what is the parameter again we proved this like 20 minutes ago the parameter by definition is the gradient at that point the gradient at that point is little q the gradient at that point is little p what does it mean when gradients are multiplying together to give negative one they're perpendicular let me show you what that looks like so yep what are we trying to find we're just exploring i'm not trying to find anything but i'm going to show you something that we just found okay here we go so don't worry too much about all of my equations over here on the left hand side that's just to make stuff work okay is that thick i can't remember if i made it thick for you yeah it's good okay now have a look at what we've just got here right um see this black line going right the way through i could have been a little more fancy to restrict it but that's the chord okay i know it's a line at the moment just picture it as an interval okay here's here's our um our chord and it's going through the focus here right so can you see this blue line is the tangent at one of like that's that's q for instance and my orange line is my tangent at p so can you see where the tangents mean do you see there's the right angle there now watch what happens as we move i think this is what i want to move there we go let's slow that down that's going way too fast let's bring it back oh i want to change direction that's what i wanted to do okay so can you see this point here see this point here that's the focus right there that's why the chord no matter where it goes it's always passing through this focus okay so you can see these tangents here are always at right angles right everywhere they go i'm going to bring it back into the middle actually it's going to come back into the middle itself there we go these tangents always meet at right angles okay so using the equation of the chord and because we know what the focus is we've proved this really interesting result that every focal chord every focal chord there it is right there right the ends that make it a focal cord they have perpendicular gradients and that's what makes their tangents perpendicular okay now this kind of geometry and the fact that it does this is why parabolas are so special just like do you remember when you did circle geometry and it's like oh my goodness look everything that's weird like this angle is equal to that angle why is that because circles are special shapes um when you've got you know a diameter and you look at that angle oh my goodness that angle is a special angle it's right angled well you get the same kinds of things on parabolas because parabolas in circles believe it or not are made of the same kind of dna |
4923 | https://oryxlearning.com/learn/adding-integers-using-integer-chips/ | Menu
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Adding Integers Using Integer Chips
Adding Integers Using Integer Chips
The term “integer” was adapted in Mathematics from Latin. Integer means intact or whole. Integers are very much like whole numbers, but they also include negative numbers among them. For example, 11, 8, 0, and −1,908 are integers whereas √5, Π are not integers. The set of integers consists of zero, the positive natural numbers, and their additive inverses.
Addition generally means to increase the value. But, in the case of integers, the addition operation might lead to an increase or decrease in the value of the given number. If we add a negative integer, the value of the given number will decrease and if we add a positive integer the value will increase.
When modeling integers, we can use colored chips to represent these integers. One color can represent a positive number and another color can represent a negative number. Here, a yellow chip will represent a positive integer and a red chip will represent a negative integer. A zero pair is the pair of the positive and negative form of the same number.
Rules
Model the addends using integer chips. If the addend is positive, use yellow chips. If the addend is negative, use red chips.2. Equal amounts of red and yellow counters create zero pairs and cancel each other out.3. Determine the amount and color of the remaining counters and write the corresponding integer.
Example
Add. -12 + 4
Solution
Model the addends using integer chips. If the addend is positive, use yellow chips. If the addend is negative, use red chips.2. Equal amounts of red and yellow counters create zero pairs and cancel each other out.3. Determine the amount and color of the remaining counters and write the corresponding integer.-12 + 4 = -8
Practice Adding Integers Using Integer Chips
Practice Problem 1
Use the counters to model the expression and then find the solution. 8 + (-3)
Practice Problem 2
Use the counters to model the expression and then find the solution. (-7) + 5
Practice Problem 3
Use the counters to model the expression and then find the solution. (-6) + 10 + (-3)
An integer is a number with no decimal or fractional part, from the set of negative and positive numbers, including zero.
Positive Integers: An integer is positive if it is greater than zero. Example: 1, 2, 3 . . .
Negative Integers: An integer is negative if it is less than zero. Example: -1, -2, -3 . . .
Zero is defined as neither negative nor positive integer. It is a whole number.
A zero pair – A pair of the positive and negative form of the same number.
Absolute value – the positive distance that a number is from 0 on a number line.
Additive inverses – two integers that are opposites.
Integer chips – Are colored chips with signs. One color represents a positive number and another color represents a negative number.
Pre-requisite Skills Integers and Absolute Value Order of Operations
Related Skills Subtracting Integers Multiplying Integers Dividing Integers Evaluating Algebraic Expressions Evaluate Algebraic Expressions Add Linear Expressions Subtract Linear Expressions Solve Complex Equations
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4924 | https://www.quora.com/In-how-many-ways-can-a-student-choose-8-questions-out-of-10-in-an-exam | In how many ways can a student choose 8 questions out of 10 in an exam? - Quora
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In how many ways can a student choose 8 questions out of 10 in an exam?
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To find the number of ways a student can choose 8 questions out of 10, we can use the combination formula, which is given by:
(n r)=n!r!(n−r)!(n r)=n!r!(n−r)!
where n n is the total number of items to choose from, r r is the number of items to choose, and !! denotes factorial.
In this case, n=10 n=10 and r=8 r=8:
(10 8)=10!8!(10−8)!=10!8!⋅2!(10 8)=10!8!(10−8)!=10!8!⋅2!
Calculating the factorials:
10!=10×9×8!10!=10×9×8!
8!8! cancels out in the numerator and denominator.
This simplifies to:
(10 8)=10×9 2!=10×9 2=90 2=45(10 8)=10×9 2!=10×9 2=90 2=45
Therefore, there are 45 ways fo
Continue Reading
To find the number of ways a student can choose 8 questions out of 10, we can use the combination formula, which is given by:
(n r)=n!r!(n−r)!(n r)=n!r!(n−r)!
where n n is the total number of items to choose from, r r is the number of items to choose, and !! denotes factorial.
In this case, n=10 n=10 and r=8 r=8:
(10 8)=10!8!(10−8)!=10!8!⋅2!(10 8)=10!8!(10−8)!=10!8!⋅2!
Calculating the factorials:
10!=10×9×8!10!=10×9×8!
8!8! cancels out in the numerator and denominator.
This simplifies to:
(10 8)=10×9 2!=10×9 2=90 2=45(10 8)=10×9 2!=10×9 2=90 2=45
Therefore, there are 45 ways for a student to choose 8 questions out of 10.
Upvote ·
Raj Yadav
B.Tech from Motilal Nehru National Institute of Technology, Allahabad (Graduated 2021)
·7y
I studied permutation and combination during preparation. It had been a long time since I solved a question on that. But might be I correct here, I think this is an easy question.
Simply, either you choose 8 questions out of 10 to answer or 2 questions out of 10 not to answer. The answer is same and it is 10C2. And this is solved as 10!/(8!×2!). The answer is 45. Always when you get a question to select r things for n things, the answer is nPr. Which is again n!/{r!×(n-r)!}.
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Henk Brozius
Studied Mathematics · Author has 3.6K answers and 2.2M answer views
·4y
Originally Answered: There are 10 questions in the exam. In how many ways can a student choose 8 questions in all if two questions are compulsory? ·
Why don't you ask directly: in how many ways can a student choose 6 questions out of 8?..
(8 6)(8 6)
Upvote ·
9 6
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More answers below
Students need to answer 8 out of 10 questions in the mathematics exam. In how many ways can a student choose 8 questions if the first three questions are mandatory?
If a student answers 8 questions out of 10, in how many ways can he select 8 questions?
A student must answer six of eight questions on an exam. If the first two questions are mandatory, how many ways can he do the exam?
A student is to answer 10 out of 12 questions in an exam, such that he must choose at least 4 from the first 5 questions. What is the number of choices available to him?
There are 10 questions in the exam. In how many ways can a student choose 8 questions in all if two questions are compulsory?
CQ
Studied The Arts&Psychology (Graduated 2015) · Author has 3.5K answers and 1.3M answer views
·1y
Originally Answered: In an examination paper, there are 12 questions. In how many different ways can a student choose eight questions in all if two questions are compulsory? ·
in an examination paper, there are 12 questions ..
how many different ways can a student choose eight questions in all if two questions that are compulsory? and this is I agree with Robert Charles Lee —— process of elimination ..
my extra 2 cents ?
skim through all the 12 questions to start ( get out of the way the stressful questions so that you can relax a bit for the rest of the paper ) —— establish the order of importance in answering the questions based on time ( usually the last questions are the most important . but not always so be careful particularly during a test )
answer the 2–3 questi
Continue Reading
in an examination paper, there are 12 questions ..
how many different ways can a student choose eight questions in all if two questions that are compulsory? and this is I agree with Robert Charles Lee —— process of elimination ..
my extra 2 cents ?
skim through all the 12 questions to start ( get out of the way the stressful questions so that you can relax a bit for the rest of the paper ) —— establish the order of importance in answering the questions based on time ( usually the last questions are the most important . but not always so be careful particularly during a test )
answer the 2–3 questions that are pertinent right off the bat while it s fresh on your mind immediately your teacher / professor has likely already alerted you in class what will be asked and how you should answer —— so practice your answers ahead of time
the remaining questions will require your basic knowledge to answer —— don t F up on this because these points are important too ..
then piece your paper together and make sure it flows from one idea to the next
remember to write a dynamic introductory paragraph about all the points in the body of your paper —— and a conclusion that sums everything up effectively
hope this helps
good luck with your question moving forward
have a great day
Upvote ·
Yogendra Sahu
Full Stack Engineer at Orion Innovation (2019–present)
·7y
45 ways
10C8 = 10! / (8! 2!)
= (10 9 8! ) / (8! 2)
= (10 9 ) / ( 2 )
= 90 / 2
= 45
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Gamini De Silva
Former Director Head of Software at JOHN KEELLS HOLDINGS PLC · Author has 2.6K answers and 725.4K answer views
·4y
Originally Answered: Students need to answer 8 out of 10 questions in the mathematics exam. In how many ways can a student choose 8 questions if the first three questions are mandatory? ·
First three are mandatory. So you have seven sums to select from. You need five more. That will be 7C5 = 7!/5!×2! = 21
Please note that your sum has nothing to do with arrangements (called permutations). It is only concerning choices disregarding how many arrangements. Hence you have to adopt combinations.
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An exam consists of ten questions, of which the student must answer eight. If the student must answer at least 4 of the first 5 questions, in how many ways can the student answer the required 8 questions?
There are 12 questions on An examination and each student must answer 8 questions including at least 4 from the first 5 questions. How many different combinations of questions could a student choose to answer?
A student is to answer a total of 10 questions from 12 available questions.If the first 6 questions are compulsory, in how many ways can this be done?
An examination requires that a student answer any 10 of 15 questions. In how many ways can the student choose which questions to answer?
A student is told to answer any 6 of 10 questions on any exam. In how many different ways can he choose the 6 questions to answer?
Jonathan Sullivan
I've passed up to Calculus III. · Author has 9.4K answers and 5.1M answer views
·6y
Originally Answered: An exam consists of ten questions, of which the student must answer eight. If the student must answer at least 4 of the first 5 questions, in how many ways can the student answer the required 8 questions? ·
There are five ways to answer four of the first five questions; answer all but the first, or second, or third, and so on.
That leaves the last five questions, again, four of which must be answered.
For each of the first five questions that isn’t answered, you have another five possible questions to not answer. 5×5=5 5×5=5
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Satvik Srivastava
Founder at BeingInquisitive.net (2020–present)
·5y
Originally Answered: If a student answers 8 questions out of 10, in how many ways can he select 8 questions? ·
To select 8 questions out of 10 means 10P8, but since we can chose them in any order hence selection becomes 10p8/8!
So, number of ways to select 8 questions out of 10 = (10! / (2! 8!)) = 45.
Therfore, there are 45 ways to select 8 questions out of 10.
Upvote ·
9 3
Venkatesan Iyer
I have studied Wittgenstein's philosophy of mathematics for my research · Author has 855 answers and 485K answer views
·7y
order of choice does not matter. It is a case of combinations.
10 c 8
10 c 8 = 10 c 2
10 9/21 = 45.
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Lance Berg
Author has 28K answers and 54.7M answer views
·6y
Originally Answered: An exam consists of ten questions, of which the student must answer eight. If the student must answer at least 4 of the first 5 questions, in how many ways can the student answer the required 8 questions? ·
There are 5 ways to answer four of the first five questions. If the student has done this, they have five remaining questions and must answer four, so again, 5 ways. Five times five makes 25 ways to answer in this manner.
It’s also possible to answer all five of the first five questions, there’s only 1 way to do this. If they do, there are 10 ways to make combinations of three from five elements. One times ten makes 10 ways to answer in this manner.
25+10=35 total ways.
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Simran Dhawan
PHP Developer (2019–present) · Author has 90 answers and 104.3K answer views
·5y
Originally Answered: If a student answers 8 questions out of 10, in how many ways can he select 8 questions? ·
It is a question of Permutations and Combinations. The formula to find various combinations is:
ⁿCᵣ = n!/((n-r)! r!)
So,
Number of ways to choose 8 questions out of 10 are = 10! / ((10–8)! 8!) =45
Therefore there are 45 ways to choose the questions.
Upvote ·
Marianne Bauer
Author has 456 answers and 326K answer views
·4y
Originally Answered: Students are to answer 8 out of 9 exam questions. In how many different ways can the questions be selected? ·
9 choose 8 = C(9,8) = 9! / (8! (9!-8!) = 9! / 8!-1! = 9 ways to select questions
1,2,3,4,5,6,7,8 (leave out 9)
1,2,3,4,5,6,7,9 (leave out 8)
1,2,3,4,5,6,8,9 (leave out 7)
1,2,3,4,5,7,8,9 (leave out 6)
1,2,3,4,6,7,8,9 (leave out 5)
1,2,3,5,6,7,8,9 (leave out 4)
1,2,4,5,6,7,8,9 (leave out 3)
1,3,4,5,6,7,8,9 (leave out 2)
2,3,4,5,6,7,8,9 (leave out 1)
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| Use the distributive property to express the sum of two whole numbers with a common factor as a multiple of a sum of two whole numbers with no common factor. Find the greatest common factor (GCF) and least common multiple (LCM) of two or more whole numbers. 1. Use factors and multiples to determine prime factorization. ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
| Arizona Academic Standards: 6.NS.B.4 ======================================== |
| Use previous understanding of factors to find the greatest common factor and the least common multiple. 1. Find the greatest common factor of two whole numbers less than or equal to 100. 2. Find the least common multiple of two whole numbers less than or equal to 12. 3. Use the distributive property to express a sum of two whole numbers 1 to 100 with a common factor as a multiple of a sum of two whole numbers with no common factor. For example, express 36 + 8 as 4(9+2). -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
| Common Core State Standards: Math.6.NS.4 or 6.NS.B.4 ======================================================== |
| Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor. For example, express 36 + 8 as 4(9+2). ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
| Georgia Math and ELA Standards: 6.PAR.6.2 ============================================= |
| Determine greatest common factors and least common multiples using a variety of strategies to make sense of applicable problems. -------------------------------------------------------------------------------------------------------------------------------- |
| Massachusetts Curriculum Frameworks: 6.NS.B.4 ================================================= |
| Use prime factorization to find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1-100 with a common factor as a multiple of a sum of two relatively prime numbers. For example, express 36 + 8 as 4(9 + 2) --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
| North Carolina - Standard Course of Study: 6.NS.4 ===================================================== |
| Understand and use prime factorization and the relationships between factors to: Find the unique prime factorization for a whole number. Find the greatest common factor of two whole numbers less than or equal to 100. Use the greatest common factor and the distributive property to rewrite the sum of two whole numbers, each less than or equal to 100. Find the least common multiple of two whole numbers less than or equal to 12 to add and subtract fractions with unlike denominators. ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
| New York State Next Generation Learning Standards: 6.NS.4 ============================================================= |
| Find the greatest common factor of two whole numbers less than or equal to 100. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor other than 1. e.g., Express 36 + 8 as 4(9 + 2). Find the least common multiple of two whole numbers less than or equal to 12. -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
| Tennessee Academic Standards: 6.NS.B.4 ========================================== |
| Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1-100 with a common factor as a multiple of a sum of two whole numbers with no common factor. For example, express 36 + 8 as 4 (9 + 2). ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
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4926 | https://en.wikipedia.org/wiki/Measurable_function | Jump to content
Measurable function
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From Wikipedia, the free encyclopedia
Kind of mathematical function
In mathematics, and in particular measure theory, a measurable function is a function between the underlying sets of two measurable spaces that preserves the structure of the spaces: the preimage of any measurable set is measurable. This is in direct analogy to the definition that a continuous function between topological spaces preserves the topological structure: the preimage of any open set is open. In real analysis, measurable functions are used in the definition of the Lebesgue integral. In probability theory, a measurable function on a probability space is known as a random variable.
Formal definition
[edit]
Let and be measurable spaces, meaning that and are sets equipped with respective -algebras and A function is said to be measurable if for every the pre-image of under is in ; that is, for all
That is, where is the σ-algebra generated by f. If is a measurable function, one writes
to emphasize the dependency on the -algebras and
Term usage variations
[edit]
The choice of -algebras in the definition above is sometimes implicit and left up to the context. For example, for or other topological spaces, the Borel algebra (generated by all the open sets) is a common choice. Some authors define measurable functions as exclusively real-valued ones with respect to the Borel algebra.
If the values of the function lie in an infinite-dimensional vector space, other non-equivalent definitions of measurability, such as weak measurability and Bochner measurability, exist.
Notable classes of measurable functions
[edit]
Random variables are by definition measurable functions defined on probability spaces.
If and are Borel spaces, a measurable function is also called a Borel function. Continuous functions are Borel functions but not all Borel functions are continuous. However, a measurable function is nearly a continuous function; see Luzin's theorem. If a Borel function happens to be a section of a map it is called a Borel section.
A Lebesgue measurable function is a measurable function where is the -algebra of Lebesgue measurable sets, and is the Borel algebra on the complex numbers Lebesgue measurable functions are of interest in mathematical analysis because they can be integrated. In the case is Lebesgue measurable if and only if is measurable for all This is also equivalent to any of being measurable for all or the preimage of any open set being measurable. Continuous functions, monotone functions, step functions, semicontinuous functions, Riemann-integrable functions, and functions of bounded variation are all Lebesgue measurable. A function is measurable if and only if the real and imaginary parts are measurable.
Properties of measurable functions
[edit]
The sum and product of two complex-valued measurable functions are measurable. So is the quotient, so long as there is no division by zero.
If and are measurable functions, then so is their composition
If and are measurable functions, their composition need not be -measurable unless Indeed, two Lebesgue-measurable functions may be constructed in such a way as to make their composition non-Lebesgue-measurable.
The (pointwise) supremum, infimum, limit superior, and limit inferior of a sequence (viz., countably many) of real-valued measurable functions are all measurable as well.
The pointwise limit of a sequence of measurable functions is measurable, where is a metric space (endowed with the Borel algebra). This is not true in general if is non-metrizable. The corresponding statement for continuous functions requires stronger conditions than pointwise convergence, such as uniform convergence.
Non-measurable functions
[edit]
Real-valued functions encountered in applications tend to be measurable; however, it is not difficult to prove the existence of non-measurable functions. Such proofs rely on the axiom of choice in an essential way, in the sense that Zermelo–Fraenkel set theory without the axiom of choice does not prove the existence of such functions.
In any measure space with a non-measurable set one can construct a non-measurable indicator function:
where is equipped with the usual Borel algebra. This is a non-measurable function since the preimage of the measurable set is the non-measurable
As another example, any non-constant function is non-measurable with respect to the trivial -algebra since the preimage of any point in the range is some proper, nonempty subset of which is not an element of the trivial
See also
[edit]
Bochner measurable function
Bochner space – Type of topological space
Lp space – Function spaces generalizing finite-dimensional p norm spaces - Vector spaces of measurable functions: the spaces
Measure-preserving dynamical system – Subject of study in ergodic theory
Vector measure
Weakly measurable function
Notes
[edit]
^ Jump up to: a b c d Strichartz, Robert (2000). The Way of Analysis. Jones and Bartlett. ISBN 0-7637-1497-6.
^ Carothers, N. L. (2000). Real Analysis. Cambridge University Press. ISBN 0-521-49756-6.
^ Folland, Gerald B. (1999). Real Analysis: Modern Techniques and their Applications. Wiley. ISBN 0-471-31716-0.
^ Royden, H. L. (1988). Real Analysis. Prentice Hall. ISBN 0-02-404151-3.
^ Dudley, R. M. (2002). Real Analysis and Probability (2 ed.). Cambridge University Press. ISBN 0-521-00754-2.
^ Aliprantis, Charalambos D.; Border, Kim C. (2006). Infinite Dimensional Analysis, A Hitchhiker's Guide (3 ed.). Springer. ISBN 978-3-540-29587-7.
External links
[edit]
Measurable function at Encyclopedia of Mathematics
Borel function at Encyclopedia of Mathematics
| v t e Measure theory | |
--- |
| Basic concepts | Absolute continuity of measures Lebesgue integration Lp spaces Measure Measure space + Probability space Measurable space/function |
| Sets | Almost everywhere Atom Baire set Borel set + equivalence relation Borel space Carathéodory's criterion Cylindrical σ-algebra + Cylinder set 𝜆-system Essential range + infimum/supremum Locally measurable π-system σ-algebra Non-measurable set + Vitali set Null set Support Transverse measure Universally measurable |
| Types of measures | Atomic Baire Banach Besov Borel Brown Complex Complete Content (Logarithmically) Convex Decomposable Discrete Equivalent Finite Inner (Quasi-) Invariant Locally finite Maximising Metric outer Outer Perfect Pre-measure (Sub-) Probability Projection-valued Radon Random Regular + Borel regular + Inner regular + Outer regular Saturated Set function σ-finite s-finite Signed Singular Spectral Strictly positive Tight Vector |
| Particular measures | Counting Dirac Euler Gaussian Haar Harmonic Hausdorff Intensity Lebesgue + Infinite-dimensional Logarithmic Product + Projections Pushforward Spherical measure Tangent Trivial Young |
| Maps | Measurable function + Bochner + Strongly + Weakly Convergence: almost everywhere of measures in measure of random variables + in distribution + in probability Cylinder set measure Random: compact set element measure process variable vector Projection-valued measure |
| Main results | Carathéodory's extension theorem Convergence theorems + Dominated + Monotone + Vitali Decomposition theorems + Hahn + Jordan + Maharam's Egorov's Fatou's lemma Fubini's + Fubini–Tonelli Hölder's inequality Minkowski inequality Radon–Nikodym Riesz–Markov–Kakutani representation theorem |
| Other results | | | | --- | | Disintegration theorem + Lifting theory Lebesgue's density theorem Lebesgue differentiation theorem Sard's theorem Vitali–Hahn–Saks theorem | | | For Lebesgue measure | Isoperimetric inequality Brunn–Minkowski theorem + Milman's reverse Minkowski–Steiner formula Prékopa–Leindler inequality Vitale's random Brunn–Minkowski inequality | |
| Applications & related | Convex analysis Descriptive set theory Probability theory Real analysis Spectral theory |
| v t e Lp spaces | |
--- |
| Basic concepts | Banach & Hilbert spaces Lp spaces Measure + Lebesgue Measure space Measurable space/function Minkowski distance Sequence spaces |
| L1 spaces | Integrable function Lebesgue integration Taxicab geometry |
| L2 spaces | Bessel's Cauchy–Schwarz Euclidean distance Hilbert space Parseval's identity Polarization identity Pythagorean theorem Square-integrable function |
| spaces | Bounded function Chebyshev distance Infimum and supremum + Essential Uniform norm |
| Maps | Almost everywhere Convergence almost everywhere Convergence in measure Function space Integral transform Locally integrable function Measurable function Symmetric decreasing rearrangement |
| Inequalities | Babenko–Beckner Chebyshev's Clarkson's Hanner's Hausdorff–Young Hölder's Markov's Minkowski Young's convolution |
| Results | Marcinkiewicz interpolation theorem Plancherel theorem Riemann–Lebesgue Riesz–Fischer theorem Riesz–Thorin theorem | | | --- | | For Lebesgue measure | Isoperimetric inequality Brunn–Minkowski theorem + Milman's reverse Minkowski–Steiner formula Prékopa–Leindler inequality Vitale's random Brunn–Minkowski inequality | |
| Applications & related | Bochner space Fourier analysis Lorentz space Probability theory Quasinorm Real analysis Sobolev space -algebra + C-algebra + Von Neumann |
| v t e | |
--- |
| History List of specific functions | |
| Types by domain and codomain | X → 𝔹 𝔹 → X 𝔹ⁿ → X X → ℤ ℤ → X X → ℝ ℝ → X ℝⁿ → X X → ℂ ℂ → X ℂⁿ → X |
| Classes/properties | Constant Identity Linear Polynomial Rational Algebraic Analytic Smooth Continuous Measurable Injective Surjective Bijective |
| Constructions | Restriction Composition λ Inverse |
| Generalizations | Relation (Binary relation) Set-valued Multivalued Partial Implicit Space Higher-order Morphism Functor |
| | |
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2006 AIME II Problems/Problem 11
Contents
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1 Problem
2 Solutions
2.1 Solution 1
2.2 Solution 2 (bash)
2.3 Solution 3 (some guessing involved)/"Engineer's Induction"
2.4 Solution 4 (if you did not know how to use the numbers given in the problem)
3 See also
Problem
A sequence is defined as follows and, for all positive integers Given that and find the remainder when is divided by 1000.
Solutions
Solution 1
Define the sum as . Since , the sum will be:
Thus , and are both given; the last four digits of their sum is , and half of that is . Therefore, the answer is .−
Solution 2 (bash)
Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:
Adding all the residues shows the sum is congruent to mod .
~ I-_-I
Solution 3 (some guessing involved)/"Engineer's Induction"
All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given and , so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some such that . From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that , at least for the first few terms. From this, we have that .
Solution by zeroman; clarified by srisainandan6
Solution 4 (if you did not know how to use the numbers given in the problem)
By the Chinese remainder theorem, each number under 1000 is uniquely determined by its mod 8 and mod 125.
We list a few terms of the sequence mod 8:
Therefore, the cycle repeats every 8 numbers, and each cycle has a sum of 4 mod 8. Therefore, the sum mod 8 is
Denote It is easy to prove that
We write the sum () of the first terms mod 125:
Therefore the desired number is
From here, we can determine the number we are looking for is -sd8
See also
2006 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10Followed by
Problem 12
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
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4928 | https://en.wikipedia.org/wiki/Collocation | Collocation - Wikipedia
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1 Expanded definition
2 In dictionaries
3 Statistically significant collocation
4 See also
5 References
6 External links
Collocation
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From Wikipedia, the free encyclopedia
Frequent occurrence of words next to each other
This article is about the corpus linguistics notion. For other uses, see Colocation (disambiguation).
In corpus linguistics, a collocation is a series of words or terms that co-occur more often than would be expected by chance. In phraseology, a collocation is a type of compositionalphraseme, meaning that it can be understood from the words that make it up. This contrasts with an idiom, where the meaning of the whole cannot be inferred from its parts, and may be completely unrelated.
There are about seven main types of collocations: adjective+noun, noun+noun (such as collective nouns), noun+verb, verb+noun, adverb+adjective, verbs+prepositional phrase (phrasal verbs), and verb+adverb.
Collocation extraction is a computational technique that finds collocations in a document or corpus, using various computational linguistics elements resembling data mining.
Expanded definition
[edit]
Collocations are partly or fully fixed expressions that become established through repeated context-dependent use. Such terms as crystal clear, middle management, nuclear family, and cosmetic surgery are examples of collocated pairs of words.
Collocations can be in a syntactic relation (such as verb–object: make and decision), lexical relation (such as antonymy), or they can be in no linguistically defined relation. Knowledge of collocations is vital for the competent use of a language: a grammatically correct sentence will stand out as awkward if collocational preferences are violated. This makes collocation a common focus for language teaching.
Corpus linguists specify a key word in context (KWIC) and identify the words immediately surrounding them, to illustrate the way words are used in practice.
The processing of collocations involves a number of parameters, the most important of which is the measure of association, which evaluates whether the co-occurrence is purely by chance or statistically significant. Due to the non-random nature of language, most collocations are classed as significant, and the association scores are simply used to rank the results. Commonly used measures of association include mutual information, t scores, and log-likelihood.
Rather than select a single definition, Gledhill proposes that collocation involves at least three different perspectives: co-occurrence, a statistical view, which sees collocation as the recurrent appearance in a text of a node and its collocates; construction, which sees collocation either as a correlation between a lexeme and a lexical-grammatical pattern, or as a relation between a base and its collocative partners; and expression, a pragmatic view of collocation as a conventional unit of expression, regardless of form. These different perspectives contrast with the usual way of presenting collocation in phraseological studies. Traditionally speaking, collocation is explained in terms of all three perspectives at once, in a continuum:
Free combination ↔ bound collocation ↔ frozen idiom
In dictionaries
[edit]
In 1933, Harold Palmer's Second Interim Report on English Collocations highlighted the importance of collocation as a key to producing natural-sounding language, for anyone learning a foreign language. Thus from the 1940s onwards, information about recurrent word combinations became a standard feature of monolingual learner's dictionaries. As these dictionaries became "less word-centred and more phrase-centred", more attention was paid to collocation. This trend was supported, from the beginning of the 21st century, by the availability of large text corpora and intelligent corpus-querying software, making it possible to provide a more systematic account of collocation in dictionaries. Using these tools, dictionaries such as the Macmillan English Dictionary and the Longman Dictionary of Contemporary English included boxes or panels with lists of frequent collocations.
There are also a number of specialized dictionaries devoted to describing the frequent collocations in a language. These include (for Spanish) Redes: Diccionario combinatorio del español contemporaneo (2004), (for French) Le Robert: Dictionnaire des combinaisons de mots (2007), and (for English) the LTP Dictionary of Selected Collocations (1997) and the Macmillan Collocations Dictionary (2010).
Statistically significant collocation
[edit]
Student's t-test can be used to determine whether the occurrence of a collocation in a corpus is statistically significant. For a bigramw 1 w 2{\displaystyle w_{1}w_{2}}, let P(w 1)=#w 1 N{\displaystyle P(w_{1})={\frac {#w_{1}}{N}}} be the unconditional probability of occurrence of w 1{\displaystyle w_{1}} in a corpus with size N{\displaystyle N}, and let P(w 2)=#w 2 N{\displaystyle P(w_{2})={\frac {#w_{2}}{N}}} be the unconditional probability of occurrence of w 2{\displaystyle w_{2}} in the corpus. The t-score for the bigram w 1 w 2{\displaystyle w_{1}w_{2}} is calculated as:
t=x¯−μ s 2 N,{\displaystyle t={\frac {{\bar {x}}-\mu }{\sqrt {\frac {s^{2}}{N}}}},}
where x¯=#w i w j N{\displaystyle {\bar {x}}={\frac {#w_{i}w_{j}}{N}}} is the sample mean of the occurrence of w 1 w 2{\displaystyle w_{1}w_{2}}, #w 1 w 2{\displaystyle #w_{1}w_{2}} is the number of occurrences of w 1 w 2{\displaystyle w_{1}w_{2}}, μ=P(w i)P(w j){\displaystyle \mu =P(w_{i})P(w_{j})} is the probability of w 1 w 2{\displaystyle w_{1}w_{2}} under the null-hypothesis that w 1{\displaystyle w_{1}} and w 2{\displaystyle w_{2}} appear independently in the text, and s 2=x¯(1−x¯)≈x¯{\displaystyle s^{2}={\bar {x}}(1-{\bar {x}})\approx {\bar {x}}} is the sample variance. With a large N{\displaystyle N}, the t-test is equivalent to a Z-test.
See also
[edit]
Linguistics portal
English collocations
Agreement (linguistics)
Cliché
Collocational restriction
Collostructional analysis
Compound noun, adjective and verb
Government (linguistics)
Idiom (language structure)
Irreversible binomial
Isocolon
Lexical item
N-gram
Phrasal verb
Phraseology
Phraseme
Sketch Engine
Statistically improbable phrase
Word sketch
References
[edit]
^Dunning, Ted (1993): "Accurate methods for the statistics of surprise and coincidenceArchived 2012-08-05 at the Wayback Machine". Computational Linguistics 19, 1 (Mar. 1993), 61–74.
^Dunning, Ted (2008-03-21). "Surprise and Coincidence". blogspot.com. Archived from the original on 2012-01-20. Retrieved 2012-04-09.
^Gledhill C. (2000): Collocations in Science WritingArchived 2023-06-29 at the Wayback Machine, Narr, Tübingen
^Firth J.R. (1957): Papers in Linguistics 1934–1951. Oxford: Oxford University Press.
^Sinclair J. (1996): "The Search for Units of Meaning", in Textus, IX, 75–106.
^Smadja F. A & McKeown, K. R. (1990): "Automatically extracting and representing collocations for language generationArchived 2015-09-06 at the Wayback Machine", Proceedings of ACL'90, 252–259, Pittsburgh, Pennsylvania.
^Hunston S. & Francis G. (2000): Pattern Grammar — A Corpus-Driven Approach to the Lexical Grammar of EnglishArchived 2023-06-29 at the Wayback Machine, Amsterdam, John Benjamins
^Hausmann F. J. (1989): Le dictionnaire de collocations. In Hausmann F.J., Reichmann O., Wiegand H.E., Zgusta L.(eds), Wörterbücher: ein internationales Handbuch zur Lexikographie. Dictionaries. Dictionnaires. Berlin/New-York: De Gruyter. 1010–1019.
^ Moon R. (1998): Fixed Expressions and Idioms, a Corpus-Based Approach. Oxford, Oxford University Press.
^Frath P. & Gledhill C. (2005): "Free-Range Clusters or Frozen Chunks? Reference as a Defining Criterion for Linguistic Units[dead link]", in Recherches anglaises et Nord-américaines, vol. 38 :25–43
^Cowie, A.P., English Dictionaries for Foreign Learners, Oxford University Press 1999:54–56
^Bejoint, H., The Lexicography of English, Oxford University Press 2010: 318
^"MED Second Edition – Key features – Macmillan". macmillandictionaries.com. Archived from the original on 2020-09-28. Retrieved 2011-08-24.
^Herbst, T. and Klotz, M. 'Syntagmatic and Phraseological Dictionaries' in Cowie, A.P. (Ed.) The Oxford History of English Lexicography, 2009: part 2, 234–243
^"Macmillan Collocation Dictionary – How it was written - Macmillan". macmillandictionaries.com. Archived from the original on 2018-12-21. Retrieved 2011-08-24.
^Manning, Chris; Schütze, Hinrich (1999). Foundations of Statistical Natural Language Processing. Cambridge, MA: MIT Press. pp.163–166. ISBN0262133601.
External links
[edit]
Look up collocation in Wiktionary, the free dictionary.
Ozdic Collocation Dictionary
A Small System Storing Spanish Collocations (Igor A. Bolshakov & Sabino Miranda-Jiménez)
Morphological characterization of collocations and semantic relationships in Spanish (Sabino Miranda-Jiménez & Igor A. Bolshakov)
Example of collocations for the word "Surgery" at wordassociations.net
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Law about the rate of chemical reactions
For other uses, see Mass action (disambiguation).
In chemistry, the law of mass action is the proposition that the rate of a chemical reaction is directly proportional to the product of the activities or concentrations of the reactants. It explains and predicts behaviors of solutions in dynamic equilibrium. Specifically, it implies that for a chemical reaction mixture that is in equilibrium, the ratio between the concentration of reactants and products is constant.
Two aspects are involved in the initial formulation of the law: 1) the equilibrium aspect, concerning the composition of a reaction mixture at equilibrium and 2) the kinetic aspect concerning the rate equations for elementary reactions. Both aspects stem from the research performed by Cato M. Guldberg and Peter Waage between 1864 and 1879 in which equilibrium constants were derived by using kinetic data and the rate equation which they had proposed. Guldberg and Waage also recognized that chemical equilibrium is a dynamic process in which rates of reaction for the forward and backward reactions must be equal at chemical equilibrium. In order to derive the expression of the equilibrium constant appealing to kinetics, the expression of the rate equation must be used. The expression of the rate equations was rediscovered independently by Jacobus Henricus van 't Hoff.
The law is a statement about equilibrium and gives an expression for the equilibrium constant, a quantity characterizing chemical equilibrium. In modern chemistry this is derived using equilibrium thermodynamics. It can also be derived with the concept of chemical potential.
History
[edit]
Two chemists generally expressed the composition of a mixture in terms of numerical values relating the amount of the product to describe the equilibrium state.
Cato Maximilian Guldberg and Peter Waage, building on Claude Louis Berthollet's ideas about reversible chemical reactions, proposed the law of mass action in 1864. These papers, in Danish, went largely unnoticed, as did the later publication (in French) of 1867 which contained a modified law and the experimental data on which that law was based.
In 1877 van 't Hoff independently came to similar conclusions, but was unaware of the earlier work, which prompted Guldberg and Waage to give a fuller and further developed account of their work, in German, in 1879. Van 't Hoff then accepted their priority.
1864
[edit]
The equilibrium state (composition)
[edit]
In their first paper, Guldberg and Waage suggested that in a reaction such as
the "chemical affinity" or "reaction force" between A and B did not just depend on the chemical nature of the reactants, as had previously been supposed, but also depended on the amount of each reactant in a reaction mixture. Thus the law of mass action was first stated as follows:
: When two reactants, A and B, react together at a given temperature in a "substitution reaction," the affinity, or chemical force between them, is proportional to the active masses, [A] and [B], each raised to a particular power
: .
In this context a substitution reaction was one such as . Active mass was defined in the 1879 paper as "the amount of substance in the sphere of action". For species in solution active mass is equal to concentration. For solids, active mass is taken as a constant. , a and b were regarded as empirical constants, to be determined by experiment.
At equilibrium, the chemical force driving the forward reaction must be equal to the chemical force driving the reverse reaction. Writing the initial active masses of A,B, A' and B' as p, q, p' and q' and the dissociated active mass at equilibrium as , this equality is represented by
represents the amount of reagents A and B that has been converted into A' and B'. Calculations based on this equation are reported in the second paper.
Dynamic approach to the equilibrium state
[edit]
The third paper of 1864 was concerned with the kinetics of the same equilibrium system. Writing the dissociated active mass at some point in time as x, the rate of reaction was given as
Likewise the reverse reaction of A' with B' proceeded at a rate given by
The overall rate of conversion is the difference between these rates, so at equilibrium (when the composition stops changing) the two rates of reaction must be equal. Hence
: ...
1867
[edit]
The rate expressions given in Guldberg and Waage's 1864 paper could not be differentiated, so they were simplified as follows. The chemical force was assumed to be directly proportional to the product of the active masses of the reactants.
This is equivalent to setting the exponents a and b of the earlier theory to one. The proportionality constant was called an affinity constant, k. The equilibrium condition for an "ideal" reaction was thus given the simplified form
[A]eq, [B]eq etc. are the active masses at equilibrium. In terms of the initial amounts reagents p,q etc. this becomes
The ratio of the affinity coefficients, k'/k, can be recognized as an equilibrium constant. Turning to the kinetic aspect, it was suggested that the velocity of reaction, v, is proportional to the sum of chemical affinities (forces). In its simplest form this results in the expression
where is the proportionality constant. Actually, Guldberg and Waage used a more complicated expression which allowed for interaction between A and A', etc. By making certain simplifying approximations to those more complicated expressions, the rate equation could be integrated and hence the equilibrium quantity could be calculated. The extensive calculations in the 1867 paper gave support to the simplified concept, namely,
: The rate of a reaction is proportional to the product of the active masses of the reagents involved.
This is an alternative statement of the law of mass action.
1879
[edit]
In the 1879 paper the assumption that reaction rate was proportional to the product of concentrations was justified microscopically in terms of the frequency of independent collisions, as had been developed for gas kinetics by Boltzmann in 1872 (Boltzmann equation). It was also proposed that the original theory of the equilibrium condition could be generalised to apply to any arbitrary chemical equilibrium.
The exponents α, β etc. are explicitly identified for the first time as the stoichiometric coefficients for the reaction.
Modern statement of the law
[edit]
The affinity constants, k+ and k−, of the 1879 paper can now be recognised as rate constants. The equilibrium constant, K, was derived by setting the rates of forward and backward reactions to be equal. This also meant that the chemical affinities for the forward and backward reactions are equal. The resultant expression
is correct even from the modern perspective, apart from the use of concentrations instead of activities (the concept of chemical activity was developed by Josiah Willard Gibbs, in the 1870s, but was not widely known in Europe until the 1890s). The derivation from the reaction rate expressions is no longer considered to be valid. Nevertheless, Guldberg and Waage were on the right track when they suggested that the driving force for both forward and backward reactions is equal when the mixture is at equilibrium. The term they used for this force was chemical affinity. Today the expression for the equilibrium constant is derived by setting the chemical potential of forward and backward reactions to be equal. The generalisation of the law of mass action, in terms of affinity, to equilibria of arbitrary stoichiometry was a bold and correct conjecture.
The hypothesis that reaction rate is proportional to reactant concentrations is, strictly speaking, only true for elementary reactions (reactions with a single mechanistic step), but the empirical rate expression
is also applicable to second order reactions that may not be concerted reactions. Guldberg and Waage were fortunate in that reactions such as ester formation and hydrolysis, on which they originally based their theory, do indeed follow this rate expression.
In general many reactions occur with the formation of reactive intermediates, and/or through parallel reaction pathways. However, all reactions can be represented as a series of elementary reactions and, if the mechanism is known in detail, the rate equation for each individual step is given by the expression so that the overall rate equation can be derived from the individual steps. When this is done the equilibrium constant is obtained correctly from the rate equations for forward and backward reaction rates.
In biochemistry, there has been significant interest in the appropriate mathematical model for chemical reactions occurring in the intracellular medium. This is in contrast to the initial work done on chemical kinetics, which was in simplified systems where reactants were in a relatively dilute, pH-buffered, aqueous solution. In more complex environments, where bound particles may be prevented from disassociation by their surroundings, or diffusion is slow or anomalous, the model of mass action does not always describe the behavior of the reaction kinetics accurately. Several attempts have been made to modify the mass action model, but consensus has yet to be reached. Popular modifications replace the rate constants with functions of time and concentration. As an alternative to these mathematical constructs, one school of thought is that the mass action model can be valid in intracellular environments under certain conditions, but with different rates than would be found in a dilute, simple environment [citation needed].
The fact that Guldberg and Waage developed their concepts in steps from 1864 to 1867 and 1879 has resulted in much confusion in the literature as to which equation the law of mass action refers. It has been a source of some textbook errors. Thus, today the "law of mass action" sometimes refers to the (correct) equilibrium constant formula,
and at other times to the (usually incorrect) rate formula.
Applications to other fields
[edit]
In plasma physics
[edit]
Main article: Saha ionization equation
In a plasma, the ionization of the atoms can be understood as a chemical equilibrium between each ionization state with the next ionization state and a freed electron:
: etc.
and accordingly a law of mass action arises for each reaction, which in the ideally dilute limit is the Saha ionization equation.
In semiconductor physics
[edit]
Main article: Mass action law (electronics)
The law of mass action also has implications in semiconductor physics. Regardless of doping, the product of electron and hole densities is a constant at equilibrium. This constant depends on the thermal energy of the system (i.e. the product of the Boltzmann constant, , and temperature, ), as well as the band gap (the energy separation between conduction and valence bands, ) and effective density of states in the valence and conduction bands. When the equilibrium electron and hole densities are equal, their density is called the intrinsic carrier density as this would be the value of and in a perfect crystal. Note that the final product is independent of the Fermi level :
Diffusion in condensed matter
[edit]
Yakov Frenkel represented diffusion process in condensed matter as an ensemble of elementary jumps and quasichemical interactions of particles and defects. Henry Eyring applied his theory of absolute reaction rates to this quasichemical representation of diffusion. Mass action law for diffusion leads to various nonlinear versions of Fick's law.
In mathematical ecology
[edit]
The Lotka–Volterra equations describe dynamics of the predator-prey systems. The rate of predation upon the prey is assumed to be proportional to the rate at which the predators and the prey meet; this rate is evaluated as xy, where x is the number of prey, y is the number of predator. This is a typical example of the law of mass action.
In mathematical epidemiology
[edit]
The law of mass action forms the basis of the compartmental model of disease spread in mathematical epidemiology, in which a population of humans, animals or other individuals is divided into categories of susceptible, infected, and recovered (immune). The principle of mass action is at the heart of the transmission term of compartmental models in epidemiology, which provide a useful abstraction of disease dynamics. The law of mass action formulation of the SIR model corresponds to the following "quasichemical" system of elementary reactions:
: The list of components is S (susceptible individuals), I (infected individuals), and R (removed individuals, or just recovered ones if we neglect lethality);
: The list of elementary reactions is
: .
: If the immunity is unstable then the transition from R to S should be added that closes the cycle (SIRS model):
: .
A rich system of law of mass action models was developed in mathematical epidemiology by adding components and elementary reactions.
Individuals in human or animal populations – unlike molecules in an ideal solution – do not mix homogeneously. There are some disease examples in which this non-homogeneity is great enough such that the outputs of the classical SIR model and their simple generalizations like SIS or SEIR, are invalid. For these situations, more sophisticated compartmental models or distributed reaction-diffusion models may be useful.
See also
[edit]
Chemical equilibrium
Chemical potential
Disequilibrium ratio
Equilibrium constant
Reaction quotient
References
[edit]
^ Péter Érdi; János Tóth (1989). Mathematical Models of Chemical Reactions: Theory and Applications of Deterministic and Stochastic Models. Manchester University Press. p. 3. ISBN 978-0-7190-2208-1.
^ Jump up to: a b Chieh, Chung. "Chemical Equilibria - The Law of Mass Action". Chemical reactions, chemical equilibria, and electrochemistry. Archived from the original on 3 October 2018. Retrieved 21 July 2019. The law of mass action is universal, applicable under any circumstance... The mass action law states that if the system is at equilibrium at a given temperature, then the following ratio is a constant.
^ Föll, Helmut. "Mass Action Law". Defects in Crystals.
^ Berthollet, C.L. (1803). Essai de statique chimique [Essay on chemical statics [i.e., equilibrium]] (in French). Paris, France: Firmin Didot. On pp. 404–407, Berthollet mentions that when he accompanied Napoleon on his expedition to Egypt, he (Berthollet) visited Lake Natron and found sodium carbonate along its shores. He realized that this was a product of the reverse of the usual reaction Na2CO3 + CaCl2 → 2NaCl + CaCO3↓ and therefore that the final state of a reaction was a state of equilibrium between two opposing processes. From p. 405: " … la décomposition du muriate de soude continue donc jusqu'à ce qu'il se soit formé assez de muriate de chaux, parce que l'acide muriatique devant se partager entre les deux bases en raison de leur action, il arrive un terme où leurs forces se balancent." ( … the decomposition of the sodium chloride thus continues until enough calcium chloride is formed, because the hydrochloric acid must be shared between the two bases in the ratio of their action [i.e., capacity to react]; it reaches an end [point] at which their forces are balanced.)
^ Levere, Trevor H. (1971). Affinity and Matter – Elements of Chemical Philosophy 1800–1865. Gordon and Breach Science Publishers. ISBN 2-88124-583-8.
^ Jump up to: a b Waage, P.; Guldberg, C.M. (1864). "Studier over Affiniteten" [Studies of affinities]. Forhandlinger I Videnskabs-selskabet I Christiania (Transactions of the Scientific Society in Christiania) (in Danish): 35–45.
^ Jump up to: a b Waage, P. (1864). "Forsøg til Bestemmelse af Lovene for Affiniteten" [Experiment for the determination of the laws of affinity]. Forhandlinger I Videnskabs-selskabet I Christiania (Transactions of the Scientific Society in Christiania) (in Danish): 92–94.
^ Jump up to: a b Guldberg, C.M. (1864). "Foredrag om Lovene for Affiniteten, specielt Tidens Indflydelse paa de kemiske Processer" [Lecture on the laws of affinity, especially the influence of time on chemical processes]. Forhandlinger I Videnskabs-selskabet I Christiania (Transactions of the Scientific Society in Christiania) (in Danish): 111–120.
^ Guldberg, C.M.; Waage, P. (1867). Études sur les affinités chimiques [Studies of chemical affinities] (in French). Christiania [Oslo], Norway: Brøgger & Christie.
^ Jump up to: a b C.M. Guldberg and P. Waage, "Experiments concerning Chemical Affinity"; German translation by Abegg in Ostwalds Klassiker der Exacten Wissenschaften, no. 104, Wilhelm Engleman, Leipzig, 1899, pp 10-125
^ McLean, Franklin C. (1938). "Application of the Law of Chemical Equilibrium (Law of Mass Action) to Biological Problems" (PDF). Physiological Reviews. 18 (4): 495–523. doi:10.1152/physrev.1938.18.4.495.
^ van 't Hoff, J.H. (1877). "Die Grenzebene, ein Beitrag zur Kenntniss der Esterbildung" [The limit plane: a contribution to our knowledge of ester formation]. Berichte der Deutschen Chemischen Gesellschaft zu Berlin (in German). 10: 669–678. doi:10.1002/cber.187701001185.
^ Jump up to: a b Guldberg, C.M.; Waage, P. (1879). "Ueber die chemische Affinität" [On chemical affinity]. Journal für praktische Chemie. 2nd series (in German). 19: 69–114. doi:10.1002/prac.18790190111. Reprinted, with comments by Abegg in Ostwalds Klassiker der Exacten Wissenschaften, no. 104, Wilhelm Engleman, Leipzig, 1899, pp 126-171
^ (Guldberg & Waage, 1879), p. 71: "Eigentlich verstehen wir unter der activen Masse nur die Menge des Stoffes innerhalb der Actionsphäre; unter sonst gleichen Umständen kann aber die Actionsphäre durch die Volumeneinheit repräsentirt werden." (Actually we understand by "active mass" only the quantity of substance inside the sphere of action; under otherwise identical conditions, however, the sphere of action can be represented by the unit volume.)
^ Guggenheim, E.A. (1956). "Textbook errors IX: More About the Laws of Reaction Rates and of Equilibrium". Journal of Chemical Education. 33 (11): 544–545. Bibcode:1956JChEd..33..544G. doi:10.1021/ed033p544.
^ Law of Mass Action
^ SC.edu
^ The Law of Mass Action
^ Leech, Mary. "Acids and Bases" (PDF). Geochemistry — GEOL 480. San Francisco State University. Archived from the original (PDF) on 2006-09-21.
^ "Recap of Fundamental Acid-Base Concepts". Chemistry 152. Washington University in St. Louis. Archived from the original on 2012-02-06.
^ Chemical Equilibria: Basic Concepts
^ "Chemical equilibrium - The Law of Mass Action". Archived from the original on 2010-02-01. Retrieved 2007-07-12.
^ "Indiana.edu" (PDF). Archived from the original (PDF) on 2007-06-15. Retrieved 2007-07-12.
^ Berkeley.edu
^ General Chemistry Online: FAQ: Acids and bases: What is the pH at the equivalence point an HF/NaOH titration?
^ law of mass action definition
^ Lab 4 – Slow Manifolds Archived 2007-11-17 at the Wayback Machine
^ Landau, Lev D.; Lifshitz, Evgeny M. (1980). "§102: The law of mass action, §104: Ionization equilibrium". Statistical Physics. Vol. 5 (3rd ed.). Butterworth-Heinemann. ISBN 978-0-7506-3372-7.
^ A.N. Gorban, H.P. Sargsyan and H.A. Wahab (2011). "Quasichemical Models of Multicomponent Nonlinear Diffusion". Mathematical Modelling of Natural Phenomena. 6 (5): 184–262. arXiv:1012.2908. doi:10.1051/mmnp/20116509. S2CID 18961678.
^ von Csefalvay, Chris (2023), "Simple compartmental models", Computational Modeling of Infectious Disease, Elsevier, pp. 19–91, doi:10.1016/b978-0-32-395389-4.00011-6, ISBN 978-0-323-95389-4, retrieved 2023-03-02
Further reading
[edit]
Studies Concerning Affinity. P. Waage and C.M. Guldberg; Henry I. Abrash, Translator.
"Guldberg and Waage and the Law of Mass Action", E.W. Lund, J. Chem. Ed., (1965), 42, 548-550.
A simple explanation of the mass action law. H. Motulsky.
The Thermodynamic Equilibrium Constant
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Equilibrium Position
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An equilibrium position is the resting point of an object attached to an elastic spring, where the spring exerts a restoring force equal to the force of gravity acting on the object.
AI generated definition based on:Introductory Differential Equations (Fifth Edition), 2018
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Chapter
Applications of Higher Order Differential Equations
2018, Introductory Differential Equations (Fifth Edition)Martha L. Abell, James P. Braselton
5.1 Simple Harmonic Motion
Suppose that an object of mass m is attached to an elastic spring that is suspended from a rigid support such as a ceiling or a horizontal rod. The object causes the spring to stretch a distance s from its natural length. The position at which it comes to rest is called the equilibrium position. According to Hooke's law, the spring exerts a restoring force in the upward (opposite) direction that is proportional to the distance s that the spring is stretched. Mathematically, this is stated as F=k s where k>0 is the constant of proportionality or spring constant. In Fig. 5.1 we see that the spring has natural length b. When the object is attached to the spring, it is stretched s units past its natural length to the equilibrium position x=0. When the system is in motion, the displacement from x=0 at time t is given by x(t).
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Figure 5.1. A spring-mass system.
By Newton's second law of motion,
F=m a=m d 2 x d t 2,
where m represents the mass of the object and a represents acceleration. If we assume that there are no forces other than the force as a result of gravity acting on the mass, we determine the differential equation that models this situation by summing the forces acting on the spring-mass system with
m d 2 x d t 2=∑(forces acting on the system)=−k(s+x)+m g=−k s−k x+m g.
At equilibrium k s=m g, so after simplification we obtain the differential equation
m d 2 x d t 2=−k x or m d 2 x d t 2+k x=0.
The two initial conditions that are used with this problem are the initial position x(0)=α and the initial velocity x′(0)=β. The function x(t) that describes the displacement of the object with respect to the equilibrium position at time t is found by solving the initial value problem
m d 2 x d t 2+k x=0,x(0)=α,x(0)=β
Although the English scientist Robert Hooke (1635–1703) discovered “Hooke's law” in 1660, he didn't formally state it until 1678. Hooke and Newton did not get along well and were not friends.
Based on the assumption made in deriving the differential equation, positive values of x(t) indicate that the mass is below the equilibrium position and negative values of x(t) indicate that the mass is above the equilibrium position. The units that are encountered in these problems are summarized in Table 5.1.
Table 5.1. Units Encountered When Solving Spring-Mass Systems Problems
| System | Force | Mass | Length | k (Spring Constant) | Time |
--- --- --- |
| English | pounds (lb) | Slugs (lbs-s 2/ft) | feet (ft) | lb/ft | seconds (s) |
| Metric | Newton (N) | kilograms (kg) | meters (m) | N/m | seconds (s) |
Example 5.1
Determine the spring constant of the spring with natural length 10 in that is stretched to a distance of 13 in by an object weighing 5 lb.
Solution: Because the object weighs 5 lb, F=5 lb, and because displacement from the equilibrium position is 13−10=3 in, s=3 in × 1 ft/12 in = 1/4 ft. Therefore,
F=k s indicates that 5=1 4 k so k=20.
Notice that the spring constant is given in the units lb/ft because F=5 lb and s=1/4 ft. □
Example 5.2
An object of weight 16 lb stretches a spring 3 in. Determine the initial value problem that models this situation if (a) the object is released from a point 4 in below the equilibrium position with an upward initial velocity of 2 ft/s; (b) the object is released from rest 6 in above the equilibrium position; (c) the object is released from the equilibrium position with a downward initial velocity of 8 ft/s.
Solution: We first determine the differential equation that models the spring mass system (we use the same equation for all three parts of the problem). The information is given in English units, so we use g=32 ft/s 2. We must convert all measurements given in inches to feet. The object stretches the spring 3 in, so s=3 in×1 ft/12 in=1/4 ft. Also, because the mass weighs 16 lb, F=16. According to Hooke's law, 16=k⋅1/4, so k=64 lb/ft. We then find the mass m of the object with F=m g to find that 16=m⋅32 or m=1/2 slug. The differential equation used to find the displacement of the object at time t is
1 2 d 2 x d t 2+64 x=0.
The initial conditions for parts (a), (b), and (c) are then found.
(a)
Because 4 in×1 ft/12 in=1/3 ft and down is the positive direction, x(0)=1/3. Notice, however, that the initial velocity is 2 ft/s in the upward (negative) direction. Hence, x′(0)=−2.
(b)
A position 6 in (or 1/2 ft) above the equilibrium position (in the negative direction) corresponds to initial position x(0)=−1/2. Being released from rest indicates that d x/d t(0)=0.
(c)
Because the mass is released from the equilibrium position, x(0)=0. Also, the initial velocity is 8 ft/s in the downward (positive) direction, so x′(0)=8. □
Example 5.3
An object weighing 60 lb stretches a spring 6 in. Determine the function x(t) that describes the displacement of the object if it is released from rest 12 in below the equilibrium position.
Solution: First, the spring constant k is determined from the supplied information. By Hooke's law, F=k s, so we have 60=k⋅0.5. Therefore, k=120 lb/ft. Next, the mass m of the object is determined using F=m g. In this case, 60=m⋅32, so m=15/8 slugs. Because k/m=64, and 12 in is equivalent to 1 ft, the initial value problem that models the situation is x″+64 x=0, x(0)=1, x′(0)=0.
The characteristic equation that corresponds to the differential equation is r 2+64=0. It has solutions r 1,2=±8 i, so a general solution of the equation is x=c 1 cos8 t+c 2 sin8 t.
To find the values of c 1 and c 2 that satisfy the initial conditions, we calculate x′=−8 c 1 sin8 t+8 c 2 cos8 t. Then x(0)=c 1=1 and x′(0)=8 c 2=0 so c 1=1, c 2=0, and x=cos8 t.
Notice that x=cos8 t indicates that the spring-mass system never comes to rest once it is set into motion. The solution is periodic, so the mass moves vertically, retracing its motion, as shown in Fig. 5.2. Motion of this type is called simple harmonic motion. □
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Figure 5.2. Simple harmonic motion.
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What is the maximum displacement of the object in Example 5.3 from the equilibrium position?
Example 5.4
An objective weighing 2 lb stretches a spring 1.5 in. (a) Determine the function x=x(t) that describes the displacement of the object if it is released with a downward initial velocity of 32 ft/s from 12 in above the equilibrium position. (b) At what value of t does the object first pass through the equilibrium position?
Solution: (a) We begin by determining the spring constant. Because the force F=2 lb stretches the spring 3/2 in×1 ft/12 in=1/8 ft, k is found by solving 2=k⋅1/8. Hence k=16 lb/ft. With F=m g, we have m=2/32=1/16 slug. The differential equation that models this situation is 1 16 x″+16 x=0 or x″+256 x=0. Because 12 in is equivalent to 1 ft, the initial position above the equilibrium (in the negative direction) is x(0)=−1. The downward initial velocity (in the positive direction) is x′(0)=32. Therefore, we must solve the initial value problem x″+256 x=0, x(0)=−1, x′(0)=32. The characteristic equation associated with x″+256 x=0 is r 2+256=0 with roots r 1,2=±16 i so a general solution of x″+256 x=0 is x=c 1 cos16 t+c 2 sin16 t with derivative x′=−16 c 1 sin16 t+16 c 2 cos16 t. Application of the initial conditions then yields x(0)=c 1 cos0+c 2 sin0=c 1=−1 and x′(0)=−16 c 1 sin0+16 c 2 cos0=16 c 2=32 so c 2=2. The position function is given by x=−cos16 t+2 sin16 t. (See Fig. 5.3 A.)
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Figure 5.3. Two plots of simple harmonic motion. (A) (on the left) is for Example 5.4 while (B) (on the right) is for Example 5.5.
(b) To determine when the object first passes through its equilibrium position, we solve the equation x=−cos16 t+2 sin16 t=0 or tan16 t=1/2. Therefore, t=1 16 tan−1(1/2)≈0.03 s, which appears to be a reasonable approximation based on the graph of x=x(t). □
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Approximate the second time the mass considered in Example 5.4 passes through the equilibrium position.
A general formula for the solution of the initial value problem
m d 2 x d t 2+k x=0,x(0)=α,d x d t(0)=β
is
(5.1)x=α cosω t+β ω−1 sinω t,where ω=k/m.
Through the use of the trigonometric identity cos(a+b)=cosa cosb−sina sinb, we can write x=x(t) in terms of a cosine function with a phase shift. First, let x=A cos(ω t−ϕ). We then apply the identity to see that x=A cosω t cosϕ+A sinω t sinϕ. Comparing the functions
x=α cosω t+β ω−1 sinω t and x=A cosω t cosϕ+A sinω t sinϕ
indicates that
A cosϕ=α and A sinϕ=β/ω.
Thus cosϕ=α/A and sinϕ=β/(A ω). Because cos 2ϕ+sin 2ϕ=1, (α/A)2+[β/(A ω)]2=1. Therefore, the amplitude of the solution is A=α 2+β 2/ω 2 and x=α 2+β 2/ω 2 cos(ω t−ϕ), where ϕ=cos−1(α α 2+β 2/ω 2) and ω=k/m. Note that the period of x=x(t) is T=2⋅π/ω=2 π m/k. In many cases, questions about the displacement function are more easily answered if the solution is written in this form.
Example 5.5
A 4-kg mass stretches a spring 0.392 m. (a) Determine the displacement function if the mass is released from 1 m below the equilibrium position with a downward initial velocity of 10 m/s. (b) What is the maximum displacement of the mass? (c) What is the approximate period of the displacement function?
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How does an increase in the magnitude (absolute value) of the initial position and initial velocity affect the amplitude of the resulting motion of the spring-mass system? From experience, does this agree with the actual physical situation?
Solution: (a) Because the mass of the object (in metric units) is m=4 kg, we use this with F=m g to determine the force. We first compute F=4⋅9.8=39.2 N. We then find the spring constant with 39.2=k⋅0.392, so k=100 N/m. The differential equation that models this spring-mass system is 4 x″+100 x=0 or x″+25 x=0. The initial position is x(0)=1, and the initial velocity is x′(0)=10. We must solve the initial value problem x″+25 x=0, x(0)=1, x′(0)=10 either directly or with the general formula obtained in Eq. (5.1). Using the general formula with α=1, β=10, and ω=100/4=5, we have
x=α 2+β 2/ω 2 cos(ω t−ϕ)=1 2+10 2/5 2 cos(5 t−ϕ)=5 cos(5 t−ϕ),
where ϕ=cos−1(1/5)≈1.11 rad, which we graph in Fig. 5.3 B.
(b) From our knowledge of trigonometric functions, we know that the maximum value of x(t)=5 cos(5 t−ϕ) is x=5. Therefore, the maximum displacement of the mass from its equilibrium position is 5≈2.236 m.
(c) The period of this trigonometric function is T=2 π/ω=2 π/5. The mass returns to its initial position every 2 π/5≈1.257 s. □
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Book2018, Introductory Differential Equations (Fifth Edition)Martha L. Abell, James P. Braselton
Chapter
CHEMICAL REACTIONS
1967, General PhysicsL.D. LANDAU, ... E.M. LIFSHITZ
§87.Chemical equilibrium
As a chemical reaction proceeds, the quantities of the original substances decrease, and reaction products accumulate. Ultimately, the reaction leads to a state in which the quantities of all the substances no longer vary. This is called a state of chemical equilibrium, a particular case of thermal equilibrium.
In chemical equilibrium there is generally present a certain quantity of the original substances as well as the products formed in the reaction. It is true that in many cases this quantity is very small, but this does not, of course, affect the principle.
The establishment of chemical equilibrium in which both initial and final substances are present occurs for the following reason. Let us consider, for example, a reaction between hydrogen and iodine gases to form hydrogen iodide:
H 2+I 2=2 HI.
As well as the formation of HI from H 2 and I 2, in a mixture of these three substances the reverse process of dissociation of HI into hydrogen and iodine will also necessarily occur: the forward reaction is always accompanied by the reverse reaction. As the quantity of HI increases and that of H 2 and I 2 decreases, the forward reaction will obviously become slower and the reverse reaction quicker, and a point is finally reached at which the rates of the two reactions become equal, with the same number of new HI molecules formed as dissociate in the same time. The quantities of all the substances thereafter remain unchanged.
Thus chemical equilibrium (and in fact other types of thermal equilibrium) is dynamic on the molecular scale; the reactions do not actually cease, but the forward and reverse reactions occur at equal rates and therefore produce no overall effect.
It is clear that, if the reaction in the above example begins from a mixture of hydrogen and iodine, the relative quantities of all three substances in the equilibrium state will be the same as in a reaction which begins with the decomposition of pure HI. The chemical equilibrium position does not depend on the side from which it is approached.
Moreover, the chemical equilibrium also does not depend on the conditions under which the reaction occurs or on the intermediate stages through which it passes. The position of equilibrium depends only on the state of the substances in equilibrium, i.e. on the temperature and pressure of the equilibrium mixture.
When the temperature changes, the position of chemical equilibrium is altered. The direction of this shift depends entirely on the heat of reaction, as is easily seen by means of Le Chatelier'sprinciple. Let us consider an exothermic reaction, such as the formation of ammonia from nitrogen and hydrogen (N 2 + 3H 2 = 2NH 3), and assume that the reaction has already reached a state of equilibrium. If the equilibrium mixture is now heated, processes must occur in it which tend to cool it: a certain quantity of ammonia must decompose, and heat is thus absorbed. This means that the chemical equilibrium is shifted in the direction such that the quantity of ammonia is decreased.
Thus the “yield” of exothermic reactions decreases when the temperature is raised; in endothermic reactions, on the other hand, the yield increases with increasing temperature.
Similarly, the dependence of the equilibrium position on the pressure is related to the change in volume accompanying the reaction (at constant pressure). Increasing the pressure lowers the yield of reactions in which the volume of the reacting mixture increases, and raises that of reactions in which the volume decreases. The latter case occurs, for instance, in the reaction of formation of gaseous ammonia: since the number of NH 3 molecules formed is less than the number of reacting N 2 and H 2 molecules, the volume of the gas mixture decreases in the reaction.
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Book1967, General PhysicsL.D. LANDAU, ... E.M. LIFSHITZ
Chapter
Applications of Higher Order Differential Equations
2018, Introductory Differential Equations (Fifth Edition)Martha L. Abell, James P. Braselton
Chapter 5 Review Exercises
1.
An object weighing 32 lb stretches a spring 6 in. If the object is lowered 4 in below the equilibrium and released from rest, determine the displacement of the object, assuming there is no damping. What is the maximum displacement of the object from equilibrium? When does the object first pass through the equilibrium position? How often does the object return to the equilibrium position?
2.
If the object in Exercise 1 is released from a point 3 in above equilibrium with a downward initial velocity of 1 ft/s, determine the displacement of the object, assuming there is no damping. What is the maximum displacement of the object from equilibrium? When does the object first pass through the equilibrium position? How often does the object return to the equilibrium position?
3.
An object of mass 5 kg is attached to the end of a spring with spring constant k=65 N/m. If the object is released from the equilibrium position with an upward initial velocity of 1 m/s, determine the displacement of the object assuming the force due to damping is F R=20 d x/d t. Find lim t→∞x(t) and the quasiperiod. What is the maximum displacement of the object from equilibrium? When does the object first pass through the equilibrium position?
4.
If the object in Exercise 3 is released from a point 1 in below equilibrium with zero initial velocity, determine the displacement. Find lim t→∞x(t) and the quasiperiod. What is the maximum displacement of the object from equilibrium? When does the object first pass through the equilibrium position?
5.
An object of mass 4 slugs is attached to a spring with spring constant k=16 lb/ft. If there is no damping and the object is subjected to the forcing function f(t)=4, determine the displacement function x(t) if x(0)=x′(0)=0. What is the maximum displacement of the object from equilibrium? When does the object first pass through the equilibrium position?
6.
If the object in Exercise 5 is subjected to the forcing function f(t)=4 cos2 t, determine the displacement. Find lim t→∞x(t) if it exists. Describe the physical phenomenon that occurs.
7.
If the object in Exercise 5 is subjected to the forcing function f(t)=4 cost, determine the displacement. Describe the physical phenomenon that occurs. Find the envelope functions.
8.
An object of mass 2 slugs is attached to a spring with spring constant k=5 lb/ft. If the resistive force is F R=6 d x/d t and the external force is f(t)=12 cos2 t, determine the displacement if x(0)=x′(0)=0. What is the steady-state solution? What is the transient solution?
9.
Find the charge and current in the L-R-C circuit if L=4 H, R=80 Ω, C=1/436 F, and E(t)=100 if Q(0)=Q′(0)=0. Find lim t→∞Q(t) and lim t→∞I(t).
10.
Find the charge and current in the L-R-C circuit in Exercise 9 if E(t)=100 sin2 t. Find lim t→∞Q(t) and lim t→∞I(t). How do these limits compare to those in Exercise 9?
11.
Find the charge and current in the L-R-C circuit if L=1 H, R=0 Ω, C=10−4 F, and E(t)=220 if Q(0)=Q′(0)=0. Find lim t→∞Q(t) and lim t→∞I(t).
12.
Find the charge and current in the L-R-C circuit in Exercise 11 if E(t)=100 sin10 t. Find lim t→∞Q(t) and lim t→∞I(t).
13.
Determine the shape of the beam of length 10 with constants E and I such that E I=1, weight distribution w(x)=x(10−x), and fixed-end boundary conditions at x=0 and x=10.
14.
Determine the shape of the beam in Exercise 13 if there are fixed-end boundary conditions at x=0 and a sliding clamped end at x=10.
15.
Determine the shape of the beam in Exercise 13 if there are fixed-end boundary conditions at x=0 and a free end at x=10.
16.
Determine the shape of the beam in Exercise 13 if there are fixed-end boundary conditions at x=0 and simple support at x=10.
17.
Use the linear approximation of the model of the simple pendulum to determine the motion of a pendulum with rod length L=1/2 ft subject to the initial conditions θ(0)=1 and d θ/d t(0)=0. What is the maximum displacement of the pendulum from the vertical position? When does the pendulum first pass through the vertical position?
18.
If the initial conditions in Exercise 17 are θ(0)=0 and θ′(0)=−1, what is the maximum displacement of the pendulum from the vertical position? When does the pendulum first return to the vertical position?
19.
How does the motion of the pendulum in Exercise 17 differ if it undergoes the damping forceF R=8 d θ/d t?
20.
Solve the model in Exercise 17 with the damping force F R=8 3 d θ/d t. How does the motion differ from that in Exercise 19?
21.
Undamped torsional vibrations (rotations back and forth) of a wheel attached to a thin elastic rod or wire satisfy the differential equationI 0 θ″+k θ=0, where θ is the angle measured from the state of equilibrium, I 0 is the polar moment of inertia of the wheel about its center, and k is the torsional stiffness of the rod. Solve this equation if k/I 0=13.69 s−2, the initial angle is 15 o≈0.2168 rad, and the initial angular velocity is 10 o s−1≈0.1745 rad⋅s−1.
22.
Determine the displacement of the spring-mass system with mass 0.250 kg, spring constant k=2.25 kg/s 2, and driving force f(t)=cost−4 sint if there is no damping, zero initial position, and zero initial velocity. For what frequency of the driving force would there be resonance?
23.
The initial value problem
y″+y={1−t 2,0⩽t<1 0,t⩾1,y(0)=y′(0)=0
can be thought of as an undamped system in which a force F acts during the interval of time 0⩽t<1. This is the situation that occurs in a gun barrel when a shell is fired. The barrel is braked with heavy springs, as indicated in the following figure.
Solve this initial value problem.
24.
Consider a buoy in the shape of a cylinder of radius r, height h, and density ρ where ρ⩽1/2 g/cm 3 (the density of water is 1 g/cm 3). Initially, the buoy sits with its base on the surface of the water. It is then released so that it is acted on by two forces: the force of gravity (in the downward direction) equal to the weight of the buoy, F 1=m g=ρ π r 2 h g; and the force of buoyancy (in the upward direction) equal to the weight of the displaced water, F 2=π r 2 x g, where x=x(t) is the depth of the base of the cylinder from the surface of the water at time t. Using Newton's second law of motion with m=ρ π r 2 h, we have the differential equation ρ π r 2 h x″=−π r 2 x g or ρ h x″+g x=0. Find the displacement of the buoy if x(0)=1 and x′(0)=0. What is the period of the solution? What is the amplitude of the solution?
25.
A cube-shaped buoy of side length ℓ and mass density ρ per unit volume is floating in a liquid of mass density ρ 0 per unit volume, where ρ 0>ρ. If the buoy is slightly submerged into the liquid and released, it oscillates up and down. If there is no damping and no air resistance, the buoy is acted on by two forces: the force of gravity (in the downward direction), which is equal to the weight of the buoy, F 1=m g=ρ ℓ 3 g; and the force of buoyancy (in the upward direction), which is equal to the weight of the displaced water, F 2=ℓ 2 x g ρ 0, where x=x(t) is the depth of the base of the buoy from the surface of the water at time t. Then by Newton's second law with m=ρ ℓ 3, we have the differential equation ρ ℓ 2 x″+g ρ 0 x=0. Determine the amplitude of the motion if ρ 0=1 g/cm 3, ρ=1/4 g/cm 3, ℓ=100 cm, g=980 cm/s 2, x(0)=25 cm and x′(0)=0.
26.
(Pursuit Models) A rabbit starts at the origin and runs with speed a due north toward a hole in a fence located at the point (0,d) on the y-axis. At the same time, a dog starts at the point (c,0) on the x-axis, running at speed b in pursuit of the rabbit. (Note: The dog runs directly toward the rabbit.) The slope of the tangent line to the dog's path is d y/d x=−(a t−y)/x, which can be written as x y′=y−a t. Differentiate both sides of this equation with respect to t to obtain x y″=−a d t/d x. If s is the length of the arc from (c,0) along the dog's path, then d s/d t=b is the dog's speed. Also, d s/d x=−1+(y′)2, where the negative sign indicates that s increases as x decreases. By the chain rule,
d t d x=d t d s d s d x=−1 b 1+(y′)2,
so substitution into the equation x y″=−a d t/d x yields the equation x y″=k 1+(y′)2, where k=a/b. Make the substitution p=y′ or y″=d p/d x to find the path of the dog with the initial condition y′(c)=0. If a<b, when does the dog catch the rabbit? Does the problem make sense if a=b? (See the following figure.)27.
(Coulomb damping) Damping that results from dry friction (as when an object slides over a dry surface) is called Coulomb damping or dry-friction damping. On the other hand, viscous damping is damping that can be represented by a term proportional to the velocity (as when an object such as a vibrating spring vibrates in air or when an object slides over a lubricated surface).
Suppose that the kinetic coefficient of friction is μ. Friction forces oppose motion. Therefore, the force as a result of friction F is shown opposite the direction of motion in Fig. 5.24. However, because F is a discontinuous function, we cannot use a single differential equation to model the motion as was done with previous damping problems. Instead, we have one equation for motion to the right and one equation for motion to the left:
Motion to right
x″+ω n 2 x=−F/m
Motion to left
x″+ω n 2 x=F/m,
where F represents the friction force, m the mass, and ω n 2=k/m.1
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Figure 5.24. A system with Coulomb (dry-friction) damping.
(a)
Find a general solution of x″+ω n 2 x=F/m.
(b)
Solve the initial value problem x″+ω n 2 x=F/m, x(0)=x 0, x′(0)=0.
(c)
Find the values of t for which the solution in (b) is valid.
(d)
At the right-hand endpoint of the interval obtained in (c), what is the displacement of the object?
(e)
What must the force F be to guarantee that the object does not move under the given initial conditions?
28.
(Self-excited vibrations) The differential equation that describes the motion of a spring-mass system with a single degree of freedom excited by the force P x′ is m x″+c x′+k x=P x′, which can be rewritten as x″+c−P m x′+k m x=0.2
(a)
Show that the roots of the characteristic equation of this equation are
r 1,2=P−c 2 m±(P−c 2 m)2−k m.
(b)
Show that if P>c the motion of the system diverges (dynamically unstable), if P=c the solution is the solution for a free undamped system, and if P<c the solution is the solution for a free damped system.
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Book2018, Introductory Differential Equations (Fifth Edition)Martha L. Abell, James P. Braselton
Chapter
SYSTEM FOR T-15 TOKAMAK POLOIDAL FIELD CONTROL
1991, Fusion Technology 1990N.D. Vasiliev, ... I.V. Mozin
1 INTRODUCTION
Plasma equilibrium pozition control is rather important and interesting problem on the most TOKAMAK-type facilities. High-temperature plasma should be confined in an equilibrium position not to contact with the wall at fast changes of the plasma and operating parameters. In this paper the description of the system an engineering realization of the plazma egualibrium control concept on a T-15 facility is given. Control is hierarchical and is realized on three levels:
-
calculation of control programs for rather long-term series of experiments;
-
correction of control law by the experimental results aimed at compensation of program errors;
-
real-time control to compensate occasional fluctuations.
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Book1991, Fusion Technology 1990N.D. Vasiliev, ... I.V. Mozin
Chapter
Tribology
2003, Encyclopedia of Physical Science and Technology (Third Edition)Andras Z. Szeri
II.D.6 Bearing Stability
In some applications both the magnitude and the direction of the external load W change in time. To respond to changing loading conditions, that is, to changes in the value of the Sommerfeld number S, the journal center will move out of its static equilibrium position. Conditions might be such that the journal will orbit with angular velocity Ω about its static position, but return to it when the load conditions once again become steady. If, on the other hand, the bearing center spirals outward from its equilibrium position as time progresses, the bearing eventually self-destroys as metal-to-metal contact will lead to excessive frictional heating and consequent bearing seizure.
The well-designed bearing is stable and the amplitude of the orbit of the journal center is small. This being the case, we are permitted to treat the instantaneous incremental lubricant force on the journal 6 π 2 ε(2+ε 2)(1−ε 2) as small perturbation of the equilibrium force, F X(0),F Y(0). This idea leads to a representation of dynamic conditions in bearings via linear springs and dampers.
(31)[d F X d F Y]=−[K X X K X Y K Y X K Y Y][X Y]−[B X X B X Y B Y X B Y Y][X˙Y˙]
Here K XX,…, K YY represent film stiffness and B XX,…, B YY represent film damping coefficients. Some of the linearized “springs” and “dampers” are shown in Fig. 20 schematically. For a finite bearing the coefficients are given by various solutions of Eq. (17). These solutions must be obtained numerically. It is found that full journal bearings are inherently unstable at small specific load P and for this reason they are rarely used on large machines. Multiple pad pivoted bearings, on the other hand, have excellent stability characteristics, especially when preloaded.
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FIGURE 20. Representation of oil film as a simple dynamic system comprising springs and dampers (cross-film springs K XY, K YX and dampers B XY, B YY are not shown). [Reprinted with permission from Raimondi, A. A., and Szeri, A. Z. Journal and thrust bearings. In Booser, E. R. (1984). “CRC Handbook of Lubrication,” CRC Press, Boca Raton, FL.]
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Reference work2003, Encyclopedia of Physical Science and Technology (Third Edition)Andras Z. Szeri
Chapter
3D force-feedback optical tweezers for experimental biology
2023, Robotics for Cell Manipulation and CharacterizationEdison Gerena, Sinan Haliyo
4.6 Calibration
The force on the trap is calculated using the optical force model :
(7.4)F o p t=K×(P l a s e r−P p r o b e)
where Plaser −Pprobe represents the displacement between the laser and the probe position as obtained from the tracking method. K is the stiffness of the trap.
Practically, to obtain the laser position, the positions of trapped probe before touching anything are recorded for a period of time and the average position is considered as the position of OTs, which is around (120, 150) in the ATIS coordinate. The stiffness is calculated experimentally using the equipartition method . For an object in a harmonic potential:
(7.5)1 2 k B T=1 2 K〈d〉2
where k B is Boltzmann's constant, T is absolute temperature, and 〈d〉2 is the displacement variance (Brownian motion) of the particle from its trapped equilibrium position. Considering 300 mW laser power it is estimated in x-axis, y-axis, and z-axis as K x = 12.3 pN/μm, K y = 12.6 pN/μm, and K z = 1.5 pN/μm, respectively, under room temperature of 25.5°C. The axial stiffness is less than the lateral stiffness, which is normal according to . This indicates that during manipulation, the loss of the trapped object is more likely caused by the reacting force in the axial direction. The same laser power will be used for the following experiments, under the same condition, that is, temperature, laser power, medium, etc. Stiffness adjustments may be required depending on the specific applications, then the calibration process will be reconducted for obtaining the OTs’ stiffness.
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Book2023, Robotics for Cell Manipulation and CharacterizationEdison Gerena, Sinan Haliyo
Chapter
Motion Artifacts Compensation in DCE-MRI Framework Using Active Contour Model
2016, Emerging Trends in Applications and Infrastructures for Computational Biology, Bioinformatics, and Systems BiologyR. Setola, ... B.B. Zobel
27.4.3 Active Contour
Once the distance map and magnitude of the gradient image are defined, we deform the initial contour by applying forces obtained by minimizing the energy functional as mentioned above (recall that the external force is given by the sum of the force calculated on the gradient image and that calculated on the distance map). The coordinates of the snake are updated at each iteration during the evolution of the algorithm. This operation is repeated until it is deemed that an equilibrium position is attained. The solutions adopted are: verify when the distance between the current contour and the previous does not exceed a predetermined limit; and operate the same analysis in terms of energy, taking into account the values assumed by the functional.
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Book2016, Emerging Trends in Applications and Infrastructures for Computational Biology, Bioinformatics, and Systems BiologyR. Setola, ... B.B. Zobel
Chapter
Physical Realization of Quantum Information Processing Systems
2012, Classical and Quantum InformationDan C. Marinescu, Gabriela M. Marinescu
Entanglement of the Motional and Internal Degrees of Freedom of Trapped Ions
The coupling of the motion of the ions inside a trap is provided by the Coulomb repulsion between ions, which is much stronger than any other interaction when the typical separations between the ions is of a few optical wavelengths. For example, the average separation between calcium ions in a linear Paul trap is about 10µm. The mutual Coulomb repulsion creates a spectrum of coupled normal modes of vibration for the trapped ions. The frequencies of different normal modes are well separated in the excitation spectrum.
The center-of-mass mode in which the ions oscillate in lock step in the harmonic potential well of the trap is the vibrational mode of lowest frequency, (ω x), along the x-axis. Initially, the ions are laser cooled at a temperature much lower than that corresponding to ω x to ensure that each vibrational mode occupies its quantum mechanical ground state. When an ion absorbs or emits a laser photon, the center of mass of the ion recoils.
The laser pulses incident on the ion can be tuned to simultaneously cause internal spin transitions and vibrational (phonon) excitations; local (atomic) internal states are thus mapped into shared (collective) phonon states. For certain choices of the laser frequency, ω L, the Hamiltonian describing the interaction between an ion and the electric field of the laser beam is resonant and the spin and the motion can be coupled efficiently (sideband coupling 6 ) . For example, for laser frequencies, ω L = ω 0 ∓ ωx, the interaction Hamiltonians take resonant forms and the spin transitions, |↓〉, → |↓〉, are accompanied by the corresponding motional mode transitions, |n〉 → |n ∓ 1〉. If the laser frequency takes the value, ω L = ω 0, the Hamiltonian is such that the spin transitions |↓〉 → |↓〉 do not change n (i.e., they are no longer accompanied by motional mode transitions).
In certain conditions, the so-called Lamb-Dicke limit, and for sufficiently low intensities, the laser beam will cause transitions that modify the state of only one of the collective modes. For example, if the laser is detuned from the resonance frequency, ω 0, by the quantity, Δω (j) L, which is equal to minus the center-of-mass frequency, ω x, i.e. Δω (j) L = –ω x (lower motional sideband), the laser excites the ion internal transition plus the center-of-mass motion mode as the only collective vibrational mode.
If the laser addresses the j-th ion in a string of N ions with a frequency tuned to the lower motional sideband of the center-of-mass mode and the equilibrium position of the ion in the linear trap coincides with the node of the laser standing wave, an interaction takes place between the internal states of ion, j, and the center-of-mass mode; the corresponding Hamiltonian for this single ion is
H j,q=η N Ω 2(|e q〉j j〈g|a e-i φ+|g〉j j〈e q|a†e i φ).
The terms, a↑ and a, are the creation and annihilation operators of center-of-mass phonons, respectively; Ω is the Rabi frequency; ψ is the laser phase; and η is the Lamb-Dicke parameter. The subscript, q = 0, 1, refers to the virtual electron level excited by the laser; the transition depends on the laser polarization. These are the intermediate (virtual, electronically excited) states of the Raman process.7 The factor, N, appears because for, N, ions the effective mass of the center-of-mass motion is NM (M is the mass of the ion), and the amplitude of the mode is proportional to 1/N M. It is assumed that (Ω/2ω x)2 η 2 ≪ 1 and in the Lamb-Dicke limit 8 η ≪ 1.
If this laser shines on the ion for a time, t=k π/(Ω η/N) (a kπ laser pulse), the evolution of the system represents a mapping of the spin qubit to the motion qubit and is described by the unitary operator
U j k,q=e-l H j,q t=exp[-i k π 2(|e q〉j j〈g|a e-i φ+h.c.)].
The operator describing the evolution of the system is unitary if the laser pulse duration is proportional to k π, with k, the laser wave vector. This transformation does not change the system state, |g〉j|0〉c, but transforms coherently the states, |g〉j|1〉c and |e q〉j|0〉c, to
|g〉j|1〉c→cos(k π 2)|g〉j|1〉c-i e i φ sin(k π 2)|e q〉j|0〉c,|e〉j|0〉c→cos(k π 2)|e q〉j|0〉c-i e i φ sin(k π 2)|g〉j|1〉c,
where |0〉c and |1〉c represent the center-of-mass states with zero phonon (vacuum), respectively, one phonon. According to the last transition, the laser pulse removes a bit of information stored as an internal excited state of the j-th ion and transfers it to the collective state of motion of all the ions. In this way, the state of motion of the i-th ion (where i ≠ j) is going to be influenced by the change of internal state of the j-th ion. The entanglement between the spin of internal atomic states and motion is evident since the final state cannot be factored out into a product of spin and motional wave functions.
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Chapter
Placement
2009, Electronic Design AutomationChris Chu
11.5 Global Placement: Analytical Approach
The basic idea of the analytical approach is to express the cost function and the constraints as analytical functions of the coordinates of the modules. Then the placement problem is transformed into a mathematical program. To illustrate this approach, an exact, but impractical, formulation is first presented in Section 11.5.1. Practical analytical placement techniques can be classified as quadratic and nonquadratic, which are presented in Sections 11.5.2 and 11.5.3, respectively. Note that analytical placement techniques generally treat both standard cells and macros in the same manner. (Sometimes, special techniques are used to handle large macros to improve wirelength or to make legalization easier, but in theory, they are not essential.) Thus, the techniques presented in this section can be considered to be for mixed-size designs.
Some notations used in this section are introduced in the following. Consider a circuit with a set of modules V and a set of nets E is to be placed in a region of width W and height H. Assume the modules in V are indexed from 1 to n. For module i ∈ V, let w i and h i be its width and height, and let x i and y i be the x- and y-coordinates of its center. For each net e ∈ E, let c e be its weight. Note that if a module i is a fixed module, then x i and y i are constants. Otherwise, they are variables. Assume for simplicity that all pins are located at the center of a module.
11.5.1 An exact formulation
It is instructive to first look at an exact formulation to the wirelength-minimized placement problem for mixed-size design. Then the rationales, pros, and cons of practical analytical placement techniques presented in Sections 11.5.2 and 11.5.3 can be better understood.
The HPWL of net e ∈ E can be written as:
HPWL e(x 1,…,x n,y 1,…,y n)=(max i∈e{x i}-min i∈e{x i})+(max i∈e{y i}-min i∈e{y i})
To express the nonoverlapping constraints among modules, it is necessary to introduce the following function:
Θ([L 1,R 1],[L 2,R 2])=[min(R 1,R 2)-max(L 1,L 2)]+
where
[z]+={z 0 if z>0 if z≤0
Θ([L 1, R 1], [L 2, R 2]) gives the length of the overlapping region of the intervals [L 1, R 1] and [L 2, R 2] as illustrated in Figure 11.9. Then the overlapping area between module i and module j is given by:
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FIGURE 11.9. The overlapping region of the intervals [L 1, R 1] and [L 2, R 2].
Overlap i j(x i,y i,x j,y j)=Θ([x i-ω i 2,x i+ω i 2],[x i-ω j 2,x j+ω j 2])Θ([y i-b i 2,y i+b i 2],[y j-b j 2,y j+b j 2])
The placement problem can be written as the following mathematical program:
Minimize∑e∈E c e×HPWL e(x 1,…,x n,y n,…,y n)Subject to Overlap i j(x i,y i,x j,y j)=0 for all i,j∈V s.t.i≠j 0≤x i-ω i 2,x i+ω i 2≤W for all i∈V 0≤y i-b i 2,y i+b i 2≤H for all i∈V
This mathematical program is extremely difficult to handle. The functions Overlap ij(x i, y i, x j, y j) for i, j ∈ V are highly nonconvex and not differentiable. The functions HPWL e (x 1, …, x n, y 1, …, y n) for e ∈ E, although convex, are not differentiable. Moreover, there are O(n 2) constraints with the number of modules n being up to millions in modern designs. Therefore, this is not a practical formulation. In practice, the wirelength is approximated by some differentiable and convex functions. The nonoverlapping constraints are usually replaced by some simpler constraints to make the module distribution roughly even. Then legalization is performed to eliminate the module overlaps, and detailed placement is applied to refine the solution with a more accurate wire-length metric. Various techniques are presented in the remainder of this section.
11.5.2 Quadratic techniques
In quadratic placement techniques, the placement problem is transformed into a sequence of convex quadratic programs. Convex quadratic program is a mathematical program with a convex and quadratic objective function and linear constraints.
11.5.2.1 Quadratic wirelength
First, the way to express the placement cost function as a quadratic function is presented. Suppose for the time being that all nets in the circuit are 2-pin nets. (Multi-pin nets will be discussed later.) Consider a net {i, j} (i.e., connecting module i and module j). Its wirelength is given by the Manhattan distance between the modules:
L{i,j}=|x i-x j|+|y i-y j|
This is usually referred to as linear wirelength. However, this function is not differentiable. So a common idea is to consider the squared Euclidean distance between the modules instead:
L˜{i,j}=(x i-x j)2+(y i-y j)2
This is usually called quadratic wirelength. To help visualize the functions, the x-components of L{i, j} and
{i, j} with a fixed value of x j = 2 are plotted as functions of x i, in Figure 11.10.
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FIGURE 11.10. Comparison of quadratic wirelength and linear wirelength.
In quadratic placement techniques, it is more convenient to set the cost function
to be half 2 of the total weighted quadratic wirelength:
L˜=1 2∑1≤i<j≤n c{i,j}×((x i-x j)2+(y i-y j)2)
can be written succinctly in matrix form in terms of the coordinates of movable modules. In the following, assume modules 1 to r are movable and modules r + 1 to n are fixed. Therefore, x 1, …, x r and y 1, …, y r are variables, and x r+1, …, x n and y r+1, …, y n are constants. Let x = (x 1 x 2 … x r)T and y = (y 1 y 2 … y r)T be the vectors of x-coordinate and y-coordinate of movable modules, respectively. Let D = (d ij)r × r be a diagonal matrix such that d ii = Σ j∈v c{i, j} for all i ∈ {1, …, r}. Let C = (c ij)r × r be the connectivity matrix among movable modules, i.e., c ij = c ji = c{i,j} for all i, j ∈ {1, …, r}. Let Q = D – C. Let d x T = (d x 1 … d x r) such that d x i = – Σ j {r + 1, …, n}c ij x j and dy = (d y 1 … d y r) such that d y 1 = – Σ j ∈ {r + 1 … n}c ij y j. Then
[L˜=1 2 x T Q x+d x x+1 2 y T Q y+d y y+constant terams]
For example, for the circuit with 3 movable modules and 3 fixed modules as represented by the graph in Figure 11.11,
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FIGURE 11.11. Connections of a circuit.
L˜=1 2(c 12((x 1-x 2)2+(y 1-y 2)2)+(c 13((x 1-x 3)2+(y 1-y 3)2)+(c 14((x 1-x 4)2+(y 1-y 4)2)+(c 24((x 2-x 4)2+(y 2-y 4)2)+(c 25((x 2-x 5)2+(y 2-y 5)2)+(c 36((x 3-x 6)2+(y 3-y 6)2)=1 2 x T Q x+d x T x+1 2 y T Q y+d y T y+1 2((c 14+c 24)x 4 2+c 25 x 5 2+c 36 x 6 2+(c 14+c 24)y 4 2+c 25 y 5 2+c 36 y 6 2
where
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It is clear that
{i, j} for all {i, j} ∈ E are convex and continuously differentiable functions. Hence, , which is a weighted sum of {i, j}'s, is also convex and differentiable. So should be easy to minimize. In particular,
∂L˜∂X=Q x+d x and∂L˜∂y=Q y+d y
Therefore, the placement with minimum wirelength is given by
(11.1)Q x+d x=0 and Q y+d y=0
In other words, if nonoverlapping constraints are ignored, the quadratic placement problem is equivalent to solving a system of linear equations. If all movable modules are connected to fixed modules either directly or indirectly, Q is positive definite and thus invertible. This implies the existence of a unique global optimal solution. The simplicity of quadratic formulation is the main reason for its popularity. Note that x and y can be solved independently. For brevity's sake, sometimes only the x-component will be discussed from now on.
11.5.2.2 Force interpretation of quadratic wirelength
The problem of quadratic wirelength minimization can also be interpreted as a classical mechanics problem of finding the equilibrium configuration for a system of objects attached to zero-length springs. Consider each circuit module as an object and each 2-pin net {i, j} as a stretched spring with spring constant c{i, j} connecting object i and object j. For the circuit represented by Figure 11.11, the corresponding spring system is shown in Figure 11.12.
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FIGURE 11.12. The spring system corresponding to the example in Figure 11.11.
The potential energy stored in spring {i, j} is:
ɛ{i,j}=1 2×c{i,j}×(Length of spring{i,j})2=1 2×c{i,j}×((x i-x j)2+(y i-j j)2)
Hence, the total potential energy of the spring system is equal to
. In other words, finding the minimum energy configuration for the spring system is equivalent to minimizing the quadratic wirelength in the quadratic placement formulation.
For a spring system, the minimum energy configuration is also the same as the force-equilibrium configuration. Note that the gradient of the total potential energy to the coordinates of an object gives the total force acting on the object. Therefore, the entries in the vectors Qx + dx and Qy + dy are the x-components and y-components of the total forces acting on the objects. In other words, another interpretation of the optimal placement conditions in Equation (11.1) is that all objects are in force equilibrium. For a nonequilibrium system (i.e., a circuit placement with suboptimal quadratic wirelength), the total force on an object provides the best way of movement for the object to minimize the total energy (i.e., the total weighted quadratic wirelength). Extra forces can also be added to influence the placement in a desirable manner (e.g., to spread out the objects). Many quadratic placement algorithms use the guidance provided by springs/extra forces to optimize a placement solution (e.g., [Quinn 1975]). Those algorithms are often called force-directed placement algorithms.
The forces exerted by the springs are given by Hooke's law. The force exerted on object i by a spring connecting objects i and j (as illustrated in Figure 11.13) is:
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FIGURE 11.13. The forces by a stretched spring connecting objects i and j.
F i j=c{i,j}×Displacement from object i to object j
Its magnitude is:
|F i j|=c{i,j}×(x i-x j)2+(y i-y j)2
To find the total force exerted by several springs on an object, it is often more convenient to decompose the force by each spring into x- and y-components:
|F x|=c{i,j}×|x j-x i|and|F y|=c{i,j}×|y j-y i|
Then the x-component and y-component of the total force are the sum of the x-component and y-component of the forces by all springs, respectively.
11.5.2.3 Net models for multi-pin nets
Circuits typically contain both 2-pin nets and multi-pin nets. To place circuits with multi-pin nets by quadratic techniques, various models have been proposed to replace each net by a group of 2-pin nets.
The traditional model is to replace each net by a clique (i.e., complete graph). For example, for the 5-pin net in Figure 11.14a, the clique model is shown in Figure 11.14b. The net weights of the 2-pin nets in the clique model should be set properly to balance the minimization of 2-pin nets and multi-pin nets. For a k-pin net with net weight c, the weight of each 2-pin net in the clique is usually set to either c/(k-1) [Vygen 1997] or 2 c/k [Kleinhans 1991; Eisenmann 1998].
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FIGURE 11.14. Clique and star models for multi-pin nets.
Another model is the star model [Vygen 1997; Mo 2000] as illustrated in Figure 11.14c. In the star model, one extra dimensionless module called the star module is introduced for each net. The star module is placed together with other movable modules during placement. Therefore, two extra variables corresponding to the x- and y-coordinates of the star module are added to the placement problem.
It is proved in [Viswanathan 2004] that the clique model and the star model are equivalent in quadratic placement if the net weights are set properly.
Specifically, consider a k-pin net connecting modules 1, …, k. In the clique model, if the weight of each 2-pin net is set to c, then the total force on module i by the 2-pin nets in the clique is given by:
F i c l i q u e=c×∑j=1 k(x j-x i)
In the star model, let x s be the x-coordinate of the star module. If the weight of each 2-pin net is set to k × c, then the total force on the star node is given by:
F s=∑j=1 k k×c×(x j-x s)
By setting F s = 0, the force-equilibrium position for the star node is:
x s 1 k∑j=1 k x j
The force on module i by the 2-pin net connecting to the star is given by:
F i s t a r=k×c×(x s-x i)=k×c×(1 k∑j=1 k x j-x i)=c×(∑j=1 k x j-k×x i)=c×∑j=1 k(x j-x i)=F i c l i u u e
Because the forces exerted are the same, the clique and the star models are equivalent, and they can be used interchangeably in quadratic placement.
On the basis of the equivalence of clique and star models, the hybrid net model [Viswanathan 2004] is a natural choice. In the hybrid net model, the clique model is used for nets with 2 to 3 pins, and the star model is used for nets with 4 or more pins. It has been shown empirically that the hybrid net model reduces the number of 2-pin nets by more than 10 × over the clique model for industrial circuits [Viswanathan 2007a]. It can also significantly reduce both the number of 2-pin nets and number of variables over the star model, because approximately 70% of nets in a typical circuit have 2 or 3 pins. Because the runtime to solve a quadratic placement problem is roughly proportional to the number of 2-pin nets and it increases slightly with the number of variables, the hybrid net model can speed up quadratic placement significantly.
11.5.2.4 Linearization methods
As shown in Figure 11.10, quadratic wirelength is a very rough approximation to linear wirelength. For small circuits, despite the inaccuracy of quadratic wire-length, quadratic placement techniques can still generate very competitive solutions. For larger circuits, however, this inaccuracy is a major bottleneck to the quality of quadratic placement solutions.
The authors in [Sigl 1991] presented a method to approximate the linear wirelength in a quadratic placement framework by iteratively adjusting the spring constant. Assume the star model is used to replace multi-pin nets. For a net e in the original circuit, let x e be the coordinate of the associated star module. Then the total linear wirelength (for the circuit after applying the star model) can be written as:
L s t a r=∑e∈E∑i∈e|x i-x e|
Consider the function:
L˜s t a r=∑e∈E∑i∈e(x i-x e)2 g i e
If g ie = |x i – x e| then
star = L star. However, g ie's are set to constants so that star would become a quadratic function. To approximate L star with star, the function star is optimized iteratively such that g ie in current iteration is set according to the coordinates of previous iteration. Intuitively, 1/g ie can be viewed as a variable spring constant that decreases with increasing spring length. The iterative process terminates when the g ie factors no longer change significantly.
Notice that even L star is just a rough approximation to the total HPWL objective function. Thus instead of setting g ie to |x i – x e|, an experimentally-verified net specific factor is used:
g i e=∑e∈E|x i-x e|for all i∈e
This choice has two advantages. First, the summation reduces the influence of nets with many connected modules and emphasizes most nets connecting only two or three modules. Second, the summation also prevents increasing the force on modules close to the star node by too much. According to HPWL, as long as a module is inside the net bounding box, it is not helpful to increase the force to pull it farther inside.
Nets becoming very short (i.e., g ie becoming very small) may cause numerical problems during the minimization of
star. Therefore, g ie is lower bounded (by the average module width for example) to ensure that g ie will never be zero.
Spindler and Johannes [Spindler 2006] introduced a BoundingBox net model, which, when combined with the preceding wirelength linearization idea, can accurately model HPWL in a quadratic placement framework. In the BoundingBox net model, a multi-pin net is transformed into only a few characteristic 2-pin nets as illustrated in Figure 11.15a. It is different from the clique model in which all possible 2-pin nets are included as shown in Figure 11.15b. Consider a k-pin net. For a given placement, suppose the modules are indexed in ascending order of their x-coordinates. Therefore, module 1 is at the left boundary and module k is at the right boundary of the net's bounding box. All connections are joined to the two boundary modules. One 2-pin net is connecting modules 1 and k. Two 2-pin nets are connecting each of the remaining k – 2 inner modules to module 1 and module k, respectively. The total number of 2-pin nets is 1 + 2(k – 2). Let N = {{1, k}, {1, 2}, {2, k}, {1, 3}, {3, k}, …, {1, k – 1}, {k – 1, k}} be the set of 2-pin nets.
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FIGURE 11.15. The BoundingBox and the clique net models for a 5-pin net.
According to the BoundingBox net model, the wirelength
BB of the k-pin net is defined as:
L˜B B=1 2∑{i,j}∈N ω{i,j}×(x i-x j)2
where
ω{i,j}=2 k-1×1 l{i,j}
If l{i, j} is set to |x i – x j| for all {i, j} ∈ N then:
L˜B B=1 2(∑{i,j}∈N 2 k-1×1|x i-x j|×(x i-x j)2)=1 k-1(∑{i,j}∈N|x i-x j|)=1 k-1(x 1-x k|+∑2≤i≤k-1(|x 1-x i|+x i-x k|))=1 k-1(|x 1-x k|+(k-2)×|x 1-x k|)=|x 1-x k|
Thus, the BoundingBox net model gives the exact HPWL if the net weights are set appropriately.
Like the linearization method in [Sigl 1991], the correct net weights w{i,j} are searched by iteratively optimizing
BB such that l{i,j} in current iteration is set to |x i – x j| of previous iteration. In each iteration, the l{i,j} factors are constants and hence BB becomes a quadratic function of module coordinates.
For simplicity, only the x-component of the model is described previously. The y-component can be constructed similarly. However, because the boundary modules and module distances may be different in x- and y-directions, the set of 2-pin nets introduced and the l{i,j} factors are most likely different. Moreover, even for a given direction, the boundary modules may change from iteration to iteration. Hence, the set of 2-pin nets and the net weights have to be updated continually. The overhead associated with maintaining two copies of connectivity matrices and the need to search for the net weights by an iterative process are the main disadvantages of the BoundingBox model over the clique, star, and hybrid models.
The BoundingBox net model has several advantages. First, it allows quadratic techniques to perform placement with the HPWL metric. Second, unlike the clique or star models, no connection is introduced among the inner modules to pull them together.3 The inner modules are able to move more freely as in the HPWL model. Third, this model introduces much fewer 2-pin nets than the clique model. It does introduce more 2-pin nets than the star/hybrid model for nets with 4 or more pins, but the difference is not very significant.
Another way to mitigate the inaccuracy of quadratic wirelength is to correct the mistakes by refining the placement solution with some linear metrics. Detailed placement can be viewed as one example of this approach. In detailed placement, as a simple problem of locally rearranging a few modules is considered, an accurate wirelength model (e.g., HPWL or even RSMT wirelength) can usually be applied. However, because corrections are restricted by the local nature and the legality requirement of module movements, the effectiveness of detailed placement in optimizing the linear cost function is limited.
A better technique called iterative local refinement (ILR) is proposed in FastPlace [Viswanathan 2004]. ILR can be applied to any global placement solution before legalization. It works in iterations. In each iteration, the placement region is divided into bins by a regular grid structure. The bin size is large at the beginning iterations and is gradually reduced to consider progressively finer module movements. After binning, the modules are examined one by one. For each module, it is tentatively moved from its original bin to its eight adjacent bins as shown in Figure 11.16. For each tentative move, one score is computed. The score is a weighted sum of a wirelength component and a density component. The wirelength component is the total change in HPWL of all nets connected to the module. The density component is a function of the module densities of the original bin and the target bin. It rewards movements from a dense bin to a sparse bin. If all eight scores are negative, the module will remain unmoved. Otherwise, the move with the highest score will be taken. This iterative process is repeated until there is no significant improvement in wirelength.
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FIGURE 11.16. Eight tentative moves of iterative local refinement.
Because ILR is not constrained by the nonoyerlapping requirement and can move modules by a relatively long distance, it is much more effective than detailed placement in correcting major problems in the placement solution. It also helps the spreading of modules. Besides, it is an extremely fast technique because of its simplicity.
11.5.2.5 Handling nonoverlapping constraints
In placement, the two primary goals are to minimize the wirelength and to avoid module overlaps. These two goals are in conflict with each other. Wirelength minimization brings modules together. Overlap avoidance requires modules to spread out. Note that if the nonoverlapping constraints are ignored, for a circuit without fixed modules, the optimal solution is to place all modules at the same location. This solution has zero wirelength but is meaningless because of the serious overlap issue. Even for a circuit with fixed modules (e.g., I/O pins at boundary), which help pulling the movable modules away from each other, the wirelength-minimized placement without considering overlaps typically has a lot of overlaps at the center of the placement region as illustrated in Figure 11.17.
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FIGURE 11.17. The placement solution for a circuit with fixed I/O pins at, boundary when quadratic wirelength is minimized and nonoverlapping constraints are ignored.
As pointed out in Section 11.5.1, instead of completely eliminating module overlaps in an analytical framework, an even distribution of modules is targeted in practice. In quadratic placement, there are basically two ways to make the module distribution more even. The first way is to add center-of-mass constraints to prevent modules from clustering together. The second way is to add forces to pull modules from dense regions to sparse regions. For both ways, the constraints/forces are added in an iterative manner to gradually spread out the modules. Note that even with the additional constraints/forces, the quadratic wirelength minimization problem can still be formulated as a convex quadratic program. Therefore, the circuit placement problem is transformed into a sequence of convex quadratic programs.
The technique of adding center-of-mass constraints is first introduced by GORDIAN [Kleinhans 1991]. Similar techniques are also used in BonnPlace [Brenner 2005] and hATP [Nam 2006]. The algorithm of GORDIAN is presented in the following. In GORDIAN, given an uneven quadratic placement solution, the module distribution can be improved by the following procedure. Assume the modules have to be spread horizontally. First, a vertical cutline is used to partition all modules into two subcircuits and the placement region into two subregions (see Figure 11.18a). Then, for each subcircuit, a constraint in the x-direction is added to force the center of mass of all its modules to be at the center of the corresponding region. Next, the placement problem with the two additional constraints is solved again. The center-of-mass constraints should pull the two subcircuits horizontally away from each other as shown in Figure 11.18b. This procedure is applied hierarchically to improve the distribution in each subregion (see Figure 11.18c) until each subcircuit contains less than a predefined number of modules. Note that at each hierarchical level, the placement of all subcircuits is considered together as a single global optimization problem.
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FIGURE 11.18. Module spreading by center-of-mass constraints in GORDIAN.
The coordinates of the center of mass are the area weighted mean values (i.e., linear functions) of the module coordinates. In other words, the center-of-mass constraints are linear equality constraints. Therefore, the global optimization problem at each hierarchical level is a convex quadratic program, which is equivalent to solving a system of linear equations.
Although the center-of-mass constraints help spreading, they hurt wirelength. For any two subcircuits belonging to the same parent in the hierarchy, the center-of-mass constraints draw them apart by a long distance. Hence the connections between them will become much longer. The wirelength impact can be minimized if the cut cost between the two subcircuits can be reduced. To avoid a large cut cost, both direction and position of the cutline are chosen carefully. To determine the cut cost of every possible partition by a vertical cutline, the cut-line is scanning from left to right and the cut cost is updated whenever the cut-line passes over a module. Only the partitions in which both subcircuits are at least 35% of the area of their parent are considered. The cut costs for horizontal cutline can be found similarly. The cutline with the smallest cut cost among all directions and positions is chosen. After that, the FM bipartitioning algorithm is optionally applied to refine the partition by moving modules that are close to the cutline. Moreover, after global optimization, if there are a lot of overlaps between two subcircuits, it indicates a bad bipartition, because many modules tend to migrate to the other region. In that case, they are repartitioned.
The technique of adding forces to spread modules in a quadratic placement framework was first introduced in Kraftwerk [Eisenmann 1998]. In Kraftwerk, density-based forces are derived to pull modules from high-density regions to low-density regions. However, constant forces are used, and the magnitude of the forces is set heuristically. As a result, the convergence is hard to control and the algorithm is not as fast as it should be. In the following, the improved technique in the new version of Kraftwerk [Spindler 2006] is presented. Note that the x-coordinates will be focused in the discussion following.
For a given placement, let x′ be the vector of current module positions and x be the vector of variables representing module positions in a new placement to be determined. The additional force for each module is separated into two components: hold force and move force. The hold force vector Fx hold is defined as:
F x=-(Q x′+d x)
It is used to counterbalance the total forces by the nets of the circuit in the current placement x′. It makes sure that if the placement problem is solved again, all modules will be held in their current positions. Hold forces do not depend on x and hence are constant forces.
The move force is used to move a module toward less dense regions. Let D(x, y) be the module density at location (x, y). It is defined as the number of modules that cover (x, y) minus the average density (Σ i w i × h i)/(W × H). The distribution D(x, y) can be viewed as a charge distribution, which creates an electrostatic potential φ based on the Poisson equation:
Δ φ=-D(x,y)
The Poisson equation can be solved efficiently by geometric multigrid solvers (e.g., [Kowarschik 2001]).
In the electrostatic formulation, the potential φ is high in regions where the distribution D(x, y) is high, and vice versa. Hence the gradient of the potential (∂φ/∂x, ∂φ/∂y)T can be used to move the modules away from high-density regions toward low-density regions and thereby reduce the overlaps among the modules. For each module i, its target position◯ i is:
x⌢i=x′-|∂Φ∂x|(x′,y′)
Move forces are added to guide modules toward their target positions. They are generated by use of the fixed-point idea proposed in FAR [Hu 2002]. Each module i is connected to its target position (i.e., a fixed point) by a spring with spring constant ĉ i. So the move force vector Fx move is given by:
F x=Q⌢(x-x⌢)
where
= diag(ĉ i). Note that spring forces rather than constant forces are used for move forces so that module movements are limited. Each module can be moved at most up to its target position. This helps the convergence of Kraft-werk significantly. The spring constants ĉ i control the tradeoff between rate of convergence and wirelength. If large ĉ i values are used, modules will be moved close to their target positions. Hence the placer will converge faster to an even density distribution. On the other hand, small ĉ i values allow module positions to be determined mostly by wirelength minimization.
In the new placement solution, the sum of net force, hold force, and move force on each module should be zero:
(Q x+d x)-(Q x′+d x)+Q⌢(x-x⌢)=0
Therefore, the new module positions x can be found by solving the following system of linear equations:
(Q+Q)(x+x′)=-Q⌢(x′-x⌢)
This spreading procedure is repeated until the placement density distribution is even enough. It can be proved that this procedure always converges to an overlap-free placement.
Note that besides this potential-based method, the target positions can also be computed by simpler heuristics like cell shifting [Viswanathan 2004] or grid warping [Xiu 2004]. All these methods try to equalize the placement density of nearby regions by locally moving modules. There is no proof of convergence for these methods, but they work well in practice.
To mitigate the negative effect of the additional forces on wirelength in force-directed spreading algorithms, a force-vector modulation technique is proposed in RQL [Viswanathan 2007b]. The technique is based on the observation that the spreading force is small for most modules but very huge for a few percent of all modules. A huge force implies that the corresponding module is pulled away from its natural position (i.e., the force-equilibrium position if spreading force is removed) by a long distance. Thus, the nets connecting to the module become very long. The force-vector modulation technique nullifies the huge forces before the next quadratic optimization iteration. As a result, modules with nullified spreading forces can return to the minimum-wirelength positions. Hence, the total wirelength can be significantly improved. Because the spreading forces of only a few percent of modules are nullified, module spreading is not seriously affected.
11.5.3 Nonquadratic techniques
Another category of analytical approach is to formulate the placement problem as a single nonlinear program as in Section 11.5.1. However, instead of exact wirelength metric and exact nonoverlapping constraints, approximations are used. In particular, the placement region is divided into bins, and the nonoverlapping constraints are replaced by bin density constraints. Let x and y be the vectors of x- and y-coordinates of the modules, respectively. Then the problem can be formulated as follows:
Minimize∑e∈E c e×WL e(x,y)Subject to D b(x,y)=T b for all bin b
WL e() is a continuously differentiable function that may be more complicated than quadratic functions but may also be more accurate in approximating HPWL. T b is the target density of bin b. D b( ) gives the density of bin b with respect to placement solution x and y. The exact bin density function is a piece-wise linear function and hence is not differentiable. D b() is a smooth version of the exact one.
Examples of placers in this category are APlace [Kahng 2004], mPL [Chan 2005], and NTUPlace [Chen 2006]. To approximate the wirelength, APlace, mPL, and NTUPlace all use the log-sum-exponential wirelength function described in a patent by [Naylor 2001]. To smooth the density function, APlace and NTUPlace use a bell-shaped function proposed also by [Naylor 2001], and mPL uses inverse Laplace transformation [Evans 2002]. The wirelength approximation and density smoothing methods are described in the following.
11.5.3.1 Log-sum-exponential wirelength function
The log-sum-exponential function is defined as:
L S E α(z 1,…,z n)=α×(log(∑i=1 n e z i/α))
It is an approximation of the maximum function:
LSE α(z 1,…,z n)≈max(z 1,…,z n)
α is a parameter controlling the accuracy of the approximation. As α converges to 0, the log-sum-exponential function converges to the maximum function. This is demonstrated in Figure 11.19, which shows for α = 0.1 and for α = 2, the log-sum-exponential function of two arguments.
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FIGURE 11.19. The log-sum-exponential function of two arguments for two different values of a.
The HPWL of a net e can be expressed in terms of the maximum function:
HPWL e(x 1,…,x n,y 1,…,y n)=(max i∈e{x i}-min i∈e{x i})+(max i∈e{y i}-min i∈e{y i})=(max i∈e{x i}+min i∈e{-x i})+(max i∈e{y i}+min i∈e{-y i})
So HPWL can be approximated by the log-sum-exponential based function as follows:
LSEWL e,α(x 1,…,x n,y 1,…,y n)=α×(log(∑i∈e e x i/α)+log(∑i∈e e-x i/α)+log(∑i∈e e y i/α)+log(∑i∈e e-y i/α))
LSEWL e,α() is strictly convex, continuously differentiable, and converges to HPWL e() as α converges to 0 [Naylor 2001].
11.5.3.2 Density constraint smoothing by bell-shaped function
To illustrate the idea, assume for now that each module is much smaller than the bins and hence is considered to be a dot. Besides, assume each module has a unit area. For a bin b, let x b be the x-coordinate of the center and w b be the width of bin b. Then the overlap function Θ x(b, i) in the x-direction between bin b and module i is shown in Figure 11.20a. This function can be approximated by a smooth bell-shaped function Θ x(b, i) as shown in Figure 11.20b. Let d x = |x i – x b|. Then:
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FIGURE 11.20. Overlap function between bin b and module i.
(11.2)Θ˜x(b,i)={1-2×d x 2/ω b 2 2×(d x-ω b)2/ω b 2 0 if 0≤d x≤ω b/2 if ω b/2≤d x≤ω b if ω b≤d x
The overlap function Θ y(b, i) in the y-direction is defined similarly.
The density function of bin b can be written as follows:
D b(x,y)=∑i∈V c i×Θ˜x(b,i)×Θ˜y(b,i)
where C i is a normalization factor so that ∑b C i×Θ˜x(b,i)×Θ˜y(b,i)=ω i×h i (i.e., area of module i).
This idea can be extended to handle large modules [Kahng 2005]. For a module i with width w i, the scope of the module in the x-direction is set to w b + w i/2 (i.e., every bin within horizontal distance w b + w i/2 from the module's center is considered to be overlapping with the module). Then:
(11.3)Θ˜x(b,i)={1-a×d 2 b×(d x-w b-w i/2)2 0 if 0≤d x≤w b/2+w i/2 if w b/2+w i/2≤d x≤+w b+w i/2 if w b+w i/2≤d x
where
a=4/((w b+w i)(2 w b+w i))b=4/(w b(2 w b+w i))
so that the function is continuous when d x = w b/2 + w i/2. Note that Equation (11.3) is the same as Equation (11.2) if w i = 0.
11.5.3.3 Density constraint smoothing by inverse Laplace transformation
Inverse Laplace transformation is a commonly used method to smooth functions. For a given placement solution x and y and for any location (x, y) in bin b, let d(x, y) be the density at (x, y), i.e., d(x, y) = D b(x, y). The smoothing operator Δ-1∈d(x, y) is defined by solving the Helmholtz equation:
(11.4){Δ ψ(x,y)-∈ψ(x,y)=d(x,y)∂ψ∂v=0(x,y)∈R(x,y)∈∂R
where ψ(x, y) is a smoothed version of d(x, y), ∈ > 0 is a parameter controlling the smoothness, R is the placement region, ∂R is the boundary of R, v is the outer unit normal vector pointing outside the boundary, and Δ = ∂2/∂x 2 + ∂2/∂y 2 is a differential operator.
The inverse operator Δ−1 d(x, y) is well defined as Equation (11.4) has a unique solution for any ɛ > 0. Because the solution of Equation (11.4) gains two more derivatives than d(x, y), ψ is at least twice differentiable.
11.5.3.4 Algorithms for nonlinear programs
For nonquadratic techniques, as the objective function and all constraints are continuously differentiable, the resulting nonlinear program can be solved by any nonlinear programming algorithms. In APlace and NTUPlace, the nonlinear program is converted by the quadratic penalty method into a sequence of unconstrained minimization problems. Each unconstrained minimization problem has the following form:
Minimize∑e∈E c e×W L e(x,y)+β×∑b(D b(x,y)-T b)2
The intuition of the quadratic penalty method is that any placement solution violating the density constraint for bin b will be charged a penalty of (D b(x, y) -T b)2. β is a parameter to specify the importance of density constraints. Its value keeps increasing in the sequence of unconstrained problems to discourage uneven placement solutions. Each unconstrained problem is solved by the conjugate gradient method [Luenberger 1984].
In mPL, the nonlinear program is solved by the Uzawa algorithm [Arrow 1958] shown in Algorithm 11.2. In the Uzawa algorithm, xk and yk are the module locations at the k-th iteration, λ b k is the Lagrange multiplier at the k-th iteration, and α is a parameter to control the rate of convergence. The nonquadratic techniques are elegant and comparable to the quadratic techniques in terms of wirelength. However, they are more complicated to implement and are usually more expensive computationally.
Algorithm 11.2
The Uzawa Algorithm
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11.5.4 Extension to multilevel
To handle large-sized problems, a multilevel scheme is commonly used in analytical placement. The application of a multilevel scheme to placement is similar to its application to partitioning presented in Section 11.3.1. It consists of three phases. First, a hierarchy of coarser netlists is constructed by clustering heavily connected modules together. Second, an initial placement of the coarsest netlist is generated. Finally, the netlist is successively unclustered, and the placement at each level is refined. The multilevel scheme can improve both the runtime and the solution quality of analytical placement algorithms. Two popular clustering techniques for netlist coarsening are introduced in the following.
11.5.4.1 First choice
The First Choice clustering technique [Karypis 1997] first represents the net-list as a weighted graph by replacing the multi-pin nets with the clique model. The weight or affinityr ij between any modules i and j in the graph is given by:
r i j=∑e∈E∧i,j∈e c e|e|-1
Then the modules are traversed in an arbitrary order. Each module i is clustered with an unclustered neighbor j with the largest r ij. After all modules are traversed, the affinity graph is updated, and the clustering process is repeated until the number of modules has reached the target. The intuition behind First Choice is that modules with high affinity should stay close together in a good placement solution.
First Choice is originally proposed for multilevel partitioning. Several modifications of First Choice targeting placement are presented in [Chan 2005]. To reduce variation in cluster size, the affinity between modules i and j is redefined as:
r i j=∑e∈E∧i,j∈e c e(|e|-1)×area(e)
where area(e) is the total area of all modules in e. In addition, modules are visited in ascending order of module area (with preference to smaller module degree to break ties). This ordering is observed to balance the area of clusters better. If a good initial placement is provided, the proximity information between modules can be incorporated into the affinity as follows:
r i j=∑e∈E∧i,j∈e c e(|e|-1)×area(e)×dist(i,j)
where dist(i, j) is the Euclidean distance between i and j.
11.5.4.2 Best choice
In the Best Choice clustering technique [Alpert 2005], the affinity is defined as:
r i j=∑e∈E∧i,j∈e c e|e|×(area(i)×area(j))
where area(i) and area(j) are the areas of modules i and j, respectively. In addition to the indirect control of the cluster size by the affinity, Best Choice imposes a hard upper limit for cluster size. Moreover, the pair of modules with the largest affinity among all pairs is clustered and, in principle, the netlist is immediately updated. In other words, Best Choice always selects the globally best pair for clustering. In practice, as the immediate update is time-consuming, a lazy updating technique is proposed to reduce the runtime. The idea is that instead of explicitly recomputing the affinities of module pairs affected by a given cluster, they are marked as invalid and are updated only after they have been selected for clustering. Because the pair selection is based on invalid affinities, lazy updating may incur some errors. However, the dramatic reduction in runtime outweighs the small errors.
As with the modified First Choice affinity, the proximity information of a good initial placement can be incorporated into the Best Choice affinity in the same way.
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PARTICLE MECHANICS
1967, General PhysicsL.D. LANDAU, ... E.M. LIFSHITZ
§13.Boundaries of the motion
If the motion of a particle is constrained so that it can move only along a certain curve, the motion is said to have one degree of freedom or to be one-dimensional. One coordinate is then sufficient to specify the position of the particle; it may be taken, for example, as the distance along the curve from a point taken as origin. Let this coordinate be denoted by x. The potential energy of a particle in one-dimensional motion is a function only of this one coordinate: U = U(x).
According to the law of conservation of energy we have
E=1 2 m v 2+U(x)=constant,
and since the kinetic energy cannot take negative values the inequality
U≤E
must hold. This implies that the particle during its motion can occupy only points where the potential energy does not exceed the total energy. If these energies are equal, we have the equation
U(x)=E,
which determines the limiting positions of the particle.
Some typical examples are the following. Let us first take a potential energy which, as a function of the coordinate x, has the form shown in Fig. 8. In order to find the boundaries of the motion of a particle in such a force field, as functions of the total energy E of the particle, we draw a straight lineU = E parallel to the x axis. This line intersects the curve of potential energy U = U(x) at two points, whose abscissae are denoted by x 1 and x 2. If the motion is to be possible it is necessary that the potential energy should not exceed the total energy. This means that the motion of a particle with energy E can occur only between the points x 1 and x 2, and a particle of energy E cannot enter the regions right of x 2 and left of x 1.
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FIG. 8.
A motion in which the particle remains in a finite region of space is called a finite motion; one in which the particle can go to any distance is called an infinite motion.
The region of finite motions depends, of course, on the energy; in the example considered here, it decreases with decreasing energy and shrinks to a single point x 0 when E = U min.
At the points x 1 and x 2 the potential energy is equal to the total energy, and therefore at these points the kinetic energy and hence the particle velocity are zero. At the point x 0 the potential energy is a minimum, and the kinetic energy and velocity have their maximum values. Since the force F is related to the potential energy F = –dU/dx, it is negative between x 0 and x 2, and positive between x 0 and x 1. This means that between x 0 and x 2 the force is in the direction of decreasing x, i.e. to the left, and between x 0 and x 1 it is to the right. Consequently, if the particle begins to move from the point x 1, where its velocity is zero, the force to the right will gradually accelerate it to a maximum velocity at the point x 0. As the particle continues to move from x 0 to x 2 under the force which is now to the left, it will slow down until it comes to rest at x 2. It will then begin to move back from x 2 to x 0. This type of motion will continue indefinitely. Thus the particle executes a periodic motion with a period equal to twice the time for the particle to go from x 1 to x 2.
At the point x 0 the potential energy is a minimum and the derivative of U with respect to x is zero; at this point the force is therefore zero, and the point x 0 is consequently a position of equilibrium of the particle. This position is evidently one of stable equilibrium, since in this case a departure of the particle from the equilibrium position causes a force which tends to return the particle to the equilibrium position. This property exists only for minima and not for maxima of the potential energy, although at the latter the force is likewise zero. If a particle is moved in either direction from a point of maximum potential energy, the resulting force in either case acts away from this point, and points where the potential energy reaches a maximum are therefore positions of unstable equilibrium.
Let us now consider the motion of a particle in a more complex field whose potential-energy curve has the form shown in Fig. 9. This curve has both a minimum and a maximum. If the particle has energy E, it can move in such a field in two regions: region I between the points x 1 and x 2, and region III to the right of the point x 3 (at these points the potential energy is equal to the total energy). The motion in the former region is of the same type as in the previous example, and is oscillatory. The motion in region III, however, is infinite, since the particle may move to any distance to the right of the point x 3. If the particle begins its motion at the point x 3, where its velocity is zero, it will continually be accelerated by the force to the right; at infinity, the potential energy is zero and the particle velocity reaches the value v∞=√(2 m E). If, on the other hand, the particle moves from infinity to the point x 3, its velocity will gradually decrease and vanish at x 3, where the particle will turn round and go back to infinity. It cannot penetrate into region I, since this is prevented by the forbidden region II lying between x 2 and x 3. This region also prevents a particle that is executing oscillations between x 1 and x 2 from entering region III, where motion with energy E is also possible. The forbidden region is called a potential barrier, and region I is called a potential well. As the particle energy increases in this case, the width of the barrier diminishes and for E ≥ U max it does not exist. The region of oscillatory motion likewise disappears, and the motion of the particle becomes infinite.
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FIG. 9.
Thus we see that the motion of a particle in a given force field may be either finite or infinite depending on the energy of the particle.
This may be illustrated also by the example of motion in a field whose potential-energy curve has the form shown in Fig. 10. In this case positive energies correspond to infinite motion, and negative energies (U min<E< 0) to finite motion.
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FIG. 10.
Whenever the potential energy is zero at infinity, motion with negative energy will necessarily be finite, since at infinity the zero potential energy exceeds the total energy, and the particle therefore cannot go to infinity.
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4931 | https://www.gauthmath.com/solution/K-4QJseckVj/Which-one-is-the-least-sweet-Sucrose-Fructose-Maltose-Lactose-A-Lactose-B-Fructo | Solved: Which one is the least sweet?Sucrose Fructose Maltose Lactose A Lactose B Fructose [Biology]
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Image 1: Which one is the least sweet?Sucrose
Fructose
Maltose
Lactose
A
Lactose
B
Fructose
C
Sucrose
D
Maltose
Which one is the least sweet?Sucrose Fructose Maltose Lactose A Lactose B Fructose C Sucrose D Maltose
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Expert Verified Solution
Answer
The least sweet carbohydrate is lactose, which has a sweetness rating of 20.
Explanation
To determine which sugar is the least sweet among sucrose, fructose, maltose, and lactose, we can refer to the relative sweetness scale where sucrose is assigned a value of 100. This scale allows us to compare the sweetness of various carbohydrates in relation to sucrose.
Identify the sweetness ratings of each carbohydrate based on the provided scale:
Sucrose: 100
Fructose: 140
Maltose: 30-50
Lactose: 20
Analyze the ratings:
Fructose has the highest sweetness rating at 140, indicating it is sweeter than sucrose.
Sucrose itself is the standard reference with a rating of 100.
Maltose has a variable sweetness rating between 30 and 50, which is less sweet than sucrose.
Lactose has the lowest sweetness rating at 20, making it the least sweet of the four carbohydrates.
Conclusion based on the ratings:
Since lactose has the lowest sweetness rating of 20, it is identified as the least sweet carbohydrate among the options provided.
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4 Easy Strategies for Comparing Fractions with Unlike Denominators
There are four easy strategies for comparing fractions. And three of the strategies make comparing fractions with unlike denominator much easier than renaming denominators. All four work for most situations. Not only are these underused strategies relatively simple to learn, but they also immerse students in deeper thinking about fractions. How I wish I was aware of these strategies at the beginning of my teaching career. What a difference they make for students!
The typical algorithm that requires renaming fractions can not only be difficult but does not promote thinking of fractions as numbers. Additionally, it is a common misconception students often hold that the only way to compare fractions is by finding common denominators. Often students know the answer to which fraction is greater if they pause for a moment to think. Instead, many students automatically grab paper and pencil to find common denominators without any reasoning.
Four strategies for comparing fractions
As you teach these four strategies consider using these language phrases to support children’s understanding of fractions.
Denominators tell us the size of the pieces. For example, frequently asking students which pieces would be larger rather than which denominators are larger, helps them think of the whole fraction as a number. When they think of a denominator as a number, rather than part of a number, it doesn’t make sense that if the value of the denominator increases the value of the fraction decreases.
Numerators tell us the number of pieces to focus on.
We can only directly compare fractions with the same size whole. This is a great time to remind students that two children that each have ¼ of a pizza don’t necessarily have the same amount of pizza. (My favorite example for students: One child may have a personal size and the other a family size. )
Students so often memorize placement (top or bottom) of the denominator and numerator without understanding what each represents. This is why I love teaching them these 4 strategies for comparing fractions.
Strategy 1: equivalent denominators – same size pieces
This strategy is by far the simplest and possibly more familiar to students. That said, it can further students’ understanding of fraction place value by consistently referring to the “like denominators” as “the same size pieces.”
How it works:
Because the size of the pieces are the same we can simply use the number of pieces to determine which fraction is greater.
For example:
Which fraction is greater: 8/12 or 10/ 12? (Younger students: 4/8 or 2/8)
We know 4/8 is greater than 2/8 because the pieces are the same size, but 4/8 represents two additional pieces.
Hints for students:
To help students think about this problem I like to carefully use the language: “We have the same size pieces.” For example, if you are sharing a quesadilla cut into 8 equal pieces your share of 4 pieces would be more than a share of 2 pieces.
Strategy 2 : equivalent numerator- same number of pieces
Often this strategy is overlooked yet it’s relatively easy for students to learn. And this method works wonders for enhancing student understanding of denominators.
How it works:
If both numerators are the same simply compare the size of the pieces assuming, of course, the same size whole. (Same size pizzas, pies, etc.)
For example:
Which fraction is greater? 2/8 or 2/6 (younger students: 2/3 or 2/4)
Eighths are larger than sixths. Because of the same number of pieces, 2/6 is greater because the pieces are larger.
Note: I like to explicitly mention to students that it’s far simpler to compare the size of pieces rather than converting 8ths and 6ths to a common denominator.
Hints for students:
To help students I explain that because we have the same number of pieces all that matters is the size of each piece. For example, would they prefer 2 pieces of a quesadilla that has been cut into 8 pieces, or 2 pieces of a quesadilla that is cut into 6 pieces? (Assuming they are hungry:))
Strategy 3: Compare the fractions to a common benchmark fraction such as ½ or 1
How it works:
Often students can recognize that a fraction is close to, greater than, or less than ½ or 1 whole. By comparing two fractions to, for example, ½ they may find that one fraction is less than ½ and one is greater than ½. That’s all they need to know to compare the two fractions. Older students may be able to use other benchmarks as well.
For example:
Which fraction is greater? 3/12 or 6/8 (younger students: 2/6 or 3/4)
3/12 is smaller than 1/2 because 6/12 is equal to 1/2
6/8 is greater than 1/2 because 4/8 is equal to 1/2
Therefore, 6/8 has to be greater than 3/12.
Hints for students:
Ask students to evaluate whether the two fractions they are comparing are greater or less than ½. If one is greater than and the other is less than ½ they know which of the two fractions is greater.
Students may find that the two fractions they are comparing are both less than ½ , but one may be obviously closer to ½ than the other. For example, 1/8 and 4/10 are both less than 1/2, but only 4/10 is close to 1/2.
Strategy 4: Missing pieces
Students analyze two fractions that are missing one piece from being equivalent to ½ or one whole.
How it works:
If both fractions are one piece short of equaling one whole, then the fraction that is missing the smaller piece is larger.
For example:
Which is greater? 8/9 or 5/6 (younger students: ⅔ or ¾)
1/9 is missing from 8/9 to equal one
1/6 is missing from 5/6 to equal one
Sixths are larger than ninths
5/6 is missing a larger piece and is less than 8/9
Or: ⅞ is farther from one whole than 9/10. (Number line placement)
Hints for students:
For example, a pie is cut into 9 pieces. Those pieces will be smaller than a pie only cut into 6 pieces. If a piece is missing from each pie then the pie cut into 9 pieces has less missing. That tells us 8/9 is the greater amount of pie remaining.
Combining strategies often works well for placing fractions in order from least to greatest
By using like denominators we know that 5/6 is greater than 2/6. To determine whether 8/9 is greater than or less than 5/6 use the missing piece strategy. 5/6 has a larger missing piece to get to one so we know 8/9 is greater than 5/6. 2/6 < 5/6 < 8/9
How might we teach the four strategies for comparing fractions?
Number talks are perfect for both teaching and practicing these strategies. Carefully choosing fractions that lend themselves to the four strategies often lead to students discovering the strategies organically. Students can share their strategies with the class. Setting pencils and paper aside during number talks will push students to develop mental strategies.
Explicitly demonstrate the strategies while asking students to explain the rationale behind the methods.
Utilize number lines and models that illustrate the strategies to enhance student understanding.
Use word problems to illustrate the strategies. This helps kiddos conceptualize and remember the strategies.
Watch for a post of teaching these four strategies with number lines and models coming soon!
How do students practice these strategies?
Number talks
Fraction sorts
Fraction cards or games: that compare fractions (which is greater than/less than) without pencils and paper
Fraction sorts: that ask students to place fractions into piles according to whether the fraction is closer to 0, ½ or 1. This helps students use reasoning to sort rather than renaming denominators.
I found that when fraction sorts are done in partners or in groups the lively discussions (constructive arguments) were amazing. What a great way to have students engage in mathematical practice standards! (MP 3)
I designed this resource specifically for students to practice the four strategies. No pencils and paper – just lively discussions!
Fraction sorts within a craftivity combine rigor and fun!
Renaming denominators is a particularly difficult skill for students
It’s a skill that can easily lead to both mistakes and frustration. Mainly, renaming denominators is an abstract algorithm that doesn’t improve students’ understanding of fractions like these four strategies do. If you haven’t read my post, Is Simplifying Fractions Important? please do. It provides more background on why it benefits students to remove the emphasis from renaming denominators.
Word Problems
Once I am sure that students understand the strategies I present them with word problems. Hopefully, at this point, they recognize the ease of thinking about the fractions, rather than automatically following the algorithm for finding common denominators. Word problems are so important to understanding fractions and understanding real-life applications.
As always, I would love to hear, in the comments, about your experiences teaching students strategies for comparing fractions!
The following packets of word problems are specifically designed for comparing fractions.
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4933 | https://brilliant.org/wiki/properties-of-real-numbers/ | Properties of Real Numbers
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Ivan Koswara,
Mehul Arora,
Rajdeep Bharati,
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Jimin Khim
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This page is under construction. You can help us by adding examples of what you think should be on this page.
This page was the subject of a [wiki collaboration party].
A real number is one whose square is positive.
( at #mathematics, on 20 September 02.30am UTC.
.999…=1
The above equation is a very common question that has misled many people over years. It even got a lengthy Wikipedia page. Is it true? They are two different numbers, no?
Surprisingly, it is true! You might have known this popular problem and hence memorized the answer. But do you know why this is true? More surprisingly, we can make this equation false! If we work on something other than the real numbers, that is. The above equation hinges on a very important property of the real numbers.
Contents
Outline dependencies
Least upper bound property
Archimedean property
Monotone convergence theorem
Geometric progression formula and .999…=1
Nested intervals theorem
Bolzano-Weierstrass theorem
Other properties
Outline dependencies
Least upper bound property
(problems)
Archimedean property
Examples
Let x∈R. Show that there exists an integer m such that m≤x<m+1 and an integer l such that x<l≤x+1.
Suppose that α,β are two real numbers satisfying α<β. Show that there exist n1,n2∈N such that α<α+n11<β and α<β−n21<β.
Monotone convergence theorem
Geometric progression formula and .999…=1
0.999…, being a number represented in decimal form, has the value 0+n=1∑∞9⋅(101)n. This is a consequence of any number that is represented in decimal: if the number is k.b1b2b3…, where k is an integer and b1,b2,b3,…∈{0,1,2,…,9}, the value of any such number is k+n=1∑∞bn⋅(101)n by definition. So we'd like to compute the value of S=n=1∑∞9⋅(101)n.
A geometric sequence is a sequence of real numbers {an}n=1∞ that satisfies the property an+1=an⋅r for some real number r and all positive integers n. It can be shown by simple induction that if {an} forms a geometric sequence, there exist real numbers a and r such that an=arn−1 for all positive integers n. (Here we take 00=1.) A geometric series (or geometric progression) is a series whose elements form a geometric sequence, i.e. a series in the form n=1∑∞arn−1. Observe that S is a geometric sequence with a=109,r=101.
To compute the value of this series, we can take a look at its partial sums n=1∑karn−1. This form is called a finite geometric series. Being the sum of a finite number of real numbers, we can apply the commutative and associative properties of addition and the distributive property of multiplication over addition, to manipulate the sum as such:
SkrSkSk−rSkSk=n=1∑karn−1=r⋅n=1∑karn−1=n=1∑k(r⋅arn−1)=n=1∑karn=n=2∑k+1arn−1=(n=1∑karn−1)−(n=2∑k+1arn−1)=ar0+n=2∑k(arn−1−arn−1)−ark=a(1−rk)=1−ra(1−rk)=1−ra−1−ra⋅rk.
By definition, the value of the series is simply the limit of its partial sums. Thus we'd like to compute k→∞limSk, or k→∞lim(1−ra−1−ra⋅rk)
It's an easy exercise to show that if {bn} is a sequence of real numbers that converges to a limit n→∞limbn and c is a constant, then n→∞lim(c+bn)=c+n→∞limbn and n→∞lim(c⋅bn)=c⋅n→∞limbn. Since 1−ra and −1−ra are constants, applying these properties gives:
k→∞limSk=k→∞lim(1−ra−1−ra⋅rk)=1−ra+k→∞lim((−1−ra)⋅rk)=1−ra+(−1−ra)⋅k→∞limrk.
Our problem has been reduced to determining k→∞limrk.
For the purpose of proving 0.999…=1, it suffices to restrict our attention to 0<r<1. It is simple induction again to show that 0rk+1, we know that the sequence {rk} is monotonically decreasing, and thus converges to a limit L by monotone convergence theorem. From the fact that rk>0, L cannot be less than 0; if L<0, then simply take ϵ=−L, and now there is no element of {rk} that lies in (L−ϵ,L+ϵ) (because L+ϵ=0N (the condition necessary for a convergent sequence to converge to the limit).
todo: show that L>0 is impossible
Thus we have shown that L<0 and L>0 lead to a contradiction. The Archimedean property tells us that L cannot be infinitesimal either, since there is no non-zero infinitesimal in the real numbers. The remaining option is thus L=0.
Hence, k→∞limSk=1−ra+(−1−ra)⋅0, or k→∞limSk=1−ra; the geometric series, for 0<r<1, gives the value 1−ra.
But S is a geometric series, with a=109,r=101. Since 0<r<1, we can apply the result we obtained above to get S=1−1/109/10=1. This proves that 0.999…=1.
Nested intervals theorem
Bolzano-Weierstrass theorem
Other properties
Density of Q in R
Cite as:
Properties of Real Numbers.
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4934 | https://numbermatics.com/n/10101/ | Number 10101 - Facts about the integer
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10101
10,101 is an odd composite number composed of four prime numbers multiplied together.
What does the number 10101 look like?
This visualization shows the relationship between its 4 prime factors (large circles) and 16 divisors.
10101 is an odd composite number. It is composed of four distinct prime numbers multiplied together. It has a total of sixteen divisors.
Prime factorization of 10101:
3 × 7 × 13 × 37
See below for interesting mathematical facts about the number 10101 from the Numbermatics database.
Quick links:
NamesFactorsDivisorsBasesRootsScalesFun
Names of 10101
Cardinal: 10101 can be written as Ten thousand, one hundred one.
Scientific notation
Scientific notation: 1.0101 × 10 4
Factors of 10101
Number of distinct prime factors ω(n): 4
Total number of prime factors Ω(n): 4
Sum of prime factors: 60
Divisors of 10101
Number of divisors d(n): 16
Complete list of divisors: 13713213739911112592734817771443336710101
Sum of all divisors σ(n): 17024
Sum of proper divisors (its aliquot sum) s(n): 6923
10101 is a deficient number, because the sum of its proper divisors (6923) is less than itself. Its deficiency is 3178
Bases of 10101
Binary: 10011101110101 2
Hexadecimal: 0x2775
Base-36: 7SL
Squares and roots of 10101
10101 squared (10101 2) is 102030201
10101 cubed (10101 3) is 1030607060301
The square root of 10101 is 100.5037312741
The cube root of 10101 is 21.6166367027
Scales and comparisons
How big is 10101?
10,101 seconds is equal to 2 hours, 48 minutes, 21 seconds.
To count from 1 to 10,101 would take you about two hours.This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
A cube with a volume of 10101 cubic inches would be around 1.8 feet tall.
Recreational maths with 10101
10101 is the same when its digits are reversed! That makes it a palindromic number.
10101 is a Harshad number.
The number of decimal digits it has is: 5
The sum of 10101's digits is 3
More coming soon!
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4935 | https://chemistry.stackexchange.com/questions/2506/acetic-acid-freezing-distilation | organic chemistry - Acetic acid freezing distilation - Chemistry Stack Exchange
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Acetic acid freezing distilation
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How can acetic acid be distilled by freezing when its freezing point is above the freezing point of water.
Examples on the web show the water portion of the vinegar as ice and the acid as a liquid but Wikipedia shows that freezing point of acetic acid is 16.6°C.
organic-chemistry
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edited Nov 5, 2012 at 22:07
Ben Norris
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1 All the sources I read confirm that acetic acid crystallizes before water upon freezing, and is separated that way. Can you include links to back your assertion (“examples on the web”)?F'x –F'x 2012-11-05 14:23:19 +00:00 Commented Nov 5, 2012 at 14:23
alchemywebsite.com/distillation_of_vinegar.htmlNccWarp9 –NccWarp9 2012-11-11 21:30:38 +00:00 Commented Nov 11, 2012 at 21:30
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The procedure you are describing sounds like fractional freezing. Acetic acid is frequently purified by this method, hence 100% acetic acid is often called "glacial" acetic acid.
As an example of how this works, let's consider a mixture of acetic acid and water.
Acetic acid has a freezing point of 16.6 o C. Water has a freezing point of 0 o. They form a eutectic at -26.7 o C (although I do not know the composition). The phase diagram probably looks something like (but is not) the following.
Phase Legend:
A = Liquid solution
B = Solid solution of acetic acid in water PLUS liquid solution
C = Solid solution of water in acetic acid PLUS liquid solution
D = Solid solution of acetic acid in water
E = Solid solution of water in acetic acid
F = Solid solution of acetic acid in water PLUS Solid solution of water in acetic acid
A mixture of acetic acid and water that is approximately 72% acetic acid is cooled until it begins to freeze (solid blue vertical line), which happens around 2 o C. The phase changes from A (liquid solution) to C (Solid solution of water in acetic acid PLUS liquid solution). The solid solution is enriched in acetic acid compared to the original liquid solution. The composition of the remaining supernatant liquid solution is enriched in water compared to the original liquid solution. The composition of the solid solution can be determined by drawing a horizontal (red) line to the phase boundary between C and E and reading down to 91% acetic acid. The composition of the remaining supernatant liquid solution can be determined by drawing a horizontal (green) line to the phase boundary between A and C and reading down to 63% acetic.
The supernatant phase is removed and discarded, and the solid phase is allowed to melt. The process is repeated. Each successive freezing yields a new solid solution that is enriched in acetic acid, until the purity is as high as you want it.
Note, that since water and acetic acid have a eutectic point, if we has started on the left side of the eutectic point, the phases would have been reverse. The solid would be enriched in water and the supernatant would have been enriched in acetic acid. The supernatant would be remove and kept, and the solid would be discarded. The supernatant would then be cooled and fractionally frozen again to further enrich the supernatant in acetic acid until the eutectic point is reached (at which point the composition of the supernatant is equivalent to the composition of the solid phase). If you want to isolate acetic acid from a solution that is mostly water, I would distil the solution until the composition is mostly acetic acid, and then fractionally freeze it.
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answered Nov 5, 2012 at 22:06
Ben NorrisBen Norris
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2 This is a great answer, just one thing - every reference I've ever read regarding the phrase 'glacial acetic acid' indicates that the name comes from the ice-like appearance of the crystals rather than the mode of separation.Richard Terrett –Richard Terrett 2012-11-06 00:26:14 +00:00 Commented Nov 6, 2012 at 0:26
The eutectic point, is it at 50% for sure? I tried to look it up, but couldn't find it.JohannesB –JohannesB 2015-01-20 21:04:29 +00:00 Commented Jan 20, 2015 at 21:04
1 A bit late, but according to this link: ddbst.com/en/EED/SLE/SLE%20Acetic%20acid%3BWater.php (which links to to primary source), the eutectic point is -27°C / 71.6% (mol/mol) / 88.3% (kg/kg).ericksonla –ericksonla 2015-06-11 19:12:47 +00:00 Commented Jun 11, 2015 at 19:12
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4936 | https://www.rgeretschlaeger.com/Folding_Questions_Proceedings_Version.pdf | Folding Questions – A Paper about Problems about Paper Robert Geretschläger; Graz, Austria Presented at WFNMC-6, Riga, Latvia, July 27th, 2010 INTRODUCTION For many years now, I have been involved in the study of the geometry of paper folding. It therefore seems like an obvious step for me to have a closer look at problems in this area that have been used as competition questions in various countries. Although the result can hardly be called complete, what follows seems to be a fair representation of the kinds of ideas that have been used in this way. In collecting these problems, I checked through various regional, national and international competitions from all around the world. I have intentionally focused on flat-folding problems, and have therefore not collected the many problems concerning nets of polyhedra or developments of cones or cylinders that can be found in the literature. The full-length version of this paper is available on-line at 1. SHAPES The most elementary paper folding problems are those concerning the shape (i.e. the outline) of a flat folded piece of paper. The medium to be folded is almost always a rectangle or a square. In most such problems, the paper is folded over only once. 1) Three shapes X, Y and Z are shown below. A sheet of A4 paper (297 mm by 210 mm) is folded once and placed flat on a table. Which of these shapes could be made? X Y Z A) Y and Z only B) Z and X only C) X and Y only D) none of them E) all of them source: UK Junior Mathematical Challenge 1999, Nr. 19. Solution: As we see in the figure, all are possible. The correct answer is therefore E. 2) A sheet of A4 size paper (297 mm x 210 mm) is folded once and then laid flat on the table. Which of these shapes could not be made? A) B) C) D) E) source: UK Intermediate Mathematical Challenge 1999, Nr. 2. Solution: As we can see in the figure, folding the rectangle along the diagonal does not yield the triangle shown in B. All other shapes are possible, and the correct answer is therefore B. 3) The diagram shows a triangular piece of paper that has been folded over to produce a shape with the outline of a pentagon. If a rectangular piece of paper is folded once, what is the smallest value of n (greater than 4) for which it is not possible to create an n-sided polygon in the same way? A) 6 B) 7 C) 8 D) 9 E) 10 source: UK Senior Mathematical Challenge 2003, Nr. 10. Solution: When the piece of paper is folded, the crease makes up one side of the resulting polygon. In addition, each of the four corners of the rectangle can contribute at most two sides to the resulting polygon. More than 9 sides are therefore certainly not possible. As we see in the figure, the other given values are possible. The correct answer is therefore E. Comments These problems are of a highly elementary type. An obvious idea for similar problems would be to consider a shape for the paper other than a rectangle. Also, any problem asking what is possible after two folds or more can become quite difficult to answer readily. Another interesting variation results from turning the situation around, and asking which shapes can be folded to make a certain shape. We can, for instance ask, which of the following shapes can be folded once in order to form a square. It may not be immediately obvious that it is possible for the hexagon on the right, but not for the one on the left, as we see below. The difference is only in the placement of the end points of the sides of the hexagons on the left and right sides of the double squares. Similar variations are possible for the points on the top and bottom, of course. Some competition problems deal with n-gons that are folded to yield m-gons. This idea can be adapted in many ways. For instance, we can ask which values of m are possible if a specific value of n is given, or vice versa. In either case, we can either specify that the polygons are convex, or not. For specific values (n, m = 3, 4, …) this yields numerous questions, with some surprising answers. Let us consider one specific example. Question: A convex n-gon is folded once. The result is a convex quadrilateral. Which values of n are possible? Answer: All values n 3 are possible. Consider the following figure: We are given polygons with n = 3, n = 4, n = 9, the last of which is exemplary for any n > 4. Each can be folded to produce a quadrilateral, as we see below. 2. FOLD AND CUT A small step leads us from considering the shapes that result from folding to shapes that result from folding, cutting, and unfolding again. One such problem is the following. 4) A square piece of paper is folded twice as shown in the figure, so that a small square results. A corner of the resulting square is cut off and the square is unfolded again. Which of the following shapes cannot result is this way? A) B) C) D) E) All shapes are possible. source: Kangaroo 2007, Écolier Solution: All four shapes are possible. In fact, they correspond to the four corners of the small square being cut. The correct answer is therefore E. Comments Problems of this type involve a bit of spatial reasoning, and there is always a certain aspect of powers of 2 involved, since any fold always doubles the number of layers of paper being considered. In essence, we are often considering simple versions of “snowflakes”, i.e. the shapes created by folding a piece of paper multiple times in such a way that a center of symmetry is created (together with the lines of symmetry resulting from the folds), making some cuts, and then unfolding again, yielding a symmetric shape. It seems like an obvious variation to consider the hexagonal shape of a real snowflake. One such question could be the following: Question: A piece of paper in the shape of a regular hexagon is folded over once and then into thirds as shown. One straight cut is made, removing a section of the folded paper, and the remaining piece is unfolded again. Which of these shapes can be the result? 1 2 3 4 A) All are possible. B) All except number 1 are possible. C) All except number 2 are possible. D) All except number 3 are possible. E) All except number 4 are possible. Answer: Because of the symmetry resulting from the folds, all but number 3 are possible. The correct answer is therefore D. 3. LENGTHS and AREA The most important aspect in most elementary questions concerning measurable attributes of flat folding, and certainly in all that involve only a single fold, is that of the resulting symmetry. „Folding over“ always implies leaving part of the paper stable on the folding surface, while reflecting the other part with respect to the folding edge. Because of the resulting symmetry, the measures of angles and of lengths are retained, and this retention allows us to calculate a number of things that may, at first glance, appear not to be uniquely determined. This aspect is key to solving such problems. An important aspect in many problems concerning lengths specifically, is the use of the Pythagorean Theorem. Right angles turn up in origami whenever a line is folded onto itself, for instance. They are therefore quite common, and calculation of lengths can therefore often be based on the sides of right triangles. Another important tool is the use of similar triangles, which also result quite naturally in folding problems because of the equal angles generated by the symmetries involved. 5) A square piece of paper ABCD is folded such that the corner A comes to lie on the mid-point M of the side BC. The resulting crease intersects AB in X and CD in Y. Show that |AX| = 5 |DY|. source: Mathematical Duel Bílovec – Chorzów – Graz – Přerov 2010 Solution: As shown in the figure, let N be the point of intersection of AM and the crease line XY. Furthermore, let the lengths of the sides of ABCD be equal to 1. Since |BM| = 2 1 , we have |AM| = 2 5 , and therefore |AN| = 2 1 |AM| = 4 5 . The right-angled triangles ABM and AXN share the angle in A, and are therefore similar, and we have |AXAN| = |AMAB| |AX|: 4 5 = 2 5 :1, and therefore |AX| = 8 5 . If I is the foot of Y on AB, the triangles YIX and ABM are congruent, since their sides are pairwise orthogonal, and |AB| = |YI| = 1 holds. We therefore have |IX| = |BM| = 2 1 , and therefore |DY| = |AXIX| = 8 5 - 2 1 = 8 1 , and we see that |AX| = 5|DY| holds as claimed. 4. ANGLES 6) A paper strip is folded three times as shown. Determine if we are given that = 70° holds. A) 140° B) 130° C) 120° D) 110° E) 100° source: Kangaroo 2010, Étudiant Solution: Since the angle 70° is folded down, the small triangle in the middle of the strip in the second figure is isosceles with angles of 70°, 70° and 40°. Folding to the left and right therefore yields angles of 140° to the left and right, which have an angle of 40° in common. the angle in question is therefore equal to 360°-(140°+140°-40°) = 120°. The correct answer is therefore C. 7) Line r passes through the corner A of a sheet of paper and makes an angle with the horizontal border, as shown in Figure 1. In order to divide into three equal parts, we proceed as follows: a) initially we mark two points B and C on the vertical border such that AB = BC; through B we draw a line s parallel to the border (Figure 2); b) after that, we fold the sheet so as to make C coincide with a point C’ on the line r and A with a point A’ on line s (Figure 3); we call B’ the point which coincides with B. r s A A A B C C B C' A' B' Figure 1 Figure 2 Figure 3 Show that lines AA’ and AB’ divide angle into three equal parts. source: 22nd Brazilian Mathematical Olympiad 2000, Nr. 1 Solution: Let P be the point in which the crease intersects the bottom edge of the paper and X the point in which the crease intersects with s. Furthermore, let = PAA’. Since AP = AP’, the triangle APA’ is isosceles and we have PA’A = AA’P = , and therefore PA’X = 2, since AP and s are parallel. Since the triangles PAX and PA’X are congruent, we therefore also have PAX = 2, and therefore XAA’ = PAA’ = . Again, since AP and s are parallel, we therefore also have AXB = 2, and thus A C B C' A' B' P X A’XB’ = 2. Since AP and s are parallel, B’ therefore lies on the extension of AX. We see that AB’ is both altitude and median in the triangle AA’C’. It is an altitude since A’B is perpendicular to AC, and therefore AB’ must be perpendicular to A’C’, and it is a median because B is the mid-point of AC, and B’ must therefore be the mid-point of A’C’. We see that AA’C’ must be isosceles, and AB’ must also be the angle bisector in A, from which we deduce B’AC’ = B’AA’ = PAA’ = , proving the claim. Note that the angle trisection described here is well established in the origami math literature, and is due to H. Abe. Comments It is not difficult to find variations on the themes of these problems. Calculating the lengths of line segments, the angles between line segments and the areas of triangles or quadrilaterals is fairly standard stuff, and not usually interesting enough to be the kind of stuff we are looking for in a competition problem, however. One possible idea for an interesting situation may be the following. Question: We are given an equilateral triangle ABC with sides of unit length. The point A is folded to the point D on BC as shown, resulting in the crease EF with E on AB and F on AC. We assume that FD is perpendicular to BC. a) Determine the angle AED. b) Determine the length of the line segment CD. c) Determine the ratio of the areas of the triangles AEF and ABC. Answer: a) AED = 90° b) |CD| = 2 - 3 c) [AEF]:[ABC] = (33 – 5):1 |
4937 | https://www.ck12.org/flexi/math-grade-7/convert-decimals-to-fractions/how-do-you-convert-improper-fractions-into-mixed-numbers/ | Flexi answers - How do you convert improper fractions into mixed numbers? | CK-12 Foundation
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Convert Decimals to Fractions
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How do you convert improper fractions into mixed numbers?
Flexi Says:
To convert an improper fraction into a mixed number, follow these steps:
Divide the numerator by the denominator: This will give you the whole number part of the mixed number.
Find the remainder: Subtract the product of the whole number and the denominator from the numerator. This remainder will be the new numerator of the fractional part.
3. Write the mixed number: Combine the whole number with the new fraction (remainder over the original denominator).
Example:
Convert 11 4 into a mixed number.
Divide: 11÷4=2 (whole number) with a remainder of 3.
Remainder: 11−(4×2)=3.
Mixed number: Combine the whole number and the fraction: 2 3 4.
So, 11 4=2 3 4.
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4938 | https://en.wikipedia.org/wiki/Waterhouse%E2%80%93Friderichsen_syndrome | Jump to content
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1 Signs and symptoms
2 Causes
3 Diagnosis
4 Prevention
5 Treatment
6 Prognosis
7 History
8 References
Waterhouse–Friderichsen syndrome
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From Wikipedia, the free encyclopedia
Adrenal gland failure from internal bleeding, often due to infection
Medical condition
| Waterhouse–Friderichsen syndrome |
| Other names | Hemorrhagic adrenalitis |
| The adrenal glands lie above the kidneys. |
| Specialty | Endocrinology |
| Causes | Bacterial infection |
Waterhouse–Friderichsen syndrome (WFS) is defined as adrenal gland failure due to hemorrhages in the adrenal glands, commonly caused by sepsis. Typically, the bacteria responsible for triggering the bleeding is Neisseria meningitidis.
The bacterial infection leads to massive bleeding into one or both adrenal glands. Bilateral adrenal gland hemorrhaging is more common. It is characterized by overwhelming bacterial infection meningococcemia leading to massive blood invasion, organ failure, coma, low blood pressure and shock, disseminated intravascular coagulation (DIC) with widespread purpura, rapidly developing adrenocortical insufficiency and death.
Signs and symptoms
[edit]
Waterhouse–Friderichsen syndrome can be caused by a number of different organisms (see below). When caused by Neisseria meningitidis, WFS is considered the most severe form of meningococcal sepsis. The onset of the illness is nonspecific with fever, rigors, vomiting, and headache. Soon a rash appears; first macular, not much different from the rose spots of typhoid, and rapidly becoming petechial and purpuric with a dusky gray color. Low blood pressure (hypotension) develops and rapidly leads to septic shock. The cyanosis of extremities can be extreme and the patient is very prostrated or comatose. In this form of meningococcal disease, meningitis generally does not occur. Low levels of blood glucose and sodium, high levels of potassium in the blood, and the ACTH stimulation test demonstrate the acute adrenal failure. Leukocytosis need not be extreme and in fact leukopenia may be seen and it is a very poor prognostic sign. C-reactive protein levels can be elevated or almost normal. Thrombocytopenia is sometimes extreme, with alteration in prothrombin time (PT) and partial thromboplastin time (PTT) suggestive of disseminated intravascular coagulation (DIC). Acidosis and acute kidney failure can be seen as in any severe sepsis. Meningococci can be readily cultured from blood or cerebrospinal fluid, and can sometimes be seen in smears of cutaneous lesions. Difficulty swallowing, atrophy of the tongue, and cracks at the corners of the mouth are also characteristic features.[citation needed]
Causes
[edit]
Multiple species of bacteria can be associated with the condition:
Meningococcus is another term for the bacterial species Neisseria meningitidis; blood infection with said species usually underlies WFS. While many infectious agents can infect the adrenals, an acute, selective infection is usually meningococcus.
Pseudomonas aeruginosa can also cause WFS.
WFS can also be caused by Streptococcus pneumoniae infections, a common bacterial pathogen typically associated with meningitis in the adult and elderly population.
Mycobacterium tuberculosis could also cause WFS. Tubercular invasion of the adrenal glands could cause hemorrhagic destruction of the glands and cause mineralocorticoid deficiency[citation needed].
Staphylococcus aureus has recently also been implicated in pediatric WFS.
It can also be associated with Haemophilus influenzae.
Adrenal hemorrhage characteristic of the Waterhouse–Friderichsen syndrome has been identified in several autopsies of patients who died of sepsis secondary to Capnocytophaga canimorsus infection.
Viruses may also be implicated in adrenal problems:
Cytomegalovirus can cause adrenal insufficiency, especially in the immunocompromised.
Ebola virus infection may also cause similar acute adrenal failure.
Diagnosis
[edit]
Diagnostic criteria are based on clinical features of adrenal insufficiency as well as identifying the causal agent. If the causal agent is suspected to be meningitis a lumbar puncture is performed. If the causal agent is suspected to be bacterial a blood culture and complete blood count is performed. An adrenocorticotropic hormone stimulation test can be performed to assess adrenal function.
Prevention
[edit]
Routine vaccination against meningococcus is recommended by the Centers for Disease Control and Prevention for all 11- to 18-year-olds and people who have poor splenic function (who, for example, have had their spleen removed or who have sickle-cell disease which damages the spleen), or who have certain immune disorders, such as a complement deficiency.
Treatment
[edit]
Fulminant infection from meningococcal bacteria in the bloodstream is a medical emergency and requires emergent treatment with vasopressors, fluid resuscitation, and appropriate antibiotics. Benzylpenicillin was once the drug of choice with chloramphenicol as a good alternative in allergic patients. Ceftriaxone is an antibiotic commonly employed today. Hydrocortisone can sometimes reverse the adrenal insufficiency. Amputations, reconstructive surgery, and tissue grafting are sometimes needed as a result of tissue necrosis (typically of the extremities) caused by the infection.[citation needed]
Prognosis
[edit]
The prognosis of Waterhouse–Friderichsen syndrome varies by severity of the illness. Around 15% of patients with significant acute bilateral adrenal bleeding experience a fatal outcome. In cases where diagnosis and appropriate treatment are delayed, the case fatality rate approaches 50%. Recovery is possible with appropriate, timely management of the illness. Mineralocorticoid and glucocorticoid treatments may be necessary depending on the recovering patient's electrolyte status and response to treatment. Research shows that people with adrenal hemorrhage can regain adrenal function to some degree.
History
[edit]
Waterhouse–Friderichsen syndrome is named after Rupert Waterhouse (1873–1958), an English physician, and Carl Friderichsen (1886–1979), a Danish pediatrician, who wrote papers on the syndrome, which had been previously described.
References
[edit]
^ "Waterhouse–Friderichsen syndrome". Genetic and Rare Diseases Information Center (GARD). Archived from the original on January 13, 2009. Retrieved 14 December 2012.
^ a b Kumar V, Abbas A, Fausto N (2005). Robins and Coltran: Pathological Basis of Disease (7th ed.). Elsevier. pp. 1214–5. ISBN 978-0-7216-0187-8.
^ Karki, Bhesh R.; Sedhai, Yub Raj; Bokhari, Syed Rizwan A. (2024), "Waterhouse-Friderichsen Syndrome", StatPearls, Treasure Island (FL): StatPearls Publishing, PMID 31855354, retrieved 2024-11-08
^ "Waterhouse-Friderichsen syndrome". MedlinePlus Medical Encyclopedia. Retrieved 2014-04-12.
^ Adem P, Montgomery C, Husain A, Koogler T, Arangelovich V, Humilier M, Boyle-Vavra S, Daum R (2005). "Staphylococcus aureus sepsis and the Waterhouse-Friderichsen syndrome in children". N Engl J Med. 353 (12): 1245–51. doi:10.1056/NEJMoa044194. PMID 16177250.
^ Morrison U, Taylor M, Sheahan DG, Keane CT (January 1985). "Waterhouse-Friderichsen syndrome without purpura due to Haemophilus influenzae group B". Postgrad Med J. 61 (711): 67–8. doi:10.1136/pgmj.61.711.67. PMC 2418124. PMID 3873065.
^ McKinney WP, Agner RC (December 1989). "Waterhouse-Friderichsen syndrome caused by Haemophilus influenzae type b in an immunocompetent young adult". South. Med. J. 82 (12): 1571–3. doi:10.1097/00007611-198912000-00029. PMID 2595428.
^ Butler, T. (July 2015). "Capnocytophaga canimorsus: an emerging cause of sepsis, meningitis, and post-splenectomy infection after dog bites". European Journal of Clinical Microbiology & Infectious Diseases. 34 (7): 1271–1280. doi:10.1007/s10096-015-2360-7. PMID 25828064. S2CID 9960310.
^ Uno, Kenji; Konishi, Mitsuru; Yoshimoto, Eiichiro; Kasahara, Kei; Mori, Kei; Maeda, Koichi; Ishida, Eiwa; Konishi, Noboru; Murakawa, Koichi; Mikasa, Keiichi (1 January 2007). "Fatal Cytomegalovirus-Associated Adrenal Insufficiency in an AIDS Patient Receiving Corticosteroid Therapy". Internal Medicine. 46 (9): 617–620. doi:10.2169/internalmedicine.46.1886. PMID 17473501.
^ "BMJ Ebola fact-sheet". British Medical Journal. Archived from the original on 2022-10-09.
^ Rosa D, Pasqualotto A, de Quadros M, Prezzi S (2004). "Deficiency of the eighth component of complement associated with recurrent meningococcal meningitis--case report and literature review" (PDF). Braz J Infect Dis. 8 (4): 328–30. doi:10.1590/S1413-86702004000400010. PMID 15565265. Archived (PDF) from the original on 2022-10-09.
^ Karki, Bhesh R.; Sedhai, Yub Raj; Bokhari, Syed Rizwan A. (2022). "Waterhouse-Friderichsen Syndrome". StatPearls. StatPearls Publishing. PMID 31855354. Retrieved 1 March 2023.
^ Waterhouse R (1911). "A case of suprarenal apoplexy". Lancet. 1 (4566): 577–8. doi:10.1016/S0140-6736(01)60988-7.
^ Friderichsen C (1918). "Nebennierenapoplexie bei kleinen Kindern". Jahrbuch für Kinderheilkunde und Physische Erziehung. 87: 109–25.
| | |
--- |
| Classification | D ICD-10: A39.1, E35.1 ICD-9-CM: 036.3 MeSH: D014884 DiseasesDB: 29316 |
| External resources | MedlinePlus: 000609 eMedicine: med/3009 |
| v t e Pseudomonadota-associated Gram-negative bacterial infections |
| α | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- | | Rickettsiales | | | | | | | | | | | | | | --- --- --- --- --- --- | | Rickettsiaceae/(Rickettsioses) | | | | --- | | Typhus | Rickettsia typhi + Murine typhus Rickettsia prowazekii + Epidemic typhus, Brill–Zinsser disease, Flying squirrel typhus | | Spottedfever | | | | --- | | Tick-borne | Rickettsia rickettsii + Rocky Mountain spotted fever Rickettsia conorii + Boutonneuse fever Rickettsia japonica + Japanese spotted fever Rickettsia sibirica + North Asian tick typhus Rickettsia australis + Queensland tick typhus Rickettsia honei + Flinders Island spotted fever Rickettsia africae + African tick bite fever Rickettsia parkeri + American tick bite fever Rickettsia aeschlimannii + Rickettsia aeschlimannii infection | | Mite-borne | Rickettsia akari + Rickettsialpox Orientia tsutsugamushi + Scrub typhus | | Flea-borne | Rickettsia felis + Flea-borne spotted fever | | | | Anaplasmataceae | Ehrlichiosis: Anaplasma phagocytophilum + Human granulocytic anaplasmosis, Anaplasmosis Ehrlichia chaffeensis + Human monocytotropic ehrlichiosis Ehrlichia ewingii + Ehrlichiosis ewingii infection | | | Hyphomicrobiales | | | | --- | | Brucellaceae | Brucella abortus + Brucellosis | | Bartonellaceae | Bartonellosis: Bartonella henselae + Cat-scratch disease Bartonella quintana + Trench fever Either B. henselae or B. quintana + Bacillary angiomatosis Bartonella bacilliformis + Carrion's disease, Verruga peruana | | |
| β | | | | | | | | | | --- --- --- --- | | Neisseriales | | | | --- | | M+ | Neisseria meningitidis/meningococcus + Meningococcal disease, Waterhouse–Friderichsen syndrome, Meningococcal septicaemia | | M− | Neisseria gonorrhoeae/gonococcus + Gonorrhea | | ungrouped: | Eikenella corrodens/Kingella kingae + HACEK Chromobacterium violaceum + Chromobacteriosis infection | | | Burkholderiales | Burkholderia pseudomallei + Melioidosis Burkholderia mallei + Glanders Burkholderia cepacia complex Bordetella pertussis/Bordetella parapertussis + Pertussis | |
| γ | | | | | | | | | | | | | | --- --- --- --- --- --- | | Enterobacteriales(OX−) | | | | --- | | Lac+ | Klebsiella pneumoniae + Rhinoscleroma, Pneumonia Klebsiella granulomatis + Granuloma inguinale Klebsiella oxytoca Escherichia coli: Enterotoxigenic Enteroinvasive Enterohemorrhagic O157:H7 O104:H4 + Hemolytic-uremic syndrome Enterobacter aerogenes/Enterobacter cloacae | | Slow/weak | Serratia marcescens + Serratia infection Citrobacter koseri/Citrobacter freundii | | Lac− | | | | --- | | H2S+ | Salmonella enterica + Typhoid fever, Paratyphoid fever, Salmonellosis | | H2S− | Shigella dysenteriae/sonnei/flexneri/boydii + Shigellosis, Bacillary dysentery Proteus mirabilis/Proteus vulgaris Yersinia pestis + Plague/Bubonic plague Yersinia enterocolitica + Yersiniosis Yersinia pseudotuberculosis + Far East scarlet-like fever | | | | Pasteurellales | | | | --- | | Haemophilus: | H. influenzae + Haemophilus meningitis + Brazilian purpuric fever H. ducreyi + Chancroid H. parainfluenzae + HACEK | | Pasteurella multocida | Pasteurellosis Actinobacillus + Actinobacillosis | | Aggregatibacter actinomycetemcomitans | HACEK | | | Legionellales | Legionella pneumophila/Legionella longbeachae + Legionnaires' disease Coxiella burnetii + Q fever | | Thiotrichales | Francisella tularensis + Tularemia | | Vibrionaceae | Vibrio cholerae + Cholera Vibrio vulnificus Vibrio parahaemolyticus Vibrio alginolyticus Plesiomonas shigelloides | | Pseudomonadales | Pseudomonas aeruginosa + Pseudomonas infection Moraxella catarrhalis Acinetobacter baumannii | | Xanthomonadaceae | Stenotrophomonas maltophilia | | Cardiobacteriaceae | Cardiobacterium hominis + HACEK | | Aeromonadales | Aeromonas hydrophila/Aeromonas veronii + Aeromonas infection | |
| ε | | | | --- | | Campylobacterales | Campylobacter jejuni + Campylobacteriosis, Guillain–Barré syndrome Helicobacter pylori + Peptic ulcer, MALT lymphoma, Gastric cancer Helicobacter cinaedi + Helicobacter cellulitis | |
| v t e Adrenal gland disorder |
| Hyperfunction | | | | --- | | Aldosterone | Hyperaldosteronism Primary aldosteronism + Conn syndrome + Bartter syndrome + Glucocorticoid remediable aldosteronism AME Liddle's syndrome 17α CAH Pseudohypoaldosteronism | | Cortisol | Cushing's syndrome + Pseudo-Cushing's syndrome Steroid-induced osteoporosis | | Sex hormones | 21α CAH 11β CAH | |
| Hypofunction | | | | --- | | Aldosterone | Hypoaldosteronism + 21α CAH + 11β CAH | | Cortisol | CAH + Lipoid + 3β + 11β + 17α + 21α | | Sex hormones | 17α CAH Inborn errors of steroid metabolism | | Adrenal insufficiency | Adrenal crisis Adrenalitis + Xanthogranulomatous Addison's disease Waterhouse–Friderichsen syndrome | |
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What percentage is grey?
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Asked 2 years, 11 months ago
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The evenly spaced lines are drawn parallel to the base of triangle. What percentage of the triangle is grey?
mathematics
geometry
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edited Oct 19, 2022 at 13:48
bobble
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asked Oct 19, 2022 at 6:21
SimdSimd
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A visual approach...
Draw nine lines parallel to each of the other 2 edges.
Now there are 100 congruent triangles, 45 of which are grey.
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answered Oct 19, 2022 at 18:23
caPNCApncaPNCApn
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22 Now this is what I call a perfect answer.Simd –Simd 2022-10-19 23:48:33 +00:00 Commented Oct 19, 2022 at 23:48
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I believe the answer to be
45%
Justification:
Observe that all of the triangles are similar to each other (base angles are equal by alternate angles). Further the n n th triangle (n n corresponds to how many bands the triangle contains) has side length ratio n:1 n:1 with the top triangle, using the fact that the bands are evenly spaced. We can use the fact that the ratio of areas of similar triangles is equal to the square of the ratio of side lengths to obtain:
T n T=n 2 1 2 T n T=n 2 1 2
Where T T is the area of the top triangle and T n T n is the area of the n n th triangle. Rearranging yields:
T n=n 2 T T n=n 2 T
To find the area of the n n th band from the top we can take the difference of T n T n and T n−1 T n−1:
A n=T n−T n−1=n 2 T−(n−1)2 A=(2 n−1)A A n=T n−T n−1=n 2 T−(n−1)2 A=(2 n−1)A
That is, the area of the n n th band is equal to 2 n−1 2 n−1 times the area of the top triangle. We can then compute the area of the shaded parts as they are simply the odd bands:
A s h a d e d=∑i∈{1,3,5,7,9}A i=∑i∈{1,3,5,7,9}(2 i−1)T=45 T A s h a d e d=∑i∈{1,3,5,7,9}A i=∑i∈{1,3,5,7,9}(2 i−1)T=45 T
We can also compute the area of all regions by summing all the shaded areas:
A a l l=∑i=1 10 A i=∑i=1 10(2 i−1)T=100 T A a l l=∑i=1 10 A i=∑i=1 10(2 i−1)T=100 T
Then taking the ratio gives:
A s h a d e d A a l l=45 T 100 T=0.45 A s h a d e d A a l l=45 T 100 T=0.45
Giving the answer of 45%
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edited Oct 19, 2022 at 23:39
Toby Speight
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answered Oct 19, 2022 at 8:59
Maxim HehirMaxim Hehir
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Obviously, the limit as the number of row-pairs grows large has to be 50%. (I.e. as the triangle grows larger and larer or, alternatively, as we slice a fixed size triangle into thinner and thinner grey-white pairs.)Kaz –Kaz 2022-10-21 21:45:51 +00:00 Commented Oct 21, 2022 at 21:45
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(1+5+9+13+17)/10 2=5∗9%=45%(1+5+9+13+17)/10 2=5∗9%=45%.
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edited Oct 19, 2022 at 7:15
Evargalo
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answered Oct 19, 2022 at 6:46
Rosie FRosie F
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1 I can see the first equation from the divisions in the visual answer, but don't understand their derivation here. Or how to get to the equivalency in the center. More detail / explanation would be appreciated.brichins –brichins 2022-10-22 22:03:21 +00:00 Commented Oct 22, 2022 at 22:03
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An approach that avoids having to count lots of tiny triangles:
Cutting the triangle into 4 subtriangles of identical shape we find that triangles marked 2,3,4 also have the same shading, whereas 1 inverts grey and white bits. Now looking at 2 and 3 (or 3 and 4) together we see that the white to grey ratio is 3:2 in triangles 2,3,4 and 2:3 in triangle 1. Total ratio is therefore (3+3+3+2):(2+2+2+3) = 11:9, i.e. 55% white, 45% grey.
Variation:
Starting with two copies of the triangle cut off subtriangle 4 from both and rearrange as shown in the bottom panel. Count white and grey stripes and take the percentage.
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edited Oct 23, 2022 at 11:19
answered Oct 22, 2022 at 17:27
loopy waltloopy walt
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The parallel lines can be seen as the bottom edges of 10 similar triangles of alternating color, all sharing a common upper corner and stacked one atop another. Since the bottom edges are all evenly spaced, if the largest triangle has height h and base b, the total area is that of the largest triangle (n n = 10, counting smallest to largest):
A 10=b h 2 A 10=b h 2
and the area of each n n th triangle is proportional to its fractional height and width of the largest triangle:
A n=(n 10 b)(n 10 h)2=(n 10)2∗b h 2=n 2 100∗A 10=n 2%A n=(n 10 b)(n 10 h)2=(n 10)2∗b h 2=n 2 100∗A 10=n 2%
The shaded areas then are the exposed bottom strip of only the gray (odd-numbered) triangles, and that strip's area is the difference between the total area of its gray triangle and the area of the next-smaller white (even-numbered) neighbor obscuring it:
A n e x p o s e d=n 2−(n−1)2 A n e x p o s e d=n 2−(n−1)2
Therefore the total shaded percentage is the sum of the exposed gray strips, or
A e x p o s e d=∑n∈{1,3,5,7,9}n 2−(n−1)2=45 A e x p o s e d=∑n∈{1,3,5,7,9}n 2−(n−1)2=45
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answered Oct 21, 2022 at 7:20
brichinsbrichins
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This visual solution is similar to @cap's but generalises with ease to any number n of stripes. n can be even or odd in either case the longest stripe shall be white.
The example has angles 45°,45°,90° for visual clarity but obviously other angles work just as well:
Using two copies of the original triangle joined at their white side we get a parallelogram that can be subdivided into smaller parallelograms as shown. Except for the all-white diagonal these smaller parallelograms are half white, half grey.
As the diagonal comprises 1/n of the total area, the grey fraction is 1/2 x (1 - 1/n) for any number of stripes n.
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answered Nov 2, 2022 at 13:12
loopy waltloopy walt
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A simple[citation-needed] answer:
Let the height of each row be h h (so the height of the triangle is 10 h 10 h).
Knowing the triangles are similar (AAA) and equally spaced, let the lengths of the bottom sides be 1 w,2 w,3 w,…,10 w 1 w,2 w,3 w,…,10 w.
Using the formula for the area of a trapezoid A=a+b 2×h A=a+b 2×h:
The area of the whole triangle is 0 w+10 w 2×10 h=50 w h 0 w+10 w 2×10 h=50 w h.
The area of the first grey row is 0 w+1 w 2×h=0.5 w h 0 w+1 w 2×h=0.5 w h.
The area of the second grey row is 2 w+3 w 2×h=2.5 w h 2 w+3 w 2×h=2.5 w h.
Continuing, we get:
area in grey total area=0.5 w h+2.5 w h+4.5 w h+6.5 w h+8.5 w h 50 w h=22.5 w h 50 w h=45 100=45%area in grey total area=0.5 w h+2.5 w h+4.5 w h+6.5 w h+8.5 w h 50 w h=22.5 w h 50 w h=45 100=45%
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answered Oct 21, 2022 at 12:59
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Another approach which starts the same way as loopy walt's answer. Divide the whole triangle into triangular quarters as in loopy walt's diagram.
Consider triangles 2 and 3. They have half the total area, and 2/5 2/5 of it is grey, which is thus 1/5 1/5 of the total area.
Consider triangles 1 and 4. Those two triangles have half the total area, and half of it is grey, which is thus 1/4 1/4 of the total area.
So the total grey area 1/5+1/4=(4+5)/20=9/20=45%1/5+1/4=(4+5)/20=9/20=45% of the total area."
My point is that you don't need to establish the white/grey proportions in each triangle, merely to see that 1 and 4 have the same division but opposite shading, so they collectively have as much grey as white. Once triangles 2 and 3 are considered together, the resulting stripes are congruent, so the 3:2 3:2 ratio is easier to see there.
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answered Oct 23, 2022 at 5:51
Rosie FRosie F
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Here is a little Scala program for arbitrary n.
```
def puzzle(n: Int): Double =
if (n == 2) {
1.0/4.0
} else if (n % 2 == 1) {
(puzzle(n - 1) (n - 1) (n - 1) + (n + (n - 1))) / (n n)
} else {
(puzzle(n - 1) (n - 1) (n - 1)) / (n n)
}
@main def main: Unit =
val height = 10
val percentage = puzzle(height)
println(s"$percentage")
```
which prints
```
run
[info] running main
0.45
```
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edited Nov 7, 2022 at 12:03
answered Nov 7, 2022 at 11:37
LucDupAtSELucDupAtSE
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I am not great with calculus and whatnot, but I think this puzzle was within my reach-- Here's what I was able to figure out, and I hope it helps anybody who struggled to understand the math lingo in the other answers:
The answer is
45%.
This can be seen as
.. a nested stack of triangles, each one 10% narrower and 10% shorter than the last. This is because if all angles are the same, a 10% reduction in height must create a 10% reduction in base width, or else the triangles would no longer be similar.
The long form, which I'm sure other answers have written in a more compact mathematical format, is as follows:
Following from the top triangle down-
((.1b.1h)/2) -- Top triangle, 1st grey area
+(((.3b.3h)/2)-((.2b.2h)/2)) --2nd grey area
+(((.5b.5h)/2)-((.4b.4h)/2)) --3rd grey area
+(((.7b.7h)/2)-((.6b.6h)/2)) --4th grey area
+(((.9b.9h)/2)-((.8b.8h)/2)) --5th grey area
We can strip out the b, h, and even the "/2" portion because these are all in proportion to each other anyway, leaving us with .1² + (.3²-.2²) + (.5²-.4²) - (.7²-.6²) + (.9²-.8²) = .45
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edited Oct 21, 2022 at 19:26
ACB
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answered Oct 21, 2022 at 19:10
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FYI
Given @cap's clever visualization solution, following recursive formulas:
number g(n)g(n) of grey triangles
g(n)=n==1 g(n)=n==1??1:g(n−1)+(2∗n−1)∗(n%2)1:g(n−1)+(2∗n−1)∗(n%2)
total number t(n)t(n) of triangles
t(n)=n==1 t(n)=n==1??1:t(n−1)+(2∗n−1)1:t(n−1)+(2∗n−1)
confirm45%45% for n=10 n=10
I found no OEIS® entry for grey, white or total series.
The highest amount n n of rows with an integer percentage p p of grey ones seems:
n=50 n=50 with p=49%p=49%
Also, if we write w(n)w(n) for number of white triangles, then following non-recursive equations hold
t(n)=g(n)+w(n)=n 2 t(n)=g(n)+w(n)=n 2
and
w(n)−g(n)=(−1)n∗n w(n)−g(n)=(−1)n∗n
And, if we write x(n)x(n) and y(n)y(n) for the alternating smallest and biggest one of w(n)w(n) and g(n)g(n) then
x(n)=n∗(n−1)/2 x(n)=n∗(n−1)/2
and
y(n)=n∗(n+1)/2 y(n)=n∗(n+1)/2
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edited Nov 4, 2022 at 1:07
answered Nov 2, 2022 at 0:32
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One other visual approach independent of n n (n=10 n=10 here)
The biggest one of both areas (white in OP case) can be seen as sum of n n areas (colored red green alternating top down here) each one equal to k k(1<=k<=n)(1<=k<=n)units. One unit being the 'top' triangle area.
But then that area equals (by Gauss formula): n∗(n+1)/2 n∗(n+1)/2, in this case: 55 55
Similarly smallest one (grey in OP case, but not given a color here) has area: n∗(n−1)/2 n∗(n−1)/2, in this case: 45 45
As a consequence this also confirms:
Total area equals n 2 n 2units, in this case: 100 100 giving both percentages 45 45 and 55 55.
Required percentages for any n n now are easy to express in terms of above for any n n.
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edited Nov 6, 2022 at 23:49
answered Nov 6, 2022 at 1:30
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4940 | https://www.siyavula.com/read/za/mathematics/grade-10/euclidean-geometry-part-1/07-euclidean-geometry-part-1-02 | Log in
7.3 Quadrilaterals
| | | |
---
| 7.2 Triangles | | 7.4 The mid-point theorem |
7.3 Quadrilaterals (EMA5X)
Quadrilateral
: A quadrilateral is a closed shape consisting of four straight line segments.
The interior angles of a quadrilateral add up to 360360°.
Mathopenref has some useful simulations on different types of quadrilaterals. Clicking on any of the
named quadrilaterals will take you to a page specific to that quadrilateral.
Parallelogram (EMA5Y)
Parallelogram
: A parallelogram is a quadrilateral with both pairs of opposite sides parallel.
Worked example 3: Properties of a parallelogram
ABCDABCD is a parallelogram with AB∥DCAB∥DC and AD∥BCAD∥BC. Show that:
AB=DCAB=DC and AD=BCAD=BC
ˆA=ˆCA^=C^ and ˆB=ˆDB^=D^
Connect ACAC to form △ABC and △CDA
Redraw the diagram and draw line AC.
Use properties of parallel lines to indicate all equal angles on the diagram
On your diagram mark all the equal angles.
Prove △ABC≡△CDA
In △ABC and △CDA:
ˆA2=ˆC3(alt ∠s; AB∥DC)ˆC4=ˆA1(alt ∠s; BC∥AD)AC(common side)∴△ABC≡△CDA(AAS)∴AB=CD and BC=DA
∴ Opposite sides of a parallelogram have equal length.
We have already shown ˆA2=ˆC3 and ˆA1=ˆC4.
Therefore,
ˆA=ˆA1+ˆA2=ˆC3+ˆC4=ˆC
Furthermore,
ˆB=ˆD(△ABC≡△CDA)
Therefore opposite angles of a parallelogram are equal.
Summary of the properties of a parallelogram:
Both pairs of opposite sides are parallel.
Both pairs of opposite sides are equal in length.
Both pairs of opposite angles are equal.
Both diagonals bisect each other.
Worked example 4: Proving a quadrilateral is a parallelogram
Prove that if both pairs of opposite angles in a quadrilateral are equal, the quadrilateral is a
parallelogram.
Find the relationship between ˆx and ˆy
In WXYZ:
ˆW=ˆY=ˆy (given)ˆZ=ˆX=ˆx (given)ˆW+ˆX+ˆY+ˆZ=360° (sum of ∠s in a
quad)∴2ˆx+2ˆy=360°∴ˆx+ˆy=180°ˆW+ˆZ=ˆx+ˆy=180°
But these are co-interior angles between lines WX and ZY. Therefore WX∥ZY.
Find parallel lines
Similarly ˆW+ˆX=180°. These are co-interior angles between lines XY
and WZ. Therefore XY∥WZ.
Both pairs of opposite sides of the quadrilateral are parallel, therefore WXYZ is a
parallelogram.
Proving a quadrilateral is a parallelogram
Prove that if both pairs of opposite sides of a quadrilateral are equal, then the
quadrilateral is a parallelogram.
Prove that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a
parallelogram.
Prove that if one pair of opposite sides of a quadrilateral are both equal and parallel, then
the quadrilateral is a parallelogram.
A quadrilateral is a parallelogram if:
Both pairs of opposite sides are parallel.
Both pairs of opposite sides are equal.
Both pairs of opposite angles are equal.
The diagonals bisect each other.
One pair of opposite sides are both equal and parallel.
Textbook Exercise 7.3
(PQRS) is a parallelogram. (PS = OS) and (QO = QR). (S\hat{O}R = 96^{\circ}) and
(Q\hat{O}R = x).
Find with reasons, two other angles equal to (x).
(S\hat{R}O = Q\hat{O}R = x) (alt (\angle)s; (SR \parallel OQ)).
(O\hat{R}Q = Q\hat{O}R = x) ((\angle)s opp equal sides).
Therefore (S\hat{R}O) and (O\hat{R}Q) are both equal to (x).
Write (\hat{P}) in terms of (x).
\begin{align}
\hat{P} &= Q\hat{R}S \qquad \text{( opp } \angle \text{s of }\parallel \text{m)} \
&= S\hat{R}O + O\hat{R}Q \
\therefore \hat{P} & = 2x
\end{align}
Calculate the value of (x).
\begin{align}
S\hat{O}R & = 96^{\circ} \qquad \text{(given)} \
S\hat{O}P & = \hat{P} \qquad \text{(}\angle\text{s opp equal sides)}\
180^{\circ} & = \hat{P} + 96^{\circ} + Q\hat{O}R \qquad \text{(sum of } \angle
\text{s on a str line)}\
84^{\circ} & = 2x + x\
3x & = 84^{\circ} \
\therefore x & = 28^{\circ}
\end{align}
Prove that the diagonals of parallelogram (MNRS) bisect one another at (P).
Hint: Use congruency.
First number each angle on the given diagram:
In (\triangle MNP) and (\triangle RSP):
\begin{align}
\hat{M_1} &= \hat{R_1} \quad \text{(alt } \angle \text{s; } MN \parallel SR\text{)} \
\hat{P_1} &= \hat{P_3} \quad \text{(vert opp } \angle \text{s } = \text{)} \
MN & = RS \text{ (opp sides of }\parallel\text{m)}
\end{align}
Therefore (\triangle MNP \equiv \triangle RSP) (AAS).
Now we know that (MP = RP) and therefore (P) is the mid-point of (MR).
Similarly, in (\triangle MSP) and (\triangle RNP):
\begin{align}
\hat{M_2} &= \hat{R_2} \quad \text{(alt } \angle \text{s; } MS \parallel NR\text{)} \
\hat{P_4} &= \hat{P_2} \quad \text{(vert opp } \angle \text{s } = \text{)} \
MS & = RN \text{ (opp sides of }\parallel\text{m)}
\end{align}
Therefore (\triangle MSP \equiv \triangle RNP) (AAS).
Now we know that (NP = SP) and therefore (P) is the mid-point of (NS).
Therefore the diagonals of a parallelogram bisect each other.
Rectangle (EMA5Z)
Rectangle
: A rectangle is a parallelogram that has all four angles equal to 90°.
A rectangle has all the properties of a parallelogram:
Both pairs of opposite sides are parallel.
Both pairs of opposite sides are equal in length.
Both pairs of opposite angles are equal.
Both diagonals bisect each other.
It also has the following special property:
Worked example 5: Special property of a rectangle
PQRS is a rectangle. Prove that the diagonals are of equal length.
Connect P to R and Q to S to form △PSR and △QRS
Use the definition of a rectangle to fill in on the diagram all equal angles and sides
Prove △PSR≡△QRS
In △PSR and △QRS:
PS=QR(opp sides of rectangle)SR(common side)PˆSR=QˆRS=90° (∠s of rectangle)∴△PSR≡△QRS (RHS)Therefore PR=QS
The diagonals of a rectangle are of equal length.
Summary of the properties of a rectangle:
Both pairs of opposite sides are parallel.
Both pairs of opposite sides are of equal length.
Both pairs of opposite angles are equal.
Both diagonals bisect each other.
Diagonals are equal in length.
All interior angles are equal to 90°
Textbook Exercise 7.4
(ABCD) is a quadrilateral. Diagonals (AC) and (BD) intersect at (T). (AC = BD),
(AT = TC), (DT = TB). Prove that:
(ABCD) is a parallelogram
(AT = TC) (given)
(\therefore DB) bisects (AC) at (T)
and (DT = TB) (given)
(\therefore AC) bisects (DB) at (T)
therefore quadrilateral (ABCD) is a parallelogram (diag of (\parallel)m)
(ABCD) is a rectangle
(AC = BD) (given).
Therefore (ABCD) is a rectangle (diags of rectangle).
Rhombus (EMA62)
Rhombus
: A rhombus is a parallelogram with all four sides of equal length.
A rhombus has all the properties of a parallelogram:
Both pairs of opposite sides are parallel.
Both pairs of opposite sides are equal in length.
Both pairs of opposite angles are equal.
Both diagonals bisect each other.
It also has two special properties:
Worked example 6: Special properties of a rhombus
XYZT is a rhombus. Prove that:
the diagonals bisect each other perpendicularly;
the diagonals bisect the interior angles.
Use the definition of a rhombus to fill in on the diagram all equal angles and sides
Prove △XTO≡△ZTO
XT=ZT (sides of rhombus)TO(common side)XO=OZ(diags of rhombus)∴△XTO≡△ZTO(SSS)∴ˆO1=ˆO4But ˆO1+ˆO4=180°(∠s on a str
line)∴ˆO1=ˆO4=90°
We can further conclude that ˆO1=ˆO2=ˆO3=ˆO4=90°.
Therefore the diagonals bisect each other perpendicularly.
Use properties of congruent triangles to prove diagonals bisect interior angles
ˆX2=ˆZ1 (△XTO≡△ZTO)and ˆX2=ˆZ2 (alt ∠s; XT∥YZ)∴ˆZ1=ˆZ2
Therefore diagonal XZ bisects ˆZ. Similarly, we can show that XZ also bisects
ˆX; and that diagonal TY bisects ˆT and ˆY.
We conclude that the diagonals of a rhombus bisect the interior angles.
To prove a parallelogram is a rhombus, we need to show any one of the following:
All sides are equal in length.
Diagonals intersect at right angles.
Diagonals bisect interior angles.
Summary of the properties of a rhombus:
Both pairs of opposite sides are parallel.
Both pairs of opposite sides are equal in length.
Both pairs of opposite angles are equal.
Both diagonals bisect each other.
All sides are equal in length.
The diagonals bisect each other at 90°
The diagonals bisect both pairs of opposite angles.
Square (EMA63)
Square
: A square is a rhombus with all four interior angles equal to 90°
OR
A square is a rectangle with all four sides equal in length.
A square has all the properties of a rhombus:
Both pairs of opposite sides are parallel.
Both pairs of opposite sides are equal in length.
Both pairs of opposite angles are equal.
Both diagonals bisect each other.
All sides are equal in length.
The diagonals bisect each other at 90°
The diagonals bisect both pairs of opposite angles.
It also has the following special properties:
All interior angles equal 90°.
Diagonals are equal in length.
Diagonals bisect both pairs of interior opposite angles (i.e. all are 45°).
To prove a parallelogram is a square, we need to show either one of the following:
It is a rhombus (all four sides of equal length) with interior angles equal to 90°.
It is a rectangle (interior angles equal to 90°).
Trapezium (EMA64)
Trapezium
: A trapezium is a quadrilateral with one pair of opposite sides parallel.
In British English a trapezium is used to indicate a quadrilateral with one pair of opposite sides
parallel while in American English a trapezium is a quadrilateral with no pairs of opposite sides
parallel. We will use the British English definition of trapezium in this book.
In British English a trapezoid is used to indicate a quadrilateral with no pairs of opposite sides
parallel while in American English a trapezoid is a quadrilateral with one pair of opposite sides
parallel.
A trapezium is sometimes called a trapezoid.
Some examples of trapeziums are given below:
Kite (EMA65)
Kite
: A kite is a quadrilateral with two pairs of adjacent sides equal.
Worked example 7: Properties of a kite
ABCD is a kite with AD=AB and CD=CB. Prove that:
AˆDC=AˆBC
Diagonal AC bisects ˆA and ˆC
Prove △ADC≡△ABC
In △ADC and △ABC:
AD=AB(given)CD=CB(given)AC(common side)∴△ADC≡△ABC(SSS)∴AˆDC=AˆBC
Therefore one pair of opposite angles are equal in kite ABCD.
Use properties of congruent triangles to prove AC bisects ˆA and ˆC
ˆA1=ˆA2(△ADC≡△ABC)and ˆC1=ˆC2(△ADC≡△ABC)
Therefore diagonal AC bisects ˆA and ˆC.
We conclude that the diagonal between the equal sides of a kite bisects the two interior angles
and is an axis of symmetry.
Summary of the properties of a kite:
Diagonal between equal sides bisects the other diagonal.
One pair of opposite angles are equal (the angles between unequal sides).
Diagonal between equal sides bisects the interior angles and is an axis of symmetry.
Diagonals intersect at 90°
Textbook Exercise 7.5
Use the sketch of quadrilateral (ABCD) to prove the diagonals of a kite are
perpendicular to each other.
First number the angles:
In (\triangle ADO) and (\triangle ABO):
[\begin{array}{rll}
AD & = AB & \text{(given)} \
AO & & \text{(common side)} \
B\hat{A}O & = D\hat{A}O & \text{(given)} \
\therefore \triangle ADO & \equiv \triangle ABO & \text{(SAS)} \
\therefore A\hat{B}O & = A\hat{D}O &
\end{array}]
In (\triangle ADB):
\begin{align}
\text{let } \hat{A_1} & = \hat{A_2} = t \
\text{and let } A\hat{D}O &= A\hat{B}O = p \
2t + 2p &= 180° \quad \text{(sum of }\angle \text{s in} \triangle \text{)} \
\therefore t + p &= 90°
\end{align}
Next we note that:
\begin{align}
\hat{O_1} & = A\hat{B}O + \hat{A_1} \text{ (ext }\angle \text{ of } \triangle \text{)}\
\hat{O_1} &= p + t \
&= 90°\
\therefore AC &\perp BD
\end{align}
Therefore the diagonals of a kite are perpendicular to each other.
Explain why quadrilateral (WXYZ) is a kite. Write down all the properties of
quadrilateral (WXYZ).
Quadrilateral (WXYZ) is a kite because is has two pairs of adjacent sides that are
equal in length.
Diagonal between equal sides bisects the other diagonal: (WP = PY).
One pair of opposite angles are equal: (\hat{W_{1}} = \hat{Y_{1}}).
Diagonal between equal sides bisects the interior angles and is an axis of
symmetry: (\hat{X_1} = \hat{X_2}).
Diagonals intersect at (90°): (WY \perp PX).
This video provides a summary of the different types of quadrilaterals and their properties.
Video: 2G77
Relationships between the different quadrilaterals
Heather has drawn the following diagram to illustrate her understanding of the relationships between
the different quadrilaterals. The following diagram summarises the different types of special
quadrilaterals.
Explain her possible reasoning for structuring the diagram as shown.
Design your own diagram to show the relationships between the different quadrilaterals and
write a short explanation of your design.
Textbook Exercise 7.6
The following shape is drawn to scale :
Give the most specific name for the shape.
We start by counting the number of sides. There are four sides in this figure and so it
is either just a quadrilateral or one of the special types of quadrilateral.
Next we ask ourselves if there are any parallel lines in the figure. You can look at the
figure to see if any of the lines look parallel or make a quick sketch of the image and
see if any pairs of opposite lines meet at a point.
Both pairs of opposite sides are parallel. This means that the figure can only be one of
the following: parallelogram, rectangle, rhombus or square.
Next we ask ourselves if all the interior angles are 90°. All the interior angles
are 90° and so this must be a square or a rectangle. Finally we check to see if all
the sides are equal in length. In this figure the sides are not equal in length and so
it is a rectangle.
Therefore this is a rectangle.
The shape is also a parallelogram and a quadrilateral. This question, however, asked for
the most specific name for the shape.
The following shape is drawn to scale :
Give the most specific name for the shape.
We start by counting the number of sides. There are four sides in this figure and so it
is either just a quadrilateral or one of the special types of quadrilateral.
Next we ask ourselves if there are any parallel lines in the figure. You can look at the
figure to see if any of the lines look parallel or make a quick sketch of the image and
see if any pairs of opposite lines meet at a point.
Both pairs of opposite sides are parallel. This means that the figure can only be one of
the following: parallelogram, rectangle, rhombus or square.
Next we ask ourselves if all the interior angles are 90°. All the interior angles
are not 90° and so this must be a parallelogram or a rhombus. Finally we check to
see if all the sides are equal in length. In this figure the sides are equal in length
and so it is a rhombus.
Therefore this is a rhombus.
The shape is also a parallelogram and a quadrilateral. This question, however, asked for
the most specific name for the shape.
Based on the shape that you see list the all the names of the shape. The figure is drawn
to scale
Both pairs of opposite sides are not parallel. This means that the figure can only be
some combination of the following: trapezium, kite, or quadrilateral.
The shape is definitely a quadrilateral because it has four sides. It does not have any
special properties: it does not have parallel sides, or right angles, or sides which are
equal in length. Therefore it is only a quadrilateral.
Based on the shape that you see list the all the names of the shape. The figure is drawn
to scale
Both pairs of opposite sides are not parallel. This means that the figure can only be
some combination of the following: trapezium, kite, or quadrilateral.
The shape is definitely a quadrilateral because it has four sides. It is also a kite
because it has two pairs of adjacent sides which are the same lengths. It cannot be a
square or a rectangle because it does not have right angles. It cannot be a
parallelogram or a trapezium because it does not have any parallel sides. And it is not
a rhombus because the four sides are not all the same length.
Therefore the correct answer is: kite and quadrilateral.
Based on the shape that you see list the all the names of the shape. The figure is drawn
to scale
Both pairs of opposite sides are parallel. That means that this shape can belong to one
or more of these groups: square, rhombus, rectangle or parallelogram.
The given shape is a square. However, it is also a rectangle. A square is also a
parallelogram, because it has parallel sides; and it is a rhombus as well, it just
happens to have right angles. A square is also a kite, a trapezium and of course a
quadrilateral.
Therefore the correct answer is: square, rectangle, rhombus, parallelogram, kite,
trapezium and quadrilateral.
Find the area of (ACDF) if (AB = 8,~ BF = 17,~ FE = EC,~ BE = ED,~ \hat{A} =
90^{\circ} ,~ C\hat{E}D = 90^{\circ})
Construct (G) such that (AC = FG)
(BCDF) is a rhombus (diagonals bisect at right angles)
Since (BCDF) is a rhombus (BC = DF). We constructed (G) such that (AC = FG).
Therefore (AB = DG).
In (\triangle ABF) and (\triangle CGD):
\begin{align}
B\hat{A}F & = C\hat{G}D = 90^{\circ} \quad \text{(given and by construction)} \
AB & = DG \quad \text{(by construction)} \
BF & = CD \quad \text{(} BCDF \text{ is a rhombus)}
\end{align}
Therefore (ABF \equiv CGD) (RHS)
Therefore (AF = CG) and so (ACGF) is a rectangle (both pairs of opposite sides equal
in length and all interior angles are 90°).
We are given the length of (AB) and (BF). Since (\triangle ABF) is right-angled we
can use the theorem of Pythagoras to find the length of (AF):
\begin{align}
BF^{2} & = AB^{2} + AF^{2} \
(17)^{2} & = (8)^{2} + AF^{2} \
AF^{2} & = 225 \
AF & = 15
\end{align}
We also know that (FD = BF = 17) and so (AC = 17 + 8 = 25).
Therefore the area of rectangle (ACGF) is:
\begin{align}
A_{\text{rectangle}} & = l \times b \
& = (25)(15) \
& = 375
\end{align}
We are almost there. We now need to calculate the area of triangle (CDG) and subtract
this from the area of the rectangle to get the area of (ACDF).
The area of triangle (CDG) is:
\begin{align}
A_{\text{triangle}} &= \frac{1}{2} DG \times CG \
& = \frac{1}{2} (8 \times 15) \
&= 60
\end{align}
Therefore the area of (ACDF) is (375 - 60 = 315).
| | | |
---
| Previous 7.2 Triangles | Table of Contents | Next 7.4 The mid-point theorem | |
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4942 | https://www.sciencedirect.com/topics/agricultural-and-biological-sciences/saprophyte | Saprophyte - an overview | ScienceDirect Topics
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Saprophyte
In subject area:Agricultural and Biological Sciences
Saprophyte is defined as an organism that does not have pathogenic properties and exists without causing harm to the host organism.
AI generated definition based on:Pratique de la thanatopraxie, 2009
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You might find these chapters and articles relevant to this topic.
Chapter
Biotic Factors Regulate Some Aspects of Plant Development
1994, Plant Growth and DevelopmentDonald E. Fosket
1.
What is meant by the term saprophyte? The organism Pseudomonas syringae causes a disease known as “bacterial canker” on almond trees. Is this consistent with its role as a saprophyte on strawberries and many other plants? Although P. syringae is a saprophyte on most plants, it can damage even these in special circumstances. How can this damage be brought about?
2.
What is a compatible reaction between a host and a plant pathogen? What are the consequences of a compatible reaction? What characteristics must the pathogen and host have for a compatible reaction to occur?
3.
What is a necrotrophic pathogen? How does a necrotrophic pathogen kill its host? Give some examples of necrotrophic pathogens.
4.
What are biotrophic pathogens? What are haustoria and what function do they serve?
5.
How does lignification help make plants resistant to a particular disease? What are lignin precursors and what are the key steps in their biosynthesis?
6.
What is the hypersensitivity response? How does an effective hypersensitivity response render a plant resistant to a particular pathogen?
7.
What are phytoalexins? What are some of the key steps in the biosynthesis of phytoalexins? What role do they play in the plant's defense response to disease organisms?
8.
What are the different pathways a pathogen might use to enter a host plant? If a fungus grows through the cell walls of a plant, digesting the components of the primary walls and using these molecules for its nutrition, what measures might the plant take to block the progress of the fungus through its tissues?
9.
What are resistance and avirulence genes? What role do they play in plant disease?
10.
What are pathogenicity genes? What types of functions do pathogenicity genes encode? What triggers the expression of pathogenicity genes?
11.
What is an elicitor? Give several examples of molecules that act as elicitors. How are elicitors produced? What function do they have in plant disease?
12.
What is the class of molecules known as “opines”? What role do opines play in crown gall disease? Where is the genetic information necessary for the synthesis and utilization of opines?
13.
What are the Ti plasmids carried by the crown gall bacterium Agrobacterium tumefaciens? What features do all Ti plasmids that cause galls have in common?
14.
What is the T-DNA of Ti plasmids? What role does it play in crown gall disease? What is the significance of the 23-bp direct repeat sequence found at either end of T-DNA?
15.
What genetic information does the T-DNA carry? Explain the effect of mutations in T-DNA genes 1,2,3, and 4 on the characteristics of tissues transformed by Ti plasmids with these mutations, both singly and when a given plasmid has mutations in all these genes.
16.
What role do plant hormones play in crown gall disease?
17.
What are the vir genes of the Ti plasmid? What role do they play in crown gall disease? What induces the expression of the vir genes? Where do these compounds come from?
18.
The overall reaction for the reduction of atmospheric N 2 shows that it takes 12 moles of ATP and 6 electrons to reduce 1 mole of N 2. Where does this ATP come from? This reaction is possible only because the legume and the bacterium are in a symbiotic relationship. Explain what each partner contributes to the reaction.
19.
How do Rhizobium bacteria find and recognize the plant species they colonize in nitrogen-fixing nodules?
20.
How do Rhizobium bacteria reach the cortical cells of the root? What is the infection thread and of what is it made? What is the significance of the peribacteroid membrane for nitrogen fixation and where does it come from?
21.
How is a bacteroid different from a free-living Rhizobium bacterium in terms of the genes it expresses? What is the difference between nif and nod genes in terms of the processes in which their products function?
22.
What are nodulins? What role do the nodulins play in nitrogen fixation?
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Book 1994, Plant Growth and DevelopmentDonald E. Fosket
Chapter
Disorders of dysbiosis
2019, The Theory of EndobiogenyKamyar M. Hedayat, Jean-Claude Lapraz
General considerations
Saprophytes adapt themselves to the metabolic demands of the organism. The greater the presentation of nutrients and calories, the greater the growth of saprophytes will be. Conversely, saprophytes can adapt the organism to its metabolic demands through the gut-brain interaction (Chapter 5).3, 4 Saprophytes participate in the metabolic response of the organism and the organism plays a part in the metabolic response of the saprophyte. This creates the endobiogenic equilibrium and determines the buffering capacity of the organism.
There is a constant interaction between the host, the flora, the environment, and the response of each and all to internal and external demands. The pathogenicity of noncommensal flora will be determined more by the endobiogenic equilibrium at the time of exposure and the quality of the adaptive response more than by the intrinsic pathogenicity of the organism.
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Book 2019, The Theory of EndobiogenyKamyar M. Hedayat, Jean-Claude Lapraz
Chapter
Facultative fungal endophytes and their potential for the development of sustainable agriculture
2021, Microbial Management of Plant StressesBarra-Bucarei Lorena PhD, ... Castro Jean Franco PhD
1.2 Endophytic fungi
Endophytic microorganisms are beneficial to plants and include fungi and bacteria. The term endophyte comes from the Greek words endon and phyton, where endon means inside and phyton means plant. It was first coined by De Bary in 1886 and is defined as an organism that spends most or all of its life cycle colonizing host plant tissues without causing obvious damage (Bascom-Slack et al., 2012). Its definition has evolved considerably, as associations with different parts of plants and the modes of association with their hosts have been taken into account (Carroll, 1988). There are approximately 300,000 species of plants on earth, each associated with one or more endophytes. These are naturally found in the ecosystem and have the capacity to occupy any plant in any environment. They are considered extremely important partners for plants (Arnold et al., 2000; Hallmann et al., 1997; Strobel and Daisy, 2003; Suryanarayanan, 2013) because they fulfill a variety of symbiotic and ecological functions (Rodriguez et al., 2009). The host range of different endophytic species is variable, ranging from single-host species to very generalist multi-host species (Sánchez Márquez et al., 2011; Stone et al., 2004). The plant receives multiple benefits from its interaction with the endophyte, in exchange for carbon-based resources (Herre et al., 2007). Within this group, entomopathogenic fungi have aroused the interest of researchers because of several characteristics that could allow them to be used with different objectives in agricultural production systems.
On the other hand, entomopathogenic fungi are a unique and highly specialized group of microbial agents that possess desirable traits that favor their development as biopesticides (Lacey et al., 2015). More than a century ago, Pasteur predicted the advantages of these fungi because of their role as pest bioregulators, acting as insect parasites that are not harmful to plants. They are constituted by species distributed around the world, which generally inhabit the soil; however, several species of entomopathogenic fungi of the genera Acremonium, Beauveria, Cladosporium, Clonostachys, Fusarium, Metarhizium, Paecilomyces, Trichoderma, and Verticillium have been reported as natural plant endophytes and have also been artificially inoculated into different cultivated plant species (Akello and Sikora, 2012; Akutse et al., 2014; Gómez-Vidal et al., 2006; Powell et al., 2009; Vega et al., 2008). Within this group, the genera Beauveria and Metarhizium stand out and have been widely studied as biological controllers. Currently, more than 700 species of arthropods are parasitized by Beauveria bassiana (Balsamo-Crivelli) Vuillemin, and another 200 by Metarhizium anisopliae (Metchnikoff), with formulations that are currently available for commercial use (Wraight et al., 2001). Thus, it is clear that the use of these endophytes as biopesticides has increased in recent decades (Feng et al., 1994; Zimmermann, 2007). These species have been widely studied and reported as natural or facultative endophytes in several plants, such as corn (Wagner and Lewis, 2000), banana (Akello et al., 2008), coffee (Posada et al., 2007; Vega et al., 2010), tomato (Barra-Bucarei et al., 2020; Leckie, 2002; Ownley et al., 2008), date palm (Gómez-Vidal et al., 2006), and chili pepper (Jaber and Ownley, 2018). Another fungus reported as entomopathogenic but better known as an antagonist of plant pathogens is Trichoderma. Studies have provided evidence that there are certain types of strains of this fungus capable of colonizing the root part of the plant and then establishing a symbiotic relationship with that particular plant. Several investigations have been carried out to demonstrate that strains of Trichoderma sp. can develop an endophytic relationship with specific plants (Han et al., 2019; Samuels et al., 2012).
Entomopathogenic endophytic fungi establish a symbiotic relationship with plants that results in increased growth, inhibition of pathogenic organisms, removal of contaminants from the soil, and increased tolerance to extreme temperatures, water availability, and salinity, among others (Ownley et al., 2010; Quesada-Moraga et al., 2009). The level of symbiosis may be mandatory, that is, the fungus can spend its entire life cycle inside plants since they depend on their metabolism to survive and can spread between plants vertically or by vector activity. It can also be facultative since they can survive outside plants at some stages of their life cycle and could be recruited by plants from the soil, mainly through the rhizosphere (Hardoim et al., 2008). This feature of their lifestyle is interesting from the standpoint of developing products for agricultural use.
1.2.1 Facultative endophytic fungi and their life inside the plant
Entomopathogenic endophytic fungi (EEFs) can to colonize plants in various ways. They can enter through its various structures and then move internally through the vascular system to establish in their tissues; in other words, they present a pattern of systemic colonization (Gómez-Vidal et al., 2006; Gurulingappa et al., 2010; Quesada-Moraga et al., 2006). Nevertheless, they can also enter and be located in certain plant parts (Wearn et al., 2012; Yan et al., 2015) or divided among plant parts (Behie et al., 2015; Zambell and White, 2015). Various studies have shown that M. anisopliae preferentially establishes in the roots (Barelli et al., 2016; Greenfield et al., 2016; Liao et al., 2014; Meyling et al., 2011), while B. bassiana has been reported as an endophyte of aboveground plant parts (Behie et al., 2015). The way EEFs enter plants could be through natural openings or through the cuticle (Landa et al., 2013), supported by the mechanical force that is exerted by the infection structure (Bidochka and Khachatourians, 1991), the enzymatic dissolution of the cuticle, or a combination of both (Kolattukudy, 1985), a mechanism similar to that used by the fungus to enter insects.
Various forms of artificial inoculation of EEF into plants have been studied, such as applications of propagules to the leaves (Rondot and Reineke, 2018; Russo et al., 2015; Tefera and Vidal, 2009), suspensions to the root (Greenfield et al., 2016; Posada and Vega, 2005; Posada et al., 2007; Qayyum et al., 2015), seed coating and immersion (Jaber, 2018; Kabaluk and Ericsson, 2007; Quesada-Moraga et al., 2009), and stem injections (Posada et al., 2007). The host range of the different species of this group of fungi is variable and ranges from single-host species to very generalist multi-host species (Sánchez Márquez et al., 2011; Stone et al., 2004).
On the other hand, several studies have shown that the transmission of EEF can be vertical, that is, it can pass from one generation to another through the seed or horizontal passing from one individual to another by dispersing its conidia (Dipali et al., 2016; Quesada-Moraga et al., 2014). There is evidence that these fungi can colonize various plant tissues for several months and can affect the physiological activities of plants, although the tissues colonized and the duration of persistence varies (Brownbridge et al., 2012; Shikano, 2017). Consequently, the establishment and persistence of endophytic fungi could be conditioned by biotic (Vicari et al., 2002) and abiotic (Bultman and Bell, 2003) factors. Among the biotic factors, those associated with the plant are especially important, such as species and variety, age, type of tissue, and endophytic microbiota (Parsa et al., 2013; Posada et al., 2010). Studies conducted by Vega et al. (2008) determined that the presence of endophytic fungi in coffee plants can affect the establishment of other endophytic fungi, such as B. bassiana. Aspects associated with the fungus, including species, strain, inoculum concentration (Parsa et al., 2013; Tefera and Vidal, 2009; Vidal and Jaber, 2015), and interaction with other antagonistic microorganisms such as fungi and bacteria (Fisher and Petrini, 1987), can also influence its establishment. There are other factors that can influence the establishment and persistence of EEFs in the plant, such as the application method, which has been confirmed in several studies (Posada et al., 2007; Quesada-Moraga et al., 2014). For example, it has been demonstrated that applications by means of foliar sprays give way to a temporary persistence of the endophyte, since it has only been possible to isolate it temporarily from the treated area, with its posterior disappearance (Batta, 2013; Biswas et al., 2012; Gurulingappa et al., 2010; Landa et al., 2013). However, this issue is still not well elucidated because in a study conducted by Tefera and Vidal (2009), it was confirmed that B. bassiana colonizes the leaves, stems, and roots of sorghum plants, regardless of the inoculation method used (foliar inoculation techniques, seeds or soil). On the other hand, among the abiotic factors affecting the establishment of EEF, the type of substrate used for the plants has proven to be especially important. For example, a sterile substrate could produce a fungistatic (Lingg and Donaldson, 1981) and antagonistic (Pereira et al., 1993) effect for the development of the endophyte. Environmental factors such as temperature, radiation, and relative humidity, to which the plants are exposed, along with their nutritional status, also have an impact (Bing and Lewis, 1991; Vidal and Jaber, 2015).
1.2.2 Facultative endophytic fungi and their life outside the plant
One of the environments where plants can live is the soil; thus, the type and amount of microorganisms present in the soil can be an indicator of its health and productive potential. Soils with more microorganisms and greater diversity are usually healthier and therefore more productive. Plant roots release part of the photosynthetically fixed carbon, in addition to a wide range of compounds, among which soluble sugars, amino acids, or secondary metabolites are of great significance; these compounds act as attractants of beneficial organisms (Badri et al., 2009). The number of species of microorganisms that can live together in the rhizosphere can vary from thousands to millions (Nihorimbere et al., 2011). Subsequently, the interactions between roots and soil microorganisms are often specialized and based on coevolutionary pressures (Duffy et al., 2004; Morrissey et al., 2004).
Cultivated plants use a diverse and complex community of rhizospheric microorganisms to maintain their health condition and primary production (Jacoby et al., 2017). Several species of EEF have been reported as plant root colonizers, providing a variety of benefits, the most important of these are phosphate solubilization, siderophore production, phytohormone production, and increased nutrient availability. Nonetheless, biocontrol activity in the soil is also imperative, and these microorganisms could be decisive in this regard for sustainable production systems. For example, Metarhizium is a ubiquitous fungus in the soil and can inhabit the plant rhizosphere (Sasan and Bidochka, 2012), providing benefits such as resistance to salt stress (Khan et al., 2012), increased biomass, and plant growth (Elena et al., 2011; Khan et al., 2012), and nitrogen acquisition. The pathway by which these fungi could make nitrogen available to plants is related to their activity as insect pathogen. The fungi can leave the plant and go to the soil while remaining connected to it through its hyphae, parasitize insects, and carry the nitrogen obtained from these to the plant (Behie et al., 2012; Behie and Bidochka, 2014). Many studies have been conducted to determine the ability of EEF to persist in the soil or to colonize plants (Fang and Leger, 2010). The persistence of these fungi in the soil would be related to several factors, including the influence of the microbial community that inhabits the soil (Scheepmaker and Butt, 2010) that could have inhibiting effects on the growth of these fungi. Studies conducted by Barelli et al. (2020) determined that strains of Metarhizium robertsii have the ability to colonize the roots of bean plants and persist in the rhizosphere of the plant, without affecting the soil microorganisms. On the other hand, Trichoderma sp. is also known as a ubiquitous fungus, and it can be found in all types of soils, including forests, agricultural soils, and even desert soils (Kour et al., 2019a, b). It has an important presence and persistence in the soil as a saprophyte, mainly due to its competitive ability that allows it to quickly colonize these environments. This fungus has a high capacity to use different types of complex substrates, such as decomposing wood. This is mainly due to heterotrophic, such as decomposition and opportunistic endophytism. Some of the functions that Trichoderma sp. performs in the soil are decomposition, bioremediation, growth promotion, and biocontrol, by fulfilling a dual function as an antagonist of pathogens and controller of insect pests (Bissett et al., 2015; Kour et al., 2019a, b; Yadav et al., 2019). One of the most widely described roles of this fungus is its ability to control soil pathogens (Verma et al., 2017).
Although soil is an environment where EEF can live and perform various functions for plants, they can also survive in a nonsoil environment called the phyllosphere. These fungi can inhabit the outside of the aerial parts of the plant, such as leaves, stems, and flowers (epiphytes or phylloplanum), forming complex interactions with the plant. Plants and microorganisms have evolved together for more than 400 million years, with the phyllosphere as an important platform for their interactions. During this time, complex interactions have developed that significantly influence plant performance, for example, by helping plants defend against certain biotic and abiotic stresses (Liu et al., 2020). As a habitat for fungi, the phyllosphere is geographically widespread and heterogeneous, forming associations that are still not well known. An important focus of study is leaf surfaces, which are very heterogeneous and provide many potential niches for the colonization of fungi (Kembel and Mueller, 2014). In the phyllosphere, fungi could reduce the infection of plant tissues due to pathogens, either directly by producing antagonistic molecules and competing for resources or indirectly by inducing the resistance response of plants. Studies by Nishi et al. (2020) determined that B. bassiana strains can grow on the surface of tomato leaves and remain outside when there are high humidity conditions, besides they also have the ability to enter the interior of the leaves, suggesting that the fungus could occupy a certain niche depending on the strain, environmental conditions, and the host plant. The environmental conditions that occur on the surface of the leaves are a major challenge to the survival of the fungus. UV-A and UV-B radiation and heat from sunlight are factors that negatively affect the growth and persistence of the fungus (Rangel et al., 2005). The ability to colonize the interior of plant tissues could allow these fungi to increase their ability to be used in agricultural production systems.
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Book 2021, Microbial Management of Plant StressesBarra-Bucarei Lorena PhD, ... Castro Jean Franco PhD
Chapter
Plant Morphology
2019, Plant Systematics (Third Edition)Michael G. Simpson
P lant L ife F orm
Plant life form denotes aspects of their structure, life cycle, and physiology. (See Raunkiaer 1934.) Life form types include: chamaephyte, an overwintering perennial with buds at or just below ground level; epiphyte, a plant growing on another plant, e.g., Tillandsia (Bromeliaceae); geophyte, a perennial herb with underground perennating rootstocks such as bulbs, corms, rhizomes; halophyte, a salt-adapted plant, e.g., Salicornia (Chenopodiaceae); mycotroph [myco-heterotroph], a usually achlorophyllous plant having an intimate contact with a fungus, from which it receives most of its nutrition; phreatophyte, a plant with a long taproot, in contact with ground water (e.g., mesquite, Prosopis); succulent, a plant with fleshy stems (stem succulents, e.g., cacti) or leaves (leaf succulents, e.g., members of Aizoaceae or Crassulaceae); therophyte, an annual plant; and xerophyte, a plant adapted to live in a dry, generally hot environment;
Plant life form types denoting nutritional physiology include: saprophyte, a heterotrophic plant living off dead organic matter; mycotroph [myco-heterotroph], a usually achlorophyllous plant having an intimate contact with a fungus, from which it receives most of its nutrition; and parasite, a plant feeding on another plant, dependent on it for all or part of its nutrition. A parasite may be further classified into a holoparasite, one that lacks photosynthesis/chloroplasts and must attach to a host to survive and reproduce (e.g., Cuscuta, Orobanche, Balanophora), or a hemiparasite, a parasitic plant that is photosynthetic during at least part of its life cycle. A hemiparasite may be either an obligate hemiparasite, requiring a host to survive and reproduce (e.g., Phoradendron, mistletoe, Viscaceae) or a facultative hemiparasite, not requiring a host connection to survive and reproduce (e.g., Krameria, Krameriaceae, or Pedicularis, lousewort, a photosynthetic member of the Orobanchaceae).
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Book 2019, Plant Systematics (Third Edition)Michael G. Simpson
Chapter
Fungi
2013, Brenner's Encyclopedia of Genetics (Second Edition)R.M. Martinez
Glossary
Eukaryotes
Cells distinguished by the presence of a true nucleus; including plants, fungi, and animals.
Hypha (pl. hyphae)
Thread-like filaments that may form branches. Hyphae produce fungal spores.
Mycelium
A mass of hyphae and spores.
Mycology
The study of fungi.
Saprophyte
Organism that lives on dead organic material.
Spore
Reproductive element that is typically easily airborne and functions in dispersal.
Yeast
Round or oval cell that reproduces by budding or fission.
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Reference work 2013, Brenner's Encyclopedia of Genetics (Second Edition)R.M. Martinez
Chapter
Plant Morphology
2019, Plant Systematics (Third Edition)Michael G. Simpson
PLANT ORGANS
The basic structural components, or organs, of plants are delimited by and strongly correlated with their specific functions. Among the liverworts, hornworts, and mosses (see Chapter 3), these organs are components of the haploid gametophyte. The gametophyte of these taxa contains rhizoids, which are uniseriate, filamentous chains of cells functioning in anchorage and water/mineral absorption. The basic body of the gametophyte can either be a flat mass of cells, termed a thallus (found in some liverworts and all hornworts) or a shoot, consisting of a generally cylindrical stem bearing leaves (found in some liverworts and all mosses; see Chapter 3). It should be noted that the shoot systems of liverworts and mosses are gametophytic tissue.
The major organs of vascular plants are sporophytic roots and shoots. Roots are present in almost all vascular plants and typically function in anchorage and absorption of water and minerals. Roots consist of an apical meristem that gives rise to a protective root cap, a central endodermis-bounded vascular cylinder, absorptive epidermal root hairs, and endogenously developed lateral roots (Figure 9.1).
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Figure 9.1. General plant structure, showing primary root and primary shoot. Note that all parts of both root and shoot are derived from cell divisions of the root or shoot apical meristem.
The sporophytic shoots of vascular plants consist of stem plus leaves (Figure 9.1). Shoots contain an apical meristem of actively dividing cells that, through continued differentiation, result in the elongation of the stem and formation of leaves and buds (see later discussion). The stem is a generally cylindrical organ that bears the photosynthetic leaves. Stems typically function in conduction of water and minerals from the roots and in support and elevation of both leaves and reproductive structures, although some stems are highly modified for other functions (see later discussion). The leaf is that organ of the shoot that is generally dorsiventrally flattened and that usually functions in photosynthesis and transpiration. Leaves are derived from leaf primordia within the shoot apex and are often variously modified. In vascular plants, leaves contain one to many vascular bundles, the veins; in some mosses, the gametophytic leaves may contain a veinlike costa, consisting of specialized (although not truly vascular) conductive tissue. Buds are immature shoot systems, typically located in the axils of leaves. Buds may grow to form lateral vegetative branches or reproductive structures (see later discussion).
Among reproductive plant organs, the sporangium is the basic spore-producing part of all land plants. The sporangium of liverworts, hornworts, and mosses is known as a capsule and typically makes up most of the sporophyte (see Chapter 3). In heterosporous plants, including all of the seed plants, sporangia are of two types: male or microsporangium and female or megasporangium (see Chapter 4). A cone, also called a strobilus, is a modified, determinate, reproductive shoot system of many nonflowering vascular plants, consisting of a stem axis bearing either sporophylls, in “simple” cones, or modified shoot systems, in “compound” cones (see Chapter 5). An ovule is a megasporangium enveloped by one or more protective integuments. A seed is the mature ovule of the seed plants, consisting of an internal embryo surrounded by nutritive tissue (comprising female gametophyte or endosperm) and enveloped by a protective seed coat (Chapter 5). The reproductive organ of angiosperms is the flower, a modified, determinate shoot bearing sporophylls called stamens and carpels, with or without outer modified leaves, the perianth (see Chapter 6). An inflorescence is an aggregate of one or more flowers, the boundaries of which generally occur with the presence of vegetative leaves. A fruit is the mature ovary of flowering plants, consisting of the pericarp (mature ovary wall), seeds, and (if present) accessory parts.
P lant H abit
Plant habit refers to the general form of a plant, encompassing a variety of components such as stem duration and branching pattern, development, or texture. Most plants can be clearly designated as an herb, vine, liana, shrub, or tree (with some subcategories; see later discussion); however, some species are difficult to accommodate into these categories. An herb is a plant in which any aboveground shoots, whether vegetative or reproductive, die back at the end of an annual growth season. Although the aboveground shoots are annual, the herb itself may be annual, biennial, or perennial, the last by means of long-lived underground rootstocks. Such perennial herbs, having a bulb, corm, rhizome, or tuber as the underground stem, are termed geophytes. A vine is a plant with elongate, weak stems, that are generally supported by means of scrambling, twining, tendrils, or roots; vines may be annual or perennial, herbaceous or woody. A liana (also spelled liane) is a vine that is perennial and woody; lianas are major components in the tree canopy layer of some tropical forests. A shrub is a perennial, woody plant with several main stems arising at ground level. A subshrub is a short shrub that is woody only at the base and that seasonally bears new, nonwoody, annual shoots above. Finally, a tree is defined as a generally tall, perennial, woody plant having one main stem (the trunk) arising at ground level. (Some plant ecologists will sometimes distinguish between shrubs and trees based primarily on an arbitrary height.)
P lant H abitat
Plant habitat refers to the general environment where the plant is growing. General habitat terms include whether the plant is terrestrial, growing on land; aquatic, growing in water; or epiphytic, growing on another plant. If aquatic, a plant can be submersed, occurring under water; floating, occurring at the water surface; or emergent, having roots or stems anchored to the substrate under water and aerial shoots growing above water. A rheophyte is a plant found along (often swiftly flowing) streams and river banks. Other aspects of the habitat include the type of substrate that the plant is growing in (e.g., whether on sandy, loam, clay, gravelly, or rocky soil). Saxicolous [epipetric] refers to a plant growing on rocks or boulders, either in or on the surface (lithophyte) or in the cracks (chasmophyte). Other habitat features include slope, aspect, elevation, moisture regime, and surrounding vegetation, community, or ecosystem (See Chapter 17, “Plant Collecting and Documentation.”)
P lant L ife F orm
Plant life form denotes aspects of their structure, life cycle, and physiology. (See Raunkiaer 1934.) Life form types include: chamaephyte, an overwintering perennial with buds at or just below ground level; epiphyte, a plant growing on another plant, e.g., Tillandsia (Bromeliaceae); geophyte, a perennial herb with underground perennating rootstocks such as bulbs, corms, rhizomes; halophyte, a salt-adapted plant, e.g., Salicornia (Chenopodiaceae); mycotroph [myco-heterotroph], a usually achlorophyllous plant having an intimate contact with a fungus, from which it receives most of its nutrition; phreatophyte, a plant with a long taproot, in contact with ground water (e.g., mesquite, Prosopis); succulent, a plant with fleshy stems (stem succulents, e.g., cacti) or leaves (leaf succulents, e.g., members of Aizoaceae or Crassulaceae); therophyte, an annual plant; and xerophyte, a plant adapted to live in a dry, generally hot environment;
Plant life form types denoting nutritional physiology include: saprophyte, a heterotrophic plant living off dead organic matter; mycotroph [myco-heterotroph], a usually achlorophyllous plant having an intimate contact with a fungus, from which it receives most of its nutrition; and parasite, a plant feeding on another plant, dependent on it for all or part of its nutrition. A parasite may be further classified into a holoparasite, one that lacks photosynthesis/chloroplasts and must attach to a host to survive and reproduce (e.g., Cuscuta, Orobanche, Balanophora), or a hemiparasite, a parasitic plant that is photosynthetic during at least part of its life cycle. A hemiparasite may be either an obligate hemiparasite, requiring a host to survive and reproduce (e.g., Phoradendron, mistletoe, Viscaceae) or a facultative hemiparasite, not requiring a host connection to survive and reproduce (e.g., Krameria, Krameriaceae, or Pedicularis, lousewort, a photosynthetic member of the Orobanchaceae).
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Book 2019, Plant Systematics (Third Edition)Michael G. Simpson
Chapter
EVOLUTIONARY ASPECTS OF SYMBIOSIS
1973, The LichensG.D. SCOTT
2 PROGRESS TOWARD ASSOCIATION
Symbiotic, parasitic, and saprophytic fungi have been labeled with different degrees of physiological efficiency according to their physical closeness to chloroplasts of the “host” organism. Such a categorization is not intended to indicate a series of evolutionary advances, nor does it indicate that one group is ecologically more successful than another. What is indicated is the existence of numerous examples of different levels of association between fungi and other organisms–mostly autotrophic plants.
The saprophyte, dependent on the previous presence of an autotroph, and the parasite, dependent on the immediate presence of an autotroph, are no less dependent than is the lichen mycobiont on the phycobiont. Both examples of association can be regarded as representing probable stages in the convergence of autotroph and heterotroph which led to the evolution of symbiotic systems.
From the least specialized saprophyte to the most highly specialized parasite, there runs a common thread of increasing dependence on specific organic substances of autotroph origin. The ultimate in specialization is complete dependence of the parasite on the metabolic products of a single species of autotroph (Quispel, 1951). At this level, a relatively stable state has been reached in which the defense-reaction mechanisms of the host are sufficiently powerful to prevent self-elimination but yet are insufficiently powerful to effect the elimination of the parasite. Physiological symbiosis, in the sense of mutual life support, is only one step further. The circumstance can easily arise, and presumably has arisen, in which the host organism through mutation develops a metabolic deficiency which is made good by the associated organism. The host becomes dependent on the parasite just as the parasite is dependent on the host, and a state of symbiosis ensues.
It is probable that the green lichens evolved by such a process, starting with the casual association of fungal hyphae and free-living green algal cells, and continuing through stages of progressive adaptation of both fungus and alga to the conditions of the new environment thus created. The present-day green lichens represent, so far as we are aware, those stages in symbiosis evolution at which the autotrophic symbiont has become fully dependent, ecologically and perhaps physiologically, on close association with the heterotrophic symbiont.
Although this may have been the general pattern of evolution of the lichen symbiosis, it is clear that not all present-day examples have reached comparable stages in the development of interdependence. The phycobionts of many blue-green lichens, for example, appear to be facultative symbionts. There is no concrete evidence that any such phycobiont cannot lead an independent existence from one generation to the next. It is evident, too, that the secondary phycobionts of cephalodial lichens, such as Stereocaulon, Peltigera (Peltidea), Nephroma, etc., are facultative associates only. Whether these blue-green lichens are to be regarded as intermediate stages in the progression to obligatory association is a matter for conjecture. The only pertinent evidence in favor appears to be the fact that Nostoc, and perhaps other blue-green phycobionts, are nitrogen fixers. So long as this attribute remains inherent to the species, there is only a slender chance of physiological adaptation to exclusive utilization of nitrogenous metabolites of the mycobiont.
The lichen symbiosis is thus a classic example of the progression from facultative to total interdependence of the symbionts. It is certain, however, despite the lack of evidence, that there is a stage of independence prior to the initiation of every individual symbiosis. This may be, and perhaps usually is, brief, but no symbiosis arising from a mycobiont ascospore can be initiated without the previous independent existence of the phycobiont. This is one instance when the lichenologist should not be confounded by the mathematical nicety that one plus one equals one—itself a fair, if not very illuminating, definition of symbiosis.
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Book 1973, The LichensG.D. SCOTT
Chapter
Dermatophytes and Dermatophytoses
1990, Diagnostic Procedure in Veterinary Bacteriology and Mycology (Fifth Edition)G.R. Carter
T. georgii
This geophilic species is most likely a soil saprophyte, but it has been isolated from the opossum (4). It does not attack hair, and the perfect state is Arthroderma ciferri. Its colonies are pale brown with a spotted brown reverse. Microaleuriospores are abundant and variable in size and shape, but no macroaleuriospores are produced.
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Book 1990, Diagnostic Procedure in Veterinary Bacteriology and Mycology (Fifth Edition)G.R. Carter
Chapter
Plant Morphology
2019, Plant Systematics (Third Edition)Michael G. Simpson
PLANT STRUCTURE
PLANT ORGANS
The basic structural components, or organs, of plants are delimited by and strongly correlated with their specific functions. Among the liverworts, hornworts, and mosses (see Chapter 3), these organs are components of the haploid gametophyte. The gametophyte of these taxa contains rhizoids, which are uniseriate, filamentous chains of cells functioning in anchorage and water/mineral absorption. The basic body of the gametophyte can either be a flat mass of cells, termed a thallus (found in some liverworts and all hornworts) or a shoot, consisting of a generally cylindrical stem bearing leaves (found in some liverworts and all mosses; see Chapter 3). It should be noted that the shoot systems of liverworts and mosses are gametophytic tissue.
The major organs of vascular plants are sporophytic roots and shoots. Roots are present in almost all vascular plants and typically function in anchorage and absorption of water and minerals. Roots consist of an apical meristem that gives rise to a protective root cap, a central endodermis-bounded vascular cylinder, absorptive epidermal root hairs, and endogenously developed lateral roots (Figure 9.1).
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Figure 9.1. General plant structure, showing primary root and primary shoot. Note that all parts of both root and shoot are derived from cell divisions of the root or shoot apical meristem.
The sporophytic shoots of vascular plants consist of stem plus leaves (Figure 9.1). Shoots contain an apical meristem of actively dividing cells that, through continued differentiation, result in the elongation of the stem and formation of leaves and buds (see later discussion). The stem is a generally cylindrical organ that bears the photosynthetic leaves. Stems typically function in conduction of water and minerals from the roots and in support and elevation of both leaves and reproductive structures, although some stems are highly modified for other functions (see later discussion). The leaf is that organ of the shoot that is generally dorsiventrally flattened and that usually functions in photosynthesis and transpiration. Leaves are derived from leaf primordia within the shoot apex and are often variously modified. In vascular plants, leaves contain one to many vascular bundles, the veins; in some mosses, the gametophytic leaves may contain a veinlike costa, consisting of specialized (although not truly vascular) conductive tissue. Buds are immature shoot systems, typically located in the axils of leaves. Buds may grow to form lateral vegetative branches or reproductive structures (see later discussion).
Among reproductive plant organs, the sporangium is the basic spore-producing part of all land plants. The sporangium of liverworts, hornworts, and mosses is known as a capsule and typically makes up most of the sporophyte (see Chapter 3). In heterosporous plants, including all of the seed plants, sporangia are of two types: male or microsporangium and female or megasporangium (see Chapter 4). A cone, also called a strobilus, is a modified, determinate, reproductive shoot system of many nonflowering vascular plants, consisting of a stem axis bearing either sporophylls, in “simple” cones, or modified shoot systems, in “compound” cones (see Chapter 5). An ovule is a megasporangium enveloped by one or more protective integuments. A seed is the mature ovule of the seed plants, consisting of an internal embryo surrounded by nutritive tissue (comprising female gametophyte or endosperm) and enveloped by a protective seed coat (Chapter 5). The reproductive organ of angiosperms is the flower, a modified, determinate shoot bearing sporophylls called stamens and carpels, with or without outer modified leaves, the perianth (see Chapter 6). An inflorescence is an aggregate of one or more flowers, the boundaries of which generally occur with the presence of vegetative leaves. A fruit is the mature ovary of flowering plants, consisting of the pericarp (mature ovary wall), seeds, and (if present) accessory parts.
P lant H abit
Plant habit refers to the general form of a plant, encompassing a variety of components such as stem duration and branching pattern, development, or texture. Most plants can be clearly designated as an herb, vine, liana, shrub, or tree (with some subcategories; see later discussion); however, some species are difficult to accommodate into these categories. An herb is a plant in which any aboveground shoots, whether vegetative or reproductive, die back at the end of an annual growth season. Although the aboveground shoots are annual, the herb itself may be annual, biennial, or perennial, the last by means of long-lived underground rootstocks. Such perennial herbs, having a bulb, corm, rhizome, or tuber as the underground stem, are termed geophytes. A vine is a plant with elongate, weak stems, that are generally supported by means of scrambling, twining, tendrils, or roots; vines may be annual or perennial, herbaceous or woody. A liana (also spelled liane) is a vine that is perennial and woody; lianas are major components in the tree canopy layer of some tropical forests. A shrub is a perennial, woody plant with several main stems arising at ground level. A subshrub is a short shrub that is woody only at the base and that seasonally bears new, nonwoody, annual shoots above. Finally, a tree is defined as a generally tall, perennial, woody plant having one main stem (the trunk) arising at ground level. (Some plant ecologists will sometimes distinguish between shrubs and trees based primarily on an arbitrary height.)
P lant H abitat
Plant habitat refers to the general environment where the plant is growing. General habitat terms include whether the plant is terrestrial, growing on land; aquatic, growing in water; or epiphytic, growing on another plant. If aquatic, a plant can be submersed, occurring under water; floating, occurring at the water surface; or emergent, having roots or stems anchored to the substrate under water and aerial shoots growing above water. A rheophyte is a plant found along (often swiftly flowing) streams and river banks. Other aspects of the habitat include the type of substrate that the plant is growing in (e.g., whether on sandy, loam, clay, gravelly, or rocky soil). Saxicolous [epipetric] refers to a plant growing on rocks or boulders, either in or on the surface (lithophyte) or in the cracks (chasmophyte). Other habitat features include slope, aspect, elevation, moisture regime, and surrounding vegetation, community, or ecosystem (See Chapter 17, “Plant Collecting and Documentation.”)
P lant L ife F orm
Plant life form denotes aspects of their structure, life cycle, and physiology. (See Raunkiaer 1934.) Life form types include: chamaephyte, an overwintering perennial with buds at or just below ground level; epiphyte, a plant growing on another plant, e.g., Tillandsia (Bromeliaceae); geophyte, a perennial herb with underground perennating rootstocks such as bulbs, corms, rhizomes; halophyte, a salt-adapted plant, e.g., Salicornia (Chenopodiaceae); mycotroph [myco-heterotroph], a usually achlorophyllous plant having an intimate contact with a fungus, from which it receives most of its nutrition; phreatophyte, a plant with a long taproot, in contact with ground water (e.g., mesquite, Prosopis); succulent, a plant with fleshy stems (stem succulents, e.g., cacti) or leaves (leaf succulents, e.g., members of Aizoaceae or Crassulaceae); therophyte, an annual plant; and xerophyte, a plant adapted to live in a dry, generally hot environment;
Plant life form types denoting nutritional physiology include: saprophyte, a heterotrophic plant living off dead organic matter; mycotroph [myco-heterotroph], a usually achlorophyllous plant having an intimate contact with a fungus, from which it receives most of its nutrition; and parasite, a plant feeding on another plant, dependent on it for all or part of its nutrition. A parasite may be further classified into a holoparasite, one that lacks photosynthesis/chloroplasts and must attach to a host to survive and reproduce (e.g., Cuscuta, Orobanche, Balanophora), or a hemiparasite, a parasitic plant that is photosynthetic during at least part of its life cycle. A hemiparasite may be either an obligate hemiparasite, requiring a host to survive and reproduce (e.g., Phoradendron, mistletoe, Viscaceae) or a facultative hemiparasite, not requiring a host connection to survive and reproduce (e.g., Krameria, Krameriaceae, or Pedicularis, lousewort, a photosynthetic member of the Orobanchaceae).
ROOTS
Roots are plant organs that function in anchorage and in absorption of water and minerals. Roots are found in all of the vascular land plants except for the psilotophytes — Psilotum and relatives (As discussed earlier, nonvascular land plants have rhizoids that assume a similar function.)
Roots, like shoots, develop by the formation of new cells within the actively growing apical meristem of the root tip. The apical meristem is covered on the outside by a rootcap, functioning both to protect the root apical meristem and to provide lubrication as the root grows into the soil. The epidermal cells away from the root tip develop hairlike extensions called root hairs; these function in greatly increasing the surface area available for water and mineral absorption. Roots of many (if not most) species of plants have an interesting symbiotic interaction with a species of fungus, known as mycorrhizae. Although the exact function of mycorrhizae is often unclear, in some species at least the fungus host aids the plant both in increasing overall surface area for absorption and in increasing the efficiency of mineral uptake, particularly phosphorus. Roots have a central vascular cylinder of conductive cells, xylem and phloem. This vascular cylinder is surrounded by a special cylinder of cells known as the endodermis. Lateral roots develop by cell divisions within the pericycle, a cylindrical layer of parenchyma cells located just inside the endodermis itself. (See Chapter 10 for more details of root anatomy.)
The first root to develop in a vascular plant is the radicle of the embryo. If the radicle continues to develop after embryo growth, it is known as the primary root. Additional roots may arise from internal tissue of either another root, the stem/shoot (often near buds), or (rarely) a leaf. Roots that arise from other roots are called lateral roots. Roots that arise from a nonroot organ (stem or leaf) are adventitious roots.
R oot T ypes (Figure 9.2)
Various modifications of roots have evolved. If the primary root becomes dominant, it is called a taproot, and the plant is described as having a taproot system. If the primary root soon withers and subsequent roots are adventitious, the plant has a fibrous root system. Several plant species, particularly those that are biennials, have storage roots in which the taproot has become greatly thickened, accumulating reservoirs of high-energy storage compounds (usually starch). Many plants that are epiphytic (grow on another plant), particularly tropical members of the monocot families Araceae and Orchidaceae, have aerial roots. These are adventitious roots that generally do not enter the soil and may absorb water and minerals from the air or from runoff from plants. Many plant species with bulbs or corms have contractile roots, roots that actually contract vertically, functioning to pull the rootstock further into the soil. Parasitic plants have specialized roots called haustoria that penetrate the tissues of a host plant. Some adventitious roots called prop roots grow from the base of the stem and function to further support the plant. Some plant species that grow in swamps or marshes have pneumatophores, roots that grow upwardly from soil to air that function to obtain additional oxygen. Buttress roots are enlarged, horizontally spreading and often vertically thickened roots at the base of trees that aid in mechanical support; they are found in certain tropical or marsh/swamp tree species.
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Figure 9.2. Root types. A. Tap root. B. Fibrous root system. C. Prop roots. D. Haustorial roots. E. Storage roots; Raphanus sativus, radish. F. Buttress roots; Ficus rubiginosa, fig. G,H. Pneumatophores; Avicennia germinans, black mangrove.
STEMS AND SHOOTS
Stems function both as supportive organs (supporting and usually elevating leaves and reproductive organs) and as conductive organs (conducting both water/minerals and sugars through the vascular tissue between leaves, roots, and reproductive organs). Structurally, stems can be distinguished from roots based on several anatomical features (see Chapter 10). As mentioned earlier, a shoot is a stem plus its associated leaves. Sporophytic shoots that are branched and bear leaves are an apomorphy for all extant vascular plants; the leafy shootlike structures of mosses and some liverworts are gametophytic and not directly homologous with shoots of vascular plants.
The first shoot of a seed plant develops from the epicotyl of the embryo (see Seeds). The epicotyl elongates after embryo growth into an axis (the stem) that bears leaves from its tip, which contains the actively dividing cells of the shoot apical meristem. Further cell divisions and growth results in the formation of a mass of tissue that develops into the immature leaf, called a leaf primordium (Figure 9.1). The point of attachment of a leaf to a stem is called the node. The region between two adjacent nodes is the internode (Figure 9.1). A bit later in development, the tissue at the upper (adaxial) junction of leaf and stem (called the axil) begins to divide and differentiate into a bud primordium. As the shoot matures, the leaves fully differentiate into an amazing variety of forms. The bud primordium matures into a bud, defined as an immature shoot system, often surrounded by protective scale leaves (see Buds). Buds have an architecture identical to the original shoot. They may develop into a lateral branch or may terminate by developing into a flower or inflorescence. Vascular strands run between stem and leaf, providing a vascular connection, composed of xylem and phloem, for water, mineral, and sugar transport. The vascular strands of leaves are termed veins.
The mostly parenchymatous tissue external to the vascular (conductive) tissue of a stem is termed the cortex. The pith is the central, mostly parenchymatous tissue, internal to the stem vasculature (e.g., in siphonosteles and eusteles). In monocots, in which there are numerous, “scattered” vascular bundles (an atactostele), the intervening parenchymatous tissue is termed ground meristem (see Chapter 10).
The stems of some vascular plants, notably the conifers and nonmonocot flowering plants, contain wood, which technically is secondary xylem tissue, derived from a vascular cambium (see Chapter 10). In these woody plants bark refers to all the tissues external to the vascular cambium, consisting of secondary phloem (inner bark), leftover cortex, and derivatives of the cork cambium (the last comprising the outer bark, or periderm; see Chapter 10).
S tem/S hoot T ypes (Figure 9.3)
Various modifications of stems and shoots have evolved, many representing specific adaptations. For example, perennial and some biennial herbs have underground stems, which are generally known as rootstocks. Rootstocks function as storage and protective organs, remaining alive underground during harsh conditions of cold or drought. When environmental conditions improve, rootstocks serve as the site of new shoot growth, sending out new adventitious roots and new aerial shoots from the apical meristem or from previously dormant buds. Different types of rootstocks have evolved in various taxonomic groups. These include the following:
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Figure 9.3. Stem types. (l.s. = longitudinal section)
1.
Bulb, in which the shoot consists of a small amount of vertical stem tissue (bearing roots below) and a massive quantity of thick, fleshy storage leaves (e.g., Allium spp., onions)
2.
Corm, in which the shoot consists mostly of generally globose stem tissue surrounded by scanty, scale-like leaves (e.g., some Iris spp., irises)
3.
Caudex, in which the rootstock consists of a relatively undifferentiated but vertically oriented stem
4.
Rhizome, in which the stem is horizontal and underground, typically with short internodes (compare stolon, below) and bearing scale-like leaves (e.g., Zingiber officinale, ginger)
5.
Tuber, which consists of a thick, underground storage stem, usually not upright, typically bearing outer buds and lacking surrounding storage leaves or protective scales (e.g., Solanum tuberosum, potato).
Rootstocks may function as reproductive structures in vegetative (clonal) propagation, either by splitting apart into separate plants or by forming proliferative structures that subsequently separate (and may even be dispersed by animals). For example, buds in the axils of the leaves of bulbs can develop into proliferative bulbels (e.g., garlic); some taxa (e.g., certain onions) can even form tiny, propagative bulbs within the aerial shoots or inflorescence of the plant, these termed bulbils. Cormose plants can, similarly from axillary buds, form proliferative corms, termed cormels. Tuberous plants typically form numerous tubers at the tips of elongate stems; these tubers can become easily separated, growing into an individual plant. Tubers can even form on aerial shoots (e.g., Dioscorea, true yams), ultimately falling off and growing into a new individual. Rhizomes frequently become highly branched; when older parts die or become broken, the separated rhizomes function as separate individuals.
A stolon or runner is a stem with long internodes that runs on or just below the surface of the ground, typically terminating in a new plantlet, as in Fragaria (strawberry). Because stolons can be underground, they are sometimes termed root-stocks and resemble narrow, elongate rhizomes. Stolons function specifically as vegetative propagative structures, however, as the terminal plantlet often becomes separated from the parent plant.
Many modified types of stems that are aerial (above-ground) also have specific functions. For example, a cladode (also termed a cladophyll or phylloclade) is a flattened, photosynthetic stem that may resemble and function as a leaf, found, e.g., in prickly-pear cacti, Asparagus, and Ruscus. Cladodes take over the primary photosynthetic function of leaves and may function to reduce water loss.
Some aerial stems may function for storage of food reserves or water. So-called succulent stems (the plants often referred to as “stem succulents”) contain a high percentage of parenchyma tissue that may store great quantities of water, allowing the plant to survive subsequent drought periods. The cacti of the New World and the stem succulent euphorbs of South Africa are classic examples of plants with succulent stems. Some of these, most notably the barrel cacti and the large columnar cacti such as saguaro’s or cardon’s, have fluted trunks that can expand rapidly following a rain, enabling the plant to store more water. Other aerial, storage stems include:
1.
A caudiciform stem, which is a low, swollen, perennial storage stem (at or above-ground level), from which arise annual or nonpersistent photosynthetic shoots (e.g., Calibanus, some Dioscorea spp.)
2.
A pachycaul, which is a woody, trunklike stem that is swollen basally, the swollen region functioning in storage (e.g., bottle trees, Brachychiton spp., and the boojum tree, Fouquieria columnaris).
Some stems or shoot types function as protective devices by deterring an herbivore from taking a bite of the plant. A thorn is a sharp-pointed stem or shoot. (A thorn is not to be confused with a spine, which is a sharp-pointed leaf or leaf part, or a prickle, which is a sharp-pointed epidermal structure found anywhere on the plant; see later discussion.) A very specialized type of shoot is the areole, a modified, reduced, nonelongating shoot apical meristem bearing leaf spines. Areoles are characteristic of the cactus family, Cactaceae.
Some stems are specialized for reproduction. For example, a scape is a “naked” (lacking vegetative leaves) peduncle (inflorescence axis), generally arising from a basal rosette of vegetative leaves and functioning to elevate flowers well above the ground. A culm refers to the flowering and fruiting stem(s) of grasses and sedges. A tiller is the general term for a proliferative grass shoot, typically growing in masses from axillary buds at the base of the stem.
Stems may have multiple or varied functions. A lignotuber or burl is largely a protective and regenerative stem following fires. Lignotubers or burls are typically swollen, woody stems, at or slightly below ground level, from which arise persistent, woody, aerial branches (e.g., some Manzanita spp.). A pseudobulb is a short, erect, aerial storage or propagative stem of certain epiphytic orchids. A short shoot or fascicle (also called a spur shoot or dwarf shoot) is a modified shoot with very short internodes from which flowers or leaves are borne. Short shoots enable the production of leaves or reproductive organs relatively quickly, with minimal stem (branch) tissue being formed. Short shoots may be found on so-called drought deciduous plants, which produce a quick flush of leaves from short shoots following a rain. Short shoots arise from the buds of more typical shoots (branches) with longer internodes, the latter termed, in contrast, long shoots. Finally, a tendril is a long, slender, coiling branch, adapted for climbing. Tendrils are typically found on weak -stemmed vines and function in support. (Note that most tendrils are leaves or leaf parts; see Leaf Structural Type.)
S tem H abit (Figure 9.4)
Stem habit is a character describing the relative position of the stem or shoot, but may also be based on stem structure, growth, and orientation. Stem habit features, like stem types, represent adaptations that enhance survival and reproduction. For example, a plant with an above-ground stem is caulescent; one that lacks an above-ground stem, other than the inflorescence axis, is termed acaulescent. Acaulescent plants bear major photosynthetic leaves only at ground level, often in a basal rosette, with the only shoot becoming aerial, being an inflorescence that eventually dies off. Acaulescent plants are often biennial herbs, in which a storage root develops in the first year and flowering (bolting) occurs in the second, or perennial herbs, in which the persistent stem remains underground and protected during extreme environmental conditions. Plants with caulescent stem habits include shrubs, trees, and herbs with aerial vegetative shoots and leaves. Some corresponding stem habit terms are arborescent, treelike in appearance and size; frutescent, having the habit of a shrub, with numerous, woody, aerial trunks; and suffrutescent, being basally woody and herbaceous apically, the habit of a subshrub. Vines are also types of caulescent plants. The stem habit of vines can be either clambering (also called scandent), sprawling across objects without specialized climbing structures, or climbing, growing upward by means of tendrils, petioles, or adventitious roots. Some plants are adapted to lying on the ground, at least in part. These include those that are prostrate (also called procumbent), trailing or lying flat, not rooting at the nodes; repent, creeping or lying flat but rooting at the nodes; or decumbent, being basally prostrate but apically ascending. Finally, some plants have a cespitose stem habit, in which multiple aerial but short-stemmed shoots arise from the base, forming a much-branched cushion. Many grasses are cespitose, these being the so-called “bunch” grasses.
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Figure 9.4. Stem habit.
S tem B ranching P attern (Figure 9.5)
The below- or above-ground stems or shoots of a plant often exhibit characteristic branching patterns. Branching pattern is determined by the relative activity of apical meristems, both the “original” shoot apical meristem derived from the seedling epicotyl and apical meristems subsequently derived from lateral buds. One major feature of branching pattern has to do with the duration of apical meristematic growth of a shoot. If a given shoot has the potential for unlimited growth, such that the apical meristem is continuously active, the growth is termed indeterminate. If instead a shoot terminates growth after a period of time, with either the abortion of the apical meristem or its conversion into a flower, inflorescence or specialized structure (such as a thorn or tendril), the growth is termed determinate. (Note that these same terms are used for inflorescence development; see later discussion.)
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Figure 9.5. Stem branching patterns.
Other terms for branching pattern center on the developmental origin of a given branch or axis. A relatively rare type of branching is dichotomous, in which a single apical meristerm divides equally into branches, e.g., Psilotum. A variant of dichotomous is branching that is pseudomonopodial, in which one branch of the initial dichotomy overtops and becomes dominant over the other. Dichotomous and pseudomonopodial branching are found primarily in lycophytes. In contrast, if a given stem axis is derived from growth of a single apical meristem, the pattern is termed monopodial. The monopodial axis may grow indefinitely and thus be indeterminate. In contrast, if a given axis (which may appear to be a single, continuous structure) is made up of numerous units that are derived from separate apical meristems, the branching pattern is sympodial. These sympodial units arise from lateral buds that are proximal to the apical meristem of the original shoot. Many rhizomes have sympodial growth.
S tem B ranching M odels (Figure 9.6)
Aside from the general stem branching patterns, more specific models of tree branching pattern have been described (Hallé et al. 1978). These are used almost exclusively for trees, but may be used with herbs as well. The models are based first on whether the tree is monoaxial, unbranched with a single (vegetative) apical meristem, or polyaxial, branched with more than one vegetative apical meristem. Additional considerations are whether the shoots are orthotropic, erect and essentially radially symmetric, the branching three-dimensional, or plagiotropic, more or less horizontal with dorsiventral symmetry, the branching two -dimensional and leaves generally in one plane (either distichous or secund). Plagiotropy may occur in two general ways. Plagiotropy by apposition is that in which extension growth of the branch is taken over by an axillary meristem, but with the original branch terminal meristem continuing growth, usually as a short shoot. Plagiotropy by substitution is that in which the original branch terminal meristem aborts or converts into a terminal inflorescence or flower, extension growth of the branch being taken over by an axillary meristem. In addition, the timing of development of a shoot can be important in plant growth. Syllepsis (or sylleptic growth) is growth of an axillary bud into a shoot without a period of rest. Prolepsis (or proleptic growth) is growth of an axillary bud into a shoot only after a period of rest.
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Figure 9.6. Stem branching pattern models, after Hallé et al. (1978). See text for explanation.
Other, related terms have to do with flowering. An indeterminate shoot that bears lateral flowers but that continues vegetative growth is termed pleonanthic. A plant with a determinate shoot that completely transforms into a flower or inflorescence is called hapaxanthic. If the entire plant flowers and fruits only once, and then dies, it is termed monocarpic; the plant itself can be an annual or perennial, but the term is usually used only for perennials, given that all annuals are monocarpic.
The following are tree growth models (after Hallé et al., 1978, illustrated in Figure 9.6), each of which is named after a botanist who contributed to our knowledge of that model. Only a very few examples of the models are listed, ones that might be familiar to the reader; see Hallé et al. (1978) for elaboration of the models and considerably more examples. An important concept, however, is that of reiteration, the growth of shoots not conforming to the parameters of the model, e.g., due to environmental stress, such as mechanical or animal damage, obscuring its normal expression. Although these models are rather specialized, they are useful in the study of tree architecture (and a challenging and intriguing exercise for the student to decipher). A given model may represent the end product of evolutionary adaptations to a given environment or life strategy and their elucidation may have taxonomic, ecological, or biomechanical significance.
Attim’s Model. Polyaxial; with a monopodial trunk with continuous growth, bearing equivalent branches, flowers always lateral. E.g., Avicennia germinans, black mangrove (Acanthaceae), Alnus incana (Betulaceae), Casuarina equisetifolia (Casuarinaceae), Euphorbia spp. (Euphorbiaceae), Eucalyptus spp. (Myrtaceae), Rhizophora mangle, red mangrove (Rhizophoraceae).
Aubréville’s Model. Polyaxial; rhythmically growing and branching, with rhythmic growth, having a monopodial trunk bearing modular whorls or pseudo-whorls of branch tiers, plagiotropic by apposition, all with a similar phyllotaxis, the inflorescences lateral E.g., Terminalia spp. (Combretaceae), Manilkara zapota (Sapotaceae).
Chamberlain’s Model. Polyaxial; having regular, sympodial branching, the modules usually orthotropic, each hypaxanthic by producing a terminal flower or inflorescence, linear growth continued by distal, lateral meristems E.g., many cycads (such as male Cycas spp.), Jatropha spp. (Euphorbiaceae), Dieffenbachia spp., Philodendron selloum (Araceae).
Champagnat’s Model. Polyaxial; with successive modular orthotropic axes with spiral leaf arrangement, each curving and becoming pendulous by its own weight, a new modular unit arising from the upper part of the curved axis E.g., Sambucus spp. (Adoxaceae), Crescentia cujete, calabash tree (Bignoniaceae), Caesalpinia pulcherrima (Fabaceae), Lagerstroemia indica, crepe myrtle (Lythraceae), Bougainvillea spectabilis (Nyctaginaceae).
Corner’s Model. Monoaxial, in which the inflorescences or sporophylls are lateral, the single stem capable of growth after flowering, not monocarpic E.g., many tree ferns, cycads (such as female Cycas spp.), many palms (Arecaceae).
Fagerlind’s Model. Polyaxial; with a monopodial, orthotropic trunk producing tiers of modular branches, each branch sympodial and plagiotropic by apposition, with spiral or decussate leaves. E.g., Hymenosporum flavum (Pittosporaceae); Magnolia grandiflora, flowering magnolia (Magnoliaceae).
Holttum’s Model. Monoaxial, with the terminal meristem developing entirely into an inflorescence, the tree dying after fruit maturation, therefore monocarpic E.g., many Agave spp. (Agavaceae), many palms (Arecaceae).
Koriba’s Model. Polyaxial; having orthotropic modules, each of which is sympodial and aborts or produces a terminal inflorescence, the modules initially equivalent, but later one becoming dominant and erect as a trunk, the others developing into branches E.g., Ochrosia spp. (Apocynaceae), Catalpa spp. (Bignoniaceae), Phytolacca dioica (Phytolaccaceae).
Leeuwenberg’s Model. Polyaxial; having equivalent, orthotropic modules, each of which is sympodial and produces a terminal inflorescence, with two or more new modules arising below it. E.g., tree Aloë spp. (Asphodelaceae), Dracaena draco (Ruscaceae), Nerium oleander, Pachypodium spp. (Apocynaceae), Pandanus spp. (Pandanaceae).
Mangenot’s Model. Polyaxial; with axes composed of modular units from a single apical meristem composed of an orthotropic proximal part (leaves often spiral), abruptly recurving into a plagiotropic distal part (the leaves often distichous), a new module orthotropically arising from the bend of the recurved section E.g., Vaccinium corymbosum, blueberry (Ericaceae), Strychnos sp., strychnine (Loganiaceae), Eugenia sp. (Myrtaceae).
Massart’s Model. Polyaxial; rhythmically growing and branching, with an orthotropic, monopodial trunk having rhythmic growth, producing regular tiers of lateral branches that are plagiotropic by leaf arrangement or symmetry, never by apposition, reproductive structure position variable E.g., Agathis spp., Araucaria spp. (Araucariaceae), Sequoia sempervirens. redwood (Cupressaceae), Diospyros spp. (Ebenaceae), Myristica fragrans (Myristicaceae), Abies spp. (Pinaceae).
Nozeran’s Model. Polyaxial; rhythmically growing and branching, with an orthotropic, sympodial trunk, each sympodial unit bearing a distal tier of monopodial or sympodial plagiotropic branches, the leaf arrangement of trunk and branches different E.g., Theobroma cacao, chocolate (Malvaceae).
Petit’s Model. Polyaxial; continuously growing and branching, with a monopodial, orthotropic trunk producing tiers of modular branches, each branch sympodial and plagiotropic by substitution (thus branch modules hapaxanthic), with spiral or decussate leaves E.g., Gossypium spp., cotton (Malvaceae); Morinda citrifolia (Rubiaceae).
Prévost’s Model. Polyaxial; rhythmically growing and branching, having two types of orthotropic modules forming trunk and branches from inception, the branch modules plagiotropic by apposition, arising sylleptically from subapical region of the trunk module, successive trunk modules proleptic and arising well below the tier of branch modules E.g., Cordia spp. (Cordiaceae), Euphorbia pulcherrima, crucifixion thorn (Euphorbiaceae).
Rauh’s Model. Polyaxial; with a monopodial trunk, rhythmically producing tiers of branches, each branch identical to trunk; flowers always lateral E.g., Ilex spp. (Aquifoliaceae), Araucaria spp. (Araucariaceae); Kalanchoe beharensis (Crassulaceae), Euphorbia spp. (Euphorbiaceae), Quercus spp. (Fagaceae), Cecropia spp./Ficus spp. (Moraceae), Fraxinus spp. (Oleaceae), most Pinus spp. (Pinaceae); Acer spp. (Sapindaceae).
Roux’s Model. Polyaxial; continuously growing and branching, with a monopodial, orthotropic trunk having spiral leaf arrangement, bearing lateral branches that are plagiotropic, but never by apposition, usually with distichous leaf arrangement, the branches inserted continuously on the trunk, reproductive structure position variable, but usually lateral on branches. E.g., Polyalthia spp. (Annonaceae), Dipterocarpus spp. (Dipterocarpaceae), Bertholletia excelsa, Brazil nut (Lecythidaceae), Durio zibethinus, durio (Malvaceae), Coffea arabica (Rubiaceae).
Scarrone’s Model. Polyaxial; rhythmically growing and branching, with a monopodial trunk bearing tiers of branches, each branch sympodial by terminal flowering, but becoming orthotropic E.g., Anacardium occidentale, cashew, Mangifera indica, mango (Anacardiaceae), Jacaranda mimosifolia (Bignoniaceae), Echium spp. (Boraginaceae), Aeonium spp. (Crassulaceae), Arbutus unedo (Ericaceae), Aesculus spp., horse-chestnut (Sapindaceae).
Schoute’s Model. Polyaxial; growth from meristems that produce orthotropic or plagiotropic trunks forking at regular, distinct intervals by equal dichotomy, otherwise with no lateral branches; inflorescences lateral. E.g., Hyphaene thebaica, Nypa fruticans (Arecaceae); Flagellaria indica (Flagellariaceae).
Stone’s Model. Polyaxial; continuously growing and branching, with a orthotropic trunk that may flower terminally, bearing orthotropic branches, with additional branching occurring sympodially below terminal inflorescences E.g., Mikania cordata (Asteraceae), Pandanus spp. (Pandanceae).
Tomlinson’s Model. Polyaxial; vegetative axes all equivalent, orthotropic, with equivalent orthotopic modules developing from basal nodes in subsequent axes. E.g., many monocots (Bromeliaceae, Cyperaceae, Poaceae, Zingiberales), Euphorbia spp. (Euphorbiaceae), Kalanchoë spp. (Crassulaceae).
Troll’s Model. Polyaxial; with all axes plagiotropic, the modules superposed upon one another, the proximal part of each module becoming erect in development, the distal part a branch. E.g., Psidium guineense (Myrtaceae), Erythroxylum coca (Erythroxylaceae), Albizzia julibrissin, Bauhinia spp. (Fabaceae), Fagus grandifolia, beech (Fagaceae), Averrhoa carambola, star-apple (Oxalidaceae).
T wigs, T runks, and B uds (Figure 9.7)
Twigs are the woody, recent-growth branches of trees or shrubs. Buds are immature shoot systems that develop from meristematic regions. In deciduous woody plants the leaves fall off at the end of the growing season and the outermost leaves of the buds may develop into protective bracts (modified leaves) known as bud scales. The bud of a twig that contains the original apical meristem of the shoot (which by later growth may result in further extension of the shoot) is called the terminal or apical bud. Buds formed in the axils of leaves are called axillary [axial] or lateral buds.
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Figure 9.7. Twigs parts and bud types (l.s. = longitudinal section).
A given bud may be vegetative, if it develops into a vegetative shoot bearing leaves; floral or inflorescence, if it develops into a flower or inflorescence; or mixed, if it develops into both flower(s) and leaves. In some species more than one axillary bud forms per node. Two or more axillary buds that are oriented sideways are called collateral buds; two or more axillary buds oriented vertically are called superposed buds. If the original terminal apical meristem of a shoot aborts (e.g., by ceasing growth or maturing into a flower), then an axillary bud near the shoot apex may continue extension growth; because this axillary bud assumes the function of a terminal bud, it is called a pseudoterminal bud.
Several scars may be identified on a woody, deciduous twig. These include the leaf scar, leaf vascular bundle scars, stipule scars (if present), and bud scale scars. Bud scale scars represent the point of attachment of the bud scales of the original terminal bud after resumption of growth during the new season. Thus, bud scale scars represent the point where the branch ceased elongation the previous growing season; the region between adjacent bud scale scars represents a single year’s growth in temperate climates, but could be shorter or longer in tropical climates.
Bark technically comprises all the tissue outside the vascular cambium of a plant with true wood (see Chapter 10). The outer bark, or periderm, are the tissues derived from the cork cambium itself. Morphologically, bark may refer to the outermost protective tissues of the stems or roots of a plant with some sort of secondary growth, whether derived from a true cork cambium or not. Bark types are often good identifying characteristics of plant taxa, particularly of deciduous trees during the time that the leaves have fallen. Various bark types include:
1.
Exfoliating, a bark that cracks or splits into large sheets
2.
Fissured, a bark split or cracked into vertical or horizontal grooves
3.
Plated, a bark split or cracked, with flat plates between the fissures
4.
Shreddy, bark coarsely fibrous
5.
Smooth, a non-fibrous bark without fissures, fibers, plates, or exfoliating sheets.
LEAVES
Leaves are the primary photosynthetic organs of plants, functioning also as the main site of transpiration. Leaves are derived from leaf primordia of the shoot apex and are, at least early in development, generally “dorsiventrally” flattened (i.e., with “dorsal” and “ventral” sides; see Position). A leaf can be gametophytic, in the leafy liverworts and mosses, or sporophytic, in the vascular plants. As mentioned earlier, sporophytic leaves characteristically are associated with buds, immature shoot systems, typically located in the axils of leaves. Buds may grow to form lateral vegetative branches or reproductive structures (see later discussion).
L eaf P arts (Figures 9.8, 9.9, 9.10)
The expanded, flat portion of the leaf, which contains the bulk of the chloroplasts, is termed the blade or lamina. Many leaves also have a proximal stalk, the petiole or (e.g., in ferns) the stipe. A leaf or leaf part (typically at the base) that partially or fully clasps the stem above the node is a leaf sheath, such as in the Poaceae (grasses) and many Apiaceae. A pseudopetiole is a petiole-like structure arising between a leaf sheath and blade, found in several monocots, such as bananas and bamboos. As mentioned earlier, leaves contain one to many vascular bundles, the veins (also sometimes called nerves); similar specialized (although not truly vascular) conductive tissue is present in mosses.
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Figure 9.8. Leaf structural types.
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Figure 9.9. Leaf structural types. A,B. Phyllode, Acacia longifolia. A. Mature. B. Young, with vestigial, caducous rachillae, representative of ancestral condition. C. Tendril, Lathyrus vestitus. D–F. Spines. D. Stipular spines, Euphorbia sp. E. Petiolar spines, Fouquieria splendens. Note mature leaf (above), dehiscence of blade and upper tissue of petiole, leaving petiolar spine (below). F. Leaf spines, cactus areole G–K. Leaf modifications of carnivorous plants. G,H. Pitcher (pitfall) leaves. G.Nepenthes sp. H.Sarracenia purpurea. I. Tentacular leaves of Drosera capensis. Note glandular trichomes (arrow). J. Snap-trap leaf, Dionaea muscipula. K. Showy flower bracts, Bougainvillea. L. Epicalyx, Lavatera bicolor. M. Bud scale, Liquidambar styraciflua. N. Unifacial leaf, Juncus phaeocephalus.
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Figure 9.10. Leaf types/parts.
Many leaves have stipules, a pair of leaflike appendages, which may be modified as spines or glands, at either side of the base of a leaf. If stipules are present, the leaves are stipulate; if absent, they are exstipulate. A specialized, scarious, sheathlike structure arising above the node in some members of the family Polygonaceae, interpreted as modified stipules, is termed an ocrea (see Polygonaceae treatment in Chapter 8). Stipels are paired leaflike structures, which may also be modified as spines or glands, at either side of the base of the leaflet of a compound leaf, as in some Fabaceae. If stipels are present, the leaves are stipellate; if absent, they are exstipellate. Stipules and stipels may, in some cases, function to protect the young, developing leaf primordia. They often are small and fall off (are “caducous”) soon after leaf maturation. In some taxa, stipules or stipels may be highly modified into spines or glands. Extreme examples are some African acacias, in which the swollen stipular spines function as a home for protective populations of ants. In the Rubiaceae the inner surface of the connate stipules (from opposite leaves) bear colleters, structures that secrete mucilage (aiding to protect young, developing shoots).
Some leaves are compound (as discussed later), i.e., divided into discrete components called leaflets. The stalk of a leaflet is termed the petiolule. Some other specialized leaf parts, restricted to certain taxa, are:
1.
Hastula, an appendage or projection at the junction of petiole and blade, as in some palms
2.
Ligule, an outgrowth or projection from the inner, top of the sheath, at its junction with the blade, as in the Poaceae
3.
Pulvinus, the swollen base of a petiole or petiolule, as in some Fabaceae.
L eaf B ehavior
The pulvinus may, in some taxa, e.g., some Fabaceae (legumes), function in thigmonasty [seismonasty] which is movement (closing) of the leaflets of a compound leaf as a response to touch, vibration, or heat (e.g., as in Mimosa pudica, a sensitive plant); a similar physiological response due to darkness (in photoperiodism) is termed nyctinasty. These physiological responses may protect the leaf from mechanical damage or help to inhibit water loss.
L eaf S tructural T ype (Figures 9.8, 9.9)
Leaf structural type (in contrast to “leaf type,” discussed later) deals with specialized modifications of leaves. One basic leaf structural type in vascular plants is whether the leaves are lycophyllous or euphyllous. Lycophylls are small, simple leaves with intercalary growth and a single, central vein that joins to the stem without a leaf gap (below). Lycophylls are found only in lycophytes and are similar to the type of leaf found in the earliest ancestors of vascular plants. Euphylls are larger, simple or compound leaves with marginal or apical growth, a leaf gap (region of parenchymatous tissue above the junction of the leaf and stem vasculature), and generally multiple veins. Euphylls are found in ferns (in the broad sense), gymnosperms, and angiosperms (see Chapter 4).
A leaf that is modified in shape and usually smaller than the major photosynthetic leaves is called a bract. In angio-sperms bracts are typically associated with flowers (flower bracts) or the axes of inflorescences (inflorescence bracts). A bractlet or bracteole (also called a prophyll or prophyllum) is a smaller or secondary bract often borne on the side of a pedicel in flowering plants. The term bract is also used for the largely nonphotosynthetic leaves that subtend the ovuliferous scales in conifer cones or that subtend the fascicles or short shoots of members of the pine family (Pinaceae). The term scale is used for a small, non-green leaf, either of a bud (bud scales), functioning to protect the delicate apical meristem and leaf primordia, or of an underground rootstock, e.g., along the internodes of a rhizome. Squamate means having or producing scales. Scales can also refer to the reduced bracts of sedge spikelets (Cyperaceae). The term cataphyll can be used to denote a scale-like, often non-green, protective leaf (e.g., in cycads or palms) or can refer to a rudimentary scale leaf found in usually hypogeous (crypto-cotylar) seedlings.
Some bractlike leaves are found in specific taxonomic groups and are given specialized names. A group of bracts resembling sepals immediately below the true calyx is termed an epicalyx (calyculus) found, e.g., in many members of the Malvaceae. Bracts subtending individual flowers of composites (Asteraceae) are collectively termed chaff or paleae (singular, palea), e.g., as found in the tribe Heliantheae of that family. The specialized bracts of the grass (Poaceae) spikelet are given different terms: glumes, the two bracts occurring at the base of a grass spikelet; lemma, the outer and lower bract at the base of the grass floret; and palea, the inner and upper bract at the base of the grass floret (See Inflorescence Type, later, and treatment of Poaceae in Chapter 7.)
A phyllary is one of the involucral bracts subtending a head (see later discussion), as in the Asteraceae. A spathe is an enlarged, sometimes colored bract subtending and usually enclosing an inflorescence, e.g., that subtending the spadix of the Araceae.
Phyllodes are leaves that consist of a flattened, bladelike petiole. Phyllodes are found in a group of mostly Australian Acacia species (the phyllodinous Acacias) and are derived from ancestrally compound leaves by loss of the rachis and leaflets. A tendril is a coiled and twining leaf or leaf part, usually a modified rachis or leaflet. (Tendril can also refer to a modified, coiling stem; see Stem Type).
A spine is a sharp-pointed leaf or leaf part. The typical spines of cacti (Cactaceae) are leaf spines, as they develop from the entire leaf primordia. A very small, deciduous leaf spine with numerous, retrorse barbs along its length is a glochidium (plural, glochidia or glochids), as found in the areoles of opuntioid cacti. Some taxa have spines that develop from a petiole, midrib, or secondary vein of a leaf, e.g., the petiolar spines of Fouquieria spp. In some palms, e.g., Phoenix, the leaflets may be modified into sharp-pointed leaflet spines. Many plants, such as the stem-succulent Euphorbias, have stipular spines; these are typically paired, at the base of a leaf.
A unifacial leaf is isobilateral, i.e., flattened side-to-side and having a left and right side, except at the base, where they are often sheathing. Some monocots belonging to several different families have unifacial leaves, notably members of the Iridaceae, the Iris family. A centric leaf is one that is cylindrical in shape, e.g., Fenestraria of the Aizoaceae. Centric leaves are sometimes a subcategory of unifacial leaves.
Some leaves are very specialized adaptations of carnivorous plants. Pitcher (or “pitfall”) leaves are those that are shaped like a container, which bears an internal fluid and functions in the capture and digestion of small animals. Several taxa have pitcher leaves, including among others Nepenthes (Nepenthaceae) and Sarracenia (Sarraceniaceae), the pitcher plants. Tentacular leaves (an “adhesive” or flypaper” type) are those bearing numerous, sticky, glandular hairs or bristles that function in capturing and digesting small animals; these are characteristics of Drosera spp. (Droseraceae), the sundews. Snap-trap leaves are those that mechanically move after being triggered, in the process capturing and digesting small animals, found in Dionaea muscipula, Venus fly trap, and Aldrovandra vesiculosa (both Droseraceae) Other specialized leaves include the “suction trap” leaves of Utricularia and the “eel trap” leaves of Genlisia (both Lentibulariaceae); see Ellison and Adamec (2018) for more details about carnivorous plant adaptations.
L eaf T ype (Figure 9.10)
The pattern of division of a leaf into discrete components or segments is termed leaf type. A simple leaf is one bearing a single, continuous blade. A compound leaf is one divided into two or more, discrete leaflets. Leaf type should not be confused with leaf division; a simple leaf may be highly divided, but as long as the divisions are not discrete leaflets, it is still technically a simple leaf; see General Terminology. For either compound or divided leaves of ferns, the first (largest) division of a leaf is termed a pinna; the ultimate divisions are termed pinnules. If the leaves are compound or divided into more than two orders, the terms “primary pinna,” “secondary pinna,” etc. can be used, with the ultimate divisions or leaflets always being pinnules.
Simple leaves were the ancestral condition in the vascular plants, as in the lycophylls of the lycopods. Simple leaves are also the norm among the psilotophytes, equisetophytes, Ginkgo, and conifers (including the Gnetales). Compound leaves are characteristic of many “ferns,” and all of the cycads. Angiosperms have the greatest diversity of leaves, ranging from simple to highly compound.
Various types of compound leaves have evolved, perhaps as a means of increasing total blade area without sacrificing structural integrity. For example, the blade tissue of a compound leaf generally may have better structural support (e.g., under windy conditions) than that of a comparably sized simple leaf. Compound leaves tend to be more common in mesic to wet environments and simple leaves in dry environments, but there are many exceptions to this and no clear trends.
Compound leaves are defined based on the number and arrangement of leaflets. A pinnately compound or pinnate leaf is one with leaflets arranged (either oppositely or alternately) along a central axis, the rachis. If a pinnate leaf has a terminal leaflet (and typically an odd number of leaflets), it is imparipinnate or odd-pinnate; if it lacks a terminal leaflet (and has an even number of leaflets), it is paripinnate or even-pinnate. A bipinnately compound or bipinnate leaf is with two orders of axes, each of which is pinnate (equivalent to a compound leaf of compound leaves). The central axis of a bipinnate leaf is still termed the rachis; the lateral axes that bear leaflets are termed rachillae (singular rachilla). Similarly, a compound leaf with three orders of axes, each pinnate, is termed tripinnately compound or tri-pinnate, etc.
A compound leaf in which four or more leaflets arise from a common point, typically at the end of the petiole, is termed palmately compound or palmate. A costapalmate leaf type is one that is essentially palmately compound to divided, but has an elongate, rachislike extension of the petiole (termed the costa), as occurs in some palms.
A compound leaf with only three leaflets is termed trifoliolate or ternately compound. (A leaf with two orders of axes, each ternately compound, is termed biternately compound. Further orders, e.g., triternately compound, can also occur). Most ternately compound leaves are palmate-ternate, in which the three leaflets join at a common point (whether petiolulate or sessile). Rarely, ternately compound leaves can be pinnate-ternate, in which the terminal leaflet arises from the tip of a rachis. Pinnate-ternate leaves are actually derived (by reduction) from an ancestral pinnately compound leaf; they are found, e.g., in some members of the Fabaceae.
Decompound is a general term for a leaf that is more than once compound, i.e., with two or more orders, being bi-, tri-, etc. pinnately, palmately, or ternately, compound. However, decompound is also used for a highly divided leaf; see Division).
A compound leaf consisting of only two leaflets is termed geminate (after Gemini, the twins, in Greek mythology). A compound leaf with two rachillae, each bearing two leaflets, is termed bigeminate. A compound leaf with two rachillae, each of these bearing a pinnate arrangement of leaflets, is termed geminate-pinnate. Finally, a very specialized type of leaf is one that appears superficially to be simple, but actually consists of a single leaflet attached to the apex of a petiole, the junction between them clearly defined. This leaf type, known as unifoliolate, is interpreted as being derived by reduction of an ancestrally compound leaf.
In some taxa, e.g., many Araceae, the leaves exhibit heteroblasty (adjective, heteroblastic), in which the juvenile leaves are distinctly different in size or shape from the adult leaves (making species identification difficult).
L eaf A ttachment (Figure 9.11)
The nature of the joining of the leaf to the stem is termed leaf attachment (sometimes treated under “Base”; see General Morphology). In general, leaves may be petiolate, with a petiole, or sessile, without a petiole. Leaflets of a compound leaf are, correspondingly, either petiolulate or sessile. (The term subsessile is sometimes used for a leaf/leaflet with a small, rudimentary petiole/petiolule.) Sessile or petiolate leaves can also have a sheathing leaf attachment, in which a flattened leaf base (the sheath) partially or wholly clasps the stem, typical of the Poaceae (grasses) and many Apiaceae. If a leaf appears to extend down the stem from the point of attachment, as if fused to the stem, the leaf attachment is decurrent (e.g., as in many Cupressaceae). A decurrent leaf base is not actually caused by later fusion of the leaf to the stem, but by extension growth of actively dividing cells of the leaf primordium at the leaf–stem junction. Last, specializations of sessile leaves may occur. If a leaf is sessile and clasps the stem most, but not all, of its circumference, the attachment is termed amplexicaul. If the leaf is sessile with the base of the blade completely surrounding the stem, it is termed perfoliolate. A special case of the latter (involving fusion of leaves) is connate-perfoliate, whereby typically two opposite leaves fuse basally such that the blade bases of the fusion product completely surrounds the stem.
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Figure 9.11. Leaf attachment.
L eaf V enation (Figures 9.12, 9.13)
The sporophytic leaves of vascular plants contain vascular bundles, known as veins, which conduct water, minerals, and sugars between the leaf and the stem The leaves of some vascular plants have only a single vein, but in most the veins are branched (termed “ramified” or “anastomosing”), sometimes in a very intricate pattern. Venation refers to this pattern of veins and vein branching. Although venation is usually described for vegetative leaves, it can also be assessed in other leaf homologues, such as bracts, sepals, petals, stamens, or carpels.
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Figure 9.12. Leaf venation, generalized terminology.
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Figure 9.13. Leaf venation, specialized terminology, redrawn from Hickey (1973), by permission.
The major vein (or veins) of a leaf, with respect to size, is termed the primary vein. From the primary vein(s), smaller, lateral veins may “branch off,” these known as secondary veins; from secondary veins, even smaller tertiary veins may arise, and so forth. [The distinctions between these vein classes can be difficult to determine in some taxa.] If a simple leaf has a single, primary vein, that vein is termed the midrib or costa (although costa may also be used for the nonvascularized conductive tissue found in the gametophytic leaves of mosses). The central, primary vein of the leaflet of a compound leaf is termed the midvein.
Venation patterns can be quite complex, and the terminology formidable (see later discussion). Four, very general venation classes are as follows (Figure 9.11):
1.
Uninervous, in which there is a central midrib with no lateral veins, e.g., as in the lycophytes, psilotophytes, and equisetophytes, as well as many conifers
2.
Dichotomous, in which veins successively branch distally into a pair of veins of equal size and orientation, e.g., in Ginkgo biloba, in which there is no actual midrib
3.
Parallel, in which the primary and secondary veins are essentially parallel to one another, the ultimate veinlets being transverse (at right angles), e.g., in most monocots
4.
Netted or reticulate, in which the ultimate veinlets form an interconnecting netlike pattern, e.g., most nonmonocot flowering plants.
Reticulate leaves can be pinnately veined (pinnate-netted), with secondary veins arising along length of a single primary vein (the midrib or, in a compound leaf, midvein); palmately veined (palmate-netted), with four or more primary veins arising from a common basal point; or ternately veined (ternate-netted), with three primary veins arising from a common basal point.
Similar to parallel venation in having transverse ultimate veinlets are penni-parallel (also called pinnate-parallel), with secondary veins arising from a single primary vein region, the former essentially parallel to one another (e.g., the Zingiberales); and palmate-parallel, with several primary veins (of leaflets or leaf lobes) arising from one point, the adjacent secondary veins parallel to these (e.g., “fan” palms).
A more detailed classification system of venation (and many other leaf features) is that of Hickey (1973) and Hickey and Wolf (1975). This system is based on the pattern of primary, secondary, and tertiary venation. The following is a summary of the terms used in this system, illustrated in Figure 9.12.
Three general venation categories are used for a basically pinnate venation: craspedodromous, in which secondary veins terminate at the leaf margin; camptodromous, in which secondary veins do not terminate at the margin; and hyphodromous, with only the primary midrib vein present or evident and secondary veins either absent, very reduced, or hidden within the leaf mesophyll.
Subcategories of craspedodromous venation include simple craspedodromous, in which all secondary veins terminate at the margin; semicraspedodromous, in which the secondary veins branch near the margin, one terminating at the margin, the other looping upward to join the next secondary vein; and mixed craspedodromous (not illustrated), with some secondary veins terminating at the margin, but with many terminating away from the margin.
Subtypes of camptodromous venation include brochidodromous, in which secondary veins form prominent upward loops near the margin, joining other, more distal, secondary veins; eucamptodromous, in which secondary veins curve upward near the margin but do not directly join adjacent secondaries; cladodromous, in which secondary veins branch toward the margin; and reticulodromous, in which secondary veins branch repeatedly, forming a very dense, netlike structure.
Parallelodromous venation is equivalent to parallel (defined earlier), in which two or more primary or secondary veins run parallel to one another, converging at the apex.
Venation is actinodromous if three or more primary veins diverge from one point (equivalent to ternate or palmate venation). Palinactinodromous is similar, but the primary veins have additional branching above the main point of divergence of the primaries.
For actinodromous and palinactinodromous types, the venation is marginal if the main, primary veins reach the blade margin, and reticulate (not to be confused with “reticulate” in the more general venation terminology) if they do not. Flabellate venation is that in which several equal, fine veins branch toward the apex of the leaf.
Campylodromous venation is that in which several primary veins run in prominent, recurved arches at the base, curving upward to converge at the leaf apex.
Finally, venation is acrodromous, if two or more primary veins (or strongly developed secondary veins) run in convergent arches toward the leaf apex (but are not recurved at the base, as in campylodromous).
For actinodromous, palinactinodromous, and acrodromous types, the venation is basal if the primaries are joined at the blade base, and suprabasal if the primaries diverge above the blade base. The venation is perfect if branching of the lateral primary veins and their branches cover at least two thirds of the leaf blade area (or reach at least two thirds of the distance toward the leaf apex), and imperfect if these veins cover less than two thirds of the leaf blade area (or reach less than two thirds of the way toward the leaf apex).
These complex venation types, along with many other details of the leaf, can be specific to certain taxonomic groups of plants. Although they are not widely used in standard morphological descriptions, their recognition can be important in identification (e.g., of many tropical and fossil plants) and classification (see Hickey and Wolf 1975).
FLOWERS
A major diagnostic feature of angiosperms is the flower. As discussed in Chapter 6, a flower is a modified reproductive shoot, basically a stem with an apical meristem that gives rise to leaf primordia. Unlike a typical vegetative shoot, however, the flower shoot is determinate, such that the apical meristem stops growing after the floral parts have formed. At least some of the leaf primordia of a flower are modified as reproductive sporophylls (leaves bearing sporangia). Flowers are unique, differing, e.g., from the cones of gymnosperms, in that the sporophylls develop either as stamens or carpels (see Chapter 6, and later discussion).
F lower P arts (Figure 9.14)
The basic parts of a flower, from the base to the apex, are as follows. The pedicel is the flower stalk. (If a pedicel is absent, the flower attachment is sessile.) Flowers may be subtended by a bract, a modified, generally reduced leaf; a smaller or secondary bract, often borne on the side of a pedicel, is termed a bracteole or bractlet (also called a prophyll or prophyllum). Bracteoles, where present, are typically paired. [In some taxa, a series of bracts, known as the epicalyx, immediately subtends the calyx (see later discussion), as in Hibiscus and other members of the Malvaceae.] The receptacle or floral receptacle (also termed a torus, although “torus” can also be used for a compound receptacle; see Inflorescence Parts) is the tissue or region of a flower to which the other floral parts are attached. The receptacle is typically at the very tip of the floral axis (derived from the original apical meristem). In some taxa the receptacle can grow significantly and assume an additional function. From the receptacle arises the basic floral parts. The perianth (also termed the perigonium) is the outermost, nonreproductive group of modified leaves of a flower. If the perianth is relatively undifferentiated, or if its components intergrade in form, the individual leaflike parts are termed tepals. In most flowers the perianth is differentiated into two groups. The calyx is the outermost series or whorl of modified leaves. Individual units of the calyx are sepals, which are typically green, leaflike, and function to protect the young flower. The corolla is the innermost series or whorl of modified leaves in the perianth. Individual units of the corolla are petals, which are typically colored (nongreen) and function as an attractant for pollination. Some flowers have a hypanthium (floral tube), a cuplike or tubular structure, around or atop the ovary, bearing along its margin the sepals, petals, and stamens.
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Figure 9.14. Flower parts, sex, and attachment.
Many flowers have a nectary, a specialized structure that secretes nectar. Nectaries may develop on the perianth parts, within the receptacle, on or within the androecium or gynoecium (below), or as a separate structure altogether. Some flowers have a disk, a discoid or doughnut-shaped structure arising from the receptacle. Disks can form at the outside and surrounding the stamens (termed an extrastaminal disk), at the base of the stamens (staminal disk), or at the inside of the stamens and/or base of the ovary (intrastaminal disk). Disks may be nectar-bearing, called a nectariferous disk.
The androecium refers to all of the male organs of a flower, collectively all the stamens. A stamen is a microsporophyll, which characteristically bears two thecae (each theca comprising a pair of microsporangia; see Chapter 6). Stamens can be leaflike (laminar), but typically develop as a stalklike filament, bearing the pollen-bearing anther, the latter generally equivalent to two fused thecae.
The gynoecium refers to all of the female organs of a flower, collectively all the carpels. A carpel is the unit of the gynoecium, at maturity enclosing one or more ovules. Carpels may form as a ring of tissue (ascidiate development) or as a modified, conduplicate female megasporophyll of a flower, (plicate or conduplicate development; see Chapter 6). A pistil is that part of the gynoecium composed of an ovary, one or more styles (which may be absent), and one or more stigmas (see later discussion).
In some taxa, e.g., Aristolochiaceae and Orchidaceae, the androecium and gynoecium are fused into a common structure, known variously as a column, gynandrium, gynostegium, or gynostemium. A stalk that bears the androecium and gynoecium is an androgynophore, e.g., Passifloraceae. A stalk-like structure that bears stamens alone is termed an androphore (e.g., some Eriocaulaceae); one that bearing one or more pistils is a gynophore or stipe (see Gynoecium, Carpel, and Pistil).
F lower S ex and P lant S ex (Figure 9.14)
Flower sex refers to the presence or absence of male and female parts within a flower. Most flowers are perfect or bisexual [monoclinous], having both stamens and carpels. Bisexual flower sex is likely the ancestral condition in angiosperms.
Many angiosperm taxa, however, have imperfect or unisexual [diclinous] flower sex. In this case, flowers are either pistillate/female, in which only carpels develop, or staminate/male, in which only stamens develop.
Plant sex refers to the presence and distribution of perfect or imperfect flowers on individuals of a species. A hermaphroditic plant is one with only bisexual flowers. A monoecious (mono, one + oikos, house) plant is one with only unisexual flowers, both staminate and pistillate on the same individual plant; e.g., Quercus spp., oaks. A dioecious (di, two + oikos, house) plant is one with unisexual flowers, but with staminate and pistillate on separate individual plants (i.e., having separate male and female individuals; e.g., Salix spp., willows).
Plant sex can vary within individuals of a species, and there may also be a combination of perfect and imperfect flowers in different individuals. (These terms are confusing, but occasionally seen in the literature.) Polygamous is a general term for a plant with both bisexual and unisexual flowers. Andromonoecious refers to a plant with both staminate and perfect flowers on the same individual, and gynomonoecious is a plant with both pistillate and perfect flowers on the same individual. Polygamomonoecious [Trimonoecious] refers to a plant with pistillate, staminate, and perfect flowers on the same individual. Androdioecious refers to a plant with staminate flowers on some individuals and perfect flowers on other individuals. Gynodioecious refers to a plant with pistillate flowers on some individuals and perfect flowers on other individuals. Polygamodioecious is a plant with staminate and perfect flowers on some individuals, pistillate and perfect flowers on other individuals. Trioecious refers to a plant with pistillate, staminate, and perfect flowers on different individuals. All of these types of nonhermaphroditic plant sex may function as a mechanism of promoting increased outcrossing between individuals of a species. (However, many hermaphroditic plants can outcross by other means; see Chapter 13.)
F lower A ttachment (Figure 9.14)
Flower attachment is pedicellate, having a pedicel; sessile, lacking a pedicel; or subsessile, having a short, rudimentary pedicel. The terms bracteate, with bracts, and ebracteate, lacking bracts, may also be used with respect to flower attachment. The adaptive significance of pedicels is likely correlated with the spatial positioning of flowers relative to pollination or eventual fruit or seed dispersal.
F lower C ycly
Flower cycly refers to the number of cycles (series or whorls) or floral parts. The two basic terms used are complete, for a flower having all four major series of parts (sepals, petals, stamens, and carpels) and incomplete, for a flower lacking one or more of the four major whorls of parts (e.g., any unisexual flower, or a bisexual flower lacking a corolla).
F lower S ymmetry (Figures 9.15, 9.16)
Flower symmetry is an assessment of the presence and number of mirror-image planes of symmetry. Actinomorphic or radial symmetry [also called polysymmetric or regular] is that in which there are three or more planes of symmetry, such that there is a repeating structural morphology when rotated less than 360° about an axis. (A variant on actinomorphic symmetry is haplomorphic, appearing radially symmetric but not having strict mirror image halves because the parts are numerous and/or spirally inserted.) Biradial symmetry [also called disymmetric] means having two (and only two) planes of symmetry. (The difference between biradial and radial symmetry is sometimes not recognized, both being termed radial symmetry or actinomorphy; however, the distinction can be useful and is recognized here.) Zygomorphic or bilateral symmetry [also called monosymmetric or irregular] is that in which there is only one plane of symmetry. An asymmetric flower lacks any plane of symmetry, usually the result of twisting of parts. Flower symmetry can sometimes be subtle and can even vary within a flower; if so, it should be separately described for calyx, corolla, androecium, and gynoecium to avoid confusion.
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Figure 9.15. Flower symmetry types.
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Figure 9.16. Flower symmetry examples. A,B. Actinomorphic/radial symmetry. A. Five planes of symmetry. B. Six planes of symmetry. C. Slight zygomorphy (bilateral symmetry), vertical plane of symmetry. D,E. Strong zygomorphy of all floral parts. F. Asymmetry, caused by twisting of floral parts.
Flower symmetry can be an important adaptation relative to pollination systems. Actinomorphic flower symmetry is likely the ancestral condition in angiosperms and is found in a large number of groups. Zygomorphy has evolved repeatedly in many groups, typically as a means of more efficiently transferring pollen to an animal (usually insect) pollinator. Zygomorphy is typically correlated with a more horizontal floral orientation, and there are many different ways that zygomorphy can come about developmentally and morphologically.
F lower M aturation
Flower maturation refers to the time of development of flowers or flower parts (see also General Terminology). Anthesis is the general time of flowering, the opening of flowers with parts available for pollination. The relative timing of development of male versus female flowers or floral parts can be an important feature in reproductive biology. Protandrous refers to stamens developing, or pollen release occurring, prior to the maturation of carpels or stigmas being receptive. Protogynous is the reverse, with carpels or stigmas developing before stamens mature or pollen is released. Both protandry and protogyny may function to promote outcrossing (and thus inhibit selfing) within individuals of a species.
Two flower maturation terms dealing with the relative direction of development of parts can be important in describing taxonomic groups. Centrifugal refers to developing from the center toward the outside or periphery, whereas centripetal is development from the outside or periphery toward the center region. Both centrifugal and centripetal can be applied to parts of the perianth, calyx, corolla, androecium, or gynoecium; the terms are often used to describe the direction of development of stamens in a multiwhorled androecium.
Finally, the term cleistogamy (adj. cleistogamous) refers to a flower in which the perianth remains closed, such that pollen produced from within the flower pollinates only the stigma(s) of that flower. Chasmogamy (adj chasmogamous) is the normal situation, in which the perianth opens and pollen may be dispersed.
PERIANTH
The perianth (or perigonium) is the outermost, nonreproductive group of modified leaves of a flower (The term perianth has also been used for components of the reproductive structures of various Gnetales, but these are not homologous.) A perianth is absent in some flowering plants, typically those taxa that have very small, reduced flowers. The perianth, where present, functions both to protect the young flowering parts and to aid in pollination.
The units of the perianth arise like leaves as primordia from the apical meristem of the flower. Typically, they may retain leaflike characters. Sepals, in fact, are usually green with stomata and veins; even petals will have veins and may have vestigial stomata. However, the perianth can undergo significant developmental changes and be highly modified (and unleaflike) at maturity.
P erianth A rrangement/C ycly/M erosity (Figures 9.17, 9.18)
A fundamental aspect of perianth structure is perianth arrangement, the position of perianth parts relative to one another. In some taxa, such as some magnolias and water lilies, the perianth parts have a spiral arrangement, i.e., spirally arranged with only one perianth part per node, not in distinct whorls. Typically, flowers with a spiral perianth arrangement have parts that are either undifferentiated (similar to one another) or that grade from an outer, sepal-like form to an inner petal-like form. In either case, the term tepal is used to describe undifferentiated or intergrading perianth parts. In most flowering plants the perianth parts have a whorled arrangement, in which the parts appear to arise from the same nodal region. (Note that, developmentally, the perianth parts may actually initiate as primordia at slightly different times and positions; however, at maturity, this is usually undetectable.)
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Figure 9.17. Perianth cycly and merosity.
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Figure 9.18. Perianth cycly A. Biseriate, homochlamydeous, with outer and inner perianth parts (tepals) similar. B. Biseriate, dichlamydeous, with distinct calyx and corolla. C. Uniseriate, with a single whorl of perianth parts, by default termed a calyx.
Cycly refers to the number of whorls (cycles, series) of parts. (See General Terminology.) Thus, perianth cycly is the number of whorls of perianth parts. The most common type of perianth cycly by far is biseriate (also called dicyclic), in which there are two discrete whorls, an outer (= lower) and inner (= upper). A less common condition in flowering plants is a uniseriate perianth cycly, with perianth parts in a single whorl. Uniseriate perianths may arise by loss or reduction of one of the whorls of an ancestrally biseriate perianth. If it is known that the calyx was evolutionarily lost, what remains should be called a corolla; if the corolla was lost, what remains should be termed a calyx. If this directionality is not known, a uniseriate perianth is usually termed a calyx by tradition (although it may simply be called a perianth). Perianths may also rarely be triseriate (or “tricyclic”) = three-whorled, tetraseriate (or “tetracyclic”) = four-whorled, etc. The term multiseriate may be used to mean “composed of three or more whorls.” Other cycly terms evaluate the similarity of the whorls of parts to one another. Dichlamydeous describes a perianth composed of a distinct outer calyx and inner corolla; in most cases, a dichlamydeous perianth is also biseriate, but it may be multiseriate (i.e., the calyx or corolla containing more than one whorl). Homochlamydeous refers to a perianth composed of similar parts, each part a tepal. Most monocots have a homochlamydeous perianth, whereas most eudicots have a dichlamydeous one. In some cases, the distinction between dichlamydeous and homochlamydeous can be difficult, as it may be difficult to assess whether outer and inner series are similar or different.
Merosity refers to the number of parts per whorl or cycle. (See General Terminology.) Thus, perianth merosity is the number of parts per whorl of the perianth. General terms for perianth merosity are isomerous, having the same number of members in different whorls (e.g., five sepals and five petals) and anisomerous, having a different number of members in different whorls (e.g., two sepals and five petals). Merosity may be described separately for each whorl of the perianth, e.g., calyx merosity and corolla merosity. It is assessed for numbers of discrete petals, sepals, and tepals, or, if perianth fusion occurs, for numbers of calyx, corolla, or perianth lobes (see later discussion). Perianth, calyx, or corolla merosity is usually designated as a simple number, although terms such as bimerous (a whorl with two members), trimerous (a whorl with three members), tetramerous (a whorl with four members), and pentamerous (a whorl with five members), etc., can be used. Terms for absence of parts include achlamydeous, lacking a perianth altogether, apetalous, having no petals or corolla, and asepalous, having no sepals or calyx.
P erianth F usion (Figure 9.19)
The term perianth fusion deals with the apparent fusion of perianth parts to one another. (This character may be treated separately as calyx or corolla fusion.) If sepals, petals, or tepals are discrete and unfused, the respective terms aposepalous [chorisepalous], apopetalous [choripetalous], and apotepalous [choritepalous] may be used. If sepals, petals, or tepals appear to be fused (even slightly at the base), the respective terms synsepalous [gamosepalous], sympetalous [gamopetalous], and syntepalous [gamotepalous] are used. The “fusion” of perianth parts does not usually occur as a separate event, e.g., petals fusing together after they are individually formed. The “fusion” is apparent, and typically results by the growth of a common floral primordium at the base of the calyx, corolla, or peranth.
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Figure 9.19. Perianth fusion
Perianth fusion results in the development of a tubelike or cuplike structure (the region of “fusion”) in the calyx, corolla, or perianth. If little fusion occurs, the tubelike region occurs only at the base and gives rise to calyx, corolla, or perianth lobes.
P erianth P arts (Figure 9.20)
Various specialized terms are used for parts of the perianth. These include the following: anterior or ventral, referring to the lower, abaxial lobe(s) or side, toward a subtending bract; beard, a tuft, line, or zone of trichomes on a perianth or perianth part (see Vestiture); claw, an abruptly narrowed base of a sepal or petal; corona, a crownlike outgrowth between stamens and corolla, which may be petaline or staminal in origin; faucal, referring to the throat of a corolla; hypanthium or floral cup, a generally tubular or cup-shaped structure at the top rim of which are attached the calyx, corolla, and androecium; labellum, a modified, typically expanded, median petal, tepal, or perianth lobe, such as in the Orchidaceae; limb, the expanded portion of usually sympetalous corolla above the tube and throat; lip, either of two variously shaped parts into which a calyx or corolla is divided, usually into upper (posterior) and/or lower (anterior) lips, such as most Lamiaceae, Orchidaceae (Note: each lip may be composed of one or more lobes); lobe, a segment of a synsepalous calyx or sympetalous corolla; petal, a corolla member or segment; a unit of the corolla; posterior or dorsal, referring to the upper, adaxial lobe(s) or side, nearest to the axis, away from the subtending bract; sepal, a calyx member or segment, a unit of the calyx; spur, a tubular, rounded or pointed projection from the calyx or corolla, functioning to contain nectar; tepal, a perianth member or segment not differentiated into distinct sepals or petals; throat, an open, expanded region of a perianth, usually of a sympetalous corolla; tube, a cylindrically shaped perianth or region of the perianth, usually of a sympetalous corolla.
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Figure 9.20. Perianth types/parts. (c. s. = cross section; l. s. = longitudinal section)
P erianth T ype (Figure 9.20)
Perianth type can include aspects of the entire perianth; however it could include aspects of only the calyx, corolla, or hypanthium (if present). Generally, perianth type is based on the structure of the corolla alone, in which case it could logically be termed corolla type. The terminology for perianth type takes into account various aspects of shape, fusion, orientation, and merosity. Perianth type is often of systematic value and may be diagnostic for certain clades of angiosperms. The perianth type typically reflects adaptive features related to pollination biology, such as attracting a pollinator or better effecting the transfer of pollen. Some perianths are highly modified for other functions, such as the lodicules of grasses, which are reduced perianth parts that, upon swelling, open up the grass floret (see Inflorescence Type, later, and Poaceae of Chapter 7).
Specific perianth types include the following: bilabiate, two-lipped, with two, generally upper and lower segments, as in many Lamiaceae; calyptrate/operculate, having calyx and corolla fused into a cap that falls off as a unit, as in Eucalyptus; campanulate, bell-shaped, with a basally rounded flaring tube about as broad as long and flaring lobes, as in Campanula (may also be used for bell-shaped apopetalous corolla or apotepalous perianth); carinate, keeled, with a sharp median fold, usually on the abaxial side; coronate, with a tubular or flaring perianth or staminal outgrowth, as in Narcissus, Asclepias spp.; cruciate, with four distinct petals in cross form, as in many Brassicaceae; cucullate/galeate, hooded, with an abaxially concave posterior lip; disk, having an actinomorphic, tubular corolla with flaring lobes, as in some Asteraceae; hypocrateriform, a corolla with a tube having abruptly spreading lobes (encompasses both rotate and salverform, below); infundibular, funnel-shaped, with a tubular base and continuously expanded apex, as in Ipomoea, morning glory; ligulate or ray, having a short, tubular corolla with a single, elongate, strap-like apical extension, as in some Asteraceae; papilionaceous, with one large posterior petal (banner or standard), two inner, lateral petals (wings), and two usually apically connate lower petals (keel), the floral structure of the Faboideae (Fabaceae); personate, two -lipped, with the upper arched and the lower protruding into the corolla throat, as in Antirrhinum, snapdragon; rotate, with a short tube, typically no longer than the calyx, and wide limbs oriented at right angles to the tube, as in Phlox; saccate, having a pouchlike evagination; salverform, trumpet-shaped; with a long tube extending beyond the calyx and flaring limbs at right angles to tube; tubular, mostly cylindrical; unguiculate, clawed, as in many Brassicaceae, Caryophyllaceae; unilabiate, one-lipped, as in many Goodeniaceae; and urceolate, urn-shaped, expanded at base and constricted at apex, as in many Ericaceae.
P erianth A estivation (Figure 9.21)
Perianth aestivation is defined by the position, arrangement, and overlapping of floral perianth parts. Aestivation can be an important systematic character for delimiting or diagnosing some flowering plant taxa. In practice, aestivation is best observed by making hand sections of mature flower buds, because after anthesis, the perianth aestivation may be obscured. For very small flowers, histological sectioning may be needed to clearly see the aestivation type.
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Figure 9.21. Perianth aestivation.
Some standard perianth aestivation terms are as follows: imbricate, general term for overlapping perianth parts; convolute or contorted, imbricate with perianth parts of a single whorl overlapping at one margin, being overlapped at the other, as in the corolla of many Malvaceae; crumpled, having a wrinkled or crinkled appearance, particularly in bud; imbricate-alternate, imbricate with the outer whorl of perianth parts (sepals or outer tepals) alternating with (along different radii) the inner whorl of perianth parts (petals or inner tepals); quincuncial, imbricate with perianth parts of a single pentamerous whorl having two members overlapping at both margins, two being overlapped at both margins, and one overlapping only at one margin; valvate, with a whorl of perianth parts meeting at the margins, not overlapping; and involute, valvate with each perianth part induplicate (folded longitudinally inward along central axis).
ANDROECIUM
The androecium consists of all the floral male (pollen -producing) reproductive organs, the units of which are stamens. Stamens are interpreted as being modified, sporangia -bearing leaves or microsporophylls. Stamens initiate as primordia from the flower apical meristem, but at maturity are attached to the receptacle, corolla (having an epipetalous stamen fusion; see below), hypanthium rim, or staminal disk, a fleshy, elevated, often nectariferous cushion of tissue.
S tamen T ype (Figure 9.22)
There are two basic stamen types: laminar and filamentous (although intermediates can occur). Laminar stamens possess a leaflike, dorsiventrally flattened structure bearing two thecae (pairs of microsporangia), these typically on the adaxial surface. Laminar stamens may represent the ancestral type in flowering plants, although they have evolved secondarily in some groups. Filamentous stamens are far more common, having a stalklike, generally terete filament with a discrete pollen-bearing part, the anther.
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Figure 9.22. Stamen types and parts.
In some taxa one or more stamens will initially form but will be nonfertile. Such a sterile stamen is termed a staminode or staminodium. Staminodes may resemble the fertile stamens and can only be identified by determining if viable pollen is released. Other staminodes may be highly modified in structure, being petaloid, clavate (clublike), nectariferous, or very reduced and vestigial. Staminodes may or may not possess an antherode, a sterile antherlike structure.
S tamen A rrangement, C ycly, P osition, and N umber
Stamen arrangement (Figure 9.23) is the placement of stamens relative to one another (see General Terminology). Two basic stamen arrangements are spiral, with stamens arranged in a spiral, and whorled, with stamens in one or more discrete whorls or series. Additional stamen arrangement types consider the relative lengths of stamens to one another: didymous, with stamens in two equal pairs; didynamous, with stamens in two unequal pairs (as in many Bignoniaceae, Lamiaceae, Scophulariaceae, etc.); and tetradynamous, with stamens in two groups of four long and two short (typical of the Brassicaceae).
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Figure 9.23. Stamen arrangement.
Stamen cycly (Figure 9.24) refers to the number of whorls or series of stamens present (applying only if the stamens are whorled to begin with). The two major types of stamen cycly are uniseriate, having a single whorl of stamens, and biseriate, with two whorls of stamens. If additional whorls are present, the terms triseriate, tetraseriate, etc., can be used.
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Figure 9.24. Stamen cycly (uniseriate or biseriate) and position.
Stamen position (Figure 9.23) is the placement of stamens relative to other, unlike floral parts, in particular to the sepals and petals. An antisepalous (also called antesepalous) stamen position is one in which the point of stamen attachment is in line with (opposite) the sepals, calyx lobes, or outer whorl of tepals; similarly, alternipetalous means having the stamens positioned between the petals or corolla lobes. Antisepalous and alternipetalous are usually synonymous because (in a biseriate perianth) petals/corolla lobes are almost always inserted between sepals/calyx lobes; however, one should describe only what is evident, such that either or both terms may be used. Antisepalous or alternipetalous stamens are very common in taxa with uniseriate stamens.
An antipetalous (also called antepetalous) stamen position is one in which the point of attachment is in line with (opposite) the petals, corolla lobes, or inner whorl of tepals; alternisepalous means that the stamens are positioned between the sepals or calyx lobes. Antipetalous and alternisepalous are usually synonymous (for the same reason cited earlier). An antipetalous/alternisepalous stamen position is relatively rare and may be diagnostic for specific groups, such as the Primulaceae and Rhamnaceae.
Other stamen position terms, that also take into account stamen cycly and number are: haplostemonous, stamens uniseriate, equal in number to the petals, and opposite the sepals (antisepalous); obhaplostemonous, stamens uniseriate, equal in number to the petals, and opposite the petals (antipetalous); diplostemonous, stamens biseriate, the outer whorl opposite the sepals and the inner whorl opposite petals; and obdiplostemonous, stamens biseriate, the outer whorl opposite the petals, the inner opposite sepals. Among taxa with a uniseriate stamen cycly, a haplostemonous position is much more common. Among those with a biseriate stamen cycly, a diplostemonous position is much more common; obdiplostemonous stamens are relatively rare, being diagnostic, e.g., for some Crassulaceae.
Stamen number is typically simply expressed as just that, a number. The term polystemonous may be used for an androecium with numerous stamens, usually many more than the number of petals
S tamen A ttachment and I nsertion (Figure 9.25)
Stamen attachment refers to the presence or absence of a stalk, being either filamentous, with a filament present, sessile, with filament absent, or subsessile, with filament very short and rudimentary. Laminar stamens are, by default, sessile.
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Figure 9.25. Stamen insertion.
Stamen insertion (Figure 9.25) can refer to either of two things. First, it can indicate whether stamens extend past the perianth or not, the two terms being exserted (also termed phanerantherous), with stamens protruding beyond the perianth, and inserted (also termed cryptantherous), with stamens included within the perianth. Insertion may also correspond to the point of insertion, which is the point of adnation of an epipetalous stamen to the corolla (see later discussion). Examples of the latter usage are “the stamens are inserted halfway up the corolla tube” or “stamens are inserted unequally” (meaning they are inserted at different levels along the length of, say, a corolla tube). Stamen insertion, by either usage, is generally indicative of an adaptation for some particular pollination mechanism, functioning to present the anthers effectively to an animal pollinator.
S tamen F usion (Figure 9.26)
Stamen fusion refers to whether and how stamens are fused. The general terms distinct (unfused to one another), connate (fused to one another), free (unfused to a different structure), and adnate (fused to a different structure) may be used (see General Terminology). Common specialized terms are apostemonous, with stamens unfused (both distinct and free); diadelphous, with two groups of stamens, each connate by filaments only, as in many Faboideae (Fabaceae), which typically have nine stamens fused most of their length and one fused only at the base or not at all; epipetalous (also called petalostemonous), with stamens adnate to (inserted on) petals or the corolla (the terms epitepalous and episepalous can be used for adnation of stamens to tepals or sepals, respectively); monadelphous, with one group of stamens connate by their filaments, as in Malvaceae; and syngenesious, with anthers connate but filaments distinct, diagnostic of the Asteraceae Stamen fusion, like stamen insertion, typically functions as a “presentation” mechanism for animal pollination.
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Figure 9.26. Stamen fusion.
A nther P arts, T ype, and A ttachment (Figure 9.27)
Anthers are discrete pollen containing units, found in the stamens of the great majority of angiosperms. Anthers typically consist of two compartments called thecae (singular theca), with each theca containing two microsporangia (the fusion product of which is a locule). (Thus, anthers are typically tetrasporangiate.) The tissue between and interconnecting the two thecae is termed the connective, to which the filament (if present) is attached. Microsporangia are the sites of production of pollen grains, the immature male gametophytes of seed plants.
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Figure 9.27. Anther types, parts, and attachment (c.s. = cross section)
Various anther types occur, as determined by their internal structure. The typical anther is dithecal, having two thecae with typically four microsporangia. In a very few taxa, such as the Cannaceae and Malvaceae, anthers are monothecal, having one theca with typically two microsporangia. Finally, an extreme type of anther is the pollinium, a typically dithecal anther in which all the pollen grains of both thecae (Orchidaceae) or of adjacent thecae (Asclepias) are fused together as a single mass. The pollinia of the Orchidaceae and Asclepias have different developmental origins and structures.
Anther attachment refers to the position or morphology of attachment of the filament to the anther. Standard anther attachment types are basifixed, anther attached at its base to apex of the filament; dorsifixed, anther attached dorsally and medially to the apex of the filament; and subbasifixed, anther attached near its base to the apex of the filament. A versatile anther attachment is one in which the anther freely pivots (“teeter -totters”) at the point of attachment with the filament; versatile anthers may be dorsifixed, basifixed, or subbasifixed.
A nther D ehiscence (Figures 9.28, 9.29)
Anther dehiscence refers to the opening of the anther in releasing pollen grains. Anther dehiscence type (Figure 9.27) is the physical mechanism of anther dehiscence. The most common, and ancestral, anther dehiscence type is longitudinal, dehiscing along a suture parallel to the long axis of the thecae. Other types are rare and specific to given groups, including poricidal, dehiscing by a pore at one end of the thecae, such as the Ericaceae; transverse, dehiscing at right angles to the long axis of the theca; and valvular, dehiscing through a pore covered by a flap of tissue, as in the Lauraceae.
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Figure 9.28. Anther dehiscence types.
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Figure 9.29. Anther dehiscence direction (c.s. = cross section)
Anther dehiscence direction (Figure 9.28) indicates the position of the anther opening relative to the center of the flower or to the ground. Anther dehiscence direction is best detected when the anthers are immature (e.g., in bud) or just beginning to open. After dehiscence, the anthers usually shrivel and twist, obscuring the original direction in which they opened. Common types of dehiscence direction are: extrorse, dehiscing outward, away from the flower center; introrse, dehiscing inward, toward the flower center; and latrorse, dehiscing laterally, to the sides. In horizontally oriented flowers, anthers may face upward or downward, relative to the ground.
One fine point of anther dehiscence direction concerns some flowers, in which at least some of the stamens have one direction early in development but become reoriented to another direction at maturity. In such a case, the dehiscence direction can be described both in the early developmental stage and in the mature stage. For example, a common condition is one in which the anthers are introrse early in development, but reorient to the top of the flower, with all the anthers facing downward. Such a dehiscence direction can be described as introrse early in development (based on observation of buds), and downward at maturity (see Figure 9.28) (In another example, the poricidal anthers of members of the Ericaceae are extrorse early in development, but introrse at maturity by inversion; see Chapter 8.)
NECTARIES
Nectaries are specialized nectar-producing structures of the flower (Figure 9.14). Nectar is a solution of one or more sugars and various other compounds and functions as an attractant (a “reward”) to promote animal pollination. Nectaries may be padlike, developing as a discrete pad of tissue extending only part-way around the base of the flower. Commonly, a floral disk, consisting of a disk-like or dough-nut-shaped mass of tissue surrounding the ovary base or top, functions as a nectary. These nectariferous disks may be inner to (intrastaminal), beneath (staminal), or outer to (extrastaminal) the androecium. A perigonal nectary is one on the perianth, usually at the base of sepals, petals, or tepals. Septal nectaries are specialized tissues embedded within the septae of an ovary, secreting nectar via a pore at the ovary base or apex.
Note that other specialized glands may secrete non-sugar compounds that function as a pollination reward, such as waxes by members of the Krameriaceae. These are not termed nectaries, but are simply called glands, e.g., wax glands.
GYNOECIUM, CARPEL, AND PISTIL (Figure 9.30)
The gynoecium refers to all female organs of a flower (Figure 9.30). The unit of the gynoecium is the carpel, defined as a modified, typically conduplicate megasporophyll that encloses one or more ovules (see Chapter 6). The carpel is one of the major features (apomorphies) that make angiosperms unique within the seed plants. Like all flower parts, a carpel is interpreted as a modified leaf, in this case a megasporophyll, defined as a reproductive leaf bearing megasporangia (which in seed plants are components of the ovules). Carpels, in fact, may develop as dorsiventrally flattened leaves that fold conduplicately, ultimately enclosing the ovules.
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Figure 9.30. Gynoecium: carpel development. (c.s. = cross section)
A pistil is that part of the gynoecium composed of an ovary, one or more styles, and/or one or more stigmas (see later discussion). The ovary is the part of the pistil containing the ovules. A style is a generally stalklike, non-ovule-bearing portion of the pistil between the stigma and ovary. Styles may be absent in some pistils. A stigma is the pollen-receptive portion of the pistil. Stigmas may be discrete structures or they may be a region (the stigmatic region) of a style or style branch, e.g., the stigmatic “lines” on the styles of Asteraceae pistils. Finally, the term stipe or gynophore is used for a basal stalk of the pistil; stipes are usually absent. [Note that stipe is also used as a synonym for a leaf petiole, especially that of ferns.]
Pistils or ovaries may be simple, composed of one carpel, or compound, composed of two or more carpels (see Carpel Number). By convention, if there is more than one ovary, style, or stigma, but if any of these appear fused in any way (e.g., three apparent ovaries fused at the base), they are all part of the same pistil. (One unique case are the Asclepiadoids, in which the gynoecium consists of two carpels made up of two distinct ovaries and styles but a single stigma joining the styles; because the stigmas of the two carpels are connate, the whole structure is termed a single pistil.)
Within the ovary, a septum (plural septa) is a partition or cross-wall. A locule is an ovary cavity, enclosed by the ovary walls and septa. Placentae (singular, placenta) are the tissues of the ovary that bear the ovules, the immature seeds. A funiculus is a stalk that may lead from the placenta to the ovule. A column is the central axis to which septae and/or placentae are attached in axile or free-central placentation (see later discussion).
G ynoecial F usion (Figures 9.31, 9.32)
Fusion of carpels is a very important systematic character, the features of which are characteristic of major taxonomic groups. An apocarpous gynoecial fusion is one in which the carpels are distinct. An apocarpous gynoecium is generally thought to be the ancestral condition in the angiosperms. In contrast, a syncarpous gynoecial fusion is one in which carpels are connate (the pistil compound) and is the most common type in flowering plants. In a syncarpous gynoecium, the degree of carpel fusion can vary considerably; from connation only at the extreme base (having a strongly lobed ovary), to fusion into one, unlobed ovary but distinct styles and/or stigmas, to complete fusion with one ovary, style, and stigma. Fusion of carpels can determine the placentation type (Figure 9.32; see later discussion). Last, if the gynoecium is composed of a single carpel (in which fusion is really inapplicable), the term unicarpellous is used.
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Figure 9.31. Gynoecial fusion, carpel number, and locule number (c.s. = cross section, l.s. = longitudinal section)
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Figure 9.32. Evolutionary sequences of carpel fusion. A. Sequence leading to axile placentation. B. Sequence leading to parietal placentation. Carpel boundaries shown with dashed lines (c.s. = cross section).
C arpel/L ocule N umber (Figure 9.31)
Carpel and locule number are important characters in angio-sperm systematics. Locule number is generally easy to determine from ovary cross- and/or longitudinal sections, being equivalent to the number of wall-enclosed chambers within the ovary. In a general sense, ovaries may be unilocular, with a single locule, or plurilocular, having two or more locules. In some angiosperms, septa may divide the ovary into chambers in one region, such as the ovary base, but not in another region, such as the ovary apex; in such a case, the chambers below are continuous with one chamber above, and the locule number is technically 1, or unilocular.
Carpel number is often critical in classification and identification of flowering plants. It is determined as follows: If the gynoecium is apocarpous, the number of carpels is equal to the number of pistils; this is because each pistil is equivalent to a single carpel in any apocarpous gynoecium. If there is a single pistil, that pistil can be equivalent to one carpel (i.e., unicarpellous) or be composed of any number of fused carpels. For one pistil the carpel number is determined (in sequence) as follows: First, carpel number is equal to the number of styles or stigmas, if either of these is greater than 1. This is true regardless of the structure of the ovary because each of the styles or stigmas is a part of a carpel or is interpreted as a vestige of an ancestral carpel. (For example, pistils of all members of the Asteraceae have two styles and stigmas, and thus carpel number is interpreted as 2. This is true even though there is but one locule, ovule, and placenta; the two styles are interpreted as ancestral vestiges of a two -carpellate pistil, which became evolutionarily reduced to a single ovuled and loculed structure.) Second, if a single pistil has only one style and stigma, the ovary must be dissected to reveal the carpel number. If the ovary is plurilocular, then locule number is generally equal to the number of carpels. Each locule, in such a case, represents the chamber of the original ancestral or developmental carpel (except in some gynobasic taxa; see later discussion). Finally, if the ovary is unilocular, the number of carpels is equal to the number of placentae. For example, a violet, with one pistil, one style/stigma, and one locule, has three carpels because of the three placentae (having parietal placentation). (Exceptions to the last two rules are the gynobasic taxa of the Lamiaceae and Boraginaceae, s.s. In both of these groups, each of the two carpels is bisected early in development by a so-called false septum, such that the mature ovary typically has four locules, each with a single placenta and ovule. Thus, in this case, the number of locules and placentae, which is four, is twice that of the number of carpels.) The term pseudomonomerous is sometimes used for a gynoecium composed of more than one carpel but that appears to be unicarpellous.
O vary A ttachment and P osition (Figures 9.31, 9.33)
Ovary attachment deals with the presence or absence of a basal stalk or stipe. A sessile ovary is one lacking a stipe and is by far the most common situation. A stipitate ovary is one having a stipe and is relatively rare (Figure 9.31).
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Figure 9.33. Ovary position and perianth/androecial position.
Ovary position (Figure 9.33) assesses the position or placement of the ovary relative to the other floral parts: hypanthium, calyx, corolla, and androecium. A superior ovary is one with sepals, petals, and stamens, and/or hypanthium attached at the base of the ovary. An inferior ovary position has sepals, petals, stamens, and/or hypanthium attached at the ovary apex. A range of intermediates between superior and inferior ovaries can occur; the term half-inferior is used for sepals, petals, stamens, and/or hypanthium attached at the middle of the ovary.
P erianth/A ndroecial P osition (Figure 9.33)
Perianth/androecial position describes placement of the perianth and androecium relative both to the ovary and to a hypanthium, if present. Although used widely, perianth/androecial position may be simply substituted with a description of both ovary position and hypanthium presence/absence.
Three perianth/androecial position terms describe a flower without a hypanthium (and are rather repetitious with ovary position). The term hypogynous is used for sepals, petals, and stamens attached at base of a superior ovary. Epigynous refers to the sepals, petals, and stamens attached at apex of an inferior ovary. Epihypogynous is used for sepals, petals, and stamens attached at middle of the ovary, the ovary being half-inferior.
Other perianth/androecial position terms denote the presence of a hypanthium, with the sepals, petals, and stamens attached to the hypanthium rim. Perigynous denotes a hypanthium attached at the base of a superior ovary. Epiperigynous denotes a hypanthium attached at the apex of an inferior ovary. (The awkward term epihypoperigynous may be used to describe a hypanthium attached at the middle of a half-inferior ovary.)
P lacentation (Figure 9.34A)
Placentation refers to the positioning of the ovules and takes into account the number and position of placentae, septa, and locules. Determining placentation requires probing or making a cross and/or longitudinal section of the ovary.
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Figure 9.34. Gynoecium. A. Placentation. (c.s. = cross section; l.s. = longitudinal section). B. Ovule types. C. Ovule position, illustrated with anatrompous ovules.
Standard placentation types are axile, with the placentae arising from the column in a compound ovary with septa, common in many flowering plants such as the Liliaceae; apical or pendulous, with a placenta at the top of the ovary; apical-axile, with two or more placentae at the top of a septate ovary, as occurs in the Apiaceae; basal, with a placenta at the base of the ovary, as occurs in the Asteraceae and Poaceae; free-central, with the placentae along the column in a compound ovary without septa, such as in the Caryophyllaceae; laminar, with ovules arising from the surface of the septae; marginal, with the placentae along the margin of a unicarpellate (simple) ovary, as in the Fabaceae; parietal, with the placentae on the ovary walls or upon intruding partitions of a unilocular, compound ovary, such as in the Violaceae; parietal-axile, with the placentae at the junction of the septum and ovary wall of a two or more loculate ovary, such as in the Brassicaceae; and parietal-septate, with placentae on the inner ovary walls but within septate locules, as in some Aizoaceae.
O vule P arts, T ype, and P osition (Figure 9.34B,C)
Ovules are immature seeds, technically consisting of a megasproangium enveloped by one or more integuments (Chapter 5). The basic parts of an ovule (see also Seeds) are the nucellus or megasporangium, within which the female gametophyte develops; integument(s); funiculus, the stalk of the ovule; micropyle, the opening in the integument through which pollen or a pollen tube enters; and the raphe, a ridge on the seed coat often present, formed from an adnate funiculus.
Ovule type is based primarily on the curvature of the funiculus and nucellus/female gametophyte. An anatropous ovule is one in which curvature during development results in displacement of the micropyle to a position adjacent to the funiculus base; this is the most common ovule type of the angiosperms and is presumed to be ancestral. An orthotropous [atropous] ovule is one in which no curvature takes place during development; the micropyle is positioned opposite the funiculus base. (An ovule somewhat intermediate in curvature between anatropous and orthotropous is sometimes termed hemitropous or hemianatropous.) A campylotropous ovule type is one in which the nucellus is bent only along the lower side. An amphitropous ovule is one in which the nucellus is bent strongly along both upper and lower sides, with a lower “basal body.” (See Chapter 11, for a more precise terminology of ovule types.)
Ovule position refers to the direction that an ovule faces relative to the floral axis, with the micropyle and raphe as regions of orientation. An epitropous ovule is one in which the micropyle points distally (toward the flower apex). This type can be further divided into epitropous-dorsal, in which the raphe is dorsal (abaxial, pointing away from the central floral or ovary axis) or epitropous-ventral, in which the raphe is ventral (adaxial, pointing toward the central floral or ovary axis). A hypotropous ovule is one in which the micropyle points proximally. This type can be further divided into hypotropous-dorsal, in which the raphe is dorsal (abaxial, pointing away from the central floral or ovary axis) or hypotropous-ventral, in which the raphe is ventral (adaxial, pointing toward the central floral or ovary axis). A final position type is a pleurotropous ovule, one in which the micropyle points to the side. This type can be further divided into pleurotropous-dorsal, in which the raphe is above or pleurotropous-ventral, in which the raphe is below (A heterotropous ovule is one that varies in orientation.)
S tyle P osition/S tructural T ype (Figure 9.35)
Style position is the placement of the style relative to the body of the ovary. A terminal or apical style position is one arising at the ovary apex; this is by far the most common type. A subapical style arises to one side, near and slightly below the ovary apex. A lateral style position is one arising at the side of an ovary, as in members of the Rosaceae, such as Fragaria. Finally, a gynobasic style arises from the base of the ovary. Gynobasic styles are characteristic of the Boraginaceae, s.s. and of most Lamiaceae, in which the style arises from the base and center of a strongly lobed ovary.
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Figure 9.35. Gynoecium: style position (left) and stigma/stigmatic region type (right).
Styles may be structurally specialized in some taxa. One specialized style structural type is a stylar beak, a persistent, extended style or basal (to subbasal) stylar region. A beak is typically accrescent and elongates during fruit formation. Beaks function in fruit dispersal, as in members of the Asteraceae (e.g., Taraxacum, dandelion) or Geraniaceae (e.g., Geranium).
S tigma/S tigmatic R egion T ypes (Figure 9.34)
The term stigma is used for a discrete structure that is receptive to pollen on the entire surface, whereas stigmatic region may be used for that portion of a larger structure (generally a style or style branch) that is receptive to pollen. General shape terms may be used to describe stigma or stigmatic region types. A few common stigma or stigmatic region types are discoid, with stigma(s) disk-shaped; globose, with stigma(s) spherical in shape; linear, with stigmas or stigmatic tissue long and narrow in shape; and plumose, stigmas with feathery, trichome-like extensions, often found in wind-pollinated taxa (e.g., in Cyperaceae, Poaceae).
INFLORESCENCES
An inflorescence is a collection or aggregation of flowers on an individual plant. Inflorescences often function to enchance reproduction. For example, the aggregation of flowers in one location will make them visually more attractive to potential pollinators. Other inflorescences are related to very specialized reproductive mechanisms, examples being the spadices and associated spathes of some Araceae or the syconia of figs (see later discussion). The structure of an inflorescence can be complicated, requiring detailed developmental study.
I nflorescence P arts
Several terms deal with leaflike structures found in the inflorescence. An inflorescence bract is one that subtends not an individual flower but an inflorescence axis or a group of flowers. (Bracts that subtend an individual flower should be termed floral bracts; however, some sources do not make the distinction or will use inflorescence bract to refer to either.) A group or cluster of bracts subtending an entire inflorescence is termed an involucre (adjective involucrate); a similar group of bracts subtending a unit of the inflorescence is an involucel. A spathe (adjective spathaceous) is an enlarged, sometimes colored bract subtending and usually enclosing an inflorescence; many Araceae are good examples of spathes, which subtend the spadix inflorescence (see later discussion). An awn is a bristlelike, apical appendage on the glumes or lemmas of grass (Poaceae) spikelets.
Other terms deal with various (stem) axes in an inflorescence. A peduncle (adjective pedunculate) is the stalk of an entire inflorescence. A compound receptacle (also called a torus, although the latter term is also used for the floral receptacle; see Flower Parts) is a mass of tissue at the apex of a peduncle that bears more than one flower. A peduncle that lacks well-developed leaves, arising from a basal rosette of vegetative leaves is termed a scape (adjective scapose), the plant habit in such a case being acaulescent. A rachis is a major, central axis within an inflorescence. However, the central axis of a grass or sedge spikelet is a rachilla. Finally, a ray is a secondary axis of a compound umbel (see later discussion).
I nflorescence P osition
There are three major inflorescence positions, defined based on where the inflorescence develops: (1) axillary, in which the entire inflorescence is positioned in the axil of the nearest vegetative leaf; (2) terminal, in which the inflorescence develops as part of a terminal shoot that gave rise to the nearest vegetative leaves; and (3) cauliflorous, in which the inflorescence grows directly from a woody trunk. Three specialized inflorescence position terms for palms are infrafoliar, in which the inflorescence arises below the crownshaft, interfoliar, in which it arises within the crown-shaft, and suprafoliar, in which it arises above the leaves of the crownshaft.
I nflorescence D evelopment
Inflorescence development is a major aspect of defining inflorescence type. The two major inflorescence developmental types are determinate and indeterminate. A determinate inflorescence is one in which the apical meristem of the primary inflorescence axis terminates in a flower; typically, the terminal flower matures first, with subsequent maturation occurring from apex to base. Determinate inflorescences are characteristic of cymes. An indeterminate inflorescence is one in which the apical meristem of the primary inflorescence axis does not develop into a flower; typically, the basal flower matures first, with maturation occurring from base to apex. Indeterminate inflorescences include a number of types, such as spikes, racemes, and panicles (see later discussion).
I nflorescence T ype (Figures 9.36–9.38)
Inflorescences that have a common development and structure with respect to presence, number, arrangement, or orientation of bracts, axes, and certain specialized structures, define an inflorescence type. One difficulty with determining inflorescence type is simply delimiting its boundaries. Generally, an inflorescence is bounded by the lowest vegetative leaf. However, there may be a gradation between lower or basal vegetative leaves and small floral bracts, such that the delimitation of the inflorescence is somewhat arbitrary. (Note that if an inflorescence consists of a single flower, it is termed solitary; a scapose inflorescence is one with one or more flowers on an essentially leafless peduncle or scape, usually arising from a basal rosette.)
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Figure 9.36. Determinate (cymose) inflorescences. A–C. Dichasial cymes. A. Simple dichasium. B. Compound dichasium. C. Compound cyme. D–G. Monochasial cymes. D. Helicoid cyme, showing derivation from compound dichasium by development of one axis on the side of the primary axis. E. Scorpioid cyme, showing derivation from compound dichasium by development of one axis on alternating sides of the primary axis. F. Rhipidium. G. Drepanium. (Terminology after and redrawn from Weberling 1989, by permission.)
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Figure 9.37. Indeterminate inflorescence types.
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Figure 9.38. Indeterminate or determinate inflorescence types.
Inflorescence types are valuable characters in systematics and are often characteristic of specific groups, such as the compound umbels of the Apiaceae, heads of the Asteraceae, and helicoid or scorpioid cymes of the Boraginaceae. Some inflorescence types are quite specialized adaptations for reproduction, such as the cyathia of Euphorbioids.
The term cyme (Figure 9.36) can be used as a general term to denote a determinate inflorescence. One type of cyme is the dichasium, one that develops along two axes, forming one or more pairs of opposite, lateral axes. A simple dichasium is a three-flowered cyme, having a single terminal flower and two, opposite lateral flowers, the pedicels of all of equal length; bracts typically subtend the two lateral flowers, although the bracts may be absent. (The term cymule may be used for a small, simple dichasium.) A compound dichasium is a many -flowered cyme of repeatedly branching simple dichasia units. In a compound dichasium, the branches are typically decussately arranged and are thus in multiple planes. Finally, a compound cyme is a branched cyme, similar to a compound dichasium but lacking a consistent dichasial branching pattern. Some compound cymes actually have the same branching pattern as a compound dichasium but with certain internodal axes being reduced or missing, yielding a more congested appearance.
A monochasium (Figure 9.36) is a cyme that develops along one axis only. (The terminology for monochasial cymes can vary from author to author, the following being just one.) A helicoid cyme or bostryx is a monochasium in which the axes develop on only one side of each sequential axis, appearing coiled at least early in development. A scorpioid cyme or cincinnus is a monochasium in which the branches develop on alternating sides of each sequential axis, typically resulting in a geniculate (zig-zag) appearance. Both helicoid cymes and scorpioid cymes have branches or axes that are in more than one plane and can be viewed as being derived by reduction from the decussate branches of a compound dichasium. Two other monochasial cymes have, by definition, axes that are in one plane. A drepanium is a monochasium in which the axes develop on only one side of each sequential axis; like a helicoid cyme, drepania typically appear coiled at least early in development. (Drepania are treated as helicoid cymes in some terminology.) A rhipidium is a monochasium in which the branches develop on alternating sides of each sequential axis; like scorpioid cymes, rhipidia typically have a geniculate (zig-zag) appearance. (Rhipidia are treated as scorpioid cymes in some terminology.) In reality, these four monochasial structures may intergrade with one another. For example, a monochasium intermediate between a helicoid cyme and a drepanium may occur. Thus, simply using the term monochasial cyme may be best in lieu of more detailed observations and descriptions.
Several indeterminate inflorescence types are recognized (Figure 9.37). All of these generally lack a flower at the top of the main axis and develop from base to apex. A spike is an indeterminate inflorescence, consisting of a single axis bearing sessile flowers. Similarly, a raceme is an indeterminate inflorescence in which the single axis bears pedicellate flowers. A panicle is like a branched raceme, defined as an indeterminate inflorescence having several branched axes bearing pedicellate flowers. Finally, a corymb is an indeterminate inflorescence consisting of a single axis with lateral axes and/or pedicels bearing flat-topped or convex flowers. Corymbs can be either simple or compound. A simple corymb is unbranched, consisting of a central axis bearing pedicellate flowers, the collection of flowers being flat-topped or convex; simple corymbs are like racemes in which the lower pedicels are much more elongate than the upper. A compound corymb is branched, consisting of two or more orders of inflorescence axes bearing flat-topped or convex, pedicellate flowers; compound corymbs are like panicles in which the lower axes and pedicles are much more elongate than the upper.
Some inflorescences may be either determinate or indeterminate (Figure 9.38). A simple umbel is a determinate or indeterminate, flat-topped or convex inflorescence with pedicels attached at one point to a peduncle. Two inflorescences in which the flowers at the point of attachment appear congested are the fascicle and glomerule. A fascicle is a racemelike or paniclelike inflorescence with pedicellate flowers in which internodes between flowers are very short. A glomerule is an inflorescence of sessile or subsessile flowers in which the internodes between flowers are very short.
In some taxa an inflorescence will appear to be one type, but (upon detailed examination) is actually a modification of another type. For example, the term pseudoumbel is used for an inflorescence appearing like a simple umbel, but actually composed of condensed, monochasial cymes, as in the Alliaceae and Amaryllidaceae.
S econdary I nflorescences (Figure 9.39)
Secondary inflorescences are defined as aggregates of unit inflorescences (also called “primary” or “partial” inflorescences); each unit inflorescence is a subunit of the secondary inflorescence that resembles an inflorescence type, per se. Examples of secondary inflorescences are “a panicle of spikelets,” “a corymb of heads,” or “a raceme of spikes.” A paracladium is a unit inflorescence resembles the secondary inflorescence, e.g., an umbel of umbels (compound umbel).
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Figure 9.39. Secondary inflorescences.
Two specific types of secondary inflorescences are the thryse and verticillaster. A thyrse is essentially a raceme of cymes, in which the main axis is indeterminate but the opposite, lateral, unit inflorescences are pedicellate cymes, typically either simple dichasia, compound dichasia, or compound cymes, occasionally monochasial cymes. A verticillaster is essentially a “spike of opposite cymes,” similar to a thyrse in having an indeterminate main axis but differing in that the lateral cymes have very reduced to absent internodal axes and pedicels, giving a congested appearance. Verticillasters are found in several members of the Lamiaceae, the mint family. A compound umbel is another secondary inflorescence in which the peduncle bears secondary axes called rays that are attached at one point and unit, simple umbels attached at the tip of the rays, as in many Apiaceae.
S pecialized I nflorescences (Figure 9.40)
Some inflorescences are quite specialized and often restricted to certain taxonomic groups. A catkin (also called an ament) is a unisexual, typically male spike or elongate axis that falls as a unit after flowering or fruiting, as in Quercus, Salix. A cyathium is an inflorescence bearing small, unisexual flowers and subtended by an involucre (frequently with petaloid glands), the entire inflorescence resembling a single flower, as in Euphorbia and relatives. An inflorescence unit, such as a cyathium, that appears as and may function like a single flower is termed a pseudanthium, typically consisting of two or more flowers fused or tightly grouped together. A head or capitulum is a determinate or indeterminate, crowded group of sessile or subsessile flowers on a compound receptacle, often subtended by an involucre, composed of involucral bracts, or phyllaries; a calyculus refers to often reduced outer bracts in the heads of the Asteraceae. Heads are typical of the Asteraceae and some other groups. (Note that some inflorescences resemble a head but lack a compound receptacle; these can be termed head-like.) A hypanthodium is an inflorescence bearing numerous flowers on the inside of a convex or involuted compound receptacle, as in Ficus. A spadix is a spike with a thickened or fleshy central axis, typically with congested flowers and usually subtended by a spathe, as in the Araceae. A spikelet literally means a “small spike” and refers to the basic inflorescence unit in the Cyperaceae, the sedges, and Poaceae, the grasses. Sedge spikelets are like a small spike, with sessile (reduced) flowers on an axis (rachilla), each flower subtended by a bract (also called a scale). A grass spikelet consists of an axis (rachilla), typically bearing two basal bracts (glumes) and one or more short lateral branch units called florets, each of which bears two bracts (lemma and palea) that subtend a terminal, reduced flower. (See family treatments of Cyperaceae and Poaceae in Chapter 7.)
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Figure 9.40. Specialized inflorescence types.
FRUITS
Fruits are the mature ovaries or pistils of flowering plants plus any associated accessory parts. Accessory parts are organs attached to a fruit but not derived directly from the ovary or ovaries, including the bracts, axes, receptacle, compound receptacle (in multiple fruits), hypanthium, or perianth. The term pericarp (rind, in the vernacular) is used for the fruit wall, derived from the mature ovary wall. The pericarp is sometimes divisible into layers: endocarp, mesocarp, and exocarp (see fleshy fruit types, discussed later).
Fruit types are based first on fruit development. The three major fruit developments are simple (derived from a single pistil of one flower), aggregate (derived from multiple pistils of a single flower, thus having an apocarpous gynoecium), or multiple (derived from many coalescent flowers; see later discussion). In aggregate or multiple fruits, the component derived from an individual pistil is called a unit fruit. The term infructescence denotes a mature inflorescence in fruit.
As mentioned in Chapter 6, the evolution of fruits was correlated with the evolution of carpels and is a significant adaptation for seed dispersal in the angiosperms.
S imple F ruit T ypes (Figures 9.41–9.43)
The simple fruit type, as well as unit fruit types of aggregate and multiple fruits, are classified based on a number of criteria, including (1) whether fleshy (succulent) or dry at maturity; (2) whether indehiscent (not splitting open at maturity) or dehiscent (splitting open along definite pores, slits, or sutures); (3) if dehiscent, the type (e.g., location, shape, and direction) of dehiscence; (4) carpel and locule number, including presence of septa; (5) seed/ovule number; (6) placentation; (7) structure of the pericarp wall; and (8) ovary position.
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Figure 9.41. Fruits: simple, dry, and indehiscent fruit types (l.s. = longitudinal section).
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Figure 9.42. Fruits: simple, dry, and dehiscent fruit types (c.s. = cross section; l.s. = longitudinal section).
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Figure 9.43. Fruits: simple, fleshy fruit types (c.s. = cross section; l.s. = longitudinal section).
One class of simple fruits are those that are dry and indehiscent at maturity (Figure 9.41). An achene is a one-seeded, dry, indehiscent fruit with seed attached to the pericarp at one point only, such as the unit fruits of sunflowers. An anthocarp or diclesium is an achene or nut, surrounded by the persistent and accrescent perianth, as in Pontederia or the Nyctaginaceae. A grain or caryopsis is a one-seeded, dry, indehiscent fruit with the seed coat adnate to pericarp wall; grains are the fruit type of all Poaceae (grasses). (The embryo of grain crops is known as germ, as in “wheat germ”; the pericarp and seed coat together are the bran.) A nut is a one -seeded, dry indehiscent fruit with a hard pericarp, usually derived from a one-loculed ovary. (Nuts and achenes may intergrade; the terms are sometimes used interchangeably.) A nutlet is a small nutlike fruit; for example, the mericarps (see schizocarp) of the Boraginaceae and Lamiaceae are termed nutlets. A samara is a winged, dry, usually indehiscent fruit, as in Acer (maple) and Ulmus (elm). A tryma is a nut surrounded by an involucre that dehisces at maturity, such as in Carya (pecan). Finally, a utricle is a small, bladdery or inflated, one-seeded, dry fruit; utricles are essentially achenes in which the pericarp is significantly larger than the mature seed, as in Atriplex (salt bush).
Other simple fruits are dry and dehiscent at maturity (Figure 9.42). Most dry, dehiscent fruits open by means of a valve, pore, or mericarp (see later discussion). However, some, of various fruit types, are explosively dehiscent, i.e., will open with force (by various mechanisms), functioning to eject the seeds, e.g., Ecballium elaterium, the squirting cucumber, or Impatiens.
A general type of dry, dehiscent fruit is the capsule. Capsules are generally dry (rarely fleshy), dehiscent fruits derived from compound (multicarpeled) ovaries. Several types of capsules can be recognized based on the type or location of dehiscence. Loculicidal capsules have longitudinal lines of dehiscence radially aligned with the locules (or between the placentae, if septa are absent). Septicidal capsules have longitudinal lines of dehiscence radially aligned with the ovary septa (or with the placentae, if septa are absent). Both loculicidal and septicidal capsules split into valves, a portion of the pericarp wall that splits off, but does not enclose the seed(s); valves may remain attached to the fruit or may fall off, depending on the taxon. A circumscissile capsule (also called a pyxis or pyxide) has a transverse (as opposed to longitudinal) line of dehiscence, typically forming a terminal lid or operculum, as in Plantago. A septifragal or valvular capsule is one in which the valves break off from the septa, as in Ipomoea, morning glory Poricidal capsules have dehiscence occurring by means of pores, as in Papaver, poppy. Other capsules can be defined by the location of dehiscence, such as acrocidal capsules, dehiscing by means of apical slits, or basicidal capsules, dehiscing by means of basal slits, as in Aristolochia spp.
Some other dry, dehiscent fruit types are really just specialized capsulelike structures. A follicle is a dry, dehiscent fruit derived from one carpel that splits along one suture, such as in the unit fruits of Magnolia. A legume is a dry, dehiscent fruit derived from one carpel that splits along two longitudinal sutures; legumes are the diagnostic fruit type of the Fabaceae, the legume family. Some legumes retain the vestige of the two, longitudinal sutures, but have become secondarily modified, such as loments, which split transversely into one-seeded segments, and indehiscent legumes, which do not split open at all (e.g., peanut). Silicles and siliques are dry, dehiscent fruits derived from a two-carpeled ovary that dehisces along two sutures but that has a persistent partition, the replum (the mature septum, generally with attached seeds). The two fruit types differ is that a silicle is about as broad or broader than long, a silique is longer than broad; both are characteristic fruit types of the Brassicaceae, the mustard family.
Finally, a schizocarp is a dry, dehiscent fruit type derived from a two or more loculed compound ovary in which the locules separate at maturity. The individual unit fruits containing each locule can be defined based on other simple fruit types. For example, a schizocarp of follicles is a fruit in which the (generally two) carpels of a pistil split at maturity, each carpel developing into a unit follicle, as in Asclepias, milkweed. A schizocarp of mericarps is one in which the carpels of a single ovary split during fruit maturation, each carpel developing into a unit mericarp, as in the Apiaceae. Mericarps are portions of the fruit that separate from the ovary as a distinct unit completely enclosing the seed(s); in the Apiaceae the two mericarps are typically attached to one another via a stalklike structure called the carpophore. Lastly, a schizocarp of nutlets is distinct in that a single ovary becomes lobed during development, the lobes developing at maturity into nutlets, which split off. Nutlets here may be viewed as specialized types of mericarps. Schizocarpic nutlets are typical of the Boraginaceae and most Lamiaceae, which have gynobasic styles attached between adjacent ovary lobes. (Note that the term eremocarp may be used for one of the fruit units developing from ovary lobes that are separate from one another at their inception, as in the Boraginaceae and Lamiaceae. By this terminology, the term schizocarp is reserved for a fruit unit that is part of a larger, single structure at inception, splitting off only at fruit maturity, as with the schizocarp of mericarps in the Apiaceae; see Hilger 2014.)
Another class of simple fruits includes those that, at maturity, are fleshy or succulent (also termed baccate or carnose; see Texture) (Figure 9.43). Fleshy fruits are general adaptations for seed dispersal by animals, the succulent pericarp being the reward (with at least some seeds either falling out or passing through the animal’s gut unharmed). Fleshy fruits are generally indehiscent, but may rarely be dehiscent, as in some Yucca spp. The pericarp of some fleshy fruits may be divided into layers. These pericarp wall layers, if present, are termed the endocarp (the innermost wall layer), mesocarp (the middle wall layer), and exocarp (the outermost wall layer); if only two layers are evident, the terms endocarp and exocarp alone are used. A berry is the general, unspecialized term for a fruit with a succulent pericarp, as in Vitis, grape. A drupe is a fruit with a hard, stony endocarp and a fleshy mesocarp, as in Prunus (peach, plum, cherry, etc.). The term pyrene can be used either for a fleshy fruit in which each of two or more seeds is enclosed by a usually bony-textured endocarp, or pyrene can refer to the seed covered by a hard endocarp unit itself, regardless of the number. A hesperidium is a septate fleshy fruit with a thick-skinned, leathery outer pericarp wall and fleshy modified trichomes (juice sacs) arising from the inner walls, as in Citrus (orange, lemon, grapefruit, etc.). A pepo is a nonseptate fleshy fruit with parietal placentation and a leathery exocarp derived from an inferior ovary, the fruit type of the Cucurbitaceae. A pome is a fleshy fruit with a cartilaginous endocarp derived from an inferior ovary, with the bulk of the fleshy tissue derived from the outer, adnate hypanthial tissue, as in Malus (apple) and Pyrus (pear). Finally, a pseudodrupe is a nut surrounded by a fleshy, indehiscent involucre, as in Juglans (walnut); thus, pseudodrupes have accessory tissue serving as the fleshy component.
A ggregate F ruit T ypes (Figure 9.44)
An aggregate fruit is one derived from two or more pistils (ovaries) of one flower. In determining the aggregate fruit type, one first identifies the unit fruit that corresponds to a single pistil. The aggregate fruit type is then indicated either as “aggregate fruit of “ the particular unit fruits or by adding the suffix “-acetum” to the unit fruit term.
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Figure 9.44. Fruits: aggregate fruit types (l.s. = longitudinal section).
An achenecetum is an aggregate fruit of achenes. A common example is Fragaria, strawberry, in which the achenes are on the surface of accessory tissue, an enlarged, fleshy receptacle. A drupecetum is an aggregate fruit of drupes, as in Rubus, raspberry or blackberry. A follicetum is an aggregate fruit of follicles, as occurs in Magnolia. A syncarp is an aggregrate fruit, typically of berries, in which the unit fruits fuse together, as in Annona. (Note that syncarps may form at the floral stage or later during fruit development; if the latter, the fruit is sometimes called a pseudosyncarp.)
M ultiple F ruit T ypes (Figure 9.45)
A multiple fruit is one derived from two or more flowers that coalesce. In determining the multiple fruit type, one may also identify the unit fruit corresponding to a single pistil of a single flower; the fruit type may be indicated as a “multiple fruit of” the particular unit fruit present.
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Figure 9.45. Fruits: multiple fruit types (l.s. = longitudinal section).
Some specialized multiple fruit types are as follows: A bur is a multiple fruit of achenes or grains surrounded by a prickly involucre, such as in Cenchrus, sandbur (Poaceae), or Xanthium, cocklebur (Asteraceae). A sorosis is a multiple fruit in which the unit fruits are fleshy berries and are laterally fused along a central axis, as in Ananas, pineapple. A syconium is a multiple fruit in which the unit fruits are small achenes covering the surface of a fleshy, inverted compound receptacle (derived from a hypanthodium), as in Ficus, fig.
SEEDS (Figure 9.46)
Aspects of seed morphology can be important systematic characters used in plant classification and identification. Some valuable aspects of seed morphology are size and shape, as well as the color and surface features of the seed coat, the outer protective covering of seed derived from the integument(s). The seed coat of angiosperms consists of two, postgenitally fused layers, an outer testa derived from the outer integument (itself sometimes divided into layers, an inner endotesta, middle mesotesta, and outer exotesta) and an inner tegmen derived from the inner integument (which can be divided into similar layers, the endotegmen, mesotegmen, and exotegmen). A seed coat that is fleshy at maturity may be termed a sarcotesta (although this may be confused with an aril, which is separate from the integuments; see later discussion). Also important in seed morphology are the shape, size, and color of the hilum, the scar of attachment of the funiculus on the seed coat, and of the raphe, a ridge on the seed coat formed from an adnate funiculus. Some seeds have an aril (adj. arillate), a fleshy outgrowth of the funiculus, raphe, or integuments (but separate from the integuments) that generally functions in animal seed dispersal. Arils may be characteristic of certain groups, such as the Sapindaceae. Similar to the aril is a caruncle (also called an elaiosome or strophiole), a fleshy outgrowth at the base of the seed; caruncles also function in animal seed dispersal, such as the carunculate seeds of Viola, violets, with regard to seed dispersal by ants.
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Figure 9.46. Seed parts, endosperm types (l.s. = longitudinal section).
Specific details of the embryo, the immature sporophyte, can be studied. These include aspects of the epicotyl (the immature shoot), radicle (the immature root; not to be confused with a “radical” position; see later discussion), hypocotyl (the transition region between the root and epicotyl), and cotyledon(s) (the first leaf/leaves of the embryo, often functioning in storage of food reserves). Some members of the Poaceae, the grass family, have the epicotyl surrounded by a protective sheath known as the coleoptile, and the radicle surrounded by a protective sheath known as the coleorhiza. Cotyledon aestivation (or ptyxis) can be a valuable systematic feature.
S eed E ndosperm T ype
All angiosperms form endosperm, the food reserve tissue derived from fusion of sperm with the polar nuclei of the female gametophyte. The typical angiosperm seed is albuminous or endospermous, having endosperm as the food reserve in mature seeds. In some angiosperms endosperm develops, but very little to none is deposited in mature seeds, a feature termed exalbuminous or nonendospermous, as in orchid seeds. Finally, some flowering plants are cotylespermous, in which the main food reserve is stored in the cotyledons. Cotylespermous seeds are typical of beans and peas.
S eed G ermination T ype
Seed germination type requires observation of young seedlings during germination and describes positioning of the cotyledons. Hypogeous [cryptocotylar] refers to a type in which the cotyledon(s) remain in the ground during germination. Epigeous [phanerocotylar] has cotyledon(s) elevated above the ground during germination. Vivipary refers to a seed that germinates into a seedling before being shed from the parent plant, e.g., Rhizophora, red mangrove.
FRUIT AND SEED DISPERSAL
The dispersal unit, or diaspore, of a plant (seeds and/or fruits, including accessory parts) often exhibits specific adaptations for dispersal from the parent plant, giving it a selective advantage. These include: anemochory, dispersal by wind (e.g., dandelion fruits, with a wind-blown pappus); autochory, self-dispersal (including ballochory, dispersal by explosive dehiscence, barochory, dispersal by gravity, or geocarpy, dispersal in which the plant pushes fruits into the ground, e.g., Arachis hypogaea, peanut); hydrochory, dispersal by water (e.g., coconuts); myrmecochory, dispersal by ants (e.g., violet seeds); and zoochory, general dispersal by (larger) animals. Two important descriptors of zoochory are: epizoochory (exozoochory), in which a fruit or seed becomes attached to and is carried away by an animal (e.g., a burr becoming attached to an animal’s fur or a human’s socks) and endozoochory, in which a fruit or seed is eaten and passes out via the animal’s feces unharmed. The absence of a specialized diaspore dispersal is termed atelochory.
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Book 2019, Plant Systematics (Third Edition)Michael G. Simpson
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Dermatophytes and Dermatophytoses
1990, Diagnostic Procedure in Veterinary Bacteriology and Mycology (Fifth Edition)G.R. Carter
T. gourvilii
This anthropophilic dermatophyte has not been isolated from infections in animals. Hair invasion is endothrix without fluorescence, and the perfect state has not been reported. Some workers regard it as a soil saprophyte (7), Colonies are folded, heaped, waxy, with a lavender to deep red pigmentation (13). Macro- and microaleuriospores are usually present in very small numbers.
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4943 | http://verso.mat.uam.es/~dmitry.yakubovich/EDA_2020_21/EDA_2020_21-H3.pdf | Ecuaciones Diferenciales y Aplicaciones (Grado de Matem´ aticas, UAM) Curso 2020/21 Hoja de problemas 3: La ecuaci´ on de Laplace 1. Demostrar que el l´ ımite uniforme de funciones arm´ onicas en un dominio es una funci´ on arm´ onica.
2. Sea F = [0, 1]. Dar ejemplos de familias L ⊂CR(F) tales que (a) L es acotada, pero no equicontinua; (b) L es equicontinua, pero no es acotada; (c) L es infinita, acotada y equicontinua.
Demostrar que, de hecho, toda familia finita L ⊂CR(F) es acotada y equicontinua.
Determinar, en cu´ ales de estos ejemplos, L es relativamente compacta en CR(F). En cada caso cuando no lo sea, encontrar una sucesi´ on de funciones {fn} ⊂L, que no tiene subsucesiones convergentes en la norma de CR(F).
3.∗Demostrar el Lema de Arzela-Ascoli (o una parte de ella) para el caso de un intervalo compacto en R.
4. Sea F es un conjunto compacto y convexo en RN tal que F ̸= ∅y F = int F. Sea L es una familia acotada de funciones en CR(F), que cumple la condici´ on |∇f(x)| ≤K, x ∈F, f ∈L, donde K es una constante absoluta. Demostrar que entonces la familia L es equicontinua.
5. Sea (uk)∞ k=1 una sucesi´ on creciente de funciones arm´ onicas definidas en un abierto conexo Ωtal que (uk(x))∞ k=1 converge para alg´ un punto x ∈Ω. Demostrar que la sucesi´ on converge uniformemente en cada subconjunto cerrado y acotado de Ωy que el l´ ımite es una funci´ on arm´ onica.
Nota: Esto da un m´ etodo alternativo para demostrar la existencia de soluciones del problema de Dirichlet para la ecuaci´ on de Laplace sin utilizar Arcel a-Ascoli.
Indicaci´ on. Usar la desigualdad de Harnack para demostrar que la sucesi´ on es de Cauchy en CR(K) para cualquier subconjunto compacto K de Ω.
6. Sean R1 > R0 > 0 y N ≥2. Consid´ erese el dominio Ω= {x ∈RN : R0 < |x| < R1}. Calcular la probabilidad de que una part´ ıcula Browniana que inicia su movimiento en un x ∈Ω, salga de Ωpor la parte interior de su frontera.
7. Se considera la funci´ on u(x, y, z) = x2(x2 + ay2 + 1) + y2(y2 + bz2 + 1) + z2(z2 + cx2 + 1).
Determinar, para qu´ e valores de par´ ametros a, b, c es esta funci´ on subarm´ onica en R3.
1 8. Sea Ωun dominio en RN. Demostrar lo siguiente.
(a) Si u1, u2, . . . , um son continuas en Ω, entonces v := m´ ax(u1, u2, . . . , um) es tambi´ en continua.
(b) Si u1, u2, . . . , um son subarm´ onicas en Ω, entonces v := m´ ax(u1, u2, . . . , um) es tambi´ en subarm´ onica.
(c) Si u es subarm´ onica en Ω, A = u(Ω) ⊂R es su imagen y ϕ : A →R es continua, creciente y convexa, entonces la composici´ on f ◦v es subarm´ onica en Ω.
Nota: Observar que A es un intervalo.
9.
(a) Sean Ωun dominio en RN, u una funci´ on arm´ onica en Ωy A := u(Ω) su imagen. Sea ϕ : A 7→R una funci´ on continua y convexa. Demostrar que v := ϕ ◦u es subarm´ onica.
(b) Demostrar que x 7→log |x| es subarm´ onica en RN \ {0} si N ≥2.
(c) Demostrar que v := |Du|2 es subarm´ onica si u es arm´ onica.
10. Sean Ω⊂RN un dominio acotado y u ∈C2(Ω) ∩C(Ω) una soluci´ on de ∆u = −1 en Ω, u|∂Ω= 0.
Demostrar que para todo x0 ∈Ωse tiene que u(x0) ≥ 1 2N m´ ın x∈∂Ω|x −x0|2.
11. (Dependencia continua) Sea u ∈C2(B1(0)) ∩C(B1(0)) una soluci´ on suave de −∆u = f en B1(0), u = g en ∂B1(0).
Demostrar que existe una constante C, independiente de u, tal que m´ ax B1(0) |u| ≤C( m´ ax ∂B1(0) |g| + m´ ax B1(0) |f|).
12. (a) Sea E ⊂RN abierto y u ∈C2(E) tal que ∆u = u en E. Demostrar que u no puede tener m´ aximos positivos ni m´ ınimos negativos en E.
(b) Sea E ⊂RN abierto, y c ∈C(E), c < 0 en E. Demostrar que el problema ∆u + cu = 0 en Ω, u = g en ∂Ω, tiene a lo sumo una soluci´ on u ∈C2(E) ∩C(E).
13. Hemos visto en la Teor´ ıa que la soluci´ on u ∈C2(Ω) ∩C(¯ Ω) del problema ( ∆u + cu = f en Ω, u = ϕ en ∂Ω (1) es ´ unica si c es una funci´ on continua en Ωy c(x) ≤0, ∀x ∈Ω.
Consideremos el caso del parallelep´ ıpedo: Ω= [0, ℓ1] × · · · × [0, ℓN], donde ℓ1, . . . , ℓN > 0. Aplicando el m´ etodo de separaci´ on de variables, demostrar que existe una sucesi´ on infinita de constantes positivas cj →+∞tales que en los correspondientes problemas (1) no hay unicidad, si ponemos c(x) ≡cj, x ∈Ω.
2 14. (a) Consideramos la semibola abierta B+ = {x ∈RN : x ∈B1(0), xN > 0}. Sea u ∈C2(B+) ∩C(B1(0) ∩{xN ≥0}) arm´ onica en U + con u = 0 en B1(0) ∩{xN = 0}. Dado x ∈B1(0), definimos v(x) := u(x) si xN ≥0, −u(x1, . . . , xN−1, −xN) si xN < 0.
Probar que v es arm´ onica en B1(0). Este resultado se conoce como Principio de reflexi´ on de Schwartz.
(b) Resu´ elvase el problema ∆u = 0 si x2 + y2 < 1, y > 0, u(x, 0) = 1 si −1 < x < 1, u(x, y) = 0 si x2 + y2 = 1, y > 0.
15. Sea E ⊂RN abierto. Demostrar que v ∈C(E) pertenece a σ(E) si y solo si para todo abierto E′ ⊂E tal que E′ ⊂E y toda funci´ on arm´ onica u tal que u = v en ∂E′ se tiene que v ≤u en E′.
16. Sea Ω⊂RN (N ≥2) un dominio suave y u ∈C2(Ω) tal que ∆u = 0 en Ω. Demostrar que cualquier esfera de radio suficientemente peque˜ no centrada en un punto donde u se anule contiene al menos un cero de u.
17. (a) (Principio de comparaci´ on) Sea Ω⊂RN abierto y acotado. Decimos que u ∈C2(Ω) ∩C(Ω) es subsoluci´ on del problema −∆u = f en Ω, u = g en ∂Ω, si −∆u ≤f en Ω, u ≤g en ∂Ω, y que es supersoluci´ on si −∆u ≥f en Ω, u ≥g en ∂Ω.
Demostrar que si u es subsoluci´ on del problema y v es supersoluci´ on, entonces u ≤v en Ω.
(b) Demostrar que si ϕ : R →R es una funci´ on continua estrictamente decreciente, entonces el principio de comparaci´ on para subsoluciones y supersoluciones es v´ alido para el problema no lineal −∆u = ϕ(u) en Ω, u = 0 en ∂Ω.
18. Sea N = 2 y sea Ωun dominio acotado en R2, cuya frontera es una uni´ on finita de curvas de clase C1.
Demostrar que todo Problema de Dirichlet para la ecuaci´ on de Laplace (PDL) en ¯ Ωtiene una ´ unica soluci´ on.
Indicaci´ on: Utilizando la expresi´ on para la soluci´ on fundamental de la ecuaci´ on de Laplace, demostrar que cada punto x∗de la frontera de Ωposee una funci´ on barrera.
19. Demostrar que lo mismo se cumple para todo dominio en R2, acotado por una curva de Jordan.
Indicaci´ on: Utilizar el Teorema de Jordan (es sobre las propiedades topol´ ogicas de las curvas de Jordan en el plano), 3 |
4944 | https://www.quora.com/To-find-the-derivative-of-10x-5-3x-4-2x-2-this-was-my-working-out-10x-5-3x-4-x-2x-2-20x-10-6x-8-dy-dx-200x-11-48x-9-why-is-this-wrong | To find the derivative of (10x^5 + 3x^4) / (2x^2) this was my working out: (10x^5 + 3x^4) x (2x^2) 20x^-10 + 6x^-8 dy/dx = 200x^-11 + 48x^-9 why is this wrong? - Quora
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To find the derivative of (10x^5 + 3x^4) / (2x^2) this was my working out: (10x^5 + 3x^4) x (2x^2) 20x^-10 + 6x^-8 dy/dx = 200x^-11 + 48x^-9 why is this wrong?
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It looks like there was a misunderstanding in your differentiation process. To find the derivative of the function y=10 x 5+3 x 4 2 x 2 y=10 x 5+3 x 4 2 x 2, you can simplify the expression first and then differentiate, or you can use the quotient rule directly. Let’s go through both methods.
Method 1: Simplification
Simplify the expression:
y=10 x 5+3 x 4 2 x 2=10 x 5 2 x 2+3 x 4 2 x 2=5 x 3+3 2 x 2 y=10 x 5+3 x 4 2 x 2=10 x 5 2 x 2+3 x 4 2 x 2=5 x 3+3 2 x 2
2. Differentiate:
d y d x=d d x(5 x 3)+d d x(3 2 x 2)=15 x 2+3 x d y d x=d d x(5 x 3)+d d x(3 2 x 2)=15 x 2+3 x
Method 2: Quotient Rule
If you want to use the quotient rule, remem
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It looks like there was a misunderstanding in your differentiation process. To find the derivative of the function y=10 x 5+3 x 4 2 x 2 y=10 x 5+3 x 4 2 x 2, you can simplify the expression first and then differentiate, or you can use the quotient rule directly. Let’s go through both methods.
Method 1: Simplification
Simplify the expression:
y=10 x 5+3 x 4 2 x 2=10 x 5 2 x 2+3 x 4 2 x 2=5 x 3+3 2 x 2 y=10 x 5+3 x 4 2 x 2=10 x 5 2 x 2+3 x 4 2 x 2=5 x 3+3 2 x 2
2. Differentiate:
d y d x=d d x(5 x 3)+d d x(3 2 x 2)=15 x 2+3 x d y d x=d d x(5 x 3)+d d x(3 2 x 2)=15 x 2+3 x
Method 2: Quotient Rule
If you want to use the quotient rule, remember that the quotient rule states that if you have a function y=u v y=u v, where u=10 x 5+3 x 4 u=10 x 5+3 x 4 and v=2 x 2 v=2 x 2, then:
d y d x=v d u d x−u d v d x v 2 d y d x=v d u d x−u d v d x v 2
Find d u d x d u d x and d v d x d v d x:
u=10 x 5+3 x 4 u=10 x 5+3 x 4 → d u d x=50 x 4+12 x 3 d u d x=50 x 4+12 x 3
v=2 x 2 v=2 x 2 → d v d x=4 x d v d x=4 x
Apply the quotient rule:
d y d x=(2 x 2)(50 x 4+12 x 3)−(10 x 5+3 x 4)(4 x)(2 x 2)2 d y d x=(2 x 2)(50 x 4+12 x 3)−(10 x 5+3 x 4)(4 x)(2 x 2)2
3. Simplify the numerator:
First term: 100 x 6+24 x 5 100 x 6+24 x 5
Second term: 40 x 6+12 x 5 40 x 6+12 x 5
Combine:
Numerator=(100 x 6+24 x 5)−(40 x 6+12 x 5)=60 x 6+12 x 5 Numerator=(100 x 6+24 x 5)−(40 x 6+12 x 5)=60 x 6+12 x 5
Combine everything:
d y d x=60 x 6+12 x 5 4 x 4=12 x 5(5 x+1)4 x 4=3 x(5 x+1)d y d x=60 x 6+12 x 5 4 x 4=12 x 5(5 x+1)4 x 4=3 x(5 x+1)
Final Answer
Thus, the derivative of y=10 x 5+3 x 4 2 x 2 y=10 x 5+3 x 4 2 x 2 is:
d y d x=15 x 2+3 x d y d x=15 x 2+3 x
or, simplified further, d y d x=3 x(5 x+1)d y d x=3 x(5 x+1).
Your original calculation seems to have some errors in applying the rules of differentiation and algebraic manipulation.
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Audley Willacey
B.Sc. in Mechanical Engineering, University of the West Indies, St. Augustine (Graduated 1976) · Author has 5K answers and 1.5M answer views
·4y
y = (10x^5 + 3x^4)/2x^2
(1)You can simplify then differentiate.
ie. y =(10x^5)/2x^2 + (3x^4)/2x^2
y = 5x^3 + (3/2)x^2
y' = 3.5x^2 + 2(3/2)x
y' = 15x^2 + 3x.
(2)If you are required to use the quotient rule
y' = (v.u' - u.v')/v^2
u = 10x^5 + 3x^4 , v = 2x^2
u' = 50x^4 + 12x^3 , v' = 4x , v^2 = 4x^4
You can see how much more involved it gets.
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Bo Jacoby
· Author has 176 answers and 106K answer views
·4y
y=10 x 5+3 x 4 2 x 2=5 x 3+3 2 x 2 y=10 x 5+3 x 4 2 x 2=5 x 3+3 2 x 2
d y d x=15 x 2+3 x d y d x=15 x 2+3 x
This is right. It is hard to tell why the other computation was wrong.
Upvote ·
Ernest Leung
B.Sc. (Hons.) in Chemistry Honors&Mathematics, The Chinese University of Hong Kong · Author has 11.9K answers and 5.8M answer views
·Jul 6
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What is the answer to (4x^5 + 6x^4 + 5x^2 - x - 10) / (2x^2 + 3) and how do you find it?
What is the answer to (4x⁵ + 6x⁴ + 5x² - x - 10) / (2x² + 3) and how do you find it?
The answer is as follows.
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What is the answer to (4x⁵ + 6x⁴ + 5x² - x - 10) / (2x² + 3) and how do you find it?
The answer is as follows.
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Ernest Leung
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How do i solve 2 x 4−13 x 3+12 x 2+17 x−10=0 2 x 4−13 x 3+12 x 2+17 x−10=0 ?
How do i solve 2x⁴ − 13x³ + 12x² + 17x − 10 = 0 ?
Let f(x) = 2x⁴ − 13x³ + 12x² + 17x − 10
f(-1) = 2(-1)⁴ − 13(-1)³ + 12(-1)² + 17(-1) − 10 = 0
f(2) = 2(2)⁴ − 13(2)³ + 12(2)² + 17(2) − 10 = 0
By Factor Theorem, (x + 1) and (x - 2) are the factors of f(x).
2x⁴ − 13x³ + 12x² + 17x − 10 = 0
(2x⁴ + 2x³) + (-15x³ - 15x²) + (27x² + 27x) + (-10x - 10) = 0
2x³(x + 1) - 15x²(x + 1) + 27x(x + 1) - 10(x + 1) = 0
(x + 1)[2x³ - 15x² + 27x - 10] = 0
(x + 1)[(2x³ - 4x²) + (-11x² + 22x) + (5x - 10)] = 0
(x + 1)[2x²(x - 2) - 11x(x - 2) + 5(x - 2)] = 0
(x + 1)(x - 2)[2x² - 11x + 5] = 0
(x + 1)(x
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How do i solve 2x⁴ − 13x³ + 12x² + 17x − 10 = 0 ?
Let f(x) = 2x⁴ − 13x³ + 12x² + 17x − 10
f(-1) = 2(-1)⁴ − 13(-1)³ + 12(-1)² + 17(-1) − 10 = 0
f(2) = 2(2)⁴ − 13(2)³ + 12(2)² + 17(2) − 10 = 0
By Factor Theorem, (x + 1) and (x - 2) are the factors of f(x).
2x⁴ − 13x³ + 12x² + 17x − 10 = 0
(2x⁴ + 2x³) + (-15x³ - 15x²) + (27x² + 27x) + (-10x - 10) = 0
2x³(x + 1) - 15x²(x + 1) + 27x(x + 1) - 10(x + 1) = 0
(x + 1)[2x³ - 15x² + 27x - 10] = 0
(x + 1)[(2x³ - 4x²) + (-11x² + 22x) + (5x - 10)] = 0
(x + 1)[2x²(x - 2) - 11x(x - 2) + 5(x - 2)] = 0
(x + 1)(x - 2)[2x² - 11x + 5] = 0
(x + 1)(x - 2)(2x - 1)(x - 5) = 0
x = -1, 2, 1/2, 5
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Will He
Lives in Michigan (2008–present) · Author has 130 answers and 24.6K answer views
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How do i solve 2 x 4−13 x 3+12 x 2+17 x−10=0 2 x 4−13 x 3+12 x 2+17 x−10=0 ?
Here, simple guess-and-checking via RRT is the best method. One should rather easily find that x=−1,1 2,2,5.x=−1,1 2,2,5.
For a purely “algebraic” method, one may try depressing this quartic and then apply Descartes’ solution. What exactly is Descartes’ solution? Here is an explanation courtesy of Wikipedia:
I don’t know about you, but I’d rather not!
Continue Reading
Here, simple guess-and-checking via RRT is the best method. One should rather easily find that x=−1,1 2,2,5.x=−1,1 2,2,5.
For a purely “algebraic” method, one may try depressing this quartic and then apply Descartes’ solution. What exactly is Descartes’ solution? Here is an explanation courtesy of Wikipedia:
I don’t know about you, but I’d rather not!
Upvote ·
9 2
Prasenjit Wakode
Maths is hobby and engineering is profession · Upvoted by
Abu Zafar
, M.Sc Naval Architecture & Mathematics, Norwegian University of Science and Technology (2017) · Author has 693 answers and 1.6M answer views
·Updated 6y
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How do I integrate 3 x 3−x 2+2 x−4√x 2−3 x+2 d x 3 x 3−x 2+2 x−4 x 2−3 x+2 d x?
I have recently seen this integral in a message on whats-app and in this message, when the limits of this integral are from 0 0 to 1 1 , it evaluates to an ATM PIN no.
So, A T M−P I N=∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x A T M−P I N=∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x
Given that, I=∫3 x 3−x 2+2 x−4√x 2–3 x+2 d x I=∫3 x 3−x 2+2 x−4 x 2–3 x+2 d x
Let, f(x)=3 x 3−x 2+2 x−4√x 2–3 x+2 f(x)=3 x 3−x 2+2 x−4 x 2–3 x+2
f(x)=(x−1)(3 x 2+2 x+4)√(x−1)(x−2)f(x)=(x−1)(3 x 2+2 x+4)(x−1)(x−2)
f(x)=\frac{(x-1)(3x f(x)=\frac{(x-1)(3x
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I have recently seen this integral in a message on whats-app and in this message, when the limits of this integral are from 0 0 to 1 1 , it evaluates to an ATM PIN no.
So, A T M−P I N=∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x A T M−P I N=∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x
Given that, I=∫3 x 3−x 2+2 x−4√x 2–3 x+2 d x I=∫3 x 3−x 2+2 x−4 x 2–3 x+2 d x
Let, f(x)=3 x 3−x 2+2 x−4√x 2–3 x+2 f(x)=3 x 3−x 2+2 x−4 x 2–3 x+2
f(x)=(x−1)(3 x 2+2 x+4)√(x−1)(x−2)f(x)=(x−1)(3 x 2+2 x+4)(x−1)(x−2)
f(x)=(x−1)(3 x 2+2 x+4)√(x−1)(x−2)(x−1)(x−2)f(x)=(x−1)(3 x 2+2 x+4)(x−1)(x−2)(x−1)(x−2)
f(x)=(3 x 2+2 x+4)√(x−1)(x−2)x−2 f(x)=(3 x 2+2 x+4)(x−1)(x−2)x−2
f(x)=(3 x+8+20 x−2)√(x−1)(x−2)f(x)=(3 x+8+20 x−2)(x−1)(x−2)
f(x)=3 x√(x−1)(x−2)+8√(x−1)(x−2)+20√(x−1)(x−2)x−2 f(x)=3 x(x−1)(x−2)+8(x−1)(x−2)+20(x−1)(x−2)x−2
f(x)=3 x√x 2–3 x+2+8√x 2–3 x+2+20√(x−1)(x−2)x−2 f(x)=3 x x 2–3 x+2+8 x 2–3 x+2+20(x−1)(x−2)x−2
f(x)=3 2(2 x−3)√x 2–3 x+2+9 2√x 2–3 x+2+8√x 2–3 x+2+20√(x−1)(x−2)x−2 f(x)=3 2(2 x−3)x 2–3 x+2+9 2 x 2–3 x+2+8 x 2–3 x+2+20(x−1)(x−2)x−2
f(x)=3 2(2 x−3)√x 2–3 x+2+25 2√x 2–3 x+2+20(x−1)(x−2)(x−2)√(x−1)(x−2)f(x)=3 2(2 x−3)x 2–3 x+2+25 2 x 2–3 x+2+20(x−1)(x−2)(x−2)(x−1)(x−2)
f(x)=3 2(2 x−3)√x 2–3 x+2+25 2√x 2–3 x+2+20(x−1)√(x−1)(x−2)f(x)=3 2(2 x−3)x 2–3 x+2+25 2 x 2–3 x+2+20(x−1)(x−1)(x−2)
f(x)=3 2(2 x−3)√x 2–3 x+2+25 2√x 2–3 x+2+20 x√(x−1)(x−2)−20√(x−1)(x−2)f(x)=3 2(2 x−3)x 2–3 x+2+25 2 x 2–3 x+2+20 x(x−1)(x−2)−20(x−1)(x−2)
f(x)=3 2(2 x−3)√x 2–3 x+2+25 2√x 2–3 x+2+20 x√x 2–3 x+2−20√x 2–3 x+2 f(x)=3 2(2 x−3)x 2–3 x+2+25 2 x 2–3 x+2+20 x x 2–3 x+2−20 x 2–3 x+2
f(x)=3 2(2 x−3)√x 2–3 x+2+25 2√x 2–3 x+2+10(2 x−3+3)√x 2–3 x+2−20√x 2–3 x+2 f(x)=3 2(2 x−3)x 2–3 x+2+25 2 x 2–3 x+2+10(2 x−3+3)x 2–3 x+2−20 x 2–3 x+2
f(x)=3 2(2 x−3)√x 2–3 x+2+25 2√x 2–3 x+2+10(2 x−3)√x 2–3 x+2+10×3√x 2–3 x+2−20√x 2–3 x+2 f(x)=3 2(2 x−3)x 2–3 x+2+25 2 x 2–3 x+2+10(2 x−3)x 2–3 x+2+10×3 x 2–3 x+2−20 x 2–3 x+2
f(x)=3 2(2 x−3)√x 2–3 x+2+25 2√x 2–3 x+2+10(2 x−3)√x 2–3 x+2+10√x 2–3 x+2 f(x)=3 2(2 x−3)x 2–3 x+2+25 2 x 2–3 x+2+10(2 x−3)x 2–3 x+2+10 x 2–3 x+2
f(x)=(3 2√x 2–3 x+2+10√x 2–3 x+2)(2 x−3)+25 2√x 2–3 x+2+10√x 2–3 x+2 f(x)=(3 2 x 2–3 x+2+10 x 2–3 x+2)(2 x−3)+25 2 x 2–3 x+2+10 x 2–3 x+2
f(x)=(3 2√x 2–3 x+2+10√x 2–3 x+2)(2 x−3)+25 2√(x−3/2)2−9/4+2+10√(x−3/2)2−9/4+2 f(x)=(3 2 x 2–3 x+2+10 x 2–3 x+2)(2 x−3)+25 2(x−3/2)2−9/4+2+10(x−3/2)2−9/4+2
f(x)=(3 2√x 2–3 x+2+10√x 2–3 x+2)(2 x−3)+25 2√(x−3/2)2−1/4+10√(x−3/2)2−1/4 f(x)=(3 2 x 2–3 x+2+10 x 2–3 x+2)(2 x−3)+25 2(x−3/2)2−1/4+10(x−3/2)2−1/4
Now, the given integral is, I=∫3 x 3−x 2+2 x−4√x 2–3 x+2 d x I=∫3 x 3−x 2+2 x−4 x 2–3 x+2 d x
I=∫f(x)d x I=∫f(x)d x
I=∫((3 2√x 2–3 x+2+10√x 2–3 x+2)(2 x−3)+(25√(x−3/2)2−1/4 2+10√(x−3/2)2−1/4))d x I=∫((3 2 x 2–3 x+2+10 x 2–3 x+2)(2 x−3)+(25(x−3/2)2−1/4 2+10(x−3/2)2−1/4))d x
I=∫(3 2√x 2–3 x+2+10√x 2–3 x+2)(2 x−3)d x+∫(25√(x−3/2)2−1/4 2+10√(x−3/2)2−1/4)d x I=∫(3 2 x 2–3 x+2+10 x 2–3 x+2)(2 x−3)d x+∫(25(x−3/2)2−1/4 2+10(x−3/2)2−1/4)d x …..(1)
In first integral above, substitute, x 2–3 x+2=p x 2–3 x+2=p
Hence, (2 x−3)d x=d p(2 x−3)d x=d p
And, in second integral substitute, x−3 2=sec q 2 x−3 2=secq 2
So, d x=sec q tan q 2 d q d x=secq tanq 2 d q
After these substitutions, integral (1) above becomes,
I=∫(3 2√p+10√p)d p+∫(25√sec 2 q 4−1/4 2+10√sec 2 q 4−1/4)sec q tan q 2 d q I=∫(3 2 p+10 p)d p+∫(25 sec 2q 4−1/4 2+10 sec 2q 4−1/4)secq tanq 2 d q
I=p 3/2+20√p+∫(25√sec 2 q−1 8+10√sec 2 q−1)sec q tan q d q I=p 3/2+20 p+∫(25 sec 2q−1 8+10 sec 2q−1)secq tanq d q
I=(20+p)√p+∫(25 tan q 8+10 tan q)sec q tan q d q I=(20+p)p+∫(25 tanq 8+10 tanq)secq tanq d q
I=(20+p)√p+25 8∫sec q tan 2 q d q+10∫sec q d q I=(20+p)p+25 8∫secq tan 2q d q+10∫secq d q
I=(20+p)√p+25 8∫sec q(sec 2 q−1)d q+10∫sec q d q I=(20+p)p+25 8∫secq(sec 2q−1)d q+10∫secq d q
I=(20+p)√p+25 8∫sec 3 q d q+55 8∫sec q d q I=(20+p)p+25 8∫sec 3q d q+55 8∫secq d q......(2)
Now, from reference to one of my answer (Reference: Prasenjit Wakode's answer to How do I evaluate [ ]∫√4 x 2+1 d x∫4 x 2+1 d x? [ ] ), it can be evaluated that, ∫sec 3 q d q=sec q tan q 2+1 2∫sec q d q∫sec 3q d q=secq tanq 2+1 2∫secq d q
Substituting this value in integral(2) , it becomes,
I=(20+p)√p+25 8(sec q tan q 2+1 2∫sec q d q)+55 8∫sec q d q I=(20+p)p+25 8(secq tanq 2+1 2∫secq d q)+55 8∫secq d q
I=(20+p)√p+25 sec q tan q 16+135 16∫sec q d q I=(20+p)p+25 secq tanq 16+135 16∫secq d q......(3)
Now, it can be proved easily that, ∫sec q d q=1 2 log(1+sin q 1−sin q)∫secq d q=1 2 log(1+sinq 1−sinq) (Reference: Prasenjit Wakode's answer to How do you integrate secx? [ ] ) . (This is a standard integral by the way.)
So, finally, integral (3) above becomes,
I=(20+p)√p+25 sec q tan q 16+135 32 log(1+sin q 1−sin q)I=(20+p)p+25 secq tanq 16+135 32 log(1+sinq 1−sinq)
This is the complete indefinite evaluation of the given integral.
To obtain the expressions in terms of ‘x x’ , consider our original substitutions as, x 2–3 x+2=p x 2–3 x+2=p and x−3 2=sec q 2 x−3 2=secq 2
Limits of integration for p :
So, when x=0 x=0 , then p=2 p=2 and when x=1 x=1 , then p=0 p=0
Limits of integration for q :
Upper Limit:
When x=1 x=1 , then q=c o s−1(−1)q=c o s−1(−1)
Hence, in this case we would get, cos(q)=−1 cos(q)=−1 , sin(q)=0 sin(q)=0 , sec q=−1 secq=−1 , tan(q)=0 tan(q)=0
Lower Limit:
When x=0 x=0 , then we get, cos(q)=−1 3 cos(q)=−1 3
Note here that, there will arise two cases now.
We get cos(q)=−1 3 cos(q)=−1 3 when sin(q)=2√2 3 sin(q)=2 2 3 or sin(q)=−2√2 3 sin(q)=−2 2 3
Case 1: When, cos(q)=−1 3 cos(q)=−1 3 and sin(q)=2√2 3 sin(q)=2 2 3
Hence, other trigonometric ratios in this case would be as follows:
sec q=−3 secq=−3 , tan(q)=−2√2 tan(q)=−2 2
Evaluation of the complete integral in this case will be as follows:
∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x=[(20+p)√p]p=0 p=2+[25 sec q tan q 16+135 32 log(1+sin q 1−sin q)]q=c o s−1(−1)q=cos−1(−1 3)∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x=[(20+p)p]p=2 p=0+[25 secq tanq 16+135 32 log(1+sinq 1−sinq)]q=cos−1(−1 3)q=c o s−1(−1)
∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x=[(20+0)√0−(20+2)√2]+[25×(−1)×(0)16+135 32 log(1+0 1−0)]−[25×(−3)×(−2√2)16+135 32 log(1+2√2 3 1−2√2 3)]∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x=[(20+0)0−(20+2)2]+[25×(−1)×(0)16+135 32 log(1+0 1−0)]−[25×(−3)×(−2 2)16+135 32 log(1+2 2 3 1−2 2 3)]
∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x=[0−22√2]+[0+0]−[75√2 8+135 32 log(3+2√2 3−2√2)]∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x=[0−22 2]+[0+0]−[75 2 8+135 32 log(3+2 2 3−2 2)]
∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x=−22√2−75√2 8−135 32 log(3+2√2 3−2√2)∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x=−22 2−75 2 8−135 32 log(3+2 2 3−2 2)
∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x=−251√2 8−135 32 log(3+2√2 3−2√2)∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x=−251 2 8−135 32 log(3+2 2 3−2 2)
∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x=−251√2 8−135 32 log((3+2√2)(3+2√2)(3−2√2)(3+2√2))∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x=−251 2 8−135 32 log((3+2 2)(3+2 2)(3−2 2)(3+2 2))
∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x=−251√2 8−135 32 log((3+2√2)2 9−8)∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x=−251 2 8−135 32 log((3+2 2)2 9−8)
∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x=−251√2 8−135 16 log(3+2√2)∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x=−251 2 8−135 16 log(3+2 2)
∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x=−251√2 8−135 16 log((1+√2)2)∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x=−251 2 8−135 16 log((1+2)2)
∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x=−251√2 8−135 8 log(1+√2)∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x=−251 2 8−135 8 log(1+2)
∫1 0 3 x 3−x 2+2 x−4√x 2–3 x+2 d x=−59.244129800...∫0 1 3 x 3−x 2+2 x−4 x 2–3 x+2 d x=−59.244129800...
Case 2: When, cos(q)=−1 3 cos(q)=−1 3 and sin(q)=−2√2 3 sin(q)=−2 2 3
Hence, other trigonometric ratios in this case would be as follows:
sec q=−3 secq=−3 , tan(q)=2√2 tan(q)=2 2
Evaluation of the complete integr...
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An Indonesian who fell in love with mathematics · Upvoted by
Michael Jørgensen
, PhD in mathematics · Author has 2.2K answers and 3.4M answer views
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Related
How do you solve this equation x 5+10 x 3+20 x−4=0 x 5+10 x 3+20 x−4=0?
Unfortunately, it has been proven the quintic formula (basically the quadratic formula, but for polynomials of degree 5) doesn’t exist, so finding the roots analytically is going to be difficult, leaving us with finding the roots numerically (finding the approximate value of the roots) as an easy way to solve it.
One thing that you can do is to graph the equation first.
As you can see, it only has one root that is between 0 and 0.5.
There are many ways to solve it numerically, but one way that I like the most is the Newton—Raphson Method. If you want to know what it is and how it works, you can w
Continue Reading
Unfortunately, it has been proven the quintic formula (basically the quadratic formula, but for polynomials of degree 5) doesn’t exist, so finding the roots analytically is going to be difficult, leaving us with finding the roots numerically (finding the approximate value of the roots) as an easy way to solve it.
One thing that you can do is to graph the equation first.
As you can see, it only has one root that is between 0 and 0.5.
There are many ways to solve it numerically, but one way that I like the most is the Newton—Raphson Method. If you want to know what it is and how it works, you can watch 3Blue1Brown’s video here between 5:55–9:42 (I highly recommend you watch the rest of the video, but it’s unrelated to the actual question right now.)
So, now we know that we can use the following formula,
x n+1=x n−f(x n)f′(x n)x n+1=x n−f(x n)f′(x n)
to numerically approximate the root in question. The function itself is f(x)=x 5+10 x 3+20 x−4 f(x)=x 5+10 x 3+20 x−4, and if you already learned calculus, you’ll see that f′(x)=5 x 4+30 x 2+20 f′(x)=5 x 4+30 x 2+20. Based on this, the formula becomes:
x n+1=x n−x 5 n+10 x 3 n+20 x n−4 5 x 4 n+30 x 2 n+20 x n+1=x n−x n 5+10 x n 3+20 x n−4 5 x n 4+30 x n 2+20
As you can see from the video, it’s best if we take a starting value close to the root to prevent it from wandering around aimlessly, so let’s take x 0=0 x 0=0.
x 0=0 x 0=0
x 1=0.2 x 1=0.2
x 2=0.196212749906 x 2=0.196212749906
x 3=0.196208655742 x 3=0.196208655742
x 4=0.196208655738 x 4=0.196208655738
x 5=0.196208655738 x 5=0.196208655738
As you can see, iterating the formula 5 times already gives us an approximate value of the root that is accurate up to 12 digits after the decimal point.
Of course, this is assuming that you are allowed to use a calculator.
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Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983) · Author has 3.9K answers and 4M answer views
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How do you solve (3x + 5) (2x - 4) ^2 = 56?
(3 x+5)(2 x−4)2=56(3 x+5)(2 x−4)2=56
(3 x+5)(4 x 2−16 x+16)=56(3 x+5)(4 x 2−16 x+16)=56
12 x 3−48 x 2+48 x+20 x 2−80 x+80−56=0 12 x 3−48 x 2+48 x+20 x 2−80 x+80−56=0
12 x 3−28 x 2−32 x+24=0 12 x 3−28 x 2−32 x+24=0
3 x 3−7 x 2−8 x+6=0 3 x 3−7 x 2−8 x+6=0
(x−3)(3 x 2+2 x−2)=0(x−3)(3 x 2+2 x−2)=0
Either:
x−3=0 x−3=0
x=3 x=3
or:
3 x 2+2 x−2=0 3 x 2+2 x−2=0
x=−2±√2 2−4(3)(−2)2(3)x=−2±2 2−4(3)(−2)2(3)
x=−2±2√7 6 x=−2±2 7 6
x=−1±√7 3 x=−1±7 3
Therefore, we find:
x=3 x=3, x=−1−√7 3 x=−1−7 3, x=−1+√7 3 x=−1+7 3
The function y=3 x 3−7 x 2−8 x+6 y=3 x 3−7 x 2−8 x+6 intersects the x x-axis, y=0 y=0, at the solutions to the given equation. A plot looks like this:
Continue Reading
(3 x+5)(2 x−4)2=56(3 x+5)(2 x−4)2=56
(3 x+5)(4 x 2−16 x+16)=56(3 x+5)(4 x 2−16 x+16)=56
12 x 3−48 x 2+48 x+20 x 2−80 x+80−56=0 12 x 3−48 x 2+48 x+20 x 2−80 x+80−56=0
12 x 3−28 x 2−32 x+24=0 12 x 3−28 x 2−32 x+24=0
3 x 3−7 x 2−8 x+6=0 3 x 3−7 x 2−8 x+6=0
(x−3)(3 x 2+2 x−2)=0(x−3)(3 x 2+2 x−2)=0
Either:
x−3=0 x−3=0
x=3 x=3
or:
3 x 2+2 x−2=0 3 x 2+2 x−2=0
x=−2±√2 2−4(3)(−2)2(3)x=−2±2 2−4(3)(−2)2(3)
x=−2±2√7 6 x=−2±2 7 6
x=−1±√7 3 x=−1±7 3
Therefore, we find:
x=3 x=3, x=−1−√7 3 x=−1−7 3, x=−1+√7 3 x=−1+7 3
The function y=3 x 3−7 x 2−8 x+6 y=3 x 3−7 x 2−8 x+6 intersects the x x-axis, y=0 y=0, at the solutions to the given equation. A plot looks like this:
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What is the result of 2x dy/dx=10x^3 y^5+y?
y 4=−x 2 4 x 5+c y 4=−x 2 4 x 5+c
Continue Reading
y 4=−x 2 4 x 5+c y 4=−x 2 4 x 5+c
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Vincent Tandya
A high school national olympiad medalist in mathematics · Upvoted by
Robby Goetschalckx
, Computer scientist for 11+ years and passionate about math since childhood. and
Michael Jørgensen
, PhD in mathematics
·Updated 8y
Related
How do you solve 6 x 4+x 3+10 x 2+6=0 6 x 4+x 3+10 x 2+6=0?
Edit: The question originally asked to solve 6 x 4+x 3+10 x 2+x+6=0 6 x 4+x 3+10 x 2+x+6=0, which is, in my opinion, a more interesting problem compared to the edited one:6 x 4+x 3+10 x 2+6=0 6 x 4+x 3+10 x 2+6=0. The solution (well, actually I can only show that it has no real solution) will be included below the original solution.
This is perhaps a more understandable solution, although it may seem like “Woah, I would never think of that!”.
Let a a be a solution. Hence, we have 6 a 4+a 3+10 a 2+a+6=0 6 a 4+a 3+10 a 2+a+6=0. It is obvious that a a is not 0 0, so we can divide both hands by a 2 a 2, giving us 6a^2+a+10+\frac{1}{a}+\frac{6}{6a^2+a+10+\frac{1}{a}+\frac{6}{
Continue Reading
Edit: The question originally asked to solve 6 x 4+x 3+10 x 2+x+6=0 6 x 4+x 3+10 x 2+x+6=0, which is, in my opinion, a more interesting problem compared to the edited one:6 x 4+x 3+10 x 2+6=0 6 x 4+x 3+10 x 2+6=0. The solution (well, actually I can only show that it has no real solution) will be included below the original solution.
This is perhaps a more understandable solution, although it may seem like “Woah, I would never think of that!”.
Let a a be a solution. Hence, we have 6 a 4+a 3+10 a 2+a+6=0 6 a 4+a 3+10 a 2+a+6=0. It is obvious that a a is not 0 0, so we can divide both hands by a 2 a 2, giving us 6 a 2+a+10+1 a+6 a 2=0 6 a 2+a+10+1 a+6 a 2=0.
Note that we can rearrange the terms to get 6(a 2+1 a 2)+(a+1 a)+10=0 6(a 2+1 a 2)+(a+1 a)+10=0. Letting b=a+1 a b=a+1 a, and realizing that a 2+1 a 2=(a+1 a)2−2=b 2−2 a 2+1 a 2=(a+1 a)2−2=b 2−2, we have a nicer equation: 6(b 2−2)+b+10=0 6(b 2−2)+b+10=0.
This can be simplified to 6 b 2+b−2=0 6 b 2+b−2=0, which is much easier to factorize. With ease, we find (3 b+2)(2 b−1)=0(3 b+2)(2 b−1)=0, or, after substituting the value of b b, (3 a+2+3 a)(2 a−1+2 a)=0(3 a+2+3 a)(2 a−1+2 a)=0.
If necessary, multiplying both hands by a a may make it look more familiar (and gives the factorization of the polynomial: (3 x 2+2 x+3)(2 x 2−x+2)(3 x 2+2 x+3)(2 x 2−x+2), a not-so-good-looking one). Finding the exact value is left as an exerc- kidding.
This trick can be applied to similar polynomials as well. That is, a polynomial with degree 4 4 and “mirroring” coefficients. When I first encountered it (the problem was asking for positive solutions of x x in the equation x 4+x 3+10 x 2+x+1=0 x 4+x 3+10 x 2+x+1=0, if I’m not mistaken), I was amazed by how nice the solution was.
(Solution for 6 x 4+x 3+10 x 2+6=0 6 x 4+x 3+10 x 2+6=0.)
There is no real solution for the above equation.
Proof: The equation above can be written as x 2(6 x 2+x+10)+6=0 x 2(6 x 2+x+10)+6=0.
Notice that 6 x 2+x+10 6 x 2+x+10 is never negative for any real number x x, because 6 x 2+x+10=5 x 2+(x 2+x+1 4)+39 4=5 x 2+(x+1 2)2+39 4 6 x 2+x+10=5 x 2+(x 2+x+1 4)+39 4=5 x 2+(x+1 2)2+39 4. The terms in this modified form is non-negative (as no square is negative), so their sum is non-negative as well.
Back to the original equation. Since x 2≥0 x 2≥0 and 6 x 2+x+10≥0 6 x 2+x+10≥0, then x 2(6 x 2+x+10)≥0 x 2(6 x 2+x+10)≥0. This gives us x 2(6 x 2+x+10)+6≥6 x 2(6 x 2+x+10)+6≥6, so there is indeed no real solution of x x.
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Ankit Sharma
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How do I integrate 3 x 3−x 2+2 x−4√x 2−3 x+2 d x 3 x 3−x 2+2 x−4 x 2−3 x+2 d x?
Divide the numerator by the term inside the square root in denominator and write it as follows:
Numerator/Denominator = Quotient + Remainder/Denominator
And then integrate both sides. Your question is quite easier to solve now and you need to know basic Integration skills to solve the question.
Continue Reading
Divide the numerator by the term inside the square root in denominator and write it as follows:
Numerator/Denominator = Quotient + Remainder/Denominator
And then integrate both sides. Your question is quite easier to solve now and you need to know basic Integration skills to solve the question.
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Bob Collier
Former EE Designed Specialized Computers for 33 Years. · Upvoted by
Dan Grubb
, Ph. D. Mathematics, Kansas State University (1986) · Author has 3.1K answers and 1.6M answer views
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Related
How do you compute for the derivative of the function below for x = 1.Unfortunatelyy=35–√x+6x2−23x3+8?
There is no trick or difficulty here.
All the terms are of the form aX^n.
What is the derivitave of aX^n? If you don’t know that, you don’t know enough to ask the question or understand the answer.
If you do know how to do that, do it 5 times and add them. That’s the answer. Since you are new at this, I’ll remind you that
35 = 35X^0 and
sqrt(X) = X^(1/2) or X^(0.5) if that makes it easier for you.
At last look, I see you have 35 & 8 - both positive. If you really believe that, I’d suggest you add them first so you have only one ‘constant’ to differentiate. There is nothing to be gained by keeping th
Continue Reading
There is no trick or difficulty here.
All the terms are of the form aX^n.
What is the derivitave of aX^n? If you don’t know that, you don’t know enough to ask the question or understand the answer.
If you do know how to do that, do it 5 times and add them. That’s the answer. Since you are new at this, I’ll remind you that
35 = 35X^0 and
sqrt(X) = X^(1/2) or X^(0.5) if that makes it easier for you.
At last look, I see you have 35 & 8 - both positive. If you really believe that, I’d suggest you add them first so you have only one ‘constant’ to differentiate. There is nothing to be gained by keeping them separate.
AFTER you have the derivative, then stick 1 in for X and do the remaining arithmetic.
That’s the answer you are seeking.
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4945 | https://high-python-ext-3-algorithms.readthedocs.io/ko/latest/chapter3.html | chapter 3: Backtrack — Python Algorithms DEV documentation
Python Algorithms
latest
chapter 1: Sorting
chapter 2: Arrays
chapter 3: Backtrack
3.1 add operators
3.2 anagram
3.3 array sum combination
3.4 combination sum
3.5 factor combinations
3.6 find words
3.7 generate abbreviations
3.8 generate parenthesis
3.9 letter combination
3.10 palindrome partioning
3.11 pattern match
3.12 Permute Unique
3.13 Permute
3.14 subsets unique
3.15 subsets
chapter 4: bfs
chapter 5: Bit manipulation
chapter 6: Calculator
chapter 7: dfs
chapter 8: Dynamic Programming
chapter 9: graph
chapter 10: Heap
chapter 11: Linked list
chapter 12: Map
chapter 13: Maths
chapter 14: Matrix
chapter 15:ml
chapter 16: queues
chapter 17: search
chapter 18: set
chapter 19: stack
chapter 20: Strings
chapter 21: tree
chapter 22: Union-find
chapter 23: Unix
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Python Algorithms
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chapter 3: Backtrack
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chapter 3: Backtrack¶
3.1 add operators¶
Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +, -, or between the digits so they prevuate to the target value.
Examples: "123", 6 -> ["1+2+3", "123"] "232", 8 -> ["23+2", "2+32"] "105", 5 -> ["10+5","10-5"] "00", 0 -> ["0+0", "0-0", "00"] "3456237490", 9191 -> []
def add_operators(num, target):
"""
:type num: str
:type target: int
:rtype: List[str]
"""
def dfs(res, path, num, target, pos, prev, multed):
if pos == len(num):
if target == prev:
res.append(path)
return
for i in range(pos, len(num)):
if i != pos and num[pos] == '0': # all digits have to be used
break
cur = int(num[pos:i+1])
if pos == 0:
dfs(res, path + str(cur), num, target, i+1, cur, cur)
else:
dfs(res, path + "+" + str(cur), num, target,
i+1, prev + cur, cur)
dfs(res, path + "-" + str(cur), num, target,
i+1, prev - cur, -cur)
dfs(res, path + "" + str(cur), num, target,
i+1, prev - multed + multed cur, multed cur)
res = []
if not num:
return res
dfs(res, "", num, target, 0, 0, 0)
return res
3.2 anagram¶
def anagram(s1, s2):
c1 = 26
c2 = 26
for c in s1:
pos = ord(c)-ord('a')
c1[pos] = c1[pos] + 1
for c in s2:
pos = ord(c)-ord('a')
c2[pos] = c2[pos] + 1
return c1 == c2
3.3 array sum combination¶
WAP to take one element from each of the array add it to the target sum. Print all those three-element combinations.
/ A = [1, 2, 3, 3] B = [2, 3, 3, 4] C = [2, 3, 3, 4] target = 7 /
Result: [[1, 2, 4], [1, 3, 3], [1, 3, 3], [1, 3, 3], [1, 3, 3], [1, 4, 2],
[2, 2, 3], [2, 2, 3], [2, 3, 2], [2, 3, 2], [3, 2, 2], [3, 2, 2]]
import itertools
from functools import partial
def array_sum_combinations(A, B, C, target):
def over(constructed_sofar):
sum = 0
to_stop, reached_target = False, False
for elem in constructed_sofar:
sum += elem
if sum >= target or len(constructed_sofar) >= 3:
to_stop = True
if sum == target and 3 == len(constructed_sofar):
reached_target = True
return to_stop, reached_target
def construct_candidates(constructed_sofar):
array = A
if 1 == len(constructed_sofar):
array = B
elif 2 == len(constructed_sofar):
array = C
return array
def backtrack(constructed_sofar=[], res=[]):
to_stop, reached_target = over(constructed_sofar)
if to_stop:
if reached_target:
res.append(constructed_sofar)
return
candidates = construct_candidates(constructed_sofar)
for candidate in candidates:
constructed_sofar.append(candidate)
backtrack(constructed_sofar[:], res)
constructed_sofar.pop()
res = []
backtrack([], res)
return res
def unique_array_sum_combinations(A, B, C, target):
"""
1. Sort all the arrays - a,b,c. - This improves average time complexity.
2. If c[i] < Sum, then look for Sum - c[i] in array a and b.
When pair found, insert c[i], a[j] & b[k] into the result list.
This can be done in O(n).
3. Keep on doing the above procedure while going through complete c array.
Complexity: O(n(m+p))
"""
def check_sum(n, nums):
if sum(x for x in nums) == n:
return (True, nums)
else:
return (False, nums)
pro = itertools.product(A, B, C)
func = partial(check_sum, target)
sums = list(itertools.starmap(func, pro))
res = set()
for s in sums:
if s is True and s not in res:
res.add(s)
return list(res)
3.4 combination sum¶
Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note: All numbers (including target) will be positive integers. The solution set must not contain duplicate combinations. For example, given candidate set [2, 3, 6, 7] and target 7, A solution set is: [
, [2, 2, 3]
]
def combination_sum(candidates, target):
def dfs(nums, target, index, path, res):
if target < 0:
return # backtracking
if target == 0:
res.append(path)
return
for i in range(index, len(nums)):
dfs(nums, target-nums[i], i, path+[nums[i]], res)
res = []
candidates.sort()
dfs(candidates, target, 0, [], res)
return res
3.5 factor combinations¶
Numbers can be regarded as product of its factors. For example,
8 = 2 x 2 x 2;= 2 x 4.
Write a function that takes an integer n and return all possible combinations of its factors.
Note: You may assume that n is always positive. Factors should be greater than 1 and less than n. Examples: input: 1 output: [] input: 37 output: [] input: 12 output: [
[2, 6], [2, 2, 3], [3, 4]
] input: 32 output: [
[2, 16], [2, 2, 8], [2, 2, 2, 4], [2, 2, 2, 2, 2], [2, 4, 4], [4, 8]
]
Iterative:
def get_factors(n):
todo, combis = [(n, 2, [])], []
while todo:
n, i, combi = todo.pop()
while i i <= n:
if n % i == 0:
combis.append(combi + [i, n//i])
todo.append((n//i, i, combi+[i]))
i += 1
return combis
Recursive:
def recursive_get_factors(n):
def factor(n, i, combi, combis):
while i i <= n:
if n % i == 0:
combis.append(combi + [i, n//i]),
factor(n//i, i, combi+[i], combis)
i += 1
return combis
return factor(n, 2, [], [])
3.6 find words¶
Given a matrix of words and a list of words to search, return a list of words that exists in the board This is Word Search II on LeetCode
board = [['o','a','a','n'], ['e','t','a','e'], ['i','h','k','r'], ['i','f','l','v'] ]
words = ["oath","pea","eat","rain"]
def find_words(board, words):
def backtrack(board, i, j, trie, pre, used, result):
'''
backtrack tries to build each words from
the board and return all words found
@param: board, the passed in board of characters
@param: i, the row index
@param: j, the column index
@param: trie, a trie of the passed in words
@param: pre, a buffer of currently build string that differs
by recursion stack
@param: used, a replica of the board except in booleans
to state whether a character has been used
@param: result, the resulting set that contains all words found
@return: list of words found
'''
if '#' in trie:
result.add(pre)
if i < 0 or i >= len(board) or j < 0 or j >= len(board):
return
if not used[i][j] and board[i][j] in trie:
used[i][j] = True
backtrack(board, i+1, j, trie[board[i][j]],
pre+board[i][j], used, result)
backtrack(board, i, j+1, trie[board[i][j]],
pre+board[i][j], used, result)
backtrack(board, i-1, j, trie[board[i][j]],
pre+board[i][j], used, result)
backtrack(board, i, j-1, trie[board[i][j]],
pre+board[i][j], used, result)
used[i][j] = False
# make a trie structure that is essentially dictionaries of dictionaries
# that map each character to a potential next character
trie = {}
for word in words:
curr_trie = trie
for char in word:
if char not in curr_trie:
curr_trie[char] = {}
curr_trie = curr_trie[char]
curr_trie['#'] = '#'
# result is a set of found words since we do not want repeats
result = set()
used = [[False]len(board) for _ in range(len(board))]
for i in range(len(board)):
for j in range(len(board)):
backtrack(board, i, j, trie, '', used, result)
return list(result)
3.7 generate abbreviations¶
given input word, return the list of abbreviations. ex) word => [1ord, w1rd, wo1d, w2d, 3d, w3 ... etc]
def generate_abbreviations(word):
def backtrack(result, word, pos, count, cur):
if pos == len(word):
if count > 0:
cur += str(count)
result.append(cur)
return
if count > 0: # add the current word
backtrack(result, word, pos+1, 0, cur+str(count)+word[pos])
else:
backtrack(result, word, pos+1, 0, cur+word[pos])
# skip the current word
backtrack(result, word, pos+1, count+1, cur)
result = []
backtrack(result, word, 0, 0, "")
return result
3.8 generate parenthesis¶
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, given n = 3, a solution set is:
["((()))", "(()())", "(())()", "()(())", "()()()"
]
def generate_parenthesis_v1(n):
def add_pair(res, s, left, right):
if left == 0 and right == 0:
res.append(s)
return
if right > 0:
add_pair(res, s + ")", left, right - 1)
if left > 0:
add_pair(res, s + "(", left - 1, right + 1)
res = []
add_pair(res, "", n, 0)
return res
def generate_parenthesis_v2(n):
def add_pair(res, s, left, right):
if left == 0 and right == 0:
res.append(s)
if left > 0:
add_pair(res, s + "(", left - 1, right)
if right > 0 and left < right:
add_pair(res, s + ")", left, right - 1)
res = []
add_pair(res, "", n, n)
return res
3.9 letter combination¶
Given a digit string, return all possible letter combinations that the number could represent.
Input:Digit string "23" Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
def letter_combinations(digits):
if digits == "":
return []
kmaps = {
"2": "abc",
"3": "def",
"4": "ghi",
"5": "jkl",
"6": "mno",
"7": "pqrs",
"8": "tuv",
"9": "wxyz"
}
ans = [""]
for num in digits:
tmp = []
for an in ans:
for char in kmaps[num]:
tmp.append(an + char)
ans = tmp
return ans
3.10 palindrome partioning¶
It looks like you need to be looking not for all palindromic substrings, but rather for all the ways you can divide the input string up into palindromic substrings. (There's always at least one way, since one-character substrings are always palindromes.)
def palindromic_substrings(s):
if not s:
return
results = []
for i in range(len(s), 0, -1):
sub = s[:i]
if sub == sub[::-1]:
for rest in palindromic_substrings(s[i:]):
results.append([sub] + rest)
return results
There's two loops. The outer loop checks each length of initial substring (in descending length order) to see if it is a palindrome. If so, it recurses on the rest of the string and loops over the returned values, adding the initial substring to each item before adding it to the results.
def palindromic_substrings_iter(s):
"""
A slightly more Pythonic approach with a recursive generator
"""
if not s:
yield []
return
for i in range(len(s), 0, -1):
sub = s[:i]
if sub == sub[::-1]:
for rest in palindromic_substrings_iter(s[i:]):
yield [sub] + rest
3.11 pattern match¶
Given a pattern and a string str, find if str follows the same pattern.
Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty substring in str.
Examples: pattern = "abab", str = "redblueredblue" should return true. pattern = "aaaa", str = "asdasdasdasd" should return true. pattern = "aabb", str = "xyzabcxzyabc" should return false. Notes: You may assume both pattern and str contains only lowercase letters.
def pattern_match(pattern, string):
"""
:type pattern: str
:type string: str
:rtype: bool
"""
def backtrack(pattern, string, dic):
if len(pattern) == 0 and len(string) > 0:
return False
if len(pattern) == len(string) == 0:
return True
for end in range(1, len(string)-len(pattern)+2):
if pattern not in dic and string[:end] not in dic.values():
dic[pattern] = string[:end]
if backtrack(pattern[1:], string[end:], dic):
return True
del dic[pattern]
elif pattern in dic and dic[pattern] == string[:end]:
if backtrack(pattern[1:], string[end:], dic):
return True
return False
return backtrack(pattern, string, {})
3.12 Permute Unique¶
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example, [1,1,2] have the following unique permutations: [
[1,1,2], [1,2,1], [2,1,1]
]
def permute_unique(nums):
perms =
for n in nums:
new_perms = []
for l in perms:
for i in range(len(l)+1):
new_perms.append(l[:i]+[n]+l[i:])
if i < len(l) and l[i] == n:
break # handles duplication
perms = new_perms
return perms
3.13 Permute¶
Given a collection of distinct numbers, return all possible permutations.
For example, [1,2,3] have the following permutations: [
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1]
]
def permute(elements):
"""
returns a list with the permuations.
"""
if len(elements) <= 1:
return elements
else:
tmp = []
for perm in permute(elements[1:]):
for i in range(len(elements)):
tmp.append(perm[:i] + elements[0:1] + perm[i:])
return tmp
def permute_iter(elements):
"""
iterator: returns a perumation by each call.
"""
if len(elements) <= 1:
yield elements
else:
for perm in permute_iter(elements[1:]):
for i in range(len(elements)):
yield perm[:i] + elements[0:1] + perm[i:]
DFS Version
def permute_recursive(nums):
def dfs(res, nums, path):
if not nums:
res.append(path)
for i in range(len(nums)):
print(nums[:i]+nums[i+1:])
dfs(res, nums[:i]+nums[i+1:], path+[nums[i]])
res = []
dfs(res, nums, [])
return res
3.14 subsets unique¶
Given a collection of integers that might contain duplicates, nums, return all possible subsets.
Note: The solution set must not contain duplicate subsets.
For example, If nums = [1,2,2], a solution is:
[, , [1,2,2], [2,2], [1,2], []
]
def subsets_unique(nums):
def backtrack(res, nums, stack, pos):
if pos == len(nums):
res.add(tuple(stack))
else:
# take
stack.append(nums[pos])
backtrack(res, nums, stack, pos+1)
stack.pop()
# don't take
backtrack(res, nums, stack, pos+1)
res = set()
backtrack(res, nums, [], 0)
return list(res)
3.15 subsets¶
Given a set of distinct integers, nums, return all possible subsets.
Note: The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,3], a solution is:
[
,
,
,
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]
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12.002 Physics and Chemistry of the Earth and Terrestrial Planets Fall 2008
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Concepts for Heat Convection
Density and Temperature
Mantle heat convection is driven by the presence of hotter rocks at great
depth than are present at shallow depth. Of course, it is not the temperature
itself, but rather the density of the rocks, that is important for convection.
Density decreases as a function of temperature and, for a given package of
rock, can be related to temperature through the coefficient of thermal
expansion, α:
ρ = ρo (1 -αT) or dρ = - ρo α dT
where ρo is the density of the material at T=0. The coefficient of thermal
expansion is generally pretty small, so the change in density due to
temperature changes in the earth or planets is very small co mpared to the
total density. For example, for the earth’s upper mantle, α is around 3 .10 -5
°C -1. For a temperature change of 2000°C, the change in density would be
around 3%.
Density depends much more strongly on pressure than on temperature. In
the eart h’s mantle, density increases downward because the increasing
pressure has a much larger effect on density than does the increasing
temperature. This leads rise to the concept of the “adiabatic gradient”.
The Mantle Adiabat
Suppose that we take a packa ge of hot, high pressure mantle from deep in
the mantle (1) and bring it to a shallower level (2) without letting it
exchange any heat with the surrounding rocks. Because the pressure drops
as the rock is uplifted through the mantle, the volume expands an d its
temperature and its density decrease. This change in temperature as a
function of depth is the adiabatic gradient.
If the adiabatic gradient and the geothermal gradient are identical, then the
uplifted rock at (2) will be at the same temperature as its surroundings before
and after uplift. Now suppose we take a second package of rock at (2) and
bring it deeper into the mantle (1), also without letting it exchange heat with
the surroundings (this is called adiabatic transport). Its temperature and it s
density will likewise increase along the same adiabatic gradient. If the 33
adiabatic gradient and the geothermal gradient are identical, then this rock
will also be at the same temperature as its surroundings having been carried
to a greater depth. This me ans that for a geothermal gradient that is equal to
the adiabatic gradient, moving material around in the mantle (without letting
it gain or lose heat) does not change the temperature or density of the mantle
as a function of depth. Thus it does not change the potential energy of the
mant le so tha t the re is no ene rgy ava ilabl e to dr ive mot ion in the mant le.
Now consider a geothermal gradient in the mantle that is greater than the
geothermal gradient. A rock uplifted from (1) to (1) via adiabatic transp ort
will decrease in temperature and density, as before. However, after
transport, it will be hotter than the surrounding rocks, but is at the same
pressure as its surroundings so it will have a lower density. Likewise a rock
transported adiabatically to greater depth, from (2) to (2), will increase in
temperature and density. But, it will be colder and denser than surrounding
rocks. Thus vertical motion of rock packages in a mantle where the thermal
gradient is greater than the adiabat will create make the mantle colder at the
bottom and hotter near to the top. It will also change the density structure so
that rocks near the bottom are denser and rocks near the top are less dense.
In doing this it decreases the potential energy of the mantle. (Remembe r
PE= ρgh Δz for a slice of material of thickness Δz and at height h above
some reference horizon.) 34
This means that if the thermal gradient is greater than the adiabatic gradient,
there is energy available to drive vertical motion of rock packages in the
mantle. This thermally -driven process of overturn is what is known as
thermal convection. It is important to notice that it is only the difference in
temperature between the geotherm (temperature profile in the earth as a
function of depth) and the adiabat (the curve whose slope is the adiabatic
gradient at all depths) that can be used to drive convection. A mantle with a
geotherm at or below the adiabat cannot convect and can only lose heat by
the slow process of conduction.
Finally, if rising or f alling of a rock package in the mantle is not
instantaneous, a rising package will lose some of its excess heat to the
surrounding mantle during ascent, and vice versa. Thus its temperature will
follow a path somewhere between the adiabat and the mantle ge otherm. If
the package rises quickly or if the diffusion of heat is slow (meaning that the
thermal diffusivity of the material, κ, is small) then it will lose only a little
heat and its path will be close to the adiabat. If it rises very slowly or has
ra pid heat diffusion, its path will lie closer to the geotherm. This means that
rapid rates of thermal conduction/diffusion in the mantle act to hinder
convection. 35
Viscous Deformation
At the high temperatures needed for convection, planetary mantle mate rial
deforms ductiley. The mathematically simplest form of ductile deformation,
which we will use here, involves “linear” or “Newtonian” viscosity. For a
layer of material with viscosity μ, undergoing simple shear with velocity u
as shown below, the stre ss (force per unit area, typically measured in units of
Pascals, or Pa, defined as N/m 2 or kg/ms 2) is linearly proportional to the
spatial gradient of velocity.
x
dz
du
z
z+dz
�
z
If the vel ocity does not change along the along the length of the channel (ie
∂u/ ∂x = 0) then stress along a horizontal plane is related to the velocity
gradient by: 36
σxz = μ ∂u/ ∂z
This relationship shows that the stress required to drive flow increases with
li nearly with the viscosity of the material. (For non -Newtonian viscosity,
there is a power -law relationship between stress and strain.)
The Rayleigh Number
The Rayleigh number is a dimensionless number that describes the vigor of
convection within a conve cting layer. It is essentially the ratio of the
parameters that increase the vigor of convection to those that inhibit
convection.
Consider a potentially convecting layer that looks something like this:
where the rolls show upwelling and downwellin g limbs of a convective
system. The thickness of the layer is d and the temperature difference across
the layer (only that in excess of the adiabat) is ΔT. The parameters that
affect subduction are:
κ thermal diffusivity (m/s 2)
α coefficient of thermal expansion (°C -1)
ΔT temperature difference across the layer (°C)
μ viscosity (Pa s = Ns/m 2 = kg/ms)
d layer thickness and approx. dimension of convection cell (m)
ρ density (kg/m 3)
Helping Convection
Convection is aided by the temperature difference from top to bottom of the
slab (after subtracting out the adiabat). The resulting density difference that
can be used to drive convection is therefore: αρΔ T. 37
Increased layer thickness d enhances convection because: (a) the potential
energy available to drive flow increases with d2 (P.E.= ρgd 2/2). (b) For a
fixed flow geometry, the spatial gradient of flow is proportional to stress.
Thus the flow velocity caused by an applied stress will scale linearly with
the dimensions of the system, giving another factor of d.
The potential energy available for convection scales linearly with the
magnitude of the gravitational field, g.
Hindering Convection
Larger viscosities mean that larger stresses are needed to cause flow at a
given rate, so convection is hindered by larger viscosity, μ.
Rapid heat diffusion hinders conduction because packets lose heat
conductively during ascent (and vice versa for sinking), reducing the density
contrast and energy to drive convection. Thus convection is hindered by
larger thermal d iffusivity, κ.
Dividing the enhancing factors by the retarding factors for convection gives
the Rayleigh number, Ra :
Ra = d3αρ gΔT / μκ
We will leave it to the homework to show that this number is dimensionless.
Experiment (laboratory and computer) show that convection will occur only
when the Rayleigh number is in excess of the range 200 -1000. The exact
value of Ra at which convection will begin depends on the exact geometry
and boundary conditions of the system, e.g rigid or stress -free upper
bou ndary, whether heat is generated internally or input from below, etc.
Because the geophysical equations that govern fluid flow are very
complicated (fourth order PDE) the Rayleigh number will give us a useful
insight into whether convection can occur and h ow vigorous it might be in
planetary interiors. |
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Middle Years
1.02 Simplifying algebraic fractions
LessonWorksheetPractice
Lesson
An algebraic fraction is a fraction where the numerator and denominator are algebraic expressions. To simplify a numeric fraction we cancel any common factors in the numerator and denominator. We use the same method to simplify algebraic fractions.
Worked example
Simplify $$4x2−16x6x−24.
Think: To simplify the fraction we want to cancel any common factors in the numerator and denominator. To find these common factors, we first want to factorise the expressions in the numerator and denominator.
Do: In $$4x2−16x there is a common factor of $$2x. In $$6x−24 there is a common factor of $$3.
| | | | |
--- --- |
| $$4x2−16x6x−24 | $$= | $$2x(2x−8)3(2x−8) | Factorising the numerator and denominator |
| | $$= | $$2x3 | Cancelling $$(2x−8) from the numerator and denominator |
Since there are no more common factors between $$2x and $$3, we conclude that $$4x2−16x6x−24=2x3.
Reflect: It's worth fully factorising the numerator and denominator first in case there are any common factors. In some cases we might have to use other factorisation techniques to do this.
Summary
An algebraic fraction is a fraction where the numerator and denominator are algebraic expressions.
To simplify an algebraic fraction, first factorise the expressions in the numerator and denominator and then cancel any common factors in the numerator and denominator.
Practice questions
Question 1
Simplify: $$64mn240m2n
Question 2
Factorise and simplify: $$5u5uv−30uw
Question 3
Factorise and simplify: $$m2−16m−4
What is Mathspace
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4948 | https://mathbooks.unl.edu/Contemporary/sec-graph-intro.html | Contemporary Mathematics: Contemporary Mathematics at Nebraska
Michelle Homp, Alyssa Seideman, Sean Gravelle, Andrew Hayes, The Mabel Elizabeth Kelly Fund
(\require{cancel}\newcommand\degree{^{\circ}} \newcommand\Ccancel[black]{\renewcommand\CancelColor{\color{#1}}\cancel{#2}} \newcommand{\blert}{\boldsymbol{\color{blue}{#1}}} \newcommand{\bluetext}{\color{blue}{#1}} \delimitershortfall-1sp \newcommand\abs{\left|#1\right|} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \definecolor{fillinmathshade}{gray}{0.9} \newcommand{\fillinmath}{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}} )
Section 15.1 Introduction to Graph Theory
In the modern world, planning efficient routes is essential for business and industry, with applications as varied as product distribution, laying new fiber optic lines for broadband internet, and suggesting new friends within social network websites like Facebook.
This field of mathematics started nearly 300 years ago as a look into a mathematical puzzle (we’ll look at it in a bit). The field has exploded in importance in the last century, both because of the growing complexity of business in a global economy and because of the computational power that computers have provided us.
Example 15.1.1.
Here is a portion of a housing development from Missoula, Montana. 1
Image source: Sam Beebe (CC-BY)
As part of her job, the development’s lawn inspector has to walk down every street in the development making sure homeowners’ landscaping conforms to the community requirements.
Naturally, she wants to minimize the amount of walking she has to do. Is it possible for her to walk down every street in this development without having to do any backtracking? While you might be able to answer that question just by looking at the picture for a while, it would be ideal to be able to answer the question for any picture regardless of its complexity.
To do that, we first need to simplify the picture into a form that is easier to work with. We can do that by drawing a simple line for each street. Where streets intersect, we will place a dot.
This type of simplified picture is called a graph.
Definition of a graph.
In graph theory, the term graph refers to an object built from vertices and edges in the following way.
A vertex in a graph is a node, often represented with a dot or a point. (Note that the singular form is vertex and the plural form is vertices.)
The edges of a graph connect pairs of vertices. We usually represent the edges as straight or curved lines.
In graph theory, a graph is a set of vertices and edges.
Two vertices are adjacent if they are connected to each other by an edge.
We often give the vertices labels (such as letters or names). Edges can be named by listing the two vertices the edge connects. For instance, the edge AB would connect two vertices labeled A and B.
In graph theory, it is very important to keep in mind that a graph is determined only by its set of vertices and set of edges. In other words, the only information we care about is which vertices are connected to each other. For example, consider the following pair of graphs:
The two diagrams in Figure 15.1.3 represent the exact same graph. In geometry, they are different shapes (a rectangle and a triangle). In graph theory, the geometry doesn’t matter; only the connections are important. Since they both include the same four vertices connected in the same way, they represent the same graph.
Because of this, graphs are is useful when we want to model a situation for which connections are important and geometry is not. For instance, the graph from Example 15.1.1 is shown below on the left. The diagram on the right represents the same graph, arranged in a different way. They are the same because they show the same vertices and the same edge connections, even though they are laid out differently.
It is also important to note that vertices only occur when a dot is explicitly placed on the graph (not necessarily where two edges cross). In our alternate drawing of the graph from Example 15.1.1 (the diagram on the right-hand side above), the edges cross at many points in the middle. However, none of the crossings on the inside of the diagram represent vertices; the vertices are only the points indicated with labels around the outside of the diagram.
Imagine a freeway overpass: the freeway and side street cross, but it is not possible to change from the side street to the freeway at that point, so there is no intersection and no vertex would be placed.
Finally, you probably already noticed that we are using the term graph differently than you may have used the term in the past to describe the graph of a mathematical function. In graph theory, the term graph always refers to these types of graphs specifically.
Now that we’ve introduced the idea of a graph, we can discuss some of their properties.
Degree of a vertex.
The degree of a vertex is the number of times it meets an edge. If a vertex is not connected to any edges, it has a degree of 0. See the diagram below for an illustration of the degrees of several vertices.
Loops.
A loop is an edge that connects a vertex to itself. In the graph below, the edge shown in red is an example of a loop. Note that vertex b has degree 4: the loop meets vertex b twice, in addition to the two other edges.
Weighted edges.
We can sometimes assign numbers to the edges of a graph, called weights. Not all graphs have weighted edges; this is only done when the weights describe something relevant to a problem the graph is being used for. Hence, what the weights represent will depend on the context of the problem. This could be anything; some possibilities are the distance between two locations, the travel time, or the travel cost. It is important to note that the weight of an edge does not necessarily correspond to the actual length of the edge when we draw it as a picture.
Example 15.1.4. The Bridges of Königsberg.
Back in the 18th century in the Prussian city of Königsberg, a river ran through the city and seven bridges crossed the forks of the river. The river and the bridges are highlighted in the picture to the right. 2
Image source: Bogdan Giuşcă
As a weekend amusement, townsfolk would see if they could find a route that would take them across every bridge once and return them to where they started.
Leonard Euler (pronounced OY-ler), one of the most prolific mathematicians ever, looked at this problem in 1735, laying the foundation for graph theory as a field in mathematics. To analyze this problem, Euler introduced edges representing the bridges:
Since the size of each land mass it is not relevant to the question of bridge crossings, each can be shrunk down to a vertex representing the location:
Notice that in this graph there are two edges connecting the north bank and island, corresponding to the two bridges in the original drawing. Depending upon the interpretation of edges and vertices appropriate to a scenario, it is entirely possible and reasonable to have more than one edge connecting two vertices.
While we haven’t answered the actual question yet of whether or not there is a route which crosses every bridge once and returns to the starting location, the graph provides the foundation for exploring this question.
Paths and Circuits.
A path is a sequence of distinct, adjacent edges. A path may be named by listing the vertices in the path in the order they are visited.
A circuit is a path that starts and ends at the same vertex. The start/end vertex should be listed at the start and end of the name of a circuit, so that we know it’s a circuit.
The length of a path is the number of edges in the path. Since a circuit is a type of path, we define the length of a circuit the same way.
Example 15.1.6. Paths and Circuits.
The following graph shows a path by highlighting the edges in red. Viewed as a path from vertex A to vertex M, we can name it ABFGHM.
The path ABFGHM has length 5, since it traverses 5 edges. If we instead view it as a path from M to A, it could be named MHGFBA. Since the graph does not specify a direction the path has to be traversed, we consider ABFGHM and MHGFBA to be the same path.
The following diagram shows another path, ABFGLKJEA, on the same graph.
The path ABFGLKJEA is also a circuit, since it starts and ends at A. Notice that we wrote A at both the start and end of the name; otherwise, we would have ABFGLKJE, which is just a path from A to E.
The circuit ABFGLKJEA has length 8, since it traverses 8 edges.
Because a circuit forms a complete loop, there are many equivalent names for any given circuit. Let’s identify some for the circuit ABFGLKJEA. For example, if we start at F and move counter-clockwise, we traverse the vertices in the order FBAEJKLGF. On the other hand, if we start at K and move clockwise, we obtain KJEABFGLK. These three names all refer to the same circuit. See if you can find other names for this circuit!
Connected and Disconnected Graphs.
We say that a graph is connected if it’s all one “piece”. Using paths, we can describe more exactly what we mean by this.
A graph is called connected if there is a path between each pair of vertices.
A graph that is not connected is called disconnected.
The different “pieces” of a disconnected graph are called components. Each component is, itself, connected, while no component is connected to another component.
A bridge in a graph is an edge whose removal would create a new component.
Exploration 15.1.1.
The graph below shows 5 cities. The weights on the edges represent the airfare for a one-way flight between the cities.
How many vertices and edges does the graph have?
Is the graph connected?
How many components does the graph have?
What is the degree of the vertex representing LA?
If you fly from Seattle to Dallas to Atlanta, is that a path? Is it a circuit? If it is a path or circuit, what is its length?
If you fly from LA to Chicago to Dallas to LA, is that a path? Is it a circuit? If it is a path or circuit, what is its length?
Solution.
The graph has 5 vertices and 10 edges.
One component. Every connected graph has a single component, consisting of the whole graph.
degree 4
If you fly from Seattle to Dallas to Atlanta, it is a path and not a circuit. Its length is 2.
If you fly from LA to Chicago to Dallas to LA, it is a path and also a circuit. Its length is 3.
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Ginkgos
Ginkgos, specifically Ginkgo biloba, are unique, deciduous trees known for their distinctive fan-shaped leaves and significant historical presence. Typically reaching heights of 80 to 100 feet, with some individuals growing even taller, ginkgos exhibit a straight trunk and a pyramidal shape that becomes denser as they mature. Their leaves emerge yellow-green in spring, transition to green in summer, and turn a striking golden color in autumn, closely resembling the fronds of the maidenhair fern. Ginkgos are dioecious, meaning male and female reproductive structures are found on separate trees; female trees produce seeds that are encased in a fleshy covering, which can emit a strong odor as it decays. Remarkably resilient, ginkgos thrive in temperate climates and are highly resistant to diseases, pests, and pollution, making them popular for urban landscaping. This species is often referred to as a "living fossil," with ancestors dating back to the Paleozoic era, showcasing its long evolutionary history. Ginkgo biloba is native to eastern China, but it has been widely cultivated in various regions around the world, contributing to its status as a significant tree in both historical and modern contexts.
Published in: 2023
By: Rhoades, Mariana Louise
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Ginkgos
Categories: Evolution; gymnosperms; paleobotany; Plantae; taxonomic groups
The ginkgo is a hairless, deciduous tree with a straight trunk and pyramid-shaped foliage usually sparsely branched when young, becoming denser with age. Leaves are fan-shaped, 2 to 3 inches (5 to 7.5 centimeters) across, sometimes divided into two lobes. The ginkgo normally reaches heights of 80 to 100 feet (24 to 30 meters) and under favorable conditions grows to 125 feet (38 meters) or more. The bark is reddish-gray and corky, with irregular, wide fissures dividing rough plates. On old trees, the bark becomes gray, rough, and deeply furrowed.
Considerable diversity in branching habit occurs, sometimes with one side of the tree having erect branching and the other side spreading limbs. Young trees send out straight branches at a skyward angle and, until maturity, the sparse branching gives the tree an erratic appearance. Upon maturity, the branches round out and become widespread, yet retain an uneven crown.
As in many conifers, the long branches (shoots) and short, spurlike shoots of Ginkgo biloba are easily distinguished. The leaves are spirally arranged on both types but widely spaced on the long shoots, with leaves in crowded, rosettelike clusters on the short shoots. Branchlets (twigs) have a horizontal or drooping habit and are occupied with short, spurlike shoots. These shorter shoots increase in length only a fraction of an inch (2.5 centimeters) per year and may produce clusters of leaves annually for many years before abruptly lengthening out into long shoots bearing scattered leaves. The fan-shaped leaf has a marked resemblance to the fronds of the maidenhair fern, thus the common name: maidenhair tree. However, in its native China it is commonly called ducks-foot tree, also based on leaf shape.
The leaves, which grow on slender stalks up to 3 inches (7.5 centimeters) long, have numerous veins radiating out from the base to an irregularly notched leaf margin. There is no central midrib vein on the somewhat leathery, textured leaf. Stomata (breathing pores in the leaves) occur on both the upper and lower surfaces of the leaves. The leaves emerge yellow-green in spring but turn green toward midsummer and become golden in autumn. Leaves on vigorous young trees can grow up to 6 to 8 inches (15 to 20 centimeters) in width. There is a morphological distinction between leaves of long branches and short shoots, with the leaves of long branches generally bilobate to four-lobed and those of short shoots only fan-shaped to bilobate.
Elliptical, naked seeds resembling a small plum appear on female trees in early spring. Seeds range from 0.75 to 1 inch long (1.9 to 2.5 centimeters) and are covered by a thin, yellowish-orange, fleshy outer wall enveloping a woody shell which contains an edible kernel in the shell interior. When falling to the ground in autumn, the seed covering begins to diminish in thickness over several months, giving off the vile odor of butanoic and hexanoic acids (butter and Romano cheese fatty acids), and is eventually lost from most seeds. Ginkgo biloba wood is light, brittle, yellowish in color, and of little value. It is used as a base wood in highly lacquered furniture and small carved items.
Reproduction
The ginkgo is dioecious: Male and female reproductive structures are borne on separate trees. The male reproductive structures appear in May and are inch-long, catkinlike structures bearing numerous paired, pollen-bearing organs. The pollen grains are similar to the elliptical grain of cycads. The pollen organs and ovules are confined to the short shoots of each ginkgo tree and arise in the leaf axils or inner bud scales.
The female reproductive structure consists of a long stalk, bearing on each side an erect, naked ovule, which is surrounded at the base with a collarlike rim. The paired ovules are borne in groups of two to ten. The three-layered ovules include a fleshy exterior layer (sarcotesta), an inner flesh, and a stony shell (sclerotesta) between the two. This three-layered structure is called the integument. The nucellus (the central cellular mass of the body of the ovule containing the embryro) is mostly free from the surrounding integument, except at its base, where it develops a pollen chamber at its apex.
Similar to the cycads, the gingko reproduces by means of flagellated sperm cells, which are carried by the wind-borne pollen to the female reproductive structures within the ovule. In the ginkgo, the vascular system is weakly developed and consists of a pair of braided bundles in the interior fleshy layer of the integument. Upon maturation of the microgametophyte (male gametophyte), pollen tubes are produced, as are large, motile sperm cells similar to those of the cycads. Megagametophyte (female gametophyte) development is similar to that in cycads as well.
Natural Regeneration
Studies into the seedling development of Ginkgo biloba reveal a unique mechanism of clonal regeneration that may help explain the species’ long survival in the natural setting. The organ of clonal regeneration in the gingko is called the basal chichi. These organs are part of aggregates of suppressed shoot buds and are located in embryonic tissue of Ginkgo biloba seedlings. When damage occurs to the seedling axis, one of these subsurface buds grows down from the tree trunk to form a woody, stemlike basal chichi. Regeneration of Ginkgo biloba by basal chichi promotes survival of the tree in the forests of China today and may have been a factor in the protracted survival of the order since the Mesozoic era.
Habitat and Range
Ginkgo biloba, a distinctive tree suited for use in singular or in group plantings on lawns or along streets, is widely cultivated in all temperate zones. It prospers in moderately moist, fertile soil in humid, temperate regions. It is extremely resistant to disease and pests, and it is highly tolerate of smoke, dust, wind, and ice.
Ginkgo biloba is apparently native in eastern China, with documented semiwild trees growing on the west peak of Tian Mu Mountain in the Tian Mu Reserve, Zhejiang Province. Ginkgo biloba is planted in the eastern United States, Europe, and along the Pacific coast.
Fossil Record
Among the plants, Ginkgo biloba is probably the best-known example of a “living fossil.” Although the ancestors to the order Ginkgoales date to the Paleozoic era, it was at the close of the Triassic period when they became a dominant part of the Mesozoic flora. During the Jurassic period, especially the middle Jurassic, Ginkgoales reached zenith numbers of species and its widest distribution.
Jurassic and Cretaceous fossil localities reveal circumpolar Ginkgophytes sites, including Alaska, Greenland, Zemlya Frantsa Iosifa (Franz Joseph Land), and Mongolia, with the Siberian locations especially productive. Southern Hemisphere Ginkgoales localities include Patagonia at the southern tip of South America, South Africa, India, Australia, and New Zealand. European fossil sites are known in England, Scotland, Germany, Italy, Hungary, Turkestan, and Afghanistan. Western Canada and the United States have Ginkgoites leaf remains from the Upper Mesozoic and Lower Tertiary deposits. The presence of Ginkgophytes in high northern latitudes during the Early Cretaceous period and its presence in southern latitudes, such as Argentina, during the Jurassic, suggests that the dispersal of the plant was from the southern to northern latitudes during the Upper Mesozoic era.
During the Tertiary period, the decline of the Ginkgophytes was evident from the presence of only two of nineteen species remaining in the fossil record. One of the remaining two species is the Ginkgo adiantoides, which declined sharply during the Oligocene period. This decline continued into the Miocene period in North America, with Ginkgo adiantoides disappearing from the fossil record at the end of the Miocene. Ginkgo adiantoides did continue into the Pliocene in Europe, however. Since the Pliocene, the fossil record indicates that Ginkgoales have been represented by the extant living fossil, Ginkgo biloba.
Researchers propose that the decline of Ginkgophyta was a result of competition from the angiosperms (flowering plants) for similar plant habitats. Also, the Ginkgophytes became more restricted to northern temperate forests in the Tertiary period. When glaciation occurred in these areas during the Pleistocene, these forests were destroyed by climate change.
Bibliography
Cleal, Christopher J., and Barry A. Thomas. Plant Fossils: The History of Land Vegetation. Suffolk, England: Boydell Press, 1999. Relates the remarkable record of earth’s floral evolution and land vegetation. Includes a generous selection of fossil plant portraits with explanatory notes.
Everett, Thomas, H. The New York Botanical Garden Illustrated Encyclopedia of Horticulture. New York: Garland, 1981. Older source, yet has good information on history and paleontology of gingko trees, physical description, and their botanical and horticultural background.
Moore, David, M., ed. Marshall Cavendish Illustrated Encyclopedia of Plants and Earth Sciences. New York: Marshall Cavendish, 1988. Covers biology of gingko tree, with readable charts on global ginkgo fossils and geological background on land plant development.
Raven, Peter H., Ray F. Evert, and Susan E. Eichhorn. Biology of Plants. 6th ed. New York: W. H. Freeman/Worth, 1999. Basic textbook covering living and extinct seed plants, including Ginkgophyta, with excellent graphs and glossary.
Steward, Wilson N., and Gar W. Rothwell. Paleobotany and the Evolution of Plants. New York: Cambridge University Press, 1993. Clearly written, reference-quality book concerning the origin and evolution of extant and extinct groups of plants as revealed by the fossil record.
Related Topics
ConifersChinaSeedsWindThe Natural by Bernard MalamudMesozoic EraSoilPaleozoic EraAlaskaGreenlandMongoliaSouth AfricaIndiaAustraliaNew ZealandEnglandScotlandGermanyItalyHungaryAfghanistanCanadaArgentinaFossil recordCompetitionAngiosperms |
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4951 | https://physics.stackexchange.com/questions/219218/why-is-the-blue-line-in-the-balmer-series-sometimes-not-visible | spectroscopy - Why is the blue line in the Balmer series sometimes not visible? - Physics Stack Exchange
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Why is the blue line in the Balmer series sometimes not visible?
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So I've conducted an experiment to find the four visible hydrogen emission spectrum lines in the Balmer series in a laboratory. I don't have any background in quantum physics.
When I looked through the eyepiece, I saw the red light, the pale blue light, and the purple light as shown in the picture below:
I've asked my lab instructor why couldn't I see the blue light and he said it is well-known that sometimes not all the spectrum is shown. He told me to look it up on the internet, since I didn't take the quantum physics course yet.
Can someone explain to me this phenomenon? Or at least refer me to an article which discusses this issue? The experiment revolved around Balmer series only. (It was my first spectroscopy experiment)
The setup looked like this:
It looked like this:
Edit: These are the wavelengths I've found:
So I think the missing wavelength is actually the 410 n m 410 n m.
What can be the reason for this?
spectroscopy
hydrogen
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edited Nov 18, 2015 at 17:54
Alexs68Alexs68
asked Nov 18, 2015 at 16:34
Alexs68Alexs68
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You haven't told us anything about either the instrument you used (aside from it's having an eyepiece) or the source (I'd guess a gas discharge tube, but you haven't said). How are we suppose to help?dmckee --- ex-moderator kitten –dmckee --- ex-moderator kitten 2015-11-18 16:47:33 +00:00 Commented Nov 18, 2015 at 16:47
I must admit I can't think of a well known reason why the 434nm should be missing. The short wavelength lines are generally a lot fainter than the red and cyan lines, but that would apply to the violet line at 410nm as well.John Rennie –John Rennie 2015-11-18 16:54:39 +00:00 Commented Nov 18, 2015 at 16:54
I've got two decent spectroscopes, one prism based, one diffraction grating based, and can confirm that the 410.1 n m 410.1 n m line is hard to see. I use a typical gas discharge tube to produce the spectrum. It probably depends on quite a few factors like intensity of source, instrument, how well you block stray light and the state of your eyes whether you will see it or not. Maybe try again after allowing your eyes to get used to the dark, as you would looking through a telescope?Gert –Gert 2015-11-18 16:54:49 +00:00 Commented Nov 18, 2015 at 16:54
Both me and my partner were in a dark room for half an hour, and we couldn't find that single line. Our instructor couldn't find it either.Alexs68 –Alexs68 2015-11-18 16:58:54 +00:00 Commented Nov 18, 2015 at 16:58
1 Can we establish what the wavelength dependence of the spectroscope plus grating is? In general, unless using specialised glasses and gratings one does expect a reduced efficiency as one heads towards the UV.ProfRob –ProfRob 2015-11-18 22:32:33 +00:00 Commented Nov 18, 2015 at 22:32
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I was looking through some NIST atomic data for the Balmer Series from here:
It lists the spontaneous emission rates for the Balmer series as follows:
λ=656 nm λ=656 nm, A 32=4.41 e+7 s−1 A 32=4.41 e+7 s−1
λ=486 nm λ=486 nm, A 42=8.42 e+6 s−1 A 42=8.42 e+6 s−1
λ=434 nm λ=434 nm, A 52=2.53 e+6 s−1 A 52=2.53 e+6 s−1
λ=410 nm λ=410 nm, A 62=9.73 e+5 s−1 A 62=9.73 e+5 s−1
λ=397 nm λ=397 nm, A 72=4.38 e+5 s−1 A 72=4.38 e+5 s−1
So, the lines should certainly get dimmer as you move toward the UV, but I don't see any reason from a quantum mechanics perspective that the blue line at λ=434 nm λ=434 nm should be dimmer than the line at λ=410 nm λ=410 nm
Based you edit above, you seem to be missing the 410 line. It might not be seen because 1. the spontaneous emission rate is lowered and 2. your eyes are not as sensitive in this part of the EM spectrum. 3. As dmckee pointed out in your comments section, being a higher excited state, it will also be less populated that lower lying levels, meaning the total spontaneous emission from this state will be even further decreased.
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edited Nov 18, 2015 at 18:06
answered Nov 18, 2015 at 17:51
tmwilson26tmwilson26
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I stand corrected, the missing wavelength was 410 n m 410 n m So the explanation is that the purple line is dimmer then the other lines? Isn't there a better answer?Alexs68 –Alexs68 2015-11-18 17:56:00 +00:00 Commented Nov 18, 2015 at 17:56
@Alexs68 I edited my answer correspondingly. Its probably do to both the lowered spontaneous emission rate and your eyes not being sensitive when you get to low wavelengths.tmwilson26 –tmwilson26 2015-11-18 17:57:56 +00:00 Commented Nov 18, 2015 at 17:57
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Why is the blue line in the Balmer series sometimes not visible?
The human eye has difficulties in distinguishing dark blue lines on a black background.
You can use "Microsoft Word" to draw a black rectangle and a few dark blue lines of different thicknesses on the rectangle. The thinner the dark blue line the less visible it is. Even the thick dark blue lines are not quite visible.
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answered Nov 18, 2015 at 17:34
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4952 | https://gueniat.fr/teaching/ENG3009_sequences_and_series_notes.pdf | Mathematics for Engineers–ENG 3009, 2018-2019 Introduction to Sequences and Series FLORIMOND GUENIAT & VIJAY VENKATESH WITH BRIAN SMITH I. Introduction In this chapter, we will look at: Arithmetic progressions Geometric progressions Infinite sequences Series and sigma notation Arithmetic and Geometric series Series and sequences are extremely powerful tools. In mathematics, they are for instance related to integrals. Sound engineering is closely related to series. In economy, interests on loans can be calcu-lated using sequences and series.
As always, please free to refer to the book [Croft and Davidson, 2016] for details.
I Introduction Def.
SEQUENCE: A sequence is a set of numbers written down in a specific order.
Sequences are very useful for many reasons. The first one is that the elements are likely related by a rule. Studying the sequences hence gives insight on this rule. It also allows to solve puzzles or understand some patterns.
Def.
TERM: an element of a sequence is called a term Sometimes we use the symbol ... to indicate that the sequence continues.
For example, 1,3,5,7,9 and −1,−2,−3,−4,−5 are both sequences of five terms.
Also, 1,2,3,...,20 is the sequence of integers from 1 to 20 inclusive.
All of the sequences given above have a finite number of terms. They are known as finite sequences.
Some sequences go on forever, and these are called infinite sequences. To indicate that a sequence might go on forever we can use the ... notation.
So when we write the sequence 1,4,7,10,13,16,19,22,25,...
It can be assumed to continue indefinitely.
It is very different from 1,4,7,10,...,22,25 which is finite, so be careful.
Page 1 of 18 ENG 3009 www.bcu.ac.uk I. Introduction Main Example Finite sequence 1,−2,4,−8,16,−32 is a sequence of 6 terms.
Its first term is 1, its second term is −2 and so on. Infinite sequence 1,−2,4,−8,16,−32,...
if this sequence seems close to the previous one, it is actually an infinite sequence.
Its first term is 1, its second term is −2 and so on.
Exercise 1.
provide the sequence of the first seven prime numbers 1.1 provide the sequence of the first 10 odd numbers 1.2 provide the sequence of bisextile years between 1900 and 2000 (included) 1.3 Very often you will be able to spot a rule that you find the next term in a sequence. For example 1,3,5,7,9,...
is a sequence of odd integers and the next term is likely to be 11. The next term in the sequence 1, 1 2, 1 4 ...
might well be 1 8.
Tip Be careful, the rule might not be obvious, so do not draw conclusions too quickly !
I a) Notation used for sequences Often, it is useful to refer to a specific term in a sequence. Consequently, a subscript notation is used to refer to different terms in a sequence.
Def.
DENOTING A TERM: The ith term of a sequence x is referred as xi For example, to denote the sequence 1,4,7,10,13 by x, the first term can be labeled x1 , the second term x2 and so on. That is, x1 = 1,x2 = 4,x3 = 7,x4 = 10,x5 = 13 and so on if the sequence is longer.
Tip Sometimes the first term in a sequence is labeled x0. When it is the case, the second is then labeled x1, and so.
Page 2 of 18 ENG 3009 www.bcu.ac.uk II. Arithmetic progressions Main Example If we have the sequence: 1,−2,4,−8,16,−32,...
we can equivalently say that we have: x1,x2,x3,...
with x1 = 1,x2 = −2,x3 = 4,..., or, even better, that xi = (−2)i−1.
We could name the sequence with s instead of x: s1,s2,s3,...
with si = (−2)i.
It is also equivalent to have: x0,x1,x2,x3,...
with x0 = 1,x1 = −2,x2 = 4,..., or that xi = (−2)i. Note that the power, in this case, is "i". So, from time to time, it is more convenient to start with a number that is not 1.
II Arithmetic progressions Def.
ARITHMETIC PROGRESSION: An arithmetic progression is a sequence where each term is found by adding a fixed quantity, called the common difference, to the previous term.
Noting d the common difference, and considering the sequence x, the i +1 th term can be deduced from the ith term by xi+1 = xi +d For example, suppose the first term is 1 and we find subsequent terms by repeatedly adding 6. We obtain 1,7,13,19,...
If we write down the first five terms of the arithmetic progression where the first term is 1 and the common difference is 3, then the second term is 4 (by adding the common difference 3, to the first term). By continuing in this way we can construct the sequence 1,4,7,10,13,...
A more general arithmetic progression has first term a and common difference d, that is the first term is a the second term is a +d the third term is a +2d the fourth term is a +3d and so on.
This leads to the following formula for the nth term.
Page 3 of 18 ENG 3009 www.bcu.ac.uk II. Arithmetic progressions Prop.
nTH TERM OF AN ARITHMETIC PROGRESSION: The nth term of an arithmetic progression denoted by x, initialized with x1 as the first term and d as the common difference, is given by: xn = x1 +(n −1)d Let’s use a proof by induction: Base case (initialization) The first term is x1, which means that x2 = x1 + d.
Or, having n = 2, we have x2 = x1 + (2 −1) × d, which correspond to the formula. The formula is correct for the initialization. The step case Let’s assume the formula is true for the rank n−1. so we have xn−1 = x1+ ¡ (n−1)−1 ¢ d.
Or, we have xn = xn−1 +d which means that, replacing xn−1: xn = x1 +(n −2)d +d and hence xn = x1 +(n −1)d The formula holds These two steps, using induction, prove that the formula is correct.
Et voilà !
Main Example We can use the formula to find the 10th term of an arithmetic progression with first term 3 and common difference 5.
a = 3,d = 5, and n = 10 The 10th term is calculated, noting s the sequence, as: s10 = 3+(101)×5 = 3+9×5 = 48 The 10th term is s10 = 48. If Pauline is saving up 100$ per month, how much will she have after 5 years ?
We have a sequence with the first term being 100 (she starts the 1st month with 100$, or we have to start at month 0), and the common difference is 100. 5 years means 60 months, hence n = 60.
She will have saved s60 = s1 +(n −1)d = 100+(60−1)∗100 = 6000 Exercise 2.
Find the nth term (in the sense xn is the nth term of a sequence) of the arithmetic progression when: x0 = 10,d = −3,n = 5 2.1 x1 = 10,d = −3,n = 5 2.2 Page 4 of 18 ENG 3009 www.bcu.ac.uk III. Geometric progressions s0 = −20,d = 9,n = 12 2.3 s3 = −4,d = 12,n = 8 2.4 v1 = 2,d = −1 2,n = 10 2.5 x1 = 1,d = −2.4,n = 11 2.6 x1 = 0,d = 5,n = 3 2.7 Exercise 3.
Does the number 203 belongs to the arithmetic progression with x1 = 3 and d = 4?
3.1 Does the number 12 belongs to the arithmetic progression with x1 = 210 and d = −13?
3.2 Exercise 4.
If Pauline is saving up 100$ per month, how long will it take for her to save 10000$?
III Geometric progressions Def.
GEOMETRIC PROGRESSION: A geometric progression is a sequence where each term is found by multiplying the previous term by a fixed quantity called the common ratio. Noting r the common ratio and considering the sequence x, the i + 1 term can be deduced from the ith term by xi+1 = r xi For example, suppose the first term is 2 and we find subsequent terms by repeatedly multiplying by 5. We obtain the sequence 2,10,50,250,...
If we write down the geometric progression whose first term is 1 and whose common ratio is 1/2, we obtain the second term by multiplying the first by the common ratio, 1/2, that is 1/2x1 = 1/2 Continuing in this way we obtain the sequence as: 1,1/2,1/4,1/8,...
A general geometric progression has first term a and common ratio r, and can therefore be written as the first term is a the second term is ar the third term is ar 2 the fourth term is ar 3 and so on.
This leads to the following formula for the nth term.
Page 5 of 18 ENG 3009 www.bcu.ac.uk III. Geometric progressions Prop.
nTH TERM OF AN GEOMETRIC PROGRESSION: The nth term of an geometric progression denoted by x, initialized with x1 as the first term and r as the common ratio, is given by: xn = x1r n−1 Let’s use a proof by induction: Base case (initialization) The first term is x1, which means that x2 = x1 ×r. Or, having n = 2, we have x2 = x1 × r 2−1, which correspond to the formula. The formula is correct for the initialization. The step case Let’s assume the formula is true for the rank n−1. so we have xn−1 = x1r n−1−1. Or, we have xn = xn−1 ×r which means that, replacing xn−1: xn = x1r n−2 ×r and hence xn = x1r n−1 The formula holds These two steps, using induction, prove that the formula is correct.
Et voilà !
Main Example Let’s use the formula to find the 7th term of a geometric progression with first term 2 and common ratio 3.
s1 = 2,r = 3, and n = 7 7th term is s7 = 2×371 = 2×36 = 1458 therefore the 7th term is 1458. If Pauline has saved up 100$ on her saving accounts, and if her saving accounts has a saving rates of 2.3%, how much will she have after 5 years ? After 10 years?
We have a sequence with the first term being 100. It is a saving account, so she get an extra 2.3%, i.e., her money is multiplied by 1.023 each year. Consequently, the common ratio is 1.023 ◦5 years means n = 5.
She will have saved s5 = s1r n−1 = 100×1.0234 = 109.52 ◦10 years means n = 10.
She will have saved s10 = s1r n−1 = 100×1.0239 = 122.71 Page 6 of 18 ENG 3009 www.bcu.ac.uk IV. Infinite sequences Exercise 5.
Find the nth term (in the sense xn is the nth term of a sequence) of the geometric progression when: x0 = 10,r = −2,n = 5 5.1 x1 = 10,r = −3,n = 5 5.2 s0 = −20,r = 0.9,n = 12 5.3 s3 = −4,r = 1.2,n = 8 5.4 v1 = 2,r = −1 2,n = 10 5.5 x1 = 1,r = −0.4,n = 11 5.6 x1 = 0,r = 15,n = 3 5.7 Exercise 6.
Does the number 48 belongs to the geometric progression with x1 = 3072 and r = 0.5?
6.1 Does the number 6072 belongs to the geometric progression with x1 = 3 and 4 = −2?
6.2 Exercise 7.
If Pauline has saved up 1000$, and have a saving rates of 4.3%, how long will it take for her to save 10000$?
IV Infinite sequences Def.
INFINITE SEQUENCE: A sequences that continues indefinitely is called an infinite sequence.
Recall that we can use the ... notation to indicate this, thus 1,1/2,1/3,1/4,1/5,...
is the infinite sequence of inverse of positive whole numbers. The sequence can be written in the ab-breviation form xk = 1/k,for k = 1,2,3,...
As k gets larger and larger and approaches infinity, the term of the sequence get closer and closer to zero. This gives us the definition of a limit: Def.
LIMIT: L is the limit of a sequence x if for any distance δ, we can find a m large enough so all the xn that follows xm are close to L up to δ.
2 4 6 8 10 12 14 L 1.5 2 m 2δ n xn It means that, when n goes to infinity, xn goes to L.
We write this concisely as: lim n→∞xn = L Page 7 of 18 ENG 3009 www.bcu.ac.uk V. Series and sigma notation In the previous example, we have: lim k→∞ 1 k = 0 lim is an abbreviation for limit, so lim k→∞means we must examine the behaviour of the sequence as k gets larger and larger. When a sequence possesses a limit it is said to converge. However, not all sequences possess a limit. The sequence defined by xk = 3k −2 which is 1,4,7,10...
is one such example. As k gets larger and larger so too do the terms of the sequence, this sequence is said to diverge.
Let’s illustrate this on the sequence xk = 3+1/k2,k = 1,2,3,...
When k gets larger, 1/k2 tends to zero, and hence lim k→∞3+ 1 k2 = 3 The sequence xk converges to the limit 3.
V Series and sigma notation Whenever the terms of a sequence are added together, we obtain what is know as a series.
Def.
SERIES: An series is a sequence where the nth term is formed with the sum of the first n terms of a given sequence. If the series is noted S and the sequence x: Sn = x1 + x2 +...+ xn For example, if we have the sequence 1,1/2,1/4,1/8, we obtain the series S, where S1 = 1 S2 = 1+ 1 2 S3 = 1+ 1 2 + 1 4 S4 = 1+ 1 2 + 1 4 + 1 8 This series ends after the 4th term and is said to be a finite series. Other series we shall meet continue indefinitely and are said to be infinite series.
Page 8 of 18 ENG 3009 www.bcu.ac.uk V. Series and sigma notation V a) Sigma notation Sigma notation, P, provides a concise and convenient way of writing long sums. The sum 1+2+3+...+10+11+12 can be written very concisely using the capital Greek letter P as k=12 X k=1 k Def.
SIGMA NOTATION P: The sigma notation, or the sum notation, appears as the symbol, P, which is the Greek upper case letter, S. The summation sign, P, instructs us to sum the elements of a sequence. A typical element of the sequence which is being summed appears to the right of the summation sign. The variable of summation, i.e. the variable which is being summed.
n P k=1 xk summation sign stopping point element to be summed index of summation starting point The P stands for for a sum, in this case the sum of all the value of k as k ranges through all whole numbers from 1 to 12. Note that the lowermost and uppermost values of k are written at the bottom and top of the sigma sign respectively. The lowermost value of k is commonly k = 1 or k = 0, but other values are certainly possible. Sometimes the sigma notation itself is abbreviated. The k = part, written at the bottom and the top of the sigma sign, can be omitted (if there is no possible confusion !) and k could be represented by other letters like n or i. k is known as a dummy variable.
For example, the following expressions are identical: k=12 X k=1 k = i=12 X i=1 i = 12 X k=1 k = 12 X 1 Page 9 of 18 ENG 3009 www.bcu.ac.uk V. Series and sigma notation Main Example For instance, if we want to write out explicitly what is meant by k=5 X k=1 k3 k ranges from 1 to 5, and if we cube each value of k, it leads to k=5 X k=1 k3 = 13 +23 +33 +43 +r 3 If we want to write out explicitly what is meant by k=4 X k=1 (−1)k2k when k = 1, (−1)k2k = (−1)121 = −2 when k = 2, (−1)k2k = (−1)222 = 4 when k = 3, (−1)k2k = (−1)323 = −8 when k = 4, (−1)k2k = (−1)424 = 16 So k=4 X k=1 (−1)k2k = −2+4−8+16 Exercise 8.
What is S = 4 P k=1 k?
8.1 What is S = 9 P k=4 k?
8.2 What is S = 4 P k=1 k3?
8.3 What is S = 5 P k=1 2k?
8.4 What is S = n P k=1 1?
8.5 V b) Arithmetic series Def.
ARITHMETIC SERIES: An arithmetic series is a series based on an arithmetic progression. In other words, an arithmetic series is a sequence where the nth term is formed with the sum of the first n terms of an arithmetic progression. If we note S the arithmetic series and x the arithmetic progression: Sn = n X k=1 xk Page 10 of 18 ENG 3009 www.bcu.ac.uk V. Series and sigma notation For example, the arithmetic progression with five terms having first term 4 and common difference 5 is 4,9,14,19,24 If these terms are added we obtain the arithmetic series S5 = 4+9+14+19+24 It is easily verified that this has sum 70. But when the given series has a larger number of terms then finding its sum by directly adding all the terms will be laborious. Fortunately there is a formula that enables us to find the sum of an arithmetic series.
Prop.
ARITHMETIC SERIES: If S is the arithmetic series formed from the arithmetic progression of first term a and common difference d, then: Sn = n ¡ 2a +(n −1)d ¢ 2 Let’s write x the arithmetic progression, with x1 = a: xn = a +(n −1)d Let’s use a proof by induction: Base case (initialization) The first term is S1 is the sum of the first terms, i.e.
S1 = x1 Or, having n = 1, we have S1 = 1 ¡ 2a+(1−1)d ¢ 2 = a, which correspond to the formula. The formula is correct for the initialization. The step case Let’s assume the formula is true for the rank n −1.
so we have Sn−1 = (n −1)× ¡ 2a +(n −1−1)d ¢ 2 . Or, we have xn = a +(n −1)d and, Sn = n P k=1 xk = n−1 P k=1 xk + xn = Sn−1 + xn So Sn = (n −1)× ¡ 2a +(n −2)d ¢ 2 + a +(n −1)d = (n −1)× ¡ 2a +(n −2)d ¢ +2× ¡ a +(n −1)d ¢ 2 = 2(n −1+1)a +(n −1)(n −2+2)d 2 = 2na +n(n −1)d 2 = n ¡ 2a +(n −1)d ¢ 2 The formula holds These two steps, using induction, prove that the formula is correct.
Et voilà !
Page 11 of 18 ENG 3009 www.bcu.ac.uk V. Series and sigma notation Main Example For finding the first 10 terms of the arithmetic series with first term 3 and common difference 4.
We use the formula Sn = n ¡ 2a +(n −1)d ¢ 2 . with n = 10,a = 3, and d = 4: S10 = 10× ¡ 2×3+(10−1)×4 ¢ 2 = 5(6+36) = 210 The sum of the first 15 terms of an arithmetic series is 165. The common difference is 2. Can we calculate the first term of the series ?
We use the formula Sn = n ¡ 2a +(n −1)d ¢ 2 , with n = 15,Sn = 165,d = 2. To find a: Sn = n ¡ 2a +(n −1)d ¢ 2 ⇔ 2Sn = n ¡ 2a +(n −1)d ¢ ⇔ 2Sn n = 2a +(n −1)d ⇔ 2Sn n −(n −1)d = 2a ⇔ a = Sn n −(n −1)d 2 And hence: a = 165 15 −(15−1)×2 2 = −3 Exercise 9.
What are the five first terms of the arithmetic series with x1 = 10,d = 2 ?
9.1 What are the seven first terms of the arithmetic series with x1 = 10,d = −2 ?
9.2 What are the five six terms of the arithmetic series with x1 = 100,d = −45 ?
9.3 V c) Geometric series Def.
GEOMETRIC SERIES: An geometric series is a series based on an geometric progression. In other words, a geometric series is a sequence where the nth term is formed with the sum of the first n terms of an geometric progression. If we note S the geometric series and x the geometric progression: Sn = n X k=1 xk For example, the geometric progression with five terms having first term 2 and common ratio 3 is 2,6,18,54,162.
If these terms are added we obtain the geometric series 2+6+18+54+162. It is easily verified that this has sum 242. Again similar to arithmetic series to find the sum of the geometric series for a larger number of terms there is a formula.
Page 12 of 18 ENG 3009 www.bcu.ac.uk V. Series and sigma notation Prop.
GEOMETRIC SERIES: Sn = a(1−r n) 1−r provided r ̸= 1.
Let’s write x the geometric progression, with x1 = a and the common ratio r: xn = ar n−1 Let’s use a proof by induction: Base case (initialization) The first term is S1 is the sum of the first terms, i.e.
S1 = x1 Or, having n = 1, we have S1 = a 1−r 1 1−r = a, which correspond to the formula. The formula is correct for the initialization. The step case Let’s assume the formula is true for the rank n −1. so we have Sn−1 = a 1−rn −1 1−r . Or, we have xn = ar n−1 and, Sn = n P k=1 xk = n−1 P k=1 xk + xn = Sn−1 + xn So Sn = a 1−r n−1 1−r + ar n−1 = a 1−r n−1 +r n−1 ×(1−r) 1−r = a 1− r n−1 + r n−1 −r n 1−r = a 1−r n 1−r The formula holds These two steps, using induction, prove that the formula is correct.
Et voilà !
The formula excludes the use of r = 1 because in this case the denominator becomes zero. If it is the case, the sequence should rather been considered as an arithmetic series with d = 0.
Main Example Let’s find the sum of the first terms of the geometric series with first term 2 and common ratio 3.
Using the formula Sn = a(1−r n) 1−r with n = 5,a = 2,r = 3: S5 = 2(1−e5) 1−3 = 242 Page 13 of 18 ENG 3009 www.bcu.ac.uk V. Series and sigma notation Going further Geometric series are pivotal in the study of fractals. Fractals are very common in nature !
For instance, snowflakes, frost, sponges and waves can be described as fractals.
wikipedia V d) Infinite geometric series When the terms of an infinity sequence are added we obtain an infinite series. It may seem strange to try to add together an infinite number of terms but under some circumstances their sum is finite and can be found. The sum of an infinite number of terms of a geometric series is denoted by S∞= a 1−r ,−1 < r < 1 Note that the common ratio should lies between -1 and 1.
Tip The ratio has to be between −1 and 1 ! In this case, the formula for the sum can be taken to the limit: lim n⇒∞Sn = lim n⇒∞ a(1−r n) 1−r Because |r| < 1, lim n⇒∞r n = 0 !
Main Example Let’s find the sum of the infinite geometric series with first term 2 and common ratio 1 3.
Using the formula S∞= a 1−r ,−1 < r < 1, with a = 2 and r = 1 3: S∞ = 2 1−1 3 = 2 2 3 = 2×3 2 = 3 Note that we can only make use of the formula because the value of r lies between -1 and 1.
Geometric series can do other cool stuff !
For instance, we can use them to convert the number m: m = 1.2121212121...
to a fraction ! m is actually: m = 1+0.21+0.0021+0.000021+...
Page 14 of 18 ENG 3009 www.bcu.ac.uk VI. Solutions to exercises which is: m = 1+ 21 100 + 21 10000 + 21 1000000 +...
or, m = 1+ 21 100 + 21 100 × µ 1 100 ¶ + 21 100 × µ 1 100 ¶2 +...
and finally, we have: m = 1+ ∞ X k=0 21 100 µ 1 100 ¶k It is the limit of the geometric series of the ratio r = 1 100 and first term 21 100. So now we can apply the formula for geometric series: m = 1+ 21 100 1 1− 1 100 = 1+ 21 100−1 = 1+ 21 99 VI Solutions to exercises Solution 1.
1,2,3,5,7,11,13 1.1 1,3,5,7,9,11,13,15,17,19 1.2 1904,1908,1912,1916,1920,1924,1928,1932,1936,1940,1944,1948, 1952,1956,1960,1964,1968,1972,1976,1980,1984,1988,1992,1996,2000 1.3 Solution 2.
x0 = 10, with d = −3 means x1 = 7. x5 = 7+(5−1)×(−3) = −5 2.1 x5 = 10+(5−1)×(−3) = −2 2.2 s0 = −20,d = 9,n = 12 means s1 = −20+9 = −11. s12 = −11+(12−1)×9 = 88 2.3 s3 = −4,d = 12,n = 8 means s3 = s1 + (3 −1) × 12 and hence s1 = s3 −(3 −1) × 12 = −28. s8 = −28+(8−1)×12 = 56.
2.4 v10 = 2+(10−1)× 1 2 = 6.5 2.5 x11 = 1+(11−1)×(−2.4) = −23 2.6 x3 = 0+(3−1)∗5 = 10 2.7 Solution 3.
xn = x1 +(n −1)d and hence xn = −1+4n. Let’s solve 4n −1 = 203, which leads to 4n = 204 and n = 51. It is an integer, hence 203 belongs to the sequence (and x51 = 203).
3.1 xn = x1 +(n −1)d and hence xn = 223−13n. Let’s solve 223−13n = 12, which leads to 13n = 211 and n = 16.23. It is not an integer, hence 12 does not belong to the sequence.
3.2 Page 15 of 18 ENG 3009 www.bcu.ac.uk VI. Solutions to exercises Solution 4.
We have a sequence with the first term being 100 (she starts the 1st month with 100$, or we have to start at month 0), and the common difference is 100. After n months, she will have saved sn = s1 +(n −1)d.
sn = s1 +(n −1)d ⇔ sn −s1 = (n −1)d ⇔ sn −s1 d = n −1 ⇔ n = 1+ sn −s1 d ⇔ n = 1+ 10000−100 100 ⇔ n = 100 So she needs 100 months, i.e., slightly more than 8 years.
Solution 5.
x0 = 10, with r = −3 means x1 = −30. x5 = x1 1−r n−1 1−r = −2430 5.1 x5 = 810 5.2 s0 = −20,r = 0.9,n = 12 means s1 = −20×9 = −180. s12 = 56.5 5.3 s3 = −4,r = 1.2,n = 8 means s3 = s1r 2 and hence s1 = s3/r 2 = −2.78. s8 = −9.96.
5.4 v10 = 2+(10−1)× 1 2 = −0.0020 5.5 x11 = 0.0001 5.6 x3 = 0 5.7 Solution 6.
xn = x1r n−1. Let’s solve 48 = 3072×(1/2)n, which leads to log 48 3072 = n log 1 2 and n = 6. It is an integer, hence 48 belongs to the sequence (and x6 = 48).
6.1 xn = x1r n−1. Let’s solve 6072 = 3 × (−2)n, which, squared, leads to log 60722 32 = 2n log2 and n = 10.98. It is not an integer, hence 6072 does not belong to the sequence.
6.2 Solution 7.
We have a sequence with the first term being 1000, and the common ratio of 1.043. After n years, she will have saved sn = s1r n−1.
sn = s1r n−1 ⇔ sn s1 = r n−1 ⇔ log µ sn s1 ¶ = log ¡ r n−1¢ ⇔ log µ sn s1 ¶ = (n −1)log(r) ⇔ log ³ sn s1 ´ log(r) = n −1 ⇔ n = 1+ log ³ sn s1 ´ log(r) ⇔ n = 1+ log ¡ 10000 1000 ¢ log(1.043) ⇔ n = 56 Page 16 of 18 ENG 3009 www.bcu.ac.uk VI. Solutions to exercises So she needs 56 years.
Solution 8.
S = 1+2+3+4 = 10 8.1 S = 39 8.2 S = 13 +23 +33 +43 = 100 8.3 S = 30 8.4 there is n terms in the sum, each being 1, hence S = n 8.5 Solution 9.
10,22,36,52,70 9.1 10,18,24,28,30,30,28 9.2 100,155,165,130,50,−75 9.3 Page 17 of 18 ENG 3009 www.bcu.ac.uk Bibliography [Croft and Davidson, 2016] Croft, A. and Davidson, R. (2016). Foundation Maths. Pearson.
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Contents
1 Introduction
2 Definition
3 Examples
3.1 Simple pendulum
3.2 Compound pendulums
3.2.1 Center of oscillation
4 Measuring moment of inertia
5 Moment of inertia of area
5.1 Sectional areas moment calculated thus
6 Motion in a fixed plane
6.1 Point mass
6.1.1 Examples
6.2 Rigid body
6.2.1 Angular momentum
6.2.2 Kinetic energy
6.2.3 Newton's laws
7 Motion in space of a rigid body, and the inertia matrix
7.1 Angular momentum
7.2 Kinetic energy
7.3 Resultant torque
7.4 Parallel axis theorem
7.5 Scalar moment of inertia in a plane
8 Inertia tensor
8.1 Definition
8.2 Alternate inertia convention
8.2.1 Determine inertia convention (principal axes method)
8.3 Derivation of the tensor components
8.4 Inertia tensor of translation
8.5 Inertia tensor of rotation
9 Inertia matrix in different reference frames
9.1 Body frame
9.2 Principal axes
9.3 Ellipsoid
10 See also
11 References
12 External links
Moment of inertia
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From Wikipedia, the free encyclopedia
Scalar measure of the rotational inertia with respect to a fixed axis of rotation
For the quantity also known as the "area moment of inertia", see Second moment of area.
| Moment of inertia |
| Flywheels have large moments of inertia to smooth out changes in rates of rotational motion. |
| Common symbols | I |
| SI unit | kg⋅m2 |
| Other units | lbf·ft·s2 |
| Derivations fromother quantities | |
| Dimension | M L2 |
| |
| Part of a series on |
| Classical mechanics |
| Second law of motion |
| History Timeline Textbooks |
| Branches Applied Celestial Continuum Dynamics Field theory Kinematics Kinetics Statics Statistical mechanics |
| Fundamentals Acceleration Angular momentum Couple D'Alembert's principle Energy + kinetic + potential Force Frame of reference Inertial frame of reference Impulse Inertia / Moment of inertia Mass Mechanical power Mechanical work Moment Momentum Space Speed Time Torque Velocity Virtual work |
| Formulations Newton's laws of motion Analytical mechanics + Lagrangian mechanics + Hamiltonian mechanics + Routhian mechanics + Hamilton–Jacobi equation + Appell's equation of motion + Koopman–von Neumann mechanics |
| Core topics Damping Displacement Equations of motion Euler's laws of motion Fictitious force Friction Harmonic oscillator Inertial / Non-inertial reference frame Motion (linear) Newton's law of universal gravitation Newton's laws of motion Relative velocity Rigid body + dynamics + Euler's equations Simple harmonic motion Vibration |
| Rotation Circular motion Rotating reference frame Centripetal force Centrifugal force + reactive Coriolis force Pendulum Tangential speed Rotational frequency Angular acceleration / displacement / frequency / velocity |
| Scientists Kepler Galileo Huygens Newton Horrocks Halley Maupertuis Daniel Bernoulli Johann Bernoulli Euler d'Alembert Clairaut Lagrange Laplace Poisson Hamilton Jacobi Cauchy Routh Liouville Appell Gibbs Koopman von Neumann |
| Physics portal Category |
| v t e |
To improve their maneuverability, combat aircraft are designed to minimize moments of inertia, while civil aircraft often are not.
The moment of inertia, otherwise known as the mass moment of inertia, angular/rotational mass, second moment of mass, or most accurately, rotational inertia, of a rigid body is defined relatively to a rotational axis. It is the ratio between the torque applied and the resulting angular acceleration about that axis.: 279: 261 It plays the same role in rotational motion as mass does in linear motion. A body's moment of inertia about a particular axis depends both on the mass and its distribution relative to the axis, increasing with mass and distance from the axis.
It is an extensive (additive) property: for a point mass the moment of inertia is simply the mass times the square of the perpendicular distance to the axis of rotation. The moment of inertia of a rigid composite system is the sum of the moments of inertia of its component subsystems (all taken about the same axis). Its simplest definition is the second moment of mass with respect to distance from an axis.
For bodies constrained to rotate in a plane, only their moment of inertia about an axis perpendicular to the plane, a scalar value, matters. For bodies free to rotate in three dimensions, their moments can be described by a symmetric 3-by-3 matrix, with a set of mutually perpendicular principal axes for which this matrix is diagonal and torques around the axes act independently of each other.
Introduction
[edit]
When a body is free to rotate around an axis, torque must be applied to change its angular momentum. The amount of torque needed to cause any given angular acceleration (the rate of change in angular velocity) is proportional to the moment of inertia of the body. Moments of inertia may be expressed in units of kilogram metre squared (kg·m2) in SI units and pound-foot-second squared (lbf·ft·s2) in imperial or US units.
The moment of inertia plays the role in rotational kinetics that mass (inertia) plays in linear kinetics—both characterize the resistance of a body to changes in its motion. The moment of inertia depends on how mass is distributed around an axis of rotation, and will vary depending on the chosen axis. For a point-like mass, the moment of inertia about some axis is given by , where is the distance of the point from the axis, and is the mass. For an extended rigid body, the moment of inertia is just the sum of all the small pieces of mass multiplied by the square of their distances from the axis in rotation. For an extended body of a regular shape and uniform density, this summation sometimes produces a simple expression that depends on the dimensions, shape and total mass of the object.
In 1673, Christiaan Huygens introduced this parameter in his study of the oscillation of a body hanging from a pivot, known as a compound pendulum. The term moment of inertia ("momentum inertiae" in Latin) was introduced by Leonhard Euler in his book Theoria motus corporum solidorum seu rigidorum in 1765, and it is incorporated into Euler's second law.
The natural frequency of oscillation of a compound pendulum is obtained from the ratio of the torque imposed by gravity on the mass of the pendulum to the resistance to acceleration defined by the moment of inertia. Comparison of this natural frequency to that of a simple pendulum consisting of a single point of mass provides a mathematical formulation for moment of inertia of an extended body.
The moment of inertia also appears in momentum, kinetic energy, and in Newton's laws of motion for a rigid body as a physical parameter that combines its shape and mass. There is an interesting difference in the way moment of inertia appears in planar and spatial movement. Planar movement has a single scalar that defines the moment of inertia, while for spatial movement the same calculations yield a 3 × 3 matrix of moments of inertia, called the inertia matrix or inertia tensor.
The moment of inertia of a rotating flywheel is used in a machine to resist variations in applied torque to smooth its rotational output. The moment of inertia of an airplane about its longitudinal, horizontal and vertical axes determine how steering forces on the control surfaces of its wings, elevators and rudder(s) affect the plane's motions in roll, pitch and yaw.
Definition
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The moment of inertia is defined as the product of mass of section and the square of the distance between the reference axis and the centroid of the section.
The moment of inertia I is also defined as the ratio of the net angular momentum L of a system to its angular velocity ω around a principal axis, that is
If the angular momentum of a system is constant, then as the moment of inertia gets smaller, the angular velocity must increase. This occurs when spinning figure skaters pull in their outstretched arms or divers curl their bodies into a tuck position during a dive, to spin faster.
If the shape of the body does not change, then its moment of inertia appears in Newton's law of motion as the ratio of an applied torque τ on a body to the angular acceleration α around a principal axis, that is: 279: 261, eq.9-19
For a simple pendulum, this definition yields a formula for the moment of inertia I in terms of the mass m of the pendulum and its distance r from the pivot point as,
Thus, the moment of inertia of the pendulum depends on both the mass m of a body and its geometry, or shape, as defined by the distance r to the axis of rotation.
This simple formula generalizes to define moment of inertia for an arbitrarily shaped body as the sum of all the elemental point masses dm each multiplied by the square of its perpendicular distance r to an axis k. An arbitrary object's moment of inertia thus depends on the spatial distribution of its mass.
In general, given an object of mass m, an effective radius k can be defined, dependent on a particular axis of rotation, with such a value that its moment of inertia around the axis is where k is known as the radius of gyration around the axis.
Examples
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See also: List of moments of inertia
Simple pendulum
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Mathematically, the moment of inertia of a simple pendulum is the ratio of the torque due to gravity about the pivot of a pendulum to its angular acceleration about that pivot point. For a simple pendulum, this is found to be the product of the mass of the particle with the square of its distance to the pivot, that is
This can be shown as follows:
The force of gravity on the mass of a simple pendulum generates a torque around the axis perpendicular to the plane of the pendulum movement. Here is the distance vector from the torque axis to the pendulum center of mass, and is the net force on the mass. Associated with this torque is an angular acceleration, , of the string and mass around this axis. Since the mass is constrained to a circle the tangential acceleration of the mass is . Since the torque equation becomes: where is a unit vector perpendicular to the plane of the pendulum. (The second to last step uses the vector triple product expansion with the perpendicularity of and .) The quantity is the moment of inertia of this single mass around the pivot point.
The quantity also appears in the angular momentum of a simple pendulum, which is calculated from the velocity of the pendulum mass around the pivot, where is the angular velocity of the mass about the pivot point. This angular momentum is given by using a similar derivation to the previous equation.
Similarly, the kinetic energy of the pendulum mass is defined by the velocity of the pendulum around the pivot to yield
This shows that the quantity is how mass combines with the shape of a body to define rotational inertia. The moment of inertia of an arbitrarily shaped body is the sum of the values for all of the elements of mass in the body.
Compound pendulums
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A compound pendulum is a body formed from an assembly of particles of continuous shape that rotates rigidly around a pivot. Its moment of inertia is the sum of the moments of inertia of each of the particles that it is composed of.: 395–396: 51–53 The natural frequency () of a compound pendulum depends on its moment of inertia, , where is the mass of the object, is local acceleration of gravity, and is the distance from the pivot point to the center of mass of the object. Measuring this frequency of oscillation over small angular displacements provides an effective way of measuring moment of inertia of a body.: 516–517
Thus, to determine the moment of inertia of the body, simply suspend it from a convenient pivot point so that it swings freely in a plane perpendicular to the direction of the desired moment of inertia, then measure its natural frequency or period of oscillation (), to obtain where is the period (duration) of oscillation (usually averaged over multiple periods).
Center of oscillation
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A simple pendulum that has the same natural frequency as a compound pendulum defines the length from the pivot to a point called the center of oscillation of the compound pendulum. This point also corresponds to the center of percussion. The length is determined from the formula, or
The seconds pendulum, which provides the "tick" and "tock" of a grandfather clock, takes one second to swing from side-to-side. This is a period of two seconds, or a natural frequency of for the pendulum. In this case, the distance to the center of oscillation, , can be computed to be
Notice that the distance to the center of oscillation of the seconds pendulum must be adjusted to accommodate different values for the local acceleration of gravity. Kater's pendulum is a compound pendulum that uses this property to measure the local acceleration of gravity, and is called a gravimeter.
Measuring moment of inertia
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The moment of inertia of a complex system such as a vehicle or airplane around its vertical axis can be measured by suspending the system from three points to form a trifilar pendulum. A trifilar pendulum is a platform supported by three wires designed to oscillate in torsion around its vertical centroidal axis. The period of oscillation of the trifilar pendulum yields the moment of inertia of the system.
Moment of inertia of area
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Moment of inertia of area is also known as the second moment of area and its physical meaning is completely different from the mass moment of inertia.
These calculations are commonly used in civil engineering for structural design of beams and columns. Cross-sectional areas calculated for vertical moment of the x-axis and horizontal moment of the y-axis .
Height (h) and breadth (b) are the linear measures, except for circles, which are effectively half-breadth derived,
Sectional areas moment calculated thus
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Source:
Square:
Rectangular: and;
Triangular:
Circular:
Motion in a fixed plane
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Point mass
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The moment of inertia about an axis of a body is calculated by summing for every particle in the body, where is the perpendicular distance to the specified axis. To see how moment of inertia arises in the study of the movement of an extended body, it is convenient to consider a rigid assembly of point masses. (This equation can be used for axes that are not principal axes provided that it is understood that this does not fully describe the moment of inertia.)
Consider the kinetic energy of an assembly of masses that lie at the distances from the pivot point , which is the nearest point on the axis of rotation. It is the sum of the kinetic energy of the individual masses,: 516–517: 1084–1085: 1296–1300
This shows that the moment of inertia of the body is the sum of each of the terms, that is
Thus, moment of inertia is a physical property that combines the mass and distribution of the particles around the rotation axis. Notice that rotation about different axes of the same body yield different moments of inertia.
The moment of inertia of a continuous body rotating about a specified axis is calculated in the same way, except with infinitely many point particles. Thus the limits of summation are removed, and the sum is written as follows:
Another expression replaces the summation with an integral,
Here, the function gives the mass density at each point , is a vector perpendicular to the axis of rotation and extending from a point on the rotation axis to a point in the solid, and the integration is evaluated over the volume of the body . The moment of inertia of a flat surface is similar with the mass density being replaced by its areal mass density with the integral evaluated over its area.
Note on second moment of area: The moment of inertia of a body moving in a plane and the second moment of area of a beam's cross-section are often confused. The moment of inertia of a body with the shape of the cross-section is the second moment of this area about the -axis perpendicular to the cross-section, weighted by its density. This is also called the polar moment of the area, and is the sum of the second moments about the - and -axes. The stresses in a beam are calculated using the second moment of the cross-sectional area around either the -axis or -axis depending on the load.
Examples
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Main article: List of moments of inertia
The moment of inertia of a compound pendulum constructed from a thin disc mounted at the end of a thin rod that oscillates around a pivot at the other end of the rod, begins with the calculation of the moment of inertia of the thin rod and thin disc about their respective centers of mass.
The moment of inertia of a thin rod with constant cross-section and density and with length about a perpendicular axis through its center of mass is determined by integration.: 1301 Align the -axis with the rod and locate the origin at its center of mass at the center of the rod, then where is the mass of the rod.
The moment of inertia of a thin disc of constant thickness , radius , and density about an axis through its center and perpendicular to its face (parallel to its axis of rotational symmetry) is determined by integration.: 1301[failed verification] Align the -axis with the axis of the disc and define a volume element as , then where is its mass.
The moment of inertia of the compound pendulum is now obtained by adding the moment of inertia of the rod and the disc around the pivot point as, where is the length of the pendulum. Notice that the parallel axis theorem is used to shift the moment of inertia from the center of mass to the pivot point of the pendulum.
A list of moments of inertia formulas for standard body shapes provides a way to obtain the moment of inertia of a complex body as an assembly of simpler shaped bodies. The parallel axis theorem is used to shift the reference point of the individual bodies to the reference point of the assembly.
As one more example, consider the moment of inertia of a solid sphere of constant density about an axis through its center of mass. This is determined by summing the moments of inertia of the thin discs that can form the sphere whose centers are along the axis chosen for consideration. If the surface of the sphere is defined by the equation: 1301
then the square of the radius of the disc at the cross-section along the -axis is
Therefore, the moment of inertia of the sphere is the sum of the moments of inertia of the discs along the -axis, where is the mass of the sphere.
Rigid body
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If a mechanical system is constrained to move parallel to a fixed plane, then the rotation of a body in the system occurs around an axis parallel to this plane. In this case, the moment of inertia of the mass in this system is a scalar known as the polar moment of inertia. The definition of the polar moment of inertia can be obtained by considering momentum, kinetic energy and Newton's laws for the planar movement of a rigid system of particles.
If a system of particles, , are assembled into a rigid body, then the momentum of the system can be written in terms of positions relative to a reference point , and absolute velocities : where is the angular velocity of the system and is the velocity of .
For planar movement the angular velocity vector is directed along the unit vector which is perpendicular to the plane of movement. Introduce the unit vectors from the reference point to a point , and the unit vector , so
This defines the relative position vector and the velocity vector for the rigid system of the particles moving in a plane.
Note on the cross product: When a body moves parallel to a ground plane, the trajectories of all the points in the body lie in planes parallel to this ground plane. This means that any rotation that the body undergoes must be around an axis perpendicular to this plane. Planar movement is often presented as projected onto this ground plane so that the axis of rotation appears as a point. In this case, the angular velocity and angular acceleration of the body are scalars and the fact that they are vectors along the rotation axis is ignored. This is usually preferred for introductions to the topic. But in the case of moment of inertia, the combination of mass and geometry benefits from the geometric properties of the cross product. For this reason, in this section on planar movement the angular velocity and accelerations of the body are vectors perpendicular to the ground plane, and the cross product operations are the same as used for the study of spatial rigid body movement.
Angular momentum
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The angular momentum vector for the planar movement of a rigid system of particles is given by
Use the center of mass as the reference point so
and define the moment of inertia relative to the center of mass as
then the equation for angular momentum simplifies to: 1028
The moment of inertia about an axis perpendicular to the movement of the rigid system and through the center of mass is known as the polar moment of inertia. Specifically, it is the second moment of mass with respect to the orthogonal distance from an axis (or pole).
For a given amount of angular momentum, a decrease in the moment of inertia results in an increase in the angular velocity. Figure skaters can change their moment of inertia by pulling in their arms. Thus, the angular velocity achieved by a skater with outstretched arms results in a greater angular velocity when the arms are pulled in, because of the reduced moment of inertia. A figure skater is not, however, a rigid body.
Kinetic energy
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The kinetic energy of a rigid system of particles moving in the plane is given by
Let the reference point be the center of mass of the system so the second term becomes zero, and introduce the moment of inertia so the kinetic energy is given by: 1084
The moment of inertia is the polar moment of inertia of the body.
Newton's laws
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Newton's laws for a rigid system of particles, , can be written in terms of a resultant force and torque at a reference point , to yield where denotes the trajectory of each particle.
The kinematics of a rigid body yields the formula for the acceleration of the particle in terms of the position and acceleration of the reference particle as well as the angular velocity vector and angular acceleration vector of the rigid system of particles as,
For systems that are constrained to planar movement, the angular velocity and angular acceleration vectors are directed along perpendicular to the plane of movement, which simplifies this acceleration equation. In this case, the acceleration vectors can be simplified by introducing the unit vectors from the reference point to a point and the unit vectors , so
This yields the resultant torque on the system as
where , and is the unit vector perpendicular to the plane for all of the particles .
Use the center of mass as the reference point and define the moment of inertia relative to the center of mass , then the equation for the resultant torque simplifies to: 1029
Motion in space of a rigid body, and the inertia matrix
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The scalar moments of inertia appear as elements in a matrix when a system of particles is assembled into a rigid body that moves in three-dimensional space. This inertia matrix appears in the calculation of the angular momentum, kinetic energy and resultant torque of the rigid system of particles.
For analysis of a spinning top, see Precession § Classical (Newtonian), and Euler's equations (rigid body dynamics).
Let the system of particles, be located at the coordinates with velocities relative to a fixed reference frame. For a (possibly moving) reference point , the relative positions are and the (absolute) velocities are where is the angular velocity of the system, and is the velocity of .
Angular momentum
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Note that the cross product can be equivalently written as matrix multiplication by combining the first operand and the operator into a skew-symmetric matrix, , constructed from the components of :
The inertia matrix is constructed by considering the angular momentum, with the reference point of the body chosen to be the center of mass : where the terms containing () sum to zero by the definition of center of mass.
Then, the skew-symmetric matrix obtained from the relative position vector , can be used to define, where defined by is the symmetric inertia matrix of the rigid system of particles measured relative to the center of mass .
Kinetic energy
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The kinetic energy of a rigid system of particles can be formulated in terms of the center of mass and a matrix of mass moments of inertia of the system. Let the system of particles be located at the coordinates with velocities , then the kinetic energy is where is the position vector of a particle relative to the center of mass.
This equation expands to yield three terms
Since the center of mass is defined by , the second term in this equation is zero. Introduce the skew-symmetric matrix so the kinetic energy becomes
Thus, the kinetic energy of the rigid system of particles is given by where is the inertia matrix relative to the center of mass and is the total mass.
Resultant torque
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The inertia matrix appears in the application of Newton's second law to a rigid assembly of particles. The resultant torque on this system is, where is the acceleration of the particle . The kinematics of a rigid body yields the formula for the acceleration of the particle in terms of the position and acceleration of the reference point, as well as the angular velocity vector and angular acceleration vector of the rigid system as,
Use the center of mass as the reference point, and introduce the skew-symmetric matrix to represent the cross product , to obtain
The calculation uses the identity obtained from the Jacobi identity for the triple cross product as shown in the proof below:
Proof
In the last statement, because is either at rest or moving at a constant velocity but not accelerated, or the origin of the fixed (world) coordinate reference system is placed at the center of mass . And distributing the cross product over the sum, we get
Then, the following Jacobi identity is used on the last term:
The result of applying Jacobi identity can then be continued as follows:
The final result can then be substituted to the main proof as follows:
Notice that for any vector , the following holds:
Finally, the result is used to complete the main proof as follows:
Thus, the resultant torque on the rigid system of particles is given by where is the inertia matrix relative to the center of mass.
Parallel axis theorem
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Main article: Parallel axis theorem
The inertia matrix of a body depends on the choice of the reference point. There is a useful relationship between the inertia matrix relative to the center of mass and the inertia matrix relative to another point . This relationship is called the parallel axis theorem.
Consider the inertia matrix obtained for a rigid system of particles measured relative to a reference point , given by
Let be the center of mass of the rigid system, then where is the vector from the center of mass to the reference point . Use this equation to compute the inertia matrix,
Distribute over the cross product to obtain
The first term is the inertia matrix relative to the center of mass. The second and third terms are zero by definition of the center of mass . And the last term is the total mass of the system multiplied by the square of the skew-symmetric matrix constructed from .
The result is the parallel axis theorem, where is the vector from the center of mass to the reference point .
Note on the minus sign: By using the skew symmetric matrix of position vectors relative to the reference point, the inertia matrix of each particle has the form , which is similar to the that appears in planar movement. However, to make this to work out correctly a minus sign is needed. This minus sign can be absorbed into the term , if desired, by using the skew-symmetry property of .
Scalar moment of inertia in a plane
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The scalar moment of inertia, , of a body about a specified axis whose direction is specified by the unit vector and passes through the body at a point is as follows: where is the moment of inertia matrix of the system relative to the reference point , and is the skew symmetric matrix obtained from the vector .
This is derived as follows. Let a rigid assembly of particles, , have coordinates . Choose as a reference point and compute the moment of inertia around a line L defined by the unit vector through the reference point , . The perpendicular vector from this line to the particle is obtained from by removing the component that projects onto . where is the identity matrix, so as to avoid confusion with the inertia matrix, and is the outer product matrix formed from the unit vector along the line .
To relate this scalar moment of inertia to the inertia matrix of the body, introduce the skew-symmetric matrix such that , then we have the identity noting that is a unit vector.
The magnitude squared of the perpendicular vector is
The simplification of this equation uses the triple scalar product identity where the dot and the cross products have been interchanged. Exchanging products, and simplifying by noting that and are orthogonal:
Thus, the moment of inertia around the line through in the direction is obtained from the calculation where is the moment of inertia matrix of the system relative to the reference point .
This shows that the inertia matrix can be used to calculate the moment of inertia of a body around any specified rotation axis in the body.
Inertia tensor
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For the same object, different axes of rotation will have different moments of inertia about those axes. In general, the moments of inertia are not equal unless the object is symmetric about all axes. The moment of inertia tensor is a convenient way to summarize all moments of inertia of an object with one quantity. It may be calculated with respect to any point in space, although for practical purposes the center of mass is most commonly used.
Definition
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For a rigid object of point masses , the moment of inertia tensor is given by
Its components are defined as where
, is equal to 1, 2 or 3 for , , and , respectively,
is the vector to the point mass from the point about which the tensor is calculated and
is the Kronecker delta.
Note that, by the definition, is a symmetric tensor.
The diagonal elements are more succinctly written as while the off-diagonal elements, also called the products of inertia, are
Here denotes the moment of inertia around the -axis when the objects are rotated around the x-axis, denotes the moment of inertia around the -axis when the objects are rotated around the -axis, and so on.
These quantities can be generalized to an object with distributed mass, described by a mass density function, in a similar fashion to the scalar moment of inertia. One then has where is their outer product, E3 is the 3×3 identity matrix, and V is a region of space completely containing the object.
Alternatively it can also be written in terms of the angular momentum operator :
The inertia tensor can be used in the same way as the inertia matrix to compute the scalar moment of inertia about an arbitrary axis in the direction , where the dot product is taken with the corresponding elements in the component tensors. A product of inertia term such as is obtained by the computation and can be interpreted as the moment of inertia around the -axis when the object rotates around the -axis.
The components of tensors of degree two can be assembled into a matrix. For the inertia tensor this matrix is given by,
It is common in rigid body mechanics to use notation that explicitly identifies the , , and -axes, such as and , for the components of the inertia tensor.
Alternate inertia convention
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There are some CAD and CAE applications such as SolidWorks, Unigraphics NX/Siemens NX and MSC Adams that use an alternate convention for the products of inertia. According to this convention, the minus sign is removed from the product of inertia formulas and instead inserted in the inertia matrix:
Determine inertia convention (principal axes method)
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If one has the inertia data without knowing which inertia convention that has been used, it can be determined if one also has the principal axes. With the principal axes method, one makes inertia matrices from the following two assumptions:
The standard inertia convention has been used .
The alternate inertia convention has been used .
Next, one calculates the eigenvectors for the two matrices. The matrix whose eigenvectors are parallel to the principal axes corresponds to the inertia convention that has been used.
Derivation of the tensor components
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The distance of a particle at from the axis of rotation passing through the origin in the direction is , where is a unit vector. The moment of inertia on the axis is
Rewrite the equation using matrix transpose: where E3 is the 3×3 identity matrix.
This leads to a tensor formula for the moment of inertia
For multiple particles, we need only recall that the moment of inertia is additive in order to see that this formula is correct.
Inertia tensor of translation
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Main article: Parallel axis theorem § Tensor generalization
Let be the inertia tensor of a body calculated at its center of mass, and be the displacement vector of the body. The inertia tensor of the translated body respect to its original center of mass is given by: where is the body's mass, E3 is the 3 × 3 identity matrix, and is the outer product.
Inertia tensor of rotation
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Let be the matrix that represents a body's rotation. The inertia tensor of the rotated body is given by:
Inertia matrix in different reference frames
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The use of the inertia matrix in Newton's second law assumes its components are computed relative to axes parallel to the inertial frame and not relative to a body-fixed reference frame. This means that as the body moves the components of the inertia matrix change with time. In contrast, the components of the inertia matrix measured in a body-fixed frame are constant.
Body frame
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Let the body frame inertia matrix relative to the center of mass be denoted , and define the orientation of the body frame relative to the inertial frame by the rotation matrix , such that, where vectors in the body fixed coordinate frame have coordinates in the inertial frame. Then, the inertia matrix of the body measured in the inertial frame is given by
Notice that changes as the body moves, while remains constant.
Principal axes
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Measured in the body frame, the inertia matrix is a constant real symmetric matrix. A real symmetric matrix has the eigendecomposition into the product of a rotation matrix and a diagonal matrix , given by where
The columns of the rotation matrix define the directions of the principal axes of the body, and the constants , , and are called the principal moments of inertia. This result was first shown by J. J. Sylvester (1852), and is a form of Sylvester's law of inertia. When the body has an axis of symmetry (sometimes called the figure axis or axis of figure) then the other two moments of inertia will be identical and any axis perpendicular to the axis of symmetry will be a principal axis.
A toy top is an example of a rotating rigid body, and the word top is used in the names of types of rigid bodies. When all principal moments of inertia are distinct, the principal axes through center of mass are uniquely specified and the rigid body is called an asymmetric top. If two principal moments are the same, the rigid body is called a symmetric top and there is no unique choice for the two corresponding principal axes. If all three principal moments are the same, the rigid body is called a spherical top (although it need not be spherical) and any axis can be considered a principal axis, meaning that the moment of inertia is the same about any axis.
The principal axes are often aligned with the object's symmetry axes. If a rigid body has an axis of symmetry of order , meaning it is symmetrical under rotations of 360°/m about the given axis, that axis is a principal axis. When , the rigid body is a symmetric top. If a rigid body has at least two symmetry axes that are not parallel or perpendicular to each other, it is a spherical top, for example, a cube or any other Platonic solid.
The motion of vehicles is often described in terms of yaw, pitch, and roll which usually correspond approximately to rotations about the three principal axes. If the vehicle has bilateral symmetry then one of the principal axes will correspond exactly to the transverse (pitch) axis.
A practical example of this mathematical phenomenon is the routine automotive task of balancing a tire, which basically means adjusting the distribution of mass of a car wheel such that its principal axis of inertia is aligned with the axle so the wheel does not wobble.
Rotating molecules are also classified as asymmetric, symmetric, or spherical tops, and the structure of their rotational spectra is different for each type.
Ellipsoid
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The moment of inertia matrix in body-frame coordinates is a quadratic form that defines a surface in the body called Poinsot's ellipsoid. Let be the inertia matrix relative to the center of mass aligned with the principal axes, then the surface or defines an ellipsoid in the body frame. Write this equation in the form, to see that the semi-principal diameters of this ellipsoid are given by
Let a point on this ellipsoid be defined in terms of its magnitude and direction, , where is a unit vector. Then the relationship presented above, between the inertia matrix and the scalar moment of inertia around an axis in the direction , yields
Thus, the magnitude of a point in the direction on the inertia ellipsoid is
See also
[edit]
Central moment
List of moments of inertia
Moment of inertia factor
Planar lamina
Rotational energy
References
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^ C. Couch and J. Mayes, Trifilar Pendulum for MOI, Happresearch.com, 2016.
^ Gracey, William, The experimental determination of the moments of inertia of airplanes by a simplified compound-pendulum method, NACA Technical Note No. 1629, 1948
^ Morrow, H. W.; Kokernak, Robert (2011). Statics and Strengths of Materials (7 ed.). New Jersey: Prentice Hall. pp. 192–196. ISBN 978-0135034521.
^ In that situation this moment of inertia only describes how a torque applied along that axis causes a rotation about that axis. But, torques not aligned along a principal axis will also cause rotations about other axes.
^ a b c d e f g h i Ferdinand P. Beer; E. Russell Johnston, Jr.; Phillip J. Cornwell (2010). Vector mechanics for engineers: Dynamics (9th ed.). Boston: McGraw-Hill. ISBN 978-0077295493.
^ Walter D. Pilkey, Analysis and Design of Elastic Beams: Computational Methods, John Wiley, 2002.
^ a b Goldstein, H. (1980). Classical Mechanics (2nd ed.). Addison-Wesley. ISBN 0-201-02918-9.
^ L. D. Landau and E. M. Lifshitz, Mechanics, Vol 1. 2nd Ed., Pergamon Press, 1969.
^ L. W. Tsai, Robot Analysis: The mechanics of serial and parallel manipulators, John-Wiley, NY, 1999.
^ David, Baraff. "Physically Based Modeling - Rigid Body Simulation" (PDF). Pixar Graphics Technologies.
^ Sylvester, J J (1852). "A demonstration of the theorem that every homogeneous quadratic polynomial is reducible by real orthogonal substitutions to the form of a sum of positive and negative squares" (PDF). Philosophical Magazine. 4th Series. 4 (23): 138–142. doi:10.1080/14786445208647087. Retrieved June 27, 2008.
^ Norman, C.W. (1986). Undergraduate algebra. Oxford University Press. pp. 360–361. ISBN 0-19-853248-2.
^ Mason, Matthew T. (2001). Mechanics of Robotics Manipulation. MIT Press. ISBN 978-0-262-13396-8. Retrieved November 21, 2014.
External links
[edit]
Wikimedia Commons has media related to Moments of inertia.
Angular momentum and rigid-body rotation in two and three dimensions
Lecture notes on rigid-body rotation and moments of inertia
The moment of inertia tensor
An introductory lesson on moment of inertia: keeping a vertical pole not falling down (Java simulation)
Tutorial on finding moments of inertia, with problems and solutions on various basic shapes
Notes on mechanics of manipulation: the angular inertia tensor
Easy to use and Free Moment of Inertia Calculator online
| v t e Classical mechanics SI units |
| | | | --- | | Linear/translational quantities | Angular/rotational quantities | | Dimensions | 1 | L | L2 | Dimensions | 1 | θ | θ2 | | T | time: ts | absement: Am s | T | time: ts | | 1 | distance: d, position: r, s, x, displacementm | area: Am2 | 1 | angle: θ, angular displacement: θrad | solid angle: Ωrad2, sr | | T−1 | frequency: fs−1, Hz | speed: v, velocity: vm s−1 | kinematic viscosity: ν,specific angular momentum: hm2 s−1 | T−1 | frequency: f, rotational speed: n, rotational velocity: ns−1, Hz | angular speed: ω, angular velocity: ωrad s−1 | | T−2 | acceleration: am s−2 | T−2 | rotational accelerations−2 | angular acceleration: αrad s−2 | | T−3 | jerk: jm s−3 | T−3 | angular jerk: ζrad s−3 | | M | mass: mkg | weighted position: M ⟨x⟩ = ∑ m x | moment of inertia: Ikg m2 | ML | | MT−1 | Mass flow rate: kg s−1 | momentum: p, impulse: Jkg m s−1, N s | action: 𝒮, actergy: ℵkg m2 s−1, J s | MLT−1 | angular momentum: L, angular impulse: ΔLkg m rad s−1 | | MT−2 | force: F, weight: Fgkg m s−2, N | energy: E, work: W, Lagrangian: Lkg m2 s−2, J | MLT−2 | torque: τ, moment: Mkg m rad s−2, N m | | MT−3 | yank: Ykg m s−3, N s−1 | power: Pkg m2 s−3, W | MLT−3 | rotatum: Pkg m rad s−3, N m s−1 | |
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高中生物专题2微生物的培养与应用第6课时土壤中分解尿素的细菌的分离与计数同步备课教学案人教版 - 百度文库
-1-/16
第
6
课时
土壤中分解尿素的细菌
的分离与计数
［
学习导航
］
1.
通过分析具体培养基的配方，归纳选择培养基的选择作用，理解筛选微生物
的原理。
2.
结合实验设计，
分析微生物数量测定实验中产生偏差的原因及解决措施，学会微
生物的计数方法，并对培养结果进行分析和评价。
［
重难点击
］
1.
研究培养基对微生物的选择作用，进行微生物数量的测定。
2.
能利用选择培
养基分离细菌，运用相关技术解决生产生活中有关微生物的计数问题。
—、研究培养基对微生物的选择作用
棊
础梳理
夯实基础突破要点
|
1.
筛选菌株
⑴自然界中目的菌株的筛选
①
原理：根据目的菌株对生存环境的要求，到相应的环境中去寻找。
②
实例：能产生耐高温的
Taq
DNA
聚合酶的
Taq
细菌就是从热泉中筛选出来的。
(2)
实验室中目的菌株的筛选
①
原理：人为提供有利于目的菌株生长的条件
(
包括营养、温度、
pH
等
)
，同时抑制或阻止其
他微生物生长。
②
实例：从土壤中筛选能分解尿素的细菌。
③
方法：利用选择培养基进行微生物的筛选。
2.
选择培养基
(1)
概念：允许特定种类的微牛物牛长，同时抑制或阻止其他种类微牛物牛长的培养基。
(2)
举例：筛选土壤中分解尿素的细菌的选择培养基是以尿素为唯一氮源，按物理性质归类为
-2-/16
1.
筛选分解尿素的细菌的选择培养基
F
面是本课题使用的培养基的配方，请分析
:
KHPQ
1.4 g
N&HPQ
2.1 g
MgSQ
・
7H
2
Q
0.2 g
葡萄糖
10.0 g
尿素
1.0 g
琼脂
15.0 g
将上述物质溶解后，用蒸馏水定容到
1 000 mL
(1)
在培养基的配方中，为微生物的生长提供碳源和氮源的分别是什么物质？
答案
碳源是葡萄糖，氮源是尿素。
(2)
分离分解尿素的细菌的选择培养基是如何进行选择的？
答案
该选择培养基的配方中，尿素是培养基中的唯一氮源，因此，只有能够利用尿素的微
生物才能够生长。
(3)
为什么只有能分解尿素的微生物才能在该培养基上生长？
答案
分解尿素的微生物能合成脲酶，而脲酶可将尿素分解成氨，从而为微生物提供氮源。
2.
设计对照实验
(1)
如何设计对照实验验证该选择培养基的确筛选到了能分解尿素的细菌？
答案
增设牛肉膏蛋白胨基础培养基并进行涂布平板操作，如果牛肉膏蛋白胨培养基上的菌
落数明显多于选择培养基上的菌落数，则说明选择培养基的确起到了筛选能分解尿素的细菌
的作用。
(2)
如何设计对照实验验证所用的选择培养基没有受到污染？
答案
设置一个未涂布平板的能筛选分解尿素的细菌的选择培养基。
归纳总结选择培养基的分类
1
利用营养缺陷型选择培养基进行的选择培养：通过控制培养基的营养成分，使营养缺陷型
微生物不能正常生长。
2
利用化学物质进行的选择培养：在完全培养基中加入某些化学物质，利用加入的化学物质
抑制某些微生物的生长，同时选择所需的微生物。
3
利用培养条件进行的选择培养：改变微生物的培养条件
如高温、特殊
pH
等，筛选特定微
生物。
固体培养基。
问题探究
理解升华
重
堆透析
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4955 | https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/15aa14b5647ea989a352a972dc4b3dfe_MIT8_01F16_pset7.pdf | 8.01x Classical Mechanics, Fall 2016 Massachusetts Institute of Technology Problem Set 7 1. Object Sliding Down an Inclined Plane An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an angle of θ = 30◦to the ground. The coefficient of kinetic friction µk = 0.2. At the bottom of the plane, the mass slides along a rough surface with a coefficient of kinetic friction µk = 0.3 until it comes to rest. The goal of this problem is to find out how far the object slides along the rough surface.
(a) What is the work done by the friction force while the mass is sliding down the inclined plane? (Is it positive or negative?) (b) What is the work done by the gravitational force while the mass is sliding down the inclined plane? (Is it positive or negative?) (c) What is the kinetic energy of the mass when it just reaches the bottom of the inclined plane?
(d) Symbolically, what is the work done by the friction force while the mass is sliding along the ground? Is this positive or negative? Express you answer in terms of some or all of the following: m, µk, g, and d where d is the distance it takes the object to stop measured from the bottom of the incline.
(e) How far from the bottom of the inclined plane does the object slide along the rough surface?
8.01x Classical Mechanics: Problem Set 7 2 2. Collision and Sliding on a Rough Surface Block A of mass mA is moving horizontally with speed vA along a frictionless surface.
It collides with block B of mass mB that is initially at rest. The two blocks stick together after the collision. At x = 0, block B enters a rough surface with a coefficient of kinetic friction that increases linearly with distance µk(x) = bx for 0 ≤x ≤d, where b is a positive constant.
At x = d, block B collides with an unstretched spring with spring constant k on a frictionless surface. The downward gravitational acceleration has magnitude g. What is the distance the spring is compressed when the blocks first comes to rest? Express you answer in terms of some or all of the following: vA, b, d, g, k, mA and mB.
8.01x Classical Mechanics: Problem Set 7 3 3. Inclined plane A body of mass m is attached to one end of a string of length R.
The other end of the string is fixed on an inclined plane making an angle φ with the horizontal as shown in the figure. The body has speed v0 at the bottom of the circle (point A). The body undergoes circular motion. There is a coefficient of sliding friction µ between the body and the plane. The downward acceleration of gravity is g. Express all answers in terms of m, φ, v0, g, µ and R as needed.
(a) How much work does the friction force do on the body as it moves from the bottom of the circle (point A) to the top of the circle (point B)?
(b) What is the tension in the string when it reaches point B? Express your answer in terms of m, φ, v0, g, µ and R as needed.
MIT OpenCourseWare 8.01 Classical Mechanics Fall 2016 For information about citing these materials or our Terms of Use, visit: |
4956 | https://math.stackexchange.com/questions/1704020/constant-functions-periodic | calculus - Constant functions periodic? - Mathematics Stack Exchange
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Constant functions periodic?
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I dont understand the meaning of this line in my book -
" sin 2 x+cos 2 x sin 2x+cos 2x is periodic but the fundamental period is not defined. "
Why is the period not defined? F(x)F(x) is 1 1 here so it is a constant function which should be periodic ?
Does this mean all constant functions are not periodic?
Please explain
calculus
functions
trigonometry
periodic-functions
constants
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edited Mar 19, 2016 at 6:24
Mikasa
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geek101geek101
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2 It is periodic, but every positive real is a period.André Nicolas –André Nicolas 2016-03-19 05:56:13 +00:00 Commented Mar 19, 2016 at 5:56
2 The fundamental period of a periodic function f(x)f(x) is the smallest T>0 T>0 (if one exists) such that f(x+T)=f(x)f(x+T)=f(x) for all x x. For example, cos(x)cos(x) the fundamental period is T=2 π T=2 π. For a constant function, this hold for any T>0 T>0 so there is no smallest positive T T and so there is no fundamental period.User8128 –User8128 2016-03-19 05:56:15 +00:00 Commented Mar 19, 2016 at 5:56
1 There are other periodic functions with no fundamental period. For example, let f(t)=0 f(t)=0 if t t is rational, and 1 1 if t t is irrational. Then every positive rational is a period.André Nicolas –André Nicolas 2016-03-19 06:09:50 +00:00 Commented Mar 19, 2016 at 6:09
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A function f:R→R f:R→R is periodic if there exists some number t>0 t>0 such that
f(x)=f(x+t)f(x)=f(x+t)
A constant function is periodic since you can take t=1,t=2 t=1,t=2, etc. (Hint: Hover over the tag "periodic-functions". What do you see?)
The fundamental period of f f is the smallest of such t t's. Since t t cannot be 0 0, you are looking for the minimum of (0,∞)(0,∞), which does not exist.
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edited Jul 23, 2016 at 20:38
hmakholm left over Monica
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answered Mar 19, 2016 at 5:57
Henricus V.Henricus V.
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The fundamental period of f f is the smallest positive number T T so that f(t±T)=f(t)f(t±T)=f(t) for all t t. Since no smallest number exists for f(t)=sin 2(t)+cos 2(t)=1 f(t)=sin 2(t)+cos 2(t)=1, there is no fundamental period.
There are different definitions of periodic (some require a fundamental period, others do not), but I prefer to exclude constant functions from being periodic.
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answered Mar 19, 2016 at 5:57
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When i googled the definition of periodic function i got-"A function f(x) is periodic if f(x) = f(x + p) for all x and some fixed p. p is called a period of f. The smallest positive p that works is called the period of f. (A constant function is periodic with any value of p as a period, so there's no such thing as "the period" of such a function." Which explains everything.
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answered Mar 19, 2016 at 6:04
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4957 | https://www.mathplanet.com/education/algebra-2/polynomial-functions/descartes-rule-of-sign | Descartes' rule of sign - Mathplanet
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Descartes' rule of sign
Do excercisesShow all 3 exercises
Descartes rules of signs I
Descartes rules of signs II
Descartes rules of signs III
Descartes' rule of sign is used to determine the number of real zeros of a polynomial function.
It tells us that the number of positive real zeros in a polynomial function f(x) is the same or less than by an even numbers as the number of changes in the sign of the coefficients. The number of negative real zeros of the f(x) is the same as the number of changes in sign of the coefficients of the terms of f(-x) or less than this by an even number.
We will show how it works with an example.
Example
Determine the number of positive and negative real zeros for the given function (this example is also shown in our video lesson):
f(x)=x 5+4 x 4−3 x 2+x−6 f(x)=x 5+4 x 4−3 x 2+x−6
Our function is arranged in descending powers of the variable, if it was not in this order we would have to rearrange the terms as our first step. Second we count the number of changes in sign for the coefficients of f(x).
Here are the coefficients of our variable in f(x):
1+4−3+1−6 1+4−3+1−6
Our variables goes from positive(1) to positive(4) to negative(-3) to positive(1) to negative(-6).
Between the first two coefficients there are no change in signs but between our second and third we have our first change, then between our third and fourth we have our second change and between our 4 th and 5 th coefficients we have a third change of coefficients. Descartes' rule of signs tells us that the we then have exactly 3 real positive zeros or less but an odd number of zeros. Hence our number of positive zeros must then be either 3, or 1.
In order to find the number of negative zeros we find f(-x) and count the number of changes in sign for the coefficients:
f(−x)=(−x)5+4(−x)4−3(−x)2+(−x)−6==−x 5+4 x 4−3 x 2−x−6 f(−x)=(−x)5+4(−x)4−3(−x)2+(−x)−6==−x 5+4 x 4−3 x 2−x−6
Here we can see that we have two changes of signs, hence we have two negative zeros or less but a even number of zeros..
In total we have 3 or 1 positive zeros or 2 or 0 negative zeros.
Do excercisesShow all 3 exercises
Descartes rules of signs I
Descartes rules of signs II
Descartes rules of signs III
More classes on this subject
Algebra 2 Polynomial functions: Basic knowledge of polynomial functionsAlgebra 2 Polynomial functions: Remainder and factor theoremsAlgebra 2 Polynomial functions: Roots and zeros
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POLYNOMIAL FUNCTIONS –Composition of functions
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4958 | https://en.wikipedia.org/wiki/Altitude_(triangle) | Jump to content
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Contents
(Top)
1 Theorems
1.1 Orthocenter
1.2 Altitude in terms of the sides
1.3 Inradius theorems
1.4 Circumradius theorem
1.5 Interior point
1.6 Area theorem
1.7 General point on an altitude
1.8 Triangle inequality
1.9 Special cases
1.9.1 Equilateral triangle
1.9.2 Right triangle
2 See also
3 Notes
4 References
5 External links
Altitude (triangle)
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From Wikipedia, the free encyclopedia
Perpendicular line segment from a triangle's side to opposite vertex
In geometry, an altitude of a triangle is a line segment through a given vertex (called apex) and perpendicular to a line containing the side or edge opposite the apex. This (finite) edge and (infinite) line extension are called, respectively, the base and extended base of the altitude. The point at the intersection of the extended base and the altitude is called the foot of the altitude. The length of the altitude, often simply called "the altitude" or "height", symbol h, is the distance between the foot and the apex. The process of drawing the altitude from a vertex to the foot is known as dropping the altitude at that vertex. It is a special case of orthogonal projection.
Altitudes can be used in the computation of the area of a triangle: one-half of the product of an altitude's length and its base's length (symbol b) equals the triangle's area: A=hb/2. Thus, the longest altitude is perpendicular to the shortest side of the triangle. The altitudes are also related to the sides of the triangle through the trigonometric functions.
In an isosceles triangle (a triangle with two congruent sides), the altitude having the incongruent side as its base will have the midpoint of that side as its foot. Also the altitude having the incongruent side as its base will be the angle bisector of the vertex angle.
In a right triangle, the altitude drawn to the hypotenuse c divides the hypotenuse into two segments of lengths p and q. If we denote the length of the altitude by hc, we then have the relation
: (geometric mean theorem; see special cases, inverse Pythagorean theorem)
For acute triangles, the feet of the altitudes all fall on the triangle's sides (not extended). In an obtuse triangle (one with an obtuse angle), the foot of the altitude to the obtuse-angled vertex falls in the interior of the opposite side, but the feet of the altitudes to the acute-angled vertices fall on the opposite extended side, exterior to the triangle. This is illustrated in the adjacent diagram: in this obtuse triangle, an altitude dropped perpendicularly from the top vertex, which has an acute angle, intersects the extended horizontal side outside the triangle.
Theorems
[edit]
The geometric altitude figures prominently in many important theorems and their proofs. For example, besides those theorems listed below, the altitude plays a central role in proofs of both the Law of sines and Law of cosines.
Orthocenter
[edit]
This section is an excerpt from Orthocenter.[edit]
The orthocenter of a triangle, usually denoted by H, is the point where the three (possibly extended) altitudes intersect. The orthocenter lies inside the triangle if and only if the triangle is acute. For a right triangle, the orthocenter coincides with the vertex at the right angle. For an equilateral triangle, all triangle centers (including the orthocenter) coincide at its centroid.
Altitude in terms of the sides
[edit]
For any triangle with sides a, b, c and semiperimeter the altitude from side a (the base) is given by
This follows from combining Heron's formula for the area of a triangle in terms of the sides with the area formula where the base is taken as side a and the height is the altitude from the vertex A (opposite side a).
By exchanging a with b or c, this equation can also used to find the altitudes hb and hc, respectively.
Any two altitudes of a triangle are inversely proportional with the sides on which they fall.
Inradius theorems
[edit]
Consider an arbitrary triangle with sides a, b, c and with corresponding altitudes ha, hb, hc. The altitudes and the incircle radius r are related by: Lemma 1
Circumradius theorem
[edit]
Denoting the altitude from one side of a triangle as ha, the other two sides as b and c, and the triangle's circumradius (radius of the triangle's circumscribed circle) as R, the altitude is given by
Interior point
[edit]
If p1, p2, p3 are the perpendicular distances from any point P to the sides, and h1, h2, h3 are the altitudes to the respective sides, then
Area theorem
[edit]
Denoting the altitudes of any triangle from sides a, b, c respectively as ha, hb, hc, and the semi-sum of the reciprocals of the altitudes as then the reciprocal of area is
General point on an altitude
[edit]
If E is any point on an altitude AD of any triangle △ABC, then: 77–78
Triangle inequality
[edit]
Since the area of the triangle is , the triangle inequality implies
: .
Special cases
[edit]
Equilateral triangle
[edit]
From any point P within an equilateral triangle, the sum of the perpendiculars to the three sides is equal to the altitude of the triangle. This is Viviani's theorem.
Right triangle
[edit]
In a right triangle with legs a and b and hypotenuse c, each of the legs is also an altitude: and . The third altitude can be found by the relation
This is also known as the inverse Pythagorean theorem.
Note in particular:
See also
[edit]
Median (geometry)
Notes
[edit]
^ Smart 1998, p. 156
^ a b Berele & Goldman 2001, p. 118
^ Andrica, Dorin; Marinescu, Dan Ştefan (2017). "New Interpolation Inequalities to Euler's R ≥ 2r" (PDF). Forum Geometricorum. 17: 149–156. Archived from the original (PDF) on 2018-04-24.
^ Johnson 2007, p. 71, Section 101a
^ Johnson 2007, p. 74, Section 103c
^ Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle," Mathematical Gazette 89, November 2005, 494.
^ Alfred S. Posamentier and Charles T. Salkind, Challenging Problems in Geometry, Dover Publishing Co., second revised edition, 1996.
^ Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle", Mathematical Gazette 89 (November 2005), 494.
^ Voles, Roger, "Integer solutions of ," Mathematical Gazette 83, July 1999, 269–271.
^ Richinick, Jennifer, "The upside-down Pythagorean Theorem," Mathematical Gazette 92, July 2008, 313–317.
References
[edit]
Altshiller-Court, Nathan (2007) , College Geometry, Dover
Berele, Allan; Goldman, Jerry (2001), Geometry: Theorems and Constructions, Prentice Hall, ISBN 0-13-087121-4
Bogomolny, Alexander. "Existence of the Orthocenter". Cut the Knot. Retrieved 2022-12-17.
Johnson, Roger A. (2007) , Advanced Euclidean Geometry, Dover, ISBN 978-0-486-46237-0
Smart, James R. (1998), Modern Geometries (5th ed.), Brooks/Cole, ISBN 0-534-35188-3
External links
[edit]
Weisstein, Eric W. "Altitude". MathWorld.
Retrieved from "
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4960 | https://www.quora.com/How-do-I-calculate-the-inverse-of-a-matrix-The-inverse-of-a-matrix-A-is-given-by-A-1-frac-1-det-A-adj-A-where-adj-A-is-the-adjugate-of-A | How to calculate the inverse of a matrix - Quora
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How do I calculate the inverse of a matrix? The inverse of a matrix A is given by [math]A^{-1} = \frac{1}{det(A)} adj(A)[/math], where adj(A) is the adjugate of A.
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Andy Baker
Works at University of Glasgow · Author has 7.3K answers and 1.7M answer views
·2y
This is the sort of question that people who know or understand little mathematics find hard to understand, but an introductory course on linear algebra will give you a lot of insight.
First, the question includes a stated formula for the inverse of a non-singular matrix, which is fine except that as soon as the matrix is large it is computationally useless! The reason is that calculating the determinant and the adjugate matrix involve enormous the matrix is numbers of steps conatins many zeros.
As you will learn when you read an introductory textbook on the subject, if you want to find the inve
Continue Reading
This is the sort of question that people who know or understand little mathematics find hard to understand, but an introductory course on linear algebra will give you a lot of insight.
First, the question includes a stated formula for the inverse of a non-singular matrix, which is fine except that as soon as the matrix is large it is computationally useless! The reason is that calculating the determinant and the adjugate matrix involve enormous the matrix is numbers of steps conatins many zeros.
As you will learn when you read an introductory textbook on the subject, if you want to find the inverse of an actually large matrix (large probably means 4 by 4 or bigger) in practice something like Gaussian elimination is the practical method. The basic idea is that for an [math]n[/math] by [math]n[/math] matrix [math]A[/math], you start with the [math]n[/math] by [math]2n[/math] matrix [math][A\mid I_n][/math] and perform elementary row operations until you get the row equivalent matrix [math][I_n\mid B][/math] in which case [math]B=A^{-1}[/math]. If at some stage you find that you get an echelon matrix with a pivot occurring in the the [math]i[/math]-th column with [math]i>n[/math] then [math]A[/math] is singular so has no inverse, so this method event tells you if no inverse exists.
Of course there are special matrices where finding inverses is something that can be done more diectly but for numerical purposes an algorithmic approach is required and that is what mathematics undergraduates learn about.
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Alexander Farrugia
Used matrices extensively in his PhD thesis. · Upvoted by
Tina Müller
, M.Sc. Mathematics · Author has 3.2K answers and 27.5M answer views
·7y
Originally Answered: How do you calculate the inverse of a matrix? ·
In high school, the adjugate method (also known as the adjoint) and the Gauss-Jordan elimination method are usually taught. Here, I shall describe a rather interesting method that uses the Cayley-Hamilton theorem.
For an [math]n\times n[/math] matrix, this method requires finding the determinant of one [math]n\times n[/math] matrix, containing an indeterminate [math]x[/math], and the product of math[/math] identical matrices.
Here is a step-by-step account of this method. We assume that the [math]n\times n[/math] matrix we require the inverse of is [math]A[/math], whose entries are math[/math].
Construct the matrix [math]M[/math] whose diagonal entries are math[/math] and whose off-
Continue Reading
In high school, the adjugate method (also known as the adjoint) and the Gauss-Jordan elimination method are usually taught. Here, I shall describe a rather interesting method that uses the Cayley-Hamilton theorem.
For an [math]n\times n[/math] matrix, this method requires finding the determinant of one [math]n\times n[/math] matrix, containing an indeterminate [math]x[/math], and the product of math[/math] identical matrices.
Here is a step-by-step account of this method. We assume that the [math]n\times n[/math] matrix we require the inverse of is [math]A[/math], whose entries are math[/math].
Construct the matrix [math]M[/math] whose diagonal entries are math[/math] and whose off-diagonal entries are [math]-a_{ij}[/math].
Find the determinant of [math]M[/math], and expand the resulting polynomial [math]p(x)[/math].
If [math]c_0[/math], the coefficient of [math]x^0[/math] in [math]p(x)[/math], is [math]0[/math], then halt. The matrix [math]A[/math] has no inverse.
Otherwise, let [math]q(x)=-\frac{1}{c_0}\left(\dfrac{p(x)-c_0}{x}\right)[/math]. The inverse of matrix [math]A[/math] is [math]q(A)[/math].
Let’s see how this method works.
Example 1: [math]A=\begin{pmatrix}1 & 2 \ 1 & -3\end{pmatrix}[/math].
Let [math]M=\begin{pmatrix}x-1 & -2 \ -1 & x+3\end{pmatrix}[/math].
[math]\det(M)=(x-1)(x+3)-(-2)(-1)=x^2+3x-x-3-2=x^2+2x-5=p(x).[/math]
[math]c_0=-5\ne 0[/math]. So inverse exists.
[math]q(x)=-\frac{1}{-5}\left(\dfrac{x^2+2x-5-(-5)}{x}\right)=\frac{1}{5}(x+2).[/math] We thus determine [math]\frac{1}{5}(A+2I).[/math]
[math]\frac{1}{5}(A+2I)=\dfrac{1}{5}\left(\begin{pmatrix}1 & 2 \ 1 & -3\end{pmatrix}+\begin{pmatrix}2 & 0 \ 0 & 2\end{pmatrix}\right)=\dfrac{1}{5}\begin{pmatrix}3 & 2 \ 1 & -1\end{pmatrix}.[/math]
Hence [math]A^{-1}=\dfrac{1}{5}\begin{pmatrix}3 & 2 \ 1 & -1\end{pmatrix}=\begin{pmatrix}3/5 & 2/5 \ 1/5 & -1/5\end{pmatrix}.[/math]
Example 2: [math]A=\begin{pmatrix}1 & 2 & 3 \ 2 & -1 & 0 \ 0 & 1 & 2\end{pmatrix}[/math].
Let [math]M=\begin{pmatrix}x-1 & -2 & -3 \ -2 & x+1 & 0 \ 0 & -1 & x-2\end{pmatrix}[/math].
Expanding from third row, [math]\det(M)=-(-1)(0-6)+(x-2)((x-1)(x+1)-4)=-6+(x-2)(x^2-1-4)=-6+(x-2)(x^2-5)=-6+x^3-5x-2x^2+10=x^3-2x^2-5x+4.[/math]
[math]c_0=4\ne 0[/math], so inverse exists.
[math]q(x)=-\frac{1}{4}(x^2-2x-5)[/math]. So we evaluate [math]-\frac{1}{4}(A^2-2A-5I)[/math].
[math]A^2=\begin{pmatrix}1 & 2 & 3 \ 2 & -1 & 0 \ 0 & 1 & 2\end{pmatrix}\begin{pmatrix}1 & 2 & 3 \ 2 & -1 & 0 \ 0 & 1 & 2\end{pmatrix}=\begin{pmatrix}5 & 3 & 9 \ 0 & 5 & 6 \ 2 & 1 & 4\end{pmatrix}.[/math]
Thus
[math]A^{-1}=-\dfrac{1}{4}\left(\begin{pmatrix}5 & 3 & 9 \ 0 & 5 & 6 \ 2 & 1 & 4\end{pmatrix}-\begin{pmatrix}2 & 4 & 6 \ 4 & -2 & 0 \ 0 & 2 & 4\end{pmatrix}-\begin{pmatrix}5 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5\end{pmatrix}\right)[/math]
[math]=-\dfrac{1}{4}\begin{pmatrix}-2 & -1 & 3 \ -4 & 2 & 6 \ 2 & -1 & -5\end{pmatrix}[/math]
[math]A^{-1}=\begin{pmatrix}1/2 & 1/4 & -3/4 \ 1 & -1/2 & -3/2 \ -1/2 & 1/4 & 5/4\end{pmatrix}.[/math]
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Alexandru Carausu
Former Former University Associate Professor at Universitatea Tehnica "Gh. Asachi" Iasi (1966–2010) · Author has 3K answers and 875.5K answer views
·2y
Originally Answered: How do I calculate the inverse of a matrix? The inverse of a square matrix A is given by A^-1 = 1/det(A) adj(A), where adj(A) is the adjugate matrix of A. ·
The answer is included in the question : the expression of A^(-1) is correctly written.
Another way to find A _(-1) is based on the Gaussian Elimination method (in fact, the Gauss-Jordan elimination), applied on the rows of a block matrix of size n-by-(2 n) :
[ A | I_n ] ~ . . . ~ [ I_n | A^(-1) . (1)
I submitted at least ten answers to similar questions, in full details. The previous formula needs the calculation of a determinant of order n , and of other n^2 determinants (minors) of order n-1. The method of (1) does not need the calculation of any determinant.
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Vishakh Rajendran
M.S. in Aerospace and Aeronautical Engineering, Nanyang Technological University · Author has 682 answers and 3M answer views
·5y
Originally Answered: How do you calculate the inverse of a matrix? ·
One very easy way to determine the inverse of a matrix is by applying the Gauss Jordan method of Matrix Inversions.
This is not to be confused with the Gauss Jordan method of solving for a system of equations using matrices. In gauss Jordan technique, you form an augumented matrix and you use elementary row transformations (Note : row only) to reduce the augumented matrix into a diagonal UNIT matrix.The details of Gauss Jordan elimination method can be found at - Vishakh Rajendran's answer to How does the Gauss-Jordan elimination method work?
So this method of determining the inverse of a matrix
Continue Reading
One very easy way to determine the inverse of a matrix is by applying the Gauss Jordan method of Matrix Inversions.
This is not to be confused with the Gauss Jordan method of solving for a system of equations using matrices. In gauss Jordan technique, you form an augumented matrix and you use elementary row transformations (Note : row only) to reduce the augumented matrix into a diagonal UNIT matrix.The details of Gauss Jordan elimination method can be found at - Vishakh Rajendran's answer to How does the Gauss-Jordan elimination method work?
So this method of determining the inverse of a matrix is also called Gauss Jordan because over here also, we form the augumented matrix and arrive at a diagonal matrix.
Principle of Gauss Jordan method to determine Inverse of a Matrix
For a matrix A, if a matrix B exists such that A B = I (Identity matrix), then B is the inverse matrix of A.
So when you convert A to I on the left of the augumented matrix, you are converting I to A inverse on the right side of the augumented matrix.
Steps
Gauss Jordan technique is valid for square matrices and which are not singular.
a. Let us assume a matrix called A as shown below-
We form the augumented matrix which is of the form A | I
b. Using elementary row operations, we convert the augumented matrix into the form I | A inverse (Note : row operations only)
So the matrix on the right side is what the Inverse is.
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4961 | https://www.youtube.com/watch?v=SGUZ-8u1OxM | "Prove" 3 = 0. Can You Spot The Mistake?
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Thanks to Lucas from Brazil who sent this problem! This is a proof that 3 = 0 using sneaky algebra that I had never seen before (this is a sneaky mistake, not a simple divide by 0 fallacy). Every step seems correct, but the conclusion is clearly false. Can you find the mistake?
My blog post for this video
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link to book from Brazil: (page 34, problem 13)
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Transcript:
Hey, this is Presh Talwalkar. Here's a portion of an email I received. Hi I'm Lucas from Brazil, and I love your videos. Here's a problem from a book with math problems that I could not solve and the book doesn't say the answer, it just said "This one we leave to the students!" I translated the problem for you, and i'm also sending you a photo of the original problem and the book link from where I got the problem I was so grateful to Lucas, I replied right away. "Thanks for writing and for translating too, I appreciate the original photo" So what was the problem? It was a false proof that 3 is equal to 0 Here's how it starts. Let x be a solution of x squared plus x plus 1 equals 0 Since x is not equal to 0 we can divide both sides by x Let's simplify this to the following form we have x plus 1, plus 1 over x is equal to 0 Now, from the first equation we have x plus 1 is equal to negative x squared We're going to substitute that into the second equation so that first x plus 1 becomes negative x squared, then we have plus 1 over x is equal to 0 We'll now rearrange this equation to get 1 over x is equal to x squared This means 1 is equal to x cubed so x is equal to 1 We now take x equals 1 and substitute it into our original equation x squared plus x plus 1 equals 0 We get 1 squared plus 1 plus 1 is equal to 0 which means 3 is equal to 0 This is clearly a nonsensical result But every single step seemed to be correct So where is the mistake in this false proof that 3 is equal to 0? Can you figure it out? Give this problem a try and when you're ready keep watching the video for the explanation. So let's take a look at this false proof Which step introduces the mistake from which the ultimate result of three equals zero originates? That would be between the second and third steps where we substitute x plus 1 is equal to negative x squared The problem with this step is that it creates an extraneous solution x equals 1, which is not a solution to the original equation, x squared plus x plus 1 equals 0 So let me explain that in a little bit more detail We start out where x is a solution of x squared plus x plus 1 equals 0, so far so good This has two solutions, which we can get from the quadratic formula We have negative 1 plus i times the square root of 3 all over 2, and we have negative 1 minus i times the square root of 3 all over 2 Now x is not equal to 0 so it is valid for us to divide both sides by x Now we have a new equation x plus 1 plus 1 over x is equal to 0 This equation again has the same two solutions So far so good Now, we're going to substitute x plus 1 is equal to negative x squared We get the equation negative x squared plus 1 over x is equal to 0. Now what happens when you solve this equation? Well you end up with another solution so you have the same two solutions as before but then you end up with the new solution of x equals 1 Notice if you substitute x equals 1 into this equation you end up with negative 1 plus 1 equals 0 So x equals 1 is a solution to this equation But it was not a solution to the previous two equations So the lesson is that when we've substituted x plus 1 is equal to negative x squared, we've got an equation that has another solution, and this is not a solution to the original equation x squared plus x plus 1 is equal to 0 So you have to be careful if a step introduces an extraneous solution, Or else you could end up with a nonsensical result like three is equal to zero Did you identify where the mistake originated? Thanks for watching this video please subscribe to my channel I make videos on math you can catch me on my blog Mind Your Decisions. If you like this video you can check out my books which are linked in the video description and you can support me on Patreon. If you have a topic suggestion or a puzzle you can email me presh [at] mindyourdecisions [dot] com And you can catch me on social media either at mindyourdecisions or @preshtalwalkar |
4962 | https://math.stackexchange.com/questions/4299243/square-free-values-of-polynomials-on-mathbbzx | number theory - Square free values of polynomials on $\mathbb{Z}[x]$ - Mathematics Stack Exchange
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Square free values of polynomials on Z[x]
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Let f(x)∈Z[x] be a separable polynomial (i.e. with no repeated roots) of positive degree. Set B:=gcd{f(n):n∈Z} and let B′ be the smallest divisor of B so that B/B′ is square-free. For each prime p we denote by p q p the largest power of p dividing B′, and by r f(p) the number of a mod p 2+q p for which f(a)/B′=0 mod p 2. We set c f=∏p(1−r f p 2+q p) which is the conjectural density of integers n for which f(n)/B′ is squarefree.
(a) For f(x)=x(x+1)(x+2)(x+3), find B f and B′f, and show that r f(p)=4 for p≠2,3. Hence c f=R∏p≠2,3(1−4 p 2). Find R.
(b) Now assume that B′=1. Show that c f>0, i.e. that r f(p)<p 2 for all primes p.
Attempt: Unfortunately I couldn't even find B f and B′f. I don't understand how should I compute gcd{f(n):n∈Z} as n varies.
So, I didn't attempted the problem after that.
I would like to attempt the other parts of problem myself. So, kindly give hints for these only. I will ask you other parts later if I am struck on it.
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edited Feb 19, 2022 at 9:28
user26857
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asked Nov 7, 2021 at 12:42
user775699 user775699
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Work “locally”, ie prime by prime. If p is a prime, what is the highest power of p that divides every n(n+1)(n+2)(n+3) for all integers n? You can also check for small values of n to eliminate most primes.Aphelli –Aphelli 2021-11-07 13:09:00 +00:00 Commented Nov 7, 2021 at 13:09
You might want to have a look at the squarefree characteristic function ∑d 2|n μ(d)TravorLZH –TravorLZH 2021-11-07 14:00:00 +00:00 Commented Nov 7, 2021 at 14:00
@TravorLZH Can you please elaborate on that?user775699 –user775699 2022-01-08 17:30:35 +00:00 Commented Jan 8, 2022 at 17:30
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Hint for computing B f: First, you know that B f divides the value f(n) for any integer n, by the definition of B f. So, start by computing the first few values, to get a sense of what B f might be. In particular, you should be able to show using this technique that B f divides the number you guess B f is.
Once you've made this guess, you need to show that it divides f(n) for every integer n. For this, it's helpful to split up the verification by prime powers. Recall, for example, the proof that 2 divides n 2+n for every integer n -- you only need to check 2 values of n here.
Here's a worked example, for the polynomial f(x)=x(x 3+1)(x+11).
First, compute f(−2)=126, f(−1)=0, f(0)=0, f(1)=24, f(2)=234. The gcd of these values is 6, so we'll guess B f is 6.
To show that 6∣n(n 3+1)(n+11) for every positive integer n, it suffices to show that 2 divides this quantity and that 3 divides this quantity. Using that a−b∣P(a)−P(b) where a and b are integers and P is a polynomial with integer coefficients, it suffices to show that 2 divides f(0) and f(1), and that 3 divides f(0), f(1), and f(2). We've already computed each of these values, so this is easy to check.
So that this is a relatively complete answer, I'll add hints for the other portions as well, but they'll be spoiler-tagged, so you can avoid looking at them if you wish.
Hints for the remainder of part (a), once B and B′ are computed:
Since these primes don't divide B, we seek the number of a modulo p 2 for which a(a+1)(a+2)(a+3) is a multiple of p 2.
The only way for this to hold is if p 2 divides one of these factors, or if p divides two of these factors. Why can the second case not happen?
One can compute r 2(f) and r 3(f) by brute force.
Hint for part (b):
If B′=1, then r p(f) is the number of a modulo p 2 for which p 2∣f(a). If r p(f)≥p 2, then all a satisfy this. What does this say about $$?
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answered Jan 10, 2022 at 8:51
Carl SchildkrautCarl Schildkraut
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4963 | https://www.studypug.com/calculus-help/monotonic-and-bounded-sequences | Home
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Monotonic and bounded sequences
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Intros
Overview:
Overview:
Monotonic Sequences
3. Overview:
Bounded Sequences
Examples
Difference between monotonic and non-monotonic sequences
Show that the following sequences
is
monotonic. Is it an increasing or decreasing sequence?
{n2}
an=3n1
{n+1n}n=1∞
{1, 1.5, 2, 2.5, 3, 3.5, ...}
Difference between bounded, bounded above, and bounded below
Determine whether the sequences are bounded below, bounded above, both, or neither
an=n(−1)n
an=n2(−1)n
an=n3
an=−n4
Convegence of sequences
Are the following sequences convergent according to
theorem 7?
{n33}n=1∞
{2(−1)2n+1}n=1∞
{n}n=4∞
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Monotonic and bounded sequences
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Notes
In this section, we will be talking about monotonic and bounded sequences. We will learn that monotonic sequences are sequences which constantly increase or constantly decrease. We also learn that a sequence is bounded above if the sequence has a maximum value, and is bounded below if the sequence has a minimum value. Of course, sequences can be both bounded above and below. Lastly, we will take a look at applying theorem 7, which will help us determine if the sequence is convergent. One important to note from the theorem is that even if theorem 7 does not apply to the sequence, there is a possibility that the sequence is convergent. It's just that the theorem will not be able to show it.
Note
Theorems:
1. A sequence is increasing if an < an+1 for every n≥1.
2. A sequence is decreasing if an > an+1 for every n≥1.
3. If a sequence is increasing or decreasing, then we call it monotonic.
4. A sequence is bounded above if there exists a number N such that an≤N for every n≥1.
5. A sequence is bounded below if there exists a number M such that an≥M for every n≥1.
6. A sequence is bounded if it is both bounded above and bounded below.
7. If the sequence is both monotonic and bounded, then it is always convergent.
Concept
Introduction
Welcome to our exploration of monotonic and bounded sequences, fundamental concepts in mathematics that play a crucial role in various fields. Our journey begins with an informative introduction video, which serves as an essential starting point for understanding these important mathematical ideas. Monotonic sequences are those that consistently increase or decrease, never changing direction. On the other hand, bounded sequences have upper and lower limits, confined within a specific range. These concepts are vital in calculus, analysis, and numerous real-world applications. Throughout this article, we'll delve deeper into the definitions, properties, and examples of both monotonic and bounded sequences. We'll examine how to identify these sequences, their significance in mathematical proofs, and their practical applications. By the end of this exploration, you'll have a solid grasp of these fundamental concepts, enabling you to tackle more advanced topics in mathematics with confidence.
FAQs
Here are some frequently asked questions about monotonic and bounded sequences:
1. What does monotonic mean in calculus?
In calculus, a monotonic sequence is one that consistently increases or decreases. A sequence is monotonically increasing if each term is greater than or equal to the previous term, and monotonically decreasing if each term is less than or equal to the previous term.
2. What is monotonic and bounded?
A sequence is monotonic and bounded if it consistently increases or decreases (monotonic) and all its terms are confined within a specific range (bounded). For example, the sequence 1, 1/2, 1/3, 1/4, ... is both monotonically decreasing and bounded between 0 and 1.
3. Is every monotonic sequence convergent?
Not every monotonic sequence is convergent. However, every bounded monotonic sequence is convergent. This is known as the Monotone Convergence Theorem. Unbounded monotonic sequences, such as 1, 2, 3, 4, ..., are divergent.
4. How do you determine if a sequence is bounded?
To determine if a sequence is bounded, you need to find an upper bound M and a lower bound m such that m an M for all terms an in the sequence. This can often be done by analyzing the general term of the sequence or by using mathematical induction.
5. What is an example of an unbounded sequence?
An example of an unbounded sequence is the natural numbers: 1, 2, 3, 4, ... This sequence is monotonically increasing but has no upper bound. Another example is the sequence n2: 1, 4, 9, 16, 25, ..., which grows without limit.
Understanding monotonic and bounded sequences is a crucial concept in advanced mathematics, particularly in calculus and analysis. However, to fully grasp this topic, it's essential to have a solid foundation in certain prerequisite areas. Two key concepts that play a significant role in comprehending monotonic and bounded sequences are upper and lower bounds and the convergence and divergence of geometric series.
Let's start with the concept of upper and lower bounds. This fundamental idea is crucial for understanding bounded sequences. When we talk about a bounded sequence, we're referring to a sequence that has both an upper and lower limit. The upper bound represents the maximum value that the sequence can approach, while the lower bound represents the minimum value. Grasping these upper and lower limits is essential for identifying whether a sequence is bounded and for determining its behavior over time.
Moving on to the convergence and divergence of geometric series, this concept is particularly relevant when dealing with monotonic sequences. A monotonic sequence is one that is either consistently increasing or consistently decreasing. Understanding the convergence of series helps in determining whether a monotonic sequence approaches a specific value (converges) or grows without bound (diverges). This knowledge is crucial for analyzing the long-term behavior of monotonic sequences and their limits.
By mastering these prerequisite topics, students can more easily grasp the intricacies of monotonic and bounded sequences. The concept of upper and lower bounds provides the framework for understanding the constraints on sequence values, while the study of convergence and divergence offers insights into the sequence's behavior as it progresses.
In conclusion, a strong foundation in these prerequisite topics is invaluable for students approaching the study of monotonic and bounded sequences. By understanding upper and lower bounds and the convergence of series, students will be better equipped to analyze, interpret, and work with these more advanced sequence concepts. This knowledge not only aids in comprehending monotonic and bounded sequences but also serves as a stepping stone for more complex mathematical ideas in calculus and analysis.
Introduction to sequences |
4964 | https://math.stackexchange.com/questions/2129103/counting-non-degenerate-triangles-in-a-square-lattice | Skip to main content
Counting non-degenerate triangles in a square lattice
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A 4 × 4 square is composed of 16 unit squares. The points formed from the unit squares are used to
create non-degenerate dank triangles. Call a triangle dank if none of its vertices lie on the same row
or column of points. How many triangles are dank?
My thoughts are first there will be 25∗16∗9 cases in total, if we dont count when the triangle is degenerated. It is the permutation so we divide by 3! to get 600. A straight line will make the triangle degenerate, so we need to minus those cases. There are 5 points on one diagonal of the square, 4 on its left and right side, and 3 on its left side. So the total case of degenerate triangles are ((53)+((43)+(33))⋅2)⋅2=40. However the answer is not 560? Can someone help me with this?
combinatorics
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By "none of the vertices in the same row or column" you mean "the vertices lie in three distinct rows and three distinct columns"?
– Parcly Taxel
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Yes I mean it. The wording is a bit ambiguous...
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There are indeed 600 dank triangles; the problem is in counting the degenerate cases. There are 52 of them and not 40:
```
. . . .
. . . .
. . . . 2 (orientation) 9 (position) = 18 cases
. . . . .
. . . . .
. . . .
. . . .
. . . . . 4 4 = 16
. . . .
. . . . .
. . . .
. . . .
. . . . . 4 1 = 4
. . . . .
. . . .
. . . .
. . . . .
. . . . 2 1 = 2
. . . . .
. . . .
. . . .
. . . .
. . . . 4 3 = 12
. . . . .
. . . . .
```
This leaves the number of non-degenerate dank triangles as 600−52=548.
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4965 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10470878/ | Research on the bearing behavior of single pile in self-weight collapsible loess areas - PMC
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. 2023 Aug 31;18(8):e0290878. doi: 10.1371/journal.pone.0290878
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Research on the bearing behavior of single pile in self-weight collapsible loess areas
Denghui Gao
Denghui Gao
1 College of Architecture and Civil Engineering, Huanghuai University, Zhumadian, Henan, China
Writing – original draft
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1,, Kuanyao Zhao
Kuanyao Zhao
1 College of Architecture and Civil Engineering, Huanghuai University, Zhumadian, Henan, China
Visualization
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1, Baohong Ma
Baohong Ma
2 Gansu Provincial Highway Aviation Tourism Investment Group Corporation Limited, Lanzhou, China
Data curation
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2, Zhiping Han
Zhiping Han
3 Shanghai Baoye Group Corporation Limited, Shanghai, China
Data curation
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3, Jifei Fan
Jifei Fan
4 School of Civil Engineering, Lanzhou University of Technology, Lanzhou, China
Writing – review & editing
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4
Editor: Malik Muhammad Akhtar 5
Author information
Article notes
Copyright and License information
1 College of Architecture and Civil Engineering, Huanghuai University, Zhumadian, Henan, China
2 Gansu Provincial Highway Aviation Tourism Investment Group Corporation Limited, Lanzhou, China
3 Shanghai Baoye Group Corporation Limited, Shanghai, China
4 School of Civil Engineering, Lanzhou University of Technology, Lanzhou, China
5 Balochistan University of Information Technology Engineering and Management Sciences, PAKISTAN
Competing Interests:The authors have declared that no competing interests exist.
✉
E-mail: dh.gao@huanghuai.edu.cn
Roles
Denghui Gao: Writing – original draft
Kuanyao Zhao: Visualization
Baohong Ma: Data curation
Zhiping Han: Data curation
Jifei Fan: Writing – review & editing
Malik Muhammad Akhtar: Editor
Received 2023 Feb 23; Accepted 2023 Aug 14; Collection date 2023.
© 2023 Gao et al
This is an open access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
PMC Copyright notice
PMCID: PMC10470878 PMID: 37651427
Abstract
The negative skin frictional caused by loess collapse will decrease the bearing capacity of single pile, which is essential to the design of pile foundations in loess areas. In this study, a method for estimating the subsidence of soil layer at any depth is firstly proposed based on the total self-weight collapse value. Secondly, a new load transfer constitutive model for pile-soil interface is developed, which considers the nonlinear stress-strain relationship and the ultimate shear strength of soil. Then, a load transfer calculation model for pile foundation is established, which can calculate the pile axial force, the pile skin frictional, neutral point position and the settlement of a single pile. The calculation results are compared with the test data that obtained from a pile foundation on-site immersion test and the effectiveness of the calculation method is verified well. This calculation method may be useful for designing pile foundations in collapsible loess regions.
Introduction
Loess is extensively distributed around the world. In northwest China, loess is widely distributed and has a large thickness, which is characterized by collapsibility and water sensitivity . Pile foundation is a common foundation type in collapsible loess areas, which can penetrate the soft or loose soil layers and transmit the structural load to competent soil layers, and providing a great bearing capacity by positive shaft resistance and pile tip resistance. But, when the settlement of soil surrounding a pile is greater than that of the pile, the positive shaft resistance will transform into the negative skin friction, which decreases the bearing capacity of single pile and increases the compressive stress in the pile shaft . Due to the wetting-sensitive of unsaturated loess, the collapsible loess may products a significant subsidence after water immersion under the combined action of soil overburden self-weight pressure and additional pressure. Then, the pile may be subjected to a significant negative skin friction and the pile capacity will be reduced or perhaps catastrophic failure . Hence, it is essential to calculate the bearing behavior of single pile with the negative skin friction in loess area.
The negative skin friction of single pile in loess area is caused by the collapsible deformation of loess after water immersion. Loess is widespread continental sediment which posses a metastable structure featured by open fabric and inter-particle bonding . The cementation materials of inter-particle bonding in natural loess are always carbonate and clays , and will be destroyed under the combined action of soil self-weight pressure and water, undergo a significant collapsible deformation. The research on the collapsible deformation calculation method of loess is mainly based on the experimental rules obtained from oedometer tests and triaxial tests, and corrected or verified with the test result of on-site immersion to ensure the applicability of the calculation method. The constitutive model of loess is established based on the triaxial test rules that can describe the stress-strain relationship of loess with different humidity under complex stress state, which is suitable for the finite element calculation of collapsible deformation [6–10]. However, many model parameters and theoretical complexity limit its wide application. The most common collapse deformation calculation method is to calculate the summation deformation of all collapsible loess layers by using the collapsibility coefficient . The collapsible coefficient can be obtained by using oedometer tests or triaxial tests [12–14]. The total collapse settlement can be calculated accurately by multiplying the summation deformation of all layers and the correction coefficient of regional collapsibility that obtained from the field immersion test . However, due to the complexity of water vapor transport in soil layers and the constraining effect of surrounding soil , the distribution rules of stress in the large thickness self-weight collapsible loess is relatively complex. The subsidence calculation of soil layer at any depth cannot be calculated accurately, which makes it difficult to calculate the relative displacement of pile-soil.
The negative skin friction of single pile caused by loess collapse deformation is usually based on empirical value, and there is no a complete calculation method. The empirical value of negative skin friction is usually determined according to the pile foundation on-site immersion test [17–19], model test [2, 20, 21] and centrifugal test [22, 23]. The empirical value is too conservative to reflect the distribution rule of negative friction along the pile length. It is necessary to establish a load transfer function that can reflect the distribution rule of the pile negative skin friction. The pile–soil load transfer function can be obtained from the soil–structural material interface tests. However, the load transfer function, such as the hyperbolic interface model , the load transfer model considering shaft resistance softening and the linearly elastic-perfectly plastic model , which are established based on the soil-structure interface tests that mainly focused on clay soil , sandy soil , and cemented soil , the interface tests few focused on loess. Hence, it is difficult to establish the pile-loess load transfer function based on the interface tests. The load acting on pile shaft is transferred to the soil in the form of shear stress through a single pile, causing shear deformation of the soil around the pile in a certain range . Hence, it is a convenient method to establish the load transfer function according to the mechanical characteristics of the soil around the pile, namely, the shear displacement method . The research on the mechanical characteristics of loess is sufficient, which is available to establish the load transfer function of pile-loess.
Accordingly, this paper aims to research the bearing behavior of single pile with the negative skin friction in loess area. To achieve this objective, the subsidence calculation method of loess layer at any depth based on the total collapse settlement is proposed, and the calculation of pile-loess relative displacement under collapsible deformation condition is realized. A new load transfer constitutive model based on the strength and deformation of the soil around the pile is developed, which considers the nonlinear stress-strain relationship and the ultimate shear strength of soil. According to the static equilibrium of pile segment, the load transfer calculation model is established by using the load transfer constitutive model and pile-soil relative displacement, which can realize the calculation of single pile bearing behavior under the collapsible deformation condition.
Subsidence calculation of soil layer at any depth
Calculation of the total self-weight collapse value
The loess standard provides a relatively simple calculation method for self-weight collapsible settlement. The total self-weight collapse value can be obtained by using the correction coefficient of regional collapsibility, the self-weight collapsible coefficient and the thickness of self-weight collapsible soil layer. In this study, we used this calculation method to estimate the total self-weight collapse value:
The self-weight collapsible coefficient δ zs is calculated as follows:
(1)
Where, h z is the stabilized height of specimen with natural humidity and structure under the action of overburden self-weight pressure (mm); is the stabilized height of saturated specimen under the same pressure after immersion (mm); and h is the initial height of specimen (mm).
The total self-weight collapse value s 0 is calculated as follows:
(2)
Where, δ zsi is the self-weight collapsible coefficient of the i th layer soil, h i is the thickness of the i th layer soil, and β 0 is the correction coefficient of regional collapsibility.
The loess layer with the self-weight collapsible coefficient δ zs less than 0.015 is regarded as the non-collapsible soil layer under self-weight stress. The deformation of non-collapsible soil layer is not included in the total self-weight collapse deformation and the depth of non-collapsible layer is the settlement calculation lower limit depth.
Calculation of stratified soil subsidence
The total self-weight collapse value can be obtained by using the correction coefficient of regional collapsibility, the self-weight collapsible coefficient and the thickness of self-weight collapsible soil layer. But, there is a large deviation between the subsidence calculation value and the measured value of stratified soil in the on-site immersion test . It can be seen from the curve of subsidence with depth that the subsidence variation law of stratified soil is similar to the Boussinesq’s vertical displacement solution [33–34]. Therefore, we can deduce the corresponding concentrated force F according to the total self-weight collapse value, and use the Boussinesq’s vertical displacement solution to calculate the stratified soil subsidence.
When the self-weight stress is small, the self-weight collapsible coefficient will be less than 0.015, and the collapsible deformation caused by soaking under self-weight stress can be ignored. The self-weight stress increases with the burial depth, and the self-weight collapsible coefficient gradually increases. When the self-weight collapsible coefficient is equal to 0.015, the collapsible deformation should be calculated, and the burial depth of the soil layer is the initial self-weight collapsible depth h 0. The initial self-weight collapsible depth h 0 is the action position of the concentrated force F in the Boussinesq’s solution. Above this depth, the self-weight collapse deformation of soil layer was 0, and stratified soil subsidence is equal to the total self-weight collapse value s 0. The subsidence of non-collapsible soil layers caused by soaking under self-weight stress is also very small, which can be ignored compared with the total settlement. Hence, the burial depth of non-collapsible soil layer is the lower limit depth h e for the calculation of collapsible deformation . The subsidence of soil layer at or below this depth was 0.
Under the action of the concentrated force F, the expression of Boussinesq’s vertical displacement at an arbitrary point is modified as follows:
(3)
Where, μ is Poisson’s ratio of the soil and E is the elastic modulus of the soil. To prevent s(z)’ from approaching infinity, the horizontal distance between the calculated point and the action point of the concentrated force R is set to be 0.8 m, which is the diameter of the pile.
The subsidence of soil layer at z = h 0 is s 0, the concentrated force F is deduced as follows:
(4)
Subsequently, Eq (4) is substituted into Eq (3) to obtain Eq (5), which can be used for calculating the subsidence of soil layer at any depth based on the total self-weight collapse value:
(5)
Because of the subsidence of soil layer at lower limit depth h e is assumed to be 0. So, the subsidence at the depth z∈[h 0,h e] is modified as follows:
(6)
When the depth of soil layer z≤h 0, its subsidence is s 0, which can be calculated using Eq (2). When the depth of soil layer z>h e, this soil layer subsidence is 0. When the depth of soil layer z between h 0 and h e, the subsidence is calculated using Eq (6). Thus, the subsidence of soil surrounding a pile at any depth can be calculated.
Pile–soil load transfer constitutive model
Principle and existing problems of shear displacement method
The shear displacement method assumes that the upper load is transferred to the soil in the form of shear stress through a single pile, causing shear deformation of the soil around the pile in a certain range , as shown in Fig 1.
Fig 1. The deformation schematic diagram of soil around pile.
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The stress analysis on the micro-unit of soil around the pile is carried out, as shown in Fig 2. According to the vertical static equilibrium differential equation of the micro-unit, the relationship between the pile skin friction and the shear stress of soil can be obtained as follows:
(7)
Fig 2. The force analysis of soil micro-unit.
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Where, τ 0 and r 0 are the pile skin friction and the radius of the pile, respectively. τ is the shear stress acting on the soil micro-unit, and r is the distance between soil micro-unit and pile axis.
According to the geometric equation of elasticity theory, and the radial deformation of the micro-unit can be omitted, then the expression of shear strain can be obtained as follows:
(8)
Where, Δ s is the vertical relative displacement of pile-soil at the distance r from the pile axis.
According to the generalized Hooke’s law, the constitutive relationship between shear stress and shear strain is obtained as follows:
(9)
Substituting Eqs (8) and (9) into Eq (7) for solution, the load transfer function of shear displacement method can be obtained as follows:
(10)
Where, G s is the shear modulus of the soil, and r m is the maximum influence radius of pile skin friction on the soil around the pile, its recommended value is r m = 10 r 0 .
Because the load transfer function of shear displacement method is derived on the basis of elastic theory, it can be seen from Eq (10) that the load transfer between piles and soil is a linear function. The greater the relative displacement of pile-soil, the greater the pile skin friction, which obviously inconsistent with the load transfer law of pile-soil measured on site. The reason is that the nonlinearity of the stress-strain relationship of soil is not fully considered during the theoretical derivation. The shear modulus of the soil changes with the shear strain, which is not a fixed value.
New load transfer model based on shear displacement method
The stress-strain curve of natural loess obtained by the triaxial shear test is approximately hyperbolic , as shown in Fig 3(A), which can be expressed as follows:
(11)
Fig 3. Schematic diagram of shear stress-strain relationship curve.
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(a) Relationship curve of τ~γ (b) Relationship curve of γ/τ~γ.
Eq (11) can be transformed into Eq (12), which is also the expression of shear modulus versus with shear strain.
(12)
Where, a and b are test parameters of soil, with a physical meaning for the inverse of the initial shear modulus G s 0 and the inverse of the ultimate shear strength τ u, respectively. In the γ/τ~γ coordinates, these test parameters can be obtained by linear fitting, as shown in Fig 3(B).
Parameters a and b are related to the stress state of soil, the parameter b can be calculated using the method in Duncan-Chang model , the calculation formula is as follows:
(13)
Where, K and n are the parameters of the model, and p atm is the standard atmospheric pressure. According to the actual shear direction of soil, the confining pressure σ 3 is regarded as the overburden self-weight stress of soil in saturated state, namely, σ 3 = γ sat z.
The inverse of parameter b is the ultimate shear strength of the soil:
(14)
Where, c and φ are the cohesion and friction angles of saturated soil, respectively. For the soil with obvious anisotropic characteristics, the strength parameters can be measured using triaxial shear tests according to the actual shear direction, or using the reduced strength parameters.
The variation of soil vertical displacement in the radial is nonlinear, as shown in Fig 4. The model for pile-soil vertical relatived is placement Δ s(z) versus radial distance γ is usually described by hyperbola and parabola . This assumption is appropriate for large scale on-site immersion test, but not applicable for local immersion condition.
Fig 4. The variation of pile-soil relative vertical displacement in the radial.
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The relational expression is defined as follows:
(15)
Where, the range of r is [r 0,r m], the range of Δ s(z) is [0,Δ S(z)], Δ S(z) is the maximum pile–soil relative displacement at depth z, and α and β are the parameters of the curve.
The boundary conditions are determined as follows:
(16)
(17)
The parameters α and β can be obtained by solving simultaneous Eqs (16) and (17).
(18)
(19)
Hence, the calculation of shear strain in Eq (8) can be obtained as follows:
(20)
The calculation equation of soil shear modulus in the radial under different pile-soil relative displacements can be obtained by substituting Eq (20) into Eq (12) as follows:
(21)
Substituting Eq (21) into Eq (9), and the constitutive relationship between shear stress and shear strain can be transformed into the following form.
(22)
Hence, the new load transfer model can be obtained by substituting Eq (22) into Eq (8):
(23)
Finally, the new load transfer model considering the nonlinear deformation of the soil around the pile is established. Compared with Eq (10), the pile skin friction in the new load transfer model approached an extreme value when the pile-soil relative displacement is large, which is consistent with the load transfer law of pile-soil measured on site.
Calculation method of single pile bearing behavior
The load transfer model of single pile is shown in Fig 5. The pile is divided into n segments along the length direction, and the calculation equation can be built according to the static equilibrium of each pile segment.
Fig 5. The load transfer model of single pile.
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(a) Schematic diagram of pile bearing behavior (b) Stress analysis of the i th pile segment.
Subsidence calculation of pile segment at any depth
As shown in Fig 5(A), the pile is divided into n segments. The subsidence of pile segment at depth z is composed of two parts: namely, the pile tip subsidence and the summation of the compression deformation of pile segments below this depth. The subsidence of pile segment at any depth can be calculated as follows:
(24)
Where, s p(Z i) is the subsidence of the i th pile segment, S b is the pile tip subsidence, ds i is the compression deformation of the i th pile segment.
It is assumed that the pile segment compression process is in the elastic stage. The forces acting on the pile segment include: the pile skin friction, the axial force acting on the upper and lower interfaces, as shown in Fig 5(B). The compression deformation of the i th pile segment can be calculated according to the following equation:
(25)
Where, A p is the cross-sectional area of pile shaft and E p is the elastic modulus of pile material.
The pile tip subsidence is given by the Boussinesq’s solution .
(26)
Where, P n+1 is the axial force at the pile tip; μ s and G s are Poisson’s ratio and the shear modulus of the bearing soil layer, respectively.
The shear modulus of the bearing soil layer G s can be obtained in accordance with its relationship with the compressive modulus E s as follows :
(27)
Hence, the subsidence of pile segment at any depth can be calculated with the pile tip subsidence.
Calculation of the pile axial force at any depth
According to the static equilibrium of pile segment in Fig 5(B), the following equation can be established:
(28)
Where, τ 0(z i) is the i th pile segment skin frictional, which can be calculated using Eq (23).
The subsidence of pile and soil can be calculated by Eqs (24) and (6), respectively. The pile–soil relative vertical displacement Δ S(z i) in Eq (23) can be calculated using the following equation:
(29)
The axial force of the pile shaft at any depth can be calculated by using Eq (26) and Eq (28), when the pile tip subsidence is given. The axial force calculation equation of pile shaft at any depth is as follows:
(30)
The solution can be obtained by following an iterative procedure. Firstly, assume a pile tip subsidence initial value S b, which is also considered as the subsidence of the n th pile segment. The axial force acting on the lower interface of the n th pile segment P n+1 can be calculated by substituting S b into Eq (26). Secondly, the pile skin frictional of the n th pile segment τ 0(z n) can be calculated as follows: the subsidence of the n th pile segment is S b, and the subsidence of soil around the pile tip s(z n) can be calculated by using Eq (6), thus, the pile-soil relative displacement of the n th pile segment Δ S(z n) can be calculated by substituting S b and s(z n) into Eq (29), and the pile skin frictional of the n th pile segment τ 0(z n) can be calculated by substituting Δ S(z n) into Eq (23). Thirdly, the axial force acting on the upper interface of the n th pile segment P n can be calculated by substituting P n+1 and τ 0(z n) into Eq (28), and the compression deformation of the n th pile segment ds n can be calculated by substituting P n and P n+1 into Eq (25). Hence, the subsidence of the (n-1)th pile segment can be calculated by substituting ds n into Eq (24). Finally, the axial force acting on the upper interface of the 1th pile segment P 1 can be calculated through successive iterations. The P 1 value is the load value acting on the pile head, which is known. Adjust the pile tip subsidence initial value S b, repeat the procedure until convergence is achieved normally (The P 1 calculated value is consistent with the pile head action load).
Calculation of the pile skin frictional and neutral point position
The pile tip subsidence value S b can be obtained when the iteration convergence. According to the iterative procedure, once the pile tip subsidence value S b is determined, the lower interface of the n th pile segment P n+1 can be calculated by substituting S b into Eq (26). Besides, the subsidence of soil around the pile tip s(z n) can be calculated by using Eq (6), the relative displacement of pile-soil Δ S(z n) can be calculated by substituting S b and s(z n) into Eq (29). Hence, the pile skin frictional of the n th pile segment τ 0(z n) can be calculated by substituting Δ S(z n) into Eq (23). Furthermore, the axial force acting on the upper interface of the n th pile segment P n can be calculated by substituting P n+1 and τ 0(z n) into Eq (28), and the compression deformation of the n th pile segment ds n can be calculated by substituting P n and P n+1 into Eq (25). Then, the subsidence of the (n-1)th pile segment s p(z i) can be calculated by substituting ds n into Eq (24). The relative displacement of pile-soil Δ S(z i) and the pile skin frictional of the n th pile segment τ 0(z i) can be calculated through successive iterations by Eq (29) and Eq (23), respectively.
The neutral point position is the depth where the relative displacement of pile-soil is 0. The subsidence of the pile s p(z i) and the subsidence of the soil around the pile s(z i) can be calculated by Eq (24) and Eq (6), respectively. When the subsidence of pile and soil is equal, namely, Δ S(z i) = 0 in Eq (29), the corresponding z value is the depth of the neutral point position.
Verification of the calculation method
To verify the reasonableness of the calculation method, the calculated results are compared with the test data of pile foundation on-site immersion test, which is carried out on the self-weight collapsible loess site in Weinan, Shaanxi, China . The lower limit depth for the calculation of collapsible deformation is 33m in this site. The test pile is 60m long and 0.8m in diameter, which is a bored pile with the pile shaft concrete strength grade of C35. The elastic modulus of pile shaft material is taken as 31.5MPa . The pile tip bearing soil layer is non-collapsible loess with the void ratio e 0 is 0.83 and the compression coefficient α 1−2 is 0.14. According to the reference [34, 40], the relevant calculation parameters of loess in this region are listed in Table 1.
Table 1. Calculation parameters of loess.
| Saturation gravity γ sat (kN/m 3) | Poisson’s ratio μ | Cohesion c(kPa) | Angle of internal friction φ (°) | Parameter K | Parameter n |
:---: :---: :---: |
| 18.9 | 0.4 | 25.4 | 23.5 | 19.3 | 0.733 |
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The calculated value is compared with the test data, as shown in Fig 6. It can be seen that the calculated value is close to the test data of the filed immersion test. The calculated depth of the neutral point position is 24.5m, which is larger than the test depth (22m) with a relative error of 11.4%. The calculated subsidence of pile head is 7.5 mm, which is smaller than the test value (8.2 mm) with a relative error of 8.5%. The calculated value of pile negative skin resistance is consistent with the test value, the calculation deviation is mainly reflected in the calculation of positive shaft resistance, and the calculated results are larger than the test values. The deviation is due to that the subsidence of soil layer below the lower limit collapse depth is assumed to be 0, which causes the calculated value of pile-soil relative displacement within this soil layer to be larger than the actual value. After analyzing the influence of model parameters on the calculation results, it is found that the cohesion of loess is the greatest influencing factor. By comparing with the test data of pile foundation on-site immersion test, the calculation method proposed in this paper can realize the calculation of single pile bearing behavior in loess site.
Fig 6. Comparison between the calculated value and test data.
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Conclusion
In this study, the calculation method of single pile bearing behavior under moistening deformation condition is established. The effectiveness of this method is verified by comparing with the test data that obtaining from a pile foundation on-site immersion test. The following conclusions can be drawn:
The subsidence calculation method of soil layer at any depth is proposed. This method defines the collapse deformation calculation range of loess layers, and the equivalent calculation between the concentrated force of Boussinesq’s solution and the total self-weight collapse settlement.
A new pile-soil load transfer constitutive model based on the strength and deformation of the soil around the pile is developed. This model takes into account the nonlinear shear deformation and ultimate shear strength of loess, and the pile skin friction under different relative displacement of pile-soil can be correctly reflect.
The calculation model of single pile bearing behavior is established. This model is established according to the static balance of the pile element, and can calculate the pile axial force, the pile skin frictional, neutral point position and the settlement of a single pile.
The calculation method is reasonable by comparison with the on-site immersion test. The calculation result show that the distribution rule of pile skin friction is similar to on-site test results, and the calculation error of neutral point position and pile settlement is within 10%.
Supporting information
S1 File. The relevant data in the manuscript.
(DOCX)
Click here for additional data file. (13.7KB, docx)
Acknowledgments
We thank the anonymous reviewers for their detailed and constructive comments.
Data Availability
All relevant data are within the manuscript and its Supporting Information files.
Funding Statement
The research described in this manuscript was financially supported by the Natural Science Foundation of China (Grant No. 42207200), the Natural Science Foundation of Gansu Province (Grant No. 22JR5RA255 and Grant No. 23JRRA768). There was no additional external funding received for this study. The Natural Science Foundation of China (Grant No. 42207200) plays an important role in the study design and preparation of the manuscript. The Natural Science Foundations of Gansu Province (Grant No. 22JR5RA255 and Grant No. 23JRRA768) play an important role in the data collection and analysis.
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Supplementary Materials
S1 File. The relevant data in the manuscript.
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Data Availability Statement
All relevant data are within the manuscript and its Supporting Information files.
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Abstract
Introduction
Subsidence calculation of soil layer at any depth
Pile–soil load transfer constitutive model
Calculation method of single pile bearing behavior
Verification of the calculation method
Conclusion
Supporting information
Acknowledgments
Data Availability
Funding Statement
References
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Key People:
: Hugo de Vries
: Robert Broom
: Charles Darwin
: Thomas Henry Huxley
: Charles Lyell
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: convergence
: punctuated equilibrium model
: molecular clock
: synthetic theory of evolution
: catastrophe theory
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evolution, theory in biology postulating that the various types of plants, animals, and other living things on Earth have their origin in other preexisting types and that the distinguishable differences are due to modifications in successive generations. The theory of evolution is one of the fundamental keystones of modern biological theory.
The diversity of the living world is staggering. More than 2 million existing species of organisms have been named and described; many more remain to be discovered—from 10 million to 30 million, according to some estimates. What is impressive is not just the numbers but also the incredible heterogeneity in size, shape, and way of life—from lowly bacteria, measuring less than a thousandth of a millimetre in diameter, to stately sequoias, rising 100 metres (300 feet) above the ground and weighing several thousand tons; from bacteria living in hot springs at temperatures near the boiling point of water to fungi and algae thriving on the ice masses of Antarctica and in saline pools at −23 °C (−9 °F); and from giant tube worms discovered living near hydrothermal vents on the dark ocean floor to spiders and larkspur plants existing on the slopes of Mount Everest more than 6,000 metres (19,700 feet) above sea level.
The virtually infinite variations on life are the fruit of the evolutionary process. All living creatures are related by descent from common ancestors. Humans and other mammals descend from shrewlike creatures that lived more than 150 million years ago; mammals, birds, reptiles, amphibians, and fishes share as ancestors aquatic worms that lived 600 million years ago; and all plants and animals derive from bacteria-like microorganisms that originated more than 3 billion years ago. Biological evolution is a process of descent with modification. Lineages of organisms change through generations; diversity arises because the lineages that descend from common ancestors diverge through time.
The 19th-century English naturalist Charles Darwin argued that organisms come about by evolution, and he provided a scientific explanation, essentially correct but incomplete, of how evolution occurs and why it is that organisms have features—such as wings, eyes, and kidneys—clearly structured to serve specific functions. Natural selection was the fundamental concept in his explanation. Natural selection occurs because individuals having more-useful traits, such as more-acute vision or swifter legs, survive better and produce more progeny than individuals with less-favourable traits. Genetics, a science born in the 20th century, reveals in detail how natural selection works and led to the development of the modern theory of evolution. Beginning in the 1960s, a related scientific discipline, molecular biology, enormously advanced knowledge of biological evolution and made it possible to investigate detailed problems that had seemed completely out of reach only a short time previously—for example, how similar the genes of humans and chimpanzees might be (they differ in about 1–2 percent of the units that make up the genes).
This article discusses evolution as it applies generally to living things. For a discussion of human evolution, see the article human evolution. For a more complete treatment of a discipline that has proved essential to the study of evolution, see the articles genetics, human and heredity. Specific aspects of evolution are discussed in the articles coloration and mimicry. Applications of evolutionary theory to plant and animal breeding are discussed in the articles plant breeding and animal breeding. An overview of the evolution of life as a major characteristic of Earth’s history is given in community ecology: Evolution of the biosphere. A detailed discussion of the life and thought of Charles Darwin is found in the article Darwin, Charles.
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Biology Bonanza
General overview
The evidence for evolution
Darwin and other 19th-century biologists found compelling evidence for biological evolution in the comparative study of living organisms, in their geographic distribution, and in the fossil remains of extinct organisms. Since Darwin’s time, the evidence from these sources has become considerably stronger and more comprehensive, while biological disciplines that emerged more recently—genetics, biochemistry, physiology, ecology, animal behaviour (ethology), and especially molecular biology—have supplied powerful additional evidence and detailed confirmation. The amount of information about evolutionary history stored in the DNA and proteins of living things is virtually unlimited; scientists can reconstruct any detail of the evolutionary history of life by investing sufficient time and laboratory resources.
Evolutionists no longer are concerned with obtaining evidence to support the fact of evolution but rather are concerned with what sorts of knowledge can be obtained from different sources of evidence. The following sections identify the most productive of these sources and illustrate the types of information they have provided.
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4967 | https://www.mometrix.com/academy/multiplying-and-dividing-fractions/ | Multiply and Divide Fractions (Review Video & Worksheets)
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[x] Transcript - [x] FAQs - [x] Practice - [x] Worksheets Calculator
Many students have a real fear of fractions. However, if you can remember what a fraction represents and a few mathematical rules on how to work with them algebraically, you will be able to face fractions with confidence. In this video, we will review how to multiply and divide fractions. Let’s get started.
We should start by defining exactly what a fraction is. A fraction represents a ratio of a “part” to a “whole,” or part over whole. The value above the division line is referred to as the numerator, and the value below the division line is the denominator.
Multiplying Fractions
To multiply fractions, simply multiply “straight across,” meaning the “numerator times the numerator” divided by the “denominator times the denominator.” Let’s look at a couple of quick examples:
2 3×2 5 2 3×2 5
Here, we want to multiply 2 3 2 3 by 2 5 2 5. As we said earlier, we’re going to multiply straight across. So we’re going to have 2 times 2 over 3 times 5. Which is equal to four over fifteen. So our answer is 4 15 4 15.
2 3×2 5=2×2 3×5=4 15 2 3×2 5=2×2 3×5=4 15
Now let’s try another one. We’re going to try 4 7 4 7 times 3 11 3 11.
4 7×3 11 4 7×3 11
Again it’s the same concept. We’re going to multiply 4 times 3, divided by 7 times 11. Which gives us 12 77 12 77.
4 7×3 11=4×3 7×11=12 77 4 7×3 11=4×3 7×11=12 77
Pretty simple, right? Now let’s take a look at dividing fractions.
Dividing Fractions
Dividing fractions involves a slightly different process. Before we jump into the mechanics of the process, let’s start by looking at an intuitive example of dividing a fraction by two. The effect of dividing by 2 is simply cutting the fraction in half, or simply multiplying the fraction by 1 over 2.
So, 4 5 4 5 divided by 2 is really the same as saying 4 5 4 5 times 1 2 1 2.
4 5 4 5÷2÷2=4 5×1 2=4 5×1 2
Then it’s going to be multiplied across just like we did before. So we have 4 times 1 is four, over 5 times 2 is 10. Which then simplifies to 2 5 2 5.
4 5 4 5÷2÷2=4 5×1 2=4 10=2 5=4 5×1 2=4 10=2 5
So in other words, 2 5 2 5 is half the size of 4 5 4 5.
Similarly, dividing a fraction by 3 would result in a fraction that is one-third the size of the original:
2 5 2 5÷3÷3=2 5×1 3=2 15=2 5×1 3=2 15
2 5 2 5 divided by 3 is the same as saying 2 5 2 5 times 1 3 1 3, which gives you 2 15 2 15.
So, 2 15 2 15 is one-third the size of 2 5 2 5.
Before we generalize this process, let’s review some important terminology. Consider the relationship between 2 and 1 2 1 2. These numbers are called reciprocals of one another, which means that the numerator of one number is the denominator of the other, and vice versa. Remember that 2 can be written as a fraction by writing it over 1, like this: 2 1 2 1. Therefore, 2 1 2 1 and 1 2 1 2 are reciprocals. The same is true of 3 and 1 3 1 3, because 3 can be written as 3 1 3 1. Therefore, 3 and 1 3 1 3 are reciprocals.
With this in mind, what pattern do you see in the process for dividing fractions?
Keep, Change, Flip
The process of dividing fractions is the same as multiplying the first fraction by the reciprocal of the second. A shorthand version of this wordy explanation that may help you remember the division process is “Keep, Change, Flip”:
Keep the first fraction as is
Change the operation from division to multiplication
Flip (or take the reciprocal of) the second fraction.
Once this adjustment is made, simply follow the rules for multiplying fractions by multiplying the numerators and dividing by the product of the denominators.
Here is an example using the “Keep, Change, Flip” process:
Say we want to divide 3 5 3 5 by 7 5 7 5. We’ll keep the first fraction as is, change the operation from division to multiplication, and flip the second number. Now we just multiply our numerators, 3 times 5 is fifteen, over 5 times 7 is 35. And then from there, we simplify to 3 7 3 7.
3 5÷7 5=3 5×5 7=15 35=3 7 3 5÷7 5=3 5×5 7=15 35=3 7
I hope this video was helpful! Thanks for watching, and happy studying!
For more help, check out our fractions calculator!
Cross Multiplying Fractions | Improper Fractions and Mixed Numbers
Frequently Asked Questions
Q
How do you multiply fractions with whole numbers?
A
Multiply fractions by whole numbers by turning the whole number into a fraction by placing it over 1. Any number divided by itself is itself, so this does not change the value of the whole number. Then, multiply across as with normal fractions.
Example:
2 3×4=2 3×4 1=8 3=2 2 3 2 3×4=2 3×4 1=8 3=2 2 3
Q
How do you multiply mixed fractions?
A
Multiply mixed fractions by first turning them into improper fractions and then multiplying across as normal. If there are common factors in the numerator and denominator, cancel those out first to simplify multiplying across. To convert the fraction back to a mixed number, divide the numerator by the denominator. The number of full divisions becomes the whole number and the remainder becomes the numerator of the fractional part over the original denominator.
Example 1:
3 1 5×2 7 9=16 5×25 9=16 1×5 9=80 9=8 8 9 3 1 5×2 7 9=16 5×25 9=16 1×5 9=80 9=8 8 9
Example 2:
3 1 5×2 7 9=16 5×25 9=16 5×5×5 9=16 1×5 9=80 9=8 8 9 3 1 5×2 7 9=16 5×25 9=16 5×5×5 9=16 1×5 9=80 9=8 8 9
Q
How do you divide fractions?
A
Divide fractions by using the phrase: “Keep, Change, Flip.” Keep the first fraction the same. Change the division sign to a multiplication sign. Flip the second fraction. Then multiply across and simplify if necessary.
Example:
4 7÷8 13=4 7×13 8=52 56=13 14 4 7÷8 13=4 7×13 8=52 56=13 14
Q
How do you divide fractions with whole numbers?
A
Divide fractions by whole numbers by first turning the whole number into a fraction and then dividing the fractions as normal by flipping the second fraction and multiplying across. Any number can be turned into a fraction by placing it over 1.
Example:
2 3÷4=2 3÷4 1=2 3×1 4=2 12=1 6 2 3÷4=2 3÷4 1=2 3×1 4=2 12=1 6
Q
How do you divide mixed fractions?
A
Divide mixed fractions by first converting them to improper fractions and then dividing the fractions as normal.
Example:
4 3 5÷2 1 2=23 5÷5 2=23 5×2 5=46 25=1 21 25 4 3 5÷2 1 2=23 5÷5 2=23 5×2 5=46 25=1 21 25
Multiplying and Dividing Fractions Problems
Question #1:
2 3×7 9=2 3×7 9=
42 9 42 9
42 47 42 47
14 27 14 27
14 9 14 9
[x] Show Answer
Answer:
The correct answer is C: 14 27 14 27. To multiply fractions, simply multiply the numerators together to get the new numerator, and multiply the denominators together to get the new denominator.
2 3×7 9=2×7 3×9=14 27 2 3×7 9=2×7 3×9=14 27
[x] Hide Answer
Question #2:
7 6÷2 3=7 6÷2 3=
7 4 7 4
9 9 9 9
14 18 14 18
7 2 7 2
[x] Show Answer
Answer:
The correct answer is A: 7 4 7 4. To divide fractions, use the phrase: Keep, Change, Flip. Keep the first fraction the same. Change the division sign to a multiplication sign. Flip the second fraction so it is its reciprocal. That process looks like this:
7 6÷2 3=7 6×3 2 7 6÷2 3=7 6×3 2
Then, multiply and simplify the fractions.
7 6×3 2=7×3 6×2=21 12=7 4 7 6×3 2=7×3 6×2=21 12=7 4
[x] Hide Answer
Question #3:
1 4×6 7÷2 9=1 4×6 7÷2 9=
7 108 7 108
47 52 47 52
12 252 12 252
27 28 27 28
[x] Show Answer
Answer:
The correct answer is D: 27 28 27 28. According to the Order of Operations (PEMDAS), multiplication and division can happen at the same time. For this example, let’s work through the multiplication and division in order from left to right. So we’ll start by multiplying 1 4 1 4 and 6 7 6 7, simplifying if necessary.
1 4×6 7=1×6 4×7=6 28=3 14 1 4×6 7=1×6 4×7=6 28=3 14
Then, divide 3 14 3 14 by 2 9 2 9.
3 14÷2 9=3 14×9 2=27 28 3 14÷2 9=3 14×9 2=27 28
Therefore, 1 4×6 7÷2 9=27 28 1 4×6 7÷2 9=27 28.
[x] Hide Answer
Question #4:
Sarah Anne is baking cookies and the recipe calls for 2 3 2 3 cups of butter. She needs a lot of cookies, so she decides to quadruple the recipe. Quickly realizing that’s a lot of cookies, she decides to back off a bit and only make the recipe 3 1 2 3 1 2 times the original. In order to do this, how many cups of butter will Sarah Anne need?
2 1 3 2 1 3 cups
4 21 4 21 cups
1 2 3 1 2 3 cups
1 7 1 7 cups
[x] Show Answer
Answer:
The correct answer is A: 2 1 3 2 1 3 cups. The first thing that needs to happen in order to solve this problem is 3 1 2 3 1 2 needs to be converted to an improper fraction.
3 1 2=3×2+1 2=7 2 3 1 2=3×2+1 2=7 2
Then, multiply the two fractions and simplify.
2 3×7 2=14 6=7 3 2 3×7 2=14 6=7 3
Finally, convert 7 3 7 3 to a mixed number.
7 3=2 1 3 7 3=2 1 3
Sarah Anne needs 2 1 3 2 1 3 cups of butter.
[x] Hide Answer
Question #5:
Antonio has 6 8 6 8 of a pizza left over and two hungry friends. If Antonio and his friends evenly split the pizza, what fraction of the pizza does each person get?
1 3 1 3
3 4 3 4
1 4 1 4
2 3 2 3
[x] Show Answer
Answer:
The correct answer is C: 1 4 1 4. This question is asking us to divide 6 8 6 8 by 3. Remember, any whole number can be turned into a fraction by placing it over 1. Here’s what the division looks like:
6 8÷3 1=6 8×1 3=6 24=1 4 6 8÷3 1=6 8×1 3=6 24=1 4
Each person gets 1 4 1 4 of the pizza.
[x] Hide Answer
Multiplying and Dividing Fraction Worksheets
Use our free printable multiplying and dividing fractions worksheets for additional practice!
Multiplying and Dividing Fractions Worksheets
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Multiplying and Dividing Fractions (Answer Key)
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Multiplying Fractions Worksheets
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Multiplying Fractions (Answer Key)
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Dividing Fractions (Answer Key)
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Half an annulus
Can you locate the point on an annulus that splits it into two areas?
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Problem Student Solutions
Problem
Image
The shaded region in the diagram, bounded by two concentric circles (circles whose centres are at the same point), is called an annulus.
The circles have radii 2 cm and 14 cm.
The dashed circle has the same centre as the other two and divides the area of this annulus into two equal areas.
What is its radius?
This problem is taken from the UKMT Mathematical Challenges.
Student Solutions
Image
The radii of the three circles are shown on the diagram, where we are trying to find r.
Using the whole annulus and the darker grey annulus
The area of the whole annulus is the area of the full circle (raduis 14 cm) without the area of the empty circle (radius 2 cm), so it is given by π 14 2−π 2 2=196 π−4 π=192 π cm 2.
The area of the darker grey annulus is the area within the dotted circle without the area in the empty circle - so π r 2−π 2 2=π r 2−4 π cm 2.
This area is equal to half of the whole annulus, so π r 2−4 π=1 2×192 π⇒π r 2−4 π=96 π.
So π r 2=100 π, so r 2=100, so r=10.
Using the whole annulus and the lighter grey annulus
The area of the whole annulus is the area of the full circle (raduis 14 cm) without the area of the empty circle (radius 2 cm), so it is given by π 14 2−π 2 2=196 π−4 π=192 π cm 2.
The area of the lighter grey annulus is the area within the full circle without the area in the dotted circle - so π 14 2−π r 2=196 π−π r 2 cm 2.
This area is equal to half of the whole annulus, so 196 π−π r 2=1 2×192 π⇒196 π−π r 2=96 π.
So 196 π−96 π=π r 2⇒π r 2=100 π, so r 2=100, so r=10.
Using the lighter grey annulus and the darker grey annulus
The area of the lighter grey annulus is the area within the full circle without the area in the dotted circle - so π 14 2−π r 2=196 π−π r 2 cm 2.
The area of the darker grey annulus is the area within the dotted circle without the area in the empty circle - so π r 2−π 2 2=π r 2−4 π cm 2.
These two areas are equal, so 196 π−π r 2=π r 2−4 π. Solving this equation for r, 196 π−π r 2=π r 2−4 π⇒196 π+4 π=π r 2+π r 2⇒200 π=2 π r 2⇒100=r 2⇒10=r
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4969 | https://www.healthdirect.gov.au/vitamin-b-deficiency | Healthdirect Free Australian health advice you can count on.
Medical problem? Call 1800 022 222. If you need urgent medical help, call triple zero immediately
healthdirect Australia is a free service where you can talk to a nurse or doctor who can help you know what to do.
beginning of content
Vitamin B deficiency
34-minute read
Listen
Key facts
Vitamin B deficiency happens when your body doesn't have enough vitamin B.
Your body needs a certain amount of each type of vitamin B to function well.
Severe vitamin B deficiency can cause diseases such as beriberi or pellagra.
People who are vitamin B deficient may feel tired, numbness or weakness, among other symptoms.
What is vitamin B deficiency?
Vitamin B deficiency happens when your body doesn't get enough vitamin B. This may happen if you don't eat enough foods containing vitamin B or because of a health condition.
Your body needs a certain amount of each type of vitamin to function well. The different types of vitamin B are all water-soluble (dissolve in water). This means that they cannot be stored in the body and must be eaten regularly to avoid deficiency.
Find out more about vitamin B and your health.
What are the symptoms of vitamin B deficiency?
Symptoms of vitamin B deficiency differ, depending on which type of vitamin B you don't have enough of.
However, symptoms are usually mild and can be treated by having more food rich in B-vitamins.
Vitamin B12, unlike other B vitamins, is stored in the liver for years. For that reason, symptoms of being deficient may take a long time to develop.
Types of vitamin B deficiency: symptoms, effects and risk factors
Thiamin
Symptoms
weight loss
short-term memory loss and confusion
muscle weakness
cardiovascular symptoms
Effects
Beriberi disease
Wernicke–Korsakoff syndrome
Risk factors
hazardous alcohol use
liver disease
malnutrition
bariatric surgery
Riboflavin
Symptoms
growth disturbances
rash
inflammation of your mouth and tongue
cracks at the corner of your mouth
anaemia
Effects
Ariboflavinosis
Risk factors
vegan diet
vegetarian athletes
pregnant and breastfeeding people and their babies
Niacin
Symptoms
mainly affects the skin
dementia
Effects
Niacin deficiency — severe disease is called pellagra
Pantothenic acid
Symptoms
irritable
restless
tiredness, apathy and malaise
sleep disturbances
nausea, vomiting and cramping
numbness and staggering gait
Effects
Pantothenic acid deficiency — only seen in individuals fed synthetic diets
Vitamin B6
Symptoms
anaemia
rash
seizures
depression and confusion
Effects
Vitamin B6 deficiency
Risk factors
older age
malnutrition
Biotin
Symptoms
rash
conjunctivitis (pink eye)
alopecia (hair loss)
central nervous system abnormalities
Effects
Biotin deficiency
Folate
Symptoms
weakness and fatigue
trouble concentrating
irritability
headaches
heart palpitations
shortness of breath
Effects
Folate deficiency
Risk factors
decreased intake — due to an eating disorder, older age, disability or isolation
conditions resulting in malabsorption
hazardous alcohol use
smoking
increased metabolic need — from rapid growth, pregnancy, burns, blood loss, damage to the gastrointestinal tract, some blood conditions
medicines — such as methotrexate and sulfasalazine
Vitamin B12
Symptoms
tired or weak
pale skin
heart palpitations
loss of appetite
weight loss
Effects
Vitamin B12 deficiency
Risk factors
autoimmune gastritis (pernicious anaemia)
gastrectomy (stomach removal)
small bowel resection
bariatric surgery
vegan diet
intestinal disorders that may affect absorption — such as coeliac disease, inflammatory bowel disease
medicines — metformin
In more severe cases, vitamin B deficiency can cause more serious problems.
What causes vitamin B deficiency?
You may be at risk of vitamin B deficiency if you don't include enough vitamin B in your diet.
The Australian Dietary Guidelines advise people with a vegan diet to take a vitamin B12 supplement. This is because vitamin B12 deficiency is more common in this group of people. If you are vegan, the only reliable sources of vitamin B12 are:
fortified plant milks
some fortified meat substitute products
some breakfast cereals
The B vitamins are water soluble. This means that they are not stored in your body, so deficiency can occur quickly if your intake is low.
Risk factors for vitamin B deficiency
You are at greater risk of vitamin B deficiency if you have changes to your gastrointestinal tract from:
conditions — such as coeliac disease or Crohn's disease
surgery to your small intestine
It can make it harder for your body to absorb vitamins from food.
Vitamin B deficiency can also be caused by alcohol abuse. Your liver needs more vitamin B to metabolise (break down) the excess alcohol.
In some cases, deficiency of one type of B vitamin can cause a deficiency in another type. This is because they need each other to function properly.
When should I see my doctor?
If you think you might be at risk of vitamin B deficiency, speak with your doctor or an accredited dietitian.
FIND A HEALTH SERVICE — The Service Finder can help you find doctors, pharmacies, hospitals and other health services.
How is vitamin B deficiency diagnosed?
Your doctor will ask you about your symptoms and may examine you. They may also want to know about your diet, whether you drink alcohol and if you smoke.
They may arrange for you to have some blood tests.
Your doctor may give you nutrition advice or refer you to an accredited dietitian.
How is vitamin B deficiency treated?
Vitamin B deficiency is treated with doses of whichever vitamin is lacking. This may be given by:
injection
tablet
liquid
nose gels or sprays
Your doctor may want to see you again after 3 months to see if the treatment has worked.
Complications of vitamin B deficiency
The complications of vitamin B deficiency depend on which B-vitamin you are lacking.
Severe anaemia can lead to tachycardia (fast heartbeat) and heart failure.
If you're pregnant, not having enough folate can increase the risk of your baby having a neural tube defect, such as spina bifida.
Folate deficiency can also increase your risk of some cancers.
Resources and support
See the Australian Guide to Healthy Eating.
The Dietitians Australia website allows you to search for an Accredited Practising Dietitian.
You can also call the healthdirect helpline on 1800 022 222 (known as NURSE-ON-CALL in Victoria). A registered nurse is available to speak with 24 hours a day, 7 days a week.
Learn more here about the development and quality assurance of healthdirect content.
Last reviewed: April 2024
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4970 | https://www.naturalspublishing.com/download.asp?ArtcID=195 | Applied Mathematics & Information Sciences 1(2)(2007), 185-194
— An International Journal
c©2007 Dixie W Publishing Corporation, U. S. A.
Common Fixed Point Theorems of Single and Set-Valued Mappings on 2-Metric Spaces
M. E. Abd El-Monsef 1, H. M. Abu-Donia 2, and Kh. Abd-Rabou 1
1
Department of Mathematics, Faculty of Science, Tanta University, Egypt
2
Department of Mathematics, Faculty of Science, Zagazig University, Egypt
Recived 11 Dec. 2006, Accepted 3 March 2007
The purpose of this paper is to study a common fixed point theorems on 2-metric spaces. Generalizations some definitions on 2-metric spaces and Fisher theorems on 2-metric spaces.
Keywords: 2-metric spaces, common fixed point, weakly compatible mappings, com-pact 2-metric spaces.
1 Introduction
The concept of a 2-metric space is a natural generalization of a metric space. It has been investigated initially by G ¨ahler . Iseki studied the fixed point theorems in 2-metric spaces. Sessa defined weak commutativity and proved common fixed point theorem for weakly commuting maps. In Jungck introduced more generalized commuting map-pings, called compatible mappings, which are more general than commuting and weakly commuting mappings. This concept has been useful for obtaining more comprehensive fixed point theorems. In [8,9] Jungck and Rhoades defined the concepts of δ-compatible and weakly compatible mappings, which extend the concept of compatible mappings in the single-valued setting on metric spaces. Several authors used these concepts to prove some common fixed point theorems (See, e.g., [13-16]). In this paper we generalized some def-initions on 2- metric spaces and studied common fixed point theorems for four mappings on 2- metric spaces.
2 Preliminaries
The concept of a 2-metric space is a natural generalization of a metric space by G ¨ahler as the following definition. 186 M. E. Abd El-Monsef, H. M. Abu-Donia, and Kh. Abd-Rabou
Definition 2.1. A 2-metric space is a set with a real-valued function on satisfying the following conditions: (1) For distinct points x, y ∈ X, there exists a point c ∈ X such that d(x, y, c ) 6 = 0 ;(2) d(x, y, c ) = 0 if at least two of x, y and c are equal equal; (3) d(x, y, c ) = d(x, c, y ) = d(c, y, x );(4) d(x, y, c ) ≤ d(x, y, z ) + d(x, z, c ) + d(z, y, c ) ∀ x, y, c, z ∈ X.
The function is called a 2-metric for the space X and the pair (X, d ) (denotes a 2-metric space. It has shown by G ¨ahler that a 2-metric d is non-negative and although d is a continuous function of any one of its three arguments, it need not be continuous in two arguments. A 2-metric space d which is continuous in all of its arguments is said to be continuous. Geometrically a 2-metric d(x, y, c ) represents the area of triangle with vertices
x, y and c .Throughout this paper, let (X, d ) be 2-metric space unless mentioned otherwise and
B(X) is the set of all nonempty bounded subset of X.
Definition 2.2. A sequence {xn} in (X, d ) is said to be convergent to a point x in X,denoted by lim n→∞ xn = x if lim n→∞ d(xn, x, c ) = 0 for all c in X. The point x is called the limit of the sequence {xn} in X.
Definition 2.3. A sequence {xn} in (X, d ) is said to be Cauchy sequence if lim n→∞
d(xm, x n, c ) = 0 , for all c in X.
Definition 2.4. The space (X, d ) is said to be complete if every Cauchy sequence in converges to an element in X.
Remark 1. We note that, in a metric space a convergent sequence is a Cauchy sequence and in a 2-metric space a convergent sequence need not be a Cauchy sequence, but every convergent sequence is a Cauchy sequence when the 2-metric d is continuous on X .
For all A, B and C in B(X), let δ(A, B, C ) be the functions defined by
δ(A, B, C ) = sup {d(a, b, c ) : a ∈ A, b ∈ B, c ∈ C},d(a, b, C ) = inf {d(x, y, c ) : c ∈ C}.
If A consists of a single point a we write δ(A, B, C ) = δ(a, B, C ), if B and C con-sist of a single point b and c respectively, we write δ(A, B, C ) = δ(a, b, c ). It follows immediately from the definition that Common Fixed Point Theorems 187
δ(A, B, C ) = δ(A, C, B ) = δ(C, B, A ) = δ(C, A, B ) = δ(B, C, A ) = δ(B, A, C ) ≥ 0.
δ(A, B, C ) ≤ δ(A, B, E ) + δ(A, E, C ) + δ(E, B, C ). For all A, B, C and E in B(X)
δ(A, B, C ) = 0 . if at least two of A, B and C consist of equal single points.
Definition 2.5. A sequence An of subsets of X is said to be convergent to a subset A of X
if (i) given a ∈ A, there is a sequence an in X such that an ∈ An for n = 1 , 2, 3, ... . {an}
converges to a.(ii) given ε > 0, there exists a positive integer N such that An ⊆ Aε for n > N where
Aε is the union of all open spheres with centers in A and radius ε.
Definition 2.6. The mappings F : X −→ B(X) and I : X −→ X are said to be weakly commuting on X if IF x ∈ B(X) and
δ(F Ix, IF x, C ) ≤ max {δ(Ix, F x, C ), δ (IF x, IF x, C )} (2.1) Note that if F is a single-valued mapping, and then the set IF x consists of a single point. Therefore, δ(IF x, IF x, C ) = d(IF x, IF x, C ) = 0 and condition (2.1) reduces to the condition given by Khan that is δ(F Ix, IF x, C ) ≤ d(Ix, F x, C ). Two commuting mapping F and I clearly weakly commuting but the converse is false.
Definition 2.7. Two single-valued mappings f and g of (X, d ) into itself are compati-ble if lim n→∞ d(f gx n, gf x n, C ) = 0 , whenever {xn} is a sequence in X such that lim n→∞ f x n
= lim n→∞ gx n = t, for some t in X.
It can be seen that two weakly commuting mappings are compatible but the converse is false.
Definition 2.8. The mappings I : X −→ X and F : X −→ B(X) are δ-compatible if
lim n→∞ δ(F Ix n, IF x n, C ) = 0 whenever {xn} is a sequence in X such that, IF x ∈ B(X),
F x n −→ { t} and Ix n −→ t for some t in X.
Definition 2.9. The mappings I : X −→ X and F : X −→ B(X) are weakly compatible if they commute at coincidence points. i.e. for each point u in X such that, F u = {Iu },we have F Iu = IF u .
Not that the equation F u = {Iu } implies that F u is singleton. It can be seen that any
δ-compatible pair {F, I } is weakly compatible but the converse is false. 188 M. E. Abd El-Monsef, H. M. Abu-Donia, and Kh. Abd-Rabou
Definition 2.10. A set-valued mapping F of X into B(X) is said to be continuous at
x ∈ X if the sequence {F x n} in B(X) converges to F x whenever {xn} is a sequence in
X converging to x in X. F is said to be continuous on X. if it is continuous at every point in X.
In , Fisher proved the following theorem:
Theorem 2.11. Let F , G be mappings of X into B(X) and I, J be mappings of X into itself satisfying:
δ(F x, Gy ) ≤ c max {d(Ix, Jy ), δ (Ix, F x ), δ (Jy, Gy )}, for all x, y ∈ X
where 0 ≤ c < 1. If F commutes with I and G commutes with J, G(x) ⊆ I(x), F (x) ⊆
J(x) and I or J is continuous, then F, G, I and J have a unique common fixed point u in
X.
Also, Fisher proved the following theorem on compact metric space:
Theorem 2.12. Let F , G be continuous mapping of a compact metric space (X, d ) into
B(X) and I, J continuous mapping of X into itself satisfying the inequality:
δ(F x, Gy ) < c max {d(Ix, Jy ), δ (Ix, F x ), δ (Jy, Gy )}, (2.2)
for all x, y ∈ X for which the right hand side of the inequality (2.2) is positive. If the mappings F and I commute, G and J commute and G(X) ⊂ I(X), F (X) ⊂ J(X), then there is a unique point u in X such that F u = Gu = {u} = {Iu } = {Ju }.
The main aim of the present paper is to prove common fixed point theorems on 2-metric spaces.
3 Some Auxiliary Lemmas and the Main Theorems
Let I, J be mappings from 2-metric space (X, d ) into itself and F, G : X −→ B(X)
set-valued mappings such that
G(x) ⊆ I(x) and F (x) ⊆ J(x) (3.1) Also, the mappings F, G, J and I satisfy the following inequality:
δ(F x, Gy, C ) ≤ k max {d(Ix, Jy, C ), δ (Ix, Gy, C ), δ (Jy, Gy, C )}, (3.2) for all x, y in X and 0 ≤ k < 1.Since F (x) ⊆ J(x), for an arbitrary point x0 in X there exists a point x1 in X such that Jx 1 ∈ F x 0. Since G(x) ⊆ I(x), for this point x1 there exist a point x2 in X such Common Fixed Point Theorems 189 that Ix 2 ∈ Gx 1 and so on. Consequently, we can define a sequence {yn} in X such that
Jx 2n+1 ∈ F x 2n = y2n and Ix 2n+2 ∈ Gx 2n+1 = y2n+1 , for all n = 1 , 2, 3, ... .In the following we introduce some auxiliary lemmas are useful in the sequel.
Lemma 3.1. Suppose that I, J be mappings from (X, d ) into itself and F, G : X −→
B(X) set-valued mappings such that conditions (3.1) and (3.2) are satisfying. Then for every n ∈ N , we have δ(yn, y n+1 , y n+2 ) = 0 .Proof. Since δ(y2n+2 , y 2n+1 , y 2n) = δ(F x 2n+2 , Gx 2n+1 , y 2n) and by using (3.2), we have
δ(y2n+2 , y 2n+1 , y 2n)= δ(F x 2n+2 , Gx 2n+1 , y 2n)
≤ c max {d(Ix 2n+2 , Ix 2n+1 , y 2n), δ (Ix 2n+2 , F x 2n+2 , y 2n), δ (Jx 2n+1 , Gx 2n+1 , y 2n)}≤ c max {δ(y2n+1 , y 2n, y 2n), δ (y2n+1 , y 2n+2 , y 2n), δ (y2n, y 2n+1 , y 2n)},
Thus, we have (1 − c)δ(y2n+2 , y 2n+1 , y 2n) ≤ 0. Then δ(y2n+2 , y 2n+1 , y 2n) = 0 .Similarly, we have δ(y2n+1 , y 2n+2 , y 2n+3 ) = 0 . Hence δ(yn, y n+1 , y n+2 ) = 0 .
Lemma 3.2. If {An} and {Bn} are sequences in B(X) converging to A and B in B(X),respectively. Then the sequence {δ(An, B n, C )} converges to δ(A, B, C ).Proof. Since An −→ A, Bn −→ B and
d(an, b n, c ) ≤ d(an, a, c ) + d(an, b n, a ) + d(a, b, c ) + d(b, b n, c ) + d(a, b n, b )
we have d(an, b n, c ) ≤ d(a, b, c ), | sup d(an, b n, c ) − sup d(a, b, c ) |= 0 .Since δ(An, B n, C ) = sup {d(an, b n, c ) : an ∈ An, b n ∈ Bn, c ∈ C}, δ(A, B, C ) = sup {d(a, b, c ) : a ∈ A, b ∈ B, c ∈ C}. Then the sequence {δ(An, B n, C )} converges to
δ(A, B, C ).Now we can introduce the main theorems:
Theorem 3.3. Let F, G be mappings of X into B(X) and I, J be mappings of X into itself satisfying:
G(x) ⊆ I(x), F (x) ⊆ J(x) (3.3)
δ(F x, Gy, C ) ≤ k max {d(Ix, Jy, C ), δ (Ix, F x, C ), δ (Jy, Gy, C )}, (3.4)
for all x, y ∈ X where 0 ≤ k < 1. If both pairs {F, I } and {G, J } are weakly compatible and one of I(X) or J(X) is complete. Then F, G, I and J have a unique common fixed point u in X.190 M. E. Abd El-Monsef, H. M. Abu-Donia, and Kh. Abd-Rabou
Proof. Let x0 be an arbitrary point in X. By (3.3), we chose a point x1 in X such that
Jx 1 ∈ F x 0. For this point x1 there exist a point x2 in X such that Ix 2 ∈ Gx 1 and so on. Inductively, we can define a sequence {yn} in X such that
Jx 2n+1 ∈ F x 2n = y2n, Ix 2n+2 ∈ Gx 2n+1 = y2n+1 (3.5) for all n = 0 , 1, 2, ... . For simplicity, we put Vn = δ(yn, y n+1 , C ), By using (3.4), we have
V2n = δ(y2n, y 2n+1 , C ) = δ(F x 2n, Gx 2n+1 , C )
≤ k max {d(Ix 2n, Jx 2n+1 , C ), δ (Ix 2n, F x 2n, C ), δ (Jx 2n+1 , Gx 2n+1 , C )}≤ k max {δ(Gx 2n−1, F x 2n, C ), δ (Gx 2n−1, F x 2n, C ), δ (F x 2n, Gx 2n+1 , C )}≤ k max {δ(y2n−1, y 2n, C ), δ (y2n, y 2n+1 , C )}≤ k max {V2n−1, V 2n}.
If V2n−1 ≤ V2n, thus (1 − k)V2n ≤ 0. Since 0 ≤ k < 1, thus V2n ≤ 0, this is a contradiction implies
V2n ≤ kV 2n−1, (3.6)
V2n+1 = δ(y2n+1 , y 2n+2 , C ) = δ(F x 2n+1 , Gx 2n+2 , C )
≤ k max {d(Ix 2n+1 , Jx 2n+2 , C ), δ (Ix 2n+1 , F x 2n+2 , C ), δ (Jx 2n+2 , Gx 2n+2 , C )}≤ k max {δ(y2n, y 2n+1 , C ), δ (y2n+1 , y 2n+2 , C )}≤ k max {V2n, V 2n+1 }.
Similarly, we obtain that
V2n+1 ≤ kV 2n, (3.7) for all n = 0 , 1, 2, · · · . By (3.6) and (3.7), we have Vn ≤ kV n−1 ≤ k2Vn−2 ≤ · · · ≤ knV0.Then
lim
n→∞
Vn = lim
n→∞
δ(yn, y n+1 , C ) = 0 . (3.8) For all n < m , we have that
δ(yn, y m, C ) ≤ δ(yn, y n+1 , y n+2 ) + δ(yn+1 , y n+2 , y n+3 ) + · · ·
δ(ym−2, y m−1, y m) + δ(ym−1, y m, C ).
By taking the limit as n, m −→ ∞ and using Lemma 3.1 and (3.8), we obtain that
lim
n→∞
δ(yn, y m, C ) = 0 .
Then {yn} is Cauchy sequence in X. Suppose that J(X) is complete. Let {xn} be the sequence defined by Jx 2n+1 ∈ F x 2n = y2n, for all n = 0 , 1, 2, · · · .Common Fixed Point Theorems 191 Since lim n,m →∞ d(Jx 2m+1 , Jx 2n+1 , C ) ≤ lim n,m →∞ δ(Jx 2m, Jx 2n, C ) = 0 , the sequence
{Jx 2n+1 } is Cauchy and hence Jx 2n+1 −→ p = Jv ∈ J(X), for some v ∈ X. But
Jx 2n ∈ Gx 2n−1 = y2n−1, so we obtain
lim
n,m →∞
d(Ix 2n, Jx 2n+1 , C ) ≤ lim
n→∞
δ(y2n−1, y 2n, C ) = 0 .
Consequently, Ix 2n −→ p. Moreover, we obtain
δ(F x 2n, p, C ) ≤ δ(F x 2n, Ix 2n, C ) + δ(Ix 2n, p, C ) + δ(F x 2n, p, Ix 2n).
Therefore, we have that lim n→∞ δ(F x 2n, p, C ) = 0 . Similarly, we have
lim
n→∞
δ(Gx 2n−1, p, C ) = 0
Since δ(F x 2n, Gv, C ) ≤ k max {d(Ix 2n, Jv, C ), δ (Ix 2n, F x 2n, C ), δ (Jv, Gv, C )},
δ(Ix 2n, Gv, C ) −→ δ(p, Gv, C ). When, Ix n −→ p, we get as n −→ ∞, (1 −
k)δ(p, Gv, C ) ≤ 0. Hence Gv = {p} = {Jv }.Since G(x) ⊆ I(x), so u ∈ X exists such that {Iu } = Gv = {Jv }.Now if F u 6 = Gv , this implies that δ(F u, Gv, C ) 6 = 0 , so that we have
δ(F u, Gv, C ) ≤ k max {d(Iu, Jv, C ), δ (Iu, F u, C ), δ (Jv, Gv, C )}.
Then F u = {p} = Gv = {Iu } = {Jv }. Since F u = {Iu } and the pair {F, I } is weakly compatible, we obtain F p = F Iu = IF u = {Ip }. By using (3.2), we obtain
δ(F p, p, C ) ≤ δ(F p, Gv, C )
≤ k max {d(Ip, Jv, C ), δ (Ip, F p, C ), δ (Jv, Gv, C )}
Then F p = {p} = {Ip }. Similarly, Gp = {p} = {Jp } if the pair {G, J } is weakly compatible, we obtain {p} = {Ip } = {Jp } = F p = Gp . Similarly, if I(X) is complete. Now, we prove the uniqueness. To see the point p is unique, suppose that w is another common fixed point of F, G, J and I with w 6 = p. Then we have
d(p, w, C ) ≤ δ(F p, Gw, C )
≤ k max {d(Ip, Jw, C ), δ (Ip, F p, C ), δ (Jw, Gw, C )}≤ kd (p, w, C ).
This implies that w = p.
Theorem 3.4. Let I, J be function of a compact 2-metric space (x, d ) into itself and F, G :
X −→ B(X) two set-valued functions with G(x) ⊆ I(x), F (x) ⊆ J(x). Suppose that the inequality:
δ(F x, Gy, C ) < max {d(Ix, Jy, C ), δ (Ix, F x, C ), δ (Jy, Gy, C )}, (3.9) 192 M. E. Abd El-Monsef, H. M. Abu-Donia, and Kh. Abd-Rabou
for all x, y ∈ X holds whenever the right hand side of the inequality (3.9) is positive. If the pairs {F, I } and {G, J } are weakly compatible, and if the function F and I are continuous, then there is a unique point u in X such that
F u = Gu = {u} = {Iu } = {Ju }.
Proof. Let η = inf
x∈X
{δ(Ix , Fx , C) }. Since X is a compact 2-metric space, there is a con-vergent sequence {xn} with limit x0 in X such that
δ(Ix n, F x n, C ) −→ η as n −→ ∞ .
Since
δ(Ix 0, F x 0, C ) ≤ d(Ix 0, Ix n, C ) + δ(Ix n, F x 0, C ) + δ(Ix 0, F x 0, Ix n)
≤ d(Ix 0, Ix n, C ) + δ(Ix n, F x n, C ) + δ(F x n, F x 0, C )+ δ(Ix 0, F x 0, F x n) + δ(Ix 0, F x 0, Ix n)
by the continuity of F and I and lim n→∞ xn = x0, we get δ(Ix 0, F x 0, C ) ≤ η and thus
δ(Ix 0, F x 0, C ) = η.
Since F (X) ⊂ J(X), there exists a point y0 in X with Jy 0 ∈ F x 0 and δ(Ix 0, Jy 0, C ) ≤
η. If η > 0. By (3.9), we obtain
δ(Jy 0, Gy 0, C ) ≤ d(F x 0, Gy 0, C )
< max {d(Ix 0, Jy 0, C ), δ (Ix 0, F x 0, C ), δ (Jx 0, Gy 0, C )}
< max {η, δ (Jx 0, Gy 0, C )},
which implies that δ(Jy 0, Gy 0, C ) ≤ η.Since G(X) ⊂ I(X), there exists a point z0 in X with Iz 0 ∈ Gy 0 and δ(Iz 0, Jy 0, C ) ≤ η.Hence
η ≤ δ(Iz 0, F z 0, C ) ≤ δ(F z 0, Gy 0, C )
< max {d(Iz 0, Jy 0, C ), δ (Iz 0, F z 0, C ), δ (Jy 0, Gy 0, C )}
< δ(Iz 0, F z 0, C ).
This contradiction demands that η = 0 . Therefore, we have Gy 0 = {Jy 0} = F x 0 =
{Ix 0} = {Iz 0}. Since F and I are weakly compatible and F x 0 = {Ix 0}, we obtain
F 2x0 = F Ix 0 = IF x 0 = {I2x0}. If I2x0 6 = Ix 0, then we have
δ(I2x0, Ix 0, C ) = δ(F 2x0, Gy 0, C )
< max {d(IF x 0, Jy 0, C ), δ (IF x 0, F 2x0, C ), δ (Jy 0, Gy 0, C )}
= δ(I2x0, Ix 0, C ).Common Fixed Point Theorems 193 So we have I2x0 = Ix 0, and hence F Ix 0 = {Ix 0} = {I2x0}. Similarly, we have
GJy 0 = {Jy 0} = {J2y0}. Let u = Ix 0 = Jy 0, thus F u = {u} = {Iu } = {Ju } = Gu .Suppose that the point y in X is a common fixed point of F, G, J and I. If either
δ(y, F y, C ) 6 = 0 or δ(y, Gy, C ) 6 = 0 , then we have that
δ(y, F y, C ) ≤ δ(F y, Gy, C )
< max {d(Iy, Jy, C ), δ (Iy, F y, C ), δ (Jy, Gy, C )}
< max {d(y, F y, C ), δ (y, Gy, C )}.
This implies that δ(y, F y, C ) < δ (y, Gy, C ). By symmetry, we have that δ(y, Gy, C ) <δ(y, F y, C ), which is impossible. So δ(y, F y, C ) = δ(y, Gy, C ), and F y = Gy = {y}.Now, we prove the uniqueness. Let y, u in X are two common fixed points of F, G, J
and I with y 6 = u. On using (3.9), we have that
d(y, u, C ) = δ(F y, Gu, C )
< max {d(Iy, Ju, C ), δ (Iy, F u, C ), δ (Ju, Gu, C )}
< d(y, u, C ).
This implies that y = u. Then F, G, J and I have a unique common fixed point in X.
References
B. Fisher, Common fixed points of mappings and set-valued mappings, Rostock. Math. Kolloq. , 18 (1981), 69-77. B. Fisher, Common fixed points of mappings and set-valued mappings on a metric spaces, Kyungpook Math. J. , 25 (1985), 35-42. B. Fisher and S. Sessa, Two common fixed point theorems for weakly commuting mappings, Period. Math. Hungar. , 20 (1989) 207-218. S. Gahler, 2-metrische Raume und ihre topologische structure, Math. Nacher. ,
26 (1963), 115-148. K. Iseki, Fixed point theorems in 2-metric spaces, Math. Sem. Notes , 3(1975), 133-136. M. D. Khan, A study of fixed point theorems , Doctoral Thesis, Aligrah Muslim Uni-versity, 1984. G. Jungck, Compatible mappings and common fixed points, Int. J. Math. math. Sci. ,
9(1986), 771-779. 194 M. E. Abd El-Monsef, H. M. Abu-Donia, and Kh. Abd-Rabou G. Jungck and B. E. Rhoades, Some fixed point theorems for compatible maps, Int. J. Math. Math. Sci. , 16 (1993), 417-428. G. Jungck and B. E. Rhoades, Fixed points for set valued functions without continuity,
Indian J. Pure Appl. Math. , 16 (1998), 227-238. P. P. Murthy, S. S. Chang, Y. J. Cho, and B. K. Sharma, Compatible mappings of type (A) and common fixed point theorem, Kyungpook Math. J. , 32 (1992), 203-216. S. V. R. Naidu and J. R. Prasad, Fixed point theorem in 2-metric spaces, Indian J. Pure Appl. Math. , 17 (1986), 974-993. H. K. Pathak, S. M. Kang, and J. H. Baek, Weak compatible mappings of type (A) and common fixed points, Kyungpook Math. J. , 35 (1995), 345-359. R. A. Rashwan and M. A. Ahmed, Common fixed points for -compatible mappings,
Southwest J. Pure Applied Math. , 1(1996), 51-61. R. A. Rashwan, Fixed points of single and set-valued mappings, Kyungpook Math. J. ,
38 (1998), 29-37. B. E. Rhoades, Common fixed points of compatible set-valued mappings, Publ. Math. Debrecen , 48 (1996), 237-240. B. E. Rhoades, S. Park, and K. B. Moon, On generalizations of the Meir-Keeler type contraction maps, J. Math. Anal. Appl. , 146 (1990), 482-494. S. Sessa, On a weak commutativity condition of mapping in fixed point considera-tions, Pub. Inst. Math. , 32 (1982), 149-153. |
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4972 | https://www.mathopenref.com/parallelogramarea.html | Area of a parallelogram
Area formula
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Recall that any of the four sides can be chosen as the base. You must use the altitude that goes with the base you choose.
The altitude (or height) of a parallelogram is the perpendicular distance
from the base to the opposite side (which may have to be extended). In the figure above, the altitude corresponding to the base CD is shown.
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Angles associated with polygons
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Back To Suma y Diferencia de PolinomiosBack
9.2
Addition and Subtraction of Polynomials
Written by:Andrew Gloag | Melissa Kramer |
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
Suppose that two cars are having a race. The distance traveled by one car after seconds is meters, while the distance traveled by the other car after seconds is meters. How far apart would the two cars be after seconds? What would you have to do to find the answer to this question?
Adding and Subtracting Polynomials
To add or subtract polynomials, you have to group the like terms together and combine them to simplify.
Let's complete the following problems:
Add and .
Subtract from .
Solving Real-World Problems Using Addition or Subtraction of Polynomials
Polynomials are useful for finding the areas of geometric objects. In the following problems, you will see this usefulness in action.
Let's write a polynomial that represents the area of each figure shown.
The blue square has the following area: .
The yellow square has the following area: .
The pink rectangles each have the following area: .
To find the area of the green region we find the area of the big square and subtract the area of the little square.
The big square has an area of .
The little square has an area of .
Area of the green region
Examples
Example 1
Earlier, you were told that in a race, the distanced traveled by one car after seconds is meters while the distance traveled by the second car after seconds is meters. How far apart would the two cars be after seconds.
To solve this question, you need to subtract the two expressions that represent distances. Let's subtract the distance of the first car from the distance of the second car and simplify.
The two cars would be meters apart after seconds.
Example 2
Subtract from .
When subtracting polynomials, we have to remember to subtract each term. If the term is already negative, subtracting a negative term is the same thing as adding:
The final answer is in standard form.
Review
Add and simplify.
Subtract and simplify.
Find the area of the following figures.
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
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4974 | https://microbenotes.com/dna-experiments-griffith-avery-mccarty-macleod-hershey-chase/ | Skip to content
Microbe Notes
DNA Experiments (Griffith & Avery, McCarty, MacLeod & Hershey, Chase)
by Prakriti Karki
DNA, deoxyribonucleic acid, is the carrier of all genetic information. It codes genetic information passed on from one generation to another and determines individual attributes like eye color, facial features, etc. Although DNA was first isolated in 1869 by a Swiss scientist, Friedrich Miescher, from nuclei of pus-rich white blood cells (which he called nuclein), its role in the inheritance of traits wasn’t realized until 1943. Miescher thought that the nuclein, which was slightly acidic and contained a high percentage of phosphorus, lacked the variability to account for its hereditary significance for diversity among organisms. Most of the scientists of his period were convinced by the idea that proteins could be promising candidates for heredity as they were abundant, diverse, and complex molecules, while DNA was supposed to be a boring, repetitive polymer. This notion was put forward as the scientists were aware that genetic information was contained within organic molecules.
Table of Contents
Toggle
Griffith’s Transformation Experiment
In 1928, a young scientist Frederick Griffith discovered the transforming principle. In 1918, millions of people were killed by the terrible Spanish influenza epidemic, and pneumococcal infections were a common cause of death among influenza-infected patients. This triggered him to study the bacteria Streptococcus pneumoniae and work on designing a vaccine against it. It became evident that bacterial pneumonia was caused by multiple strains of S. pneumoniae, and patients developed antibodies against the particular strain with which they were infected. Hence, serum samples and bacterial isolates used in experiments helped to identify DNA as the hereditary material.
He used two related strains of S. pneumoniae and mice and conducted a series of experiments using them.
When type II R-strain bacteria were grown on a culture plate, they produced rough colonies. They were non-virulent as they lacked an outer polysaccharide coat. Thus, when RII strain bacteria were injected into a mouse, they did not cause any disease and survived.
When type I S-strain bacteria were grown on a culture plate, they produced smooth, glistening, and white colonies. The smooth appearance was apparent due to a polysaccharide coat around them that provided resistance to the host’s immune system. It was virulent and thus, when injected into a mouse, resulted in pneumonia and death.
In 1929, Griffith experimented by injecting mice with heat-killed SI strain (i.e., SI strain bacteria exposed to high temperature ensuing their death). But, this failed to harm the mice, and they survived.
Surprisingly, when he mixed heat-treated SI cells with live RII cells and injected the mixture into the mice, the mice died because of pneumonia. Additionally, when he collected a blood sample from the dead mouse, he found that sample to contain live S-strain bacteria.
Conclusion of Griffith’s Transformation Experiment
Based on the above results, he inferred that something must have been transferred from the heat-treated S strain into non-virulent R strain bacteria that transformed them into smooth coated and virulent bacteria. Thus, the material was referred to as the transforming principle.
Following this, he continued with his research through the 1930s, although he couldn’t make much progress. In 1941, he was hit by a German bomb, and he died.
Avery, McCarty, and MacLeod Experiment
During World War II, in 1943, Oswald Avery, Maclyn McCarty, and Colin MacLeod working at Rockefeller University in New York, dedicated themselves to continuing the work of Griffith in order to determine the biochemical nature of Griffith’s transforming principle in an in vitro system. They used the phenotype of S. pneumoniae cells expressed on blood agar in order to figure out whether transformation had taken place or not, rather than working with mice. The transforming principle was partially purified from the cell extract (i.e., cell-free extract of heat-killed type III S cells) to determine which macromolecule of S cell transformed type II R-strain into the type III S-strain. They demonstrated DNA to be that particular transforming principle.
Initially, type III S cells were heat-killed, and lipids and carbohydrates were removed from the solution.
Secondly, they treated heat-killed S cells with digestive enzymes such as RNases and proteases to degrade RNA and proteins. Subsequently, they also treated it with DNases to digest DNA, each added separately in different tubes.
Eventually, they introduced living type IIR cells mixed with heat-killed IIIS cells onto the culture medium containing antibodies for IIR cells. Antibodies for IIR cells were used to inactivate some IIR cells such that their number doesn’t exceed the count of IIIS cells. that help to provide the distinct phenotypic differences in culture media that contained transformed S strain bacteria.
Observation of Avery, McCarty, and MacLeod Experiment
The culture treated with DNase did not yield transformed type III S strain bacteria which indicated that DNA was the hereditary material responsible for transformation.
Conclusion of Avery, McCarty, and MacLeod Experiment
DNA was found to be the genetic material that was being transferred between cells, not proteins.
Hershey and Chase Experiment
Although Avery and his fellows found that DNA was the hereditary material, the scientists were reluctant to accept the finding. But, not that long afterward, eight years after in 1952, Alfred Hershey and Martha Chase concluded that DNA is the genetic material. Their experimental tool was bacteriophages-viruses that attack bacteria which specifically involved the infection of Escherichia coli with T2 bacteriophage.
T2 virus depends on the host body for its reproduction process. When they find bacteria as a host cell, they adhere to its surface and inject its genetic material into the bacteria. The injected hereditary material hijacks the host’s machinery such that a large number of viral particles are released from them. T2 phage consists of only proteins (on the outer protein coat) and DNA (core) that could be potential genetic material to instruct E. coli to develop its progeny. They experimented to determine whether protein or DNA from the virus entered into the bacteria.
Bacteriophage was allowed to grow on two of the medium: one containing a radioactive isotope of phosphorus(32P) and the other containing a radioactive isotope of sulfur (35S).
Phages grown on radioactive phosphorus(32P) contained radioactive P labeled DNA (not radioactive protein) as DNA contains phosphorus but not sulfur.
Similarly, the viruses grown in the medium containing radioactive sulfur (35S) contained radioactive 35S labeled protein (but not radioactive DNA) because sulfur is found in many proteins but is absent from DNA.
E. coli were introduced to be infected by the radioactive phages.
After the progression of infection, the blender was used to remove the remains of phage and phage parts from the outside of the bacteria, followed by centrifugation in order to separate the bacteria from the phage debris.
Centrifugation results in the settling down of heavier particles like bacteria in the form of pellet while those light particles such as medium, phage, and phage parts, etc., float near the top of the tube, called supernatant.
Observation of Hershey and Chase Experiment
On measuring radioactivity in the pellet and supernatant in both media, 32P was found in large amount in the pellet while 35S in the supernatant that is pellet contained radioactively P labeled infected bacterial cells and supernatant was enriched with radioactively S labeled phage and phage parts.
Conclusion of Hershey and Chase Experiment
Hershey and Chase deduced that it was DNA, not protein which got injected into host cells, and thus, DNA is the hereditary material that is passed from virus to bacteria.
References
Fry, M. (2016). Landmark Experiments in Molecular Biology. Academic Press.
About Author
Prakriti Karki
Prakriti Karki completed her B.Sc. in the field of Microbiology. She is interested in working in the interface of immunology, microbiology, synthetic biology, bioinformatics, and open science. She has worked as a project lead at Media Lab Nepal, as a research associate in the BMSIS program, and as an awareness community member at the iGEM WiSTEM initiative.
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4975 | https://evokewellness.com/blog/z-drug-mechanism-action-inducing-sleep/ | Z-Drug Mechanism of Action For Inducing Sleep? | Evoke Wellness
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Z-Drug Mechanism of Action For Inducing Sleep?
Neil McKinnellJuly 19, 2021Drug Abuse
Z-Drugs are a class of psychoactive drugs known as sedative-hypnotics. They are also sometimes known as nonbenzodiazepines or nonbenzodiazepine hypnotics because they are benzodiazepine-like. These drugs are prescribed for sleep problems. They have a very quick onset of action and help improve sleep quality; however, they do not improve the length of sleep.
The Effects of Z-Drugs on Human Performance and Driving
The National Institute of Health “In the Zzz Zone: The Effects of Z-Drugs on Human Performance and Driving” states that:
The Z-drugs, zolpidem, zopiclone, and zaleplon, were developed as hypnotics with improved pharmacokinetics compared to benzodiazepines, the traditional treatments for insomnia. Their pharmacology and toxicology have been previously reviewed. Zolpidem, zaleplon, and eszopiclone are the three Z-drugs currently approved by the US Food and Drug Administration (FDA) to treat insomnia. Like benzodiazepines, Z-drugs are GABAA receptor agonists; however, their clinically attractive properties include short duration of action, non-disturbance of overall sleep architecture, and diminished residual effects during daytime hours. While they have been studied in the elderly, there is increasing interest in Z-drug effectiveness and residual effects in shift workers, pilots, and military personnel. (NIH)
Z-Drugs were developed as a safer alternative to benzodiazepines, but they come with dangers, including dependence. Z-Drugs are considered controlled substances and are Schedule IV drugs according to the DEA. Schedule IV drugs are listed as low risk for dependence and low potential for abuse.
Z-Drugs Mechanism of Action
Z-Drugs work by slowing down brain activity. They work on the central nervous system and directly affect the GABA receptors. GABA receptors produce natural sedative-like effects, so Z-Drugs enhance the effects of GABA transmission, like benzodiazepines. This slows down activity in the brain and central nervous system and induces sleep. Z-Drugs are great for short-term sleep issues. Due to the risk of abuse and dependence, they should only be prescribed on a short-term basis.
Side Effects of Z-Drugs
Z-Drugs do slow down the central nervous system, so slowed breathing and heart rate can occur. Also, misusing any of these drugs can cause respiratory failure, unconsciousness, and even death. Some of the other complications that z-drugs can cause are sleepwalking nocturnal eating and sensory disturbances. Ambien has been associated with falls in the elderly as well.
More About Side Effects of Z-Drugs
Most people tolerate Z-Drugs well, but they do have some adverse effects. One of the biggest and most disturbing side effects reported with drugs is visual hallucinations. Some people have also reported amnesia and unusual or inappropriate behavior. Some of the other side effects can include:
Delusions
Extreme sleepiness
Clumsiness
Aggression
Dizziness
Loss of balance
Loss of consciousness
Nausea and vomiting
Some users have reported feeling tired, confused, sluggish, and foggy-headed after taking a z-drug. Ambien can also make symptoms of depression and anxiety worse. And if an individual takes the drug for a long-term period, it can even start causing issues with their sleep (rebound insomnia). Combining z-drugs with alcohol or other benzodiazepines is extremely dangerous. These are all central nervous system depressants, and when they are combined, the risk of respiratory arrest and overdose death is dramatically increased.
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Tagged addiction, drugs, insomnia, mechanisms, side effects, sleeping disorder, Z-drugs
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4976 | https://math.stackexchange.com/questions/218421/what-are-the-practical-applications-of-the-taylor-series | Skip to main content
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Related: Motivating Infinite Series.
– Mike Spivey
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Dec 19, 2012 at 17:40
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One reason is that we can approximate solutions to differential equations this way: For example, if we have
y′′−x2y=ex
To solve this for y would be difficult, if at all possible. But by representing y as a Taylor series ∑anxn, we can shuffle things around and determine the coefficients of this Taylor series, allowing us to approximate the solution around a desired point.
It's also useful for determining various infinite sums. For example:
11−x=∑n=0∞xn
11+x=∑n=0∞(−1)nxn
Integrate:
ln(1+x)=∑n=0∞(−1)nxn+1n+1
Substituting x=1 gives
ln2=1−12+13−14+15−16⋯
There are also applications in physics. If a system under a conservative force (one with an energy function associated with it, like gravity or electrostatic force) is at a stable equilibrium point x0, then there are no net forces and the energy function is concave upwards (the energy being higher on either side is essentially what makes it stable). In terms of taylor series, the energy function U centred around this point is of the form
U(x)=U0+k1(x−x0)2+k2(x−x0)3⋯
Where U0 is the energy at the minimum x=x0. For small displacements the high order terms will be very small and can be ignored. So we can approximate this by only looking at the first two terms:
U(x)≈U0+k1(x−x0)2⋯
Now force is the negative derivative of energy (forces send you from high to low energy, proportionally to the energy drop). Applying this, we get that
F=ma=mx′′=−2k1(x−x0)
Rephrasing in terms of y=x−x0:
my′′=−2k1y
Which is the equation for a simple harmonic oscillator. Basically, for small displacements around any stable equilibrium the system behaves approximately like an oscillating spring, with sinusoidal behaviour. So under certain conditions you can replace a potentially complicated system by another one that's very well understood and well-studied. You can see this in a pendulum, for example.
As a final point, they're also useful in determining limits:
limx→0sinx−xx3
limx→0x−16x3+1120x5⋯−xx3
limx→0−16+1120x2⋯
−16
which otherwise would have been relatively difficult to determine. Because polynomials behave so much more nicely than other functions, we can use taylor series to determine useful information that would be very difficult, if at all possible, to determine directly.
EDIT: I almost forgot to mention the granddaddy:
ex=1+x+12x2+16x3+124x4⋯
eix=1+ix−12x2−i16x3+124x4⋯
=1−12x2+124x4⋯+ix−i16x3+i1120x5⋯
=cosx+isinx
eix=cosx+isinx
Which is probably the most important equation in complex analysis. This one alone should be motivation enough, the others are really just icing on the cake.
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edited Jan 28, 2016 at 15:38
answered Oct 22, 2012 at 1:54
Robert MastragostinoRobert Mastragostino
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Granddaddy is used to approximate exp, sin and cos, right?
– krlmlr
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Oct 22, 2012 at 11:37
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@user946850 without the imaginary part, yes. Taylor series can also be used to approximate these functions in computers to pretty high accuracy. sinx≈x−16x3 has an error of at most 8%. Adding the next term reduces that to less than 0.5% and using this in conjunction with sin(π/2−x)=cosx and cosx≈1−12x2+124x4 can lower this even further. There are many other approximations that exist as well (the CORDIC algorithm, Chebyshev approximation, etc.) but this is sometimes used in practice.
– Robert Mastragostino
Commented
Oct 22, 2012 at 15:43
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@RobertMastragostino what x range did you use to determine that the error is at most 8%? For x∈R the error is unbounded.
– Ruslan
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Feb 12, 2014 at 13:10
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@Ruslan you can use the symmetries and periodicity of sin(x) to restrict your calculation to [0,π/2]. It's this range that has the maximum error of 8%.
– Robert Mastragostino
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Feb 12, 2014 at 14:41
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In the calculator era, we often don't realize how deeply nontrivial it is to get an arbitrarily good approximation for a number like e, or better yet, esin(2√). It turns out that in the grand scheme of things, ex is not a very nasty function at all. Since it's analytic, i.e. has a Taylor series, if we want to compute its values we just compute the first few terms of its Taylor expansion at some point.
This makes plenty of sense for computing, say, e1/2:1+1/2+1/2!(1/2)2+1/3!(1/2)3+... is obviously going to converge very quickly: 1/4!24<1/100 and 1/5!25<1/1000, so we know for instance we can get e1/2 to 2 decimal places by summing the first 5 terms of the Taylor expansion.
But why should this work for computing something like e100? Now the expansion looks like 1+100+1002/2+1003/3!+..., and initially it blows up incredibly fast. This is where analytic functions really show how special they are: the denominators n! grow so fast that it doesn't matter what xn we have in the numerators, before too long the series will converge. That's the essence of the Taylor approximation: analytic functions are those that are unreasonably close to polynomials.
There are much faster methods for getting approximations like the one for e√, in theory: using Newton's method to solve x2−e=0 will give you an approximation to e√ accurate to a number of places that goes like the square of the number of iterations you've done. But how do we apply Newton's method here? The first formula is
x1=x0−2x0x20−e
So, if we want a decimal expansion of e√, we'd better be able to get one of x20−e. And how are we going to get that?
The Taylor series.
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answered Oct 22, 2012 at 1:57
Kevin CarlsonKevin Carlson
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Great answere here. Could you please provide your analogical insights into my query there?
– bonCodigo
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Nov 7, 2014 at 4:21
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Also, Newton-Raphson itself uses Taylor series approximation (first order) to find the root.
– Anindya Mahajan
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Jun 8, 2020 at 21:39
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You might wanna read this Taylor Series as Definitions.
Taylor Series are studied because polynomial functions are easy and if one could find a way to represent complicated functions as series (infinite polynomials) then one can easily study the properties of difficult functions.
Evaluating definite Integrals:
Some functions have no antiderivative which can be expressed in terms of familiar functions. This makes evaluating definite integrals of these functions difficult because the Fundamental Theorem of Calculus cannot be used. If we have a polynomial representation of a function, we can oftentimes use that to evaluate a definite integral.
Understanding asymptotic behaviour: Sometimes, a Taylor series can tell us useful information about how a function behaves in an important part of its domain.
Understanding the growth of functions
Solving differential equations
I'm pretty sure this is not all but with a little research you can find as many as possible.
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edited Jun 12, 2020 at 10:38
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answered Oct 22, 2012 at 1:54
user31280user31280
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The applications of Taylor series is mainly to approximate ugly functions into nice ones(polynomials)!
Example: Take f(x)=sin(x2)+ex4. This is not a nice function, but it can be approximated to a polynomial using Taylor series.
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answered Oct 22, 2012 at 1:38
ILoveMathILoveMath
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That's correct, but maybe add a bit more detail?
– MrChessR
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A good example of Taylor series and, in particular, the Maclaurin series, is in special relativity, where the Maclaurin series are used to approximate the Lorrentz factor γ. Taking the first two terms of the series gives a very good approximation for low speeds. You can actually show that at low speeds, special relativity reduces to classical (Newtonian) physics. For example, in special relativity, the momentum is given by p=γmv, and at low speeds γ≈1, so p≈mv, which is the (linear) momentum in classical mechanics.
Also, the most famous equation in physics, E=mc2, is actually an approximation for low velocities, which, again, can be derived using Taylor series.
By the way,
γ=11−v2/c2−−−−−−−−√,
where v is the velocity and c is the speed of light.
Another example is again from physics: When we first study pendulum motion, we often begin with an assumption sinθ≈θ, which also comes from Taylor series because
sinθ=θ−θ33!+θ55!−θ77!+⋯
Moreover, any software that graphs various functions actually uses very good Taylor approximations.
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edited Nov 13, 2013 at 5:53
answered Oct 22, 2012 at 1:57
glebovgglebovg
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"Moreover, any software that graphs various functions actually uses very good Taylor approximations". I don't think that's always true, do you have a source for that claim?
– R R
Commented
Dec 14, 2013 at 21:05
I am pretty sure there should be an article about this. Do you honestly think a computer can plot ex, sinx, cosx, etc. with infinite precision? Almost all real numbers are transcendental (because algebraic numbers are countable), and therefore irrational, but all numbers in a computer are rational.
– glebovg
Commented
Dec 16, 2013 at 7:15
2
I'm not claiming that at all and I'm not sure how you got that impression. You're claiming that computers always use taylor approximations, I'm thinking taylor approximations are not always efficient or accurate and that there are probably other computational tricks for computing tricky functions. One example off the top of my head is the CORDIC system.
– R R
Commented
Dec 16, 2013 at 17:43
1
Can you expand on your comment that E=mc2 is an approximation for low velocities?
– littleO
Commented
Feb 3, 2015 at 20:42
@littleO Have a look at Einstein's famous energy-momentum relation, which is not very difficult to derive. For a particle at rest, the momentum p is zero, so we get E=mc2. If p≈0, then E≈mc2.
– glebovg
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Feb 23, 2015 at 21:50
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Taylor series provide the basic method for computing transcendental functions
such as ex, sinx, and cosx.
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answered Oct 22, 2012 at 1:49
Matt EMatt E
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No one's mentioned the combinatorial side of things, so I'll be the first to say it: generating functions. We use generating functions to pass hard discrete counting problems to the continuous, where things are easy. Generating functions are a central tool in combinatorics (counting, graph theory, etc.) and probability (where we have moment generating functions). Taylor series is the fundamental idea behind all of these. Read: for details, and take a combinatorics or mathematical probability class to learn more.
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answered Dec 19, 2012 at 16:23
Gyu Eun LeeGyu Eun Lee
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All of computational science is built on Taylor's theorem.
The biggest hammer by far is Newton's method, which is fragile in its raw form but serves as the basis of many efficient and practical algorithms for solving equations
f(x)=0
for horribly nonlinear functions f. It's hard to get more general than that!
Let me mention one other specific application: simulating physics, using Newton's laws. Suppose you have some object with position x(t) that is being acted upon by several possibly complicated nonlinear forces. The second law says
F=ma
or
F=md2xdt2.
Typically F is a function of x: for instance, the gravitational force of one object acting on another obeys an inverse square law, which depends on x. This gives us the second-order ODE
F(x)=md2xdt2.
If F is complicated enough, there is no hope of solving this equation analytically. But suppose we know the initial position x(0)=x0 and initial velocity dxdt(0)=v0, and we want to know what the position and velocity will be at time h. We can Taylor-expand x(t):
x(h)=x(0)+hdxdt(0)+h22d2xdt2(0)+O(h3).
If h is small, we can ignore the higher-order terms, and plug in the above into Newton's law to get
F(x(0))≈2mx(h)−x(0)−hdxdt(0)h2
or
x(h)≈x0+hv0+h22mF(x0),
and similarly
v(h)≈v0+hF(x0).
Once you have the position and velocity at time h, you can predict them at time 2h, by using the above calculations and replacing x0 with xh and v0 with vh. Repeating this process, you can get a good approximation for x(t) for all t!
The error of the each step above will depend on the error you incurred by truncating the Taylor series, which depends on h. But you know that the error should scale roughly like h3/h2=h, so that halving h halves the error at each step. Even more accurate methods can be developed along this vein, where by taking into account more terms of the Taylor series, you have less error at every step, at the cost of more expensive computation each step.
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answered Nov 13, 2013 at 6:37
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We can also use Taylor series to approximate integrals that are impossible with the other integration techniques.
A classic example is ∫sin(x2)dx.
We can't actually integrate this, but using the taylor series for sin(x) we can substitute x2 in for x at each term of the series, and then integrate each term individually. After doing so, we can write a new sum.
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edited Oct 22, 2012 at 7:07
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answered Oct 22, 2012 at 3:02
Nick NicoliniNick Nicolini
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In physics you often approximate a complicated function by taking the first few terms in the Taylor series (the Taylor polynomial). For small values of the independent variable, you often assume linearity, which can allow you to get a closed form solution. For example, if you take an introductory physics class then you usually study the motion of the pendulum by approximating sin(θ) by θ for small angles.
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answered Dec 19, 2012 at 16:34
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Someone already mentioned the usefulness of Taylor's series in relativity, I would like to spent a few words to further explore this point because relativity is a good arena to test the very important role of Taylor series in solving practical problems in physics. Let's consider relativistic kinetic energy formula
EK=mc2⎛⎝⎜11−v2c2−−−−−√−1⎞⎠⎟
Taylor series says that for v≪c the kinetic energy is about
EK≈mv22+3mv48c2
and this allow you to evaluate the relativistic value when you are in classical regime, and then to get an idea of how relativistic corrections are far from our daily experience. At first glance you can say "who care about EK≈ blah blah...? We are in XXI century, I don't need approximate formulae to simplify pen & paper calculations, I simply can take my computer and insert numbers to see what happens". Well, things are not so simple. Let's consider this exercise I took from a Taha Sochi's book: we are evaluating kinetic energy of a 1 kg body moving at 100 m/s.
Classical mechanics says 5000 J but what is the relativistic answer? The book's answer is completely wrong and it is very instructive to see what happened here. I think the author used 3.33⋅10−7 in place of vc, he took his computer, he entered the value, and he found an absurd 4996 J: a relativistic energy lower than classical one! You could think this is a problem related to the very bad habit to do big roundings in intermediate steps. You could say: "I can correct easily this naive mistake: let's use some more digits in vc value!". The idea of using so many digits until the result stabilizes seem reasonable. You can do calculations by exploiting Spreadsheets, WolframAlpha, or simply Google search cell, probably you will find (try!) 5009 J (or 5016 J if you use the approximate value 3⋅108 for c used by the author). You may feel satisfied and feel that the result is right, after all it is just a bit greater than classical. But wait a minute! Is it plausible that for a 1 kg ball moving at the ridiculously low speed of a fast car or slow airplane, the relativistic correction is of some joule? This would be decidedly huge: the second answer too is completely wrong. The problem is that computers usually works with a very limitate number of digits, and from something like 1,0000000(...small)−1 you can get zero or any other strange results! The only way to solve this problem, as far as I know, is using Taylor formula (unless you know how to force computer using more digits, it is possible do that with some programming language, but probably this would be a more complicated and less sure way to solve the problem). Using Taylor formula written before (adding more Taylor terms the change is negligible) you simply got the correct relativistic kinetic energy of our moving ball: about 5000.000000000417 J (3mv48c2≈4.17⋅10−10 J). So in this case classical and relativistic results differ of about 0.00000000001%. All this show that Taylor series are not only illuminating and useful, but sometimes practically indispensable.
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answered Sep 5, 2020 at 16:25
Fausto VezzaroFausto Vezzaro
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The Taylor Series is used in power flow analysis of electrical power systems (Newton-Raphson method).
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answered Oct 22, 2012 at 12:12
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Multivariate Taylor series is used in different optimization techniques; that is you approximate your function as a series of linear or quadratic forms, and then successively iterate on them to find the optimal value.
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answered Oct 22, 2012 at 12:51
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Say you were navigating or orienteering and had plenty of time: One could use the law of sines (and the Taylor series) to evaluate lengths of triangles on maps ( SineA/A = SineB/B = SineC/C). Thus distances could be done with incredible accuracy. Three terms of the series would be plenty. It allows you to calculate sine without a calculator. This is obviously a little ridiculous but ever so slightly useful. If you you didn't have paper you could compute in the sand!
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answered Sep 29, 2014 at 22:11
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One of the main tools in statistical sciences (that you can find almost in every research in Social sciences, Economics and Medicine) is regression analysis. One of the justification of validity of such analysis is that linear regression can be viewed as a linear approximation to some unknown function f(x). Namely, you have a data set that is {xi,yi)}ni=1 and you assume that your data come from some process
Yi=f(Xi)+ϵi,
where
E[Y|X]=f(x)=f(0)+f′(0)x+R1(x)=β0+β1x+R1(x)
namely, you can approximate the data generating process by estimating
yi=β0+β1xi+ϵi.
In such a case you can use some pretty simple methods to estimate the parameters, however in a non-linear models one can use the Newton-Raphson method that uses a linear approximation (first order Taylor expansion) to estimate the parameters. The same logic holds for multiple regression models, where the linear regression is just a first order Taylor expansion (models with interactions and quadratic terms can be viewed as second-order Taylor expansions).
Another useful application is the result that is called the Delta-method. In the context of statistical inference and parameter estimation. Let θ be a parameter of interest and Xn its estimator, then if
n−−√(Xn−θ)−→DN(0,σ2)
and let g some function where g′(θ) exists and not equals zero, then
n−−√(Xn−θ)−→DN(0,g′(θ)2σ2).
This result a straight-forward consequence of Taylor expansion of g(Xn) at θ. This result (and its multivariate) analog allows us to compute asymptotically-correct confidence intervals to various parameters, including for the aforementioned regression parameters.
Using the same basic logic, Taylor expansion allows to approximate variance of complicated configurations (functions) where the explicit variance is too complicated for precise analytical calculations.
The bottom line that in computational sciences where the basic tools are models and the main goals are approximations of (unknown) functions, Taylor series is maybe one of the most fundamental tools to start with.
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answered Mar 8, 2019 at 9:22
V. VancakV. Vancak
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Discrete Mathematics and Its Applications
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"Discrete Mathematics and Its Applications" by Kenneth H. Rosen offers a comprehensive and accessible exploration of fundamental mathematical concepts, tailored to meet the needs of diverse fields such as mathematics, computer science, and engineering. This esteemed, best-selling text stands as a market leader, providing flexibility and effectiveness as a teaching resource for educators. The book also integrates with McGraw-Hill Education's Connect, a cutting-edge digital learning platform designed to enhance student engagement and success. Connect adapts to individual learning needs, streamlining the assignment and grading process for instructors while offering randomized problems and multi-step solutions that support student comprehension and foster independent problem-solving skills.
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Author : Kenneth H. Rosen
Dr. Kenneth H. Rosen holds a B.S. in Mathematics from the University of Michigan, Ann Arbor (1972), and a Ph.D. from the Massachusetts Institute of Technology (1976). Renowned for his contributions to number theory and mathematical modeling, he has published numerous articles in professional journals. Dr. Rosen is also the author of several textbooks, including the widely used "Elementary Number Theory and Its Applications," currently in its fifth edition with Addison-Wesley, as well as "Discrete Mathematics and Its Applications."
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Chapter 1 | 1 The Foundations: Logic and Proofs
Chapter 1: The Foundations: Logic and Proofs
1.1 Propositional Logic
- Logic underpins mathematical reasoning and has significant applications in computer science.
- A proposition is a declarative statement that can be true or false but not both.
- The domain of logic includes basic components like propositions, logical operators, and truth tables.
1.1.1 Introduction to Propositions
- Propositions serve as the fundamental building blocks of logical reasoning, where true values (T) and false values (F) are assigned.
1.1.2 Compound Propositions
- Logical operators like negation (¬), conjunction (∧), and disjunction (∨) are used to create new propositions.
- Truth tables display the truth values of compound propositions.
1.1.3 Conditional and Biconditional Statements
- Conditional statements (→) express a dependency between propositions.
- Biconditional statements (↔) assert that two propositions are true under the same conditions.
1.2 Applications of Propositional Logic
- Propositional logic is essential in various fields, ranging from software design to artificial intelligence.
- Logic is crucial for constructing clear, unambiguous specifications.
1.3 Propositional Equivalences
- Logical equivalences are essential for simplifying and manipulating propositions.
- Truth tables help establish conditions under which propositions are equivalent.
1.4 Predicates and Quantifiers
- Predicates allow the expression of statements involving variables.
- Quantifiers (∀ for "for all," ∃ for "there exists") provide a way to generalize statements over a domain.
1.5 Nested Quantifiers
- Nested quantifiers allow for more complex logical structures, reflecting relationships where one quantifier may be contained within another.
1.6 Rules of Inference
- These are tools for argumentation that establish valid conclusions based on premises.
- Examples include Modus Ponens and Modus Tollens, which form the basis of many logical arguments.
1.7 Introduction to Proofs
- A proof is a formal way to demonstrate the truth of a statement, relying on axioms and previously established results.
- Different methods include direct proofs, proofs by contraposition, and proofs by contradiction.
1.8 Proof Methods and Strategy
- A systematic approach to constructing proofs is essential, involving clear definitions and considerations of all cases.
- The use of existing proofs can provide insights and strategies for proving new theorems.
Summary of Main Concepts:
- Logic serves as the foundation for mathematical reasoning and various applications.
- Propositions, predicates, and quantifiers form the basis of logical expressions.
- Understanding and applying logical laws and inference rules is key to constructing valid arguments and proofs.
- Various methods for proving theorems include direct methods, contraposition, contradiction, and case analysis.
- Counterexamples are vital for testing the truth of conjectures and statements in logic.
This structured overview encapsulates the themes and components of Chapter 1, reflecting the importance of logic in mathematics and computer science.
| Section | Content |
--- |
| Chapter 1 | The Foundations: Logic and Proofs |
| 1.1 | Propositional Logic: Foundation of logic and mathematical reasoning; includes propositions, logical operators, and truth tables. |
| 1.1.1 | Introduction to Propositions: Propositions are foundational building blocks with true (T) and false (F) values. |
| 1.1.2 | Compound Propositions: Created using logical operators (¬, ∧, ∨); truth tables illustrate truth values. |
| 1.1.3 | Conditional and Biconditional Statements: Conditional statements (→) show dependency; biconditional (↔) indicates mutual truth. |
| 1.2 | Applications of Propositional Logic: Important in software design, AI, and creating clear specifications. |
| 1.3 | Propositional Equivalences: Simplification and manipulation of propositions using logical equivalences and truth tables. |
| 1.4 | Predicates and Quantifiers: Predicates express variable statements; quantifiers (∀, ∃) generalize statements. |
| 1.5 | Nested Quantifiers: Enable complex logical structures with relationships between quantifiers. |
| 1.6 | Rules of Inference: Tools for valid conclusions based on premises, e.g., Modus Ponens, Modus Tollens. |
| 1.7 | Introduction to Proofs: Formal demonstrations of truth using axioms and methods like direct proof, contraposition, contradiction. |
| 1.8 | Proof Methods and Strategy: Systematic approaches for proofs, involving definitions and consideration of all cases. |
| Summary of Main Concepts | Logic is foundational for reasoning; propositions, predicates, and quantifiers are critical; understanding logical laws is essential; various proof methods exist; counterexamples are key for testing conjectures. |
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Key Point : Understanding propositions and logical operators is essential for effective problem-solving.
Example : Imagine you're debating whether to stay in or go out based on the weather. You might form a proposition like, 'If it rains, then I will stay home.' Understanding this conditional statement allows you to navigate your plans logically: if you check the forecast and it predicts rain, premises lead you to decide to stay in. This illustrates how propositional logic forms the basis for clear reasoning, helping you make informed decisions in everyday situations.
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Key Point : The foundational role of logic in mathematics and computer science is emphasized, but its interpretation may vary.
Critical Interpretation : While Rosen highlights propositional logic as critical for clear reasoning, it's important to recognize that the field of logic is expansive and interpretations differing among scholars can lead to contrasting applications and understandings of logic. Other sources, such as "Logic: A Very Short Introduction" by Graham Priest, also discuss the philosophical perspectives that challenge Rosen's viewpoint, suggesting there are alternative frameworks, or even critiques regarding the limits of classical logic, such as paraconsistent or intuitionistic logic frameworks, that offer valuable insights into how different logics can lead to different conclusions.
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Chapter 2 | 2 Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
Chapter Summary: Basic Structures: Sets, Functions, Sequences, Sums, and Matrices
1. Introduction to Discrete Mathematics
Discrete mathematics revolves around discrete structures that represent collections of objects, where sets play a critical role. The chapter introduces foundational concepts including sets, functions, sequences, and matrices, which are essential for various applications in mathematics and computer science.
2. Sets
- Definition: A set is an unordered collection of distinct elements.
- Notation: Elements are denoted with lowercase letters, and sets with uppercase letters.
- Methods of Description: Sets can be described using roster or set builder notation.
- Operations: Key operations on sets include union, intersection, difference, and Cartesian product.
- Venn Diagrams: These visual tools help illustrate relationships between sets.
- Subsets: A subset includes elements of another set, while proper subsets exclude the set itself.
- Cardinality: The size of a set is referred to as its cardinality, with concepts of finite and infinite sets introduced.
3. Functions
- Definition: A function assigns each element of one set (domain) to exactly one element of another set (codomain).
- Types of Functions: Includes one-to-one (injective), onto (surjective), and bijective functions.
- Inverse Functions: For bijections, inverses can be defined, reversing the mapping of the original function.
- Function Composition: Composition of functions involves applying one function to the result of another.
4. Sequences and Summations
- Sequences: Defined as ordered lists of elements where the index is typically drawn from integers.
- Types: Sequences can be arithmetic or geometric progressions.
- Recurrence Relations: These express terms in a sequence based on previous terms.
- Summation Notation: Utilizes the sigma symbol (∑) to represent the sum of sequence terms, with key properties and formulas presented.
5. Cardinality of Sets
- Countable vs. Uncountable: Countable sets can be listed in a sequence, while uncountable sets cannot.
- Examples: The set of positive integers is countable, while the set of real numbers is uncountable, demonstrated through Cantor's diagonalization argument.
6. Matrices
- Matrix Definition: A matrix is a rectangular array of numbers defined by its rows and columns.
- Matrix Operations: Involves addition, multiplication, transposes, and properties of identity matrices.
- Zero-One Matrices: Used in Boolean algebra to represent discrete structures, leading to operations such as the join and meet.
7. Additional Concepts
- The chapter concludes with applications of the discussed concepts in various fields and presents proofs for important results, such as the Schröder-Bernstein theorem, which relates to cardinalities.
Through established definitions, examples, and exercises, this chapter lays a foundation for further study in discrete mathematics, providing the essential tools for analyzing discrete structures and their interactions.
| Section | Summary |
--- |
| 1. Introduction to Discrete Mathematics | Focuses on discrete structures like sets, functions, sequences, and matrices, essential for mathematics and computer science. |
| 2. Sets | Definition: Unordered collection of distinct elements. Notation: Lowercase for elements, uppercase for sets. Methods of Description: Roster and set builder notation. Operations: Union, intersection, difference, Cartesian product. Venn Diagrams: Visualize relationships between sets. Subsets: Include elements of another set; proper subsets exclude the set itself. Cardinality: Size of a set; distinguishes between finite and infinite sets. |
| 3. Functions | Definition: Assigns each element of one set to one element of another. Types of Functions: Includes injective, surjective, and bijective. Inverse Functions: Defined for bijections, reversing original mapping. Function Composition: Applying one function to the result of another. |
| 4. Sequences and Summations | Sequences: Ordered lists typically drawn from integers. Types: Arithmetic and geometric progressions. Recurrence Relations: Define terms based on previous terms. Summation Notation: Uses sigma (∑) to represent sums, with properties and formulas. |
| 5. Cardinality of Sets | Countable vs. Uncountable: Countable can be listed; uncountable cannot. Examples: Positive integers are countable; real numbers are uncountable (Cantor's diagonalization). |
| 6. Matrices | Matrix Definition: Rectangular array of numbers defined by rows and columns. Matrix Operations: Addition, multiplication, transposes, identity matrices. Zero-One Matrices: Represent discrete structures in Boolean algebra. |
| 7. Additional Concepts | Applications in various fields, including proofs like the Schröder-Bernstein theorem relating to cardinalities. |
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Chapter 3 | 3 Algorithms
3 Chapter Algorithms
3.1 Algorithms
Algorithms are defined as finite sequences of precise instructions for performing computations or solving problems. The chapter explores various algorithms, including searching and sorting algorithms, emphasizing brute-force and greedy algorithms. The growth of functions and computational complexity, measured through big-O, big-Omega, and big-Theta notation, are discussed to understand the efficiency of algorithms.
3.1.1 Introduction
Algorithms can be initiated through modeling discrete problems using mathematical structures. A typical algorithm example is finding the maximum integer in a sequence.
3.1.2 Searching Algorithms
Searching algorithms locate an element in a list. The linear search method sequentially looks for an element, while the binary search method divides the list into halves to find an element. The complexities are analyzed, pointing out that linear search is O(n) and binary search is O(log n).
3.1.3 Sorting
Sorting algorithms organize list elements. Key sorting strategies include bubble sort and insertion sort, both of which have O(n^2) complexity.
3.1.4 String Matching
String matching algorithms, crucial in fields such as bioinformatics, examine occurrences of one string within another. The naive string matcher is highlighted as a foundational method.
3.1.5 Greedy Algorithms
Greedy algorithms build solutions step-by-step, making the best choice at each stage. The cashier's algorithm for making change serves as an illustration.
3.1.6 The Halting Problem
The halting problem is presented, demonstrating that there is no algorithmic way to determine if a program halts for a given input.
3.2 The Growth of Functions
This section introduces big-O notation to represent the growth of functions, used to analyze algorithm efficiency without considering constant factors.
3.2.1 Big-O Notation
Big-O notation simplifies the analysis of algorithms by estimating the growth of functions in a way that abstracts away constant factors.
3.2.3 Big-O, Omega, and Theta Notation
Big-Omega provides a lower bound for functions, while big-Theta combines both upper and lower bounds. Together, they help understand algorithm efficiency.
3.3 Complexity of Algorithms
This section discusses time complexity, typically measured through the number of operations an algorithm performs.
3.3.1 Time Complexity
Time complexity can be detailed in terms of worst-case and average-case scenarios. Algorithms can vary widely in efficiency, often analyzed using big-O notation.
3.3.4 Algorithmic Paradigms
Various algorithmic paradigms, such as brute force, are introduced, setting the stage for more complex approaches like dynamic programming and divide-and-conquer strategies.
3.3.5 Understanding the Complexity of Algorithms
Practical considerations regarding algorithm efficiency relate to technological improvements in computing resources, impacting both tractable and intractable problems.
In summary, Chapter 3 of "Discrete Mathematics and Its Applications" delves deeply into algorithm theory, complexity analysis, and practical implications in computing, equipping readers with a fundamental understanding of algorithmic performance. Understanding discrete probability is essential for analyzing algorithms that rely on stochastic processes. This concept applies when evaluating the expected performance of algorithms, especially in cases of randomized algorithms. By integrating discrete probability with algorithm analysis, one can better predict outcomes and enhance the efficiency of computing solutions.
| Section | Content |
--- |
| 3 Chapter Algorithms | Explores algorithms, including searching and sorting, and discusses computational complexity through big-O, big-Omega, and big-Theta notation. |
| 3.1 Algorithms | Algorithms are finite sequences of precise instructions for performing computations or solving problems. |
| 3.1.1 Introduction | Initiation of algorithms through modeling discrete problems, e.g., finding the maximum integer in a sequence. |
| 3.1.2 Searching Algorithms | Methods to locate an element in a list. Linear search (O(n)) vs. binary search (O(log n)). |
| 3.1.3 Sorting | Organizing list elements using algorithms like bubble sort and insertion sort, both with O(n^2) complexity. |
| 3.1.4 String Matching | Examines occurrences of one string within another, highlighting the naive string matcher. |
| 3.1.5 Greedy Algorithms | Solutions built step-by-step with the best choice at each stage, exemplified by the cashier's algorithm. |
| 3.1.6 The Halting Problem | Demonstrates that no algorithm can determine if a program halts for given input. |
| 3.2 The Growth of Functions | Introduces big-O notation to analyze algorithm efficiency without constant factors. |
| 3.2.1 Big-O Notation | Simplifies analysis of algorithms by estimating function growth, abstracting constant factors. |
| 3.2.3 Big-O, Omega, and Theta Notation | Big-Omega provides lower bounds; big-Theta combines upper and lower bounds for efficiency understanding. |
| 3.3 Complexity of Algorithms | Discusses time complexity, analyzing the number of operations algorithms perform. |
| 3.3.1 Time Complexity | Explains worst-case and average-case scenarios for time complexity analysis using big-O notation. |
| 3.3.4 Algorithmic Paradigms | Introduces various paradigms like brute force, dynamic programming, and divide-and-conquer. |
| 3.3.5 Understanding the Complexity of Algorithms | Practical implications of algorithm efficiency relate to technological progress affecting problem tractability. |
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Key Point : Understanding Algorithm Efficiency
Example : Imagine you're organizing your bookshelf. You could either search each book one-by-one from left to right, which takes longer as your collection grows (linear search, O(n)), or you could sort them first and then find a specific title quickly by dividing the remaining possibilities in half (binary search, O(log n)). This illustrates how knowing the efficiency of different algorithms can save you time, especially as your list expands.
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Chapter 4 | 4 Number Theory and Cryptography
Chapter 4: Number Theory and Cryptography
Overview
This chapter introduces number theory, emphasizing integer properties significant for computer science. It covers divisibility, modular arithmetic, integer representations, primes, greatest common divisors (gcd), linear congruences, and applications in cryptography.
4.1 Divisibility and Modular Arithmetic
Divisibility: An integer ( a ) divides ( b ) if there exists an integer ( c ) such that ( b = ac ).
Modular Arithmetic: Focuses on remainders when integers are divided by a modulus. Key operations include addition, multiplication, and the modulus function.
Division Algorithm: For integers ( a ) and a positive integer ( d ), there are unique integers ( q ) and ( r ) such that ( a = dq + r ), with ( 0 \leq r < d ).
4.2 Integer Representations and Algorithms
Base Representations: Integers can be expressed in any base ( b > 1 ), including binary, octal, and hexadecimal.
Algorithms for Arithmetic Operations: Introduces algorithms for addition, multiplication, and modular exponentiation, crucial for computer arithmetic.
4.3 Primes and Greatest Common Divisors
Prime Definitions: A prime number has exactly two distinct positive divisors: 1 and itself. Theorem: Every integer greater than 1 can be uniquely factored into prime numbers (fundamental theorem of arithmetic).
Trial Division: Method for determining primality through divisors up to the square root of the number.
Greatest Common Divisor: Defined, with methods for calculation including the Euclidean algorithm and its properties, like Bézout's Theorem, which provides a linear combination representation for gcd.
4.4 Solving Congruences
Linear Congruences: Solutions can be found using modular inverses.
Chinese Remainder Theorem: Provides conditions under which a system of linear congruences has a unique solution.
4.5 Applications of Congruences
Hashing Functions: Used in assigning memory locations and for error detection in data transmission.
Pseudorandom Number Generation: Methods for generating integers that mimic randomness are crucial in computer science.
Check Digits: Techniques for detecting errors in identification numbers through congruences.
4.6 Cryptography
Classical Cryptography: Discusses methods like Caesar and affine ciphers. Emphasizes encryption and decryption processes.
Public Key Cryptography: Introduces the RSA cryptosystem, allowing secure communication via public and private keys.
Protocols: Describes key exchange methods like the Diffie-Hellman protocol and digital signatures for message authenticity.
Key Concepts and Definitions:
- Modular Arithmetic (a mod b): Remainder of division.
- GCD and LCM: Fundamental properties and relationships.
- Pseudoprime: A composite number that appears prime under certain tests.
- Primitive Roots and Discrete Logarithms: Key concepts in number theory related to modular arithmetic.
- RSA Cryptosystem: Based on the difficulty of factoring large primes, crucial for secure communication.
Exercises and Applications: Engage with problems related to prime factorization, solving congruences, and cryptographic techniques.
This summary captures the essential topics of Chapter 4, laying a foundation for both theoretical aspects and practical applications related to number theory and cryptography.
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Key Point : Understanding Modular Arithmetic is Fundamental to Cryptography
Example : Imagine you're sending a secret message to a friend. Using modular arithmetic, you encode your message into numbers, where the operations wrap around upon reaching a certain number, say 26 for the alphabet. By assigning numbers to letters (A=0, B=1, ..., Z=25), adding a key number allows the sender to transform the message into a seemingly random string of numbers. This method not only secures communication but also ensures that only those with the key can decode it, illustrating the importance of modular arithmetic in modern cryptography.
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Key Point : Importance of Cryptography in Computer Science
Critical Interpretation : While the chapter emphasizes the critical role of cryptography in ensuring data security through precise mathematical principles, it is essential to recognize that the author's view may overly simplify the complexities involved in real-world applications. Cybersecurity risks continue to evolve, and reliance on foundational cryptographic techniques can introduce vulnerabilities. Scholars like Bruce Schneier argue in works such as 'Secrets and Lies: Digital Security in a Networked World' that cryptographic measures are often only as strong as their implementation, highlighting the necessity of continual adaptation beyond theoretical frameworks. Readers are encouraged to explore alternative perspectives on cryptography and security beyond those presented here.
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Chapter 5 | 5 Induction and Recursion
Chapter 5: Induction and Recursion
5.1 Mathematical Induction
Mathematical induction is a proof technique for establishing that a property holds true for all positive integers. The process consists of two steps: the basis step, which involves proving the property for the initial integer (usually 1), and the inductive step, where one shows that if the property holds for an integer ( k ), it must also hold for ( k + 1 ).
5.1.1 Introduction
The essence of mathematical induction is to establish that a statement about natural numbers is true beginning from a base case and demonstrating that this truth carries through the integers. Examples include properties of factorials, divisibility, and combinatorial identities.
5.1.2 Mathematical Induction
A proposition ( P(n) ) is proven for all positive integers ( n ) by showing:
- ( P(1) ) is true (basis step).
- For all ( k ), if ( P(k) ) is true, then ( P(k + 1) ) is true (inductive step).
5.1.3 Why Mathematical Induction is Valid
Induction is valid based on the well-ordering property: every non-empty subset of positive integers has a least element. If a property does not hold for some integer, it must hold for the smallest integer in the subset, leading to a contradiction.
5.1.4 Choosing the Correct Basis Step
The basis step may vary; one can establish properties for integers starting from any integer ( b ), provided the induction appropriately reflects the initial condition.
5.1.5 Guidelines for Proofs by Mathematical Induction
To construct proofs effectively, express statements clearly, verify the basis step, and systematically use the inductive hypothesis in the inductive step.
5.1.6 The Good and the Bad of Mathematical Induction
Induction can affirm conjectures but is not a method for discovery. Preferred proofs are those illuminating why theorems are true, as induction proofs tend to lack insight.
5.1.7 Examples of Proofs by Mathematical Induction
Various examples illustrate the breadth of induction, covering summation formulas, inequalities, divisibility, and behavioral claims about algorithms and computer programs.
5.2 Strong Induction and Well-Ordering
Strong induction generalizes mathematical induction, allowing one to assume the property holds for all integers ( j ) up to ( k ) in proving it for ( k + 1 ). It is especially helpful in cases where direct induction may fail.
5.2.1 Introduction
Using strong induction simplifies assertions about structured sets, making use of all previous results to prove the next.
5.2.2 Strong Induction
To prove ( P(n) ) using strong induction, establish the basis for initial cases, then show the result holds for ( k + 1 ) under the assumption that it holds for all smaller integers.
5.2.3 Examples of Proofs Using Strong Induction
- Example: Show every integer can be expressed as the product of primes.
- Example: Prove a winning strategy exists in variants of games through induction.
- Example: Show properties of postage can be achieved with specific coin denominations.
5.2.4 Using the Well-Ordering Property
The well-ordering property forms the foundation for proving theorems not easily established via simple induction, such as in establishing the division algorithm.
5.3 Recursive Definitions and Structural Induction
Recursive definitions allow for objects or sequences to be defined through smaller versions of themselves, which opens the door to proofs via structural induction.
5.3.1 Introduction
Recursive definitions apply rules for creating complex structures out of simpler ones, useful in areas ranging from mathematics to computer science.
5.3.2 Recursively Defined Functions
Functions can be defined recursively, and their values computed through established initial cases and subsequent rules.
5.3.3 Recursively Defined Sets and Structures
Sets can be constructed through recursive rules; examples include well-formed formulas, strings, and trees.
5.4 Recursive Algorithms
Algorithms can exploit recursion for function evaluation and sorting, as shown in merge sort or calculating Fibonacci numbers. The complexity of recursive solutions often contrasts sharply with their iterative counterparts.
5.5 Program Correctness
To ensure a program functions correctly, proofs of partial correctness and termination must be established using appropriate assertions and rules of inference. Loop invariants are crucial in this analysis.
In summary, this chapter provides a comprehensive overview of induction, recursion, and their applications in mathematical proofs and algorithm verification, highlighting their importance in computer science and discrete mathematics.
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Chapter 6 | 6 Counting
6 Counting
6.1 The Basics of Counting
- Introduction: Counting is a fundamental part of combinatorics and applied in various fields including computer science and biology.
- Basic Counting Principles: Introduces product and sum rules. The product rule states if a procedure has two tasks, the total ways to complete the procedure are the product of the ways to each task. The sum rule states if a task can be performed in one of two ways, the total is the sum of the ways.
- Examples of Counting: Practical examples include assigning offices, labeling chairs, and counting telephone numbers.
6.2 The Pigeonhole Principle
- The pigeonhole principle states if more objects are placed into fewer boxes, at least one box must contain more than one object.
- Generalized Pigeonhole Principle: If N objects are put into k boxes, at least one box contains at least ⌈N/k⌉ objects.
- Applications: Proved through examples such as birthdays in a group or students sharing scores.
6.3 Permutations and Combinations
- Introduction: Counting problems often require arranging or selecting distinct elements.
- Permutations: Calculation of r-permutations (order matters) is shown through examples with sets. The formula is given as P(n, r) = n! / (n - r)!.
- Combinations: Focuses on choosing subsets where order doesn’t matter, denoted as C(n, r) and calculated using C(n, r) = n! / (r! (n - r)!).
- Example Applications: Events like seating arrangements, committee formation, and various practical scenarios.
6.4 Binomial Coefficients and Identities
- The Binomial Theorem: Describes the expansion of (x + y)ⁿ using binomial coefficients.
- Properties and theorems include Pascal's identity, Vandermonde’s identity, and combinatorial proofs for identities involving binomial coefficients.
6.5 Generalized Permutations and Combinations
- Permutations with repetition: P(n, r) = n^r.
- Combinations with repetition: Described using the formula C(n + r - 1, r).
- Distribution of objects into distinguishable versus indistinguishable boxes is examined, with specific formulas and examples for both.
6.6 Generating Permutations and Combinations
- Generating Methods: Algorithms for generating permutations and combinations in lexicographic order.
- Next Permutation and Combination: Detailed procedures for obtaining the next permutation or combination following a defined order provided with algorithmic descriptions.
Key Terms and Results
- Combinatorics, Permutations, Combinations, Binomial Coefficients: Defined along with relevant formulae and identities central to counting problems.
Exercises and Applications: Includes various exercises that illustrate the application of principles discussed, such as counting methods, generating permutations, and solving related real-world problems using combinatorial reasoning. Discrete mathematics plays a crucial role in understanding the principles of counting. It provides the foundational concepts needed for topics such as combinatorics and algorithms, essential for computer science and other technical fields. Mastery of discrete mathematics not only enhances problem-solving skills but also prepares students for advanced studies in mathematics and related disciplines.
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Chapter 7 | 7 Discrete Probability
7 Discrete Probability
7.1 An Introduction to Discrete Probability
7.1.1 Introduction
Probability theory has roots in combinatorics and was developed over 300 years ago, initially to analyze gambling games. It is widely applied in genetics and computer science and is useful for solving problems through probabilistic algorithms.
7.1.2 Finite Probability
An experiment is a procedure yielding outcomes from a sample space. The probability of an event is defined using Laplace's definition for finite, equally likely outcomes. Probability values range from 0 to 1.
7.1.3 Probabilities of Complements and Unions of Events
The probability of complements and unions of events can be calculated using Theorems that establish relationships between these probabilities.
7.1.4 Probabilistic Reasoning
Probabilistic reasoning helps determine which of two competing events (such as winning a game) is more likely, exemplified by the Monty Hall problem.
7.2 Probability Theory
7.2.1 Introduction
Conditions under which probabilities are defined reflect scenarios where outcomes are not necessarily equally likely, and involves concepts such as conditional probability.
7.2.2 Assigning Probabilities
Probabilities must assign a nonnegative value between 0 and 1 for each outcome, summing to 1 across the sample space.
7.2.3 Probabilities of Complements and Unions of Events
Similar principles apply for different definitions of events and established methods remain consistent for calculating these probabilities.
7.2.4 Conditional Probability
Conditional probability measures the likelihood of an event occurring, given that another event has occurred.
7.2.5 Independence
Events are independent if the occurrence of one does not affect the probability of the other. This leads to useful conclusions regarding the joint probabilities.
7.2.6 Bernoulli Trials and the Binomial Distribution
Bernoulli trials have two outcomes, and the probability of k successes in n trials follows the binomial distribution.
7.2.7 Random Variables
Random variables assign values to outcomes, and follow rules that govern expectations and variances.
7.2.8 The Birthday Problem
The Birthday problem explores the probability that at least two individuals in a group share the same birthday.
7.2.9 Monte Carlo Algorithms
Probabilistic algorithms, like Monte Carlo methods, rely on random choices to solve problems, with applications in decision-making under uncertainty.
7.2.10 The Probabilistic Method
The probabilistic method is a nonconstructive proof strategy used to show that certain objects exist within a set.
7.3 Bayes’ Theorem
7.3.1 Introduction
Bayes' Theorem allows for the computation of conditional probabilities and has applications in various fields including medicine and spam filtering.
7.3.2 Bayes’ Theorem
Bayes' Theorem integrates prior knowledge and evidence to calculate updated probabilities for event occurrences.
7.3.3 Bayesian Spam Filters
Spam filters utilize Bayes' Theorem to classify emails based on the presence of certain words, adjusting probabilities accordingly.
7.4 Expected Value and Variance
7.4.1 Introduction
Expected value facilitates the assessment of central tendencies within random variables, while variance reflects the spread of values around this mean.
7.4.2 Expected Values
The concept and calculation of expected values leverage prior examples and results highlighting linearity.
7.4.3 Linearity of Expectations
The linear nature of expected values simplifies calculations for sums of random variables, greatly aiding computational complexity considerations.
7.4.4 Average-Case Computational Complexity
Average-case complexity employs expected values to analyze algorithms under stochastic input conditions.
7.4.5 The Geometric Distribution
The geometric distribution and its expected values, providing insights into occurrences over trials until a designated outcome arises, are detailed.
7.4.6 Independent Random Variables
Independence among random variables ensures that joint outcomes can be simply calculated using their individual distributions.
7.4.7 Variance
Variance and standard deviation measure the extent of variability in random variables, with relationships derived from expected values central to analysis.
7.4.8 Chebyshev’s Inequality
Chebyshev's Inequality provides bounds on the probability that a random variable deviates from its expected value by a specified margin, highlighting variability.
This summary outlines the critical concepts in discrete probability, including the foundational theories, theorems, applications like spam filtering, and statistical measures of mean and variability.
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Key Point : The Role of Probability in Decision-Making
Critical Interpretation : Probability theory serves as a fundamental framework for analyzing uncertainty in diverse fields, which sparks discussion about its applicability and potential biases. While Kenneth H. Rosen presents a structured approach to probability and its foundational role in algorithms and decision-making, it is essential for readers to critically evaluate how probabilistic models may oversimplify complex real-world scenarios. For instance, relying heavily on probabilistic outcomes can sometimes mislead decision-makers if underlying assumptions are flawed. Doubts about the effectiveness of probabilistic reasoning in various contexts can be supported by scholarly critiques, such as those found in 'Against the Gods: The Remarkable Story of Risk' by Peter L. Bernstein, which explores the limitations of probability in assessing risk and decision-making.
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Chapter 8 | 8 Advanced Counting Techniques
Chapter 8: Advanced Counting Techniques
8.1 Applications of Recurrence Relations
- Many counting problems are complex and require recurrence relations.
- Example: Number of bit strings of length n with no consecutive zeros follows the recurrence relation ( a_{n+1} = a_n + a_{n-1} ).
- Discusses dynamic programming and divide-and-conquer paradigms in algorithm analysis.
- Illustrates counting problems modeled by recurrence relations (e.g., population growth, Tower of Hanoi).
8.2 Solving Linear Recurrence Relations
- Defines recursive definitions and recurrence relations.
- Explains linear homogeneous recurrence relations with constant coefficients.
- Introduces techniques for solving such relations systematically.
- Applies theorems for different cases of recurrence relations, including examples such as Fibonacci numbers and others involving distinct roots.
8.3 Divide-and-Conquer Algorithms and Recurrence Relations
- Explains the divide-and-conquer paradigm, reducing problems into smaller subproblems.
- Sets up recurrence relations based on algorithm characteristics (e.g., binary search, merge sort).
- Provides the Master Theorem for analyzing the complexity of divide-and-conquer recurrences.
8.4 Generating Functions
- Introduces generating functions as a way to encode sequences and solve counting problems.
- Discusses properties of power series and operations on generating functions.
- Provides examples illustrating how to derive series using the extended binomial theorem and calculate coefficients.
8.5 Inclusion–Exclusion
- Explains the principle of inclusion-exclusion to count elements in unions of sets.
- Derives formulas for the number of elements in unions of two, three, and n sets.
- Employs examples to illustrate practical applications in counting problems (e.g., student enrollments, integers with certain properties).
8.6 Applications of Inclusion–Exclusion
- Demonstrates the use of inclusion-exclusion in estimating the number of primes and onto functions.
- Discusses specific combinatorial problems (e.g., seating arrangements, derangements).
- Provides derivations and examples to show the effectiveness of the principle.
Key Terms and Results
- Recurrence Relation: A formula that expresses terms of a sequence as functions of previous terms.
- Dynamic Programming: An algorithmic strategy for solving problems by breaking them down into simpler overlapping subproblems.
- Derangement: A permutation of objects where no object appears in its original position.
- Principle of Inclusion-Exclusion: A method to calculate the size of unions of multiple sets by considering intersecting sizes.
These advanced counting techniques serve as powerful tools for solving complex combinatorial problems across various applications in mathematics and computer science.
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Chapter 9 | 9 Relations
9 C H A P T E R Relations
9.1 Relations and Their Properties
- Relationships among elements of sets are studied through binary relations, subsets of Cartesian products.
- Properties of binary relations include reflexivity, symmetry, antisymmetry, and transitivity.
- n-ary relations represent relationships among elements of more than two sets.
9.2 n-ary Relations and Their Applications
- n-ary relations are used in relational databases to manage complex data structures.
- Key operations include selection, projection, and join.
- SQL facilitates querying n-ary relations, with selection filtering records and projection reducing the dataset.
9.3 Representing Relations
- Relations can be represented through matrices or directed graphs.
- Zero-one matrices denote the presence of relationships, while digraphs visually depict the connections between elements.
- Specific properties (reflexivity, symmetry, antisymmetry) can be identified through these representations.
9.4 Closures of Relations
- The reflexive closure adds pairs (a, a) to ensure reflexivity.
- The symmetric closure involves including reverse pairs (b, a).
- The transitive closure includes all pairs that can be connected through paths in a directed graph.
- Warshall's algorithm efficiently computes the transitive closure.
9.5 Equivalence Relations
- An equivalence relation is reflexive, symmetric, and transitive, thus partitioning a set into equivalence classes.
- Examples include congruence relations and shared characteristics like graduating from the same high school.
- The equivalence classes form a partition of the set, where each element belongs to exactly one class.
9.6 Partial Orderings
- A partial ordering is reflexive, antisymmetric, and transitive, exemplified by divisibility and inclusion relations.
- Compatible total orders can be derived from partial orders, known as topological sorting.
- Hasse diagrams illustrate the structure of posets, allowing for easy identification of maximal and minimal elements.
Key Terms and Results
- Binary Relation: A subset of A × B.
- Equivalence Relation: A relation that is reflexive, symmetric, and transitive.
- Partial Order: A relation that is reflexive, antisymmetric, and transitive.
- Lattice: A poset where every two elements have a least upper bound and a greatest lower bound.
Exercises
- Various exercises are provided to deepen understanding of concepts like reflexive closures, symmetric closures, equivalence relations, and the construction and analysis of posets.
This content format effectively organizes the key concepts and definitions from Chapter 9, as well as providing structured insights into exercises that reinforce learning objectives within the study of relations in discrete mathematics.
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Chapter 10 | 10 Graphs
Chapter 10: Graphs
10.1 Graphs and Graph Models
Graphs are discrete structures consisting of vertices connected by edges, which may be directed or undirected. Graphs are widely used as models in various disciplines, including ecology, organizational structures, tournament outcomes, and social networks. A basic definition of a graph G = (V, E) includes a set of vertices V and edges E, where edges may connect two different vertices or may consist of loops. Different types of graphs such as simple graphs, multigraphs, pseudographs, and directed graphs arise based on the characteristics of the edges and connections.
10.2 Graph Terminology and Special Types of Graphs
Basic terminology includes concepts like adjacency, degree of vertices, and isolated/pseudo vertices. We introduce types of graphs such as complete graphs (K_n), cycles (C_n), wheels (W_n), and bipartite graphs (K_m,n). Different properties of graphs, like connectivity and the presence of cut vertices and edges, influence graph characteristics and behavior.
10.3 Representing Graphs and Graph Isomorphism
Graph representation can be done using edge lists, adjacency lists, and adjacency matrices. Isomorphic graphs have the same structure, meaning a one-to-one correspondence can be made between their vertices preserving adjacency. Isomorphism can be proven through invariants like vertex count, edge count, and degree sequence.
10.4 Connectivity
Connected graphs allow a path between any two vertices, whereas disconnected graphs do not. We introduce concepts like connected components, cut vertices, edge connectivity, and vertex connectivity to characterize graph robustness. Algorithms help to analyze connectivity and find properties like cut edges or vertices within a connected graph.
10.5 Euler and Hamilton Paths
Euler circuits allow each edge of a graph to be traversed once, and conditions for their existence depend on vertex degree (all vertices must have even degree). The Hamilton circuit visits every vertex exactly once, and no simple necessary conditions currently exist. However, sufficient conditions like Dirac's and Ore's theorems help determine the existence of Hamilton circuits.
10.6 Shortest-Path Problems
Weighted graphs model real-world problems like transportation and communication costs. Dijkstra's algorithm efficiently finds the shortest path between two vertices in a connected graph where edge weights are positive. The traveling salesperson problem seeks the shortest Hamilton circuit visiting all vertices, known to be NP-complete.
10.7 Planar Graphs
A planar graph can be drawn in the plane without edge crossings. K_5 and K_3,3 are examples of nonplanar graphs. Kuratowski’s theorem states that a graph is nonplanar if it contains a subgraph homeomorphic to K_5 or K_3,3. Euler's formula relates vertices, edges, and regions in planar graphs.
10.8 Graph Coloring
Graph coloring involves assigning colors to vertices so that adjacent vertices receive different colors. The chromatic number (𝜒(G)) indicates the minimum number of colors needed. The four color theorem states that any planar graph can be colored with no more than four colors. Applications include scheduling exams and optimizing circuit designs.
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Key Point : The Complexity of Graph Problems and Their Real-World Applications
Critical Interpretation : The chapter emphasizes the utility of graph theory in modeling real-world problems; however, the author's assertion that algorithms like Dijkstra's unequivocally provide optimal solutions may not account for the intricacies of some scenarios. Not all practical applications yield easily solvable problems, as exemplified by the traveling salesperson problem which remains NP-complete. Readers are encouraged to reflect on whether reliance on algorithmic solutions limits understanding of underlying complexities in real-world phenomena. Sources regarding algorithmic complexity can be found in 'Computers and Intractability' by Garey and Johnson.
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Chapter 11 | 11 Trees
11 Trees
11.1 Introduction to Trees
Trees are a type of connected graph without simple circuits. They have widespread applications in various fields, notably in computer science for efficient algorithms like searching and sorting. Trees can also model decision processes and enhance computational complexity studies.
11.1.1 Rooted Trees
Rooted trees are a specific type of tree where one vertex is designated as the root. Each edge in a rooted tree can be directed away from the root. Key terminology includes parent, child, sibling, ancestor, descendant, internal vertex, and leaf. Subtrees can be created from any vertex.
11.1.2 Trees as Models
Trees are useful in various applications from modeling chemical compounds to structuring organizations and file systems.
11.2 Applications of Trees
11.2.1 Introduction
We explore how trees can be utilized in data storage, decision processes, and character encoding.
11.2.2 Binary Search Trees
Binary search trees efficiently locate items in a sorted manner. They are constructed by recursively placing items based on defined ordering.
11.2.3 Decision Trees
Decision trees model problems where outcomes depend on decisions, allowing for systematic evaluations.
11.2.4 Prefix Codes
Prefix codes, important in data compression, ensure no encoded character prefix overlaps with another, maintaining unique representation.
11.2.5 Game Trees
Game trees analyze decision-making processes in games, involving strategies and outcomes for optimal play.
11.3 Tree Traversal
Traversal of trees is essential for accessing and listing vertices systematically. Common methods include preorder, inorder, and postorder traversals, each serving specific purposes in relation to tree structure.
11.4 Spanning Trees
A spanning tree connects all vertices in a graph with minimal edges. Depth-first search and breadth-first search are common methods to construct these trees, each with its algorithms.
11.5 Minimum Spanning Trees
Minimum spanning trees are crucial for optimizing connection costs in networks. Prim’s and Kruskal’s algorithms are the primary methods for finding them, employing greedy strategies to minimize edge weights in a spanning tree while ensuring connectivity.
Key Terms and Results
- Trees: Connected undirected graphs without circuits.
- Forests: Collections of trees.
- Rooted Trees: Trees with a designated root vertex.
- Spanning Trees: Connected subgraphs encompassing all vertices of a graph.
- Minimum Spanning Trees: Trees with the minimal sum of edge weights.
Important Algorithms
- Prim’s Algorithm: Iteratively adds the cheapest edge connected to the tree.
- Kruskal’s Algorithm: Adds the cheapest edge connecting different components without forming circuits.
This chapter emphasizes the versatile role trees play in algorithmic design, problem-solving, and optimization across disciplines in discrete mathematics.
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Chapter 12 | 12 Boolean Algebra
12 Boolean Algebra
12.1 Boolean Functions
- Introduction: Boolean algebra operates with the binary set {0, 1}, which is fundamental in electronic circuits where inputs and outputs are binary. Key operations are complementation, Boolean sum (OR), and Boolean product (AND).
- Boolean Expressions and Functions: Boolean variables represent values 0 or 1, and a Boolean function of degree n is a mapping from B^n to B. Boolean expressions recursively define valid expressions.
- Identities of Boolean Algebra: Various identities govern operations in Boolean algebra, including idempotent laws, domination laws, commutative laws, and distributive laws.
- Duality: Each identity has a corresponding dual obtained by interchanging operations and values of 0 and 1. This principle aids in deriving new identities.
- Abstract Definition: Boolean algebras can be abstractly defined based on properties that define their operations and include elements 0 and 1.
12.2 Representing Boolean Functions
- Sum-of-Products: Every Boolean function can be expressed as a sum of products of literals, known as the sum-of-products expansion or disjunctive normal form.
- Functional Completeness: Sets of operations (e.g., {AND, OR, NOT}) can fully represent Boolean functions. Reduction to fewer operators, such as NAND and NOR, is possible.
12.3 Logic Gates
- Introduction to Gates: Basic gates (inverters, OR, AND) perform Boolean operations in circuits. Multiple input configurations are possible for AND and OR gates.
- Combinations of Gates: Complex circuits can be constructed from combinations of simple gates, keeping track of input for shared connections.
- Examples of Circuits: Applications include designing voting systems and circuits for controlling fixtures with multiple switches.
12.4 Minimization of Circuits
- Introduction to Minimization: The goal is to design circuits with minimal gates and complexity, crucial for efficiency and reliability.
- Karnaugh Maps: Visual methods simplify Boolean expressions for functions with fewer than six variables. They help identify groups of minterms for combination.
- Quine–McCluskey Method: A systematic, algorithmic approach allows simplification for functions with any number of variables, providing a structured way to minimize expressions efficiently.
- Don’t Care Conditions: Certain input combinations can be ignored in function design, allowing flexibility in circuit construction.
- Practical Applications: Summarizes the importance of simplification in circuit design relating to reliability, cost, and functionality, with insights on historical context and modern methodologies.
This chapter provides foundational insights into Boolean algebra's principles, applications, and significance in digital circuit design.
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Chapter 13 | 13 Modeling Computation
Chapter 13: Modeling Computation
13.1 Languages and Grammars
- Introduction: Discussion of how languages are generated through grammars, which can help in understanding valid sentence structures in both natural and programming languages.
- Phrase-Structure Grammars: Definition of grammars, including vocabularies, terminals, nonterminals, and productions. This section introduces the formal structure of different types of grammars (type 0 to type 3) used to define languages.
13.2 Finite-State Machines with Output
- Introduction: Overview of finite-state machines (FSMs) as models for computation, capable of producing outputs based on input and state transitions.
- Formal Definition: Explanation of FSMs characterized by states, input alphabet, output alphabet, transition functions, and a starting state.
- Examples: Practical applications of FSMs in modeling various systems (e.g., vending machines) and introduction to Mealy and Moore machines.
13.3 Finite-State Machines with No Output
- Introduction: Different types of FSMs designed for language recognition, specifically finite-state automata (FSAs) that accept languages by reaching designated final states.
- Set of Strings: Definitions and concepts related to how FSAs recognize input strings based on state transitions and final states.
- Designing Automata: Techniques for constructing FSMs that recognize specific languages and exploring the properties of reachability and state equivalence.
13.4 Language Recognition
- Introduction: Explanation of regular sets recognized by FSMs, finite-state grammars, and the relationship between them.
- Kleene's Theorem: Establishes that regular sets are those recognized by FSMs, and details are provided on constructing respective automata from regular expressions and grammars.
- Regular Sets and Regular Grammars: Discussion on how every regular grammar generates a regular set and vice versa.
13.5 Turing Machines
- Introduction: Turing machines as a more powerful model of computation that expands upon FSMs by allowing for an infinite tape and more intricate computations, including the ability to perform complex algorithms.
- Formal Definition: Detailed description of Turing machines including their structure, operation, and how they recognize sets.
- Computing Functions: Overview of how Turing machines can model number-theoretic functions and other computation tasks.
Key Concepts:
- Finite-state machines can be classified into those with and without output, with various applications in computer science.
- Language recognition is fundamental for understanding how algorithms can be constructed and how various language types correspond to different computational models.
- Turing machines provide a robust framework for exploring computability and complexity in algorithms, revealing the limits of what can be computed effectively.
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4978 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9672860/ | Gaussian graphical models with applications to omics analyses - PMC
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As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health.
Learn more: PMC Disclaimer | PMC Copyright Notice
Stat Med
. Author manuscript; available in PMC: 2023 Nov 10.
Published in final edited form as: Stat Med. 2022 Sep 26;41(25):5150–5187. doi: 10.1002/sim.9546
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Gaussian graphical models with applications to omics analyses
Katherine H Shutta
Katherine H Shutta
1 Department of Biostatistics and Epidemiology, University of Massachusetts - Amherst, Amherst, Massachusetts, USA
2 Department of Biostatistics, Harvard T.H. Chan School of Public Health, Boston, Massachusetts, USA
3 Channing Division of Network Medicine, Department of Medicine, Brigham and Women’s Hospital and Harvard Medical School, Boston, Massachusetts, USA
Find articles by Katherine H Shutta
1,2,3, Roberta De Vito
Roberta De Vito
4 Department of Biostatistics and Data Science Initiative, Brown University, Providence, Rhode Island, USA
Find articles by Roberta De Vito
4, Denise M Scholtens
Denise M Scholtens
5 Division of Biostatistics, Department of Preventive Medicine, Northwestern University Feinberg School of Medicine, Chicago, Illinois, USA
Find articles by Denise M Scholtens
5, Raji Balasubramanian
Raji Balasubramanian
1 Department of Biostatistics and Epidemiology, University of Massachusetts - Amherst, Amherst, Massachusetts, USA
Find articles by Raji Balasubramanian
1
Author information
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Copyright and License information
1 Department of Biostatistics and Epidemiology, University of Massachusetts - Amherst, Amherst, Massachusetts, USA
2 Department of Biostatistics, Harvard T.H. Chan School of Public Health, Boston, Massachusetts, USA
3 Channing Division of Network Medicine, Department of Medicine, Brigham and Women’s Hospital and Harvard Medical School, Boston, Massachusetts, USA
4 Department of Biostatistics and Data Science Initiative, Brown University, Providence, Rhode Island, USA
5 Division of Biostatistics, Department of Preventive Medicine, Northwestern University Feinberg School of Medicine, Chicago, Illinois, USA
✉
Correspondence: Katherine H. Shutta, Department of Biostatistics, Harvard T.H. Chan School of Public Health, 677 Huntington Ave., Boston, MA 02115, USA. kshutta@hsph.harvard.edu
Issue date 2022 Nov 10.
PMC Copyright notice
PMCID: PMC9672860 NIHMSID: NIHMS1830332 PMID: 36161666
The publisher's version of this article is available at Stat Med
Abstract
Gaussian graphical models (GGMs) provide a framework for modeling conditional dependencies in multivariate data. In this tutorial, we provide an overview of GGM theory and a demonstration of various GGM tools in R. The mathematical foundations of GGMs are introduced with the goal of enabling the researcher to draw practical conclusions by interpreting model results. Background literature is presented, emphasizing methods recently developed for high-dimensional applications such as genomics, proteomics, or metabolomics. The application of these methods is illustrated using a publicly available dataset of gene expression profiles from 578 participants with ovarian cancer in The Cancer Genome Atlas. Stand-alone code for the demonstration is available as an RMarkdown file at
Keywords: Gaussian graphical models, graphical lasso, network medicine, omics, partial correlation networks
1 |. INTRODUCTION
With the onset of the omics revolution over the last two decades, methods for analyzing complex datasets with many variables have become widely necessary. Graphical modeling with biological networks is one popular way to analyze this type of data. Biological networks are frameworks in which nodes represent individual units such as genes, proteins, or metabolites and edges reflect a relationship between node pairs such as co-expression or physical interaction. Network medicine proposes that the state of a biological system is a manifestation of complex interdependencies between biological molecules, and that network-level disruptions of these interdependencies are hallmarks of complex disease.1 Estimation and analysis of biological networks is a foundational component of network medicine.
Network models have been successfully applied to study the organization and structure of several classes of biomolecules in the cell. Classic examples of biological networks in humans include protein-protein interaction networks,2 gene regulatory networks,3 and gene coexpression networks.4 Biological networks have been crucial in identifying disease susceptibility genes and in identifying potential therapeutic drug targets.5
Topological properties of networks such as connected components, sparsity, degree centrality, and degree distribution provide a perspective from which to search for signatures of a disease state or a complex biological process. Because biological networks can be estimated in an agnostic manner, this perspective can provide insights into biological functions that may not yet have been identified using traditional laboratory approaches. For example, module-finding in biological networks is a tool for de novo pathway discovery.6 In addition to identifying novel pathways, biological networks can provide reinforcement for, or rationale to challenge, previously discovered pathways in an organism.
Gaussian graphical model (GGM) estimation is one approach to estimating biological networks. In a GGM, edges represent pairwise conditional dependencies: two nodes in a GGM are connected with an edge if and only if they are conditionally dependent given the other nodes in the network.7 Modeling biological networks with GGMs enables a biologically meaningful interpretation of the network as a roadmap containing information about which molecules are directly associated within a given state.
The utility and popularity of GGMs have inspired a range of new methods and tutorial papers in recent years,8–10 yet there is still a need for practical instruction to bridge the gap between efforts of statisticians in developing new GGM technologies and efforts of epidemiologists and clinical research scientists in applying these technologies in practice, particularly for omics and other high-dimensional data applications. Here, we address this gap by providing an overview of GGMs and straightforward guidance for their use in health sciences applications. The goal of this tutorial is to prepare the reader to independently conduct GGM-based analyses and interpretation.
We begin by discussing the theory behind GGMs, focusing on intuitive interpretation of GGM edge weights (partial correlations). We compare GGMs to correlation networks, noting that correlation networks represent marginal associations between variables, and so may be quite dense and difficult to interpret; in contrast, GGMs represent conditional associations between variables, are likely to be more sparse, and have a straightforward interpretation in terms of conditional dependence given the state of the measured biological system. We discuss the recently derived pair-path subscore, which demonstrates that the correlation between two nodes in a GGM can be expressed as a linear combination of partial correlations (ie, edge weights) along the network paths connecting them.11
As in many omics analyses, data dimensionality demands attention in GGM estimation. We review a range of methodological advances that have enabled fast estimation of sparse GGMs through penalized likelihood methods. These regularized methods are especially relevant in high-dimensional settings, where the number of samples n is less than the number of variables p; such approaches are also useful for estimating sparse GGMs in low-dimensional settings (n ≥ p).12
We discuss two key areas for using GGMs in practice. The first is precision matrix estimation, which enables estimation of partial correlations between all variables in the network and a resultant weighted graph in which edge weights correspond to these partial correlations. The second is inference on the edge set of the graphical model to test the null hypothesis that an edge does not exist between two nodes (zero partial correlation). This method can be used to produce an unweighted graph in which the adjacency matrix is binary. We review a variety of approaches for GGM estimation and inference, illustrating their use on publicly available ovarian cancer gene expression data13 in R. To support individual exploration, an RMarkdown document containing the code demonstration is available on Github at We conclude with a discussion of the state of the field and areas for future work.
2 |. PRELIMINARIES
2.1 |. Gaussian graphical models
GGMs are a special case of the more general concept of an undirected graphical model. An undirected graphical model is a representation of a multivariate random vector as a graph 𝒢(V, E), where V is a set of nodes representing the p components of and E is a set of edges between these nodes, representing pairwise relationships. A useful representation of E is as a binary adjacency matrix A, where A ij = 1 if there is an edge between X i and X j and zero otherwise. An extension of this representation is a weighted adjacency matrix where A ij represents the edge’s weight between X i and X j. A zero entry in a weighted adjacency matrix indicates the absence of an edge, as in the binary case. A glossary of these and other terms related to graphical modeling is available in the Supplement (Appendix S1).
A conditional independence graph is one type of undirected graphical model. Let A, B, and C denote disjoint node sets in V. C is said to separate A and B when there is no path between A and B that does not pass through C. In a conditional independence graph, the absence of an edge between two nodes A and B corresponds to conditional independence of the random vectors A, B given the separating set C.7 Conditional independence may be conceptualized by differentiating between direct and indirect associations. A direct association between two variables cannot be removed by conditioning on any other set of variables, while an indirect association can be removed by conditioning on another set of variables. Conditional independence is a statement about the lack of direct association between the sets A and B: although there may be indirect association between A and B, it is fully explained by variables in C. An example of a conditional independence graph and its conditional independence properties are shown in Figure 1.
FIGURE 1.
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An example of an undirected graphical model on a multivariate random variable . For each of the sets A, B, and C described in (i) to (iv), the conditional independence property A⊥B|C holds. (i) A = X 1 , B = {X 4 , X 5}, and C = {X 2 , X 3} (ii) A = X 2 , B = {X 3 , X 5}, and C = {X 1 , X 4} (iii) A = {X 1 , X 2 , X 3}, B = X 5, and C = X 4 (iv) A = X 1 , B = X 5, and C = X 4. Note that these are not the only conditional independence properties implied by the graph: for example, it is also true that X 2 ⊥X 5|X 4
The GGM is a special case of an conditional independence graph. In a GGM, we assume follows the multivariate normal distribution:
(1)
where is a p-dimensional vector of means and Σ is the p × p covariance matrix of .
The edge weights of a GGM correspond to partial correlations between the variables in . The partial correlation between variables X i and X j is a measure of their conditional association, given the remaining elements of . Let X−i,−j denote the set of all variables in except for X i and X j. Using this notation, the partial correlation is defined as:
(2)
In the special case of the Gaussian distribution, conditional independence between X i and X j given the remaining variables X−i,−j is equivalent to zero partial correlation between the two variables. The edge set of a GGM is therefore defined by the set of all pairs (X i , X j) with nonzero partial correlation:
(3)
The multivariate normal distribution can be parameterized in terms of the precision matrix Θ = Σ−1. This is a useful parameterization in the context of GGMs, because partial correlations can be expressed in terms of the entries of this precision matrix,{θ ij}.7 Specifically, θ ij = 0 if and only if there is zero partial correlation between X i and X j. Thus the sparsity pattern of Θ contains the conditional independence relations encoded in the corresponding GGM, and the problem of estimating a GGM is equivalent to the problem of estimating Θ.
2.2 |. Intuiting partial correlation in the Gaussian setting
In the Gaussian setting, partial correlation can be intuited in two main ways. The first is in the context of the bivariate conditional distributions of the multivariate normal distribution, and the second is in the context of multiple linear regression. The first approach allows us to derive a straightforward formula for the partial correlation as a function of the precision matrix Θ, and the second shows us that multiple linear regression coefficients are directly proportional to partial correlation coefficients.
2.2.1 |. Bivariate conditionals
Under the multivariate normal assumption on X 1, … , X p, it is well known that the conditional distribution of X i and X j (i.e., the distribution X i , X j|X−i,−j) is a bivariate normal distribution.7 Denote the covariance of this distribution as the 2 × 2 matrix , and its inverse as . Let θ ij represent the (i, j) entry of the p × p matrix Θ. In the Supplement (Appendix S2), we demonstrate that the entries of the 2 × 2 matrix in terms of the entries of Θ, the precision matrix of the full dataset X 1, … , X p, are:
(4)
The interpretation of Equation (4) is that the entries of the inverse of the covariance matrix of the bivariate conditional distribution of X i and X j, that is, the bivariate conditional precision matrix, are the same as the corresponding entries in the overall precision matrix Θ. This is where the connection between the precision matrix and the partial correlations becomes apparent. One can obtain , the conditional covariance matrix of X i and X j, by inverting Equation (4) using the usual formula for the inverse of a 2 × 2 matrix:
(5)
Converting this conditional covariance matrix to a conditional correlation matrix by scaling to the diagonal and noting that θij = θji by symmetry yields:
(6)
The off-diagonal elements of represent the conditional correlation between X i and X j, conditioned on X−i,−j, the rest of the variables in , that is, the partial correlation between X i and X j. This yields the useful relationship between the partial correlation and corresponding elements of the precision matrix:
(7)
Note the negative coefficient of θ ij indicating that the sign of the off-diagonal elements of Θ changes when converting from the precision matrix to a matrix of partial correlations.
2.2.2 |. Multiple linear regression
Next, consider the problem of regressing the i th variable X i on the remainder of the variables X−i. Let [β−i]j be the coefficient of X j in this regression.7 In the Supplement (Appendix S3), we show that:
(8)
The main conclusion is that we can interpret the partial correlation as (i) the conditional correlation described in Equation (7), and (ii) as directly proportional to [β−i]j, the regression coefficient described in Equation (8). A corollary is that [β−i]j = 0 if and only if θ ij = 0, a relationship exploited in early GGM estimation methods based on neighborhood selection (Section 3.1).14
2.3 |. GGMs vs correlation networks
A correlation network is another type of undirected graphical model that is commonly seen in the literature. A correlation network also consists of a graph 𝒢(V, E) where vertices in V still represent a multivariate random variable , but the edge definition is different. In a correlation network, an edge exists between X i and X j if and only if the marginal correlation between the two is nonzero:
(9)
The marginal correlation (also referred to as simple correlation, bivariate correlation, or Pearson correlation) measures the association between two variables, not adjusting for the effects of any other variables. Marginal correlation, therefore, is a quantitative measure that combines both direct association and indirect association. To understand the interpretation of a correlation network, it is informative to consider the meaning of a non-edge in this context. Suppose the edge (i, j) is not in E. This means that the marginal correlation between X i and X j is zero: these two variables have no direct or indirect associations. In the context of biological data, this is a very strong condition. Consequently, correlation networks are likely to be quite dense, and it may be difficult to gather insights into underlying patterns in the data based on this approach.
The practice of thresholding a network involves removing any edges with edge weight less than a fixed threshold. Thresholding emphasizes the heaviest-weighted edges and imposes sparsity on the network structure. While thresholding is a practical analysis choice, it can be problematic in that thresholds are often chosen in an arbitrary way and interpretation of the resulting network is difficult. In this context, GGMs have another advantage over correlation networks: because they are typically more sparse, it is often easier to avoid thresholding when conducting analyses on GGMs vs correlation networks.
To highlight the difference between a correlation network and GGM, we consider a simple example of three genes: Gene 1, Gene 2, and Gene 3. Let represent the expression levels of these three genes and let be the random error present in the experiment. Suppose that the true relationship between the three is as follows:
Gene 1 is expressed with a mean at some fixed level, say 3, and is independent of Gene 2 or Gene 3:
(10)
Gene 2 is positively regulated by Gene 1 and expressed according to the model:
(11)
Gene 3 is negatively regulated by Gene 1, and expressed according to the model:
(12)
Under this setup, the covariance matrix can be found algebraically (Appendix S4):
(13)
To find the partial correlations, we first find the precision matrix by inverting the covariance matrix:
(14)
From Equation (14), we can see from the sparsity pattern (zero entries) of Θ that X 2 and X 3 are conditionally independent given X 1. We can go further and calculate all the pairwise partial correlations using Equation (7):
(15)
(16)
(17)
In this example, we see that the correlation network and GGM are quite different (Table 1, Figure 2). Because Gene 2 and Gene 3 both depend on the constitutive expression of Gene 1, they have a substantial marginal association (correlation of −0.632). This is reflected in the correlation network’s edge between X 2 and X 3. However, the marginal association is entirely due to the relationship of Gene 2 and Gene 3 with Gene 1. Therefore, there is no remaining association (zero partial correlation) between Gene 2 and Gene 3 after conditioning on Gene 1. This is reflected in the absence of an edge between X 2 and X 3 in the GGM.
TABLE 1.
Correlations and partial correlations between variables in the 3-gene example
| (A) Correlation network adjacency matrix | (B) GGM (partial correlation network) adjacency matrix |
:--- |
| | X 1 | X 2 | X 3 | | X 1 | X 2 | X 3 |
| X 1 | 0 | 0.894 | −0.707 | X 1 | 0 | 0.816 | −0.408 |
| X 2 | 0.894 | 0 | −0.632 | X 2 | 0.816 | 0 | 0 |
| X 3 | −0.707 | −0.632 | 0 | X 3 | −0.408 | 0 | 0 |
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FIGURE 2.
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Structure of the correlation network vs partial correlation network in the 3-gene example. (A) Correlation network and (B) Gaussian graphical model
Importantly, conditional independence does not imply marginal independence, nor does marginal independence imply conditional independence. The example above shows the first tenet; in the Supplement (Appendix S4), we present an example of the second. Neither of these independence relations is stronger or “better” than the other; rather, they are simply two different perspectives from which we can analyze the relationships between sets of random variables.
In a biological context such as genomics or metabolomics, the correlation network may be quite dense. Many genes or metabolites play a similar role as Gene 1 in the thought experiment above. Because of the complex, interrelated nature of biochemical pathways, such features are typically strongly associated with many other features in the network. These associations can induce further marginal associations that contribute to the correlation network, resulting in a correlation network that is difficult to interpret. The GGM, on the other hand, can be both more sparse and easier to interpret than the correlation network because its edges represent direct associations rather than indirect associations (correlation network). A summary of these differences is shown in Table 2.
TABLE 2.
Key characteristics of GGMs vs correlation networks
| Graphical model | Non-edge independence assertion | Edge interpretation | Typical density |
:--- :--- |
| GGM (partial correlation network) | (i,j) ∉ E ⇔ X i⊥X j⊥X−i,−j | Direct dependence | More sparse |
| Correlation network | (i,j) ∉ E ⇔ X i⊥X j | Direct and indirect dependence | More dense |
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Based on the above motivations, we focus on GGMs rather than correlation networks in this tutorial. We note, however, that an elegant connection between correlation networks and GGMs in the context of network topology has been recently derived.11 Gill et al11 demonstrate that the correlation between two nodes in a GGM can be expressed as a linear combination of the product of partial correlations (edge weights) along the set of network paths connecting them. This relationship enables the decomposition of indirect dependence into contributions from various sources of direct dependence using the pair-path subscore (PPS), providing a mechanism for interrogating path-specific components of correlation. The utility of the PPS is demonstrated on metabolomics data, which are well-known to demonstrate high between-metabolite correlations that can make interpretation particularly challenging.15 Keeping this advance in mind, we note that our focus on GGMs in this tutorial does not preclude, but rather can help to facilitate, correlation-based analyses.
Finally, it is essential to note that the structure used in this illustrative example is a directed acyclic graph (DAG), while we have modeled it with GGMs and correlation networks, both of which are undirected graphs. Learning the structure of DAGs from observational data is an exciting and active area of research that is beyond the scope of this article (see, eg, References 16 and 17). At the time of writing of this tutorial, it is often a more practical choice to model omics data with an undirected graphical model such as a GGM or a correlation network for several reasons. First, it is common for omics data to contain feedback loops that cannot be modeled with a DAG. Second, methods for learning DAGs rely on an assumption of causal sufficiency that specifies that there cannot be any unmeasured common confounders, that is, that all variables affecting multiple nodes in the graph are themselves included as nodes in the graph.18 This is an unrealistic assumption in the case of omics data (and most observational data). For example, myriad other variables such as chromatin remodeling proteins or microRNAs may affect the expression of multiple genes in a gene expression network. As a third example, we note that even if a directed graph is acyclic and causal sufficiency is met, a single observational dataset may have conditional independence properties that can be satisfied by multiple distinct directed graphs, a phenomenon known as Markov equivalence.19 While undirected graphical models themselves have limitations, they comprise a more straightforward framework for modeling high-dimensional data that has been well-studied in omics contexts.
2.4 |. Deriving insight from GGMs in omics applications
The result of GGM estimation is an undirected network with weighted edges, a product that enables the use of various network science tools for practical interpretation. Strategies for interpretation can include node-based and network-level properties. Here, we discuss several strategies for both types of analysis.
Several node-based measures exist to assess the importance of each node within the framework of its network. A first measure is the number of edges connected to a node, known as its degree. A weighted version of degree is sometimes used in graphs with weighted edges. Hub centrality, betweenness centrality, and closeness centrality are three other node-based measures we will use in our example below. Hub centrality is a measure of how much influence a node has in a network based on its connections in the context of the overall network structure; for a more precise definition, see the work of Kleinberg. Betweenness centrality is a measure of how many of the shortest paths through the network travel through a node.20,21 Closeness centrality is a measure of the average shortest path length from a node to each node in the network.20 These three centrality measures provide complementary views of the influence of each node within the overall network.
On the network level, measures such as degree distribution and sparsity provide structural characterization. Investigating clusters of highly connected nodes called communities can also provide an important network-level perspective.22,23 A community in a GGM represents a set of nodes that are highly conditionally dependent, given the rest of the network. Knowledge of these conditionally dependent sets can provide insight into specific biologic functions: a community of metabolites, for example, may be indicative of a metabolic process, or a community of genes may be indicative of shared functional regulation. A range of different community detection algorithms have been developed;24 some options implemented in igraph include a fast greedy clustering algorithm25 (cluster_fast_greedy), a method based on random walks26 (cluster_walktrap), and a method drawing on spin glass models from statistical physics27–29 (cluster_spinglass). Importantly, different community detection methods can produce variable results. A researcher should take the time to understand the basics of how a particular community detection method works prior to using it. This topic is beyond the scope of the current tutorial; see, for example, Fortunato and Hric.23
In the case where a researcher has a high-dimensional dataset such as a gene expression dataset coupled with an outcome of interest such as disease status, algorithms to detect subnetworks enriched for association with the outcome can be applied. BioNet30 and the jActiveModules tool of the Cytoscape suite6,31 are two examples of such tools.
3 |. FREQUENTIST METHODS FOR GGM ESTIMATION AND INFERENCE
In this section, we review frequentist methods for two key tasks: GGM estimation and GGM inference. GGM estimation is the task of determining the sparsity pattern (binary adjacency matrix) or precision matrix (weighted adjacency matrix) from an input dataset. GGM inference involves studying the variability of these estimates and conducting hypothesis tests accordingly. This section provides theoretical and historical context for the techniques that will be applied in Section 6.
3.1 |. GGM estimation
3.1.1 |. Precision matrix estimation with standard maximum likelihood methods
A first maximum likelihood approach for GGM estimation is to invert the maximum likelihood estimator (MLE) of the covariance matrix, that is, set . Matrix inversion, when possible, is a continuous transformation; inverting S thus yields the MLE for the precision matrix.32,33 However, S is not always invertible. One important such case occurs in high-dimensional settings where the number of predictors p is larger than the sample size n. Regularized methods that impose a penalty on the size of the estimated precision matrix can be applied in these cases; rapid increases in high-dimensional data and applications over the last few decades have spurred the development of several such approaches. However, there is also a case for regularized methods in low-dimensional problem settings. rarely has entries equal to exactly zero, meaning the corresponding estimated GGM is a completely connected graph. Regularized methods can shrink the entries of , inducing exact zeros. These zeros, then, translate into the absence of an edge in the GGM, yielding a sparse network that may have a more interpretable topology. Regularized methods are thus an attractive option even for low-dimensional data where S can be inverted.
One of the earliest regularized methods was developed by Meinshausen and Bühlmann.14 These authors focused on estimating the graphical structure by solving the related problem of neighborhood selection. Consider the undirected graph 𝒢(V, E). For a node a ∈ V, the neighborhood of a is denoted as ne a; ne a is the set of all nodes b such that (a, b) ∈ E. Meinshausen and Bühlmann showed that, in the case where is distributed as multivariate normal, the neighborhood selection problem is equivalent to selecting the nonzero coefficients in a linear regression of X a on the remaining variable set X−a. The rationale for this approach is rooted in the relationship between partial correlations and corresponding regression coefficients in a multiple linear model presented in Equation (8).
Notably, the Meinshausen-Bühlmann approach estimates only the sparsity pattern of the precision matrix, not the actual values of the nonzero matrix entries, providing an approximate rather than exact solution to the problem of estimating the parameters of a GGM.
In 2007 and 2008, several groups made advances on the exact problem of estimating Θ.12,34,35 Around the same time, Yuan and Lin,12 Banerjee et al,34 and Friedman et al35 separately developed methodology for estimating a sparse precision matrix by maximizing the ℓ 1 penalized likelihood of a GGM with respect to the space of non-negative definite precision matrices Θ:
(18)
where ||Θ||1 = ∑i,j |θ ij| or ∑i≠j |θ ij| (depending on if one wishes to penalize the diagonal elements of Θ), S is the sample covariance matrix, and λ ≥ 0 is a non-negative tuning parameter. Equation (18) represents a tradeoff between optimizing the unpenalized Gaussian likelihood and establishing a sparse precision matrix: larger values of λ increasingly bias the entries of the estimate to zero, corresponding to sparser estimated precision matrices.
Regularized approaches have two key characteristics that are helpful for the estimation of high-dimensional GGMs. First, formulating the objective function with a penalty on the magnitude of the elements of the precision matrix leads to an estimated graph with fewer edges; such a graph may be more feasible to interpret in high-dimensional cases. Second, the algorithm may be used in settings where p ≫ n, that is, the MLE for the precision matrix does not exist.
Based on the framework above, the graphical lasso was published in 2008 by Friedman et al.35 The graphical lasso was a transformative advance that forms the foundation for much of the current research in GGM estimation. In 2011, two different research teams36,37 independently discovered that the sample covariance matrix contains all the information necessary to determine the block diagonal structure of the precision matrix estimate for a fixed value of λ. The problem of estimating the full GGM for a specific λ is thus reducible to the problem of solving Equation 18 on each block, a task that is computationally simpler and easily parallelized. This beautiful property enabled massive computational improvements: graphical lasso problems became feasible in a few minutes for sparse graphs on p ~ 20 000 predictors, vs a previous limit of p ~ 2000.36,37 Further details of this insight are presented in the Supplement (Appendix S5).
3.1.2 |. Other penalties used in GGM estimation
While the ℓ 1 penalty in Equation (18) is commonly used, an important limitation of this penalty is that variables with larger coefficients are subject to higher bias.38 This is true of ℓ 1 penalized regression (i.e., LASSO) in general as well as in the graphical lasso. While it is desirable to bias small coefficients towards zero to achieve the goal of variable selection, it may also be important to reduce the bias in estimates of larger coefficient values. Zou39 introduced the adaptive LASSO for this purpose in the context of regression. The adaptive LASSO penalty is a weighted ℓ 1 penalty that is smaller for larger coefficients, simultaneously accomplishing the desired shrinkage of small coefficients while mitigating bias in the estimation of larger coefficients.39 The adaptive LASSO penalty can also be applied in Equation (18) to obtain unbiased estimates of large partial correlations while still biasing smaller partial correlations to zero to obtain a sparse graph.40 Another penalty that also achieves this goal is the smoothly clipped absolute deviation (SCAD) penalty proposed by Fan and Li.38 Fan and Li present a detailed comparison of GGM estimation with the LASSO, adaptive LASSO, and SCAD penalties. In simulation settings, GGMs estimated with the adaptive LASSO and SCAD penalties outperform those estimated with the LASSO penalty according to sensitivity, specificity, entropy loss, and quadratic loss; the SCAD penalty typically results in a higher Type 1 error rate and a lower Type 2 error rate than the adaptive LASSO penalty.40 A third alternative penalty is the minimax concave penalty (MCP) proposed by Zhang.41 Unlike the Lasso penalty, the MCP does not require the strong irrepresentable condition for model consistency. This condition is based on the covariance of the predictors and assumes that the set of predictors not in the model are not highly correlated with those in the true model.42 Further, while the adaptive LASSO requires an initial estimate of the coefficients that is within a certain distance from the true values and assumes that the minimum nonzero coefficient is a certain distance away from zero, MCP does not require either. Williams43,44 provides a survey of the SCAD, MCP, and other nonconvex penalties in the context of GGMs; they also implement GGM estimation with these penalties in the R package GGMncv. In the remainder of this tutorial, we focus on methods based on the ℓ 1 penalty, that is, methods based on the graphical lasso.
3.1.3 |. Methods for tuning parameter selection
Selecting the tuning parameter λ is critical for GGM estimation. Higher values of λ lead to more sparse graphs, while lower values lead to denser graphs. Using a value of λ that is too large will result in a graph that is overly sparse and may be missing critical edges that are important to the characterization of the data. Using a value of λ that is too small will result in a graph that is overly dense and may contain false positive edges that mask true conditional independence patterns. As is the general case for tuning parameters, we recommend that the optimal value of λ be selected in a data-driven manner. Wysocki and Rhemtulla45 present an overview of four common data-driven tuning parameter selection approaches, which we briefly summarize here.
Cross-validation involves partitioning available data into training and test sets, estimating a precision matrix from the training data, and assessing the out-of-sample likelihood of using the test data.45 The value of λ that maximizes the out-of-sample likelihood is chosen as optimal by this criterion. Cross-validation may involve a simple train-test split of the data, K-fold cross-validation, or leave-one-out cross-validation.46
The rotation information criterion (RIC) is based on an approximate null distribution of the precision matrix estimates. First, each row of the original n × p dataset is permuted to generate a data matrix in which all true associations between variables have been disrupted.45,47 Next, the smallest value of λ such that the estimated precision matrix is diagonal, that is, no edges are observed in the estimated GGM, is determined. This procedure is repeated over a user-defined number of permutations, and the smallest λ obtained in these repeated procedures is selected as optimal. Intuitively, this approach finds a tuning parameter that one would expect not to select any false positives in a truly empty graph. The RIC may tend to underselect edges in practice, giving a graph that is sparse but has many false negatives.48
The stability approach to regularization selection (StARS) is a subsampling-based approach.49 For a dataset with n observations, N subsamples of size b (1 < b < n) are generated by sampling without replacement.45,49 Next, for a particular λ, N different GGMs are estimated. Stability of the estimated GGMs is evaluated by comparing the edge sets across these N different GGMs. A preferable λ is one for which there is high agreement in the edge sets across all the subsamples.
The extended Bayesian information criterion (eBIC) optimizes a BIC-like criterion.45,50 For a GGM consisting of E edges, n samples, and p predictors, the eBIC is given by:
(19)
where γ is a non-negative tuning parameter, with higher values of γ encouraging a sparser graphical model.50 Note that when γ = 0, this equation corresponds to the usual BIC.51 To perform model selection with the eBIC, for each candidate λ, the corresponding precision matrix is estimated from the full n × p dataset. The eBIC is then calculated using . The value of λ resulting in the smallest value of the eBIC is then chosen as the best.
3.1.4 |. Incorporating non-Gaussian data via the nonparanormal transformation
The assumption of multivariate normal data can be limiting in practice. To relax this assumption, Liu et al52 introduce the nonparanormal transformation. For a random variable X = (X 1, … , X p), the nonparanormal transformation seeks to find a set of monotonic functions f 1, … , f p such that the distribution of f(X) = (f 1(X 1), … , f p(X p)) is multivariate normal. Because each function f i only depends on the i th variable, the distribution of f(X) has the same factorization as the distribution of X, meaning that the transformation preserves independence and conditional independence relationships. Consequently, the sparsity pattern of the graph based on the transformed data f(X) will reflect these properties in the original data. It is important to note that edge weights in this graph do not represent partial correlations between, for example, X i and X j, but rather between their transformed versions f i(X i) and f j(X j). For details on the nonparanormal transformation and the estimation of f, we refer the reader to Liu et al.52 Liu et al53 also present a procedure for estimating f with improved efficiency. Finally, we note that the nonparanormal transformation is a special case of a Gaussian copula and refer readers to Dobra and Lenkoski54 for background in this area.
3.2 |. GGM inference
In addition to the problem of estimating a GGM, the challenge of performing inference on the estimated network has been an active area of study in recent years. While a broad range of inferential questions may arise based on an estimated network, here we primarily focus on edge-level inferential questions such as the presence or absence of an edge or the variability of an estimated edge weight.
In the GGM estimation tools described above, the goal of the problem is either to select a set of unweighted edges which comprise the network (as in the Meinshausen-Bühlmann neighborhood selection approach), or to estimate the precision matrix as in the graphical lasso, thereby estimating the partial correlations and a corresponding edge-weighted network. We turn our attention now to quantifying the uncertainty of the selected edge set or estimated partial coefficients. First, we discuss several methods that use large-sample theory to assess variability (Section 3.2.1). Second, we discuss two methods that use bootstrap approaches to approximate the sampling distributions of relevant test statistics (Section 3.2.2). Both approaches provide scope for hypothesis testing and the construction of confidence intervals.
3.2.1 |. Methods based on large-sample theory
Early work in the area of GGM inference for the low-dimensional setting was done by Drton and Perlman,55 who first introduce the following null and alternative hypotheses for testing the partial correlation between variables X i and X j:
(20)
(21)
For a single hypothesis test of this form, Drton and Perlman propose the use of the Fisher-transformed sample partial correlation coefficient,56
where r ij is the sample partial correlation between (X i , X j) and where z ij is asymptotically normal under with mean 0 and variance equal to . When the sample covariance matrix is invertible, a straightforward application of Equation (7) results in the desired test statistic. Drton and Perlman frame edge set selection as a collection of the p(p − 1)/2 hypothesis tests of this form, for which p(p − 1)/2 corresponding test statistics can be calculated. The GGM edge set is proposed to include any edge (i, j) for which is rejected in favor of . We provide further details of this approach in Appendix S6.
In the high-dimensional setting where regularized methods must be used to estimate the GGM, inferential approaches cannot be based on the methods of Drton and Perlman as inference must be undertaken considering the complexities of regularization. Liu57 begins by noting that estimating a GGM in a high-dimensional setting requires selecting a tuning parameter, and establishes the focus of assessing the relationship between the value of the tuning parameter and the number of falsely discovered edges. Liu develops a method called “GGM estimation with FDR control” (G by developing test statistics that can be used when p > n. Beginning with a multivariate normal variable X = (X 1, … , X p), for variables X i and X j, let X−i and X−j represent the remaining variables with i or j removed, respectively. Consider the linear models
(22)
(23)
Liu notes that covariance between the residuals from these two models have the property:
(24)
where θ ij is the (i, j)th entry of the precision matrix Θ = Σ−1. Test statistics that can be used when p > n are then derived from the distributions of these residuals. We refer interested readers to the original publication for technical details.57 Ren et al58 also introduce a method that uses linear regression models, in the spirit of Liu. This contribution differs in that where other methods have focused on regressing one predictor against all the others, their method uses a bivariate linear regression.58 For two predictors X i and X j, Ren et al consider the bivariate regression of (X i , X j) against the remaining predictors X−i,−j:
(25)
Under the usual multivariate normal assumption, the covariance between the residuals of this regression is
(26)
Ren et al58 propose to directly estimate the values of θ ij from the residuals of this “two-vs-rest” regression. A scaled lasso estimator with an ℓ 1 penalty is used for these regressions.
Janková and van de Geer59–61 published a series of works regarding a de-biased graphical lasso estimator and corresponding inferential techniques such as hypothesis testing and the construction of confidence intervals. In the first work published in 2015, the authors demonstrate that it is possible to invert the Karush-Kuhn-Tucker conditions used in the original graphical Lasso estimator to obtain a de-biased precision matrix estimator and derive the asymptotic normality of this estimator under certain conditions on the true precision matrix.59 The resulting distribution facilitates element-wise or simultaneous inference on the entries of the estimated precision matrix, leading to an edge selection procedure with useful large-sample probability guarantees.59 One key assumption is an irrepresentability condition similar to that required for model selection consistency when using LASSO regression for variable selection.42 In the context of the LASSO, the irrepresentable condition essentially states that the set of variables that are in the true model not be too highly correlated with the set of variables that are not.42 LASSO and the graphical lasso are both examples of a broader class of estimators called regularized M-estimators, and such an irrepresentability condition is an important part of proving model selection consistency for this class.62 Janková and van de Geer note that this condition is, in practice, difficult to verify. In the second paper, the authors present a different estimator that achieves similar results but does not require the irrepresentability condition.60 The estimator of the 2017 work is a variant of the Meinshausen-Bühlmann algorithm, as opposed to the estimator of the 2015 work, which is a variant of the graphical Lasso algorithm. In both papers, a variety of further details such as convergence rates and edge selection performance are addressed.59,60 A notable advantage of both methods over other work is that the usual multivariate normal assumption is relaxed to an assumption of sub-Gaussian tails on the marginal distributions. The R package SILGGM implements both methods for straightforward use and comparison.63 In the third paper, Janková and van de Geer present an overview of methods for inference in GGMs as well as directed graphs.
3.2.2 |. Bootstrap-based network inference
Bootstrap Inference for Network Construction (BINCO)
Li et al64 note that determining the right level of sparsity is an outstanding challenge in GGM model selection and propose a bootstrap-based approach to identifying the optimal regularization parameter in network estimation. They frame the problem in the context of balancing the tradeoff between FDR control and statistical power to detect edges. This work focuses on network estimation as an edge selection problem, meaning that estimating an unweighted network is the goal. For an edge selection procedure A(λ) and dataset Y, Li et al define the selection probability pij of each edge (i, j) as the estimand of interest. The authors generate a large number of estimated networks by resampling the dataset Y and repeatedly applying procedure A(λ). The empirical selection probability (selection frequency) Xij of edge (i, j) among these bootstrap networks is used as an estimator for pij. Having defined this estimator, all that remains is to choose some threshold c ∈ [0, 1] and select all edges such that Xij ≥ c. Simulation studies demonstrated that BINCO is largely resilient to changes in network properties such as network topology, dimensionality, and edge strength. An R package BINCO was previously available on the CRAN repository, but was removed in 2013. An archived version is available.
Bootnet
Epskamp et al9 also present a bootstrap-based approach to network analysis in their R package bootnet. The authors emphasize that their software can be used to develop estimates of edge weights along with corresponding confidence intervals based on bootstrap estimates of the sampling variability in the network, but that these cannot be used to validly assess whether or not an edge weight differs significantly from zero due to the LASSO component of the estimation process.9 Instead, Epskamp et al suggest using the confidence intervals to gain some idea of network accuracy or to test the hypothesis that two edge weights significantly differ. Epskamp et al9 also address the issue of multiple hypothesis testing that arises naturally from testing hypotheses about many edges in a graph and suggest it is an area for future research. The scope of the work also includes hypothesis testing for centrality measures, which provides an important perspective on the sampling variability of the network structure.
4 |. BAYESIAN APPROACHES TO GGM ESTIMATION AND INFERENCE
Bayesian approaches for GGM estimation have also been developed. While a thorough review of these methods is beyond the scope of this tutorial, we briefly highlight some advances in this area.
4.1 |. The Bayesian graphical lasso and Bayesian adaptive graphical lasso
The coefficients estimated by LASSO regression have a Bayesian interpretation as the posterior mode obtained by considering the usual linear regression model Y = Xβ + ϵ, ϵ ~ N(0 , σ 2) with where the g i are independent double exponential densities.46 Similarly, the precision matrix estimated by graphical lasso has a Bayesian interpretation as the mode of a posterior distribution based on coupling the multivariate normal data model with a particular prior on the precision matrix entries.65 While this prior is again based on the double exponential, it does not take exactly that form due to the requirement that the precision matrix must be positive definite; we refer readers to Wang65 for details.
The Bayesian graphical lasso suffers the same limitation as the graphical lasso in terms of over-shrinking large nonzero precision matrix entries; to remedy this, Wang65 also introduces a Bayesian analogue of the adaptive graphical lasso. In simulation studies on p = 30 and p = 100 predictors across six different graph topologies, the Bayesian adaptive graphical lasso often performed best according to Stein’s loss; the Bayesian graphical lasso, adaptive graphical lasso, or the SCAD method were runners-up in most cases; the case of a star topology with 100 predictors was an exception, here the original graphical lasso had the best performance, followed by the adaptive graphical lasso.
Importantly, the work of Wang focuses on small to moderate (p 10–100) dimensional problems; application of the Bayesian graphical lasso to omics datasets will therefore require some other type of variable selection or dimensionality reduction.
4.2 |. The Wishart, G-Wishart, spike-and-slab, and horseshoe prior distributions
The Wishart distribution is a conjugate prior for the precision matrix Θ in the multivariate normal distribution N(0, Θ−1).66 An attractive feature of the Wishart conjugate prior is that it admits a closed-form solution for the posterior mode and variance.67 However, GGM estimates based on the corresponding posterior distribution are not sparse.67
An alternative prior that does impose sparsity is the G-Wishart distribution. The G-Wishart distribution is a relative of the Wishart distribution in which specific off-diagonal matrix entries are fixed to zero; it is also a conjugate prior for Θ.68–72 However, sampling from and approximating the normalizing constant of the corresponding G-Wishart posterior can be computationally challenging, and much work in the field has addressed these problems.72,73 Wang74 proposes mitigating the challenge of sampling from the G-Wishart by using a spike-and-slab prior on the off-diagonal elements of the precision matrix. While the G-Wishart prior fixes certain entries exactly to zero, the spike-and-slab prior is a mixture distribution of two normal distributions, both centered at zero but with distinct variances. The mixture component with the smaller variance can be interpreted as the distribution of the “non-edges,” while the mixture component with the larger variance can be interpreted as the distribution of the true edges.
The spike-and-slab prior introduced by Wang74 depends hierarchically on a binomial distribution which represents the idea that the edge set is a set of Bernoulli random variables with success probability (edge presence probability) π. The spike-and-slab graphical lasso enables substantial computational improvements over other Bayesian graphical lasso approaches by introducing a more efficient updating scheme and avoiding approximating the normalizing constant; however, a limitation of adopting the spike-and-slab prior is that the posterior estimated edge weight for non-edges is larger than it is for the G-Wishart prior. If the goal is to classify each edge as present or absent, this limitation is irrelevant; however, if the goal is to estimate the partial correlation, estimates may be too high.
Li et al75 introduce the graphical horseshoe estimator (GHE), which takes a different approach by assigning two distinct types of priors to different precision matrix elements. Diagonal elements of the precision matrix are given an uninformative prior, while each off-diagonal element ω ij has its own horseshoe prior parameterized by a global hyperparameter τ and a local hyperparameter λ ij, both of which follow a half-Cauchy distribution.75 In this framework, the induced marginal prior for each off-diagonal element of the precision matrix is horseshoe-shaped, with heavy mass at both 0 and 1 and little mass in between. In this way, the GHE of Li et al allows global shrinkage to control the sparsity of the precision matrix via the shared hyperparameter τ while reducing the bias of nonzero off-diagonal elements via λ ij.
4.3 |. Unregularized Bayesian GGM estimation
Recent work in psychometrics has focused on the use of the Wishart distribution as a conjugate prior for unregularized GGM estimation in the case where p < n.67 To impose sparsity on the estimated GGM, a decision rule based on posterior probabilities is used to assess the evidence that an entry is or is not zero.67 This method can be extended to include regularization.67 Williams and Mulder76 additionally present Bayesian hypothesis tests for single partial correlations as well as a set of multiple partial correlations with order constraints. These methods are available in the R package BGGM.77
5 |. GGM EXTENSIONS
In recent years, many researchers have built on the foundations of GGM theory and methodology to develop extensions suited to problems such as estimating scale-free graphs and GGM meta-analysis. We address several such extensions below.
5.1 |. Extensions incorporating structural characteristics of graphs
Graphical models can be characterized by structural characteristics such as hubs or connected components. The original graphical lasso does not incorporate these properties into estimation; two extensions that have been developed for this purpose are the hub graphical lasso78 and the cluster graphical lasso (CGL).79
In developing the hub graphical lasso, Tan et al78 observe that a key limitation of the ℓ 1 penalty is that it treats all edges in the graph as independent and equally likely to be nonzero. In an Erdös-Renyi random graph, this assumption is true; however, it is well-known that many real-world networks do not have this topology.80–82 One topology of particular interest is scale-free networks, which (in contrast to random graphs) are characterized by hub-and-spoke structures and have been identified in a wide variety of real-world settings.81 Tan et al modify the ℓ 1 penalty in the original graphical lasso objective function in order to encourage a hub structure in the estimated network. In addition to the hub graphical lasso, Tan et al78 also present similar approaches for estimating Gaussian covariance graph models and binary Ising graphical models. Software is available in the R package hglasso.83
The CGL is based on the realization that sparse inverse covariance estimation with the graphical lasso is equivalent to a two-phase estimation approach of (1) performing single linkage hierarchical clustering (SLHC) to identify connected components of the network, followed by (2) performing the graphical lasso individually within each connected component.79 Tan et al79 point out that SLHC has widely known shortcomings and propose the CGL as an alternative. The CGL is also a two-phase approach, but is more general than the graphical lasso in that it allows the use of any clustering method in the first phase. There are two key advantages to the CGL. The first advantage is the flexibility of using any clustering method, which helps to address some of the issues with SLHC. The second advantage relates to tuning. One property of the graphical lasso is that the number of edges and connected components is effectively tuned by the same tuning parameter. CGL improves this by allowing clustering (ie, selection of number of the connected components) in the first step followed by shrinkage in the second step, meaning that the two aims can be decoupled and addressed sequentially.
5.2 |. Differential network analysis with GGMs
The comparison of GGMs across multiple datasets or conditions, often referred to as differential network analysis, is an important task for the practical utility of GGMs in answering research questions. In this context, a balance is typically sought between improving power to detect a shared network signal by borrowing strength across conditions via joint estimation and identifying condition-specific network signals. Shojaie84 presents a thorough overview of differential network analysis based on GGMs and other types of graphical models. While such a discussion is beyond the scope of this tutorial, we briefly mention two such methods. The joint graphical lasso85 (JGL) adds an additional penalty term to the optimization problem in Equation (18) that encourages similarity between precision matrices. While the form of the penalty is generally left flexible, two specific penalties are proposed: the fused graphical lasso, which addresses model selection by encouraging similarity in sparsity patterns, and the group graphical lasso, which addresses parameter estimation by encouraging similarity in coefficient values. A different approach is taken by Ha et al86 in their tool DiNGO (Differential Network Analysis in Genomics). Ha et al86 model each condition-specific GGM as the convolution of a shared global GGM with a local component. This multi-stage process involves fitting a global GGM to the pooled data, calculating the residuals from this fit, and regressing these residuals on an indicator variable of condition. A key difference between the JGL and DiNGO is that while both methods estimate condition-specific networks, DiNGO directly estimates a global GGM while the JGL does not.
5.3 |. Graphical modeling on mixed data types
Mixed graphical models incorporate multiple data types in the same graphical framework. In mixed graphical models, the absence of an edge between variables still indicates their conditional independence as in the GGM setting; however, the underlying data need not have a Gaussian distribution, and multiple different types of distributions can be accommodated in the same graph. Importantly, edges in mixed graphical models must be interpreted strictly as conditional independence relationships, not partial correlations. This framework is particularly important in the context of multi-omic integration with GGMs, as many types of omics data have distributions that are not Gaussian. For example, genetic variation may be measured in terms of single nucleotide polymorphisms (SNPs), which can be represented as ordinal random variables taking values in the set {0, 1, 2} based on whether a locus has zero, one, or two copies of a minor allele.87 RNA-seq, used to assess gene expression, is a count variable that is typically modeled as Poisson or negative binomial.88
One approach to incorporating discrete and continuous variables into a mixed graphical model is based on the conditional Gaussian distribution, which assumes that the conditional distribution of each continuous variable, conditioned on all possible values of the discrete variables, is normal.7,89 A limitation of this approach is that the exponential nature of this conditioning quickly becomes prohibitive in practice as the number of discrete variables and the size of their support grows. Cheng et al90 present an extension for high-dimensional applications that reduces the size of the problem by incorporating sparsity and a group lasso penalty, demonstrating their approach on a mixed graphical model with 118 discrete variables and 16 continuous variables. Similar approaches are presented by Fellinghauer et al91 and by Lee and Hastie.92 Yang et al93 take a different approach by introducing mixed exponential Markov random fields, a graphical modeling framework that considers the joint distribution of a set of random variables in the exponential family. This family includes the binomial, Poisson, and Gaussian distributions, all of which have been used for modeling various omics data types.
5.4 |. Meta-analysis of GGMs
Meta-analysis of GGMs is another important extension. Because the parameter space for GGM estimation grows quadratically with the number of nodes (predictors) in the graphical model, small sample sizes can be problematic even for estimation methods that accommodate high-dimensional data. Integrating multiple studies into a single GGM via meta-analysis is one way that researchers may be able to mitigate this problem. To this end, Epskamp et al94 present meta-analytic Gaussian network aggregation (MAGNA), a modeling framework that permits meta-analyzing multiple estimated GGMs from different studies. Fixed-effects MAGNA assumes that there is not between-study heterogeneity, while random-effects MAGNA directly models between-study heterogeneity, breaking down variation in estimated partial correlations into variation due to sampling error and variation due to between-study differences.94 These tools are available in the R package psychonetrics.95 A limitation of MAGNA is that it is very computationally intensive and cannot currently handle a large number of predictors; the demonstrated example focuses on a set of 16 post-traumatic stress disorder symptoms, whereas omics applications typically involve analysis of hundreds or even thousands of predictors.
6 |. DEMONSTRATION
6.1 |. Introduction to the motivating dataset and software choices
The demonstration in this tutorial focuses on ovarian cancer gene expression; in this context, constructing a GGM provides us with a graphical model of conditional independence patterns between gene expression levels in individuals with ovarian cancer. Generally, network-based analyses of gene expression can help generate hypotheses about the disease’s regulatory patterns. In ovarian cancer in particular, gene expression has been effectively used to classify tumors, and differences in gene expression between tumor types have implicated specific biological functions in disease etiology and progression,96 and network-based analyses have been applied to identify systems-level patterns.97,98 In addition to the existing literature linking gene expression with ovarian cancer phenotypes, a practical reason for the study of ovarian cancer gene expression data is the high-quality publicly available data in the curatedOvarianData R package.
In this demonstration, we use R packages huge for GGM estimation and SILGGM and bootnet for GGM inference. We also use GGMncv to compare graphs estimated with alternative penalties (adaptive graphical lasso, SCAD, MCP). We selected these packages because they are well-documented, user-friendly, and incorporate a broad range of methods and options. An overview of some additional R software that is available for GGM-related tasks is shown in Table 3.
TABLE 3.
An overview of publicly available R packages for Gaussian graphical models
| Package | Description |
:--- |
| BGGM77 | Suite of Bayesian methods for GGM estimation and inference. Includes features to support exploratory and confirmatory hypothesis testing, GGM comparison, and assessment of node-wise predictability based on the posterior distribution. |
| bootnet9 | Bootstrap methods for estimating accuracy and stability of network structures and summary measures such as degree centrality. Incorporates a comprehensive set of R packages (including huge and qgraph) for undirected graphical modeling in a user-friendly interface, facilitating easy comparison of results from different methods. |
| GGMncv43 | Implementation of GGM estimation with the SCAD, MCP, and other nonconvex alternatives to the ℓ 1 penalty. |
| glasso14,35,36 | Original package implementing the graphical lasso as proposed by Friedman et al, with the computational advances proposed by Witten et al; additionally implements the Meinshausen-Bühlmann algorithm for approximate solutions. |
| hglasso78,83 | Implementation of the hub graphical lasso, including a BIC-type criterion (hglassoBIC) to aid in model selection. |
| huge47,48 | Functions for GGM estimation with correlation thresholding, the Meinshausen-Buhlmann algorithm, and glasso; GGM model selection with the RIC, StARS, and eBIC (glasso only) scoring criteria; optional preprocessing of data using the nonparanormal transformation for non-Gaussian input and/or lossy screening for computational efficiency. |
| igraph99 | Comprehensive suite of tools for generating and plotting graphs, analyzing graph properties such as centrality measures, and performing community detection. |
| qgraph100 | GGM estimation with the graphical lasso, including model selection with the eBIC; tools for visualization and analysis of estimated networks, including computation of centrality measures, mutual information, and small-worldness index. |
| SILGGM57–60,63 | Efficient methods for local and global inference on large Gaussian graphical models; implements the methods of Liu et al (2013), Ren et al (2015), Jankova and van de Geer (2015, 2017). Supports export formats that facilitate downstream analysis in Cytoscape. |
| visNetwork101 | Visualization tool that allows a user to easily produce interactive network visualizations by interfacing with the JavaScript tool “vis.js.” |
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6.2 |. Preparing the R workspace
6.2.1 |. Package installation
First, users should install and load the following libraries (available from CRAN at the time of writing this tutorial).
# Libraries specific to GGM estimation and
# inference
library(bootnet) # Epskamp 2018
library(GGMncv) # Williams 2020
library(glasso) # Friedman, Hastie, Tibshirani 2007
library(huge) # Zhao et al 2012
library(SILGGM) # Zhang et al 2018
# (implements Ren et al 2015, Janková and van de
# Geer 2015, Janková and van de Geer 2017, and
# Liu 2013)
# Other utilities
library(dplyr)
library(igraph)
library(msigdbr)
library(MVN)
To access and process the ovarian cancer data, we will need the curatedOvarianData package, which can be installed via Bioconductor.13,102
if (!requireNamespace("BiocManager", quietly = TRUE)) {
install.packages("BiocManager")
}
# BiocManager∷install('curatedOvarianData')
1ibrary(curatedOvarianData)
6.2.2 |. Loading and previewing the data
The data used in this tutorial are from a The Cancer Genome Atlas (TCGA) study of ovarian cancer available in the curatedOvarianData R package.13 We begin by demonstrating how to load the data. The data are stored in an ExpressionSet object, helpful documentation for which can be found the vignette for the Bioconductor’s ExpressionSet class.103 We use the exprs function to extract the array data from the ExpressionSet.
1ibrary(curatedOvarianData)
data(TCGA_eset)
ocData = exprs(TCGA_eset)
The resulting object is a 13 104 × 578 matrix where rows correspond to genes, columns correspond to participant identifiers, and cells correspond to gene expression levels that have been preprocessed as described in Ganzfried et al.13
dim(ocData)
13104 578
ocData[1:5, 1:5]
TCGA.20.0987 TCGA.23.1031 TCGA.24.0979 TCGA.23.1117
A1CF 2.923522 3.052169 2.846371 3.002209
A2M 10.353008 11.635772 7.954542 9.971500
A4GNT 3.321405 3.666463 3.258038 3.596212
AAAS 4.608010 5.142133 5.025422 5.139928
AACS 7.279213 7.048869 7.750161 6.206031
TCGA.23.1021
A1CF 3.062993
A2M 8.971334
A4GNT 3.388706
AAAS 5.256831
AACS 7.835422
6.3 |. Data preprocessing
6.3.1 |. Standardizing the data
An important starting point in GGM estimation is to consider standardizing the predictors. Because many methods for GGM estimation employ shrinkage methods, the size of the entries in the covariance matrix, and consequently the precision matrix, will affect results. In applications such as metabolomics that output relative abundance of individual metabolites, the absolute measurements have no practical interpretation; in these settings, standardizing the predictors ahead of time is important. Some packages handle this internally, while others do not; therefore, we recommend standardizing predictors prior to beginning analysis unless you deliberately wish to study the variance of the absolute measurements. Here, we also center for convenience.
standardize = function(x) {
return((x − mean(x))/sd(x))
}
ocDataStd = apply(ocData, 1, standardize)
Using the apply function has the additional advantage of transposing the array data so that rows (n) contain participant identifiers and columns (p) contain gene names and the data frame is in the more familiar n × p format which is standard input for most GGM estimation software.
We can check to make sure the standardization worked as expected:
summary(apply(ocDataStd, 2, mean))
Min. 1st Qu. Median Mean 3rd Qu.
-3.645e-15 -3.545e-16 -2.456e-18 -3.340e-18 3.495e-16
Max.
3.334e-15
summary(apply(ocDataStd, 2, sd))
Min. 1st Qu. Median Mean 3rd Qu. Max.
1 1 1 1 1 1
The resulting centered, standardized predictors are now prepared for GGM estimation.
6.3.2 |. Identifying genes for network analysis
Graphical lasso can handle tens of thousands of predictors. However, a frequent practice is the use of some prior knowledge to reduce the set of predictors being considered. In this tutorial, we will leverage two gene sets from the Molecular Signatures Database (MolSigDB).104 Both gene sets come from a study by Wamunyokoli et al105 profiling gene expression in ovarian cancer. The first set corresponds to genes that were downregulated in mucinous ovarian tumors, and the second corresponds to genes that were upregulated. In both cases, the baseline for comparison was normal ovarian epithelial cells. We access both gene sets using the msigdbr R package (
human_gene_sets = msigdbr(species = "Homo sapiens",
category = "C2", subcategory = "CGP")
ocDown = human_gene_sets[human_gene_sets$gs_name ==
"WAMUNYOKOLI_OVARIAN_CANCER_GRADES_1_2_DN", ]$gene_symbol
ocUp = human_gene_sets[human_gene_sets$gs_name == "WAMUNYOKOLI_OVARIAN_CANCER_GRADES_1_2_UP",
]$gene_symbol
ocGenes = union(ocDown, ocUp)
length(ocGenes)
212
head(ocGenes)
"ANXA8" "ARMCX1" "ATP2B4" "ATP2C1" "BASP1" "BNIP3"
Of the 212 genes contained in both the “down” and “up” gene sets, 156 are available for analysis in the TCGA data. Extracting these genes yields a 578 × 156 dataset for further investigation:
ocDataSmall = data.frame(ocDataStd[, which(colnames(ocDataStd) %in%
ocGenes)])
dim (ocDataSmall)
578 156
6.4 |. GGM estimation and tuning parameter selection
We begin fitting a GGM with the glasso function from the glasso package. When using this function, the tuning parameter must be specified. In this package, the tuning parameter is denoted by rho, vs the notation λ used earlier in this article. We demonstrate over a series of values of rho. Plotting the GGM requires converting from the results of glasso, which correspond to estimates of the precision matrix, to the matrix of partial correlations, which is accomplished with the code -cov2cor(glassoMod$wi) (see Equation 7).
par(mfrow = c(1, 4), mar = c(0, 1, 0, 1))
for (rho in c(0.2, 0.3, 0.4, 0.5)) {
glassoMod = glasso(cov(ocDataSmall), rho = rho,
nobs = nrow(ocDataSmall))
glassoGGM = graph_from_adjacency_matrix(-cov2cor(glassoMod$wi),
weighted = T, diag = F, mode = “undirected”)
plot(glassoGGM, vertex.size = 0, vertex.label = NA,
layout = layout_with_graphopt)
title(paste(“glasso: rho =”, rho), line = -8)
}
mtext(“GGMs for Ovarian Carcinoma Gene Set”, side = 3,
line = -6, outer = T
These graphs demonstrate the graphical lasso property that higher values of the tuning parameter result in a sparser graph.
Rather than make a choice of tuning parameter based on inspection of the graph, a more principled approach involves the use of a scoring criterion such as those described in the review above. The huge.select function of the R package huge has implemented three such criteria: the extended Bayesian information criterion (eBIC), the RIC, and the stability approach to regulation selection (StARS). We begin by looking at the eBIC.
The eBIC itself has a non-negative tuning parameter gamma, where smaller values of gamma result in a more dense graph and higher values result in a sparser graph.50 To use the eBIC to select a model, we first fit the GGM over a range of values of λ using the huge function and then use the huge.select function to select the optimal λ from the results of the first step, while setting γ to 0:
hugeResults = huge(as.matrix(ocDataSmall), method = “glasso”)
hugeSelectResults = huge.select(hugeResults, criterion = “ebic”,
ebic. gemma = 0)
Conducting extended Bayesian information criterion (ebic) selection.…done
We can view the set of λ considered:
hugeResults$lambda
0.69424965 0.53753229 0.41619173 0.32224214 0.24950039
0.19317909 0.14957155 0.11580782 0.08966579 0.06942497
as well as the optimal λ selected by huge.select:
hugeSelectResults$opt.lambda
0.06942497
Note that huge.select has selected the smallest value of λ that was considered in the model fitting process. huge.select supplies a plot function that lets us visualize the relationship between λ and the sparsity of the resulting graph:
plot(hugeSelectResults)
We see a dense GGM is favored by the eBIC scoring criterion with the ebic.gamma parameter set to 0. To understand the impact of ebic.gamma, we set ebic.gamma=0.5 (which is the default option). Note that we don’t need to refit the models to do this, we simply run huge.select again:
hugeSelectResults_0.5 = huge.select(hugeResults, criterion = “ebic”,
ebic.gamma = 0.5)
Conducting extended Bayesian information criterion (ebic) selection.…done
Now the optimal λ is the largest value of λ that was considered in the original model fitting:
hugeSelectResults_0.5$opt.lambda
0.6942497
and plotting the results lets us see that in this case, a completely empty graph has been selected:
plot(hugeSelectResults_0.5)
The selection of ebic.gamma is somewhat of a Goldilocks situation—the package documentation states that the ebic.gamma “can only be tuned by experience.”47 Because neither the very dense graph nor the empty graph is practically very useful, we can try a value of ebic.gamma that is halfway:
hugeSelectResults_0.25 = huge.select(hugeResults, criterion = “ebic”,
ebic.gamma = 0.25)
Conducting extended Bayesian information criterion (ebic) selection.…done
The optimal λ for ebic.gamma=0.25 is in between the two earlier values that we had obtained:
hugeSelectResults_0.25$opt.lambda
0.1931791
and the plot shows us that with this criterion, the optimal value of λ is near the middle of the range of λ values that were considered:
plot(hugeSelectResults_0.25, cex.main = 0.5)
Next, we investigate two other scoring criteria provided by huge.select. The RIC (ric) requires no criterion-specific tuning parameter. The stability approach to regulation selection (StARS; stars) criterion requires the selection of a variability threshold (stars.thres) and a subsampling ratio (stars.subsample.ratio).
hugeSelectResults_ric = huge.select(hugeResults, criterion = “ric”)
The optimal λ selected by the ric scoring criterion is similar to that selected by the eBIC with ebic.gamma=0.25:
hugeSelectResults_ric$opt.lambda
0.1650767
Next, we run model selection with the stars criterion and its default tuning parameters, stars.thres=0.1 and stars.subsample.ratio calculated as where n = 578.
hugeSelectResults_stars = huge.select(hugeResults,
criterion = “stars”)
The value of λ selected by StARS is larger than that selected by the RIC or by the eBIC with ebic.gamma=0 or ebic.gamma=0.25:
hugeSelectResults_stars$opt.lambda
0.3222421
It is evident that the optimal value of λ can be highly sensitive to the definition of “optimality” (ie, choice of criterion). We explore this further and propose solutions in Reference 106.
One important consideration in the selection of the tuning parameter is the range of λ considered in the first estimation step. Most methods have a default range of λ that is based on theoretical properties of the graphical lasso model, computed from the number of λ values of interest and the minimum value of λ that will result in an empty graph. A user may also define their own sequence of λ values if they wish to search in a more refined region. For example, the previous λ selected by stars is about 0.322, but the next closest values evaluated were 0.416 and 0.250:
hugeSelectResults_stars$lambda
0.69424965 0.53753229 0.41619173 0.32224214 0.24950039
0.19317909 0.14957155 0.11580782 0.08966579 0.06942497
We can try a larger number of λ values resulting in a finer grid by using the nlambda argument in the huge function to see if there is a better fit:
hugeResults_smallLambda = huge(as.matrix(ocDataSmall),
method = “glasso”, nlambda = 100)
hugeSelectResults_stars_smallLambda = huge.select(hugeResults_smallLambda,
criterion = “stars”)
When we do this, we find the optimal value of λ is much smaller:
hugeSelectResults_stars_smallLambda$opt.lambda
0.2553714
For the purposes of this tutorial, we will proceed using the value of λ = 0.322 determined by the StARS criterion with the coarse-grained sequence of λ values: it is the largest tuning parameter we found that produces a non-empty graph, and this sparsity facilitates easier demonstration of plotting techniques.
6.5 |. Visualizing and interpreting the estimated GGM
6.5.1 |. Graphing the GGM
To transform the estimated GGM into an object that can be visualized, we extract the 156 × 156 precision matrix from the results of huge.select:
precMat = hugeSelectResults_stars$opt.icov
dim(precMat)
156 156
Next, we convert the precision matrix to a GGM by combining the cov2cor function from base R and the graph_from_adjacency_matrix function from igraph (see Equation 7 for rationale).
ggm = graph_from_adjacency_matrix(-cov2cor(precMat),
weighted = T, mode = “undirected”, diag = F)
The summary function tells us our GGM has 156 nodes and 184 edges, along with an attribute for edge weight:
summary(ggm)
IGRAPH 52aff6b U-W- 156 187 --
+ attr: weight (e/n)
We can add a name attribute corresponding to our genes. The vertices of the GGM have the same ordering as the original data used to fit the model, so we can use the column names from ocDataSmall as vertex names:
V(ggm)$name = names(ocDataSmall)
summary(ggm)
IGRAPH 52aff6b UNW- 156 187 --
+ attr: name (v/c), weight (e/n)
We use the plot function from igraph to visualize the network. Here, for simplicity, we look at only nodes that have at least one edge in the network. We have set vertex size to be proportional to the degree of each node, labeled vertices with degree > 6, assigned edge width to be proportional to the magnitude of each edge weight, and assigned edge color to be dark red for positive partial correlations and blue for negative partial correlations. Finally, we select a graph layout. Several options are provided in igraph, each with advantages and disadvantages for particular applications. We recommend trying several. Here, we use layout_with_graphopt, which is a force-directed layout based on treating the network as a mass-and-spring system. Because there is a random component to the layout function, it is a good idea to set a seed before running it.
V(ggm)$name = names(ocDataSmall) # assign names to nodes
ggmConnected = delete_vertices(ggm, which(degree(ggm) ==
0))
set.seed(2021)
thisLayout = layout_with_graphopt(ggmConnected)
plot(ggmConnected, vertex.size = degree(ggmConnected),
vertex.label = ifelse(degree(ggmConnected) > 6,
V(ggmConnected)$name, NA), vertex.label.cex = 0.5,
edge.width = 20 abs(E(ggmConnected)$weight),
edge.color = ifelse(E(ggmConnected)$weight > 0,
“darkred”, “blue”), layout = thisLayout)
title(“Ovarian Cancer Gene Expression (Wamunyokoli et al.)”)
6.5.2 |. Interpretation with quantitative measures
Once a legible plot has been obtained, a natural next step is to begin making conclusions about the important genes in the network. For example, it appears from our plot above that EPCAM is an important gene in this network—and indeed, the literature also indicates that EPCAM is an important player in ovarian cancer.107 However, while some conclusions can be easily made from inspection of the network, we caution this approach: visualization tools for graphical models can be deceptive, and a simple qualitative approach to analyzing the GGM opens the door for researcher bias to influence model interpretation.
Several quantitative methods for network analysis can be applied to derive insights from an estimated GGM. Above, we described four such measures: degree, hub centrality, betweenness centrality, and closeness centrality.20,21,108 We can easily calculate all of these using igraph, with a few minor processing steps.
First, betweenness centrality is only well-defined for connected graphs; since our graph above is almost connected, we remove the two outlying vertices that do not belong to the large central connected component. If there are multiple non-negligible connected components, a better approach would be to analyze each one individually.
ggmConnComp = delete_vertices(ggmConnected, which(components(ggmConnected)$membership ==
2))
ocDegree = degree(ggmConnComp)
ocHub = hub_score(ggmConnComp)$vector
Next, it is important to know that betweenness centrality and closeness centrality interpret edge weights as distances (ie, dissimilarities), where larger values indicate more dissimilar nodes. In contrast, in a GGM, a high edge weight magnitude (high absolute value of partial correlation) indicates closeness/similarity. To calculate betweenness centrality and closeness centrality, we therefore first transform the partial correlations in a GGM to function as a distance measure for which large values indicate dissimilarity as shown below:
transformedWeights = 1/abs(E(ggmConnComp)$weight)
ocCloseness = closeness(ggmConnComp, weights = transformedWeights)
ocBetweenness = betweenness(ggmConnComp, weights = transformedWeights)
We assemble a final data frame containing all four measures:
ocGraphMeasures = data.frame(gene = V(ggmConnComp)$name,
degree = ocDegree, hubScore = ocHub, closeness = ocCloseness,
betweenness = ocBetweenness)
ocGraphMeasures = ocGraphMeasures[order(-ocGraphMeasures$degree),
]
Because absolute numbers can be hard to interpret, we can look at ranks for each of these measures, along with an average rank across all four:
ocGraphRanks = data.frame(gene = ocGraphMeasures$gene,
degreeRank = order(-ocGraphMeasures$degree), hubScoreRank = order(-ocGraphMeasures$hubScore),
closenessRank = order(-ocGraphMeasures$closeness),
betweennessRank = order(-ocGraphMeasures$betweenness))
ocGraphRanks$meanRank = 1/4 (ocGraphRanks$degreeRank +
ocGraphRanks$hubScoreRank + ocGraphRanks$closenessRank +
ocGraphRanks$betweennessRank)
ocGraphRanks = ocGraphRanks[order(ocGraphRanks$meanRank),
]
head(ocGraphRanks)
gene degreeRank hubScoreRank closenessRank
14 SEC22B 14 7 3
1 EPCAM 1 3 16
10 ABCD3 10 27 5
15 COX7A2 15 28 1
12 GOLPH3 12 29 6
2 DAB2 2 20 42
betweennessRank meanRank
14 6 7.50
1 16 9.00
10 3 11.25
15 2 11.50
12 11 14.50
2 9 18.25
We see that SEC22B, while not easy to identify based on visual inspection of the network, ranks highly in terms of all four of the measures considered. Indeed, based on TCGA data (find out if this is the same exact data used in tutorial), SEC22B is an important prognostic marker for ovarian cancer, with high levels of SEC22B corresponding to favorable ovarian cancer outcomes.109,110 As another example, FLRT2, the gene with the third best average rank, was recently identified as a novel methylated tumor suppressor gene.111 These examples highlight the importance of performing quantitative analyses of network measures as part of GGM interpretation.
6.6 |. Inference
It is important to recognize that the network estimated and analyzed above is a point estimate—at this stage in the tutorial, we have not yet performed any statistical inference on the presence or absence of edges in the GGM. Here, we will present methods for edge-wise inference using the SILGGM package and methods for understanding variability of edge weight estimates and comparing two different edge weights using the bootnet package.
6.6.1 |. With SILGGM
The four different methods described above for GGM inference57–60 are all implemented in the R package SILGGM. Each of the algorithms described here returns an undirected, unweighted network corresponding to an underlying GGM. We can easily apply all of them with similar syntax. Note that for the method of Liu,57 we have two versions: one using regular Lasso and one using scaled Lasso. We refer to these as liu2013 and liu2013scaled in the code below.
fdr = 0.05
liu2013 = SILGGM(x = as.matrix(ocDataSmall), method = “GFC_L”,
alpha = fdr)
liu2013scaled = SILGGM(x = as.matrix(ocDataSmall),
method = “GFC_SL”, alpha = fdr)
ren2015 = SILGGM(x = as.matrix(ocDataSmall), method = “B_NW_SL”,
alpha = fdr, global = T)
jankova2015 = SILGGM(x = as.matrix(ocDataSmall), method = “D-S_GL”,
alpha = fdr, global = T)
jankova2017 = SILGGM(x = as.matrix(ocDataSmall), method = “D-S_NW_SL”,
alpha = fdr, global = T)
A quick look at the five different binary networks estimated with these functions shows there is some variation between the results:
Each of the methods presented here provides theoretical properties about edge selection that are based on detailed assumptions. Researchers should refer to the original papers describing these methods in order to determine if assumptions are plausible in their research setting; however, as some of the authors note, it is often difficult or impossible to verify an assumption in practice. Here, we proceed with the results of Jankova 2017, which does not require an irrepresentability condition on the edge set.
Although edges in the binary network have meaningful statistical interpretations, they do not offer any sense of magnitude or direction of partial correlations in the way that the weighted edges in the weighted GGM do. The SILGGM implementations of the Ren 2015, Jankova 2015, and Jankova 2017 methods do estimate a precision matrix, but these estimates are not sparse (no entries are exactly zero). Consequently, it can be difficult to identify patterns of interest in the corresponding GGMs.
To overcome this shortcoming, we propose using SILGGM in practice to validate patterns observed in the sparse GGM obtained from one of the regularized methods. For example, the gene WT1 appears be a hub in an interesting subnetwork of the weighted network estimated above with huge:
contNbhd = neighborhood(ggmConnected, nodes = which(V(ggmConnected)$name ==
“WT1”))
contGraph = induced_subgraph(ggmConnected, contNbhd)
plot(contGraph, vertex.size = 40, vertex.label.cex = 0.5,
edge.width = 20 abs(E(contGraph)$weight), edge.color = ifelse(E(contGraph)$weight >
0, “darkred”, “blue”), layout = layout_with_graphopt)
If we want to make an inferential statement about this subnetwork, we can look at the results of the edgewise hypothesis tests in the results from SILGGM. For consistency, we use the Jankova 2015 graphical lasso-based method and rerun SILGGM with the value of lambda that we selected for the weighted GGM:
jankova2015 = SILGGM(x = as.matrix(ocDataSmall), method = “D-S_GL”,
alpha = 0.05/(156 155/2), global = T, lambda = 0.322)
Use method ′ “D-S_GL” ′
In this case, lambda = 0.322
Center each column.
Pre-calculate inner product matrixes.
Calculate graphical Lasso.
Perform global inference.
Use prespecified level(s): 4.13565e-06
True graph is not available.
Next, we extract the neighborhood of WT1 in the Jankova 2015 inference graph. We do this by thresholding based on edge-wise Bonferroni-corrected p-values presented in the estimation results, consistent with the application shown in Reference 59:
inferenceGraph = graph_from_adjacency_matrix(jankova2015$p_precision <
0.05/(156 155/2), mode = “undirected”, diag = F)
V(inferenceGraph)$name = names(ocDataSmall)
wt1Nbhd = neighborhood(inferenceGraph, nodes = which(V(inferenceGraph)$name ==
“WT1”))
wt1Graph = induced_subgraph(inferenceGraph, wt1Nbhd)
plot(wt1Graph, vertex.size = 40, vertex.label.cex = 0.5,
layout = layout_with_graphopt)
Although Jankova 2015 uses the graphical lasso as a foundation, due to the de-sparsifying procedure, the neighborhood of WT1 in this binary network may be quite different from that from the weighted GGM. To consider them together, we take their union using the %u% operator from igraph:
overlayGraph = wt1Graph %u% contGraph
bothNames = intersect(V(wt1Graph)$name, V(contGraph)$name)
binaryOnly = setdiff(V(wt1Graph)$name, V(contGraph)$name)
weightedOnly = setdiff(V(contGraph)$name, V(wt1Graph)$name)
plot(overlayGraph, vertex.size = 40, vertex.label.cex = 0.5,
vertex.color = ifelse(V(overlayGraph)$name %in%
bothNames, "purple", ifelse(V(overlayGraph)$name %in%
binaryOnly, "red", "blue")), vertex.label.color = "black",
layout = layout_with_graphopt)
legend("topleft", legend = c("Weighted and Binary",
"Binary Only"), col = c("purple", "red"), pch = 20,
cex = 0.5)
This graph shows that all of the six genes in the neighborhood of WT1 in the weighted GGM have statistically significant partial correlation with WT1 according to the Jankova 2015 method (SH3BP5, C3orf52, ABCC3, PTGS1, RNF128). There are also several genes which the binary network reports to have a statistically significant partial correlation with WT1 that did not appear in the weighted GGM, for the reasons mentioned above.
We again emphasize that the inference here is only valid under the assumptions presented in each of the corresponding papers. These assumptions can be difficult to verify in practice, so inferential results should be interpreted with caution.
6.6.2 |. With bootnet
The methods presented in SILGGM are based on theoretical derivations and principles of FDR control, and lead to conclusions about edge presence or absence. The bootnet package takes a different approach that utilizes the bootstrap to assess variability of estimated edge weights. Via resampling, bootnet is able to estimate quantile-based or Wald-type confidence intervals for the edges in the GGM. The authors of the bootnet package note that these confidence intervals should not be used to determine presence or absence of an edge, but that they can be used to generally understand the magnitude of edge weight variability and to compare partial correlations between two edges.9
Here, we apply bootnet with the EBICglasso method to explore such questions. The code chunk below is time-consuming. If you would like to try a quicker version, change the nBoots variable to something smaller for exploration; as with any bootstrap process, large values of nBoots should be used in actual analyses.
bn = bootnet∷bootnet(as.matrix(ocDataSmall), nBoots = 100,
default = "EBICglasso", nlambda = 30, lambda.min.ratio = 0.1,
tuning = 0.25, type = "nonparametric", model = "GGM")
| |
| |
The results from bootnet enable us to construct bootstrap-based confidence intervals for a partial correlation using the bootTable attribute of the output. This table stores the values of each edge weight over the bootstrap samples conducted, and includes the necessary information to estimate standard deviation of the edge weight sampling distribution. Below, we focus on the edge between genes ESRP1 and EPCAM:
esrp1_to_epcam = bn$bootTable %>%
filter(node1 == "EPCAM" & node2 == "ESRP1")
sd(esrp1_to_epcam$value)
0.06442792
We can use this information to construct approximate 95% quantile or Wald-type confidence intervals; here, we show both:
ciLowerQuantile = quantile(esrp1_to_epcam$value, 0.025)
ciUpperQuantile = quantile(esrp1_to_epcam$value, 0.975)
ciLowerQuantile
2.5%
0.04190299
ciUpperQuantile
97.5%
0.2819411
pointEstimate = bn$sampleTable %>%
filter(node1 == "EPCAM" & node2 == "ESRP1")
ciLowerWald = pointEstimate$value − 1.96 sd(esrp1_to_epcam$value)
ciUpperWald = pointEstimate$value + 1.96 sd(esrp1_to_epcam$value)
ciLowerWald
0.08661881
ciUpperWald
0.3391762
Note that if many confidence intervals are to be investigated, multiple hypothesis testing should be considered. While a simple adjustment such as a Bonferroni correction can be implemented, there are difficulties related to the dependence between the multiple tests and this is an area for future work.9
bootnet can also help us compare partial correlations by using the distribution of edge weights in the bootstrap samples. For example, the gene ESRP1 has several strong positive connections in our network, one to EPCAM, one to TPD52, and one to ERBB3.
It appears from the graph that the ESRP1-EPCAM and ERSP1-TPD52 edges are of similar magnitude, and that there are also some weaker connections such as the ESRP1 connection to F11R. We can formally test hypotheses like these using the differenceTest function in bootnet. First, we show that there is no significant difference between the weight of the ESRP-TPD52 edge and the ESRP1-ERBB3 edge:
differenceTest(bn, x = "ESRP1", x2 = "TPD52", y = "ERBB3",
y2 = "ESRP1", measure = "edge")
id1 id2 measure lower upper
1 ESRP1--TPD52 ERBB3--ESRP1 edge -0.1866988 0.07393816
significant
1 FALSE
Next, when we compare ESRP1-F11R and ESRP1-ERBB3, two edges that appear quite different in edge weight on the graph, we see that a statistically significant difference is detected:
differenceTest(bn, x = "ESRP1", x2 = "F11R", y = "ERBB3",
y2 = "ESRP1", measure = "edge")
Expected significance level given number of bootstrap samples is approximately:
0.05
id1 id2 measure lower upper
1 ESRP1-F11R ERBB3--ESRP1 edge 0.02863092 0.1920666
significant
1 TRUE
6.7 |. Community detection
Community detection, as discussed earlier in this article, is one useful tool for interpreting a graph once it has been estimated. We note that community detection is a general technique that can be used on a broad range of graphs, not solely GGMs. The igraph package provides an easy interface for several different community detection methods. Here, we demonstrate the cluster_walktrap algorithm,26 labeling nodes with degree > 6.
ggmComms = cluster_walktrap(ggmConnected, weights = abs(E(ggmConnected)$weight))
plot(ggmComms, ggmConnected, vertex.size = degree(ggmConnected),
vertex.label = ifelse(degree(ggmConnected) > 6,
V(ggmConnected)$name, NA), vertex.label.cex = 0.5,
vertex.label.color = "black", layout = thisLayout)
We can see that EPCAM is part of a small dark blue community. If we want to look into its community further, we can extract the community identifier using the membership and names attributes of the community detection object ggmComms:
ggmComms$membership[which(ggmComms$names == "EPCAM")]
5
This shows us that Community 5 is the identifier for EPCAM’s community in this network. Next, we extract the induced subgraph of the genes in Community 5 (ie, the subgraph formed by all of these genes and any edges between them):
epcamCommunity = induced_subgraph(ggmConnected, V(ggmConnected)[which(ggmComms$membership ==
5)])
plot(epcamCommunity, vertex.size = 40, vertex.label.cex = 0.5,
edge.width = 20 abs(E(ggmConnected)$weight),
edge.color = ifelse(E(ggmConnected)$weight > 0,
"darkred", "blue"), layout = layout_with_graphopt)
Looking at the community level can be a helpful way to understand which nodes in the network are most closely associated with a particular node (in this case, gene) of interest, which can lead to further biological insights. It is important to note that community detection algorithms can produce highly variable results, so we encourage the user to undertake a thorough sensitivity analysis. This topic is beyond the scope of the current tutorial; see, for example, Fortunato and Hric.23
6.8 |. Comparison with the adaptive graphical lasso, SCAD, and MCP alternative penalties using GGMncv
Finally, we demonstrate the difference in estimated GGMs with the use of penalties other than the ℓ 1 (LASSO) penalty, as discussed in Section 3.1.2. The GGMncv package provides the ggmncv function which allows us to easily fit several different penalties.
First, as a benchmark, we fit a GGM with the LASSO penalty using the same tuning parameter selected above using the StARS criterion (λ = 0.322). (While ggmncv also provides an option for selecting the tuning parameter using a generalized information criterion, here our focus is on comparing penalties with the same value of λ and so we do not explore this functionality; interested readers should refer to Williams44 for details.)
lassoGGM = ggmncv(cor(ocDataSmall), n = nrow(ocDataSmall),
penalty = "lasso", # choice of penalty function
progress = FALSE, # don’t show a progress bar
Y = ocDataSmall, # input data
lambda = 0.322) # penalty parameter
Next, we fit GGMs with the adaptive graphical lasso, the SCAD penalty, and the MCP. Again, we use the same value of λ = 0.322 for purposes of comparison between these approaches; in practice, the optimal value of λ is specific to each penalty type and the appropriate penalty must be used in tuning parameter selection.
adaptiveGGM = ggmncv(cor(ocDataSmall), n = nrow(ocDataSmall),
penalty = "adapt", progress = FALSE, Y = ocDataSmall,
lambda = 0.322)
scadGGM = ggmncv(cor(ocDataSmall), n = nrow(ocDataSmall),
penalty = "scad", progress = FALSE, Y = ocDataSmall,
lambda = 0.322)
mcpGGM = ggmncv(cor(ocDataSmall), n = nrow(ocDataSmall),
penalty = "mcp", progress = FALSE, Y = ocDataSmall,
lambda = 0.322)
A pairwise comparison of coefficients shows that, among coefficients that are nonzero in at least one of the pair of models, coefficients with larger magnitude are typically shrunk less by the SCAD and the MCP penalties but the same phenomenon is not observed as strongly in the adaptive graphical lasso:
It is worth mentioning again that the optimal λ is a function of the penalty type and that any cross-validation or other approaches for selecting λ should be performed using the desired penalty of the final model. The ggmncv function’s built-in select option is a convenient tool for performing such selection.44
7 |. DISCUSSION
The GGM is an important tool for studying conditional dependence relationships in complex datasets. Because the precision matrix of a multivariate normal distribution can be directly used to determine the corresponding GGM, estimating a GGM is equivalent to the problem of estimating the precision matrix. While inverting the sample covariance is an option in low-dimensional cases, such an inverse does not always exist and it is often necessary to use regularized approaches to estimate a precision matrix and its corresponding GGM. A common strategy for regularization is the use of an ℓ 1 penalty on the estimated precision matrix, such as in the graphical lasso. Known limitations in the ℓ 1 penalty have inspired other penalty choices such as the SCAD and MCP penalties discussed above. Regularized estimation requires tuning parameter selection, which should be done in a data-driven manner. Cross-validation, the RIC, the StARS, and the eBIC are four examples of criteria for tuning parameter selection.
GGM estimation does not focus on inferential questions such as measures of uncertainty and hypothesis testing. These questions have been approached from two broad viewpoints: asymptotic theory based on large-sample properties of precision matrix estimators and bootstrap methods. Asymptotic theory yields straightforward calculations, but often requires conditions that are difficult or impossible to verify in practice. Bootstrap-based approaches typically don’t require such conditions, but can be prohibitively time-consuming. While neither approach is perfect, it is not uncommon to see networks presented in literature without any treatment of uncertainty. As with all estimates, uncertainty is key to interpretation of estimated GGMs. We urge researchers to report measures of uncertainty and inferential results when publishing network-based analyses.
In Section 6, we have shown a basic GGM workflow using a variety of freely available R packages, including estimation with huge, tuning parameter selection with huge.select, inference with SILGGM and bootnet, and visualization with igraph. The software that we have demonstrated has been chosen for its ease of use and its accessibility: all R packages were available on CRAN at the time of this writing.112 Although the goal of our Demonstration here is to present one complete workflow for analysis, it is important to note that many methods exist for GGM estimation, and within each method, the researcher needs to make further decisions regarding model selection (range of tuning parameters to consider, criterion for selecting the tuning parameters). Recent work has shown that the estimated GGM can be highly sensitive to these choices.113,114 Using the default parameters in a given software package may not be the right fit for a particular research situation, and it is critical that researchers try various approaches for model selection and develop a rationale for the chosen approach for the final model. To facilitate such exploration from a machine-learning perspective, our group has recently introduced an ensemble approach that uses a likelihood-based loss function to learn a consensus network from a user-defined library of GGM estimation tools.106
Visualization of a GGM may be helpful for generating hypotheses about the relationships between the variables included in the network. Tools such as igraph99 and vis.js101 provide options for graph layouts that can help the researcher identify aspects such as hub nodes, connected components, and clusters. With large numbers of variables, it may become difficult to see these kinds of structures through visual inspection. In this case, it may be helpful to use tools such as the three node importance measures introduced above (hub centrality, betweenness centrality, and closeness centrality) or to calculate network-level measures (such as sparsity and degree distribution) in order to interpret the network and generate further hypotheses.
As accessibility, affordability, and scalability of omics technologies improve, it is becoming increasingly common to obtain more complex omics datasets across a variety of conditions. There is a commensurate need for the development of new analysis techniques. For example, Ni et al115 demonstrate the use of matrix-variate GGMs to analyze a complex set of proteomic data consisting of the expression of 50 proteins measured across 31 ovarian cancer cell lines over 3 conditions. The level of each protein is a 31 × 3 matrix-valued random variable, and the matrix-variate GGM models conditional dependencies between protein levels based on the sparsity patterns of the row and column covariance matrices in the corresponding matrix-valued Gaussian distribution. Many modern lines of investigation consider the collection of multiple omics types on the same sample set, followed by the application of integrative methods for analysis (“multi-omic integration”). In one approach to this setting, Weighill et al116 estimate multi-omic GGMs using a method called DRAGON (Determining Regulatory Associations using Graphical Models on Multi-Omic Networks). A third interesting setting is the case of multi-level data, where one wishes to construct a graphical model of variables that exist in a hierarchy. An example of multi-level data in an omics context is genes and pathways; each gene belongs to one or more pathways, and one can estimate a two-level graphical model consisting of within-pathway edges between genes and between-pathway edges between pathways.117,118
We close by noting that while GGMs are not a particularly new concept, their utility has been re-emphasized as technology has evolved over the last several decades. In the early 21st century, advances in omics and other big data outpaced existing statistical methods; the development of regularization-based approaches such as the graphical lasso was a transformative step forward. As the breadth and nature of omics data continues to expand, further innovations in GGM methodology have been inspired. The complexity of the data and methodology, however, need not be matched by the complexity of the analysis: we have demonstrated in this tutorial that a basic GGM workflow can be accomplished with a few lines of open-source code, and user-friendly tools are rapidly being developed to address more advanced questions. As the development of new technologies and data-mining methodologies continue to reinforce each other in the coming years, GGMs hold promise as a key tool for turning the growing wealth of complex omics data into meaningful biological discoveries.
Supplementary Material
supinfo
NIHMS1830332-supplement-supinfo.pdf (534.1KB, pdf)
FUNDING INFORMATION
Research reported in this publication was supported by the U.S. Natural Library of Medicine of the National Institutes of Health under award number R01LM013444-01.
U.S. National Library of Medicine, Grant/Award Number: 1R01LM013444-01
Abbreviations:
DAG
directed acyclic graph
DiNGO
differential network analysis in genomics
eBIC
extended Bayesian information criterion
FDR
false discovery rate
FWER
family-wise error rate
GGM
Gaussian graphical model
GHE
graphical horseshoe estimator
JGL
joint graphical lasso
LASSO
least absolute shrinkage and selection operator
MAGNA
meta-analytic Gaussian network aggregation
MCP
minimax concave penalty
PPS
pair-path subscore
RIC
rotation information criterion
SCAD
smoothly clipped absolute deviation
SNP
single nucleotide polymorphism
StARS
stability approach to regularization selection
TCGA
The Cancer Genome Atlas
Footnotes
The content is solely the responsibility of the authors and does not necessarily represent the official views of the National Institutes of Health.
Here, we use our notation for the partial correlation coefficient as introduced in Equation (2), which differs slightly from that introduced in Drton and Perlman55 but has the same interpretation.
SUPPORTING INFORMATION
Additional supporting information can be found online in the Supporting Information section at the end of this article.
DATA AVAILABILITY STATEMENT
The data used in this work are publicly available in the R package curatedOvarianData13 available for download from Bioconductor at
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4979 | https://www.iknowit.com/lessons/e-dividing-whole-numbers-and-unit-fractions.html | Interactive - 5th Grade Math Lesson | Dividing Whole Numbers and Unit Fractions
Dividing Whole Numbers and Unit Fractions
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Understanding Division with Whole Numbers and Unit Fractions
Dividing whole numbers by unit fractions, and vice versa, is a key concept in 5th grade math that helps students deepen their understanding of how division and fractions work together. A unit fraction has a numerator of 1, and learning how to divide by or into these fractions teaches students to think critically about part-whole relationships and the meaning of reciprocal values. This concept has practical applications in real life, such as dividing ingredients in recipes or understanding how many portions a whole can be split into. Mastering this skill prepares students for more advanced fraction operations in middle school.
How This Topic Is Taught in the Classroom
In the 5th grade classroom, teachers introduce the division of whole numbers and unit fractions through models like fraction strips, number lines, and visual representations. Students explore how dividing by a fraction results in a larger number, which can be a surprising and important discovery. Lessons often include word problems, hands-on practice, and opportunities to discuss strategies for solving these problems. Teachers reinforce this skill through class discussions, guided practice, and regular homework or classwork assignments that challenge students to apply their knowledge in different ways.
Reinforce Learning with Interactive Math Practice
Online learning tools like I Know It offer a powerful way to reinforce this important 5th grade skill through engaging and interactive lessons. This educational game-style practice provides immediate feedback, helpful hints, and animated encouragement to keep students motivated. Whether used for in-class review, homework, or independent study, this interactive lesson supports elementary teaching goals and helps students build confidence in dividing whole numbers and unit fractions. By combining digital tools with classroom instruction, learners gain a deeper and more lasting understanding of complex fraction and division concepts.
Level
This interactive math lesson is categorized as Level E and will help 5th grade students learn and grow advanced division skills.
Common Core Standard
5.NF.7, MA.5.FR.2.4, MA.5.AR.1.3, 5.3J
Number And Operations - Fractions
Apply And Extend Previous Understandings Of Multiplication And Division To Multiply And Divide Fractions.
Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.1
You might also be interested in...
Adding Mixed Numbers with Unlike Denominators (Level E)
Try out this fifth grade level math lesson for adding mixed numbers with unlike denominators practice with your class today!
Dividing Whole Numbers and Fractions (Level E)
Try out this fifth grade level math lesson for dividing whole numbers and fractions practice with your class today!
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Tools & Reference>Nephrology
Acute Pyelonephritis Treatment & Management
Updated: Aug 20, 2025
Author: Ahmed I Kamal, MD, MSc, PhD, FISN; Chief Editor: Vecihi Batuman, MD, FASN more...;)
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Acute Pyelonephritis
Sections Acute Pyelonephritis
Overview
Background
Pathophysiology
Etiology
Epidemiology
Prognosis
Patient Education
Show All
Presentation
History
Physical Examination
Complications
Show All
DDx
Workup
Approach Considerations
Collection of Urine Specimens
Urinalysis
Urine and Blood Cultures
Indications for Imaging Studies
Computed Tomography
Magnetic Resonance Imaging
Ultrasonography
Scintigraphy
CT and MR Urography
Diagnosis of Papillary Necrosis
Histologic Findings
Show All
Treatment
Approach Considerations
Antibiotic Selection
Outpatient Treatment
Inpatient Treatment
Abscesses
Calculi-Related Infections
Other Complicated Infections
Pregnant Patients
Pediatric Patients
Consultations
Diet
Activity
Prevention
Show All
Guidelines
NICE Guidelines
IDSA Guidelines
Show All
Medication
Medication Summary
Sulfonamides
Fluoroquinolones
Cephalosporins, Second Generation
Cephalosporins, Third Generation
Cephalosporins, 4th Generation
Penicillins, Amino
Penicillins, Extended-Spectrum
Carbapenems
Aminoglycosides
Glycopeptides
Monobactams
Urinary Analgesics
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Media Gallery;)
Tables)
References;)
Approach Considerations
Ambulatory younger women who present with signs and symptoms of uncomplicated acute pyelonephritis may be candidates for outpatient therapy. They must be otherwise healthy and must not be pregnant. In addition, they must be treated initially in the emergency department (ED) with vigorous oral or intravenous (IV) fluids, antipyretic pain medication, and a dose of parenteral antibiotics. Studies have shown that outpatient therapy for selected patients is as safe as inpatient therapy for a comparable group of patients and is much less expensive.
Use analgesics as needed. Early in the course of the illness, parenteral analgesics are often necessary to reduce morbidity from symptoms. Nonsteroidal anti-inflammatory drugs and narcotics are complementary; do not assume that one class is better than the other.
Admission is usually appropriate for patients who are severely ill, pregnant, or elderly or who have comorbid disorders that increase the complexity of management or the complication rate (eg, diabetes mellitus, chronic lung disease, congenital or acquired immunodeficiency). Admission may also be advisable for patients whose social situation is unstable, because of the possibility of poor compliance or poor follow-up.
Emergency surgery may be indicated in a patient with fever or positive blood culture results persisting longer than 48 hours; in a patient whose condition deteriorates; or in a patient who appears toxic for longer than 72 hours. These patients may have an abscess, emphysematous pyelonephritis, or an obstructing calculus. The etiology may not be immediately evident, but an unexpected change in the clinical picture warrants immediate evaluation for potential surgical intervention.
After recovery from the acute infection, patients may be candidates for elective surgery to reverse conditions that predispose to recurrent kidney infections and damage, such as the following:
Congenital anomalies
Fistulae involving the urogenital tract
Prostatic hyperplasia
Renal calculi
Vesicoureteral reflux
Next:
Antibiotic Selection
Antibiotic selection is typically empirical, because the results of blood or urine cultures are rarely available by the time a decision must be made. Initial selection should be guided by local antibiotic resistance patterns. Culture results from specimens collected before the initiation of therapy should be checked in 48 hours to determine antibiotic efficacy.
The pathogen in community-acquired infections is usually Escherichia coli or other Enterobacteriaceae. Acceptable regimens may include fluoroquinolones, cephalosporins, penicillins, extended-spectrum penicillins, carbapenems, and aminoglycosides. [22, 23] If enterococci are suggested on the basis of Gram stain results, ampicillin or vancomycin can replace the fluoroquinolone. If any doubt exists as to the diagnosis, coverage of both Enterobacteriaceae and enterococci is acceptable.
There is a higher incidence of enterococcal infections in hospitalized and other institutionalized patients. Ampicillin or amoxicillin should be included in the regimen. If the patient is allergic to penicillin, vancomycin should be substituted.
In choosing an empirical antibiotic regimen, consideration should include the local antibiogram and drug-resistance rates. For example, in a community with growing fluoroquinolone resistance, agents in that class may not be an ideal first-line choice. In light of increasing resistance, short courses of treatment are preferred. In one clinical trial, a 7-day course of oral ciprofloxacin was shown to be a safe and successful treatment for acute pyelonephritis in women, including older women and those with more severe infection.
For uncomplicated pyelonephritis, the American College of Physicians (ACP) recommends administering a short course of fluoroquinolones (5 to 7 days) or trimethoprim-sulfamethoxazole (TMP-SMX; 14 days), based on antibiotic susceptibility.
Patient characteristics should also be considered. For example, patients who have been frequently exposed to antibiotics (eg, solid-organ transplant and hematopoietic transplant patients) or are from institutional facilities are at a greater risk for infection with drug-resistant pathogens, such as extended-spectrum beta-lactamaseproducing or carbapenemase-producing organisms.
Oral versus parenteral administration
Growing data suggest that oral antibiotic therapy, parenteral antibiotic therapy, and initial parenteral antibiotic therapy followed by oral antibiotic therapy are equally effective regimens, although most of the studies have been small. [26, 27] Some clinicians believe that initiating therapy with an intravenous or intramuscular dose of medication reduces the risk of therapeutic failure; other clinicians believe that a completely oral course is sufficient. Data exist to support both assertions.
One conventional regimen comprises levofloxacin, 500 mg/day, given intravenously and then orally for 7-14 days. A short-course regimen of intravenous levofloxacin at 750 mg/day for 5 days proved non-inferior to that conventional regimen in a prospective, open-label, randomized, controlled clinical trial in 317 Chinese patients with complicated urinary tract infections and acute pyelonephritis.
To be considered for oral therapy, patients must meet several prerequisites. They must, of course, be able to tolerate oral medication. In addition, they must have no indication for admission, and close monitoring to ensure good compliance must be possible.
One prospective study supports using oral therapy alone in patients who can tolerate oral intake, lack signs of sepsis, and do not show signs of obstruction or kidney suppuration on urinary tract ultrasonography. Another study (prospective, randomized, unblinded) in a controlled hospital setting found no difference in efficacy between oral and intravenous ciprofloxacin for the initial management of severe pyelonephritis.
The 2010 guidelines from the Infectious Diseases Society of America (IDSA) recommend that women with pyelonephritis who require hospitalization be treated initially with an intravenous antimicrobial regimen. The choice of antimicrobial agents should be based on local resistance data, with the regimen tailored on the basis of susceptibility results.
Although the guidelines recommend either 14 days of TMP-SMX or 7 days of ciprofloxacin for the treatment of pyelonephritis, a study in 272 women with susceptible E coli pyelonephritis reported similar clinical outcomes with 7 days of TMP-SMX therapy compared with 7 days of ciprofloxacin.
Because of the high rate of resistance of E coli, the empirical use of TMP-SMX should be avoided in patients who require hospitalization. If beta-lactam drugs and fluoroquinolones are contraindicated, administer aztreonam parenterally. As such, the patient will need to be admitted.
Regimens for complicated cases
With complicated acute pyelonephritis, treat patients parenterally until defervescence and improvement in the clinical condition warrants changing to oral antibiotics. Complete the course of therapy with an oral agent selected on the basis of culture results. Acceptable regimens include the following:
Ampicillin and an aminoglycoside
Cefepime
Imipenem
Meropenem
Piperacillin/tazobactam
Cefepime/enmetazobactam
Ticarcillin/clavulanate
Ceftazidime/avibactam [30, 31]
Meropenem/vaborbactam
If the patient is allergic to penicillin, vancomycin should be substituted. Vancomycin or linezolid are options if enterococci are a consideration.
Also, see Guidelines for recommendations from the Infectious Diseases Society of America.
Cefepime/enmetazobactam (Exblifep) is indicated for treatment of adults with complicated urinary tract infections UTIs), including pyelonephritis, caused by E coli, Klebsiella pneumoniae, Pseudomonas aeruginosa, Proteus mirabilis, and Enterobacter cloacae complex. Approval of cefepime/enmetazobactam was based on the phase 3 ALLIUM trial, in which 79.1% (273/345) of patients receiving cefepime/enmetazobactam achieved clinical cure and microbiologic eradication, compared with 58.9% (196/333) of those receiving piperacillin/tazobactam.
Meropenem/vaborbactam (Vabomere) is a combination of the carbapenem antibiotic meropenem with the beta-lactamase inhibitor vaborbactam. Specifically, vaborbactam inhibits bacterial production of the K pneumoniae carbapenemase (KPC) enzyme, which confers resistance to carbapenems. The US Food and Drug Administration (FDA) has approved meropenem/vaborbactam for adults aged 18 years and older with pyelonephritis and other complicated UTIs caused by designated susceptible Enterobacteriaceae: E coli, K pneumoniae, and Enterobacter cloacae species complex.
The safety and efficacy of meropenem/vaborbactam were demonstrated in a study of more than 500 adults with complicated UTI, including pyelonephritis, in which cure or improvement in symptoms and a negative urine culture were seen in about 98% of patients treated with meropenem/vaborbactam, versus about 94% of patients treated with piperacillin/tazobactam. Approximately 7 days after completing treatment, about 77% of patients treated with meropenem/vaborbactam demonstrated resolution of symptoms and a negative urine culture, compared with about 73% of patients treated with piperacillin/tazobactam.
Previous
Next:
Outpatient Treatment
Antibiotic therapy
Patients presenting with acute pyelonephritis can be treated with a single dose of a parenteral antibiotic followed by oral therapy, provided they are monitored within the first 48 hours. In a study of febrile, nonpregnant women presenting with symptoms of acute pyelonephritis, 25% were hospitalized; of nonhospitalized patients, 80% were treated with a single parenteral dose of ceftriaxone or gentamicin, followed by oral therapy (usually TMP-SMX). Twelve percent returned with persistent symptoms, most in the first day; most of those patients were admitted.
Acute pyelonephritis has customarily been treated with 14 days of antibiotics, and the IDSA guidelines maintain this recommendation for TMP-SMX and beta-lactam agents. However, evidence suggests that in young, healthy women who are receiving a fluoroquinolone, including ciprofloxacin, the course of treatment can be shortened to 7 days. Levofloxacin, 750 mg/day, can be given for 5 days. [34, 35] Young, healthy males should complete a 14-day course.
Outpatient treatment is appropriate for patients who have an uncomplicated infection that does not warrant hospitalization. Oral antibiotics are used to treat patients with mild to moderate illness. (See Table 2, below, for a description of outpatient treatments for pyelonephritis.)
Table 2. Outpatient Treatment for Pyelonephritis (Open Table in a new window))
| |
| First-line therapy |
| Ciprofloxacin (Cipro) 500 mg PO BID for 7d or Ciprofloxacin extended-release (Cipro XR) 1000 mg PO daily for 7d or Levofloxacin (Levaquin) 750 mg PO daily for 5d If local fluoroquinolone resistance is thought to be > 10%, administer a single dose of ceftriaxone (Rocephin) 1g IV or a consolidated 24-hour dose of an aminoglycoside (gentamicin 7 mg/kg IV ortobramycin 7 mg/kg IV or amikacin 20 mg/kg IV) |
| Second-line therapy |
| Trimethoprim/sulfamethoxazole 160 mg/800 mg (Bactrim DS, Septra DS) 1 tablet PO BID for 14d If trimethoprim/sulfamethoxazole is used when the susceptibility is not known, an initial single IV dose of the following may also be given: ceftriaxone 1 g IV or a consolidated 24-h dose of an aminoglycoside (gentamicin 7 mg/kg IV or tobramycin 7 mg/kg IV or amikacin 20 mg/kg IV) |
| Alternative therapy |
| Oral beta-lactams are not as effective for treating pyelonephritis; however, if they are used, administer a single dose of ceftriaxone 1 g IV or a consolidated 24-h dose of an aminoglycoside (gentamicin 7 mg/kg IV or tobramycin 7 mg/kg IV or amikacin 20 mg/kg IV) Amoxicillin-clavulanate (Augmentin) 500 mg/125 mg PO BID for 14d or Amoxicillin-clavulanate 250 mg/125 mg PO TID for 3-7d or Cefaclor 500 mg PO TID for 7d Should generally be avoided in elderly patients because of the risk of affecting kidney function. |
For female patients suspected of having acute pyelonephritis, the IDSA guidelines recommend sending urine for culture and susceptibility testing and then starting empirical antibiotic therapy.
The fluoroquinolones are well tolerated and quite effective. They are probably the outpatient antibiotic treatment of choice for pyelonephritis. They (along with TMP-SMX) do not eradicate the protective lactobacilli from the vagina. An important caveat for the use of fluoroquinolones in the elderly is their potential to cause aortic aneurysm or dissection; hypoglycemia; and a variety of neuropsychiatric symptoms, ranging from seizures to worsening of dementia.
In communities where the prevalence of resistance in uropathogens to fluoroquinolones is not known to be greater than 10%, the IDSA guidelines advise that patients who do not require hospitalization be treated with oral ciprofloxacin, 500 mg twice daily for 7 days, with or without an initial 400-mg dose of intravenous ciprofloxacin. If the uropathogens fluoroquinolone resistance is greater than 10%, an initial intravenous dose of a long-acting parenteral antimicrobial (eg, ceftriaxone, 1 g) is recommended.
On an individual basis, for persons with community-onset UTI and fever, a case-control study found that the risk of fluoroquinolone-resistant E coli infection increased if the patient had undergone recent hospitalization or urinary catheterization or if the patient had used a fluoroquinolone within the past 6 months.
Intravenous fluids
If oral intake is not tolerated, intravenous hydration is warranted. Intravenous fluids should include 1 L of 5% dextrose in saline to reverse any existing ketosis, regardless of whether ketones are detected in the urine. Additional intravenous hydration is accomplished with normal saline. Exercise caution regarding conditions that might be adversely affected by improper amounts of fluid, saline, or glucose.
Follow-up
If the patient is not admitted at the time of diagnosis, follow-up reevaluation is important in 1-2 days to be sure the patient is progressing properly. A good rule based on common sense is that if the managing clinician is concerned that the patient may not respond well to outpatient management but still thinks the patient deserves a trial at home, the initial follow-up visit should take place in 24 hours. If the clinician thinks that the patient will do quite well with outpatient management, the initial follow-up visit can take place in 48 hours.
If the patient thinks that he or she is not progressing well or is getting worse, the patient should be evaluated emergently for consideration for admission and intravenous antibiotics.
Continue supportive care by prescribing antiemetics, antipyretics, analgesics, and urinary tract analgesics as needed. Have the patient complete a 14-day course of oral antibiotics. Evidence suggests that when treating a young, healthy female, the course of treatment can be shortened to 7 days from 14 days, if the antibiotic being used is a fluoroquinolone. Healthy young males should complete a 14-day course.
Obtain follow-up urine culture results in any patient with a complicated UTI, a complicated course, or increased risk of infection. Urine cultures are generally not indicated in healthy, nonpregnant women with resolved symptoms. All patients with a complicated UTI should be considered for outpatient follow-up imaging of the urinary tract to identify abnormalities that predispose to further infections.
Rest is essential for recovery. Activity should be minimal. The patient should not return to work for 2 weeks, so as to allow time for the infection to be eliminated and for the patient to recover physical strength. Temper this recommendation depending on the physical condition of the patient and the presence of comorbid conditions.
Previous
Next:
Inpatient Treatment
The decision regarding admission of a patient with acute pyelonephritis depends on the following:
Age
Systemic factors such as immunocompromising chemotherapy or chronic diseases
Known urinary tract structural abnormalities
Renal calculi
Recent hospitalization or urinary tract instrumentation
Response to ED therapy
Admit all patients with complicated UTI. Complicating factors include the following:
Structural urinary tract abnormalities (eg, calculi, tract anomalies, indwelling catheter, obstruction)
Metabolic disease (eg, diabetes, kidney insufficiency)
Impaired host defenses (eg, HIV, current chemotherapy, underlying active cancer)
Pregnancy
Admission is also indicated for patients with apparent clinically uncomplicated pyelonephritis who have any of the following:
Inability to maintain adequate oral hydration
Evidence of vasomotor instability
Unrelenting fever despite antipyretic therapy
Debilitating pain or dehydration that cannot be corrected promptly in the ED
Inadequate home care or resources to fill prescriptions or comply with the medical regimen (eg, homeless patients, adolescents, elderly patients in an acute illness setting who are at risk for clouded judgment, patients with substance abuse issues or other issues that will prevent adequate compliance)
Inpatient care includes the following:
Continue supportive care
Monitor urine and blood culture results
Monitor comorbid conditions for deterioration
Maintain hydration status with IV fluids until hydration can be maintained with oral intake
Continue IV antibiotics until defervescence and significant symptomatic improvement occur; convert to an oral regimen tailored to urine or blood culture results
In patients with acute pyelonephritis who require hospitalization, treatment begins with IV antibiotics. IV therapy should be given for 24-48 hours or until severe symptoms improve. A systematic review of 8 randomized, controlled trials in hospitalized patients with acute pyelonephritis found that early switching to oral antibiotic treatment appears to be as effective and safe as exclusively IV regimens.
Selection of a regimen should be based on local resistance data, and the regimen should be tailored on the basis of susceptibility results. Recommended regimens are listed in Table 3, below. However, a multinational study in 184 patients found that cefazolin is non-inferior to ceftriaxone for the empirical treatment of acute pyelonephritis in hospitalized patients. At 72 hours, defervescence or normalization of white blood cell count had occurred in 87.0% of the cefazolin group versus 85.9% of the ceftriaxone group; in addition, no difference was observed between the two groups for length of stay or 30-day readmission for cystitis or pyelonephritis.
Table 3. Inpatient Treatment for Acute Pyelonephritis (Open Table in a new window))
| |
| First-line therapy |
| Ciprofloxacin (Cipro) 400 mg IV q12h for 10-14d or Levofloxacin (Levaquin) 250 mg IV q24h for 10d or Levofloxacin (Levaquin) 750 mg IV q24h for 5d |
| Second-line therapy |
| Extended-spectrum cephalosporins or penicillins: Ampicillin/sulbactam (Unasyn) 1.5 g IV q6h or Piperacillin/tazobactam (Zosyn) 3.375 g IV q6h or Cefepime/enmetazobactam 2.5 g IV q8h or Ticarcillin/clavulanate (Timentin) 3.1 g IV 4-6h or Cefotaxime (Claforan) 1-2 g IV q8h or Ceftriaxone (Rocephin) 1 g IV q24h or Ceftazidime (Fortaz, Tazicef) 2 g IV q8h All of the above can be administered with or without an aminoglycoside (except in pregnant patients); see Aminoglycosides, below Carbapenems: Meropenem (Merrem) 500 mg IV q8h or Ertapenem (Invanz) 1 g IV q24h or Monobactam (for patients with penicillin allergy): Aztreonam (Azactam) 1 g IV q8-12h |
| Alternative therapy |
| Aminoglycosides (because of their potential nephrotoxicity, aminoglycoside antibiotics should be reserved for patients with serious and potentially life-threatening infections, and the dosage and blood levels should be carefully monitored to minimize the risk of nephrotoxicity): Gentamicin 3 mg/kg/day IV/IM in 3 divided doses or 7 mg/kg/day pulsed dosing or Tobramycin 3 mg/kg/day IV/IM in 3 divided doses or 7 mg/kg/day pulsed dosing or Amikacin 10 mg/kg/day IV/IM in 3 divided doses or 20 mg/kg/day pulsed dosing or Plazomicin 15 mg/kg IV q24hr infused over 30 minutes |
However, an Israeli medical center that instituted a program of initial empirical treatment of pyelonephritis with aminoglycosides reported higher rates of in vitro activity and lower overall mortality than with non-aminoglycoside antibiotic therapy, without excess nephrotoxicity. In the study, which included 2026 patients, 715 treated with aminoglycosides and 1311 treated with non-aminoglycoside drugs, aminoglycosides were more likely to have in vitro activity against clinical isolates (odds ratio 2.0; P < 0.001); death at 30 days occurred in 55 (7.6%) versus 145 (11%) patients treated with aminoglycosides and non-aminoglycoside drugs, respectively (adjusted hazard ratio 0.78; P = 0.013). The incidence of acute kidney injury was 2.5% versus 2.9%, respectively; P = 0.6).
Duration of therapy
Duration of therapy should be at least 10-14 days, inclusive of initial IV therapy. Patient circumstances may warrant more prolonged therapy.
The PROGRESS (Procalcitonin-guided Antimicrobial Therapy to Reduce Long-Term Sequelae of Infections) study found that in septic patients, procalcitonin levels can be used to guide discontinuation of antibiotics. This multicenter randomized trial in 266 patients, 95 of whom had acute pyelonephritis, used an ¥80% reduction in procalcitonin level or any procalcitonin level of ¤0.5 μg/L at Day 5 or later as a criterion for discontinuing antibiotics.
In PROGRESS, procalcitonin-guided discontinuation, compared with standard of care, resulted in a shorter median length of antibiotic therapy (5 versus 10 days), a lower rate of infection-associated adverse events such as infections by multidrug-resistant organisms and Clostridioides difficile (7.2% versus 15.3%), and lower 28-day mortality (15.2% versus 28.2%). The cost of hospitalization was also reduced in the procalcitonin arm.
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Abscesses
For renal cortical abscesses (renal carbuncles), surgical drainage was once the only treatment. However, modern antibiotics alone often are curative. A semisynthetic penicillin, cephalosporin, fluoroquinolone, or vancomycin is recommended. Generally, parenteral antibiotics should be administered for 10-14 days, followed by oral therapy for 2-4 weeks. Fever should resolve within 5-6 days, and pain should resolve within 24 hours. If parenteral antibiotic therapy is successful, oral therapy is instituted for an additional 2-4 weeks.
If patients do not respond within 48 hours, percutaneous (or open) drainage should be performed. Other surgical options are enucleation of the carbuncle or nephrectomy.
Treatment of renal corticomedullary abscess includes parenteral antibiotics for 48 hours (usually successful), incision and drainage, and nephrectomy. If antibiotic therapy is successful, oral therapy is instituted for an additional 2 weeks.
Mortality associated with perinephric abscesses is 20-50%, but this rate can be decreased with early recognition, surgical drainage, and antimicrobial therapy (not adequate alone). Antibiotics should include an aminoglycoside and an antistaphylococcal agent. If Pseudomonas species are considered or demonstrated, an antipseudomonal beta-lactam antibiotic should be added. For enterococci, an aminoglycoside and ampicillin are recommended. Other organisms that have been reported include tuberculosis and fungi. Nephrectomy may be necessary.
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Calculi-Related Infections
A major challenge with calculi-related UTI is that the organisms can survive within the calculus. In the presence of acute infection, calculi must be removed immediately using cystoscopy or open surgical procedure. In a review of adult patients with obstructive pyelonephritis due to a ureteral stone or kidney stone with hydronephrosis (n=311,100), delayed decompression (2 or more days after admission) increased the odds of in-hospital death by 29% (odds ratio 1.29, 95% confidence interval 1.03-1.63, P = 0.032).
The preferred method of treatment is surgical. Options include extracorporeal shockwave lithotripsy (ESWL), endoscopic methods, percutaneous methods, and open surgery. For staghorn calculus, the treatment of choice is to remove the whole stone. Fragments left behind remain infected and will grow again. Antibiotic therapy should be used in conjunction.
Although food and vitamin supplements that are rich in phosphorus and magnesium are advisable, remember that magnesium (and other divalent cations) can chelate fluoroquinolones, preventing their absorption from the gut.
Acidifying agents have been used. Ascorbic acid does not significantly decrease urinary pH, and ammonium chloride provides only temporary acidification. This approach is of little clinical usefulness. Urease inhibitors are effective in reducing stone formation, but long-term use is fraught with neurosensory, hematologic, and dermatologic adverse effects.
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Other Complicated Infections
Treatment for renal papillary necrosis is admission for intravenous antibiotic therapy. The agents chosen should cover E coli and Enterobacter, Proteus, and Klebsiella species. Treatment for more serious infections also should cover Pseudomonas and Enterococcus species.
Parenteral agents that may be used empirically pending the result of a urine culture include the following:
Gentamicin
Cefotaxime
Ceftriaxone
Ceftazidime
Cefepime
Piperacillin-tazobactam
Imipenem-cilastatin
Meropenem
Ciprofloxacin
Parenteral therapy should be continued until the fever and other symptoms resolve. Duration of therapy generally is 14 days. If the response is not good, CT-guided drainage or surgical drainage with debridement is indicated.
The treatment of choice for xanthogranulomatous pyelonephritis is nephrectomy after the patient is stabilized. See Xanthogranulomatous Pyelonephritis for more information.
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Pregnant Patients
Physiologic changes in the urinary tract predispose pregnant women to an increased risk of cystitis and pyelonephritis, which may lead to preterm labor and kidney damage. Hydroureter of pregnancy develops around the seventh week and progresses throughout the remainder of pregnancy; it resolves by 8 weeks post partum. The ureters may dilate sufficiently to contain 200 mL or more of urine. In addition, the kidneys enlarge and bladder capacity may double. The left kidney is more affected than the right.
The prevalence of bacteriuria in pregnancy is 2-25%, depending on the study criteria. Symptomatic UTI occurs in 1-3% of all pregnancies and leads to premature labor in 20-50% of cases.
A screening urine culture is recommended for all pregnant women at 16 weeks' gestation. If the results are negative for a UTI, no additional cultures are indicated. If the patient has a history of recurrent UTIs, further cultures and other screening techniques (eg, nitrite dipstick or urine Gram stain) may be needed to detect the development of asymptomatic bacteriuria.
Successful treatment of bacteriuria prevents pyelonephritis. Fluoroquinolones and aminoglycosides should be avoided in pregnancy. Accepted regimens for treating asymptomatic bacteriuria include the following:
Amoxicillin 250 mg orally 3 times a day for 3 or 7 days; or 3 g in a single dose
Cephalexin 2 or 3 g in a single dose
Nitrofurantoin 200 mg in a single dose; or 100 mg 4 times a day for 3 or 7 days
Inpatient admission is warranted for any pregnant patient with pyelonephritis. The treatment of choice during pregnancy includes the use of beta-lactam antibiotics. Intravenous antibiotics should be administered until the patient is afebrile for 24 hours and symptomatically improved. It is recommended to avoid fluoroquinolones in pregnant patients. Aminoglycosides should also be avoided due to potential risk of ototoxicity following prolonged fetal exposure. See Table 4, below, for regimens for pyelonephritis in pregnant patients.
Table 4. Treatment of Pyelonephritis During Pregnancy (Open Table in a new window))
| |
| Mild to moderate pyelonephritis |
| Ceftriaxone (Rocephin) 1 g IV q24h or Cefepime (Maxipime) 1 g IV q12h or Cefotaxime (Claforan) 1-2 g IV q8h or Ceftazidime (Fortaz, Tazicef) 2 g IV q8h or Ampicillin 1-2 g IV q6h plus gentamicin IV 1.5 mg/kg q8h |
| Severe pyelonephritis |
| If the patient is immunocompromised and/or has incomplete urinary drainage: Ticarcillin-clavulanate (Timentin) 3.1 g IV q6h or Ampicillin-sulbactam (Unasyn) 1.5 g IV q6h or Piperacillin-tazobactam (Zosyn) 3.375 g IV q6h |
Once patients are afebrile for at least 48 hours, they can be switched to oral antibiotics, with the regimen choice guided by results from susceptibility studies, and discharged to complete 10-14 days of treatment. Obtain an additional urine culture 1-2 weeks after the completion of therapy to verify eradication of the infection. Obtain monthly urine cultures until delivery to monitor for recurrent infection.
Postcoital therapy with cephalexin or nitrofurantoin is recommended for prophylaxis against recurrent infection. If the initial infection requires a second agent for clearing the infection or a recurrent infection occurs, suppressive therapy until delivery is indicated with nitrofurantoin (50 mg or 100 mg at bedtime). Recurrent infection or persistent bacteriuria is an indication for urologic evaluation 3-6 months after delivery.
Caution should be used when prescribing nitrofurantoin for prophylaxis, as adverse effects of this medication include peripheral neuropathy, pulmonary toxicity, and hepatotoxicity. Nitrofurantoin-induced liver injury and death have been reported. Hepatotoxicity is reversible if recognized early and nitrofurantoin is stopped, so hepatic enzymes should be monitored in women at higher doses or with prolonged use.
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Pediatric Patients
In pediatric patients with acute pyelonephritis, indications for immediate urologic referral include the following:
Abnormal electrolyte values associated with acidosis
Elevated blood urea nitrogen level
Hypertension
A palpable bladder
Voiding difficulty (dribbling, poor stream, straining)
Aside from the effects of acute infection, the overriding concern is progressive deterioration in function of an already compromised kidney (hypoplastic or dysplastic) secondary to scarring from recurrent pyelonephritis with or without associated obstruction. Infants and preschool-aged children are at greatest risk. Initial management varies with patient age and presentation.
Close follow-up, regardless of whether the patient is initially admitted, is essential to ensure recovery. Immediate reevaluation is encouraged for any recurrence of symptoms. Neither treatment of asymptomatic bacteriuria nor long-term suppressive therapy has been found to be effective.
Urologic evaluation is necessary to establish the presence of any urologic abnormality. The preferred imaging study for the diagnosis of acute pyelonephritis is technetium-99m dimercaptosuccinic acid (99m Tc-DMSA) scintigraphy. Ultrasonography is the imaging study of choice for the diagnosis of urinary tract structural abnormalities. (See Workup/Scintigraphy).
Age-related data adapted from Harwood-Nuss and colleagues and Hansson and colleagues are presented below in Table 5. [43, 44]
Table 5. Pediatric Urinary Tract Infections (Open Table in a new window))
| | | | | |
--- ---
| | Neonates | Infants 6 weeks to 3 years of age | Children 3-6 years of age | Children 6-11 years of age |
| UTI frequency (%) | 1 | 1.5-3 | 1.5-3 | 1.2 |
| Female-to-male ratio | 1:1.5 | 10:1 | 10:1 | 30:1 |
| Route of infection | Blood | Ascending | Ascending | Ascending |
| Signs and symptoms | Failure to thrive, fever, hypothermia, irritability, jaundice, poor feeding, sepsis, vomiting | Diarrhea, failure to thrive, fever, irritability, poor feeding, strong-smelling urine, vomiting | Abdominal pain, dysuria, enuresis, fever, gross hematuria, meningismus, strong-smelling urine, urinary urgency, urinary frequency, vomiting | Dysuria, enuresis, fever, flank pain or tenderness, urinary urgency, urinary frequency |
| Predominant organism | Klebsiella species | E coli | E coli, Proteus species in older boys | E coli |
| Management | Admit for intravenous ampicillin and gentamicin and further evaluation | Admit for intravenous ampicillin and gentamicin and further evaluation | Follow adult guidelines, but avoid fluoroquinolones, which are theoretically contraindicated due to potential effects on the musculoskeletal system | Follow adult guidelines, but avoid fluoroquinolones, which are theoretically contraindicated due to potential effects on the musculoskeletal system |
See also Pediatric Pyelonephritis.
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Consultations
Consultation is indicated if the infection is complicated. Most cases of acute pyelonephritis occur in adult women and are readily managed without consultation. The following are reasons for consulting various subspecialists:
A urologist may be consulted regarding patients with ureteral or urethral obstruction, urinary stones, urogenital abnormality, or recurrent pyelonephritis, or for the first episode of pyelonephritis in an infant or child.
A nephrologist may be consulted regarding patients with acute kidney injury or advanced chronic kidney disease or for neonates or infants.
An infectious disease specialist may be consulted regarding patients with an unusual or resistant pathogen, those who are immunocompromised, patients with persisting fever (> 48 hours) or toxicity (> 72 hours), or patients whose blood culture results are positive for more than 48 hours.
An obstetrician may be consulted for patients who are pregnant.
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Diet
A regular diet is permitted as tolerated. Special diets, such as those for patients with diabetes mellitus, should be honored. Maintaining hydration is very important. If admission is not indicated and the patient will be monitored in an outpatient setting, hydration status should be normalized with intravenous fluids; the physician should not assume that the patient can or will accomplish this with oral hydration alone.
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Activity
Rest is essential for recovery. Activity should be minimized. Patients who are treated in an outpatient setting should not return to work for 2 weeks in order to allow time for the infection to be eliminated. This time also allows the patient to recover physical strength. This recommendation can be tempered in special circumstances as warranted by the clinician.
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Prevention
Prevention of pyelonephritis involves identifying clinical situations that could lead to pyelonephritis and developing a strategy to decrease that likelihood. These strategies may include a change in contraceptive behavior, administration of prophylactic antibiotics, or early identification and treatment of cystitis.
If these strategies do not eliminate infection, recurrence of infection, or relapse (reinfection less than 14 days after completing an appropriate regimen), the patient needs to undergo systematic evaluation for predispositional anatomic, functional, or structural abnormalities.
For more information, see Prevention of Urinary Tract Infection (UTI).
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Guidelines
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Media Gallery
Nonobstructing distal left ureteral calculus 2 X 1 X 2 cm.
Multiple abscesses, upper pole of left kidney.
Bilateral hydronephrosis.
of 3
Tables
Table 1. Bacterial Etiology of Urinary Tract Infections;)
Table 2. Outpatient Treatment for Pyelonephritis;)
Table 3. Inpatient Treatment for Acute Pyelonephritis;)
Table 4. Treatment of Pyelonephritis During Pregnancy;)
Table 5. Pediatric Urinary Tract Infections;)
Table 1. Bacterial Etiology of Urinary Tract Infections
| | | |
---
| Bacteria | % Uncomplicated | % Complicated |
| Gram negative | | |
| Escherichia coli | 70-95 | 21-54 |
| Proteus mirabilis | 1-2 | 1-10 |
| Klebsiella spp | 1-2 | 2-17 |
| Citrobacter spp | < 1 | 5 |
| Enterobacter spp | < 1 | 2-10 |
| Pseudomonas aeruginosa | < 1 | 2-19 |
| Other | < 1 | 6-20 |
| Gram positive | | |
| Coagulase-negative staphylococci | 5-10 | 1-4 |
| Enterococci | 1-2 | 1-23 |
| Group B streptococci | < 1 | 1-4 |
| Staphylococcus aureus | < 1 | 1-23 |
| Other | < 1 | 2 |
| Adapted from Hooton TM. The current management strategies for community-acquired urinary tract infection. Infect Dis Clin North Am. Jun 2003;17(2):303-32. [QxMD MEDLINE Link]. S saprophyticus |
Table 2. Outpatient Treatment for Pyelonephritis
| |
| First-line therapy |
| Ciprofloxacin (Cipro) 500 mg PO BID for 7d or Ciprofloxacin extended-release (Cipro XR) 1000 mg PO daily for 7d or Levofloxacin (Levaquin) 750 mg PO daily for 5d If local fluoroquinolone resistance is thought to be > 10%, administer a single dose of ceftriaxone (Rocephin) 1g IV or a consolidated 24-hour dose of an aminoglycoside (gentamicin 7 mg/kg IV ortobramycin 7 mg/kg IV or amikacin 20 mg/kg IV) |
| Second-line therapy |
| Trimethoprim/sulfamethoxazole 160 mg/800 mg (Bactrim DS, Septra DS) 1 tablet PO BID for 14d If trimethoprim/sulfamethoxazole is used when the susceptibility is not known, an initial single IV dose of the following may also be given: ceftriaxone 1 g IV or a consolidated 24-h dose of an aminoglycoside (gentamicin 7 mg/kg IV or tobramycin 7 mg/kg IV or amikacin 20 mg/kg IV) |
| Alternative therapy |
| Oral beta-lactams are not as effective for treating pyelonephritis; however, if they are used, administer a single dose of ceftriaxone 1 g IV or a consolidated 24-h dose of an aminoglycoside (gentamicin 7 mg/kg IV or tobramycin 7 mg/kg IV or amikacin 20 mg/kg IV) Amoxicillin-clavulanate (Augmentin) 500 mg/125 mg PO BID for 14d or Amoxicillin-clavulanate 250 mg/125 mg PO TID for 3-7d or Cefaclor 500 mg PO TID for 7d Should generally be avoided in elderly patients because of the risk of affecting kidney function. |
Table 3. Inpatient Treatment for Acute Pyelonephritis
| |
| First-line therapy |
| Ciprofloxacin (Cipro) 400 mg IV q12h for 10-14d or Levofloxacin (Levaquin) 250 mg IV q24h for 10d or Levofloxacin (Levaquin) 750 mg IV q24h for 5d |
| Second-line therapy |
| Extended-spectrum cephalosporins or penicillins: Ampicillin/sulbactam (Unasyn) 1.5 g IV q6h or Piperacillin/tazobactam (Zosyn) 3.375 g IV q6h or Cefepime/enmetazobactam 2.5 g IV q8h or Ticarcillin/clavulanate (Timentin) 3.1 g IV 4-6h or Cefotaxime (Claforan) 1-2 g IV q8h or Ceftriaxone (Rocephin) 1 g IV q24h or Ceftazidime (Fortaz, Tazicef) 2 g IV q8h All of the above can be administered with or without an aminoglycoside (except in pregnant patients); see Aminoglycosides, below Carbapenems: Meropenem (Merrem) 500 mg IV q8h or Ertapenem (Invanz) 1 g IV q24h or Monobactam (for patients with penicillin allergy): Aztreonam (Azactam) 1 g IV q8-12h |
| Alternative therapy |
| Aminoglycosides (because of their potential nephrotoxicity, aminoglycoside antibiotics should be reserved for patients with serious and potentially life-threatening infections, and the dosage and blood levels should be carefully monitored to minimize the risk of nephrotoxicity): Gentamicin 3 mg/kg/day IV/IM in 3 divided doses or 7 mg/kg/day pulsed dosing or Tobramycin 3 mg/kg/day IV/IM in 3 divided doses or 7 mg/kg/day pulsed dosing or Amikacin 10 mg/kg/day IV/IM in 3 divided doses or 20 mg/kg/day pulsed dosing or Plazomicin 15 mg/kg IV q24hr infused over 30 minutes |
Table 4. Treatment of Pyelonephritis During Pregnancy
| |
| Mild to moderate pyelonephritis |
| Ceftriaxone (Rocephin) 1 g IV q24h or Cefepime (Maxipime) 1 g IV q12h or Cefotaxime (Claforan) 1-2 g IV q8h or Ceftazidime (Fortaz, Tazicef) 2 g IV q8h or Ampicillin 1-2 g IV q6h plus gentamicin IV 1.5 mg/kg q8h |
| Severe pyelonephritis |
| If the patient is immunocompromised and/or has incomplete urinary drainage: Ticarcillin-clavulanate (Timentin) 3.1 g IV q6h or Ampicillin-sulbactam (Unasyn) 1.5 g IV q6h or Piperacillin-tazobactam (Zosyn) 3.375 g IV q6h |
Table 5. Pediatric Urinary Tract Infections
| | | | | |
--- ---
| | Neonates | Infants 6 weeks to 3 years of age | Children 3-6 years of age | Children 6-11 years of age |
| UTI frequency (%) | 1 | 1.5-3 | 1.5-3 | 1.2 |
| Female-to-male ratio | 1:1.5 | 10:1 | 10:1 | 30:1 |
| Route of infection | Blood | Ascending | Ascending | Ascending |
| Signs and symptoms | Failure to thrive, fever, hypothermia, irritability, jaundice, poor feeding, sepsis, vomiting | Diarrhea, failure to thrive, fever, irritability, poor feeding, strong-smelling urine, vomiting | Abdominal pain, dysuria, enuresis, fever, gross hematuria, meningismus, strong-smelling urine, urinary urgency, urinary frequency, vomiting | Dysuria, enuresis, fever, flank pain or tenderness, urinary urgency, urinary frequency |
| Predominant organism | Klebsiella species | E coli | E coli, Proteus species in older boys | E coli |
| Management | Admit for intravenous ampicillin and gentamicin and further evaluation | Admit for intravenous ampicillin and gentamicin and further evaluation | Follow adult guidelines, but avoid fluoroquinolones, which are theoretically contraindicated due to potential effects on the musculoskeletal system | Follow adult guidelines, but avoid fluoroquinolones, which are theoretically contraindicated due to potential effects on the musculoskeletal system |
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Contributor Information and Disclosures
Ahmed I Kamal, MD, MSc, PhD, FISN Assistant Professor in Transplant Nephrology, Medical University of South Carolina College of Medicine; Assistant Professor of Nephrology, Mansoura Urology and Nephrology Center, Mansour University Faculty of Medicine, EgyptAhmed I Kamal, MD, MSc, PhD, FISN is a member of the following medical societies: American Society of Nephrology, Egyptian Society of Nephrology and Transplantation, International Society of NephrologyDisclosure: Nothing to disclose.
Tibor Fulop, MD, PhD, FACP, FASN Professor of Medicine, Department of Medicine, Division of Nephrology, Medical University of South Carolina College of Medicine; Attending Physician, Medical Services, Ralph H Johnson VA Medical CenterTibor Fulop, MD, PhD, FACP, FASN is a member of the following medical societies: American Academy of Urgent Care Medicine, American College of Physicians, American Society of Hypertension, American Society of Nephrology, International Society for Apheresis, International Society for Hemodialysis, Magyar Orvosi Kamara (Hungarian Chamber of Medicine)Disclosure: Nothing to disclose.
Specialty Editor Board
Eleanor Lederer, MD, FASN Professor of Medicine, Chief, Nephrology Division, Director, Nephrology Training Program, Director, Metabolic Stone Clinic, Kidney Disease Program, University of Louisville School of Medicine; Consulting Staff, Louisville Veterans Affairs HospitalEleanor Lederer, MD, FASN is a member of the following medical societies: American Association for the Advancement of Science, American Society for Bone and Mineral Research, American Society of Nephrology, American Society of Transplantation, International Society of Nephrology, Kentucky Medical Association, National Kidney FoundationDisclosure: Serve(d) as a director, officer, partner, employee, advisor, consultant or trustee for: American Society of Nephrology
Received income in an amount equal to or greater than $250 from: Healthcare Quality Strategies, Inc.
Chief Editor
Vecihi Batuman, MD, FASN Professor of Medicine, Section of Nephrology-Hypertension, Deming Department of Medicine, Tulane University School of MedicineVecihi Batuman, MD, FASN is a member of the following medical societies: American College of Physicians, American Society of Hypertension, American Society of Nephrology, Southern Society for Clinical InvestigationDisclosure: Nothing to disclose.
Amy J Behrman, MD Associate Professor, Department of Emergency Medicine, Director, Division of Occupational Medicine, University of Pennsylvania School of Medicine
Amy J Behrman, MD is a member of the following medical societies: American College of Occupational and Environmental Medicine
Disclosure: Nothing to disclose.
Christopher Edwards, MD Resident Physician, Department of Emergency Medicine, University of Pennsylvania School of Medicine
Christopher Edwards, MD is a member of the following medical societies: American College of Emergency Physicians and Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Judith Green-McKenzie, MD, MPH Associate Professor, Director of Clinical Practice, Occupational Medicine Residency Director, University of Pennsylvania School of Medicine
Judith Green-McKenzie, MD, MPH is a member of the following medical societies: American College of Occupational and Environmental Medicine, American College of Physicians, American College of Preventive Medicine, National Medical Association, and Society of General Internal Medicine
Disclosure: Nothing to disclose.
Eleanor Lederer, MD Professor of Medicine, Chief, Nephrology Division, Director, Nephrology Training Program, Director, Metabolic Stone Clinic, Kidney Disease Program, University of Louisville School of Medicine; Consulting Staff, Louisville Veterans Affairs Hospital
Eleanor Lederer, MD is a member of the following medical societies: American Association for the Advancement of Science, American Federation for Medical Research, American Society for Biochemistry and Molecular Biology, American Society for Bone and Mineral Research, American Society of Nephrology, American Society of Transplantation, International Society of Nephrology, Kentucky Medical Association, National Kidney Foundation, and Phi Beta Kappa
Disclosure: Dept of Veterans Affairs Grant/research funds Research
Chike Magnus Nzerue, MD Associate Dean for Clinical Affairs, Vice-Chairman of Internal Medicine, Meharry Medical College
Chike Magnus Nzerue, MD is a member of the following medical societies: American Association for the Advancement of Science, American College of Physicians, American College of Physicians-American Society of Internal Medicine, American Society of Nephrology, and National Kidney Foundation
Disclosure: Nothing to disclose.
Suzanne Moore Shepherd, MD, MS, DTM&H, FACEP, FAAEM Associate Professor, Education Officer, Department of Emergency Medicine, Hospital of the University of Pennsylvania; Director of Education and Research, PENN Travel Medicine
Suzanne Moore Shepherd, MD, MS, DTM&H, FACEP, FAAEM is a member of the following medical societies: Alpha Omega Alpha, American Academy of Emergency Medicine, American Society of Tropical Medicine and Hygiene, International Society of Travel Medicine, Society for Academic Emergency Medicine, and Wilderness Medical Society
Disclosure: Nothing to disclose.
William H Shoff, MD, DTM&H Director, PENN Travel Medicine; Associate Professor, Department of Emergency Medicine, Hospital of the University of Pennsylvania, University of Pennsylvania School of Medicine
William H Shoff, MD, DTM&H is a member of the following medical societies: American College of Physicians, American Society of Tropical Medicine and Hygiene, International Society of Travel Medicine, Society for Academic Emergency Medicine, and Wilderness Medical Society
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4981 | https://t5k.org/prove/prove4_2.html | Primality Proving 4.2: Elliptic curves and the ECPP test
Finding primes & proving primality 4.2: Using elliptic curves and the ECPP test Home>Primality Proving>Chapter Four> Elliptic Curves
Elliptic Curves
What is the next big leap in primality proving? To switch from Galois groups to some other, perhaps easier to work with groups--in this case the points on Elliptic Curves modulo n. An Elliptic curve is a curve of genus one, that is a curve that can be written in the form
E(a,b): y_2= _x_3+ _ax+ b(with 4_a_3+ 27_b_2not zero)
They are called "elliptic" because these equations first arose in the calculation of the arc-lengths of ellipses.
The rational points on such a curve form a group with addition defined using the "chord and tangent method:" That is, if the two points P 1 and P 2 are rational (have rational coefficients), then the line through P 1 and P 2 intersects the curve again in a third rational point which we call -(P 1+P 2) (the negative is to make the associative law work out). Reflect through the x-axis to get P 1+P 2. (If P 1 and P 2 are not distinct, then use the tangent line at P 1.)
If we then reduce this group modulo a prime p we get a small group E(a,b)/p whose size can be used in roughly the way we use the size of (Z/pZ) in the first of the classical tests. Let |E| be the order (the size) of the group E:
Theorem:|E(a,b)/p| lies in the interval (p+1-2sqrt(p),p+1+2sqrt(p)) and the orders are fairly uniformly distributed (as we vary a_and _b).
Obviously we are again getting out of our depth, but perhaps you see that we now have replaced the groups of order n-1 and n+1 used in the classical test with a far larger range of group sizes. We can keep switching curves until we find one we can "factor." This improvement comes at the cost of having to do a great deal of work to find the actual size of these groups.
About 1986, S. Goldwasser & J. Kilian [GK86] and A. O. L. Atkin [Atkin86] introduced elliptic curve primality proving methods. Atkin's method, ECPP, was implemented by a number of mathematicians, including Atkin & Morain [AM93]. François Morain's C-code (discussed in [AM93]) is available on the web for many platforms. For Windows based platforms the Primo implementation is easier to use.
Heuristically, the best version of ECPP is O((log n)4+eps) for some eps> 0 [LL90] (see also D. J. Bernstein's page It has been proven to be polynomial time for almost all choices of inputs.
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4982 | https://r-knott.surrey.ac.uk/fibonacci/fibformula.html | Finding a Formula for the Fibonacci Numbers
The calculators on this page require JavaScript but you appear to have switched JavaScript off
(it is disabled). Please go to the Preferences for this browser and enable it if you want to use the
calculators, then Reload this page.
Is there a formula for the n-th Fibonacci number F(n) in terms of only n?
How many digits does F(n) have?
Given one Fibonacci number can we compute the next directly using only that number?
How can I tell if a given number is a Fibonacci number?
All these questions are answered on this page, using no more maths than you meet at school (UK, up to age 16).
Contents of this page
The icon means there is a
You do the maths... section of questions to start your own investigations.
The calculator icon
indicates that there is a live interactive calculator in that section.
1Binet's Formula for the nth Fibonacci number
1.1
A Calculator for F(n) using Binet's Formula
1.2 An approximation is good enough
1.2.1
An Approximating Calculator
1.2.2
You do the maths...
1.3 Historical Note - Binet's, de Moivre's or Euler's Formula?
2 How many digits does a number have?
2.1 Using the display on your calculator
2.2
Using the LOG button on your calculator
2.3 How many digits are there in Fib(n)?
2.4 What are the first few digits of Fib(n)?
2.4.1
Number of Digits in F(n) Calculator
3 Finding Fibonacci Numbers Fully
4Calculating the next Fibonacci number directly
4.1 An example
4.2 But there's a problem....
4.3Proving that this formula is correct
5 Detecting when N is a Fibonacci Number
5.1
Is N a Fibonacci Number Calculator
6 Finding the Fibonacci index number i for a given n
6.1
Nearest Fibonacci Number ≤ n Calculator
7
Binet's Formula for negative n?
8 Binet's formula for non-integer values of n?
8.1 Complex Numbers
8.2
Applications of Complex numbers
8.3 Argand Diagrams
8.4 Plotting functions on an Argand Diagram
8.5 References on Complex Numbers
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..
1 Binet's Formula for the nth Fibonacci number
We have only defined the nth Fibonacci number in terms of the
two before it:
the n-th Fibonacci number is the sum of the (n-1)th and the (n-2)th.
So to calculate the 100th Fibonacci number, for instance, we need to compute
all the 99 values before it first - quite a task, even with a calculator!
A natural question to ask therefore is:
Can we find a formula for F(n) which involves
only n and does not need any other (earlier) Fibonacci values?
Yes! It involves our golden section number Phi and its reciprocal phi:
Here it is:
| | | | |
--- --- |
| Fib(n) = | Phin − (−Phi)−n | = | Phin − ( −phi)n |
| √5 | √5 |
On these pages we use
| | | | |
--- --- |
| Phi = | √5 + 1 | = | 1·61803 39887 49894 84820 ... |
| 2 |
| | | | | |
--- ---
| phi = Phi −1 = | 1 | = | √5 −1 | = 0·61803 39887 49894 84820 ... |
| Phi | 2 |
Since phi is the name we use for 1/Phi on these pages, then we can remove the
fraction in the numerator here and make it simpler, giving the second form of
the formula at the start of this section.
We can also write this in terms of 5 by using the values for Phi and phi above.
| | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- |
| Fib(n) = | Phin – (–phi)n | = | Phin–(–phi)n | = | 1 | ( ( | 1 + √5 | ) | n | – | ( | 1 – √5 | ) | n | ) |
| Phi – (–phi) | √5 | √5 | 2 | 2 |
If you prefer values in your formulae, then here is another form:-
| | |
--- |
| Fib(n) = | 1.6180339..n – (–0.6180339..)n |
| 2.236067977.. |
This is a surprising formula since it involves
square roots and powers of Phi (an irrational number)
but it always gives an integer for all (integer) values of n!
Here's how it works: Phi =(1·618..) and -phi=-0·618..:
Try some other values of n here, but remember that this calculator has some built-in limits
for the number of decimal places it can accurately compute. Sometimes
these inaccuracies will make numbers round to the wrong values.
You should be suspicious of large numbers that end with 0
because this could indicate a loss of some of the final digits. Only the left-hand digits of a large number
are correct -- the question is just how many!
But the mathematics does tell us that the resulting Fib(n) really should
be an integer for all values of n! We will return to this problem in the next section.
1.1 A Calculator for F(n) using Binet's Formula
C A L C U L A T O R
Fib(n) using Binet's Formula with n=
R E S U L T S:
| | | | | | |
--- --- --- |
Binet's Formula | Approximating | F(n)'s Number of Digits | Is N a Fibonacci number? | Nearest Fibonacci Number ≤ n |
You might want to look at two ways to prove this formula:
the first way is very simple and the second is more advanced and is
for those who are already familiar with matrices.
1.2 An approximation is good enough
Since phi is less than one in size, its powers decrease rapidly. We can use this
to derive the following simpler formula for the n-th Fibonacci number F(n):
F(n) = round( Phin / √5 ) provided n ≥ 0
where the round function gives the nearest integer to its argument.
Notice how, as n gets larger, the value of Phin/√5
is almost an integer.
If n<0 then the powers of (-phi) become larger than the same power of Phi, so then we use
F(n) = round( (-phi)n / √5 ) provided n < 0
1.2.1 An Approximating Calculator
C A L C U L A T O R
rounded
powers of Phi and (-phi) when n=
R E S U L T S:
| | | | | | |
--- --- --- |
Binet's Formula | Approximating | F(n)'s Number of Digits | Is N a Fibonacci number? | Nearest Fibonacci Number ≤ n |
1.2.2 You do the maths...
What is F(100) according to Binet's formula?
The first calculator will give you
some of the initial digits, but the right-hand digits will be wrong.
You may choose to write
a computer program for this, or use a package (such as Mathematica or Maple)
which lets you work out very long integers exactly.
How many digits does F(100) have? (the approximate value on the first calculator above
should tell you).
Check your answer with the list of Fibonacci numbers.
Look at the following table:
| n: | Phin=X: | (-Phi)-n=Y: | X-Y: | (X-Y)/(5): |
--- ---
| 1 | 1·618033989 | -0·61803399 | 2·23606798 | 1 |
You might nave noticed that we didn't ADD the X and Y values to get 1·618..-0·618..=1
directly but instead in Binet's Formula, we subtracted and divided by sqrt(5).
Let's see what happens if we
do just ADD the X and Y columns:
(a) Add a new column to the table above which is X+Y. Fill it in and you'll
notice something very surprising - another integer series
that is not the Fibonacci numbers!! These numbers are called the
Lucas Numbers and they also have some similar properties to the
Fibonacci numbers and are covered in another page at this site (see
Fibonacci Home page).
(b) Can you spot the rule whereby the latest two Lucas numbers are used to
generate the next Lucas number?
1.3 Historical Note - Binet's, de Moivre's or Euler's Formula?
Many authors say that this formula was discovered by
J. P. M. Binet (1786-1856) in 1843 and so call it Binet's Formula.
Graham, Knuth and Patashnik in Concrete Mathematics
(2nd edition, 1994) mention that Euler had already published this formula in 1765.
But Don Knuth in The Art of Computer Programming, Volume 1
Fundamental Algorithms, section 1.2.8, traces it back even further, to
A de Moivre (1667-1754). He had written about "Binet's" formula in 1730
and had indeed found a method for finding formulae for any general series of
numbers formed in a similar way to the Fibonacci series.
Like many results in Mathematics, it is often not the original discoverer who gets
the glory of having their name attached to the result, but someone later!
Concrete Mathematics
(2nd edition, 1994) by Graham, Knuth and Patashnik, Addison-Wesley.
The Art of Computer Programming D E Knuth,
Volume 1: Fundamental Algorithms (now in its Third Edition, 1997).
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..
2 How many digits does a number have?
2.1 Using the display on your calculator
One of the questions above asks you to use your calculator to
find out how many digits are in a number.
When the number gets too big for the calculator's
display, it shows the first few digits and a little "exponent" which says how to move
the decimal place from where it is shown to it true place -
negative means move it to the left, otherwise
move it to the right from where it is shown in the display.
So Phi20/sqrt(5) on my calculator is 6765·000029 and Fib(20)=6765.
But Phi60/sqrt(5) shows as 1·548008755 12 where the little figures
at end are the "exponent", that is, the true value is 1·548008755x1012.
If we move the decimal point 12 places to the right
(putting in 0s for the missing digits), we get:
and the correct value for Fib(60) is
1548008755920
So the exponent, when positive, has told us how many digits there are in the
number calculated, showing just the first few of the digits if not all of them
will fit into the display window!
Similarly, phi60 is just 1/Phi60 which we've just calculated.
Using the "1/x" button
on my calculator when it is showing the value above gives:
6·459911784-13 meaning 6·459911784x10-13. This time we must
move the decimal place to the left since the exponent is negative and we must
move it 13 places.
This gives
0·00000000000064511784 as the value for phi60 - quite small!
2.2 Using the LOG button on your calculator
But how can we calculate the number of digits in a given whole number?
This section shows how to use the LOG button on your calculator to find out
how long a number is.
Returning to the investigation above where you calculated F(100),
this number is usually too big for most calculators to compute, but we can find how
long it is as follows, using the simplified formula:
Fib(n) = round( Phin/√5 )
[This very nearly gives the correct value of F(n) since the part of the formula
we have omitted is very small indeed for large n.]
The LOG button on your calculator can be used to compute how long a number is,
that is, how many decimal digits it has.
This is the "logarithm to base 10". Another
button, usually labelled LN is the "logarithm to base e".
Take the LOG of any 3-digit number
and the answer should be "2 point something".
Try with any 4-digit number and you get a LOG of "3.something". So,
the number of digits in any integer is just 1+ the whole-part of its LOG.
LOGs have useful properties such as:
+ if we ADD LOGS we find the length of the PRODUCT of the original numbers;
e.g. the LOG of x2 is just 2 times LOG x and
the LOG of x3 = 3 LOG x
if we SUBTRACT LOGS we find the length of the QUOTIENT (DIVISION):
e.g. the LOG of √x = (LOG x)/2
2.3 How many digits are there in Fib(n)?
So, now you have enough information to answer the question:
How many digits has F(1000)?
Computing LOG (Phi1000/√5) is the same as computing
1000LOG(Phi) - (LOG √5) = 1000LOG Phi - (LOG 5)/2
= 208.638155.
So 1+the whole number part of your answer is the number of
digits in F(1000), i.e. 209 digits.
In fact, you can find the first few digits by using
the rest of the LOG answer as we'll see in the next section.
2.4 What are the first few digits of Fib(n)?
The whole number part of the log tells us how many digits are in the number. The rest of the log
of a number tells us what those initial digits are. This is because if the log of the
number is xxxx.yyyyyy
then the number itself is 10xxxx.yyyyyy=10xxxx+0.yyyyyy=10xxxx x 100.yyyyyy
10xxxx is just a power of ten so it tells us how to move the decimal point in 100.yyyyyy.
Since 0.yyyy is between 0 and 1 and 100=1 and 101=10 then
100.yyyyyy has a single non-zero digit before its decimal point.
The part after the decimal point in number r is often called the fractional part of r
and written as frac(r).
So the first few digits of a number r are given by 10frac(log10(r)).
2.4.1 Number of Digits in F(n) Calculator
C A L C U L A T O R
| | |
--- |
| | i= up to |
| | |
R E S U L T S
| | | | | | |
--- --- --- |
Binet's Formula | Approximating | F(n)'s Number of Digits | Is N a Fibonacci number? | Nearest Fibonacci Number ≤ n |
There is a
PUMAS
Practical Uses of Maths and Science page by
Kim Aaron of the Jet Propulsion Lab, entitled
"Just what is a log anyway?"
It shows how Kim has found many practical uses of
logarithms as a working engineer.
This page is designed for middle school students, but teachers will also
find it well worth checking out too!
3 Finding Fibonacci Numbers Fully
If we want to study the full and exact integer form of a Fibonacci number, how do we do it?
We will assume that our calculator or computer can add, subtract and multiply whole
numbers of any length (such as Maple or
Mathematica). [If not, it is an interseting and easy programming
exercise to represent integers of any length as an array of single digits together with
functions that manipluate them to perform
addition, subtraction and multiplication using the rules that we all learned at school.]
We have the basic recursive definition, of course (where recursive just means that
we are defining one value of F in terms of two other, simpler, values of F):
F(0)=0
F(1)=1
F(n) = F(n-1) + F(n-2), n>1
but the disadvantage is that to calculate F(1000), for example,
we need to compute all 999 numbers before it.
Alternatively, using Binet's formula above means we need to
compute many decimal places of √5.
So the question we have is:
Can we find a quicker method using only integers?
One method was proposed by the late
Prof. Edsgar W Dijkstra around 1978,
described in note EWD654 "In honor of Fibonacci" (PDF, 38K; download the free Acrobat Reader to view it if your browser will not display it).
Note that Dijkstra's series begins F(1)=0 and F(2)=1 so, using our index system
which has F(0)=0 and F(1)=1, we have:
F(2n-1) = F(n-1)2 + F(n)2
F(2n) = ( 2 F(n-1) + F(n) ) F(n)
which need only F(n) and F(n-1) to compute both F(2n) and F(2n-1).
Although the formula in Dijkstra's note
were "well-known" in 1978, his method of using these two formula in this way as
an efficient algorithm for computing big Fibonacci numbers may be original.
So, to compute F(1000) we would use the two formulae as follows:
F(1000) needs F(500) and F(499)
F(500) and
F(499) need F(250) and F(249)
F(250) and
F(249) need F(124) and F(125)
F(124) needs F(61) and F(62)
F(125) needs F(62) and F(63)
F(63) needs F(32) and F(31)
F(62) and
F(61) needs F(31) and F(30)
F(32) and
F(31) need F(16) and F(15)
F(30) needs F(15) and F(14)
F(16) and
F(15) need F(8) and F(7)
F(14) needs F(7) and F(6)
F(8) and
F(7) need F(4) and F(3)
F(6) needs F(3) and F(2)
F(4) and
F(3) need F(2) and F(1)
F(2) needs F(1) and F(0)
F(1) and
F(0) are 1 and 0 by definition
So there are very few (22) F values needed to compute F(1000) and it takes
much less work
than using the method of "add the previous two F values to get the next one".
See also Computers use the Rabbit Sequence
on the Golden String page at this site for another insight
into computers and computing Fibonacci numbers.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..
4 Calculating the next Fibonacci number directly
Suppose we have evaluated Fib(100) and we want to know the
next value: Fib(101). Do we have to use Binet's formula again?
Well we could do, of course, but here is a short-cut.
There is also a formula that,
given one Fibonacci number, returns the next Fibonacci number directly,
calculating
it in terms only of the previous value (ie not needing the value before as well).
F(n+1) = round( F(n) Phi )
The round function applies to a number (whole or decimal) and changes it to an integer.
round(x) means "the integer nearest to x".
If we apply it to a number ending with ·5 then we will round up
eg round(3.5) is 4.
If we apply the round function to a value which is already a whole number then it does not change it.
4.1 An example
Here's an example of our "next Fibonacci" formula using a small value of n:
Since F(4)=3 then
F(5)
= round( 3 Phi )
= round( 3x1·618... )
= round( 4·854... )
= 5 which is correct!
4.2 But there's a problem....
F(1)=1 and the next Fibonacci number is F(2)=1.
ALSO, F(2)=1 but the next Fibonacci number is F(3)=2.
But we cannot have two different values for "the next Fibonacci number
after 1"!
If we put F(n)=1 in the formula, we have round(1 x Phi), or the nearest integer to Phi = 1.618..., which is 2.
So we cannot apply this formula when n is 1 (and you can check that it also fails if n is 0).
So we have an important restriction on the formula above:
the formula only applies when n is bigger than 1.
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| F(n+1) = round( F(n) Phi ) for all n > 1 |
4.3 Proving that this formula is correct
You can easily evaluate F(2) from F(1)=1 and F(3) from F(2)=2
by this formula and see that they give F(2)=3 and F(3)=5.
Then, if you are familiar with proof by induction
you can show that, supposing the formula is true for F(n-1) and F(n) then
it must also be true for F(n+1) by showing that adding the formula's
expressions for F(n) and F(n-1) gives the formula's expression for F(n+1).
Other ways of proving it involve results about recurrence relations
and how to solve them, which are very like solving differential equations, except
that they deal with integer values not real number values.
This is often included in university level courses on
Pure or Discrete Mathematics.
[ For the university level mathematician, there is an interesting
HAKMEM note
on a fast way of computing Fibonacci numbers and its applications.]
5 Detecting when N is a Fibonacci Number
Given a number N it is not easy to spot if it is a Fibonacci number or not.
Is there a simple test to see if N is a Fibonacci number?
I Gessel solved this in 1972 with a surprisingly simple test.
N is a Fibonacci number if and only if 5 N2 + 4 or
5 N2 – 4 is a square number.
For instance,
3 is a Fibonacci number since 5x32+4 is 49 which is 72
5 is a Fibonacci number since 5x52–4 is 121 which is 112
4 is not a Fibonacci number since neither 5x42+4=84 nor 5x42–4=76
are pefect squares.
It is easy to test if a whole number is square on a calculator by taking its square root and
checking that it has nothing after the decimal point.
In a computer programming
language, take the square root, round it to the nearest integer and then
square the result. If this is the same as the original whole
number then the original was a perfect square.
5.1 Is N a Fibonacci Number Calculator
C A L C U L A T O R
if
is a Fibonacci number
R E S U L T S
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Binet's Formula | Approximating | F(n)'s Number of Digits | Is N a Fibonacci number? | Nearest Fibonacci Number ≤ n |
Problem H-187: n is a Fibonacci number if and only if 5n2+4 or 5n2-4 is a square posed and solved by I Gessel in Fibonacci Quarterly
(1972) vol 10, page 417.
The method above needs to square the number n being tested and then has to check the new number 5 n2 ± 4 is a square number.
If n is large, this can be a problem as n2 has twice as many digits as n.
Another approach is to find the index number i of n or, to put it another way,
to find the (index number of the) closest number equal to or less than n that is a Fibonacci number:
so that F(i)≤n but no larger i will have this property.
This involves using logs but can cope with larger numbers that the approach of this current section
because the log to base 10 of a number is "the number of digits in a number" even when it is not an exact power of 10.
We examine it in the next section.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..
6 Finding the Fibonacci index number i for a given n
It is sometime useful to find the index number i in Fib(i)=n for a given value of n, or to find which index number is close.
Here again Binet's Formula comes in handy - we met it above
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| Fib(i) = | Phii − ( −phi)i |
| √5 |
But since phi=0.618 and phi2= 0.382, the powers of phi quickly get very small and have a smaller and smaller effect on Phii.
By ignoring the small term we find a simpler formula for Fib(i) from which we can find a formula for the index number i:
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| Fib(i) ≈ | Phii |
| √5 |
so that
√5 Fib(i) ≈ Phii
Taking logs of both sides gives:
log(Fib(i) √5) ≈ log( Phii)
and using the log rule that log(a b)=log(a)+log(b) we have:
log(Fib(i)) + log(√5) ≈ log( Phii)
Also we have √a = a1/2 and
log(ab) ≈ b log(a) and applying these rules gives:
log(Fib(i)) + log(5)/2 ≈ i log(Phi)
Rearranging we have our formula for the index number i in terms of Fib(i):
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--- |
| i ≈ | log(Fib(i)) + log(5)/2 |
| log(Phi) |
If we are given a number N and we wish to find the nearest Fibonacci number below or equal to it, then we compute
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| i ≈ | log( N ) + log(5)/2 |
| log(Phi) |
If N is not an actual Fibonacci number then this value for i will be a fractional value.
If we round it down to the nearest whole number then
we will have the Fibonacci index-number i where Fib(i) is below or equal to N; i.e. the largest value of i for which Fib(i) ≤ N.
In fact, using either Round(i) (the nearest integer to i) or Floor(i) (the nearest integer below i)
on the computed value of i will sometimes result in an error for small values of N
which are just below or equal to a Fibonacci number because in these cases the −(−phi)i
term that we ignored actually does affect a few results when N is small.
In practice we can use Floor(i) (round i down to the nearest integer)
and then test this value and also test (i + 1) to be sure.
In the following Calculator, we indicate where a rounding UP has been necessary in the Results.
We can also precompute log(5)/2 and log(Phi) using logs to base 10 or base e so
long as you use the same base for all the logs in the formula. With logs to base 10 to 30 dps we have:
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| log(√5) | = log(5)/2 | = 0.80471 89562 17050 18730 03796 66613 ... |
| log(Phi) | = log((√5+1)/2) | = 0.48121 18250 59603 44749 77589 13424 ... |
6.1 Nearest Fibonacci Number ≤ n Calculator
C A L C U L A T O R : Nearest Fibonacci number ≤ n
the nearest Fibonacci index number i where Fib(i) ≤
R E S U L T S
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Binet's Formula | Approximating | F(n)'s Number of Digits | Is N a Fibonacci number? | Nearest Fibonacci Number ≤ n |
7 Binet's Formula for negative n?
Earlier on this page we looked at Binet's formula for the Fibonacci numbers:
Fib(n) = { Phi n - (-phi) n } / √5
Here Phi=1·6180339... and
phi = 1/Phi = Phi-1 = (√5-1)/2 = 0·6180339... .
We only used this formula for positive whole values of n and found - surprisingly - it
only gives integer results. Well perhaps it was not so surprising really since the
formula is supposed to be define the Fibonacci numbers which are integers;
but it is surprising in that this formula involves
the square root of 5,
Phi and phi which are all irrational numbers i.e. cannot be expressed exactly as the ratio
of two whole numbers.
Suppose we try negative whole numbers for n in Binet's formula.
The formula extends the definition of the Fibonacci
numbers F(n) to negative n.
In fact, if we try to extend the Fibonacci series backwards, still
keeping to the rule that a Fibonacci number is the sum of the two numbers on its LEFT,
we get the following:
| | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- |
| n : | ... | -6 | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | ... |
| Fib(n): | ... | -8 | 5 | -3 | 2 | -1 | 1 | 0 | 1 | 1 | 2 | 3 | 5 | 8 | ... |
and this is consistent with Binet's formula for negative whole values of n.
So we can think of Fib(n) being defined an all integer values of n, both positive
and negative and the Fibonacci series extending infinitely far in both the positive and
negative directions.
8 Binet's formula for non-integer values of n?
This section is optional and at an advanced level
i.e. post 16 years education.
Take me back to the Fibonacci Home page now
or learn about square roots of negative numbers in what follows!
Well now we've tried negative values for n, why not try fractional or other non-whole
values for n?
This doesn't make sense in terms of numbers in a series (there is a 2nd
and a 3rd term and even perhaps a -2nd term
but what can we possibly mean by
a 2·5th term for instance)??
However, this could give us some interesting
(mathematical) insights into the whole-number terms which are our familiar Fibonacci series.
8.1 Complex Numbers
The trouble is that in Binet's formula:
Fib(n) = { Phi n - (-phi) n }/√5
the second term (-phi)n means we have to find the n-th power of
a negative number: -phi and n is not a whole number. If n
was 0·5 for instance, meaning sqrt(-phi), then we are taking the square-root of a
negative value which is "impossible".
Mathematicians have already extended the real-number system to
cover such "imaginary" numbers. They are called complex numbers and have
two parts A and B, both normal real numbers:
a real part, A, and an imaginary part, B. The imaginary part is a multiple
of the basic "imaginary" quantity √-1, denoted i.
So complex numbers are written
as x + i y or x + y i or sometimes as x + I y or
even more simply as (x,y).
8.2 Applications of Complex numbers
To me it is still surprising that such
"imaginary" numbers - or numbers involving the imaginary quantity that is the square
root of a negative number - have very practical applications in the real world.
For instance, electrical engineers have already found many applications for such
"imaginary" or complex numbers.
Whereas resistance can be described by a real number often
measured in ohms, complex numbers are used for the inductance
and capacitance,
so they have very practical uses!
Electrical engineers tend to use j rather than i when writing
complex numbers.
Mathematicians find uses for complex numbers in solving equations:
Every equation of the form Ax+B=0 has a solution which is a fraction:
namely X=-B/A if A and B are integers.
These are called linear equations where A and B are, in general, any real
numbers.
Every equation of the form Ax2 + Bx + C=0 has either
one or two solutions IF we allow complex numbers for x. (Here A is not zero or
we just get a linear equation.)
For instance x2=2 has two solutions:
+sqrt(2) and -sqrt(2)
but x2=0 has just one solution namely x=0.
Note that x2=-2 has two solutions too:
x=sqrt(-2)=isqrt(2) and x=-sqrt(-2)=-isqrt(2)
Every equation of the form Ax3 + Bx2 + Cx + D = 0
has at most 3 solutions again allowing x to be a complex number if necessary.
This leads to a beautiful theorem about solving equations
which are sums of (real number multiples of) powers of x,
called polynomials in x:
If the highest power of x in a polynomial is n
then there are at most n different complex or real number solutions
which make the polynomial's value zero
Complex Numbers and Their Applications by
F J Budden, Longman's 1968,
is now out of print but is an excellent introduction
to the fascinating subject of complex numbers and their applications.
8.3 Argand Diagrams
Writing (x,y) for a complex numbers suggests we might be able to plot complex
numbers on a graph, the x distance being the real part of a complex number
and the y height being its complex part.
Such plots are called Argand diagrams after
J. R. Argand (1768-1822).
We can plot an individual point such as 1 - 2i as the point (1,-2).
Numbers which are real have zero as their complex part so the real number 3 is the
same as the complex number 3 + 0 i and has "coordinates" (3,0).
The real number -1·5
is the same as -1·5 + 0 i or (-1·5,0).
In general, the real number r is the complex number r + 0 i and
is plotted at (r,0) on the Argand diagram.
In fact, all the real values are already in the graph along the x axis
also called the real axis.
Numbers which are purely imaginary have a real-part of zero
and so are of the
form 0+yi always lying exactly on the y axis ( the imaginary axis).
8.4 Plotting functions on an Argand Diagram
We can plot a complex function on an Argand diagram,
that is, a function whose values
are complex numbers. This is where Binet's formula comes in since it will give us
complex numbers as n varies over the real numbers.
So what happens if we plot a graph of F(n) described by Binet's formula,
plotting the results on an Argand diagram?
The blue plot
is for positive values of n from 0 to 6. Note how this curve crosses the
x axis (representing the "real part of the complex number")
at the Fibonacci numbers, 0, 1, 2, 3, 5 and 8.
But there is a loop so it crosses the axis twice at x=1, and
we really do get the whole Fibonacci sequence 0,1,1,2,3,5,8..
as the crossing points.
The red plot is of negative values of n from -6 to 0.
It also crosses the
x axis at the values -8, 5, -3, 2, -1, 1 and 0 corresponding to the Fibonacci numbers
F(-6), F(-5), F(-4), F(-3), F(-2), F(-1) and F(0).
Spirals?
Note that the red spiral for negative values of n
is NOT an equiangular or logarithmic spiral that we found in
sea-shells on the Fibonacci in Nature page.
This is because the curve crosses the x axis at 1 and next at 2,
so the distance from the origin has doubled,
but the next crossing is not at 4 (which would mean another doubling as required for
a logarithmic spiral) but at 5.
If you adjust the width of your browser window, you should be able
to see both curves side by side.
Now it looks as if the two curves are made from the same 3-dimensional
spiral spring-shape,
a bit like the spiral bed-springs in cartoons,
getting narrower towards one end.
The red curve seems to be looking down the centre
of the three-dimensional spring and the blue one looking at the same spring shape
but from the side.
Comparing the two diagrams we can see that even the heights of the loops are the same!
I haven't yet found an explanation for this - can you find one? [Let me know if you
do!]
The plots were produced using parametric plotting function with Maple's built-in
"plot" function:
``
Phi:=(sqrt(5)+1)/2;phi:=(sqrt(5)-1)/2;
f:=n->(Phi^n-(-phi)^n)/sqrt(5);
plot([Re(f(n)),Im(f(n)),n=-6..0],color=RED,
title=Fib(n),-6≤n≤0, Argand diagram);
plot([Re(f(n)),Im(f(n)),n=0..6],color=BLUE,
title=Fib(n),0≤n≤6, Argand diagram`);
```
Kurt Papke has
a Web page with a Java applet to show the two Argand diagrams but animating
the formula that f(n)=f(n-1)+f(n-2) for any real value n. The complex numbers
f(n), f(n-1) and f(n-2) can be illustrated as vectors, and so the formula
f(n)=f(n-1)+f(n-2) becomes a vector equation showing that the vector f(n-1) added to
(followed by) the vector f(n-2) gives the same length-and-direction-movement as the
vector f(n).
Kurt has an excellent
3D version of the spiral that
you can rotate on the screen (using a Java applet) AND one also for the Lucas numbers formula!
For a different complex function based on Binet's formula, see the following two articles
where they both introduce the factor
ei n
which is 1 when n is an integer:
Argand Diagrams of Extended Fibonacci and Lucas Numbers, F J Wunderlich, D E Shaw,
M J Hones Fibonacci Quarterly, vol 12 (1974), pages 233 - 234;
An Extension of Fibonacci's Sequence P J deBruijn, Fibonacci Quarterly
vol 12 (1974) pages 251-258.
8.5 References on Complex Numbers
Complex Numbers are included in some (UK based) Mathematics
syllabuses at Advanced level (school/college examinations taken at about age 17).
Here are some books relating to different Advanced level Examination Boards
syllabus entries on Complex Numbers:
GCE A level Maths: Complex Numbers A. Nicolaides,ISBN: 1872684270, 1995.
Nuffield Advanced Mathematics: Complex Numbers and Numerical Analysis
June 1994, Longman, ISBN: 0582099846.
School Maths Project 16-19: Complex Numbers Cambridge, 1992, ISBN: 0521426529.
| | | |
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| The Puzzling World of the Fibonacci Numbers | Fibonacci Home Page The Mathematical Magic of the Fibonacci Numbers WHERE TO NOW??? Fibonacci bases and other ways of representing integers | The next Topic is... The Golden Section - the Number and Its Geometry |
© 1996-2016 Dr Ron Knott
Created 1996, Updated: 29 August 2023 |
4983 | https://physlab.tamucc.edu/Phys1/06%20Atwood/Atwood%20Instructions-Sum24.pdf | Atwood Machine Equipment • Computer, PASCO Interface • Table Clamp • Double pulley apparatus (one smart pulley) • Smart Pulley Data Cable • String • Two Mass Hangers • One Mass Set (1×500, 2×200, 1×100, 1×50, 2×20, 1×10, 1×5, 2×2, 1×1) • Styrofoam Pad Objectives • To verify Newton’s Second Law as applied to an Atwood Machine. Introduction According to Newton’s 2nd Law the acceleration of an object depends on the net applied force and the object’s mass. ∑𝐹 ⃗= 𝑚𝑎 ⃗ (1) In an Atwood Machine, there are two objects, each one is a mass hanging from one end of a string. The string is hung over a pulley so that both masses are hanging. The weight of one mass pulls the system in one direction, while the weight of the other mass pulls in the other direction. Figure 1. Individual free body diagrams of each of the hanging masses. In the above free body diagrams, 𝑇 is the tension in the string, Assuming that the pulley is massless and frictionless, and the string has no mass and doesn’t stretch, the tension T will be the same throughout the string. The weights are 𝐹 𝑔1 = 𝑚1𝑔 and 𝐹 𝑔2 = 𝑚2𝑔 where 𝑔 is the acceleration due to gravity. Taking the convention that up is positive and down is negative, the net force equations for 𝑚1 and 𝑚2 are: ∑𝐹 1𝑦= 𝑇−𝑚1𝑔= 𝑚1𝑎1 (2) ∑𝐹 2𝑦= 𝑇−𝑚2𝑔= 𝑚2𝑎2 (3) The key to solving this is to realize that in the system of 𝑚1 and 𝑚2, one mass accelerates upward while the other accelerates downward. So, for mass 𝑚2, we flip the coordinates and then the accelerations of 𝑚1 and 𝑚2 are the same and we can just call them both 𝑎. ∑𝐹 1𝑦= 𝑇−𝑚1𝑔= 𝑚1𝑎 (4) ∑𝐹 2𝑦= −𝑇+ 𝑚2𝑔= 𝑚2𝑎 (5) We can add equations (4) and (5) to obtain: 𝑚2𝑔−𝑚1𝑔= 𝑚1𝑎+ 𝑚2𝑎 (𝑚2 −𝑚1)𝑔= 𝑚tot𝑎 (6) (7) Solving for a, the acceleration of the system of both masses, the theoretical acceleration is g times the difference in mass divided by the total mass: 𝑎= 𝑔((𝑚2 −𝑚1) 𝑚tot ) (8) If 𝑚2 > 𝑚1, then 𝐹 𝑔2 is stronger, and the net force on the system accelerates 𝑚2 downward and 𝑚1 upward. Treating the Atwood Machine as a System. Notice that Eq. 7 looks a lot like Newton’s 2nd Law (Eq. 1). The “object” is the pair of masses attached to either end of a string. On the right side of Eq. 7, the mass of the system is the total mass of the two hangers. The acceleration of the system is directly proportional to the exerted force and inversely proportional to the mass of the system. On the left of Eq. 7, we have the net force on the system ∑F = (M −m)𝑔. The net force acting on the system of both masses is just the difference in weight between two hanging masses. This net force accelerates both of the hanging masses. The heavier mass is accelerated downward, and the lighter mass is accelerated upward. Experimental Plan Newton’s 2nd Law can be verified by setting up the Atwood Machine under various conditions. By carefully selecting the masses, we can graph: (1) ∑F vs. 𝑎 while keeping the system mass ∑m constant. (2) 𝑎 vs. 1/∑𝑚 while keeping the net force, ∑F = (M-m)𝑔, constant. In graph (1), the result should be linear with the slope of the graph equal to the total mass. In graph (2), the result should be linear with the slope of the graph equal to the exerted net force (M −m)𝑔. If this is true, then Newton’s 2nd Law is verified. Experimental Setup We will use the Photogate/pulley system to measure the motion of both masses as one moves upward and the other moves downward. Capstone software will record the changing speed of the masses as they move. The recorded data will be displayed as a velocity vs. time graph, where the slope of the graph is the acceleration of the system. Capstone file “Atwood’s Machine” is posted on Blackboard in a corresponding folder and should be open before starting the experiment. Hardware Setup 1. Mount the universal clamp to the edge of the table. 2. Insert the connecting rod of the pulley into the clamp. 3. Plug the photogate into digital channel 1of the interface. You can tell it’s working by spinning the pulley. A little red light near the pulley should blink. 4. Use a piece of string just long enough so that when one mass is on the floor/table, the other is a few centimeters below its pulley. 5. Hang mass hangers from each end of the string, and hang the string over the pulleys, as shown in Figure 2. Figure 2. Schematic representation of the Atwood Machine. (Image source: PASCO) 6. Make sure the string isn’t rubbing against anything, such as the data cable. m M Part 1: Constant Total Mass 1. Choose the amount of mass for Run 1. With the mass kits we have, this is a good amount: • 𝑀 (large mass) = 0.36 kg = 360 grams. Holder (50 g) + 200 g + 100 g + 10 g • 𝑚 (small mass) = 0.34 kg = 340 grams. Holder (50 g) + 200 g + 50 g + 20 g + 20 g • The total mass should be 0.7 kg = 700 grams. 2. Position the masses so that 𝑀 is hanging near the top and 𝑚 is on or near the Styrofoam pad. 3. When you are ready to start, press Start in Capstone and release system. 4. When 𝑀 reaches the bottom, press Stop. 5. Fill in the columns of the Table 1. • Run #: Record the run number from Capstone, in case you should come back to it. • 𝑚, and 𝑀: The individual masses including the hanger. • 𝑎: the experimental acceleration recorded by Capstone. • ∑𝐹= (M −m)𝑔: net force acting on the system. DO NOT USE the measured acceleration for this column! • ∑𝑚: the total mass. Run # m (kg) 𝑴 (kg) a (m/s2) ∑F (N) ∑𝒎 (kg) 0.34 0.36 0.33 0.37 0.32 0.38 0.31 0.39 0.30 0.40 Table 1. Example table for recording run parameters, results, and calculations for Part 1. 6. Perform 4 more trials. For each trial, shift mass back and forth so the total is always 0.7 kg = 700 grams. (Sometimes you have to move 20 g over and move 10 g back.).Use the collected data to fill in Table 1. 7. Plot ∑𝐹 vs. 𝑎 and fit it into linear trendline. Record your results in Table 2. To validate 2nd Law of Motion, the slope of the linear trendline should be equal to the total mass of the system in kilograms. Compare the two by calculating % Error = ABS(Slope – ∑Mass)/∑Mass Total Mass (kg) Slope of ∑𝐹 vs. a (kg) % Error Table 2. Experimental and expected total mass. Part 2: Constant Net Force This time, the data runs will always have the same mass difference, hence the same net force. 1. Choose the masses for the first run. With our mass sets, this is a good amount: • 𝑀 (large mass) = 0.30 kg = 300 grams: Holder (50 g) + 200 g + 50 g • 𝑚 (small mass) = 0.25 kg = 250 grams: Holder (50 g) + 200 g 2. Position the masses so that 𝑀 is hanging near the top and 𝑚 is on or near the Styrofoam pad. 3. When you are ready to start, press Start in Capstone and release system. 4. When 𝑀 reaches the bottom, press Stop. 5. Fill in the columns of the Table 3. • Run #: Record the run number from Capstone, in case you should come back to it. • 𝑚, and 𝑀: The individual masses including the hanger. • 𝑎: the experimental acceleration recorded by Capstone. • ∑𝐹= (M −m)𝑔: net force acting on the system. DO NOT USE the measured acceleration for this column! • ∑𝑚: the total mass. Run # m (kg) 𝑴 (kg) 𝟏/∑𝒎 (kg) a (m/s2) ∑F (N) 0.25 0.30 0.26 0.31 0.27 0.32 0.28 0.33 0.30 0.35 Table 3. Example table for recording run parameters, results, and calculations for Part 2. 6. Perform 4 more trials, adding another 10 g to both hangers for each trial (one 10g set comes from smaller weights: 5+2+2+1). Use the collected data to fill in Table 3. (Note: You cannot make M=0.34 kg. Just skip to the next point.) 7. Plot 𝑎 vs. 1 ∑𝑚 and fit it into linear trendline. Record your results in Table 4.To validate 2nd Law of Motion, the slope of the linear trendline should be equal to the net force acting on the system. Compare the two values by calculating % Error = ABS(Slope – ∑F)/∑F Net Force (N) Slope of 𝑎 vs. 1 ∑𝑚 (N) % Error Table 4. Experimental and expected net force. Requirements for the Report (also consult the rubric): Save your Excel files through your Blackboard Group File Exchange • The abstract section must contain the following explanations in paragraph form: o How the data was collected and calculated for Table 1 o How the data from Table 1 was analyzed including interpretation of the trendline and comparison of values in Table 2 o How the data was collected and calculated for Table 3 o How the data from Table 3 was analyzed including interpretation of the trendline and comparison of values in Table 4 o A general statement based on Tables 2 and 4 about how the experimental results support Second Law of Motion (∑F = ∑ma) • The data section must include o 4 Tables (labeled and captioned) o 2 Graphs (title, axis labels, units, labeled and captioned) ▪ ΣF vs aexp ▪ aexp vs 1 ∑𝑚 (Inverse Mass) |
4984 | https://www.ncbi.nlm.nih.gov/sites/books/NBK574574/ | Standard Deviation - 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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Standard Deviation
Samy El Omda; Shane R. Sergent.
Author Information and Affiliations
Authors
Samy El Omda 1; Shane R. Sergent 2.
Affiliations
1 Royal Liverpool Hospital
2 Michigan State University College of OM
Last Update: November 25, 2024.
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Definition/Introduction
The standard deviation (SD) measures the extent of scattering in a set of values, typically compared to the mean value of the set.The calculation of the SD depends on whether the dataset includes the entire target population or merely samples it. Ideally, studies obtain data from an entire population, which defines the population parameter. However, this is rarely possible in medical research, so a population sample is often used.
The sample SD is determined through the following steps:
Calculate the deviation of each observation (each data point) from the mean.
Square each of these results to remove negative values.
Add these new values together.
Subtract 1 from the total number of observations, then divide this number by the sum obtained above; this result is the sample variance.
Find the square root of the sample variance; this number will be the sample SD.
The process of determining the sample SD of a data set is outlined in the example below, which will use the following data:4, 5, 5, 5, 7, 8, 8, 8, 9, 10
First,calculate the mean of the data set by adding the value of each observation together and then dividing by the number of observations. The sum of our example values is 69, which is then divided by 10 to calculate a mean of 6.9.
Follow the same steps as above to work out the standard deviation:
Subtract the mean from each observation to obtain the following values: -2.9, -1.9, -1.9, -1.9, 0.1, 1.1, 1.1, 1.1, 2.1, 3.1
Each value is squared, resulting in new values: 8.41, 3.61, 3.61, 3.61, 0.01, 1.21, 1.21, 1.21, 4.41, 9.61
The sum of these new values is 36.9.
The sample variance is calculated by dividing this sum by 10 minus 1: 36.9/9 = 4.1.
Finally, the square root of the sample variance is the sample standard deviation, which is 2.02 (to 2 decimal places).
For the population SD, the only difference is during step 4; the sample variance is divided by the total number of observations (without subtracting 1).
Using the same data set to calculate the SD (assuming the data set was the total population):
Subtract the mean from each observation to obtain the following values: -2.9, -1.9, -1.9, -1.9, 0.1, 1.1, 1.1, 1.1, 2.1, 3.1
Each value is squared to remove any negative values, resulting in new values: 8.41, 3.61, 3.61, 3.61, 0.01, 1.21, 1.21, 1.21, 4.41, 9.61
The sum of these new values is 36.9.
The population variance is calculated by dividing this sum by the total number of observations: 36.9/10 = 3.69
Finally, the square root of the sample variance is the population standard deviation, which is 1.92 (to 2 decimal places).
The greater the deviation of each value from the data set's mean, the greater the standard deviation; a smaller SD indicates the values are all relatively close to the mean.
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Issues of Concern
The method of calculating sample SD described above is highly accurate. However, this method becomes increasingly inefficient as the data set values increase, as the difference from the mean of every value must be calculated. This step can be bypassed using the following steps:
List the observed values of the data set in a column, or 'x.'
Find the 'sum of x' by adding all the values in this column together.
Square the value of the 'sum of x.'
Divide this result by the total number of observations to find a new value, or'y'.
Square each value in the 'x' column.
Find the sum of these squared values.
Subtract 'y' from this sum.
If calculating the sample SD, divide this value by the total number of observations minus 1. If calculating the population SD, divide the value by the total number of observations.
The square root of this result is the sample (or population) SD.
This method can quickly calculate the sample SD of a large data set,particularly if paired with a memory function-enabled calculator or an electronic data analysis program.
One mistake researchers sometimes make involves choosing the SD or the standard error of the mean (SEM) to be reported alongside the mean. The distinction between the SD and SEM is crucial but often overlooked. Authors may report the incorrect variable in their data. While the SD refers to the scatter of values around the sample mean, the SEM refers to the accuracy of the sample mean itself. The SEM is a measurement of the precision of the sample mean compared to the total population mean. In contrast to the SD, the SEM does not provide information on the scatter of the sample.
Despite this difference, the SEM and SD are often incorrectly used as interchangeable variables.There have been many hypotheses for this,including a lack of understanding of the meaning of these statistical concepts, leading authors to report what they have seen other authors report.Consequently, multiple articles may publish the SEM in an improper context. Another reason is that the SEM is smaller than the SD; if presented alongside the mean,a smaller value erroneously suggests a higher precision.
This error can also be observed in the figures generated from the same analysis (eg, using the SEM will shorten the error bars).This can even confuse experienced readers who typically expect the SD to be paired with the mean.In response to these issues, some journals only allow authors to present SD to remove any chance of an author mistakenly using the SEM in an inappropriate context.
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Clinical Significance
Using SD allows for a quick overview of a population that is distributed along a normal (Gaussian) pattern. In these populations, 1 SD covers 68% of observations, 2 SD covers 95%, and 3 SD covers 99.7%.This adds greater context to the mean value reported in many studies.
A hypothetical example may help clarify this point. For example, a study looking at the effect of a new chemotherapy agent on life expectancy concludes that the agent increases life expectancy by 5 years.
Widely different data sets could be analyzed to obtain a mean of 5 years. For example, everyone in the sample could have shown an increase in life expectancy between 4 to 6 years. However, another data set that would satisfy this mean would be if half the sample showed no statistical increase in life expectancy while the other half experienced an increase of 10 years. Including an SD can help readers quickly resolve this ambiguity, as the former case would have a small SD while the latter would have a large SD. This allows readers to interpret the results of the study more accurately.
Furthermore, finding a “normal” range of measurements within medicine is often critical information. One example could be a range of laboratory values (such as complete blood count) expected if the measurement was conducted in healthy individuals. Often, this is the 95% reference range, where 2.5% of values will be below the reference range, and 2.5% of values will be greater than the reference range. This makes the SD extremely useful in data sets that follow a normal distribution because it can quickly calculate the range in which 95% of values lie.
Another advantage of reporting the SD is that it reports the scatter within the data in the same units as the data itself. This is in contrast to the variance of the data set, which is equivalent to the square of the SD; hence, its units are the units of the data squared. Therefore, although the variance can be useful in certain scenarios, it is generally not used when describing data.By sharing the same unit, the SD allows the data to be more easily interpreted.
Clinicians should also know the disadvantages of using standard deviation. The main issue is in data sets where there are extreme values or severe skewness, as these results can influence the mean and SD by a significant amount. Consequently, in scenarios where the data set does not follow a normal (Gaussian) distribution, other measures of dispersion are often used. Most commonly, the interquartile range (IQR) is used alongside the median of the dataset. This is due to the IQR being significantly more resistant to extreme values, as only the data between the first and third quartiles are factored in when calculating the IQR. The data between the first and third quartile represents the middle 50% of values, so any unusually high or low values will not affect the calculation of the IQR. This gives a more accurate picture of the data set’s distribution than the SD.
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Nursing, Allied Health, and Interprofessional Team Interventions
Evidence-based medicine plays a significant role in patient care and relies on integrating individual clinical expertise with the best available literature.Keeping up to date with the current literature is vital for all members of an interprofessional team to ensure they are acting in the best interest of a patient. To do this, all members must understand descriptive statistics (eg, mean, median, mode, standard deviation). Without this knowledge, it can be challenging to interpret the conclusions of research articles accurately.
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Review Questions
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Comment on this article.
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References
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Wissing DR, Timm D. Statistics for the nonstatistician: Part I. South Med J. 2012 Mar;105(3):126-30. [PubMed: 22392207]
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Andrade C. Understanding the Difference Between Standard Deviation and Standard Error of the Mean, and Knowing When to Use Which. Indian J Psychol Med. 2020 Jul;42(4):409-410. [PMC free article: PMC7746895] [PubMed: 33402813]
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Barde MP, Barde PJ. What to use to express the variability of data: Standard deviation or standard error of mean? Perspect Clin Res. 2012 Jul;3(3):113-6. [PMC free article: PMC3487226] [PubMed: 23125963]
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Bartko JJ. Rationale for reporting standard deviations rather than standard errors of the mean. Am J Psychiatry. 1985 Sep;142(9):1060. [PubMed: 4025621]
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Swinscow TD. Statistics at square one. III--Standard deviation. Br Med J. 1976 Jun 05;1(6022):1393-4. [PMC free article: PMC1640090] [PubMed: 1276702]
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Hazra A, Gogtay N. Biostatistics Series Module 7: The Statistics of Diagnostic Tests. Indian J Dermatol. 2017 Jan-Feb;62(1):18-24. [PMC free article: PMC5286748] [PubMed: 28216720]
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Driscoll P, Lecky F, Crosby M. An introduction to everyday statistics--2. J Accid Emerg Med. 2000 Jul;17(4):274-81. [PMC free article: PMC1725424] [PubMed: 10921817]
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Hohmann E, Wetzler MJ, D'Agostino RB. Research Pearls: The Significance of Statistics and Perils of Pooling. Part 2: Predictive Modeling. Arthroscopy. 2017 Jul;33(7):1423-1432. [PMC free article: PMC7017839] [PubMed: 28457678]
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Madadizadeh F, Ezati Asar M, Hosseini M. Common Statistical Mistakes in Descriptive Statistics Reports of Normal and Non-Normal Variables in Biomedical Sciences Research. Iran J Public Health. 2015 Nov;44(11):1557-8. [PMC free article: PMC4703239] [PubMed: 26744717]
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Sackett DL, Rosenberg WM, Gray JA, Haynes RB, Richardson WS. Evidence based medicine: what it is and what it isn't. BMJ. 1996 Jan 13;312(7023):71-2. [PMC free article: PMC2349778] [PubMed: 8555924]
Disclosure:Samy El Omda declares no relevant financial relationships with ineligible companies.
Disclosure:Shane Sergent declares no relevant financial relationships with ineligible companies.
Definition/Introduction
Issues of Concern
Clinical Significance
Nursing, Allied Health, and Interprofessional Team Interventions
Review Questions
References
Copyright © 2025, StatPearls Publishing LLC.
This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.
Bookshelf ID: NBK574574 PMID: 34662088
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Definition/Introduction
Issues of Concern
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4985 | https://www.youtube.com/watch?v=Xyl1M40zmbs | Solving f(x)+f(y)=f[(x+y)/(1-xy)]
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Transcript:
Intro hello everyone in this video we're going to be solving a functional equation a nice one f is differentiable everywhere and this function is from r to r and of course there are some exceptions here like you can't really use all the values for x and y so you don't want to make x y equal to 1 because that's going to be undefined anyways so f is differentiable so we're going to use differentiation to solve this functional equation i'll be Partial Derivatives presenting two methods and let's start with the first one all right so i'm going to be taking the partial derivatives here since we have two variables x and y first of all i'm going to differentiate this the with the you know the del operator uh this is del over tau x anyways some people just say partial derivative so i'm going to be taking the partial derivative with respect to x first let's go ahead and do that partial derivative with respect to x so that means y is going to be considered a constant so if you differentiate f of x that's going to be f prime of x f of y is constant therefore it's just going to be 0 and on the right hand side i'm going to have to use chain rule so it's just going to be f prime of this multiply by the derivative of the inside but the inside is a quotient so i kind of have to consider that so if you differentiate x plus y it is going to be the derivative of x is one multiply by one minus x y right and then uh it's going to be minus the derivative of 1 minus x y in this case and y is kind of like a constant here therefore the derivative of negative x y is just going to be negative y so that's going to be plus y times the first thing and all of that is divided by 1 minus xy quantity squared so we use the quotient rule here let's go ahead and simplify the right hand side a little bit oops i was supposed to write it a little differently okay so that's like f prime of a quotient multiply by this okay let's see what this looks like 1 minus xy plus xy plus y squared so the numerator is going to be y squared plus 1 or 1 plus y squared i guess i could write it that way it's a little better 1 plus y squared and that is going to be divided by 1 minus xy quantity squared now similarly you can definitely you know just you don't have to go through the same thing it's going to be the same here first and then the only difference is going to be you know y is kind of being replaced with x so it's going to be instead of 1 plus y squared it's going to be 1 plus x squared and everything else will be the same awesome so we have these two things now and why not divide these equations side by side and the motivation behind that division is we have the same thing like a lot of things look similar so when we divide lots of things will cancel out and we're going to end up with a simple expression okay after dividing these we're going to get something like this f prime of x divided by f prime of y equals now these two expressions are going to cancel out so we don't have to worry about it these two expressions are going to cancel out we don't have to worry about it so we're going to end up with 1 plus y squared divided by 1 plus x squared which is awesome great so and we want to cross multiply here so if you cross multiply we get 1 plus x squared multiply by f prime and 1 plus y squared multiply by f prime of y and obviously we have some sort of symmetry here and this is really cool now what is this equation supposed to mean we're trying to solve for f and so f of x can be expressed as a function of x or in terms of x and f of y can be expressed in terms of y right but we have a function of x on one side and function of y on the other side it's only possible if both of these are equal to a constant otherwise you can't have a function of x only equal to a function of y only therefore from here we get something real nice 1 plus x squared multiplied by f prime equals a constant so from here i can isolate f prime and write it as c over 1 plus x squared i mean you don't have to go through the d y over the x thing if you want you can but if you differentiate both sides consider the fact that the integral of f prime d x is f of x plus c or some constant k let's say we can just go ahead and integrate both sides and that's going to give us f of x so on the left hand side we get f of x let's save the constant for the right hand side what is the integral of this guy over here and that is basically arc tangent some people call it our tangent or you can call it tan inverse so that's the inverse function inverse of tangent x and so it's we can write it as c times tan inverse of x and plus of course i have to use a constant how about k okay great so this is f of x but it's not done yet because i still want to find if possible c or k or both and to find it i'm going to use the value for f of 0 what is f of 0 in our original problem in our original equation if you replace y with 0 and x with 0 at the same time but i want to replace y with 0 first because there's a reason behind that you're going to get f of x plus f of 0 equals f of x so i replace y with 0. and from here we're going to be getting f of x is going to cancel out we're going to get f of 0 equals 0. this is nice right okay cool but then we're going to simply um plug this in uh if we actually we didn't need to replace x with zero anyways f of zero is equal to zero we can go ahead and plug it in here f of zero and the right hand side is gonna give me c times tan inverse of zero plus k but tan inverse of zero is zero therefore this is k k equals zero and that means f of x can be written as c times tan inverse of x plus zero so i don't have to write it and this brings us to the end of the Second Method first method great let's go ahead and talk about the second method second method also use uses derivatives but in an interesting way so our second method let me rewrite the original problem now before i start differentiating i'm going to do a little bit of functional equation stuff so let's replace y with negative x if we do we get f of x plus f of negative x here notice that we're going to get 0 and 0 divided by something which is by the way 1 plus x squared which is non-zero is going to give us f of 0. now this is significant because we know that f of 0 is 0 do we well so pretend that you don't know it so let's go ahead and replace x with 0 here that gives us f of 0 plus f of 0 equals f of 0 which means 2 f of 0 equals f of 0 which means f of 0 equals 0. well we found it with the first method but this is the second method okay great now having said that let's go ahead and plug that in here so we now have f of x plus f of negative x equals f of 0 which is 0 and f of negative x becomes negative f of x which means f is odd awesome f is an odd function let's go ahead and use that fact in an interesting way so since f is differentiable everywhere i can write f prime by using the limit definition of derivative this is equal to limit as h approaches zero of f of x plus h minus f of x divided by h great now let's save this for future now we have the fact that f is odd how am i going to use it right well i have a difference here but my original problem is given as a sum so why don't we just convert that to a difference right let's go ahead and take a look at the following and we're going to go back to the definition of the derivative okay so let's consider f of x plus f of negative y which is which can be written as f of x plus negative y which is x minus y divided by 1 minus x times negative y which is 1 plus x y great so this comes from the original equation i didn't use the fact that f is odd yet but i am going to since f is odd f of negative y can be written as negative f of y and this is the critical part because now we have an equation for a difference awesome beautiful now let's go ahead and use that definition or identity here so f prime becomes limit as h approaches 0 of f of x plus h minus f of x so it's kind of like a difference of 2 to f's and if i use the identity that i have here i'm going to get the following this is going to give me limit as h approaches 0 of f of it was f of x plus h minus f of x therefore it's going to be f of x plus h minus x divided by 1 plus x times x plus h hopefully you get the idea and that is divided by h so i just use the limit definition and just use this identity for f of x plus h minus f of x great let's go ahead and simplify this x cancels out unfortunately it doesn't simplify a whole lot but don't worry we'll take care of that and now we can basically write this as limit as h approaches 0 of f of h over x squared plus x h plus 1 and then divided by h but here's what i would like to do so this is what we would like to do uh i want to get something um you know this you see the stuff inside the parentheses i want to get the same thing at the bottom so let's go ahead and divide by this but we're dividing the denominator so it's kind of equivalent to multiplying so we kind of have to divide again i can do the following right hopefully now here's what i would like you to see this limit here is kind of special and it's equivalent to something like this so let's go ahead and for example if i call this h over x squared plus xh plus 1. let's go ahead and call that k if h approaches 0 k also approaches 0 obviously right so we can write this as limit as k approaches 0 of f of k over k and so that's what it is this is what it is and it's going to be multiplied by something of course but don't worry about that we'll take care of that with the limit and as h approaches 0 it's going to be 1 over x squared plus 1 by the way so we can kind of write this whole thing include that as well right and we can write it like that okay since h x and h are different we're just replacing h with 0 here okay cool so this is the uh f prime can be basically written as follows f prime of x can be written as limit as k approaches 0 of f of k over k times 1 over x squared plus 1. okay great now how am i going to use this information right well here's the thing if you consider f prime at 0 by using the definition again and remember f of x plus h it's just going to be f of oops i forgot to write the limit limit as h approaches 0 or k doesn't really matter same thing you're going to write it as f of x plus h which is f of h minus f of x which is f of 0 divided by h but f of 0 is 0 remember we found it so this can be written as limit as h approaches 0 of f of h over h which is the same as this limit by the way therefore that is equivalent to f prime at 0. so in other words we can write f prime of x as f prime of 0 which is this one here or this one here multiplied by one over x squared plus one now this is the super duper critical part what is f prime at zero it is a constant yay so we can set it equal to c and now write f prime of x as c over x squared plus one and you know the rest of the story from here by way of integrating you get c times tan inverse of x the cool thing about the second method is that we don't have to deal with the plus k that comes from integration because we didn't use integration we only used the definition of the derivative and this brings us to the end of this video i thank you for watching i hope you enjoyed it please let me know don't forget to comment like and subscribe i'll see you tomorrow with another video until then be safe take care and bye bye |
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The molar mass of hydrogen gas (H2) is approximately 2.02 g/mol. This is calculated by adding the molar masses of the two hydrogen atoms in a molecule of hydrogen gas (1.01 g/mol each).
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4987 | https://www.medicalnewstoday.com/articles/what-is-crest-syndrome | What is crest syndrome? Symptoms and treatment
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What is CREST syndrome? All you need to know
Medically reviewed by Stella Bard, MD — Written by Zia Sherrell, MPH on December 23, 2022
What it is
Vs. scleroderma
Symptoms
Causes
Diagnosis
Treatment
Outlook
Summary
CREST syndrome is an autoimmune condition that affects the connective tissues. The word CREST is an acronym for the conditions five main features.
CREST syndrome affects various systems in the body, and the symptoms a person experiences can depend on the affected organ. Doctors use a variety of tests to diagnose CREST syndrome.
Due to advancements in treatment, the outlook for those with the condition is improving.
This article explores how CREST syndrome affects the body, and its causes, diagnosis, and treatments.
What is CREST syndrome?
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Henrik Sorensen/Getty Images
CREST syndrome is a form of systemic sclerosis (SSc), a condition that causes the hardening and tightening of the skin. Other names for this condition are cutaneous systemic sclerosis or limited scleroderma.
In CREST, a person has at least three of the five clinical elements:
Calcinosis: The deposit of calcium in the skin and other organs.
Raynaud’s phenomenon: Spasms and narrowing of blood vessels that cause decreased blood flow to fingers and toes.
Esophageal dysmotility: Difficulty swallowing due to the muscles lining the esophagus losing function.
Sclerodactyly: Thickening and tightening of the skin on the fingers and toes due to collagen deposition.
Telangiectasia: Dilated capillaries — a network of small blood vessels — that form red spots on the skin.
Learn more about Raynaud’s phenomenon.
Scleroderma vs. CREST syndrome
CREST syndrome is a type of scleroderma that is slow to progress. Scleroderma refers to a group of diseases affecting the connective tissue that supports the skin and organs.
Doctors consider CREST syndrome a milder form of scleroderma with a better outlook.
Although CREST follows the same process as typical scleroderma, organ involvement comes slower and later in the disease course.
What body systems does scleroderma affect?
In scleroderma, the skin and blood vessels typically become thickened and hard, and people may have swelling or pain in their muscles and joints.
There are 2 main types of scleroderma: localized and systemic.
The localized type only affects the skin, whereas the systemic type affects the skin and other body parts. These might include underlying tissues, blood vessels, and major organs, such as the heart, lungs, and kidneys.
Symptoms of CREST syndrome
The symptoms of CREST syndrome can vary depending on the affected organs.
Aside from the five clinical elements mentioned above, common symptoms include:
swelling and stiffness in the joints
muscle weakness, pain, or stiffness
swelling or inflammation in areas such as the hands or face
scarring of the lungs that affects breathing (pulmonary fibrosis)
heart problems
kidney problems
Causes of CREST syndrome
Experts are unsure of the exact cause of scleroderma and CREST syndrome. However, it appears to be an autoimmune condition. This means that the immune system attacks the body’s connective tissue.
Several factors likely interact to cause CREST syndrome and scleroderma. These might include:
atypical immune activity, which affects the way the immune system reacts to body cells
environmental triggers — scleroderma has associations with silica dust, organic solvents, and L-tryptophan
genetics, where possessing certain genes may increase the likelihood of having the disease
CREST syndrome does not appear to pass from parent to child.
Diagnosing CREST syndrome
Doctors diagnose CREST syndrome by considering a person’s medical history and physical exams.
Various tests, either routine or specialized, can also be beneficial.
The process can include:
checking the skin for swelling, thickening, and tightening
blood tests to check for signs of inflammation and autoimmunity
serum creatinine tests to measure kidney function
creatinine kinase tests to measure muscle involvement
imaging tests such as MRI, CT, or X-ray, to assess for any atypical growths
lung function tests to assess breathing
electromyography (EMG) and nerve conduction studies
Treatment for CREST syndrome
Depending on their symptoms, a person may require a combination of treatments and/or other management strategies.
The primary goals of treatment are to:
reduce inflammation
limit the progression of the disease
improve quality of life
Treatment options may include:
Anti-inflammatory drugs:Aspirin, indomethacin (Indocin), and naproxen (Naprosyn) can reduce inflammation.
Immunosuppressants: Steroids may slow disease progression and limit flare-ups.
D- penicillamine and colchicine: These medications can help control the hardening of the skin and internal organs.
Vasodilators: These medications ease Raynaud’s symptoms and prevent lung and kidney damage.
Laxatives or heartburn medication: These medications can reduce digestive symptoms.
People should work with their doctor for individualized treatment plans to control their symptoms. Surgery may sometimes be needed to reduce organ damage or improve blood flow.
Outlook
People with limited sclerosis may live with pain and discomfort. However, they can often lead normal lives.
However, those with more severe or systemic symptoms may require ongoing monitoring. Severe forms of SSc also have a higher mortality rate because of the effects on the heart and lungs.
The outlook for those with SSc and CREST syndrome is improving. The 10-year survival rate for CREST syndrome, which has slow and late organ involvement, is over 90%.
However, people with SSc have an increased risk of malignancies, especially lung cancer. This means that regular checkups are necessary.
Summary
CREST syndrome is a type of scleroderma that affects the skin, small blood vessels, joints, lungs, and other organs.
It can cause symptoms such as Raynaud’s phenomenon, swelling and stiffness in the joints, thickening and hardening of the skin, difficulty swallowing, and other symptoms.
Treatment options vary depending on the severity of the symptoms. However, doctors may prescribe anti-inflammatory and immunosuppressant drugs.
The outlook for those with CREST syndrome is improving due to advances in treatment, and people can often lead normal lives. However, regular checkups are still beneficial because of the increased risk of malignancies.
Body Aches
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How we reviewed this article:
Sources
Medical News Today has strict sourcing guidelines and relies on peer-reviewed studies, academic research institutions, and medical journals and associations. We only use quality, credible sources to ensure content accuracy and integrity. You can learn more about how we ensure our content is accurate and current by reading oureditorial policy.
Adigun, R., et al. (2022). Systemic Sclerosis.
Bobeica, C., et al. (2022). CREST syndrome in systemic sclerosis patients – Is dystrophic calcinosis a key element to a positive diagnosis?
CREST syndrome - autoimmune association. (2021).
Diffuse cutaneous systemic sclerosis. (2021).
Peng, H. et al.(2020). Association between systemic sclerosis and risk of lung cancer: results from a pool of cohort studies and Mendelian randomization analysis.
Scleroderma affects skin and more. (2018).
Scleroderma. (2021).
Sobolewski, P., et al. (2019). Systemic sclerosis – multidisciplinary disease: clinical features and treatment.
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Medically reviewed by Stella Bard, MD — Written by Zia Sherrell, MPH on December 23, 2022
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4988 | https://chemistry.stackexchange.com/questions/81370/correlation-between-resonance-and-coordinate-covalent-bond | Correlation between resonance and coordinate covalent bond - Chemistry Stack Exchange
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Correlation between resonance and coordinate covalent bond [closed]
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So I was listening to an online chemistry lecture the other day, and the theme was "bond".
In the 'coordinate covalent bond' part, the teacher said "Examples of the coordinate covalent bond can be S O X 2 S O X 2, H N O X 3 H N O X 3, etc." and I could figure out that both molecules have resonance structures: between N N and O O atoms in H N O X 3 H N O X 3, between O O and S S atoms in S O X 2 S O X 2. Also I am aware of the resonance structures in the C X 6 H X 6 C X 6 H X 6(benzene) molecule, so I looked up the Wikipedia page and could find this sentence:
"..., while the VB description involves a superposition of resonance structures."
So I became curious about this: Is there any close correlation between resonance and coordinate covalent bond? In other words, is it safe to view a molecule as having coordinate covalent bond when it has resonance structure in it?
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asked Aug 17, 2017 at 12:58
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Correlation? Safe? Your question doesn't make much sense...Mithoron –Mithoron 2017-08-17 13:54:03 +00:00 Commented Aug 17, 2017 at 13:54
I think what you mean by resonance is formal charges. Not resonance. I once thought of this too.Tan Yong Boon –Tan Yong Boon 2017-08-17 22:53:12 +00:00 Commented Aug 17, 2017 at 22:53
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Coordinate covalent bonds and Resonance are very very vastly different things. You may have coordinate compounds without resonance, and you may have resonance involving compounds without any coordinate covalent bonds. There's no connection.
Coordinate Covalent Bonds (a.k.a Dative Bonds):
These bonds are just like any other covalent bond. They involve sharing of electrons in between the two atoms that make up the bond. The only difference between a regular covalent bond and a coordinate one is that both the electrons belong to one atom in case of coordinate covalent bonds. Head over to Wikipedia's page on Coordinate Covalent Bonds to learn more, if you're interested.
Resonance:
This is, as I mentioned, vastly different from dative bonds. This is simply a delocalization of π π-electrons along a larger stretch of the molecule as opposed to being confined to being in-between two atoms. Resonance occurs when two π π-systems are in conjugation with each other.
For example, consider butadiene:
There are two π π-systems that are in conjugation with each other. As a result, the two systems kind of overlap and become one bigger system.
You might wonder, why do these molecules undergo resonance anyway? Well it's all based on a very simple fact: Like charges repel. Electrons in the π π molecular orbitals are no exception either. As opposed to being confined to being between two atoms, they prefer to spread over the molecule when given the chance. If you put a 100 misanthropes into a large hall, would they all form two clumps or evenly spread throughout the hall?
Anyway, now you can see, there isn't any relationship between dative bonds and resonance. There are compounds which have no dative bonds, but possess resonance, eg. benzene, butadiene, pyridine, et cetera. There are compounds that have dative bonds, and yet no resonance, eg. ammonia-boron adduct, tetraaquacopper(II), et cetera.
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edited Jun 11, 2020 at 10:20
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answered Aug 17, 2017 at 13:38
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Thanks for the detailed explanation! As many pinpointed my question as unclear, I also felt the same and you fixed it.PenPoint –PenPoint 2017-08-18 02:09:24 +00:00 Commented Aug 18, 2017 at 2:09
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Based on the examples you have given, that does appear to be the case but if you look for more examples you'll find that this is not particularly true. Two well known compounds can be used to demonstrate:
(1) Carbon Monoxide: Coordinate Covalent Bond from O -> C, but no resonance structure
(2) Benzene: Resonance Structure but no Coordinate Covalent Bonds
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answered Aug 17, 2017 at 13:31
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4989 | https://www.youtube.com/watch?v=RsEhAUNI9Ls | 利用坐標求面積｜測量員公式｜鞋帶公式
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4990 | https://perso.telecom-paristech.fr/decreuse/_downloads/c22155fef582344beb326c1f44f437d2/rudin.pdf | REAL AND COMPLEX ANALYSIS REAL AND COMPLEX ANALYSIS Third Edition Walter Rudin Professor of Mathematics University of Wisconsin, Madison ) McGraw-Hill Book Company New York St. Louis San Francisco Auckland Bogota Hamburg London Madrid Mexico Milan Montreal New Delhi Panama Paris Sao Paulo Singapore Sydney Tokyo Toronto REAL AND COMPLEX ANALYSIS INTERNATIONAL EDITION 1987 Exclusive rights by McGraw-Hill Book Co., Singapore for manufacture and export. This book cannot be re-exported from the country to which it is consigned by McGraw-Hill. 0123 45678920BJE9876 Copyright © 1987, 1974, 1966 by McGraw-Hill, Inc. All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or a retrieval system, without the prior written permission of the publisher. This book was set in Times Roman. The editor was Peter R. Devine. The p roduction supervisor was Diane Renda. Library of Congress Cataloging-in-Publication Data Rudin, Walter, 1921 -Real and complex analysis. Bibliography: p. Includes index. 1. Mathematical analysis. I. Title. QA300.R82 1987 515 86-7 ISBN 0-07-054234-1 When ordering this title use ISBN 0-07-100276-6 Printed in Singapore ABOUT THE AUTHOR Walter Rudin is the author of three textbooks, Principles of Mathematical Analysis, Real and Complex Analysis, and Functional Analysis, whose widespread use is illustrated by the fact that they have been translated into a total of 13 languages. He wrote the first of these while he was a C.L.E. Moore Instructor at M.I.T., just two years after receiving his Ph.D. at Duke University in 1949. Later he taught at the University of Rochester, and is now a Vilas Research Professor at the University of Wisconsin-Madison, where he has been since 1959. He has spent leaves at Yale University, at the University of California in La Jolla, and at the University of Hawaii. His research has dealt mainly with harmonic analysis and with complex vari ables. He has written three research monographs on these topics, Fourier Analysis on Groups, Function Theory in Polydiscs, and Function Theory in the Unit Ball ofC0• CONTENTS Preface Xlll Prologue: The Exponential Function 1 Chapter 1 Abstract Integration Set-theoretic notations and terminology The concept of measurability Simple functions Elementary properties of measures Arithmetic in [0, oo] Integration of positive functions Integration of complex functions The role played by sets of measure zero Exercises Chapter 2 Positive Borel Measures Vector spaces Topological preliminaries The Riesz representation theorem Regularity properties of Borel measures Lebesgue measure Continuity properties of measurable functions Exercises Chapter 3 LP-Spaces Convex functions and inequalities The LP -spaces Approximation by continuous functions Exercises 5 6 8 15 16 18 19 24 27 31 33 33 35 40 47 49 55 57 61 61 65 69 71 vii viii CONTENTS Chapter 4 Elementary Hilbert Space Theory Inner products and linear functionals Orthonormal sets Trigonometric series Exercises Chapter 5 Examples of Banach Space Techniques Banach spaces Consequences of Baire's theorem Fourier series of continuous functions Fourier coefficients of L 1-functions The Hahn-Banaeh theorem An abstract ap6roach to the Poisson integral Exercises Chapter 6 Complex Measures Total variation Absolute continuity Consequences of the Radon-Nikodym theorem Bounded linear functionals on LP The Riesz representation theorem Exercises Chapter 7 Differentiation Derivatives of measures The fundamental theorem of Calculus Differentiable transformations Exercises Chapter 8 Integration on Product Spaces Measurability on cartesian products Product measures The Fubini theorem Completion of product measures Convolutions Distribution functions Exercises Chapter 9 Fourier Transforms Formal properties The inversion theorem The Plancherel theorem The Banach algebra L 1 Exercises 76 76 82 88 92 95 95 97 1 00 103 1 04 1 08 112 1 1 6 1 1 6 120 124 126 129 1 32 135 135 1 44 1 50 1 56 160 160 1 63 1 64 1 67 1 70 1 72 1 74 1 78 178 1 80 185 1 90 1 93 Chapter 10 Elementary Properties of Holomorphic Functions Complex differentiation Integration over paths The local Cauchy theorem The power series representation The open mapping theorem The global Cauchy theorem The calculus of residues Exercises Chapter 11 Harmonic Functions The Cauchy-Riemann equations The Poisson integral The mean value property Boundary behavior of Poisson integrals Representation theorems Exercises Chapter 12 The Maximum Modulus Principle Introduction The Schwarz lemma The Phragmen-Lindelof method An interpolation theorem A converse of the maximum modulus theorem Exercises Chapter 13 Approximation by Rational Functions Preparation Runge's theotem The Mittag-Leffler theorem Simply connected regions Exercises Chapter 14 Conformal Mapping Preservation of angles Linear fractional transformations Normal families The Riemann mapping theorem The class f/ Continuity at the boundary Conformal mapping of an annulus Exercises CONTENTS ix 1 96 1 96 200 204 208 21 4 21 7 224 227 231 231 233 237 239 245 249 253 253 254 256 260 262 264 266 266 270 273 274 276 278 278 279 281 282 285 289 291 293 X CONTENTS Chapter 15 Zeros of Holomorphic Functions Infinite products The Weierstrass factorization theorem An interpolation problem Jensen's formula Blaschke products The Mtintz-Szasz theorem Exercises Chapter 16 Analytic Continuation Regular points and singular points Continuation along curves The monodro_y theorem Construction of a modular function The Picard theorem Exercises Chapter 17 HP-Spaces Subharmonic functions The spaces HP and N The theorem of F. and M. Riesz Factorization "theorems The shift operator Conjugate functions Exercises Chapter 18 Elementary Theory of Banach Algebras Introduction The invertible elements Ideals and homomorphisms Applications Exercises Chapter 19 Holomorphic Fourier Transforms Introduction Two theorems of Paley and Wiener Quasi-analytic classes The Denjoy-Carleman theorem Exercises Chapter 20 Uniform Approximation by Polynomials Introduction Some lemmas Mergelyan's theorem Exercises 298 298 301 304 307 31 0 31 2 315 31 9 319 323 326 328 331 332 335 335 337 341 342 346 350 352 356 356 357 362 365 369 371 371 372 377 380 383 386 386 387 390 394 CONTENTS xi Appendix: Hausdorff's Maximality Theorem 395 Notes and Comments 397 Bibliography 405 List of Special Symbols 407 Index 409 PREFACE This book contains a first-year graduate course in ·which the basic techniques and theorems of analysis are presented in such a way that the intimate connections between its various branches are strongly emphasized. The traditionally separate subjects of " real analysis " and " complex analysis " are thus united; some of the basic ideas from functional aࡨalysis are also included. Here are some examples of the way in which these connections are demon strated and exploited. The Riesz representation theorem and the Hahn-Banach theorem allow one to " guess " the Poisson integral formula. They team up in the proof of Runge's theorem. They combine with Blaschke's theorem on the zeros of bounded holomorphic functions to give a proof of the Miintz-Szasz theorem, which concerns approximation on an interval. The fact that 13 is a Hilbert space is used in the proof of the Radon-Nikodym theorem, which leads to the theorem about differentiation of indefinite integrals, which in turn yields the existence of radial limits of bounded harmonic functions. The theorems of Plancherel and Cauchy combined give a theorem of Paley and Wiener which, in turn, is used in the Denjoy-Carleman theorem about infinitely differentiable functions on the real line. The maximum modulus theorem gives information about linear transform ations on I! -spaces. Since most of the results presented here are quite classical (the novelty lies in the arrangement, and some of the proofs are new), I have not attempted to docu ment the source of every item. References are gathered at the end, in Notes and Comments. They are not always to the original sources, but more often to more recent works where further references can be found. In no case does the absence of a reference imply any claim to originality on my part. The prerequisite for this book is a good course in advanced calculus (settheoretic manipulations, metric spaces, uniform continuity, and uniform convergence). The first seven chapters of my earlier book " Principles of Mathe matical Analysis " furnish sufficient preparation. xiii xiv PREFACE Experience with the first edition shows that fir·st-year graduate students can study the first 15 chapters in two semesters, plus some topics from 1 or 2 of the remaining 5. These latter are quite independent of each other. The first 15 should be taken up in the order in which they are presented, except for Chapter 9, which can be postponed. The most important difference between this third edition and the previous ones is the entirely new chapter on differentiation. The basic facts about differen tiation are now derived froࡩ the existence of Lebesgue points, which in turn is an easy consequence of the so-called " weak type " inequality that is satisfied by the maximal functions of measures on euclidean spaces. This approach yields strong theorems with minimal effort. Even more important is that it familiarizes stu dents with maximal functions,· since these have become increasingly useful in several areas of analysis. One of these is the study of the boundary behavior of Poisson integrals. A related one concerns HP-spaces. Accordingly, large parts of Chapters 11 and 17 were rewritten and, I hope, simplified in the process. I have also made several smaller changes in order to improve certain details: For example, parts of Chapter 4 have been simplified; the notions of equi continuity and weak convergence are presented in more detail ; the boundary behavior of conformal maps is studied by means of Lindelof's theorem about asymptotic valued of bounded holomorphic functions in a disc. Over the last 20 years, numerous students and colleagues have offered com ments and criticisms concerning the content of this book. I sincerely appreciated all of these, and have tried to follow some of them. As regards the present edition, my thanks go to Richard Rochberg for some useful last-minute suggestions, and I especially thank Robert Burckel for the meticulous care with which he examined the entire manuscript. Walter Rudin PROLOGUE THE EXPONENTIAL FUNCTION This is the most important function in mathematics. It is defined, for every com plex number z, by the formula 00 zn exp (z) = L --,. n=O n. (1) The series (1) converges absolutely. for every z and converges uniformly on every bounded subset of the complex plane. Thus exp is a continuous function. The absolute convergence of (1) shows that the computation 00 ak 00 bm 00 1 n n ! 00 (a + b )n L -k, L -, = L -, L kl( k)l akbn-k = L I k=O • m=O m. n=O n. k=O • n-· n=O n. is correct. It gives the important addition formula exp (a) exp (b) = exp (a + b), valid for all complex numbers a and b. (2) We define the number e to be exp (1), and shall usually replace exp (z) by the customary shorter expression ez. Note that e0 = exp (0) = 1, by (1). Theorem (a) For every complex z we have ez #= 0. (b) exp is its own derivative: exp' (z) = exp (z). 1 2 REAL AND COMPLEX ANALYSIS (c) The restriction of exp to the real axis is a monotonically increasing positive function, and exr 0 as xr - 00 . (d) There exists a positive number n such that e1ti/l = i and such that ez = 1 if and only if z/(2ni) is an integer. (e) exp is a periodic function, with period 2ni. (f) The mapping t r eit maps the real axis onto the unit circle. (g) Jfw is a complex number and w # 0, then w = ez for some z. PROOF By (2), ez . e-z = ez-z = e0 = 1. This implies (a). Next, , ( ) 1. exp (z + h) - exp (z) ( ) 1. exp (h) - 1 ( ) exp z = tm h = exp z tm h = exp z . hƤo hƥo The first of the above equalities is a matter of definition, the second follows from (2), and the third from (1), and (b) is proved. That exp is monotonically increasing on the positive real axis, and that exr oo as x r oo , is clear from ( 1 ). The other assertions of (c) are conse quences of ex . e-x = 1. For any real number t, (1) shows that e-it is the complex conjugate of eit. Thus or (t real). (3) In other words, if t is real, eit lies on the unit circle. We define cos t, sin t to be the real and imaginary parts of eit : cos t = Re [eit], If we differentiate both sides of Euler's identity eit = cos t + i sin t, which is equivalent to (4), and if we apply (b), we obtain (t real). cos' t + i sin' t = ieit = - sin t + i cos t, so that cos' = - sin, sin' = cos. The power series ( 1) yields the representation t2 t4 t6 cos t = 1 - - + - - - + .. . 2 ! 4! 6! . (4) (5) (6) (7) PROLOGUE: THE EXPONENTIAL FUNCTION 3 Take t = 2. The terms of the series (7) then decrease in absolute value (except for the first one) and their signs alternate. Hence cos 2 is less than the sum of the first three terms of (7), with t = 2; thus cos 2 < - ࡦ. Since cos 0 = 1 and cos is a continuous real function on the real axis, we conclude that there is a smallest positive number t0 for which cos t0 = 0. We define n = 2t0 • (8) It follows from (3) and (5). that sin t0 = + 1. Since sin' (t) = cos t > 0 on the segment (0, t0) and since sin 0 = 0, we have sin t0 > 0, hence sin t0 = 1, and therefore e1ti/2 = i. (9) It follows that e1ti = i2 = - 1, e21ti = (- 1)2 = 1, and then e21tin = 1 for every integer n. Also, (e) follows immediately : (10) If z =X + iy, X and y real, then ez = exeiY; hence I ez I =ex. If ez = 1, we there fore must have ex= 1, so that x = 0; to prove that y/2n must be an integer, it is enough to show that eiy # 1 if 0 < y < 2n, by (10). Suppose 0 < y < 2n, and eiY/4 = u + iv (u and v real). (1 1) Since 0 < y/4 < n/2, we have u > 0 and v > 0. Also eiy = (u + iv)4 = u4 - 6u2v2 + v4 + 4iuv(u2 - v2). (12) The right side of (12) is real only if u2 = v2 ; since u2 + v2 = 1, this happens only when u2 = v2 = ࡧ, and then (12) shows that eiy = - 1 # 1. This completes the proof of (d). We already know that tÒ eit maps the real axis into the unit circle. To prove (f), fix w so that I w I = 1 ; we shall show that w = eit for some real t. Write w = u + iv, u and v real, and suppose first that u > 0 and v > 0. Since u < 1, the definition of n shows that there exists a t, 0 < t < n/2, such that cos t = u; then sin2 t =+= 1 - u2 = v2, and since sin t > 0 if 0 < t < n/2, we have sin t = v. Thus w = eit. If u < 0 and v > 0, the preceding conditions are satisfied by - iw. Hence -- iw = eit for some real t, and w = ei. Finally, if v < 0, the preceding two cases show that - w = eit for some real t, hence w = ei. This com pletes the proof of (f). If w # 0, put a.= w/ I w 1. Then w = I w I a.. By (c), there is a real x such that I w I = ex. Since I a. I = 1, (f) shows that a. = eiy for some real y. Hence w = ex+iy. This proves (g) and completes the theorem. /Ill 4 REAL AND COMPLEX ANALYSIS We shall encounter the integral of (1 + x2)- 1 over the real line. To evaluate it, put <p (t) = sin tjcos t in (-n/2, n/2). By (6), <p ' = 1 + <p 2• Hence <pis a mono tonically increasing mapping of ( -n/2, n/2) onto ( - oo, oo ), and we obtain J oo dx J n/2 <p '(t) dt J n/2 2 = 2 = dt = n. - oo 1 + X -n/2 1 + (/) (t) -n/2 CHAPTER ONE ABSBfRACT INTEGRATION Toward the end of the nineteenth century it became clear to many mathemati cians that the Riemann integral (about which one learns in calculus courses) should be replaced by some other type of integral, more general and more flex ible, better suited for dealing with limit processes. Among the attempts made in this direction, the most notable ones were due to Jordan, Borel, W. H. Young, and Lebesgue. It was Lebesgue's construction which turned out to be the most successful. In brief outline, here is the main idea: The Riemann integral of a function f over an interval [a, b] can be approximated by sums of the form n L f(ti)m(Ei) i= 1 where E1, • • • , En are disjoint intervals whose union is [a, b], m(Ei) denotes the length of Ei, and ti E Ei for i = 1, . . . , n. Lebesgue discovered that a completely satisfactory theory of integration results if the sets Ei in the above sum are allowed to belong to a larger class of subsets of the line, the so-called " measurable sets," and if the class of functions under consideration is enlarged to what he called " measvrable functions." The crucial set-theoretic properties involved are the following: The union and the intersection of any countable family of measurable sets are measurable; so is the complement of every measur able set ; and, most important, the notion of·" length " (now called " measure ") can be extended to them in such a way that s 6 REAL AND COMPLEX ANALYSIS for every countable collection { Ei} of pairwise disjoint measurable sets. This pro perty of m is called countable additivity. The passage from Riemann's theory of integration to that of Lebesque is a process of completion (in a sense which will appear more precisely later). It is of the same fundamental importance in analysis as is the construction of the real number system from the rationals. The above-mentioned measure m is of course intimately related to the geometry of the real line. In this chapter we shall present an abstract (axiomatic) version of the Lebesgue integral, relative to any countably additive measure on any set. (The precise definitions follow.) This abstract theory is not in any way more difficult than the special case of the real line; it shows that a large part of integration theory is independent of any geometry (or topology) of the underlying space; and, of course, it gives us a tool of much wider applicability. The existence of a large class of measures, among them that of Lebesgue, will be established in Chap. 2. Set-Theoretic Notations and Terminology 1.1 Some sets can be described by listing their members. Thus {x h . . . , xn} is the set whose members are x1, • • • , xn; and {x} is the set whose only member is x. More often, sets are described by properties. We write {x: P} for the set of all elements x which have the property P. The symbol 0 denotes the empty set. The words collection, family, and class will be used synonymously with set. We write x e A if x is a member of the set A; otherwise x ¢ A. If B is a subset of A, i.e., if x e B implies x e A, we write· B c A. If B c A and A c B, then A = B. If B c A and A =F B, B is a proper subset of A. Note that 0 c A for every set A. A u B and A n B are the union and intersection of A and B, respectively. If {A«} is a collection of sets, where a, runs through some index set I, we write for the union and intersection of {A«}: U A«= {x: x e A« for at least one a, e I} cxel n A«= {x: X E A« for every a, E I}. ael If I is the set of all positive integers, the customary notations are n= l n= l ABSTRACT INTEGRATION 7 If no two members of {Acx} have an element in common, then {Acx} is a disjoint collection of sets: We write A- B = {x: x e A, x Ӷ B}, and denote the complement of A by Ac whenever it is clear from the context with respect to which larger set the com plement is taken. The cartesian product A 1 x · · · x An of the sets A 1, • • . , An is the set of all ordered n-tuples (a1, • • • , an) where a; e A; for i = 1, . . . , n. The real line (or real number system) is R 1, and Rk = R1 X ... X R 1 ( k factors). The extended real number system is R1 with two symbols, oo and -oo, adjoined, and with the obvious ordering. If -oo <a< b < oo, the interval [a, b] and the segment (a, b) are defined to be [a, b] = {x: a< x < b}, (a, b) = { x : a < x < b} . We also write [a, b) = { x: a < x < b}, (a, b] = { x: a < x < b}. If E c [-oo, oo] and E # 0, the least upper bound (supremum) and great est lower bound (infimum) of E exist in [-oo, oo] and are denoted by sup E and inf E. Sometimes (but only when sup E e E) we write max E for sup E. The symbol f: X Ò Y means thatf is a function (or mapping or transformation) of the set X into the set Y; i.e., f assigns to each x e X an element f(x) e Y. If A c X and B c Y, the image of A and the inverse image (or pre-image) of B are f(A) = {y: y =f(x) for some x e A}, f - 1(B) = {x:f(x) e B}. Note that/ - 1(B) may be empty even when B =1= 0. The domain off is X. The range offisf(X). lff(X) = Y,fis said to map X onto Y. We write f-1(y), instead off - 1( {y} ), for every y e Y. Iff -1(y) consists of at most one point, for each y e Y, f is said to be one-to-one. Iff is one-to-one, then f - 1 is a function with domainf(X) and range X. Iff: X ˰ [-oo, oo] and E c X, it is customary to write supx e E f(x) rather than sup f(E). Iff: X˰ Y and g: Y Ò Z, the composite function g of: X Ò Z is defined by the formula (g o f)(x) = g(f(x)) (x e X). 8 REAL AND COMPLEX ANALYSIS If the range off lies in the real line (or in the complex plane), then f is said to be a real function (or a complex function). For a complex function/, the statement "!> 0 " means that all valuesf(x) off are nonnegative real numbers. The Conµept of Measurability The class of measurable functions plays a fundamental role in integration theory. It has some basic properties in common with another most important class of functions, namely, the continuous ones. It is helpful to keep these similarities in mind. Our presentation is therefore organized in such a way that the analogies between the concepts topological space, open set, and continuous function, on the one hand, and measurable space, measurable set, and measurable function, on the other, are strongly emphasized. It seems that the relations between these concepts emerge most clearly when the setting is quite abstract, and this (rather than a desire for mere generality) motivates our approach to the subject. 1.2 Definition (a) A collection r of subsets of a set X is said to be a topology in X if r has the following three properties : (i) 0 E r and X E r. ( ii) If ࡤ E r f 0 r i = 1' . . . ' n, then vl n v2 n . . . n v,. E r. (iii) If { ƚ} is an arbitrary collection of members of r (finite, countable, or uncountable), then Ua ƚ E r. (b) If r is a topology in X, then X is called a topological space, and the members of r are called the open sets in X. (c) I f X and Y are topological spaces and iff is a mapping of X into Y, then f is said to be continuous provided that f- 1( V) is an open set in X for every open set V in Y. 1.3 Definition (a) A collection 9J1 of subsets of a set X is said to be a a-algebra in X if 9J1 has the following properties : (i) X E 9J1. (ii) If A E 9J1, then Ac E IDl, where Ac is the complement of A relative to X. (iii) If A = Uł= 1 An and if An E 9J1 for n = 1, 2, 3, . . . , then A E IDl. (b) If 9J1 is a a-algebra in X, then X is called a measurable space, and the members of 9J1 are called the measurable sets in X. . (c) If X is a measurable s pace, Y is a topological space, and f is a mapping of X into Y, then f is said to be measurable provided that f - 1(V) is a measurable set in X for every open set V in Y. ABSTRACT INTEGRATION 9 It would perhaps be more satisfactory to apply the term " measurable space " to the ordered pair (X, 9Jl), rather than to X. After all, X is a set, and X has not been changed in any way by the fact that we now also have a a-algebra of its subsets in mind. Similarly, a topological space is an ordered pair (X, rf. B ut if this sort of thing were systematically done in all mathematics, the terminology would become awfully cumbersome. We shall discuss this again at somewhat greater length in Sec. 1.21. 1.4 Comments on Definition 1.2 The most familiar topological spaces are the metric spaces. We shall assume some familiarity with metric spaces but shall give the basic definitions, for the sake of completeness. A metric space is a set X in which a distance function (or metric) p is defined, with the following properties : (a) 0 < p(x, y) < oo for all x andy E X. (b) p(x, y) = 0 if and only if x = y. (c) p(x, y) = p(y, x) for all x and y E X. (d) p(x, y) < p(x, z) + p(z, y) for all x, y, and z E X. Property (d) is called the triangle inequality. If x E X and r > 0, the open ball with center at x and radius r is the set {y E X : p(x, y) < r}. If X is a metric space and if r is the collection of all sets E c X which are arbitrary unions of open balls, then r is a topology in X. This is not hard to verify; the intersection property depends on the fact that if x E B 1 n B2 , where B1 and B2 are open balls, then x is the center of an open ball B c B1 n B2 • We leave this as an exercise. For instance, in the real line R 1 a set is open if and only if it is a union of open segments (a, b). In the plane R2, the open sets are those which are unions of open circular discs. Another topological space, which we shall encounter frequently, is the extended real line [-oo, oo]; its topology is defined by declaring the following sets to be open: (a, b), [-oo, a), (a, oo ], and any union of segments of this type. The definition of continuity given in Sec. 1.2(c) is a global one. Frequently it is desirable to define continuity locally : A mapping f of X into Y is said to be continuous at the point x0 E X if to every neighborhood V of f(x0) there corre sponds a neighborhood W of x0 such thatf(W) c V. (A neighborhood of a point x i ࡥ, by definition, an open set which contains x.) When X and Y are metric spaces, this local definition is of course the same as the usual epsilon-delta definition, and is equivalent to the requirement that limf(xn) = f(x0) in Y whenever lim xn = x0 in X. The following easy proposition relates the local and global definitions of con tinuity in the expected manner: 1.5 Proposition Let X and Y be topological spaces. A mapping f of X into Y is continuous if and only iff is continuous at every point of X. 10 REAL AND COMPLEX ANALYSIS PROOF Iff is continuous and x0 E X, then f - 1( V) is a neighborhood of x0 , for every neighborhood V of f(x0). Since /(/-1(V)) c V, it follows that f is continuous at x0 • Iff is continuous at every point of X and if V is open in Y, every point x E f - 1(V) has a neighborhood ˲ such that f(W x) c V. Therefore ˲ c f-1(V). It follows that f - 1(V) is the union of the open sets˲' so f- 1(V) is itself open. Thusfis continuous. /Ill 1.6 Comments on Definition 1.3 Let 9Jl be a a-algebra in a set X. Referring to Properties (i) to (iii) of Definition 1.3(a), we immediately derive the following facts. (a) Since 0 = xc, (i) and (ii) imply that 0 E IDl. (b) Taking An + l = An +l = · · · = 0 in (iii), we see that A1 u A2 u · · · u An E 9Jl if Ai E 9Jl for i = 1, ... , n. (c) Since 9Jl is closed under the formation of countable (and also finite) intersec tions. (d) Since A - B =Be n A, we have A - B E 9Jl if A E 9Jl and B E IDl. The prefix u refers to the fact that (iii) is required to hold for all countable unions of members of IDl. If (iii) is required for finite unions only, then 9Jl is called an algebra of sets. 1.7 Theorem Let Y and Z be topological spaces, and let g: Y Ò Z be contin uous. (a) I f X is a topological space, iff: X Ò Y is continuous, and if h = g of, then h: X Ò Z is continuous. (b) I f X is a measurable space, iff: XÒ Y is measurable, and if h = g of, then h: X Ò Z is measurable. Stated informally, continuous functions of continuous functions are contin uous; continuous functions of measurable functions are measurable. PROOF If V is open in Z, then g- 1(V) is open in Y, and If/is continuous, it follows that h- 1(V) is open, proving (a). Iff is measurable, it follows that h - 1(V) is measurable, proving (b). /Ill ABSTRACT INTEGRATION 1 1 1.8 Theorem Let u and v be real measurable functions on a measurable space X, let be a continuous mapping of the plane into a topological space Y, and define h(x) = (u(x), v(x)) for x EX. Then h: X Ò Y is measurable. PROOF Put f(x) = (u(x), v(x)). Then f maps X into the plane. Since h = of, Theorem 1.7 shows that it is enough to prove the measurability off If R is any open rectangle in. the plane, with sides parallel to the axes, then R is the cartesian product of two segments I1 and I2 , and which is measurable, by our assumption on u and v. Every open set V in the plane is a countable union of such rectangles Ri , and since f -1(V) is measurable. /Ill 1.9 Let X be a measurable space. The following propositions are corollaries of Theorems 1. 7 and 1.8: (a) Iff= u + iv, where u and v are real measurable functions on X, then f is a complex measurable function on X. This follows from Theorem 1.8, with (z) = z. (b) Iff= u + iv is a complex measurable function on X, then u, v, and If I are real measurable functions on X. This follows from Theorem 1.7, with g(z) = Re (z), Im (z), and I z 1. (c) Iff and g are complex measurable functions on X, then so aref + g andfg. For real/ and g this follࡢws from Theorem 1.8, with ( s' t) = s + t and (s, t) = st. The complex case then follows from (a) and (b). (d) If E is a measurable set in X an.d if then XE is a measurable function. if X E E if X ¢ E This is obvious. We call XE the characteristic function of the set E. The letter x will be reserved for characteristic functions throughout this book. (e) Iff is a complex measurable function on X,- there is a complex measurable function C( on X such that I C( I = 1 andf = C( I fl. 12 REAL AND COMPLEX ANALYSIS PROOF Let E = {x:f(x) = 0}, let Y be the complex plane with the origin removed, define qJ(z) = zj I z I for z e Y, and put cx(x) = <p(f(x) + XE(x)) (x E X). If x E E, cx(x) = 1; if x ¢ E, cx(x) = f(x)j I f(x) 1 . Since <p is continuous on Y and since E is measurable (why?), the measurability of ex follows from (c), (d), and Theorem 1.7. //// We now show that a-algebras exist in great profusion. 1.10 Theorem I f ff is any collection of subsets of X, there exists a smallest a-algebra IDl in X such that ff c IDl. This IDl is sometimes called the a-algebra generated by ff. PROOF Let Q be the family of all a-algebras 9Jl in X which contain ff. Since the collection of all subsets of X is such a a-algebra, Q is not empty. Let IDl be the intersection of all 9Jl E Q. It is clear that ff c IDl and that IDl lies in every a-algebra in X which contains ff. To complete the proof, we have to show that IDl is itself a a-algebra. If An E IDl for n = 1, 2, 3, . . . , and if 9Jl E Q, then An E IDl, so U An E IDl, since 9Jl is a a-algebra. Since U An E 9Jl for every 9Jl e Q, we conclude that U An e IDl. The other two defining properties of a a-algebra are verified in the same manner. //// 1.11 Borel Sets Let X be a topological space. By Theorem 1.10, there exists a smallest a-algebra ˳ in X such that every open set in X belongs to ˳. The members of˳ are called the Borel sets of X. In particular, closed sets are Borel sets (being, by definition, the complements of open sets), and so are all countable unions of closed sets and all countable intersections of open sets. These last two are called Fa's and G,'s, respectively, and play a considerable role. The notation is due to Hausdorff. The letters F and G were used for closed and open sets, respectively, and a refers to union (Summe), b to intࡣrsection (Durchschnitt). For example, every half-open interval [a, b) is a G, and an Fa in R 1. Since 81 is a a-algebra, we may now regard X as a measurable space, with the Borel sets playing the role of the measurable sets; more concisely, we consider the measurable space (X, P1). Iff: X -4 Y is a continuous mapping of X, where Y is any topological space, then it is evident from the definitions that f -1(V) E 81 for every open set V in Y. In other words, every continuous mapping o f X is Borel \ measurable. Borel measurable mappings are often called Borel mappings, or Borel func tions. ABSTRACT INTEGRATION 13 1.12 Theorem Suppose 9Jl is a a-algebra in X, and Y is a topological space. Let f map X into Y. (a) If Q is the collection of all sers E c Y such that f - 1(£) E IDl, then Q is a a-algebra in Y. (b) Iffis measurable and Eisa Borel set in Y, thenf - 1(E) E IDl. (c) If Y = [- oo, oo] and f - 1((cx, oo ]) E 9Jl for every real ex, then f is measur able. (d) Iff is measurable, if Z is a topological space, if g: Y Ò Z is a Borel mapping, and if h = g of, then h: X Ò Z is measurable. Part (c) is a frequently used criterion for the measurability of real-valued functions. (See also Exercise 3.) Note that (d) generalizes Theorem 1.7(b). PROOF (a) follows from the relations f - 1(Y) = X, f - 1(Y - A) = X -f - 1(A), and f - 1(A1 u A2 u · · ·) =f - 1(A1) uf-1(A2) u · · · . To prove (b), let n be as in (a); the measurability of f implies that Q contains all open sets in Y, and since Q is a a-algebra, Q contains all Borel sets in Y. To prove (c), let Q be the collection of all E c [- oo, oo] such that f -1(E) E IDl. Choose a real number cx, and choose cxn < cx so that cxn | cx as n | oo. Since ( cxn , oo] E Q for each n, since 00 00 [ - oo, cx) = U [- oo, cxnJ = U (cxn, oo]c, n =1 n=l and since (a) shows that Q is a a-algebra, we see that [ - oo, cx) E Q. The same is then true of (ex, fJ) = [ - oo, fJ) n ( cx, oo]. Since every open set in [ - oo, oo] is a countable union of segments of the above types, Q contains every open set. Thusfis measurable. To prove (d), let V c Z be open. Then g - 1(V) is a Borel set of Y, and stnce h- 1 ( V) = f - 1 (g - 1 ( V) ), (b) shows that h - 1(V) E IDl. /Ill 1.13 Definition Let {an} be a sequence in [ - oo, oo], and put ( k = 1, 2, 3, . . . ) (1) 14 REAL AND COMPLEX ANALYSIS and fJ = inf {bh b2, b3, . . . }. (2) We call fJ the upper limit of {an}, and write fJ = lim sup an . (3) The following properties are easily verified: First, b1 > b2 > b3 > · · · , so that bk----+ fJ as k----+ oo ; secondly, there is a subsequence { anJ of {an} such that ani----+ {3 as i----+ oo, and fJ is the largest number with this property. The lower limit is defined analogously: simply interchange sup and inf in (1) and (2). Note that lim inf an = -lim sup (-an>· (4) If {an} converges, then evidently lim sup an= lim inf an= lim an. (5) Suppose {fn} is a sequence of extended-real functions on a set X. Then sup fn and lim sup fn are the functions defined on X by If ( sàp fn}x) = sßp (f,(x)), (tiÞ ... ÜPfn) 0, and h > 0, then!+ < g andf- <h. PROOF f < g and 0 < g clearly implies max {f, 0} <g. Ill/ Simple Functions 1.16 Definition A complex function s on a measurable space X whose range consists of only finitely many points will be called a simple function. Among these are the nonnegative simple functions, whose range is a finite subset of [0, oo ). Note that we explicitly exclude oo from the values of a simple func tion. If (X1, .
.
• , (Xn are the distinct values of a simple function s, and if we set Ai = { x: s(x) = (Xi}, then clearly n s = L (Xi XAi' i= 1 where XAi is the characteristic function of Ai, as defined in Sec. 1.9(d). It is also clear that s is measurable if and only if each of the sets Ai is measurable. 1.17 Theorem Let f: X----+ [0, oo] be measurable. There exist simple measur able functions sn on X such that (a) 0 < s 1 < s2 < · · · <f. (b) sn(x)----+ f(x) as n----+ oo,for every x e X. PROOF Put Ȝn = 2 -n. To each positive integer n and each real number t cor responds a unique integer k = kn(t) that satisfies kȜn < t < (k + 1 )Ȝn. Define ifOSt<n (1) -if n S t S oo. 16 REAL AND COMPLEX ANAJ... YSIS Each CfJn is then a Borel function on [0, oo ], if O<t<n, (2) 0 < cp1 < cp2 < · · · < t, and CfJn(t)--+ t as n--+ oo, for every t e [0, oo ]. It follows that the functions satisfy (a) and (b); they are measurable, by Theorem 1.12(d). Elementary Properties of Measures 1.18 Definition (3) Ill/ (a) A positive measure is a function Jl, defined on a a-algebra 9Jl, whose range is in [0, oo] and which is countably additive. This means that if {A;} is a disjoint countable collection of members of 9Jl, then (1) To avoid trivialities, we shall also assume that Jl(A) < oo for at least one A E 9Jl. (b) A measure space is a measurable space which has a positive measure defined on the a-algebra of its measurable sets. (c) A complex measure is a complex-valued countably additive function defined on a a-algebra. Note: What we have called a positive measure is frequently just called a measure; we add the word " positive" for emphasis. If J.l(E) = 0 for every E e 9Jl, then Jl is a positive measure, by our definition. The value oo is admissible for a positive measure; but when we talk of a complex measure Jl, it is understood that Jl(E) is a complex number, for every E e 9Jl. The real measures form a subclass of the complex ones, of course. 1.19 Theorem Let Jl be a positive measure on a a-algebra 9Jl. Then (a) J.l(0) = 0. (b) Jl(A1 u · · · u An) = Jl(A1) + · · · + Jl(An) if A1, • • • , An are pairwise disjoint members of9Jl. (c) A c B implies Jl(A) < Jl(B) if A e 9Jl, B e 9Jl. (d) Jl(An)--+ Jl(A) as n--+ oo if A = U:'= 1 An , An e 9Jl, and and Jl(A 1) is finite. ABSTRACT INTEGRATION 17 As the proof will show, these properties, with the exception of (c), also hold for complex measures; (b) is called finite additivity; (c) is called monotonicity. PROOF (a) Take A e 9Jl so that Jl(A) < oo, and take A1 = A and A2 = A3 = · · · = 0 in 1.18(1). (b) Take An+ 1 = An+l = · · · = 0 in 1.18(1). (c) Since B = A u (B-A) and A n (B-A) = 0, we see that (b) implies Jl(B) = Jl(A) + Jl(B-A) > Jl(A) . (d) Put B1 = A1, and put Bn =An-An-1 for n = 2, 3, 4, . . . . Then Bn E IDl, B; n Bi = 0 ifi :f=j, An= B1 u · · · u Bn, and A= U?:: 1 B;. Hence n oo Jl(An) = L Jl(B;) and Jl(A) = L Jl(B;) . i= 1 i= 1 Now (d) follows, by the definition of the sum of an infinite series. (e) Put en= A1 -An. Then e1 c: e2 c: e3 c: · · ·, A1-A= U en, and so (d) shows that Jl(A 1) -Jl(A) = Jl(A 1 -A) = lim Jl( en) = Jl(A 1) - lim Jl(An) . n-+ oo n-+ oo This implies (e). Ill/ 1.20 Examples The construction of interesting measure spaces requires some labor, as we shall see. However, a few simple-minded examples can be given immediately: (a) For any E c: X, where X is any set, define Jl(E) = oo if E is an infinite set, and let Jl(E) be the number of points in E if E is finite. This Jl is called the counting measure on X. (b) Fix x0 e X, define Jl(E) = 1 if x0 e E 'and Jl(E) = 0 if x0 ¢ E, for any E c: X. This Jl may be called the unit mass concentrated at x0 • (c) Let Jl be the counting measure on the set {1, 2, 3, ... }, let An= {n, n + 1, n + 2, . . . }. Then n An= 0 but Jl(AJ = oo for n = 1, 2, 3, . . .. This shows that the hypothesis is not superfluous in Theorem 1.19(e). 1.21 A Comment on Terminology One frequently sees measure spaces referred to as " ordered triples" (X, IDl, Jl) where X is a set, IDl is a a-algebra in X, and Jl is a measure defined on IDl. Similarly, measurable spaces are " ord.et;eq࡞ pࡠr$f'-ࡡ (X, IDl). 18 REAL AND COMPLEX ANALYSIS This is logically all right, and often convenient, though somewhat redundant. For instance, in (X, 9Jl) the set X is merely the largest member of 9Jl, so if we know 9Jl we also know X. Similarly, every measure has a a-algebra for its domain, by definition, so if we know a measure J1 we also know the a-algebra 9Jl on which J1 is defined and we know the set X in which 9Jl is a a-algebra. It is therefore perfectly legitimate to use expressions like " Let J1 be a measure " or, if we wish to emphasize the a-algebra or the set in question, to say " Let J1 be a measure on 9Jl" or " Let J1 be a measure on X." What is logically rather meaningless but customary (and we shall often follow mathematical custom rather than logic) is to say " Let X be a measure space "; the emphasis should not be on the set, but on the measure. Of course, when this wording is used, it is tacitly understood that there is a measure defined on some a-algebra in X and that it is this measure which is really under dis-. CUSSIOn. Similarly, a topological space is an ordered pair (X, r), where r is a topology in the set X, and the significant data are contained in r, not in X, but " the topological space X " is what one talks about. This sort of tacit convention is used throughout mathematics. Most mathe matical systems are sets with some class of distinguished subsets or some binary operations or some relations (which are required to have certain properties), and one can list these and then describe the system as an ordered pair, triple, etc., depending on what is needed. For instance, the real line may be described as a quadruple (R1, +, · , <), where +, ·, and < satisfy the axioms of a complete archimedean ordered field. But it is a safe bet that very few mathematicians think of the real field as an ordered quadr\lple. Arithmetic in [0, oo] 1.22 Throughout mtegration theory, one inevitably encounters oo. One reason is that one wants to be able to integrate over sets of infinite measure; after all, the real line has infinite length. Another reason is that even if one is primarily inter ested in real-valued functions, the lim sup of a sequence of positive real functions or the sum of a sequence of positive real functions may well be oo at some points, and much of the elegance of theorems like 1.26 and 1.27 would be lost if one had to niake some special provisions whenever this occurs. Let us define a + oo = oo + a = oo if 0 < a < oo, and a · oo = oo · a={: ifO<a< oo if a= 0; sums and products of real numbers are of course defined in the usual way. It may seem strange to define 0 · oo = 0. Howe-ver, one verifies without diffi culty that with this definition the commutative, associative, and distributive laws hold in [0, oo] without any restriction. ABSTRACT INTEGRATION 19 The cancellation laws have to be treated with some care : a + b = a + c implies b = c only when a < oo, and ab = ac implies b = c only when 0 <a < oo. Observe that the following useful proposition holds : If we combine this with Theorems 1.17 and 1.14, we see that sums and pro ducts of measurable functions into [0, oo] are measurable. Integration of Positive Functions In this section, 9Jl will be a a-algebra in a set X and J1 will be a positive measure on 9Jl. 1.23 Definition If s: X Ò [0, oo) is a measurable simple function, of the form n s = L (Xi XAi' i= 1 (1) where rx1, • • • , rxn are the distinct values of s (compare Definition 1.16), and if E E m, we define (2) The convention 0 · oo = 0 is used here; it may happen that rxi = 0 for some i and that Jl(Ai n E) = oo. Iff: XÒ [0, oo] is measurable, and E E 9Jl, we define Lf dJ1. = sup Ls dJ1., (3) the supremum being taken over all simple measurable functions s such that O<s<f The left member of (3) is called the Lebesgue integral off over E, with respect to the measure Jl. It is a number in [0, oo ]. Observe that we apparently have two definitions for JE f dJ1 iff is simple, namely, (2) and (3). However, these assign the same value to the integral, since f is, in this case, the largest of the functions s which occur on the right of (3). 1.24 The following propositions are immediate consequences of the definitions. The functions and sets occurring in them are assumed to be measurable: (a) I fO < f < g, then JE f dJ1 < JE g dJ1. (b) I f A c B andf> 0, then JA f dJ1 < JB f dJ1. 20 REAL AND COMPLEX ANALYSIS (c) Iff> 0 and cis a constant, 0 < c < oo , then (d) Iff(x) = Ofor all x E E, then'JE f dJ1 = 0, even if J1(E) = oo . (e) If J1(E) = 0, then JE f dJ1 = 0, even iff(x) = oo for every x E E. (f) Iff> 0, then JE f dJ1 = Jx XEf dJ1. This last result shows that we could have restricted our definition of integra tion to integrals over all of X, without losing any generality. If we wanted to integrate over subsets, we could then use (f) as the definition. It is purely a matter of taste which definition is preferred. One may also remark here that every measurable subset E of a measure space X is again a· measure space, in a perfectly natural way: The new measur able sets are simply those measurable subsets of X which lie in E, and the measure is unchanged, except that its domain is restricted. This shows again that as soon as we have integration defined over every measure space, we automati cally have it defined over every measurable subset of every measure space. 1.25 Proposition Let s and t be nonnegative measurable simple functions on X. For E E IDl, define (1) Then <p is a measure on IDl. Also (2) (This proposition contains provisional forms of Theorems 1.27 and 1.29.) PROOF If s is as in Definition 1.23, and if E1, E2, • • • are disjoint members of 9Jl whose union is E, the countable additivity of J1 shows that n n oo <p(E) = L 0i Jl(Ai n E) = L 0i L ,u(Ai n Er) i= 1 i=1 r=1 oo n oo = L L 0i Jl(Ai n Er) = L qJ(Er). r=1 i=1 r= 1 Also, ({J(0) = 0, so that <p is not identically oo . ABSTRACT INTEGRATION 21 Next, let s be as before, let {31, • • • , 13m be the distinct values of t, and let Bi = {x: t(x) = Pi}. If Eii = Ai n Bi, then and L? + t) dJl. = (cxi + {Ji) Jl(Eii) L,js dJ1. + 1./ dJ1. = cxi Jl.(Eii) + {Ji Jl.(Eii) . Thus (2) holds with Eii in place of X. Since X is the disjoint union of the sets Eii (1 < i < n, 1 < j < m), the first half of our proposition implies that (2) holds. II II We now come to the interesting part of the theory. One of its most remark able features is the ease with which it handles limit operations. 1.26 Lebesgue's Monotone Convergence Theorem Let {fn} be a sequence of measurable functions on X, and suppose that" (a) 0 <f1(x) <f2(x) < · · · < oo for every x E X, (b) fn(x)r f(x) as nr oo,for every x E X. Thenfis measurable, and as nr oo. PROOF Since J fn < J fn + h there exists an a. E [0, oo] such that l.r.. dJ1.-. ex as nr oo. (1) By Theorem 1.14, f is measurable. Sincef n </, we have J fn < J f for every n, so (1) implies (2) Let s be any simple measurable function such that 0 < s < f, let c be a constant, 0 < c < 1, and define (n = 1, 2, 3, . . . ). (3) Each En is measurable, E 1 c E2 c E3 c · · · , and X = U En. To see this equality, consider some x E X. If f(x) = 0, then x E E1; if f(x) > 0, then cs(x) <f(x), since c < 1; hence x E En for some n. Also r fn djl. > r f.. djl. > c r s djl. Jx JEn JEn (n = 1, 2, 3, . . . ) . (4) 22 REAL AND COMPLEX ANALYSIS Let nŠ oo, applying Proposition 1.25 and Theorem 1.19(d) to the last inte gral in ( 4 ). The result is Since (5) holds for every c < 1, we have ex> Is dJJ. for every simple measurable s satisfying 0 < s < f, so that ex> If dJJ.. The theorem follows from ( 1 ), (2), and (7). 1.27 Theorem Iff,.: Xr [0, oo] is measurable, for n = 1, 2, 3, ... , and 00 f(x) = L f n(x) (x EX), n = l then If djJ. = n̖ l I f, djJ.. (5) (6) (7) Ill/ (1) (2) PROOF First, there are sequences {sa' { s˂'} of simple measurable functions such that sͥ r f1 and s˂' Š /2 , as in Theorem 1.17. If si = sͥ + s˂', then sir f1 + /2, and the monotone convergence theorem, combined with Propo sition 1.25, shows that (3) Next, put gN = /1 + · · · + fN. The sequence {gN} converges monotoni cally to f, and if we apply induction to (3) we see that (4) Applying the monotone convergence theorem once more, we obtain (2), and the proof is complete. /Ill If we let J1 be the counting measure on a countable set, Theorem 1.27 is a statement about double series of nonnegative real numbers (which can of course be proved by more elementary means): ABSTRACT INTEGRATION 23 Corollary If aii > 0 fori and j = 1, 2, 3, ... , then 00 00 00 00 L L aij = L L aij . i=l j=l j=l i=l 1.28 Fa too's Lemma If fn: Xr [0, oo] is measurable, for each positive integer n, then i (lim inf fn) dp, < lim inf l f n dp,. X n - oo n - oo X Strict inequality can occur in (1); see Exercise 8. PROOF Put gk(x) = inf };(x) Then gk < fk, so that i ǁk (k = 1, 2, 3, . . . ; X E X). 1 gk dp. < lit dp. (k = 1, 2, 3, ... ). (1) (2) (3) Also, 0 < g1 < g2 < · · · , each gk is measurable, by Theorem 1.14, and gk(x)r lim inff n(x) as k Ò oo, by Definition 1.13. The monotone convergence theorem shows therefore that the left side of (3) tends to the left side of (1), as kr oo. Hence (1) follows from (3). /Ill 1.29 Theorem Suppose f: X -x [0, oo] is measurable, and l{J(E) = 1ġ dp. (E e 9Jl). (1) Then cp is a measure on 9Jl, and (2) for every measurable g on X with range in [0, oo ]. PROOF Let E1, E2, E3, Observe that be disjoint members of IDl whose union is E. and that 00 XEf = L XEjf j= 1 (3) (4) 24 REAL AND COMPLEX ANALYSIS It now follows from Theorem 1.27 that 00 qJ(E) = L qJ(E j). (5) j=l Since ({J(0) = 0, ( 5) proves that qJ is a measure. Next, (1) shows that (2) holds whenever g = XE for some E e 9Jl. Hence (2) holds for every simple measurable function g, and the general case follows from the monotone convergence theorem. /Ill Remark The second assertion of Theorem 1.29 is sometimes written in the form dqJ = f djJ.. (6) We assign no independent meaning to the symbols dqJ and djJ.; (6) merely means that (2) holds for every measurable g > 0. Theorem 1.29 has a very important converse, the Radon-Nikodym theorem, which will be proved in Chap. 6. Integration of Complex Functions As before, J.l will in this section be a positive measure on an arbitrary measurable space X. 1.30 Definition We define L1(JJ.) to be the collection of all complex measur able functions/ on X for which 11!1 dJl < 00 . Note that the measurability off implies that of If I, as we saw in Propo sition 1.9(b); hence the above integral is defined. The members of L1(JJ.) are called Lebesgue integrable functions (with respect to JJ.) or summable functions. The significance of the exponent 1 will become clear in Chap. 3. 1.31 Definition Iff = u + iv, where u and v are real measurable functions on X, and iff E L1 (JJ. ), we define ifdJl = iu+ dJl-Lu- dJl + i iv+ dJl-i iv- dJl (1) for every measurable set E. Here u + and u- are the positive and negative parts of u, as defined in Sec. 1.15 ; v + and v - are similarly obtained from v. These four functions are measurable, real, and nonnegative; hence the four integrals on the right of (1) exist, by Definition 1.23. Furthermore, we have u + < I u I < If I, etc., so that ABSTRACT INTEGRATION 25 each of these four integrals is finite. Thus (1) defines the integral on the left as a complex number. Occasionally it is desirable to define the integral of a measurable func tion/with range in [ - oo, oo] to be (2) provided that at least one of the integrals on the right of (2) is finite. The left side of (2) is then a number in [ - oo, oo ]. 1.32 Theorem Suppose f and g E L1(J.t) and C( and P are complex numbers. Then C(f + pg E L1(J.t), and (1) PROOF The measurability of C(f + pg follows from Proposition 1.9(c). By Sec. 1.24 and Theorem 1.27, l l oc f + pg I dJl < l ( l oc I I f I + I P I I g I ) dJl = l oc l l l / 1 d11 + I P i l l g l dJl < oo. Thus C(f + pg E I!(J.l). To prove (1), it is clearly sufficient to prove fxu + g) d11 = fx! d11 + fxg d11 and l (oc f) dJl = oc If dJl, and the general case of (2) will follow if we prove (2) for real f and g in L1(J.t). Assuming this, and setting h = f + g, we have or By Theorem 1.27, h + - h - =!+ - f - + g + - g -and since each of these integrals is finite, we may transpose and obtain (2). (2) (3) (4) (5) 26 REAL AND COMPLEX ANALYSIS That (3) holds if C( > 0 follows from Proposition 1.24(c). It is easy to verify that (3) holds if C( = - 1, using relations like ( -u) + = u-. The case C( = i is also easy: Iff= u + iv, then fan= f (iu-v) = f (-v) + if u = -f v +i f u = {f u +if v) =iff Combining these cases with (2), we obtain (3) for any complex C(. /Ill 1.33 Theorem Iff E L1(Jl), then fxt dJ.L < ll/1 dJ.L. PROOF Put z = Jx f dJl. Since z is a complex number, there is a complex number C(, with I C( I = 1, such that C(Z = I z 1. Let u be the real part of C(f Then u < I C(f I = I f I . Hence If dJl = oc If dJl = Locf dJ.L = l u dJl < l1 f I dJl. The third of the above equalities holds since the preceding ones show that J C(f dJl is real. /Ill We conclude this section with another important convergence theorem. 1.34 Lebesgue's Dominated Convergence Theorem Suppose {fn} is a sequence of complex measurable functions on X such that f(x) = lim fn(x) exists for every x E X. If there is a function g E L1(Jl) such that and I fn(x) I < g(x) (n = 1, 2, 3, ... ; x E X), lim I If. -fl d11 = o, n- 00 Jx lim r f. dJ.L = r f dJl. n - oo Jx Jx (1) (2) (3) (4) ABSTRACT INTEGRATION 27 PROOF Since If I < g and f is measurable, f E L1(J1.). Since I fn -f I < 2g, Fatou's lemma applies to the functions 2g - I f n -f I and yields I 2g dp. < lim inf I (2g - I fn -f I ) d J1. Jx n - oo Jx = I 29 dp. + lim inf ( - I I !. -! I dJi.) Jx n - oo Jx = I 2g dp. - lim sup I I f. -! I dJi.. Jx n - oo Jx Since J 2g dJl is finite, we may subtract it and obtain lim sup I 1 /. -/ I dJi. < o. n - 00 Jx (5) If a sequence of nonnegative real numbers fails to converge to 0, then its upper limit is positive. Thus ( 5) implies (3). By Theorem 1.33, applied to fn -f, (3) implies (4). //// The Role Played by Sets of Measure Zero 1.35 Definition Let P be a property which a point x may or may not have. For instance, P might be the property "f(x) > 0" iff is a given function, or it might be " {fn(x)} converges " if {I n} is a given sequence of functions. If Jl is a measure on a a-algebra 9Jl and if E E 9Jl, the statement " P holds almost everywhere on E" (abbreviated to " P holds a.e. on E ") means that there exists an N E 9Jl such that Jl(N) = 0, N c E, and P holds at every point of E -N. This concept of a.e. depends of course very strongly on the given measure, and we shall write " a.e. [Jl.] " whenever clarity requires that the rneasure be indicated. For example, iff and g are measurable functions and if . J.l( {X : f (X) # g( X)}) = 0, (1) we say that/= g a.e. [Jl.] on X, and we may write/"' g. This is easily seen to be an equivalence relation. The transitivity (/"' g and g "' h implies f"' h) is a consequence of the fact that the union of two sets of measure 0 has measure 0. Note that iff"' g, then, for every E E IDl, (2) To see this, let N be the set which appears in (1); then E is the union of the disjoint sets E - N and E n N; on E -N,f= g, and J.l(E n N) = 0. 28 REAL AND COMPLEX ANALYSIS Thus, generally speaking, sets of measure 0 are negligible in integration. It ought to be true that every subset of a negligible set is negligible. But it may happen that some set N E 9Jl with J.l(N) = 0 has a subset E which is not a member of IDl. Of course we can define J.l(E) = 0 in this case. But will this extension of J.l still be a measure, i.e., will it still be defined on a a-algebra? It is a pleasant fact that the answer is affirmative : 1.36 Theorem Let (X, 9Jl, J.l) be a measure space, let IDl be the collection of all E c X for which there exist sets A and B E 9Jl such that A c E c B and J1(B - A) = 0, and define J.l(E) = Jl(A) in this situation. Then 9Jl is a a-algebra, and J.l is a measure on 9Jl. This extended measure J.l is called complete, since all subsets of sets of measure 0 are now measurable ; the a-algebra IDl is called the Jl-completion of IDl. The theorem says that every measure can be completed, so, whenever it is conve nient, we may assume that any given measure is complete; this just gives us more measurable sets, hence more measurable functions. Most measures that one meets in the ordinary course of events are already complete, but there are excep tions; one of these will occur in the proof of Fubini's theorem in Chap. 8 . PROOF We begin by checking that J.l is well defined for every E E IDl. Suppose A c E c B, A1 c E c B1, and J.l(B - A) = J.l(B1 -A1) = 0. (The letters A and B will denote members of 9Jl throughout this proof.) Since A -A1 c E-A1 c B1 -A1 we have J.l(A -A1) = 0, hence J.l(A) = J.l(A n A1). For the same reason, J.l(A1) = J.l(A 1 n A) . We conclude that indeed J1(A1) = J.l(A). Next, let us verify that IDl has the three defining properties of a a algebra. (i) X E IDl, because X E 9Jl and 9Jl c IDl. (ii) If A c E c B then Be c Ee c Ae. Thus E E IDl implies Ee E IDl, because Ae -Be :._ Ae n B = B -A. (iii) If Ai c Ei c Bi, E = U Ei, A = U Ai, B = U Bi, then A c E c B and 00 00 B -A = U (Bi -A) c U (Bi -Ai). 1 1 Since countable unions of sets of measure zero have measure zero, it follows that E E IDl if Ei e IDl for i = 1, 2, 3, .... Finally, if the sets Ei are disjoint in step (iii), the same is true of the sets Ai, and we conclude that 00 00 J.l(E) = J.l(A) = L J.l(Ai) = L J.l(Ei). 1 1 This proves that J.l is countably additive on IDl. Ill/ ABSTRACT INTEGRATION 29 1.37 The fact that functions which are equal a.e. are indistinguishable as far as integration is concerned suggests that our definition of measurable function might profitably be enlarged. Let us call a function f defined on a set E E 9Jl measurable on X if Jl(Ec) = 0 and if/ - 1(V) n E is measurable for every open set V. If we define f(x) = 0 for x E Ec, we obtain a measurable function on X, in the old sense. If our measure happens to be complete, we can define f on Ec in a perfectly arbitrary manner, and we still get a measurable function. The integral of f over any set A E 9Jl is independent of the definition off on Ec ; therefore this definition need not even be specified at all. There are many situations where this occurs naturally. For instance, a func tion f on the real line may be differentiable only almost everywhere (with respect to Lebesgue measure), but under certain conditions it is still true that f is the integral of its derivative ; this will be discussed in Chap. 7. Or a sequence {In} of measurable functions on X may converge only almost everywhere; with our new definition of measurability, the limit is still a measurable function on X, and we do not have to cut down to the set on which convergence actually occurs. To illustrate, let us state a corollary of Lebesgue's dominated convergence theorem in a form in which exceptional sets of measure zero are admitted: 1.38 Theorem Suppose {fn} is a sequence of omplex measurable functions defined a.e. on X such that Then the series 00 f(x) = L fn(x) n=l converges for almost all x,f E L1(J1,), and r t djl = f r t. djl. Jx n= 1 Jx (1) (2) (3) PROOF Let Sn be the set on whichf n is defined, so that Jl(Sӵ) = 0. Put <p(x) = L I fn(x) I' for X E s = n sn . Then Jl(Sc) = 0. By (1) and Theorem 1.27, f. <p djl < 00 . (4) If E = {xES: <p(x) < oo }, it follows from (4) that Jl(Ec) = 0. The series (2) converges absolutely for every x E E, and if f(x) is defined by (2) for x E E , then I f(x) I < <p(x) on E, so that f E L1(J1,) on E, by (4). If gn = /1 + · · · + fn, then I gn I < 1/n}, n = 1, 2, 3, . . . , then ! .u(A.> < i 1 d,u < r 1 d,u = o, n An JE so that Jl(An) = 0. Since { x E E: f(x) > 0} = U An , (a) follows. (b) Put f-= u + iv, let E = {x: u(x) > 0}. The real part of JE f dJ1 is then JE u+ dJ1. Hence JE u+ dJ1 = 0, and (a) implies that u+ = 0 a.e. We con clude similarly that - + - 0 U =V =V = a.e. (c) Examine the proof of Theorem 1.33. Our present assumption implies that the last inequality in the proof of Theorem 1.33 must actually be an equality. Hence J (If I -u) dJ1 = 0. Since If I -u > 0, (a) shows that If I = u a.e. This says that the real part of rxf is equal to I rxf I a.e., hence rxf = I rxf I = If I a.e ., which is the desired conclusion. /Ill 1.40 Theorem Suppose J1(X) < oo, f e L1(J1), S is a closed set in the complex plane, and the averages lie inS for every E E 9Jl with J1(E) > 0. Thenf(x) e S for almost all x EX. ABSTRACT INTEGRATION 31 PROOF Let n be a closed circular disc (with center at C( and radius r > 0, say) in the complement of S. Since sc is the union of countably many such discs, it is enough to prove that Jl.(E) = 0, where E = f-1(n). If we had Jl.( E) > 0, then I AJ.f) -ex I = Jl(Ġ i (/-ex) dJl < Jl(Ġ i I!- C( I dJi. < r, which is impossible, since AJf) E S. Hence Jl.(E) = 0. Ill/ 1.41 Theorem Let {Ek} be a sequence of measurable sets in X, such that (1) Then almost all x eX lie in at most finitely many ofthe sets Ek. PROOF If A is the set of all x which lie in infinitely many Ek, we have to prove that J.l(A) = 0. Put 00 g(x) = L XEk(x) (x EX). (2) k=l For each x, each term in this series is either 0 or 1. Hence x e A if and only if g(x) = oo. By Theorem 1.27, the integral of g over X is equal to the sum in (1). Thus g e L 1(J1.) , and so g(x) < oo a.e. /Ill Exercises 1 Does there exist an infinite a-algebra which has only countably many members? l Prove an analogue of Theorem 1.8 for n functions. 3 Prove that iff is a real function on a measurable space X such that { x: f(x) > r} is measurable for every rational r, then f is measurable. 4 Let {a,.} and {b,.} be sequences in [ - oo, oo], and prove the following assertions : (a) lim sup (-a,.) = - lim inf a,. . n-+ oo n -+ oo (b) lim sup (a,. + b,.) < lim sup a,. + lim sup b,. provided none of the sums is of the form oo - oo. (c) If a,. r b,. for all n, then n -+ oo lim inf a,. < lim inf b,. . n -+ oo n -+ oo Show by an example that strict inequality can hold in (b). n -+ oo 32 REAL AND COMPLEX ANALYSIS 5 (a) Suppose f: X y [ - oo, oo] and g: Xy [ - oo, oo] are measurable. Prove that the sets {x : f(x) < g(x)}, {x: f(x) = g(x)} are measurable. (b) Prove that the set of points at which a sequence of measurable real-valued functions con verges (to a finite limit) is measurable. 6 Let X be an uncountable set, let 9Jl be the collection of all sets E c: X such that either E or Ec is at most countable, and define JJ.(E) = 0 in the first case, JJ.(E) = 1 in the second. Prove that 9Jl is a a-algebra in X and that Jl is a measure on 9Jl. Describe the corresponding measurable functions and their integrals. 7 Suppose !, : X y [0, oo] is measurable for n = 1, 2, 3, .
. . , /1 > /2 > /3 > · · · > 0, f ,(x) y f(x) as ny oo, for every x e X, and /1 e L1(JJ.). Prove that then lim r !
, djJ. = r f djl n-+ oo Jx Jx and show that this conclusion does not follow if the condition "/1 e L1(Jl) " is omitted. 8 Put !, = XE if n is odd, !, = 1 - XE if n is even. What is the relevance of this example to Fatou's lemma? 9 Suppose Jl is a positive measure on X,f: Xy [0, oo] is measurable, Jx f dJl = c, where 0 < c < oo, and IX is a constant. Prove that 00 lim r n log [1 + (f /n)«] djJ. = c n-+ oo Jx 0 if 0 < IX < 1, if IX = 1, if 1 < IX < 00 . Hint: If IX > 1, the integrands are dominated by IX j. If IX < 1, Fatou's lemma can be applied. 10 Suppose JJ.(X) < oo, {!,} is a sequence of bounded complex measurable functions on X, and!, ź f uniformly on X. Prove that I lim r !
, dJl = r f djl, n= oo Jx Jx and show that the hypothesis " JJ.(X) < oo " cannot be omitted. 11 Show that 00 00 A = n U Ek n = 1 k =n in Theorem 1.41, and hence prove the theorem without any reference to integration. 12 Suppose f e E(JJ.). Prove that to each E > 0 there exists a {J > 0 such that JE I f I dJl < E whenever Jl(E) < b. 13 Show that proposition 1.24(c) is also true when c = oo. CHAPTER TWO POSITIVE BOREL MEASURES Vector Spaces 2.1 Definition A complex vector space (or a vector space over the complex field) is a set V, whose elements are called vectors and in which two oper ations, called addition and scalar multiplication, are defined, with the follow ing familiar algebraic properties : To every pair of vectors x and y there corresponds a vector x + y, in such a way that x + y = y + x and x + (y + z) = (x + y) + z; V contains a unique vector 0 (the zero vector or origin of V) such that x + 0 = x for every x E V; and to each x E V there corresponds a unique vector -x such that x+(-x)=O. To each pair (C(, x), where x E V and C( is a scalar (in this context, the word scalar means complex number), there is associated a vector C(X E V, in such a way that lx = x, C((f3x) = (C(f3)x, and such that the two distributive laws C((X + y) = C(X + C(y, (C( + {3)x = C(X + {3x (1) hold. , A linear transformation of a vector space V into a vector space V1 is a mapping A of V into V1 such that A(C(X + {3y) = C(AX + {3Ay (2) for all x and y E V and for all scalars rx and {3. In the special case in which V1 is the field of scalars (this is the simplest example of a vector space, except for the trivial one consisting of 0 alone), A is called a linear functional. A linear functional is thus a complex function on V which satisfies (2). Note that one often writes Ax, rather than A(x), if A is linear. 33 34 REAL AND COMPLEX ANALYSIS The preceding definitions can of course be made equally well with any field whatsoever in place of the complex field. Unless the contrary is explicitly stated, however, all vector spaces occurring in this book will be complex, with one notable exception : the euclidean spaces Rk are vector spaces over the real field. 2.2 Integration as a Linear Functional Analysis is full of vector spaces and linear transformations, and there is an especially close relationship between integration on the one hand and linear functionals on the other. For instance, Theorem 1.32 shows that L1(f.l) is a vector space, for any posi tive measure f.l, and that the mapping (1) is a linear functional on L1(J1.). Similarly, if g is any bounded measurable function, the mapping t-lfg djJ (2) is a linear functional on L 1(J1.); we shall see in Chap. 6 that the functionals (2) are, in a sense, the only interesting ones on L1(f.l). For another example, let C be the set of all continuous complex functions on the unit interval I = [0, 1]. The sum of the two continuous functions is contin uous, and so is any scalar multiple of a continuous function. Hence C is a vector space, and if Af= rf(x) dx (f E C), (3) the integral being the ordinary Riemann integral, then A is clearly a linear func tional on C; A has an additional interesting property: it is a positive linear func tional. This means that Af > 0 whenever f > 0. One of the tasks which is still ahead of us is the construction of the Lebesgue measure. The construction can be based on the linear functional (3), by the fol lowing observation : Consider a segment (a, b) c I and consider the class of all f e C such that 0 <! < 1 on I and f(x) = 0 for all x not in (a, b). We have Af < b -a for all such f, but we can choose f so that Af is as close to b -a as desired. Thus the length (or measure) of (a, b) is intimately related to the values of the functional A. The preceding observation, when looked at from a more general point of view, leads to a remarkable and extremely important theorem of F. Riesz: To every positive linear functional A on C corresponds a finite positive Borel measure f.l on I such that (f E C). (4) POSITIVE BOREL MEASURES 35 [The converse is obvious : if J1 is a finite positive Borel measure on I and if A is defined by (4), then A is a positive linear functional on C.] It is clearly of interest to replace the bounded interval I by R1• We can do this by restricting attention to those continuous functions on R1 which vanish outside some bounded interval. (These functions are Riemann integrable, for instance.) Next, functions of several variables occur frequently in analysis. Thus we ought to move from R 1 to R". It turns out that the proof of the Riesz theorem still goes through, with hardly any changes. Moreover, it turns out that the euclidean properties of R" (coordinates, orthogonality, etc.) play no role in the proof; in fact, if one thinks of thetn too much they just get in the way. Essential to the proof are certain topological properties of R". (Naturally. We are now dealing with continuous functions.) The crucial property is that of local com pactness: Each point of R" has a neighborhood whose closure is compact. We shall therefore establish the Riesz theorem in a very general setting (Theorem 2.14). The existence of Lebesgue measure follows then as a special case. Those who wish to concentrate on a more concrete situation may skip lightly over the following section on topological preliminaries (Urysohn's lemma is the item of greatest interest there; see Exercise 3) and may replace locally compact Hausdorff spaces by locally compact metric spaces, or even by euclidean spaces, without missing any of the principal ideas. It should also be mentioned that there are situations, especially in probability theory, where measures occur naturally on spaces without topology, or on topo logical spaces that are not locally compact. An example is the so-called Wiener measure which assigns numbers to certain sets of continuous functions and which is a basic tool in the study of Brownian motion. These topics will not be dis cussed in this book. Topological Preliminaries 2.3 Definitions Let X be a topological space, as defined in Sec. 1.2. (a) A set E c X is closed if its complement Ec is open. (Hence 0 and X are closed, finite unions of closed sets are closed, and arbitrary intersections of closed sets are closed.) (b) The closure E of a set E c X is the smallest closed set in X which con tains E. (The following argument proves the existence of E: The collec tion Q of all closed subsets of X which contain E is not empty, since X e Q; let E be the intersection of all members of Q.) (c) A set K c X is compact if every open cover of K contains a finite sub cover. More explicitly, the requirement is that if { Œ} is a collection of open sets whose union contains K, then the union of some finite sub collection of { Ya} also contains K. In particular, if X is itself compact, then X is called a compact space. (d) A neighborhood of a point p e X is any open subset of X which contains p. (The use of this term is not quite standardized; some use 36 REAL AND COMPLEX ANALYSIS " neighborhood of p" for any set which contains an open set containing p.) (e) X is a Hausdorff space if the following is true: If p E X, q E X, and p =I= q, then p has a neighborhood U and q has a neighborhood V such that U n V = 0. (f) X is locally compact if every point of X has a neighborhood whose closure is compact. Obviously, every compact space is locally compact. We recall the Heine-Borel theorem : The compact subsets of a euclidean space R" are precisely those that are closed and bounded (,t Theorem 2.41). From this it follows easily that R" is a locally compact Hausdorff space. Also, every metric space is a Hausdorff space. 2.4 Theorem Suppose K is compact and F is closed, in a topological space X. IfF c K, then F is compact. PROOF If { ƚ} is an open cover of F and W = Fe, then W u Ua ƚ covers X; hence there is a finite collection { ƚJ such that K c W u ƚ1 u · · · u V«n . Then F c ƚ1 u · · · U ƚn • Ill/ Corollary If A c B and if B has compact closure, so does A. 2.5 Theorem Suppose X is a Hausdorff space, K c X, K is compact, and p E Kc. Then there are open sets U and W such that p E U, K c W, and U n W = 0. PROOF If q E K, the Hausdorff separation axiom implies the existence of dis joint open sets U q and ( , such that p E U q and q E ( . Since K is compact, there are points qb . . . , qn E K such that K c (1 u · · · u Vq" . Our requirements are then satisfied by the sets Corollaries U = Uq1 n · · · n Uq" and W = (1 U · · · u V qn · //// (a) Compact subsets of Hausdorff spaces are closed. (b) If F is closed and K is compact in a Hausdorff space, then F n K zs compact. Corollary (b) follows from (a) and Theorem 2.4. t Numbers in brackets refer to the Bibliography. POSITIVE BOREL MEASURES 37 2.6 Theorem I f { Ka} is a collection of compact subsets of a Hausdorff space and if na Ka = 0, then some finite subcollection of { Ka} also has empty inter section. PROOF Put ɓ = K. Fix a member K1 of {Ka}· Since no point of K1 belongs to every Ka, { Œ} is an open cover of K 1. Hence K 1 c ɓ u · · · u ɓ" for some finite collection { ɓJ. This implies that Ill/ 2.7 Theorem Suppose U is open in a locally compact Hausdorff space X, K c U, and K is compact. Then there is an open set V with compact closure such that K c V c V c U. PROOF Since every point of K has a neighborhood with compact closure, and since K is covered by the union of finitely many of these neighborhoods, K lies in an open set G with compact closure. If U =X, take V = G. Otherwise, let C be the complement of U. Theorem 2.5 shows tha!!o each p E C there corresponds an open set W, such that K c W, and p ¢ WP. Hence { C n G n W,}, where p ranges over C, is a collection of compact sets with empty intersection. By Theorem 2.6 there are points Pb . . . , Pn E C such that The set V= G n W n ··· n W Pl Pn then has the required properties, since V C G (\ WPl (\ • • • (\ WPn • Ill/ 2.8 Definition Let f be a real (or extended-real) function on a topological space. If {x:f(x) > ex} is open for every real ex, f is said to be lower semicontinuous. If {x:f(x) <ex} is open for every real ex, f is said to be upper semicontinuous. A real function is obviously continuous if and only if it is both upper and lower semicontinuous. The simplest examples of semicontinuity are furnished by characteristic func tions: 38 REAL AND COMPLEX ANALYSIS (a) Characteristic functions of open sets are lower semicontinuous. (b) Characteristic functions of closed sets are upper semicontinuous. The following property is an almost immediate consequence of the defini tions: (c) The supremum of any collection of lower semicontinuous functions is lower semicontinuous. The infimum of any collection of upper semicontinuous func tions is upper semicontinuous. 2.9 Definition The support of a complex function f on a topological space X is the closure of the set {x:f(x) =F 0}. The collection of all continuous complex functions on X whose support is compact is denoted by Cc(X). Observe that Cc(X) is a vector space. This is due to two facts: (a) The support off + g lies in the union of the support off and the support of g, and any finite union of compact sets is compact. (b) The sum of two continuous complex functions is continuous, as are scalar multiples of continuous functions. (Statement and proof of Theorem 1.8 hold verbatim if " measurable function " is replaced by " continuous function," " measurable space " by " topological space "; take q,(s, t) = s + t, or s, t) = st, to prove that sums and products of contin uous functions are continuous.) 2.10 Theorem Let X and Y be topological spaces, and let f: X---+ Y be contin uous. If K is a compact subset of X, thenf(K) is compact. PROOF If { Œ} is an open cover of f(K), then {f- 1( Œ)} is an open cover of K, hence K c f - 1(ɓ1) u · · · u f -t(ŒJ for some cx1, .
.
• , cxn , and therefore f ( K) c Œ 1 u · · · u Œ" . I I I I Corollary The range of any f e Cc(X) is a compact subset of the complex plane. In fact, if K is the support off e Cc(X), then f(X) c f(K) u { 0}. If X is not compact, then 0 e f(X), but 0 need not lie inf(K), as is seen by easy examples. 2.1 1 Notation In this chapter the following conventions will be used. The notation K<.f (1) POSITIVE BOREL MEASURES 39 will mean that K is a compact subset of X, that/ E Cc(X), that 0 <f(x) < 1 for all x E X, and thatf(x) = 1 for all x E K. The notation !« v (2) will mean that V is open, that/ E Cc(X), 0 <!< 1, and that the support off lies in V. The notation K «fˁ V (3) will be used to indicate that both (1) and (2) hold. 2.12 Urysohn's Lemma Suppose X is a locally compact Hausdorff space, V is open in X, K c V, and K is compact. Then there exists an f E Cc(X), such that K «f c:< V. (1) In terms of characteristic functions, the conclusion asserts the existence of a continuous function f which satisfies the inequalities XK < f < Xv . Note that it is easy to find semicontinuous functions which do this ; examples are XK and Xv . PROOF Put r 1 = 0, r2 = 1, and let r3, r 4 , r5 , • • • be an enumeration of the rationals in (0, 1). By Theorem 2.7, we can find open sets V 0 and then V1 such that V 0 is compact and (2) Suppose n > 2 and V,.1 , • • • , V,." have been chosen in such a manner that ri < ri implies -v,.i c V,.i . Then one of the numbers rb ... , rn, say ri, will be the largest one which is smaller than rn+ 1 , and another, say ri , will be the smal lest one larger than r n + 1 . Using Theorem 2. 7 again, we can find V,." + 1 so that -v,.j C V,.n+ 1 C -v,.n + 1 C V,.i • Continuing, we obtain a collection { V,.} of open sets, one for every rational r E [0, 1], with the following properties: K c Vb V 0 c V, each -v,. is compact, and Define fr(x)={p . and s > r implies ʉ c V,.. if X E V,., otherwise, f = sup/,. , r g.(x) = {! s if X E ʉ ' otherwise, (3) (4) (5) The remarks following Definition 2.8 show that f is lower semi continuous and that g is upper semicontinuous. It is clear that 0 <! < 1, that 40 REAL AND COMPLEX ANALYSIS f(x) = 1 if x e K, and that f has its support in V 0 . The proof will be com pleted by showing thatf = g. The inequality fr(x) > gs(x) is possible only if r > s, x e V,., and x ¢ ʉ . But r > s implies V,. c Ӵ . Hence/,. < gs for all r and s, so f < g. Suppose f(x) < g(x) for some x. Then there are rationals r and s such that f(x) < r < s < g(x). Since f(x) < r, we have x ¢ V,.; since g(x) > s, we have x e ʉ . By (3), this is a contradiction. Hencef =g. /Ill 2.13 Theorem Suppose Vb . . . , V, are open subsets of a locally compact H aus dorff space X, K is compact, and K c V 1 u · · · u Vn . Then there exist functions hiƙ f!i (i = 1, ... , n) such that (x E K). (1) Because of (1), the collection { hb . . . , hn} is called a partition of unity on K, subordinate to the cover { V 1, • • • , V,}. PROOF By Theorem 2.7, each x e K has a neighborhood x with compact closure x c V; for some i (depending on x). There are points xb ... , xm such that x1 u · · · u xm :::J K. If 1 < i < n, let Hi be the union of those W xi which lie in V; . By Urysohn's lemma, there are functions gi such that Hiƙ gi ƙ V; . Define h1 = g1 h2 = (1 -g1)g2 hn = (1 -g1)(1 -g2) · · · (1 -gn- l)gn · Then hiƙ V; . It is easily verified, by induction, that (2) h 1 + h 2 + · · · + h n = 1 -( 1 -g 1 )( 1 -g 2) • • • ( 1 -g n)• ( 3) Since K c H 1 u · · · u Hn, at least one gi(x) = 1 at each point x e K ; hence (3) shows that ( 1) holds. I I I I The Riesz Representation Theorem 2.14 Theorem Let X be a locally compact Hausdorff space, and let A be a positive linear functional on Cc(X). Then there exists a a-algebra 9Jl in X which contains all Borel sets in X, and there exists a unique positive measure J.l on 9Jl which represents A in the sense that (a) Af = Jx f dJ.lfor every f E Cc(X), and which has the following additional properties: POSITIVE BOREL MEASURES 41 (b) Jl(K) < oo for every compact set K c X. (c) For every E E 9Jl, we have Jl(E) = inf {!1( V): E c V, V open}. (d) The relation 11(E) = sup {Jl(K): K c E, K compact} holds for every open set E, and for every E e 9Jl with Jl(E) < oo. (e) If E E 9Jl, A c E, and Jl(E) = 0, then A e IDl. For the sake of clarity, let us be more explicit about the meaning of the word "positive " in the hypothesis: A is assumed to be a linear functional on the complex vector space Cc(X), with the additional property that Afis a nonnegative real number for every f whose range consists of nonnegative real numbers. Briefly, iff(X) c [0, oo) then Af e [0, oo ). Property (a) is of course the one of greatest interest. After we define 9Jl and Jl, (b) to (d) will be established in the course of proving that 9Jl is a a-algebra and that 11 is countably additive. We shall see later (Theorem 2. 18) that in "reasonable " spaces X every Borel measure which satisfies (b) also satisfies (c) and (d) and that (d) actually holds for every E E 9Jl, in those cases. Property (e) merely says that (X, 9Jl, 11) is a complete measure space, in the sense of Theorem 1.36. Throughout the proof of this theorem, the letter K will stand for a compact subset of X, and V will denote an open set in X. Let us begin by proving the uniqueness of Jl. If 11 satisfies (c) and (d), it is clear that 11 is determined on 9Jl by its values on compact sets. Hence it suffices to prove that /11 (K) = Jl2(K) for all K, whenever 111 and 112 are measures for which the theorem holds. So, fix K and E > 0. By (b) and (c), there exists a V :::J K with Jl2(V) < Jl2(K) + E; by Urysohn's lemma, there exists an f so that K -<..
f-<.. V; hence Jll(K) = I XK dJ11 < If dJ11 = Af = If dJ12 < I Xv dJ12 = J12(V) < J12(K) + E. Thus /1 1 (K) < Jl2(K). If we interchange the roles of 111 and 112, the opposite inequality is obtained, and the uniqueness of 11 is proved. Incidentally, the above computation shows that (a) forces (b). Construction of 11 and 9Jl For every open set V in X, define Jl(V) = sup { Af: f-<.. V}. (1) 42 REAL AND COMPLEX ANALYSIS If V1 c V2, it is clear that (1) implies Jl(Vd < Jl(V2). Hence Jl(E) = inf {Jl(V): E c V, V open}, (2) if E is an open set, and it is consistent with ( 1) to define Jl(E) by (2), for every E c X. Note that although we have defined Jl(E) for every E c X, the countable additivity of Jl will be proved only on a certain a-algebra 9Jl in X. Let IDlF be the class of all E c X which satisfy two conditions: Jl(E) < oo, and Jl(E) = sup {Jl(K): K c E, K compact}. (3) Finally, let 9Jl be the class of all E c X such that E n K e IDlF for every compact K. Proof that Jl and 9Jl have the required properties It is evident that Jl is monotone, i.e., that Jl(A) < Jl(B) if A c B and that Jl(E) = 0 implies E e IDlF and E e IDl. Thus (e) holds, and so does (c), by defini tion. Since the proof of the other assertions is rather long, it will be convenient to divide it into several steps. Observe that the positivity of A implies that A is monotone: f < g implies Af < Ag. This is clear, since Ag = Af + A(g -f) and g -f > 0. This monot onicity will be used in Steps II and X. STEP I If E1, E2, E3, • • • are arbitrary subsets of X, then (4) PROOF We first show that (5) if V1 and V2 are open. Choose g ƙ V1 u V2• By Theorem 2.13 there are func tions h 1 and h2 such that hi ƙ V; and h 1 (x) + h2(x) = 1 for all x in the support of g. Hence hi g ƙ V;, g = h1g + h2 g, and so (6) Since (6) holds for every g ƙ V1 u V2, (5) follows. If Jl(Ei) = oo for some i, then ( 4) is trivially true. Suppose therefore that Jl(Ei) < oo for every i. Choose E > 0. By (2) there are open sets V; => Ei such that ( i = 1' 2, 3' . . . ). POSITIVE BOREL MEASURES 43 Put V = Uf Ⱦ, and choose/ V. Sincefhas compact support, we see that f V1 u · · · u V,. for some n. Applying induction to (5), we therefore obtain 00 Af < ,u( v1 u . . . u Vn) < ,u( v1) + . . . + ,u( Vn) < L ,u( E i) + E. i= 1 Since this holds for every f V, and since U Ei c V, it follows that .u(91 E) < ,u(V) < J/(Ei) + E, which proves (4), since E was arbitrary. /Ill STEP II If K is compact, then K e 9RF and ,u(K) = inf {Af: K ˁf}. (7) This implies assertion (b) of the theorem. PROOF If K ȿ and 0 < rx < 1, let Œ = {x: f(x) > rx}. Then K c Œ, and rxg < f whenever g Œ. Hence JJ(K) < Jl(Œ) = sup {Ag: g Œ} < rx-1 Af Let rx ׳ 1, to conclude that Jl(K) < Af. (8) Thus Jl(K) < oo. Since K evidently satisfies (3), K e 9RF. If E > 0, there exists V :::J K with Jl(V) < Jl(K) + E. By Urysohn's lemma, K ˁ V for some f Thus Af < Jl(V) < Jl(K) + E, which, combined with (8), gives (7). Ill/ STEP III Every open set satisfies (3). Hence 9RF contains every open set V with Jl(V) < oo . PROOF Let rx be a real" number such that rx < Jl(V). There exists anf V with rx < Af If W is any open set which contains the support K off, then f W, hence Af < Jl(W). Thus Af < Jl(K). This exhibits a compact K c V with rx < JJ(K), so that (3) holds for V. /Ill STEP IV Suppose E = u% 1 Ei, where E1, E2' E3' ... are'pairwise disjoint members of9RF. Then (9) i= 1 If, in addition, Jl(E) < oo, then also E e 9RF. 44 REAL AND COMPLEX ANALYSIS PROOF We first show that (10) if K 1 and K 2 are disjoint compact sets. Choose E > 0. By Urysohn's lemma, there exists f E Cc(X) such that f(x) = 1 on K 1, f(x) = 0 on K2, and 0 < f < 1. By Step II there exists g such that K1 u K2-< g and Ag < Jl(K1 u K2) +E. Note that K 1 -<fg and K2-< (1 -f)g. Since A is linear, it follows from (8) that Since E was arbitrary, (10) follows now from Step I. If J1(E) = oo, (9) follows from Step I. Assume therefore that J1(E) < oo , and choose E > 0. Since Ei E IDlF, there are compact sets Hi c Ei with (i = 1, 2, 3, ... ). Putting Kn = H1 u · · · u Hn and using induction on (10), we obtain n n Jl(E) > J1(Kn) = L J1(Hi) > L J1(Ei)-E. i=1 i=l (1 1) (12) Since ( 12) holds for every n and every E > 0, the left side of (9) is not smaller than the right side, and so (9) follows from Step I. But if J1(E) < oo and E > 0, (9) shows that N J1(E) < L J1(Ei) + E (13) i = 1 for some N. By (12), it follows that Jl(E) < J1(KN) + 2E, and this shows that E satisfies (3); hence E E IDlF. I I I I STEP v If E E IDlF and E > 0, there is a compact K and an open V such that K c E c V and J1(V -K) < E. PROOF Our definitions show that there exist K c E and V :::J E so that E E J1(V) - 2 < Jl(E) < J1(K) + 2 . Since V -K is open, V -K E IDlF, by Step III. Hence Step IV implies that J1(K) + J1(V-K) = J1(V) < J1(K) +E. /Ill STEP VI If A E IDlF and B E IDlF, then A - B, A u B, and A n B belong to IDlF. POSITIVE BOREL MEASURES 45 PROOF If E > 0, Step V shows that there are sets Ki and- V; such that K1 c A c V1, K2 c B c V2 , and ,u(V; -Ki) < E, for i = 1, 2. Since A - B c V1 - K2 c (V1 - K1) u (K1 - V2) u (V2 - K2), Step I shows that (14) Since K 1 - V2 is a compact subset of A - B, (14) shows that A - B satisfies (3), so that A -B E IDlF . Since A u B = (A - B) u B, an application of Step IV shows that A u B E IDlp . Since A n B = A - (A -B), we also have A n BE IDlp . /Ill STEP vn 9Jl is a a-algebra in X which contains all Borel sets. PROOF Let K be an arbitrary compact set in X. If A E IDl, then Ac n K = K - (A n K), so that Ac n K is a difference of two members of Jt F . Hence Ac n K E IDlF , and we conclude: A E 9Jl implies Ac E 9Jl. Next, suppose A = U'f Ai , where each Ai E IDl. Put B1 = A 1 n K, and (n = 2, 3, 4, . . . ). (15) Then { Bn} is a disjoint sequence of members of IDlF , by Step VI, and A n K = U 'f Bn. It follows from Step IV that A n K E IDlF . Hence A E IDl. Finally, if C is closed, then C n K is compact, hence C n K E IDlF , so C E IDl. In particular, X E IDl. We have thus proved that 9Jl is a a-algebra in X which contains all closed subsets of X. Hence 9Jl contains all Borel sets in X. /Ill STEP VIII IDlF consists of precisely those sets E E 9Jl for which J.l(E) < oo . This implies assertion (d) of the theorem. PROOF If E E IDlF , Steps II and VI imply that E n K E IDlF for every compact K, hence E E IDl. Conversely, suppose E E 9Jl and ,u(E) < oo , and choose E > 0. There is an open set V ::::> E with ,u(V) < oo ; by III and V, there is a compact K c V with ,u(V - K) <E. Since E n K E IDlp , there is a compact set H c E n K with ,u(E n K) < ,u(H) + E. Since E c (E n K) u (V - K), it follows that ,u(E) < J.l(E n K) + ,u(V - K) < J.l(H) + 2E, which implies that E E IDlF . Ill/ STEP IX ,u is a measure on IDl. 46 REAL AND COMPLEX ANALYSIS PROOF The countable additivity of J.l on 9Jl follows immediately from Steps IV and VIII. I II I STEP X For every f E Cc(X), Af = Jx f dj.J,. This proves (a), and completes the theorem. PROOF Clearly, it is enough to prove this for real f Also, it is enough· to prove the inequality (16) for every real f E Cc(X). For once (16) is establiɀɁhed, the linearity of A shows that -Af= A(-f) < l(-/) dJ.l = -lfdp., which, together with (16), shows that equality holds in (16). Let K be the support of a real f E Cc(X), let [a, b] be an interval which contains the range off (note the Corollary to Theorem 2.10), choose E > 0, and choose Yi, for i = 0, 1, ... , n, so that Yi - Yi- 1 < E and Yo <a < Y1 < · · · < Yn = b. (17) Put Ei = {x: Yi- 1 <f(x) < Yi} n K (i = 1, ... , n). (18) Since f is continuous, f is Borel measurable, and the sets Ei are therefore disjoint Borel sets whose union is K. There are open sets V; => Ei such that (i = 1, ... , n) (19) and such that f(x) < Yi + E for all x E V; . By Theorem 2.13, there are func tions hi Ѣ V; such that L hi = 1 on K. Hence f = L hi f, and Step II shows that POSITIVE BOREL MEASURES 47 Since h;f < (Y; + e)h;, and since Y; -E <f(x) onE;, we have n n Af = L A(h;f) < L (Y; + e)Ah; i= 1 i= l n n = L (I a I + Y; + e)Ah; - I a I L Ah; i=l i= l n < L (I a I + Y; + e)[JJ(E;) + eln] - I a I JJ(K) i= 1 n € n = L (y; -e)JJ(E;) + 2eJJ(K) + - L (I a I + Yi + e) i= l n i= 1 < LfdJt + E[2t(K) + Ia I + b + E]. Since e was arbitrary, (16) is established, and the proof of the theorem is complete. I I I I Regularity Properties of Borel Measures 2.15 Definition A measure Jl defined on the a--algebra of all Borel sets in a locally compact Hausdorff space X is called a Borel measure on X. If Jl is positive, a Borel set E c X is outer regular or inner regular, respectively, if E has property (c) or (d) of Theorem 2.14. If every Borel set in X is both outer and inner regular, Jl is called regular. In our proof of the Riesz theorem, outer regularity of every set E was built into the construction, but inner regularity was proved only for the open sets and for those E e IDl for which JJ(E) < oo. It turns out that this flaw is in the nature of things. One cannot prove regularity of JJ under the hypothesis of Theorem 2 .. 14; an example is described in Exercise 17. However, a slight strengthening of the hypotheses does give us a regular measure. Theorem 2.17 shows this. And if we specialize a little more, Theorem 2.18 shows that all regularity problems neatly disappear. . 2.16 Definition A set E in a topological space is called u-compact if E is a countable union of compact sets. A set E in a measure space (with measure JJ) is said to have u-finite measure if E is a countable union of sets E; with JJ(E;) < oo. For example, in the situation described in Theorem 2.14, every a- compact set has a--finite measure. Also, it is easy to see that if E e IDl and E has u-finite measure, then E is inner regular. 48 REAL AND COMPLEX ANALYSIS 2.17 Theorem Suppose X is a locally compact, a-compact Hausdorff space. If 9J1 and J1 are as described in the statement of Theorem 2.14, then 9Jl and J1 have the following properties: (a) If E E 9Jl and E > 0, there is a closed set F and an open set V such that F c E c V and Jl(V - F) < E. (b) J1 is a regular Borel measure on X. (c) If E E 9Jl, there are sets A and B such that A zs an Fa , B is a GtS, A c E c B, and J1(B - A) = 0. As a corollary of (c) we see that every E E 9Jl is the union of an Fa and a set of measure 0. PROOF Let X = K 1 u K 2 u K 3 u · · · , where each Kn is compact. If E E 9Jl and E > 0, then Jl(Kn n E) < oo , and there are open sets V, :::J Kn n E such that (n = 1, 2, 3, ... ). (1) If V = U V,, then V-E c U (V,-(Kn n E)), so that E J1(V-E)< 2. Apply this to Ec in place of E: There is an open set W :::J Ec such that J1(W -Ec) < E/2. If F = we , then F c E, and E - F = W -Ec. Now (a) follows. Every closed set F c X is a-compact, because F = U (F n Kn). Hence (a) implies that every set E E 9Jl is inner regular. This proves (b). If we apply (a) with E = 1/j U = 1, 2, 3, . . . ), we obtain closed sets Fi and open sets l-j such that Fi c E c l-j and Jl(l-} - Fi) < 1/j. Put A = U Fi and B = n l-j . Then A c E c B, A is an Fa , B is a G tS, and J1(B -A) = 0 since B-A c l-j - Fi for j = 1, 2, 3, .... This proves (c). /Ill 2.18 Theorem Let X be a locally compact Hausdorff space in which every open set is a-compact. Let A be any positive Borel measure on X such that A(K) < oo for every compact set K. Then A is regular. Note that every euclidean space Rk satisfies the present hypothesis, since every open set in Rk is a countabie union of closed balls. POSITIVE BOREL MEASURES 49 PROOF Put Af = Jx fdA., for f e Cc(X). Since A.(K) < oo for every compact K, A is a positive linear functional on Cc(X), and there is a regular measure Jl, satisfying the conclusions of Theorem 2.17, such that (1) We will show that A. = Jl. Let V be open in X. Then V = U Ki, where Ki is compact, i = 1, 2, 3, . . . . By Urysohn's lemma we can choose_h so that Ki <..
.h <.. V. Let 9n = max (/b . . . , fn). Then 9n E Cc(X) and 9n(x) increases to Xv(x) at every point x e X. Hence (1) and the monotone convergence theorem imply A(V) = lim f 9n dA = lim f 9n dp. = p.(V) . n- oo Jx n- oo Jx (2) Now let E be a Borel set in X, and choose E > 0. Since J.l satisfies Theo rem 2.17, there is a closed set F and an open set V such that F c E c V and Jl(V-F) < E. Hence Jl(V) < Jl(F) + E < Jl(E) + E. Since V -F is open, (2) shows that A.(V -· F) < E, hence A.(V) ѡ A.(E) + E. Consequently and A.(E) < A.(V) = Jl(V) < Jl(E) + E Jl(E) < Jl(V) = A.(V) < A.(E) + E ' so that I A.( E) -Jl(E) I < E for every E > 0. Hence A.( E) = JL(E). /Ill In Exercise 18 a compact Hausdorff space is described in which the com plement of a certain point fails to be a-compact and in which the conclusion of the preceding theorem is not true. Lebesgue Measure 2.19 Euclidean Spaces Euclidean k-dimensional space Rk is the set of all points x = (· 1, .
. 0 , ·k) whose coordinates ·i are real numbers, with the following alge braic and topological structure: If x = (·1, • .
. , ĉk), y = (171, ... , 11k), and ex is a real number, x + y and cxx are defined by (1) This makes Rk into a real vector space. If x · y = L ·; 1'/; and I xI = (x · x)112, the Schwarz inequality I x 0 y I < I x I I y ! leads to the triangle inequality lx-yl <lx-zl + lz-yl; (2) hence we obtain a metric by setting p(x, y) = I x -y 1 . We assume that these facts are familiar and shall prove them in greater generality in Chap. 4. 50 REAL AND COMPLEX ANALYSIS If E c Rk and x e Rk, the translate of E by x is the set E +X= {y + x: y E E}. A set of the form W = {X: (Xi < =i < {3i , 1 < i < k}, (3) (4) or any set obtained by replacing any or all of the < signs in ( 4) by < , is called a k-cell; its volume is defined to be k vol (W) = fl (f3i - rxi). (5) i= 1 If a e Rk and b > 0, we shall call the set Q( a; b) = { x: rxi < = i < rxi + b, 1 < i < k} (6) the b-box with corner at a. Here a = ( rx b . . . , rxk). For n = 1, 2, 3, . . . , we let P n be the set of all x e Rk whose coordinates are integral multiples of 2 -n, and we let nn be the collection of all 2 -n boxes with corners at points of Pn . We shall need the following four properties of {!ln}· The first three are obvious by inspection. (a) Ifn isfixed, each x e Rk lies in one and only one member of !ln. (b) If Q' E nn ' Q" E Qr ' and r < n, then either Q' c Q" or Q' n Q" = 0. (c) If Q E nr , then vol (Q) = 2-rk; and if n > r, the set pn has exactly 2(n-r)k points in Q. (d) Every nonempty open set in Rk is a countable union of disjoint boxes belonging to 01 u 02 u 03 u · · · . PROOF OF (d) If V is open, every x e V lies in an open ball which lies in V; hence x E Q c V for some Q belonging to some Qn . In other words, V is the union of all boxes which lie in V and which belong to some Qn . From this collection of boxes, select those which belong to Qb and remove those in 02, 03 , . . . which lie in any of the selected boxes. From the remaining collection, select those boxes of 02 which lie in V, and remove those in 03, 04, • • • which lie in any of the selected boxes. If we proceed in this way, (a) and (b) show that (d) holds. /Ill 2.20 Theorem There exists a positive complete measure m defined on a u algebra 9Jl in Rk, with the following properties: (a) m(W) = vol (W)for every k-cell W. (b) 9Jl contains all Borel sets in Rk; more precisely, E e 9Jl if and only if there are sets A and B c Rk such that AcE c B, A is an Fa, B is a G״, and m(B - A) = 0. Also, m is regular. POSITIVE BOREL MEASURES 51 (c) m is translation-invariant, i.e., m(E + x) = m(E) for every E e 9Jl and every x e Rk. (d) If J1 is any positive translation-invariant Borel measure on Rk such that J1(K) < oo for every compact set K, then there is a constant c such that J1(E) = cm(E)for all Borel sets E c Rk. (e) To every linear transformation T of Rk into Rk corresponds a real number A(T) such that m(T(E)) = A(T)m(E) for every E e IDl. In particular, m(T(E)) = m(E) when Tis a rotation. The members of 9Jl are the Lebesgue measurable sets in Rk; m is the Lebesgue measure on Rk. When clarity requires it, we shall write mk in place of m. PROOF Iff is any complex function on Rk, with compact support, define An f= 2 -nk L f(x) ( n = 1 , 2, 3, . . . ), (1) x e Pn where P n is as in Sec. 2.19. Now suppose f e Cc(Rk), f is real, W is an open k-cell which contains the support of f, and E > 0. The uniform continuity of f(, Theorem 4.19) shows that there is ɂn integer N and that there are functions g and h with support in W, such that (i) g and h are constant on each box belonging to QN, (ii) g <f< h, and (iii) h-g N, Property 2.19(c) shows that (2) Thus the upper and lower limits of {An/} differ by at most E vol (W), and since E was arbitrary, we have proved the existence of Af= lim Anf (3) It is immediate that A is a positive linear functional on Cc(Rk). (In fact, Af is precisely the Riemann integral off over Rk. We went through the preceding construction in order not to have to rely on any theorems about Riemann integrals in several variables.) We define m and 9Jl to be the measure and a-algebra associated with this A as in Theorem 2.14. Since Theorem 2.14 gives us a complete measure and since Rk is u compact, Theorem 2.17 implies assertion (b) of Theorem 2.20. 52 REAL AND COMPLEX ANALYSIS To prove (a), let W be the open cell 2.19(4), let Er be the union of those boxes belonging to Qr whose closures lie in W, choose /,. so that Er-< /,.-< W, and put gr = max {/1, • .
. ,!,.}. Our construction of A shows that (4) As rr oo , vol (Er)r vol (W), and Ag, =I g, dm---+ m(W) (5) by the monotone convergence theorem, since gr(x)r Xw(x) for all x e Rk. Thus m(W) = vol (W) for every open cell W, and since every k-cell is the intersection of a decreasing sequence of open k-cells, we obtain (a). The proofs of (c), (d), and (e) will use the following observation: If A. is a positive Borel measure on Rk and A.(E) = m(E) for all boxes E, then the same equality holds for all open sets E, by property 2.19(d), and therefore for all Borel sets E, since A. and m are regular (Theorem 2.18). To prove (c), fix x e Rk and define A.(E) = m(E + x). It is clear that A. is then a measure; by (a), A.(E) = m(E) for all boxes, hence m(E + x) = m(E) for all Borel sets E. The same equality holds for every E e IDl, because of (b). Suppose next that J1 satisfies the hypotheses of (d). Let Q0 be a 1-box, put c = J1(Q0). Since Q0 is the union of 2nk disjoint 2 -n boxes that are translates of each other, we have 2"kJ1(Q) = J1(Q0) = cm(Q0) = c · 2nkm(Q) for every 2 -"-box Q. Property 2.19(d) implies now that J1(E) =em( E) for all open sets E c Rk. This proves (d). To prove (e), let T: RkŠ Rk be linear. If the range of T is a subspace Y of lower dimension, then m(Y) = 0 and the desired conclusion holds with A(T) = 0. In the other case, elementary linear algebra tells us that T is a one-to-one map of Rk onto Rk whose inverse is also linear. Thus T is a homeomorphism of Rk onto Rk, so that T(E) is a Borel set for every Borel set E, and we can therefore define a positive Borel measure J1 on Rk by J1(E) = m(T(E)). The linearity of T, combined with the translation-invariance of m, gives J1(E + x) = m(T(E + x)) = m(T(E) + Tx) = m(T(E)) = J1(E). Thus J1 is translation-invariant, and the first assertion of (e) follows from (d), first for Borel sets E, then for all E e IDl by (b). To find A(T), we merely need to know m(T(E))/m(E) for one set E with 0 < m(E) < oo . If T is a rotation, let E be the unit ball of Rk; then T(E) = E, and A(T) = 1. //// POSITIVE BOREL MEASURES 53 2.21 Remarks If m is the Lebesgue measure on Rk, it is customary to write L1(Rk) in place of L1(m). If E is a Lebesgue measurable subset of Rk, and if m is restricted to the measurable subsets of E, a new measure space is obtained in an obvious fashion. The phrase "f E L1 on E" or "f e L1(E)" is used to indicate that f is integrable on this measure space. If k = 1, if I is any of the sets (a, b), (a, b ], [a, b), [a, b ], and iff e L1(I), it is customary to write r f(x) dx in place of if dm. Since the Lebesgue measure of any single point is 0, it makes no difference over which of these four sets the integral is extended. Everything learned about integration in elementary Calculus courses is still useful in the present context, for iff is a continuous complex function on [a, b ], then the Riemann integral off and the Lebesgue integral off over [a, b] coincide. This is obvious from our construction iff(a) = f(b) = 0 and iff(x) is defined to be 0 for x < a and for x > b. The general case follows without difficulty. Actually the same thing is true for every Riemann integrable f on [a, b]. Since we shall have no occasion to discuss Riemann integrable functions in the sequel, we omit the proof and refer to Theorem 11.33 of . Two natural questions may have occurred to some readers by now: Is every Lebesgue measurable set a Borel set? Is every subset of Rk Lebesgue measurable? The answer is negative in both cases, even when k = 1. The first question can be settled by a cardinality argument which we sketch briefly. Let c be the cardinality of the continuum (the real line or, equivalently, the collection of all sets of integers). We know that Rk has a countable base (open balls with rational radii and with centers in some countable dense subset of Rk), and that &lk (the collection of all Borel sets of Rk) is the Ƀ-algebra generated by this base. It follows from this (we omit the proof) that PAk has cardinality c. On the other hand, there exist Cantor sets E c R1 with m(E) = 0. (Exercise 5.) The completeness of m implies that each of the 2c subsets of E is Lebesgue measur able. Since 2c > c, most subsets of E are not Borel sets. The following theorem answers the second question. 2.22 Theorem If A c R 1 and every subset of A is Lebesgue measurable then m(A) = 0. Corollary Every set of positive measure has nonmeasurable subsets. PROOF We shall use the fact that R1 is a group, relative to addition. Let Q be the subgroup that consists of the rational numbers, and let E be a set that contains exactly one point from each coset of Q in R 1. (The assertion that 54 REAL AND COMPLEX ANALYSIS there is such a set is a direct application of the axiom of choice.) Then E has the following two properties. (a) (E + r) n (E + s) = 0 if r e Q, s e Q, r =1= s. (b) Every x e R1 lies in E + r for some r e Q. To prove (a), suppose x e (E + r) n (E + s). Then x = y + r = z + s for some y e E, z e E, y =1= z. But y - z = s -r e Q, so that y and z lie in the same coset of Q, a contradiction. To prove (b), let y be the point of E that lies in the same coset as x, put r = x - y. Fix t e Q, for the moment, and put At = A n (E + t). By hypothesis, At is measurable. Let K c At be compact, let H be the union of the translates K + r, where r ranges over Q n [0, 1]. Then H is bounded, hence m(H) < oo. Since K c E + t, (a) shows that the sets K + r are pairwise disjoint. Thus m(H) = Lr m(K + r). But m(K + r) = m(K). It follows that m(K) = 0. This holds for every compact K c At . Hence m(At) = 0. Finally, (b) shows that A = U At , where t ranges over Q. Since Q is countable, we conclude that m(A) = 0. /Ill 2.23 Determinants The scale factors ɠ(T) that occur in Theorem 2.20(e) can be interpreted algebraically by means of determinants. Let { e 1, • • . , ek} be the standard basis for Rk: the ith coordinate of ei is 1 if i = j, 0 if i =1= j. If T: Rkr Rk is linear and k Te . = """rx .
. e. J ơ l) l i= 1 (1 < j < k) (1) then det T is, by definition, the determinant of the matrix [T] that has rxii in row i and columnj. We claim that A( T) = I det T I . (2) If T = T1 T2, it is clear that ɠ(T) = ɠ(T1)ɠ(T2). The multiplication theorem for determinants shows therefore that if (2) holds for T1 and T2 , then (2) also holds for T. Since every linear operator on Rk is a product of finitely many linear operators of the following three types, it suffices to establish (2) for each of these: (I) {Te1, ... , Tek} is a permutation of {e1, .
.
• , ek}· (II) Te1 = rxe1, Tei = e; for i = 2, .
. . , k. (III) Te1 = e1 + e2, Te; = e; fori = 2, . . . , k. Let Q be the cube consisting of all x = ( = 1, . . . , k) with 0 < =; < 1 for i = 1, . . . ' k. If T is of type (1), then [T] has exactly one 1 in each row and each column and has 0 in all other places. So det T = + 1. Also, T(Q) = Q. So A(T) = 1 = l det T l . POSITIVE BOREL MEASURES 55 If T is of type (II), then clearly A(T) = I r:t. l = I det T 1. If T is of type (III), then det T = 1 and T(Q) is the set of all points L ĉi ei whose coordinates satisfy ĉ1 < ĉ2 < ĉ1 + 1, 0 < ĉ i < 1 if i =1= 2. (3) If S1 is the set of points in T(Q) that have ĉ2 < 1 and if S2 is the rest of T(Q), then (4) and S1 n (S2 -e2) is empty. Hence A(T) = m(S1 u S2) = m(S1) + m(S2-e2) = m(Q) = 1, so that we again have A(T) = I det T 1 . Continuity Properties of Measurable Functions Since the continuous functions played such a prominent role in our construction of Borel measures, and of Lebesgue measure in particular, it seems reasonable to expect that there are some interesting relations between continuous functions and measurable functions. In this section we shall give two theorems of this kind. We shall assume, in both of them, that J1 is a measure on a locally compact I -I ausdorff space X which has the properties stated in Theorem 2.14. In particular, J1 could be Lebesgue measure on some Rk. 2.24 Lusin's Theorem Suppose f is a complex measurable function on X, Jl(A) < oo,f(x) = 0 if x ¢ A, and E > 0. Then there exists a g E Cc(X) such that Jl( {X : f (X) # g( X)}) < €. (1) Furthermore, we may arrange it so that sup I g(x) I < sup I f(x) 1 . (2) x e X x e X PROOF Assume first that 0 < f < 1 and that A is compact. Attach a sequence {sn} to f, as in the proof of Theorem 1.17, and put t1 = s1 and tn = sn-sn-l for n = 2, 3, 4, . .. . Then 2ntn is the characteristic function of a set T, .c A, and 00 f(x) = L tn(x) (x EX). (3) n=1 Fix an open set V such that A c V and V is compact. There are compact sets Kn and open sets V,. such that Kn c T, c V,. c V and J1(V,. -Kn) < 2 -n €. By Urysohn's lemma, there are functions hn such that Kn-< hn -< V,. Define 00 g(x) = L 2-nhn(x) (x EX). (4) n=1 This series converges uniformly on X, so g is continuous. Also, the support of g lies in V. Since 2 -nhn(x) = tn(x) except in V,. -Kn, we have g(x) = f(x) 56 REAL AND COMPLEX ANALYSIS except in U (V,. -Kn), and this latter set has measure less than E. Thus (1) holds if A is compact and 0 <! < 1. It follows that (1) holds if A is compact and f is a bounded measurable function. The compactness of A is easily removed, for if Jl(A) < oo then A contains a compact set K with Jl(A -K) smaller than any preassigned posi tive number. Next, if f is a complex measurable function and if Bn = {x: I f(x) I > n}, then n Bn = 0, so Jl(Bn)r 0, by Theorem 1.19(e). Since f coincides with the bounded function (1 -XB) · f except on Bn, (1) follows in the general case. Finally, let R = sup { I f(x) I : x E X}, and define <p(z) = z if I z I < R, R. Then <p is a continuous mapping of the complex plane onto the disc of radius R. If g satisfies (1) and g1 = <p o g, then g1 satisfies (1) and (2). /Ill Corollary Assume that the hypotheses of Lusin 's theorem are satisfied and that If I < 1. Then there is a sequence {gn} such that gn E Cc(X), I gn I < 1, and f(x) = lim gn(x) a.e. (5) PROOF The theorem implies that to each n there corresponds a gn E Cc(X), with I gn I < 1, such that Jl(En) < 2 -n, where En is the set of all x at which f(x) =I= gn(x). For almost every x it is then true that x lies in at most finitely many of the sets En (Theorem 1.41). For any such x, it follows that f(x) = gn(x) for all large enough n. This gives (5). /Ill 2.25 The Vitali-Caratheodory Theorem Suppose f E L1(Jl), f is real-valued, and E > 0. Then there exist functions u and v on X such that u < f < v, u is upper semicontinuous and bounded above, v is lower semicontinuous and bounded below, and 1 (v -u) dJl < E. (1) PROOF Assume first that f > 0 and that f is not identically 0. Since f is the pointwise limit of an increasing sequence of simple functions sn, f is the sum Of the simple functions tn = Sn - Sn _ 1 (taking So = 0), and since tn is a linear combination of characteristic functions, we see that there are measurable sets Ei (not necessarily disjoint) and constants ci > 0 such that 00 f(x) = L ci XE;(x) (x EX). (2) i= 1 POSITIVE BOREL MEASURES 57 Since (3) the series in (3) converges. There are compact sets Ki and open sets V; such that Ki c Ei c V; and Put 00 v = L CiXVi' i= 1 (i = 1, 2, 3, . . . ). N u = L ci XKi' i= 1 where N is chosen so that Then v is lower semicontinuous, u is upper semicontinuous, u < f < v, and N oo v - u = L ci(Xvi -XKi) + L ci X vi i= l N+ l 00 00 < L ci(Xvi- XKi) + L ci XEi i=1 N+1 so that (4) and (6) imply (1). (4) (5) (6) In the general case, write f = f + -! -, attach u1 and v1 to f +, attach u2 and v2 tof-, as above, and put u = u1 - v2 , v = v1 - u2 • Since -v2 is upper semicontinuous and since the sum of two upper semicontinuous functions is upper semicontinuous (similarly for lower semicontinuous; we leave the proof of this as an exercise), u and v have the desired properties. I I I I Exercises 1 Let {f n} be a sequence of real nonnegative functions on R 1, and consider the following four state ments: (a) If/1 and/2 are upper semicontinuous, then /1 + /2 is upper semicontinuous. (b) lf/1 and/2 are lower semicontinuous, then f1 + /2 is lower semicontinuous. (c) If eachf n is upper semicontinuous, then Lf f n is upper semicontinuous. (d) If each/,. is lower semicontinuous, then Lf f.. is lower semicontinuous. Show that three of these are true and that one is false. What happens if the word " nonnegative " is omitted ? Js the truth of the statements affected if R 1 is replaced by a general topological space? 2 Letfbe an arbitrary complex function on R1, and define q>(x, b) = sup { I f(s) -f(t) I : s, t e (x - b, x + b)}, q>(x) = inf {q>(x, b): b > 0}. Prove that q> is upper semicontinuous, that f is continuous at a point x if and only if q>(x) = 0, and hence that the set of points of continuity of an arbitrary complex function is a G6 • Formulate and prove an analogous statement for general topological spaces in place of R 1• 58 REAL AND COMPLEX ANALYSIS 3 Let X be a metric space, with metric p. For any nonempty E c X, define pJ.x) = inf {p(x, y): y e E}. Show that PE is a uniformly continuous function on X. If A and B are disjoint nonempty closed subsets of X, examine the relevance of the function to· U rysohn 's lemma. f(x) = PA(x) PA(x) + pa(x) 4 Examine the proof of the Riesz theorem and prove the following two statements : (a) If E 1 c V1 and E2 c V2, where V1 and V2 are disjoint open sets, then JJ.(E 1 u E2) = f.l(E d + J.t(E2), even if E 1 and E2 are not in IDl (b) If E e IDlF, then E = N u K1 u K2 u · · · , where {K;} is a disjoint countable collection of compact sets and J.J.(N) = 0. . In Exercises 5 to 8, m stands for Lebesgue measure on R 1• S Let E be Cantor's familiar " middle thirds " set. Show that m(E) = 0, even though E and R1 have the same cardinality. 6 Construct a totally disconnected compact set K c R 1 such that m(K) > 0. (K is to have no con nected subset consisting of more than one point.) If v is lower semicontinuous and v < XK , show that actually v < 0. Hence XK cannot be approx imated from below by lower semicontinuous functions, in the sense of the Vitali-Caratheodory theorem. 7 If 0 < E < 1, construct an open set E c [0, 1] which is dense in [0, 1], such that m(E) = E. (To say that A is dense in B means that the closure of A contains B.) 8 Construct a Borel set E c R 1 such that 0 < m(E n J) < m(I) for every nonempty segment I. Is it possible to have m(E) < oo for such a set? 9 Construct a sequence of continuous function /,. on [0, 1] such that 0 <!, < 1, such that lim 11 /,.(x) dx = 0, n -+ CX> Jo but such that the sequence {fix)} converges for no x e [0, 1]. 10 If {f n} is a sequence of continuous functions on [0, 1] such that 0 <f..< 1 and such thatf..
(x) -+ 0 as n -+ oo, for every x e [0, 1], then lim rl f,(x) dx = 0. n .... CX> Jo Try to prove this without using any measure theory or any theorems about Lebesgue integration. (This is to impress you with the power of the Lebesgue integral. A nice proof was given by W. F. Eberlein in Communications on Pure and Applied Mathematics, vol. X, pp. 357-360, 1957.) 1 1 Let J1. be a regular Borel measure on a compact Hausdorff space X; assume J.J.(X) = 1. Prove that there is a compact set K c X (the carrier or support of f.l) such that J.J.(K) = 1 but Jl(H) < 1 for every proper compact subset H of K. Hint : Let K be the intersection of all compact K« with J.J.(K«) = 1 ; show that every open set V which contains K also contains some K«. Regularity of J1. is needed; compare Exercise 18. Show that Kc is the largest open set in X whose measure is 0. 12 Show that every compact subset of R 1 is the support of a Borel measure. POSITIVE BOREL MEASURES 59 13 Is it true that every compact subset of R 1 is the support of a continuous function? If not, can you describe the class of all compact sets in R 1 which are supports of continuous functions? Is your description valid in other topological spaces? 14 Let f be a real-valued Lebesgue measurable function on Rk. Prove that there exist Borel functions g and h such that g(x) = h(x) a.e. [m], and g(x) < f(x) < h(x) for every x E Rk. IS It is easy to guess the limits of as n---+ oo. Prove that your guesses are correct. 16 Why is m(Y) = 0 in the proof of Theorem 2.20(e)? 17 Define the distance between points (x1, y1) and (x2 , y2) in the plane to be Show that this is indeed a metric, and that the resulting metric space X is locally compact. Iff e Cc(X), let x1, • • • , xn be those values of x for whichf(x, y) =1- 0 for at least one y (there are only finitely many such x !), and define Let f.J. be the measure associated with this A by Theorem 2.14. If E is the x-axis, show that f.J.(E) = oo although f.J.(K) = 0 for every compact K c E. 18 This exercise requires more set-theoretic skill than the preceding ones. Let X be a well-ordered uncountable set which has a last element wb such that every predecessor of w1 has at most countably many predecessors. (" Construction " : Take any well-ordered set which has elements with uncountably many predecessors, and let w1 be the first of these; w1 is called the first uncountable ordinal.) For IX E X, let P«[S«] be the set of all predecessors (successors) of IX, and call a subset of X open if it is a P« or an Sp or a P« n Sp or a union of such sets. Prove that X is then a compact Hausdorff space. (Hint : No well-ordered set contains an infinite decreasing sequence.) Prove that the complement of the point w1 is an open set which is not u-compact. Prove that to every f E C(X) there corresponds an IX =1- w1 such thatfis constant on S«. Prove that the intersection of every countable collection { Kn} of uncountable compact subsets of X is uncountable. (Hint : Consider limits of increasing countable sequences in X which intersect each Kn in infinitely many points.) Let 9Jl be the collection of all E c X such that either E u {w1} or Ec u {w1 } contains an uncountable compact set; in the first case, define A.( E) = 1 ; in the second case, define A.( E) = 0. Prove that 9Jl is a u-algebra which contains all Borel sets in X, that A. is a measure on 9Jl which is not regular (every neighborhood of w1 has measure 1), and that for every f E C(X). Describe the regular f.J. which Theorem 2.14 associates with this linear functional. 19 Go through the proof of Theorem 2.14, assuming X to be compact (or even compact metric) rather than just locally compact, and see what simplifications you can find. 20 Find continuous functions /, : [0, 1] ---+ [0, oo) such that f n(x)---+ 0 for all x E [0, 1] as n ---+ oo, Jӳ f n(x) dx---+ 0, but sup" !, is not in I.J. (This shows that the conclusion of the dominated convergence theorem may hold even when part of its hypothesis is violated.) 21 If X is compact and f: X---+ ( - oo, oo) is upper semicontinuous, prove that f attains its maximum at some point of X. 60 REAL AND COMPLEX ANALYSIS 22 Suppose that X is a metric space, with metric d, and that f: X---+ [0, oo] is lower semicontinuous, f(p) < oo for at least one p E X. For n = 1, 2, 3, . . . , x E X, define and prove that (i) I gn(x) - giy) I < nd(x, y), (ii) 0 < g1 < g2 < . . . < f, gix) = inf {f(p) + nd(x, p): p E X} (iii) gix)---+ f(x) as n ---+ oo, for all x E X. Thus f is the pointwise limit of an increasing sequence of continuous functions. (Note that the con verse is almost trivial.) 23 Suppose V is open in Rk and J1 is a finite positive Borel measure on Rk. Is the function that sends x to p(V + x) necessarily continuous? lower semicontinuous? upper semicontinuous? 24 A step f unction is, by definition, a finite linear combination of characteristic functions of bounded intervals in R 1• Assume f E lJ(R 1 ), and prove that there is a sequence {gn} of step functions so that ÙÚ': L: I f(x) - g.(x) I dx = 0. 25 (i) Find the smallest constant c such that (ii) Does log ( 1 + e1) < c + t (0 < t < oo). lim -log { 1 + e"f} dx 1 11 n .... CX> n 0 exist for every realf E L1 ? If it exists, what is it? CHAPTER THREE LP-SPACES Convex Functions and Inequalities Many of the most common inequalities in analysis have their origin in the notion of convexity. 3.1 Definition A real function qJ defined on a segment (a, b), where - oo <a < b < oo, is called convex if the inequality ({J((1 -A)x + Ay) < (1 -A)qJ(x) + AqJ(y) (1) holds whenever a < x < b, a < y < b, and 0 < A < 1. Graphically, the condition is that if x < t < y, then the point (t, qJ(t)) should lie below or on the line connecting the points (x, qJ(x)) and (y, qJ(y)) in the plane. Also, (1) is equivalent to the requirement that ({J(ͤ-qJ(ײ ({J(U) -({J(ͤ ----<----whenever a< s < t < u <b. t-s - u-t (2) The mean value theorem for differentiation, combined with (2), shows imme diately that a real differentiable function qJ is convex in (a, b) if and only if a < s < t < b implies qJ'(s) < qJ'(t), i.e., if and only if the derivative qJ' is a mono tonically increasing function. For example, the exponential function is convex on (- oo, oo ). 3.2 Theorem If qJ is convex on (a, b) then qJ is continuous on (a, b). 61 62 REAL AND COMPLEX ANALYSIS PROOF The idea of the proof is most easily conveyed in geometric language. Those who may worry that this is not "rigorous " are invited to transcribe it in terms of epsilons and deltas. Suppose a < s < x < y < t < b. Write S for the point (s, qJ(s)) in the plane, and deal similarly with x, y, and t. Then X is on or below the line S Y, hence Y is on or above the line through S and X; also, Y is on or below X T. As yr x, it follows that Y rX, i.e., qJ(y)Š qJ(x). Left-hand limits are handled in the same manner, and the continuity of <p follows. I I I I Note that this theorem depends on the fact that we are working on an open segment. For instance, if <p(x) = 0 on [0, 1) and qJ(1) = 1, then (/) satisfies 3.1(1) on [0Ʉ 1] without being continuous. 3.3 Theorem (Jensen's Inequality) Let J1 be a positive measure on a a-algebra 9Jl in a set Q, so that ,u(Q) = 1. Iff is a real function in IJ(J1), if a <f(x) < b for all x e n, and if <p is convex on (a, b), then (1) Note: The cases a= - oo and b = oo are not excluded. It may happen that (/) of is not in L1(,u); in that case, as the proof will show, the integral of (/) of exists in the extended sense described in Sec. 1.31, and its value is + oo. PROOF Put t = fn f dJ1. Then a < t < b. If P is the supremum of the quotients on the left of 3.1(2), where a< s < t, then p is no larger than any of the quotients on the right of 3.1(2), for any u e (t, b). It follows that <p(t) + p(s -t) (a< s <b). (2) Hence qJ(f(x)) - 0 (3) for every x e n. Since <p is continuous, (/) of is measurable. If we integrate both sides of (3) with respect to J1, (1) follows from our choice of t and the assumption ,u(Q) = 1. 1111 To give an example, take qJ(x) =ex. Then (1) becomes If Q is a finite set, consisting of points p1, • • • , Pn , say, and if (4) If-SPACES 63 (4) becomes exp {! (x1 + ··· + x,)} < !(ex' + ··· + ex•), (5) for real xi . Putting Yi = exi, we obtain the familiar inequality between the arith metic and geometric means of n positive numbers: (6) Going back from this to (4), it should become clear why the left and right sides of exp {ilog g dJ1.} < Lg dJ1. (7) are often called the geometric and arithmetic means, respectively, of the positive function g. If we take Jl( {pi}) = rxi > 0, where L rxi = 1, then we obtain Ylltyll2 . . . yCXn < (X y + (X y + ... + (X y 1 2 n -1 1 2 2 n n (8) in place of (6). These are just a few samples of what is contained in Theorem 3.3. For a con verse, see Exercise 20. 3.4 Definition If p and q are positive real numbers such that p + q = pq, or equivalently 1 1 - + - = 1, p q (1) then we call p and q a pair of conjugate exponents. It is clear that (1) implies 1 < p < oo and 1 < q < oo. An important special case is p = q = 2. As p-+ 1, (1) forces q-+ oo. Consequently 1 and oo are also regarded as a pair of conjugate exponents. Many analysts denote the exponent conjugate to p by p', often without saying so explicitly. 3.5 Theorem Let p and q be conjugate exponents, 1 < p < oo. Let X be a measure space, with measure Jl. Let f and g be measurable functions on X, with range in [0, oo]. Then (1) and (2) The inequality (1) is Holder's; (2) is Minkowski's. If p = q = 2, (1) is known as the Schwarz inequality. 64 REAL AND COMPLEX ANALYSIS PROOF Let A and B be the two factors on the right of (1). If A = 0, thenf = 0 a.e. (by Theorem 1 .39); hencefg = 0 a.e., so (1) holds. If A > 0 and B = oo, (1) is again trivial. So we need consider only the case 0 < A < oo, 0 < B < oo. Put This gives f F= A' l FP dJl = l Gq dJl = 1. (3) (4) If x e X is such that 0 < F(x) < oo and 0 < G(x) < oo , there are real numbers s and t such that F(x) = 3IP, G(x) = etfq. Since 1/p + ljq = 1, the convexity of the exponential function implies that (5) It follows that (6) for every x e X. Integration of (6) yields · lFG dJl < p-1 + q -1 = 1, (7) by (4); inserting (3) into (7), we obtain (1). Note that (6) could also have been obtained as a special case of the inequality 3.3(8). To prove (2), we write (f + g)P =f. (f + g)p-1 + g . (f + g)P -1. (8) Holder's inequality gives If. (f + g)p-1 < {f fP riP {f (f + g)(p-l)qr'q. (9) Let (9') be the inequality (9) with f and g interchanged. Since (p- 1)q = p, addition of (9) and (9') gives Clearly, it is enough to prove (2) in the case that the left side is greater than 0 and the right side is less than oo. The convexity of the function tP for 0 < t < oo shows that I!'-SPACES 65 Hence the left side of (2) is less than oo, and (2) follows from ( 1 0) if we divide by the first factor on the right of (10), bearing in mind that 1 - 11q = 1lp. This completes the proof. I I I I It is sometimes useful to know the conditions under which equality can hold in an inequality. In many cases this information may be obtained by examining the proof of the inequality. For instance, equality holds in (7) if and only if equality holds in (6) for almost every x. In (5), equality holds if and only if s = t. Hence "FP = Gq a.e." is a necessary and sufficient condition for equality in (7), if (4) is assumed. In terms of the original functions f and g, the following result is then obtained: Assuming A < oo and B < oo, equality holds in (1) if and only if there are constants C( and {3, not both 0, such that C(f P = {3gq a.e. We leave the analogous discussion of equality in (2) as an exercise. The I! -spaces In this section, X will be an arbitrary measure space with a positive measure JL 3.6 Definition If 0 < p < oo and iff is a complex measurable function on X, define Jl / ll p = {f)fiP djJ.} lfp and let IJ'(J1) consist of allffor which We call II fliP the IJ'-norm off 11 / ll p < 00. (1) (2) If J1 is Lebesgue measure on Rk, we write IJ'(Rk) instead of IJ'(J1), as in Sec. 2.21. If J1 is the counting measure on a set A, it is customary to denote the corresponding I! -space by tP(A), or simply by tP, if A is countable. An element of tP may be regarded as a complex sequence x = { n}, and 3.7 Definition Suppose g: Xr [0, oo] is measurable. Let S be the set of all real C( such that Jl(g-1(( C(, 00 ])) = 0. (1) If S = 0, put {3 = oo. If S =1= 0, put {3 = inf S. Since g-l((p, oo])= n91 g-l((P+ !• oo]} (2) 66 REAL AND COMPLEX ANALYSIS and since the union of a countable collection of sets of measure 0 has measure 0, we see that {3 e S. We call {3 the essential supremum of g. Iff is a complex measurable function on X, we define llfll oo to be the essential supremum of I f I , and we let L00(J1) consist of all f for which II f II oo < oo. The members of L00(J1) are sometimes called essentially bounded measurable functions on X. It follows from this definition that the inequality I f(x) I < A holds for almost all x if and only if A > llfll oo . As in Definition 3.6, L00(Rk) denotes the class of all essentially bounded (with respect to Lebesgue measure) functions on Rk, and t00(A) is the class of all bounded functions on A. (Here bounded means the same as essentially bounded, since every nonempty set has positive measure!) 3.8 Theorem If p and q are conjugate exponents, 1 < p < oo, and iff e l!(J1) and g e /J(J1), then fg e L1(J1), and (1) PROOF For 1 < p < oo, (1) is simply Holder's inequality, applied to If I and lgl. lfp = oo, note that I f(x)g(x) I < II f II oo I g(x) I (2) for almost all x; integrating (2), we obtain (1). If p = 1, then q = oo, and the same argument applies. //// 3.9 Theorem Suppose 1 < p < oo , and f e l!(J1), g E I!(J1). Then f + g E I!(J1), and (1) PROOF For 1 < p < oo, this follows from Minkowski's inequality, since For p = 1 or p = oo, (1) is a trivial consequence of the inequality 1/+gl <l/l +lgl. /Ill 3.10 Remarks Fix p, 1 < p < oo. Iff e I!'(J1) and a. is a complex number, it is clear that a.f e I!(J1). In fact, II a.f II p = I a. Ill f II p • (1) In conjunction with Theorem 3.9, this shows that I!(J1) is a complex vector space. I! -SPACES 67 Suppose f, g, and h are in I!(J1). Replacing f by f-g and g by g -h in Theorem 3.9, we obtain 11/-hllp < 11/-gllp + llg-hllp· (2) This suggests that a metric may be introduced in I!(J1) by defining the dis tance between f and g to be II f-g II P. Call this distance d(f, g) for the moment. Then 0 < d(f, g) < oo, d(f, f) = 0, d(f, g) = d(g, f), and (2) shows that the triangle inequality d(f, h) < d(f, g) + d(g, h) is satisfied. The only other property which d should· have to define a metric space is that d(f, g) = 0 should imply that f = g. In our present situation this need not be so; we have d(f, g)= 0 precisely whenf(x) = g(x)for almost all x. Let us write f g if and only if d(f, g) = 0. It is clear that this is an equivalence relation in I!(J1) which partitions I!(J1) into equivalence classes; each class consists of all functions which are equivalent to a given one. If F and G are two equivalence classes, choose f e F and g e G, and define d(F, G) = d(f, g); note that/ /1 and g g1 implies d(f, g) = d(fl, gl), so that d(F, G) is well defined. With this definition, the set of equivalence classes is now a metric space. Note that it is also a vector space, since f /1 and g g1 implies f + g !1 + g1 and rxf rxft· When I!(J1) is regarded as a metric space, then the space which is really under consideration is therefore not a space whose elements are functions, but a space whose elements are equivalence classes of functions. For the sake of simplicity of language, it is, however, customary to relegate this distinction to the status of a tacit understanding and to continue to speak of I!(J1) as a space of functions. We shall follow this custom. If {fn} is a sequence in I!(J1), iff e I!(J1), and if limn-+oo 11/, -fliP= 0, we say that {In} converges to fin l!(J1) (or that {In} converges to fin the mean of order p, or that {In} is /!-convergent to f). If to every E > 0 there corre sponds an integer N such that 11/n -fmllp < E as soon as n > N and m > N, we call {In} a Cauchy sequence in I!(J1). These definitions are exactly as in any metric space. It is a very important fact that I!(J.l) is a complete metric space, i.e., that every Cauchy sequence in I!(J1) converges to an element of I!(J1): 3.11 Theorem I!(J1) is a complete metric space, for 1 < p < oo and for every positive measure Jl. PROOF Assume first that 1 < p < oo. Let {In} be a Cauchy sequence in I!(J1). There is a subsequence {fnJ, n1 < n2 < · · ·, such that (i = 1, 2, 3, . . . ). (1) 68 REAL AND COMPLEX ANALYSIS Put k g k = L I fni + 1 -fni I ' i= 1 00 g = L lfni+l -fni 1. (2) i= 1 Since (1) holds, the Minkowski inequality shows that llgkiiP < 1 for k = 1, 2, 3, . . . . Hence an application of Fatou's lemma to {gf} gives llgiiP < 1. In particular, g(x) < oo a.e., so that the series 00 fnt(x) + L (fni+ 1(x) -fni(x)) i= 1 (3) converges absolutely for almost every x E X. Denote the sum of (3) by f(x), for those x at which (3) converges; put f(x) = 0 on the remaining set of measure zero. Since k-1 fnt + L (fni+l -fni) = /,.k' (4) i= 1 we see that f(x) = lim fnJx) a.e. (5) i- oo Having found a function f which is the pointwise limit a.e. of {fnJ, we now have to prove that this f is the I!'-limit of {In}. Choose E > 0. There exists an N such that 11/n -fm liP< E if n > N and m > N. For every m > N, Fatou's lemma shows therefore that (6) We conclude from (6) that f-fm E I!'(Jl.), hence that /E I!'(Jl.) [since f = (f-fm) + fm], and finally that 11/-/m liPr 0 as mr oo. This completes the proof for the case 1 < p < oo. In L00(J1.) the proof is much easier. Suppose {in} is a Cauchy sequence in L00(J1.), let Ak and Bm, n be the sets where I fk(x) I > 11/k II oo and where I fn(x) -fm(x) I > II/,. -fm II oo , and let E be the union of these sets, for k, m, n = 1, 2, 3, . . . . Then J.l(E) = 0, and on the complement of E the sequence {In} converges uniformly to a bounded function! Definef(x) = 0 for x E E. Then f E L00(J1.), and 11/n -/II 00 | 0 as nr oo. //// The preceding proof contains a result which is interesting enough to be stated separately: 3.12 Theorem If 1 < p < oo and if {fn} is a Cauchy sequence in I!'(J.l), with limit f, then {fn} has a subsequence which converges pointwise almost every where to f(x). I! -SPACES 69 The simple functions play an interesting role in I!(Jl): 3.13 Theorem Let S be the class of all complex, measurable, simple functions on X such that Jl( {X : s( X) # 0}) < 00 . (1) If 1 < p < oo, then S is dense in I!(p,). PROOF First, it is clear that S c I!(Jl). Suppose f > 0, f e I!(p,), and let { sn} be as in Theorem 1.17. Since 0 < sn < f, we have Sn E I!(p,), hence sn E S. Since If -sn JP < fP, the dominated convergence theorem shows that II ! -sn liPr 0 as nr oo. Thus f is in the I!-closure of S. The general case (f complex) follows from this. /Ill Approximation by Continuous Functions So far we have considered I!(p,) on any measure space. Now let X be a locally compact Hausdorff space, and let J1 be a measure on a a-algebra 9Jl in X, with the properties stated in Theorem 2.14. For example, X might be R\ and J1 might be Lebesgue measure on Rk. Under these circumstances, we· have the following analogue of Theorem 3.13: 3.14 Theorem For 1 < p < oo, Cc(X) is dense in l!(J1). PROOF Define S as in Theorem 3.13. If s e S and E > 0, there exists a g e Cc(X) such that g(x) = s(x) except on a set of measure < E, and lgl < llslloo (Lusin's theorem). Hence Since S is dense in I!(Jl), this completes the proof. (1) Ill/ 3.15 Remarks Let us discuss the relations between the spaces I!(Rk) (the I! spaces in which the underlying measure is Lebesgue measure on Rk) and the space Cc(Rk) in some detail. We consider a fixed dimension k. For every p E [1, oo] we have a metric on Cc(Rk); the distance between/ and g is 11/-gllp· Note that this is a genuine metric, and that we do not have to pass to equivalence classes. The point is that if two continuous func tions on Rk are not identical, then they differ on some nonempty open set V, and m(V) > 0, since V contains a k-cell. Thus if two members of Cc(Rk) are equal a.e., they are equal. It is also of interest to note that in Cc(Rk) the essential supremum is the same as the actual supremum: for f e Cc(Rk) 11/11 oo = sup I f(x) f . (1) x e Rk 70 REAL AND COMPLEX ANALYSIS If 1 < p < oo, Theorem 3.14 says that Cc(Rk) is dense in I!(Rk), and Theorem 3.11 shows that I!'(Rk) is complete. Thus I!'(Rk) is the completion of the metric space which is obtained by endowing Cc(Rk) with the 1!-metric. The cases p = 1 and p = 2 are the ones of greatest interest. Let us state once more, in different words, what the preceding result says if p = 1 and k = 1 ; the statement shows that the Lebesgue integral is indeed the "right" generalization of the Riemann integral: If the distance between two continuous functions f and g, with compact supports in R1, is defined to be L:l f(t)-g(t) I dt, (2) the completion of the resulting metric space consists precisely of the Lebesgue integrable functions on R 1, provided we identify any two that are equal almost everywhere. Of course, every metric space S has a completion S whose elements may be viewed abstractly as equivalence classes of Cauchy sequences in S (see , p. 82). The important point in the present situation is that the various I!' completions of Cc(Rk) again turn out to be spaces of functions on Rk. The case p = oo differs from the cases p < oo. The L00-completion of Cc(Rk) is not L00(Rk), but is C0(Rk), the space of all continuous functions on Rk which "vanish at infinity," a concept which will be defined in Sec. 3.16. Since (1) shows that the L00-norm coincides with the supremum norm on Cc(Rk), the above assertion about C0(Rk) is a special case of Theorem 3.17. 3.16 Definition A complex function f on a locally compact Hausdorff space X is said to vanish at infinity if to every E > 0 there exists a compact set K c X such that I f(x) I < E for all x not in K. The class of all continuous f on X which vanish at infinity is called C0(X). It is clear that Cc(X) c C0(X), and that the two classes coincide if X is compact. In that case we write C(X) for either of them. 3.17 Theorem If X is a locally compact Hausdorff space, then C0(X) is the completion of Cc(X), relative to the metric defined by the supremum norm 11/11 = sup I f(x) 1 . (1) x_e X PROOF An elementary verification shows that C0(X) satisfies the axioms of a metric space if the distance between/ and g is taken to be II!-gil- We have to show that (a) Cc(X) is dense in C0(X) and (b) C0(X) is a complete metric space. 1!-SPACES 71 Given f e C0(X) and E > 0, there is a compact set K so that I f(x) I < e outside K. Urysohn's lemma gives us a function g e Cc(X) such that 0 < g < 1 and g(x) = 1 on K. Put h = fg. Then h e Cc(X) and 11/- hll < E. This proves (a). To prove (b), let {In} be a Cauchy sequence in C0(X), i.e., assume that {In} converges uniformly. Then its pointwise limit function f is continuous. Given e > 0, there exists an n so that 11/n -/II < E/2 and there is a compact set K so that I fn(x) I < e/2 outside K. Hence I f(x) I < E outside K, and we have proved that/vanishes at infinity. Thus C0(X) is complete. · /Ill Exercises 1 Prove that the supremum of any collection of convex functions on (a, b) is convex on (a, b) (if it is finite) and that pointwise limits of sequences of convex functions are convex. What can you say about upper and lower limits of sequences of convex functions? 2 If cp is convex on (a, b) and if t/1 is convex and nondecreasing on the range of cp, prove that t/1 o cp is convex on (a, b). For cp > 0, show that the convexity of log cp implies the convexity of cp, but not vice versa. 3 Assume that cp is a continuous real function on (a, b) such that Łjx + y) . 1 1 '\ 2 < 2 cp(x) + 2 cp(y) for all x and y E (a, b). Prove that cp is convex. (The conclusion does not follow if continuity is omitted from the hypotheses.) 4 Suppose/is a complex measurable function on X, J1 is a positive measure on X, and q>(p) = ll fiP dJl = »»ǹ»»Ǻ (0 < p < oo). Let E = {p: cp(p) < oo }. Assume llfll oo > 0. (a) If r < p < s, r E E, and s E E, prove that p E E. (b) Prove that log cp is convex in the interior of E and that cp is continuous on E. (c) By (a), E is connected. Is E necessarily open? Closed? Can E consist of a single point? Can E be any connected subset of (0, oo )? (d) If r < p < s, prove that II/IlP < max (llfllr, IIIIIs). Show that this implies the inclusion :(11) n e(11) c I!'(Jl). (e) Assume that llfllr < oo for some r < oo and prove that II/IlP-+ llflloo as p-+ oo. S Assume, in addition to the hypotheses of Exercise 4, that p(X) = 1. (a) Prove that llfllr < II fils if 0 < r < s < oo. (b) Under what conditions does it happen that 0 < r < s < oo and llfllr = llflls < oo ? (c) Prove that ͞(p) => e(p) if 0 < r < s. Under what conditions do these two spaces contain the same functions? (d) Assume that llfllr < oo for some r > 0, and prove that if ex p { - oo} is defined to be 0. lim II fliP = exp { [ log I f I dp} p .... o Jx 72 REAL AND COMPLEX ANALYSIS 6 Let m be Lebesgue measure on [0, 1 ], and define II f II P with respect to m. Find all functions ci> on [0, oo) such that the relation (lim 11/ llp) = f\ci> of) dm p .... o Jo holds for every bounded, measurable, positive f Show first that (x > 0, 0 < c < 1). Compare with Exercise 5(d). 7 For some measures, the relation r < s implies e(JJ.) c l.!(JJ.); for others, the inclusion is reversed ; and there are some for which e(JJ.) does not contain l.!(JJ.) if r =1- s. Give examples of these situations, and find conditions on J1. under which these situations will occur. 8 If g is a positive function on (0, 1) such that g(x)---+ oo as x ---+ 0, then there is a convex function h on (0, 1) such that h < g and h(x)---+ oo as x---+ 0. True or false? Is the problem changed if (0, 1) is replaced by (0, oo) and x---+ 0 is replaced by x---+ oo ? 9 Suppose f is Lebesgue measurable on (0, 1), and not essentially bounded. By Exercise 4(e), 11/ II P---+ oo as p ---+ oo . Can 11 /IIP tend to oo arbitrarily slowly? More precisely, is it true that to every positive function ci> on (0, oo) such that Cl>(p) ---+ oo as p---+ oo one can find an f such that II f II P ---+ oo as p ---+ oo , but II f II P < Cl>(p) for all sufficiently large p? 10 Suppose f n E IJ'(JJ. ), for n = 1, 2, 3, . . . , and II!, -f II P ---+ 0 and!, ---+ g a. e., as n ---+ oo . What relation exists between f and g? · 11 Suppose JJ.(!l) = 1, and suppose f and g are positive measurable functions on n such that f g > 1. Prove that 12 Suppose JJ.(!l) = 1 and h : 0---+ [0, oo] is measurable. If prove that If J1. is Lebesgue measure on [0, 1] and if h is continuous, h = f', the above inequalities have a simple geometric interpretation. From this, conjecture (for general Q) under what conditions on h equality can hold in either of the above inequalities, and prove your conjecture. 13 Under what conditions onf and g does equality hold in the conclusions of Theorems 3.8 and 3.9? You may have to treat the cases p = 1 and p = oo separately. 14 Suppose 1 < p < oo,f E lJ' = IJ'((O, oo )), relative to Lebesgue measure, and 1 ix F(x) = -f(t) dt X 0 (0 < x < oo). (a) Prove Hardy's inequality which shows that the mapping/---+ F carries lJ' into IJ'. (b) Prove that equality holds only iff= 0 a.e. (c) Prove that the constant p/(p - 1) cannot be replaced by a smaller one. (d) Iff > 0 andf E I1, prove that F ¢ L1• /!-SPACES 73 Suggestions: (a) Assume first thatf > 0 andf E Cc((O, oo )). Integration by parts gives f" FP(x) dx = -p f" FP- 1(x)xF'(x) dx. Note that xF' = f - F, and apply Holder's inequality to J FP- 1f Then derive the general case. (c) Takef(x) = x - t tp on [1, A],f(x) = 0 elsewhere, for large A. See also Exercise 14, Chap. 8. 15 Suppose {an} is a sequence of positive numbers. Prove that oo ( 1 N )p ( p )p oo L - L an < L a: N = 1 N n= 1 P - 1 n= 1 if 1 < p < oo . Hint : If an > an+ 1, the result can be made to follow from Exercise 14. This special case implies the general one. 16 Prove Egoroff's theorem : If J.l.(X) < oo, if {fn} is a sequence of complex measurable functions which converges pointwise at every point of X, and if E > 0, there is a measurable set E c X, with J.l.(X - E) < E, such that {fn} converges uniformly on E. (The conclusion is that by redefining the/, on a set of arbitrarily small measure we can convert a pointwise convergent sequence to a uniformly convergent one; note the similarity with Lusin's theorem.) Hint : Put S(n, k) = . (I {x: I f.{xJ -f,{x) I < o}' r, J>n k show that J.l.(S(n, k)). Jl(X) as n . oo , for each k, and hence that there is a suitably increasing sequence { nk} such that E = n S(nk , k) has the desired property. Show that the theorem does not extend to a-finite spaces. Show that the theorem does extend, with e͟sentially the same proof, to the situation in which the sequence {fn} is replaced by a family {ft}, where t ranges over the positive reals; the assumptions are now that, for all x E X, (i) lim ft(x) = f(x) and t-+ 00 (ii) t _. ft(x) is continuous. 17 (a) If 0 < p < oo, put Yp = max (1, 2P-1), and show that for arbitrary complex numbers IX and p. (b) Suppose Jl is a positive measure on X, 0 < p < oo , f E l!(J.l.),!,. E l!(Jl), f,(x) . f(x) a.e., and 11/,.IIP. II! liP as n. oo . Show that then lim II!-f n liP = 0, by completing the two proofs that are sketched below. (i) By Egoroff's theorem, X = A u B in such a way that JA I f IP < E, Jl(B) < oo , and f n . f uniformly on B. Fatou's lemma, applied to J8 I !,. IP, leads to lim sup f... I f.. IP dJl < E. (ii) Put hn = yp( I f IP + I !,. IP) - I f-!, IP, and use Fatou's lemma as in the proof of Theorem 1.34. (c) Show that the conclusion of (b) is false if the hypothesis llfnllp___. 11/IIP is omitted, even if J.l.( X) < oo . 74 REAL AND COMPLEX ANALYSIS 18 Let Jl be a positive measure on X. A sequence {/,.} of complex measurable functions on X is said to converge in measure to the measurable function/if to every E > 0 there corresponds an N such that Jl( {X: I /,.(X) -f (X) I > E}) < E for all n > N. (This notion is of importance in probability theory.) Assume JL(X) < oo and prove the following statements: (a) If !,(x) -4 f(x) a.e., then!, -4 f in measure. (b) Iff, E l!(JL) and II/ ,. -!IIP-4 0, thenf,.-4 f in measure; here 1 < p < oo. (c) Iff, -4 f in measure, then {!,} has a subsequence which converges to f a.e. Investigate the converses of (a) and (b). What happens to (a), (b), and (c) if J,J.(X) = oo, for instance, if Jl is Lebesgue measure on R 1 ? 19 Define the essential range of a function f E L00(JL) to be the set R 1 consisting of all complex numbers w such that JL( { x: I f (x) - w I < E}) > 0 for every E > 0. Prove that R 1 is compact. What relation exists between the set R 1 and the number 11/ll oo? Let A 1 be the set of all averages where E E 9Jl and JL(E) > 0. What relations exist between A 1 and R 1? Is A 1 always closed? Are there measures Jl such that A 1 is convex for every f E L00(Jl)? Are there measures Jl such that A 1 fails to be convex for some f E L00(JL)? How are these results affected if L00(f.J.) is replaced by I.J(f.J.), for instance? 20 Suppose cp is a real function on R1 such that for every real bounded measurable f Prove that cp is then convex. 21 Call a metric space Y a completion of a metric space X if X is dense in Y and Y is complete. In Sec. 3.1 5 reference was made to "the " completion of a metric space. State and prove a uniqueness theorem which justifies this terminology. 22 Suppose X is a metric space in which every Cauchy sequence has a convergent subsequence. Does it follow that X is complete? (See the proof of Theorem 3.1 1.) 23 Suppose Jl is a positive measure on X, f.J.(X) < oo,f E L00(Jl), II f II oo > 0, and ( n = 1, 2, 3, . .
. ). Prove that 24 Suppose Jl is a positive measure,/ E lJ'(f.J.), g E lJ'(f.J.). (a) If 0 < p < 1, prove that that ll(f, g) = J I f- g IP df.J. defines a metric on lJ'(f.J.), and that the resulting metric space is complete. (b) If 1 < p < oo and II/IlP < R, llgiiP < R, prove that f l if l" - l g i• j dJl < 2pR"- 1 1if- gil •. Hint: Prove first, for x > 0, y > 0, that if 0 < p < 1, if 1 < p < 00. I!'-SPACES 75 Note that (a) and (b) establish the continuity of the mapping/-+ I f IP that carries l!(JJ.) into L1(JJ.). 25 Suppose J1. is a positive measure on X and f: X ---+ (0, oo) satisfies J x f dJJ. = 1. Prove, for every E c: X with 0 < JJ.(E) < oo, that and, when 0 < p < 1, i(logf) dJJ. < JJ.(E) log -1 -E JJ.(E) if• dJl < Jl(E)t -•. 26 Ifjis a positive measurable function on [0, 1], which is larger, f f(x) log f(x) dx or f f(s) ds flog f(t) dt? CHAPTER FOUR ELEMENTARY HILBERT SPACE THEORY Inner Products and Linear Functionals 76 4.1 Definition A complex vector space H is called an inner product space (or unitary space) if to each ordered pair of vectors x and y e H there is associ ated a complex number (x, y), the so-called "inner product " (or "scalar product ") of x and y, such that the following rules hold: (a) (y, x) = (x, y). (The bar denotes complex conjugation.) (b) (x + y, z) = (x, z) + (y, z) if x, y, and z e H. (c) (C(X, y) = C((X, y) if x and y e H and C( is a scalar. (d) (X, X) > 0 for all X E H. (e) (x, x) = 0 only if x = 0. (f) Let us list some immediate consequences of these axioms: (c) implies that (0, y) = 0 for all y E H. (b) and (c) may be combined into the statement: For every y e H, the mapping x ׯ (x, y) is a linear functional on H. (a) and (c) show that (x, C(y) = a(x, y). (a) and (b) imply the second distributive law: (z, x + y) = (z, x) + (z, y). By (d), we may define llxll, the norm of the vector x E H, to be the non negative square root of (x, x). Thus llxll2 = (x, x). ELEMENTARY HILBERT SPACE THEORY 77 4.2 The Schwarz Inequality The properties 4.1 (a) to (d) imply that I (x, y) I < llxll IIYII for all x and y E H. PROOF Put A= llxll2, B = l(x, y)l, and C = IIYII2• There is a complex number C( such that I C( I = 1 and C((y, x) = B. For any real r, we then have (x -rC(y, x -rC(y) = (x, x) -rC((y, x) -r&(x, y) + r2(y, y). (1) The expression on the left is real and not negative. Hence A -2Br + Cr2 > 0 (2) for every real r. If C = 0, we must have B = 0; otherwise (2) is false for large positive r. If C > 0, take r = B/C in (2), and obtain B2 < AC. /Ill 4.3 The Triangle Inequality For x andy E H, we have llx + Yll < llxll + IIYII-PROOF By the Schwarz inequality, llx + Yll2 = (x + y, x + y) = (x, x) + (x, y) + (y, x) + (y, y) < llxll2 + 2llxii iiYII + IIYII2 = (llxll + llyll)2• Ill/ 4.4 Definition It follows from the triangle inequality that llx-zll < llx -Yll + IIY -zll (x, y, z E H). (1) If we define the distance between x and y to be llx -yll, all the axioms for a .metric space are satisfied; here, for the first time, we use part (e) of Definition 4.1. Thus H is now a metric space. If this metric space is complete, i.e., if every Cauchy sequence converges in H, then H is called a Hilbert space. Throughout the rest of this chapter, the letter H will denote a Hilbert space. 4.5 Examples (a) For any fixed n, the set C" of all n-tuples x = ( e 1 , • • . , B ,), where B b . . . , e, are complex numbers, is a Hilbert space if addition and scalar multiplication are defined componentwise, as usual, and if n (x, y) = L Biiji j= 1 78 REAL AND COMPLEX ANALYSIS (b) If J1 is any positive measure, I3(J1) is a Hilbert space, with inner product (f, g) = Lf{j djJ.. The integrand on the right is in L1(J1), by Theorem 3.8, so that (f, g) is well defined. Note that 11 ! 11 = (f./)112 = {LI/12 dJ1.r12 = llfll2· The completeness of I3(J1) (Theorem 3.11) shows that /J(J1) is indeed a Hilbert space. [We recall that L2(J1) should be regarded as a space of equivalence classes of functions; compare the discussion in Sec. 3.10.] For H = I3(J1), the inequalities 4.2 and 4.3 turn out to be special cases of the inequalities of Holder and Minkowski. Note that Example (a) is a special case of (b). What is the measure in (a)? (c) The vector space of all continuous compiex functions on [0, 1] is an inner product space if (f, g) = r f(t)g(t) dt but is not a Hilbert space. 4.6 Theorem For any fixed y e H, the mappings x---+ (x, y), x---+ (y, x), x---+ llxll are continuous functions on H. PROOF The Schwarz inequality implies that I (xt, y)-(x2, y) I = I (xl -x2, y) I < llx1 -x2II IIYII, which proves that x---+ (x, y) is, in fact, uniformly continuous, and the same is true for x---+ (y, x). The triangle inequality II x 1 11 < II x 1 -x2ll + II x21װ yields llxtll - llx2ll <·llxt -x2ll, and if we interchange x1 and x2 we see that l llxtll - llx2ll I < llxt -x2ll for all x1 and x2 e H. thus X --+ llxll is also uniformly continuous. /Ill 4.7 Subspaces A subset M of a vector space V is called a subspace of V if M is itself a vector space, relative to the addition and scalar multiplication ·which are defined in V. A necessary and sufficient condition for a set M c V to be a sub space is that x + y e M and C(X e M whenever x and y e M and C( is a scalar. ELEMENTARY HILBERT SPACE THEORY 79 In the vector space context, the word " subspace" will always have this meaning. Sometimes, for emphasis, we may use the term "linear subspace " in place of subspace. For example, if V is the vector space of all complex functions on a set S, the set of all bounded complex functions on S is a subspace of V, but the set of all f e V with I f(x) I < 1 for all x e S is not. The real vector space R3 has the follow ing subspaces, and no others: (a) R3, (b) all planes through the origin 0, (c) all straight lines through 0Ʌ and (d) {0}. A closed subspace of H is a subspace that is a closed set relative to the topol ogy induced by the metric of H. Note that if M is a subspace of H, so is its closure M. To see this, pick x and y in M and let (X be a scalar. There are sequences {xn} and {Yn} in M that converge to x and y, respectively. It is then easy to verify that xn + Yn and (XXn converge to x + y and (XX, respectively. Thus x + y e M and (XX e M. 4.8 Convex Sets A set E in a vector space V is said to be convex if it has the following geometric property: Whenever x e E, y e E, and 0 < t < 1, the point zt = (1 -t)x + ty also lies in E. As t runs from 0 to 1, one may visualize zt as describing a straight line segment in V, from x to y. Convexity requires that E contain the segments between any two of its points. It is clear that every subspace of V is convex. Also, if E is convex, so is each of its translates E +X= {y + x: y E E}. 4.9 Orthogonality If (x, y) = 0 for some x and y e H, we say that x is orthogonal to y, and sometimes write x l_ y. Since (x, y) = 0 implies (y, x) = 0, the relation l_ is symmetric. Let x.l denote the set of all y e H which are orthogonal to x; and if M is a subspace of H, let M.l be the set of all y e H which are orthogonal to every X E M. Note that xj_ is a subspace of H, since x l_ y and x l. y' implies x i (y + y') and x _i (Xy. Also, xJ. is precisely the set of points where the continuous function y---+ (x, y) is 0. Hence x.l is a closed subspace of H. Since M J. is an intersection of closed subs paces, and it follows that M J. is a closed subspace of H. 4.10 Theorem Every nonempty, closed, convex set E in a Hilbert space H con tains a unique element of smalle$t norm. 80 REAL AND COMPLEX ANALYSIS In other words, there is one and only one x0 E E such that llx0ll < llxll for every x E E. PROOF An easy computation, using only the properties listed in Definition 4.1, establishes the identity (x and y E H). (1) This is known as the parallelogram law: If we interpret llx ll to be the length of the vector x, (1) says that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides, a familiar propo sition in plane geometry. Let - = inf { llxll : x E E}. For any x and y E E, we apply (1) to .x and .y and obtain X + y 2 lllx - Yll 2 = .llxll 2 + /IIYII 2 -2 Since E is convex, (x + y)/2 E E. Hence llx - Y ll 2 < 211xll 2 + 211YII 2 - 4-2 (x and y E E). (2) (3) If also llx ll = IIYII = ð' then (3) implies x = y, and we have proved the unique ness assertion of the theorem. The definition of - shows that there is a sequence {Yn} in E so' that llYn II Ɇ ð as nr oo. Replace x and y in (3) by Yn and Ym. Then, as nr oo and mr oo, the right side of (3) will tend to 0. This shows that {Yn} is a Cauchy sequence. Since H is complete, there exists an x0 E H so that Yn r x0 , i.e., llYn - x0 ll | 0, as nr oo. Since Yn E E and E is closed, x0 E E. Since the norm is a continuous function on H (Theorem 4.6), it follows that llxo ll = lim IIYnll = ð-n-+ oo 4.1 1 Theorem Let M be a closed subspace of a Hilbert space H. (a) Every x E H has then a unique decomposition x = Px + Qx into a sum of Px EM and Qx E M1.. (b) Px and Qx are the nearest points toxin M and in M1., respectively. (c) The mappings P: H r M and Q : H | M.l. are linear. (d) llxll 2 = II Pxll 2 + 11 Qxll 2 • . Corollary If M "# H, then there exists y E H, y "# 0, such that y _i M. P and Q are called the orthogonal projections of H onto M and M1.. Ill/ ELEMENTARY HILBERT SPACE THEORY 81 PROOF As regards the uniqueness in (a), suppose that x' + y' = x" + y" for some vectors x', x" in M and y', y" in M l.. Then x' -x" = y" -y'. _Since x' -x" e M, y" - y' e M1., and M n M.l = {0} [an immediate conse quence of the fact that (x, x) = 0 implies x = 0], we have x" = x', y" = y'. To prove the existence of the decomposition, note that the set x + M = {x + y: y eM} is closed and convex. Define Qx to be the element of smallest norm tn x + M; this exists, by Theorem 4.10. Define Px = x- Qx. Since Qx e x + M, it is clear that Px e M. Thus P maps H into M. To prove that Q maps H into Ml. we show that (Qx, y) = 0 for all y e M. Assume IIYII = 1, without loss of generality, and put z = Qx. The minimizing property of Qx shows that (z, z) = llzll2 < liz-flYII2 = (z-fly, z-fly) for every scalar fl. This simplifies to 0 < -fl(y, z) -ti(z, y) + flti. With fX = (z, y), this gives 0 < -I (z, y) 12, so that (z, y) = 0. Thus Qx e M1.. We have already seen that Px eM. If y EM, it follows that llx-Yll2 = IIQx + (Px-y)ll2 = 11Qxll2 + IIPx-Yll2 which is obviously minimized when y = Px. We have now proved (a) and (b). If we apply (a) to x, to y, and to fXX + py, we obtain P(fXX + py) -fXPX -pPy = fXQX + PQy -Q(fXX + py). The left side is in M, the right side in M1.. Hence both are 0, so P and Q are linear. Since Px _i Qx, (d) follows from (a). To prove the corollary, take x e H, x ¢ M, and put y = Qx. Since Px eM, x =I= Px, hence y = x-Px =I= 0. /Ill We have already observed that xr (x, y) is, for each y e H, a continuous linear functional on H. It is a very important fact that all continuous linear functionals on H are of this type. 4.12 Theorem If L is a continuous linear functional on H, then there is a unique y e H such that Lx = (x, y) (x E H). (1) 82 REAL AND COMPLEX ANALYSIS PROOF If Lx = 0 for all x, take y = 0. Otherwise, define M = { x: Lx = 0}. (2) The linearity of L shows that M is a subspace. The continuity of L shows that M is closed. Since Lx =1= 0 for some x e H, Theorem 4.11 shows that M.L does not consist of 0 alone. Hence there exists z e M.L, with liz II = 1. Put u = (Lx)z -(Lz)x. (3) Since Lu = (Lx)(Lz)-(Lz)(Lx) = 0, we have u e M. Thus (u, z) = 0. This gtves Lx = (Lx)(z, z) = (Lz)(x, z). (4) Thus (1) holds with y = rxz, where (X = Lz. The uniqueness of y is easily proved, for if (x, y) = (x, y') for all x e H, set z = y- y'; then (x, z) = 0 for all x e H; in particular, (z, z) = 0, hence z = 0. /Ill Orthonormal Sets 4.13 Definitions If V is a vector space, if xb ... , xk E V, and if c1, ... , ck are scalars, then c1x1 + · · · + ck xk is called a linear combination of x1, ... , xk. The set { x 1 , . . . , xk} is called independent if c 1 x 1 + · · · + ck xk = 0 implies that c 1 = · · · = ck = 0. A set S c V is independent if every finite subset of S is independent. The set [S] of all linear combinations of all finite subsets of S (also called the set of all finite linear combinations of members of S) is clearly a vector space; [S] is the smallest subspace of V which contains S; [S] is called the span of S, or the space spanned by S. A set of vectors ua in a Hilbert space H, where rx runs through some index set A, is called orthonormal if it satisfies the orthogonality relations (ua, up) = 0 for all rx =1= {3, rx e A, and {3 e A, and if it is normalized so that lluall = 1 for each rx e A. In other words, {ua} is orthonormal provided that {1 if (X = {3, (u .. ' up) = 0 if a ¥= {J. (1) If {ua: rx e A} is orthonormal, we associate with each x e H a complex function x on the index set A, defined by (rx E A). (ɇ) One sometimes calls the numbers .X(rx) the Fourier coefficients of x, relative to the set { ua}. We begin with some simple facts about finite orthonormal sets. ELEMENTARY HILBERT SPACE THEORY 83 4.14 Theorem Suppose that {ua : C( e A} is an orthonormal set in Hand that F is a finite subset of A. Let M F be the span of { ua : C( e F}. (a) If <p is a complex function on A that is 0 outside F, then there is a vector y e M F, namely (1) that has y(C() = <p(C()for every C( E A. Also, IIYII2= L lqJ(C()I2• (2) aeF (b) Ifx e Hand sp(x) = L X(C()Ua (3) aeF then llx -sp(x)ll < llx -sll (4) for every s e M F, except for s = sp(x), and L I X(C() 12 < llxll2• (5) aeF PROOF Part (a) is an immediate consequence of the orthogonality relations 4.13(1). In the proof of (b), let us write sF in place of sp(x), and note that sp(C() = x(C() for all C( e F. This says that (x -sF) j ua if C( e F, hence (x -sp) j_ (sF -s) for every s E M F, and therefore llx-sll2 = ll(x-Sp) +(sF-s)ll2 = llx-sFII2 + llsF-sll2• (6) This gives (4). With s = 0, (6) gives llsFII2 < llxll2, which is the same as (5), because of (2). //// The inequality (4) states that the "partial sum " sp(x) of the " Fourier series " L x(C()Ua of x is the unique best approximation to x in M F, relative to the metric defined by the Hilbert space norm. 4.15 We want to drop the finiteness condition that appears in Theorem 4.14 (thus obtaining Theorems 4.17 and 4.18) without even restricting ourselves to sets that are necessarily countable. For this reason it seems advisable to clarify the meaning of the symbol. La e 4 qJ(C() when C( ranges over an arbitrary set A. Assume 0 < <p(C() < oo for each C( e A. Then L qJ (1) aeA denotes the supremum of the set of all finite sums qJ(C(1) + · · · + qJ(C(n), where C(1, ... , C(n are distinct members of A. 84 REAL AND COMPLEX ANALYSIS A moment's consideration will show that the sum (1) is thus precisely the Lebesgue integral of qJ relative to the counting measure J1 on A. In this context one usually writes tP(A) for l!(J1). A complex function qJ with domain A is thus in t2(A) if and only if (2) Example 4.5(b) shows that t2(A) is a Hilbert space, with inner product ( (/J, t/1 > = L ({J( rx )t/1( Ct ). (3) cx e A Here, again, the sum over A stands for the integral of qJtfr with respect to the counting measure; note that qJtfr e t1(A) because qJ and t/J are in t2(A). Theorem 3.13 shows that the functions qJ that are zero except on some finite subset of A are dense in t2(A). Moreover, if qJ E t2(A), then { rx e A : qJ(rx) =1= 0} is at most countable. For if An is the set of all (J. where I qJ(rx) I > 1/n, then the number of elements of A is at most ex E An cx e A Each An (n = 1, 2, 3, . . . ) is thus a finite set. The following lemma about complete metric spaces will make it easy to pass from finite orthonormal sets to infinite ones. 4.16 Lemma Suppose that (a) X and Yare metric spaces, X is complete, (b) f: X Ò Y is continuous, (c) X has a dense subset X0 on whichfis an isometry, and (d) f(X 0) is dense in Y. Thenfis an isometry of X onto Y. The most important part of the conclusion is thatj'maps X onto all of Y. Recall that an isometry is simply a mapping that preserves distances. Thus, by assumption, the distance betweenf(x1) andf(x2) in Y is equal to that between x1 and x2 in X, for all points x1, x2 in X 0 . PROOF The fact that f is an isometry on X is an immediate consequence of the continuity off, since X 0 is dense in X. Pick y e Y. Since f(X 0) is dense in Y, there is a sequence { xn} in X 0 such thatf(xn) Ò y as nr oo. Thus {f(xn)} is a Cauchy sequence in Y. Sincefis an isometry on X 0 , it follows that { xn} is also a Cauchy sequence. The com pleteness of X implies now that { xn} converges to some x e X, and the con tinuity 9ffshows thatf(x) = lim f(xn) = y. //// ELEMENTARY HILBERT SPACE THEORY 85 4.17 Theorem Let { ua: a. e A} be an orthonormal set in H, and let P be the space of all finite linear combinations of the vectors ua. The inequality L I x(a.) 12 < llxll2 (1) a e A holds then for every x e H, and xr xis a continuous linear mapping of H onto t2(A) whose restriction to the closure P of P is an isometry of Ponto t2(A). PROOF Since the inequality 4.14(5) holds for every finite set F c A, we have (1), the so-called Bessel inequality. Define f on H by f(x) = x. Then (1) shows explicitly that f maps H into t2(A). The linearity off is obvious. If we apply (1) to x-y we see that llf(y) -f(x)ll2 = II.Y-xll2 < IIY-xll. Thus f is continuous. Theorem 4.14(a) shows that f is an isometry of P onto the dense subspace of t2(A) consisting ࡚f those functions whose support is a finite set F cA. The theorem follows therefore from Lemma 4.16, applied with X= P, X0 = P, Y = t2(A); note that P, being a closed subset of the complete metric spate H, is itself complete. /Ill The fact that the mapping xŠ x carries H onto t2(A) is known as the Riesz Fischer theorem. 4.18 Theorem Let {ua: a. e A} be an orthonormal set in H. Each of the follow ing four conditions on { ua} implies the other three: (i) { ua} is a maximal orthonormal set in H. (ii) The set P of all finite linear combinations of members of { ua} is dense in H. (iii) The equality holds for every x e H. (iv) The equality L I x(a.) 12 = llxll2 a e A L x(a.).Y(a.) = (x, y) a e A holds for all x e H and y e H. The last formula is known as Parseval 's identity. Observe that x and y are in t2(A), hence xy is in t1(A), so that the sum in (iv) is well defined. Of course, (iii) is the special case x = y of (iv). Maximal orthonormal sets are often called complete orthonbrmal sets or orthonormal bases. 86 REAL AND COMPLEX ANALYSIS PROOF To say that {u«} is maximal means simply that no vector of H can be adjoined to { u«} in such a way that the resulting set is still orthonormal. This happens precisely when there is no x =F 0 in H that is orthogonal to every u«. We shall prove that (i) Ò (ii) Ò (iii)Ò (iv) Ò (i). If P is not dense in H, then its closure P is not all of H, and the corollary to Theorem 4.1 1 implies that pl. contains a nonzero vector. Thus { u«} is not maximal when P is not dense, and (i) implies (ii). If (ii) holds, so does (iii), by Theorem 4.17. The implication (iii) Ò (iv) follows from the easily proved Hilbert space identity (sometimes called the " polarization identity ") 4(x, y) = llx + Yll2- llx-Yll2 + illx + iyll2-illx- iyll2 which expresses the inner product (x, y) in terms of norms and which is equally valid with .X, y in place of x, y, simply because t2(A) is also a Hilbert space. (See Exercise 19 for other identities of this type.) Note that the sums in (iii) and (iv) are II .XII ࡛ and (.X, y), respectively. Finally, if (i) is false, there exists u =F 0 in H so that (u, u«) = 0 for all C( e A. If x = y = u, then (x, y) = llull2 > 0 but .X(C() = 0 for all C( e A, hence (iv) fails. Thus (iv) implies (i), and the proof is complete. I I I I 4.19 Isomorphisms Speaking informally, two algebraic systems of the same nature are said to be isomorphic if there is a one-to-one mapping of one onto the other which preserves all relevant properties. For instance, we may ask whether two groups are isomorphic or whether two fields are isomorphic. Two vector spaces are isomorphic if there is a one-to-one linear mapping of one onto the other. The linear mappings are the ones which preserve the relevant concepts in a vector space, namely, addition and scalar multiplication. In the same way, two Hilbert spaces H 1 and H 2 are isomorphic if there is a one-to-one linear mapping A of H 1 onto H 2 which also preserves inner products : (Ax, Ay) = (x, y) for all x and y e H1• Such a A is an isomorphism (or, more specifically, a Hilbert space isomorphism) of-H 1 onto H 2 • Using this terminology, Theorems 4.17 and 4.18 yield the following statement : I f {u«: C( e A} is a maximal orthonormal set in a Hilbert space H, and ifx(C() = (x, u«), then the mapping xŠ xis a Hilbert space isomorphism of H onto t2(A). One can prove (we shall omit this) that t2(A) and t2(B) are isomorphic if and only if the sets A and B have the same cardinal number. But we shall 'Prove that every nontrivial Hilbert space (this means that the space does not consist of 0 alone) is isomorphic to some t2(A), by proving that every such space contains a maximal orthonormal set (Theorem 4.22). The proof will depend on a property of partially ordered sets which is equivalent to the axiom of choice. 4.20 Partially Ordered Sets A set f!/J is said to be partially ordered by a binary relation < if (a) a < b and b < c implies a < c. (b) a < a for every C( E f!JJ . (c) a < b and b < a implies a = b. ELEMENTARY HILBERT SPACE THEORY 87 A subset f2 of a partially ordered set f!JJ is said to be totally ordered (or lin early ordered) if every pair a, b E f2 satisfies either C( < b or b < a. For example, every collection of subsets of a given set is partially ordered by the inclusion relation c . To give a more specific example, let f!JJ be the collection of all open subsets of the plane, partially ordered by set inclusion, and let f2 be the collection of all open circular discs with center at the origin. Then f2 c f!JJ , f2 is totally ordered by c , and f2 is a maximal totally ordered subset of 9. This means that if any member of f!JJ not in f2 is adjoined to fl, the resulting collection of sets is no longer totally ordered by c . 4.21 The Hausdorff Maximality Theorem Every nonempty partially ordered set contains a maximal totally ordered subset. This is a consequence of the axiom of choice and is, in fact, equivalent to it ; another (very similar) form of it is known as Zorn's lemma. We give the proof in the Appendix. If now H is a nontrivial Hilbert space, then there exists a u E H with llull = 1, so that there is a nonempty orthonormal set in H. The existence of a maximal orthonormal set is therefore a consequence of the following theorem : 4.22 Theorem Every orthonormal set B in a Hilbert space H is contained in a maximal orthonormal set in H. PROOF Let f!JJ be the class of all orthonormal sets in H which contain the given set B. Partially order f!JJ by set inclusion. Since B E f!JJ , f!JJ =1= 0. Hence f!JJ contains a maximal totally ordered class n. Let S be the union of all members of n. It is clear that B c S. We claim that S is a maximal orthonor mal set : If u1 and u2 e S, then u1 E A1 and u2 e A2 for some A1 and A2 E n. Since n is total ordered, Al c A2 (or A2 c At), so that ul E A2 and u2 E A2 . Since A2 is orthonormal, (u1, u2) = 0 if u1 =1= u2 , (u1, u2) = 1 if u1 = u2 . Thus S is an orthonormal set. Suppose S is not maximal. Then S is a proper subset of an orthonormal set S. Clearly, S ¢ n, and S contains every member of n. Hence we may adjoin S to Q and still have a total order. This contradicts the maximality n. 1m 88 REAL AND COMPLEX ANALYSIS Trigonometric Series 4.23 Definitions Let T be the unit circle in the complex plane, i.e., the set of all complex numbers of absolute value 1. If F is any function on T and iff is defined on R 1 by (1) thenfis a periodic function of period 2n. This means thatf(t + 2n) =f(t) for all real t. Conversely, iff is a function on R1, with period 2n, then there is a function F on T such that (1) holds. Thus we may identify functions on T with 2n-periodic functions· on R 1; and, for simplicity of notation, we shall sometimes write f(t) rather than f(eit), even if we think off as being defined on T. With these conventions in mind, we define I.!'(T), for 1 < p < oo, to be the class of all complex, Lebesgue measurable, 2n-periodic functions on R 1 for which the norm is finite. { 1 f1t }1/p 11/IIP= 2n ,}f(t)IPdt (2) In other words, we are looking at I.!'(J1), where J1 is Lebesgue measure on [0, 2n] (or on T), divided by 2n. L00(T) will be the class of all 2n-periodic members of L00(R1), with the essential supremum norm, and C(T) consists of all continuous complex functions on T (or, equivalently, of all continuous, complex, 2n-periodic functions on R1), with norm 11/11 oo = sup I f(t) I, t (3) The factor 1/(2n) in (2) simplifies the formalism we are about to develop. For instance, the 1.!'-norm of the constant function 1 is 1. A trigonometric polynomial is a finite sum of the form N f(t) = a0 + L (an cos nt + bn sin nt) (4) n=l where a0, ab ... , aN and bb . . . , bN are complex numbers. On accoqnt of the Euler identities, (4) can also be written in the form N f(t) = L c" eint (5) n= -N \vhich is more convenient for most purposes. It is clear that every trigono metric polynomial has period 2n. ELEMENTARY HILBERT SPACE THEORY 89 We shall denote the set of all integers (positive, zero, and negative) by Z, and put (n E Z). If we define the inner product in L2( T) by 1 J 1t (f, g) = ln _ /(t)g(t) dt [note that this is in agreement with (2)], an easy computation shows that (un ' um) = ! f1t ei 0 there is a trigonometric poly nomial P such that II f- P 112 < E. Since II g 112 < II g II oo for every g E C( T), the esti mate II f-P 112 < E will follow from II f- P II oo < E, and it is this estimate which we shall prove. Suppose we had trigonometric polynomials Q1, Q2 , Q3 , • .
• , with the follow ing properties : (a) (b) 1 f1t -2 Qk(t) dt = 1. 1t - 1t (c) If YJk(t5) = sup { Qk(t): t5 < It I < n }, then lim YJk(t5) = 0 k-+oo for every t5 > 0. Another way of stating (c) is to say : for every t5 > 0, Qk(t)r 0 uniformly on [ - n, -t5] u [b, n]. To eachf E C(T) we associate the functions Pk defined by ( k = 1' 2, 3' . .. ). (1) 90 REAL AND COMPLEX ANALYSIS If we replace s by -s (using Theorem 2.20(e)) and then by s - t, the periodicity of f and Qk shows that the value of the integral is not affected. Hence ( k = 1' 2, 3, ... ). (2) Since each Qk is a trigonometric polynomial, Qk is of the form Nk Q (t) = " a eint k 1-.J n, k ' (3) and if we replace t by t -s in (3) and substitute the result in (2), we see that each P k is a trigonometric polynomial. Let E > 0 be given. Since f is uniformly continuous on T, there exists a ç > 0 such that I f(t) -f(s) I < E whenever It-s I < ç- By (b), we have 1 J1t Pk(t) -f(t) = 2n _ , {f(t- s) -f(t)}Qk(s) ds, and (a) implies, for all t, that 1 J1t I Pk(t) -f(t) I < 2n _ ,1 f(t- s) -f(t) I Qk(s) ds = A1 + A2 , where A1 is the integral over [-Ī, Ī] and A2 is the integral over [-n, - Ī] u [£5, n]. In Ab the integrand is less than EQk(s), so A1 < E, by (b). In A2 , we have Qk(s) < 'lk(ç), hence (4) for sufficiently large k, by (c). Since these estimates are independent of t, we have proved that lim II!-Pklloo = 0. (5) k-+ 00 It remains to construct the Qk. This can be done in many ways. Here is a simple one. Put {1 + cos t}k Qk(t) = ck 2 , (6) where ck is chosen so that (b) holds. Since (a) is clear, we only need to show (c). Since Qk is even, (b) shows that 1 ck i1t {1 + cos t}kd ck i1t {1 + cos t}k . d 2ck = -t > -SID t t = . n o 2 n 0 2 n( k + 1) Since Qk is decreasing on [0, n], it follows that Qit) < Qk(l5) < n(k; 1) c +nOS 15y (0 < ç < I t I < n). (7) ELEMENTARY HILBERT SPACE THEORY 91 This implies (c), since 1 + cos ç < 2 if 0 < ç < n. We have proved the following important result : 4.25 Theorem I ff e C(T) and E > 0, there is a trigonometric polynomial P such that I f(t) -P(t) I < E for every real t. A more precise result was proved by Fejer (1904): The arithmetic means of the partial sums of the Fourier series of any f e C(T) converge uniformly to f. For a proof (quite similar to the above) see Theorem 3.1 of [ 45], or p. 89 of , vol. I. 4.26 Fourier Series For any f e L1(T), we define the Fourier coefficients off by the formula A 1 f1t • f(n) = 2n ,/(t)e-"'1 dt (n E Z), (1) where, we recall, Z is the set of all integers. We thus associate with each/ e L1(T) a function/on Z. The Fourier series of/is (2) and its partial sums are N sױt) = L /(n)eint (N = 0, 1, 2, ... ). (3) -N Since L2(T) c L1(T), (1) can be applied to every f e I3(T). Comparing the defi nitions made in Sees. 4.23 and 4.13, we can now restate Theorems 4.17 and 4.18 in concrete terms: The Riesz-Fischer theorem asserts that if {en} is a sequence of complex numbers such that (4) n= - ao then there exists anf e I3(T) such that 1 f 1t . en = -f(t)e - lnt dt 2n _ 1t (n e Z). (5) The Parseval theorem asserts that (6) 92 REAL AND COMPLEX ANALYSIS whenever f E L2(T) and g E L2(T); the series on the left of (6) converges absolu tely; and if sN is as in (3), then lim II!- sN ll 2 = 0, (7) N-+oo since a special case of ( 6) yields llf- sN IIC = L l /(n) l2• (8) lni >N Note that (7) says that every f E I!(T) is the 13-limit of the partial sums of its Fourier series ; i.e., the Fourier series off converges to f, in the L2-sense. Pointwise convergence presents a more delicate problem, as we shall see in Chap. 5. The Riesz-Fischer theorem and the Parseval theorem may be summarized by saying that the mappingf +--+ /is a Hilbert space isomorphism of L2(T) onto t2(Z). The theory of Fourier series in other function spaces, for instance in L1(T), is much more difficult than in I!(T), and we shall touch only a few aspects of it. Observe that the crucial ingredient in ·the proof of the Riesz-Fischer theorem is the fact that 13 is complete. This is so well recognized that the name " Riesz Fischer theorem " is sometimes given to the theorem which asserts the complete ness of L2, or even of any /J'. Exercises In this set of exercises, H always denotes a Hilbert space. 1 If M is a closed subspace of H, prove that M = (M.i).i. Is there a similar true statement for sub spaces M which are not necessarily closed? 2 Let { xn : n = 1, 2, 3, . . . } be a linearly independent set of vectors in H. Show that the following construction yields an orthonormal set {un} such that {x1, • • • , xN} and {u1, • • • , uN} have the same span for all N. Put u1 = xtfllx1 ll . Having u1, . . . , un _ 1 define n - 1 Vn = xn - L (xn , uJui , i = 1 Note that this leads to a proof of the existence of a maximal orthonormal set in separable Hilbert spaces which makes no appeal to the Hausdorff maximality principle. (A space is separable if it contains a countable dense subset.) 3 Show that I!(T) is separable if 1 < p < oo, but that L00(T) is not separable. 4 Show that H is separable if and only if H contains a maximal orthonormal system which is at most countable. 5 If M = {x: Lx = 0}, where L is a continuous linear functional on H, prove that M.i is a vector space of dimension 1 (unless M = H). 6 Let { un} (n = 1, 2, 3, . . . ) be an orthonormal set in H. Show that this gives an example of a closed and bounded set which is not compact. Let Q be the set of all x E H of the form (where I c. I <m). Prove that Q is compact. (Q is called the Hilbert cube.) ELEMENTARY HILBERT SPACE THEORY 93 More generally, let { <5,.} be a sequence of positive numbers, and let S be the set of all x E H of the form Prove that S is compact if and only if Lf <>; < oo . Prove that H is not locally compact. 7 Suppose {a,.} is a sequence of positive numbers such that L a,. b,. < oo whenever b,. > 0 and L b; < oo . Prove that L a; < oo . Suggestion: IfL a; = oo then there are disjoint sets Ek (k = 1, 2, 3, .
. . ) so that L a; > 1. ,. E EƦc Define bn so that b,. = ck a,. for n E Ek . For suitably chosen ck , L a,. b,. = oo although L b; < oo . 8 If H 1 and H 2 are two Hilbert spaces, prove that one of them is isomorphic to a subspace of the other. (Note that every closed subspace of a Hilbert space is a Hilbert space.) 9 If A c: [0, 2n] and A is measurable, prove that lim f cos nx dx = lim f sin nx dx = 0. n-+oo A n-+oo A 10 Let n 1 < n2 < n3 < · · · be positive integers, and let E be the set of all x E [0, 2n] at which {sin nk x} converges. Prove that m(E) = 0. Hint : 2 sin2 IX = 1 - cos 21X, so sin nk x ___. + 1/.Ji a.e. on E, by Exercise 9. 1 1 Find a nonempty closed set E in I3(T) that contains no element of smallest norm. 12 The constants ck in Sec. 4.24 were shown to be such that k- 1ck is bounded. Estimate the relevant integral more precisely and show that 0 < lim k- 112ck < oo . k-+ 00 13 Suppose f is a continuous function on R 1, with period 1. Prove that lim - L f(nǭX) = f(t) dt 1 N il N-+oo N n= l 0 for every irrational real number IX. Hint : Do it first for 14 Compute and find f(t) = exp (2nikt), k = 0, + 1, + 2, . . . . min J 1 I x3 - a - bx - cx2 12 dx a, b, c -1 where g is subject to the restrictions 94 REAL AND COMPLEX ANALYSIS IS Compute State and solve the corresponding maximum problem, as in Exercise 14. 16 If x0 E H and M is a closed linear subspace of H, prove that min {llx - x0ll : x E M} = max {l(x0 , y)l : y E MJ., IIYII = 1}. 17 Show that there is a continuous one-to-one mapping y of [0, 1] into H such that y(b) - y(a) is orthogonal to y(d) -- y(c) whenever 0 < a < b r c < d < 1. (y may be called a " curve with orthogonal increments.") Hint : Take H = 13, and consider characteristic functions of certain subsets of [0, 1]. 18 Define us<t) = eisr for all s E R 1, t E R 1 • Let X be the complex vector space consisting of all finite linear combinations of these functions us . Iff E X and g E X, show that 1 fA -(f, g) = lim -2 f(t)g(t) dt A .... oo A -A exists. Show that this inner product makes X into a unitary space whose completion is a non separable Hilbert space H. Show also that {us: s E R 1} is a maximal orthonormal set in H. 19 Fix a positive integer N, put ro = e21Ci/N, prove the orthogonality relations ! i (J)nk = { 1 if k = 0 N n = 1 0 if 1 < k < N - 1 and use them to derive the identities 1 N (x, Y) = - L llx + ro"yll2ro" N n = 1 that hold in every inner product space if N > 3. Show also that (x, y) = ŀ f1C llx + ei8yll2ei8 dO. 2n 1C CHAPTER FIVE EXAMPLES OF BANACH SPACE TECHNIQUES Banach Spaces 5.1 In the preceding chapter we saw how certain analytic facts about trigonomet ric series can be made to emerge from essentially goemetric considerations about general Hilbert spaces, involving the notions of convexity, subspaces, orthog onality, and completeness. There are many problems in analysis that can be attacked with greater ease when they are placed within a suitably chosen abstract framework. The theory of Hilbert spaces is not always suitable since orthogonality is something rather special. The class of all Banach spaces affords greater variety. In this chapter we shall develop some of the basic properties of Banach spaces and illustrate them by applications to concrete problems. 5.2 Definition A complex vector space X is said to be a normed linear space if to each x e X there is associated a nonnegative real number llxll, called the norm of x, such that (a) llx + Yll < llxll + IIYII for all x and y e X, (b) II axil = I C( I llxll if x e X and C( is a scalar, (c) llxll = 0 implies x = 0. By (a), the triangle inequality llx -Yll < llx -zll + liz -Yll (x, y, z e X) holds. Combined with (b) (take C( = 0, C( = - 1) and (c) this shows that every normed linear space may be regarded as a metric space, the distance between x and y being llx-Yll. A Banach space is a normed linear space which is complete in the metric defined by its norm. 96 REAL AND COMPLEX ANALYSIS For instance, every Hilbert space is a Banach space, so is every IJ'(J1) normed by 11/IIP (provided we identify functions which are equal a.e.) if 1 < p < oo, and so is C0(X) with the supremum norm. The simplest Banach space is of course the complex field itself, normed by llxll = I x l -One can equally well discuss real Banach spaces; the definition is exactly the same, except that all scalars are assumed to be real. 5.3 Definition Consider a linear transformation A from a normed linear space X into a normed linear space Y, and define its norm by IIAII = sup .{II Axil: x e X, llxll < 1}. (1) If II A II < oo, then A is called a bounded linear transformation. In (1), llxll is the norm of x in X, IIAxll is the norm of Ax in Y; it will frequently happen that several norms occur together, and the context will make it clear which is which. Observe that we could restrict ourselves to unit vectors x in (1), i.e., to x with llxll = 1, without changing the supremum, since IIA(C(x)ll = IIC(Axll = I C( Ill Axil-(2) Observe also that II All is the smallest number such that the inequality IIAxll < IIAII IIxll (3) holds for every x e X. The following geometric picture is helpful: A maps the closed unit ball in X, i.e., the set {x e X : llxll < 1}, (4) into the closed ball in Y with center at 0 and radius II All-An important special case is obtained by taking the complex field for Y; in that case we talk about bounded linear functionals. 5.4 Theorem For a linear transformation A of a normed linear space X into a normed linear space Y, each of the following three conditions implies the other two: (a) A is bounded. (b) A is continuous. (c) A is continuous at one point of X. PROOF Since IIA(x1 -x2)11 < 11AII IIx1 -x2ll, it is clear that (a) implies (b), and (b) implies (c) trivially. Suppose A is continuous at x0• To each E > 0 one can then find a b > 0 so that llx-x0ll < b implies II Ax-Ax0ll < E. In other words, II x II < b implies IIA(x0 + x)-Ax0ll < E. EXAMPLES OF BANACH SPACE TECHNIQUES 97 But then the linearity of A shows that II Axil < E. Hence II All < E/b, and (c) implies (a). /Ill Consequences of Haire's Theorem 5.5 The manner in which the completeness of Banach spaces is frequently exploited depends on the following theorem about complete metric spaces, which also has many applications in other parts of mathematics. It implies two of the three most important theorems which make Banach spaces useful tools in analysis, the Banach-Steinhaus theorem and the open mapping theorem. The third is the Hahn-Banach extension theorem, in which completeness plays no role. 5.6 Baire's Theorem If X is a complete metric space, the intersection of every countable collection of dense open subsets of X is dense in X. In particular (except in the trivial case X = 0), the intersection is not empty. This is often the principal significance of the theorem. PROOF Suppose vl, v2' v3' . . . are dense and open in X. Let w be any open set in X. We have to show that n V,. has a point in W if W =1= 0. Let p be the metric of X; let us write S(x, r) = {y E X : p(x, y) < r} (1) and let S(x, r) be the closure of S(x, r). [Note: There exist situations in which S(x, r) does not contain all y with p(x, y) < r !] Since V1 is dense, W n V1 is a nonempty open set, and we can therefore find x 1 and r 1 such that (2) If n > 2 and xn _ 1 and r n _ 1 are chosen, the denseness of V,. shows that V,. n S(xn _ 1, rn _ 1) is not empty, and we can therefore find xn and rn such that (3) By induction, this process produces a sequence { xn} in X. If i > n and j > n, the construction shows that xi and xi both lie in S(xn, rn), so that p(xi' X j) < 2r n < 2/n, and hence { xn} is a Cauchy sequence. Since X is com plete, there is a point x E X such that x = lim xn . Since xi lies in the closed set S(xn, r n) if i > n, it follows that x lies in each S(xn, r n), and (3) shows that x lies in each V,. . By (2), x E W. This completes the proof. I I I I Corollary In a complete metric space, the intersection of any countable collec tion of dense G ª ' s is again a dense G ª . 98 REAL AND COMPLEX ANALYSIS This follows from the theorem, since every Gd is the intersection of a count able collection of open sets, and since the union of countably many countable sets is countable. 5.7 Baire's theorem is sometimes called the category theorem, for the following reason. Call a set E c X nowhere dense if its closure E contains no nonempty open subset of X. Any countable union of nowhere dense sets is called a set of the first category; all other subsets of X are of the second category (Baire's terminology). Theorem 5.6 is equivalent to the statement that no complete metric space is of the first category. To see this, just take complements in the statement of Theorem 5.6. 5.8 The Banach-Steinhaus Theorem Suppose X is a Banach space, Y is a normed linear space, and {A«} is a collection of bounded linear transformations of X into Y, where 1X ranges over some index set A. Then either there exists an M < oo such that (1) for every C( e A, or sup IIAaxll = oo (2) « e A for all x belonging to some dense Gd in X. In geometric terminology, the alternatives are as follows: Either there is a ball B in Y (with radius M and center at 0) such that every Aa maps the unit ball of X into B, or there exist x e X (in fact, a whole dense Gd of them) such that no ball in Y contains Aa x for all C(. The theorem is sometimes referred to as the uniform boundedness principle. PROOF Put and let n} (x EX) (n = 1, 2, 3, .
. . ). (3) (4) Since each Aa is continuous and since the norm of Y is a continuous function on Y (an immediate consequence of the triangle inequality, as in the proof of Theorem 4.6), each function X --+ IIAaxll is continuous on X. Hence cp is lower semicontinuous, and each V,. is open. EXAMPLES OF BANACH SPACE TECHNIQUES 99 If one of these sets, say VN , fails to be dense in X, then there exist an x0 e X and an r > 0 such that llxll < r implies x0 + x rJ VN; this means that <p(x0 + x) < N, or (5) for all C( e A and all x with llxll < r. Since x = (x0 + x)-x0, we then have (6) and it follows that (1) holds with M = 2Nir. The other possibility is that every V,. is dense in X. In that case, n V,. is a dense G, in X, by Baire's theorem; and since (/)(X)= 00 for every X E n v,. ' the proof is complete. I I I I 5.9 The Open Mapping Theorem Let U and V be the open unit balls of the Banach spaces X and Y. To every bounded linear transformation A of X onto Y there corresponds a ç > 0 so that A(U) => ç V. (1) Note the word " onto " in the hypothesis. The symbol (J V stands for the set y: y e V}, i.e., the set of all y e Y with IIYII < ç. It follows from (1) and the linearity of A that the image of every open ball in , with center at x0, say, contains an open ball in Y with center at Ax0• Hence e image of every opt!n set is open. This explains the name of the theorem. Here is another way of stating (1): To every y with IIYII < ç there corresponds x with llxll < 1 so that Ax = y. PROOF Given y e Y, there exists an x e X such that Ax = y; if llxll < k, it follows that y e A(kU). Hence Y is the union of the sets A(kU), for k = 1, 2, 3, . .
.. Since Y is complete, the Baire theorem implies that there is a nonempty open set W in the closure of some A(kU). This means that every point of W is the limit of a sequence {Ax;}, where X; e kU; from now on, k and W are fixed. Choose y0 e W, and choose 11 > 0 so that Yo + y e W if IIYII < 11· For any such y there are sequences { xa, { xț'} in kU such that Axț---+ Yo, Axț'---+ Yo + y (i---+ oo ). (2) Setting xi = xț' -xț, we have II X; II < 2k and Ax;---+ y. Since this holds for every y with II y II < 11, the linearity of A shows that the following is true, if (J = 1112k: To each y e Y and to each E > 0 there corresponds an x e X such that llxll < CJ-1IIYII and IIY-Axil < E. (3) This is almost the desired conclusion, as stated just before the start of the proof, except that there we had E = 0. 100 REAL AND COMPLEX ANALYSIS Fix y e bV, and fix E > 0. By (3) there exists an x1 with llx111 < 1 and IIY -Ax1ll < ȚbE. Suppose x1, • • • , x,. are chosen so that (4) (5) Use (3), with y replaced by the vector on the left side of ( 5), to obtain an x,. + 1 so that (5) holds with n + 1 in place of n, and (n = 1, 2, 3, . . . ). (6) If we set sn = x1 + · · · + xn, (6) shows that { s,.} is a Cauchy sequence in X. Since X is complete, there exists an x e X so that s,.r x. The inequality II x 111 < 1, together with ( 6), shows that II x II < 1 + E. Since A is continuous, As,.r Ax. By (5), AsnŠ y. Hence Ax= y. We have now proved that A((1 + E)U) :::> b V, (7) or A( U) :::> ( 1 + E) -1 b v' (8) for every E > 0. The union of the sets on the right of (8), taken over all E > 0, is bV. This proves (1). //// 5.10 Theorem If X and Y are Banach spaces and if A is a bounded linear transformation of X onto Y which is also one-to-one, then there is a b > 0 such that II Ax II > b II x II (x E X). In other words, A -1 is a bounded linear transformation of Y onto X. (1) PRooF If b is chosen as in the statement of Theorem 5.9, the conclusion of that theorem, combined with the fact that A is now one-to-one, shows that IIAxll < b implies llxll < 1. Hence llxll > 1 implies IIAxll > b, and (1) is proved. The transformation A-1 is defined on Y by the requirement that A -1 y = x if y = Ax. A trivial verification shows that A -1 is linear, and (1) implies that IIA -111 < 1/b. /Ill Fourier Series of Continuous Functions 5.1 1 A Convergence Problem Is it true for every f e C( T) that the Fourier series of f converges to f(x) at every point x? EXAMPLES OF BANACH SPACE TECHNIQUES 101 Let us recall that the nth partial sum of the Fourier series off at the point x is given by ( n = 0, 1, 2, . . . ), (1) n where Dn(t) = L eikt. (2) k= -n This follows directly from formulas 4.26(1) and 4.26(3). The problem is to determine whether lim sn(f; x) = f(x) (3) n -+ oo for every f e C(T) and for every real x. We observed in Sec. 4.26 that the partial sums do converge to f in the L2-norm, and Theorem 3.12 implies therefore that each f e I3(T) [hence also each f e C(T)] is the pointwise limit a.e. of some sub sequence of the full sequence of the partial sums. But this does not answer the present question. We shall see that the Banach-Steinhaus theorem answers the question nega tively. Put s(f; x) = sup I sn(f; x) 1 . (4) n To begin with, take x = 0, and define (fe C(T); n = 1, 2, 3, .. . ). (5) We know that C(T) is a Banach space, relative to the supremum norm 11/11 oo . It follows from (1) that each An is a bounded linear functional on C(T), of norm (6) We claim that as n ----+ oo . (7) This will be proved by showing that equality holds in (6) and that as n ----+ oo . (8) Multiply (2) by eit/2 and by e -it/2 and subtract one of the resulting two equa tions from the other, to obtain D ( ) = sin (n + t)t n t sin (t/2) · (9) 102 REAL AND COMPLEX ANALYSIS Since I sin x I < I x I for all real x, (9) shows that 2 i1t ( 1) dt 2 i(n + 112)1t dt 11Dnll1 > -sin n + - t - = -lsin t I-1t 0 2 t 1t 0 t which proves (8). 2 n 1 ik1t 4 n 1 > - L -I sin t I dt = 2 L ----+ oo, 7t k = 1 kn (k -1 )1t 7t k = 1 k Next, fix n, and put g(t) = 1 if Dn(t) > 0, g(t) = -1 if Dn(t) < 0. There exist jj e C(T) such that -1 <jj < 1 andf,{t)---+ g(t) for every t, asj---+ oo. By the domi nated convergence theorem, Thus equality holds in (6), and we have proved (7). Since (7) holds, the Banach-Steinhaus theorem asserts now that s(f; 0) = oo for every fin some dense G«S-set in C(T). We chose x = 0 just for convenience. It is clear that the same result holds for every other x : To each real number x there corresponds a set Ex c C(T) which is a dense G«S in C(T), such that s(f; x) = oo for every f e Ex . In particular, the Fourier series of each f e Ex diverges at x, and we have a negative answer to our question. (Exercise 22 shows the answer is positive if mere continuity is replaced by a somewhat stronger smoothness assumption.) It is interesting to observe that the above result can be strengthened by another application of Baire's theorem. Let us take countably many points xi , and let E be the intersection of the corresponding sets EXi c C(T). By Baire's theorem, E is a dense G«S in C(T). Every f e E has s(f; xi) = oo at every point xi. For each/, s(f; x) is a lower semicontinuous function of x, since (4) exhibits it as the supremum of a collection of continuous functions. Hence { x: s(f; x) = oo} is a G «S in R 1, for each f If the above points xi are taken so that their union is dense in (-n, n), we obtain the following result: 5.12 Theorem There is a set E c C(T) which is a dense G«S in C(T) and which has the following property: For each f e E, the set Q1 = {x: s(f; x) = oo} is a dense G«S in R1• EXAMPLES OF BANACH SPACE TECHNIQUES 103 This gains in interest if we realize that E, as well as each Q1 , is an uncount able set : 5.13 Theorem In a complete metric space X which has no isolated points, no countable dense set is a G˱ . PROOF Let xk be the points of a countable dense set E in X. Assume that E is a G ˱ . Then E = n V,. , where each V,. is dense and open. Let n W, = V,. - U {xk}· k=l Then each W, is still a dense open set, but n W,. = 0, in contradiction to Baire's theorem. II II Note: A slight change in the proof of Baire's theorem shows actually that every dense G˱ contains a perfect set if X is as above. Fourier Coefficients of L 1-functions 5.14 As in Sec. 4.26, we associate to every f e L1(T) a function Jon Z defined by /(n) = -f(t)e- lnt dt 1 f1t . 2n -1t (n E Z). (1) It is easy to prove that /(n)---+ 0 as In I-+ oo, for every f e L1• For we know that C(T) is dense in L1(T) (Theorem 3.14) and that the trigonometric polynomials are dense in C(T) (Theorem 4.25). If E > 0 and f e L1(T), this says that there is a g e C(T) and a trigonometric polynomial P such that II f-g 111 < E and llg -Pll oo < E. Since llg-Pll1 < llg-Plloo if follows that II/-Pll1 < 2e; and if In I is large enough (depending on P), then 1 f1t 1 /(n) I = 21!: -x {f(t)-P(t)}e-int dt < 11/-Pll1 < 2E. (2) Thus/(n)---+ 0 as n---+ + oo. This is known as the Riemann-Lebesgue lemma. The question we wish to raise is whether the converse is true. That is to say, if {an} is a sequence of complex numbers such that an -+ 0 as n---+ + oo , does it follow that there is anf e L1(T) such that/(n) = an for all n e Z? In other words, is something like the Riesz-Fischer theorem true in this situation? This can easily be answered (negatively) with the aid of the open mapping theorem. 104 REAL AND COMPLEX ANALYSIS Let c0 be the space of all complex functions <p on Z such that <p(n)----+ 0 as n----+ + oo , with the supremum norm II <p II oo = sup { I <p( n) I : n e Z}. (3) Then c0 is easily seen to be a Banach space. In fact, if we declare every subset of Z to be open, then Z is a locally compact Hausdorff space, and c0 is nothing but C0(Z). The following theorem contains the answer to our question: " 5.15 Theorem The mapping f----+ f is a one-to-one bounded linear transformation of L1(T) into (but not onto) c0. PROOF Define A by Af = J It is clear that A is linear. We have just proved that A maps L1(T) into c0, and formula 5.14(1) shows that I /(n) I < 11 / 11 1, so that II A II < 1. (Actually, II A II = 1 ; to see this, take f = 1.) Let us now prove that A is one-to-one. Suppose/ e L1(T) and/(n) = 0 for every n e Z. Then f/(t)g(t) dt = 0 (1) if g is any trigonometric polynomial. By Theorem 4.25 and the dominated convergence theorem, (1) holds for every g e C(T). Apply the dominated con vergence theorem once more, in conjunction with the Corollary to Lusin's theorem, to conclude that (1) holds if g is the characteristic function of any measurable set in T. Now Theorem 1.39(b) shows that/= 0 a.e. If the range of A were all of c0, Theorem 5.10 would imply the existence of a b > 0 such that 11/lloo > bll/llt (2) for every f e L1(T). But if Dn(t) is defined as in Sec. 5.1 1, then Dn e L1(T), liD nil oo = 1 for n = 1, 2, 3, . . . , and IIDnll1----+ oo as n----+ oo . Hence there is no b > 0 such that the inequalities (3) hold for every n. This completes the proof. /Ill The Hahn-Banach Theorem 5.16 Theorem If M is a subspace of a normed linear space X and iff is a bounded linear functional on M, then f can be extended to a bounded linear functional F on X so that II F II = II f 11. Note that M need not be closed. EXAMPLES OF BANACH SPACE TECHNIQUES 105 Before we turn to the proof, some comments seem called for. First, to say (in the most general situation) that a function F is an extension off means that the domain of F includes that off and that F(x) = f(x) for all x in the domain off Second, the norms II F II and II f II are computed relative to the domains of F and f; explicitly, 11/11 = sup {1/(x)l: x E M, llxll < 1}, IIFII = sup {IF(x)l: x E X, llxll < 1}, The third comment concerns the field of scalars. So far everything has been stated for complex scalars, but the complex field could have been replaced by the real field without any changes in statements or proofs. The Hahn-Banach theorem is also true in both cases; nevertheless, it appears to be essentially a " real " theorem. The fact that the complex case was not yet proved when Banach wrote his classical book " Operations lineaires " may be the main reason that real scalars are the only ones considered in his work. It will be helpful to introduce some temporary terminology. Recall that V is a complex (real) vector space if x + y E V for x and y E V, and if ax E V for all complex (real) numbers a.. It follows trivially that every complex vector space is also a real vector space. A complex function <p on a complex vector space V is a complex-linear functional if <p(x + y) = <p(x) + <p(y) and <p(a.x) = a.<p(x) (1) for all x and y E V and all complex a.. A real-valued function <p on a complex (real) vector space V is a real-linear functional if (1) holds for all real a.. If u is the real part of a complex-linear functional f, i.e., if u(x) is the real part of the complex number f(x) for all x E V, it is easily seen that u is a real-linear functional. The following relations hold between/ and u: 5.17 Proposition Let V be a complex vector space. (a) If u is the real part of a complex-linear functional! on V, then f(x) = u(x) -iu(ix) (x E V). (1) (b) If u is a real-linear functional on V and iff is defined by (1), then f is a complex-linear functional on V. (c) If V is a normed linear space and f and u are related as in (1), then 11/11 = llull. PROOF If a. and {3 are real numbers and z = a. + i/3, the real part of iz is - {3. This gives the identity z = Re z -i Re (iz) for all complex numbers z. Since Re (if(x)) = Re f(ix) = u(ix), (1) follows from (2) with z = f(x). (2) (3) 106 REAL AND COMPLEX ANALYSIS Under the hypotheses (b), it is clear that f(x + y) = f(x) + f(y) and that f(rxx) = rxf(x) for all real rx. But we also have f(ix) = u(ix) -iu(-x) = u(ix) + iu(x) = if(x), (4) which proves thatfis complex-linear. Sinࡖe lu(x)l < 1/(x)l, we have llull < 11/11. On the other hand, to every x E V there corresponds a complex number rx, I rxl = 1, so that rxf(x) = I f(x) 1. Then I f(x) I = f(rxx) = u(rxx) < llull · llrxxll = llull · llxll, which proves that 11/11 < llull. (5) Ill/ 5.18 Proof of Theorem 5.16 We first assume that X is a real normed linear space and, consequently, that f is a real-linear bounded functional on M. If II f II = 0, the desired extension is F = 0. Omitting this case, there is no loss of generality in assuming that II f II = 1. Choose x0 E X, x0 ¢ M, and let M 1 be the vector space spanned by M and x0 • Then M 1 consists of all vectors of the form x + Ax0, where x E M and A is a real scalar. If we define /1 (x + Ax0) = f(x) + Arx, where rx is any fixed real number, it is trivial to verify that an extension of f to a linear functional on M 1 is obtained. The problem is to choose rx so that the extended functional still has norm 1. This will be the case provided that I f(x) + Arxl < II x + AXo II (x E M, A real). (1) Replace x by -Ax and divide both sides of (1) by I AI. The requirement is then that lf(x)-rxl < llx-xoll (x E M), (2) i.e., that Ax < rx < Bx for all x E M, where Ax = f(x)- llx-x011 and Bx = f(x) + llx-Xoll· (3) There exists such an rx if and only if all the intervals [Ax, Bx] have a common point, i.e., if and only if (4) for all x and y E M. But f(x) -f(y) =f(x-y) < llx-Yll < llx-Xoll + IIY-Xoll, (5) and so ( 4) follows from (3). We have now proved that there exists a norm-preserving extension/1 off on M1. Let f!/J be the collection of all ordered pairs (M', /'), where M' is a sub space of X which contains M and where f' is a real-linear extension off to M', with 11/'11 = 1. Partially order f!/J by declaring (M',f') < (M",/") to mean that M' c M" and f"(x) = f'(x) for all x e M'. The axioms of a partial order EXAMPLES OF BANACH SPACE TECHNIQUES 107 are clearly satisfied, fJJ is not empty since it contains (M,f), and so the Haus dorff maximality theorem asserts the existence of a maximal totally ordered subcollection Q of &J. Let $ be the collection of all M' such that (M', f') e Q. Then $ is totally ordered, by set inclusion, and therefore the union M of all members of $ is a subspace of X. (Note that in general the union of two subspaces is not a subspace. An example is two planes through the origin in R3.) If x e M, then x e M' for some M' e ; define F(x) = f'(x), where f' is the function which occurs in the pair (M', f') e n. Our definition of the partial order in Q shows that it is immaterial which M' e $ we choose to define F(x), as long as M' contains x. It is now easy to check that F is a linear functional on M, with IIFII = 1. If M were a proper subspace X, the first part of the proof would give us a further extension of F, and this would contradict the maximality of Q. Thus M = X, and the proof is complete for the case of real scalars. If now f is a complex-linear functional on the subspace M of the complex normed linear space X, let u be the real part off, use the real Hahn-Banach theorem to extend u to a real-linear functional U on X, with II Ull = !lull, and define F(x) = U(x) -iU(ix) (x E X). (6) By Propositi0n 5.17, F is a complex-linear extension off, and II F II = II U II = II u II = II f II. This completes the proof. /Ill Let us mention two important consequences of the Hahn-Banach theorem : 5.19 Theorem Let M be a linear subspace of a normed linear space X, and let x0 E X. Then x0 is in the closure M of M if and only if there is no bounded linear functional f on X such that f(x) = 0 for all x E M but f(x0) -# 0. PROOF If x0 e M, f is a bounded linear functional on X, and f(x) = 0 for all x e M, the continuity off shows that we also have f(x0) = 0. Conversely, suppose x0 ¢ M. Then there exists a ç > 0 such that llx - x0 ll > Ī for all x e M. Let M' be the subspace generated by M and x0 , and definef(x + Ax0) = A if x e M and A is a scalar. Since Ī I A I < I A l ii Xo + A-1 x II = II AXo + x II, we see that f is a linear functional on M' whose norm is at most Ī -1. Also f(x) = 0 on M,f(x0) = 1. The Hahn-Banach theorem allows us to extend this /from M' to X. /Ill 5.20 Theorem If X is a normed linear space and if x0 E X, x0 -# 0, there is a bounded linear functional[ on X, ofnorm 1, so thatf(x0) = llxoll-108 REAL AND COMPLEX ANALYSIS PROOF Let M = {A.x0}, and definef(A.x0) = A.llx0ll. Thenfis a linear function al of norm 1 on M, and the Hahn-Banach theorem can again be applied. //// 5.21 Remarks If X is a normed linear space, let X be the collection of all bounded linear functionals on X. If addition and scalar multiplication of linear functionals are defined in the obvious manner, it is easy to see that X is again a normed linear space. In fact, X is a Banach space ; this follows from the fact that the field of scalars is a complete metric space. We leave the verification of these properties of X as an exercise. One of the consequences of Theorem 5.20 is that X is not the trivial ·vector space (i.e., X consists of more than 0) if X is not trivial. In fact, X separates points on X. This means that if x 1 =I= x2 in X there exists an f e X such that f(x1) =I= f(x2). To prove this, merely take x0 = x2 -x1 in Theorem 5.20. Another consequence is that, for x e X, llxll = sup { 1/(x)l:fe X, 11/11 = 1}. Hence, for fixed x E X, the mapping f O f(x) is a bounded linear functional on X, of norm llxll. This interplay between X and X (the so-called " dual space " of X) forms the basis of a large portion of that part of mathematics which is known as functional analysis. An Abstract Approach to the Poisson Integral 5.22 Successful applications of the Hahn-Banach theorem to concrete problems depend of course on a knowledge of the bounded linear functionals on the normed linear space under consideration. Sࡗ far we have only determined the bounded linear functionals on a Hilbert space (where a much simpler proof of the Hahn-Banach theorem exists ; see Exercise 6), and we know the positive linear functionals on Cc(X). We shall now describe a general situation in which the last-mentioned func tionals occur naturally. Let K be a compact Hausdorff space, let H be a compact subset of K, and let A be a subspace of C(K) such that 1 E A (1 denotes the function which assigns the number 1 to each x e K) and such that 11/IIK = 11/IIH (f E A). (1) Here we used the notation 11/IIE = sup {lf(x)l: x E E}. (2) Because of the example discussed in Sec. 5.23, H is sometimes called a bound ary of K, corresponding to the space A. EXAMPLES OF BANACH SPACE TECHNIQUES 109 Iff E A and x E K, (1) says that lf(x)l < 11/IIH· (3) In particular, if f(y) = 0 for every y E H, then f(x) = 0 for all x E K. Hence if f1 and /2 E A and /1 (y) = /2(y) for every y E H, then /1 = /2; to see this, put f = !t -!2. Let M be the set of all functions on H that are restrictions to H of members of A. It is clear that M is a subspace of C(H). The preceding remark shows that each member of M has a unique extension to a member of A. Thus we have a natural one-to-one correspondence between M and A, which is also norm preserving, by (1). Hence it will cause no confusion if we use the same letter to designate a member of A and its restriction to H. Fix a point x E K. The inequality (3) shows that the mapping fr f(x) is a bounded linear functional on M, of norm 1 [since equality holds in (3) iff= 1]. By the Hahn-Banach theorem there is a linear functional A on C(H), of norm 1, such that Af=f(x) (fe M). (4) We claim that the properties A1 = 1 ' IIAII = 1 (5) imply that A is a positive linear functional on C(H). To prove this, suppose fe C(H), 0</< 1, put g=2f- 1, and put Ag = C( + i/3, where C( and {3 are real. Note that -1 < g < 1, so that I g + ir 12 < 1 + r2 for every real constant r. Hence (5) implies that (6) Thus {32 + 2rf3 < 1 for every real r, which forces {3 = 0. Since llgll8 < 1, we have I a I < 1 ; hence Af= ȚA(1 +g)= !{1 + C() > 0. (7) Now Theorem 2.14 can be applied. It shows that there is a regular positive Borel measure Jlx on H such that (f E C(H)). (8) In particular, we get the representation formula (f E A). (9) What we have proved is that to each x E K there corresponds a positive measure Jlx on the ɟɟboundary " H which ɟɟrepresents " x in the sense that (9) holds for every f E A. 1 10 REAL AND COMPLEX ANALYSIS Note that A determines Jlx uniquely; but there is no reason to expect the Hahn-Banach extension to be unique. Hence, in general, we cannot say much about the uniqueness of the representing measures. Under special circumstances we do get uniqueness, as we shall see presently. . . 5.23 To see an example of the preceding situation, let U = { z : I z I < 1} be the open unit disc in the complex plane, put K = 0 (the closed unit disc), and take for H the boundary T of U. We claim that every polynomial/, i.e., every function of the form N f(z) = L an zn, (1) n = O where a0 , ... , aN are complex numbers, satisfies the relation llfllu = llfllr· (2) (Note that the continuity off shows that the supremum of If I over U is the same as that over 0.) Since 0 is compact, there exists a z0 e 0 such that I f(z0) I > I f(z) I for all z E 0. Assume z0 E U. Then N f(z) = L bn(z - z0)n, (3) n = O so that b1 = b2 = · · · = bN = 0; i.e., f is constant. Thus z0 e T for every noncon stant polynomial f, and this proves (2). (We have just proved a special case of the maximum modulus theorem; we shall see later that this is an important property of all holomorphic functions.) 5.24 The Poisson Integral Let A be any subspace of C(U) (where 0 is the closed unit disc, as above) such that A contains all polynomials and such that llfllu = llfllr (1) holds for every f E A. We do not exclude the possibility that A consists of precisely the polynomials, but A might be larger. The general result obtained in Sec. 5.22 applies to A and shows that to each z e U there corresponds a positive Borel measure Jlz on T such that f(z) = J/ dJlz (f E A). (2) (This also holds .for z e T, but is then trivial : Jl.z is simply the unit mass concen trated at the point z.) EXAMPLES OF BANACH SPACE TECHNIQUES 1 1 1 We now fix z e U and write z = rei8, 0 < r < 1, (J real. If un(w) = w", then un E A for n = 0, 1, 2, . . . ; hence (2) shows that ( n = 0, 1, 2, ... ). (3) Since u_n = un on T, (3) leads to ( n = 0, + 1, + 2, . . . ). (4) This suggests that we look at the real function 00 Pr((J-t) = L r1nlein(8-t) (t real), (5) n=- oo . stnce (n = 0, + 1, + 2, . .. ). (6) Note that the series (5) is dominated by the convergent geometric series L rlnl, so that it is legitimate to insert the series into the integral (6) and to integrate term by term, which gives (6). Comparison of (4) and (6) gives r f dp.z = 21 I" f(eiˀP,(() -t) dt JT n - 1t (7) for f = un, hence for every trigonometric polynomial f, and Theorem 4.25 now implies that (7) holds for every f e C(T). [This shows that Jlz was uniquely deter mined by (2). Why?] In particular, (7) holds iff e A, and then (2) gives the representation 1 J1t f(z) = 2n _J(e;ˀP,(() - t) dt (f E A). The series ( 5) can be summed explicitly, since it is the real part of Thus 2 x ( _it)× _ eit + z _ 1 -r2 + 2ir sin ( (J - t) 1 + Ɵ ze - it -1 1 -it 12 • 1 e - z -ze p r( (J -t) = 1 -2r cos ( (J -t) + r2 • This is the so-called " Poisson kernel." Note that Pr((J - t) > 0 if 0 < r < 1. We now summarize what we have proved: (8) (9) 1 12 REAL AND COMPLEX ANALYSIS 5.25 Theorem Suppose A is a vector space of continuous complex functions on the closed unit disc 0. If A contains all polynomials, and if sup I f(z)l = sup I f(z) I z e U z e T (1) for every f e A (where Tis the unit circle, the boundary of U), then the Poisson integral representation 1 f1t 1 -r2 . f(z) = -2 1 2 (fJ ) 2 f(elˀ dt n _ 1t - r cos -t + r is valid for every f e A and every z e U. Exercises (2) 1 Let X consist of two points a and b, put f.J.( {a}) = f.J.( { b}) = t, and let IJ'(f.J.) be the resulting real IJ'-space. Identify each real function/ on X with the point (f(a),f(b)) in the plane, and sketch the unit balls of IJ'(f.J.), for 0 < p < oo. Note that they are convex if and only if 1 < p < oo. For which p is this unit ball a square? A circle? If f.J.({a}) =F f.J.(b), how does the situation differ from the preceding one? 2 Prove that the unit ball (open or closed) is convex in every normed linear space. 3 If 1 < p < oo , prove that the unit ball of IJ'(f.J.) is strictly convex; this means that if f=F g, h = t<f + g), then II h II P < 1. (Geometrically, the surface of the ball contains no straight lines.) Show that this fails in every I.J(f.J.), in every L00(f.J.), and in every C(X). (Ignore trivialities, such as spaces consisting of only one point.) 4 Let C be the space of all continuous functions on [0, 1], with the supremum norm. Let M consist of allf E C for which r112 f dt - rl f dt = 1. Jo J112 Prove that M is a closed convex subset of C which contains no element of minimal norm. S Let M be the set of allf E L1([0, 1]), relative to Lebesgue measure, such that f f(t) dt = 1. Show that M is a closed convex subset of L1([0, 1]) which contains infinitely many elements of minimal norm. (Compare this and Exercise 4 with Theorem 4.10.) 6 Let f be a bounded linear functional on a subspace M of a Hilbert space H. Prove that f has a unique norm-preserving extension to a bounded linear functional on H, and that this extension van ishes on M1.. 7 Construct a bounded linear functional on some subspace of some L1(f.J.) which has two (hence infinitely many) distinct norm-preserving linear extensions to L1(f.J.). 8 Let X be a normed linear space, and let X be its dual space, as defined in Sec. 5.21, with the norm 11/ 11 = sup { l f(x) l : llxll < 1}. (a) Prove that X is a Banach space. (b) Prove that the mapping /---+ f(x) is, for each x E X, a bounded linear functional on X, of norm llxll. (This gives a natural imbedding of X in its " second dual " X, the dual space of X.) EXAMPLES OF BANACH SPACE TECHNIQUES 1 13 (c) Prove that {llxnll} is bounded if {xn} is a sequence in X such that {f(xn)} is bounded for every f E X. 9 Let Co ' t1' and t00 be the Banach spaces consisting of all complex sequences X = { eJ, i = 1, 2, 3, . . . , defined as follows: X E t1 if and only if llxll t = L I ei I < 00. X E too if and only if llxll 00 = sup I ei I < 00. Co is the subspace of t00 consisting of all X E t00 for which ei _. 0 as i . 00. Prove the following four statements. (a) If y = { '7;} E t1 and Ax = L ei 17; for every X E Co ' then A is a bounded linear functional on c0 , and IIAII = IIYII 1. Moreover, every A E (c0) is obtained in this way. In brief, (c0) = t1• (More precisely, these two spaces are not equal; the preceding statement exhibits an isometric vector space isomorphism between them.) (b) In the same sense, (t1) = t00• (c) Every y E t1 induces a bounded linear functional on t00, as in (a). However, this does not give all of (t00), since (t00) contains nontrivial functionals that vanish on all of c0 • (d) c0 and t1 are separable but too is not. 10 If L (Xi e i con verges for every sequence ( e J such that e i _. 0 as i . 00' prove that L I rx; I < 00. 11 For 0 < rx < 1, let Lip rx denote the space of all complex functions/on [a, b] for which I f(s) -f(t) I M 1 = sup « < oo. s :l= t l s - t l Prove that Lip rx is a Banach space, if II f II = I f(a) I + M 1; also, if 11! 11 = M1 + sup l f(x) l . X (The members of Lip rx are said to satisfy a Lipschitz condition of order rx.) 12 Let K be a triangle (two-dimensional figure) in the plane, let H be the set consisting of the vertices of K, and let A be the set of all real functions/ on K, of the form f(x, y) = rxx + py + y ( rx, p, and y real). Show that to each (x0 , y0) E K there corresponds a unique measure J1. on H such that f(xo , Yo) = lf diJ.. (Compare Sec. 5.22.) Replace K by a square, let H again be the set of its vertices, and let A be as above. Show that to each point of K there still corresponds a measure on H, with the above property, but that uniqueness is now lost. Can you extrapolate to a more general theorem? (Think of other figures, higher dimensional spaces.) 13 Let {In} be a sequence of continuous complex functions on a (nonempty) complete metric space X, such thatf(x) = lim !,.(x) exists (as a complex number) for every x E X. (a) Prove that there is an open set V =1- 0 and a number M < oo such that I fix) I < M for all x E V and for n = 1, 2, 3, . . . . (b) If E > 0, prove that there is an open set V =1- 0 and an integ͝r N such that I f(x) -fn(x) I < E if x E V and n > N. Hint for (b): For N = 1, 2, 3, . . . , put AN = {x: l fm(x) -fn(x)l < E if m > N and n > N}. Since X = U AN , some AN has a nonempty interior. 114 REAL AND COMPLEX ANALYSIS 14 Let C be the space of all real continuous functions on I = [0, 1] with the supremum norm. Let X,. be the subset of C consisting of those f for which there exists a t € I such that I f(s) -f(t) I s; n I s - t I for all s E /. Fix n and prove that each open set in C contains an open set which does not intersect X,. . (Each f E C can be uniformly approximated by a zigzag function g with very large slopes, and if II g - h II is small, h ¢ X,. .) Show that this implies the existence of a dense G tJ in C which consists entirely of nowhere differentiable functions. IS Let A = (a;} be an infinite matrix with complex entries, where i, j = 0, 1, 2, . . . . A associates with each sequence {sJ a sequence {0';}, defined by CX> u; = L a;i si j = O (i = 1, 2, 3, . . . ), provided that these series converge. Prove that A transforms every convergent sequence { sJ to a sequence { u;} which converges to the same limit if and only if the following conditions are satisfied : and (a) (b) (c) lim aii = 0 i .... CX> CX> sup; L laii I < oo. j= O CX> lim L aii = 1. i-+ oo j= O for each j. The process of passing from { si} to { u;} is called a summability method. Two examples are : 1 i + 1 if 0 < j < i, 0 if i < j, 0 < r; < 1, Prove that each of these also transforms some divergent sequences {sJ (even some unbounded ones) to convergent sequences { 0';}. 16 Suppose X and Y are Banach spaces, and suppose A is a linear mapping of X into Y, with the following property : For every sequence {x,.} in X for which x = lim x,. and y = lim Ax,. exist, it is true that y = Ax. Prove that A is continuous. This is the so-called " closed graph theorem." Hint: Let X 9 Y be the set of all ordered pairs (x, y), x E X and y E Y, with addition and scalar multiplication defined componentwise. Prove that X 9 Y is a Banach space, if ll(x, y) ll = llx ll + IIYII . The graph G of A is the subset of X 9 Y formed by the pairs (x, Ax), x E X. Note that our hypothesis says that G is closed ; hence G is a Banach space. Note that (x, Ax) __. x is continuous, one-to-one, and linear and maps G onto X. Observe that there exist nonlinear mappings (of R1 onto R1, for instance) whose graph is closed although they are not continuous :f(x) = 1/x if x =F O,f(O) = 0. 17 If J1 is a positive measure, each f E L 00(J1) defines a multiplication operator M 1 on 13(11 ) into J!(p), such that Mj(g) = f g. Prove that II M111 < ll f ll oo · For which measures J1 is it true that II M111 = ll f ll oo for allf E L 00(p)? For which/ E L 00(J1) does M 1 map J!(p) onto J!(p)? 18 Suppose {A,.} is a sequence of bounded linear transformations from a normed linear space X to a Banach space Y, suppose IIA,.Il < M < oo for all n, and suppose there is a dense set E c X such that {A,. x} converges for each x E E. Prove that {A,. x} converges for each x E X. EXAMPLES OF BANACH SPACE TECHNIQUES 1 15 19 If sn is the nth partial sum of the Fourier series of a function f E C(T), prove that sn/log n---+ 0 uniformly, as n---+ oo , for each/ E C(T). That is, prove that lim II s" II oo = o. n-+oo log n On the other hand, if A.Jlog n---+ 0, prove that there exists an f E C(T) such that the sequence {sif; 0)/A.n} is unbounded. Hint: Apply the reasoning of Exercise 18 and that of Sec. 5.1 1, with a better estimate of 11Dnll 1 than was used there. 20 (a) Does there exist a sequence of continuous positive functions f.. on R 1 such that {fix)} ts unbounded if and only if x is rational ? (b) Replace " rational " by " irrational " in (a) and answer the resulting question. (c) Replace " {fix)} is unbounded " by "!,(x)---+ oo as n---+ oo " and answer the resulting ana logues of (a) and (b). 21 Suppose E c R1 is measurable, and 'm(E) = 0. Must there be a translate E + x of E that does not intersect E? Must there be a homeomorphism h of R1 onto R1 so that h(E) does not intersect E? 22 Suppose f E C( T) and f E Lip tX for some tX > 0. (See Exercise 1 1.) Prove that the Fourier series off converges to f(x), by completing the following outline : It is enough to consider the case x = 0, f(O) = 0. The difference between the partial sums sn(f; 0) and the integrals 1 fn sin nt -f(t) dt n -n t tends to 0 as n---+ oo. The functionf(t)/t is in L 1(T). Apply the Riemann-Lebesgue lemma. More careful reasoning shows that the convergence is actually uniform on T. CHAPTER SIX COMPLEX MEASURES Total Variation 6.1 Introduction Let 9Jl be a a-algebra in a set X. Call a countable collection { EJ of members of 9Jl a partition of E if Ei n Ei = 0 whenever i # j, and if E = U Ei. A complex measure Jl on 9Jl is then a complex function on 9Jl such that (E E IDl) (1) i= 1 for every partition { EJ of E. Observe that the convergence of the series in (1) is now part of the require ment (unlike for positive measures, where the series could either converge or diverge to oo ). Since the union of the sets Ei is not changed if the subscripts are permuted, every rearrangement of the series (1) must also converge. Hence (, Theorem 3.56) the series actually converges absolutely. Let us consider the problem of finding a positive measure A which dominates a given complex measure Jl on IDl, in the sense that I p,(E) I < A( E) for every E e IDl, and let us try to keep A as small as we can. Every solution to our problem (if there is one at all) must satisfy 00 00 A(E) = L A(Ei) > L I p,(Ei) I' i= 1 1 (2) for every partition { Ei} of any set E e IDl, so that A( E) is at least equal to the supremum of the sums on the right of (2),. taken over all partitions of E. This suggests that we define a set function I Jll on 9Jl by 00 I .u I (E) = sup L I p,( Ei) I i= 1 (E E 9J1), the supremum being taken over all partitions { Ei} of E. 1 16 (3) COMPLEX MEASURES 1 17 This notation is perhaps not the best, but it is the customary one. Note that I J11 (E) > I J1(E) I, but that in general I J1 1 (E) is not equal to I J1(E) 1 . It turns out, as will be proved below, that I J11 actually is a measure, so that our problem does have a solution. The discussion which led to (3) shows then clearly that I J11 is the minimal solution, in the sense that any other solution A. has the property A.( E) > I J11 (E) for all E e IDl. The set function I J11 is called the total variation of Jl, or sometimes, to a void misunderstanding, the total variation measure. The term " total variation of J1" is also frequently used to denote the number I J11 (X). If J1 is a positive measure, then of course I J11 = Jl. Besides being a measure, I J11 has another unexpected property: I J11 (X) < oo. Since I J1(E) I < I J11 (E) < I J11 (X), this implies that every complex measure J1 on any a-algebra is bounded : If the range of J1 lies in the complex plane, then it actually lies in some disc of finite radius. This property (proved in Theorem 6.4) is sometimes expressed by saying that J1 is of bounded variation. 6.2 Theorem The total variation I J11 of a complex measure J1 on ID1 is a positive measure on 9Jl. PROOF Let { EJ be a partition of E e IDl. Let t; be real numbers such that t; < I J11 (E;). Then each E; has a partition { Aii} such that L I Jl(A;j) I > t; j (i = 1' 2, 3, . . . ). Since { Aii} (i, j = 1, 2, 3, . . . ) is a partition of E, it follows that L t; < L I Jl(Aij) I < I J1l (E). i i, j (1) (2) Taking the supremum of the left side of (2), over all admissible choices of { t;}, we see that L I J11 ( E ;) < I J11 (E). (3) i To prove the opposite inequality, let {Ai} be any partition of E. Then for any fixed j, {A i n EJ is a partition of A i, and for any fixed i, {A i n EJ is a partition of E;. Hence L I J-l(A)I = L L Jl(Ai n E;) j j i j i i j (4) i 1 18 REAL AND COMPLEX ANALYSIS Since (4) holds for every partition { Ai} of E, we have I Ill (E) < L I Ill ( E i). i By (3) and (5), I Jll is countably additive. Note that the Corollary to Theorem 1.27 was used in (2) and (4). (5) That I Jll is not identically oo is a trivial consequence of Theorem 6.4 but can also be seen right now, since I Jll (0) = 0. I I I I 6.3 Lemma If z 1 , . .. , z N are complex numbers then there is a subset S of { 1, ... , N} for which PROOF Write zk = I zk I eiak. For - n < fJ < n, let S(fJ) be the set of all k for which cos ( rxk -lJ) > 0. Then N L zk = L e-iozk > Re L e-iozk = L I zk I cos+ (rxk-fJ). S(8) S(8) S(8) k = 1 Choose 00 so as to maximize the last sum, and put S = S(lJ0). This maximum is at least as large as the average of the sum over [ - n, n], and this average is n-1 L I zk I , because for every rx. -cos+ (rx-lJ) dlJ = -1 fn 1 2n - n 1t 6.4 Theorem If Jl is a complex measure on X, then I ,u I (X) < oo. Ill/ PROOF Suppose first that some set E E IDl has I Jll (E) = oo. Put t = n(1 + I Jl(E) I). Since I Jll (E) > ·t, there is a partition { Ei} of E such that i= 1 for some N. Apply Lemma 6.3, with zi = Jl(Ei), to conclude that there is a set A c E (a union of some of the sets Ei) for which I Jl(A) I > t/n > 1. Setting B = E -A, it follows that t I Jl( B) I = I Jl( E) -Jl( A) I > I Jl( A) I - I Jl( E) I > -- I Jl( E) I = 1. 1t COMPLEX MEASURES 1 19 We have thus split E into disjoint sets A and B with I Jl(A) I > 1 and I Jl(B) I > 1. Evidently, at least one of I Jll (A) and I Jll (B) is oo, by Theorem 6.2. Now if I Jl l (X) = oo, split X into A1, Bh as above, with I Jl(A1) I > 1, I Jl l (B1) = oo . Split B1 into A2, B2, with I Jl(A2) I > 1, I Jll (B2) = oo. Contin uing in this way, we get a countably infinite disjoint collection {Ai}, with I Jl(Ai) I > 1 for each i. The countable additivity of Jl implies that But this series cannot converge, since Jl(Ai) does not tend to 0 as i-+ oo. This contradiction shows that I Jl l (X) < oo. /Ill 6.5 If Jl and A are complex measures on the same a-algebra IDl, we define Jl + A and CJl by (Jl + A)(E) = Jl(E) + A(E) (cJl)(E) = CJl(E) (E E 9Jl) (1) for any scalar c, in the usual manner. It is then trivial to verify that Jl + A and CJl are complex measures. The collection of all complex measures on IDl is thus a vector space. If we put IIJlll = I Il l (X), (2) it is easy to verify that all axioms of a normed linear space are satisfied. 6.6 Positive and Negative Variations Let us now specialize and consider a real measure Jl on a a-algebra IDl. (Such measures are frequently called signed mea sures.) Define I Jl l as before, and define Jl + = !< I Il l + Jl ), (1) Then both Jl + and Jl- are positive measures on IDl, and they are bounded, by Theorem 6.4. Also, (2) The measures Jl + and Jl- are called the positive and negative variations of Jl, respectively. This representation of Jl as the difference of the positive measures Jl + and Jl- is known as the Jordan decomposition of Jl. Among all representations of Jl as a difference of two positive measures, the Jordan decomposition has a certain minimum property which will be established as a corollary to Theorem 6.14. 120 REAL AND COMPLEX ANALYSIS Absolute Continuity 6.7 Definitions Let J1 be a positive measure on a a-algebra IDl, and let A. be an arbitrary measure on IDl; A. may be positive or complex. (Recall that a complex measure has its range in the complex plane, but that our usage of the term " positive measure " includes oo as an admissible value. Thus the positive measures do not form a subclass of the complex ones.) We say that A. is absolutely continuous with respect to Jl, and write (1) if A.(E) = 0 for every E e 9Jl for which J1(E) = 0. If there is a set A e 9Jl such that A.(E) = A.(A n E) for every E e IDl, we say that A. is concentrated on A. This is equivalent to the hypothesis that A.( E) = 0 whenever E n A = 0. Suppose At and A.2 are measures on IDl, and suppose there exists a pair of disjoint sets A and B such that At is concentrated on A and A.2 is concen trated on B. Then we say that At and A.2 are mutually singular, and write (2) Here are some elementary properties of these concepts. 6.8 Proposition Suppose, Jl, A., A.b and A.2 are measures on a a-algebra IDl, and J1 is positive. (a) If A. is concentrated on A, so is I A. I. (b) If A.1 l. A.2 , then I A.t I l. I A.2 l · (c) If At l. J1 and A.2 l. J1, then At + A.2 l. Jl. (d) If At J1 and A.2 J1, then At + A.2 Jl. (e) If A. J1, then I A. I Jl. (f) If At J1 and A.2 l. J1, then At l. A.2 . (g) If A. Jl and A. l. Jl, then A. = 0. PROOF (a) If E n A = 0 and { Ei} is any partition of E, then A.(Ei) = 0 for all j. Hence I A. I (E) = 0. (b) This follows immediately from (a). (c) There are disjoint sets A1 and Bt such that At is concentrated on At and J1 on B1, and there are disjoint sets A2 and B2 such that A.2 is concen trated on A2 and J1 on B2• Hence At + A.2 is concentrated on A = A1 u A2, J1 is concentrated on B = Bt n B2, and A n B = 0. (d) This is obvious. (e) Suppose Jl(E) = 0, and { Ei} is a partition of E. Then Jl(Ei) = 0; and since A. Jl, A.(Ei) = 0 for allj, hence L I A.(Ei) I = 0. This implies I A. I (E) = 0. COMPLEX MEASURES 121 (f) Since A.2 .l J1, there is a set A with J1(A) = 0 on which A.2 is concentrated. Since At J1, A.t(E) = 0 for every E c A. So At is concentrated on the complement of A. (g) By (f), the hypothesis of (g) implies, that A. .l A., and this clearly forces A. = 0. Ill/ We come now to the principal theorem about absolute continuity. In fact, it is probably the most important theorem in measure theory. Its statement will involve a-finite measures. The following lemma describes one of their significant properties. 6.9 Lemma If J1 is a positive a-finite measure on a a-algebra 9Jl in a set X, then there is afunction wE Lt(J1) such that 0 < w(x) < 1for every x EX. PROOF To say that J1 is a-finite means that X is the union of countably many sets En E 9Jl (n = 1, 2, 3, . . . ) for which J1(En) is finite. Put wn(x) = 0 if x E X-En and put if X E En. Then w = Lr wn has the required properties. Ill/ The point of the lemma is that J1 can be replaced by a finite measure jJ, (namely, djl = w dJ1) which, because of the strict positivity of w, has precisely the same sets of measure 0 as J1. 6.10 The Theorem of Lebesgue-Radon-Nikodym Let J1 be a positive a-finite measure on a a-algebra 9Jl in a set X, and let A. be a complex measure on 9Jl. (a) There is then a unique pair of complex measures A.a and A.s on 9Jl such that If A. is positive and finite, then so are A. a and As. (b) There is a unique h E Lt(J1) such that A.a(E) = 1 h d,u for every set E E IDl. (1) (2) The pair (A.a, A.s) is called the Lebesgue decomposition of A. relative to Jl. The uniqueness of the decomposition is easily seen, for if (A., A.) is another pair which satisfies ( 1 ), then (3) A. -A.a Jl, and A.s -A. .l J1, hence both sides of (3) are 0; we have used 6.8(c), 6.8(d), and 6.8(g). 122 REAL AND COMPLEX ANALYSIS The existence of the decomposition is the significant part of (a). Assertion (b) .is known as the Radon-Nikodym theorem. Again, uniqueness of h is immediate, from Theorem 1.39(b). Also, if h is any member of L1(J1.), the integral in (2) defines a measure on 9Jl (Theorem 1.29) which is clearly absolutely continuous with respect to J.l. The point of the Radon-Nikodym theorem is the converse: Every A. J1. (in which case A.a = A.) is obtained in this way. The function h which occurs in (2) is called the Radon-Nikodym derivative of A.a with respect to J.l. As noted after Theorem 1.29, we may express (2) in the form dA.a = h dJ1., or even in the form h = dA.a/dJ1.. The idea of the following proof, which yields both (a) and (b) at one stroke, is due to von Neumann. PROOF Assume first that A. is a positive bounded measure on IDl. Associate w to J1. as in Lemma 6.9. Then d<p = dA. + w dJ1. defines a positive bounded measure qJ on IDl. The definition of the sum of two measures shows that (4) for f = XE , hence for simple f, hence for any nonnegative measurable f If f E /3( <p), the Schwarz inequality gives Since <p(X) < oo , we see that (5) is a bounded linear functional on L2(<p). We know that every bounded linear functional on a Hilbert space H is given by an inner product with an element of H. Hence there exists a g E I3(<p) such that (6) for every f e /3( qJ ). Observe how the completeness of L2(<p) was used to guarantee thࡘ exis tence of g. Observe also that although g is defined uniquely as an element of L2( <p ), g is determined only a.e. [ <p] as a point function on X. Put f = XE in (6), for any E E 9Jl with 0. The left side of (6) is then A.( E), and since 0 < A. < <p, we have 1 f A.(E) 0 < ({J(E) J/ dqJ = qJ(E) < 1. (7) COMPLEX MEASURES 123 Hence g(x) e [0, 1] for almost all x (with respect to <p), by Theorem 1.40. We may therefore assume that 0 < g(x) < 1 for every x e Xࡕ without affecting (6), and we rewrite (6) in the form l (1 -g)f dA = lfgw dJl. (8) Put A = {x: O<g(x)< l}, B = {X: g( X) = 1}, (9) and define measures A.a and A.s by A.s(E) = A.(B n E), (10) for all E e 9Jl Iff= XB in (8), the left side is 0, the right side is JB w dJ1. Since w(x) > 0 for all x, we conclude that J1(B) = 0. Thus A.s l. Jl. Since g is bounded, (8) holds iff is replaced by (1 + g + ... + g")XE for n = 1, 2, 3, ... , E e 9Jl. For such f, (8) becomes 1(1 -g"+1) d).= Lg(l + g + · · · + g ")w dJl. (1 1) At every point of B, g(x) = 1, hence 1 -g"+ 1(x) = 0. At every point of A, g"+1(x)rO monotonically. The left side of (1 1) converges therefore to A.( A n E) = A.a(E) as n r oo . The integrands on the right side of (1 1) increase monotonically to a non negative measurable limit h, and the monotone convergence theorem shows that the right side of (1 1) tends to JE h dJ1 as nr oo . We have thus proved that (2) holds for every E e 9Jl. Taking E = X, we see that h E L1(J1), since A.a(X) < oo . Finally, (2) shows that A.a Jl,. and the proof is complete for positive A.. If A. is a complex measure on IDl, then A. = .A.1 + iA.2, with A.1 and A.2 real, and we can apply the preceding case to the positive and negative variations of A.1 and A.2• /Ill If both J1 and A. are positive and a-finite, most of Theorem 6.10 is still true. We can now write X= U Xn, where J1(Xn) < oo and A.(Xn) < oo, for n = 1, 2, 3, . . . . The Lebesgue decompositions of the measures A.(E n Xn) still give us a Lebesgue decomposition of A., and we still get a function h which satisfies Eq. 6.10(2); however, it is no longer true that h e L1(J1), although h is " locally in L1 ," i.e., J Xn h dJ1 < oo for each n. Finally, if we go beyond a-finiteness, we meet situation࡙ where the two theo rems under consideration actually fail. For example, let J1 be Lebesgue measure on (0, 1), and let A. be the counting measure on the a-algebra of all Lebesgue 124 REAL AND COMPLEX ANALYSIS measurable sets in (0, 1). Then A has no Lebesgue decomposition relative to J.l, and although J.l A and J.l is bounded, there is no h e L1(A) such that dJ.l = h dA. We omit the easy proof. The following theorem may explain why the word " continuity " is used in connection with the relation A J.l. 6.11 Theorem Suppose J.l and A are measures on a a-algebra IDl, J.l is positive, and A is complex. Then the following two conditions are equivalent: (a) A JL (b) To every E > 0 corresponds a b > 0 such that IA(E) I < E for all E e 9Jl with J.l(E) < b. Property (b) is sometimes used as the definition of absolute continuity. However, (a) does not imply (b) if A is a positive unbounded measure. For instance, let J.l be Lebesgue measure on (0, 1 ), and put for every Lebesgue measurable set E c (0, 1 ). PROOF Suppose (b) holds. If J.l(E) = 0, then J.l(E) < b for every b > 0, hence lA( E) I < E for every E > 0, so A( E) = 0. Thus (b) implies (a). Suppose (b) is false. Then there exists an E > 0 and there exist sets En e 9Jl (n = 1, 2, 3, ... ) such that J.l(En) < 2 -n but lA( En) I > E. Hence IAI (En) > E. Put 00 A = n An. (1) n=l Then J.l(An) < 2 -n+ 1, An ::) An+ b and so Theorem 1.19(e) shows that J.l(A) = 0 and that since IAI (An) > IAI (En). It follows that we do not have IAI J.l, hence (a) is false, by Proposition 6.8(e). //// Consequences of the Radon-Nikodym Theorem 6.12 Theorem Let J.l be a complex measure on a a-algebra 9Jl in X. Then there is a measurable function h such that I h(x) I = 1 for all x e X and such that dJ.l = h d I J.ll. (1) COMPLEX MEASURES 125 By analogy with the representation of a complex number as the product of ts absolute value and a number of absolute value 1, Eq. (1) is sometimes referred o as the polar representation (or polar decomposition) of JL PROOF It is trivial that J1 I J11, and therefore the Radon-Nikodym theorem guarantees the existenee of some h E L1( I J11) which satisfies (1). Let Ar = {x: I h(x) I < r}, where r is some positive number, and let {Ei} be a partition of Ar. Then l IJL(Ei)l = l Lih d iJLI < l riJLI(E) = riJLI(A,), so that I J11 (Ar) < r I J11 (Ar). If r < 1, this forces I J11 (Ar) = 0. Thus I hI > 1 a.e. On the other hand, if I J11 (E) > 0, (1) shows that 1 ih dl I - IJ1(E)I < I J11 (E) E J1 - I J11 (E) - 1. We now apply Theorem 1.40 (with the closed unit disc in place of S) and conclude that I h I < 1 a.e. Let B = {x E X : I h(x) I #- 1}. We have shown that I J11 (B)= 0, and if we redefine h on B so that h(x) = 1 on B, we obtain a function with the desired properties. I I I I 6.13 Theorem Suppose J1 is a positive measure on IDl, g E L1(J1), and (E E 9Jl). (1) Then (E E 9Jl). (2) PROOF By Theorem 6.12, there is a function h, of absolute value 1, such that dA = h d I A 1. By hypothesis, dA = g dJ1. Hence h d I A I = g dJ1. This gives d I A I = hg dJ1. (Compare with Theorem 1.29.) Since I A I > 0 and J1 > 0, it follows that hg > 0 a.e. [J1], so that fig = I g I a.e. [J1]. //// 6.14 The Hahn Decomposition Theorem Let J1 be a real measure on a u algebra 9Jl in a set X. Then there exist sets A and B E 9Jl such that 126 REAL AND COMPLEX ANALYSIS A u B = X, A n B = 0, and such that the positive and negative variations J1 + and Jl- of J1 satisfy Jl+(E) = Jl(A n E), J1-(E) = -Jl(B n E) (E E IDl). (1) In other words, X is the union of two disjoint measurable sets A and B, such that "A carries all the positive mass of J1" [since (1) implies that Jl(E) > 0 if E c A] and " B carries all the negative mass of J1" [since J1(E) < 0 if E c B]. The pair (A, B) is called a Hahn decomposition of X, induced by Jl. PROOF By Theorem 6.12, dJ1 = h d I J11, where I hI = 1. Since J1 is real, it fol lows that h is real (a.e., and therefore everywhere, by redefining on a set of measure 0), hence h = + 1. Put A = {X: h( X) = 1}' Since J1 + = !( I J11 + J1 ), and since B= {x: h(x)= -1}. !{1 +h)={̕ we have, for any E e IDl, on A, on B, 1 i I' J1 +(E) = 2 (1 + h) d I J11 = J h d I J11 = J1(E (\ A). E E '"' A (2) (3) (4) Since J1(E) = J.t(E n A) + Jl(E n B) and since J1 = Jl+ - Jl-, the second half of (1) follows from the first. II II Corollary If J1 = A.1 -A.2, where A.1 and A.2 are positive measures, then A.1 > J1 + and A.2 > Jl-. This is the minimum property of the Jordan decomposition which was men tioned in Sec. 6.6. PROOF Since J1 < A.1, we have /Ill Bounded Linear Functionals on lJ' 6.15 Let J1 be a positive measure, suppose 1 < p < oo , and let q be the exponent conjugate to p. The Holder inequality (Theorem 3.8) shows that if g e JJ(J1) and if 9 is defined by (1) COMPLEX MEASURES 127 then <1>9 is a bounded linear functional on I.!'(Jl.), of norm at most llgllq. The ques tion naturally arises whether all bounded linear functionals on I!'(Jl.) have this form, and whether the representation is unique. For p = oo, Exercise 13 shows that the answer is negative: L 1(m) does not furnish all bounded linear functionals on L00(m). For 1 < p < oo, the answer is affirmative. It is also affirmative for p = 1, provided certain measure-theoretic pathologies are excluded. For a-finite measure spaces, no difficulties arise, and we shall confine ourselves to this case. 6.16 Theorem Suppose 1 < p < oo, J1. is a a-finite positive measure on X, and џ is a bounded linear functional on I.!'(J.l). Then there is a unique g e JJ(Jl.), where q is the exponent conjugate to p, such that (f E I.!'(Jl. }). (1) Moreover, if<!> and g are related as in (1), we have (2) In other words, JJ(Jl.) is isometrically isomorphic to the dual space of I!'(Jl), under the stated conditions. PROOF The uniqueness of g is clear, for if g and g' satisfy (1), then the integral of g -g' over any measurable set E of finite measure is 0 (as we see by taking XE for f), and the a-finiteness of J1. implies therefore that g -g' = 0 a.e. Next, if (1) holds, Holder's inequality implies (3) So it remains to prove that g exists and that equality holds in (3). If II<!> II = 0, (1) and (2) hold with g = 0. So assume 11<1>11 > 0. We first consider the case Jl.(X) < oo. For any measurable set E c X, define Since <!> is linear, and since XA u 8 = XA + Xs if A and B are disjoint, we see that A is additive. To prove countable additivity, suppose E is the union of countably many disjoint measurable sets Ei , put Ak = E 1 u · · · u Ek, and note that (k O oo); (4) the continuity of <!> shows now that A(Ak) Ơ A( E). So A is a complex measure. [In (4) the assumption p < oo was used.] It is clear that A(E) = 0 if J.t(E) = Ŀ' 128 REAL AND COMPLEX ANALYSIS since then IIXEIIP = 0. Thus A. J1, and the Radon-Nikodym theorem ensures the existence of a function g e L1(J1) such that, for every measurable E c X, q,(XE) = l g djJ. = l XE g djJ.o (5) By linearity it follows that (6) holds for every simple measurable f, and so also for every f E L00(J1), since every f e L00(J1) is a uniform limit of simple functions_h . Note that the uniform con vergence of .h to f implies II h -f II P---+ 0, hence ( h)---+ ( f), as i---+ oo . We want to conclude that g E /J(J1) and that (2) holds; it is best to split the argument into two cases. CASE 1 p = 1. Here (5) shows that lo dJJ. < llq,ll 0 llxEIIl = llq,ll 0 JJ.(E) for every E E 9Jl. By Theorem 1.40, I g(x) I < 11<1>11 a.e., so that II gil oo < 11<1>11. CASE 2 1 < p < oo . There is a measurable function ex, I ex I = 1, such that cxg = I g I [Proposition 1.9(e)]. Let En = {x: I g(x) I < n} , and define f = XEn I g lq-1cx. Then If IP = I g lq on En ,f e L00(J1), and (6) gives l.1o1q djJ. = fxto djJ. = q,(f) < llq,ll{l.lolqf'p, so that l XE.I g lq djJ. < llq,llq ( n = 1 , 2, 3, . . . ). (7) If we apply the monotone convergence theorem to (7), we obtain llgllq < 11<1>11. Thus (2) holds and g e /J(J1). It follows that both sides of (6) are contin uous functions on I!(J1). They coincide on the dense subset L00(J1) of I!(J1); hence they coincide on all of I!(J1), and this completes the proof if J1(X) < oo . If J1(X) = oo but J1 is a-finite, choose w E L1(J1) as in Lemma 6.9. Then djl = w dJ1 defines a finite measure on IDl, and (8) is a linear isometry of I!(jl) onto I!(J1), because w(x) > 0 for every x E X. Hence 'P(F) = (w11PF) (9) defines a bounded linear functional 'P on I!(jl), with II 'I'll = 11<1>11. COMPLEX MEASURES 129 The first part of the proof shows now that there exists G e JJ(jJ,) such that '¥(F) = LFG djl Put g = w11qG. (If p = 1, g = G.) Then (F E I!(jl)). i1 g lq dp. = i1 G lq djl = ll'¥11q = llэllq (10) (11) if p > 1, whereas llgll oo = II GII oo = ll\1' 11 = 11<1>11 if p = 1. Thus (2) holds, and since G djl = w11Pg dJl, we finally get <ll (f) = '¥(w- 11P f) = i w- 11"fG djl = fxto dp. (12) for every f e l!(Jl). Ill/ 6.17 Remark We have already encountered the special case p = q = 2 of Theorem 6.16. In fact, the proof of the general case was based on this special case, for we used the knowledge of the bounded linear functionals on 1.3(/1) in the proof of the Radon-Nikodym theorem, and the latter was the key to the proof of Theorem 6.16. The special case p = 2, in turn, depended on the completeness of I.3(Jl), on the fact that 1.3(/1) is therefore a Hilbert space, and on the fact that the bounded linear functionals on a Hilbert space are given by inner products. We now turn to the complex version of Theorem 2.14. The Riesz Representation Theorem ö.18 Let X be a locally compact Hausdorff space. Theorem 2.14 characterizes the oositive linear functionals on Cc(X). We are now in a position to characterize the ?ounded linear functionals on Cc(X). Since Cc(X) is a dense subspace of C0(X), relative to the supremum norm, every such has a unique extension to a bounded linear functional on C0(X). Hence we may as well assume to begin with that we are dealing with the Banach--space C0(X). If Jl is a complex Borel measure, Theorem 6.12 asserts that there is a complex Borel function h with I h I = 1 such that df.l = h d I J.l l . It is therefore reasonable to iefine integration with respect to a complex measure Jl by the formula I f dp. = I jh d I P. l · (1) The relation J XE dJl = Jl(E) is a special case of (1). Thus i XE d(p. + A) = (p. + A.)(E) = p.(E) + A(E) = i XE dp. + L XE dA (2) 130 REAL AND COMPLEX ANALYSIS whenever J1 and A. are complex measures on 9Jl and E e IDl. This leads to the addition formula (3) which is valid (for instance) for every bounded measurable! We shall call a complex Borel measure J1 on X regular if I J11 is regular in the sense of Definition 2.15. If J1 is a complex Borel measure on X, it is clear that the mapptng (4) is a bounded linear functional on C0(X), whose norm is no larger than I J11 (X). That all bounded linear functionals on C0(X) are obtained in this way is the content of the Riesz theorem : 6.19 Theorem If X is a locally compact Hausdorff space, then every bounded linear functional on C0(X) is represented by a unique regular complex Borel measure Jl, in the sense that (1) for every f e C0(X). Moreover, the norm of is the total variation of J1: 11<1>11 = IJ11(X). (2) PROOF We first settle the uniqueness question. Suppose J1 is a regular complex Borel measure on X and J f dJ1 = 0 for all f e C0(X). By Theorem 6.12 there is a Borel function h, with I hI = 1, such that dJ1 = h d I J11. For any sequence {fn} in C0(X) we then have I Il l (X) = 1 (ii -fn)h d I 111 < 11 ii -In I d I J1l, (3) and since Cc(X) is dense in L1( I J11) (Theorem 3.14), {fn} can be so chosen that the last expression in (3) tends to 0 as n r oo. Thus I J11 (X) = 0, and J1 = 0. It is easy to see that the difference of two regular complex Borel measures on X is regular. This shows that at most one J1 corresponds to each <1>. Now consider a given bounded linear functional on C0(X). Assume 11<1>11 = 1, without loss of generality. We shall construct a positive linear func tional A on Cc(X), such that I (f ) I < A( I f I ) < II f II where II f II denotes the supremum norm. (4) COMPLEX MEASURES 131 Once we have this A, we associate with it a positive Borel measure A., as in Theorem 2.14. The conclusion of Theorem 2.14 shows that A. is regular if A.(X) < oo . Since A.(X) = sup {Af: 0 <f < 1,fe Cc(X)} and since I Af I < 1 if II f II < 1, we see that actually A.( X) < 1. We also deduce from (4) that 1ͣ/)1 < A(l/1) = il/1 dA = 11/ ll t (5) The last norm refers to the space L1(A.). Thus is a linear functional on Cc(X) of norm at most 1, with respect to the L1(A.)-norm on Cc(X). There is a norm preserving extension of to a linear functional on L1(A.), and therefore Theorem 6. 16 (the case p = 1) gives a Borel function g, with I g I < 1, such that (6) Each side of (6) is a continuous functional on C0(X), and Cc(X) is dense in C0(X). Hence (6) holds for allf e C0(X), and we obtain the representation (1) with dJl = g dA.. Since 11<1>11 = 1, (6) shows that ilgl dA > sup {lͣf)l:fe C0(X), 11/ 11 < 1} = 1. (7) We also know that A.(X) < 1 and I g I < 1. These facts are compatible only if A.(X) = 1 and I g I = 1 a.e. [A.]. Thus d I Jll = I g I dA. = dA., by Theorem 6. 13, and I Jll (X)= A.(X) = 1 = 11<1>11, (8) which proves (2). So all depends on finding a positive linear functional A that satisfies ( 4). Iff e cc+(X) [the class of all nonnegative real members of Cc(X)], define Af = sup {I (h) I : h E Cc(X), I hI 0, A satisfies (4), 0 </1 </2 implies A/1 < A/2, and A(cf) = cAfif c is a positive constant. We have to show that A(f + g) = Af + Ag (10) and we then have to extend A to a linear functional on Cc(X). Fix f and g e cc+(X). If E > 0, there exist h1 and h2 e Cc(X) such that I h1l <f, I h2l < g, and Af < I (hl) I + E, Ag < I (h2) I + E. (11) 132 REAL AND COMPLEX ANALYSIS There are complex numbers rxb I rxi I = 1, so that rxi <f)( hi) = I <f)( hi) I, i = 1, 2. Then Af + Ag < I <f)(hl) I + I <f)(h2) I + 2E = <f)(rx.l h1 + fX2 h2) + 2E <A( I htl + I h2l) + 2E < A(f + g) + 2E, so that the inequality > holds in (10). Next, choose h e Cc(X), subject only to the condition I hI s/ + g, let V = {x:f(x) + g(x) > 0}, and define h.(x) = f(x)h(x) , h2(x) = g(x)h(x) f(x) + g(x) f(x) + g(x) h1(x) = h2(x) = 0 (x ¢ V). (x E V), (12) It is clear that h1 is continuous at every point of V. If x0 ¢ V, then h(x0) = 0; since h is continuous and since I h1(x) I < I h(x) I for all x e X, it follows that x0 is a point of continuity of h1• Thus h1 e Cc(X), and the same holds for h2• Since h1 + h2 = h and I h11 </,I h21 < g, we have I ( h) I = I (h1) + (h2) I < I <f)(ht) I + I <f)(h2) I < Af + Ag. Hence A(/+ g)< Af + Ag, and we have proved (10). Iff is now a real function, f e Cc(X), then 2/ + = If I + f, so that f + e cc+(X) ; likewise,/ - E cc+(X) ; and since/=/ + -f-, it is natural to define (f e Cc(X), f real) (13) and A(u + iv) = Au + iAv. (14) Simple algebraic manipulations, just like those which occur in the proof of Theorem 1.32, show now that our extended functional A is linear on Cc(X). This completes the proof. I I I I Exercises 1 If f.J. is a complex measure on a a-algebra 9Jl, and if E E 9Jl, define the supremum being taken over all finite partitions { Ei} of E. Does it follow that A. = I f.J. I ? 2 Prove that the example given at the end of Sec. 6. 10 has the stated properties. 3 Prove that the vector space M(X) of all complex regular Borel measures on a locally compact Hausdorff space X is a Banach space if llf.J.II = I f.J. I (X). Hint: Compare Exercise 8, Chap. 5. [That the difference of any two members of M(X) is in M(X) was used in the first paragraph of the proof of Theorem 6.19 ; supply a proof of this fact.] COMPLEX MEASURES 133 4 Suppose 1 < p < oo , and q is the exponent conjugate to p. Suppose J.t is a positive a-finite measure and g is a measurable function such thatf g E IJ(p) for every f E I.!'(p). Prove that then y E IJ(J.t). S Suppose X consists of two points a and b; define p({a}) = 1, p({b}) = p(X) = oo , and J.t(0) = 0. Is it true, for this J.t, that L00(J.t) is the dual space of I!(J.t)? 6 Suppose 1 < p < oo and prove that /J(f.J.) is the dual space of I.!'(J.t) even if J.t is not a-finite. (As usual, 1/p + 1/q = 1.) 7 Suppose J.t is a complex Borel measure on [0, 27t) (or on the unit circle T), and define the Fourier coefficients of J.t by [l(n) = f e-••• d!l(t) (n = 0, + 1, + 2, . . . ). Assume that jJ.(n)---+ 0 as n ---+ + oo and prove that then jJ.(n)---+ 0 as n ---+ - oo . Hint : The assumption also holds with f df.J. in place of df.J. iff is any trigonometric polynomial, hence iff is continuous, hence iff is any bounded Borel function, hence if dJ.t is replaced by d I J.t 1 . 8 In the terminology of Exercise 7, find all J.t such that jl is periodic, with period k. [This means that jl(n + k) = jl(n) for all integers n; of course, k is also assumed to be an integer.] 9 Suppose that {g,.} is a sequence of positive continuous functions on I = [0, 1], that J.t is a positive Borel measure on I, and that (i) lim,.. oo gix) = 0 a.e. [m], (ii) fi g,. dm = 1 for all n, (iii) lim,._. oo f I f g,. dm = f I f dJ.t for every f E C(/). Does it follow that J.t .l m? 10 Let (X, 9Jl, J.t) be a positive measure space. Call a set c L1(J.1.) unif ormly integrable if to each € > 0 corresponds a b > 0 such that whenever f E and J.l.(E) < b. (a) Prove that every finite subset of I!(J.l.) is uniformly integrable. (b) Prove the following convergence theorem of Vitali : If (i) J.t(X) < oo , (ii) {In} is unif ormly integrable, (iii) / ,.(x)---+ f(x) a.e. as n ---+ oo , and (iv) I f(x) I < oo a.e., then f E L1(J.t) and Suggestion : Use Egoroff's theorem. (c) Show that (b) fails if J.1. is Lebesgue measure on ( - oo, oo), even if { 11/,.11 1} is assumed to be bounded. Hypothesis (i) can therefore not be omitted in (b). (d) Show that hypothesis (iv) is redundant in (b) for some J.t (for instance, for Lebesgue measure on a bounded interval), but that there are finite measures for which the omission of (iv) would make (b) false. (e) Show that Vitali's theorem implies Lebesgue's dominated convergence theorem, for finite measure spaces. Construct an example in which Vitali's theorem applies although the hypotheses of Lebesgue's theorem do not hold. (f) Construct a sequence {/,.}, say on [0, 1], so thatf ,.(x) ---+ 0 for every x, f !
,. ---+ 0, but {In} is not uniformly integrable (with respect to Lebesgue measure). (g) However, the following converse of Vitali's theorem is true : I f J.t(X) < oo ,fn E L1(J.1.), and exists f or every E E 9Jl, then {f,.} is unif ormly integrable. 134 REAL AND COMPLEX ANALYSIS Prove this by completing the following outline. Define p(A, B) = I I X.t - XB I dJ.t. Then (rol, p) is a complete metric space (modulo sets of measure 0), and E-+ IE !,. dJ.t is continuous for each n. If £ > 0, there exist E0 , ў' N (Exercise 13, Chap. 5) so that 1 (f. -fN) dJl < E if p(E, E0) < b, n > N. () If J.t(A) < , () holds with B = E0 - A and C = E0 u A in place of E. Thus () holds with A in place of E and 2£ in place of €. Now apply (a) to {/1, . . . , fN}: There exists {l > 0 such that Lf. dJl < 3E if Jl(A) < b', n = 1, 2, 3, . . . . l l Suppose J.l is a positive measure on X, J.t(X) < oo,f, e I!(p.) for n = 1, 2, 3, ... ,f,(x)-+ f(x) a.e., and there exists p > 1 and C < oo such that f x I / ,. IP dp < C for all n. Prove that Hint: {/,} is uniformly integrable. lim [ I f-!,. I dJ.t = o. n - co Jx 12 Let 9Jl be the collection of all sets E in the unit interval [0, 1] such that either E or its complement is at most countable. Let J.t be the counting measure on this a-algebra ln. If g(x) = x for 0 s; x s; 1, show that g is not rol-measurable, although the mapping !-+ L xf(x) = f f g dJl makes sense for every f e I!(p) and defines a bounded linear functional on lJ(p). Thus (L 1) '# L eo in this situation. 13 Let L00 = L00(m), where m is Lebesgue measure on I = [0, 1]. Show that there is a bounded linear functional A '# 0 on L 00 that is 0 on C(J), and that therefore there is no g e L 1(m) that satisfies A f = f 1 f g dm for every f E L00• Thus (L 00) '# I!. CHAPTER SEVEN DIFFERENTIATION In elementary Calculus we learn that integration and differentiation are inverses of each other. This fundamental relation persists to a large extent in the context of the Lebesgue integral. We shall see that some of the most important facts about differentiation of integrals and integration of derivatives can be obtained with a minimum of effort by first studying derivatives of measures and the associ ated maximal functions. l"he Radon-Nikodym theorem and the Lebesgue decom position will play a prominent role. Derivatives of Measures We begin with a simple, theorem whose main purpose is to motivate the defini tions that follow. 7.1 Theorem Suppose J.l is a complex Borel measure on R1 and f(x) = J.l((- oo, x)) (1) If x E R1 and A is a complex number, each of the following two statements implies the other: (a) f is diff erentiable at x and f'(x) = A. (b) To every e > 0 corresponds a ç > 0 such that J.l(l) -A m(J) < € (2) 135 136 REAL AND COMPLEX ANALYSIS for every open segment I that contains x and whose length is less than Ł. Here m denotes Lebesgue measure on R1• 7.2 Definitions Theorem 7.1 suggests that one might define the derivative of J.l at x to be the limit of the quotients J.l(I)/m(I), as the segments I shrink to x, and that an analogous definition might be appropriate in several variables, i.e., in Rk rather than in R 1. Accordingly, let us fix a dimension k, denote the open ball with center x E Rk and radius r > 0 by B(x, r) = {y E Rk: I y-xI < r} (1) (the absolute value indicates the euclidean metric, as in Sec. 2.19), associate to any complex Borel measure J.l on Rk the quotients (Q, JLXx) = JL(B(x, r)), m(B(x, r)) (2) where m = mk is Lebesgue measure on Rk, and define the symmetric derivative of J.l at x to be (Dtt)(x) = lim (Qr J.l}(x) (3) r- o at those points x E Rk at which this limit exists. We shall study DJ.l by means of the maximal function MJ.l. For J.l > 0, this is defined by (MJ.l)(x) = sup (Qr J.l}(x), (4) O < r < oo and the maximal function of a complex Borel measure J.l is, by definition, that of its total variation I J.ll. The functions M J.l: Rk = [0, oo] are lower semicontinuous, hence measur able. To see this, assume J.l > 0, pick A. > 0, let E = { M J.l > A.}, and fix x E E. Then there is an r > 0 such that J.l(B(x, r)) = tm(B(x, r)) (5) for some t > A., and there is a Ī > 0 that satisfies (6) If I y -xI < Ł' then B(y, r + Ł) ::) B(x, r), and therefore J.l(B(y, r + Ł)) > tm(B(x, r)) = t[r/(r + Ł)]km(B(y, r + Ł)) > A.m(B(y, r + Ł)). Thus B(x, Ł) c E. This proves that E is open. Our first objective is the " maximal theorem " 7.4. The following covering lemma will be used in its proof. DIFFERENTIATION 137 7.3 Lemma If W is the union of a finite collection of balls B(xi , ri), 1 < i < N, then there is a setS c { 1, ... , N} so that (a) the balls B(xi , ri) with i E S are disjoint, (b) W c U B(xh 3ri), and i e S (c) m(W) < 3k 2: m(B(xi , ri)). i e S PROOF Order the balls Bi = B(xi , ri) so that r1 > r2 > · · · > rN. Put i1 = 1. Discard all Bi that intersect Bi1• Let Bil be the first of the remaining Bi, if there are any. Discard all Bi with j > i2 that intersect Bi2 , let Bi3 be the first of the remaining ones, and so on, as long as possible. This process stops after a finite number of steps and gives S = { i b i2, • • • } • It is clear that (a) holds. Every discarded Bi is a subset of B(xi, 3ri) for some i E S, for if r' < r and B(x', r') intersects B(x, r), then B(x', r') c B(x, 3r). This proves (b), and (c) follows from (b) because m(B(x, 3r)) = 3km(B(x, r)) /Ill The following theorem says, roughly speaking, that the maximal function of a measure cannot be large on a large set. 7.4 Theorem If J1 is a complex Borel measure on Rk and A. is a positive number, then m{MJ1 >A.} < 3kA. - 1 IIJ111. (1) . Here II J111 = I J1 1 (Rk), and the left side of (1) is an abbreviation for the more cumbersome expression m({x E Rk: (MJ1)(x) >A.}). (2) We shall often simplify notation in this way. PROOF Fix J1 and A.. Let K be a compact subset of the open set {MJ1 >A.}. Each x E K is the center of an open ball B for which I J1 1 (B) > A.m(B). Some finite collection of these B's covers K, and Lemma 7.3 gives us a dis joint subcollection, say { B 1, ... , Bn}, that satisfies n n m(K) < 3k 2: m(Bi) < 3kA. - 1 2: I J1 1 (Bi) < 3kA. - 1 IIJ111. 1 1 The disjointness of {Bh . . . , Bn} was used in the last inequality. Now (1) follows by taking the supremum over all compact K c {MJ1 >A.}. //// 138 REAL AND COMPLEX ANALYSIS 7.5 Weak L1 Iff e I!(Rk) and A. > 0, then m { I f I > A.} < A.-1 II f li t because, putting E = {If I > A.}, we have Am(E) < [ 1 ! 1 dm < [ I ! I dm = 11/ll t· JE JRk Accordingly, any measurable function/for which A. · m{lfi >A.} is a bounded function of A. on (0, oo) is said to belong to weak L1. (1) (2) (3) Thus weak L1 contains L1. That it is actually larger is shown most simply by the function 1/x on (0, 1). We associate to each f e I!(Rk) its maximal function Mf: Rk = [0, oo ], by setting (Mf)(x) = sup / ) [ 1 ! 1 dm. 0 <r< 00 m Br JB(x, r) (4) [We wrote Br in place of B(x, r) because m(B(x, r)) depends only on the radius r.] If we identify f with the measure J1 given by dJ1 = f dm, we see that (4) agrees with the previously defined M Jl. Theorem 7.4 states therefore that the "maximal operator " M sends L1 to weak I!, with a bound (namely 3k) that depends only on the space Rk: For every f e L1(Rk) and every A.> 0, (5) 7.6 Lebesgue points Iff e L1(Rk), any x e Rk for which it is true that lim / [ lf(y) -f(x) f dm(y) = 0 r- 0 m Br) JB(x, r) (1) is called a Lebesgue point off For example, (1) holds iff is continuous at the point x. In general, (1) means that the averages of I f-f(x) I are small on small balls centered at x. The Lebesgue points off are thus points where f does not oscillate too much, in an average sense. It is probably far from obvious that every f e L1 has Lebesgue points. But the following remarkable theorem shows that they always exist. (See also Exercise 23.) 7.7 Theorem Iff e L1(Rk), then almost every x e Rk is a Lebesgue point off DIFFERENTIATION 139 PROOF Define (T,./Xx) = (k) I 1/-f(x)l dm m r JB(x, r) (1) for x E R\ r > 0, and put (Tf)(x) = lim sup (T,. f)(x). (2) r-o We have to prove that Tf = 0 a.e. [m]. Pick y > 0. Let n be a positive integer. By Theorem 3.14, there exists g E C(Rk) so that 11/-gll1 < 11n: Put h =! -g. Since g is continuous, T g = 0. Since we have ( T,. h)( X) < / ) r I h I dm + I h(x) I m Br JB(x, r) Th < Mh + I hi. Since T,. f < T,. g + T,. h, it follows that Tf< Mh + I hi. Therefore {Tf> 2y} c {Mh > y} u {I hi > y}. (3) (4) (5) (6) Denote the union on the right of (6) by E(y, n). Since llhll1 < 1ln, Theorem 7.4 and the inequality 7.5(1) show that m(E(y, n)) < (3k + 1 )l(yn). (7) The left side of (6) is independent of n. Hence 00 {Tf> 2y} c n E(y, n). (8) n = l This intersection has measure 0, by (7), so that { Tf > 2y} is a subset of a set of measure 0. Since Lebesgue measure is complete, {Tf> 2y} is Lebesgue measurable, and has measure 0. This holds for every positive y. Hence Tf = 0 a.e. [ m J. I I I I Theorem 7.7 yields interesting information, with very little effort, about topics such as (a) differentiation of absolutely continuous measures, (b) differentiation using sets other than balls, (c) differentiation of indefinite integrals in R 1, (d) metric density of measurable sets. 140 REAL AND COMPLEX ANALYSIS We shall now discuss these topics. 7.8 Theorem Suppose Jl is a complex Borel measure on Rk, and Jl m. Let f be the Radon-Nikodym derivative of Jl with respect to m. Then DJ1 = f a.e. [m], and (1) for all Borel sets E c Rk. In other words, the Radon-Nikodym derivative can also be obtained as a limit of the quotients Q, Jl. PROOF The Radon-Nikodym theorem asserts that (1) holds with/in place of DJ1. At any Lebesgue point x off, it follows that f(x) = lim 1 i f dm = lim JL(B(x, r)) . r - o m(B,) B(x, r) r - o m(B(x, r)) (2) Thus (DJl)(x) exists and equals f(x) at every Lebesgue point off, hence a.e. [m]. Ill/ 7.9 Nicely shrinking sets Suppose x E Rk. A sequence {Ei} of Borel sets in Rk is said to shrink to x nicely if there is a number C( > 0 with the following property: There is a sequence of balls B(x, ri), with lim ri = 0, such that Ei c B(x, ri) and (1) for i = 1, 2, 3, . . . . Note that it is not required that x E Ei, nor even that x be in the closure of Ei. Condition (1) is a quantitative version of the requirement that each Ei must occupy a substantial portion of some spherical neighborhood of x. For example, a nested sequence of k-cells whose longest edge is at most 1,000 times as long as its shortest edge and whose diameter tends to 0 shrinks nicely. A nested sequence of rectangles (in R2) whose edges have lengths 1/i and (1/i)2 does not shrink nicely. 7.10 Theorem Associate to each x E Rk a sequence {Ei(x)} that shrinks to x nicely, and let f E L1(Rk). Then f(x) = lim 1 i f dm i- oo m(Ei(x)) Ei(x) at every Lebesgue point off, hence a.e. [m]. (1) DIFFERENTIATION 141 PROOF Let x be a Lebesgue point off and let C((x) and B(x, ri) be the positive number and the balls that are associated to the sequence {Ei(x)}. Then, because Ei(x) c B(x, ri), a(x) r lf-f(x)l dm< 1 r 1!-f(x)l dm. m(E;(x)) JE;(x) m(B(x, ri)) JB(x, ri) The right side converges to 0 as ir oo, because rir 0 and x is a Lebesgue point off Hence the left side converges to 0, and ( 1) follows. I I I I Note that no relation of any sort was assumed to exist between { Ei(x)} and { Ei(Y)}, for different points x and y. Note also that Theorem 7 .. 10 leads to a correspondingly stronger form of Theorem 7.8. We omit the details. 7.11 Theorem Iffe L1(R1) and F(x) =roof dm (-oo <x < oo), then F'(x) = f(x) at every Lebesgue point off, hence a.e. [m]. (This is the easy half of the fundamental theorem of Calculus, extended to Lebesgue integrals.) PROOF Let {<5i} be a sequence of positive numbers that converges to 0. Theorem 7.10, with Ei(x) = [x, x + <5J, shows then that the right-hand deriv ative of F exists at all Lebesgue points of x off and that it is equal to f(x) at these points. If we let Ei(x) be [x -bi, x] instead, we obtain the same result for the left-hand derivative of F at x. I I I I 7.12 Metric density Let E be a Lebesgue measurable subset of Rk. The metric density of E at a point x e Rk is defined to be I. m(E n B(x, r)) tm ----r-+ o m(B(x, r)) provided, of course, that this limit exists. (1) If we let f be the characteristic function of E and apply Theorem 7.8 or Theorem 7.10, we see that the metric density of E is 1 at almost every point of E, and that it is 0 at almost every point of the complement of E. Here is a rather striking consequence of this, which should be compared with Exercise 8 in Chap. 2: If E > 0, there is no set E c R 1 that satisfies m(E n I) 1 E < m(I) < -E (2) for every segment I. 142 REAL AND COMPLEX ANALYSIS Having dealt with differentiation of absolutely continuous measures, we now turn to those that are singular with ·respect to m. 7.13 Theorem Associate to each x E Rk a sequence { E;(x)} that shrinks to x nicely. If J1 is a complex Borel measure and J1 i m, then (1) PROOF The Jordan decomposition theorem shows that it suffices to prove (1) under the additional assumption that J1 > 0. In that case, arguing as in the proof of Theorem 7.10, we have Hence ( 1) is a consequence of the special case which will now be proved. (DJ1)(x) = 0 a.e. [m], The upper derivative DJ1, defined by (DJ1)(x) = lim [ sup (Qr J1)(x)J n-+ oo 0 < r < 1/n (2) (3) is a Borel function, because the quantity in brackets decreases as n increases and is, for each n, a lower semicontinuous function of x; the reasoning used in Sec. 7.2 proves this. Choose A > 0, E > 0. Since J1 i m, J1 is concentrated on a set of Lebesgue measure 0. The regularity of J1 (Theorem 2.18) shows therefore that there is a compact set K, with m(K) = 0, J1(K) > IIJ111 - E. Define J11(E) = Jl(K n E), for any Borel set E c Rk, and put J12 = J1-J11• Then II J12ll < E, and, for every x outside K, (4) Hence (5) and Theorem 7.4 shows that (6) Since (6) holds for every E > 0 and for every A > 0, we conclude that DJ1 = 0 a.e. [m], i.e., that (2) holds. /Ill Theorems 7.10 and 7.13 can be combined in the following way: DIFFERENTIATION 143 7.14 Theorem Suppose that to each x e Rk is associated some sequence {Eרx)} that shrinks to x nicely, and that Jl is a complex Borel measure on Rk. Let dp, = f dm + dp,s be the Lebesgue decomposition of Jl with respect to m. Then . p,(Ei(x)) Ö1m (E, )) = f(x) a.e. [m]. , .... oo m i\x In particular, Jl l m if and only if(Dp,)(x) = 0 a.e. [m]. The following result contrasts strongly with Theorem 7.13: 7.15 Theorem If Jl is a positive Borel measure on Rk and Jl _i m, then (Dp,)(x) = oo a.e. [p,]. (1) PROOF There is a Borel set S c: Rk with m(S) = 0 and p,(Rk -S) = 0, and there are open sets Jj => S with m(Jj) < 1/j, for j = 1, 2, 3, . . . . For N = 1, 2, 3, . . . , let EN be the set of all xeS to which correspond radii ri = ri(x), with lim ri = 0, such that (2) Then (1) holds for every xeS- U EN. N Fix N and j, for the moment. Every x e EN is then the center of a ball Bx c: Jj that satisfies (2). Let Px be the open ball with center x whose radius is 1/3 of that of Bx. The union of these balls Px is an open set K-}, N that con tains EN and lies in J'} . We claim that (3) To prove (3), let K c: K-}, N be compact. Finitely many Px cover K. Lemma 7.3 shows therefore that there is a finite set F c: EN with the follow ing properties : (a) {Px: x e F} is a disjoint collection, and (b) K c: U Bx. x e F Thus This proves (3). x e F x e F x e F Now put nN = n K-}, N· Then EN c: nN, nN is a G., , p,(ON) = 0, and j (Dp,)(x) = oo at every point of S - U nN . Ill/ N 144 REAL AND COMPLEX ANALYSIS The Fundamental Theorem of Calculus 7.16 This theorem concerns functions defined on some compact interval [a, b] in R1• It has two parts. The first asserts, roughly speaking, that the derivative of the indefinite integral of a function is that same function. We dealt with this in Theorem 7.1 1. The second part goes the other way: one returns to the original function by integrating its derivative. More precisely f(x) -f(a) = ix f'(t) dt (a< x <b). (1) In the elementary version of this theorem, one assumes that f is differentiable at every point of [a, b] and that/' is a continuous function. The proof of (1) is then easy. In trying to extend (1) to the setting of the Lebesgue integral, questions such as the following come up naturally: Is it enough to assume that f' e L1, rather than that f' is continuous? If f is continuous and differentiable at almost all points of [a, b ], must (1) then hold? Before proving any positive results, here are two examples that show how (1) can fail. (a) Put f(x) = x2 sin (x-2) if x "# 0, f(O) = 0. Then f is differentiable at every point, but flf'(t)l dt = oo, (2) sof' ¢ L1• If we interpret the integral in (1) (with [0, 1] in place of [a, b ]) as the limit, as e = 0, of the integrals over [e, 1], then (1) still holds for thisf More complicated situations can arise where this kind of passage to the limit is of no use. There are integration processes, due to Denjoy and Perron (see , ), which are so designed that (1) holds whenever f is differentiable at every point. These fail to have the property that the integrability off implies that of If I , and therefore do not play such an important role in analysis. (b) Suppose f is continuous on [a, b ], f is differentiable at almost every point of [a, b], and/' E L1 on [a, b]. Do these assumptions imply that (1) holds? Answer: No. Choose {<5n} so that 1 =<50 > <51 > <52 > · · · , bnŠ 0. Put E0 = [0, 1]. Suppose n > 0 and En is constructed so that En is the union of 2n disjoint closed intervals, each of length 2-nbn. Delete a segment in the center of each of these 2n intervals, so that each of the remaining 2n + 1 intervals has length DIFFERENTIATION 145 2-n-l£5n+1 (this is possible, since £5n+1 < £5n), and let En + 1 be the union of these 2n + 1 intervals. Then E 1 :::J E2 :::J • • · , m(En) = £5n, and if 00 E = n En, (3) n=1 then E is compact and m(E) = 0. (In fact, E is perfect.) Put (n = 0, 1, 2, . . . ). (4) Then ln(O) = 0, ln(1) = 1, and each In is a monotonic function which is con stant on each segment in the complement of En . If I is one of the 2n intervals whose union is En, then (5) It follows from (5) that (6) and that (7) Hence {In} converges uniformly to a continuous monotonic function f, with f(O) = 0, 1(1) = 1, and l'(x) = 0 for all x rJ E. Since m(E) = 0, we have I' = 0 a.e. Thus (1) fails. If £5n = (2/3)n, the set E is Cantor's " middle thirds " set. Having seen what can go wrong, assume now that I' e L1 and that (1) does hold. There is then a measure J.l, defined by dJ.l =I' dm. Since J.l m, Theorem 6.1 1 shows that there corresponds to each E > 0 a £5 > 0 so that I J.ll (E) < E whenever E is a union of disjoint segments whose total length is less than £5. Since l(y) -l(x) = J.l((x, y)) if a < x < y < b, it follows that the absolute continuity off, as defined below, is necessary for (1). Theorem 7.20 will show that this necessary condition is also sufficient. 7.17 Definition A complex function f, defined on an interval I = [a, b ], is said to be absolutely continuous on I (briefly, I is AC on I) if there corre sponds to every E > 0 a £5 > 0 so that n . L I l(f3i) -l(rxi) I < E i= 1 (1) 146 REAL AND COMPLEX ANALYSIS for any n and any disjoint collection of segments (cx1, {31), • • • , (cxn, Pn) in I whose lengths satisfy n 'L ({3i -cxi) < b. (2) i= 1 Such anfis obviously continuous: simply taken= 1. In the following theorem, the implication (b)r (c) is probably the most interesting. That (a)r (c) without assuming monotonicity off is the content of Theorem 7.20. 7.18 Theorem Let I= [a, b], let f: IŠ R1 be continuous and nondecreasing. Each of the following three statements about f implies the other two: (a) fis AC on I. (b) fmaps sets of measure 0 to sets of measure 0. (c) f is diff erentiable a.e. on I,f' e L1, and f(x) -f(a) = lx f'(t) dt (ex < x 0. Associate b > 0 to f and E, as in Definition 7. 17. There is then an open set V with m(V) < b, so that E c V c I. Let (cxb {3i) be the disjoint segments whose union is V. Then 'L ({3i -cxi) < b, and our choice of b shows that therefore 'L (f(f3i) -f(cxi)) < E. (2) i [Definition 7.17 was stated in terms of finite sums; thus (2) holds for every partial sum of the (possibly) infinite series, hence (2) holds also for the sum of the whole series, as stated.] Since E c V,f(E) c U [f(cxi),f(f3i)]. The Lebesgue measure of this union is the left side of (2). This says that f(E) is a subset of Borel sets of arbitrarily small measure. Since Lebesgue measure -is complete, it follows thatf(E) e 9Jl and m(f(E)) = 0. We have now proved that (a) implies (b). DIFFERENTIATION 147 Assume next that (b) holds. Define g(x) = x + f(x) (a< x <b). (3) If the /-image of some segment of length '1 has length rJ', then the g-image of that same segment has length 11 + 17'. From this it follows easily that g satisfies (b), since f does. Now suppose E c I, E e 9Jl. Then E = E1 u E0 where m(E0) = 0 and E1 is an Fa (Theorem 2.20). Thus E 1 is a countable union of compact sets, and so is g(E1), because g is continuous. Since g satisfies (b), m(g(E0)) = 0. Since g(E) = g(E1) u g(E0), we conclude: g(E) e 9Jl Therefore we can define Jl( E) = m(g( E)) (E c I, E E 9Jl). (4) Since g is one-to-one (this is our reason for working with g rather than f), disjoint sets in I have disjoint g-images. The countable additivity of m shows therefore that J1 is a (positive, bounded) measure on 9Jl. Also, J1 m, because g satisfies (b). Thus dJ1 = h dm for some h e L1(m), by the Radon-Nikodym theorem. If E = [a, x], then g(E) = [g(a), g(x)], and (5) gives g(x) -g(a) = m(g(E)) = JL(E) = L h dm = ix h(t) dt. If we now use (3), we conclude that f(x) -f(a) = r [h(t) - 1] dt Thusf'(x) = h(x)- 1 a.e. [m], by Theorem 7.1 1. We have now proved that (b) implies (c). (C( < X < b). (5) (6) The discussion that preceded Definition 7.17 showed that (c) implies (a). 7.19 Theorem Supposef: Ir R1 is AC, I = [a, b]. Define N F(x) = sup L I f(ti) -f(ti-1) I i= 1 (a< x <b) where the supremum is taken over all N and over all choices of { t;} such that a= t0 < t1 < · · · < tN = x. The functions F, F + f, F -!are then nondecreasing and AC on I. Ill/ (1) (2) 148 REAL AND COMPLEX ANALYSIS [F is called the total variation function off. Iff is any (complex) function on I, AC or not, and F(b) < oo, then f is said to have bounded variation on I, and F(b) is the total variation off on I. Exercise 13 is relevant to this.] PROOF If (2) holds and x < y < b, then N F(y) > I f(y) -f(x) I + L I f(ti) -f(ti-1) 1 . (3) i= 1 Hence F(y) > I f(y) -f(x) I + F(x). In particular F(y) > f(y) -f(x) + F(x) and F(y) > f(x) -f(y) + F(x). (4) This proves that F, F + f, F -fare nondecreasing. Since sums of two AC functions are obviously AC, it only remains to be proved that F is AC on I. If (a, p) c I then n F(p) -F(rx) = sup L I f(ti) -. f(ti-1) I, (5) 1 the supremum being taken over all { tJ that satisfy rx = t0 < · · · < tn = p. Note that L (ti-ti-1) = P- rx . Now pick E > 0, associate b > 0 to f and E as in Definition 7.17, choose disjoint segments (rxi, P) c I with L (pi-rx) < b, and apply (5) to each (rxi, pi) . It follows that L (F(pi) -F(rxi)) < E, j by our choice of b. Thus F is AC on I. We have now reached our main objective: (6) Ill/ 7.20 Theorem Iff is a complex function that is AC on I= [a, b], then f is diff erentiable at almost all points of I,f' E L1(m), and f(x) -f(a) = r f'(t) dt (a< x <b). (1) PROOF It is of course enough to prove this for real f. Let F be its total variation function, as in Theorem 7.19, define ft = !(F +f), f2 = !(F -f), and apply the implication (a)r (c) of Theorem 7.18 to f1 and f2• Since f=f1 -f2 this yields ( 1 ) . (2) (3) /Ill DIFFERENTIATION 149 The next theorem derives (1) from a different set of hypotheses, by an entirely different method of proof. 7.21 Theorem Iff: [a, b] } R1 is diff erentiable at every point of [a, b] and f' e L1 on [a, b ], then f(x) -f(a) = r f'(t) dt (a< x 0. Theorem 2.25 ensures the existence of a lower semicontinuous function g on [a, b] such that g > f' and r g(t) dt < r f'(t) dt + E. (2) Actually, Theorem 2.25 only gives g > f', but since m([a, b]) < oo, we can add a small constant to g without affecting (2). For any 11 > 0, define F,(x) = r g(t) dt -f(x) + f(a) + rJ(X-a) (a< x 0 such that g(t) > f'(x) and f(t) -f(x) f'(x). For any such t we therefore have F,(t)-F,(x) = fg(s) ds- [f(t) -f(x)] + rJ(t-x) > (t -x)f'(x) -(t -x)[f'(x) + 17] + rJ(t -x) = 0. Since F ,(a) = 0 and F, is continuous, there is a last point x e [a, b] at which F ,(x) = 0. If x < b, the preceding computation implies that F ,(t) > 0 for t e (x, b]. In any case, F,(b) > 0. Since this holds for every 11 > 0, (2) and (3) . now gtve f(b) -f(a) < r g(t) dt < r f'(t) dt + E, (5) and since E was arbitrary, we conclude that f(b) -f(a) < r f'(t) dt. (6) 150 REAL AND COMPLEX ANALYSIS If f satisfies the hypotheses of the theorem, so does -!; therefore ( 6) holds with -fin place off, and these two inequalities together give (1). /Ill Differentiable Transformations 7.22 Definitions Suppose V is an open set in Rk, T maps V into R\ and x E V. If there exists a linear operator A on Rk (i.e., a linear mapping of Rk into R\ as in Definition 2.1) such that lim I T(x + h) - T(x) - Ah I = 0 h-+0 lhl (1) (where, of course, hE Rk), then we say that T is diff erentiable at x, and define T'(x) =A. (2) The linear operator T'(x) is called the derivative of T at x. (One shows easily that there is at most one linear A that satisfies the preceding require ments; thus it is legitimate to talk about the derivative of T.) The term diff er ential is also often used for T'(x). The point of (1) is of course that the difference T(x + h)- T(x) is approximated by T'(x)h, a linear function of h. Since every real number ex gives rise to a linear operator on R 1 (mapping h to cxh), our definition of T'(x) coincides with the usual one when k = 1. When A: RkŠ Rk is linear, Theorem 2.20(e) shows that there is a number L(A) such that m(A(E)) = A(A)m(E) for all measurable sets E c Rk. Since A'(x) = A (3) (4) and since every differentiable transformation T can be locally approximated by a constant plus a linear transformation, one may conjecture that m(T(E))"' L(T'(x)) m(E) (5) for suitable sets E that are close to x. This will be proved in Theorem 7.24, and furnishes the motivation for Theorem 7.26. Recall that A( A) = I det A I was proved in Sec. 2.23. When T is differen tiable at x, the determinant of T'(x) is called the Jacobian of T at x, and is denoted by J T(x). Thus A(T'(x)) = I J T(x) 1 . (6) The following lemma seems geometrically obvious. Its proof depends on the Brouwer fixed point theorem. One can avoid the use of this theorem by imposing DIFFERENTIATION 151 stronger hypotheses on F, for example, by assuming that F is an open mapping. But this would lead to unnecessarily strong assumptions in Theorem 7.26. 7.23 Lemma Let S = { x: I xI = 1} be the sphere in Rk that is the boundary of the open unit ball B = B(O, 1). IfF: B----+ Rk is continuous, 0 < E < 1, and I F(x)-xl < E for all x e S, then F(B) ::) B(O, 1 - e). (1) PROOF Assume, to reach a contradiction, that some point a e B(O, 1 - e) is not in F(B). By (1), I F(x) I·> 1 -E if x e S. Thus a is not in F(S), and there fore a =t= F(x), for every x e B. This enables us to define a continuous map G: B----+ B by a-F(x) G(x) = I a -F(x) 1 · If x e S, then x · x = I x 12 = 1, so that (2) x · (a -F(x)) = x · a + x · (x -F(x)) - 1 < I a I + E - 1 < 0. (3) This shows that x · G(x) < 0, hence x =1= G(x). If x e B, then obviously x =1= G(x), simply because G(x) e S. Thus G fixes no point of B, contrary to Brouwer's theorem which states that every continuous map of .8 into B has at least one fixed point. /Ill A proof of Brouwer's theorem that is both elementary and simple may be found on pp. 38-40 of " Dimension Theory " by Hurewicz and Wallman, Princeton University Press, 1948. 7.24 Theorem If (a) V is open in Rk, (b) T: V----+ Rk is continuous, and (c) T is diff erentiable at some point x e V, then lim m(T(B(x, r))) = .1(T'(x)). r-+ O m(B(x, r)) (1) Note that T(B(x, r)) is Lebesgue measurable; in fact, it is a-compact, because B(x, r) is a-compact and T is continuous. PRooF Assume, without loss of generality, that x = 0 and T(x) = 0. Put A= T'(O). The following elementary fact about linear operators on finite dimensional vector spaces will be used: A linear operator A on Rk is one-to-152 REAL AND COMPLEX ANALYSIS one if and only if the range of A is all of Rk. In that case, the inverse A -1 of A is also linear. Accordingly, we split the proof into two cases. CASE 1 A is one-to-one. Define (x E V). (2) Then F'(O) = A -1 T'(O) = A-lA =I, the identity operator. We shall prove that lim m(F(B(O, r))) = 1. r- o m(B(O, r)) Since T(x) = AF(x), we have m(T(B)) = m(A(F(B))) = A(A)m(F(B)) for every ball B, by 7.22(3). Hence (3) will give the desired result. (3) (4) Choose E > 0. Since F(O) = 0 and F'(O) = I, there exists a ç > 0 such that 0 < I xI < ç implies I F(x) -xI < E I x 1 . (5) We claim that the inclusions B(O, (1 -E)r) c F(B(O, r)) c B(O, (1 + E)r) (6) hold if 0 < r < Ī. The first of these follows from Lemma 7.23, applied to B(O, r) in place of B(O, 1 ), because I F(x) -x I < Er for all x with I x I = r. The second follows directly from (5), since I F(x) I < (1 + E) I x 1 . It is clear that (6) implies and this proves (3). (1 _ )k < m(F(B(O, r))) < (1 + )k E - m(B(O, r)) -E (7) CASE n A is not one-to-one. In this case, A maps Rk into a subspace of lower dimension, i.e., into a set of measure 0. Given E > 0, there is therefore an '1 > 0 such that m(E,) < E if E, is the seࡔ of all points in Rk whose distance from A(B(O, 1)) is less than '7· Since A = T'(O), there is a ç > 0 such that I x I < ç implies I T(x) -Ax I < '7 1 x 1 . (8) If r < ç' then T(B(O, r)) lies therefore in the set E that consists of the points whose distance from A(B(O, r)) is less than rJr. Our choice of '1 shows that m(E) < Erk. Hence m(T(B(O, r))) < Erk (0 < r < ç). (9) DIFFERENTIATION, 153 Since rk = m(B(O, r))/m(B(O, 1 )), (9) implies that lim m(T(B(O, r))) = O. r- o m(B(O, r)) This proves (1), since A(T'(O)) = A(A) = 0. 7.25 Lemma Suppose E c Rk, m(E) = 0, T maps E into Rk, and I. I T(y)- T(x) I tm sup I I < oo y -x for every x e E, as y tends to x within E. Then m(T(E)) = 0. (10) Ill/ PROOF Fix positive integers n and p, let F = F n , P be the set of all x e E such that I T(y)- T(x)l < nly-xl for all y e B(x, 1/p) n E, and choose E > 0. Since m(F) = 0, F can be covered by balls Bi = B(xi, ri), where xi e F, ri < 1/p, in such a way that L m(Bi) < e. (To do this, cover F by an open set W of small measure, decompose W into disjoint boxes of small diameter, as in Sec. 2.19, and cover each of those that intersect F by a ball whose center lies in the box and in F.) If x e F n Bi then I xi -xI < ri < 1/p and xi e F. Hence I T(xi) - T(x) I < n I xi -xI < nri so that T(F n Bi) c B(T(xi), nri). Therefore T(F) c U B(T(xi), nri). i The measure of this union is at most L m(B(T(xi), nri) = nk L m(Bi) < nke. i i Since Lebesgue measure is complete and e was arbitrary, it follows that T(F) is measurable and m(T(F)) = 0. To complete the proof, note that E is the union of the countable collec-tion { F n. p}. I I I I Here is a special case of the lemma: If V is open in Rk and T: V } Rk is diff erentiable at every point of V, then T maps sets of measure 0 to sets of measure 0. We now come to the change-of-variables theorem. 7.26 Theorem Suppose that (i) X c V c Rk, Vis open, T: V } Rk is continuous; 154 REAL AND COMPLEX ANALYSIS (ii) X is Lebesgue measurable, T is one-to-one on X, and T is diff erentiable at every point of X; (iii) m(T(V-X))= 0. Then, setting Y = T(X), ifdm = l(fo T)llrl dm (1) for every measurable f: Rk----+ [0, oo ]. The case X = V is perhaps the most interesting one. As regards condition (iii), it holds, for instance, when m( V -X) = 0 and T satisfies the hypotheses of Lemma 7.25 on V-X. The proof has some elements in common with that of the implication (b)--+ (c) in Theorem 7.18. It will be important in this proof to distinguish between Borel sets and Lebesgue measurable sets. The a-algebra consisting of the Lebesgue measurable subsets of Rk will be denoted by IDl. PROOF We break the proof into the following three steps: (I) If E e 9Jl and E c V, then T(E) e IDl. (II) For every E E IDl, m(T(E n X))= lxEilrl dm. (Ill) For every A e IDl, ixA dm = L<xA o T)llrl dm. If E0 e IDl, E0 c V, and m(E0) = 0, then m(T(E0 -X)) = 0 by (iii), and m(T(E0 n X)) = 0 by Lemma 7.25. Thus m(T(E0)) = 0. If E 1 c V is an F (1 , then E 1 is a-compact, hence T(E 1) is a-compact, because T is continuous. Thus T(E 1) e IDl. Since every E E 9Jl is the union of an F (1 and a set of measure 0 (Theorem 2.20), (I) is proved. To prove (II), let n be a positive integer, and put V,. = {x E V: I T(x)l < n}, Because of (1), we can define Jln(E) = m(T(E n X n)) (E E IDl). (2) (3) Since T is one-to-one on X n, the countable additivity of m shows that Jln is a measure on IDl. Also, Jln is bounded (this was the reason for replacing X temporarily by Xn), and Jln m, by another application of Lemma 7.25. DIFFERENTIATION 155 Theorem 7.8 tells us therefore that (DJ.ln)(x) exists a.e. [m], that DJ.ln e L1(m), and that p..(E) = 1 (Dp..) dm (E E IDl). (4) We claim next that (DJ.ln)(x) = I J T(x) I (5) To see this, fix x e X n, and note that B(x, r) c V, for all sufficiently small r > 0, because V, is open. Since V, -X n c V -X, hypothesis (iii) enables us to replace X n by V, in (3) without changing J.ln(E). Hence, for small r > 0, J.ln(B(x, r)) = m(T(B(x, r))). (6) If we divide both sides of (6) by m(B(x, r)) and refer to Theorem 7.24 and formula 7.22(6), we obtain (5). Since (3) implies that J.ln(E) = J.ln(E n X n), it follows from (3), (4), and (5) that m(T(E n X.))= r XEIJTi dm Jxn (E E IDl). (7) If we apply the monotone convergence theorem to (7), letting n ---+ oo , we obtain (II). We begin the proof of (III) by letting A be a Borel set in Rk. Put E = T-1(A) = {x e V: T(x) e A}. (8) Then XE = XA o T. Since XA _is a Borel function and T is continuous, XE is a Borel function (Theorem 1.12), hence E e IDl. Also T(E n X)= A n Y (9) which implies, by (II), that lx...t dm = m(T(E n X))= l(X...t o T)IJTI dm. (10) Finally, if N e 9Jl and m(N) = 0, there is a Borel set A => N with m(A) = 0. For this A, (10) shows that (XA o T) I J T I = 0 a.e. [m]. Since 0 < XN < XA, it follows that both integrals in (10) are 0 if A is replaced by N. Since every Lebesgue measurable set is the disjoint union of a Borel set and a set of measure 0, ( 1 0) holds for every A e IDl. This proves (Ill). Once we have (III), it is clear that (1) holds for every nonnegative Lebesgue measurable simple function! Another application of the monotone convergence theorem completes the proof. I I I I 156 REAL AND COMPLEX ANALYSIS Note that we did not prove that f o T is Lebesgue measurable for all Lebesgue measurable f. It need not be; see Exercise 8. What the proof does estab lish is the Lebesgue measurability of the product (f o T) I J T 1 . Here is a special case of the theorem: Suppose cp: [a, b] } [C(, PJ is AC, monotonic, cp(a) = C(, 0 is Lebesgue measurable. Then r f(t) dt = r f(<p(x))q/(x) dx. (1 5) To derive this from Theorem 7.26, put V =(a, b), T = <p, let Q be the union of the maximal segments on which <p is constant (if there are any) and let X be the set of all x E V -Q where <p'(x) exists (and is finite). Exercises 1 Show that I f(x) I < (M f)(x) at every Lebesgue point ofjiff E lJ(Rk). 2 For > 0, let /() be the segment ( -, ) c: R1• Given ct and p, 0 < ct < p < 1, construct a measur able set E c: R 1 so that the upper and lower limits of m(E n /()) 2 are P and ct, respectively, as ---+ 0. (Compare this with Section 7.12.) 3 Suppose that E is a measurable set of real numbers with arbitrarily small periods. Explicitly, this means that there are positive numbers Pi , converging to 0 as i---+ oo, so that E + Pi = E (i = 1, 2, 3, . . . ). Prove that then either E or its complement has measure 0. Hint: Pick ct E R1, put F(x) = m(E n [ct, x]) for x > ct, show that if ct + Pi < x < y. What does this imply about F'(x) if m(E) > 0? 4 Call t a period of the function f on R 1 if f(x + t) = f(x) for all x E R 1. Suppose f is a real Lebesgue measurable function with periods s and t whose quotient s/t is irrational. Prove that there is a con stant c such thatf(x) = c a.e., but thatfneed not be constant. Hint: Apply Exercise 3 to the sets {!> A.}. S If A c: R1 and B c: R1, define A + B = {a + b: a e A, b E B}. Suppose m(A) > 0, m(B) > 0. Prove that A + B contains a segment, by completing the following outline. There are points a0 and b0 where A and B have metric density 1. Choose a small > 0. Put c0 = a0 + b0 • For each €, positive or negative, define Bl to be the set of all c0 + € - b for which b E B and I b - b0 I < . Then Bl c: (a0 + € - , a0 + € + ). If was well chosen and I € I is sufficiently small, it follows that A intersects Bl , so that A + B :::> (c0 - €0, c0 + €0) for some €0 > 0. Let C be Cantor's " middle thirds " set and show that C + C is an interval, although m( C) = 0. (See also Exercise 19, Chap. 9.) 6 Suppose G is a subgroup of R 1 (relative to addition), G =F R 1, and G is Lebesgue measurable. Prove that then m( G) = 0. Hint: Use Exercise 5. DIFFERENTIATION 157 7 Construct a continuous monotonic function f on R 1 so that f is not constant on any segment althoughf'(x) = 0 a.e. 8 Let V = (a, b) be a bounded segment in R 1• Choose segments W,. c: V in such a way that their union W is dense in V and the set K = V - W has m(K) > 0. Choose continuous functions cp" so that cpix) = 0 outside W,., 0 < cpix) < 2 -n in W,.. Put qJ = L qJ" and define T(x) = r q>(t) dt (a < x < b). Prove the following statements: (a) T satisfies the hypotheses of Theorem 7.26, with X = V. (b) T' is continuous, T'(x) = 0 on K, m(T(K)) = 0. (c) If E is a nonmeasurable subset of K (see Theorem 2.22) and A = T(E), then XA is Lebesgue measurable but XA o T is not. (d) The functions cp" can be so chosen that the resulting T is an in finitely diff erentiable homeo morphism of V onto some segment in R 1 and (c) still holds. 9 Suppose 0 < ct < 1. Pick t so that ta. = 2. Then t > 2, and the construction of Example (b) in Sec. 7.16 can be carried out with b" = (2/t)". Show that the resulting function f belongs to Lip ct on [0, 1]. 10 Iff E Lip 1 on [a, b], prove thatfis absolutely continuous and thatf' E L00• 1 1 Assume that 1 < p < oo , f is absolutely continuous on [a, b ], f' E 1!, and ct = 1/q, where q is the exponent conjugate to p. Prove that f E Lip ct. 12 Suppose cp : [a, b] ---+ R 1 is nondecreasing. (a) Show that there is a left-continuous nondecreasing f on [a, b] so that {f #- qJ} is at most countable. [Left-continuous means: if a < x < b and E > 0, then there is a b > 0 so that l f(x) -f(x - t) l < E whenever 0 < t < b.] (b) Imitate the proof of Theorem 7.18 to show that there is a positive Borel measure J1 on [a, b] for which f(x) -f(a) = Jl([a, x)) (a < x < b). (c) Deduce from (b) thatf'(x) exists a.e. [m], thatf' E L1(m), and that f(x) -f(a) = r f'(t) dt + s(x) (a < x < b) where s is nondecreasing and s'(x) = 0 a.e. [m]. (d) Show that Jl _L m if and only if f'(x) = 0 a.e. [m], and that J1 ¥ m if and only iff is AC on [a, b ]. (e) Prove that cp'(x) = f'(x) a.e. [m]. 13 Let BV be the class of all f on [a, b] that have bounded variation on [a, b], as defined after Theorem 7.19. Prove the following statements. (a) Every monotonic bounded function on [a, b] is in BV. (b) Iff E BV is real, there exist bounded monotonic functions f1 and f2 so that f = /1 -f2 • Hint : Imitate the proof of Theorem 7.19. (c) If f E BV is left-continuous then f1 and f2 can be chosen in (b) so as to be also left continuous. (d) Iff E BV is left-continuous then there is a Borel measure Jl on [a, b] that satisfies f(x) -f(a) = J,L([a, x)) J1 ¥ m if and only iff is AC on [a, b]. (a < x < b); (e) Every f E BV is differentiable a.e. [m], andf' E V(m). 14 Show that the product of two absolutely continuous functions on [a, b] is absolutely continuous. Use this to derive a theorem about integration by parts. 158 REAL AND COMPLEX ANALYSIS IS Construct a monotonic function f on R 1 so that f'(x) exists (finitely) for every x E R 1, but f' is not a continuous function. 16 Suppose E c [a, b], m(E) = 0. Construct an absolutely continuous monotonic function! on [a, b] so thatf'(x) = oo at every x E E. Hint: E c n V,. , V,. open, m(V,.) c 2-n. Consider the sum of the characteristic functions of these sets. 17 Suppose {,un} is a sequence of positive Borel measures on Rk and ,U( E) = L J.ln( E). n= 1 Assume ,u(Rk) < oo. Show that ,u is a Borel measure. What is the relation between the Lebesgue decompositions of the J.ln and that of ,u? Prove that 00 (D,u)(x) = L (D,u,.)(x) a.e. [m]. rt= 1 Derive corresponding theorems for sequences {fn} of positive nondecreasing functions on R 1 and their sums f = L !,. . 18 Let q>0(t) = 1 on [0, 1), q>0(t) = - 1 on [1, 2), extend q>0 to R 1 so as to have period 2, and define q>n(t) = q>0(2nt), n = 1, 2, 3, . . . . Assume that L I en 12 < oo and prove that the series () converges then f or almost every t. Probabilistic interpretation : The series L ( +en) converges with probability 1. Suggestion : { q>,.} is orthonormal on [0, 1 ], hence () is the Fourier series of some f E L 2• If a = j · 2-N, b = U + 1) · 2-N, a < t < b, and sN = c1q>1 + · · · + cN q>N, then, for n > N, 1 f.b 1 f.b SN(t) = SN dm = Sn dm, b - a b - a a a and the last integral converges to J! f dm, as n-+ oo . Show that () converges to f(t) at almost every Lebesgue point off 19 Supposefis continuous on R1,f(x) > 0 if O < x < 1,f(x) = 0 otherwise. Define hc(x) = sup {ncf(nx): n = 1, 2, 3, . . . }. Prove that (a) he is in IJ(R1) if O < c < 1, (b) h1 is in weak IJ but not in L1(R1), (c) he is not in weak IJ if c > 1. 20 (a) For any set E c R2, the boundary aE of E is, by definition, the closure of E minus the interior of E. Show that E is Lebesgue measurable whenever m(aE) = 0. (b) Suppose that E is the union of a (possibly uncountable) collection of closed discs in R2 whose radii are at least 1. Use (a) to show that E is Lebesgue measurable. (c) Show that the conclusion of (b) is true even when the radii are unrestricted. (d) Show that some unions of closed discs of radius 1 are not Borel sets. (See Sec. 2.21.) (e) Can discs be replaced by triangles, rectangles, arbitrary polygons, etc., in all this? What is the relevant geometric property? DIFFERENTIATION 159 21 Iff is a real function on [0, 1] and y(t) = t + lf(t), the length of the graph off is, by definition, the total variation of y on [0, 1]. Show that this length is finite if and only iff E BV. (See Exercise 13.) Show that it is equal to f J 1 + (f'(t)]2 dt iff is absolutely continuous. ll (a) Assume that both f and its maximal function M f are in L1(Rk). Prove that then f(x) = 0 a.e. [m]. Hint: To every other f e I!(Rk) corresponds a constant c = c(f) > 0 such that (M f)(x) > c l x l -k whenever I x I is sufficiently large. (b) Definef(x) = x- 1(log x)-2 if O < x < f,f(x) = 0 on the rest of R1• Thenf e I!(R1). Show that (Mf)(x) ƣ l 2x log (2x) 1 - 1 (0 < X < 1/4) so that Jȩ (M f)(x) dx = oo. 23 The definition of Lebesgue points, as made in Sec. 7.6, applies to individual integrable functions, not to the equivalence classes discussed in Sec. 3.10. However, if F e I!(Rk) is one of these equivalence classes, one may call a point x e Rk a Lebesgue point o f F if there is a complex number, let us call it (SF)(x), such that for one (hence for every) f e F. lim 1 l I f- (SF)(x) I dm = 0 r-+0 m(Br) B(:x, r) Define (SF)(x) to be 0 at those points x e Rk that are not Lebesgue points of F. Prove the following statement: Iff e F, and x is a Lebesgue point off, then x is also a Lebesgue point of F, andf(x) = (SF)(x). Hence SF E F. Thus S " selects " a member of F that has a maximal set of Lebesgue points. CHAPTER EIGHT INTEGRATION ON PRODUCT SPACES This chapter is devoted to the proof and discussion of the theorem of Fubini concerning integration of functions of two variables. We first present the theorem in its abstract form. Measurability on Cartesian Products 160 8.1 Definitions If X and Y are two sets, their cartesian product X x Y is the set of all ordered pairs (x, y), with x e X and y e Y. If A c X and B c Y, it follows that A x B c X x Y. We call any set of the form A x B a rectangle in X x Y. Suppose now that (X, 9') and (Y, 5'") are measurable spaces. Recall that this simply means that 9' is a a-algebra in X and 5'" is a a-algebra in Y. A measurable rectangle is any set of the form A x B, where A e 9' and B E 5'". If Q = R1 u · · · u Rn , where each R; is a measurable rectangle and R; n Ri = 0 for i # j, we say that Q e 8, the class of all elementary sets. 9' x 5'" is defined to be the smallest a-algebra in X x Y which contains every measurable rectangle. A monotone class IDl is a collection of sets with the following properties : If A; e IDl, B; e IDl, A; c A;+ 1, B; ::) B;+ 1, for i = 1, 2, 3, . .
. , and if 00 A = U A· l ' (1) i = 1 then A e IDl and B e 9Jl. INTEGRATION ON PRODUCT SPACES 161 If E c X x Y, x e X, y e Y, we define Ex = {y : (x, y) E E}, EY = {X: ( x, y) E E}. (2) We call Ex and EY the x-section and y-section, respectively, of E. Note that Ex C Y, EY C X. 8.2 Theorem If E e !/ x §", then Ex e §" and EY e !/,for every x e X and y E Y. PROOF Let Q be the class of all E e !/ x §" such that Ex e §" for every x e X. If E = A x B, then Ex = B if x e A, Ex = 0 if x ¢A. Therefore every measurable rectangle belongs to Q. Since §" is a a-algebra, the following three statements are true. They prove that Q is a a-algebra and hence that Q =!/ X ff : (a) X X y E n. (b) If E E n, then (Ec)x = (ExY, hence Ec E n. (c) If Ei E Q (i = 1, 2, 3, . . . ) and E = U Eb then Ex = U (Ei)x , hence E e Q. The proof is the same for EY. Ill/ 8.3 Theorem !/ x ff is the smallest monotone class which contains all elemen tary sets. PROOF Let 9Jl be the smallest monotone class which contains 8; the proof that this class exists is exactly like that of Theorem 1. 10. Since !/ x f7 is a monotone class, we have 9Jl c !/ x f/. The identities (A1 x B1) n (A2 x B2) = (A1 n A2) x (B1 n B2), (A1 x B1)-(A2 x B2) = [(A1 - A2) x B1] u [(A1 n A2) x (B1 -B2)] show that the intersection of two measurable rectangles is a measurable rec tangle and that their difference is the union of two disjoint measurable rec tangles, hence is an elementary set. If P e 8 and Q e 8, it follows easily that P n Q E C and P - Q E C. Since P u Q = (P - Q) u Q and (P - Q) n Q = 0, we also have P u Q E 8. For any set P c X x Y, define Q(P) to be the class of all Q c X x Y such that P - Q e IDl, Q - P e IDl, and P u Q E IDl. The following properties are obvious: (a) Q e Q(P) if and only if P e !l(Q). (b) Since 9Jl is a monotone class, so is each Q(P). 162 REAL AND COMPLEX ANALYSIS Fix P e C. Our preceding remarks about 8 show that Q e Q(P) for all Q e C, hence 8 c Q(P), and now (b) implies that 9J1 c Q(P). Next, fix Q e 9Jl. We just saw that Q e Q(P) if P e 8. By (a), P e Q(Q), hence C c Q(Q), and if we refer to (b) once more we obtain 9J1 c Q(Q). Summing up : If P and Q e IDl, then P -Q e 9J1 and P u Q e IDl. It now follows that 9J1 is a a-algebra in X x Y : (i) X x Y e 8. Hence X x Y e IDl. (ii) If Q e IDl, then Qc e IDl, since the difference of any two members of 9Jl is in IDl. (iii) If Pi e 9Jl for i = 1, 2, 3, .
.. , and P = U Pi, put Since 9Jl is closed under the formation of finite unions, Qn e IDl. Since Qn c Qn + 1 and P = U Qn, the mono tonicity of 9Jl shows that p E 9Jl. Thus 9Jl is a a-algebra, 8 c 9Jl c !/ x :T, and (by definition) !/ x :T is the smallest a-algebra which contains 8. Hence 9Jl = !/ x :T. /Ill 8.4 Definition With each function f on X x Y and with each x e X we associate a functionfx defined on Y by fx(Y) = f(x, y). Similarly, if y e Y,fY is the function defined on X by fY(x) = f(x, y). Since we are now dealing with three a-algebras, !/, :T, and !/ x :T, we shall, for the sake of clarity, indicate in the sequel to which of these three a-algebras the word " measurable " refers. 8.5 Theorem Let f be an (!/ x :T)-measurable function on X x Y. Then (a) For each x e X,fx is a !T-measurablefunction. (b) For each y e Y,fY is an !/-measurable/unction. PROOF For any open set V, put Q = {(x, y): f(x, y) e V}. Then Q e !/ x !T, and Qx = {y: f x(Y) E V}. Theorem 8.2 shows that Qx e :T. This proves (a); the proof of (b) is similar. I I I I INTEGRATION ON PRODUCT SPACES 163 Product Measures 8.6 Theorem Let (X, !/, J1) and (Y, !/, A.) be a-finite measure spaces. Suppose Q E !/ X !/.If t/J(y) = Jl(QY) (1) for every x E X and y E Y, then dp. = I"' dA, (2) Notes: The assumptions on the measure spaces are, more explicitly, that J1 and A. are positive measures on !/ and !/, respectively, that X is the union of countably many disjoint sets X n with Jl(X n) < oo, and that Y is the union of countably many disjoint sets Y m with A.(Y m) < oo . Theorem 8.2 shows that the definitions (1) make sense. Since A(Qx) = I XQ(x, y) dA(y) (x E X), with a similar statement for Jl(QY), the conclusion (2) can be written in the form l dp.(x) l XQ(x, y) dA(y) = l dA(y) l XQ(x, y) dp.(x). (3) (4) PROOF Let Q be the class of all Q E !/ x ff for which the conclusion of the theorem holds. We claim that Q has the following four properties : (a) Every measurable rectangle belongs to !l. (b) If Q1 c Q2 c Q3 c · · · , if each Qi E Q, and if Q = U Qb then Q E n. (c) If { QJ is a disjoint countable collection of members of Q, and if Q = u Qi , then Q E Q. (d) If J1(A) < 00 and A.(B) < 00, if A X B ::::) Ql ::::) Q2 ::::) Q3 ::::) . . . ' if Q = n Qi and Qi E Q for i = 1, 2, 3, . . . , then Q E Q. If Q = A x B, where A E !/, B E!/, then (5) and therefore each of the integrals in (2) is equal to ,u(A)A.(B). This gives (a). To prove (b), let <pi and t/Ji be associated with Qi in the way in which (1) associates <p and t/1 with Q. The countable additivity of J1 and A. shows that (ir oo), (6) the convergence being monotone increasing at every point. Since cpi and t/1 i are assumed to satisfy the conclusion of the theorem, (b) follows from the monotone convergence theorem. 164 REAL AND COMPLEX ANALYSIS For finite unions of disjoint sets, (c) is clear, because the characteristic function of a union of disjoint sets is the sum of their characteristic functions. The general case of (c) now follows from (b). The proof of (d) is like that of (b), except that we use the dominated convergence theorem in place of the monotone convergence theorem. This is legitimate, since Jl(A) < oo and A.(B) < oo. Now define Qmn = Q n (X n X Y m) ( m, n = 1' 2, 3, .. . ) (7) and let 9Jl be the class of all Q e !/ x ff such that Qmn e Q for all choices of m and n. Then (b) and (d) show that 9Jl is a monotone class; (a) and (c) show that & c IDl; and since IDl c !/ x ff, Theorem 8.3 implies that 9Jl = !/ x ff. Thus Qmn E Q for every Q E !/ x ff and for all choices of m and n. Since Q is the union of the sets Qmn and since these sets are disjoint, we conclude from (c) that Q E n. This completes the proof. Ill I 8.7 Definition If (X, !/, Jl) and (Y, !/, A.) are as in Theorem 8.6, and if Q e !/ x !/, we define (p x A)(Q) = l A(QJ dp(x) = lp(QY) dA(y). (1) The equality of the integrals in (1) is the content of Theorem 8.6. We call Jl x A. the product of the measures Jl and A.. That Jl x A. is really a measure (i.e., that Jl x A. is countably additive on !/ x 5'") follows immediately from Theorem 1.27. Observe also that Jl x A. is a-finite. The Fubini Theorem 8.8 Theorem Let (X, !/, Jl) and ( Y, !/, A.) be a-finite measure spaces, and let f be an(!/ x !f)-measurable function on X x Y. (a) IfO <f < oo , and if rft(y) = fxr dp (x e X, y e Y), then <p is !/-measurable, t/1 is !/-measurable, and r (/) dp = r 1 d(p x A) = r"' dA. Jx JxxY Jy (b) Iff is complex and if qJ(x) = l1 fix dA and l qJ dp < oo, then f e L1(Jl x A.). (1) (2) (3) INTEGRATION ON PRODUCT SPACES 165 (c) Iff E L1(Jl x A.), thenfx E L1(A.)for almost all x E X,fY E L1(Jl)for almost all y E Y; the functions qJ and t/1, defined by (1) a.e., are in L1(Jl) and L1(A.), respectively, and (2) holds. Notes: The first and last integrals in (2) can also be written in the more usual form l dJl(x) 1/(x, y) dA(y) = 1 dA(y) Lf(x, y) dJl(x). (4) These are the so-called " iterated integrals " off. The middle integral in (2) is often referred to as a double integral. The combination of (b) and (c) gives the following useful result : Iff is (!/ x §)-measurable and if l dJl(x) 1 I f(x, y) I dA(y) < oo, (5) then the two iterated integrals (4) are finite and equal. In other words, " the order of integration may be reversed " for (!/ x §") measurable functions f whenever f > 0 and also whenever one of the iterated integrals of If I is finite. PROOF We first consider (a). By Theorem 8.5, the definitions of qJ and t/1 make sense. Suppose Q E !/ x fT and f = XQ. By Definition 8. 7, (2) is then exactly the conclusion of Theorem 8.6. Hence (a) holds for all nonnegative simple (!/ x §")-measurable functions s. In the general case, there is a sequence of such functions sn, such that 0 < s 1 < s2 < · · · and sn(x, y)----+ f(x, y) at every point of X x Y. If lfJn is associated with sn in the same way in which qJ was associated to f, we have r (/),. dJl = r s,. d(Jl x A) Jx Jxx Y (n = 1, 2, 3, ... ). (6) The monotone convergence theorem, applied on (Y, §", A.), shows that lfJn(x) increases to qJ(x), for every x EX, as n----+ oo . Hence the monotone con vergence theorem applies again, to the two integrals in ( 6), and the first equality (2) is obtained. The second half of (2) follows by interchanging the roles of x and y. This completes (a). If we apply (a) to If I, we see that (b) is true. Obviously, it is enough to prove (c) for real L1(Jl x A.); the complex case then follows. Iff is real, (a) applies to f + and to f-. Let ({J1 and ({J2 corre spond to f + and f- as qJ corresponds to f in (1). Since f E L1(Jl x A.) and 166 REAL AND COMPLEX ANALYSIS f + < If I, and since (a) holds for f +, we see that cp1 e L1(J1.). Similarly, cp2 e L1(J1.). Since (7) we have fx e L1(A) for every x for which cp1(x) < oo and cp2(x) < oo ; since cp1 and cp2 are in L1(J1.), this happens for almost all x; and at any such x, we have cp(x) = ·cp1(x)-<p2(x). Hence cp e L1(J1.). Now (2) holds with cp1 and f + and with <p2 and/ -, in place of cp and/; if we subtract the resulting equations, we obtain one half of (c). The other half is proved in the same manner, with fY and t/1 in place of fx and <p. I I I I 8.9 Counterexamples The following three examples will show that the various hypotheses in Theorems 8.6 and 8.8 cannot be dispensed with. (a) Let X = Y = [0, 1], J1. = A = Lebesgue measure on [0, 1]. Choose {<5n} so that 0 = <51 < <52 < <53 < · · · , <5n----+ 1, and let gn be a real continuous function with support in (£5n, <5n + 1), such that JA gn(t) dt = 1, for n = 1, 2, 3, ... . Define 00 f(x, y) = L [gn(x)-gn + l(x)Jgn(y). n=l Note that at each point (x, y) at most one term in this sum is different from 0. Thus no convergence problem arises in the definition off. An easy computa tion shows that f dx f f(x, y) dy = 1 "I= 0 = f dy f f(x, y) dx, so that the conclusion of the Fubini theorem fails, although both iterated integrals exist. Note that f is continuous in this example, except at the point (1, 1), but that fax f lf(x, y)l dy = oo. (b) Let X = Y = [0, 1], J1. = Lebesgue measure on [0, 1], A = counting measure on Y, and putf(x, y) = 1 if x = y,f(x, y) = 0 if x #= y. Then lf(x, y) dJl(x) = 0, If(x, y) dA(y) = 1 for all x and y in [0, 1 ], so that 1 dA.(y) lf(x, y) dJl(x) = 0 "I= 1 = l dJl(x) If(x, y) dA.(y). This time the failure is due to the fact that A is not a-finite. Observe that our function f is (9' x §')-measurable, if 9' is the class of all Lebesgue measurable sets in [0, 1] and ff consists of all subsets of [0, 1]. INTEGRATION ON PRODUCT SPACES 167 To see this, note that f = Xn , where D is the diagonal of the unit square. Given n, put I --[j - 1 j] i -n 'n and put Qn = (I 1 X I 1) U (I 2 X I 2) U · · · U (In X In). Then Qn is a finite union of measurable rectangles, and D = n Qn . (c) In examples (a) and (b), the failure of the Fubini theorem was due to the fact that either the function or the space was " too big." We now turn to the role played by the requirement that f be measurable with respect to the a-algebra f/ X fT. To pose the question more precisely, suppose ,u(X) = A.( Y) = 1, 0 <! < 1 (so that " bigness " is certainly avoided); assume f x is :T -measurable and fY is !/-measurable, for all x and y; and assume cp is f/ -measurable and t/1 is :T -measurable, where cp and t/1 are defined as in 8.8(1). Then 0 < cp < 1, 0 < t/1 < 1, and both iterated integrals are finite. (Note that no reference to product measures is needed to define iterated integrals.) Does it follow that the two iterated integrals of fare equal? The (perhaps surprising) answer is no. In the following example (due to Sierpinski), we take (X, !/, ,u) = (Y, :T, A.) = [0, 1] with Lebesgue measure. The construction depends on the continuum hypoth esis. It is a consequence of this hypothesis that there is a one-to-one mapping j of the unit interval [0, 1] onto a well-ordered set W such that j(x) has at most countably many predecessors in W, for each x e [0, 1]. Taking this for granted, let Q be the set of all (x, y) in the unit square such that j(x) precedes j(y) in W. For each x e [0, 1], Qx contains all but countably many points of [0, 1] ; for each y e [0, 1], QY contains at most countably many points of [0, 1]. Iff= XQ , it follows thatfx andfY are Borel measurable and that q>(x) = r f(x, y) dy = 1, t/l(y) = r f(x, y) dx = 0 for all x and y. Hence r dx r f(x, y) dy = 1 =I= 0 = r dy r f(x, y) dx. Completion of Product Measures 8.10 If (X, f/, ,u) and (Y, :T, A.) are complete measure spaces, it need not be true that (X x Y, !/ x :T, ,u x A.) is complete. There is nothing pathological about 168 REAL AND COMPLEX ANALYSIS this phenomenon: Suppose that there exists an A e !/, A =1= 0, with ,u(A) = 0; and suppose that there exists a B c Y so that B ¢ ff. Then A x B c A x Y, (,u x A.)(A x Y) = 0, but A x B ¢ !/ x S"'. (The last assertion follows from Theorem 8.2.) For instance, if ,u = A. = m1 (Lebesgue measure on R1), let A consist of any one point, and let B be any nonmeasurable set in R1• Thus m1 x m1 is not a complete measure; in particular, m1 x m1 is not m2 , since the latter is complete, by its construction. However, m2 is the completion of m1 x m1• This result gener alizes to arbitrary dimensions : 8.1 1 Theorem Let mk denote Lebesgue measure on Rk. If k = r + s, r > 1, s > 1, then mk is the completion of the product measure mr x ms. PROOF Let f!4 k and IDlk be the a-algebras of all Borel sets and of all Lebesgue measurable sets in Rk, respectively. We shall first show that (1) Every k-cell belongs to IDlr x IDls. The a-algebra generated by the k-cells is PAk. Hence &lk c IDlr x IDls. Next, suppose E E IDlr and FE IDls. It is easy to see, by Theorem 2.20(b), that both E x Rs and Rr x F belong to IDlk. The same is true of their intersection E x F. It follows that IDlr x IDls c IDlk. Choose Q e IDlr x IDls. Then Q e IDlk, so there are sets P 1 and P 2 e 31 k such that P 1 c Q c P 2 and mk(P 2 -P 1) = 0. Both mk and mr x ms are trans lation invariant Borel measures on Rk. They assign the same value to each k-cell. Hence they agree on f!4 k, by Theorem 2.20(d). In particular, (mr X ms)(Q - P1) < (mr X ms)(P2 -P1) = mk(P2- P1) = 0 and therefore So mr X ms agrees with mk on IDlr X IDls. It now follows that IDlk is the (mr x ms)-completion of IDlr x IDls, and this is what the theorem asserts. /Ill We conclude this section with an alternative statement of Fubini's theorem which is of special interest in view of Theorem 8.1 1. 8.12 Theorem Let (X,!/, ,u) and (Y, ff, A.) be complete a-finite measure spaces. Let(!/ x ff) be the completion of!/ x ff, relative to the measure ,u x A.. Let f be an (!/ x §')-measurable function on X x Y. Then all conclusions of Theorem 8.8 hold, the only diff erence being as follows: The §'-measurability of fx can be asserted only for almost all x eX, so that q>(x) is only defined a.e. [,u] by 8.8(1); a similar statement holds for fY and t/J. INTEGRATION ON PRODUCT SPACES 169 The proof depends on the following two lemmas : Lemma 1 Suppose v is a positive measure on a a-algebra IDl, IDl is the com pletion of 9Jl relative to v, and f is an IDl-measurable function. Then there exists an IDl-measurable function g such that f = g a.e. [ v ]. (An interesting special case of this arises when v is Lebesgue measure on Rk and 9Jl is the class of all Borel sets in Rk.) Lemma 2 Let h be an (g x §')-measurable function on X x Y such that h = 0 a.e. [Jl x A.]. Then for almost all x E X it is true that h(x, y) = 0 for almost all y E Y; in particular, hx is !T -measurable for almost all x E X. A similar statement holds for hY. If we assume the lemmas, the proof of the theorem is immediate: Iff is as in the theorem, Lemma 1 (with v = Jl x A.) shows that f = g + h, where h = 0 a.e. [Jl x A.] and g is (g x !f)-measurable. Theorem 8.8 applies to g. Lemma 2 shows that fx = 9x a.e. [A.] for almost all x and that fY = gY a.e. [Jl] for almost all y. Hence the two iterated integrals off, as well as the double integral, are the same as those of g, and the theorem follows. PROOF OF LEMMA 1 Suppose f is IDl-measurable and f > 0. There exist IDl measurable simple functions 0 = s0 < s1 < s2 < · · · such that sn(x)r f(x) for each x EX, as nr oo. Hence f = L (sn+ 1 -sn). Since sn+ 1 -sn is a finite linear combination of characteristic functions, it follows that there are con stants ci > 0 and sets Ei E IDl such that 00 f(x) = L ci XEi(x) (x EX). i= 1 The definition of IDl (see Theorem 1.36) shows that there are sets Ai E IDl, Bi E IDl, such that Ai c Ei c Bi and v(Bi-Ai) = 0. Define 00 g(x) = L ci XAi(x) (x EX). i= 1 Then the function g is IDl-measurable, and g(x) = f(x), except possibly when x E U (Ei -Ai) c U (Bi -Ai). Since v(Bi -Ai) = 0 for each i, we conclude that g = fa.e. [v]. The general case (freal or complex) follows from this. /Ill PROOF OF LEMMA 2 Let P be the set of all points in X x Y at which h(x, y) =1= 0. Then P E (9' x !T) and (Jl x A.)(P) = 0. Hence there exists a Q E g x !T such that P c Q and (Jl x A.)(Q) = 0. By Theorem 8.6, I A(Qx) dJl(X) = 0. (1) 170 REAL AND COMPLEX ANALYSIS Let N be the set of all x eX at which A.(Qx) > 0. It follows from (1) that J.l(N) = 0. For every x £1 N, A.(Qx) = 0. Since P x c Qx and (Y, ff, A.) is a com plete measure space, every subset of P x belongs to ff if x ¢ N. If y ¢ P x, then hx(Y) = 0. Thus we see, for every x £1 N, that hx is ff -measurable and that hx(Y) = 0 a.e. [A.]. //// Convolutions 8.13 It happens occasionally that one can prove that a certain set is not empty by proving that it is actually large. The word " large " may of course refer to various properties. One of these (a rather crude one) is cardinality. An example is furnished by the familiar proof that there exist transcendental numbers : There are only countably many algebraic numbers but uncountably many real numbers, hence the set of transcendental real numbers is not empty. Applications of Baire's theorem are based on a topological notion of largeness: The dense G a's are " large " subsets of a complete metric space. A third type of largeness is measure-theoretic: One can try to show that a certain set in a measure space is not empty by showing that it has positive measure or, better still, by showing that its complement has measure zero. Fubini's theorem often occurs in this type of argument. For example, let f and g e IJ(R1), assume f> 0 and g > 0 for the moment, and consider the integral h(x) = L:f(x -t)g(t) dt ( - 00 < X < 00 ). (1) For any fixed x, the integrand in (1) is a measurable function with range In [0, oo ], so that h(x) is certainly well defined by (1), and 0 < h(x) < oo. But is there any x for which h(x) < oo ? Note that the integrand in (1) is, for each fixed x, the product of two members of Ll, and such a product is not always in L1. [Example: f(x) = g(x) = 1/Jx if 0 < x < 1, 0 otherwise.] The Fubini theorem will give an affirmative answer. In fact, it will show that h e L1(R 1 ), hence that h(x) < oo a.e. 8.14 Theorem Supposefe L1(R1), g e L1(R1). Then L: I f(x -y)g(y) I dy < oo for almost all x. For these x, define h(x) = L:f(x -y)g(y) dy. (1) (2) (3) INTEGRATION ON PRODUCT SPACES 171 where 11/llt = I: lf(x) I dx. (4) We call h the convolution off and g, and write h = f g. PROOF There exist Borel functions /0 and g0 such that /0 = f a.e. and g0 = g a. e. The integrals ( 1) and (2) are unchanged, for every x, if we replace f by fo and g by g0• Hence we may assume, to begin with, that f and g are Borel functions. To apply Fubini's theorem, we shall first prove that the function F defined by F(x, y) = f(x -y)g(y) (5) is a Borel function on R2• Define <p: R2r R1 and t/J : R2r R1 by <p(x, y) = x - y, t/J(x, y) = y. (6) Thenf(x - y) = (f o <p)(x, y) and g(y) = (g o t/J)(x, y). Since <p and t/1 are Borel functions, Theorem 1.12(d) shows that f o <p and g o t/1 are Borel functions on R 2• Hence so is their product. Next we observe that J_0000 dy I-: I F(x, y) I dx = J_0000 I g(y) I dy L: I f(x - y) I dx = II f lltllg lit, (7) smce L: l f(x -y)l dx = 11/llt (8) for every y e R 1, by the translation-in variance of Lebesgue measure. Thus F e L1(R2), and Fubini's theorem implies that the integral (2) exists for almost all x e R1 and that he L1(R1). Finally, II hil t = L: I h(x) I dx < L: dx L: I F(x, y) I dy = I: dy J_: I F(x, y) I dx = II/ lltllgll1, by (7). This gives (3), and completes the proof. Convolutions will play an important role in Chap. 9. /Ill 172 REAL AND COMPLEX ANALYSIS Distribution Functions 8.15 Definition Let Jl be a a-finite positive measure on some a-algebra in some set X. Let f: Xr [0, oo] be measurable. The function that assigns to each t E [0, oo) the number Jl { f > t} = Jl( {X E X : f (X) > t}) (1) is called the distribution function off It is clearly a monotonic (nonincreasing) function of t and is therefore Borel measurable. One reason for introducing distribution functions is that they make it possible to replace integrals over X by integrals over [0, oo); the formula . (2) is the special case qJ(t) = t of our next theorem. This will then be used to derive an I!'-property of the maximal functions that were introduced in Chap. 7. 8.16 Theorem Suppose that f and Jl are as above, that qJ: [0, oo] r [0, oo] is monotonic, absolutely continuous on [0, T] for every T < oo, and that qJ(O) = 0 and qJ(t) ͢ qJ( oo) as t r oo. Then (1) PROOF Let E be the set of all (x, t) e X x [0, oo) where f(x) > t. When f is simple, then E is a union of finitely many measurable rectangles, and is there fore measurable. In the general case, the measurability of E follows via the standard approximation off by simple functions (Theorem 1. 17). As in Sec. 8.1, put Et = {X E X : ( x, t) E E} (0 < t < oo). (2) The distribution function off is then J-L(E1) = l XE(x, t) dJ-L(x). (3) The right side of (1) is therefore f" J-L(E1)ql(t) dt = l dJ-L(x) f" XE(x, t)ql(t) dt, (4) by Fubini's theorem. INTEGRATION ON PRODUCT SPACES 173 For each x e X, XE(x, t) = 1 if t f(x). The inner integral in (4) is therefore rf(x) Jo ql(t) dt = cp(f(x)) (5) by Theorem 7.20. Now (1) follows from (4) and (5). Ill/ 8.17 Recall now that the maximal function Mf lies in weak L1 when f e L1(Rk) (Theorem 7.4). We also have the trivial estimate IIM/IIoo < 11/lloo (1) valid for all f e L00(Rk). A technique invented by Marcinkiewicz makes it possible to " interpolate " between these two extremes and to prove the following theorem of Hardy and Littlewood (which fails when p = 1 ; see Exercise 22, Chap. 7). 8.18 Theorem If 1 < p < oo and f e I!'(Rk) then Mf e I!'(Rk). PROOF Since Mf = M( I f I ) we may assume, without loss of generality, that f> 0. Theorem 7.4 shows that there is a constant A, depending only on the dimension k, such that A m{Mg > t} <- llgllt t (1) for every g e IJ(Rk). Here, and in the rest of this proof, m = mk , the Lebesgue measure on Rk. Pick a constant c, 0 < c < 1, which will be specified later so as to mini mize a certain upper bound. For each t e (0, oo ), split/into a sum where ( ) _ {f(x) 9t X - 0 iff(x) > ct if f(x) <ct. Then 0 < ht(x) < ct for every x e Rk. Hence ht e L00, Mht < ct, and Mf< Mgt+ Mht t for some x, it follows that (Mgt)(x) > (1 -c)t. Setting Et = {/> ct}, (5), (1), and (3) imply that m{Mf> t} < m{Mgt > (1 -c)t} < (l i c)t llg,llt = (1 j c)t L/ dm. (2) (3) (4) (5) 174 REAL AND COMPLEX ANALYSIS We now use Theorem 8.16, with X= Rk, J1 = m, qJ(t) = tP, to calculate [ (M.f)P dm = p [00m{M.f> t}tp-1 dt < Ap [00tp-2 dt [ fdm JRk Jo 1 -c Jo JEr = fdm tP-2 dt= fPdm. Ap i if/c Apcl-p i 1 - C Rk 0 ( 1 - C )(p - 1) Rk This proves the theorem. However, to get a good constant, let us choose c so as to minimize that last expression. This happens when c = (p- 1)/p = 1/q, where q is the exponent conjugate to p. For this c, c 1 -P = 1 + < e, ( 1 )p-1 p- 1 and the preceding computation yields (6) Ill/ Note that cPÒ 1 as pÒ oo, which agrees with formula 8.17(1), and that cpr.oo as pÒ 1. Exercises 1 Find a monotone class 9Jl in R 1 which is not a u-algebra, even though R 1 E 9Jl and R 1 - A E 9Jl for every A E IDl 2 Suppose f is a Lebesgue measurable nonnegative real function on R 1 and A(f) is the ordinate set of f This is the set of all points (x, y) E R2 for which 0 < y < f(x). (a) Is it true that A(f) is Lebesgue measurable, in the two-dimensional sense? (b) If the answer to (a) is affirmative, is the integral off over R1 equal to the measure of A(f)? (c) Is the graph off a measurable subset of R2? (d) If the answer to (c) is affirmative, is the measure of the graph equal to zero? 3 Find an example of a positive continuous function fin the open unit square in R2, whose integral (relative to Lebesgue measure) is finite but such that cp(x) (in the notation of Theorem 8.8) is infinite for some x E (0, 1 ). 4 Suppose 1 < p < oo,f E L1(R1), and g E IJ'(R1). (a) Imitate the proof of Theorem 8.14 to show that the integral defining (f • g)(x) exists for almost all x, that f • g E I!(R 1 ), and that (b) Show that equality can hold in (a) if p = 1 and if p = oo, and find the conditions under which this happens. (c) Assume 1 < p < oo, and equality holds in (a). Show that then either f = 0 a.e. or g = 0 a.e. (d) Assume 1 < p < oo, E > 0, and show that there existf E L1(R1) and g E IJ'(R1) such that INTEGRATION ON PRODUCT SPACES 175 S Let M be the Banach space of all complex Borel measures on R 1• The norm in M is ll,u II = 1 ,u I (R 1 ). Associate to each Borel set E c R 1 the set E2 = {(x, y): x + y E E} c R2• If ,u and A. e M, define their convolution ,u • A. to be the set function given by for every Borel set E c R 1 ; ,u x A. is as in Definition 8.7. (a) Prove that ,u • A. E M and that ll,u • A.ll < ll,ull IIA.II. (b) Prove that ,u • A. is the unique v e M such that J f dv = JJ f(x + y) dJl(X) dA(y) for every f e C0(R 1 ). (All integrals extend over R 1.) (c) Prove that convolution in M is commutative, associative, and distributive with respect to addition. (d) Prove the formula (Jl d)(E) = J Jl(E - t) dA(t) for every ,u and A. E M and every Borel set E. Here E - t = {x - t: x e E}. (e) Define ,u to be discrete if ,u is concentrated on a countable set; define ,u to be continuous if ,u( { x}) = 0 for every point x e R 1 ; let m be Lebesgue measure on R 1 (note that m ¢ M). Prove that ,u • A. is discrete if both ,u and A. are discrete, that ,u • A. is continuous if ,u is continuous and A. e M, and that ,u • A. ¥ m if ,u ¥ m. (f) Assume d,u = f dm, dA. = g dm, f e L1(R 1 ), and g E IJ(R 1 ), and prove that d(,u • A.) = (f • g) dm. (g) Properties (a) and (c) show that the Banach space M is what one calls a commutative Banach algebra. Show that (e) and (f) imply that the set of all discrete measures in M is a subalgebra of M, that the continuous measures form an ideal in M, and that the absolutely continuous measures (relative to m) form an ideal in M which is isomorphic (as an algebra) to IJ(R 1 ). (h) Show that M has a unit, i.e., show that there exists a e M such that • ,u = ,u for all ,u E M. (i) Only two properties of R1 have been used in this discussion : R1 is a commutative group (under addition), and there exists a translation invariant Borel measure m on R 1 which is not identi cally 0 and which is finite on all compact subsets of R 1• Show that the same results hold if R 1 is replaced by Rk or by T (the unit circle) or by Tk (the k-dimensional torus, the cartesian product of k copies of T), as soon as the definitions are properly formulated. 6 (Polar coordinates in Rk.) Let Sk _ 1 be the unit sphere in Rk, i.e., the set of all u e Rk whose distance from the origin 0 is 1. Show that every x e Rk, except for x = 0, has a unique representation of the form x = ru, where r is a positive real number and u E Sk- t· Thus Rk - {0} may be regarded as the cartesian product (0, 00) X Sk- 1• Let mk be Lebesgue measure on Rk, and define a measure uk_ 1 on Sk_ 1 as follows: If A c Sk- 1 and A is a Borel set, let A be the set of all points ru, where 0 < r < 1 and u e A, and define 176 REAL AND COMPLEX ANALYSIS Prove that the formula is valid for every nonnegative Borel function f on Rk. Check that this coincides with familiar results when k = 2 and when k = 3. Suggestion : If 0 < r1 < r2 and if A is an open subset of Sk_ 1, let E be the set of all ru with r 1 < r < r 2 , u E A, and verify that the formula holds for the characteristic function of E. Pass from these to characteristic functions of Borel sets in Rk. 7 Suppose (X, !/', J.J.) and (Y, !/, A.) are a-finite measure spaces, and suppose t/1 is a measure on !/' x ff such that t/1( A x B) = JJ.( A )A.( B) whenever A E !/' and B E ff. Prove that then t/f(E) = (JJ. x A.)(E) for every E E !/' x ff. 8 (a) Suppose f is a real function on R2 such that each sectionf x is Borel measurable and each section fY is continuous. Prove thatfis Borel measurable on R2• Note the contrast between this and Example 8.9(c). (b) Suppose g is a real function on Rk which is continuous in each of the k variables separately. More explicitly, for every choice of x2 , . . . , xk , the mapping x1 ---+ g(x1, x2 , . . . , xk) is continuous, etc. Prove that g is a Borel function. Hint: If (i - 1)/n = ai- l < x < ai = i/n, put 9 Suppose E is a dense set in R1 andfis a real function on R2 such that (a)fx is Lebesgue measurable for each x E E and (b) fY is continuous for almost all y E R 1• Prove that f is Lebesgue measurable on R2. 10 Suppose f is a real function on R2, f x is Lebesgue measurable for each x, and fY is continuous for each y. Suppose g : R 1 ---+ R 1 is Lebesgue measurable, and put h(y) = f(g(y), y). Prove that h is Lebesgue measurable on R 1. Hint : Define f.. as in Exercise 8, put hn(Y) = f..
(g(y), y), show that each hn is measurable, and that hn(Y)---+ h(y). 1 1 Let Blk be the a-algebra of all Borel sets in Rk. Prove that Blm+n = 31m x Bin . This is relevant in Theorem 8.14. 12 Use Fubini's theorem and the relation to prove that 1 iCX> - = e -xt dt X 0 (x > 0) i A sin x n lim dx = -. A-+oo 0 X 2 13. If J1. is a complex measure on a a-algebra IDl, show that every set E E 9Jl has a subset A for which 1 I J.J.(A) I > - I J.J. I (E). 1C INTEGRATION ON PRODUCT SPACES 177 Suggestion : There is a measurable real function () so that df.J. = ei8 d I J.J. I . Let A« be the subset of E where cos (() - a) > 0, show that Re (e -••JJ(AJ] = 1 cos+ (8 - tX) d I Il L and integrate with respect to a (as in Lemma 6.3). Show, by an example, that 1/n is the best constant in this inequality. 14 Complete the following proof of Hardy's inequality (Chap. 3, Exercise 14): Suppose f> 0 on (0, oo),f E I!, 1 < p < oo, and 1 ix F(x) = -f(t) dt. X 0 Write xF(x) = JӲ f(t)t«t -« dt, where 0 -:::: a < 1/q, use Holder's inequality to get an upper bound for F(x)P, and integrate to obtain f" P(x) dx < (1 - tXq)1 -"(tXp) - 1 f" fP(t) dt. Show that the best choice of ct yields r P(x) dx < c Õ lr r JP(t) dt. IS Put q>(t) = 1 - cos t if 0 < t < 2n, q>(t) = 0 for all other real t. For - oo < x < oo, define f(x) = 1, g(x) = q>'(x), h(x) =f.., qJ(t) dt. Verify the following statements about convolutions of these functions: (i) (f • g)(x) = 0 for all x. (ii) (g • h)(x) = (q> • q>)(x) > 0 on (0, 4n). (iii) Therefore (f • g) • h = 0, whereas! • (g • h) is a positive constant. But convolution is supposedly associative, by Fubini's theorem (Exercise 5(c)). What went wrong? 16 Prove the following analogue of Minkowski's inequality, forf > 0: {ff J<x. y) d.?.(y)T dJl(x)} 1,. < f[f f"<x. y) dJJ(x)J' . d.?.(y). Supply the required hypotheses. (Many further developments of this theme may be found in .) CHAPTER NINE FOURIER TRANSFORMS Formal Properties 9.1 Definitions In this chapter we shall depart from the previous notation and use the letter m not for Lebesgue measure on R 1 but for Lebesgue measure divided by fo,. This convention simplifies the appearance of results such as the inversion theorem and the Plancherel theorem. Accord ingly, we shall use the notation foo 1 Joo _ oo f(x) dm(x) = fo, _ oo f(x) dx, where dx refers to ordinary Lebesgue measure, and we define and {fro rjp II/IlP = -oolf(x)IP dm(x) (f g)(x) = L:f(x-y)g(y) dm(y) /(t) = L:f(x)e-ixt dm(x) (1 < p < oo), (x E R1), (t E R1 ). (1) (2) (3) (4) Throughout this chapter, we shall write I! in place of I!(R 1 ), and C0 will denote the space of all continuous functions on R 1 which vanish at infinity. If/ E L\ the integral (4) is well defined for every real t. The function/is called the Fourier transform off Note that the term " Fourier transform " is also applied " to the mapping which takes f to f 178 FOURIER TRANSFORMS 179 The formal properties which are listed in Theorem 9.2 depend intimately on the translation-invariance of m and on the fact that for each real ex the mapping x q eiax is a character of the additive group R 1. By definition, a function qJ is a character of R 1 if I ({J( t) I = 1 and if ({J(S + t) = qJ(S)qJ(t) (5) for all real s and t. In other words, qJ is to be a homomorphism of the additive group R1 into the multiplicative group of the complex numbers of absolute value 1. We shall see later (in the proof of Theorem 9.23) that every continuous charac ter of R1 is given by an exponential. 9.2 Theorem Suppose f E L1, and ex and A. are real numbers. (a) If g(x) = f(x)eiax, then g(t) = /(t -ex). (b) If g(x) = f(x-ex), then g(t) = /(t)e-iat. (c) If g E L1 and h = f g, then h(t) = /(t)g(t). Thus the Fourier transform converts multiplication by a ,character into translation, and vice versa, and it converts convolutions to pointwise products. (d) If g(x) = f(-x), then g(t) = f(t). (e) If g(x) = f(x/ A.) and A. > 0, then g(t) = A./(A.t). (f) Ifg(x) = -ixf(x) and g E L1, then/is diff erentiable and/'(t) = g(t). PROOF (a), (b), (d), and (e) are proved by direct substitution into formula 9.1(4). The proof of (c) is an application of Fubini's theorem (see Theorem 8.14 for the required measurability proof): h(t) = s_: e-itx dm(x) s_: f(x -y)g(y) dm(y) = s_:g(y)e-ity dm(y) s_:f(x-y)e-it(x-y) dm(x) = s_:g(y)e-ity dm(y) s_:f(x)e-itx dm(x) = g(t)/(t). Note how the translation-invariance of m was used. To prove (f), note that /(s) -/(t) = foo f(x)e-ixtqJ(X, s-t) dm(x) s- t -oo (s ¥= t), (1) 180 REAL AND COMPLEX ANALYSIS where cp(x, u) = (e-ixu- 1)/u. Since I cp(x, u) I < I xI for all real u # 0 and since cp(x, u)r -ix as ur 0, the dominated convergence theorem applies to (1), if s tends to t, and we conclude that /'(t) = -i I-: xf(x)e-ixt dm(x). (2) /Ill 9.3 Remarks (a) In the preceding proof, the appeal to the dominated convergence theorem may seem to be illegitimate since the dominated convergence theorem deals only with countable sequences of functions. However, it does enable us to conclude that 1. /(sn) -/(t) . Joo f( ) -ixt d (t) tm = - z x x e m n - 00 sn -t - 00 for every sequence { sn} which converges to t, and this says exactly that 1. /(s) -/(t) . Joo f( ) -ixt d (t) 1m = - z x x e m . s-t S -t -oo We shall encounter similar situations again, and shall apply con vergence theorems to them without further comment. (b) Theorem 9.2(b) shows that the Fourier transform o£ [f(x + ex) -f(x)]/cx . IS " eia.t - 1 f(t) . C( This suggests that an analogue of Theorem 9.2(/) should 'be true under " certain conditions, namely, that the Fourier transform of f' is itf(t). If f e I!, f' e L1, and iff is the indefinite integral off', the result is easily established by an integration by parts. We leave this, and some related results, as exercises. The fact that the Fourier transform converts differ entiation to multiplication by ti makes the Fourier transform a useful tool in the study of differential equations. ,.fhe Inversion Theorem 9.4 We have just seen that certain operations on functions correspond nicely to operations on their Fourier transforms. The usefulness and interest of this corre spondence will of course be enhanced if there is a way of returning from the transforms to the functions, that is to say, if there is an inversion formula. FOURIER TRANSFORMS 181 Let us see what such a formula might look like, by analogy with Fourier series. If 1 f1t . en=- f(x)e-lnX dx 2n 1t (n E Z), (1) then the inversion formula is 00 t = L en einx. (2) - oo We know that (2) holds, in the sense of !!-convergence, iff e I!(T). We also know that (2) does not necessarily hold in the sense of pointwise convergence, even iff is continuous. Suppose now thatf e L1(T), that {en} is given by (1), and that 00 L lcnl < 00 . (3) - oo Put 00 g(x) = L en einx. (4) - oo By (3), the series in (4) converges uniformly (hence g is continuous), and the Fourier coefficients of g are easily computed : 1 f1t . 1 f1t { 00 • } • - g(x)e-lkx dx = -L ene'"x e-lkX dx 2n - 1t 2n - 1t n=- 00 00 1 f1t . -L en- e'x dx n=- 00 2n -1t (5) Thusfand g have the same Fourier coefficients. This implies/= g a.e., so the Fourier series off converges to f(x) a.e. The analogous assumptions in the context of Fourier transforms are that f E L1 and/ E L1, and we might then expect that a formula like f(x) = s: /(t)eitx dm(t) (6) is valid. Certainly, iff e Ll, the right side of (6) is well defined; call it g(x); but if we want to argue as in (5), we run into the integral f ao ei(t -s)x dx, -- oo (7) which is meaningless as it stands. Thus even under the strong assumption that / e L1, a proof of (6) (which is true) has to proceed over a more devious route. 182 REAL AND COMPLEX ANALYSIS [It should be mentioned that (6) may hold even if J ¢ L1, if the integral over ( -oo, oo) is interpreted as the limit, as A --+ oo, of integrals over (-A, A). (Analogue: a series may converge without converging absolutely.) We shall not go in to this.] 9.5 Theorem For any function f on R1 and every y e R1, let fY be the translate off defined by fy(x) = f(x -y) (1) If 1 < p < oo and iff e IJ', the mapping y-+fy (2) is a uniformly continuous mapping of R1 into IJ'(R1). PROOF Fix E > 0. Since f e IJ', there exists a continuous function g whose support lies in a bounded interval [-A, A], such that ll f- gllp < € (Theorem 3.14). The uniform continuity of g shows that there exists a b e (0, A) such that Is -t I < b implies I g(s) -g(t) I < (3A) - l tp€. If I s -t I < b, it follows that r: g(x -s) -g(x -t) IP dx < (3A) - lEP(2A + b) < EP, so that llgs - gtiiP < €. ࡓote that IJ'-norms (relative to Lebesgue measure) are translation-invariant : II f li P = ll .fsllp · Thus ll .fs -.ftllp < ll .fs·- gsllp + llgs - gtll p + llgt -.ftll p = ll(f- g)sll p + llgs - gtllp + ll(g -f)tllp < 3€ whenever I s -t I < b. This completes the proof. 9.6 Theorem Iff e L1, then J e C0 and II! II oo < II f ll 1. PROOF The inequality (1) is obvious from 9.1(4). If tn-+ t, then Ill/ (1) (2) FOURIER TRANSFORMS 183 The integrand is bounded by 2 1 f(x) I and tends to 0 for every x, as n----+ oo . Hence /(tn)----+ /(t), by the dominated convergence theorem. Thus/ is contin uous. Since eni = - 1 , 9.1(4) gives /(t) = -L:f(x)e-it(x+nJt> dm(x) = -L:f(x -n/t)e-itx dm(x). (3) Hence 2/(t) = L: {f(x) -!( x- ;) }e-itx dm(x), so that which tends to 0 as t----+ + oo , by Theorem 9.5. (4) (5) /Ill 9.7 A Pair of Auxiliary Functions In the proof of the inversion theorem it will be convenient to know a positive function H which has a positive Fourier transform whose integral is easily calculated. Among the many possibilities we choose one which is of interest in connection with harmonic functions in a half plane. (See Exercise 25, Chap. 1 1.) Put H(t) = e-ltl (1) and define h;,(x) = L: H(A.t)eitx dm(t) (A> 0). (2) A simple computation gives (3) and hence L: h;,(x) dm(x) = 1. (4) Note also that 0 < H(t) < 1 and that H(At)----+ 1 as A-+ 0. 9.8 Proposition Iff e L1, then (f h;,Xx) = L: H(A.t)/(t)eixt dm(t). 184 REAL AND COMPLEX ANALYSIS PROOF This is a simple application of Fubini's theorem. (f h.,)(x) = I:f(x -y) dm(y) I: H(A.t)eity dm(t) = I: H(A.t) dm(t) I:f(x -y)eity dm(y) =I: H(A.t) dm(t) I:f(y)eit(x-y) dm(y) = I: H(A.t)](t)eitx dm(t). 9.9 Theorem If g E L00 and g is continuous at a point x, then lim (g h;.)(x) = g(x) . .A.-o PROOF On account of9.7(4), we have (g h.A.)(x) -g(x) = I: [g(x -y) -g(x)]h.A.(y) dm(y) =I: [g(x-y)-g(x)]r1h{h) dm(y) =I: [g(x-A.s)-g(x)]h1(s) dm(s). Ill/ (1) The last integrand is dominated by 211911 oo h1(s) and converges to 0 pointwise for every s, as A -+-0. Hence (1) follows from the dominated convergence theorem. I I I I 9.10 Theorem If 1 < p < oo and f E I!, then lim llf h.A.-fllp=O. (1) .A.-o The cases p = 1 and p = 2 will be the ones of interest to us, but the general case is no harder to prove. PROOF Since h.A. E IJ, where q is the exponent conjugate to p, (f h.A.)(x) is defined for every x. (In fact, f h.A. is continuous ; see Exercise 8.) Because of 9.7(4) we have (f h.A.)(x) -f(x) = I: [f(x -y) -f(x)]h.A.(Y) dm(y) (2) FOURIER TRANSFORMS 185 and Theorem 3.3 gives I (f h;.)(x) -f(x) IP < L: I f(x -y) -f(x) IPh;.(y) dm(y). (3) Integrate (3) with respect to x and apply Fubini's theorem: llf h;. -/11ʿ < s_:llf, -/llʿh;.(Y) dm(y). (4) If g(y) = 11/y -/IIʿ, then g is bounded and continuous, by Theorem 9.5, and g(O) = 0. Hence the right side of (4) tends to 0 as A.r 0, by Theorem 9.9. /Ill 9.11 The Inversion Theorem Iff E L1 and J E L1, and if. g(x) = 1: /(t)eixt dm(t) then g e C0 andf(x) = g(x) a.e. PROOF By Proposition 9.8, (f hA)(x) = L: H(A.t)/(t)eixt dm(t). (1) (2) The integrands on the right side of (2) are bounded by 1/(t) I, and since H(A.t) ͢ 1 as A.r 0, the right side of (2) converges to g(x), for every x e R 1, by the dominated convergence theorem. If we combine Theorems 9.10 and 3.12, we see that there is a sequence { A.n} such that. Anr 0 and lim (/ h;.)(x) = f(x) a.e. (3) Hencef(x) = g(x) a.e. That g e C0 follows from Theorem 9.6. Ill/ 9.12 The Uniqueness Theorem If fe L1 and /(t) = 0 for all t E R1, then f(x) = 0 a.e. PROOF Since J = 0 we have J e L1, and the result follows from the inversion theorem. I I I I The Plancherel Theorem Since the Lebesgue measure of R 1 is infinite, 13 is not a subset of L1, and the definition of the Fourier transform by formula 9.1(4) is therefore not directly applicable to every f e L2• The definition does apply, however, iff E I! n 13, and it turns out that then J E L2• In fact, 11/112 = II f 112! This isometry of L1 n 13 into L2 extends to an isometry of L2 onto 13, and this extension defines the Fourier 186 REAL AND COMPLEX ANALYSIS transform (sometimes called the Plancherel transform) of every f E I3. The resulting L2-theory has in fact a great deal more symmetry than is the case in L1. In L2,fand/play exactly the same role. 9.13 Theorem One can associate to each f E L2 a function J E L2 so that the following properties hold: (a) Jffe L1 n 13, then/is the previously defined Fourier transform off (b) For everyfe L2, ll/ ll 2 = ll f ll 2 · (c) The mappingfr /is a Hilbert space isomorphism of L2 onto L2• (d) The following symmetric relation exists between! and/: If <pit) = f_AAf(x)e-ixt dm(x) and r/t A(x) = f_AA/(t)eixt dm(t), then ll <fJ A -/ll 2 r 0 and ll t/1 A -f 11 2 r 0 as Ar oo . Note: Since L1 n 13 is dense in 13, properties (a) and (b) determine the mapping fr /uniquely. Property (d) may be called the L2 inversion theorem. PROOF Our first objective is the relation (1) We fix/ E L1 n L2, put /(x) = f(-x), and define g = f J Then g(x) = I-:f(x-y)f(-y) dm(y) = I-:f(x + y)f(y) dm(y), (2) or g(x) = (f-x,f), (3) where the inner product is taken in the Hilbert space L2 and f-x denotes a translate off, as in Theorem 9.5. By that theorem, xr f-x is a continuous mapping of R1 into L2, and the continuity of the inner product (Theorem 4.6) therefore implies that g is a continuous function. The Schwarz inequality shows that I g( X) I < II f-X 11 2 II f 11 2 = II f II C so that g is bounded. Also, g E L1 since f E L1 and J E L1• Since g E L1, we may apply Proposition 9.8 : (g h;.XO) = I: H(A.t)l dm(t). Since g is continuous and bounded, Theorem 9.9 shows that lim (g h;J(O) = g(O) = II f II C . .A. - o (4) (5) (6) FOURIER TRANSFORMS 187 Theorem 9.2(d) shows that g = 1/12 > 0, and since H(At) increases to 1 as Ar 0, the monotone convergence theorem gives (7) Now (5), (6), and (7) shows that J E L2 and that (1) holds. This was the crux of the proof. Let Y be the space of all Fourier transforms J of functions f E L1 n I3. By (1), Y c /3. We claim that Y is dense in !.3, i.e., that y1. = {0}. The functions xq eiaxH(Ax) are in L1 n L2, for all real C( and all A> 0. Their Fourier transforms h;.(C(-t) = L: eiru:H(A.x)e-ixt dm(x) are therefore in Y. If w E L2, w E Y l., it follows that (h;. W)(1X) = L: h;.(IX -t)w(t) dm(t) = 0 for all C(. Hence w = 0, by Theorem 9.1 0, and therefore Y is dense in L2• " (8) (9) Let us introduce the temporary notation f for f. From what has been proved so far, we see that is an !.3-isometry from one dense subspace of 1.3, namely L1 n L2, onto another, namely Y. Elementary Cauchy sequence argu ments (compare with Lemma 4.16) imply therefore that extends to an isometry a> of L2 onto L2• If we write /for d'> f, we obtain properties (a) and (b). Property (c) follows from (b), as in the proof of Theorem 4.18. The Par seval formula L:f(x)g(x) dm(x) = L:/(t)[J(t) dm(t) holds therefore for all f E L2 and g E L2• (10) To prove (d), let kA be the characteristic function of [-A, A]. Then kAf E L1 n L2 iff E L2, and (/)A = (k A f)''' . Since II f-k A f 112 q 0 as Aq oo, it follows from (b) that 11/-qJ A 112 = II(/-k A f) ק112 } 0 as A q oo . The other half of (d) is proved the same way. 9.14 Theorem Iff e L2 and/ e IJ, then . f(x) = L:/(t)eixt dm(t) a.e. (11) (12) /Ill 188 REAL AND COMPLEX ANALYSIS PROOF This is corollary of Theorem 9. 13(d). Ill/ 9.15 Remark Iff E L1, formula 9.1(4) defines/(t) unambiguously for every t. If fe L2, the Plancherel theorem defines/ uniquely as an element of the Hilbert space L2, but as a point function /(t) is only determined almost everywhere. This is an important difference between the theory of Fourier transforms in L1 and in L2• The indeterminacy of /(t) as a point function will cause some difficulties in the problem to which we now turn. 9.16 Translation-Invariant Subspaces of L2 A subspace M of L2 is said to be translation-invariant iff E M implies that fa E M for every real ex, where fa(x) = f(x - ex). Translations have already played an important part in our study of Fourier transforms. We now pose a problem whose solution will afford an illus tration of how the Plancherel theorem can be used. (Other applications will occur in Chap. 19.) The problem is : Describe the closed translation-invariant sub spaces of L2• Let M be a closed translation-invariant subspace of L2, and let M be the image of M under the Fourier transform. Then M is closed (since the Fourier transform is an 13-isometry). If fa is a translate off, the Fourier transform of fa is lea, where ea(t) = e-iat; we proved this for fE I! in Theorem 9.2; the result extends to /3, as can be seen from Theorem 9.13(d). It follows that M is invariant under multiplication by ea ,for all ex E R 1. Let E be any measurable set in R 1. If M is the set of all <p E e which vanish a.e. on E, then M certainly is a subspace of L2, which is invariant under multipli cation by all ea (note that I ea I = 1, so <pea E L2 if <p E L2), and M is also closed. Proof: <p E M if and only if <p is orthogonal to every t/1 E L2 which vanishes a.e. on the complement of E. If M is the inverse image of this M, under the Fourier transform, then M is a space with the desired properties. One may now conjecture that every one of our spaces M is obtained in this manner, from a set E c R1• To prove this, we have to show that to every closed translation-invariant M c L2 there corresponds a set E c R 1 such that f E M if " and only if f(t) = 0 a.e. on E. The obvious way of constructing E from M is to " associate with each f E M the set E 1 consisting of all points at whichf(t) = 0, and to define E as the intersection of these sets E 1 . But this obvious attack runs into a serious difficulty: Each E 1 is defined only up to sets of measure 0. If { AJ is a countable collection of sets, each determined up to sets of measure 0, then n Ai is also determined up to sets of measure 0. But there are uncountably many f E M, so we lose all control over n E f . This difficulty disappears entirely if we think of our functions as elements of the Hilbert space 13, and not primarily as point functions. .. We shall now prove the conjecture. Let M be the image of a closed translation-invariant subspace M c L2, under the Fourier transform. Let P be the FOURIER TRANSFORMS 189 orthogonal projection of I3 onto M (Theorem 4.1 1): To each f e J3 there corre sponds a unique P f eM such that/- P fis orthogonal to M. Hence f- P f l_ Pg (1) and since M is invariant under multiplication by ea , we also have f - P f l_ (Pg)ea (2) If we recall how the inner product is defined in /3, we see that (2) is equivalent to (3) and this says that the Fourier transform of (f- P f) · Pg (4) is 0. The function (4) is the product of two E-functions, hence is in L1, and the uniqueness theorem for Fourier transforms shows now that the function (4) is 0 a.e. This remains true if Pg is replaced by Pg. Hence f · Pg = (P f) · (Pg) (5) Interchanging the roles off and g leads from (5) to f · Pg = g · P f (6) Now let g be a fixed positive function in L2 ; for instance, put g(t) = e -ltl. Define ( ) _ (Pg)(t) (/) t -g(t) . (7) (Pg)(t) may only be defined a.e. ; choose any one determination in (7). Now (6) becomes P f= cp · f (8) If f e M, then P f = f This says that P2 = P, and it follows that cp2 = <p, because <p2 . g = <p . Pg = p2g = Pg = <p . g. (9) Since <p2 = <p, we have <p = 0 or 1 a.e., and if we let E be the set of all t where <p(t) = 0, then M consists precisely of those f e L2 which are 0 a.e. on E, since f e M if and only if/= P f= <p · f We therefore obtain the following solution to our problem. 190 REAL AND COMPLEX ANALYSIS 9.17 Theorem Associate to each measurable set E c R1 the space ME of all f E I.3 such that/= 0 a. e. on E. Then ME is a closed translation-invariant sub space of I.3. Every closed translation-invariant subspace of I3 is ME for some E, and MA = M8 if and only if m((A -B) u (B -A)) = 0. The uniqueness statement is easily proved; we leave the details to the reader. The above problem can of course be posed in other function spaces. It has been studied in great detail in L1• The known results show that the situation is infinitely more complicated there than in /3. The Banach Algebra L1 9.18 Definition A Banach space A is said to be a Banach algebra if there is a multiplication defined in A which satisfies the inequality llxyll < llxll IIYII (x and y e A), (1) the associative law x(yz) = (xy)z, the distributive laws x(y + z) = xy + xz, (y + z)x = yx + zx (x, y, and z e A), (2) and the relation (cxx)y = x(cxy) = cx(xy) (3) where ex is any scalar. 9.19 Examples (a) Let A = C(X), where X is a compact Hausdorff space, with the supr.emum norm and the usual pointwise multiplication of functions: (fg)(x) = f(x)g(x). This is a commutative Banach algebra (fg = gf) with unit (the constant function 1). (b) C0(R1) is a commutative Banach algebra without unit, i.e., without an element u such that uf=ffor allfe C0(R1). (c) The set of all linear operators on Rk (or on any Banach space), with the operator norm as in Definition 5.3, and with addition and multiplication defined by (A + B)(x) = Ax + Bx, (AB)x = A(Bx), is a Banach algebra with unit which is not commutative when k > 1. (d) L1 is a Banach algebra if we define multiplication by convolution; since the norm inequality is satisfied. The associative law could be verified directly (an application of Fubini's theorem), but we can proceed as FOURIER TRANSFORMS 191 follows : We know that the Fourier transform off g is / · g, and we know that the mapping/-+ /is one-to-one. For every t e Rl, /(t)[g(t)h(t)] = [/(t)g(t)]h(t), by the associative law for complex numbers. It follows that f (g h) = (f g) h. In the same way we see immediately that f g = g f. The remaining requirements of Definition 9.18 are also easily seen to hold in L1• Thus L1 is a commutative Banach algebra. The Fourier transform is an algebra isomorphism of L1 into C0• Hence there is no f e L1 with J = 1, and therefore L1 has no unit. 9.20 Complex Homomorphisms The most important complex functions on a Banach algebra A are the homomorphisms of A into the complex field. These are precisely the linear functionals which also preserve multiplication, i.e., the _func tions cp such that cp(iJ.x + f3y) = iJ.cp(x) + f3cp(y), cp(xy) = cp(x)cp(y) for all x and y e A and all scalars iJ. and {3. Note that no boundedness assumption is made in this definition. It is a very interesting fact that this would be redundant: 9.21 Theorem If cp is a complex homomorphism on a Banach algebra A, then the norm of cp, as a linear functional, is at most 1. PROOF Assume, to get a contradiction, that I cp(x0) I > II x0 II for some x0 e A. Put A.= cp(x0), and put x = x0/A.. Then llxll < 1 and cp(x) = 1. Since llxnll < llxlln and llxll < 1, the elements 2 n S = -X-X -··· -X n (1) form a Cauchy sequence in A. Since A is complete, being a Banach space, there exists a y e A such that IIY -snll-+ 0, and it is easily seen that x + sn = xsn -1, so that x + y = xy. (2) Hence cp(x) + cp(y) = cp(x)cp(y), which is impossible since cp(x) = 1. /Ill 9.22 The Complex Homomorphisms of L1 Suppose cp is a complex homo morphism of L1, i.e., a linear functional (of norm at most 1, by Theorem 9.21) which also satisfies the relation cp(f g) = cp(f)cp(g) (1) 192 REAL AND COMPLEX ANALYSIS By Theorem 6.16, there exists a p e L00 such that <p(f) = t: f(x)P(x) dm(x) (2) We now exploit the relation (1) to see what else we can say about p. On the one hand, <p(f g) = t: (f g)(x)P(x) dm(x) On the other hand, = t:p(x) dm(x) f_0000j(x -y)g(y) dm(y) = f: g(y) dm(y) L:fy(x)p(x) dm(x) = t: g(y)<p(fy) dm(y). <p(f)<p(g) = <p(f) t: g(y)p(y) dm(y). (3) (4) Let us now assume that cp is not identically 0. Fix f e L1 so that cp(f) =F 0. Since the last integral in (3) is equal to the right side of (4) for every g e L1, the uniqueness assertion of Theorem 6.16 shows that cp(f)p(y) = cp(fy) (5) for almost all y. But y-+ /y is a continuous mapping of R1 into L1 (Theorem 9.5) and <p is continuous on L1• Hence the right side of (5) is a continuous function of y, and we may assume [by changing p(y) on a set of measure 0 if necessary, which does not affect (2)] that P is continuous. If we replace y by x + y and then f by fx in ( 5), we obtain cp(f)P(x + Y) = cp(fx + y) = cp((fx)y) = cp(fx)P(y) = 0 such that Ip(y) dy = c ¥= o. (7) FOURIER TRANSFORMS 193 Then ,,«S [«S ix + tS cP(x) = Jo p(y)p(x) dy = Jo p(y + x) dy = x p(y) dy. (8) Since p is continuous, the last integral is a differentiable function of x; hence (8) shows that p is differentiable. Differentiate (6) with respect to y, then put y = 0; the result is P'(x) = AP(x), A = P'(O). (9) Hence the derivative of p(x)e-Ax is 0, and since p(O) = 1, we obtain (10) But p is bounded on R1• Therefore A must be pure imaginary, and we conclude: There exists a t e R 1 such that p(x) = e-itx. (1 1) We have thus arrived at the Fourier transform. 9.23 Theorem To every complex homomorphism 0. Prove that 1 /(y) I < /(0) for every y #- 0. 2 Compute the Fourier transform of the characteristic function of an interval. For n = 1, 2, 3, ... , let gn be the characteristic function of [ -n, n], let h be the characteristic function of [ - 1, 1], and compute gn • h explicitly. (The graph is piecewise linear.) Show that gn • h is the Fourier transform of a function!,. E L1 ; except for a multiplicative constant, . . SID X Sln nx f,.(x) = __ x _ 2 __ Show that 11fnll 1 ---+ oo and conclude that the mapping/---+ /maps IJ into a proper subset of C0 • Show, however, that the range of this mapping is dense in C0 • 3 Find where A. is a positive constant. fA sin A.t . lim e•tx dt A .... oo -A t ( - oo < x < oo) 4 Give examples off e I3 such thatf¢ L1 but/ e L1• Under what circumstances can this happen? 194 REAL AND COMPLEX ANALYSIS S If f E I! and J I t./(t) I dm(t) < oo , prove that f coincides a.e. with a differentiable function whose derivative is i 1: tf(t)e'"' dm(t). 6 Suppose f E L1, f is differentiable almost Ȩverywhere, and f' E I!. Does it follow that the Fourier transform off' is tlf(t)? 7 Let S be the class of all functions f on R 1 which have the following property : f is infinitely differen tiable, and there are numbers Amif) < oo , for m and n = 0, 1, 2, . . . , such that Here D is the ordinary differentiation operator. Prove that the Fourier transform maps S onto S. Find examples of members of S. 8 If p and q are conjugate exponents, f E I!, g E I!J, and h = f • g, prove that h is uniformly contin-uous. If also 1 < p < oo , then h E C0 ; show that this fails for some f E L1, g E L 00• • 9 Suppose 1 < p < oo,f E I!, and lx+ l g(x) = x f(t) dt. Prove that g E C0 • What can you say about g iff E L00 ? 10 Let coo be the class of all infinitely differentiable complex functions on R 1, and let CC: consist of all g E coo whose supports are compact. Show that CC: does not consist of 0 alone. Let L1uc be the class of all f which belong to I! locally; that is, f E L1vc provided that f is measur able and J1 I f I < oo for every bounded interval /. Iff E Lwc and g E CC:, prove thatf • g E C00• Prove that there are sequences {gn} in CC: such that as n-+ oo , for every f E L1• (Compare Theorem 9.10.) Prove that {gn} can also be so chosen that (f • gn)(x)-+ f(x) a.e., for every f E L11oc ; in fact, for suitable {gn} the convergence occurs at every point x at whichfis the derivative of its indefinite integral. Prove that (f • h;,)(x)-+f(x) a.e. iff e I!, as A. -+ 0, and that f • h;. E coo, although h;. does not have compact support. (h;. is defined in Sec. 9.7.) 11 Find conditions onfand/or /which ensure the correctness of the following formal argument: If and q>(t) = -f(x)e-ux dx 1 J oo . 2n _ oo 00 F(x) = L f(x + 2kn) k= - 00 then F is periodic, with period 2n, the nth Fourier coefficient of F is q>(n), hence F(x) = L q>(n)einx. In particular, More generally, 00 00 L f(2kn) = L q>(n). k= - oo n= - oo 00 00 L f(kP> = ex L q>(ncx) k=·- oo n= -oo if et > 0, P > 0, etP = 2n. () FOURIER TRANSFORMS 195 What does () say about the limit, as IX ---+ 0, of the right-hand side (for " nice " functions, of course)? Is this in agreement with the inversion theorem? [() is known as the Poisson summation formula.] 12 Take f(x) = e -lxl in Exercise 1 1 and derive the identity e21ra. + 1 1 oo IX e21ta. - 1 = n n=; CX> rx.l + n2 • 13 If 0 < c < oo, define .fc(x) = exp (-cx2). (a) Computefc . Hint: If cp = fc , an integration by parts gives 2ccp'(t) + tcp(t) = 0. (b) Show that there is one (and only one) c for which fc = fc . (c) Show that .fa • !, = yfc ; find y and c explicitly in terms of a and b. (d) Take f = fc in Exercise 1 1. What is the resulting identity? 14 The Fourier transform can be defined for f E I!(Rk) by J(y) = r f(x)e-ix·y dmk(x) JRk where X . y = L ei '1i if X = (e1, . . . , ek), y = ('71, . . . , '1k), and mk is Lebesgue measure on Rk, divided by (2n)kfl for convenience. Prove the inversion theorem and the Plancherel theorem in this context, as well as the analogue of Theorem 9.23. 15 Iff E L1(Rk), A is a linear operator on Rk, and g(x) = f(Ax), how is g related to j? Iff is invariant under rotations, i.e., iff(x) depends only on the euclidean distance of x from the origin, prove that the same is true of J 16 The Laplacian of a function f on Rk is provided the partial derivatives exist. What is the relation between/ and g if g = L\ f and all necessary integrability conditions are satisfied? It is clear that the Laplacian commutes with translations. Prove that it also commutes with rotations, i.e., that L(f o A) = (Af) o A whenever fhas continuous second derivatives and A is a rotation of Rk. (Show that it is enough to do this under the additional assumption thatfhas compact support.) 17 Show that every Lebesgue measurable character of R1 is continuous. Do the same for Rk. (Adapt part of the proof of Theorem 9.23.) Compare with Exercise 18. 18 Show (with the aid of the Hausdorff maximality theorem) that there exist real discontinuous func tions f on R 1 such that f(x + y) = f(x) + f(y) for all x and y e R 1• Show that if (1) holds andfis Lebesgue measurable, thenfis continuous. Show that if (1) holds and the graph off is not dense in the plane, thenfis continuous. Find all continuous functions which satisfy (1). (1) 19 Suppose A and B are measurable subsets of R 1, having finite positive measure. Show that the convolution XA • XB is continuous and not identically 0. Use this to prove that A + B contains a segment. (A different proof was suggested in Exercise 5, Chap. 7.) CHAPTE R TEN ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS Complex Differentiation We shall now study complex functions defined in subsets of the complex plane. It will be convenient to adopt some standard notations which will be used through out the rest of this book. 196 10.1 Definitions If r > 0 and a is a complex number, D( a; r) = { z: I z -a I < r} (1) is the open circular disc with center at a and radius r. D(a; r) is the closure of D(a; r), and D'(a; r) = {z: 0 < I z -a I < r} (2) is the punctured disc with center at a and radius r. A set E in a topological space X is said to be not connected if E is the union of two nonempty sets A and B such that A n B = 0 = A n B. (3) If A and B are as above, and if V and W are the complements of A and B, respectively, it follows that A c W and B c V. Hence E c V u W, E n V =F 0, E n W =F 0, E n V n W = 0. (4) Conversely, if open sets V and W exist such that (4) holds, it is easy to see that E is not connected, by taking A = E n W, B = E n V. If E is closed and not connected, then (3) shows that E is the union of two disjoint nonempty closed sets; for if A c A u B and A n B = 0, then A = A. ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 197 If E is open and not connected, then (4) shows that E is the union of two disjoint nonempty open sets, namely E n V and E n W. Each set consisting of a single point is obviously connected. If x e E, the family x of all connected subsets of E that contain x is therefore not empty. The union of all members of x is easily seen to be connected, and to be a maximal connected subset of E. These sets are called the components of E. Any two components of E are thus disjoint, and E is the union of its components. By a region we shall mean a nonempty connected open subset of the complex plane. Since each open set Q in the plane is a union of discs, and since all discs are connected, each component of Q is open. Every plane open set is thus a union of disjoint regions. The letter Q will from now on denote a plane open set. 10.2 Definition Suppose f is a complex function defined in Q. If z0 e Q and if 1. f(z) -f(zo) tm ---ࡎ (1) z - z0 exists, we denote this limit by f'(z0) and call it the derivative off at z0 • If/'(z0) exists for every z0 e Q, we say that f is holomorphic (or analytic) in Q. The class of all holomorphic functions in Q will be denoted by H(Q). To be quite explicit, f'(z0) exists if to every E > 0 there corresponds a ç > 0 such that f(z) -f(zo) -/'(zo) < E z - z0 for all z e D'(z0; Ī). (2) Thus f'(z0) is a complex number, obtained as a limit of quotients of complex numbers. Note thatfis a mapping of Q into R2 and that Definition 7.22 associates with such mappings another kind of derivative, namely, a linear operator on R2• In our present situation, if (2) is satisfied, this linear operator turns out to be multiplication by f'(z0) (regarding R2 as the complex field). We leave it to the reader to verify this. 10.3 Remarks If f e H(Q) and g e H(Q), then also f + g e H(Q) and fg e H(Q), so that H(Q) is a ring; the usual differentiation rules apply. More interesting is the fact that superpositions of holomorphic functions are holomorphic: Iffe H(Q), iff(!l) c 01, ifg e H(!l1), and ifh = g o f, then h e H(Q), and h' can be computed by the chain rule h'(z0) = g'(f(z0))/'(z0) (z0 E Q). To prove this, fix z0 e !l, and put w0 = f(z0). Then f(z) -f(z0) = /'(z0) + E(z), g(w) - g(w0) = g'(w0) + 17(w), (1) (2) (3) 198 REAL AND COMPLEX ANALYSIS where e(z)----+ 0 as z----+ z0 and 17(w)----+ 0 as w----+ w0• Put w = f(z), and substitute (2) into (3): If z i= z0, h(z)-h(zo) = [g'(f(zo)) + 11(/(z))][f'(zo) + €(z)]. z-z0 (4) The differentiability off forces f to be continuous at z0• Hence (1) follows from (4). 10.4 Examples For n = 0, 1, 2, . . . , zn is holomorphic in the whole plane, and the same is true of every polynomial in z. One easily verifies directly that 1/z is holomorphic in {z: z i= 0}. Hence, taking g(w) = 1/w in the chain rule, we see that iff1 andf2 are in H(Q) and 00 is an open subset of Q in whichf2 has no zero, thenf1/f2 e H(Qo). Another example of a function which is holomorphic in the whole plane (such functions are called entire) is the exponential function defined in the Prologue. In fact, we saw there that exp is differentiable everywhere, in the sense of Definition 10.2, and that exp' (z) = exp (z) for every complex z. 10.5 Power Series From the theory of power series we shall assume only one fact as known, namely, that to each power series (1) there corresponds a number R e [0, oo] such that the series converges absolutely and uniformly in D(a; r), for every r < R, and diverges if z ¢ D(a; R). The " radius of convergence " R is given by the root test: (2) Let us say that a function f defined in Q is representable by power series in Q if to every disc D(a; r) c Q there corresponds a series (1) which ࡏonverges to f(z) for all z e D(a; r). 10.6 Theorem Iff is representable by power series in Q, then f e H(Q) and f' is also representable by power series in 0. In fact, if 00 f(z) = L cn(z -a)n (1) n=O for z e D(a; r), then for these z we also have 00 f'(z) = L ncn(z-a)n-l. (2) n=l ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 199 PROOF If the series (1) converges in D(a; r), the root test shows that the series (2) also converges there. Take a = 0, without loss of generality, denote the sum of the series (2) by g(z), fix wE D(a; r), and choose p so that I w I < p < r. If z =I= w, we have f(z) -f(w) ( ) x [zn-wn ... .n-1] -g w = L..J en -nw . z- w n=1 z-w The expression in brackets is 0 if n = 1, and is n-1 (z-w) L kwk-1zn-k-1 k=1 if n > 2. If I z I < p, the absolute value of the sum in ( 4) is less than n(n- 1) n-2 2 p so (3) (4) (5) (6) Since p < r, the last series converges. Hence the left side of ( 6) tends to 0 as z-+ w. This says thatf'(w) = g(w), and completes the proof. /Ill Corollary Since f' is seen to satisfy the same hypothesis as f does, the theorem can be applied to f'. It follows that f has derivatives of all orders, that each derivative is representable by power series inn, and that 00 f(k)(z) = L n(n - 1) · · · (n-k + 1)cn(z -a)n-k (7) n=k if ( 1) holds. Hence ( 1) implies that k !ck = f<k)(a) ( k = 0, 1' 2, . . . ), (8) SO that for each a E Q there is a unique sequence {en} for which ( 1) holds. We now describe a process which manufactures functions that are represent able by power series. Special cases will be of importance later. 10.7 Theorem Suppose J1 is a complex (finite) measure on a measurable space X, cp is a complex measurable function on X, Q is an open set in the plane which does not intersect cp(X), and (z E 0). (1) Then f is representable by power series in Q. 200 REAL AND COMPLEX ANALYSIS PROOF Suppose D(a; r) c n. Since z-a lz-al 1 < < r (()-a)n+ 1 = <p(() - Z (2) (3) converges uniformly on X, for every fixed z E D(a; r). Hence the series (3) may be substituted into (1), andf(z) may be computed by interchanging sum mation and integration. It follows that 00 /(z) = L cn(z - a)n (z E D(a; r)) (4) 0 where i dJ1(() C n = X ( <p( () -a )n + 1 (n = 0, 1, 2, . . . ). (5) Ill/ Note: The convergence of the series (4) in D(a; r) is a consequence of the proof. We can also derive it from (5), since (5) shows that lcnl < f5) ( n = 0, 1, 2, . . . ). (6) Integration over Paths Our first major objective in this chapter is the converse of Theorem 10.6: Every f E H(Q) is representable by power series in 0. The quickest route to this is via Cauchy's theorem which leads to an important integral representation of holo morphic functions. In this section the required integration theory will be devel oped; we shall keep it as simple as possible, and shalࡐ regard it merely as a useful tool in the investigation of properties of holomorphic functions. 10.8 Definitions If X is a topological space, a curve in X is a continuous mapping y of a compact interval [et, fJ] c R1 into X; here et < fJ. We call [ et, fJ] the parameter interval of y and denote the range of y by y. Thus y is a mapping, and y is the set of all points y(t), for et < t < fJ. If the initial point y(et) of y coincides with its end point y(fJ), we call y a closed curve. A path is a piecewise continuously differentiable curve in the plane. More explicitly, a path with parameter interval [a, fJ] is a continuous complex function y on [ et, fJ], such that the following holds : There are finitely many ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 201 points s i, C( = s0 < s 1 < · · · < sn = {3, and the restriction of y to each interval [si_1, si] has a continuous derivative on [si_ b si]; however, at the points sb ... , sn _1 the left- and right-hand derivatives of y may differ. A closed path is a closed curve which is also a path. Now suppose y is a path, and f is a continuous function on y. The integral off over y is defined as an integral over the parameter interval [C(, fJ] of y: if(z) dz = r f(y(t))y'(t) dt. (1) Let <p be a continuously differentiable one-to-one mapping of an interval [C(1, fJ1] onto [C(, fJ], such that <p(C(1) = C(, (t)))y'(q>(t))q/(t) dt = .. f(y(s))y'(s) ds, so that our " reparametrization " has not changed the integral : r f(z) dz = r f(z) dz. JYl Jy (2) Whenever (2) holds for a pair of paths y and y 1 (and for all f), we shall regard y and y 1 as equivalent. It is convenient to be able to replace a path by an equivalent one, i.e., to choose parameter intervals at will. For instance, if the end point of y1 coin cides with the initial point of y2, we may locate their parameter intervals so that y1 and y2 join to form one path y, with the property that rf = r f+ r f Jy JYl JY2 (3) for every continuous f on y = yf u y! . However, suppose that [0, 1] is the parameter interval of a path y, and y1(t) = y(1 -t), 0 < t < 1. We call y1 the path opposite to y, for the following reason : For anyfcontinuous on yf = y, we have r f(Yt(t))yJ.(t) dt = -r f(y(l - t))y'(l -t) dt = -r f(y(s))y'(s) ds, so that r ! = _ r 1 JYl Jy (4) 202 REAL AND COMPLEX ANALYSIS From (1) we obtain the inequality if(z) dz < 11!11 oo r I y'(t) I dt, (5) where II f II oo is the maximum of Ill on y and the last integral in (5) is (by definition) the length of y. 10.9 Special Cases (a) If a is a complex number and r > 0, the path defined by y(t) = a + reit (0 < t < 2n) is called the positively oriented circle with center at a and radius r; we have if(z) dz = ir 12" f(a + rei8)ei8 d(), and the length of y is 2nr, as expected. (b) If a and b are complex numbers, the path y given by y( t) = a + ( b -a )t (0 < t < 1) is the oriented interval [a, b] ; its length is I b -a I , and If l f(z) dz = (b -a) 11 f[a + (b -a)t] dt. [a, b] Jo ( ) a(f3 - t) + b( t -ex) }' t t = ---{3 --_ex __ _ (ex < t < {3), (1) (2) (3) (4) (5) we obtain an equivalent path, which we still denote by [a, b]. The path opposite to [a, b] is [b, a]. (c) Let {a, b, c} be an ordered triple of complex numbers, let A = A(a, b, c) be the triangle with vertices at a, b, and c (A is the smallest convex set which contains a, b, and c), and define r f= r f+ r f+l f, Jaa J[a, b] J[b, c] [c, a] (6) for any f continuous on the boundary of A. We may regard (6) as the defini tion of its left side. Or we may regard oA as a path obtained by joining [a, b] to [b, c] to [c, a], as outlined in Definition 10.8, in which case (6) is easily proved to be true. ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 203 If {a, b, c} is permuted cyclically, we see from (6) that the left side of (6) is unaffected. If {a, b, c} is replaced by {a, c, b}, then the left side of ( 6) changes . stgn. We now come to a theorem which plays a very important role in function theory. 10.10 Theorem Let y be a closed path, let Q be the complement of y (relative to the plane), and define 1 i d( Indy (z) = -2 . ( 'Ttl - z y (z e Q). (1) Then Indy is an integer-valued function on Q which is constant in each com ponent of!l and which is 0 in the unbounded component of!l. We call IndY (z) the index of z with respect to y. Note that y is compact, hence y lies in a bounded disc D whose complement De is connected; thus ·De lies in some component of n. This shows that Q has precisely one unbounded com ponent. PROOF Let [ex, fJ] be the parameter interval of }', fix z E n, then 1 f.fJ y'(s) Indy (z) = -2 . ( ) ds. 'Ttl tX }' s - z (2) Since wf2ni is an integer if and only if ew = 1, the first assertion of the theorem, namely, that Indy (z) is an integer, is equivalent to the assertion that lfJ(fJ) = 1, where {f.t y'(s) } <p(t) = exp ex y(s) - z ds (ex < t < fJ). (3) Differentiation of (3) shows that qJ'(t) -({J(t) y'( t) (4) y(t) - z except possibly on a finite set S where y is not differentiable. Therefore qJj(y - z) is a continuous function on [ex, fJ] whose derivative is zero in [ex, fJ] -S. Since S is finite, qJj(y - z) is constant on [ex, fJ]; and since qJ(ex) = 1, we obtain <p(t) = y(t) - z y(ex) - z (ex < t < fJ). (5) We now use the assumption that y is a closed path, i.e., that y(fJ) = y(ex); (5) shows that lfJ(fJ) = 1, and this, as we observed above, implies that Indy (z) is an integer. 204 REAL AND COMPLEX ANALYSIS By Theorem 10.7, (1) shows that Indy e H(Q). The image of a connected set under a continuous mapping is connected (, Theorem 4.22), and since Indy is an integer-valued function, Indy must be constant on each component of !l. Finally, (2) shows that I IndY (z) I < 1 if I z I is sufficiently large. This implies that Indy (z) = 0 in the unbounded component of Q. //// Remark : If A.(t) denotes the integral in (3), the preceding proof shows that 2n Indy (z) is the net increase in the imaginary part of A.(t), as t runs from a. to {3, and this is the same as the net increase of the argument of y(t) -z. (We have not defined " argument " and will have no need for it.) If we divide this increase by 2n, we obtain " the number of times that y winds around z," and this explains why the term " winding number " is frequently used for the index. One virtue of the preceding proof is that it establishes the main properties of the index without any reference to the (multiple-valued) argu ment of a complex number. 10.1 1 Theorem If y is the positively oriented circle with center at a and radius r, then Ind1 (z) = {ץ if I z-a I < r, if I z-a I > r. PROOF We take y as in Sec. 10.9(a). By Theorem 10.10, it is enough to compute Indy (a), and 10.9(2) shows that this equals -. = - (relt)-1eltdt= l. 1 i dz r l21t • • 2nz Y z-a 2n 0 The Local Cauchy Theorem /Ill There are several forms of Cauchy's theorem. They all assert that if y is a closed path or cycle in Q, and if y and Q satisfy certain topological conditions, then the integral of every f e H(Q) over y is 0. We shall first derive a simple local version of this (Theorem 10.14) which is quite sufficient for many applications. A more general global form will be established later. 10.12 Theorem Suppose F e H(Q) and F' is continuous in Q. Then iF'(z) dz = 0 for every closed path y in Q. ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 205 PROOF If [C(, fJ] is the parameter interval of y, the fundamental theorem of calculus shows that i F'(z) dz = r F'(y(t))y'(t) dt = F(y(p)) -F(y( C()) = 0, since y(fJ) = y( C( ). Ill/ Corollary Since zn is the derivative of zn+ 1 /(n + 1) for all integers n =I= -1, we have iz" dz = 0 for every closed path y if n = 0, 1, 2, ... , and for those closed paths y for which 0 ¢ y if n = -2, - 3, -4, .... The case n = -1 was dealt with in Theorem 10.10. 10.13 Cauchy's Theorem for a Triangle Suppose A is a closed triangle in a plane open set n, p E !l,fis continuous on n, andf E H(Q- {p}). Then r /(z) dz = 0. Jaa (1) For the definition of oA we refer to Sec. 10.9(c). We shall see later that our hypothesis actually implies that f E H(Q), i.e., that the exceptional point p is not really exceptional. However, the above formulation of the theorem will be useful in the proof of the Cauchy formula. · PROOF We assume first that p ¢ A. Let a, b, and c be the vertices of A, let a', b', and c' be the midpoints of [b, c], [c, a], and [a, b], respectively, and con sider the four triangles Ai formed by the ordered triples {a, c', b'}, {b, a', c'}, { c, b', a'}, {a', b', c'}. If J is the value of the integral (1), it follows from 10.9(6) that J = itl 111/(z) dz. (2) (3) The absolute value of at least one of the integrals on the right of (3) is there fore at least I J/4 1 . Call the corresponding triangle A1, repeat the argument with A 1 in place of A, and so forth. This generates a sequence of triangles An 206 REAL AND COMPLEX ANALYSIS such that A => A1 => A2 => • • · , such that the length of oAn is 2-nL, if L is the length of oA, and such that I J I < 4n i f(z) dz oAn ( n = 1 , 2, 3, . . . ). (4) There is a (unique) point z0 which the triangles An have in common. Since A is compact, z0 e A, so /is differentiable at z0• Let E > 0 be given. There exists an r > 0 such that I f(z) -f(zo) -/'(zo)(z -zo) I < E I z -Zo I (5) whenever I z -z0 I < r, and there exists an n such that I z -z0 I < r for all z e An. For this n we also have lz-z0 I < 2-nL for all z e An. By the Corol lary to Theorem 1 0.12, ľ j(z) dz = i [f(z) -f(zo) -j'(zo)(z- Zo)J dz, JoAn oAn so that (5) implies and now (4) shows that I J I < eL2• Hence J == 0 if p ¢ A. (6) (7) Assume next that p is a vertex of A, say p = a. If a, b, and c are collinear, then (1) is trivial, for any continuous f If not, choose points x e [a, b] and y e [a, c], both close to a, and observe that the integral off over oA is the sum of the integrals over the boundaries of the triangles {a, x, y}, {x, b, y}, and { b, c, y}. The last two of these are 0, since these triangles do not contain p. Hence the integral over oA is the sum of the integrals over [a, x], [x, y], and [y, a], and since these intervals can be made arbitrarily short and f is boundeࡑ on A, we again obtain (1). Finally, if p is an arbitrary point of A, apply the preceding result to {a, b, p}, { b, c, p}, and { c, a, p} to complete the proof. I I I I 10.14 Cauchy's Theorem in a Convex Set Suppose Q is a convex open set, pEn, f is continuous on n, and f E H(Q- {p}). Then f = F' for some F e H(O.). Hence if(z) dz = 0 (1) for every closed path y in n. ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 207 PROOF Fix a E 0. Since 0 is convex, 0 contains the straight line interval from a to z for every z e n, so we can define F(z) = f f@ dĉ J[a, z] (z e 0). (2) For any z and z0 · e 0, the triangle with vertices at a, z0, and z lies in 0; hence F(z)-F(z0) is the integral off over [z0, z], by Theorem 10.13. Fixing z0, we thus obtain F(z) -F(zo) -f(zo) = 1 f [f(e) -f(zo)] dĉ, (3) Z -Zo z -Zo J[zo, z] if z =F z0 • Given E > 0, the continuity off at z0 shows that there is a Ī > 0 such that lf(ĉ)-f(z0)1 < E ifl ࡒ-z0 1 < ç; hence the absolute value of the left side of (3) is less than E as soon as I z-z0 I < Ī. This proves that f = F'. In particular, Fe H(O). Now (1) follows from Theorem 10.12. /Ill 10.15 Cauchy's Formula in a Convex Set Suppose y is a closed path in a convex open set 0, and f e H(O). If z e 0 and z ¢ y, then 1 i f(ĉ) f(z) · Ind1 (z) = -2 . ĉ dĉ. 'Ttl - z }' The case of greatest interest is, of course, Indy (z) = 1. PRooF Fix z so that the above conditions hold, and define f(ĉ) -f(z) g(ĉ) = ĉ-z if ĉ e o, ĉ =F z, f'(z) ifĉ = z. Then g satisfies the hypotheses of Theorem 10.14. Hence _ 21. ig(ĉ) dĉ = 0. 'Ttl }' If we substitute (2) into (3) we obtain (1 ). (1) (2) (3) Ill/ The theorem concerning the representability of holomorphic functions by power series is an easy consequence of Theorem 1 0.15, if we take a circle for y: 10.16 Theorem For every open set 0 in the plane, every f e H(O) is represent able by power series in 0. 208 REAL AND COMPLEX ANALYSIS PROOF Suppose f e H(Q) and D(a; R) c Q. If y is a positively oriented circle with center at a and radius r < R, the convexity of D(a; R) allows us to apply Theorem 10.15; by Theorem 10.11, we obtain f(z) = Ó i f(e> de 2nz Y ĉ - z (z e D(a; r)). (1) But now we can apply Theorem 10.7, with X = [0, 2n], <p = y, and dJ1(t) = f(y(t))y'(t) dt, and we conclude that there is a sequence { c,.} such that 00 f(z) = L c,.(z -a)" (z e D(a; r)). (2) n = O The uniqueness of {c,.} (see the Corollary to Theorem 10.6) shows that the same power series is obtained for every r < R (as long as a is fixed). Hence the representation (2) is valid for every z e D(a; R), and the proof is complete. Ill/ Corollary Iff e H(Q), thenf' e H(!l). PROOF Combine Theorems 10.6 and 10.16. Ill/ The Cauchy theorem has a useful converse: 10.17 Morera's Theorem Suppose f is a continuous complex function in an open set Q such that r f(z) dz = 0 Jaa for every closed triangle A c Q. Thenf e H(Q). PROOF Let V be a convex open set in Q. As in the proof of Theorem 10.14, we can construct Fe H(V) such that F' =f. Since derivatives of holomorphic functions are holomorphic (Theorem 10.16), we have f e H(V), for every convex open V c n, hence f e H(Q). I I I I The Power Series Representation The fact that every holomorphic function is locally the sum of a convergent power series has a large number of interesting consequences. A few of these are developed in this section. 10.18 Theorem Suppose Q is a region,/ e H(O), and Z(f) = {a e O:f(a) = 0}. (1) ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 209 Then either Z(f) = Q, or Z(f) has no limit point in Q. In the latter case there corresponds to each a e Z(/) a unique positive integer m = m(a) such that f(z) = (z - a)mg(z) (z e Q), (2) where g e H(Q) and g(a) =I= O;furthermore, Z(f) is at most countable. (We recall that regions are connected open sets.) The integer m is called the order of the zero which f· has at the point a. Clearly, Z(/) = Q if and only if/is identically 0 in n. We call Z(/) the zero set of f Analogous results hold of course for the set of ex-points off, i.e., the zero set of f- ex, where ex is any complex number. PROOF Let A be the set of all limit points of Z(f) in n. Since fis continuous, A c Z(f). Fix a e Z(f), and choose r > 0 so that D(a; r) c Q. By Theorem 10.16, 00 f(z) = L cn(z - a)" (z e D(a; r)). (3) n=O There are now two possibilities. Either all en are 0, in which case D(a; r) c A and a is an interior point of A, or there is a smallest integer m [necessarily positive, sincef(a) =OJ such that em =1= 0. In that case, define g(z) = {(z - a)-m f(z) em (z E Q - {a}), (z = a). Then (2) holds. It is clear that g e H(Q - {a}). But (3) implies 00 g(z) = L cm+k(z - a)k (z e D(a; r)). k=O Hence g e H(D(a; r)), so actually g e H(Q). (4) (5) Moreover, g(a) =1= 0, and the continuity of g shows that there is a neigh borhood of a in which g has no zero. Thus a is an isolated point of Z(f), by (2). If a e A, the first case must therefore occur. So A is open. If B = Q - A, it is clear from the definition of A as a set of limit points that B is open. Thus Q is the union of the disjoint open sets A and B. Since Q is connected, we have either A = n, in which case Z(/) = n, or A = 0. In the latter case, Z(f) has at most finitely many points in each compact subset of !l, and since Q is a-compact, Z(f) is at most countable. I I I I Corollary I ff and g are holomorphic functions in a region Q and if f(z) = g(z) for all z in some set which has a limit point in Q, thenf(z) = g(z)for all z E !l. 210 REAL AND COMPLEX ANALYSIS In other words, a holomorphic function in a region n is determined by its values on any set which has a limit point in n. This is an important uniqueness theorem. Note: The theorem fails if we drop the assumption that n is connected : If n = no u nl, and no and nl are disjoint open sets, put f = 0 in no and f = 1 in nl. 10.19 Definition If a e n and f e H(n - {a}), then f is said to have an iso lated singularity at the point a. Iff can be so defined at a that the extended function is holomorphic in n, the singularity is said to be removable. 10.20 Theorem Supposefe H(n - {a}) andfis bounded in D'(a; r),for some r > 0. Thenf has a removable singularity at a. Recall that D' (a; r) = { z : 0 < I z -a I < r}. PROOF Define h(a) = 0, and h(z) = (z-a)2f(z) in n- {a}. Our boundedness assumption shows that h'(a) = 0. Since h is evidently differentiable at every other point of n, we have h E H(O.), so 00 h(z) = L cn(z-a)" (z e D(a; r)). n=2 We obtain the desired holomorphic extension off by setting f(a) = c2, for then 00 f(z) = Len+ 2(z -a)" (z E D(a; r)). Ill/ n=O 10.21 Theorem If a e n and f e H(n - {a}), then one of the following three cases must occur: (a) f has a removable singularity at a. (b) There are complex numbers c1, .
• . , em , where m is a positive integer and em =I= 0, such that has a removable singularity at a. (c) Ifr > 0 and D(a; r) c n, thenf(D'(a; r)) is dense in the plane. In case (b),f is said to have a pole of order m at a. The function m L ck(z-a) -k, k=1 ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 21 1 a polynomial in (z -a)-1, is called the principal part off at a. It is clear in this situation that I f(z) Ir oo as zr a. In case (c), f is said to have an essential singularity at a. A statement equiva lent to (c) is that to each complex number w there corresponds a sequence { zn} such that znr a andf(zn)r w as nŠ oo . PROOF Suppose (c) fails. Then there exist r > 0, b > 0, and a complex number w such that I f(z) - w I > b in D'(a; r). Let us write D for D(a; r) and D' for D'(a; r). Define 1 g(z) = f(z) - w (z e D'). (1) Then g e H(D') and I g I < 1lb. By Theorem 10.20, g extends to a holo morphic function in D. If g(a) # 0, (1) shows thatf is bounded in D'(a; p) for some p > 0. Hence (a) holds, by Theorem 10.20. If g has a zero of order m > 1 at a, Theorem 10.18 shows that (zED), (2) where g1 e H(D) and g1(a) # 0. Also, g1 has no zero in D', by (1). Put h = 1lg1 in D. Then he H(D), h has no zero in D, and f(z)- w = (z-a)-mh(z) (z E D'). (3) But h has an expansion of the form 00 h(z) = L bn(z -a)" (zED), (4) n = O with b0 # 0. Now (3) shows that (b) holds, with ck = bm -k , k = 1, . . . , m. This completes the proof. I I I I We shall now exploit the fact that the restriction of a power series L cn(z -a)" to a circle with center at a is a trigonometric series. 10.22 Theorem If 00 f(z) = L cn(z -a)" (z e D(a; R)) (1) n = O and ifO < r < R, then (2) 212 REAL AND COMPLEX ANALYSIS PROOF We have 00 f(a + rei8) = L en rnein8. n=O For r < R, the series (3) converges uniformly on [ - n, n]. Hence 1 J1t .8 . 8 en rn = - f(a + re' )e-'n df} 2n _1t ( n = 0, 1, 2, . .. ), and (2) is seen to be a special case of Parseval's formula. Here are some consequences : 10.23 Liouville's Theorem Every bounded entire function is constant. Recall that a function is entire if it is holomorphic in the whole plane. (3) (4) /Ill PROOF Suppose f is entire, I f(z) I < M for all z, and f(z) = L en zn for all z. By Theorem 10.22, 00 L lcn12r2n < M2 n=O for all r, which is possible only if en = 0 for all n > 1. Ill/ 10.24 The Maximum Modulus Theorem Suppose Q is a region, f e H(Q), and D(a; r) c n. Then I f(a) I < max I f(a + rei8) 1. 8 Equality occurs in (1) if and only iff is constant in Q. (1) Consequently, If I has no local maximum at any point of n, unless f is con stant. PROOF Assume that I f(a + rei8) I < I f(a) I for all real 9. In the notation of Theorem 10.22 it follows then that / 00 L lcnl2r2n< lf(a)l2 = lcol2. n=O Hence c1 = c2 = c3 = · · · = 0, which implies that f(z) = j(a) in D(a; r). Since Q is connected, Theorem 10.18 shows thatfis constant in Q. /Ill Corollary Under the same hypotheses, iff has no zero in D(a; r). I f(a) I > min I f(a + rei8) I 8 (2) ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 213 PROOF If f(a + rei8) = 0 for some (} then (2) is obvious. In the other case, there is a region 00 that contains D( a; r) and in which f has no zero; hence (1) can be applied to 1/ f, and (2) follows. /Ill 10.25 Theorem Ifn is a positive integer and P( ) n n - 1 z = z + an - 1 z + 0 0 0 + a 1 z + ao ' where a0, • • • , an _ 1 are complex numbers, then P has precisely n zeros in the plane. Of course, these zeros are counted according to their multiplicities: A zero of order m, say, is counted as m zeros. This theorem contains the fact that the complex field is algebraically closed, i.e., that every nonconstant polynomial with complex coefficients has at least one complex zero. PROOF Choose r > 1 + 21 a0 I + I a 11 + · · · + I an _ 11 . Then I P(rei8) I > I P(O) I (0 < (} < 2n). If P had no zeros, then the function f = 1/P would be entire and would satisfy I f(O) I > I f(re;8) I for all 0, which contradicts the maximum modulus theorem. Thus P(z1) = 0 for some z1• Consequently, there is a polynomial Q, of degree n- 1, such that P(z) = (z - z1)Q(z). The proof is completed by induction on n. I I I I 10.26 Theorem (Cauchy's Estiõates) Iff e H(D(a; R)) and I f(z) I < M for all z e D(a; R), then (n = 1, 2, 3, . . . ). (1) PROOF For each r < R, each term of the series 10.22(2) is bounded above by M2• Ill/ If we take a= 0, R = 1, and f(z) = z", then M = 1, J(O) = n!, and we see that (1) cannot be improved. 10.27 Definition A sequence {.fj} of functions in Q is said to converge to f uniformly on compact subsets of Q if to every compact K c Q and to every E > 0 there corresponds an N = N(K, E) such that I h{z) -f(z) I < E for aJl z E K ifj > N. For instance, the sequence {zn} converges to 0 uniformly on compact subsets of D(O; 1), but the convergence is not uniform in D(O; 1). 214 REAL AND COMPLEX ANALYSIS It is uniform convergence on compact subsets which arises most natu rally in connection with limit operations on holomorphic functions. The term " almost uniform convergence " is sometimes used for this concept. 10.28 Theorem Suppose jj e H(O.), for j = 1, 2, 3, ... , and jjr f uniformly on compact subsets of Q. Then f e H(Q), and f)r f' uniformly on compact subsets ofO.. PROOF Since the convergence is unifor.m on each compact disc in 0., f is continuous. Let A be a triangle in Q. Then A is compact, so r f(z) dz = lim r fiz) dz = 0, JaA i- oo JaA by Cauchy's theorem. Hence Morera's theorem implies thatf e H(O.). Let K be compact, K c Q. There exists an r > 0 such that the union E of the closed discs D(z; r), for all z e K, is a compact subset of 0.. Applying Theorem 10.26 tof-jj , we have (z E K), where II / liE denotes the supremum of If I on E. SincejjÒ funiformly on E, it follows that f)r f' uniformly on K. /Ill Corollary Under the same hypothesis,fנn>rf uniformly, asjr oo , on every compact set K c Q, and for every positive integer n. Compare this with the situation on the real line, where sequences of infinitely differentiable functions can converge uniformly to nowhere differentiable func tions ! The Open Mapping Theorem JfQ is a region andf e H(O.), thenf(O.) is either a region or a point. This important property of holomorphic functions will be proved, in more detailed form, in Theorem 10.32. 10.29 Lemma Iffe H(Q) and g is defined in Q x Q by f(z) -f(w) if w =I= z, g(z, w) = z- w f'(z) if w = z, then g is continUOUS in Q X Q. ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 215 PROOF The only points (z, w) E n x Q at which the continuity of g is poss ibly in doubt have z = w. Fix a E Q. Fix E > 0. There exists r > 0 such that D(a ; r) c Q and I/'(') -f'(a) I < E for all ' E D(a ; r). If z and w are in D(a ; r) and if '(t) = (1 -t)z + tw, then '(t) E D(a ; r) for 0 < t < 1, and g(z, w) -g(a, a) = r [f'(((t)) -f'(a)] dt. The absolute value of the integrand is <E, for every t. Thus I g(z, w) -g(a, a) I <E. This proves that g is continuous at (a, a). /Ill 10.30 Theorem Suppose cp E H(Q), z0 E Q, and cp'(z0) i= 0. Then Q contains a neighborhood V of z0 such that (a) <p is one-to-one in V, (b) W = <p(V) is an open set, and (c) ift/1: W / Vis defined by t/J(cp(z)) = z, then t/1 E H(W). Thus <p: V / W has a holomorphic inverse. PROOF Lemma 10.29, applied to ȚI cp'(zo) I I Z1 -z2l if z 1 E V and z2 E V. Thus (a) holds, and also \ 0 so that D(a, r) c V. By (1) there exists c > 0 such that I <p( a + rei8) - 2c (-n < (} < n ). If A E D( <p(a) ; c), then I A -<p(a) I < c, hence (3) implies min I A-cp(a + rei8) I > c. 8 (3) (4) By the corollary. to Theorem 1 0.24, A -cp must therefore have a zero in D(a ; r). Thus A = <p(z) for some z E D(a ; r) c V. This proves that D(<p(a) ; c) c <p(V). Since a was an arbitrary point of V, 0, then nm(z) = w if and only if z = r11mei<B + 2kn:)fm, k = 1, . . . , m. Note also that each nm is an open mapping: If V is open and does not contain 0, then nm(V) is open by Theorem 10.30. On the other hand, nm(D(O; r)) = D(O; rm). Compositions of open mappings are clearly open. In particular, nm o <p is open, by Theorem 10.30, if <p' has no zero. The following theorem (which contains the more detailed version of the open mapping theorem that was mentioned prior to Lemma 10.29) states a converse : Every nonconstant holo morphic function in a region is locally of the form nm o <p, except for an additive constant. 10.32 Theorem Suppose Q is a region, f E H(Q), f is not constant, z0 E Q, and w0 = f(z0). Let m be the order of the zero which the function f-w0 has at z0• Then there exists a neighborhood V of z0, V c Q, and there exists <p E H(V), such that (a) f(z) = w0 + [<p(z)]m for all z E V, (b) <p' has no zero in V and <p is an invertible mapping of V onto a disc D(O; r). Thus/ -w0 = nm o <p in V. It follows that f is an exactly m-to-1 mapping of V -{ z0} onto D'(w0; rm), and that each w0 E f(Q) is an interior point of f(!l). Hencej(Q) is open. PROOF Without loss of generality we may assume that Q is a convex neigh borhood of z0 which is so small that f(z) =I= w0 if z E Q - { z0}. Then (z E Q) (1) for some g E H(Q) which has no zero in Q. Hence g'/g E H(Q). By Theorem 10.14, g'/g = h' for some hE H(Q). The derivative of g · exp (-h) is 0 in Q. If h is modified by the addition of a suitable constant, it follows that g = exp (h). Define h(z) <p(z) = (z -z0) exp -m Then (a) holds, for all z E Q. (z E Q). (2) Also, <p(z0) = 0 and 1, so that f would be m-to-1 in some deleted neighborhood of z0 • Now apply part (c) of Theorem 10.30. /Ill Note that the converse of TĽeorem 10.33 is false : If f(z) = ez, then f'(z) =I= 0 for every z, butfis not one-to-one in the whole complex plane. The Global Cauchy Theorem Before we state and prove this theorem, which will remove the restriction to convex regions that was imposed in Theorem 10.14, it will be convenient to add a little to the integration apparatus which was sufficient up to now. Essentially, it is a matter of no longer restricting ourselves to integrals over single paths, but to consider finite " sums " of paths instead. A simple instance of this occurred already in Sec. 10.9(c). 10.34 Chains and Cycles Suppose y1, • • • , Yn are paths in the plane, and put K = yf u · · · u }':. Each }'; induces a linear functional Yi on the vector space C(K), by the formula (1) Define (2) Explicitly, f(/) = y1(f) + · · · + Yn(f) for all f e C(K). The relation (2) suggests that we introduce a " formal sum " r = Y 1 -f- · · · -f- Yn (3) and define Lf(z) dz = f(f). (4) Then (3) is merely an abbreviation for the statement Lf(z) dz = it. Lf(z) dz (f E C(K)). (5) Note that (5) serves as the definition of its left side. 218 REAL AND COMPLEX ANALYSIS The objects r so defined are called chains. If each y i in (3) is a closed path, then r is called a cycle. If each y i in (3) is a path in some open set Q, we say that r is a chain in Q. If (3) holds, we define r = Yt u · · · u y:. If r is a cycle and rx ¢ r, we define the index of rx with respect to r by 1 l dz Indr (rx) = -2 . , nz r z -rx just as in Theorem 10.10. Obviously, (3) implies n Indr ( rx) = 2: IndYi ( rx ). i = 1 (6) (7) (8) If each Yi in (3) is replaced by its opposite path (see Sec. 10.8), the resulting chain will be denoted by - r. Then f f(z) dz = - r f(z) dz -r Jr (f e C(r)). (9) In particular, Ind_r (rx) = - Indr (rx) if r is a cycle and rx ¢ r. Chains can be added and subtracted in the obvious way, by adding or sub tracting the corresponding functionals: The statement r = r 1 .f- r 2 means r f(z) dz = r f(z) dz + r f(z) dz Jr Jr1 Jr2 for every f e C(rt u r!). (10) Finally, note that a chain may be represented as a sum of paths in many ways. To say that means simply that for every f that is continuous on yf u · · · u }': u <5f u · · · u <5t. In particular, a cycle may very well be represented as a sum of paths that are not closed. 10.35 Cauchy's Theorem Suppose f e H(Q), where Q is an arbitrary open set in the complex plane. I jr is a cycle in Q that satisfies Indr ( rx) = 0 for every rx not in Q, (1) ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 219 then and then 1 i f(w) f(z) · Indr (z) = -. dw 2nz r w-z Lf(z) dz = 0. If r 0 and r 1 are cycles in Q such that for Z E Q -- r for every C( not in n, r f(z) dz = r f(z) dz. Jro Jr1 PROOF The function g defined in n X n by f(w) -f(z) if w =I= z, g(z, w) = w-z f'(z) if w = z, is continuous in n X n (Lemma 10.29). Hence we can define h(z) = -2 1 . r g(z, w) dw 1tl Jr (z e Q). (2) (3) (4) (5) (6) (7) For z E n - r' the Cauchy formula (2) is clearly equivalent to the assertion that h(z) = 0. (8) To prove (8), let us first prove that h e H(Q). Note that g is uniformly continuous on every compact subset of n X n. If z E n, Zn E n, and Zn = z, it follows that g(zn, w) = g(z, w) uniformly for w e r (a compact subset of Q). Hence h(zn) = h(z). This proves that h is continuous in Q. Let A be a closed triangle in Q. Then r h(z) dz = -2 1 . r ( r g(z, w) dz) dw. JaA 7tl Jr JaA (9) For each w E n, zס g(z, w) is holomorphic in n. (The singularity at z = w is removable.) The inner integral on the right side of (9) is .therefore 0 for every w E r. Morera's theorem shows now that h E H(Q). 220 REAL AND COMPLEX ANALYSIS Next, we let Q1 be the set of all complex numbers z for which Indr (z) = 0, and we define h.(z) = Ò r f(w) dw 2nz Jr w-z (z e 01). (10) If z E n (\ n1, the definition of nl makes it clear that hl(z) = h(z). Hence there is a function <p e H(Q u 01) whose restriction to Q is h and whose restriction to n1 is h1. Our hypothesis (1) shows that !11 contains the complement of n. Thus <p is an entire function. 01 also contains the unbounded component of the com plement of r, since Indr (ӱ) is 0 there. Hence lim <p(z) = lim h1(z) = 0. lzl - oo lzl - oo (1 1) Liouville's theorem implies now that <p(z) = 0 for every z. This proves (8), and hence (2). To deduce (3) from (2), pick a E n - r and define F(z) = (z -a)f(z). Then 1 i 1 i F(z) -. f(z) dz = -. dz = F(a) · Indr (a) = 0, 2nz r 2nz r z -a because F(a) = 0. (12) Finally, (5) follows from (4) if (3) is applied to the cycle r = r 1 - r 0 • This completes the proof. I I I I 10.36 Remarks (a) If y is a closed path in a convex region Q and if C( ¢ n, an application of Theorem 10.14 to f(z) = (z -C() - 1 shows that Indy (C() = 0. Hypothesis (1) of Theorem 10.35 is therefore satisfied by every cycle in Q if Q is convex. This shows that Theorem 10.35 generalizes Theorems 10.14 and 10.15. (b) The last part of Theorem 10.35 shows under what circumstances integra tion over one cycle can be replaced by integration over another, without changing the value of the integral. For example, let Q be the plane with three disjoint closed discs Di removed. If r, y1, y2 , y3 are positively oriented circles in Q such that r surrounds D1 u D2 u D3 and Yi sur rounds Di but not Di for j "# i, then Lf(z) dz = itt 1/(z) dz for every f e H(Q). (c) In order to apply Theorem 10.35, it is desirable to have a reasonably efficient method of finding the index of a point with respect to a closed path. The following theorem does this for all paths that occur in practice. ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 221 It says, essentially, that the index increases by 1 when the path is crossed " from right to left." If we recall that Indy (C() = 0 if C( is in the unbounded component of the complement W of y, we can then successively deter mine Indy (C() in the other components of W, provided that W has only finitely many components and that y traverses no arc more than once. 10.37 Theorem Suppose y is a closed path in the plane, with parameter interval [C(, fJ]. Suppose C( < u < v < fJ, a and bare complex numbers, I b I = r > 0, and (i) y(u) = a -b, y(v) = a + b, (ii) I y(s) -a I < r if and only if u < s < v, (iii) I y(s) -a I = r if and only if s = u or s = v. Assume furthermore that D(a; r) -y is the union of two regions, D + and D _ , labeled so that a + bi E i5 + and a -bi E i5 _ . Then Indy (z) = 1 + Indy (w) if x E D + and w E D _ . As y(t) traverses D(a; r) from a -b to a + b, D _ is " on the right " and D + is " on the left " of the path. PROOF To simplify the writing, reparametrize y so that u = 0 and v = n. Define C(s) = a-beis f(s) = {C(s) y(2n -s) {y(s) g(s) = C(s) h(s) = { cde) (0 < s < 2n) (0 < s < n) (n < s < 2n) (0 < s < n) (n < s < 2n) (C( < s < 0 or n < s < fJ) (0 < s < n). Since y(O) = C(O) and y(n) = C(n),f, g, and h are closed paths. If E c D(a; r), I ( -a I = r, and ( ¢ E, then E lies in the disc D(2a -(; 2r) which does not contain (. Apply this to E = g, ( = a-bi, to see [from Remark 10.36(a)] that Indg (a-bi) = 0. Since i5 _ is connected and D _ does not intersect g, it follows that Indg (w) = 0 ifw ED_. (1) The same reasoning shows that lnd1 (z) = 0 (2) 222 REAL AND COMPLEX ANALYSIS We conclude that Indy (z) = Indh (z) = Indh (w) = Indc (w) + IndY (w) = 1 + Indy (w). The first of these equalities follows from (2), since h = y -+- f The second holds because z and w lie in D(a; r), a connected set which does not intersect h. The third follows from (1), since h-+-g = C -+- y, and the fourth is a conse quence of Theorem 10.1 1. This completes the proof. /Ill We now turn to a brief discussion of another topological concept that is relevant to Cauchy's theorem. 10.38 Homotopy Suppose y0 and y1 are closed curves in a topological space X, both with parameter interval I = [0, 1]. We say that y0 and y1 are X-homotopic if there is a continuous mapping H of the unit square I2 = I x I into X such that H(s, 0) = y0(s), H(s, 1) = y1(s), H(O, t) = H(1, t) (1) for all s E I and t e I. Put yis) = H(s, t). Then (1) defines a one-parameter family of closed curves Yt in X, which connects y0 and y1• Intuitively, this means that y0 can be continuously deformed to y1, within X. If y0 is X -homotopic to a constant mapping y1 (i.e., if yf consists of just one point), we say that y0 is null-homotopic in X. If X is connected and if every closed curve in X is null-homotopic, X is said to be simply connected. For example, every convex region Q is simply connected. To see this, let y0 be a closed curve in Q, fix z 1 e Q, and define H(s, t) = (1 -t)y0(s) + tz 1 (0 <: s < 1, 0 < t < 1). (2) Theorem 10.40 will show that condition (4) of Cauchy's theorem 10.35 holds whenever r 0 and r 1 are Q-homotopic closed paths. As a special case of this, note that condition (1) of Theorem 10.35 holds for every closed path r in Q if!l is simply connected. 10.39 Lemma I f y0 and y1 are closed paths with parameter interval [0, 1], if rx is a complex number, and if I Y 1 (s) - Yo(s) I < I rx -Yo(s) I (0 < s < 1) . (1) PROOF Note first that (1) implies that rx ¢ yv and rx ¢ yf. Hence one can define y = ( y 1 - rx )/( y 0 - rx ). Then y' yw yx Y Y1 - rx Yo - rx (2) ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 223 and 1 1 - y I < 1, by (1). Hence y c D(1 ; 1), which implies that Indy (0) = 0. Integration of (2) over [0, 1] now gives the desired result. I I I I 10.40 Theorem If r 0 and r 1 are 0.-homotopic closed paths in a region 0., and if C( ¢ 0., then Indr1 (C() = Indr0 (C(). PROOF By definition, there is a continuous H : /2 / 0. such that H(s, 0) = r 0(s), H(s, 1) = r 1(s), H(O, t) = H(1, t). Since /2 is compact, so is H(/2). Hence there exists E > 0 such that I C( - H ( s, t) I > 2E if Since H is uniformly continuous, there is a positive integer n such that I H(s, t) - H(s', t') I < E if I s - s' I + I t - t' I < 1 ln. Define,polygonal closed paths Yo , . . . , Yn by Yk(s) = H(: , ̒}ns + 1 - i) + Heb 1 , a)(i - ns) if i - 1 < ns < i and i = 1, . . . , n. By (4) and (5), I Yk(s) - H(s, kin) I < E (k = 0, . . . , n; 0 < s < 1 ). In particular, taking k = 0 and k = n, I Yo(s) - r o(s) I < E, By ( 6) and (3), (k = 0, . . . , n; 0 < s < 1). On the other hand, (4) and (5) also give ( k = 1, . . . , n; 0 < s < 1 ). (1) (2) (3) (4) (5) (6) (7) (8) (9) Now it follows from (7), (8), (9), and n + 2 applications of Lemma 10.39 that C( has the same index with respect to each of the paths r 0, y0, y 1, • • • , Yn , r 1• This proves the theorem. I I I I Note: If rt(s) = H(s, t) in the preceding proof, then each rt is a closed curve, but not necessarily a path, since H is not assumed to be differentiable. The paths Yk were introduced for this reason. Another (and perhaps more satisfactory) way to circumvent this difficulty is to extend the definition of index to closed curves. This is sketched in Exercise 28. 224 REAL AND COMPLEX ANALYSIS The Calculus of Residues 10.41 Definition A function f is said to be meromorphic in an open set Q if there is a set A c Q such that (a) A has no limit point in !l, (b) f e H(Q - A), (c) fhas a pole at each point of A. Note that the possibility A = 0 is not excluded. Thus every f e H(Q) is meromorphic in n. Note also that (a) implies that no compact subset of Q contains infinitely many points of A, and that A is therefore at most countable. Iff and A are as above, if a e A, and if m Q(z) = 2: ck(z - a)-k (1) k= 1 is the principal part off at a, as defined in Theorem 10.21 (i.e., iff- Q has a removable singularity at a), then the number c1 is called the residue off at a: c1 = Res (f; a). If r is a cycle and a ¢ r, (1) implies Ó j Q(z) dz = c1 Indr (a) = Res (Q; a) lndr (a). 2nz Jr This very special case of the following theorem will be used in its proof. (2) (3) 10.42 The Residue Theorem Suppose f is a meromorphic function in Q. Let A be the set o f points in Q at which f has poles. If r is a cycle in Q - A such that then lndr (et) = 0 for all -2 1 . r f(z) dz = L Res (f; a) Indr (a). 1t l Jr a E A (1) (2) PROOF Let B = {a e A : Indr (a) "# 0}. Let W be the complement of r. Then lndr (z) is constant in each component V of W. If- V is unbounded, or if V intersects nc, (1) implies that Indr (z) = 0 for every z E v. Since A has no limit point in !l, we conclude that B is a finite set. The sum in (2), though formally infinite, is therefore actually finite. Let a1, • • • , an be the points of B, let Q1, . . . , Qn be the principal parts off at ah . . . , an , and put g = f- (Q1 + · · · + Qn). (If B = 0, a possibility which is not excluded, then g =f.) Put 00 = Q - (A -B). Since g has removable ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 225 singularities at a1, • • • , an , Theorem 10.35, applied to the function g and the open set 0.0 , shows that Lg(z) dz = 0. (3) Hence and sincefand Qk have the same residue at ak , we obtain (2). Ill/ We conclude this chapter with two typical applications of the residue theorem. The first one concerns zeros of holomorphic functions, the second is the evaluation of a certain integral. 10.43 Theorem Suppose y is a closed path in a region Q, such that Indy (C() = 0 for every C( not in 0.. Suppose also that Indy ( C() = 0 or 1 for every C( E Q - y, and let 01 be the set of all C( with IndY (C() = 1. For any f E H(Q) let N 1 be the number o f zeros off in Q 1, counted accord ing to their multiplicities. (a) Iff e H(Q) andfhas no zeros on y then 1 r f'(z) N 1 = 2ni J1 f(z) dz = Indr (0) where r = f 0 y. (b) If also g E H(Q) and I f ( z) - g( z) I < I f ( z) I then Ng = N1 . for all z e y (1) (2) Part (b) is usually called Rouche's theorem. It says that two holomorphic functions have the same number of zeros in 0.1 if they are close together on the boundary of 0.1, as specified by (2). PROOF Put qJ =f'/ f, a meromorphic function in 0.. If a e Q andfhas a zero of order m = m(a) at a, then f(z) = (z - a)mh(z), where h and 1/h are holo morphic in some neighborhood V of a. In V - {a}, Thus f'(z) m h'(z) <p(z) = f(z) = z - a + h( z) · Res (({J; a) = m(a). (3) (4) 226 REAL AND COMPLEX ANALYSIS Let A = {a e Q1 : f(a) = 0}. If our assumptions about the index of y are combined with the residue theorem one obtains 1 i f'(z) -2 . f( ) dz = L Res (<p ; a) = L m(a) = N1 . 'Ttl y z a e A a e A This proves one half of (1). The other half is a matter of direct computation : 1 i dz 1 i21t r'(s) Indr (0) = -- = -ds 2ni r z 2ni 0 r(s) 1 i21t f'(y(s)) , 1 i f'(z) = -y (s) ds = -dz. 2ni 0 f(y(s)) 2ni Y f(z) The parameter interval of y was here taken to be [0, 2n]. Next, (2) shows that g has no zero on y. Hence (1) holds with g in place off Put r 0 = g o y. Then it follows from (1), (2), and Lemma 10.39 that Ng = Indr0 (0) = Indr (0) = N1 . 10.44 Problem For real t,find the limit, as A q oo, of f A • Sin X . t d -- e'x X. -A X Ill/ (1) SoLUTION Since z- 1 · sin z · eitz is entire, its integral over [-A, A] equals that over the path r A obtained by going from -A to - 1 along the real axis, from - 1 to 1 along the lower half of the unit circle, and from 1 to A along the real axis. This follows from Cauchy's theorem. r A a voids the origin, and we may therefore use the identity 2i sin z = eiz - e-iz to see that (1) equals <p A(t + 1) - <p A(t - 1), where 1 l i eisz - <pA(s) = -. - dz. n 2nz r A z (2) Complete r A to a closed path in two ways : First, by the semicircle from A to - Ai to - A ; secondly, by the semicircle from A to Ai to -A. The function eisz /z has a single pole, at z = 0, where its residue is 1. It follows that and 1 1 f o . - <p A(s) = -2 exp (isAe'9) df} n n _1t 1 1 i1t . - <p A(s) = 1 - -2 exp (isAe'8) dO. n n 0 (3) (4) ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 227 Note that I exp (isAei6) I = exp (-As sin 0), (5) and that this is < 1 and tends to 0 as Aq oo if s and sin (} have the same sign. The dominated convergence theorem shows therefore that the integral in (3) tends to 0 if s < 0, and the one in ( 4) tends to 0 if s > 0. Thus lim 0, if s < 0, and if we apply (6) to s = t + 1 and to s = t - 1, we get 1. fA sin x itx d {n tm e x = A-+oo -A X 0 if - 1 < t < 1, if It I > 1. Since 0 such that I IX - P I > b for all ct E A and P E B. Prove this, with an arbitrary metric space in place of the plane. 2 Suppose thatfis an entire function, and that in every power series 00 f(z) = L cn<z - a)" n=O at least one coefficient is 0. Prove thatfis a polynomial. Hint: n! en = J<">(a). 3 Suppose f and g are entire functions, and I f(z) I < I g(z) I for every z. What conclusion can you draw? 4 Supposefis an entire function, and I f(z) I < A + B I z lk for all z, where A, B, and k are positive numbers. Prove thatfmust be a polynomial. S Suppose {In} is a uniformly bounded sequence of holomo"rphic functions in n such that {fn(z)} converges for every z e n. Prove that the convergence is uniform on every compact subset of n. Hint: Apply the dominated convergence theorem to the Cauchy formula for fn -f m . 6 There is a region n that exp (!l) = D(l ; 1). Show that exp is one-to-one in n, but that there are many such n. Fix one, and define log z, for I z - l l < 1, to be that w E n for which ew = z. Prove that log' (z) = 1/z. Find the coefficients an in 1 z n=O 228 REAL AND COMPLEX ANALYSIS and hence find the coefficients en in the expansion CX> log z = L cn X roCX> dx Jc 1 + x" (n = 2, 3, 4, . . . ). [For even n, the method of Exercise 8 can be used. However, a different path can be chosen, which simplifies the computation and which also works for odd n: from 0 to R to R exp (2ni/n) to 0.] Answer: (n/n)/sin (n/n). 14 Suppose 0.1 and 0.2 are plane regions, f and g are nonconstant complex functions defined in 0.1 and 0.2 , respectively, and /(0.1) c 0.2 • Put h = g of Iff and g are holomorphic, we know that h is holomorphic. Suppose we know that f and h are holomorphic. Can we conclude anything about g? What if we know that g and h are holomorphic? IS Suppose n is a region, ({J E H(Q), qJ' has no zero in n, f E H(({J(Q)), g = f 0 qJ, Zo E n, and Wo = qJ(z0). Prove that iff has a zero of order m at w0 , then g also has a zero of order m at z0 • How is this modified if qJ' has a zero of order k at z0? ELEMENTARY PROPERTIES OF HOLOMORPHIC FUNCTIONS 229 16 Suppose J1 is a complex measure on a measure space X, n is an open set in the plane, cp is a bounded function on n x X such that cp(z, t) is a measurable function of t, for each z E n, and cp(z, t) is holomorphic in n, for each t € X. Define f(z) = L q>(z, t) dJl(t) for z E n. Prove thatf E H(!l). Hint: Show that to every compact K c: n there corresponds a constant M < oo such that cp(z, t) - cp(z0 , t) < M (z and z0 E K, t E X). z - z0 17 Determine the regions in which the following functions are defined and holomorphic : f(z) = f 1 : tz ' ioo e'z g(z) = 2 dt, 0 1 + t fl e'z h(z) = 2 dt. - 1 1 + t Hint : Either use Exercise 16, or combine Morera's theorem with Fubini's. 18 Suppose f E H(!l), D(a; r) c: n, y is the positively oriented circle with center at a and radius r, and f has no zero on y. For p = 0, the integral _ 1 i f'(z) zP dz 2ni. , f(z) is equal to the number of zeros off in D(a; r). What is the value of this integral (in terms of the zeros off) for p = 1, 2, 3, . . . ? What is the answer if zP is replaced by any cp E H(!l)? 19 Suppose! E H(U), g E H(U), and neither fnor g has a zero in U. If (n = 1, 2, 3, . . . ) find another simple relation between! and g. 20 Suppose n is a region,!,. E H(!l) for n = 1, 2, 3, . . . ' none of the functions!, has a zero in n, and {!,.} converges to funiformly on compact subsets of n. Prove that either fhas no zero in n or f(z) = 0 for all Z E !l. If !l' is a region that contains every !,(!l), and iff is not constant, prove thatf(!l) c: !l'. 21 Suppose f E H(!l), n contains the closed unit disc, and I f(z) I < 1 if I z I = 1. How many fixed points mustfhave in the disc? That is, how many solutions does the equationf(z) = z have there? 22 Suppose! E H(!l), n contains the closed unit disc, I f(z) I > 2 if I z I = 1, andf(O) = 1. Mustfhave a zero in the unit disc? 23 Suppose P ,.(z) = 1 + z/1 ! + · · · + z" /n !, Q,.(z) = P ,.(z) - 1, where n = 1, 2, 3, . . . . What can you say about the location of the zeros of P,. and Q,. for large n? Be as specific as you can. 24 Prove the following general form of Rouche's theorem : Let n be the interior of a compact set K in the plane. Suppose f and g are continuous on K and holomorphic in n, and I f(z) - g(z) I < I f(z) I for all z E K - n. Then f and g have the same number of zeros in n. 25 Let A be the annulus {z: r1 < I z I < r2}, where r1 and r2 are given positive numbers. (a) Show that the Cauchy formula f(z) = Ñ (i + i ) f(() d( 2nz ,. ,2 C - z is valid under the following conditions:/ E H(A), 230 REAL AND COMPLEX ANALYSIS and (0 < t r 21t). (b) Show by means of (a) that every f E H(A) can be decomposed into a sumf = f1 + f2 , wheref1 is holomorphic outside D(O; r1) and f2 e H(D(O; r2)); the decomposition is unique if we require that /1(z)-+ 0 as I z I -+ oo . (c) Use this decomposition to associate with eachf e H(A) its so-called " Laurent series " which converges to fin A. Show that there is only one such series for eachf Show that it converges to f uniformly on compact subsets of A. (d) Iff E H(A) andfis bounded in A, show that the components /1 and /2 are also bounded. (e) How much of the foregoing can you extend to the case r1 = O (or r2 = oo, or both)? (f) How much of the foregoing can you extend to regions bounded by finitely many (more than two) circles? 26 It is required to expand the function CX> in a series of the form L en z". - oo 1 1 --2 + --1 - z 3 - z How many such expansions are there? In which region is each of them valid? Find the coeffi-cients en explicitly for each of these expansions. 27 Suppose n is a horizontal strip, determined by the inequalities a < y < b, say. Suppose f E H(!l), andf(z) = f(z + 1) for all z E n. Prove thatfhas a Fourier expansion in n, CX> f = L en e21tinz, - oo which converges uniformly in {z : a + € < y < b - €}, for every € > 0. Hint: The map z -+ e21tiz con verts/to a function in an annulus. Find the integral formulas by means of which the coefficients en can be computed from f. 28 Suppose r is a closed curve in the plane, with parameter interval [0, 21t]. Take ct ¢ r. Approx imate r uniformly by trigonometric polynomials rn . Show that Indr .. (ct) = lndr"' (ct) if m and n are sufficiently large. Define this common value to be Indr (ct). Prove that the result does not depend on the choice of {rn}; prove that Lemma 10.39 is now true for closed curves, and use this to give a different proof of Theorem 10.40. 29 Define 1 il f1t d(} f (z) = -r dr iB • 1t 0 _ 1t re + z Show thatf(z) = z if I z I < 1 and thatf(z) = 1/z if I z I > 1. Thus f is not holomorphic in the unit disc, although the integrand is a holomorphic function of z. Note the contrast between this, on the one hand, and Theorem 10.7 and Exercise 16 on the other. Suggestion : Compute the inner integral separately for r < I z I and for r > I z 1 . 30 Let n be the plane minus two points, and show that some closed paths r in n satisfy assumption ( 1) of Theorem 10.35 without being null-homotopic in n. CHAPTER ELEVEN HARMONIC FUNCTIONS The Cauchy-Riemann Equations 11.1 The Operators o and a Suppose f is a complex function defined in a plane open set Q. Regard/ as a transformation which maps Q into R2, and assume that f has a differential at some point z0 e n, in the sense of Definition 7.22. For simplicity, suppose z0 = f(z0) = 0. Our differentiability assumption is then equiv alent to the existence of two complex numbers a. and {3 (the partial derivatives off with respect to x and y at z0 = 0) such that f(z) = a.x + {3y + rJ(z)z (z = x + iy), where rJ(z)q 0 as zq 0. Since 2x = z + z and 2iy = z - z, (1) can be rewritten in the form a. - i/3 a. + i/3 f ( z) = 2 z + 2 z + 17( z )z. This suggests the introduction of the differential operators Now (2) becomes _ f( z) = (Of)(O) + (i3f)(O) · Z + rt(z) z z (z =1= 0). (1) (2) (3) (4) For real z, z/z = 1 ; for pure imaginary z, z/z = - 1. Hence f(z)jz has a limit at 0 if and only if (Jf)(O) = 0, and we obtain the following characterization of holomorphic functions : 231 232 REAL AND COMPLEX ANALYSIS 1 1.2 Theorem Suppose f is a complex function in 0 that has a diff erential at every point o fO.. Thenf E H(Q) if and only if the Cauchy-Riemann equation (8/)(z) = 0 (1) holds for every z E 0.. In that case we have f'(z) = (8/)(z) (z E 0). (2) Iff = u + iv, u and v real, ( 1) splits into the pair of equations where the subscripts refer to partial differentiation with respect to the indicated variable. These are the Cauchy-Riemann equations which must be satisfied by the real and imaginary parts of a holomorphic function. 1 1.3 The Laplacian Let f be a complex function in a plane open set 0., such that f xx and / yy exist at every point of 0.. The Laplacian off is then defined to be AJ = fxx + / yy · (1) Iff is continuous in n and if Af= O (2) at every point of 0., thenfis said to be harmonic in n. Since the Laplacian of a real function is real (if it exists), it is clear that a complex function is harmonic in n if and only if both its real part and its imaginary part are harmonic in 0.. Note that Af= 488! (3) provided that f xy = / yx , and that this happens for all f which have continuous second-order derivatives. Iff is holomorphic, then 8! = O,fhas continuous derivatives of all orders, and therefore (3) shows : 1 1.4 Theorem H olomorphic functions are harmonic. We shall now turn our attention to an integral representation of harmonic functions which is closely related to the Cauchy formula for holomorphic func tions. It will show, among· other things, that every real harmonic function is locally the real part of a holomorphic function, and it will yield information about the boundary behavior of certain classes of holomorphic functions in open discs. HARMONIC FUNCTIONS 233 The Poisson Integral 11.5 The Poisson Kernel This is the function 00 Pr(t) = L r'"'eint (0 < r < 1, t real). (1) - oo We may regard Pr(t) as a function of two variables r and t or as a family of functions of t, indexed by r. If z = rei8 (0 < r < 1, (} real), a simple calculation, made in Sec. 5.24, shows that [eit + z] 1 - r2 P (fJ - t) = Re . = . r elt -z 1 - 2r cos ( (} - t) + r2 From (1) we see that 1 f1t -2 Pr(t) dt = 1 n -1t (0 < r < 1). From (2) it follows that Pr(t) > 0, Pr(t) = Pr(- t), that P r( t) < P r( b) (0 < b < I t I < n ), and that (0 < b < n). (2) (3) (4) (5) These properties are reminiscent of the trigonometric polynomials Qk(t) that were discussed in Sec. 4.24. The open unit disc D(O; 1) will from now on be denoted by U. The unit circle - the boundary of U in the complex plane - will be denoted by T. Whenever it js convenient to do so, we shall identify the spaces I.!'(T) and C(T) with the corre sponding spaces of 2n-periodic functions on R 1, as in Sec. 4.23. One can also regard Pr(fJ - t) as a function of z = rei8 and eit. Then (2) becomes for z E U, eit E T. it 1 -I z 12 P(z, e ) = I it 12 e -z 11.6 The Poisson Integral Iff E L1(T) and 1 f1t F(reiÔ = l:n: , P,(fJ - t)f(t) dt, (6) (1) then the function F so defined in U is called the Poisson integral off We shall sometimes abbreviate the relation (1) to F = P[f]. (2) 234 REAL AND COMPLEX ANALYSIS Iffis real, formula 1 1.5(2) shows that P[f] is the real part of 1 In eit + z -2 it f(t) dt, n -n e -z (3) which is a holomorphic function of z = rei8 in U, by Theorem 10.7. Hence P[f] is harmonic in U. Since linear combinations (with constant coefficients) of harmo nic functions are harmonic, we see that the following is true : 11.7 Theorem Iff e L1(T) then the Poisson integral P[f] is a harmonic func tion in U. The following theorem shows that Poisson integrals of continuous functions behave particularly well near the boundary of U. 11.8 Theorem Iff e C(T) and if H fis de fined on the closed unit disc 0 by if r = 1, if 0 < r < 1, (1) then Hf e C(O). PROOF Since P,(t) > 0, formula 1 1.5(3) shows, for every g e C(T), that (0 < r < 1), (2) so that IIHgll u = llgii T (g E C(T)). (3) (As in Sec. 5.22, we use the notation llgiiE to denote the supremum of I g I on the set E.) If n= -N is any trigonometric polynomial, it follows from 1 1.5( 1) that so that Hg e C(O). N (H g)(rei8) = L e n r1nlein8, n= -N (4) (5) Finally, there are trigonometric polynomials gk such that llgk -f II Tq 0 as k q oo . (See Sec. 4.24.) By (3), it follows that IIHgk - Hfll u = IIH(gk -f)ll uq 0 (6) as kࡍ oo . This says that the functions Hgk e C(O) converge, uniformly on 0, to H f Hence H f e C(U). /Ill HARMONIC FUNCTIONS 235 Note: This theorem provides the solution of a boundary value problem (the Dirichlet problem): A continuous functionfis given on T and it is required to find a harmonic function F in U " whose boundary values are f." The theorem exhibits a solution, by means of the Poisson integral of f, and it states the relation between f and F more precisely. The uniqueness theorem which corresponds to this existence theorem is contained in the following result. 11.9 Theorem Suppose u is a continuous real f unction on the closed unit disc 0, and suppose u is harmonic in U. Then (in U) u is the Poisson integral of its restriction to T, and u is the real part of the holomorphic function 1 f1t eit + z . f(z) = -2 it u(e't) dt n 1t e - z (z e U). (1) PRooF Theorem 10.7 shows that f e H(U). If u1 = Re f, then (1) shows that u1 is the Poisson integral of the boundary values of u, and the theorem will be proved as soon as we show that u = u1• Put h = u - u1• Then h is continuous on 0 (apply Theorem 1 1.8 to u1), h is harmonic in U, and h = 0 at all points of T. Assume (this will lead to a contradiction) that h(z0) > 0 for some z0 e U. Fix E so that 0 < E < h(z0), and define g(z) = h(z) + Ei z 12 (z e 0). (2) Then g(z0) > h(z0) > E. Since g e C(U) and since g = E at all points of T, there exists a point z 1 E U at which g has a local maximum. This implies that gxx < 0 and gYY < 0 at z 1• But (2) shows that the Laplacian of g is 4E > 0, and we have a contradiction. Thus u - u1 < 0. The same argument shows that u1 - u < 0. Hence u = u 1, and the proof is complete. I I I I 11.10 So far we have considered only the unit disc U = D(O; 1). It is clear that the preceding work can be carried over to arbitrary circular discs, by a simple change of variables. Hence we shall merely summarize some of the results : If u is a continuous real function on the boundary of the disc D(a; R) and if u is defined in D(a; R) by the Poisson integral 1 f1t R2 2 ifJ - r it u(a + re ) = -2 R2 2R ((} ) 2 u(a + Re ) dt n _ 1t -r cos - t + r then u is continuous on D(a; R) and harmonic in D(a; R). (1) If u is harmonic (and real) in an open set Q and if D(a; R) c Q, then u satisfies (1) in D(a; R) and there is a holomorphic function f defined in D(a; R) whose real part is u. This f is uniquely defined, up to a pure imaginary additive constant. For if two functions, holomorphic in the same region, have the same real part, their difference must be constant (a corollary of the open mapping theorem, or the Cauchy-Riemann equations). 236 REAL AND COMPLEX ANALYSIS We may summarize this by saying that every real harmonic function is locally the real part of a holomorphic function. Consequently, every harmonic function has continuous partial derivatives of all orders. The Poisson integral also yields information about sequences of harmonic functions : 11.1 1 Harnack's Theorem Let { un} be a sequence o f harmonic f unctions in a region n. (a) /funr u uniformly on compact subsets of!l, then u is harmonic in Q. (b) If u1 < u2 < u3 < · · · , then either { un} converges uniformly on compact subsets o jQ, or Un(z)r 00 f or every Z E Q. PROOF To prove (a), assume D(a; R) c n, and replace u by un in the Poisson integral 1 1.10(1). Since un Ò u uniformly on the boundary of D(a; R), we con clude that u itself satisfies 1 1.10(1) in D(a; R). In the proof of (b), we may assume that u1 > 0. (If not, replace un by un - u1.) Put u = sup un , let A = {z E Q: u(z) < oo}, and B = Q - A. Choose D(a; R) c Q. The Poisson kernel satisfies the inequalities R - r R2 - r2 R + r --< < --R + r - R2 - 2rR cos (0 - t) + r - R - r for 0 < r < R. Hence The same inequalities hold with u in place of un. It follows that either u(z) = oo for all z e D(a; R) or u(z) < oo for all z e D(a; R). Thus both A and B are open; and since Q is connected, we have either A = 0 (in which case there is nothing to prove) or A = n. In the latter case, the monotone convergence theorem shows that the Poisson formula holds for u in every disc in Q. Hence u is harmonic in Q. Whenever a sequence of continuous functions converges monotonically to a continuous limit, the con vergence is uniform on compact sets (, Theorem 7.13). This completes the proof. //// HARMONIC FUNCTIONS 237 The Mean Value Property 11.12 Definition We say that a continuous function u in an open set n has the mean value property if to every z E Q there corresponds a sequence {rn} such that rn > 0, rnq 0 as nq oo, and 1 f1t u(z) = -2 u(z + rn eit) dt n -1t ( n = 1, 2, 3, ... ). (1) In other words, u(z) is to be equal to the mean value of u on the circles of radius r n and with center at z. Note that the Poisson formula shows that (1) holds for every harmonic function u, and for every r such that D(z; r) c n. Thus harmonic functions satisfy a much stronger mean value property than the one that we just defined. The following theorem may therefore come as a surprise : 11.13 Theorem If a continuous function u has the mean value property in an open set n, then u is harmonic in n. PROOF It is enough to prove this for real u. Fix D(a; R) c n. The Poisson integral gives us a continuous function h on D(a; R) which is harmonic in D(a; R) and which coincides with u on the boundary of D(a; R). Put v = u - h, and let m = sup { v(z}: z E D(a; R)}. Assume m > 0, and let E be the set of al1 z E D( a; R) at which v(z) = m. Since v = 0 on the boundary of D(a; R), E is a compact subset of D(a; R). Hence there exists a z0 E E such that I Zo - a I > I z - a I for all z E E. For all small enough r, at least half the circle with center z0 and radius r lies outside E, so that the corresponding mean values of v are all less than m = v(z0). But v has the mean value property, and we have a contradic tion. Thus m = 0, so v < 0. The same reasoning applies to - v. Hence v = 0, or u = h in D(a; R), and since D(a; R) was an arbitrary closed disc in n, u is harmonic in n. I I I I Theorem 1 1.13 leads to a reflection theorem for holomorphic functions. By the upper half plane n + we mean the set of all z = x + iy with y > 0; the lower half plane n- consists of all z whose imaginary part is negative. 1 1.14 Theorem (The Schwarz reflection principle) Suppose L is a segment o f the real axis, n + is a region in n +, and every t E L is the center o f an open disc Dt such that n + n Dt lies in n+. Let n- be the re flection o f O.+ : n- = {z: .z E n+}. (1) 238 REAL AND COMPLEX ANALYSIS Suppose f = u + iv is holomorphic in Q +, and (2) f or every sequence { zn} in Q + which converges to a point o f L. Then there is a function F, holomorphic in n+ u L u n-, such that F(z) = f(z) in n+ ; this F satis fies the relation F(z) = F(z) (z e n + u L u n-). (3) The theorem asserts that f can be extended to a function which is holo morphic in a region symmetric with respect to the real axis, and (3) states that F preserves this symmetry. Note that the continuity hypothesis (2) is merely imposed on the imaginary part off PROOF Put n = n+ u L u n-. We extend v to n by defining v(z) = 0 for z E L and v(z) = -v(z) for z E n-. It is then iJnmediate that v is continuous and that v has the mean value property in n, no that v is harmonic in n, by Theorem 1 1.13. Hence v is locally the imaginary part of a holomorphic function. This means that to each of the discs Dt there corresponds an ft E H(Dt) such that Im ft = v. Each ft is determined by v up to a real additive constant. If this constant is chosen so that ft(z) = f(z) for some z e Dt n n +, the same will hold for all z E Dt n n +' since f-It is constant in the region Dt n n +. We assume that the functions ft are so adjusted. The power series expansion of ft in powers of z - t has only real coeffi cients, since v = 0 on L, so that all derivatives of ft are real at t. It follows that ft(z) = ft(z) (4) Next, assume that Ds n Dt =I= 0. Then ft =f=!s in Dt n Ds n n + ; and since Dt n Ds is connected, Theorem 10.18 shows that ft(z) = fs(z) Thus it is consistent to define f(z) F(z) = ft(z) f(z) for z e n+ for z e Dt for Z E g-(5) (6) and it remains to show that F is holomorphic in n-. If D(a; r) c: n-, then D(a; r) c: g+, so for every z E D(a; r) we have 00 f(z) = L c"<z - a)". (7) n = O HARMONIC FUNCTIONS 239 Hence 00 F(z) = L cn(z - a)n (z e D(a; r)). (8) n=O This completes the proof. Ill/ Boundary Behavior of Poisson Integrals 11.15 Our next objective is to find analogues of Theorem 1 1.8 for Poisson inte grals of If-functions and measures on T. Let us associate to any function u in U a family of functions ur on T, defined by (0 < r < 1). (1) Thus ur is essentially the restriction of u to the circle with radius r, center 0, but we shift the domain of ur to T. Using this terminology, Theorem 1 1.8 can be stated in the following form: If f e C(T) and F = P[f], then Frqf uniformly on T, as r-+ 1. In other words, lim IIFr -/11 oo = 0, (2) r-1 which implies of course that lim Fr(ei8) = f(ei9) (3) r-1 at every point of T. As regards (2), we shall now see (Theorem 1 1.16) that the corresponding norm-convergence result is just as easy in I!'. Instead of confining ourselves to investigating radial limits, as in (3), we shall then study nontangential limits of Poisson integrals of measures and If-functions; the differentiation theory devel oped in Chap. 7 will play an essential role in that study. 1 1.16 Theorem If 1 < p < oo ,f e I!'(T), and u = P[f], then (0 < r < 1). (1) I/ 1 s p < oo , then lim llur - fli P = 0. (2) r-1 PROOF If we apply Jensen's inequality (or Holder's) to 1 J1t u,(eiÔ = l:n: , /(t)P,(() - t) dt (3) 240 REAL AND COMPLEX ANALYSIS we obtain (4) If we integrate (4) with respect to fJ, over [ -n, n] and use Fubini's theorem, we obtain (1 ). Note that formula 1 1.5(3) was used twice in this argument. To prove (2), choose E > 0, and choose g E C(T) so that llg -fliP < E (Theorem 3.14). Let v = P[g]. Then Ur -f = (ur - vr) + (vr - g) + (g -f). By (1), llur - vriiP = ll(u - v)riiP < 11/- giiP < E. Thus llur -fliP < 2E + llvr - giiP (5) (6) for all r < 1. Also, llvr - giiP < llvr - gil 00 , and the latter converges to 0 as rq 1, by Theorem 1 1.8. This proves (2). /Ill 11.17 Poisson Integrals of Measures If J1 is a complex measure on T, and if we want to replace integrals over T by integrals over intervals of length 2n in R 1, these intervals have to be taken half open, because of the possible presence of point masses in J1. To avoid this (admittedly very minor) problem, we shall keep integration on the circle in what follows, and will write the Poisson integral u = P[dJ1] of J1 in the form u(z) = J/(z, eif) dJL(eiь (z E U) where P(z, eit) = (1 -I z 12)/ I eit -z 12, as in formula 1 1.5(6). (1) The reasoning that led to Theorem 1 1.7 applies without change to Poisson integrals of measures. Thus u, de fined by (1), is harmonic in U. Setting IIJ111 = I J1 1 (T), the analogue of the first half of Theorem 1 1.16 is (2) To see this, replace J1 by I J1 1 in (1), apply Fubini's theorem, and refer to formula 1 1.5(3). 1 1.18 Approach Regions For 0 < C( < 1, we define Qa to be the union of the disc D(O; C() and the line segments from z = 1 to points of D(O; C(). In other words, Qa is the smallest convex open set that contains D(O; C() and has the point 1 in its boundary. Near z = 1, Qa is an angle, bisected by the radius of U that terminates at 1, of opening 2fJ, where C( = sin 0. Curves that approach 1 within Qa cannot be tangent to T. Therefore Qa is called a nontangential approach region, with vertex 1. HARMONIC FUNCTIONS 241 The regions Qa expand when C( increases. Their union is U, their intersection is the radius [0, 1 ). Rotated copies of Qa, with vertex at eit, will be denoted by eitga. 11.19 Maximal Functions If 0 < C( < 1 and u is any complex function with domain U, its nontangential maximal function N a u is defined on T by (1) Similarly, the radial maximal function of u is (Mrad u)(ei' = sup { I u(reit) I : 0 < r < 1 }. (2) If u is continuous and A is C1 positive number, then the set where either of these maximal functions is <A is a closed subset of T. Consequently, N a u and M rad u are lower semicontinuous on T; in particular, they are measurable. Clearly, M rad u < N a u, and the latter increases with C(. If u = P[dJ.L], Theorem 1 1.20 will show that the size of N a u is, in turn, controlled by the maximal func tion MJ.L that was defined in Sec. 7.2 (taking k = 1). However, it will simplify the notation if we replace ordinary Lebesgue measure m on T by u = mj2n. Then u is a rotation-invariant positive Borel measure on T, so normalized that u(T) = 1. Accordingly, MJ.L is now defined by (M )( i8) I J.L I (I) J.l e = sup u(J) . (3) The supremum is taken over all open arcs I c T whose centers are at ei8, includ ing T itself (even though T is of course not an arc). Similarly, the derivative DJ.L of a measure J.L on T is now .8 • J.L(l) (DJL)(e' ) = hm u(J) , (4) as the open arcs I c T shrink to their center ei8, and ei8 is a Lebesgue point of f E L1(T) if (5) where {I} is as in (4). If dJ.L = f du + dJ.Ls is the Lebesgue decomposition of a complex Borel measure J.L on T, where f E L1(T) and J.ls i a, Theorems 7.4, 7.7, and 7.14 assert that (6) that almost every point of T is a Lebesgue point off, and that DJ.L = f, DJ.Ls = 0 a.e. [ u ]. 242 REAL AND COMPLEX ANALYSIS We will now see, for any complex Borel measure 11 on T, that the non tangential and radial maximal functions of the harmonic function P[dJJ,] are con trolled by M 11· In fact, if any one of them is finite at some point of T, so are the others; this can be seen by combining Theorem 1 1.20 with Exercise 19. 1 1.20 Theorem Assume 0 < a. < 1. Then there is a constant ca > 0 with the f ollowing property: I f 11 is a positive finite Borel measure on T and u = P[dJJ,] is its Poisson integral, then the inequalities (1) hold at every point ei6 E T. PROOF We shall prove ( 1) for () = 0. The general case follows then if the special case is applied to the rotated measure JJ.8(E) = JJ.(ei8E). Since u(z) = J T P(z, eit) dJJ,(eit), the first inequality in (1) will follow if we can show that Ca P(z, eit) < P( I z I, eit) (2) holds for all z E Qa and all eit E T. By formula 1 1.5(6), (2) is the same as (3) where r = I z 1 . The definition of n(% shows that I z - r I /(1 - r) is bounded in n(% ' say by Ya . Hence 1 eit - r I < I eit -z I + I z - r I < I eit -z I + }'a( 1 - r) < ( 1 + }'a) I eit -z I so that (3) holds with CCX = ( 1 + }'a) - 2• This proves the first half of ( 1 ). For the second half, we have to prove that (0 < r < 1). (4) Fix r. Choose open arcs Ii c T, centered at 1, so that I 1 c I 2 c · · · c In_1, put In = T. For 1 < j < n, let Xi be the characteristic function of Ii, and let hi be the largest positive number for which hi Xi< Pr on T. Define n K = L (hi- hi+1)Xi i=1 (5) where hn+ 1 = 0. Since Pr(t) is an even function of t that decreases as t increases from 0 to n, we see that hi-hi+ 1 > 0, that K = hi on Ii-Ii_1 (putting I 0 = 0), and that K < P r . The definition of M JJ. shows that JJ,(I j) < (M JJ,)(1 )a( I j). (6) HARMONIC FUNCTIONS 243 Hence, setting (MJl)(1) = M, i K dJ1 = it(hi - hi+ 1)J1(li) < M J1(hi - hi+ 1)u(Ii) = M i K du < M I TP, du = M. (7) Finally, if we choose the arcs Ii so that their endpoints form a sufficiently fine partition of T, we obtain step functions K that converge to Pr , uniformly on T. Hence (4) follows from (7). /Ill 11.21 Nontangential Limits A function F, defined tn U, ts said to have non tangential limit A at ei8 e T if, for each C( < 1, lim F(zi) = A j- oo for every sequence { zi} that converges to ei8 and that lies in ei8Qa. 11.22 Theorem I f Jl is a positive Borel measure on T and (DJl)(ei8) = 0 for some fJ, then its Poisson integral u = P[dJl] has nontangential limit 0 at ei8• PROOF By definition, the assumption (DJl)(ei8) = 0 means that lim Jl(I)/u(I) = 0 (1) as the open arcs I c T shrink to their center ei8• Pick E > 0. One of these arcs, say I 0 , is then small enough to ensure that Jl(I) < Ea(I) (2) for every I c I 0 that has ei8 as center. Let Jlo be the restriction of J1 to I 0 , put Jlt = J1 -Jlo, and let ui be the Poisson integral of Jli (i = 0, 1). Suppose zi converges to ei8 within some region ei8Qa. Then zi stays at a positive distance from T - I 0 • The inte grands in (3) converge therefore to 0 as j / oo , uniformly on T - I 0 • Hence lim u1(zi) = 0. (4) j- oo Next, use (2) together with Theorem 1 1.20 to see that (5) 244 REAL AND COMPLEX ANALYSIS lim sup u0(z i) < E/Crz . (6) i- oo Since u = u0 + u1 and E was arbitrary, (4) and (6) give lim u(zi) = 0. (7) j- oo /Ill 11.23 Theorem Iff e L1(T), then PUJ has nontangential limit f(ei8) at every Lebesgue point ei8 off PROOF Suppose ei8 is a Lebesgue point off By subtracting a constant from/ we may assume, without loss of generality, thatf(ei8) = 0. Then lim cr:I) i I f I da = 0 (1) as the open arcs I c T shrink to their center ei8• Define a Borel measure J1 on T by (2) Then (1) says that (DJ1}(ei8) = 0; hence P[dJ1] has nontangential limit 0 at ei8, by Theorem 1 1.22. The samࡌ is true of P[f], because I P[f] I < P[ I f I] = P[dJ1]. The last two theorems can be combined as follows. (3) Ill/ 11.24 Theorem I f dJ1 = f du + dJ1s is the Lebesgue decomposition o f a complex Borel measure J1 on T, where f e L1(T), Jls l u, then P[dJ1] has nontangential limit f(f!i9) at almost all points o f T. PROOF Apply Theorem 1 1.22 to the positive and negative variations of the real and imaginary parts of Jls , and apply Theorem 1 1.23 tDf //// Here is another consequence of Theorem 1 1.20. 11.25 Theorem For 0 < C( < 1 and 1 < p < oo, there are constants A( C(, p) < oo with the following properties: HARMONIC FUNCTIONS 245 (a) If J1 is a complex Borel measure on T, and u = P[dJ1], then (b) If 1 < p < oo,f E I!(T), and u = P[f], then (0 < A. < oo). II N a ull p < A( ex, p)llfll p . PROOF Combine Theorem 1 1.20 with Theorem 7.4 and the inequality (7) in the proof of Theorem 8.18. I I I I The nontangential maximal functions Na u are thus in weak L1 if u = P[dJ1], and they are in I.!'(T) if u = P[f] for some f E I!(T), p > 1. This latter result may be regarded as a strengthened form of the first part of Theorem 1 1.16. Representation Theorems 11.26 How can one tell whether a harmonic function u in U is a Poisson integral or not? The preceding theorems (1 1.16 to 1 1.25) contain a number of necessary conditions. It turns out that the simplest of these, the I! -boundedness of the family { ur: 0 < r < 1} is also sufficient ! Thus, in particular, the boundedness of 11urll 1, as rq 1, implies the existence of nontangential limits a.e. on T, since, as we will see in Theorem 1 1.30, u can then be represented as the Poisson integral of a measure. This measure will be obtained as a so-called " weak limit " of the functions ur . Weak convergence is an important topic in functional analysis. We will approach it through another important concept, called equicontinuity, which we will meet • again later, in connection with the so-called " normal families " of holomorphic functions. 11.27 Definition Let ] be a collection of complex functions on a metric space X with metric p. We say that ] is equicontinuous if to every E > 0 corresponds a b > 0 such that I f(x) -f(y) I < E for every f E ] and for all pairs of points x, y with p(x, y) < b. (In particular, every f E §" is then uniformly continuous.) We say that ] is pointwise bounded if to every x E X corresponds an M(x) < oo such that I f(x) I < M(x) for every f E ]-11.28 Theorem (Arzela-Ascoli) Suppose that ] is a pointwise bounded equi continuous collection o f complex functions on a metric space X, and that X contains a countable dense subset E. Every sequence {fn} in ff has then a subsequence that converges uniformly on every compact subset of X. 246 REAL AND COMPLEX ANALYSIS PROOF Let xb x2 , x3 , .
.
• be an enumeration of the points of E. Let S0 be the set of all positive integers. Suppose k > 1 and an infinite set Sk _ 1 c S0 has been chosen. Since {fn(xk): n E Sk_ 1} is a bounded sequence of complex numbers, it has a convergent subsequence. In other words, there is an infinite set sk c sk - 1 so that lim fn(xk) exists as n----+ 00 within sk . Continuing in this way, we obtain infinite sets S0 :::J S 1 :::J S2 :::J • • • with the property that lim fn(x) exists, for 1 < j < k, if n----+ 00 within sk . Let rk be the kth term of Sk (with respect to the natural order of the positive integers) and put For each k there are then at most k - 1 terms of S that are not in Sk . Hence lim fn(x) exists, for every x E E, as n----+ oo within S. (The construction of S from { Sk} is the so-called diagonal process.) Now let K c X be compact, pick E > 0. By equicontinuity, there is a Ī > 0 so that p(p, q) < Ī implies I fn(p) -fn(q) I < E, for all n. Cover K with open balls B1, • • • , BM of radius Ī/2. Since E is dense in X there are points Pi E Bi n E for 1 < i < M. Since Pi E E, lim fn(Pi) exists, as n----+ oo within S. Hence there is an integer N such that for i = 1, . . . , M, if m > N, n > N, and m and n are in S. To finish, pick x E ע- Then x E Bi for some i, and p(x, Pi) < Ī- Our choice of Ī and N shows that < E + E + E = 3€ if m > N, n > N, m E S, n E S. 11.29 Theorem Suppose that (a) X is a separable Banach space, (b) {An} is a sequence o f linear functionals on X, (c) supiiAnll = M < oo . n Then there is a subsequence {An;} such that the limit Ax = lim An; X i-ao exists for every x EX. Moreover, A is linear, and II All < M. Ill/ (1) (In this situation, A is said to be the weak limit of {An;} ; see Exercise 18.) HARMONIC FUNCTIONS 247 PROOF To say that X is separable means, by definition, that X has a count able dense subset. The inequalities show that {An} is pointwise bounded and equicontinuous. Since each point of X is a compact set, Theorem 1 1.28 implies that there is a subsequence { AnJ such that {An; x} converges, for every x e X, as i q oo. To finish, define A by (1). It is then clear that A is linear and that II All < M. /Ill Let us recall, for the application that follows, that C(T) and I.!'(T) (p < oo) are separable Banach spaces, because the trigonometric polynomials are dense in them, and because it is enough to confine ourselves to trigonometric polynomials whose coefficients lie in some prescribed countable dense subset of the complex field. 11.30 Theorem Suppose u is harmonic in U, 1 < p < oo, and sup lluriiP = M < oo. (1) O<r< 1 (a) If p = 1, it f ollows that there is a unique complex Borel measure J1 on T so that u = P[dJ1]. (b) If p > 1, it follows that there is a unique f e I.!'(T) so that u = f[f]. (c) Every positive harmonic function in U is the Poisson integral o f a unique positive Borel measure on T. PROOF Assume first that p = 1. Define linear functionals Ar on C(T) by (0 < r < 1). (2) By (1), IIArll < M. By Theorems 1 1.29 and 6.19 there is a measure J1 on T, with IIJ111 < M, and a sequence riq 1, so that for every g e C(T). lim r gu,j da = r g djJ. j- oo JT JT (3) Put h1{z) = u(ri z). Then hi is harmonic in U, continuous on 0, and is therefore the Poisson integral of its restriction to T (Theorem 1 1.9). Fix z e U, and apply (3) with (4) 248 REAL AND COMPLEX ANALYSIS Since hieit) = urieit), we obtain u(z) = lim u(ri z) = lim hiz) j j = lim j P(z, eiŭhiei1) da(eiŭ j JT = LP(z, ei1) dJL(ei1) = PdJL. If 1 < p < oo, let q be the exponent conjugate to p. Then I!!
(T) is separa ble. Define Ar as in (2), but for all g E I!!
(T). Again, II Arll < M. Refer to Theo rems 6.16 and 1 1.29 to deduce, as above, that there is an f E I.!'(T), with II fliP < M, so that (3) holds, with / du in place of dJ1, for every g E I!!
(T). The rest of the proof is as it was in the case p = 1. This establishes the existence assertions in (a) and (b). To prove unique ness, it suffices to show that P[dJ1] = 0 implies J1 = 0. Pick f E C(T), put u = P[f], v = P[dJ1]. By Fubini's theorem, and the symmetry P(rei8, eit) = P(reit, ei8), fu, dJL = fvJ da (0 < r < 1). (5) When v = 0 then vr = 0, and since u,q f uniformly, as rq 1, we conclude that Lġ djL = o (6) for every f E C(T) if P[dJ1] = 0. By Theorem 6.19, (6) implies that J1 = 0. Finally, (c) is a corollary of (a), since u > 0 implies (1) with p = 1 : L I u, I da = L u, da = u(O) (0 < r < 1) (7) by the mean value property of harmonic functions. The functionals A, used in the proof of (a) are now positive, hence J1 > 0. /Ill 1 1.31 Since holomorphic functions are harmonic, all of the preceding results (of which Theorems 1 1.16, 1 1.24, 1 1.25, 1 1.30 are the most significant) apply to holo morphic functions in U. This leads to the study of the HP-spaces, a topic that will be taken up in Chap. 17. At present we shall only give one application, to functions in the space H00• This, by definition, is the space of all bounded holomorphic functions in U ; the norm 11!11 oo = sup { lf(z) I: z E U} turns H00 into a Banach space. HARMONIC FUNCTIONS 249 As before, L00(T) is the space of all (equivalence classes of) essentially bounded functions on T, normed by the essential supremum norm, relative to Lebesgue measure. For g E L00(T), II gil oo stands for the essential supremum of I g 1 . 1 1.32 Theorem To every f E H00 corresponds a function f E L00(T), de fined almost everywhere by f ( e;8) = lim f (re;8). The equality 11/11 oo = II/ II oo holds. r- 1 (1) If f(ei8) = 0 for almost all ei8 on some arc I c T, then f(z) = 0 for every Z E U. (A considerably stronger uniqueness theorem will be obtained later, in Theorem 15.19. See also Theorem 17.18 and Sec. 17.19.) PROOF By Theorem 1 1.30, there is a unique g E L00(T) such that/= P[g]. By Theorem 1 1.23, (1) holds with / = g. The inequality 11/ll oo < 11/ ll oo follows from Theorem 1 1. 16(1); the opposite inequality is obvious. In particular, iff = 0 a.e., then II! II 00 = 0, hence 11/11 00 = 0, hencef = 0. Now choose a positive integer n so that the length of I is larger than 2njn. Let C( = exp { 2ni/n} and define n F(z) = [1 f(C(kz) (z E U). (2) k= 1 Then F E H00 and F = 0 a.e. on T, hence F(z) = 0 for all z E U. If Z(/), the zero set off in U, were at most countable, the same would be true of Z(F), since Z(F) is the union of n sets obtained from Z(f) by rotations. But Z(F) = U. Hence/ = 0, by Theorem 10.18. //// Exercises 1 Suppose u and v are real harmonic functions in a plane region n. Under what conditions is uv harmonic? (Note that the answer depends strongly on the fact that the question is one about real functions.) Show that u2 cannot be harmonic in n, unless u is constant. For which f E H(!l) is I f 12 harmonic? 2 Suppose f is a complex function in a region n, and both f and f2 are harmonic in n. Prove that either for fis holomorphic in n. 3 If u is a harmonic function in a region n, what can you say about the set of points at which the gradient of u is 0? (This is the set on which ux = uY = 0.) 4 Prove that every partial derivative of every harmonic function is harmonic. Verify, by direct computation, that Pr((} - t) is, for each fixed t, a harmonic function of rei8• Deduce (without referring to holomorphic functions) that the Poisson integral P[d,u] of every finite Borel measure ,u on T is harmonic in U, by showing that every partial derivative of P[d,u] is equal to the integral of the corresponding partial derivative of the kernel. 5 Suppose f E H(!l) and f has no zero in n. Prove that log I f I is harmonic in n, by computing its Laplacian. Is there an easier way? 250 REAL AND COMPLEX ANALYSIS 6 Suppose! E H(U), where U is the open unit disc,fis one-to-one in U, Q = f(U), andf(z) = L cn z". Prove that the area of Q is Hint: The Jacobian off is I f' 12. 7 (a) Iff E H(Q),f(z) =I= 0 for z E n, and - 00 < tX < oo, prove that by proving the formula in which t/J is twice differentiable on (0, oo) and q>(t) = tt/J"(t) + t/J'(t). (b) Assume f E H(Q) and is a complex function with domain f(Q), which has continuous second-order partial derivatives. Prove that L[<1> o JJ = [(L\<1>) o JJ · I f' 1 2. Show that this specializes to the result of (a) if (w) = ( I w 1). 8 Suppose Q is a region, !, E H(Q) for n = 1, 2, 3, . . . , un is the real part of!, , { un} converges uni formly on compact subsets of Q, and {f n(z)} converges for at least one z E Q. Prove that then {!,} converges uniformly on compact subsets of Q. 9 Suppose u is a Lebesgue measurable function in a region Q, and u is locally in L1• This means that the integral of I u I over any compact subset of Q is finite. Prove that u is harmonic if it satisfies the following form of the mean value property : u(a) = 11:: 2 JJ u(x, y) dx dy D(a; r) whenever .i5(a; r) c: n. 10 Suppose I = [a, b] is an interval on the real axis, q> is a continuous function on I, and Show that f(z) = -. dt 1 lb q>(t) 2nz a t - z lim [f(x + iE) -f(x - iE)] exists for every real x, and find it in terms of q>. (z ¢ /). (E > 0) How is the result affected if we assume merely that qJ E I! ? What happens then at points x at which qJ has right- and left-hand limits? 1 1 Suppose th(lt I = [a, b ], Q is a region, I c: Q, f is continuous in Q, and f E H(Q - /). Prove that actually f E H(Q). Replace I by some other sets for which the same conclusion can be drawn. 12 (Harnack's Inequalities) Suppose Q is a region, K is a compact subset of n, Zo E n. Prove that there exist positive numbers a. and {3 (depending on z0 , K, and Q) such that for every positive harmonic function u in Q and for all z E K. HARMONIC FUNCTIONS 251 If {un} is a sequence of positive harmonic functions in n and if un(zo)---+ 0, describe the behavior of { un} in the rest of Q. Do the same if un(z0)---+ oo. Show that the assumed positivity of { un} is essential for these results. 13 Suppose u is a positive harmonic function in U and u(O) = 1. How large can u(!) be? How small? Get the best possible bounds. 14 For which pairs of lines L1, L2 do there exist real functions, harmonic in the whole plane, that are 0 at all points of L1 u L2 without vanishing identically? 15 Suppose u is a positive harmonic function in U, and u(rei8)---+ 0 as r---+ 1, for every ei8 =1- 1. Prove that there is a constant c such that 16 Here is an example of a harmonic function in U which is not identically 0 but all of whose radial limits are 0: Prove that this u is not the Poisson integral of any measure on T and that it is not the difference of two positive harmonic functions in U. 17 Let be the set of all positive harmonic functions u in U such that u(O) = 1. Show that is a convex set and find the extreme points of <1>. (A point x in a convex set is called an extreme point of if x lies on no segment both of whose end points lie in and are different from x.) Hint: If C is the convex set whose members are the positive Borel measures on T, of total variation 1, show that the extreme points of C are precisely those f.J. E C whose supports consist of only one point of T. 18 Let X be the dual space of the Banach space X. A sequence {An} in X is said to converge weakly to A E X if Anx---+ Ax as n---+ oo , for every x E X. Note that An---+ A weakly whenever An---+ A in the norm of X. (See Exercise 8, Chap. 5.) The converse need not be true. For example, the functionals f---+ j(n) on I3(T) tend to 0 weakly (by the Bessel inequality), but each of these functionals has norm 1. Prove that {IIAnll} must be bounded if {An} converges weakly. 19 (a) Show that bPr(b) > 1 if b = 1 - r. (b) If f.J. > 0, u = P[df.J.], and /ı c: T is the arc with center 1 and length 2b, show that f.J.(/ ı) < bu( 1 - b) and that therefore ( M f.J. )( 1) < n( M rad U )( 1 ). (c) If, furthermore, f.J. _i m, show that a.e. [JJ.]. Hint: Use Theorem 7. 15. 20 Suppose E c: T, m(E) = 0. Prove that there is anf E H00, withf(O) = 1, that has lim f(rei8) = 0 r-+ 1 at every ei6 E E. Suggestion: Find a lower semicontinuous t/1 E L1(T), t/1 > 0, t/1 = + oo at every point of E. There is a holomorphic g whose real part is P[t/1]. Letf = 1/g. 21 Define f E H(U) and g E H(U) by f(z) = exp {(1 + z)/(1 - z)}, g(z) = (1 - z) exp { -f(z)}. Prove that g(ei6) = lim g(re;6) r .... l exists at every e;6 E T, that g E C(T), but that g is not in H00• 252 REAL AND COMPLEX ANALYSIS Suggestion: Fix s, put t + is - 1 z = ----t t + is + 1 (0 < t < 00 ). For certain values of s, I g(zr) I . oo as t . oo . 22 Suppose u is harmonic in U, and { u,: 0 < r < 1} is a uniformly integrable subset of I.J(T). (See Exercise 10, Chap. 6.) Modify the proof of Theorem 11.30 to show that u = P[f] for some f E I.J(T). 23 Put (}n = 2 -n and define 00 u(z) = L n-2 { P(z, ei6") - P(z, e-i9")}, n = l for z E U. Show that u is the Poisson integral of a measure on T, that u(x) = 0 if - 1 < x < 1, but that u(1 - € + i€) is unbounded, as € decreases to 0. (Thus u has a radial limit, but no nontangential limit, at 1.) Hint: If € = sin (} is small and z = 1 - € + i€, then P(z, ei9) - P(z, e-i9) > 1/€. 24 Let Dn(t) be the Dirichlet kernel, as in Sec. 5.11, define the Fejer kernel by 1 K N = ( D 0 + D 1 + . . . + D N), N + 1 put LN(t) = min (N, n2/Nt2). Prove that and that J T LN du < 2. 1 1 - cos Nt KN- l(t) = - . < LN(t) N 1 - cos t Use this to prove that the arithmetic means So + sl + . . . + SN (1 - -----N -N + 1 of the partial sums sn of the Fourier series of a function! E L1(T) converge to f(ei9) at every Lebesgue point of f. (Show that sup I u N I is dominated by M f, then proceed as in the proof of Theorem 11.23.) 25 If 1 < p < oo and f E I!(R1), prove that (f • h;.)(x) is a harmonic function of x + iA. in the upper half plane. (h;. is defined in Sec. 9.7 ; it is the Poisson kernel for the half plane.) CHAPTER TWELVE THE MAXIMUM MODULUS PRINCIPLE Introduction 12.1 The maximum modulus theorem (10.24) asserts that the constants are the only homomorphic functions in a region Q whose absolute values have a local maximum at any point of n. Here is a restatement : I f K is the closure of a bounded region n, iff is contin uous on K and holomorphic in Q, then I f ( z) I < II f II an for every z E n. I f equality holds at one point z E n, then f is constant. [The right side of (1) is the supremum of I ! I on the boundary an of n.] (1) For if I f(z) I > II/ II an at some z E n, then the maximum of I f I on K (which is attained at some point of K, since K is compact) is actually attained at some point of n, so fis constant, by Theorem 10.24. · The equality 11 /ll oo = 11/ ll oo , which is part of Theorem 1 1.32, implies that I /(z) I < 11 /11 oo (z E U, f E H00(U)). (2) This says (roughly speaking) that I f(z) I is no larger than the supremum of the boundary values off, a statement similar to (1). But this time boundedness on U is enough ; we do not need continuity on 0. This chapter contains further generalizations of the maximum modulus theorem, as well as some rather striking applications of it, and it concludes with a theorem which shows that the maximum property " almost " characterizes the class of holomorphic functions. 253 254 REAL AND COMPLEX ANALYSIS The Schwarz Lemma This is the name usually given to the following theorem. We use the notation established in Sec. 1 1.31. 12.2 Theorem Supposef e H00, 11/lloo < 1, andf(O) = 0. Then I f(z) I < I z I I /'(0) I < 1 ; (z e U), (1) (2) if equality holds in (1) for one z e U - {0}, or if equality holds in (2), then f(z) = A.z, where A. is a constant, I A. I = 1. In geometric language, the hypothesis is that f is a holomorphic mapping of U into U which keeps the origin fixed; part of the conclusion is that either f is a rotation or/moves each z e U - {0} closer to the origin than it was. PROOF Since f(O) = 0, f(z)lz has a removable singularity at z = 0. Hence there exists g e H(U) such thatf(z) = zg(z). If z E U and I z I < r < 1, then I ( ) I I f(reiÔ I 1 g z < max <-. 8 r r Letting r----+ 1, we see that I g(z) I < 1 at every z e U. This gives (1). Since /'(0) = g(O), (2) follows. If I g(z) I = 1 for some z e U, then g is constant, by another application of the maximum modulus theorem. I I I I Many variants of the Schwarz lemma can be obtained with the aid of the following mappings of U onto U: 12.3 Definition For any C( e U, de fine Z-C( cp«(z) = 1 - . -C(Z 12.4 Theorem Fix C( e U. Then <p« is a one-to-one mapping which carries T onto T, U onto U, and C( to 0. The inverse of cp« is cp -« . We have (1) PROOF <p« is holomorphic in the whole plane, except for a pole at 11ࡋ which lies outside U. Straightforward substitution shows that (2) THE MAXIMUM MODULUS PRINCIPLE 255 Thus CfJa is one-to-one, and <p -a is its inverse. Since, for real t, eit - C( I eit - C( I -- 1 1 - &.eit - I e-it - &. I -(3) (z and z have the same abs.olute value), CfJa maps T into T; the same is true of <fJ-a; hence <fJa(T) = T. It now follows from the maximum modulus theorem that CfJa(U) c U, and consideration of cp-a shows that actually CfJa(U) = U. /Ill 12.5 An Extremal Problem Suppose C( and P are complex numbers, I C( I < 1, and I P I < 1. How large can 1 /'(C() I be iffis subject to the conditionsfe H00, llfll oo < 1, andf(C() = P? To solve this, put g = (/)p o f o CfJ -a • (1) Since CfJ -a and (/)p map U onto U, we see that g E H00 and llgll oo < 1; also, g(O) = 0. The passage from f to g has reduced our problem to the Schwarz lemma, which gives I g'(O) I < 1. By (1 ), the chain rule gives g'(O) = cp{lP)f'(C()cp'a(O). If we use Eqs. 12.4(1), we obtain the inequality , < 1 - I P l2 I f (C() I - 1 _ I C( 12 • (2) (3) This solves our problem, since equality can occur in (3). This happens if and only if I g'(O) I = 1, in which case g is a rotation (Theorem 12.2), so that (z E U) (4) for some constant A. with I A. I = 1. A remarkable feature of the solution should be stressed. We imposed no smoothness conditions (such as continuity on 0, for instance) on the behavior of f near the boundary of U. Nevertheless, it turns out that the functions f which maximize I /'( C() I under the stated restrictions are actually rational functions. Note also that these extremal functions map U onto U (not just into) and that they are one-to-one. This observation may serve as the motivation for the proof of the Riemann mapping theorem in Chap. 14. At present, we shall merely show how this extremal problem can be used to characterize the one-to-one holomorphic mappings of U onto U. 12.6 Theorem Suppose f e H(U), f is one-to-one, f(U) = U, C( e U, and f(C() = 0. Then there is a constant A., I A. I = 1, such that (z e U). (1) In other words, we obtain/by composing the mapping CfJa with a rotation. 256 REAL AND COMPLEX ANALYSIS PROOF Let g be the inverse off, defined by g(f(z)) = z, z E U. Since f is one to-one, f' has no zero in U, so g E H(U), by Theorem 10.33. By the chain rule, g'(O)f'((X) = 1. (2) The solution of 12.5, applied to f and to g, yields the inequalities 1 I f'((X) I < 1 _ I (X 12 , (3) By (2), equality must hold in (3). As we observed in the preceding problem (with {3 = 0), this forcesfto satisfy (1). /Ill The Phragmen-Lindelof Method 12.7 For a bounded region Q, we saw in Sec. 12.1 that iff is continuous on the closure of Q and iff E H(Q), the maximum modulus theorem implies llflln = 11/ llan · For unbounded regions, this is no longer true. To see an example, let (1) n = {z = x + iy: - ; < y < ;}; (2) Q is the open strip bounded by the parallel lines y = + n/2; its boundary an is the union of these two lines. Put f(z) = exp (exp (z)). (3) For real x, 1( x + Ğ) = exp ( + iex) (4) since exp (ni/2) = i, so I f(z) I = 1 for z E an. Butf(z)---+ oo very rapidly as x---+ oo along the positive real axis, which lies in Q. "Very " is the key word in the preceding sentence. A method developed by Phragmen and Lindelof makes it possible to prove theorems of the following kind: Iff E H(Q) and if I f I < g, where g(z)---+ oo " slowly " as z ---+ oo in Q (just what " slowly " means depends on Q), then f is actually bounded in Q, and this usually implies further conclusions about/, by the maximum modulus theorem. Rather than describe the method by a theorem which would cover a large number of situations, we shall show how it works in two cases. In both, Q will be a strip. In the first, f will be assumed to be bounded, and the theorem will improve the bound; in the second, a growth condition will be imposed onfwhich just excludes the function (3). In view of later applications, Q will be a vertical strip in Theorem 12.8. THE MAXIMUM MODULUS PRINCIPLE 257 First, however, let us mention another example which also has this general flavor: Supposefis an entire function and I f(z) I < 1 + I z 1112 (5) for all z. Thenfis constant. This follows immediately from the Cauchy estimates 10.26, since they show that f(O) = 0 for n = 1, 2, 3, . . . . 12.8 Theorem Suppose Q = {x + iy: a < x < b}, Q = {x + iy: a < x < b}, (1) f is continuous on n, f E H(Q), and suppose that I f(z) I < B for all z E Q and somefixed B < oo . If M(x) = sup { I f(x + iy) I : - oo < y < oo} then we actually have (a < x < b) (a < x < b). (2) (3) Note: The conclusion (3) implies that the inequality I f I < B can be replaced by I f I < max (M(a), M(b)), so that I f I is no larger in Q than the supremum of I f I on the boundary of Q. If we apply the theorem to strips bounded by lines x = C( and x = p, where a < C( < p < b, the conclusion can be stated in the following way: Corollary Under the hypotheses of the theorem, log M is a convex function on (a, b). PROOF We assume first that M(a) = M(b) = 1. In this case we have to prove that I f(z) I < 1 for all z E n. For each E > 0, we define an auxiliary function 1 hiz) = -1 _ + E ( z---a -) (z E 0). (4) Since Re {1 + E(Z - a)} = 1 + E(X - a) > 1 in n, we have I h£ I < 1 in n, so that I f(z)h£(z) I < 1 Also, 1 1 + E(z - a) I > E I y I , so that B I f (z )h,(z) I < E I y I (z E iJQ). (5) (z = X + iy E Q). (6) Let R be the rectangle cut off from Q by the lines y = + B/E. By (5) and (6), I .fh£ I < 1 on oR, hence I .fh£ I < 1 on R, by the maximum modulus theorem. B.ut (6) shows that I .fh£ I < 1 on the rest of n. Thus I f(z)hl(z) I < 1 258 REAL AND COMPLEX ANALYSIS for all z e Q and all E > 0. If we fix z e Q and then let E / 0, we obtain the desired result I f(z) I < 1. We now turn to the general case. Put g(z) = M(a)(b-z)/(b-a)M(b)(z-a)/(b-a), where, for M > 0 and w complex, Mw is defined by Mw = exp (w log M), and log M is real. Then g is entire, g has no zero, 1/g is bounded in 0, I g(a + iy) I = M(a), I g(b + iy) I = M(b), (7) (8) (9) and hence f /g satisfies our previous assumptions. Thus I f /g I < 1 in n, and this gives (3). (See Exercise 7.) //// 12.9 Theorem Suppose n = {x + iy: l y l < ;} (1) Suppose f is continuous on n, f E H(Q), there are constants C( < 1, A < oo , such that l f(z) l < exp {A exp (C( I x l )} (z = X + iy E Q), (2) and f(x + Ğ) < 1 ( - 00 < X < 00 ). (3) Then l f(z) l < 1 for all z E 0. Note that the conclusion does not follow if C( = 1, as is shown by the function exp (exp z). PROOF Choose {3 > 0 so that rx < {3 < 1. For E > 0, define h£(z) = exp { - E(ePz + e-Pz)}. (4) For z E n, Re [ePz + e-Pz] = (ePx + e-Px) cos {3y > b(ePx + e-Px) (5) where b = cos (f3n/2) > 0, since I /3 1 < 1. Hence l h£(z) l < exp { - eb(ePx + e-Px)} < 1 (z e Q). (6) It follows that I jh£ I < 1 on an and that I f(z)hiz) I < exp {Aealxl - Eb(ePx + e-Px)} (z E. Q). (7) THE MAXIMUM MODULUS PRINCIPLE 259 Fix E > 0. Since Eb > 0 and P > C(, the exponent in (7) tends to - oo as x---+ + oo. Hence there exists an x0 so that the right side of (7) is less than 1 for all x > x0 . Since I fhf. l < 1 on the boundary of the rectangle whose ver tices are + x0 + (ni/2), the maximum modulus theorem shows that actually I fhf. l < 1 on this rectangle. Thus 1 ./hf. l < 1 at every point of Q, for every E > 0. As E---+ 0, h((z)---+ 1 for every z, so we conclude that I f(z) I < 1 for all z E Q. Ill/ Here is a slightly different application of the same method. It will be used in the proof of Theorem 14. 18. 12.10 Lindelof's Theorem Suppose r is a curve, with parameter interval [0, 1], such that I r(t) I < 1 if t < 1 and r(1) = 1. I f g E H00 and lim g(r(t)) = L, (1) t-l then g has radial limit L at 1. (It follows from Exercise 14, Chap. 14, that g actually has nontangential limit L at 1.) PROOF Assume I g I < 1, L = 0, without loss of generality. Let E > 0 be given. There exists t0 < 1 so that, setting r0 = Re r(t0), we have 1 I g(r(t)) I < E and Re r(t) > ro > 2 as soon as t0 < t < 1. Pick r, r 0 < r < 1. Define h in Q = D(O; 1) n D(2r; 1) by h(z) = g(z)g(z)g(2r - z)g(2r - z). Then h E H(Q) and I h I < 1. We claim that I h(r) I < E. Since h(r) = I g(r) 14, the theorem follows from (4). (2) (3) (4) To prove (4), let £1 = r([t1, 1]), where t1 is the largest t for which Re r(t) = r, let E2 be the reflection of E 1 in the real axis, and let E be the union of £1 u £2 and its reflection in the line x = r. Then (2) and (3) imply that Pick c > 0, define I h(z) I < E if z e Q n E. hc(z) = h(z)(1 - zY(2r - 1 - z)c (5) (6) 260 REAL AND COMPLEX ANALYSIS for z e Q, and put hc(1) = hc(2r - 1) = 0. If K is the union of E and the bounded components of the complement of E, then K is compact, he is con tinuous on K, holomorphic in the interior of K, and (5) implies that I he I < E on the boundary of K. Since the construction of E shows that r e K, the maximum modulus theorem implies that I hc(r) I < E. Letting c ----+ 0, we obtain (4). /Ill An Interpolation Theorem 12.11 The convexity theorem 12.8 can sometimes be used to prove that certain linear transformations are bounded with respect to certain If-norms. Rather than discuss this in full generality, let us look at a particular situation of this kind. I Let X be a measure space, with a positive measure J1, and suppose { t/1 n} (n = 1, 2, 3, . . . ) is an orthonormal set of functions in /J(J1); we recall what this means: if m = n, if m =F n. (1) Let us also assume that { t/1 n} is a bounded sequence in L00(J1): There exists an M < oo such that (n = 1, 2, 3, . . . ; x e X). Then for any f e I!(J1), where 1 < p < 2, the integrals ( n = 1, 2, 3, .. . ) exist and define a function Jon the set of all positive integers. There are now two very easy theorems: For f e L1(J1), (2) gives and for f e L2(J1), the Bessel inequality gives . where the norms are defined as usual : II! li P = [J I f lp dp. J'p, and 11/1100 = supn 1/(n) 1 . 11/llq = [L 1/(n) lq]lfq, (2) (3) (4) (5) (6) Since (1, oo) and (2, 2) are pairs of conjugate exponents, one may conjecture that 11/11 q is finite whenever f e I!'(J1) and 1 < p < 2, q = p/(p - 1 ). This is indeed true and can be proved by " interpolation " between the preceding trivial cases p = 1 and p = 2. THE MAXIMUM MODULUS PRINCIPLE 261 12.12 The Hausdorff-Young Theorem Under the above assumptions, the inequality IIlii q < M(2-p)/p II f II p holds if 1 < p < 2 and iff E I!(J1). PROOF We first prove a reduced form of the theorem. (1) Fix p, 1 < p < 2. Letf be a simple complex function such that II/IlP = 1, and let bh . . . , bN be complex numbers such that L I bn IP = 1. Our objective is the inequality N L bn/(n) < M<2-p)/P. (2) n= l Put F = If IP, and put Bn = I bn IP. Then there is a function cp and there are complex numbers {31, • • • , PN such that and f = F11Pcp, b = B11Pf3 n n n ' lcpl = 1, 1 Pn 1 = 1, IF dp. = 1, (3) N L Bn = 1. (4) n= l If we use these relations and the definition of/(n) given in Sec. 12.11, we obtain .. tl b,.l(n) = ..tl Bש!Pp,. l pl!Pq>,ji" dp.. (5) Now replace 1/p by z in (5), and define Cl)(z) = J1 Bת P,. 1Fzq>,ji,. dJl. (6) for any complex number z. Recall that Az = exp (z log A) if A > 0; if A = 0, we agree that Az = 0. Since F is simple, since F > 0, and since Bn > 0, we see that is a finite linear combination of such exponentials, so is an entire function which is bounded on {z: a < Re (z) < b} for any finite a and b. We shall take a = t and b = 1, shall estimate on the edges of this strip, and shall then apply Theorem 12.8 to estimate <1>(1/p). For - oo < y < oo, define c,.(y) = 1 pl/2 pillq>,ji,. dJl.. (7) 262 REAL AND COMPLEX ANALYSIS The Bessel inequality gives f. I c,.(y) 12 < [ IF112 FiYq> 12 dJJ. = [ I F I dJJ. = 1, n= l Jx Jx and then the Schwarz inequality shows that (8) I q,(Ț + iy) I = ..t̑ B!'2 B!ifJ .. c.. < ttl B .. .
.. tl I c .. 12 r,2 < 1. (9) The estimate 1<1>(1 + iy)l < M ( - oo < y < oo) (10) follows trivially from (3), (4), and (6), since 111/!nll 00 < M. We now conclude from (9), (10), and Theorem 12.8 that l(x + iy)l < M2x-l (Ț < x < 1, - oo < y < oo). (1 1) With x = 1/p and y = 0, this gives the desired inequality (2). The proof is now easily completed. Note first that Lt. I /(n) lqr/q = sup .. t. b,/(n) ' (12) the supremum being taken over all {b1, ... , bN} with L I bn IP = 1, since the I!J norm of any function on any measure space is equal to its norm as a linear functional on I!. Hence (2) shows that { N }1/q 11צ11/(n) lq < M<2-pl/PIIf11P for every simple complex f e I!(p,). (13) If now f e I!(p, ), there are simple functions jj such that II jj -f II P---+ 0 as j---+ oo . Then/ in) -+ /(n) for every n, because 1/Jn e JJ(JL). Thus since (13) holds for each.fj , it also holds forf Since N was arbitrary, we finally obtain (1). /Ill A Converse of the Maximum Modulus Theorem We now come to the theorem which was alluded to in the introduction of the present chapter. The letter j will denote the identity function: j(z) = z. The function which assigns the number 1 to each z e 0 will be denoted by 1. 12.13 Theorem Suppose M is a vector space of continuous complex functions on the closed unit disc 0, with the following properties: (a) 1 eM. (b) Iff e M, then also if e M. (c) Iffe M, then llfll u = ll ffi T · Then every f e M is holomorphic in U. THE MAXIMUM MODULUS PRINCIPLE 263 Note that (c) is a rather weak form of the maximum modulus principle; (c) asserts only that the overall maximum of I f I on 0 is attained at some point of the boundary T, but (c) does not a priori exclude the existence of local maxima of l f l in U. PROOF By (a) and (b), M contains all polynomials. In conjunction with (c), this shows that M satisfies the hypotheses of Theorem 5.25. Thus every f e M is harmonic in U. We shall use (b) to show that every f e M actually satisfies the Cauchy-Riemann equation. Let a and a be the differential operators introduced in Sec. 1 1.1. The product rule for differentiation gives (aJ)(fg) =f . (aJg) + (af) . (Jg) + (J f) . (ag) + (aaj) . g. Fix f e M, and take g = j. Then jj e M. Hence f and jj are harmonic, so aaj = 0 and (aJ)(jj) = 0. Also, Jj = 0 and aj = 1. The above identity therefore reduces to J f= 0. Thusfe H(U). //// This result will be used in the following proof. 12.14 Rado's Theorem Assume f e C(U), n is the set of all z E U at whichf(z) =F 0, andfis holomorphic in n. Thenfis holomorphic in U. In particular, the theorem asserts that U - n is at most countable, unless n = 0. PROOF Assume n =I= 0. We shall first prove that n is dense in u. If not, there exist C( e n and {3 e U - Q such that 2 1 {3 - C( I < 1 - I /3 1 . Choose n so that 2n I /( C() I > II f II T . Define h(z) = (z - {3)-"f(z), for z E n. If z E u n an, then f(z) = 0, hence h(z) = 0. If z E T n an, then This contradicts the maximum modulus theorem. Thus n is dense in U. Next, let M be the vector space of all g e C(O) that are holomorphic in n. Fix g E M. For n = 1, 2, 3, . . . , f gn = 0 on U n an. The maximum modulus theorem implies therefore, for every C( E n, that I f(C() I I g(C() In < 11/gnllan = 11 /gnll T < II/ II Tllgll} · If we take nth roots and then let nq oo , we see that I g(C() I < II gil T, for every C( E n. Since n is dense in U, II gil u = II gil T . It follows that M satisfies the hypotheses of Theorem 12.13. Since f e M, f is holomorphic in U. I I I I 264 REAL AND COMPLEX ANALYSIS Exercises 1 Suppose L\ is a closed equilateral triangle in the plane, with vertices a, b, c. Find max ( I z - a I I z - b I I z - c I ) as z ranges over L. 2 Suppose f E H(II + ), where II + is the upper half plane, and I f I < 1. How large can I f'(i) I be? Find the extremal functions. (Compare the discussion in Sec. 12.5.) 3 Suppose f E H(O.). Under what conditions can I f I have a local minimum in 0.? 4 (a) Suppose Q is a region, D is a disc, i5 c: Q, f E H(Q),f is not constant, and I f I is constant on the boundary of D. Prove thatfhas at least one zero in D. (b) Find all entire functions/such that l f(z) l = 1 whenever l z l = 1. 5 Suppose Q is a bounded region, {f n} is a sequence of continuous functions on fi which are holo morphic in Q, and {fn} converges uniformly on the boundary of 0.. Prove that {f n} converges uni formly on fi. 6 Suppose f E H(Q), r is a cycle in n such that Indr (IX) = 0 for all (X ¢ 0., I /(() I < 1 for every ' E r' and Indr (z) =F 0. Prove that I f(z) I < 1. 7 In the proof of Theorem 12.8 it was tacitly assumed that M(a) > 0 and M(b) > 0. Show that the theorem is true if M(a) = 0, and that thenf(z) = 0 for all z E 0.. 8 If 0 < R1 < R2 < oo, let A(R1, R2) denote the annulus {z: R1 < l z l < R2}. There is a vertical strip which the exponential function maps onto A(R1, R2). Use this to prove Hadamard's three-circle theorem: Iff E H(A(R1, R2)), if M(r) = max I f(rei6) I 6 and if R 1 < a < r < b < R 2 , then log (b/r) log (r/a) log M(r) < log M(a) + log M(b). log (b/a) log (b/a) [In other words, log M(r) is a convex function of log r.] For which f does equality hold in this inequality? 9 Let n be the open right half plane (z E n if and only if Re z > 0). Suppose f is continuous on the closure of n (Re z > O),f E H(TI), and there are constants A < oo and a < 1 such that I f(z) I < A exp ( I z I«) for all z E n. Furthermore, I f(iy) I < 1 for all real y. Prove that I f(z) I < 1 in n. Show that the conclusion is false for IX = 1. How does the result have to be modified if n is replaced by a region bounded by two rays through the origin, at an angle not equal to n? 10 Let n be the open right half plane. Suppose that f E H(TI), that I f(z) I < 1 for all z E n, and that there exists IX, -n/2 < IX < n/2, such that Prove thatf = 0. log I f(rei«) I --------+- - oo as r----+-oo. r Hint: Put gn(z) = f(z)e"z, n = 1, 2, 3, . . . . Apply Exercise 9 to the two angular regions defined by - n/2 < (} < IX, (X < (} < n/2. Conclude that each gn is bounded in n, and hence that I gn I < 1 in n, for all n. 1 1 Suppose r is the boundary of an unbounded region n, f E H(O.), f is continuous on Q u r, and there are constants B < oo and M < oo such that I f I < M on r and I f I < B in Q. Prove that we then actually have I f I < M in n. THE MAXIMUM MODULUS PRINCIPLE 265 Suggestion: Show that it involves no loss of generality to assume that U n Q = 0. Fix z0 E Q, let n be a large integer, let V be a large disc with center at 0, and apply the maximum modulus theorem to the functionfn(z)/z in the component of V n Q which contains z0 . 12 Letfbe an entire function. If there is a continuous mapping y of [0, 1) into the complex plane such that y(t)---+ oo and f(y(t)) ----. ct as t---+ 1, we say that ct is an asymptotic value off [In the complex plane, " y( t)---+ oo as t ---+ 1 " means that to each R < oo there corresponds a t R < 1 such that I y( t) I > R if tR < t < 1.] Prove that every nonconstant entire function has oo as an asymptotic value. Suggestion: Let En = { z : I f(z) I > n }. Each component of En is unbounded (proof?) and contains a component of En+ 1 , by Exercise 1 1. 13 Show that exp has exactly two asymptotic values : 0 and oo . How about sin and cos? Note: sin z and cos z are defined, for all complex z, by eiz - e-iz sln z = ---2i eiz + e-iz cos z = ---2 14 Iff is entire and if ct is not in the range off, prove that ct is an asymptotic value off unless f is constant. IS Suppose f E H(U). Prove that there is a sequence { zn} in U such that I zn 1 ---+ 1 and {f(zn)} is bounded. 16 Suppose Q is a bounded region,/ E H(Q), and lim sup I f(zn) I < M n .... oo for every sequence { zn} in Q which converges to a boundary point of Q. Prove that I f(z) I < M for all Z E Q. 17 Let be the set of allf E H(U) such that 0 < I f(z) I < 1 for z e U, and let <>c be the set of allf E that have f(O) = c. Define M(c) = sup { l f'(O) I : f e <>c}, M = sup { l f'(O) I : f e <1>}. Find M, and M(c) for 0 < c < 1. Find anf e withf'(O) = M, or prove that there is no suchf Suggestion: log fmaps U into the left half plane. Compose logfwith a properly chosen map that . takes this half plane to U. Apply the Schwarz lemma. CHAPTER THIRTEEN APPROXIMATIONS BY RATIONAL FUNCTIONS Preparation 13.1 The Riemann Sphere It is often convenient in the study of holomorphic functions to compactify the complex plane by the adjunction of a new point called oo. The resulting set S2 (the Riemann sphere, the union of R2 and { oo }) is topologized in the following manner. For any r > 0, let D'( oo ; r) be the set of all complex numbers z such that I z I > r, put D( oo ; r) = D'( oo ; r) u { oo },_ and declare a subset of S2 to be open if and only if it is the union of discs D(a; r), where the a's are arbitrary points of S2 and the r's are arbitrary positive numbers. On S2 - { oo }, this gives of course the ordinary topology of the plane. It is easy to see that S2 is homeomorphic to a sphere (hence the notation). In fact, a homeo morphism <p of S2 onto the unit sphere in R3 can be explicitly exhibited: Put · <p( oo) = (0, 0, 1 ), and put .9 (2r cos fJ 2r sin fJ r2 - 1) rn(re' ) -.., -r2 + 1 ' r2 + 1 ' r2 + 1 (1) for all complex numbers rei9• We leave it to the reader to construct the geometric picture that goes with (1). Iff is holomorphic in D'( oo ; r), we say thatfhas an isolated singularity at oo. The nature of this singularity is the same as that which the function j, defined in D'(O; 1/r) by /(z) = f(1/z), has at 0. Thus if f is bounded in D'( oo ; r), then limz-oo f(z) exists and is a complex number (as we see if we apply Theorem 10.20 to /), we define f(oo) to be this limit, and we thus obtain a function in D( oo ; r) which we call holomorphic: note that this is defined in terms of the behavior of J near 0, and not in terms of differentiability off at oo. 266 APPROXIMATIONS BY RATIONAL FUNCTIONS 267 If J has a pole of order m at 0, then f is said to have a pole of order m at oo ; the-principal part off at oo is then an ordinary polynomial of degree m (compare Theorem 10.21), and if we subtract this polynomial from f, we obtain a function with a removable singularity at oo . Finally, if]has an essential singularity at 0, then/is said to have an essential singularity at oo . For instance, every entire function which is not a polynomial has an essential singularity at oo . Later in this chapter we shall encounter the condition " S2 - Q is connected," where Q is an open set in the plane. Note that this is not equivalent to the condition " the complement of Q relative to the plane is connected." For example, if Q consists of all complex z = x + iy with 0 < y < 1, the complement of Q rela tive to the plane has two components, but S2 - Q is connected. 13.2 Rational Functions A rational function f is, by definition, a quotient of two polynomials P and Q :f = P/Q. It follows from Theorem 10.25 that every noncon stant polynomial is a product of factors of degree 1. We may assume that P and Q have no such factors in common. Thenfhas a pole at each zero of Q (the pole off has the same order as the zero of Q). If we subtract the corresponding prin cipal parts, we obtain a rational function whose only singularity is at oo and which is therefore a polynomial. Every rational function/ = P/Q has thus a representation of the form k f(z) = A 0(z) + L Ai(z - a)- 1) (1) j = 1 where A0 , Ab . . . , Ak are polynomials, Ab . . . , Ak have no constant term, and ab . . . , ak are the distinct zeros of Q; ( 1) is called the partial fractions decomposition off We turn to some topological considerations. We know that every open set in the plane is a countable union of compact sets (closed discs, for instance). However, it will be convenient to have some additional properties satisfied by these com pact sets : 13.3 Theorem Every open set Q in the plane is the union o f a sequence {Kn}, n = 1, 2, 3, . . . , of compact sets such that (a) Kn lies in the interior of Kn + 1,for n = 1, 2, 3, . . . . (b) Every compact subset of!l lies in some Kn . (c) Every component of S2 - Kn contains a component of S2 - !l,for n = 1, 2, 3, . . . . Property (c) is, roughly speaking, that Kn has no holes except those which are forced upon it by the holes in Q. Note that Q is not assumed to be connected. The interior of a set E is, by definition, the largest open subset of E. 268 REAL AND COMPLEX ANALYSIS PROOF For n = 1, 2, 3, . . . , put V,. = D(oo ; n) u U D(a; !) a ¢ 0 n (1) and put Kn = S2 - V,. . [Of course, a =1= oo in (1).] Then Kn is a closed and bounded (hence compact) subset of Q, and Q = U Kn . If z e Kn and r = n- 1 - (n + 1)- 1, one verifies easily that D(z; r) c Kn + 1. This gives (a). Hence Q is the union of the interiors W, of Kn . If K is a compact subset of Q, then K c w1 u . . . u W N for some N, hence K c KN . Finally, each of the discs in (1) intersects S2 - Q; each disc is connected; hence each component of .V, intersects S2 - Q; since v,. ::::> S2 - Q, no com ponent of S2 - n can intersect two components of v,. . This gives (c). /Ill 13.4 Sets of Oriented Intervals Let be a finite collection of oriented intervals in the plane. For each point p, let m1(p)[mE(p)] be the number of members of that have initial point [end point] p. If m1(p) = mE(P) for every p, we shall say that is balanced. If is balanced (and nonempty), the following construction can be carried out. Pick }'1 = [a0 , a1] e <1>. Assume k > 1, and assume that distinct members y1, . . . , Yk of have been chosen in such a way that }'; = [a;_1, a;] for 1 < i < k. If ak = a0 , stop. If ak =I= a0 , and if precisely r of the intervals y 1, • • • , Yk have ak as end point, then only r - 1 of them have ak as initial point; since is balanced, contains at least one other interval, say Yk + h whose initial point is ak . Since is finite, we must return to a0 eventually, say at the nth step. Then }' 1, . . . , Yn join (in this order) to form a closed path. The remaining members of still form a balanced collection to which the above construction can be applied. It follows that the members of can be so numbered that they form finitely many closed paths. The sum of these paths is a cycle. The following conclusion is thus reached. If = {y1, .
• • , YN} is a balanced collection of oriented intervals, and if r = y 1 + · · · + YN then r is a cycle. 13.5 Theorem I f K is a compact subset of a plane open set Q ( =I= 0), then there is a cycle r in Q - K such that the Cauchy formula f(z) = Ð I f(C) dC 2nz Jr ' - z holds for every f e H(Q) and for every z e K. (1) PROOF Since K is compact and n is open, there exists an '1 > 0 such that the distance from any point of K to any point outside n is at least 21]. Construct APPROXIMATIONS BY RATIONAL FUNCTIONS 269 a grid of horizontal and vertical lines in the plane, such that the distance between any two adjacent horizontal lines is 1'/, and likewise for the vertical lines. Let Q1, . .. , Qm be those squares (closed 2-cells) of edge 11 which are formed by this grid and which intersect K. Then Qr c Q for r = 1, . . . , m. If ar is the center of Qr and ar + b is one of its vertices, let Yrk be the oriented interval and define [ ·kb ·k + lb] y rk = ar + l ' ar + l JQr = Yr1 .f- Yr2 .f- Yr3 .f- Yr4 (r = 1, . . . , m). (2) (3) It is then easy to check (for example, as a special case of Theorem 10.37, or by means of Theorems 10.1 1 and 1 0.40) that IndvQ. (oc) = g if a. is in the interior of Qr , if a. is not in Qr . (4) Let I: be the collection of all Yrk (1 < r < m, 1 < k < 4). It is clear that I: is balanced. Remove those members of I: whose opposites (see Sec. 10.8) also belong to I:. Let be the collection of the remaining members of I:. Then is balanced. Let r be the cycle constructed from <1>, as in Sec. 13.4. If an edge E of some Qr intersects K, then the two squares in whose boundaries E lies intersect K. Hence I: contains two oriented intervals which are each other's opposites and whose range is E. These intervals do not occur in <1>. Thus r is a cycle in Q -K. The construction of from I: shows also that m Indr (a.) = L IndaQr (a.) r= 1 if a. is not in the boundary of any Qr . Hence (4) implies if a. is in the interior of some Qr , if a. lies in no Qr . (5) (6) If z E K, then z ¢ r, and z is a limit point of the interior of some Qr . Since the left side of (6) is constant in each component of the complement of r, (6) gives lndr (z) = {̏ if z E K, if z ¢ n. Now (1) follows from Cauchy's theorem 10.35. (7) /Ill 270 REAL AND COMPLEX ANALYSIS Runge's Theorem The main objective of this section is Theorem 13.9. We begin with a slightly different version in which the emphasis is on uniform approximation on one com pact set. 13.6 Theorem Suppose K is a compact set in the plane and { C(i} is a set which contains one point in each component of S2 - K. IfO. is open, n ::::) K,f E H(O.), and E > 0, there exists a rational function R, all of whose poles lie in the pre scribed set { C(i}, such that l f(z) - R(z) l < E (1) for every z E K. Note that S2 - K has at most countably many components. Note also that the preassigned point in the unbounded component of S2 - K may very well be oo ; in fact, this happens to be the most interesting choice. PROOF We consider the Banach space C(K) whose members are the contin uous complex functions on K, with the supremum norm. Let M be the sub space of C(K) which consists of the restrictions to K of those rational functions which have all their poles in { C(i}. The theorem asserts that f is in the closure of M. By Theorem 5.19 (a consequence of the Hahn-Banach theorem), this is equivalent to saying that every bounded linear functional on C(K) which vanishes on M also vanishes at /, and hence the Riesz represent ation theorem (Theorem 6.19) shows that we must prove the following asser tion : If J1 is a complex Borel measure on K such that 1 R dp. = 0 (2) for every rational function R with poles only in the set { C(i}, and iff E H(O.), then we also have (3) So let us assume that J1 satisfies (2). Define (z E S2 - K). (4) By Theorem 10.7 (with X = K, cp(() = (), h E H(S2 - K). APPROXIMATIONS BY RATIONAL FUNCTIONS 271 Let Jt] be the component of S2 - K which contains a.i, and suppose D(a.i; r) c Jt]. If a.i =F oo and if z is fixed in D(a.i; r), then 1 _ 1. ļ (z - a.)" - Im Ɵ n + 1 ( - Z N -+ oo n = O (( - a.j) (5) uniformly for ( e K. Each of the functions on the right of (5) is one to which (2) applies. Hence h(z) = 0 for all z e D(a.i; r). This implies that h(z) = 0 for all z e Jt] , by the uniqueness theorem 1 0.18. If a.i = oo , (5) is replaced by 1 N -- = - lim L z -" - 1(" ( - Z N-+ oo n = O ( C E K, I z I > r), (6) which implies again that h(z) = 0 in D( oo ; r), hence in Jt]. We have thus proved from (2) that h(z) = 0 (z E S2 - K). (7) Now choose a cycle r in Q - K, as in Theorem 13.5, and integrate this Cauchy integral representation of f with respect to JL An application of Fubini's theorem (legitimate, since we are dealing with Borel measures and continuous functions on compact spaces), combined with (7), gives The last equality depends on the fact that r c Q -K, where h(w) = 0. Thus (3) holds, and the proof is complete. I I I I The following special case is of particular interest. 13.7 Theorem Suppose K is a compact set in the plane, S2 - K is connected, and f e H(Q), where Q is some open set containing K. Then there is a sequence { P ,.} o f polynomials such that P ,.(z) / f(z) uniformly on K. PROOF Since S2 - K has now only one component, we need only one point a.i to apply Theorem 13.6, and we may take a.i = oo. I I I I 13.8 Remark The preceding result is false for every compact K in the plane such that S2 - K is not connected. For in that case S2 - K has a bounded component V. Choose a. e V, put f(z) = (z - a.)- 1, and put 272 REAL AND COMPLEX ANALYSIS m = max { I z - ex I : z e K}. Suppose P is a polynomial, such that I P(z) -f(z) I < 1/m for all z e K. Then I ( z - ex )P( z) - 1 I < 1 (z E K). (1) In particular, ( 1) holds if z is in the boundary of V ; since the closure of V is compact, the maximum modulus theorem shows that (1) holds for every z e V; taking z = ex, we obtain 1 < 1. Hence the uniform approximation is not possible. The same argument shows that none of the cxi can be dispensed with in Theorem 13.6. We now apply the preceding approximation theorems to approximation in open sets. Let us emphasize that K was not assumed to be connected in Theorems 13.6 and 13.7 and that Q will not be assumed to be connected in the theorem which follows. 13.9 Theorem Let Q be an open set in the plane, let A be a set which has one point in each component of S2 - Q, and assume f e H(Q). Then there is a sequence { Rn} o f rational functions, with poles only in A, such that Rn----+-f uni-formly on compact subsets of!l. In the special case in which S2 - Q is connected, we may take A = { oo} and thus obtain polynomials Pn such that Pn-+-f uniformly on compact subsets o f!l. Observe that S2 - Q may have uncountably many components; for instance, we may have S2 - Q = { oo} u C, where C is a Cantor set. PROOF Choose a sequence of compact sets Kn in n, with the properties speci fied in Theorem 13.3. Fix n, for the moment. Since each component of S2 - Kn contains a component of S2 - Q, each component of S2 - Kn con tains a point of A, so Theorem 13.6 gives us a rational function Rn with poles in A such that 1 I Rn(z) -f(z) I < -n (1) If now K is any compact set in n, there exists an N such that K c Kn for all n > N. It follows from (1) that 1 I Rn(z) -f(z) I < -n which completes the proof. (z e K, n > N), (2) Ill/ APPROXIMATION;S BY RATIONAL FUNCTIONS 273 The Mittag-Leffler Theorem Runge's theorem will now be used to prove that meromorphic functions can be constructed with arbitrarily preassigned poles. 13.10 Theorem Suppose Q is an open set in the plane, A c Q, A has no limit point in Q, and to each C( E A there are associated a positive integer m(C() and a rational June tion m(«) P (%< z) = L c j. (%< z - C() -j. j= 1 Then there exists a meromorphic f unction f in Q, whose principal part at each C( E A is P « and which has no other poles in Q. PROOF We choose a sequence {Kn} of compact sets in Q, as in Theorem 13.3 : For n = 1, 2, 3, . . . , Kn lies in the interior of Kn + 1, every compact subset of Q lies in some Kn , and every component of S2 - Kn contains a component of S2 - Q. Put A1 = A n K1, and An = A n (Kn - Kn-. 1) for n = 2, 3, 4, . . . . Since An c Kn and A has no limit point in Q (hence none in Kn), each An is a finite set. Put Qn(z) = L P «(z) ( n = 1, 2, 3, . . . ). (1) « e An Since each An is finite, each Qn is a rational function. The poles of Qn lie in Kn - Kn _ 1, for n > 2. In particular, Qn is holomorphic in an open set con taining Kn_ 1• It now follows from Theorem 13.6 that there exist rational functions Rn , all of whose poles are in S2 - Q, such that (2) We claim that 00 /(z) = Q1(z) + L (Qn(z) - Rn(z)) (z e !l) (3) n=2 has the desired properties. Fix N. On KN, we have N oo f = Q1 + L (Qn - Rn) + L (Qn - Rn). (4) n=2 N+ 1 By (2), each term in the last sum in (4) is less than 2 -n on KN; hence this last series converges uniformly on KN , to a function which is holomorphic in the interior of K N . Since the poles of each Rn are outside n, 274 REAL AND COMPLEX ANALYSIS is holomorphic in the interior of K N. Thus f has precisely the prescribed principal parts in the interior of KN, and hence in Q, since N was arbitrary. /Ill Simply Connected Regions We shall now summarize some properties of simply connected regions (see Sec. 1 0.38) which illustrate the important role that they play in the theory of holomorphic functions. Of these properties, (a) and (b) are what one might call internal topological properties of Q; (c) and (d) refer to the way in which Q is embedded in S2 ; properties (e) to (h) are analytic in character; U) is an algebraic statement about the ring H(Q). The Riemann mapping theorem 14.8 is another very important property of simply connected regions. In fact, we shall use it to prove the implication U)q (a). 13.1 1 Theorem For a plane region Q, each of the following nine conditions implies all the others. (a) Q is homeomorphic to the open unit disc U. (b) Q is simply connected. (c) Indy (C() = Ofor every closed path y in Q and for every C( Ӯ S2 - Q. (d) S2 - Q is connected. (e) Every f E H(Q) can be approximated by polynomials, uniformly on compact subsets o f!l. (f) For every f E H(Q) and every closed path }' in Q, 1/(z) dz = 0. (g) To every f E H(Q) corresponds an F E H(Q) such that F' =f. (h) Iff E H(Q) and 1/ f E H(Q), there exists a g E H(Q) such thatf = exp (g). U) Iff E H(Q) and 1/ f E H(Q), there exists a <p E H(O.) such thatf = <p2• The assertion of (h) is that f has a " holomorphic logarithm " g in 0.; U) asserts that f has a " holomorphic square root " 0 such that D(a; r) c Q. For z E D(a; r) we can compute F(z) by integrating/ over a path r(a), followed by the interval [a, z]. Hence, for z E D'(a; r), F(z) - F(a) = 1 l f(() d(, z - a z - a r 1 a , z (2) and the continuity off at a implies now that F'(a) = f(a), as in the proof of Theorem 10. 14. (g) implies (h). Iff E H(Q) andfhas no zero in Q, thenf'l f E H(Q), and (g) implies that there exists a g E H(Q) so that g' =!'I f We can add a constant to g, so that exp {g(z0)} = f(z0) for some z0 E Q. Our choice of g shows that tbe derivative offe-g is 0 in Q, hence fe-g is constant (since Q is connected), and it follows that f = eg. (h) implies U). By (h),f = eg. Put qJ = exp (ыg). U) implies (a). If Q is the whole plane, then Q is homeomorphic to U: map z to zl(1 + I z I ). If Q is a proper subregion of the plane which satisfies U), then there actually exists a holomorphic homeomorphism of Q onto U (a confo_rmal mapping). This assertion is the Riemann mapping theorem, which is the main objective of the next chapter. Hence the proof of Theorem 13.1 1 will be com plete as soon as the Riemann mapping theorem is proved. (See the note following the statement of Theorem 14.8.) I I I I 276 REAL AND COMPLEX ANALYSIS The fact that (h) holds in every simply connected region has the following consequence (which can also be proved by quite elementary means): 13.12 Theorem I ff e H(Q), where Q is any open set in the plane, and iff has no zero in Q, then log I f I is harmonic in Q. PROOF To every disc D c Q there corresponds a function g e H(D) such that f = eg in D. If u = Re g, then u is harmonic in D, and I f I = e". Thus log I f I is harmonic in every disc in Q, and this gives the desired conclusion. I I I I Exercises 1 Prove that every meromorphic function on S2 is rational. 2 Let Q = { z: I z I < 1 and 1 2z - 1 1 > 1}, and suppose f E H(Q). (a) Must there exist a sequence of polynomials Pn such that Pn---+ f uniformly on compact subsets of Q? (b) Must there exist such a sequence which converges to/uniformly in Q? (c) Is the answer to (b) changed if we require more off, namely, that f be holomorphic in some open set which contains the closure of Q? 3 Is there a sequence of polynomials Pn such that Pn(O) = 1 for n = 1, 2, 3, . . . , but Pn(z)---+ 0 for every z =F 0, as n ---+ oo ? 4 Is there a sequence of polynomials P n such that lim Pn(z) = n-+ ex> 1 0 - 1 if Im z > 0, if z is real, if lm z < 0? S For n = 1, 2, 3, . . . , let An be a closed disc in U, and let Ln be an arc (a homeomorphic image of [0, 1]) in U - An which intersects every radius of U. There are polynomials Pn which are very small on An and more or less arbitrary on Ln. Show that {An}, { Ln}, and { P n} can be so chosen that the series!= I: Pn defines a function! E H(U) which has no radial limit at any point of T. In other words, for no real 8 does limr-+ 1f(rei6) exist. 6 Here is another construction of such a function. Let { nk} be a sequence of integers such that n1 > 1 and nk + 1 > 2knk . Define CX> h(z) = L 5kzn". k=l Prove that the series converges if I z I < 1 and prove that there is a constant c > 0 such that I h(z) I > c · sm for all z with I z I = 1 - (1/nm). [Hint: For such z the mth term in the series defining h(z) is much larger than the sum of all the others.] Hence h has no finite radial limits. Prove also that h must have infinitely many zeros in U. (Compare with Exercise 15, Chap. 12.) In fact, prove that to every complex number IX there correspond infinitely many z E U at which h(z) = IX. 7 Show that in Theorem 13.9 we need not assume that A intersects each component of S2 - Q. It is enough to assume that the closure of A intersects each component of S2 - Q. 8 Prove the Mittag-Leffier theorem for the case in which Q is the whole plane, by a direct argument which makes no appeal to Runge's theorem. 9 Suppose Q is a simply connected region, f E H(Q), f has no zero in Q, and n is a positive integer. Prove that there exists a g E H(Q) such that gn = f APPROXIMATIONS BY RATIONAL FUNCTIONS 277 10 Suppose Q is a region, f E H(Q), and f ¢ 0. Prove that f has a holomorphic logarithm in n if and only if/has holomorphic nth roots in Q for every positive integer n. 1 1 Suppose that f n E H(Q) (n = 1, 2, 3, . . . ), f is a complex function in n, and f(z) = limn-+ ex> f n(z) for every z E Q. Prove that Q has a dense open subset V on which f is holomorphic. Hint: Put cp = sup I fn i . Use Baire's theorem to prove that every disc in Q contains a disc on which cp is bounded. Apply Exercise 5, Chap. 10. (In general, V #- Q. Compare Exercises 3 and 4.) 12 Suppose, however, that f is any complex-valued measurable function defined in the complex plane, and prove that there is a sequence of holomorphic polynomials Pn such that limn-oo Piz) = f(z) for almost every z (with respect to two-dimensional Lebesgue measure). CHAPTER FOURTEEN CONFORMAL MAPPING Preservation of Angles 14.1 Definition Each complex number z =1= 0 determines a direction from the origin, defined by the point z A[z] = -1 z I (1) on the unit circle. Suppose f is a mapping of a region Q into the plane, z0 E Q, and z0 has a deleted neighborhood D'(z0; r) c Q in which f(z) =I= f(z0). We say that f pre serves angles at z0 if lim e-iB A[f(z0 + rei8) -f(z0)] (r > 0) (2) r- o exists and is independent of 0. In less precise language, the requirement is that for any two rays £ and £', starting at z0 , the angle which their images/(£) andf(£') make atf(z0) is the same as that made by £ and £', in size as well as in orientation. The property of preserving angles at each point of a region is character istic of holomorphic functions whose derivative has no zero in that region. This is a corollary of Theorem 14.2 and is the reason for calling holomorphic functions with nonvanishing derivative conformal mappings. 14.2 Theorem Let f map a region Q into the plane. If f'(z0) exists at some z0 E Q and f'(z0) #= 0, then f preserves angles at z0 • Conversely, if the diff eren tial off exists and is diff erent from 0 at z0 , and iff preserves angles at z0 , then f'(z0) exists and is diff erent f rom 0. 278 CONFORMAL MAPPING 279 Here f'(z0) = lim [f(z) -f(z0)]1(z - z0), as usual. The differential off at z0 is a linear transformation L of R2 into R2 such that, writing z0 = (x0 , y0), f(xo + x, Yo + y) = f(xo , Yo) + L(x, y) + (x2 + y2)11217(x, y), (1) where 17(x, y) Ò 0 as x Ò 0 and y Ò 0, as in Definition 7.22. PROOF Take z0 = f(z0) = 0, for simplicity. If /'(0) = a =1= 0, then it is imme diate that (rƈ 0), (2) so f preserves angles at 0. Conversely, if the differential off exists at 0 and is different from 0, then (1) can be rewritten in the form f(z) = rxz + {3z + I z I 17(z), (3) where 17(z) Ò 0 as z Ò 0, and rx and {3 are complex numbers, not both 0. Iff also preserves angles at 0, then . . rx + pe- 2i8 lim e-'8 A[f(re'8)] = . r-o I rx + pe- 2l8 I (4) exists and is independent of 0. We may exclude those fJ for which the denominator in (4) is 0; there are at most two such (} in [0, 2n). For all other fJ, we conclude that rx + {3e- 2i8 lies on a fixed ray through 0, and this is possible only when {3 = 0. Hence rx =1= 0, and (3) implies that /'(0) = rx. I I I I Note: No holomorphic function preserves angles at any point where its derivative is 0. We omit the easy proof of this. However, the differential of a transformation may be 0 at a point where angles are preserved. Example:f(z) = I z I z, z0 = 0. Linear Fractional Transformations 14.3 If a, b, c, and d are complex numbers such that ad - be =1= 0, the mapping az + b zÒ cz + d (1) is called a linear fractional trans formation. It is convenient to regard (1) as a mapping of the sphere S2 into S2, with the obvious conventions concerning the point oo . For instance, -die maps to oo and oo maps to ale, if c =1= 0. It is then easy to see that each linear fractional transformation is a one-to-one mapping of S2 onto S2• Furthermore, each is obtained by a superposition of transformations of the following types: (a) Translations : zÒ z + b. (b) Rotations: z ƈ az, I a I = 1. 280 REAL AND COMPLEX ANALYSIS (c) Homotheties : z-+ rz, r > 0. (d) Inversion : z-+ ·1/z. If c = 0 in ( 1 ), this is obvious. If c =F 0, it follows from the identity az + b a A ---= - + ' cz + d c cz + d (2) The first three types evidently carry lines to lines and circles to circles. This is not true of (d). But if we let §" be the family consisting of all straight lines and all circles, then !F is preserved by (d), and hence we have the important result that §" is preserved by every linear fractional transformation. [It may be noted that when §" is regarded as a family of subsets of S2, then §" consists of all circles on S2, via the stereographic projection 13.1(1); we shall not use this property of §" and omit its proof.] The proof that !F is preserved by inversion is quite easy. Elementary analytic geometry shows that every member of !F is the locus of an equation шzz + Pz + iJz + y = 0, (3) where Ŕ and y are real constants and p is a complex constant, provided that piJ > Ŕy. If Ŕ =F 0, (3) defines a circle; Ƣ = 0 gives the straight lines. Replacement of z by 1/z transforms (3) into Ŕ + Pz + Pz + yzz = 0, (4) which is an equation of the same type. Suppose a, b, and c are distinct complex numbers. We construct a linear fractional transformation <p which maps the ordered triple {a, b, c} into {0, 1, oo }, namely, ( ) (b - c)(z - a) <p z = (b - a)(z - c) (5) There is only one such <p. For if <p(a) = 0, we must have z - a in the numer ator; if <p(c) = oo, we must have z - c in the denominator; and if <p(b) = 1, we are led to (5). If a or b or c is oo , formulas analogous to (5) can easily be written down. If we follow (5) by the inverse of a transformation of the same type, we obtain the following result : For any two ordered triples {a, b, c} and {a', b', c'} in S2 there is one and only one linear fractional transformation which maps a to a', b to b', and c to. c'. (It is of course assumed that a # b, a # c, and b # c, and likewise for a', b', and c'.) We conclude from this that every circle can be mapped onto every circle by a linear fractional transformation. Of more interest is the fact that every circle can be mapped onto every straight line (if oo is regarded as part of the line), and hence that every open disc can be con formally mapped onto every open half plane. CONFORMAL MAPPING 281 Let us discuss one such mapping more explicitly, namely, 1 + z qJ(z) = 1 . - z (6) This qJ maps { - 1, 0, 1} to {0, 1, oo} ; the segment ( - 1, 1) maps onto the positive real axis. The unit circle T passes through - 1 and 1 ; hence qJ(T) is a straight line through ({J( - 1) = 0. Since T rnakes a right angle with the real axis at - 1, qJ(T) makes a right angle with the real axis at 0. Thus qJ(T) is the imaginary axis. Since qJ(O) = 1, it follows that qJ is a conformal one-to-one mapping o f the open unit disc onto the open right half plane. The role of linear fractional transformations in the theory of conformal mapping is also .well illustrated by Theorem 12.6. 14.4 Linear fractional transformations make it possible to transfer theorems con cerning the behavior of holomorphic functions near straight lines to situations where circular arcs occur instead. It will be enough to illustrate the method with an informal discussion of the reflection principle. Suppose Q is a region in U, bounded in part by an arc L on the unit circle, andfis continuous on Q, holomorphic in n, and real on L. The function Z - l t/J(z) = . z + z (1) maps the upper half plane onto U. If g = f o t/1, Theorem 11.14 gives us a holo morphic extension G of g, and then F = G o t/1- 1 gives a holomorphic extension F offwhich satisfies the relation f(z) = F(z), where z = 1/z. The last assertion follows from a property of t/1: If w = t/J(z) and w1 = t/J(z), then w1 = w, as is easily verified by computation. Exercises 2 to 5 furnish other applications of this technique. Normal Families The Riemann mapping theorem will be proved by exhibiting the mapping func tion as the solution of a certain extremum problem. The existence of this solution depends on a very useful compactness property of certain families of holo morphic functions which we now formulate. 14.5 Definition Suppose ʈ c H(Q), for some region n. We call ʈ a normal family if every sequence of members of ʈ contains a subsequence which con verges uniformly on compact subsets of n. The limit function is not required to belong to ʈ. 282 REAL AND COMPLEX ANALYSIS (Sometimes a wider definition is adopted, by merely requiring that every sequence in §' either converges or tends to oo , uniformly on compact subsets of Q. This is well adapted for dealing with meromorphic functions.) 14.6 Theorem Suppose §' c H(Q) and §' is uniformly bounded on each compact subset of the region Q. Then §' is a normal family. PROOF The hypothesis means that to each compact K c: Q there corresponds a number M(K) < oo such that I f(z) I < M(K) for allf E §' and all z E K. Let { Kn} be a sequence of compact sets whose union is Q, such that Kn lies in the interior of Kn+ 1 ; such a sequence was constructed jn Theorem 13.3. Then there exist positive numbers Łn such that (1) Consider two points z' and z" in Kn , such that I z' - z" I < Łn, let }' be the positively oriented circle with center at z' and radius 2Ł", and estimate I f(z') -f(z") I by the Cauchy formula. Since we have 1 1 z' - z" ( - z' ( - z" ( ( - z')( ( - z") ' , , z' - z" i f ( () f(z ) -f(z ) = 2ni Y (C - z')(( - z") dC, (2) and since I ( - z' I = 2Ł" and I ( - z" I > Łn for all ( E y, (2) gives the inequal ity I f(z') -f(z") I < M(`n+.) I z' - z" I , n (3) valid for all f E §' and all z' and z" E Kn, provided that I z' - z" I < Łn. This was the crucial step in the proof: We have proved, for each Kn, that the restrictions of the members of §' to Kn form an equicontinuous family. If Jj E ff, for j = 1, 2, 3, . . . , Theorem 1 1.28 implies therefore that there are infinite sets sn of positive integers, s 1 ::) s2 ::) s3 ::) . . . ' so that {Jj} con verges uniformly on Kn as j Ò oo within Sn. The diagonal process yields then an infinite set S such that {Jj} converges uniformly on every Kn (and hence on every compact K c: Q) asjÒ oo within S. /Ill The Riemann Mapping Theorem 14.7 Conformal Equivalence We call two regions Q1 and Q2 conformally equiva lent if there exists a <p E H(Q1) such that <p is one-to-one in Q1 and such that <p(Q1) = Q2 , i.e., if there exists a conformal one-to-one mapping of Q1 onto Q2 . CONFORMAL MAPPING 283 Under these conditions, the inverse of <p is holomorphic in Q2 (Theorem 10.33) and hence is a conformal mapping of n2 onto n1. It follows that conformally equivalent regions are homeomorphic. But there is a much more important relation between conformally equivalent regions : If <p is as above, fq f o <p is a one-to-one mapping of H(Q2) onto H(Q1) which pre serves sums and products, i.e., which is a ring isomorphism of H(Q2) onto H(Q 1). If Q 1 has a simple structure, problems about H(Q2) can be transferred to prob lems in H(Q1) and the solutions can be carried back to H(Q2) with the aid of the mapping function <p. The most important case of this is based on the Riemann mapping theorem (where 02 is the unit disc U), which reduces the study of H(Q) to the study of H(U), for any simply connected proper subregion of the plane. Of course, for explicit solutions of problems, it may be necessary to have rather precise information about the mapping function. 14.8 Theorem Every simply connected region Q in the plane (other than the plane itself) is conformally equivalent to the open unit disc U. Note: The case of the plane clearly has to be excluded, by Liouville's theorem. Thus the plane is not conformally equivalent to U, although the two regions are homeomorphic. The only property of simply connected regions which will be used in the proof is that every holomorphic function which has no zero in such a region has a holomorphic square root there. This will furnish the conclusion " U) implies (a) " in Theorem 13.1 1 and will thus complete the proof of that theorem. PROOF Suppose Q is a simply connected region in the plane and let w0 be a complex number, w0 rJ n. Let l: be the class of all t/1 E H(Q) which are one to-one in Q and which map Q into U. We have to prove that some t/1 E l: maps n onto u. We first prove that l: is not empty. Since Q is simply connected, there exists a <p E H(Q) so that <p2(z) = z - w0 in Q. If <p(z 1) = <p(z 2), then also <p2(z 1) = <p2(z 2), hence z 1 = z 2 ; thus <p is one-to-one. The same argument shows that there are no two distinct points z 1 and z 2 in Q such that <p(z 1) = - <p(z2). Since <p is an open mapping, <p(Q) contains a disc D(a; r), with 0 < r < I a 1 . The disc D(- a; r) therefore fails to intersect <p(Q), and if we define t/1 = r/( <p + a), we see that t/1 E l:. The next step consists in showing that if t/1 E l:, if tjJ(Q) does not cover all o f U, and if Zo E n, then there exists a t/lt E l: with It will be convenient to use the functions <fJcx defined by Z - r:t. <fJcx(z) = 1 - . - r:t.Z 284 REAL AND COMPLEX ANALYSIS For C( E U, <fJa is a one-to-one mapping of U onto U; its inverse is <p - a (Theorem 12.4). Suppose t/1 E 1:, C( E U, and C( ¢ t/1(0.). Then <fJa o t/1 E I:, and <fJa o t/1 has no zero in 0.; hence there exists a g E H(Q) such that g2 = <fJa o t/J. We see that g is one-to-one (as in the proof that I: =I= 0), hence g E I:; and if t/11 = <pp o g, where f3 = g(z0), it follows that t/11 E I:. With the notation w2 = s(w), we now have t/J = <fJ - a o S o g = <fJ - a o S o <fJ - (J o t/J1. Since t/11 (z0) = 0, the chain rule gives where F = <fJ - a o s o <fJ - p · We see that F(U) c U and that F is not one-to one in U. Therefore I F'(O) I < 1, by the Schwarz lemma (see Sec. 12.5), so that I t/J'(z0) I < lt/1'1(z0) 1 . [Note that t/J'(z0) =I= 0, since t/1 is one-to-one in 0..] Fix Zo E n, and put 11 = sup { I t/J'(z0) I : t/1 E I:}. The foregoing makes it clear that any h E I: for which I h'(z0) I = 11 will map 0. onto U. Hence the proof will be completed as soon as we prove the existence of such an h. Since I t/f(z) I < 1 for all t/1 E I: and z E 0., Theorem 14.6 shows that I: is a normal family. The definition of 11 shows that there is a sequence { t/1 n} in I: such that I t/Jъ(z0) I & 17, and by normality of I: we can extract a subsequence (again denoted by { t/ln}, for simplicity) which converges, uniformly on compact subsets of 0., to a limit h E H(O.). By Theorem 10.28, I h'(z0) I = 11· Since I: =I= 0, 11 > 0, so h is not constant. Since tPn(Q) c U, for n = 1, 2, 3, . . . , we have h(O.) c 0, but the open mapping theorem shows that actually h(O.) c U. So all that remains to be shown is that h is one-to-one. Fix distinct points z1 and z2 E 0.; put C( = h(z1) and C(n = t/ln(z1) for n = 1, 2, 3, . . . ; and let i5 be a closed circular disc in 0. with center at z 2 , such that z 1 ¢ i5 and such that h - C( has no zero on the boundary of D. This is possible, since the zeros of h - C( have no limit point in 0.. The functions t/ln - C(n converge to h - C(, uniformly on D ; they have no zero in D since they are one-to-one and have a zero at z 1 ; it now follows from Rouche's theorem that h - C( has no zero in D; in particular, h(z2) =I= h(z 1). Thus h E I:, and the proof is complete. I I I I A more constructive proof is outlined in Exercise 26. CONFORMAL MAPPING 285 14.9 Remarks The preceding proof also shows that h(z0) = 0. For if h(z0) = {3 and {3 =F 0, then qJ p o h e l:, and I (<pp o h)'(zo) I = I I h'(zo) 1 . It is interesting to observe that although h was obtained by maximizing I t/l'(z0) I for t/1 e l:, h also maximizes I /'(z0) I iff is allowed to range over the class consisting of all holomorphic mappings of n into U (not necessarily one-to-one). For if f is such a function, then g = f o h - l maps U into U, hence I g'(O) I < 1, with equality holding (by the Schwarz lemma) if and only if g is a rotation, so the chaiQ rule gives the following result : I ff E H(Q), f(Q) c U, and Zo E n, then I /'(zo) I < I h'(zo) I . Equality holds if and only iff(z) = Ah(z),for some constant A with I A I = 1. The Class !/ 14.10 Definition !/ is the class of all/ e H(U) which are one-to-one in U and ·which satisfy /(0) = 0, /'(0) = 1. (1) Thus every f e !/ has a power series expansion 00 /(z) = z + L an z" (z e U). (2) n = 2 The class !/ is not closed under addition or multiplication, but has many other interesting properties. We shall develop only a few of these in this section. Theorem 14. 1 5 will be used in the proof of Mergelyan's theorem, in Chap. 20. 14.11 Example If I C( I < 1 and then fa E !/. z 00 1 (z) -ѝ nC(" - 1 z" Jrz = (1 - C(Z)2 nGl For if fa(z) = frz(w), then (z - w)(1 - C(2zw) = 0, and the second factor is not 0 if I z I < 1 and I w I < 1. When I C( I = 1,/a is called a Koebe function. We leave it as an exercise to find the regionsf a(U). 14.12 Theorem (a) Iff e !/, I C( I = 1, and g(z) = Ct.f(C(z), then g e !/. (b) I ff e !/ there exists a g e !/ such that (z e U). (1) 286 REAL AND COMPLEX ANALYSIS PROOF (a) is clear. To prove (b), write f(z) = z<p(z). Then cp e H(U), cp(O) = 1, and cp has no zero in U, sincefhas no zero in U - {0}. Hence there exists an h e H(U) with h(O) = 1, h2(z) = cp(z). Put (z e U). (2) Then g.2(z) = z2h2(z2) = z2cp(z2) = f(z2), so that (1) holds. It is clear that g(O) = 0 and g'(O) = 1. We have to show that g is one-to-one. Suppose z and w e U and g(z) = g(w). Since f is one-to-one, (1) implies that z2 = w2• So either z = w (which is what we want to prove) or z = - w. In the latter case, (2) shows that g(z) = -g(w); it follows that g(z) = g(w) = 0, and since g has no zero in U - {0}, we have z = w = 0. /Ill 14.13 Theorem IfF e H(U - {0}), F is one-to-one in U, and then 1 00 F(z) = - + L rxn zn Z n=O 00 L n I rxn 12 < 1. n= l (z e U), (1) (2) This is usually called the area theorem, for reasons which will become apparent in the proof. PROOF The choice of rx0 is clearly irrelevant. So assume rx0 = 0. Neither the hypothesis nor the conclusion is affected if we replace F(z) by A.F(A.z) ( I A. I = 1). So we may assume that rx1 is real. Put Ur = {z: l z l < r}, Cr = {z : l z l = r}, and V,. = {z : r < l z l < 1}, for 0 < r < 1. Then F(Ur) is a neighborhood of oo (by the open mapping theorem, applied to 1/F); the sets F(Ur), F(Cr), and F(V,.) are disjoint, since F is one-to-one. Write F = u + iv, and 1 F(z) = - + rx1 z + cp(z) z For z = rei8, we then obtain (z e U), (3) (4) u = A cos () + Re cp and v = -B sin () + Im <p. (5) CONFORMAL MAPPING 287 Divide Eqs. (5) by A and B, respectively, square, and add : - + - -+ e r n + -Im r n + u2 v2 1 2 cos fJ R (Re (/))2 2 sin (} (Im 0 such that, for all sufficiently small r, (6) This says that F( Cr) is in the interior of the ellipse Er whose semiaxes are Aj1 + rJr3 and Bj1 + 1Jr3, and which therefore bounds an area nAB(l + '7r3) = n(! + rt1r)(! - rt1r)(l + 'fr3) < ; (1 + '7r3). (7) Since F(Cr) is in the interior of Er, we have Er c F(Ur); ·hence F(V,.) is in the interior of Er , so the area of F(V,.) is no larger than (7). The Cauchy Riemann equations show that the Jacobian of the mapping (x, y)q (u, v) is I F' 12• Theorem 7.26 therefore gives the following result: ; (1 + 'lr3) > f f I F' 12 v, 00 = n { r-2 - 1 + L n I an 12( 1 - r2n)}. 1 If we divide (8) by n and then subtract r-2 from each side, we obtain N L n I an 12( 1 - r2n) < 1 + rJr n= l (8) (9) for all sufficiently small r and for all positive integers N. Let r v 0 in (9), then let N v oo. This gives (2). II I I Corollary Under the same hypothesis, I a 1 I < 1. That this is in fact best possible is shown by F(z) = (11z) + az, I a I = 1, which is one-to-one in U. 288 REAL AND COMPLEX ANALYSIS 14.14 Theorem Iff e !/, and 00 f(z) = z + L an zn, n=2 then (a) I a2 1 < 2, and (b) f(U) ::) D(O; !). The second assertion is thatf(U) contains all w with I w I < !. PROOF By Theorem 14.12, there exists a g e !/ so that g2(z) = f(z2). If G = 11g, then Theorem 14.13 applies to G, and this will give (a). Since we have and hence f(z2) = zף(l + a2 z2 + · · ·), 1 1 2 1 a2 G( z) = - ( 1 - 2 a 2 z + · · ·) = - -- z + · · · . z z 2 The Corollary to Theorem 14.13 shows now that I a2 1 < 2. To prove (b), suppose w ¢f(U). Define f(z) h(z) = 1 -f(z)lw · Then h e H(U), h is one-to-one in U, and h(z) = (z + a2 z2 + · · ·{ 1 + : + . . -) = z + (a 2 + ] )z2 + . . · , so that h e !/. Apply (a) to h: We have I a2 + (11w) I < 2, and since I a2 1 < 2, we finally obtain 1 11w I < 4. So I w I > ! for every w ¢ f(U). This completes the proof. I I I I Example 14. 1 1 shows that both (a) and (b) are best possible. Moreover, given any C( =F 0, one can find entire functions f, with f(O) = 0, f'(O) = 1, that omit the value C(. For example, f(z) = C(( 1 - e-zfa). Of course, no such/ can be one-to-one in U if I C( I < !. 14.15 Theorem Suppose F e H(U - {0}), F is one-to-one in U, F has a pole o f order 1 at z = 0, with residue 1, and neither w1 nor w2 are in F(U). CONFORMAL MAPPING 289 PROOF Iff = 1/(F - w 1), then f e !/, hencef(U) :::J D(O, !), so the image of U under F - w1 contains all w with I w I > 4. Since w2 - w1 is not in this image, we have I w2 - w 1 I < 4. /Ill Note that this too is best possible: If F(z) = z- 1 + z, then F(U) does not contain the points 2, - 2. In fact, the complement of F(U) is precisely the interval [ - 2, 2] on the real axis. Continuity at the Boundary Under certain conditions, every conformal mapping of a simply connected region Q onto U can be extended to a homeomorphism of its closure Q onto U. The nature of the boundary of Q plays a decisive role here. 14.16 Definition A boundary point P of a simply connected plane region Q will be called a simple boundary point o f!l if P has the following property: To every sequence { C(n} in n such that C(n q P as n q oo there corresponds a curve y, with parameter interval [0, 1], and a sequence {tn}, 0 < t1 < t2 < · · · , tnq 1, such that y(tn) = C(n (n = 1, 2, 3, . . . ) and y(t) E Q for 0 < t < 1. In other words, there is a curve in Q which passes through the points C(n and which ends at p. 14.17 Examples Since examples of simple boundary points are obvious, let us look at some that are not simple. If n is U - {x: 0 < x < 1}, then Q is simply connected; and if O < p < 1, p is a boundary point of Q which is not simple. To get a more complicated example, let no be the interior of the square with vertices at the points 0, 1, 1 + i, and i. Remove the intervals [ 1 1 n - 1 ] [ 1 i 1 J 2n ' 2n + n i and 2n + 1 + n ' 2n + 1 + i from no . The resulting region n is simply connected. If 0 < y < 1, then iy is a boundary point which is not simple. 14.18 Theorem Let Q be a bounded simply connected region in the plane, and letfbe a conformal mapping ofO. onto U. (a) I f P is a simple boundary point o f Q, then f has a continuous extension to Q u {P}. Iffis so extended, then l f(P) I = 1. (b) If P1 and P2 are distinct simple boundary points o f!l and iff is extended to n u {Pt} u {P2} as in (a), thenf(Pt> =1= f(P2). PROOF Let g be the inverse off Then g E H{U), by Theorem 10.33, g(U) = n, g is one-to-one, and g E H00' since n is bounded. 290 REAL AND COMPLEX ANALYSIS Suppose (a) is false. Then there is a sequence { C(n} in Q such that C(n q p, f(C(2n) q w1, f(C(2n+ 1)щ w2 , and w1 #- w2 . Choose }' as in Definition 14.16, and put r(t) = f(y(t)), for 0 < t < 1. Put Kr = g(D(O; r)), for 0 < r < 1. Then Kr is a compact subset of Q. Since y(t) q P as t q 1, there exists a t < 1 (depending on r) such that y(t) ¢ Kr if t < t < 1. Thus I r(t) I > r if t < t < l. This says that l r(t) l ˰ 1 as tq l. Since r(t2n)q w1 and r(t2n + 1)q w2 , we also have I w1 1 = I w2 1 = 1. It now follows that one of the two open arcs J whose union is T- ({w1} u {w2}) has the property that every radius of U which ends at a point of J intersects the range of r in a set which has a limit point on T. Note that g(r(t)) = y(t) for 0 < t < 1 and that g has radial limits a.e. on T, since g E H00• Hence lim g(reit) = p (a.e. on J), (1) r- 1 since g(r(t)) q p as t q 1. By Theorem 1 1.32, applied to g - p, (1) shows that g is constant. But g is one-to-one in U, and we have a contradiction. Thus w1 = w2 , and (a) is proved. Suppose (b) is false. If we multiply f by a suitable constant of absolute value 1, we then have P1 #- P2 butf(p1) = f(P2) = 1. Since P1 and /32 are simple boundary points of Q, there are curves Yi with parameter interval [0, 1] such that yl[O, 1)) c Q for i = 1 and 2 and Yi(1) = pi . Put ri(t) = f(yi(t)). Then ri([O, 1)) c U, and r 1(1) = r 2(1) = 1. Since g(ri(t)) = Yi(t) on [0, 1), we have lim g(ri(t)) = Pi (i = 1, 2). (2) t- 1 Theorem 12.10 implies therefore that the radial limit of g at 1 is P1 as well as P 2 . This is impossible if P1 "# P2 • I I I I 14.19 Theorem If Q is a bounded simply connected region in the plane and if every boundary point o f!l is simple, then every conformal mapping o f!l onto U extends to a homeomorphism o fO onto 0. PROOF Suppose f e H(Q), f(Q) = U, and f is one-to-one. By Theorem 14.18 we can extend f to a mapping of Q into 0 such that f(C(n) q f(z) whenever { C(n} is a sequence in Q which converges to z. If { zn} is a sequence in Q which converges to z, there exist points C(n E Q such that I C(n - zn I < 11n and I f(C(n) -f(zn) I < 11n. Thus C(n q z, hence f(C(n)q f(z), and this shows that f(zn)ƈ f(z). We have now proved that our extension off is continuous on Q. Also U cf(Q) c 0. The compactness of 0 implies that f(O) is compact. Hence f(O) = 0. CONFORMAL MAPPING 291 Theorem 14.18(b) shows that f is one-to-one on Q. Since every contin uous one-to-one mapping of a compact set has a continuous inverse (, Theorem 4. 17), the proof is complete. I I I I 14.20 Remarks (a) The preceding theorem has a purely topological corollary: I f every bound ary point o f a bounded simply connected plane region Q is simple, then the boundary o jQ is a Jordan curve, and n is homeomorphic to 0. (A Jordan curve is, by definition, a homeomorphic image of the unit circle.) The converse is true, but we shall not prove it : If the boundary of Q is a Jordan curve, then every boundary point of Q is simple. (b) Supposefis as in Theorem 14.19, a, b, and c are distinct boundary points of Q, and A, B, and C are distinct points of T. There is a linear fractional transformation ({J which maps the triple {f(a), f(b), f(c)} to {A, B, C} ; suppose the orientation of {.t4, B, C} agrees with that of {f(a),f(b),f(c)} ; then ({J(U) = U, and the function g = cp o f is a' homeomorphism of Q onto 0 which is holomorphic in Q and which maps {a, b, c} to pre scribed values {A, B, C}. It follows from Sec. 14.3 that g is uniquely determined by these requirements. (c) Theorem 14.19, as well as the above remark (b), extends without difficulty to simply connected regions Q in the Riemann sphere S 2 , all of whose boundary points are simple, provided that S2 - Q has a nonempty inte rior, for then a linear fractional transformation brings us back to the case in which Q is a bounded region in the plane. Likewise, U can be replaced, for instance, by a half plane. (d) More generally, if/1 and/2 map 01 and 02 onto U, as in Theorem 14.19, then f = f 2 1 0 !1 is a homeomorphism of nl onto n2 which is holo-morphic in 01• . Conformal Mapping of an Annulus 14.21 It is a consequence of the Riemann mapping theorem that any two simply connected proper subregions of the plane are conformally equivalent, since each of them is conformally equivalent to the unit disc. This is a very special property of simply connected regions. One may ask whether it extends to the next simplest situation, i.e., whether any two annuli are conformally equivalent. The answer is negative. For 0 < r < R, let A(r, R) = { z: r < I z I < R} (1) be the annulus with inner radius r and outer radius R. If A. > 0, the mapping zq A.z maps A(r, R) onto A(A.r, A.R). Hence A(r, R) and A(r1, R1) are conformally equivalent if Rlr = R 1/r 1. The surprising fact is that this sufficient condition is 292 REAL AND COMPLEX ANALYSIS also necessary; thus among the annuli there is a different conformal type associ ated with each real number greater than 1. 14.22 Theorem A(r1, R1) and A(r2 , R2) are conformally equivalent if and only if R1/r 1 = R2/r2 · PROOF Assume r1 = r2 = 1, without loss of generality. Put (1) and assume there exists fe H(A1) such that f is one-to-one and f(A1) = A2 . Let K be the circle with center at 0 and radius r = .JR;. Since f - 1 : A2 q A 1 is also holomorphic,f - 1(K) is compact. Hence (2) for some e > 0. Then V = f(A(l, 1 + e)) is a connected subset of A2 which does not intersect K, so that V c A(l, r) or V c A(r, R2). In the latter case, replace f by R2/ f. So we can assume that V c A(l, r). If 1 < I zn I < 1 + e and I zn I q 1, then f(zn) E V and {f(zn)} has no limit point in A2 (since f- 1 is continuous); thus I f(zn) I q 1. In the same manner we see that I f(zn) I q R2 if l zn i q Rl. Now define (3) and u( z) = 2 log I f ( z) I - 2cx log I z I (4) Let o be one of the Cauchy-Riemann operators. Since of= 0 and o f= f', the chain rule gives o(2 log I f I ) = o(log (!J)) =!'I f, (5) so that (Ou)(z) = ̗̍: -: (6) Thus u is a harmonic function in A1 which, by the first paragraph of this proof, extends to a continuous function on A1 which is 0 on the boundary of A1• Since nonconstant harmonic functions have no local maxima or minima, we conclude that u = 0. Thus f'(z) ex - -f(z) z (7) CONFORMAL MAPPING 293 Put y(t) = ft: eit ( -n < t < n); put r = f o y. As in the proof of Theorem 1 0.43, (7) gives 1 r f'(z) IX = 21ti JY f(z) dz = lndr (0). (8) Thus C( is an integer. By (3), C( > 0. By (7), the derivative of z-'i(z) is 0 in A1• Thus f(z) = cz«. Since f is one-to-one in A 1, C( = 1. Hence R2 = R 1. I I I I Exercises 1 Find necessary and suffic͚ent conditions which the complex numbers a, b, c, and d have to satisfy so that the linear fractional transformation z---+ (az + b)/(cz + d) maps the upper half plane onto itself. 2 In Theorem 1 1.14 the hypotheses were, in simplified form, that n c: n +, L is on the real axis, and Im f(z)---+ 0 as z---+ L. Use this theorem to establish analogous reflection theorems under the following hypotheses: (a) n c: n +' L on real axis, I f(z) 1 ---+ 1 as z---+ L. (b) n c: U, L c: T, I f(z) 1 ---+ 1 as z---+ L. (c) !l c: U, L c: T, lmf(z)---+ 0 as z---+ L. In case (b), iff has a zero at ct E n, show that its extension has a pole at 1/Ǭ. What are the analogues of this in cases (a) and (c)? 3 Suppose R is a rational function such that I R(z) I = 1 if I z I = 1. Prove that k z - ct R(z) = czm 0 _ " n= 1 1 - ctn z where c is a constant, m is an integer, and ct1, • • • , ctk are complex numbers such that ctn "# 0 and I ctn I "# 1. Note that each of the above factors has absolute value 1 if I z I = 1. 4 Obtain an analogous description of those rational functions which are positive on T. Hint: Such a function must have the same number of zeros as poles in U. Consider products of factors of the form where I ct I < 1 and I P I < 1. (z - ct){1 - Ǭz) (z - P)(1 - Pz) 5 Suppose f is a trigonometric polynomial, n f(O) = L k= -n andf(O) > 0 for all real 0. Prove that there is a polynomial P(z) = c0 + c1z + · · · + cn z" such that (8 real). Hint: Apply Exercise 4 to the rational function I:ak zl'. Is the result still valid if we assume·f(O) > 0 instead ofj(O) > 0? 6 Find the fixed points of the mappings cp« (Definition 12.3). Is there a straight line which cp« maps to itself? 7 Find all complex numbers ct for whichf: is one-to-one in U, where Describe f «( U) for all these cases. z f «(z) = 2 • 1 + ctz 294 REAL AND COMPLEX ANALYSIS 8 Suppose f(z) = z + (1/z). Describe the families of ellipses and hyperbolas onto which f maps circles with center at 0 and rays through 0. 9 (a) Suppose n = {z: - 1 < Re z < 1}. Find an explicit formula for the one-to-one conformal mapping/ of n onto U for whichf(O) = 0 andf'(O) > 0. Computef'(O). (b) Note that the inverse of the function constructed in (a) has its real part bounded in U, whereas its imaginary part is unbounded. Show that this implies the existence of a continuous real function u on 0 which is harmonic in U and whose harmonic conjugate v is unbounded in U. [v is the function which makes u + iv holomorphic in U; we can determine v uniquely by the requirement v(O) = 0.] (c) Suppose g E H(U), I Re g I < 1 in U, and g(O) = 0. Prove that Hint: See Exercise 10. . 2 1 + r I g(re'8) I < - log . 1t 1 - r (d) Let n be the strip that occurs in Theorem 12.9. Fix a point ct + ip in n. Let h be a conformal one-to-one mapping of n onto n that carries ct + ip to 0. Prove that I h'(ct + iP) I = 1/cos p. 10 Supposefand g are holomorphic mappings of U into !l,fis one-to-one,f(U) = n, andf(O) = g(O). Prove that g(D(O; r)) c f(D(O; r)) (0 < r < 1). 1 1 Let n be the upper half of the unit disc U. Find the conformal mapping/ of n onto U that carries { - 1, 0, 1} to { - 1, - i, 1}. Find z E n such thatf(z) = 0. Findf(i/2). Hint:f = cp o s o t/J, where cp and t/1 are linear fractional transformations and s(.A.) = .A. 2• 12 Suppose n is a convex region,/ E H(!l), and Re f'(z) > 0 for all z E n. Prove thatfis one-to-one in n. Is the result changed if the hypothesis is weakened to Re f'(z) > 0? (Exclude the trivial case f = constant.) Show by an example that " convex " cannot be replaced by " simply connected." 13 Suppose n is a region,!,. E H(!l) for n = 1, 2, 3, . . . ' each!,. is one-to-one in n, and!,.---+ f uniformly on compact subsets of n. Prove that f is either constant or one-to-one in n. Show that both cases can occur. 14 Suppose n = {x + iy: - 1 < y < 1 },f E H(!l), I f I < 1, andf(x)-+ 0 as X-+ 00 . Prove that lim f(x + iy) = 0 ( - 1 < y < 1) x -+ ex> and that the passage to the limit is uniform if y is confined to an interval [ - ct, ct], where ct < 1. Hint: Consider the sequence {f..
}, where f n(z) = z + n, in the square I x I < 1, I y I < 1 . What does this theorem tell about the behavior of a function g E H00 near a boundary point of U at which the radial limit of g exists? IS Let .9' be the class of all f E H( U) such that Re f > 0 and f(O) = 1. Show that §" is a normal family. Can the condition '}(0) = 1 " be omitted? Can it be replaced by " I f(O) I < 1 "? 16 Let !F be the class of all f E H( U) for which Is this a normal family? fJ l f(z) l2 dx dy < 1. u 17 Suppose n is a region,fn E H(!l) for n = 1, 2, 3, . . . ,fn---+ f uniformly on compact subsets of n, and f is one-to-one in n. Does it follow that to each compact K c n there corresponds an integer N(K) such thatf n is one-to-one on K for all n > N(K)? Give proof or counterexample. CONFORMAL MAPPING 295 18 Suppose n is a simply connected region, z0 E n, and f and g are one-to-one conformal mappings of n onto U which carry z0 to 0. What relation exists between f and g? Answer the same question if f(z0) = g(z0) = a, for some a E U. 19 Find a homeomorphism of U onto U which cannot be extended to a continuous function on 0. 20 Iff E !/ (Definition 14.10) and n is a positive integer, prove that there exists a g e !/ such that gn(z) = f(zn) for all z E U. 21 Find allf E !/ such that (a)f(U) T U, (b) f(U) T 0, (c) I a2 1 = 2. 22 Suppose f is a one-to-one conformal mapping of U onto a square with center at 0, and f(O) = 0. Prove that f(iz) = if(z). If f(z) = I:cn zn, prove that en = 0 unless n - 1 is a multiple of 4. Generalize this: Replace the square by other simply connected regions with rotational symmetry. 23 Let n be a bounded region whose boundary consists of two nonintersecting circles. Prove that there is a one-to-one conformal mapping of n onto an annulus. (This is true for every region n such that S2 - n has exactly two components, each of which contains more than one point, but this general situation is harder to handle.) 24 Complete the details in the following proof of Theorem 14.22. Suppose 1 < R2 < R1 and f is a one-to-one conformal mapping of A(1, R1) onto A(1, R2). Definef1 = f and f n = f o !, 1• Then a sub sequence of {fn} converges uniformly on compact subsets of A(1, R1) to a function g. Show that the range of g cannot contain any nonempty open set (by the three-circle theorem, for instance). On the other hand, show that g cannot be constant on the circle { z : I z 12 = R 1}. Hence f cannot exist. 25 Here is yet another proof of Theorem 14.22. If f is as in 14.22, repeated use of the reflection principle extends f to an entire function such that I f(z) I = 1 whenever I z I = 1. This implies f(z) = rlzn, where I r1 l = 1 and n is an integer. Complete the details. 26 Iteration of Step 2 in the proof of Theorem 14.8 leads to a proof (due to Koebe) of the Riemann mapping theorem which is constructive in the sense that it makes no appeal to the theory of normal families and so does not depend on the existence of some unspecified subsequence. For the final step of the proof it is convenient to assume that n has property (h) of Theorem 13.1 1. Then any region conformally equivalent to n will satisfy (h). Recall also that (h) implies U), trivially. By Step 1 in Theorem 14.8 we may assume, without loss of generality, that 0 E n, n c: U, and n =1- U. Put n = 00 • The proof consists in the construction of regions 01, 02 , 03 , • • • and of func tions/1,/2 ,/3 ' . . . ' so thatf n(nn- 1) = nn and so that the functionsfn ° f n- 1 ° . . . 0 !2 ° !1 converge to a conformal mapping of n on to U. Complete the details in the following outline. (a) Suppose nn 1 is constructed, let rn be the largest number such that D(O; rn) c: nn_ 1, let rln be a boundary point of nn _ 1 with 1 r1n 1 = r n , choose Pn so that p; = - r1n , and put (The notation is as in the proof of Theorem 14.8.) Show that Fn has a holomorphic inverse Gn in nn_ 1, and put f n = An Gn , where An = I c I /c and c = GĮ(O). (This !, is the Koebe mapping associated with nn_ 1. Note that !, is an elementary function. It involves only two linear fractional transformations and a square root.) (b) Compute thatfĮ(O) = (1 + rn)/2Jr:. > 1. . (c) Put t/10(z) = z and t/ln(z) = f it/Jn_ 1(z)). Show that t/ln is a one-to-one mapping of n onto a region nn c: u' that { t/1Į(0)} is bounded, that and that therefore rn͜ 1 as nǸ oo. n 1 + r "'Į(o) = n Jf. . k= 1 2 rk (d) Write t/ln(z) = zhn(z), for z E n, show that I hn I < I hn+ 1 I ' apply Harnack's theorem and Exer cise 8 of Chap. 1 1 to {log 1 hn 1 } to prove that { t/1 n} converges uniformly on compact subsets of n, and show that lim t/Jn is a one-to-one mapping of n onto U. 296 REAL AND COMPLEX ANALYSIS 27 Prove that L:'= 1 (1 - rn)2 < oo, where {rn} is the sequence which occurs in Exercise 26. Hint: 1 + r (1 - Jr)2 -- = 1 + . 2Jr 2Jr 28 Suppose that in Exercise 26 we choose (Xn E U - Qn_ 1 without insisting that I an I = rn . For example, insist only that 1 + rn I (X" I < 2 · Will the resulting sequence { .Pn} still converge to the desired mapping function? 29 Suppose Q is a bounded region, a E O.,f e H(O.),j(Q) c: 0., andf(a) = a. (a) Put /1 = f and f n = f o f n _ 1 , compute fĮ(a), and conclude that I f'(a) I < 1. (b) Iff'( a) = 1, prove thatf(z) = z for all z E n. Hint: If compute the coefficient of (z - a)m in the expansion of !,(z). (c) If I f'(a) I = 1, prove thatfis one-to-one and thatf(O.) = 0.. Hint : If y = f'(a), there are integers nk---. oo such that y"" ---. 1 and fn" ---. g. Then g'(a) = 1, g(Q) c: Q (by Exercise 20, Chap. 10), hence g(z) = z, by part (b). Use g to draw the desired conclusions aboutf 30 Let A be the set of all linear fractional transformations. If {(X, p, y, } is an ordered quadruple of distinct complex numbers, its cross ratio is defined to be (IX - p)(y - ) [(X, p, y, J = ((X - )(y - P> " If one of these numbers is oo, the definition is modified in the obvious way, by continuity. The same applies if (X coincides with p or y or . (a) If cp(z) = [z, (X, p, y ], show that cp E A and cp maps {(X, p, y} to {0, 1, oo }. (b) Show that the equation [w, a, b, c] = [z, (X, p, y] can be solved in the form w = q>(z); then cp E A maps {(X, p, y} to {a, b, c}. r _tne. (c) If cp E A, show that [cp((X), cp(p), cp(y), cp()] = [(X, p, y, ]. (d) Show that [(X, p, y, ] is real if and only if the four points lie on the same circle or straight (e) Two points z and z are said to be symmetric with respect to the circle (or straight line) C through (X, p, and y if [z, (X, p, y] is the complex conjugate of [z, (X, p, y ]. If C is the unit circle, find a simple geometric relation between z and z. Do the same if C is a straight line. (f) Suppose z and z are symmetric with respect to C. Show that <p(z) and cp(z) are symmetric with respect to cp( C), for every cp E A. 31 (a) Show that A (see Exercise 30) is a group, with composition as group operation. That is, if <p E A and t/1 E A, show that cp o .p E A and that the inverse cp - 1 of <p is in A. Show that A is not commutative. (b) Show that each member of A (other than the identity mapping) has either one or two fixed points on S2• [A fixed point of cp is a point (X such that cp(a) = (X.] (c) Call two mappings cp and cp 1 E A conjugate if there exists a t/1 E A such that cp 1 = t/1 - 1 o cp o t/1. Prove that every cp E A with a unique fixed point is conjugate to the mapping z-. z + 1. Prove that every cp E A with two distinct fixed points is conjugate to the mapping z---. az, where (X is a complex number; to what extent is (X determined by cp? CONFORMAL MAPPING 297 (d) Let IX be a complex number. Show that to every <p e A which has IX for its unique fixed point there corresponds a p such that 1 1 --- --- + P . <p(z) - IX z - ct Let Ga. be the set of all these <p, plus the identity transformation. Prove that Ga. is a subgroup of A and that Ga. is isomorphic to the additive group of all complex numbers. (e) Let IX and p be distinct complex numbers, and let Ga., fJ be the set of all <p e A which have IX and p as fixed points. Show that every <p e Ga., fJ is given by <p(z) - ct z - ct = }' . <p( z) - p z - p ' where y is a complex number. Show th͛t Grx, fJ is a subgroup of A which is isomorphic to the multipli cative group of all nonzero complex numbers. (f) If <p is as in (d) or (e), for which circles C is it true that <p(C) = C? The answer should be in terms of the parameters IX, p, and y. 32 For z e 0, z2 =1- 1, define f(z) = exp i log , { 1 + z} 1 - z choosing the branch of log that has log 1 = 0. Describe f(E) if E is (a) U, (b) the upper half of T, (c) the lower half of T, (d) any circular arc (in U) from - 1 to 1, (e) the radius [0, 1), (f) any disc { z: I z - r I < 1 - r}, 0 < r < 1. (g) any curve in U tending to 1. 33 If <prx is as in Definition 12.3, show that (a) Ļ i I ({Jӯ 12 dm = 1, 1t u 1 i I 1 - I ct 12 1 (b) -I <pa. l dm = 2 log 2• 1t u I IX I 1 - l ct l Here m denotes Lebesgue measure in R 2• CHAPTER FIFTEEN ZEROS OF HOLOMORPHIC FUNCTIONS Infinite Products 15.1 So far we have met only one result concerning the zero set Z(f) of a non constant holomorphic function/ in a region n, namely, Z(f) has no limit point in n. We shall see presently that this is all that can be said about Z(f), if no other conditions are imposed on f, because of the theorem of Weierstrass (Theorem 15.1 1) which asserts that every A c Q without limit point in Q is Z(f) for some f E H(Q). If A = { C(n}, a natural way to construct such an f is to choose functions f n E H(Q) so that fn has only one zero, at C(n , and to consider the limit of the products P n = /1 /2 · · · fn ' as n / oo. One has to arrange it so that the sequence {Pn} converges to some f E H(Q) and so that the limit function f is not 0 except at the prescribed points C(n . It is therefore advisable to begin by studying some general properties of infin ite products. 298 15.2 Definition Suppose { un} is a sequence of complex numbers, and p = limn-oo Pn exists. Then we write 00 p = n (1 + un)· n= l (1) (2) The Pn are the partial products of the infinite product (2). We shall say that the infinite product (2) converges if the sequence {Pn} converges. ZEROS OF HOLOMORPHIC FUNCTIONS 299 In the study of infinite series :r-an it is of significance whether the an approach 0 rapidly. Analogously, in the study of infinite products it is of interest whether the factors are or are not close to 1. This accounts for the above notation : 1 + un is close to 1 if un is close to 0. 15.3 Lemma I f ub . . . , uN are complex numbers, and if N N PN = fl (1 + un), Pʑ = fl (1 + I Un I ), (1) n= 1 n= 1 then (2) and l PN - 1 1 < Pʑ - 1. (3) PROOF For x > 0, the inequality 1 + x < ex is an immediate consequence of the expansion of ex in powers of x. Replace x by I u1 I , . . . , I uN I and multiply the resulting inequalities. This gives (2). For N = 1, (3) is trivial. The general case follows by induction : For k = 1, . . . , N - 1, Pk+ 1 - 1 = Pk(1 + uk+ 1) - 1 = (pk - 1)(1 + uk+ 1) + uk+ b so that if (3) holds with k in place of N, then also I Pk + 1 - 1 1 < (pt - 1 )( 1 + I uk + 1 I ) + I uk + 1 I = Pt + 1 - 1. I I I I 15.4 Theorem Suppose { un} is a sequence o f bounded complex functions on a set S, such that :r, I un(s) I converges uniformly on S. Then the product 00 f(s) = fl (1 + un(s)) (1) n= l converges uniformly on S, and f(s0) = 0 at some s0 e S if and only if un(s0) = - 1 for some n. Furthermore, if {n1, n2 , n3 , .
.
• } is any permutation o f { 1, 2, 3, . . . }, then we also have 00 f(s) = fl (1 + unk(s)) (s e S). (2) k = 1 PROOF The hypothesis implies that :r, I un(s) I is bounded on S, and if PN denotes the Nth partial product of (1), we conclude from Lemma 15.3 that there is a constant C < oo such that I PN(s) I < C for all N and all s. 300 REAL AND COMPLEX ANALYSIS Choose E, 0 < E < > . There exists an N 0 such that 00 L I un(s) I < E (s E S). (3) n=No Let {n1, n2 , n3 , • • . } be a permutation of {1, 2, 3, . . . }. If N > N0 , if M is so large that { 1, 2, . . . , N} c { n b n 2 , .
.
• , n M}, and if qM(s) denotes the Mth partial product of (2), then qM - PN = PN{n (1 + unk) - 1}. (4) (5) The nk which occur in (5) are all distinct and are larger than N 0 • Therefore (3) and Lemma 15.3 show that (6) If nk = k (k = 1, 2, 3, . . . ), then qM = p M , and ( 6) shows that {PN} con verges uniformly to a limit function! Also, (6) shows that I PM - PNo I < 2 1 PNo I E so that I PM I > (1 - 2E) I PNo 1 . Hence I f(s) I > (1 - 2E) I PNo(s) I (M > N0), (s E S), which shows that f(s) = 0 if and only if PN0(s) = 0. (7) (8) Finally, (6) also shows that {qM} converges to the same limit as {PN}· /Ill 15.5 Theorem Suppose 0 < un < 1. Then 00 if and only if L un < oo. n= l PROOF If PN = (1 - u1) · · • (1 - uN), then p1 > p2 > · · · , PN > 0, hence p = lim PN exists. If чun < co , Theorem 15.4 implies p > 0. On the other hand, N p < PN = n (1 - un) < exp { - ul - u2 -.
. . - uN}, 1 and the last expression tends to 0 as N ̐ oo, if l:un = oo. We shall frequently use the following consequence of Theorem 15.4: /Ill 15.6 Theorem Suppose fn E H(Q) for n = 1, 2, 3, . . . , no fn is identically 0 in any component of!l, and 00 L 1 1 -fn(z) l (1) n= l ZEROS OF HOLOMORPHIC FUNCTIONS 301 converges uniformly on compact subsets ofO.. Then the product 00 f(z) = n fn(z) (2) n= l converges uniformly on compact subsets of 0.. Hence f E H(O.). Furthermore, we have 00 m(f; z) = L m(fn; z) (z E 0.), (3) n= l where m(f; z) is de fined to be the multiplicity o f the zero off at z. [If f(z) i= 0, then m(f; z) = 0.] PROOF The first part follows immediately from Theorem 15.4. For the second part, observe that each z E 0. has a neighborhood V in which at most finitely many of the fn have a zero, by (1). Take these factors first. The product of the remaining ones has no zero in V, by Theorem 15.4, and this gives (3). Inci dentally, we see also that at most finitely many terms in the series (3) can be positive for any given z E n. Ill/ The Weierstrass Factorization Theorem 15.7 Definition Put E0(z) = 1 - z, and for p = 1, 2, 3, . . . , { z2 zP} EP(z) = (1 - z) exp z + 2 + · · · + p . These functions, introduced by Weierstrass, are sometimes called elementary factors. Their only zero is at z = 1. Their utility depends on the fact that they are close to 1 if I z I < 1 and p is large, although EP(1) = 0. 15.8 Lemma For I z I < 1 and p = 0, 1, 2, . . . , 1 1 - E p( z) I < I z I p + 1. PROOF For p = 0, this is obvious. For p > 1, direct computation shows that { z2 zP} - EǙ(z) = zP exp z + 2 + · · · + p . So - EǙ has a zero of order p at z = 0, and the expansion of - EǙ in powers of z has nonnegative real coefficients. Since 1 - EP(z) = - ( EǙ(w) dw, J[o, z] 302 REAL AND COMPLEX ANALYSIS 1 - EP has a zero of order p + 1 at.z = 0, and if ( ) _ 1 - EP(z) qJ z -p + 1 ' z then 0. Hence lqJ(z) I < qJ( 1 ) = 1 if I z I < 1, and this gives the assertion of the lemma. I I I I 15.9 Theorem Let { zn} be a sequence of complex numbers such that zn =I= 0 and I zn Ir oo as n r oo. If {Pn} is a sequence of nonnegative integers such that oo ( r)l + pn L -< oo n= 1 rn for every positive r (where r n = I zn I), then the infinite product (1) (2) defines an entire function P which has a zero at each point zn and which has no other zeros in the plane. More precisely, if C( occurs m times in the sequence { zn}, then P has a zero of order m at C(. Condition (1) is always satisfied if Pn = n - 1,for instance. PROOF For every r, rn > 2r for all but finitely many n, hence rlrn < > for these n, so (1) holds with 1 + Pn = n. Now fix r. If I z I < r, Lemma 15.8 shows that ( z ) z 1 + Pn ( r) 1 + Pn 1 - E -< -< -Pn --ћ Zn ќ if rn > r, which holds for all but finitely many n. It now follows from (1) that the series oo ( z) L 1 - E -n= 1 Pn Zn converges uniformly on compact sets in the plane, and Theorem 15.6 gives the desired conclusion. I Ill Note: For certain sequences {rn}, (1) holds for a constant sequence {Pn}. It is of interest to take this constant as small as possible; the resulting function (2) is then called the canonical product corresponding to {zn}· For instance, if l: 11r n < oo, we can take Pn = 0, and the canonical product is simply n 1 - -oo ( z) " = 1 Zn . ZEROS OF HOLOMORPHIC FUNCTIONS 303 If 1:1/rn = oo but 1:1/r; < oo , the canonical product is Canonical products are of great interest in the study of entire functions of finite order. (See Exercise 2 for the definition.) We now state the Weierstrass factorization theorem. 15.10 Theorem Let f be an entire function, suppose f(O) "# 0, and let z 1 , z2 , z3 , . . . be the zeros off, listed according to their multiplicities. Then there exist an entire function g and a sequence {Pn} of nonnegative integers, such that (1) Note: (a) Iff has a zero of order k at z = 0, the preceding applies to f(z)jzk. (b) The factorization (1) is not unique; a unique factorization can be associated with thosefwhose zeros satisfy the condition required for the convergence of a canon ical product. PROOF Let P be the product in Theorem 15.9, formed with the zeros of f Then f /P has only removable singularities in the plane, hence is (or can be extended to) an entire function. Also, f / P has no zero, and since the plane is simply connected,f /P = eg for some entire function g. /Ill The proof of Theorem 15.9 is easily adapted to any open set : 15.11 Theorem Let Q be an open set in S2, Q "# S2• Suppose A c Q and A has no limit point in Q. With each a. E A associate a positive integer m( a.). Then there exists an f E H(Q) all of whose zeros are in A, and such that f has a zero of order m(a.) at each a. E A. PROOF It simplifies the argument, and causes no loss of generality, to assume that 00 E n but 00 ¢ A. (If this is not so, a linear fractional transformation will make it so.) Then S2 - Q is a nonempty compact subset of the plane, and oo is not a limit point of A. If A is finite, we can take a rational function for f If A is infinite, then A is countable (otherwise there would be a limit point in Q). Let { a.n} be a sequence whose terms are in A and in which each a. E A is listed precisely m(a.) times. Associate with each a.n a point Pn E S2 - n such that I Pn - a.n I < I {3 - an I for all {3 E S2 - n; this is possible since S2 - Q is compact. Then 304 REAL AND COMPLEX ANALYSIS as n O oo ; otherwise A would have a limit point in Q. We claim that has the desired properties. f(z) = fJ E,.((X" - Pn) n= 1 Z - /3n Put rn = 2 I rxn - f3n 1 . Let K be a compact subset of Q. Since rn O 0, there exists an N such that I z - Pn I > rn for all z E K and all n > N. Hence (Xn - f1n 1 <-z - f1n -2' which implies, by Lemma 15.8, that _ ((Xn - f1n) < (!)"+ 1 1 En z - f1n -2 (z E K, n > N), and this again completes the proof, by Theorem 15.6. Ill/ As a consequence, we can now obtain a characterization of meromorphic functions (see Definition 10.41): 15.12 Theorem Every meromorphic function in an open set Q is a quotient o f two functions which are holomorphic in Q. The converse is obvious: If g E H(Q), h E H(Q), and h is not identically 0 in any component of Q, then gfh is meromorphic in Q. PROOF Supposefis. meromorphic in Q; let A be the set of all poles of fin Q; and for each rx E A, let m(rx) be the order of the pole off at rx. By Theorem 15. 1 1 there exists an h E H(Q) such that h has a zero of multiplicity m(rx) at each rx E A, and h has no other zeros. Put g = jh. The singularities of g at the points of A are removable, hence we can extend g so that g E H(Q). Clearly, f= g/h in !l - A. Ill/ An Interpolation Problem The Mittag-Leffier theorem may be combined with the Weierstrass theorem 15.11 to give a solution of the following problem: Can we take an arbitrary set A c Q, without limit point in !l, and find a function f E H(Q) which has pre scribed values at every point of A? The answer is affirmative. In fact, we can do even better, and also prescribe finitely many derivatives at each point of A : 15.13 Theorem Suppose Q is an open set in the plane, A c Q, A has no limit point in Q, and to each rx E A there are associated a nonnegative integer m(rx) ZEROS OF HOLOMORPHIC FUNCTIONS 305 and complex numbers wn, a , 0 < n < m(ex). Then there exists an f E H(Q) such that (ex E A, 0 < n < m(ex)). (1) PROOF By Theorem 15.1 1, there exists a g E H(Q) whose only zeros are in A and such that g has a zero of order m(ex) + 1 at each ex E A. We claim we can associate to each ex E A a function P a of the form 1 + m(a) P a(z) = L ci. a(z - ex)-i j = 1 such that gPa has the power series expansion (2) g(z)Pa(z) = Wo, a + w1, a(z - ex) + . .
. + wm(a), a(z - ex)m(a) + . . . (3) in some disc with center at ex. To simplify the writing, take ex = 0 and m(ex) = m, and omit the subscripts ex. For z near 0, we have where b 1 =I= 0. If then P(z) = C z - 1 + · · · + c z-m- 1 1 m+ 1 ' (4) (5) g(z)P(z) = (em+ 1 + em z + · · · + c 1zm) (b1 + b2 z + b3 z2 + · · ·). (6) The b's are given, and we want to choose the c's so that (7) If we compare the coefficients of 1, z, . . . , zm in (6) and (7), we can solve the resulting equations successively for cm+ l' em , . . . , c1, since b1 =I= 0. In this way we obtain the desired Pa's. The Mittag-Leffier theorem now gives us a meromorphic h in Q whose principal parts are these Pa's, and if we putf = gh we obtain a function with the desired properties. /Ill The solution of this interpolation problem can be used to determine the structure of all finitely generated ideals in the rings H(Q). 15.14 Definition The ideal [gb . . . , gn] generated by thefunctions g1, . . . , gn E H(Q) is the set of all functions of the form "£ /; gi , where /; E H(O.) for i = 1, . . . , n A principal ideal is one that is generated by a single function. Note that = H(O.). Iff E H(O.), ex E 0., and f is not identically 0 in a neighborhood of ex, the multiplicity of the zero off at ex will be denoted by m(f; ex). If /(ex) =I= 0, then m(f; ex) = 0, as in Theorem 15.6. 306 REAL AND COMPLEX ANALYSIS 15.15 Theorem Every finitely generated ideal in H(Q) is principal. More explicitly: If g1, .
• . , gn E H(Q), then there exist functions g,h , hi E H(Q) such that n g = L h gi and gi = hi g i = 1 (1 < i < n). PROOF We shall assume that Q is a region. This is done to avoid problems posed by functions that are identically 0 in some components of Q but not in all. Once the theorem is proved for regions, that case can be applied to each component of an arbitrary open set Q, and the full theorem can be deduced. We leave the details of this as an exercise. Let P(n) be the following proposition : Ifg1, • • • , gn E H(O.), ifno gi is identically 0, and ifno point o jQ is a zero of every g; , then [gb . . . , gnJ = . P( 1) is trivial. Assume that n > 1 and that P(n - 1) is true. Take gb . . . , gn E H(O.), without common zero. By the Weierstrass theorem 1 5.1 1 there exists qJ e H(O.) such that m( qJ ; Ŕ) = min { m(g i ; Ŕ): 1 < i < n - 1} (Ŕ E Q). (1) The functionsh = gJqJ (1 < i < n - 1) are in H(O.) and have no common zero in Q. Since P(n - 1) is true, [fb . . . , fn _ 1] = . Hence (2) Moreover, our choice of qJ shows that gn(Ŕ) =I= 0 at every point of the set A = {Ŕ E Q: ({J(Ŕ) = 0}. Hence it follows from Theorem 15.13 that there exists h E H(Q) such that (ct E Q). (3) Such an h is obtained by a suitable choice of the prescribed values of h(Ŕ) for Ŕ E A and for 0 < k < m(qJ ; Ŕ). By (3), (1 - hgn)/({J has removable singularities. Thus (4) for somef E H(Q). By (2) and (4), 1 E [g1, .
.
• , gn]. We have shown that P(n - 1) implies P(n). Hence P(n) is true for all n. Finally, suppose G1, . . . , Gn E H(O.), and no Gi is identically 0. (This involves no loss of generality.) Another application of Theorem 15.1 1 yields qJ e H(O.) with m(qJ ; Ŕr = min m(Gi; Ŕ) for all Ŕ e Q. Put gi = G/qJ. Then gi e H(Q), and the functions g1, . . . , gn have no common zeros in 0.. By P(n), [g1, . . . , gn] = . Hence [G1, . . . , GnJ = [({J]. This completes the proof. //// ZEROS OF HOLOMORPHIC FUNCTIONS 307 Jensen's Formula 15.16 As we see from Theorem 15.11, the location of the zeros of a holomorphic function in a region Q is subject to no restriction except the obvious one concern ing the absence of limit points in Q. The situation is quite different if we replace H(Q) by certain subclasses which are defined by certain growth conditions. In those situations the distribution of the zeros has to satisfy certain quantitative conditions. The basis of most of these theorems is Jensen's formula (Theorem 15.18). We shall apply it to certain classes of entire functions and to certain sub classes of H(U). The following lemma affords an opportunity to apply Cauchy's theorem to the evaluation of a definite integral. 1 i21t 15.17 Lemma l:n: 0 log 1 1 - ei8 1 dfJ = 0. PROOF Let Q = {z: Re z < 1}. Since 1 - z =F 0 in Q and Q is simply con nected, there exists an h e H(Q) such that exp {h(z)} = 1 - z in Q, and this h is uniquely determined if we require that h(O) = 0. Since Re (1 - z) > 0 in Q, we then have Re h(z) = log 1 1 - z I , n I Im h(z) I < 2 (z e Q). (1) For small b > 0, let r be the path (b < t < 2n - b), (2) and let y be the circular arc whose center is at 1 and which passes from eiʒ to e-iʒ within U. Then 2\ rx-c!
log 1 1 - ei8 1 d() = Re [ 2[i L h(z) d: J = Re [ 2Zi i h(z) Y z J. (3) The last equality depended on Cauchy's theorem; note that h(O) = 0. The length of y is less than nb, so (1) shows that the absolute value of the last integral in (3) is less than Cb log (1/b), where C is a constant. This gives the result if br 0 in (3). /Ill 15.18 Theorem Suppose Q = D(O; R), f e H(f;l), f(O) =F 0, 0 < r < R, and C(1, • • • , C(N are the zeros o ff in D(O; r), listed according to their multiplicities. Then N r { 1 J1t } I /(0) I ll I = exp -2 log I f(rei8) 1 d8 . " = 1 I C(" n _ 1t (1) 308 REAL AND COMPLEX ANALYSIS This is known as Jensen'sformula. The hypothesisf(O) =1= 0 causes no harm in applications, for iff has a zero of order k at 0, the formula can be applied to f(z)fzk. PROOF Order the points cxi so that a.1, . . . , cxm are in D(O; r) and I a.m+ 1 I = · · · = I a.N I = r. (Of course, we may have m = N or m = 0.) Put m r2 _ њ z N (X g(z) =f(z) n n n n n= 1 r(cxn - z) n=m+ 1 (Xn - z (2) Then g e H(D), where D = D(O; r + e) for some E > 0, g has no zero in D, hence log I g I is harmonic in D (Theorem 13.12), and so By (2), 1 J1t log I g(O) I = 2n _ _ }og I g(reiÔ I dO. m I g(O) I = I f(O) I TI r n = 1 I (Xn I . (3) (4) For 1 < n < m, the factors in (2) have absolute value 1 if I z I = r. If a.n = rei8" for m < n < N, it follows that N log I g(rei8) I = log I f(reiÔ I -L log 11 - ei<B - Bn) 1 . (5) n=m+1 Lemma 15.17 shows therefore that the integral in (3) is unchanged if g is replaced by f. Comparison with (4) now gives (1). /Ill Jensen's formula gives rise to an inequality which involves the boundary values of bounded holomorphic functions in U (we recall that the class of these functions has been denoted by H00): 15.19 Theorem Iff e H00 ,f not identically 0, de fine 1 J1t Jl.,(f) = 2n }og I f(reiÔ I d() (0 < r < 1) and JJ.(f) = 2X f..tog I f(ei8) I d() where f is the radial limit function off, as in Theorem 1 1.32. Then J.lr(f) < f.ls(f) if 0 < r < s < 1' J.lr(f)-+ log I f(O) I as r-+ 0, (1) (2) (3) (4) ZEROS OF HOLOMORPHIC FUNCTIONS 309 and Jlr(f) < Jl(f) if 0 < r < 1. (5) Note the following consequence: One can choose r so thatf(z) =1= 0 if I z I = r; then Jlr(f) is finite, and so is Jl(f), by (5). Thus log I f I e L1(T), and f(ei8) =1= 0 at almost every point of T. PROOF There is an integer m > 0 such that f(z) = zmg(z), g e H00, and g(O) =1= 0. Apply Jensen's formula 15.18(1) to g in place of f. Its left side obviously cannot decrease if r increases. Thus Jlr(g) < Jls(g) if r < s. Since we have proved (3). Jlr(f) = Jlr(g) + m log r, Let us now assume, without loss of generality, that I f I < 1. Write f,.(ei8) in place of f(rei8). Then f,.---+ /(0) as r---+0, and /,.--+/ a.e. as r---+1. Since log (1/ I /,. I ) > 0, two applications of Fatou's lemma, combined with (3), give (4) and (5). /Ill 15.20 Zeros of Entire Functions Suppose f is an entire function, M(r) = sup I f(rei8) I (0 < r < oo), (1) 8 and n(r) is the number of zeros of f in D(O; r). Assume f(O) = 1, for simplicity. Jensen's formula gives { 1 J1t } n(2r) 2r n(r) 2r M(2r) > exp -log I f(2rei8) I dO = ll > ll > 2n, 2n - 1t n = 1 I (Xn I n = 1 I (Xn I if { (Xn} is the sequence of zeros off, arranged so that I (X1 1 < I (X2 1 < · · · . Hence n(r) log 2 < log M(2r). (2) Thus the rapidity with which n(r) can increase (i.e., the density of the zeros of f) is controlled by the rate of growth of M(r). Suppose, to look at a more specific situation, that for large r M(r) < exp { Ar"} where A and k are given positive numbers. Then (2) leads to I . log n(r) k tm sup 1 < . r-+oo og r For example, if k is a positive integer and f(z) = 1 - ezk, (3) (4) (5) 310 REAL AND COMPLEX ANALYSIS then n(r) is about n- 1 krk, so that lim log n(r) = k. r-+ 00 log r This shows that the estimate (4) cannot be improved. Blaschke Products (6) Jensen's formula makes it possible to determine the precise conditions which the zeros of a nonconstantf e H00 must satisfy. 15.21 Theorem If { 1Xn} is a sequence in U such that 1Xn =I= 0 and 00 L ( 1 - I (Xn I ) < 00, n= l if k is a nonnegative integer, and if ( ) k [1 00 1Xn - Z I 1Xn I B z = z n = 1 1 - ʐn Z 1Xn (z e U), (1) (2) then B e H00, and B has no zeros except at the points 1Xn (and at the origin, if k > 0). We call this function B a Blaschke product. Note that some of the 1Xn may be repeated, in which case B has multiple zeros at those points. Note also that each factor in (2) has absolute value 1 on T. The term " Blaschke product " will also be used if there are only finitely many factors, and even if there are none, in which case B(z) = 1. PROOF The nth term in the series 00 L n= l IS 1Xn - Z I 1Xn I 1 -. --1 - ʐn Z 1Xn if I z I < r. Hence Theorem 15.6 shows that B e H(U) and that B has only the prescribed zeros. Since each factor in (2) has absolute value less than 1 in U, it follows that I B(z) I < 1, and the proof is complete. /Ill 15.22 The preceding theorem shows that 00 L ( 1 _ · I (Xn I ) < 00 (1) n= l ZEROS OF HOWMORPHIC FUNCTIONS 31 1 is a su fficient condition for the existence of an f e H00 which has only the presc ribed zeros { cxn}. This condition also turns out to be necessary: Iff e H00 and f is not identically zero, the zeros off must satis fy (1). This is a special case of Theorem 1 5.23. It is interesting that (1) is a necessary condition in a much larger class of functions, which we now describe. For any real number t, define log+ t = log t if t > 1 and log+ t = 0 if t < 1. We let N (for Nevanlinna) be the class of allf e H(U) for which 1 f1t sup -2 log+ I f(reiÔ I d(} < oo. O < r < l 1t -1t (2) It is clear that H00 c N. Note that (2) imposes a restriction on the rate of growth of I f(z) I as I z I --+ 1, whereas the boundedness of the integrals (3) imposes no such restriction. For instance, (3) is independent of r iff= e9 for any g e H(U). The point is that (3) can stay small because log I f I assumes large nega tive values as well as large positive ones, whereas log+ I f I > 0. The class N will be discussed further in Chap. 17. 15.23 Theorem Suppose f e N,f is not identically 0 in U, and cx1, cx2 , cx3 , • • • are the zeros off, listed according to their multiplicities. Then 00 L ( 1 - I CXn I ) < 00 • (1) n= l (We tacitly assume that f has infinitely many zeros in U. If there are only finitely many, the above sum has only finitely many terms, and there is nothing to prove. Also, I ex" I < lcxn + 1 1 .) PROOF Iff has a zero of order m at the origin, and g(z) = z-m f(z), then g e N, and g has the same zeros as f, except at the origin. Hence we may assume, without loss of generality, that f(O) =F 0. Let n(r) be the number of zeros off in D(O; r), fix k, and take r < 1 so that n(r) > k. Then Jensen's formula implies that n(r) r { 1 f1t } I f(O) I }I 1 an 1 = exp 2n " log 1 J(reiÔ 1 de (2) (3) 312 REAL AND COMPLEX ANALYSIS Our assumption that f e N is equivalent to the existence of a constant C < oo which exceeds the right side of (3) for all r, 0 < r < 1. It follows that k f1 I C(n I > c - 1 I f (0) I rk. (4) n= 1 The inequality persists, for every k, as r----+ 1. Hence 00 f1 I C(n I > c - 1 I f (0) I > o. (5) n= 1 By Theorem 15.5, (5) implies (1). Ill/ Corollary Iff e Hao (or even iff e N), if C(h C(2 , C(3 , • • • are the zeros off in U, and if љ(1 - I C(n I ) = oo, thenf(z) = Ofor all z E U. For instance, no nonconstant bounded holomorphic function in U can have a zero at each of the points (n - 1)/n (n = 1, 2, 3, . . . ). We conclude this section with a theorem which describes the behavior of a Blaschke product near the boundary of U. Recall that as a member of H00, B has radial limits B(e;8) at almost all points of T. 15.24 Theorem If B is a Blaschke product, then I B(ei8) I = 1 a.e. and 1 J1t lim -2 log I B(re;8) I ĺlJ = 0. r- 1 n -1t (1) PROOF The existence of the limit is a consequence of the fact that the integral is a monotonic function of r. Suppose B(z) is as in Theorem 15.21, and put BN(z) = fi CXn --:: Z • I CXn I n = N 1 - C(n Z C(n (2) Since log ( I B/BN I ) is continuous in an open set containing T, the limit (1) is unchanged if B is replaced by B N . If we apply Theorem 15.19 to B N we therefore obtain 1 J1t . 1 J1t . log I BN(O) I < !Wrr: 2:n: }og I B(re'Ô I d() < 2:n: }og I B(e'Ô I d() < 0. (3) As N ----+ oo, the first term in (3) tends to 0. This gives (1), and shows that J log I B I = 0. Since log I B I < 0 a.e., Theorem 1.39(a) now implies that log I B I = 0 a.e. /Ill The Miintz-Szasz Theorem 15.25 A classical theorem of Weierstrass (, Theorem 7.26) states that the polynomials are dense in C(I), the space of all continuous complex functions on ZEROS OF HOLOMORPHIC FUNCTIONS 313 the closed interval I = [0, 1], with the supremum norm. In other words, the set of all finite linear combinations of the functions 1 2 3 ' t, t ' t ' . . . (1) is dense in C(I). This is sometimes expressed by saying that the functions (1) span C(I). This suggests a question: If 0 < A1 < A2 < A.3 < · · · , under what conditions is it true that the functions 1 tAl tA2 tA3 ' ' ' ' . . . (2) span C(I)? It turns out that this problem has a very natural connection with the problem of the distribution of the zeros of a bounded holomorphic function in a half plane (or in a disc; the two are conformally equivalent). The surprisingly neat answer is that the functions (2) span C(I) if and only if 1:1/ An = oo. Actually, the proof gives an even more precise conclusion : 15.26 Theorem Suppose 0 < A1 < A2 < A3 < · · · qnd let X be the closure in C(I) of the set o f all .finite linear combinations o f the functions 1 tAl tA.2 tA3 ' ' ' ' . . . . (a) I fl:1/An = oo, then X = C(I). (b) I f 1:1/ An < oo, and if A ¢ {An}, A =F 0, then X does not contain the function tA.. PROOF It is a consequence of the Hahn-Banach theorem (Theorem 5.19) that cp e C(I) but cp ¢ X if and only if there is a bounded linear functional on C(I) which does not vanish at cp but which vanishes on all of X. Since every bounded linear functional on C(I) is given by integration with respect to a complex Borel measure on I, (a) will be a consequence of the following pro position : I fl:1/An = oo and if J.l is a complex Borel measure on I such that (n = 1, 2, 3, . . . ), (1) then also fcc dJ-t(t) = o (k = 1, 2, 3, . . . ). (2) For if this is proved, the preceding remark shows that X contains all functions t"; since 1 e X, all polynomials are then in X, and the Weierstrass theorem therefore implies that X = C(I). 314 REAL AND COMPLEX ANALYSIS So assume that (1) holds. Since the integrands in (1) and (2) vanish at 0, we may as well assume that Jl is concentrated on (0, 1]. We associate with Jl the function (3) For t > 0, tz = exp (z log t), by definition. We claim that f is holo morphic in the right half plane. The continuity off is easily checked, and we can then apply Morera's theorem. Furthermore, if z = x + iy, if x > 0, and if 0 < t < 1, then I tz I = tx < 1. Thus f is bounded in the right half plane, and (1) says thatj(An) = 0, for n = 1, 2, 3, . . . . Define (1 + z) g(z) = f 1 - z (z e U). (4) Then g e H00 and g(rxn) = 0, where rxn = (An - 1)/(An + 1). A simple computa tion shows that 1:(1 - I rxn I ) = oo if 1:1/An = oo. The Corollary to Theorem 15.23 therefore tells us that g(z) = 0 for all z E U. Hence f = 0. In particular, f(k) = 0 for k = 1, 2, 3,. , . . , and this is (2). We have thus proved part (a) of the theorem. To prove (b) it will be enough to construct a measure Jl on I such that (3) defines a function f which is holomo'rphic in the half plane Re z > - 1 (anything negative would do here), which is 0 at 0, A1, A2 , A3 , • • • and which has no other zeros in this half plane. For the functional induced by this measure Jl will then vanish on X but will not vanish at any function t;. if A # 0 and A ¢ {An}· We begin by constructing a function f which has these prescribed zeros, and we shall then show that this f can be represented in the form (3). Define Since f(z) = z fi A. - z (2 + z)3 n= 1 2 + An + Z A - z 1 -n 2 + An + Z 2z + 2 (5) the infinite product in (5) converges uniformly on every compact set which contains none of the points - An - 2. It follows thatfis a meromorphic func tion in the whole plane, with poles at - 2 and - An - 2, and with zeros at 0, A1, A2 , A3 , • .
.
. Also, each factor in the infinite product (5) is less than 1 in absolute value if Re z > - 1. Thus I f(z) I < 1 if Re z > - 1. The factor , (2 + z)3 ensures that the restriction offto the line Re z = - 1 is in L1• Fix z so that Re z > - 1, and consider the Cauchy formula for f(z), where the path of integration consists of the semicircle with center at - 1, radius R > 1 + I z I , from - 1 - iR to - 1 + R to - 1 + iR, followed by the ZEROS OF HOLOMORPHIC FUNCTIONS 315 interval from - 1 + iR to - 1 - iR. The integral over the semicircle tends to 0 as R = oo, so we are left with But f(z) = - ! foo f(- 1 .+ is) ds 2n _ 00 - 1 + zs - z (Re z > - 1). _ _ 1 _ _ = ll tz - is dt 1 + z - is Jo (Re z > - 1). Hence (6) can be rewritten in the form f(z) = [ \z{! foo /( - 1 + is)e- is log t ds} dt. Jo 2n - oo (6) (7) (8) The interchange in the order of integration was legitimate: If the integrand in (8) is replaced by its absolute value, a finite integral results. Put g(s) = f(- 1 + is). Then the inner integral in (8) is.g(log t), where g is the Fourier transform of g. This is a bounded continuous function on (0, 1], and if we set dJ1(t) = g(Iog t) dt we obtain a measure which represents fin the desired form (3). This completes the proof. I I I I 15.27 Remark The theorem implies that whenever { 1, tA1, tA2, • • • } spans C(J), then some infinite subcollection of the tAi can be removed without altering the span. In particular, C(J) contains no minimal spanning sets of this type. This is in marked contrast to the behavior of orthonormal sets in a Hilbert space : if any element is removed from an orthonormal set, its span is dimin ished. Likewise, if { 1, tA1 , tA2 • • • } does not span C(J), removal of any of its elements will diminish the span; this follows from Theorem 15.26(b). Exercises 1 Suppose {an} and { bn} are sequences of complex numbers such that I: I an - bn I < oo. On what sets will the product converge uniformly? Where will it define a holomorphic function? 2 Suppose f is entire, A. is a positive number, and the inequality I f(z) I < exp ( I z 1;.) holds for all large enough I z 1 . (Such functions f are said to be of finite order. The greatest lower bound of all A. for which the above condition holds is the order of f) If f(z) = I:an z", prove that the inequality (eA.)"';. lan l < -n 316 REAL AND COMPLEX ANALYSIS holds for all large enough n. Consider the functions exp (i'), k = 1, 2, 3, . . . , to determine whether the above bound on I a" I is close to best possible. 3 Find all complex z for which exp (exp (z)) = 1. Sketch them as points in the plane. Show that there is no entire function of finite order which has a zero at each of these points (except, of course,/= 0). 4 Show that the function e1tiz + e-1riz n cot nz = ni -. -- e1t'z _ e-1tiz has a simple pole with residue 1 at each integer. The same is true of the function 1 00 2z N 1 f(z) = - + L 2 2 = lim L · z n = 1 z - n N -+ oo n = - N z - n Show that both functions are periodic [f(z + 1) = f(z)], that their difference is a bounded entire function, hence a constant, and that this constant is actually 0, since i oo dt lim f(iy) = - 2i 2 = -ni. y -+ oo 0 1 + t This gives the partial fractions decomposition 1 00 2z n cot nz = - + L 2 2 • z 1 z - n (Compare with Exercise 12, Chap. 9.) Note that n cot nz is (g'/gX.z) if g(z) = sin nz. Deduce the product representation sin nz = Il (1 _ z:). nz n = l n S Suppose k is a positive integer, { zn} is a sequence of complex numbers such that I: I zn 1-k- 1 < oo , and f(z) = Il Ek(.:)· n = l zn (See Definition 15.7.) What can you say about the rate of growth of M(r) = max I f(rei8) I ? 8 6 Suppose f is entire, f(O) #- 0, I f(z) I < exp ( I z IP) for large I z I , and { zn} is the sequence of zeros off, counted according to their multiplicities. Prove that I: I zn 1-p-t < oo for every € > 0. (Compare with Sec. 15.20.) 7 Suppose/is an entire function,f(.Jn) = 0 for n = 1, 2, 3, . . . , and there is a positive cortstant ct such that I f(z) I < exp ( I z I«) for all large enough I z 1 . For which ct does it follow that f(z) = 0 for all z? [Consider sin (nz 2).] 8 Let { zn} be a sequence of distinct complex numbers, zn #- 0, such that Zn---+ oo as n---+ oo , and let {mn} be a sequence of positive integers. Let g be a meromorphic function in the plane, which has a simple pole with residue mn at each zn and which has no other poles. If z ¢ { zn}, let y(z) be any path from 0 to z which passes through none of the points zn , and define f(z) = exp { r g(() d(}. Jy(z) ZEROS OF HOLOMORPHIC FUNCTIONS 317 Prove that f(z) is independent of the choice of y(z) (although the integral itself is not), that f is holomorphic in the complement of {zn}, that f has a removable singularity at each of the points zn , and that the extension off has a zero of order mn at zn . The existence theorem contained in Theorem 15.9 can thus be deduced from the Mittag-Leffier theorem. 9 Suppose 0 < ct < 1, 0 < P < l,f E H(U),f(U) c: U, andf(O) = ct. How many zeros canfhave in the disc D(O; P)? What is the answer if (a) ct = t, P = t; (b) ct = i, P = t; (c) ct = j, P = i; (d) ct = 1/1,000 , p = 1/10? 10 For N = 1, 2, 3, . . . , define oo ( z2) grv(z) = n 1 - 2 n=N n Prove that the ideal generated by {gN} in the ring of entire functions is not a principal ideal. 1 1 Under what conditions on a sequence of real numbers Yn does there exist a bounded holomorphic function in the open right half plane which is not identically zero but which has a zero at each point 1 + iyn? In particular, can this happen if (a) Yn = log n, (b) Yn = Jn, (c) Yn = n, (d) Yn = n2? 12 Suppose 0 < I ctn I < 1, I:(1 - I ctn I ) < oo, and B is the Blaschke product with zeros at the points ctn . Let E be the set of all points 1-,(iXn and let n be the complement of the closure of E. Prove that the product actually converges uniformly on every compact subset of n, so that B E H(!l), and that B has a pole at each point of E. (This is of particular interest in those cases in which n is connected.) 13 Put ctn = 1 - n-2, for n = 1, 2, 3, . . . , let B be the Blaschke product with zeros at these points ct, and prove that lim, .... 1 B(r) = 0. (It is understood that 0 < r < 1.) More precisely, show that the estimate is valid if ctN 1 < r < ctN . N- 1 r - et N- 1 ct - ct I B(r) I < n n < n N n < 2e -N /3 1 1 - ctn r 1 1 - ctn 14 Prove that there is a sequence { ctn} with 0 < ctn < 1, which tends to 1 so rapidly that the Blaschke product with zeros at the points ctn satisfies the condition lim sup I B(r) I = 1. r-+1 Hence this B has no radial limit at z = 1. IS Let cp be a linear fractional transformation which maps U onto U. For any z E U define the cp-orbit of z to be the set { cpiz)}, where cp0(z) = z, cpn(z) = cp(cpn_ 1(z)), n = 1, 2, 3, . . . . Ignore the case cp(z) = z. (a) For which cp is it true that the cp-orbits satisfy the Blaschke condition I:(1 - I cpiz) I ) < oo? [The answer depends in part on the location of the fixed points of cp. There may be one fixed point in U, or one fixed point on T, or two fixed points on T. In the last two cases it is advantageous to transfer the problem to (say) the upper half plane, and to consider transformations on it which either leave only oo fixed or leave 0 and oo fixed.] (b) For which cp do there exist nonconstant functions f E H00 which are invariant under cp, i.e., which satisfy the relationf(cp(z)) = f(z) for all z E U? 16 Suppose I ct1 I < I ct2 1 < lct3 1 < · · · < 1, and let n(r) be the number of terms in the sequence { cti} such that I cti I < r. Prove that 17 If B(z) = I:ck ј is a Blaschke product with at least one zero off the origin, is it possible to have ck > 0 for k = 0, 1, 2, . . . ? 318 REAL AND COMPLEX ANALYSIS 18 Suppose B is a Blaschke product all of whose zeros lie on the segment (0, 1) and f(z) = (z - 1)2 B(z). Prove that the derivative off is bounded in U. 19 Putf(z) = exp [ -(1 + z)/(1 - z)]. Using the notation of Theorem 15.19, show that lim p.,(/) < p.(f), although/ e H00• Note the contrast with Theorem 15.24. 20 Suppose A1 > A2 > · · · , and An---+ 0 in the Miintz-Szasz theorem. What is the conclusion of the theorem, under these conditions? 21 Prove an analogue of the Miintz-Szasz theorem, with I!( I) in place of C(J). 22 Putf,(t) = t"e-r (0 < t < oo , n = 0, 1, 2, . . . ) and prove that the set of all finite linear combinations of the functions fn is dense in 13(0, oo ). Hint: If g E 13(0, oo) is orthogonal to each /, and if F(z) = f' e-•• g(t) dt (Re z > 0), then all derivatives of F are 0 at z = 1. Consider F(1 + iy). 23 Suppose n ;:::, 0, f E H(!l), I f(e;6) I > 3 for all real 8, /(0) = 0, and A1, A2 , . . . , AN are the zeros of 1 -fin U, counted according to their multiplicities. Prove that I AtA2 " ' AN I < !. Suggestion: Look at B/(1 -f), where B is a certain Blaschke product. CHAPTER SIXTEEN ANALYTIC CONTINUATION In this chapter we shall be concerned with questions which arise because func tions which are defined and holomorphic in some region can frequently be extended to holomorphic functions in some larger region. Theorem 10.18 shows that these extensions are uniquely determined by the given functions. The exten sion process is called analytic continuation. It leads in a very natural way to the consideration of functions which are defined on Riemann surfaces rather than in plane regions. This device makes it possible to replace " multiple-valued functions " (such as the square-root function or the logarithm) by functions. A systematic treatment of Riemann surfaces would take us too far afield, however, and we shall restrict the discussion to plane regions. Regular Points and Singular Points 16.1 Definition Let D be an open circular disc, suppose f e H(D), and let p be a boundary point of D. We call p a regular point off if there exists a disc D1 with center at P and a function g e H(D 1) such that g(z) = f(z) for all z e D n D 1 • Any boundary point of D which is not a regular point off is called a singular point off It is clear from the definition that the set of all regular points off is an open (possibly empty) subset of the boundary of D. In the following theorems we shall take the unit disc U for D, without any loss of generality. 319 320 REAL AND COMPLEX ANALYSIS 16.2 Theorem Supposef e H(U), and the power series 00 /(z) = L an zn (z e U) (1) n=O has radius o f convergence 1. Then f has at least one singular point on the unit circle T. PROOF Suppose, on the contrary, that every point of T is a regular point off The compactness of T implies then that there are open discs D1, . . . , Dn and functions gi E H(D) such that the center of each Di is on T, such that T c D1 u · · · u Dn , and such that giz) = f(z) in Di n U, for j = 1, . . . , n. If Di n Di "# 0 and V;i = Di n Di n U, then V;i "# 0 (since the centers of the Di are on T), and gi = f = gi in V;i . Since Di n Di is connected, it follows from Theorem 10.18 that gi = gi in Di n Di . Hence we may define a function h in Q = U u D1 u · · · u Dn by h(z) = {/(z) gi(z) (z E U), (z e Di), (2) Since Q => 0 and Q is open, there exists an E > 0 such that the disc D(O; 1 + E) c Q. But h e H(Q), h(z) is given by (1) in U, and now Theorem 10.16 implies that the radius of convergence of (1) is at least 1 + E, contrary to our assumption. I I I I 16.3 Definition Iff e H(U) and if every point of T is a singular point of f, then T is said to be the natural boundary off In this case, f has no holo morphic extension to any region which properly contains U. 16.4 Remark It is very easy to see that there exist f e H(U) for which T is a natural boundary. In fact, if Q is any region, it is easy to find an f e H(Q) which has no holomorphic extension to any larger region. To see this, let A be any countable set in Q which has no limit point in Q but such that every boundary point of Q is a limit point of A. Apply Theorem 15.1 1 to get a function f e H(Q) which is 0 at every point of A but is not identically 0. If g e H(Q1), where Q1 is a region which properly contains Q, and if g =fin Q, the zeros of g would have a limit point in 01, and we have a contradiction. A simple explicit example is furnished by 00 f(z) = L z2" = z + z2 + z4 + z8 + · · · (z e U). (1) n=O This f satisfies the functional equation /(z2) = /(z) - z, (2) from which it follows (we leave the details to the reader) that f is unbounded on every radius of U which ends at exp {2nikl2n}, where k and n are positive ANALYTIC CONTINUATION 321 integers. These points form a dense subset of T; and since the set of all singular points off is closed,fhas T as its natural boundary. That this example is a power series with large gaps (i.e., with many zero coefficients) is no accident. The example is merely a special case of Theorem 16.6, due to Hadamard, which we shall derive from the following theorefo of Ostrowski: 16.5 Theorem Suppose A., Pk , and qk are positive integers, P 1 < P2 < P3 < . . . , and (k = 1, 2, 3, . . . ). (1) Suppose 00 f(z) = L anzn (2) n=O has radius o f convergence 1, and an = 0 whenever Pk < n < qk for some k. If sp(z) is the pth partial sum of (2), and if f3 is a regular point o ff on T, then the sequence {sPk(z)} converges in some neighborhood of {3. Note that the full sequence {sp(z)} cannot converge at any point outside 0. The gap condition (1) ensures the existence of a subsequence which converges in a neighborhood of [3, hence at some points outside U. This phenomenon is called overconvergence. PROOF If g(z) = f(f3z), then g also satisfies the gap condition. Hence we may assume, without loss of generality, that f3 = 1. Then f has a holomorphic extension to a region Q which contains U u { 1}. Put (3) and define F(w) = f(qJ(w)) for all w such that qJ(w) E n. If I w I < 1 but w =F 1, then I qJ(w) I < 1, since 1 1 + w I < 2. Also, qJ(1) = 1. It follows that there exists an E > 0 such that qJ(D(O; 1 + e)) c n. Note that the region qJ(D(O; 1 + e)) contains the point 1. The series 00 F(w) = L bm wm (4) m=O converges if I w I < 1 + e. 322 REAL AND COMPLEX ANALYSIS The highest and lowest powers of w in [<p(w)]n have exponents (A. + 1)n and A.n. Hence the highest exponent in [<p(w)]Pk is less than the lowest expo nent in [<p(w)]qk, by (1). Since 00 F(w) = L an[<p(w)]n ( I w I < 1), n=O the gap condition satisfied by {an} now implies that Pk (A.+ l)Pk L an[<p(w)]n = L bm Wm n=O m=O ( k = 1 ' 2, 3' . . . ). (5) (6) The right side of (6) converges, as k---+ oo, whenever I w I < 1 +E. Hence {sPk(z)} converges for all z E <p(D(O; 1 +e)). This is the desired conclusion. //// Note: Actually, {sPk(z)} converges uniformly in some neighborhood of {3. We leave it to the reader to verify this by a more careful examination of the preceding proof. 16.6 Theorem Suppose A. is a positive integer, {Pk} is a sequence of positive integers such that ( k = 1' 2, 3, . . . ), (1) and the power series 00 f(z) = L ck zPk (2) k= 1 has radius o f convergence 1. Thenfhas T-as its natural boundary. PROOF The subsequence {sPk} of Theorem 16.5 is now the same (except for repetitions) as the full sequence of partial sums of (2). The latter cannot con verge at any point outside 0 ; hence Theorem 16.5 implies that no point of T can be a regular point off /Ill 16.7 Exa!J!Pie Put an = 1 if n is a power of 2, put an = 0 otherwise, put '7n = exp (-vn), and define Since 00 f(z) = L an '1n zn. n=O lim sup 1 an '1n 11/n = 1, n-+ oo (1) (2) ANALYTIC CONTINUATION 323 the radius of convergence of (1) is 1. By Hadamard's theorem, f has T as its natural boundary. Nevertheless, the power series of each derivative off, 00 f(z) = L n(n - 1) · · · (n - k + 1)a n '7n zn -k, (3) n=k converges uniformly on the closed unit disc. Each f is therefore uniformly continuous on 0, and the restriction off to T is infinitely differentiable, as a function of fJ, in spite of the fact that T is the natural boundary off The example demonstrates rather strikingly that the presence of singularities, in the sense of Definition 16.1, does not imply the presence of discontinuities or (stated less precisely) of any lack of smoothness. This seems to be the natural place to insert a theorem in which continuity does preclude the existence of singularities : 16.8 Theorem Suppose Q is a region, L is a straight line or a circular arc, Q - L is the union of two regions 01 and 02 , f is continuous in Q, and f is holomorphic in 01 and in 02 • Thenfis holomorphic in Q. PROOF The use of linear fractional transformations shows that the general case follows if we prove the theorem for straight lines L. By Morera's theorem, it is enough to show that the integral off over the boundary an is 0 for every triangle n in Q. The Cauchy theorem implies that the integral off vanishes over every closed path y in ࡊ n Q1 or in A n Q2 • The continuity of f shows that this is still true if part of y is in L, and the integral over an is the sum of at most two terms of this sort. I I I I Continuation along Curves 16.9 Definitions A function element is an ordered pair (/, D), where D is an open circular disc and f e H(D). Two function elements (f0 , D0) and (f1, D1) are direct continuations of each other if two conditions hold : D0 n D 1 =F 0, andf0(z) =f1(z) for all z E D0 n D1. In this case we write (1) A chain is a finite sequence CC of discs, say CC = {D0 , D1, • • • , Dn}, such that D;- 1 n D; =F 0 for i = 1, . . . , n. If (/0 , D0) is given and if there exist elements (/; , D;) such that (/; 1, D; 1) "' (/; , D;) for i = 1, . . . , n, then (f n , Dn) is said to be the analytic continuation of(f0 , D0) along CC. Note thatfn is uniquely deter mined by fo and CC (if it exists at all). To see this, suppose (1) holds, and suppose (1) also holds with g1 in place of/1. Then g1 =fo =!1 in D0 n D1 ; and since D 1 is connected, we have g 1 = f1 in D 1. The uniqueness of f n now follows by induction on the number of terms in CC. 324 REAL AND COMPLEX ANALYSIS If (fn , Dn) is the continuation of (/0 , D0) along CC, and if Dn n D0 =F 0, it need not be true that (/0 , D 0) "' (f n , D n); in other words, the relation "' is not transitive. The simplest example of this is furnished by the square-root func tion : Let D0 , D1, and D2 be discs of radius 1, with centers 1, ro, and ro2, where ro3 = 1, choose jj e H(D) so that fJ(z) = z and so that (/0 , D0) e (fl, D1), (fh D1) e (/2 , D2). In Do n D2 we have/2 = -fo =Ffo · A chain CC = { D 0 , . . . , D n} is said to cover a curve y with parameter inter val [0, 1] if there are numbers 0 = s0 < s 1 < · · · < sn = 1 such that y(O) is the center of D0 , y(1) is the center of Dn , and (i = 0, . . . , n - 1 ). (2) If (f0 , D0) can be continued along this CC to (f,, Dn), we call (fn , Dn) an analytic continuation o f (f0 , D0) along y (uniqueness will be proved in Theorem 16.1 1); (/0 , D0) is then said to admit an analytic continuation along y. Although the relation (1) is not transitive, a restricted form of transitivity does hold. It supplies the key to the proof of Theorem 16.1 1. 16.10 Proposition Suppose that D0 n D1 n D2 =F 0, (D0 ,/0) e (D1,/1), and (D 1' /1) e (D2 , /2). Then (Do , fo) e (D2 , /2). PROOF By assumption, /0 = f1 in D0 n D1 and /1 = /2 in D1 n D2 • Hence fo = /2 in the nonempty open set D0 n D1 n D2 . Since fo and /2 are holo morphic in D0 n D2 and D0 n D2 is connected, it follows that /0 = /2 in Do n D2 . //// 16.11 Theorem If (f, D) is a function element and if y is a curve which starts at the center o f D, then (f, D) admits at most one analytic continuation along y. Here is a more explicit statement of what the theorem asserts : If y is covered by chains CC1 = {A0 , A1, • • • , Am} and CC2 = {B0 , B1, • • • Bn}, where A0 = B0 = D, if (f, D) can be analytically continued along CC 1 to a function element (gm , Am), and if (f, D) can be analytically continued along CC 2 to (hn , Bn), then gm = hn in Am n Bn . Since Am and Bn are, by assumption, discs with the same center y( 1 ), it follows that gm and hn have the same expansion in powers of z - y(1), and we may as well replace Am and Bn by whichever is the larger one of the two. With this agreement, the conclusion is that gm = hn . PROOF Let CC 1 and CC 2 be as above. There are numbers 0 = s0 < S 1 < · · · < S m = 1 = Sm + 1 and 0 = u0 < u1 < · · · < an = 1 = un+ l such that y([si , si + 1]) c: Ai , (0 < i < m, 0 < j < n). (1) ANALYTIC CONTINUATION 325 There are function elements (gi , Ai) Ů (gi+ t' Ai+ 1) and (hi, Bi) Ů (hi+ 1, Bi+ 1), for 0 < i < m - 1 and 0 < j < n - 1. Here g0 = h0 = f We claim that if 0 < i < m and 0 < j < n, and if [s;, si+ 1] intersects [ui, ui+ 1], then (gi , Ai) Ů (hi, Bi). Assume there are pairs (i, j) for which this is wrong. Among them there is one for which i + j is minimal. It is clear that then i + j > 0. Suppose si > ui. Then i > 1, and since [s;, si+ 1] intersects [ui, ui+ 1], we see that (2) The minimality of i + j shows that (gi_ 1, Ai_ 1) Ů (hi, B); and since (gi_ 1, Ai_ 1) Ů (gi , Ai), Proposition 16.10 implies that (gi , Ai) Ů (hi, B). This contradicts our assumption. The possibility si < ui is ruled out in the same way. So our claim is established. In particular, it holds for the pair (m, n), and this is what we had to prove. /Ill 16.12 Definition Suppose (X and f3 are points in a topological space X and cp is a continuous mapping of the unit square I2 = I x I (where I = [0, 1]) into X such that cp(O, t) = (X and cp(1, t) = f3 for all t e I. The curves Yt defined by Yt(s) = cp(s, t) (s E I, t E I) (1) are then said to form a one-parameter family {Yt} of curves from (X to f3 in X. We now come to a very important property of analytic continuation: 16.13 Theorem Suppose {Yt} (0 < t < 1) is a one-parameter family of curves f rom (X to f3 in the plane, D is a disc with center at (X, and the function element (f, D) admits analytic continuation along each }'p to an element (gt, Dt). Then g1 = go · The last equality is to be interpreted as in Theorem 16.1 1 : (gt, D1) Ů (go ' Do), and D0 and D1 are discs with the same center, namely, {3. PROOF Fix t e I. There is a chain ct = { A0 , • • • , An} which covers Yt , with A0 = D, such that (gt , Dt) is obtained by continuation of (f, D) along ct. There are numbers 0 = s0 < · · · < sn = 1 such that ( i = 0, 1, . . . , n - 1 ) . (1) There exists an e > 0 which is less than the distance from any of the compact sets Ei to the complement of the corresponding open disc Ai . The uniform continuity of cp on I2 (see Definition 16.12) shows that there exists a Ȝ > 0 such that if s E I' u E I' I u - t I < Ȝ-(2) 326 REAL AND COMPLEX ANALYSIS Suppose u satisfies these conditions. Then (2) shows that CC covers Yu , and therefore Theorem 16.1 1 shows that both gt and gu are obtained by continua tion of (f, D) along this same chain CC. Hence gt = gu . Thus each t e I is covered by a segment Jt such that gu = gt for all u e I n Jt. Since I is compact, I is covered by finitely many Jt; and since I is connected, we see in a finite number of steps that g1 = g0 • /Ill Our next item is an intuitively obvious topological fact. 16.14 Theorem Suppose r 0 and r 1 are curves in a topological space X, with common initial point ex and common end point p. If X is simply connected, then there exists a one-parameter family {Yt} (0 < t < 1) o f curves from ex to P in X, such that Yo = r 0 and y1 = r 1• PROOF Let [0, n] be the parameter interval of r 0 and r 1• Then r(s) = {r o(s) r1(2n - s) (0 < s < n) (n < s < 2n) (1) defines a closed curve in X. Since X is simply connected, r is null-homotopic in X. Hence there is a continuous H: [0, 2n] x [0, 1] Ò X such that H(s, 0) = r(s), H(s, 1) = c e X, H(O, t) = H(2n, t). If : 0 Ơ X is defined by (reiÔ = H(O, 1 - r) (0 < r < 1, 0 < (J < 2n), (2) implies that is continuous. Put and Yt(O) = <1>[(1 - t)ei6 + te-i8] (0 < (J < n, 0 < t < 1). Since ( ei8) = H( 0, 0) = r( 0), it follows that Yt(O) = <1>(1) = r(O) = ex (0 < t < 1), Yt(n) = ( - 1) = r(n) = p (0 < t < 1), (0 < (J < n) (0 < (J < n). This completes the proof. The Monodromy Theorem (2) Ill/ The preceding considerations have essentially proved the following important theorem. ANALYTIC CONTINUATION 327 16.15 Theorem Suppose Q is a simply connected region, (f, D) is a function element, D c n, and (f, D) can be analytically continued along every curve in Q that starts at the center of D. Then there exists g E H(Q) such that g(z) = f(z) for all z E D. PROOF Let r 0 and r 1 be two curves in n from the center C( of D to some point {3 E n. It follows from Theorems 16.13 and 16.14 that the analytic con tinuations of (f, D) along r 0 and r 1 lead to the same element (g fJ , D p), where Dp is a disc with center at {3. If Dp1 intersects Dp , then (gpl' Dp1) can be obtained by first continuing (/, D) to {3, then along the straight line from {3 to {31• This shows that gp1 = gp in Dp1 n Dp . The definition g(z) = gp(z) (z E Dp) is therefore consistent and gives the desired holomorphic extension off I I I I 16.16 Remark Let Q be a plane region, fix w ¢ Q, let D be a disc in Q. Since D is simply connected, there exists f E H(D) such that exp [f(z)] = z - w. Note that f'(z) = (z - w) - 1 in D, and that the latter function is holomorphic in all of n. This implies that (f, D) can be analytically continued along every path y in Q that starts at the center C( of D: If y goes from C( to {3, if D p = D(f3; r) c Q, if (z E Dp) (1) and if (z E Dp), (2) then (g fJ , D p) is the continuation of (f, D) along y. Note that gJ,(z) = (z - w) - 1 in Dp . Assume now that there exists g E H(Q) such that g(z) = f(z) in D. Then g'(z) = (z - w) - 1 for all z E n. If r is a closed path in n, it follows that Indr (w) = -2 1 . j g'(z) dz = 0. 1tl Jr (3) We conclude (with the aid of Theorem 13.1 1) that the monodromy theorem fails in every plane region that is not simply connected. 328 REAL AND COMPLEX ANALYSIS Construction of a Modular Function 16.17 The Modular Group This is the set G of all linear fractional transform ations qJ of the form ( ) az + b qJ z = cz + d (1) where a, b, c, and d are integers and ad - be = 1. Since a, b, c, and d are real, each qJ e G maps the real axis onto itself (except for oo ). The imaginary part of qJ(i) is (c2 + d2)- 1 > 0. Hence (({J E G), where ll + is the open upper half plane. If qJ is given by (1), then _ 1( ) dw - b qJ w = - cw + a so that qJ - 1 e G. Also cp o t/1 e G if qJ e G and t/1 e G. (2) (3) Thus G is a group, with composition as group operation. In view of (2) it is customary to regard G as a group of transformations on n + . The transformations z-+ z + 1 (a = b = d = 1, c = 0) and z-+ - 1/z (a = d = 0, b = - 1, c = 1) belong to G. In fact, they generate G (i.e., there is no proper subgroup of G which contains these two transformations). This can be proved by the same method which will be used in Theorem 16.19(c). A modular function is a holomorphic (or meromorphic) function f on n + which is invariant under G or at least under some nontrivial subgroup r of G. This means that/ o qJ =/for every qJ e r. 16.18 A Subgroup We shall take for r the group generated by a and -r, where z a(z) = 2z + 1 ' -r(z) = z + 2. (1) One of our objectives is the construction of a certain function A which is invari ant under r and which leads to a quick proof of the Picard theorem. Actually, it is the mapping properties of A which are important in this proof, not its invari ance, and a quicker construction (using just the Riemann mapping theorem and the reflection principle) can be given. But it is instructive to study the action of r on n + , in geometric terms, and we shall proceed along this route. Let Q be the set of all z which satisfy the following four conditions, where Z = X + iy: y > 0, - 1 <X< 1, 1 2z + 1 1 > 1, 1 2z - 1 1 > 1. (2) Q is bounded by the vertical lines x = - 1 and x = 1 and is bounded below by two semicircles of radius . , with centers at -! and at t. Q contains those of its boundary points which lie in the left half of TI +. Q contains no point of the real axis. ANALYTIC CONTINUATION 329 We claim that Q is afundamental domain ofr. This means that statements (a) and (b) of the following theorem are true. 16.19 Theorem Let r and Q be as above. (a) If cp1 and <p2 e r and <p 1 =F E r cp(Q) = rr + • (c) r contains all trans formations qJ E G of the form ( ) az + b qJ z = cz + d for which a and d are odd integers, b and c are even. (1) PROOF Let r 1 be the set of all cp e G described in (c). It is easily verified that r 1 is a subgroup of G. Since (J E r 1 and 't E r 1' it follows that r c r 1• To show that r = r 1 ' i.e., to prove (c), it is enough to prove that (a') and (b) hold, where (a') is the statement obtained from (a) by replacing r by r 1 . For if (a') and (b) hold, it is clear that r cannot be a proper subset of r 1. We shall need the relation Im z Im q>(z) = l cz + d l2 (2) which is valid for every cp e G given by ( 1 ). The proof of (2) is a matter of straightforward computation, and depends on the relation ad - be = 1. We now prove (a'). Suppose cp1 and cp2 e r 1, cp1 =1= cp2 , and define <p = <tJ1 1 o cp2 • If z e <p1 (Q) n cp2(Q), then cp1 1(z) e Q n 1 for all z e Q. Otherwise, the disc D(-djc; 1/ I c I ) would intersect Q. The description of Q shows that if a. =F -ĺ is a real number and if D(a.; r) intersects Q, then at least one of the points - 1, 0, 1 lies in D(a.; r). Hence I cw + d I < 1, for w = - 1 or 0 or 1. But for these w, cw + d is an odd integer whose absolute value cannot be less than 1. So I cz + d I > 1, and it now follows from (2) that Im cp(z) < Im z for 330 REAL AND COMPLEX ANALYSIS every z e Q. If it were true for some z e Q that q>(z) E Q, the same argument would apply to q>- 1 and would show that Im z = Im q> - 1(q>(z)) < Im q>(z). (4) This contradiction shows that (3) holds. Hence (a') is proved. To prove (b), let l: be the union of the sets q>(Q), for q> e r. It is clear that l: c ll +. Also, l: contains the sets 'tn(Q), for n ·= 0, + 1, + 2, . . . , where rn(z) = z + 2n. Since u maps the circle 1 2z + 1 1 = 1 onto the circle 1 2z - 1 1 = 1, we see that l: contains every z e n + which satisfies all inequal ities 1 2z - (2m + 1) I > 1 ( m = 0, + 1, + 2, . . . ). (5) Fix w E n +. Since Im w > 0, there are only finitely many pairs of inte gers c and d such that I cw + d I lies below any given bound, and we can choose q>0 e r so that I cw + d I is minimized. By (2), this means that Im q>(w) < Im q>0(w) Put z = q>0(w). Then (6) becomes ((z) < Im z ( = ur-n and to q> = u- 1r-n. Since z - 2n (ur -n)(z) = , 2z - 4n + 1 it follows from (2) and (7) that _ 1 -n z - 2n (u r )(z) = - 2z + 4n + 1 ' (6) (7) (8) 1 2z - 4n + 1 1 > 1, 1 2z - 4n - 1 1 > 1 (n = 0, + 1, + 2, . . . ). (9) Thus z satisfies (5), hence z e l:; and since w = = A/or every q> e r. (b) A is one-to-one on Q. (c) The range Q of A [which is the same as A(Q), by (a)], is the region consist ing of all complex numbers diff erent f rom 0 and 1. (d) A has the real axis as its natural boundary. ANALYTIC CONTINUATION 331 PROOF Let Q0 be the right half of Q. More precisely, Q0 consists of all z E II + such that 0 < Re z < 1, 1 2z - 1 1 > 1. (1) By Theorem 14.19 (and Remarks 14.20) there is a continuous function h on Q0 which is one-to-one on Q0 and holomorphic in Q0 , such that h(Q 0) = II +, h(O) = O, h(1) = 1, and h(oo) = oo. The reflection principle (Theorem 1 1.14) shows that the formula h(-x + iy) = h(x + iy) (2) extends h to a continuous function on the closure Q of Q which is a confor mal mapping of the interior of Q onto the complex plane minus the non negative real axis. We also see that h is one-to-one on Q, that h(Q) is the region Q described in (c), that h(- 1 + iy) = h(1 + iy) = h(-r(- 1 + iy)) (0 < y < oo), (3) and that h(- ĺ + ĺei9) = h(ĺ + ĺei(1t-9)) = h( u(- ĺ + ĺei9)) (0 < () < n). (4) Since h is real on the boJndary of Q, (3) and (4) follow from (2) and the definitions of u and r. 1 We now define the function A: (z E (/)( Q), <p E r). (5) By Theorem 16.19, each z E n + lies in qJ(Q) for one and only one <p E r. Thus (5) defines A(z) for z E II +, and we see immediately that A has properties (a) to (c) and that A is holomorphic in the interior of each of the sets qJ(Q). It follows from (3) and ( 4) that A is continuous on Q u !- l(Q) u 0' - l(Q), hence on an open set V which contains Q. Theorem 16.8 now shows that A is holomorphic in V. Since II + is covered by the union of the sets <p( V), 0, and the radius of convergence of the series is 1. Prove that f has a singularity at z = 1. Hint: Expand fin powers of z - ! . If 1 were a regular point off, the new series would converge at some x > 1. What would this imply about the original series? 2 Suppose (f, D) and (g, D) are function elements, P is a polynomial in two variables, and P(f, g) = 0 in D. Suppose f and g can be analytically continued along a curve y, to / 1 and g 1 • Prove that ANALYTIC CONTINUATION 333 P(/1, g1) = 0. Extend this to more than two functions. Is there such a theorem for some class of functions P which is larger than the polynomials? 3 Suppose n is a simply connected region, and u is a real harmonic function in n. Prove that there exists an f E H(!l) such that u = Ref. Show that this fails in every region which is not simply con nected. 4 Suppose X is the closed unit square in the plane,fis a continuous complex function on X, andfhas no zero in X. Prove that there is a continuous function g on X such that f = e9• For what class of spaces X (other than the above square) is this also true? S Prove that the transformations z---+ z + 1 and z---+ - 1/z generate the full modular group G. Let R consist of all z = x + iy such that I x I < ! , y > 0, and I z I > 1, plus those limit points which have x < 0. Prove that R is a fundamental domain of G. 6 Prove that G is also generated by the transformations qJ and t/1, where 1 qJ(z) = - -, z z - 1 tjl(z) = --z Show that qJ has period 2, t/1 has period 3. 7 Find the relation between composition of linear fractional transformations and matrix multiplica tion. Try to use this to construct an algebraic proof of Theorem 16.19(c) or of the first part of Exercise 5. 8 Let E be a compact set on the real axis, of positive Lebesgue measure, let n be the complement of E, relative to the plane, and define f(z) = l dt E t - Z (z E !l). Answer the following questions: (a) Is/ constant? (b) Canfbe extended to an entire function? (c) Does lim z f(z) exist as z---+ oo? If so, what is it? (d) Doesfhave a holomorphic square root in !l? (e) Is the real part of/bounded in !l? (/) Is the imaginary part off bounded in n? [If " yes " in (e) or (f), give a bound.] (g) What is J, f(z) dz if y is a positively oriented circle which has E in its interior? (h) Does there exist a bounded holomorphic function qJ in n which is not constant? 9 Check your answers in Exercise 8 against the special case E = [ - 1, 1]. 10 Call a compact set E in the plane removable if there are no nonconstant bounded holomorphic functions in the complement of E. (a) Prove that every countable compact set is removable. (b) If E is a compact subset of the real axis, and m(E) = 0, prove that E is removable. Hint : E can be surrounded by curves of arbitrarily small total length. Apply Cauchy's formula, as in Exercise 25, Chap 10. (c) Suppose E is removable, n is a region, E c: n, f E H(!l - E), and f is bounded. Prove that f can be extended to a holomorphic function in n. (d) Formulate and prove an analogue of part (b) for sets E which are not necessarily on the real aXIS. (e) Prove that no compact connected subset of the plane (with more than one point) is remov able. 334 REAL AND COMPLEX ANALYSIS 1 1 For each positive number a., let r« be the path with parameter interval ( - oo, oo) defined by - t - ni ( - 00 < t < - a.), ( - a. < t < a.), t + ni (a. < t < 00 ). Let Q« be the component of the complement of r: which contains the origin, and define 1 l exp (ew) d fiz) = -. w 2nl r2 w - z (z E OJ. Prove thatf11 is an analytic continuation off« if a. < p. Prove that therefore there is an entire function! whose restriction to Q« is f « . Prove that r-+ ex> for every ei6 =F 1. (Here r is positive and (} is real, as usual.) Prove that f is not constant. [Hint: Look atf(r).] If prove that for every ei6• g = f exp (-f), r-+ ex> Show that there exists an entire function h such that 12 Suppose lim h(nz) = {1 n -+ ex> 0 if z = 0, if z =F 0. f(z) = I ( z z2) 3' - f: a. z". k = l 2 n = l Find the regions in which the two series converge. Show that this illustrates Theorem 16.5. Find the singular point off which is nearest to the origin. 13 Let n = {z: ! < I z I < 2}. For n = 1, 2, 3, . .
. let xn be the set of all f E H(Q) that are nth deriv atives of some g E H(Q). [In other words, X n is the range of the differential operator D" with domain H(Q).] (a) Show thatf E X 1 if and only if J, f(z) dz = 0, where y is the positively oriented unit circle. (b) Show that f E X n for every n if and only iff extends to a holomorphic function in D(O; 2). 14 Suppose Q is a region, p E n, R < oo . Let ơ be the class of all f e H(Q) such that I f(p) I < R and f(Q) contains neither 0 nor 1. Prove that !F is a normal family. 15 Show that Theorem 16.2 leads to a very simple proof of the special case of the monodromy theorem (16.15) in which Q and D are concentric discs. Combine this special case with the Riemann mapping theorem to prove Theorem 16.15 in the generality in which it is stated. CHAPTER SEVENTEEN This chapter is devoted to the study of certain subspaces of H(U) which are defined by certain growth conditions; in fact, they are all contained in the class N defined in Chap. 15. These so-called HP-spaces (named for G. H. Hardy) have a large number of interesting properties concerning factorizations, boundary values, and Cauchy-type representations in terms of measures on the boundary of U. We shall merely give some of the highlights, such as the theorem of F. and M. Riesz on measures J.l whose Fourier coefficients jJ,(n) are 0 for all n < 0, Beurling's classification of the invariant subspaces of H2, and M. Riesz's theorem on conju gate functions. A convenient approach to the subject is via subharmonic functions, and we begin with a brief outline of their properties. Subharmonic Functions 17.1 Definition A function u defined in an open set 0 in the plane is said to be sub harmonic if it has the following four properties. (a) - 00 < u(z) < 00 for all Z E 0. (b) u is upper semicontinuous in n. (c) Whenever D(a; r) c Q, then 1 J1t u(a) < -2 u(a + reiő dO. 1t - n (d) None of the integrals in (c) is -oo. 335 336 REAL AND COMPLEX ANALYSIS Note that the integrals in (c) always exist and are not + oo, since (a) and (b) imply that u is bounded above on every compact K c Q. [Proo f: If Kn is the set of all z e K at which u(z) > n, then K => K 1 => K 2 • • · , so either Kn = 0 for some n, or n Kn =1= 0, in which case u(z) = oo for some z e K.] Hence (d) says that the integrands in (c) belong to L1(T). Every real harmonic function is obviously subharmonic. 17.2 Theorem I f u is subharmonic in n, and if cp is a monotonically increasing convex function on R 1, then cp o u is sub harmonic. [To have cp o u defined at all points of n, we put cp(- oo) = lim cp(x) as x ----+ - oo.] PROOF First, cp o u is upper semicontinuous, since cp is increasing and contin uous. Next, if D(a; r) c n, we have The first of these inequalities holds since cp is increasing and u is sub harmonic; the second follows from the convexity of cp, by Theorem 3.3. /Ill 17.3 Theorem If Q is a region, f e H(Q), and f is not identically 0, then log I f I is sub harmonic in !l, and so are log+ I f I aud I f IP (0 < p < oo ). PROOF It is understood that log I f(z) I = - oo if f(z) = 0. Then log I f I is upper semicontinuous in n, and Theorem 15.19 implies that log I f I is sub harmonic. The other assertions follow if we apply Theorem 17.2 to log I f I in place of u, with cp(t) = max (0, t) and cp(t) = ePt. Ill/ 17.4 Theorem Suppose u is a continuous subharmonic function in Q, K is a compact subset of !l, h is a continuous real function on K which is harmonic in the interior V of K, and u(z) < h(z) at all boundary points of K. Then u(z) < h(z) for all z E K. This theorem accounts for the term " subharmonic." Continuity of u is not necessary here, but we shall not need the general case and leave it as an exercise. PROOF Put u1 = u - h, and assume, to get a contradiction, that u1(z) > 0 for some z e V. Since u1 is continuous on K, u1 attains its maximum m on K ; and since u1 < 0 on the boundary of K, the set E = {z e K : u1(z) = m} is a nonempty compact subset of V. Let z0 be a boundary point of E. Then for HP-SPACES 337 some r > 0 we have D(z0; r) c V, but some subarc of the boundary of D(z0 ; r) lies in the complement of E. Hence and this means that u1 is not subharmonic in V. But if u is subharmonic, so is u - h, by the mean value property of harmonic functions, and we have our contradiction. //// 17.5 Theorem Suppose u is a continuous subharmonicfunction in U, and 1 J1t m(r) = -2 u(reiő d(} 1t -1t (0 < r < 1). (1) PROOF Let h be the continuous function on D(O; r2) which coincides with u on the boundary of D(O; r2), and which is harmonic in D(O; r2). By Theorem 17.4, u < h in D(O; r2). Hence 1 J1t 1 J1t m(r1) < -2 h(r1ei6) d(} = h(O) = -2 h(r2 ei6) dfJ = m(r2). 1t -1t 1t -1t The Spaces HP and N 17.6 Notation As in Sees 1 1.15 and 1 1.19, we define!,. on T by (0 < r < 1) Ill/ (1) if f is any continuous function with domain U, and we leĹ u denote Lebesgue measure on T, so normalized that u(T) = 1. Accordingly, If-norms will refer to I!(u). In particular, ll.friiP = {11 fr IP da} l/p (0 < p < oo), 11/,.11 oo = sup I f(re!ő I , tJ and we also introduce 11 /,.110 = exp Llog+ I f.. I da. 17.7 Definition Iff e H(U) and 0 < p < oo, we put 11 / IIP = sup { II!,.IIP: 0 < r < 1}. (2) (3) (4) (1) 338 REAL AND COMPLEX ANALYSIS If 0 < p < oo , HP is defined to be the class of all f E H(U) for which II f II P < oo. (Note that this coincides with our previously introduced termi nology in the case p = oo.) The class N consists of all f E H( U) for which II f II 0 < oo . It is clear that H00 c HP c Hs c N if 0 < s < p < oo. 17.8 Remarks (a) When p < oo , Theorems 17.3 and 17.5 show that ll f,. II P is a nondecreasing function of r, for every f E H(U); when p = oo, the same follows from the maximum modulus theorem. Hence II f II p = lim II /,.II p • r- 1 (1) (b) For 1 < p < oo , II / IlP satisfies the triangle inequality, so that HP is a normed linear space. To see this, note that the Minkowski inequality gives (2) if O < r < 1. As rq 1, we obtain (3) (c) Actually, HP is a Banach space, if 1 < p < oo : To prove completeness, suppose {In} is a Cauchy sequence in HP, I z I < r < R < 1, and apply the Cauchy formula to f n -f m , integrating around the circle of radius R, center 0. This leads to the inequalities from which we conclude that {In} converges uniformly on compact subsets of U to a functionf e H(U). Given E > 0, there is an m such that 11 / n -f mii P < E for all n > m, and then, for every r < 1, (4) n- oo This gives 11 /-/ mllpA 0 as mA oo . (d) For p < 1, HP is still a vector space, but the triangle inequality is no longer satisfied by II f II P . We saw in Theorem 15.23 that the zeros of any f e N satisfy the Blaschke condition l:(1 - I C(n I ) < oo . Hence the same is true in every HP. It is interesting that the zeros of any f e HP can be divided out without increasing the norm: 17.9 Theorem Supposefe N,f¥= 0, and B is the Blaschke productformed with the zeros off Put g = f /B. Then g E N and llg ll o = 11 / 11 0 • Moreover, iffe HP, then g E HP and llg iiP = 11 / II P (O < p < oo). HP-SPACES 339 PROOF Note first that I g( z) I > I f ( z) I (z E U). (1) In fact, strict inequality holds for every z E U, unless f has no zeros in U, in which case B = 1 and g = f If s and t are nonnegative real numbers, the inequality log+ (st) < log+ s + log+ t (2) holds since the left side is 0 if st < 1 and is log s + log t if st > 1. Since I g I = I f Ill B I , (2) gives 1 log+ l o l < log+ 1 ! 1 + log I B I ' (3) By Theorem 15.24, (3) implies that II g II 0 < II f II 0 , and since (1) holds, we actually have llgllo = 11/ llo · Now suppose f E HP for some p > 0. Let Bn be the finite Blaschke product formed with the first n zeros off (we arrange these zeros in some sequence, taking multiplicities into account). Put gn = f l Bn . For each n, 1 Bn(rei8) 1 q 1 uniformly, as rƈ 1. Hence llgniiP = 11/ llp · As n q oo, l gn l increases to I g I , so that (0 < r < 1), (4) by the monotone convergence theorem. The right side of (4) is at most 11/IIP' for all r < 1. If we let rq 1, we obtain llgiiP < 11/ llp · Equality follows now from ( 1 ), as before. I I I I 17.10 Theorem Suppose 0 < p < oo, f E HP, f =/= 0, and B is the Blaschke product formed with the zeros off Then there is a zero{ree function h E H2 such that (1) In particular, every f E H1 is a product f= gh (2) in which both factors are in H2• PROOF By Theorem 17.9, / IB E HP; in fact, 11/ IBIIP = 11/llp · Since f iB has no zero in U and U is simply connected, there exists cp E H(U) so that exp (cp) = f iB (Theorem 13.1 1). Put h = exp (pcpl2). Then h E H(U) and I h 12 = I f iB IP, hence h E H2, and (1) holds. In fact, II h II ĸ = II f IIх . To obtain (2), write (1) in the form/ = (Bh) · h. II II 340 REAL AND COMPLEX ANALYSIS We can now easily prove some of the most important properties of the HP spaces. 17.11 Theorem IfO < p < oo andf e HP, then (a) the nontangential maximal functions N « fare in I!'(T),for all C( < 1 ; (b) the nontangential limitsf(e;8) exist a.e. on T, andf e I!'(T); (c) limr-1 11/ -f,.IIP = 0, and (d) 11/ llp = 11 /llp · Iff e H1 thenfis the Cauchy integral as well as the Poisson integral off. PROOF We begin by proving (a) and (b) for the case p > 1. Since holomorphic functions are harmonic, Theorem 1 1.30(b) shows that every f e HP is then the Poisson integral of a function (call it /) in I!'(T). Hence Naf e I!(T), by Theorem 1 1.25(b), and f(e;8) is the nontangential limit off at almost every e;8 e T, by Theorem 1 1.23. If 0 < p < 1 and f e HP, use the factorization f= Bh21P (1) given by Theorem 17.10, where B is a Blaschke product, h e H2, and h has no zero in U. Since I f I < I h I21P in U, it follows that -(Naf)P < (N« h)2, (2) so that Nafe I!(T), because Nrz h e I3(T). Similarly, the existence of B and h a.e. on T implies that the non tangential limits off(call them/) exist a.e. Obviously, I f I < Nrzfwherever f exists. Hence/ e I.!'(T). This proves (a) and (b), for 0 < p < oo. Since /,. / f a.e. and I /,. I < N a f, the dominated convergence theorem gives (c). If p > 1, (d) follows from (c), by the triangle inequality. If p < 1, use Exer cise 24, Chap. 3, to deduce (d) from (c). Finally, if/ e H1, r < 1, andf,.(z) = f(rz), then/,. e H(D(O, 1/r)), and there fore/,. can be represented in U by the Cauchy formula f.(z) = .!. f1t f,.(eit) . dt r 2n - 1t 1 - e -ltz and by the Poisson formula /,.(z) = 2 1 J1t P(z, eit)f,.(eit) dt. 1t - 1t (3) (4) For each z E U, 1 1 - e-itz l and P(z, eit) are bounded functions on T. The case p = 1 of (c) leads therefore from (3) and (4) to and 1 J1t f ( eit) f(z) = -. dt 2n _ 1t 1 - e-ltz 1 J1t . . f(z) = -2 P(z, elt)f(e'ŭ dt. n - 1 (5) (6) Ill/ The space H2 has a particularly simple characterization in terms of power series coefficients: 17.12 Theorem Supposefe H(U) and Thenf E H2 if and only ifLц I an 12 < oo. PROOF By Parseval's theorem, applied to/,. with r < 1, /Ill The Theorem of F. and M. Riesz 17.13 Theorem If J.J. is a complex Borel measure on the unit circle T and (1) for n = - 1, -2, - 3, . . . , then J.J. is absolutely continuous with respect to Lebesgue measure. PROOF Put/ = P[dJ.J.]. Then/satisfies II /,.ll 1 < II J.J.II (0 < r < 1). (2) (See Sec. 1 1.1 7.) Since, setting z = rei8, 00 P(z, eit) = L r1nlein8 e-int, (3) - oo 342 REAL AND COMPLEX ANALYSIS as in Sec. 1 1.5, the assumption (1), which amounts to saying that the Fourier coefficients jl( n) are 0 for all n < 0, leads to the power series 00 f(z) = L jl(n)zn (z e U). (4) 0 By (4) and (2), fe H1. Hence f= P[f], by Theorem 17.1 1, where f e I!(T). The uniqueness of the Poisson integral representation (Theorem 1 1.30) shows now that dJl = f du. /Ill The remarkable feature of this theorem is that it derives the absolute contin uity of a measure from an apparently unrelated condition, namely, the vanishing of one-half of its Fourier coefficients. In recent years the theorem has been extended to various other situations. Factorization Theorems We already know from Theorem 17.9 that every f e HP (except f = 0) can be factored into a Blaschke product and a function g e HP which has no zeros in U. There is also a factorization of g which is of a more subtle nature. It concerns, roughly speaking, the rapidity with which g tends to 0 along certain radii. 17.14 Definition An inner function is a function M e H00 for which I M I = 1 a.e. on T. (As usual, M denotes the radial limits of M.) If <p is a positive measurable function on T such that log <p e L1(T), and if { 1 f n eit + z . } Q(z) = c exp -2 it _ log qJ(e't) dt n - n e z for z e U, then Q is called an outer function. Here c is a constant, I c I = 1. (1) Theorem 15.24 shows that every Blaschke product is an inner function, but there are others. They can be described as follows. 17.15 Theorem Suppose c is a constant, I c I = 1, B is a Blaschke product, Jl is a finite positive Borel measure on T which is singular with respect to Lebesgue measure, and { I n eit + z } M(z) = cB(z) exp -it _ dJl(t) - n e z (z e U). Then M is an inner function, and every inner function is of this f orm. (1) PROOF If (1) holds and g = M/B, then log l g l is the Poisson integral of - dJl, hence log I g I < 0, so that g e H00, and the same is true of M. Also DJ1 = 0 a.e., since Jl is singular (Theorem 7.13), and therefore the radial limits of HP-sPACES 343 log I g I are 0 a.e. (Theorem 1 1.22). Since I B I = 1 a.e., we see that M is an inner function. Conversely, let B be the Blaschke product formed with the zeros of a given inner function M and put g = MIB. Then log I g I is harmonic in U. Theorems 15.24 and 17.9 show that I g I < 1 in U and that I g I = 1 a.e. on T. Thus log I g I < 0. We conclude from Theorem i 1.30 that log I g I is the Poisson integral of -dJ.l, for some positive measure Jl on T. Since log I g I = 0 a.e. on T, we have DJ.l = 0 a.e. on T, so Jl is singular. Finally, log I g I is the real part of f1t eit + z h(z) = -it dJ.l(t), _ 1t e - z and this implies that g = c exp (h) for some constant c with I c I = 1. Thus M is of the form ( 1 ). This completes the proof. I I I I The simplest example of an inner function which is not a Blaschke product is the following: Take c = 1 and B = 1, and let Jl be the unit mass at t = 0. Then {z + 1} M(z) = exp z _ 1 , which tends to 0 very rapidly along the radius which ends at z = 1. 17.16 Theorem Suppose Q is the outer function related to cp as in De finition 17.14. Then (a) log I Q I is the Poisson integral of log cp. (b) limr- 1 1 Q(reiő I = cp(ei6) a.e. on T. (c) Q e HP if and only if cp e I!'(T). In this case, II Q liP = llcpllp · PROOF (a) is clear by inspection and (a) implies that the radial limits of log I Q I are equal to log cp a.e. on T, which proves (b). If Q e HP, Fatou's lemma implies that II QIIP < IIQIIP' so llcpiiP < IIQIIP ' by (b). Conversely, if cp e I!'( T), then by the inequality between the geometric and arithmetic means (Theorem 3.3), and if we integrate the last inequality with respect to (} we find that II Q II P < llcpiiP if p < oo. The case p = oo is trivial. Ill/ 344 REAL AND COMPLEX ANALYSIS 17.17 Theorem Suppose 0 < p < oo, f e HP, and f is not identically 0. Then log I f I e L1(T), the outer function { 1 J1t eit + z . } Q1(z) = exp -2 it _ log I f(eu) I dt n 1t e z is in HP, and there is an inner function M 1 such that Furthermore, 1 J1t log I /(0) I < 27t , !og I f(e;ŭ I dt. Equality holds in (3) if and only if M 1 is constant. (1) (2) (3) The functions M 1 and Q1 are called the inner and outer factors off, respec tively; Q1 depends only on the boundary values of I f 1 . PROOF We assume first thatfe H1• If B is the Blaschke product formed with the zeros of f and if g = fiB, Theorem 17.9 shows that g e H1 ; and since I g I = I f I a.e. on T, it suffices to prove the theorem with g in place off So let us assume that f has no zero in U and that f(O) = 1. Then log I f I is harmonic in U, log I f(O) I = 0, and since log = log+ - log-, the mean value property of harmonic functions implies that 1 J1t . 1 J1t . 2n _ , !og- I f(re'Ô I dO = 2n _ , !og + I f(re'Ô dO < II f II o < II fli t (4) for 0 < r < 1. It now follows from Fatou's lemma that both log+ I f I and log- I f I are in L1(T), hence so is log I f 1 . This shows that the definition (1) makes sense. By Theorem 17.16, Q1 e H1• Also, I Qj I = I f I =1= 0 a.e., since log I f I e L1(T). If we can prove that (z e U), (5) thenf/Q1 will be an inner function, and we obtain the factorization (2). Since log I Q1 I is the Poisson integral of log I f I , (5) is equivalent to the inequality log I f I < P[log I ! I ], (6) which we shall now prove. Our notation is as in Chap. 1 1 : P[h] ts the Poisson integral of the function h e L1 ( T). For I z I < 1 and 0 < R < 1, putfR(z) = f(Rz). Fix z e U. Then log I fR(z) I = Plog+ I fR I - Plog- I fR I . (7) Since I log+ u - log+ v I < I u - v I for all real numbers u and v, and since 11/R -! 11 1 ---+ 0 as R ---+ 1 (Theorem 17.1 1), the the first Poisson integral in (7) converges to P[log + I f I ], as R---+ 1. Hence Fatou's lemma gives P[log- I / I ] < lim inf P[log- I fR I ] = P[log+ I / I ] - log I f I , (8) R-+1 which is the same as (6). We have now established the factorization (2). If we put z = 0 in (5) we obtain (3); equality holds in (3) if and only if I f(O) I = I Q1(0) I , i.e., if and only if I M1(0) 1 = 1 ; and since IIM111 oo = 1, this happens only when M1 is a con stant. This completes the proof for the case p = 1. , If 1 < p < oo, then HP c H1, hence all that remains to be proved is that Q1 e HP. But iff e HP, then I f I e IJ'(T), by Fatou's lemma; hence Q1 e HP, by Theorem 17.16(c). Theorem 17.10 reduces the case p < 1 to the case p = 2. /Ill The fact that log I f I e L1(T) has a consequence which we have already used in the proof but which is important enough to be stated separately: 17.18 Theorem IfO < p < oo,f e HP, andf is not identically 0, then at almost all points of T we have f(ei8) =F 0. PROOF Iff = 0 then log I f I = - oo, and if this happens on a set of posi tive measure, then f..tog I f(eiŭ I dt = - oo. Ill/ Observe that Theorem 17.18 imposes a quantitative restriction on the loca tion of the zeros of the radial limits of an f e HP. Inside U the zeros are also quantitatively restricted, by the Blaschke condition. As usual, we can rephrase the above result about zeros as a uniqueness theo rem : Iff e HP, g e HP, and f(eiő = g(ei8) on some subset o f T whose Lebesgue measure is positive, thenf(z) = g(z)for all z e U. 17.19 Let us take a quick look at the class N, with the purpose of determining how much of Theorems 17.17 and 17.18 is true here. Iff e N and f ӭ 0, we can divide by a Blaschke product and get a quotient g which has no zero in U and which is in N (Theorem 17.9). Then log I g I is harmonic, and since I log I g I I = 2 log+ I g I - log I g I (1) 346 REAL AND COMPLEX ANALYSIS and 1 f1t 27t _}og I g(re;9) I d() = log I g(O) 1 , (2) we see that log I g I satisfies the hypotheses of Theorem 1 1.30 and is therefore the Poisson integral of a real measure JL Thus {i eit + z } f(z) = cB(z) exp it dJ1(t) , r e -z (3) where c is a constant, I c I = 1, and B is a Blaschke product. Observe how the assumption that the integrals of log+ I g I are bounded (which is a quantitative formulation of the statement that I g I does not get too close to oo) implies the boundedness of the integrals of log- I g I (which says that I g I does not get too close to 0 at too many places). If J1 is a negative measure, the exponential factor in (3) is in H00 • Apply the Jordan decomposition to Jl. This shows: To every f e N there correspond two functions b1 and b2 e H00 such that b2 has no zero in U and f = b1/b2 • Since b! =1= 0 a.e., it follows thatfhas finite radial limits a.e. Also, f =1= 0 a.e. Is log I f I e L1(T)? Yes, and the proof is identical to the one given in Theorem 17.17. However, the inequality (3) of Theorem 17.17 need no longer hold. For example, if {1 + z} f(z) = exp 1 -z ' then 11/llo = e, I / I = 1 a.e., and 1 f1t log I /(0) I = 1 > 0 = 2n }og I /(e;ŭ I dt. The Shift Operator (4) (5) 17.20 Invariant Subspaces Consider a bounded linear operator S on a Banach space X ; that is to say, S is a bounded linear transformation of X into X. If a closed subspace Y of X has the property that S(Y) c Y, we call Y an S-invariant subspace. Thus the S-invariant subspaces of X are exactly those which are mapped into themselves by S. The knowledge of the invariant subspaces of an operator S helps us to visual ize its action. (This is a very general-and hence rather vague-principle: In studying any transformation of any kind, it helps to know what the transform ation leaves fixed.) For instance, if S is a linear operator on an n-dimensional vector space X and if S has n linearly independent characteristic vectors x1, • • • , xn , the one-dimensional spaces spanned by any of these xi are S-invariant, and we obtain a very simple description of S if we take {xh . . . , xn} as a basis of X. We shall describe the invariant subs paces of the so-called " shift operator " S on t2• Here t2 is the space of all complex sequences (1) for which ll xll = Lt̓ğnl2f'2 < oo, (2) and S takes the element x e t2 given by (1) to Sx = { 0, eo , e 1, e 2 , • • ·} • (3) It is clear that S is a bounded linear operator on t2 and that liS II = 1. A few S-invariant subspaces are immediately apparent : If ࡃ is the set of all x e t2 whose first k coordinates are 0, then lk is S-invariant. To find others we make use of a Hilbert space isomorphism between t2 and H2 which converts the shift operator S to a multiplication operator on H2• The point is that this multiplication operator is easier to analyze (because of the - richer structure of H2 as a space of holomorphic functions) than is the case in the original setting of the sequence space t2• We associate with each x e t2, given by (1), the function 00 f(z) = L ğnzn (z e U). n=O By Theorem 17.12, this defines a linear one-to-one mapping of t2 onto H2• If n=O and if the inner product in H2 is defined by (f, g) = ! I1t f(eiB)g(eiÔ dO, 2n - 1r (4) (5) (6) the Parsevai theorem shows that (f, g) = (x, y). Thus we have a Hilbert space isomorphism of t2 onto H2, and the shift operator S has turned into a multiplica tion operator (which we still denote by S) on H2: (S f)(z) = zf(z) (fe H2, z e U). (7) The previously mentioned invariant subspaces lk are now seen to consist of all f e H2 which have a zero of order at least k at the origin. This gives a clue: For any finite set { (Xh ••• , (Xk} c: U, the space Y of all f e H2 such that f( (X 1) = · · · = f( (Xk) = 0 is S-in variant. If B is the finite Blaschke product with zeros at (Xh ••• , (Xk , then/ e Y if and only iff/B e H2• Thus Y = BH2• 348 REAL AND COMPLEX ANALYSIS This suggests that infinite Blaschke products may also give rise to S-invariant subspaces and, more generally, that Blaschk.e products might be replaced by arbi trary inner functions cp. It is not hard to see that each cpH2 is a closed S-invariant subspace of H2, but that every closed S-invariant subspace of H2 is of this form is a deeper result. 17.21 Beurling's Theorem (a) For each inner function cp the space is a closed S-invariant subspace o f H2• (1) (b) If cp 1 and cp2 are inner functions and if cp 1 H2 = cp2 H2, then cp 1/ cp2 is constant. (c) Every closed S-invariant subspace Y of H2, other than {0}, contains an inner function cp such that Y = cpH2• PROOF H2 is a Hilbert space, relative to the norm (2) If cp is an inner function, then I cp I = 1 a.e. The mapping f-+ cpf is there fore an isometry of H2 into H2; being an isometry, its range cpH2 is a closed subspace of H2• [Proof: If cpf n-+ g in H2, then .{cpf n} is a Cauchy sequence, hence so is {In}, hence fn-+ f E H2, so g = cpf e cpH2 .] The S-invariance of cpH2 is also trivial, since z · cpf = cp · z f Hence (a) holds. If cp1H2 = cp2 H2, then cp1 = cp2 f for some f E H2, hence cp1/cp2 E H2• Similarly, cp2/cp1 E H2• Put cp = cp1/cp2 and h = cp + (1/cp). Then h e H2, and since I cp I = 1 a.e. on T, h is real a.e. on T. Since h is the Poisson integral of h, it follows that h is real in U, hence h is constant. Then cp must be con stant, and (b) is proved. The proof of (c) will use a method originated by Helson and Lowdensla ger. Suppose Y is a closed S-invariant subspace of H2 which does not consist of 0 alone. Then there is a smallest integer k such that Y contains a function/ . of the form 00 /(z) = L en zn, (3) n=k Then f fJ z Y, where we write z Y for the set of all g of the form g(z) = z f(z),fe Y. It follows that zY is a proper closed subspace of Y [closed by the argument used in the proof of (a)], so Y contains a nonzero vector which is orthogonal to zY (Theorem 4.1 1). HP-SPACES 349 So there exists a cp E Y such that llcpll 2 = 1 and cp .l zY. Then <p .l z"<p, for n = 1, 2, 3, . . . . By the definition of the inner product in H2 [see 17.20(6)] this means that ( n = 1, 2, 3, . . . ). (4) These equations are preserved if we replace the left sides by their complex conjugates, i.e., if we replace n by -n. Thus all Fourier coefficients of the function I <p 12 E L1(T) are 0, except the one corresponding to n = 0, which is 1. Since L1-functions are determined by their Fourier coefficients (Theorem 5.15), it follows that I <p I = 1 a.e. on T. But cp e H2, so <p is the Poisson integral of cp, and hence I <p I < 1. We conclude that cp is an inner function. Since cp E Y and Y is S-invariant, we have cpzn E Y for all n > 0, hence cpP e Y for every polynomial P. The polynomials are dense in H2 (the partial sums of the power series of any f E H2 converge to f in the H2-norm, by Parseval's theorem), and since Y is closed and I cp I < 1, it follows that cpH2 c Y. We have to prove that this inclusion is not proper. Since cpH2 is closed, it is enough to show that the assumptions h E Y and h .l <pH2 imply h = 0. If h .l cpH2, then h .l cpzn for n = 0, 1, 2, . . . , or ! f1t h(ei6)cp(ei8)e-in8 d(} = 0 2n _1t ( n = 0, 1, 2, . . . ). (5) If h E Y, then znh E z Y if n = 1, 2, 3, . . . , and our choice of <p shows that znh .l <p, or ( n = - 1, - 2, - 3, . . . ). (6) Thus all Fourier coefficients of hcp are 0, hence hcp = 0 a.e. on T; and since I <p I = 1 a.e., we have h = 0 a.e. Therefore h = 0, and the proof is complete. I I I I 17.22 Remark If we combine Theorems 17.15 and 17.21, we see that the S-invariant subs paces of H2 are characterized by the following data: a sequence of complex numbers { C(n} (possibly finite, or even empty) such that I C(n I < 1 and 1:(1 - I C(n I ) < oo, and a positive Borel measure J.l on T, singular with respect to Lebesgue measure (so DJ.l = 0 a.e.). It is easy (we leave this as an exercise) to find conditions, in terms of { C(n} and J.l, which ensure that one S-invariant subspace of H2 contains another. The partially ordered set of all S-invariant subspaces is thus seen to have an extremely complicated struc ture, much more complicated than one might have expected from the simple definition of the shift operator on t2• We conclude the section with an easy consequence of Theorem 17.21 which depends on the factorization described in Theorem 17.17. 350 REAL AND COMPLEX ANALYSIS 17.23 Theorem Suppose M 1 is the inner factor of a function f e H2, and Y is the smallest closed S-invariant subspace of H2 which contains f. Then Y = M1 H2• In particular, Y = H2 if and only iff is an outer function. (1) PROOF Let f = M 1 Q1 be the factorizࡄtion of f into its inner and outer factors. It is clear thatf e M1 H2 ; and since M1 H2 is closed and S-invariant, we have Y c M1 H2• On the other hand, Theorem 17.21 shows that there is an inner function qJ such that Y = qJH2• Since/ e Y, there exists an h = Mh Qh e H2 such that (2) Since inner functions have absolute value 1 a.e. on T, (2) implies that Q1 = Qh , hence M 1 = qJM h e Y, and therefore Y must contain the smallest S in variant closed subspace which contains M 1 . Thus M 1 H2 c Y, and the proof is complete. I I I I It may be of interest to summarize these results in terms of two questions to which they furnish answers. Iff E H2, which functions g e H2 can be approximated in the H2-norm by functions of the form fP, where P runs through the polynomials? Answer: Pre cisely those g for which gl M 1 E H2• For which/ e H2 is it true that the set {JP} is dense in H2 ? Answer: Precisely for those/for which Conjugate Functions 1 f1t log I /(0) I = 2n }og I f(eiŭ I dt. 17.24 Formulation of the Problem Every real harmonic function u in the unit disc U is the real part of one and only one f e H( U) such that f (0) = u(O). Iff= u + iv, the last requirement can also be stated in the form v(O) = 0. The function v is called the harmonic conjugate of u, or the conjugate function of u. Suppose now that u satisfies (1) for some p. Does it follow that (1) holds then with v in place of u? Equivalently, does it follow that/ e HP? The answer (given by M. Riesz) is affirmative if 1 < p < oo. (For p = 1 and p = oo it is negative; see Exercise 24.) The precise statement is given by Theorem 17.26. HP-SPACES 351 Let us recall that every harmonic u that satisfies (1) is the Poisson integral of a function u e I!(T) (Theorem 1 1.30) if 1 < p < oo. Theorem 11.11 suggests therefore another restatement of the problem: If 1 < p < oo, and if we associate to each h e I!(T) the holomorphic function 1 I1t eit + z . (t/Jh)(z) = -2 it h(elļ dt (z e U), n 1t e - z do all ofthesefunctions t/lh lie in HP? Exercise 25 deals with some other aspects of this problem. (2) 17.25 Lemma If 1 < p < 2, ç = n/(1 + p), fX = (cos ç)- 1, and p = fXP(1 + fl), then 1 < p(cos ep)P - fX cos pep (-1t<rn<1t) 2 - 'f' - 2 " PROOF If ç < I ep I < n/2, then the right side of ( 1) is not less than - (X cos pep > -(X cos pç = (X cos ç = 1, and it exceeds p(cos ç)P - fX = 1 if I ep I < ç-(1) /Ill 17.26 Theorem If 1 < p < oo, then there is a constant AP < oo such that the inequality llt/lhiiP < APIIhiiP (1) holds for every h e I!(T). More explicitly, the conclusion is that t/Jh (defined in Sec. 17.24) is in HP, and that (2) where du = dfJj2n is the normalized Lebesgue measure on T. Note that h is not required to be a real function in this theorem, which asserts that t/1 : I!----+ HP is a bounded linear operator. PROOF Assume first that 1 < p < 2, that h E I!(T), h > 0, h =/= 0, and let u be the real part off= t/Jh. Formula 11.5(2) shows that u = P[h], hence u > 0 in U. Since U is simply connected and f has no zero in U, there is a g e H(U) such that g = fP, g(O) > 0. Also, u = I f I cos ep, where ep is a real function with domain U that satisfies I ep I < n/2. If fX = fXP and p = PP are chosen as in Lemma 17.25, it follows that for 0 < r < 1. II fr IP da < P I (u,)P da - IX i1 f..IP cos (pqJ,) da (3) 352 REAL AND COMPLEX ANALYSIS Note that I f IP cos p 0. Hence (0 < r < 1) 11 fr iP da < P 1hP da because u = P[h] implies lluriiP < llhllp · Thus llt/Jhllp < P11pllhllp if h E JJ'(T), h > 0. (4) (5) If h is an arbitrary (complex) function in I.!(T), the preceding result applies to the positive and negative parts of the real and imaginary parts of h. This proves (2), for 1 < p < 2, with A P = 4p11P. To complete the proof, consider the case 2 < p < oo. Let w E IJ(T), where q is the exponent conjugate to p. Put w(ei9) = w(e-iő. A simple compu tation, using Fubini's theorem, shows for any h E I.!(T) that f/1/!h), W da = 1 (1/!w)) i da (0 < r < 1). (6) Since q < 2, (2) holds with w and q in place of h and p, so that (6) leads to 1 (I/! h), W da < Aq II wllq [I hiiP. (7) Now let w range over the unit ball of IJ(T) and take the supremum on the left side of (7). The result is (0 < r < 1). (8) Hence (2) holds again, with AP < Aq . Ill/ (If we take the smallest admissible values for AP and Aq , the last calculation can be reversed, and shows that AP = Aq .) Exercises 1 Prove Theorems 17.4 and 17.5 for upper semicontinuous subharmonic functions. 2 Assume! E H(O.) and prove that log (1 + If I) is subharmonic in n. 3 Suppose 0 < p < oo and f E H(U). Prove that f E HP if and only if there is a harmonic function u in U such that I f(z) IP < u(z) for all z E U. Prove that if there is one such harmonic majorant u of I f IP, then there is at least one, say u 1. (Explicitly, I f IP < u 1 and u 1 is harmonic; and if If IP < u and u is harmonic, then u1 < u.) Prove that II fliP = u,.(0)11P. Hint: Consider the harmonic functions in D(O; R), R < 1, with boundary values I f IP, and let R ---+ 1. 4 Prove likewise thatf E N if and only if log+ I f I has a harmonic majorant in U. HP-SPACES 353 S Suppose f E HP, cp E H(U), and cp(U) c: U. Does it follow that f o cp e HP? Answer the same ques tion with N in place of HP. 6 If 0 < r < s < oo, show that Hs is a proper subclass of Hr. 7 Show that H00 is a proper subclass of the intersection of all HP with p < oo. 8 Iff E H1 andf E I.!'(T), prove thatf e HP. 9 Supposefe H(U) and f(U) is not dense in the plane. Prove thatf has finite radial limits at almost all points of T. 10 Fix IX e U. Prove that the mapping !---+ f(a) is a bounded linear functional on H2• Since H2 is a Hilbert space, this functional can be represented as an inner product with some g e H2• Find this g. 1 1 Fix a E U. How large can l f'(a) l be if 11/112 < 1 ? Find the extremal functions. Do the same for J(n)(IX). 12 Suppose p > 1,/ E HP, andf is real a.e. on T. Prove thatfis then constant. Show that this result is false for every p < 1. 13 Suppose f E H(U), and suppose there exists an M < oo such that f maps every circle of radius r < 1 and center 0 onto a curve Yr whose length is at most M. Prove thatfhas a continuous extension to 0 and that the restriction offto T is absolutely continuous. 14 Suppose J1. is a complex Borel measure on T such that (n = 1, 2, 3, . . . ). Prove that then either J1. = 0 or the support of J1. is all of T. IS Suppose K is a proper compact subset of the unit circle T. Prove that every continuous function on K can be uniformly approximated on K by polynomials. Hint: Use Exercise 14. 16 Complete the proof of Theorem 17.17 for the case 0 < p < 1. 17 Let cp be a nonconstant inner function with no zero in U. (a) Prove that 1/cp ¢ HP if p > 0. (b) Prove that there is at least one ei6 E T such that limr .... 1 cp(rei8) = 0. Hint: log I cp I is a negative harmonic function. 18 Suppose cp is a nonconstant inner function, I a I < 1, and IX ¢ cp(U). Prove that limr .... 1 cp(rei8) = a for at least one ei8 e T. 19 Suppose! E H1 and 1/f E H1• Prove thatfis then an outer function. 20 Suppose! E H1 and Re [f(z)] > 0 for all z e U. Prove thatfis an outer function. 21 Prove thatf E N if and only iff= g/h, where g and h e H00 and h has no zero in U. 22 Prove the following converse of Theorem 15.24: Iff E H(U) and if ÎÏ f .
l log I f(re'8) I I d(J = 0, thenfis a Blaschke product. Hint: () implies lim fn log+ I f(rei6) I d(J = 0. r-+1 -n () Since log+ If I > 0, it follows from Theorems 17.3 and 17.5 that log+ I f I = 0, so If I < 1. Now f = Bg, g has no zeros, I g I < 1, and () holds with 1/g in place off By the first argument, 1 1/g I < 1. Hence I g I = 1. 23 Find the conditions mentioned in Sec. 17.22. 354 REAL AND COMPLEX ANALYSIS 24 The conformal mapping of U onto a vertical strip shows that M. Riesz's theorem on conjugate functions cannot be extended to p = oo . Deduce that it cannot be extended to p = 1 either. 25 Suppose 1 < p < oo, and associate with eachf E IJ'(T) its Fourier coefficients (n) = -f(ear)e-mr dt l 1 f n . . 2n - n (n = 0, + 1, + 2, . . . ). Deduce the following statements from Theorem 17.26: (a) To each f E IJ'(T) there corresponds a function g E IJ'(T) such that g(n) = ](n) for n > 0 but g(n) = 0 for all n < 0. In fact, there is a constant C, depending only on p, such that The mapping/-+ g is thus a bounded linear projection of IJ'(T) into IJ'(T). The Fourier series of g is obtained from that offby deleting the terms with n < 0. (b) Show that the same is true if we delete the terms with n < k, where k is any given integer. (c) Deduce from (b) that the partial sums s, of the Fourier series of any f e IJ'(T) form a bounded sequence in IJ'(T). Conclude further that we actually have , .... 00 (d) Iff E IJ'(T) and if 00 F(z) = L z", n = O then F e HP, and every F e HP is so obtained. Thus the projection mentioned in (a) may be regarded as a mapping of IJ'(T) onto HP. 26 Show that there is a much simpler proof of Theorem 17.26 if p = 2, and find the best value of A2 • 27 Suppose f(z) = Lj a, z" in U and L I a, I < oo . Prove that for all 8. 28 Prove that the following statements are correct if { n"} is a sequenȦ of positive integers which tends to oo sufficiently rapidly. If 00 z"" f(z) = L -k = 1 k then I f'(z) I > nJ(lOk) for all z such that 1 1 1 - - < l z l < 1 - -. n" 2n" Hence f I f'(re1'} I dr = oo for every 8, although lim [R f'(rei8) dr R .... l Jo fiP-SPACES 355 exists (and is finite) for almost all 8. Interpret this geometrically, in terms of the lengths of the images under f of the radii in U. 29 Use Theorem 17.1 1 to obtain the following characterization of the boundary values of HP functions, for 1 < p < oo : A function g E I!'(T) isf (a.e.) for somef E HP if and only if for all negative integers n. -g(e'')e-•nt dt = 0 1 In . . 2n -n CHAPTER EIGHTEEN ELEMENTARY THEORY OF BANACH ALGEBRAS Introduction 356 18.1 Definitions A complex algebra is a vector space A over the complex field in which an associative and distributive multiplication is defined, i.e., x(yz) = (xy)z, (x + y)z = xz + yz, x(y + z) = xy + xz (1) for x, y, and z e A, and which is related to scalar multiplication so that (xy) = x(y) = (x)y (2) for x and y e A, a scalar. If there is a norm defined in A which makes A into a normed linear space and which satisfies the multiplicative inequality llxyll < llxii iiYII (x and y e A), (3) then A is a normed complex algebra. If, in addition, A is a complete metric space relative to this norm, i.e., if A is a Banach space, then we call A a Banach algebra. The inequality (3) makes multiplication a continuous operation. This means that if xnË x and Yn¢ y, then Xn Yn¢ xy, which follows from (3) and the identity Xn Yn - xy = (xn - x)yn + x(yn - y). (4) Note that we have not required that A be commutative, i.e., that xy = yx for all x and y e A, and we shall not do so except when explicitly stated. ELEMENTARY THEORY OF BANACH ALGEBRAS 357 However, we shall assume that A has a unit. This is an element e such that xe =ex= x (x E A). (5) It is easily seen that there is at most one such e (e' = e'e = e) and that II ell > 1, by (3). We shall make the additional assumption that llell = 1. (6) An element x e A will be called invertible if x has an inverse in A, i.e., if there exists an element x- 1 e A such that (7) Again, it is easily seen that no x e A has more than one inverse. If x and y are invertible in A, so are x- 1 and xy, since (xy)- 1 = y - 1 x- 1 • The invertible elements therefore form a group with respect to multiplication. The spectrum of an element x e A is the set of all complex numbers A. such that x - A.e is not invertible. We shall denote the spectrum of x by a(x). 18.2 The theory of Banach algebras contains a great deal of interplay between algebraic properties on the one hand and topological ones on the other. We already saw an example of this in Theorem 9.21, and shall see others. There are also close relations between Banach algebras and holomorphic functions : The easiest proof of the fundamental fact that a(x) is never empty depends on Liouville's theorem concerning entire functions, and the spectral radius formula follows naturally from theorems about power series. This is one reason for re stricting our attention to complex Banach algebras. The theory of reࡅl Banach algebras (we omit the definition, which should be obvious) is not so satisfactory. The Invertible Elements In this section, A will be a complex Banach algebra with unit e, and G will be the set of all invertible elements of A. 18.3 Theorem I fx e A and llxll < 1, then e + x. e G, 00 (e + x)- 1 = L (- 1)nxn, n=O and - 1 < llxll 2 ll(e + x) -e + xll _ 1 _ llxll " PROOF The multiplicative inequality 18.1(3) shows that llxnll < llxlln. If SN פe-X+ X2- • • • + (-1)NXN, (1) (2) (3) 358 REAL AND COMPLEX ANALYSIS it follows that {sN} is a Cauchy sequence in A, hence the series in (1) con verges (with respect to the norm of A) to an element y e A. Since multiplica tion is continuous and (e + x)sN = e + (- 1)NxN+ 1 = sĽe + x), (4) we see that (e + x)y = e = y(e + x). This gives (1), and (2) follows from (5) /Ill 18.4 Theorem Suppose x e G, II x- 1 11 = 1/rx, h e A, and II h II = P < rx. Then x + h e G, and II ( h)- 1 - 1 - 1 h - 1 11 p2 x + - x + x x < rx2(rx - p) . (1) PROOF llx- 1hll < {3/rx < 1, hence e + x- 1h e G, by Theorem 18.3; and since x + h = x(e + x- 1h), we have x + h e G and Thus (2) (x + h)- 1 - x- 1 + x- 1hx- 1 = [(e + x- 1h)- 1 - e + x- 1h]x- 1, (3) and the inequality (1) follows from Theorem 18.3, with x- 1h in place of x. /Ill Corollary 1 G is an open set, and the mapping x----+ x- 1 is a homeomorphism of G onto G. For if x e G and llhll ----+0, (1) implies that ll(x + h)- 1 - x- 1 11 ----+0. Thus x----+ x- 1 is continuous; it clearly maps G onto G, and since it is its own inverse, it is a homeomorphism. Corollary 2 The mapping x----+ x- 1 is diff erentiable. Its diff erential at any x e G is the linear operator which takes h e A to -x- 1 hx- 1• This can also be read off from (1). Note that the notion of the differential of a transformation makes sense in any normed linear space, not just in Rk, as in Definition 7.22. If A is commutative, the above differential takes h to -x-2h, which agrees with the fact that the derivative of the holomorphic function z - 1 is -2 -z . Corollary 3 For every x e A, u(x) is compact, and I A. I < llxll if A. e a(x). ELEMENTARY THEORY OF BANACH ALGEBRAS 359 For if I A I > llxll, then e - A -1x e G, by Theorem 18.3, and the same is true of x - Ae = -A.(e - A - 1x); hence A. ¢ a(x). To prove that a(x) is closed, observe (a) A e a(x) if and only if x - Ae ¢ G ; (b) the complement of G is a closed subset of A, by Corollary 1 ; and (c) the mapping A---+ x - Ae is a continuous mapping of the complex plane into A. 18.5 Theorem Let be a bounded linear functional on A,fix x e A, and define (A. ¢ a(x)). (1) Thenf is holomorphic in the complement o f a(x), andf(A.)---+ 0 as A.--+ oo. PROOF Fix A ¢ a(x) and apply Theorem 18.4 with x - A.e in place of x and with (A - Jl)e in place of h. We see that there is a constant C, depending on x and A, such that ll(x - Jl.e)- 1 - (x - A.e)- 1 + (A. - Jl}(x - Ae)-2 11 < C I J1 - A l2 (2) for all J1 which are close enough to A. Thus ( x - Jle)- 1 - ( x - Ae)- 1 ( ǀ ) _ 2 _; ;, ; _ ; ;.... ....;.. _ ---+ X - Ae Jl - A (3) as J1---+ A, and if we apply to both sides of (3), the continuity and linearity of show that f(Jl) -f(A)---+ q,[(x - A.e)-2]. Jl - A (4) So f is differentiable and hence holomorphic outside a(x). Finally, as A---+ oo we have Aj(A) = [A(x - Ae) - 1] = q,[ C- e) - 1 ]---+ľ-e), (5) by the continuity of the inversion mapping in G. /Ill 18.6 Theorem For every x e A, a(x) is compact and not empty. PROOF We already know that a(x) is compact. Fix x e A, and fix A.0 ¢ a(x). Then (x - Ao e) - 1 #= 0, and the Hahn-Banach theorem implies the existence of a bounded linear functional on A such that f(A0) =F 0, where f is defined as in Theorem 18.5. If a(x) were empty, Theorem 18.5 would imply ࡆhat f is an entire function which tends to 0 at oo, hence f(A.) = 0 for every A., by Liouville's theorem, and this contradictsf(A.0) "# 0. So a(x) is not empty. /Ill 18.7 Theorem (Gelfand-Mazur) I f A is a complex Banach algebra with unit in which each nonzero element is invertible, then A is (isometrically isomorphic to) the complex field. 360 REAL AND COMPLEX ANALYSIS An algebra in which each nonzero element is invertible is called a division algebra. Note that the commutativity of A is not part of the hypothesis; it is part of the conclusion. PROOF If x e A and .A1 =1= .A.2 , at least one of the elements x - .A1 e and x - .A2 e must be invertible, since they cannot both be 0. It now follows from Theorem 18.6 that u(x) consists of exactly one point, say .A(x), for each x e A. Since x - ) .. (x)e is not invertible, it must be 0, hence x = .A(x)e. The mapping xH .A(x) is therefore an isomorphism of A onto the complex field, which is also an isometry, since I .A(x) I = II.A(x)ell = llxll for all x e A. /Ill 18.8 Definition For any x e A, the spectral radius p(x) of x is the radius of the smallest closed disc with center at the origin which contains u(x) (sometimes this is also called the spectral norm of x; see Exercise 14): p( x) = sup { I .A I : .A e a( x)}. 18.9 Theorem (Spectral Radius Formula) For every x e A, lim II x" 11 11" = p(x). (1) (This existence of the limit is part of the conclusion.) PROOF Fix x e A, let n be a positive integer, .A a complex number, and assume .A" ¢ u(x"). We have (x" - .A"e) = (x - .A.e)(x"- 1 + .Ax"-2 + · · · + .A n- 1e). (2) Multiply both sides of (2) by (x" - .A"e)- 1• This shows that x - .A.e is invert ible, hence .A ¢ u(x). So if .A e u(x), then .A" e u(x") for n = 1, 2, 3, . . . . Corollary 3 to Theorem 18.4 shows that I .A" I < llx"ll, and therefore I .A I < llx"ll 11". This gives p(x) < lim inf llx"ll 1'"· Now if I .A I > llxll, it is easy to verify that 00 (.A.e - x) L .A.-"- 1x" = e. n=O (3) (4) The above series is therefore -(x - .Ae)- 1. Let be a bounded linear func tional on A and define/as in Theorem 18.5. By (4), the expansion 00 /(.A) = - L (x").A -n- 1 n=O (5) ELEMENTARY THEORY OF BANACH ALGEBRAS 361 is valid for all A such that I A I > llxll. By Theorem 18.5, f is holomorphic outside a(x), hence in the set {A: I A. I > p(x)}. It follows that the power series (5) converges if I A I > p(x). In particular, ( I A I > p( X)) (6) n for every bounded linear functional on A. It is a consequence of the Hahn-Banach theorem (Sec. 5.21) that the norm of any element of A is the same as its norm as a linear functional on the dual space of A. Since (6) holds for every <1>, we can now apply the Banach-Steinhaus theorem and conclude that to each A with I A. I > p(x) there corresponds a real number C(A) such that (n = 1, 2, 3, . . . ). Multiply (7) by I A In and take nth roots. This gives II Xn ll 11n < I A I [ C(A)] 11n (n = 1, 2, 3, . . . ) if I A I > p(x), and hence lim sup II xn ll 11n < p(x). The theorem follows from (3) and (9). 18.10 Remarks (7) (8) (9) Ill/ (a) Whether an element of A is or is not invertible in A is a purely algebraic property. Thus the spectrum of x, and likewise the spectral radius p(x), are defined in terms of the algebraic structure of A, regardless of any metric (or topological) considerations. The limit in the statement of Theorem 18.9, on the other hand, depends on metric properties of A. This is one of the remarkable features of the theorem: It asserts the equality of two quantities which arise in entirely different ways. (b) Our algebra may be a subalgebra of a larger Banach algebra B (an example follows), and then it may very well happeࡇ that some x e A is not invertible in A but is invertible in B. The spectrum of x therefore depends on the algebra; using the obvious notation, we have a A(x) => a8(x), and the inclusion may be proper. The spectral radius of x, however, is unaffected by this, since Theorem 18.9 shows that it can be expressed in terms of metric properties of powers of x, and these are independent of anything that happens outside A. 18.11 Example Let C(T) be the algebra of all continuous complex functions on the unit circle T (with pointwise addition and multiplication and the supremum norm), and let A be the set of all f e C(T) which can be extended to a continuous function F on the closure of the unit disc U, such that F is holomorphic in U. It is easily seen that A is a subalgebra of C(T). lffn e A 362 REAL AND COMPLEX ANALYSIS and {fn} converges uniformly on T, the maximum modulus theorem forces the associated sequence {Fn} to converge uniformly on the closure of U. This shows that A is a closed subalgebra of C(T), and so A is itself a Banach algebra. Define the functionf0 by f0(ei8) = ei8• Then F0(z) = z. The spectrum off0 as an element of A consists of the closed unit disc; with respect to C(T), the spectrum of fo consists only of the unit circle. In accordance with Theorem 18.9, the two spectral radii coincide. Ideals and Homomorphisms From now on we shall deal only with commutative algebras. 18.12 Definition A subset I of a commutative complex algebra A is said to be an ideal if (a) I is a subspace of A (in the vector space sense) and (b) xy e I whenever x e A and y e I. If I =1= A, I is a proper ideal. Maximal ideals are proper ideals which are not contained in any larger proper ideals. Note that no proper ideal contains an invertible element. If B is another complex algebra, a mapping qJ of A into B is called a homomorphism if qJ is a linear mapping which also preserves multiplication : qJ(x)qJ(y) = qJ(xy) for all x and y e A. The kernel (or null space) of qJ is the set of all x e A such that qJ(x) = 0. It is trivial to verify that the kernel of a homomorphism is an ideal. For the converse, see Sec. 18.14. 18.13 Theorem If A is a commutative complex algebra with unit, every proper ideal of A is contained in a maximal ideal. I f, in addition, A is a Banach algebra, every maximal ideal o f A is closed. PROOF The first part is an almost immediate consequence of the Hausdorff maximality principle (and holds in any commutative ring with unit). Let I be a proper ideal of A. Partially order the collection &J of all proper ideals of A which contain I (by set inclusion), and let M be the union of the ideals in some maximal linearly ordered subcollection f2 of &J. Then M is an ideal (being the union of a linearly ordered collection of ideals), I c M, and M =1= A, since no member of &J contains the unit of A. The maximality of f2 implies that M is a maximal ideal of A. If A is a Banach algebra, the closure M of M is also an ideal (we leave the details of the proof. of this statement to the reader). Since M contains no invertible element of A and since the set of all invertible elements is open, we --have M =I= A, and the maximality of M therefore shows that M = M. /Ill 18.14 Quotient Spaces and Quotient Algebras Suppose J is a subspace of a vector space A, and associate with each x e A the coset ({J(X) = X + J = {x + y: y E J}. (1) ELEMENTARY THEORY OF BANACH ALGEBRAS 363 If x1 - x2 e J, then <p(x1) = qJ(x2). If x1 - x2 ¢ J, qJ(x1) r\ qJ(x2) = 0. The set of all cosets of J is denoted by A/ J; it is a vector space if we define <p(x) + qJ(y) = <p(x + y), A<p( X) = (/)(AX) (2) for x and y e A ࡈand scalars A.. Since J is a vector space, the operations (2) are well defined ; this means that if <p( x) = qJ( x') and qJ(y) = qJ(y'), then <p( x) + qJ(y) = qJ( x') + qJ(y'), (3) Also, <p is clearly a linear mapping of A onto A/ J; the zero element of A/ J is <p(O) = J. Suppose next that A is not merely a vector space but a commutative algebra and that J is a proper ideal of A. If x' - x e J and y' - y e J, the identity x'y' - xy = (x' - x)y' + x(y' - y) (4) shows that x'y' - xy e J. Therefore multiplication can be defined in A/J in a consistent manner: qJ(x)<p(y) = 0. It is clear that IIA.qJ(x)ll = I A. l llqJ(x)ll . If x1 and x2 E A and E > 0, there exist y1 and y2 E J so that (i = 1, 2). (2) Hence ll<p(xl + x2)ll < llx1 + X2 + Y1 + Y2ll < llqJ(xl)ll + ll qJ(x2)11 + 2e, (3) which gives the triangle inequality and proves (a). Suppose A is complete and {<p(xn)} is a Cauchy sequence in A/J. There is a subsequence for which ( i = 1 ' 2, 3' . . . ), (4) and there exist elements zi so that zi - xn. E J and llzi - zi+ 1 11 < 2 -i. Thus { zi} is ' . a Cauchy sequence in A; and since A is complete, there exists z E A such that llzi - zll / 0. It follows that qJ(xn) converges to qJ(z) in A/J. But if a Cauchy sequence has a convergent subsequence, then the full sequence converges. Thus A/J is complete, and we have proved (b). To prove (c), choose x1 and x2 E A and E > 0, and choose y1 and y2 E J so that (2) holds. Note that (x1 + y1)(x2 + y2) E x1x2 + J, so that ll <p(xlx2)11 < ll(xl + Y1)(x2 + Y2)ll < llx1 + Y1 ll llx2 + Y2ll· (5) Now (2) implies (6) Finally, if e is the unit element of A, take x1 ¢ J and x2 = e in (6); this gives II qJ(e)ll > 1. But e E <p(e), and the definition of the quotient norm shows that ll<p(e)ll < llell = 1. So ll<p(e)ll = 1, and the proof is complete. 18.16 Having dealt with these preliminaries, we are now in a position to derive some of the key facts concerning commutative Banach algebras. Suppose, as before, that A is a commutative complex Banach algebra with unit element e. We associate with A the set A of all complex homomorphisms of A; these are the homomorphisms of A onto the complex field, or, in different terminology, the multiplicative linear functionals on A which are not identically 0. As before, u(x) denotes the spectrum of the element x E A, and p(x) is the spectral radius of x. Then the following relations hold: 18.17 Theorem (a) Every maximal ideal M o f A is the kernel o f some h E A. (b) A. E u(x) if and only if h(x) = A. for some h E A. (c) x is invertible in A if and only if h(x) "# Ofor every h e A. (d) h(x) e u(x)for every x e A and h e A. (e) I h(x) I < p(x) < llxll for every x e A and h E A. ELEMENTARY THEORY OF BANACH ALGEBRAS 365 PROOF If M is a maximal ideal of A, then A/ M is a field; and since M is closed (Theorem 18.13), A/M is a Banach algebra. By Theorem 18.7 there is an isomorphism j of A/ M onto the complex field. If h = j o cp, where cp is the homomorphism of A onto A/ M whose kernel is M, then h e A and the kernel of h is M. This proves (a). If A. e a(x), then x - A.e is not invertible; hence the set of all elements (x - A.e)y, where y e A, is a proper ideal of A, which lies in a maximal ideal (by Theorem 18.13), and (a) shows that there exists an h e A such that h(x - A.e) = 0. Since h(e) = 1, this gives h(x) = A.. On the other hand, if A. ¢ u(x), then (x - A.e)y = e for some y e A. It follows that h(x - A.e)h(y) = 1 for every h e A, so that h(x - A.e) =1= 0, or h(x) =1= A.. This proves (b). Since x is invertible if and only if 0 ¢ a(x), (c) follows from (b). Finally, (d) and (e) are immediate consequences of (b). /Ill Note that (e) implies that the norm of h, as a linear functional, is at most 1. In particular, each h e A is continuous. This was already proved earlier (Theorem 9.21). Applications We now give some examples of theorems whose statements involve no algebraic concepts but which can be proved by Banach algebra techniques. 18.18 Theorem Let A(U) be the set of all continuous functions on the closure 0 of the open unit disc U whose restrictions to U are holomorphic. Suppose fh . . . , fn are members o f A(U), such that I ft(z) I + · · · + I fn(z) I > o (1) f or every z e 0. Then there exist gb . . . , gn e A(U) such that n L h(z)gi(z) = 1 (z e 0). (2) i= 1 PROOF Since sums, products, and uniform limits of holomorphic functions are holomorphic, A(U) is a Banach algebra, with the supremum norm. The set J of all functions "£_h gi , where the gi are arbitrary members of A(U), is an ideal of A(U). We have to prove that J contains the unit element 1 of A(U). By Theorem 18.13 this happens if and only if J lies in no maximal ideal of A( U). By Theorem 18.17( a) it is therefore enough to prove that there is no homomorphism h of A(U) onto the complex field such that h(_h) = 0 for every i (1 < i < n). 366 REAL AND COMPLEX ANALYSIS Before we determine these homomorphisms, let us note that the poly nomials form a dense subset of A(U). To see this, suppoࡉe f e A(U) and E > 0; since f is uniformly continuous on 0, there exists an r < 1 such that I f(z) -f(rz) I < E for all z e 0; the expansion of f(rz) in powers of z con verges if I rz I < 1, hence converges to f(rz) uniformly for z e 0, and this gives the desired approximation. Now let h be a complex homomorphism of A(U). Put f0(z) = z. Then /0 e A(U). It is obvious that a(/0) = 0. By Theorem 18.17(d) there exists an ex e 0 such that h(/0) = ex. Hence h(fÌ) = exn =/Ì(ex), for n = 1, 2, 3, . . . , so h(P) = P(ex) for every polynomial P. Since h is continuous and since the poly nomials are dense in A(U), it follows that h(f) = f(ex) for every f e A(U). Our hypothesis (1) implies that I .fr{ex) I > 0 for at least one index i, 1 < i < n. Thus h(/i) =1= 0. We have proved that to each h e L\ there corresponds at least one of the given functions h such that h(i'i) =I= 0, and this, as we noted above, is enough to prove the theorem. I I I I Note: We have also determined all maximal ideals of A(U), in the course of the preceding proof, since each is the kernel of some h e L\ : If ex e U and if M ex is the set o f all f e A(U) such that f(ex) = 0, then Mcx is a maximal ideal of A(U), and all maximal ideals o f A(U) are obtained in this way. A( U) is often called the disc algebra. 18.19 The restrictions of the members of A(U) to the unit circle T form a closed subalgebra of C(T). This is the algebra A discussed in Example 18.1 1. In fact, A is a maximal subalgebra of C(T). More explicitly, if A c B c C(T) and B is a closed (relative to the supremum norm) subalgebra of C(T), then either B = A or B = C(T). It is easy to see (compare with Exercise 29, Chap. 17) that A consists pre cisely of those f e C(T) for which A 1 J 1t •9 • (J f(n) = -f(el )e -ln d(J = 0 2n _ 1t (n = - 1, - 2 , - 3, . . . ). (1) Hence the above-mentioned maximality theorem can be stated as an approx imation theorem: 18.20 Theorem Suppose g e C(T) and g(n) =1= 0 for some n < 0. Then to every f e C(T) and to every E > 0 there correspond polynomials such that m(n) p (ei8\ = ї a eik8 n J ơ n, k k=O N (n = 0, . . . , N) I f(eiÔ - L Pn(ei8)gn(eiÔ I < E (ei8 e T). n=O (1) (2) ELEMENTARY THEORY OF BANACH ALGEBRAS 367 PROOF Let B be the closure in C = C(T) of the set of all functions of the form (3) The theorem asserts that B = C. Let us assume B #= C. The set of all functions (3) (note that N is not fixed) is a complex algebra. Its closure B is a Banach algebra which contains the function f o , where f 0(ei6) = ei8• Our assumption that B :I= C implies that 1/!0 ¢ B, for otherwise B would contain f"o for all integers n, hence all trigonometric polynomials would be in B; and since the trigonometric polynomials are dense in C (Theorem 4.25) we should have B = C. So fo is not invertible in B. By Theorem 18. 17 there is a complex homo morphism h of B such that h(/ 0) = 0. Every homomorphism onto the complex field satisfies h(1) = 1 ; and since h(/0) = 0, we also have h(f"o) = [h(f o)Jn = 0 ( n = 1 , 2, 3, . . . ) . (4) We know that h is a linear functional on B, of norm at most 1. The Hahn-Banach theorem extends h to a linear functional on C (still denoted by h) of the same norm. Since h(l) = 1 and ll hll < 1, the argument used in Sec. 5.22 shows that h is a positive linear functional on C. In particular, h(f) is real for real j; hence h(j) = h(f). Since f 0 n is the complex conjugate of f"o , it follows that (4) also holds for n = - 1, -2, - 3, . . .
. Thus h(f"o) = { 1 Íf n = 0, (S) 0 tf n =/= 0. Since the trigonometric polynomials are dense in C, there is only one bounded linear functional on C which satisfies (5). Hence h is given by the formula (f E C). (6) Now if n is a positive integer, gf0 e B; and since h is multiplicative on B, (6) gives 1 f1t "6 . 6 g(-n) = -g(e' )e'n df) = h(gf"o) = h(g)h(f"o) = 0, 2n _1t by (5). This contradicts the hypothesis of the theorem. We conclude with a theorem due to Wiener. 18.21 Theorem Suppose - ao c ein6 n ' - ao (7) Ill/ (1) 368 REAL AND COMPLEX ANALYSIS and f(ei8) =/= 0 for every real 0. Then 1 v . 8 v f(e;Ô = w y, e'" with w I y, I < oo. (2) PROOF We let A be the space of all complex functions f on the unit circle which satisfy (1 ), with the norm 00 II f II = L I en 1 . - oo (3) It is clear that A is a Banach space. In fact, A is isometrically isomorphic to t1, the space of all complex functions on the integers which are integrable with respect to the counting measure. But A is also a commutative Banach algebra, under pointwise multiplication. For if g e A and g(eiő = Ybn einB, then and hence (4) llfgll = L L Cn-k bk < L l bk l L l cn-k l = 11/11 · llgll. (5) n k k n Also, the function 1 is the unit of A, and 11 1 11 = 1. Putf0(eiő = ei6, as before. Thenf0 e A, 1/!0 e A, and lifo II = 1 for n = 0, + 1, + 2, . . . . If h is any complex homomorphism of A and h(/0) = A., the fact that II h II < 1 implies that I A.n l = l h(fo) l < 11/oll = t (n = 0, + 1, + 2, . . . ). (6) Hence I A. I = 1. In other words, to each h corresponds a point eia e T such that h(/0) = eia, so ( n = 0, + 1, + 2, . . . ). (7) Iffis given by (1), then/= Ŀcnfo · This series converges in A; and since h is a continuous linear functional on A, we conclude from (7) that (f E A). (8) Our hypothesis that f vanishes at no point of T says therefore that f is not in the kernel of any complex homomorphism of A, and now Theorem 18.17 implies that f is invertible in A. But this is precisely what the theorem asserts. I I I I ELEMENTARY THEORY OF BANACH ALGEBRAS 369 Exercises 1 Suppose B(X) is the algebra of all bounded linear operators on the Banach space X, with if A, A1, and A2 e B(X). Prove that B(X) is a Banach algebra. 2 Let n be a positive integer, let X be the space of all complex n-tuples (normed in any way, as long as the axioms for a normed linear space are satisfied), and let B(X) be as in Exercise 1. Prove that the spectrum of each member of B(X) consists of at most n complex numbers. What are they? 3 Take X = 13( - oo, oo ), suppose cp e L00(- oo, oo ), and let M be the multiplication operator which takes f e I3 to cp f Show that M is a bounded linear operator on I3 and that the spectrum of M is equal to the essential range of cp (Chap. 3, Exercise 19). 4 What is the spectrum of the shift operator on t2? (See Sec. 17.20 for the definition.) S Prove that the closure of an ideal in a Banach algebra is an ideal. 6 If X is a compact Hausdorff space, find all maximal ideals in C(X). 7 Suppose A is a commutative Banach algebra with unit, which is generated by a single element x. This means that the polynomials in x are dense in A. Prove that the complement of u(x) is a con nected subset of the plane. Hint: If A. ¢ u(x), there are polynomials P, such that P n<x)---+ (x - A.e)- 1 in A. Prove that Piz)---+ (z - A.)-1 uniformly for z E u(x). 8 Suppose L I c, I < oo, f(z) = L c, z", I f(z) I > 0 for every z e 0, and 1/f(z) = L a, z". Prove that L I a, I < oo. 9 Prove that a closed linear subspace of the Banach algebra L1(R1) (see Sec. 9.19) is translation invariant if and only if it is an ideal. 10 Show that L1(T) is a commutative Banach algebra (without unit) if multiplication is defined by 1 J1t (f g)(t) = -f(t - s)g(s) ds. 2n - n Find all complex homomorphisms of L1(T), as in Theorem 9.23. If E is a set of integers and if IE is the set of all f e IJ(T) such that /(n) = 0 for all n e E, prove that IE is a closed ideal in L1(T), and prove that every closed ideal in IJ(T) is obtained in this manner. 1 1 The resolvent R(A., x) of an element x in a Banach algebra with unit is defined as R(A., x) = (A.e - x) -1 for all complex A. for which this inverse exists. Prove the identity R(A., x) - R(Jl, x) = (Jl - A.)R(A.)R(Jl) and use it to give an alternative proof of Theorem 18.5. 12 Let A be a commutative Banach algebra with unit. The radical of A is defined to be the intersec tion of all maximal ideals of A. Prove that the following three statements about an element x e A are equivalent: (a) x is in the radical of A. (b) lim llx" 1111" = 0. n -+ oo (c) h(x) = 0 for every complex homomorphism of A. 13 Find an element x in a Banach algebra A (for instance, a bounded linear operator on a Hilbert space) such that x" #: 0 for all n > 0, but lim, .... 0 llx"ll 11" = 0. 370 REAL AND COMPLEX ANALYSIS 14 Suppose A is a commutative Banach algebra with unit, and let L\ be the set of all complex homo morphisms of A, as in Sec. 18.16. Associate with each x E A a function x on L\ by the formula x(h) = h(x) (h E A). x is called the Gelfand transform of x. Prove that the mapping x y x is a homomorphism of A onto an algebra A of complex functions on A, with pointwise multiplication. Under what condition oq A is this homomorphism an iso morphism? (See Exercise 12.) Prove that the spectral radius p(x) is equal to llxlloo = sup { l x(h)l: h E A}. Prove that the range of the function x is exactly the spectrum u(x). 15 If A is a commutative Banach algebra without unit, let A 1 be the algebra of all ordered pairs (x, A.), with x E A and A. a complex number; addition and multiplication are defined in the " obvious " way, and ll(x, A.) II = llxll + I A. I . Prove that A1 is a commutative Banach algebra with unit and that the mapping x y (x, 0) is an isometric isomorphism of A onto a maximal ideal of A1. This is a standard embedding of an algebra without unit in one with unit. 16 Show that H00 is a commutative Banach algebra with unit, relative to the supremum norm and pointwise addition and multiplication. The mapping fy f(a) is a complex homomorphism of H00, whenever I a I < 1. Prove that there must be others. 17 Show that the set of all functions (z - 1)2/, where f E H00, is an ideal in H00 which is not closed. Hint: if I z I < 1, € > o. 18 Suppose cp is an inner function. Prove that { cp f: f E H00} is a closed ideal in H00• In other words, prove that if {!,.} is a sequence in H00 such that cp f, y g uniformly in U, then g/cp E H00• CHAPTER NINETEEN HOLOMORPHIC FOURIER TRANSFORMS Introduction 19.1 In Chap. 9 the Fourier transform of a function / on R1 was defined to be a function / on R 1. Frequently / can be extended to a function which is holo morphic in a certain region of the plane. For instance, if/(t) = e- ltl, then/(x) = (1 + x2) - l, a rational function. This should not be too surprising. For each real t, the kernel eitz is an entire function of z, so one should expect that there are " conditions onfunder which/will be holomorphic in certain regions. We shall describe two classes of holomorphic functions which arise in this manner. For the first one, let F be any function in /3(- oo, oo) which vanishes on (- oo, 0) [i.e., take F e /3(0, oo )] and define f(z) = f" F(t)eitz dt (1) where n + is the set of all z = X + iy with y > 0. If z E n+' then I eitz I = e-ty, which shows that the integral in (1) exists as a Lebesgue integral. If Im z > ƒ > 0, Im zn > Ī' and zn----+ z, the dominated convergence theorem shows that lim i ""I exp (itz,J - exp (itz) 12 dt = 0 n-oo 0 because the integrand is bounded by the L1-function 4 exp ( - 2ƒt) and tends to 0 for every t > 0. The Schwarz inequality implies therefore that f is continuous in n+. The theorems of Fubini and Cauchy show that Jr f(z) dz = 0 for every closed path y in ll +. By Morera's theorem,/ e H(ll +). 371 372 REAL AND COMPLEX ANALYSIS Let us rewrite (1) in the form f(x + iy) = ro F(t)e-tyeitx dt, regard y as fixed, and apply Plancherel's theorem. We obtain -l f(x + iy) l2 dx = I F(t) l2e- 2tY dt < I F(t) l2 dt 1 Joo 1oo 1oo 2n _ 00 0 o (2) (3) for every y > 0. [Note that our notation now differs from that in Chap. 9. There the underlying measure was Lebesgue measure divided by fo,. Here we just use Lebesgue measure. This accounts for the factor 1/(2n) in (3).] This shows: (a) Iff is of the form (1), thenfis holomorphic in n+ and its restrictions to horizon tal lines in n + form a bounded set in 13(- 00 ' 00 ). Our second class consists of allf of the form f(z) = L.4 A F(t)eitz dt (4) where 0 < A < oo and F e 13(- A, A). These functions f are entire (the proof is the same as above), and they satisfy a growth condition: I /(z) I < f A A I F(t) I e-ty dt < eAIYI f_ A A I F(t) I dt. (5) If C is this last integral, then C < oo, and (5) implies that I f(z) I < CeAizl. (6) [Entire functions which satisfy (6) are said to be of exponential type.] Thus: (b) Every f of the form (4) is an entire function which satis fies (6) and whose restriction to the real axis lies in 13 (by the Plancherel theorem). It is a remarkable fact that the converses of (a) and (b) are true. This is the content of Theorems 19.2 and 19.3. Two Theorems of Paley and Wiener 19.2 Theorem Supposef e H(ll+) and 1 Joo sup -2 l f(x + iy) l2 dx = C < oo. O<y<oo 7r - oo Then there exists an F e 13(0, oo) such that f(z) = fo F(t)eitz dt (1) (2) HOLOMORPHIC FOURIER TRANSFORMS 373 and (3) Note: The function F we are looking for is to have the property that f(x + iy) is the Fourier transform of F(t)e-yt (we regard y as a positive constant). Let us apply the inversion formula (whether or not this is correct does not matter; we are trying to motivate the proof that follows): The desired F should be of the form 1 Joo . 1 f . F(t) = ety · -f(x + iy)e-,tx dx = -f(z)e-ltz dz. 2n _ oo 2n (4) The last integral is over a horizontal line in n +, and if this argument is correct at all, the integral will not depend on the particular line we happen to choose. This suggests that the Cauchy theorem should be invoked. PROOF Fix y, 0 < y < oo. For each (X > 0 let ra be the rectangular path with vertices at + (X + i and + (X + iy. By Cauchy's theorem r f(z)e-itz dz = 0. Jr« (5) We consider only real values of t. Let <P(/3) be the integral off(z)e-itz over the straight line interval from f3 + i to f3 + iy ([3 real). Put I = [y, 1] if y < 1, I = [ 1, y] if 1 < y. Then I fll (PW = 1f(p + iu)e-it(/l+iu) du 2 < 11 f(P + iuW du 1e2tu du. (6) Put A(p) = 11 f(P + iu) 12 du. Then (1) shows, by Fubini's theorem, that 1 Joo 2n _ oo A(p) dp < Cm(/). Hence there is a sequence {(Xi} such that (Xi-+ oo and A((X) + A( - (Xi)-+ 0 By (6), this implies that U-+ oo). as j-+ oo. (7) (8) (9) (10) 374 REAL AND COMPLEX ANALYSIS Note that this holds for every t and that the sequence {(Xi} does not depend on t. Let us define 1 frJ.j . oiY, t) = ln ,.f(x + iy)e-•tx dx. Then we deduce from (5) and (10) that ( - 00 < t < 00 ). j-+ 00 (1 1) (12) Write /y(x) for f(x + iy). Then f, e 13(- oo, oo ), by hypothesis, and the Plancherel theorem asserts that JUV s:l /,(t) - giy, t) 12 dt = 0, (13) where/y is the Fourier transform off , . A subsequence of {giy, t)} converges therefore pointwise to /,(t), for almost all t (Theorem 3.12). If we define (14) it now follows from (12) that (15) Note that (14) does not involve y and that (15) holds for every y e (0, oo ). Plancherel's theorem can be applied to ( 15): f oo f oo 1 f oo - oo e-Zty i F(tW dt = -ool /,(t) l2 dt = ln _ 00l /y(x) l2 dx < C. If we let yË oo, (16) shows that F(t) = 0 a.e. in (- oo, 0). If we let y ¢ 0, ( 16) shows that 1 oo l F(t) 12 dt < C. (16) (17) It now follows from (15) that/, e L1 if y > 0. Hence Theorem 9.14 gives fy(x) = s_: /,(t)eitx dt (18) or f(z) = 100 F(t)e-yteitx dt = 100 F(t)eitz dt (19) This is (2), and now (3) follows from (17) and formula 19.ŀ {3). Ill/ HOLOMORPHIC FOURIER TRANSFORMS 375 19.3 Theorem Suppose A and C are positive constants and f is an entire func tion such that I f(z) I < CeAizl for all z, and I:l f(xW dx < oo. Then there exists an F e 13(- A, A) such that f(z) = f_A A F(t)e11z dt f or all z. PROOF Put.f:(x) = f(x)e-£1xl, for E > 0 and x real. We shall show that lim f ao f.(x)e-itx dx = 0 £-0 - 00 ( t real, I t I > A). (1) (2) (3) (4) Since 11 .1: -/11 2 q 0 as e Ѡ o, the Plancherel theorem implies that the Fourier transforms off: converge in J3 to the Fourier transform F off(more precisely, of the restriction off to the real axis). Hence (4) will imply that F vanishes outside [-A, A}, and then Theorem 9.14 shows that (3) holds for almost every real z. Since each side of (3) is an entire function, it follows that (3) holds for every complex z. Thus (4) implies the theorem. For each real (X, let r a be the path defined by (0 < s < oo), (5) put (6) and if W E lla , define q,"(w) = r f(z)e-wz dz = e1" reo f(se1") exp (- wseia) ds. (7) Jr« Jo By (1) and (5), the absolute value of the integrand is at most C exp { - [Re (weia) - A]s}, and it follows (as in Sec. 19.1) that a is holomorphic in the half plane lla . However, more is true if C( = 0 and if C( = n : We have q,0(w) = Lao f(x)e-wx dx (Re w > 0), (8) 376 REAL AND COMPLEX ANALYSIS 1t(w) = -s: 00/(x)e-wx dx (Re w < 0). (9) <1>0 and 1t are holomorphic in the indicated half planes because of (2). The significance of the functions a to ( 4) lies in the easily verified rela tion s_: f.(x)e-itx dx = $0( E + it) - 1t(- E + it) (t real). (10) Hence we have to prove that the right side of (10) tends to 0 as E ---+ 0, if t > A and if t < -A. We shall do this by showing that any two of our functions a agree in the intersection of their domains of definition, i.e., that they are analytic con tinuations of each other. Once this is done, we can replace <1>0 and 1t by 1t12 in ( 1 0) if t < -A, and by 1t12 if t > A, and it is then obvious that the difference tends to 0 as E ---+ 0. So suppose 0 < {3 - C( < n. Put If w = I w l e-ir, then C( + /3 }' = 2 ' {3 - C( 11 = cos 2 > 0. (1 1) Re (weia) = 11 1 w I = Re (weiP) (12) so that w E na n llp as soon as I w I > A/1]. Consider the integral Lf(z)e-wz dz (13) over the circular arc r given by r(t) = reit, C( < t < {3. Since Re(-wz)= - l w l r cos (t - y) < - l w l rY/, the absolute value of the integrand in (13) does not exceed C exp {(A - I w l 17)r}. If I w I > A/17 it follows that (13) tends to 0 as r---+ oo. (14) We now apply the Cauchy theorem. The integral of f(z)e-wz over the interval [0, reiP] is equal to the sum of (13) and the integral over [0, reia]. Since (13) tends to 0 as r---+ oo, we conclude that <>a(w) = p(w) if w = I w I e-iy and I w I > A/1], and then Theorem 10.18 shows that <>a and p coincide in the intersection of the half planes in which they were originally defined. This completes the proof. /Ill 19.4 Remarks Each of the two preceding proofs depended on a typical appli cation of Cauchy's theorem. In Theorem 19.2 we replaced integration over one horizontal line by integration over another to show that 19.2(15) was HOLOMORPHIC FOURIER TRANSFORMS 377 independent of y. In Theorem 19.3, replacement of one ray by another was used to construct analytic continuations; the result actually was that the functions a are restrictions of one function which is holomorphic in the complement of the interval [- Ai, Ai]. The class of functions described in Theorem 19.2 is the half plane ana logue of the class H2 discussed in Chap. 17. Theorem 19.3 will be used in the proof of the Denjoy-Carleman theorem (Theorem 19.1 1). Quasi-analytic Classes 19.5 If Q is a region and if z0 e Q, every f e H(Q) is uniquely determined by the numbers f(z0), f'(z0), f"(z0), • • • . On the other hand, there exist infinitely differen tiable functions on R1 which are not identically 0 but which vanish on some interval. Thus we have here a uniqueness property which holomorphic functions possess but which does not hold in coo (the class of all infinitely differentiable complex functions on R1). Iff e H(Q), the growth of the sequence { I J(z0) I } is restricted by Theorem 10.26. It is therefore reasonable to ask whether the above uniqueness property holds in suitable subclasses of coo in which the growth of the derivatives is subject to some restrictions. This motivates the following definitions ; the answer to our question is given by Theorem 19.1 1. 19.6 The Classes C{Mn} If M0 , M1, M2 , • • • are positive numbers, we let C{Mn} be the class of allf e coo which satisfy inequalities of the form ( n = 0, 1, 2, . . . ). (1) Here D0f = f, D"f is the nth derivative off if n > 1, the norm is the supremum norm over R1, and {31 and B1 are positive constants (depending on f, but not on n). Iff satisfies ( 1 ), then . {II D"f II }ttn hm sup oo < B1 . n-oo Mn (2) This shows that B 1 is a more significant quantity than {3 1 . However, if {3 1 were omitted in (1), the case n = 0 would imply II f II oo < M 0 , an undesirable restriction. The inclusion of {31 makes C{Mn} into a vector space. Each C { M n} is invariant under a ffine transformations. More explicitly, suppose f e C{ M n} and g(x) = f(ax + b). Then g satisfies (1), with /3g = {3 1 and Bg = aB 1 . We shall make two standing assumptions on the sequences {Mn} under con sideration : Mo = 1. M; < Mn- tMn+ t ( n = 1, 2, 3, . . . ). Assumption (4) can be expressed in the form: {log Mn} is a convex sequence. (3) (4) 378 REAL AND COMPLEX ANALYSIS These assumptions will simplify some of our work, and they involve no loss of generality. [One can prove, although we shall not do so, that every class --C{Mn} is equal to a class C{Mn}, where {Mn} satisfies (3) and (4).] The following result illustrates the utility of (3) and ( 4): 19.7 Theorem Each C { M n} is an algebra, with respect to pointwise multiplica tion. PROOF Supposefand g E C{Mn}, and p1 , B1, pg , and Bg are the correspond ing constants. The product rule for differentiation shows that (1) Hence I Dn(fg) l < pf pg t (̌)B}B:-jMjMn-j• j=O } (2) The convexity of {log Mn}, combined with M0 = 1, shows that MiMn-i < M n for 0 < j < n. Hence the binomial theorem leads from (2) to IID"(f g)ll oo < pf pg(Bf + Bg)"Mn SO thatfg E C{Mn}. ( n = 0, 1, 2, . . . ), 19.8 Definition A class C { M n} is said to be quasi-analytic if the conditions (D"f)(O) = 0 (for n = 0, 1, 2, . . . ) imply thatf(x) = 0 for all x E R1• (3) Ill/ (1) The content of the definition is of course unchanged if (D"f)(O) is replaced by (D" f)(x0), where x0 is any given point. The quasi-analytic classes are thus the ones which have the uniqueness property we mentioned in Sec. 19.6. One of these classes is very intimately related to holomorphic functions : 19.9 Theorem The class C{ n !} consists of all f to which there corresponds a b > 0 such that f can be extended to a bounded holomorphic function in the strip de fined by I lm(z) I < b. Consequently C{ n !} is a quasi-analytic class. PROOF Suppose f E H(Q) and I f(z) I < P for all z e !l, where Q consists of all z = x + iy with I y I < b. It follows from Theorem 10.26 that (n = 0, 1, 2, . . . ) (1) for all real x. The restriction off to the real axis therefore belongs to C { n!}. HOLOMORPHIC FOURIER TRANSFORMS 379 Conversely, suppose f is defined on the real axis and f E C{n !}. In other words, II D"f II oo < PB"n ! (n = 0, 1, 2, . . . ). We claim that the representation f(x) = R (D"f)(a) (x - a)" n=O n! (2) (3) is valid for all a E R1 if a - B- 1 < x < a + B- 1• This follows from Taylor's formula f(x) = Tf.1 (Dq!(a) (x - a)i + ( S 1)' fx (x - t)"- 1(D".f)(t) dt, (4) ) = 0 } • n . Ja which one obtains by repeated integrations by part. By (2) the last term in (4) (the " remainder ") is dominated by nPB" i x (x - t)"- 1 dt = P I B(x - a) 1". (5) If I B(x - a) I < 1, this tends to 0 as n q oo, and (3) follows. We can now replace x in (3) by any complex number z such that I z - a I < 1/B. This defines a holomorphic function Fa in the disc with center at a and radius 1/B, and Fa(x) = f(x) if x is real and I x - a I < ljB. The various functions Fa are therefore analytic continuations of each other; they form a holomorphic extension F off in the strip I y I < 1/ B. lf O < ç < 1/B and z = a + iy, I Y I < Ī' then I F(z) I = I Fa(z) I = f (D "fQ(a) (iy)" < P f (Bb)" = p . n = o n · n = .o 1 - BĪ This shows that F is bounded in the strip I y I < Ī' and the proof is complete. Ill/ 19.10 Theorem The class C{Mn} is quasi-analytic if and only if C{Mn} con tains no nontrivial function with compact support. PROOF If C{Mn} is quasi-analytic, iff E C{Mn}, and iff has compact support, then evidently f and all its derivatives vanish at some point, hence f(x) = 0 for all x. Suppose C{M,.} is not quasi-analytic. Then there exists an f e C{Mn} such that (D" f)(O) = 0 for n = 0, 1, 2, . . . , butf(x0) #- 0 for some x0 • We may assume x0 > 0. If g(x) = f(x) for x > 0 and g(x) = 0 for x < 0, then g E C{Mn}. Put h(x) = g(x)g(2x0 - x). By Theorem 19.7, h E C{Mn}. Also, h(x) = 0 if x < 0 and if x > 2x0 • But h(x0) = f2(x0) =1= 0. Thus h is a non trivial member of C{Mn} with compact support. //// 380 REAL AND COMPLEX ANALYSIS We are now ready for the fundamental theorem about quasi-analytic clases. The Denjoy-Carleman Theorem 19.11 Theorem Suppose M0 = 1, M; < Mn 1Mn+ 1for n = 1, 2, 3, . . . , and 00 xn Q(x) = L -, n=O Mn xn q(x) = sup - , n;;::O Mn for x > 0. Then each o f the following five conditions implies the other four: (a) (b) (c) (d) (e) C { M n} is not quasi-analytic. 100 dx log Q(x) 1 2 < oo. o + x 100 dx log q(x) 1 2 < oo. o + x 00 ( 1 y'· L -< oo. n= l Mn f M.- t n= l Mn < 00 . Note: If M n Ò oo very rapidly as n ƈ oo, then Q(x) tends to infinity slo\vly as xÒ oo. Thus each of the five conditions says, in its own way, that Mnƈ oo rapidly. Note also that Q(x) > 1 and q(x) > 1. The integrals in (b) and (c) are thus always defined. It may happen that Q(x) = oo for some x < oo. In that case, the integral (b) is + oo, and the theorem asserts that C{Mn} is quasi-analytic. If M n = n !, then M n _ 1/ M n = 1/n, hence (e) is violated, and the theorem asserts that C { n !} is quasi-analytic, in accordance with Theorem 19.9. PROOF THAT (a) IMPLIES (b) Assume that C{Mn} is not quasi-analytic. Then C{Mn} contains a nontrivial function with compact support (Theorem 19.10). An affine change of variable gives a function F e C{Mn}, with support in some interval [0, A], such that (n = 0, 1, 2, . .. ) and such that F is not identically zero. Define and f(z) = 1.4 F(t)eitz dt c - iw) g(w) = f 1 + w . (1) (2) (3) HOLOMORPHIC FOURIER TRANSFORMS 381 Then f is entire. If Im z > 0, the absolute value of the integrand in (2) is at most I F(t) 1 . Hence f is bounded in the upper half plane; therefore g is bounded in U. Also, g is continuous on U, except at the point w = -1. Since f is not identically 0 (by the uniqueness theorem for Fourier transforms) the same is true of g, and now Theorem 15.19 shows that 1 J1t -2 log I g(ei8) I dfJ > - oo. n - 1t (4) If x = i(1 -ei8)/(1 + ei8) = tan (fJ/2), then dfJ = 2(1 + x2)-1 dx, so (4) is the same as 1 Joo dx -log l f(x) l 1 2 > -oo . n _ 00 + x On the other hand, partial integration of (2) gives f(z) = (iz)-n LA(DnF)(t)eitz dt (z =1= 0) (5) (6) since F and all its derivatives vanish at 0 and at A. It now follows from (1) and (6) that (x real, n = 0, 1, 2, ... ). Hence Q(x) i f(x) i = f xn l f(x) l < 2A n=O Mn (x > 0), and (5) and (8) imply that (b) holds. PROOF THAT (b) IMPLIES (c) q(x) < Q(x). (7) (8) /Ill /Ill PROOF THAT (c) IMPLIES (d) Put an= M!1n. Since M0 = 1 and since M; < Mn-1M n+ 1' it is easily verified that an < an+ 1' for n > 0. If X > ean' then xn/Mn >en, so xn log q(x) > log -> log en = n. Mn Hence e f."" tog q(x) · dÌ > e f n f.ean+1X-2 dx + e f.oo (N + 1)x-2 dx ea 1 X n = 1 ean ea N + 1 N ( 1 1 ) N + 1 N+ 1 1 - L n - -+ = L -n = 1 an an + 1 aN+ 1 n = 1 an for every N. This shows that (c) implies (d). (9) (10) /Ill 382 REAL AND COMPLEX ANALYSIS PROOF THAT (d) IMPLIES (e) Put A = Mn- 1 n M n (n = 1, 2, 3, . . . ). (1 1) (an An)n < Mn · A1A2 · · · An = 1. (12) Thus An < 1/an , and the convergence of 1:(1/an) implies that of l:An . //// PROOF THAT (e) IMPLIES (a) The assumption now is that l:An < oo, where An is given by (1 1). We claim that the function /(z) =(sin z)2 fi sin An z z n= 1 An z (13) is an entire function of exponential type, not identically zero, which satisfies the inequalities (sin x)2 I x'}'(x) I < M k x (x real, k = 0, 1, 2, . . . ). (14) Note first that 1 - z- 1 sin z has a zero at the origin. Hence there is a constant B such that It follows that so that the series . stn z 1 -< B i z l ( I z I < 1). z 00 L n= 1 sin An z 1 ----1 An Z (15) (16) (17) converges uniformly on compact sets. (Note that 1/ An----+ oo as n----+ oo, since l:An < oo.) The infinite product (13) therefore defines an entire function f which is not identically zero. Next, the identity -- = -e' z t sin z 1 f1 .t d z 2 - 1 shows that I z- 1 sin z I < eiYI if z = x + iy. Hence 00 with A = 2 + L An . n= 1 (18) (19) HOLOMORPHIC FOURIER TRANSFORMS 383 For real x, we have I sin x I < I x I and I sin x I < 1. Hence I x'}'(x) I < I xk I (sin x)2 n sin A.,. X X n= 1 An X < (si: xy(A.1 • • • A.1) 1 = M{si: xy. This gives (14), and if we integrate (14) we obtain 1 fao 7t _ 001 x" f(x) I dx < Mk (k = 0, 1, 2, . . . ). (20) (21) We have proved that f satisfies the hypotheses of Theorem 19.3. The Fourier transform off, 1 fao . F(t) = -2 f(x)e-,tx dx 1t -ao (t real) (22) is therefore a function with compact support, not identically zero, and (21) shows that F e coo and that (DkF)(t) = -( - ix)'1(x)e -,tx dx, 1 fao . 2n _ 00 (23) by repeated application of Theorem 9.2(/). Hence IIDkFII ao < Mk , by (21), which shows that F e C{Mn}· Hence C{Mn} is not quasi-analytic, and the proof is complete. //// Exercises 1 Supposefis an entire function of exponential type and (/J(y) = J :l f(x + iyW dx. Prove that either q>(y) = oo for all real y or q>(y) < oo for all real y. Prove that f = 0 if q> is a bounded function. 2 Suppose f is an entire function of exponential type such that the restriction off to two nonparallel lines belongs to I!. Prove that f = 0. 3 Suppose f is an entire function of exponential type whose restriction to two nonparallel lines is bounded. Prove thatfis constant. (Apply Exercise 9 of Chap. 12.) 4 Suppose f is entire, I f(z) I < C exp (A I z I ), and f(z) = l:a, z". Put oo n! a, (w) = L + 1 " n = O w" Prove that the series converges if I w I > A, that f(z) = Ë [ (w)ewz dw 2nz Jr 384 REAL AND COMPLEX ANALYSIS if r(t) = (A + €)eir, 0 < t < 2n, and that is the function which occurred in the proof of Theorem 19.3. (See also Sec. 19.4.) S Suppose f satisfies the hypothesis of Theorem 19.2. Prove that the Cauchy formula (0 < € < y) () holds; here z = x + i y. Prove that f(x) = lim f(x + iy) y-+0 exists for almost all x. What is the relation between f and the function F which occurs in Theorem 19.2? Is () true with € = 0 and withf in place off in the integrand? 6 Suppose cp e L2(- oo, oo) and cp > 0. Prove that there exists anfwith If I = cp such that the Fourier transform off vanishes on a half line if and only if foo dx log cp(x) 2 > - oo. _ 00 1 + X Suggestion: Consider f, as in Exercise 5, where f = exp (u + iv) and 1 foo Y u(z) = -2 2 log q>(t) dt. n _ oo (x - t) + y 7 Letfbe a complex function on a closed set E in the plane. Prove that the following two conditions onf are equivalent: (a) There is an open set Q => E and a function F E H(Q) such that F(z) = f(z) for z e E. (b) To each ct E E there corresponds a neighborhood Ķ of ct and a function F« E H(V J such that Fiz) = f(z) in Ķ n E. (A special case of this was proved in Theorem 19.9.) 8 Prove that C{n! } = e{n"}. 9 Prove that there are quasi-analytic classes which are larger than C{n ! }. 10 Put An = Mn_ tfMn , as in the proof of Theorem 19. 1 1. Pick g0 E Cc(R1), and define (n = 1, 2, 3, . . . ). Prove directly (without using Fourier transforms or holomorphic functions) that g = lim gn is a func tion which demonstrates that (e) implies (a) in Theorem 19.11. (You may choose any g0 that is conve nient.) 11 Find an explicit formula for a function cp e coo, with support in [ -2, 2], such that cp(x) = 1 if - 1 < x < l. --12 Prove that to every sequence {ctn} of complex numbers there corresponds a function f E coo such that (D"l')(O) = ctn for n = 0, 1, 2, . . . . Suggestion: If cp is as in Exercise 1 1, if Pn ·= ctJn !, if gn(x) = Pn x"cp(x), and if then II Dk !, II 00 < 2-" for k = 0, . . . , n - 1, provided that An is large enough. Take f = "£ f n . 13 Construct a function/ E coo such that the power series oo (D"l')(a) L (x - a)" n=O n! HOLOMORPHIC FOURIER TRANSFORMS 385 has radius of convergence 0 for every a E R 1 . Suggestion: Put CX> f(z) = L ck ei)."x, k= 1 where { ck} and { A.k} are sequences of positive numbers, chosen so that I:ck A.IJ < oo for n = 0, 1, 2, .. . and so that en A.: increases very rapidly and is much larger than the sum of all the other terms in the series I:ck A.IJ . For instance, put ck = ;.: -k, and choose { A.k} so that k- 1 A.k > 2 L ciA.' and A.k > k2k. j = 1 14 Suppose C{Mn} is quasi-analytic, f E C{Mn}, and f(x) = 0 for infinitely many x E [0, 1]. What follows? IS Let X be the vector space of all entire functions f that satisfy I f(z) I < Cexlzl for some C < oo , and whose restriction to the real axis is in 13. Associate with each f E X its restriction to the integers. Prove thatfȧ {f(n)} is a linear one-to-one mapping of X onto (2• 16 Assumef is a measurable function on ( - oo , oo) such that l f(x) l < e-lxl for all x. Prove that its Fourier transform] cannot have compact support, unlessf(x) = 0 a.e. CHAPTER TWENTY UNIFORM APROXIMATION BY POLYNOMIALS Introduction 20.1 Let K0 be the interior of a compact set K in the complex plane. (By defini tion, K0 is the union of all open discs which are subsets of K; of course, K0 may be empty even if K is not.) Let P(K) denote the set of all functions on K which are uniform limits of polynomials in z. Which functions belong to P(K)? Two necessary conditions come to mind immediately : If f E P(K), then f E C(K) and/ E H(K0). The question arises whether these necessary conditions are also sufficient. The answer is negative whenever K separates the plane (i.e., when the com plement of K is not connected). We saw this in Sec. 13.8. On the other hand, if K is an interval on the real axis (in which case K0 = 0), the Weierstrass approx imation theorem asserts that P(K) = C(K). So the answer is positive if K is an interval. Runge's theorem also points in this direction, since it states, for compact sets K which do not separate the plane, that P(K) contains at least all those f E C(K) which have holomorphic extensions to some open set Q ::) K. In this chapter we shall prove the theorem of Mergelyan which states, without any superfluous hypotheses, that the above-mentioned necessary condi tions are also sufficient if K does not separate the plane. The principal ingredients of the proof are : Tietze's extension theorem, a smoothing process invoving convolutions, Runge's theorem, and Lemma 20.2, whose proof depends on properties of the class !/ which was introduced in Chap. 14. 386 UNIFORM APPROXIMATION BY POLYNOMIALS 387 Some Lemmas 20.2 Lemma Suppose D is an open disc of radius r > 0, E c D, E is compact and connected, Q = S2 - E is connected, and the diameter of E is at least r. Then there is afunction g e H(Q) and a constant b, with thef ollolving property: If the inequalities Q((, z) = g(z) + (( - b)g2(z), 100 I Q((, z) l < -r 1 1 000r2 Q(,, z) - z - C < I; - ' 13 hold for all z e Q and for all ' e D. (1) (2) (3) We recall that S2 is the Riemann sphere and that the diameter of E is the supremum of the numbers I z 1 - z2 1, where z 1 e E and z2 e E. PROOF We assume, without loss of generality, that the center of D is at the origin. So D = D(O; r). The implication (d)-+ (b) of Theorem 13.1 1 shows that Q is simply con nected. (Note that oo e Q.) By the Riemann mapping theorem there is there fore a conformal mapping F of U onto Q such that F(O) = oo . F has an expansion of the form We define a oo F(w) = - + L en wn W n=O 1 g( z) = - F - 1( z) a (w e U). (z e Q), where F - 1 is the mapping of Q onto U which inverts F, and we put b = -2 1 . r zg(z) dz, 1tl Jr where r is the positively oriented circle with center 0 and radius r. (4) (5) (6) By (4), Theorem 14.15 can be applied to Fja. It asserts that the diameter of the complement of (F/a)(U) is at most 4. Therefore diam E < 4 1 a 1 . Since diam E > r, it follows that (7) 388 REAL AND COMPLEX ANALYSIS Since g is a conformal mapping of Q onto D(O; 1/ I a I), (7) shows that 4 I g(z) I < -r (z e Q) and since r is a path in n, of length 2nr, (6) gives I b I < 4r. If ' e D, then I ' I < r, so (1), (8), and (9) imply 4 (16) 100 I Q I < r + 5r -;I < -r - . This proves (2). Fix ' e D. (8) (9) (10) If z = F(w), then zg(z) = wF(w)ja; and since wF(w)---+ a as w---+ 0, we have zg(z)---+ 1 as z---+ oo. Hence g has an expansion of the form 1 A2(') A3(') g( z) = z - ' + ( z - ,)2 + ( z - ,)3 + . . . ( I z - ' I > 2r). (1 1) Let r 0 be a large circle with center at 0; (1 1) gives (by Cauchy's theorem) that (12) Substitute this value of A2(') into (1 1). Then (1) shows that the function (13) is bounded as z---+ oo. Hence cp has a removable singularity at oo. If z e Q n D, then l z - 'I < 2r, so (2) and (13) give I cp(z) I < 8r3 1 Q(,, z) I + 4r2 < 1,000r2• (14) By the maximum modulus theorem, (14) holds for all z e Q. This proves (3). /Ill 20.3 Lemma Suppose f e Cʎ(R2), the space of all continuously diff erentiable functions in the plane, with compact support. Put a = !(Ô + i ̋). 2 ax ay Then the following ʏʏCauchy formula " holds: f(z) = - !_I ' r (13/X() dŐ d'1 1t JR2 ' - Z (1) (2) UNIFORM APPROXIMATION BY POLYNOMIALS 389 PROOF This may be deduced from , Green's theorem. However, here is a sim pie direct proof: Put 0, (J real. If ( = z + rei8, the chain rule gives -1 iB [a i a J (o f)(() = 2 e or + ; ofJ cp(r, fJ). The right side of (2) is therefore equal to the limit, as E---+ 0, of - -- + - - d(J dr. 1 100 i21t (0(/J i 0(/J) 2n £ 0 or r ofJ (3) (4) For each r > 0, <p is periodic in 0, with period 2n. The integral of ocpjofJ is therefore 0, aࡂd (4) becomes 1 i21t i 00 0(/J 1 i21t - 2n o d() • Or dr = 2n o qJ(E, 0) dO. As E---+ 0, cp(E, fJ)---+ f(z) uniformly. This gives (2). (5) Ill/ We shall establish Tietze's extension theorem in the same setting in which we proved Urysohn's lemma, since it is a fairly direct consequence of that lemma. 20.4 Tietze's Extension Theorem Suppose K is a compact subset of a locally compact Hausdorff space X, and f e C(K). Then there exists an F e Cc(X) such that F(x) = f(x)for all x e K. (As in Lusin's theorem, we can also arrange it so that II F II x = II f II K .) PROOF Assume f is real, - 1 <! < 1. Let W be an open set with compact closure so that K c W. Put K+ = {x E K: f(x) > ƺ}, K- = {x E K:f(x) < - ƺ}. (1) Then K + and K- are disjoint compact subsets of W. As a consequence of Urysohn's lemma there is a function /1 e Cc(X) such that f1(x) = ƺ on K +, /1(x) = - ƺ on K-, - ƺ </1(x) < ƺ for all x e X, and the support of/1 lies in W. Thus 1 !-ft l < ϕ on K, I /1 1 < ƺ on X. (2) Repeat this construction with f-!1 in place of f: There exists an /2 e Cc(X), with support in W, so that I /2 I < ƺ · ϕ on X. (3) In this way we obtain functions fn e Cc(X), with supports in W, such that l f-ft - · · · -fn l < (ϖ)" on K, I fn l < ƺ · (ϖ)"- 1 on X. (4) 390 REAL AND COMPLEX ANALYSIS Put F = f1 + f2 + f3 + · · · . By (4), the series converges to f on K, and it converges uniformly on X. Hence F is continuous. Also, the support of F lies in W. //// Mergelyan's Theorem 20.5 Theorem If K is a compact set in the plane whose complement is con nected, iff is a continuous complex function on K which is holomorphic in the interior of K, and if E > 0, then there exists a polynomial P such that I f(z) - P(z) I < efor all z e K. If the interior of K is empty, then part of the hypothesis is vacuously satis fied, and the conclusion holds for every f e C(K). Note that K need not be con nected. PROOF By Tietze's theorem, f can be extended to a continuous function in the plane, with compact support. We fix one such extension, and denote it again by f. For any Ī > 0, let w(ç) be the supremum of the numbers I f(z2) -f(z 1) I where z 1 and z 2 are subject to the condition I z 2 - z 1 I < ç- Since f is uni formly continuous, we have lim w(ç) = 0. (1) <-o From now on, ç will be fixed. We shall prove that there is a polynomial P such that I f(z) - P(z) I < 10,000 w(ç) (z e K). (2) By (1), this proves the theorem. Our first objective is the construction of a function e Cʎ(R2), such that for all z and I f(z) - (z) I < w(ç), I (8ct>)(z) I < 2i15) , (z) = -!II ({jct>XO dŐ d'1 n , _ z X (3) (4) (5) UNIFORM APPROXIMATION BY POLYNOMIALS 391 where X is the set of all points in the support of whose distance from the complement of K does not exceed b. (Thus X contains no point which is "far within " K.) We construct as the convolution off with a smoothing function A. Put a(r) = 0 if r > b, put 3 ( r2)2 a(r) = n@2 1 - @2 (0 < r < @), (6) and define A( z) = a( I z I ) (7) for all complex z. It is clear that A e Cт(R2). We claim that (8) R2 (9) (10) The constants are so adjusted in (6) that (8) holds. (Compute the integral in polar coordinates.) (9) holds simply because A has compact support. To compute (10), express aA in polar coordinates, as in the proof of Lemma 20.3, and note that oAjo() = 0, I oAjor I = -a'(r). Now define (z) =II f(z - C)A(() dŐ drt =II A(z - ()/(() dŐ d17. (1 1) R2 R2 Since f and A have compact support, so does <1>. Since (z) -/(z) =II [f(z - () -f(z)]A(() de d17 R2 (12) and A(() = 0 if I ( I > @' (3) follows from (8). The difference quotients of A converge boundedly to the corresponding partial derivatives of A, since 392 REAL AND COMPLEX ANALYSIS A E CÛ(R2). Hence the last expression in (1 1) may be differentiated under the integral sign, and we obtain (afl>Xz> = J J (aA)(z - ()/(() de d11 R2 =If f(z - ()(8A)(() dŐ d17 R2 = JJ f(z - () -f(z)(() de dYf. R2 (13) The last equality depends on (9). Now (10) and (13) give (4). If we write (13) with x and y in place of a, we see that has continuous partial derivatives. Hence Lemma 20.3 applies to <1>, and (5) will follow if we can show that a = 0 in G, where G is the set of all z E K whose distance from the complement of K exceeds b. We shall do this by showing that (z) = f(z) (z E G) ; (14) note that aj = 0 in G, since f is holomorphic there. (We r.ecall that a is the Cauchy-Riemann operator defined in Sec. 1 1.1.) Now if z E G, then z - ( is in the interior of K for all ( with I ( I < b. The mean value property for harmo nic functions therefore gives, by the first equation in (1 1), for all z E G. f«S f21t <D(z) = Jo a(r)r dr Jo f(z - reiÔ d(} = 2nf(z) f a(r)r dr = /(z) f f A = /(z) R2 We have now proved (3), (4), and (5). (15) The definition of X shows that X is compact and that X can be covered by finitely many open discs D1, • • • , Dn , of radius 2b, whose centers are not in K. Since S2 - K is connected, the center of each Di can be joined to oo by a polygonal path in S2 - K. It follows that each Di contains a compact con nected set Ei , of diameter at least 2b, so that S2 - Ei is connected and so that K n Ei = 0. We now apply Lemma 20.2, with r = 2£5. There exist functions UNIFORM APPROXIMATION BY POLYNOMIALS 393 gi e H(S2 - Ei) and constants bi so that the inequalities 50 I Qj(, z) l < F, 1 4,000 <52 Qi(, z) - z - ( < I z - ( 13 hold for z ¢ Ei and ( e Di , if (16) (17) (18) Let Q be the complement of E1 u · · · u En . Then Q is an open set which contains K. Put X1 = X n D1 and Xi = (X n D) - (X1 u · · · u Xi 1), for 2 < j < n. Define (19) and F(z) = : f f (8q,X()R((, z) dŐ drJ (z e Q). (20) X Since F(z) = J. : f f (8q,)(0Qj((, z) dŐ drJ, (21) Xj (18) shows that F is a finite linear combination of the functions gi and gJ. Hence F e H(Q). By (20), (4), and (5) we have 2w(<5) If 1 I F(z) - (z) I < nb R((, z) - z _ ( dŐ d17 X (z e Q). (22) Observe that the inequalities (16) and (17) are valid with R in place of Qi if ( e X and z e n. For if ( e X then ( e Xi for some j, and then R((, z) = Qj(, z) for all z e Q. Now fix z E n, put ' = z + pei6, and estimate the integrand in (22) by (16) if p < 4<5, by (17) if 4<5 < p. The integral in (22) is then seen to be less than the sum of 14d (50 1) 2n 0 ķ + p p dp = 808m5 (23) 394 REAL AND COMPLEX ANALYSIS and Hence (22) yields i 00 4,000<52 2n 3 p dp = 2,000 n<5. 4«S p I F(z) - (z) I < 6,000 ro(<5) (z e Q). (24) (25) Since F e H(Q), K c Q, and S2 - K is connected, Runge's theorem shows that F can be uniformly approximated on K by polynomials. Hence (3) and (25) show that (2) can be satisfied. This completes the proof. I I I I One unusual feature of this proof should be pointed out. We had to prove that the given function f is in the closed subspace P(K) of C(K). (We use the terminology of Sec. 20.1.) Our first step consisted in approximating f by <1>. But this step took us outside P(K), since was so constructed that in general will not be holomorphic in the whole interior of K. Hence is at some positive distance from P(K). However, (25) shows that this distance is less than a constant multiple of ro( <5). [In fact, having proved the theorem, we know that this distance is at most ro(<5), by (3), rather than 6,000 ro(<5).] The proof of (25) depends on the inequality (4) and on the fact that 8<1> = 0 in G. Since holomorphic functions <p are characterized by J<p = 0, ( 4) may be regarded as saying that is not too far from being holomorphic, and this interpretation is confirmed by (25). Exercises 1 Extend Mergelyan's theorem to the case in which S2 - K has finitely many components: Prove that then every f E C(K) which is holomorphic in the interior of K can be uniformly approximated on K by rational functions. 2 Show that the result of Exercise 1 does not extend to arbitrary compact sets K in the plane, by verifying the details of the following example. For n = 1, 2, 3, .
.
. , let Dn = D(ctn; rn) be disjoint open discs in U whose union V is dense in U, such that (rn < oo. Put K = 0 - V. Let r and Yn be the paths 0 < t < 2n, and define L(f) = If(z) dz -J. i.f(z) dz (f E C(K)). Prove that L is a bounded linear functional on C(K), prove that L(R) = 0 for every rational function R whose poles are outside K, and prove that there exists anf E C(K) for which L(f) '# 0. 3 Show that the function g constructed in the proof of Lemma 20.2 has the smallest supremum norm among allf E H(O.) such that zf(z)-+ 1 as z-+ oo. (This motivates the proof of the lemma.) Show also that b = c0 in that proof and that the inequality I b I < 4r can therefore be replaced by I b I < r. In fact, b lies in the convex hull of the set E. APPENDIX HAUSDORFF'S MAXIMALITY THEOREM We shall first prove a lemma which, when combined with the axiom of choice, leads to an almost instantaneous proof of Theorem 4.21. If §" is a collection of sets and c §", we call a subchain of §" provided that is totally ordered by set inclusion. Explicitly, this means that if A e and B e <1>, then either A c B or B c A. The union of all members of will simply be referred to as the union of . Lemma Suppose §" is a nonempty collection o f subsets o f a set X such that the union of every subchain of !F belongs to §". Suppose g is a function which associates to each A e §" a set g(A) e §" such that A c g(A) and g(A) - A consists o f at most one element. Then there exists an A e !F for which g(A) = A. PROOF Fix A0 e §". Call a subcollection §"' of §" a tower if §"' has the fol-_ lowing three properties : (a) A0 e §"'. (b) The union of every subchain of !F' belongs to §"'. (c) If A e §"', then also g(A) e !F'. The family of all towers is nonempty. For if §" 1 is the collection of all A E §" such that A0 c A, then §" 1 is a tower. Let §" 0 be the intersection of all towers. Then §" 0 is a tower (the verification is trivial), but no proper subcollection of !F 0 is a tower. Also, A0 c A if A e !F 0 . The idea of the proof is to show that !F 0 is a su bchain of !F. 395 396 REAL AND COMPLEX ANALYSIS Let r be the collection of all C e §" 0 such that every A e §" 0 satisfies either A c C or C c A. For each C e r, let ( C) be the collection of all A e §" 0 such that either A c C or g( C) c A. Properties (a) and (b) are clearly satisfied by r and by each (C). Fix C e r, and suppose A e (C). We want to prove that g(A) e (C). If A e ( C), there are three possibilities : Either A c C and A =I= C, or A = C, or g(C) c A. If A is a proper subset of C, then C cannot be a proper subset of g(A), otherwise g(A) - A would contain at least two elements; since C e r, it follows that g(A) c C. If A = C, then g(A) = g(C). If g(C) c A, then also g(C) c g(A), since A c g(A). Thus g(A) e ( C), and we have proved that ( C) is a tower. The minimality of §" 0 implies now that ( C) = §" 0 , for every c e r. In other words, if A e !F 0 and C e r, then either A c C or g(C) c A. But this says that g(C) e r. Hence r is a tower, and the minimality of §" 0 shows that r = !F 0 • It follows now from the definition of r that !F 0 is totally ordered. Let A be the union of §" 0 • Since §" 0 satisfies (b), A e §" 0 • By (c), g(A) e !F 0 • Since A is the largest member of §" 0 and since A c g(A), it follows that A = g(A). /Ill Definition A choice function for a set X is a function f which associates to each nonempty subset E of X an element of E:f(E) e E. In more informal terminology, f " chooses " an element out of each non empty subset of X. The Axiom of Choice For every set there is a choice function. Hausdorff's Maximality Theorem Every nonempty partially ordered set P contains a maximal totally ordered subset. PROOF Let §" be the collection of all totally ordered subsets of P. Since every subset of P which consists of a single element is totally ordered, !F is not empty. Note that the union of any chain of totally ordered sets is totally ordered. Letfbe a choice function for P. If A e §", let A be the set of all x in the complement of A such that A u {x} e §". If A =I= 0, put g(A) = A u {/(A)}. If A = 0, put g(A) = A. By the lemma, A = 0 for at least one A e §", and any such A is a maximal element of §". /Ill NOTES AND COMMENTS Chapter 1 The first book on the modern theory of integration and differentiation is Lebesgue's "Le͗ons sur }'integration," published in 1904. For an illuminating history of the earlier attempts that were made toward the construction of a satisfactory theory of integration, see [ 42]t, which contains interesting discussions of the difficulties that even first-rate mathematicians had with simple set-theoretic concepts before these were properly defined and well understood. The approach to abstract integration presented in the text is inspired by Saks . Greater generality can be attained if a-algebras are replaced by a-rings (Axioms: U A; e 9t and A1 - A2 e 9t if Ai E 9t for i = 1, 2, 3, . . . ; it is not required that X E at), but at the expense of a necessarily fussier definition of measurability. See Sec. 18 of . In all classical applications the measurability of X is more or less automatic. This is the reason for our choice of the somewhat simpler theory based on a-algebras. Sec. 1.1 1. This definition of Bl is as in . In the Borel sets are defined as the a-ring generated by the compact sets. In spaces which are not a-compact, this is a smaller family than ours. Chapter 2 Sec. 2.12. The usual statement of Urysohn's lemma is : If K0 and K 1 are disjoint closed sets in a normal Hausdorff space X, then there is a continuous function on X which is 0 on K0 and 1 on K 1• The proof is exactly as in the text. Sec. 2.14. The original form of this theorem, with X = [0, 1], is due to F. Riesz (1909). See , pp. 373, 380-381, and , pp. 134-135 for its further history. The theorem is here presented in the same generality as in . The set function J1. which is defined for all subsets of X in the proof of Theorem 2.14 is called an outer measure because of its countable subadditivity (Step I). For systematic exploitations (originated by Caratheodory) of this notion, see and . Sec. 2.20. For direct constructions of Lebesgue measure, along more classical lines, see ͘ , and . t Numbers in brackets refer to the Bibliography. 397 398 REAL AND COMPLEX ANALYSIS Sec. 2.21. A proof that the cardinality of every countably generated a-algebra is s;c may be found on pp. 133-134 of [ 44]. That this cardinality is either finite or > c should be clear after doing Exercise 1 of Chap. 1. Sec. 2.22. A very instructive paper on the subject of nonmeasurable sets in relation to measures invariant under a group is : J. von Neumann, Zur allgemeinen Theorie des Masses, Fundament a Math., vol. 13, pp. 73-1 16, 1929. See also Halmos's article in the special (May, 1958) issue of Bull. Am. Mat h. Soc. devoted to von Neumann's work. Sec. 2.24. , p. 72. Sec. 2.25. , p. 75. There is another approach to the Lebesgue theory of integration, due to Daniell (Ann. Math., vol. 19, pp. 279-294, 1917-1918) based on extensions of positive linear function als. When applied to Cc(X) it leads to a construction which virtually turns the Vitali-Caratheodory theorem into the definition of measurability. See and, for the full treatment, . Exercise 8. Two interesting extensions of this phenomenon have appeared in Amer. Math. Monthly: see vol. 79, pp. 884-886, 1972 (R. B. Kirk) and vol. 91, pp. 564-566, 1984 (F. S. Cater). Exercise 17. This example appears in A Theory of Radon Measures on Locally Compact Spaces, by R. E. Edwards, Acta Mat h., vol. 89, p. 160, 1953. Its existence was unfortunately over looked in . Exercise 18. , p. 231 ; originally due to Dieudonne. Chapter 3 -The best general reference is . See also Chap. 1 of . Exercise 3. Vol. 1 (1920) of Fundamenta Math. contains three papers relevant to the parentheti cal remark. Exercise 7. A very complete answer to this question was found by A. Villani, in Am. Math. Monthly, vol. 92, pp. 485-487, 1985. Exercise 16. , p. 18. When t ranges over all positive real numbers, a measurability problem arises in the suggested proof. This is the reason for including (ii) as a hypothesis. See W. Walter, Am. Math. Monthly, vol. 84, pp. 1 18-1 19, 1977. Exercise 17. The second suggested proof of part (b) was published by W. P. Novinger, in Proc. Am. Math. Soc., vol. 34, pp. 627-628, 1972. It was also discovered by David Hall. Exercise 18. Convergence in measure is a natural concept in probability theory. See , Chap. IX. Chapter 4 There are many books dealing with Hilbert space theory. We cite and as main references. See also , , and . The standard work on Fourier series is . For simpler introductions, see , , , and . Exercise 2. This is the so-called Gram-Schmidt orthogonalization process. Exercise 18. The functions that are uniform limits (on R 1) of members of X are called almost periodic. Chapter 5 The classical work here is . More recent treatises are , , and . See also and . Sec. 5. 7. The relations between measure theory on the one hand and Baire's theorem on the other are discussed in great detail in [ 48]. Sec. 5.1 1. Although there are continuous functions whose Fourier series diverge on a dense G͙ , the set of divergence must have measure 0. This was proved by L. Carleson (Acta Math., vol. 1 16, pp. 135-157, 1966) for all fin I3(T); the proof was extended to IJ'(T), p > 1, by R. A. Hunt. See also , especially Chap. II. Sec. 5.22. For a deeper discussion of representing measures see Arens and Singer, Proc. Am. Math. Soc., vol. 5, pp. 735-745, 1954. Also , . NOTES AND COMMENTS 399 Chapter 6 Sec. 6.3. The constant 1/n is best possible. See R. P. Kaufman and N. W. Rickert, Bull. Am. Math. Soc., vol. 72, pp. 672-676, 1966, and (for a simpler treatment) W. W. Bledsoe, Am. Math. Monthly, vol. 77, pp. 180-182, 1970. Sec. 6.10. von Neumann's proof is in a section on measure theory in his paper: On Rings of Operators, III, Ann. Math., vol. 41, pp. 94-161, 1940. See pp. 124-130. Sec. 6.1 5. The phenomenon L00 #- (C) is discussed by J. T. Schwartz in Proc. Am. Math. Soc., vol. 2, pp. 270-275, 1951, and by H. W. Ellis and D. 0. Snow in Can. Math. Bull., vol. 6, pp. 21 1-229, 1963. See also , p. 131, and , p. 36. Sec. 6.19. The references to Theorem 2.14 apply here as well. Exercise 6. See , p. 43. Exercise 10(g). See , vol. I, p. 167. Chapter 7 Sec. 7.3. This simple covering lemma seems to appear for the first time in a paper by Wiener on the ergodic theorem (Duke Math. J., vol. 5, pp. 1-18, 1939). Covering lemmas play a central role in the theory of differentiation. See , , and, for a very detailed treatment, . Sec. 7.4. Maximal functions were first introduced by Hardy and Littlewood, in Acta Math., vol. 54, pp. 81-1 16, 1930. That paper contains proofs of Theorems 8.18, 1 1.25(b), and 17.1 1. Sec. 7.21. The same conclusion can be obtained under somewhat weaker hypotheses; see , Theorems 260-264. Note that the proof of Theorem 7.21 uses the existence and integrability of only the right-hand derivative of f, plus the continuity of f For a further refinement, see P. L. Walker, Amer. Math. Monthly, vol. 84, pp. 287-288, 1977. Sees. 7.25, 7.26. This treatment of the change of variables formula is quite similar to D. Varberg's in Amer. Math. Monthly, vol. 78, pp. 42-45, 1971. Exercise 5. A very simple proof, due to K. Stromberg, is in Proc. Amer. Mat h. Soc., vol. 36, p. 308, 1972. Exercise 12. For an elementary proof that every monotone function (hence every function of bounded variation) is differentiable a.e., see , pp. 5-9. In that work, this theorem is made the starting point of the Lebesgue theory. Another, even simpler, proof by D. Austin is in Proc. Amer. Math. Soc., vol. 16, pp. 220--221, 1965. Exercise 18. These functions cpn are the so-called Rademacher functions. Chap. V of con tains further theorems about them. Chapter 8 Fubini's theorem is developed here as in and . For a different approach, see . Sec. 8.9(c) is in Fundamenta Math., vol. 1, p. 145, 1920. Sec. 8.18. This proof of the Hardy-Littlewood theorem (see the reference to Sec. 7.4) is essen tially that of a very special case of the Marcinkiewicz interpolation theorem. A full discussion of the latter may be found in Chap. XII of . See also . Exercise 2. Corresponding to the idea that an integral is an area under a curve, the theory of the Lebesgue integral can be developed in terms of measures of ordinate sets. This is done in . Exercise 8. Part (b), in even more precise form, was proved by Lebesgue in J. M athematiques, ser. 6, vol. 1, p. 201, 1905, and seems to have been forgotten. It is quite remarkable in view of another example of Sierpinski (Fundamenta Math., vol. 1, p. 1 14, 1920) : There is a plane set E which is not Lebesgue measurable and which has at most two points on each straight line. Iff= XE , then f is not Lebesgue measurable, although all of the sections f x and f' are upper semicontinuous; in fact, each has at most two points of discontinuity. (This example depends on the axiom of choice, but not on the continuum hypothesis.) 400 REAL AND COMPLEX ANALYSIS Chapter 9 For another brief introduction, see , chap. XVI. A different proof of Plancherel's theorem is in . Group-theoretic aspects and connections with Banach algebras are discussed in , , and . For more on invariant subspaces (Sec. 9. 16) see [1 1] ; the corresponding problem in IJ is described in , Chap. 7. Chapter 10 General references : , , , , , and . Sec. 10.8. Integration can also be defined over arbitrary rectifiable curves. See , vol. I, Appendix C. Sec. 10.10. The topological concept of index is applied in and is fully utilized in . The computational proof of Theorem 10.10. is as in , p. 93. Sec. 10.13. Cauchy published his theorem in 1825, under the additional assumption that f' is continuous. Goursat showed that this assumption is redundant, and stated the theorem in its present form. See , p. 163, for further historical remarks. Sec. 10.16. The standard proofs of the power series representation and of the fact that f E H(O) implies f' E H(O.) proceed via the Cauchy integral formula, as here. Recently proofs have been con structed which use the winding number but make no appeal to integration. For details see . Sec. 10.25. A very elementary proof of the algebraic completeness of the complex field is in , p. 170. Sec. 10.30. The proof of part (b) is as in . Sec. 10.32. The open mapping theorem and the discreteness of Z(f) are topological properties of the class of all nonconstant holomorphic functions which characterize this class up to homeomorp hisms. This is Stoilov's theorem. See [34 ]. Sec. 10.35. This strikingly simple and elementary proof of the global version of Cauchy's theorem was discovered by John D. Dixon, Proc. Am. Math. Soc., vol. 29, pp. 625-626, 1971. In the proof is based on the theory of exact differentials. In the first edition of the present book it was deduced from Runge's theorem. That approach was used earlier in , p. 177. There, however, it was applied in simply connected regions only. Chapter 1 1 General references : , Chap. 5; , Chap. 1. Sec. 1 1.14. The reflection principle was used by H. A. Schwarz to solve problems concerning conformal mappings of polygonal regions. See Sec. 17.6 of . Further results along these lines were obtained by Caratheodory; see , vol. II, pp. 88-92, ,and Commentarii Mathematici Helvetici, vol. 19, pp. 263-278, 1946-1947. Sees. 1 1.20, 1 1.25. This is the principal result of the Hardy-Littlewood paper mentioned in the reference to Sec. 7 .4. The proof of the second inequality in Theorem 1 1.20 is as in [ 40], p. 23. Sec. 1 1.23. The first theorems of this type are in Fatou's thesis, Series trigonometriques et series de Taylor, Acta Mat h., vol. 30, pp. 335-400 , 1906. This is· the first major work in which Lebesgue's theory of integration is applied to the study of holomorphic functions. Sec. 1 1.30. Part (c) is due to Herglotz, Leipziger Berichte, vol. 63, pp. 501-51 1, 191 1. Exercise 14. This was suggested by W. Ramey and D. Ullrich. Chapter 12 Sec. 12.7. For further examples, see , pp. 176-186. Sec. 12.1 1. This theorem was first proved for trigonometric series by W. H. Young (1912; q = 2, 4, 6, . . . ) and F. Hausdorff (1923 ; 2 < q < oo). F. Riesz (1923) extended it to uniformly bounded NOTES AND COMMENTS 401 orthonormal sets, M. Riesz (1926) derived this extension from a general interpolation theorem, and G. 0. Thorin (1939) discovered the complex-variable proof of M. Riesz's theorem. The proof of the text is the Calder6n-Zygmund adaptation (1950) of Thorin's idea. Full references and discussions of other interpolation theorems are in Chap. XII of . Sec. 12.13. In slightly different form, this is in Duke Math. J., vol. 20, pp. 449-458, 1953. Sec. 12. 14. This proof is essentially that of R. Kaufman (Math. Ann., vol. 169, p. 282, 1967). E. L. Stout (Math. Ann., vol. 177, pp. 339-340, 1968) obtained a stronger result. Chapter 13 Sec. 13.9. Runge's theorem was published in Acta Math., vol. 6, 1885. (Incidentally, this is the same year in which the Weierstrass theorem on uniform approximation by polynomials on an interval was published ; see Mathematische Werke, vol. 3, pp. 1-37.) See , pp. 171-177, for a proof which is close to the original one. The functional analysis proof of the text is known to many analysts and has probably been independently discovered several times in recent years. It was communicated to me by L. A. Rubel. In , vol. II, pp. 299-308, attention is paid to the closeness of the approximation if the degree of the polynomial is fixed. Exercises 5, 6. For yet another method, see D. G. Cantor, Proc. Am. Math. Soc., vol. 15, pp. 335-336, 1964. Chapter 14 General reference : . Many special mapping functions are described there in great detail. Sec. 14.3. More details on linear fractional transformations may be found in , pp. 22-35 ; in , pp. 46-57 ; in ; and especially in Chap. 1 of L. R. Ford's book ŵ'Automorphic Functions," McGraw-Hill Book Company, New York, 1929. Sec. 14.5. Normal families were introduced by Montel. See Chap. 15 of . Sec. 14.8. The history of Riemann's theorem is discussed in , pp. 320-32 1, and in , p. 230. Koebe's proof (Exercise 26) is in J. f ur reine und angew. Math., vol. 145, pp. 177-223, 1915; doubly connected regions are also considered there. Sec. 14.14. Much more is true than just I a2 I < 2 : In fact, I an I < n for all n > 2. This was conjec tured by Bieberbach in 1916, and proved by L. de Branges in 1984 [Acta Math., vol. 154, pp. 137-152, (1985)]. Moreover, if I an I = n for just one n > 2, then f is one of the Koebe functions of Example 14.1 1. Sec. 14.19. The boundary behavior of conformal mappings was investigated by Caratheodory in Math. Ann., vol. 73, pp. 323-370, 1913. Theorem 14.19 was proved there for regions bounded by Jordan curves, and the notion of prime ends was introduced. See also , vol. II, pp. 88-107. Exercise 24. This proof is due to Y. N. Moschovakis. Chapter IS Sec. 15.9. The relation between canonical products and entire functions of finite order is discussed in Chap. 2 of , Chap. VII of , and Chap. VIII of . Sec. 15.25. See Szasz, Math. Ann., vol. 77, pp. 482-496, 1916, for further results in this direction. Also Chaj). II of . Chapter 16 The classical work on Riemann surfaces is . (The first edition appeared in 1913.) Other references : Chap. VI of , Chap. 10 of , Chap. VI of , and . Sec. 16.5. Ostrowski's theorem is in J. London Math. Soc., vol. 1, pp. 251-263, 1926. See J.-P. Kahane, Lacunary Taylor and Fourier Series, Bull. Am. Math. Soc., vol. 70, pp. 199-213, 1964, for a more recent account of gap series. Sec. 16.15. The approach to the monodromy theorem was a little simpler in the first edition of this book. It used the fact that every simply connected plane region is homeomorphic to a convex 402 REAL AND COMPLEX ANALYSIS one, namely U. The present proof is so arranged that it applies without change to holomorphic functions of several complex variables. (Note that when k > 2 there exist simply connected open sets in Rk which are not homeomorphic to any convex set; spherical shells furnish examples of this.) Sec. 16.17. Chap. 13 of , Chap. VIII of , and part 7 of . Sec. 16.21. Picard's big theorem is proved with the aid of modular functions in part 7 of . " Elementary " proofs may be found in [3 1], pp. 277-284, and in Chap. VII of . Exercise 10. Various classes of removable sets are discussed by Ahlfors and Beurling in Confor mal Invariants and Function-theoretic Null-Sets, Acta Math., vol. 83, pp. 101-129, 1950. Chapter 17 The classical reference here is . See also , Chap. VII. Although deals mainly with the unit disc, most proofs are so constructed that they apply to other situations which are described there. Some of these generalizations are in Chap. 8 of . Other recent books on these topics are , [ 40], and [ 46]. Sec. 17.1. See for a thorough treatment of subharmonic functions. Sec. 17.13. For a different proof, see , or the paper by Helson and Lowdenslager in Acta Math., vol. 99, pp. 165-202, 1958. An extremely simple proof was found by B. K. 0ksendal, in Proc. Amer. Math. Soc., vol. 30, p. 204, 1971. Sec. 17.14. The terms " inner function " and " outer function " were coined by Beurling in the paper in which Theorem 17.21 was proved : On Two Problems Concerning Linear Transformations in Hilbert Space, Acta Math., vol. 81, pp. 239-255, 1949. For further developments, see [1 1]. Sees. 17.25, 17.26. This proof of M. Riesz's theorem is due to A. P. Calderon. See Proc. Am. Math. Soc., vol. 1, pp. 533-535, 1950. See also , vol. I, pp. 252-262. Exercise 3. This forms the basis of a definition of HP -spaces in other regions. See Trans. Am. Mat h. Soc., vol. 78, pp. 46-66, 1955. Chapter 18 General references : , , and ; also . The theory was originated by Gelfand in 1941. Sec. 18.18. This was proved in elementary fashion by P. J. Cohen in Proc. Am. Math. Soc., vol. 12, pp. 159-163, 1961. · Sec. 18.20. This theorem is Wermer's, Proc. Am. Math. Soc., vol. 4, pp. 866-869, 1953. The proof of the text is due to Hoffman and Singer. See [1 5], pp. 93-94, where an extremely short proof by P. J. Cohen is also given. (See the reference to Sec. 18.1 8.) Sec. 18.21. This was one of the major steps in Wiener's original proof of his Tauberian theorem. See , p. 91. The painless proof given in the text was the first spectacular success of the Gelfand theory. Exercise 14. The set L\ can be given a compact Hausdorff topology with respect to which the function x is continuous. Thus x .. x is a homomorphism of A into C(A). This representation of A as an algebra of continuous functions is a most important tool in the study of commutative Banach algebras. Chapter 19 Sees. 19.2, 19.3 : , pp. 1-1 3. See also , where functions of exponential type are the main subject. Sec. 19.5. For a more detailed introduction to the classes C{M,.}, see S. Mandelbrojt, " Series de Fourier et classes quasi-analytiques," Gauthier-Villars, Paris, 1935. Sec. 19.1 1. In , the proof of this theorem is based on Theorem 19.2 rather than on 19.3. Exercise 4. The function is called the Borel transform of f. See , Chap. 5. · Exercise 12. The suggested proof is due to H. Mirkil, Proc. Am. Math. Soc., vol. 7, pp. 65Ŷ52, 1956. The theorem was proved by Borel in 1895. NOTES AND COMMENTS 403 Chapter 20 See S. N. Mergelyan, Uniform Approximations to Functions of a Complex Variable, Uspehi Mat. Nauk (N.S.) 7, no. 2 (48), 31-122, 1952 ; Amer. Math. Soc. Translation No. 101, 1954. Our Theorem 20.5 is Theorem 1.4 in Mergelyan's paper. A functional analysis proof, based on measure-theoretic considerations, has recently been published by L. Carleson in Mat h. Scandinavica, vol. 1 5, pp. 167-17 5, 1964. Appendix The maximality theorem was first stated by Hausdorff on p. 140 of his book " Grundziige der Mengenlehre," 19 14. The proof of the text is patterned after Sec. 16 of Halmos's book . The idea to choose g so that g(A) - A has at most one element appears there, as does the term " tower." The proof is similar to one of Zermelo's proofs of the well-ordering theorem; see Mat h. Ann., vol. 65, pp. 107-128, 1908. BIBLIOGRAPHY 1. L. V. Ahlfors : " Complex Analysis," 3d ed., McGraw-Hill Book Company, New York, 1978. 2. S. Banach : Theorie des Operations lineaires, " Monografie Matematyczne," vol. 1, Warsaw, 1932. 3. R. P. Boas : " Entire Functions," Academic Press Inc., New York, 1954. 4. C. Caratheodory : " Theory of Functions of a Complex Variable," Chelsea Publishing Company, New York, 1954. 5. N. Dunford and J. T. Schwartz : " Linear Operators," Interscience Publishers, Inc., New York, 1958. 6. P. R. Halmos : " Introduction to Hilbert Space and the Theory of Spectral Multiplicity," Chelsea Publishing Company, New York, 1951. 7. P. R. Halmos : " Measure Theory," D. Van Nostrand Company Inc., Princeton, N. J., 1950. 8. P. R. Halmos : " Naive Set Theory," D. Van Nostrand Company, Inc., Princeton, N. J., 1960. 9. G. H. Hardy, J. E. Littlewood, and G. P6lya : " Inequalities," Cambridge University Press, New York, 1934. 10. G. H. Hardy and W. W. Rogosinski : " Fourier Series," Cambridge Tracts no. 38, Cambridge, London, and New York, 1950. 1 1. H. Helson : " Lectures on Invariant Subspaces," Academic Press Inc., New York, 1964. 12. E. Hewitt and K. A. Ross : "Abstract Harmonic Analysis," Springer-Verlag, Berlin, vol. I, 1963 ; vol. II, 1970. 1 3. E. Hille : "Analytic Function Theory," Ginn and Company, Boston, vol. I, 1959; vol. II, 1962. 14. E. Hille and R. S. Phillips : " Functional Analysis and Semigroups," Amer. Math. Soc. Colloquium Publ. 3 1, Providence, 1957. 15. K. Hoffman : " Banach Spaces of Analytic Functions," Prentice-Hall, Inc., Englewood Cliffs, N. J., 1962. 16. H. Kestelman : " Modern Theories of Integration," Oxford University Press, New York, 1937. 17. L. H. Loomis : "An Introduction to Abstract Harmonic Analysis," D. Van Nostrand Company, Inc., Princeton, N. J., 1953. 18. E. J. McShane : " Integration," Princeton University Press, Princeton, N. J. 1944. 19. M. A. Naimark : " Normed Rings," Erven P. Noordhoff, NV, Groningen Netherlands, 1959. 20. Z. Nehari : " Conformal Mapping," McGraw-Hill Book Company, New York 1952. 21. R. E. A. C. Paley and N. Wiener: " Fourier Transforms in the Complex Domain," Amer. Math. Soc. Colloquium Publ. 19, New York, 1934. 22. T. Rado : Subharmonic Functions, Ergeb. Math., vol. 5, no. 1, Berlin, 1937. 23. C. E. Rickart : " General Theory of Banach Algebras," D. Van Nostrand Company, Inc., Prince ton, N. J., 1960. 405 406 BIBLIOGRAPHY 24. F. Riesz and B. Sz.-Nagy : " Leŷons d'Analyse Fonctionnelle," Akademiai Kiad6, Budapest, 1952. 25. H. L. Royden : " Real Analysis," The Macmillan Company, New York, 1963. 26. W. Rudin : " Principles of Mathematical Analysis," 3d ed., McGraw-Hill Book Company, New York, 1976. 27. W. Rudin : " Fourier Analysis on Groups," Interscience Publishers, Inc., New York, 1962. 28. S. Saks : " Theory of the Integral," 2d ed., " Monografie Matematyczne," vol. 7, Warsaw, 1937. Reprinted by Hafner Publishing Company, Inc., New York. 29. S. Saks and A. Zygmund : "Analytic Functions," " Monografie Matematyzcne," vol. 28, Warsaw, 1952. 30. G. Springer : " Introduction to Riemann Surfaces," Addison-Wesley Publishing Company, Inc., Reading, Mass., 1957. 31. E. C. Titchmarsh : " The Theory of Functions," 2d ed., Oxford University Press, Fair Lawn, N. J., 1939. 32. H. Weyl : " The Concept of a Riemann Surface," 3d ed., Addison-Wesley Publishing Company, Inc., Reading, Mass., 1964. 33. N. Wiener : " The Fourier Integral and Certain of Its Applications," Cambridge University Press, New York, 1933. Reprinted by Dover Publications, Inc., New York. 34. G. T. Whyburn : " Topological Analysis," 2d ed., Princeton University Press, Princeton, N. J., 1964. 35. J. H. Williamson : " Lebesgue Integration," Holt, Rinehart and Winston, Inc., New York, 1962. 36. A. Zygmund : " Trigonometric Series," 2d ed., Cambridge University Press, New York, 1959. Supplementary References 37. R. B. Burckel : "An Introduction to Classical Complex Analysis," Birkhauser Verlag, Basel, 1979. 38. P. L. Duren : " Theory of HP Spaces," Academic Press, New York, 1970. 39. T. W. Gamelin : " Uniform Algebras," Prentice-Hall, Englewood Cliffs, N. J., 1969. 40. J. B. Garnett : " Bounded Analytic Functions," Academic Press, New York, 1981. 41. M. de Guzman : " Differentiation of Integrals in R"," Lecture Notes in Mathematics 481, Springer Verlag, Berlin, 197 5. 42. T. Hawkins : " Lebesgue's Theory of Integration," University of Wisconsin Press, Madison, 1970. 43. H. Helson : "Harmonic Analysis," Addison-Wesley Publishing Company, Inc., Reading, Mass., 1983. 44. E. Hewitt and K. Stromberg : " Real and Abstract Analysis," Springer-Verlag, New York, 1965. 45. Y. Katznelson : "An Introduction to Harmonic Analysis," John Wiley and Sons, Inc., New York, 1968. 46. P. Koosis : " Lectures on H P Spaces," London Math. Soc. Lecture Notes 40, Cambridge Uni-versity Press, London, 1980. 47. R. Narasimhan : " Several Complex Variables," University of Chicago Press, Chicago, 1971. 48. J. C. Oxtoby : " Measure and Category," Springer-Verlag, New York, 1971. 49. W. Rudin : " Functional Analysis," McGraw-Hill Book Company, New York, 1973. 50. E. M. Stein : " Singular Integrals and Differentiability Properties of Functions," Princeton Uni-versity Press, Princeton, N. J., 1970. · 51. E. M. Stein and G. Weiss : " Introduction to Fourier Analysis on Euclidean Spaces," Princeton University Press, Princeton, N. J., 1971. 52. E. L. Stout : " The Theory of Uniform Algebras," Bogden and Quigley, Tarrytown-on-Hudson, 1971. 53. R. L. Wheeden and A. Zygmund : " Measure and Integral," Marcel Dekker Inc., New York, 1977. exp (z) 'l' 9Jl XE lim sup lim inf f +,f -Lt(Jl) a.e. E Cc(X) K --<f--< V IDlp m, mk A(T) L1(Rk), L1(E) II/IlP' 11/lloo I!'(Jl), l!'(Rk), {P Loo(Jl), Loo(Rk), too C0(X), C(X) (x, y) II X II X l_ y, Ml. LIST OF SPECIAL SYMBOLS AND ABBREVIA TIONSt 1 x(C() 82 8 T 88 8 I!'(T), C(T) 88 1 1 z 89 " 14 f(n) 91 14 Co 104 15 11/IIE 108 24 u 1 10 27 Pr(f) - t) 1 1 1 35 Lip C( 1 13 38 I 11 1 (E) 1 16 39 Jl+, Jl-1 19 42 A у Jl 120 51 Al l_ A2 120 51, 150 B(x, r) 136 53 Qr Jl 136 65, 66 DJ1 136, 241 65 MJl 136, 241 66 M f 138 70 AC 145 76 T'(x) 150 76 JT 150 79 BV 157 t The standard set-theoretic symbols are described on pages 6 to 8 and are not listed here. 407 408 LIST OF SPECIAL SYMBOLS AND ABBREVIATIONS Ex , £Y 161 P[dJl] 240 fx , fY 162 Q(% 240 Jl X A 164 Na 241 f g 171 Mrad 241 Jl A 175 (J 241 " f(t) 178 Hoo 248 coo 194 f(eit) 249 coo c 194 ({Ja(z) 254 D(a; r), D '(a; r), D(a; r) 196 s2 266 Q 197 !/ 285 H(Q) 197 Ep(z) 301 y, y 200 log+t 311 a A 202 N 31 1 Indy(z) 203 (fo , Do) "' (/1, D1) 323 Z(f) 208 HP 338 . + 217 Mf , Qf 344 I2 = I X I 222 u(x) 357 a, a 231 p(x) 360 A 232 C{Mn} 377 P[f] 233 P(K) 386 n + n -' 237 Cф(R2) 388 Absolute continuity, 120 of functions, 145 Absolute convergence, 1 16 Absolutely convergent Fourier series, 367 Addition formula, 1 Affine transformation, 377 Ahlfors, L. V ., 402 Algebra, 356 of measures, 17 5 of sets, 10 Algebraically closed field, 213 Almost everywhere, 27 Almost periodic function, 94 Almost uniform convergence, 214 Analytic continuation, 323, 377, 379 Analytic function, 197 Annulus, 229, 264, 292 Approach region, 240 Area, 250 Area theorem, 286 Arens, R., 398 Argument, 204 Arithmetic mean, 63 Arzela-Ascoli theorem, 245 Associative law, 18 Asymptotic value, 265 Austin, D., 399 Average, 30 Axiom of choice, 396 I Baire's theorem, 97 Balanced collection, 268 INDEX Ball, 9 Banach, S,, 105 Banach algebra, 190, 356 Banach space, 9 5 Banach-Steinhaus theorem, 98 Bessel's inequality, 85, 260 Beurling, A., 335, 402 Beurling's theorem, 348 Bieberbach conjecture, 401 Blaschke product, 3 10, 3 17, 318, 338, 353 Bledsoe, W. W., 399 Borel, E., 5, 402 Borel function, 12 Borel measure, 4 7 Borel set, 12 Borel transform, 402 Boundary, 108, 202, 289 Bounded function, 66 Bounded linear functional, 96, 1 13, 127, 130 Bounded linear transformation, 96 Bounded variation, 1 17, 148, 157 Box, 50 Brouwer's fixed point theorem, 151 Calderon, A. P ., 401, 402 Cancellation law, 19 Canonical product, 302 Cantor, D. G., 401 Cantor set, 58, 145 Caratheodory, C., 397, 400 , 401 Carleson, L., 398 Carrier, 58 409 410 INDEX Cartesian product, 7, 160 Category theorem, 98 Cater, F. S., 398 Cauchy, A., 400 Cauchy formula, 207, 219, 229, 268, 341, 384, 388 Cauchy-Riemann equations, 232 Cauchy sequence, 67 Cauchy theorem, 205, 206, 207, 218 Cauchy's estimates, 213 Cell, 50 Chain, 218 Chain rule, 197 Change of variables, 153, 156 Character, 179 Characteristic function, 1 1 Choice function, 396 Class, 6 Closed curve, 200 Closed graph theorem, 1 14 Closed path, 201 Closed set, 13, 35 Closed subspace, 78 Closure, 35 Cohen, P. J., 402 Collection, 6 Commutative algebra, 356 Commutative law, 18 Compact set, 35 Complement, 7 Complete measure, 28 Complete metric space, 67 Completion : of measure space, 28, 168 of metric space, 70, 74 Complex algebra, 356 Complex field, 213, 359 Complex homomorphism, 191, 362 Complex-linear functional, 105 Complex measure, 16, 1 16 Complex vector space, 33 Component, 197 Composite function, 7 Concentrated, 1 19 Conformal equivalence, 282 Conjugate exponents, 63 Conjugate function, 350 Connected set, 196 Continuity, 8, 9 Continuous function, 8 Continuous linear functional, 81, 96, 1 1 3, 127, 130 Continuous measure, 175 Continuum hypothesis, 167 Con vergence : dominated, 26 in IJ, 61 in measure, 7 4 monotone, 21 almost uniform, 214 uniform on compact subsets, 214 weak, 245 Convex function, 61 Convex sequence, 410 Convex set, 79 Convexity theorem, 257 Convolution, 170, 175, 178 Coset, 362 Cosine, 2, 265 Countable additivity, 6, 16 Counting measure, 17 Cover, 35, 324 Covering lemma, 137, 399 Curve, 200 with orthogonal increments, 94 Cycle, 218 Daniell, P. J ., 398 de Branges, L., 401 Denjoy, A., 144 Denjoy-Carleman theorem, 380 Dense set, 58 Derivative, 135, 197 of Fourier transform, 179 of function of bounded variation, 1 57 of integral, 141 of measure, 136, 142, 143, 241 symmetric, 1 36 of transformation, 1 50 Determinant, 54 Diagonal process, 246 Dieudonne, J., 398 Differentiable transformation, 150 Differential, 1 50 Direct continuation, 323 Dirichlet kernel, 101 Dirichlet problem, 235 Disc, 9, 196 Discrete measure, 17 5 Disjoint sets, 7 Distance function, 9 Distribution function, 172 Distributive law, 18 Division algebra, 360. Dixon, J. D., 400 Domain, 7 Dominated convergence theorem, 26, 29, 180 Double integral, 165 Dual space, 108, 1 12, 127, 130 Eberlein, W. F., 58 Edwards, R. E., 398 Egoroff's theorem, 73 Elementary factor, 301 Elementary set, 161 Ellipse, 287 Ellis, H. W., 399 Empty set, 6 End point, 200 Entire function, 198 Equicontinuous family, 245 Equivalence classes, 67 Equivalent paths, 201 Essential range, 7 4 Essential singularity, 21 1, 267 Essential supremum, 66 Euclidean space, 34, 49 Euclid's inequality, 77 Euler's identity, 2 Exponential function, 1, 198 Exponential type, 372, 382, 383 Extended real line, 7, 9 Extension, 105 Extensiop theorem, 105 Extremal function, 255 Extreme point, 25 1 Fa-set, 12 Factorization, 303, 338, 344 Family, 6 Fatou, P., 400 Fatou's lemma, 23, 68, 309, 344 Fatou's theorem, 249 Fejer kernel, 252 Fejer's theorem, 91 Field, 394 Finite additivity, 17 First category, 98 Fixed point, 151, 229, 247, 293, 297, 314, 318, 395 Ford, L. R., 40 1 Fourier coefficients, 82, 91 of measure, 1 33 Fourier series, 83, 91 Fourier transform, 178 Fubini's theorem, 164, 168 Function, 7 absolutely continuous., 145 analytic, 197 Borel, 1 3 bounded, 66 of bounded variation, 148 complex, 8 continuous, 8 convex, 61 entire, 198 essentially bounded, 66 exponential, 1 of exponential type, 372, 383 harmonic, 232 holomorphic, 197 Lebesgue integrable, 24 left-continuous, 157 locally integrable, 194 lower semicontinuous, 37 measurable, 8, 29 meromorphic, 224 modular, 328 nowhere differentiable, 1 14 rational, 267 real, 8 simple, 15 subharmonic, 335 summable, 24 upper semicontinuous, 37 Function element, 323 Functional : bounded, 96 on C0 , 1 30 complex-linear, 105 continuous, 96 on Hilbert space, 81 on I!, 127 linear, 33 multiplicative, 191, 364 positive, 34 real-linear, 105 Functional analysis, 108 Fundamental domain, 329 INDEX 41 1 Fundamental theorem of calculus, 144, 148 G"-set, 12 Gap series, 276, 32 1, 334, 354, 385 Gelfand, I., 402 Gelfand-Mazur theorem, 359 Gelfand transform, 370 Geometric mean, 63 Goursat, E., 400 Graph, 1 14, 174 Greatest lower bound, 7 Green's theorem, 389 Hadamard, J., 264, 321 Hahn-Banach theorem, 104, 107, 270, 313, 359 Hahn decomposition, 126 Hall, D., 398 412 INDEX Halmos, P. R., 398, 403 Hardy, G. H., 173, 335, 399 Hardy's inequality, 72, 177 Harmonic conjugate, 350 Harmonic function, 232 Harmonic majorant, 352 Harnack's theorem, 236, 250 Hausdorff, F., 12, 400 Hausdorff maximality theorem, 87, 107, 195, 362, 396 Hausdorff separation axiom, 36 Hausdorff space, 36 Hausdorff-Young theorem, 261 Heine-Borel theorem, 36 Helson, H., 348 Herglotz, G., 400 Hilbert cube, 92 Hilbert space, 77 isomorphism, 86 Hoffman, K., 402 Holder's inequality, 63, 66 Holomorphic function, 197 Homomorphism, 179, 191, 364, 402 Homotopy, 222 Hunt, R. A., 398 Ideal, 175, 305, 362 Image, 7 Independent set, 82 Index : of curve, 223, 230 of cycle, 218 of path, 203 Infimum, 7 Infinite product, 298 Initial point, 200 Inner factor, 344 Inner function, 342 Inner product, 76, 89 Inner regular set, 4 7 Integral, 19, 24 Integration, 19 over cycle, 218 of derivative, 146, 148, 149 over measurable set, 20 by parts, 157 over path, 200 with respect to complex measure, 129 Interior, 267 Interpolation, 173, 260, 304 Intersection, 6 Interval, 7 Invariant subspace, 188, 346 Inverse image, 7 Inverse mapping, 7, 215 Inversion, 280 Inversion formula, 181 Inversion theorem, 185, 186 Invertible element, 357 Isolated singularity, 210, 266 Isometry, 84 Isomorphism, 86 Iterated integral, 165 Jacobian, 150, 250 Jensen's formula, 307 Jensen's inequality, 62 Jordan, C., 5 Jordan curve, 291 Jordan decomposition, 1 19 Kahane, J. P., 401 Kaufman, R. P., 399, 437 Kirk, R. B., 398 Koebe function, 285, 401 Laplace equation, 232 Laplacian, 19 5 Laurent series, 230 LeaŸ upper bound, 7 Lebesgue, H. J., 5, 21, 397, 399 Lebesgue decomposition, 121 Lebesgue integrable function, 24 Lebesgue integral, 19 Lebesgue measurable set, 51 Lebesgue measure, 51 Lebesgue point, 138, 159, 241 Left-continuous function, 1 57 Left-hand limit, 157 Length, 159, 202 Limit, pointwise, 14 in mean, 67 of measurable functions, 14 in measure, 7 4 Lindelof's theorem, 259 Linear combination, 82 Linear fractional transformation, 279, 296 Linear independence, 82 Linear transformation, 33 Linearly ordered set, 87 Liouville's theorem, 212, 220, 359 Lipschitz condition, 1 1 3 Littlewood, J. E., 173, 399 Locally compact space, 36 Locally integrable function, 194 Logarithm, 227, 274 Lowdenslager, D., 348 Lower half plane, 237 Lower limit, 14 Lower semicontinuous function, 37 Lusin's theorem, 55 Mandelbrojt, S., 402 Mapping, 7 continuous, 8 one-to-one, 7 open, 99, 214 (See also Function) Marcinkiewicz, J ., 173 Maximal function, 136, 138, 241 Maximal ideal, 362, 364 Maximal orthonormal set, 85 Maximal subalgebra, 366 Maximality theorem, 87, 396 Maximum modulus theorem, 1 10, 212 Mean value property, 237 Measurable function, 8, 29 Measurable set, 8 Measurable space, 8 Measure, 17 absolutely continuous, 120 Borel, 47 complete, 28 complex, 1 6 continuous, 175 counting, 18 discrete, 1 7 5 Lebesgue, 51 positive, 16 real, 16 regular, 47 representing, 109 a-finite, 4 7 signed, 1 19 singular, 120 translation-invariant, 51 Measure space, 16 Mergelyan's theorem, 390, 394 Meromorphic function, 224, 304 Metric, 9 Metric density, 141 Metric space, 9 Minkowski's inequalitŹ, 63, 1 77 Mirkil, H., 402 Mittag-Leffier theorem, 273 Modular function, 328 Modular group, 328 Monodromy theorem, 327 Monotone class, 160 Monotone convergence theorem, 21 Monotonicity, 17, 42 Morera's theorem, 208 Moschovakis, Y. N., 401 Multiplication operator, 1 14, 34 7 Multiplicative inequality, 356 Multiplicative linear functional, 364 Multiplicity of a zero, 209 Miintz-Szasz theorem, 313, 318 Natural boundary, 320, 330 Negative part, 15 Negative variation, 1 19 Neighborhood, 9, 35 Neumann, J. von, 122, 398, 399 Nevanlinna, R., 31 1 INDEX 413 Nicely shrinking sequence, 140 Nonmeasurable set, 53, 1 57, 167 Nontangential approach region, 240 Nontangential limit, 243, 340 Nontangential maximal function, 241, 340 Norm, 65, 76, 95, 96 Norm-preserving extension, 106 Normal family, 28 1 Normed algebra, 356 Normed linear space, 95 Novinger, W. P., 398 Nowhere dense, 98 Nowhere differentiable function, 1 14 Null-homotopic curve, 222 Null space, 362 One-to-one mapping, 7 One-parameter family, 222, 326 Onto, 7 Open ball, 9 Open cover, 35 Open mapping theorem, 99, 214, 216 Open set, 8 Opposite path, 201 Orbit, 317 Order : of entire function, 3 1 5 of pole, 210 of zero, 209 Ordinal, 59 Ordinate set, 174, 399 Oriented interval, 202 Orthogonal projection, 80 Orthogonality, 79 Orthogonality relations, 82 Orthonormal basis, 85 Orthonormal set, 82 Ostrowski, A., 321 Outer factor, 344 Outer function, 342 Outer measure, 397 414 INDEX Outer regular set, 47 Overconvergence, 321 Paley-Wiener theorems, 372, 375 Parallelogram law, 80 Parameter interval, 200 Parseval's identity, 85, 91, 187, 212 Partial derivative, 23 1 Partial fractions, 267 Partial product, 298 Partial sum of Fourier series, 83, 91, 101, 354 Partially ordered set, 86 Partition : of set, 1 16 of unity, 40 Path, 200 Periodic function, 2, 88, 93, 1 56 Perron, 0., 144 Phragmen-Lindelof method, 256 n, 3 Picard theorem, 332 Plancherel theorem, 186 Plancherel transform, 186 Pointwise limit, 14 Poisson integral, 1 1 2, 233, 240, 252 Poisson kernel, 1 1 1, 233 Poisson summation formula, 195 Polar coordinates, 175 Polar representation of measure, 125 Pole, 210, 267 Polynomial, 1 10 Positive linear functional, 34, 40, 109 Positive measure, 16 Positive part, 15 Positive variation, 1 19 Positively oriented circle, 202 Power series, 198 Pre-image, 7 Preservation of angles, 278 Prime end, 401 Principal ideal, 305 Principal part, 2 1 1 Product measure, 164 Projection, 80 Proper subset, 6 Punctured disc, 196 Quasi-analytic class, 378 Quotient algebra, 363 Quotient norm, 363 Quotient space, 363 Rademacher functions, 158 Radial limit, 239 Radial maximal function, 241 Radical, 369 Radius of convergence, 198 Radon-Nikodym derivative, 122, 140 Radon-Nikodym theorem, 121 Rado's theorem, 263 Ramey, W., 400 Range, 7 Rational function, 267, 276 Real line, 7 Real-linear functional, 105 Real measure, 1 6 Rectangle, 160 Reflection principle, 237 Region, 197 Regular Borel measure, 4 7 Regular point, 319 Removable set, 333 Removable singularity, 210 Representable by power series, 198 Representat:on theorems, 40, 81, 130 Representing measure, 109, 398 Residue, 224 Residue theorem, 224 Resolvent, 369 Restriction, 109 Rickert, N. W., 399 Riemann integral, 5, 34 Riemann-Lebesgue lemma, 103 Riemann mapping theorem, 283, 295 Riemann sphere, 266 Riesz, F., 34, 341, 397, 400 Riesz, M., 341, 350, 401 Riesz-Fischer theorem, 85, 91 Riesz representation theorem, 34, 40, 130 Right-hand derivative, 399 Root test, 198 Rotation, 279 Rotation-invariance, 51, 195 Rouche's theorem, 225, 229 Rubel, L. A., 401 Runge's theorem, 270, 272, 387 Saks, S., 397 Scalar, 33 Scalar product, 76 Schwartz, J. T ., 399 Schwarz, H. A., 400 Schwarz inequality, 49, 63, 77 Schwarz lemma, 254 Schwarz reflection principle, 237 Second category, 98 Section, 161 Segment, 7 Separable space, 92, 247 Set, 6 Borel, 13 closed, 13, 35 compact, 36 connected, 196 convex, 79 dense, 58 elementary, 161 empty, 6 Fa , 12 G6 , 12 inner regular, 47 measurable, 8, 51 nonmeasurable, 53, 157, 1 67 open, 7 outer regular, 47 partially ordered, 86 strictly convex, 1 12 totally disconnected, 58 totally ordered, 87 Shift operator, 347 Sierpinski, W., 167, 399 a-algebra, 8 a-compact set, 47 a-finite measure, 47 a-ring, 397 Signed measure, 1 19 Simple boundary point, 289 Simple function, 15 Simply connected, 222, 274 Sine, 2, 265, 3 16 Singer, I. M., 398, 402 Singular measure, 120 Singular point, 3 19 Snow, D. 0., 399 Space : Banach, 95 compact, 35 complete metric, 67 dual, 108, 1 12, 127, 1 30 Hausdorff, 36 Hilbert, 77 inner product, 76 locally compact, 36 measurable, 8 metric, 9 normed linear, 95 separable, 92 topological, 8 unitary, 76 vector, 33 Span, 82 Spectral norm, 360 Spectral radius, 360 Spectrum, 357 Square root, 274 Stoilov's theorem, 400 Stout, E. L., 401 Strictly convex set, 1 12 Stromberg, K., 399 Subadditivity, 397 Subchain, 395 Subharmonic function, 335 Subset, 6 Subspace, 78 Sum of paths, 217 Summability method, 1 14 Summable function, 24 Supremum, 7 Supremum norm, 70 Support, 38, 58 Symmetric derivative, 136 Szasz, 0 ., 401 Tauberian theorem, 402 Taylor's formula, 379 Thorin, G. 0., 401 Three-circle theorem, 264 Tietze's extension theorem, 389 Topological space, 8 Topology, 8 Total variation, 1 17, 148 Totally disconnected set, 58 Totally ordered set, 87 Tower, 395 Transcendental number, 170 Transformation : affine, 377 bounded linear, 96 differentiable, 150 linear, 33 linear fractional, 279, 296 (See also Function) Transitivity, 324 Translate : of function, 182 of set, 50 Translation-invariance, 51 Translation-invariant measure, 51 Translation-invariant subspace, 188 Triangle, 202 Triangle inequality, 9, 49, 77 Trigonometric polynomial, 88 Trigonometric system, 89 Ullrich, D., 400 Uniform boundedness principle, 98 INDEX 415 416 INDEX Uniform continuity, 51 Uniform integrability, 133 Union, 6 Unit, 357 Unit ball, 96 Unit circle, 2 Unit disc, 1 10 Unit mass, 17 Unit vector, 96 Unitary space, 7 6 Upper half plane, 237 Upper limit, 14 Upper semicontinuous function, 37 Urysohn's lemma, 39 Vanish at infinity, 70 Varberg, D. E., 399 Vector space, 33 Villani, A., 398 Vitali-Caratheodory theorem, 56 Vitali's theorem, 133 Volume, 50 von Neumann, J., 122 Walker, P. L., 399 Weak convergence, 245, 246, 25 1 Weak L1, 138 Weierstrass, K., 301, 332 Weierstrass approximation theorem, 312, 387 Weierstrass factorization theorem, 303 Wermer, J., 402 Wiener, N., 367 Winding number, 204 Young, W. H., 5, 400 Zermelo, E., 403 Zero set, 209 Zorn's lemma, 87 Zygmund, A., 401 |
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Binomial Expansion
This page details the more advanced use of binomial expansion. You should be familiar with all of the material from the more basic Binomial Expansion page first.
Recall that the first formula provided in the Edexcel formula booklet is:
for and where . However, this formula is only valid for positive integer .
In addition to this, the booklet also provides a second formula for negative and fractional powers:
The first formula is only valid for positive integer but this formula is valid for all . This includes negative and fractional powers. Note, however, the formula is not valid for all values of . As stated, the values must be between -1 and 1.
Range of Validity for Binomial Expansions
As stated above, the second formula for binomial expansion in the Edexcel Formula Booklet is only valid for . This is because, unlike for positive integer , these expansions have an infinite number of terms (as indicted by the … in the formula). Subsequently, we require the series to converge as the powers of become large. For this to happen, we must have . Questions may ask you to find the binomial expansion without explicitly stating the value of and ask you to identify the values for which the expansion is valid. See the Identifying the Power Example.
Also notice that in this second formula there is a very specific format inside the brackets – it must be 1 plus something. Therefore, if there is something other than 1 inside these brackets, the coefficient must be factored out. Do this by first writing . Then find the expansion of using the formula. Do this by replacing all with . This inevitably changes the range of validity. It follows that this expansion will be valid for or . See the Factoring Out Example.
At more advanced levels, questions may ask you to use partial fractions first. See the Using Partial Fractions question.
Examples
Identifying the Power
Find the first four terms in ascending powers of of the binomial expansion of .
State the range of validity for your expansion.
Solution:
Firstly, write the expression as . The power is negative and so we must use the second formula. We can then find the expansion by setting and replacing all with :
We find the range of validity by replacing with in the expression to give . It follows that the expansion is valid for . We can also write this as .
Factoring out
Find the first three terms, in ascending powers of , of the expansion of . State the range of validity of your expansion and use it to find an approximation to .
Solution:
Begin by writing the expression as . The power is fractional so we must use the second formula. However, this first requires us to remove a factor of 4:
Note that the expansion of is given by:
Hence, multiplying by the factor of gives:
This expansion is valid for , that is .
Finally, by setting , we can find an approximation to :
to 4 decimal places. Note that we can do this since . The exact value of to 4 decimal places, which is a reasonable approximation.
Using Partial Fractions
Express as partial fractions.
Show that the quadratic approximation to is given by .
State the range of values of for which this approximation is valid.
Solution:
Firstly, revise partial fractions by clicking here. Note that . Hence, and . It follows that and and so .
From part 1, we can write . Now, take binomial expansions for each term: Hence,
as given.
The first expansion is valid for (or ) while the second expansion is valid for (or . In order for both of these inequalities to be satisfied, we must have .
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4992 | https://www.aafp.org/pubs/afp/issues/2020/0101/p34.html | Lymphoma: Diagnosis and Treatment
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WILLIAM D. LEWIS, MD, SETH LILLY, PharmD, BCPS, AND KRISTIN L. JONES, PA-C
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Am Fam Physician. 2020;101(1):34-41
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Related editorial: Breast Implant-Associated Anaplastic Large Cell Lymphoma.
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Patient information: See related handout on lymphoma, written by the authors of this article.
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Author disclosure: No relevant financial affiliations.
Lymphoma is a group of malignant neoplasms of lymphocytes with more than 90 subtypes. It is traditionally classified broadly as non-Hodgkin or Hodgkin lymphoma. Approximately 82,000 new U.S. patients are diagnosed with lymphoma annually. Any tobacco use and obesity are major modifiable risk factors, with genetic, infectious, and inflammatory etiologies also contributing. Lymphoma typically presents as painless adenopathy, with systemic symptoms of fever, unexplained weight loss, and night sweats occurring in more advanced stages of the disease. An open lymph node biopsy is preferred for diagnosis. The Lugano classification system incorporates symptoms and the extent of the disease as shown on positron emission tomography/computed tomography to stage lymphoma, which is then used to determine treatment. Chemotherapy treatment plans differ between the main subtypes of lymphoma. Non-Hodgkin lymphoma is treated with CHOP (cyclophosphamide, doxorubicin, vincristine, and prednisone) with or without rituximab (R-CHOP), bendamustine, and lenalidomide. Hodgkin lymphoma is treated with combined chemotherapy with ABVD (doxorubicin, bleomycin, vinblastine, and dacarbazine), Stanford V (a chemotherapy regimen consisting of mechlorethamine, doxorubicin, vinblastine, vincristine, bleomycin, etoposide, and prednisone), or BEACOPP (bleomycin, etoposide, doxorubicin, cyclophosphamide, vincristine, procarbazine, and prednisone) with radiotherapy. Subsequent chemotherapy toxicities include neuropathy, cardiotoxicity, and secondary cancers such as lung and breast, and should be considered in the shared decision-making process to select a treatment regimen. Once remission is achieved, patients need routine surveillance to monitor for complications and relapse, in addition to age-appropriate screenings recommended by the U.S. Preventive Services Task Force. Patients should receive a 13-valent pneumococcal conjugate vaccine followed by a 23-valent pneumococcal polysaccharide vaccine at least eight weeks later with additional age-appropriate vaccinations because lymphoma is an immunosuppressive condition. Household contacts should also be current with their immunizations.
Lymphoma represents a heterogeneous group of malignant neoplasms of lymphocytes, which can involve lymphatic tissue, bone marrow, or extranodal sites. The World Health Organization’s classification system identifies more than 90 different subtypes (Table 1).1,2 The initial stratification is derived from B-cell, T-cell, or natural killer cell origin. Further classification of distinct lymphoma subtypes is beyond the scope of this article; however, they are ultimately each defined by morphology, immunopheno-type, genetic, molecular, and clinical features.1,3 This article will focus on the types of lymphoma traditionally classified as non-Hodgkin or Hodgkin.
SORT: KEY RECOMMENDATIONS FOR PRACTICE
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| Clinical recommendation | Evidence rating | Comments |
---
| Open lymph node biopsy should be used to definitively diagnose lymphoma.14,15 | C | Expert opinion and clinical review articles |
| Positron emission tomography/computed tomography should be used to determine the staging of the lymphoma.19 | C | Expert opinion and clinical review article |
| Patients with lymphoma should have intensive follow-up surveillance for the first two years following remission.40 | C | Expert opinion and clinical review article |
| A 13-valent pneumococcal conjugate vaccine (Prevnar 13), followed by a 23-valent pneumococcal polysaccharide vaccine (Pneumovax 23) at least eight weeks later and then again at least five years later, should be administered following lymphoma treatment.44,45 | C | Expert opinion and guidelines |
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, go to
TABLE 1.
Common Lymphoma Subtypes with Incidence and Five-Year Survival
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| Lymphoma subtype | Incidence per 100,000 | Five-year survival |
---
| Hodgkin | 2.8 | 85.7% |
| Non-Hodgkin B-cell lymphomas | | |
| Burkitt | 0.4 | 64.1% |
| Diffuse large B cell | 7.2 | 63.2% |
| Follicular | 3.5 | 88.4% |
| Marginal zone | 2.2 | 90.3% |
| Precursor B cell | 1.5 | 68.9% |
| Non-Hodgkin T-cell and natural killer cell lymphomas | | |
| Mycosis fungoides | 0.6 | 90.9% |
| Peripheral T-cell | 1.2 | 58.4% |
Information from references 1 and 2.
Epidemiology
More than 82,000 new patients are projected to be diagnosed with lymphoma in 2019, representing 4.7% of all new cancer cases in the United States. The current five-year survival rate for non-Hodgkin lymphoma is 72.0%, and for Hodgkin lymphoma it is 86.6%. Almost 21,000 people are projected to die from lymphoma in 2019, representing 3.5% of all cancer deaths. Incidence of non-Hodgkin lymphoma is higher in men and whites, and it increases with age. The median age of patients at diagnosis of non-Hodgkin lymphoma is 67 years, and the median age at death is 76. Hodgkin lymphoma is most commonly diagnosed at 20 to 34 years of age; however, the median age at death is 68 because of the higher survival rate among younger patients.2,4
Risk Factors
Genetic, infectious, and inflammatory etiologies increase the risk of lymphoma. First-degree relatives of patients with non-Hodgkin lymphoma and Hodgkin lymphoma have a respective 1.7-fold and 3.1-fold increased risk of developing lymphoma. A family history of a specific subtype of lymphoma is associated with developing that same subtype.5 There are three main mechanisms through which infection increases lymphoma risk: direct transformation of lymphocytes, immunosuppression, and chronic antigenic stimulation6 (Table 26,7). Rheumatoid arthritis, systemic lupus erythematosus, Sjögren syndrome, dermatomyositis, and celiac disease are inflammatory conditions that increase the risk of lymphoma through disease-specific causes and the chronic use of immunosuppressive medications.8
TABLE 2.
Lymphoma-Related Infections
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| Mechanism | Infection | Lymphoma type |
---
| Direct lymphocyte transformation | Epstein-Barr virus | Burkitt, non-Hodgkin, Hodgkin |
| Human T-lymphotropic virus type 1 | T-cell leukemia |
| Immunosuppression | HIV | Hodgkin, non-Hodgkin |
| Chronic antigenic stimulation | Helicobacter pylori Chlamydia psittaci Campylobacter jejuni Campylobacter coli Borrelia burgdorferi | Non-Hodgkin (mucosa-associated lymphoid tissue) |
| Hepatitis C | Splenic marginal zone |
Information from references 6 and 7.
Modifiable risk factors include current or former tobacco use9 and obesity (body mass index of 30 kg per m2 or higher).10 Breast implants and long-term pesticide exposure have also been associated with non-Hodgkin lymphoma.11–13
Clinical Presentation
Lymphoma commonly presents as painless adenopathy. Adenopathy can wax and wane over years in indolent presentations or involve rapidly progressive adenopathy in more aggressive subtypes. Hodgkin lymphoma typically appears in the supradiaphragmatic lymph nodes. Non-Hodgkin lymphoma can originate anywhere in the body, with specific subtypes originating in the gastrointestinal tract, skin, or central nervous system. Systemic symptoms of fever, unexplained weight loss, and night sweats occur in a subset of patients with more advanced disease. Lymphoma spreads to extranodal sites by direct invasion or by hematogenous spread to the spleen, liver, lungs, or bone marrow.14,15 High-grade lymphomas can present as oncologic emergencies because of the structural compression from the enlarging tumor, including superior vena cava syndrome, malignant epidural spinal cord compression, or malignant pericardial effusion.16 Paraneoplastic syndromes are rare with lymphoma, occurring as paraneoplastic cerebellar degeneration in Hodgkin lymphoma and as dermatomyositis and polymyositis in Hodgkin and non-Hodgkin lymphomas.17
Diagnosis
The diagnosis of lymphoma is made using an open lymph node biopsy, based off morphology, immunohistochemistry, and flow cytometry.3 Although fine-needle aspiration and core needle biopsy are often part of the initial evaluation of any adenopathy, neither will provide adequate tissue for the diagnosis of lymphoma because of the need to verify Hodgkin lymphoma via the presence of Reed-Sternberg cells.15,18
Staging
The Ann Arbor staging system was initially developed in 1971 for Hodgkin lymphoma, and was later adapted for non-Hodgkin lymphoma. The Lugano classification system further modified staging by incorporating positron emission tomography/computed tomography (PET-CT) results to determine the staging of the lymphoma (Table 319). PET-CT is used for fluorodeoxyglucose-avid lymphoma subtypes, with symptoms alone being used for staging the remaining subtypes. The new staging system incorporates two symptom-based classifications: A (absence of symptoms) and B (presence of fever, weight loss, and night sweats) for Hodgkin lymphoma. A bone marrow biopsy is now recommended only for diffuse large B-cell lymphoma with a negative PET-CT result.19
TABLE 3.
Lugano Classification for Staging Lymphoma
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| Stage | Description of disease from positron emission tomography/computed tomography results |
--- |
| I | Single nodal group or single extralymphatic lesion |
| II† | Multiple nodal groups on same side of diaphragm or with limited contiguous extralymphatic involvement |
| III | Multiple nodal groups on both sides of the diaphragm; may involve the spleen |
| IV | Noncontiguous extralymphatic involvement |
—Staging for Hodgkin lymphoma is further subdivided for systemic symptoms; A for absence of symptoms or B for fevers > 101.3°F (38.5°C), drenching night sweats, or 10% (of body weight) unintentional weight loss over the past six months.
†—Stage II may also be classified as bulky disease (> 10-cm mass), which may be treated as limited or advanced disease based on several prognostic factors.
Adapted with permission from Cheson BD, Fisher RI, Barrington SF, et al.; Alliance, Australasian Leukaemia and Lymphoma Group; Eastern Cooperative Oncology Group; European Mantle Cell Lymphoma Consortium, et al. Recommendations for initial evaluation, staging, and response assessment of Hodgkin and non- Hodgkin lymphoma: the Lugano classification. J Clin Oncol. 2014;32(27):3062.
Prognosis
The International Prognostic Index is used broadly for all subtypes of non-Hodgkin lymphoma, and the International Prognostic Score is used for Hodgkin lymphoma20,21 (Table 422,23).
TABLE 4.
Comparison of Prognostic Indices in Lymphoma
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| Non-Hodgkin | Hodgkin |
--- |
| International Prognostic Index | International Prognostic Score |
| Criteria | Criteria |
| Age > 60 | Age > 45 |
| Elevated serum lactate dehydrogenase | Male sex |
| Eastern Cooperative Oncology | Serum albumin concentration < 4.0 g per dL (40 g per L) |
| Group performance status ≥ 2 | Hemoglobin concentration < 10.5 g per dL (105 g per L) |
| Ann Arbor stage III or IV disease† |
| Extranodal sites > 1 | Ann Arbor stage IV disease† Leukocytosis (≥ 15,000 μL [15 × 109 white blood cells per L]) Lymphopenia (< 600 lymphocytes per μL [0.6 × 109 per L], or < 8% of total white blood cell count) |
| Total score____ | Total score____ |
| Five-year overall survival rate based on number of criteria from International Prognostic Index/Score | |
| Score 0 or 1 = 73% | Score 0 = 89% |
| Score 2 = 51% | Score 1 = 90% |
| Score 3 = 43% | Score 2 = 81% |
| Score 4 or 5 = 26% | Score 3 = 78% |
| | Score 4 = 61% |
| | Score ≥ 5 = 56% |
Note: Each criterion = 1 point.
—Eastern Cooperative Oncology Group performance status: 0 = fully active, 1 = ambulatory but restricted to light work, 2 = ambulatory but unable to carry out activities, 3 = limited self-care only, 4 = bedridden, 5 = dead.
†—Ann Arbor stage III = multiple nodal groups on both sides of the diaphragm, may involve the spleen; stage IV = noncontiguous extralymphatic involvement.
Information from references 22 and 23.
Treatment
Treatment of lymphoma consists of chemotherapy alone or in combination with radiotherapy.24 Radiotherapy alone is not recommended.25 Toxicity from radiotherapy can lead to serious long-term complications such as secondary cancers in the irradiated area, including breast or lung cancers.25 Additionally, patients receiving chemotherapy can subsequently develop breast or lung cancers, melanoma, or acute myeloid leukemia.26,27 Patients who are older than 60 years at diagnosis have worse outcomes, regardless of the staging. The National Comprehensive Cancer Network (NCCN) recommends avoiding certain chemotherapeutic agents in patients older than 60 years. The physician should focus on shared decision-making when discussing treatment options with all patients, but particularly for those older than 60 years, including whether the patient should pursue treatment.25
The standard treatment for Hodgkin lymphoma is ABVD (doxorubicin [Adriamycin], bleomycin, vinblastine [Velban], and dacarbazine), but other regimens such as the Stanford V (doxorubicin, vinblastine, mechlorethamine, etoposide [Toposar], vincristine, bleomycin, and prednisone) and escalated-BEACOPP (bleomycin, etoposide, doxorubicin, cyclophosphamide, vincristine, procarbazine [Matulane], and prednisone) can be used.24–28 Treatment for non-Hodgkin lymphoma varies depending on the histology, but often uses treatments such as CHOP (cyclophosphamide, doxorubicin, vincristine, and prednisone) with or without rituximab (Rituxan; R-CHOP), a monoclonal antibody specific for CD20-positive B lymphocytes.29 Other medications such as bendamustine (Bendeka), an alkylating agent, and lenalidomide (Revlimid) are also used in many non-Hodgkin lymphoma treatments.30,31 Common complications of these therapies are listed in Table 5.25–27,29–36
TABLE 5.
Common Chemotherapy Regimens and Complications
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| Therapy | Regimen | Short-term complications | Long-term complications |
--- --- |
| Hodgkin lymphoma | | | |
| ABVD | Doxorubicin (Adriamycin) | Nausea/vomiting Alopecia Neutropenia Neuropathy Bleomycin-induced pulmonary toxicity | Cardiotoxicity (heart failure) Neuropathy Pulmonary fibrosis Increased risk of myocardial infarction |
| Bleomycin |
| Vinblastine (Velban) |
| Dacarbazine |
| Stanford V | Doxorubicin | Nausea/vomiting Fatigue Pulmonary toxicity Neuropathy | Neuropathy Pulmonary fibrosis Cardiotoxicity Rarely solid secondary malignancies of breast, lung, and skin |
| Vinblastine |
| Mechlorethamine |
| Etoposide (Toposar) |
| Vincristine |
| Bleomycin |
| Prednisone |
| Escalated-BEACOPP | Bleomycin | Anemia Leukopenia Thrombocytopenia Nausea/vomiting Infection Disulfiram reaction between ethanol and procarbazine | Acute myeloid leukemia Sterility/infertility |
| Etoposide |
| Doxorubicin (Adriamycin) |
| Cyclophosphamide |
| Vincristine (Oncovin) |
| Procarbazine (Matulane) |
| Prednisone |
| Non-Hodgkin lymphoma | | | |
| CHOP | Cyclophosphamide | Heart failure Constipation Hyperglycemia Neuropathy | Cardiomyopathy Myelosuppression Neuropathy |
| Doxorubicin |
| (Hydroxydaunorubicin) |
| Vincristine (Oncovin) |
| Prednisone |
| R-CHOP | Rituximab (Rituxan) + CHOP | Reactivate hepatitis B infection | Progressive multifocal leukoencephalopathy |
Information from references 25–27, and 29–36.
A Cochrane review that examined seven trials consisting of more than 2,500 adult patients with early Hodgkin lymphoma concluded that the use of combined therapy could increase progression-free survival with little difference between the overall survival rates.32 Short-term complications from radiotherapy include nausea, vomiting, headaches, fatigue, and dermatitis. Radiotherapy can also lead to long-term complications, including cardiac and pulmonary toxicity, hypothyroidism, or breast or lung cancers.24–32 Radiotherapy can be avoided in patients with stage IA or IIA lymphoma without bulky disease25 (Table 319).
Interim Reassessment
PET-CT scans, and subsequent Deauville scoring (Table 621), should be used to assess the response to chemotherapy in non-Hodgkin and Hodgkin lymphoma.25,30,31,33 A score of 3 or less is considered complete remission in non-Hodgkin lymphoma and should conclude the current treatment course. A score of 4 or 5 is an indicator to consider escalating therapy.25 Patients with Hodgkin lymphoma with a Deauville score of 1 or 2 have been shown to have similar progression and mortality outcomes between radiotherapy and no further treatment.32 Patients who receive a score of 3 or 4 should receive additional chemotherapy and/or radiotherapy, and a score of 5 indicates the need for a biopsy (excisional or core needle) in addition to chemotherapy and radiotherapy.25 A positive biopsy should be considered refractory disease.25
TABLE 6.
Deauville Score for Assessing PET-CT Scans
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| PET-CT finding | Score |
--- |
| No FDG uptake related to lymphoma | 1 |
| FDG uptake at lymphoma site is ≤ mediastinum FDG uptake | 2 |
| FDG uptake at lymphoma site is > mediastinum FDG uptake but < liver FDG uptake | 3 |
| FDG uptake at lymphoma site is > liver FDG uptake at any site | 4 |
| FDG uptake at lymphoma site is substantially > liver FDG uptake or new FDG uptake sites found | 5 |
FDG = fluorodeoxyglucose; PET-CT = positron emission tomography/computed tomography.
Adapted with permission from Armitage JO, Gascoyne RD, Lunning MA, et al. Non-Hodgkin lymphoma. Lancet . 2017;390(10091):302.
Relapse
Relapse rates for non-Hodgkin lymphoma are variable and based on the specific subtype. The most common subtype, diffuse large B-cell lymphoma, has a 40% lifetime relapse rate.37 Lifetime relapse in Hodgkin lymphoma occurs in 10% to 15% of patients with early stage disease and 40% of patients with advanced stage disease.38
Surveillance
Patients who have achieved remission need routine surveillance to monitor for complications and relapse, as well as age-appropriate screenings recommended by the U.S. Preventive Services Task Force.39 Complications of lymphoma treatment include secondary malignancies (e.g., breast, lung, skin, colon), cardiac disease, infertility, and endocrine, neurologic, and psychiatric dysfunctions. Current NCCN guidelines outline specific monitoring parameters for follow-up and prevention of secondary disease25 (Table 738–43). The extent and frequency of follow-up specifically depend on the histologic subtype of lymphoma. Patients should follow up with an oncologist every three to six months for the first two years, every six to 12 months until year 3, then annually thereafter. After five years of being cancer free, the patient can be transitioned to a primary care physician.40
TABLE 7.
Lymphoma Surveillance for up to Five Years Posttreatment
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| Cancer screening | Laboratory screening | Cardiac screening | Counseling | Immunizations |
--- ---
| Breast: annual screening mammography starting at age 40; history of chest or axilla radiation: start eight to 10 years after treatment or at age 40, whichever comes first; consider annual breast magnetic resonance imaging if chest radiation was received between ages 10 and 30; consider referral to breast subspecialist to discuss possible chemoprevention | Complete blood count, fasting blood glucose, and comprehensive metabolic panel annually | Annual blood pressure screening, lifestyle modification, and treatment of obesity, hypertension, and tobacco use | Annual depression screening | Age-appropriate immunizations per the Centers for Disease Control and Prevention schedule, including annual influenza vaccine; resume live vaccines at least three months after completion of chemotherapy |
| Routine surveillance tests for cervical, colorectal, lung, and prostate cancers per the USPSTF guidelines | Lipid profile per the USPSTF guidelines | Consider stress test and/or echocardiography at 10-year intervals (frequency of testing based on findings and other associated risk factors) | Neurocognitive impairment screening for any patient who is high risk (e.g., history of brain radiation or intrathecal treatment) | PCV13 (Prevnar 13), followed by PPSV23 (Pneumovax 23) at least eight weeks later and again at least five years later |
| | Thyroid-stimulating hormone annually if neck irradiation | Carotid ultrasonography every 10 years if neck irradiation | Infertility: consider reproductive endocrinologist referral | Haemophilus influenzae type b: three doses following hematopoietic stem cell transplantation |
PCV13 = 13-valent pneumococcal conjugate vaccine; PPSV23 = 23-valent pneumococcal polysaccharide vaccine; USPSTF = U.S. Preventive Services Task Force.
Information from references 38–43.
If a patient is asymptomatic, routine surveillance imaging does not improve outcomes or provide a clinical benefit.40,41 Surveillance imaging should be used in patients who have reported symptoms or who are at high risk of relapse in a place that would not be easily examined, and who would be candidates for treatment. However, NCCN imaging guidelines for lymphoma surveillance state that it is acceptable to perform chest radiography or CT of the chest every six to 12 months for the first two years and then yearly for the next three to five years posttreatment.41 Surveillance imaging with PET-CT scans following complete remission is not recommended.40,41 Disease marker research is ongoing, examining minimal residual disease measurements, a polymerase chain reaction–based method that looks at identifying tumor-specific DNA sequences.41
Immunizations
All patients with lymphoma should receive pneumococcal vaccination initially with a 13-valent pneumococcal conjugate vaccine (Prevnar 13), followed at least eight weeks later by a 23-valent pneumococcal polysaccharide vaccine (PPSV23; Pneumovax 23) and then another PPSV23 at least five years later.44 Patients receiving anti–B-cell antibodies should not receive annual influenza vaccination, and administration of live vaccines is contraindicated during chemotherapy. Routine vaccinations recommended by the Centers for Disease Control and Prevention (CDC) should resume, including any recommended inactivated or live vaccines three months after chemotherapy or six months after anti–B-cell antibody therapy.43,45 Patients receiving a hematopoietic stem cell transplant should receive a series of three doses of Haemophilus influenzae type b vaccine starting six to 12 months after a successful transplant. Household contacts should receive appropriate CDC-recommended immunizations.43
This article updates a previous article on this topic by Glass.46
Data Sources: A PubMed search was completed using combinations of the key terms lymphoma, non-Hodgkin, Hodgkin, presentation, diagnosis, staging, treatment, and follow up. The search included meta-analyses, randomized controlled trials, clinical trials, and reviews. Search dates: April 18, May 17, and May 31, 2018, and August 30, 2019. We also searched the Agency for Healthcare Research and Quality evidence reports, UpToDate, the Cochrane database, Essential Evidence Plus, the National Comprehensive Cancer Network, and the Surveillance, Epidemiology, and End Results database. Search dates: April 18, 2018, and August 30, 2019.
Research reported in this article was supported by the National Institute of General Medical Sciences of the National Institutes of Health under award number 5U54GM104942-03. The content is solely the responsibility of the authors and does not necessarily represent the official views of the National Institutes of Health.
WILLIAM D. LEWIS, MD, FAAFP, is an associate professor in the Department of Family Medicine at West Virginia University Eastern Division and the West Virginia Clinical and Translational Science Institute, Harpers Ferry, and is codirector of the West Virginia Practice-Based Research Network, Morgantown.
SETH LILLY, PharmD, BCPS, is an assistant professor of clinical pharmacy at West Virginia University Eastern Division.
KRISTIN L. JONES, PA-C, is a physician assistant in the Department of Family Medicine at West Virginia University Eastern Division.
Address correspondence to William D. Lewis, MD, West Virginia University, 171 Taylor St., Harpers Ferry, WV 25425 (email: lewisw@wvumedicine.org). Reprints are not available from the authors.
Author disclosure: No relevant financial affiliations.
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Edwards-Bennett SM, Jacks LM, Moskowitz CH, et al. Stanford V program for locally extensive and advanced Hodgkin lymphoma: the Memorial Sloan-Kettering Cancer Center experience. Ann Oncol. 2010;21(3):574-581.
Swerdlow AJ, Higgins CD, Smith P, et al. Second cancer risk after chemotherapy for Hodgkin’s lymphoma: a collaborative British cohort study. J Clin Oncol. 2011;29(31):4096-4104.
Filippi AR, Levis M, Parikh R, et al. Optimal therapy for early-stage Hodgkin’s lymphoma: risk adapting, response adapting, and role of radiotherapy. Curr Oncol Rep. 2017;19(5):34.
Pfreundschuh M, Trümper L, Osterborg A, et al.; MabThera International Trial Group. CHOP-like chemotherapy plus rituximab versus CHOP-like chemotherapy alone in young patients with good-prognosis diffuse large-B-cell lymphoma: a randomised controlled trial by the MabThera International Trial (MInT) Group. Lancet Oncol. 2006;7(5):379-391.
National Comprehensive Cancer Network. NCCN guidelines & clinical resources. B-cell lymphomas guideline. Accessed June 16, 2018.
National Comprehensive Cancer Network. NCCN guidelines & clinical resources. T-cell lymphoma. Accessed June 16, 2018.
Blank O, von Tresckow B, Monsef I, et al. Chemotherapy alone versus chemotherapy plus radiotherapy for adults with early stage Hodgkin lymphoma. Cochrane Database Syst Rev. 2017(4):CD007110.
Van Heertum RL, Scarimbolo R, Wolodzko JG, et al. Lugano 2014 criteria for assessing FDG-PET/CT in lymphoma: an operational approach for clinical trials. Drug Des Devel Ther. 2017;11:1719-1728.
Radford J, Illidge T, Counsell N, et al. Results of a trial of PET-directed therapy for early-stage Hodgkin’s lymphoma. N Engl J Med. 2015;372(17):1598-1607.
van Nimwegen FA, Ntentas G, Darby SC, et al. Risk of heart failure in survivors of Hodgkin lymphoma: effects of cardiac exposure to radiation and anthracyclines. Blood. 2017;129(16):2257-2265.
Conway JL, Connors JM, Tyldesley S, et al. Secondary breast cancer risk by radiation volume in women with Hodgkin lymphoma. Int J Radiat Oncol Biol Phys. 2017;97(1):35-41.
Sarkozy C, Sehn LH. Management of relapsed/refractory DLBCL. Best Pract Res Clin Haematol. 2018;31(3):209-216.
Bröckelmann PJ, Goergen H, Kohnhorst C, et al. Late relapse of classical Hodgkin lymphoma: an analysis of the German Hodgkin study group HD7 to HD12 trials. J Clin Oncol. 2017;35(13):1444-1450.
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In simple terms, what does it mean to parameterize something in mathematics, particularly calculus?
·
In mathematics, particularly in calculus, to parameterize something means to express a mathematical object, like a curve or surface, using one or more parameters. This involves defining the object in terms of variables that can take on different values.
For example:
Curves: Instead of describing a curve using a traditional equation (like y=f(x)), you might use a parameter t. A curve in 2D can be parameterized as:
x(t)=t
y(t)=t2
Here, t is the parameter that varies, and as t changes, the coordinates (x(t),y(t)) trace out the curve.
Surfaces: Similarly, a surface in 3D can be parameterized usi
In mathematics, particularly in calculus, to parameterize something means to express a mathematical object, like a curve or surface, using one or more parameters. This involves defining the object in terms of variables that can take on different values.
For example:
Curves: Instead of describing a curve using a traditional equation (like y=f(x)), you might use a parameter t. A curve in 2D can be parameterized as:
x(t)=t
y(t)=t2
Here, t is the parameter that varies, and as t changes, the coordinates (x(t),y(t)) trace out the curve.
Surfaces: Similarly, a surface in 3D can be parameterized using two parameters, say u and v:
x(u,v)=u
y(u,v)=v
z(u,v)=u2+v2
This approach allows for more flexibility in describing complex shapes and is particularly useful for integration, differentiation, and analyzing geometric properties.
Leo Cutter
Studied Mathematics at Bishop Gorman High School (Graduated 2020)
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Author has 356 answers and 4.1M answer views
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Updated 6y
To parametrize something in mathematics means to write (usually a non-bijective function) in terms of what is called a parameter. So here’s an example of that:
This right here is the equation for a parabola:
y=−x2+5x−1
And when we parameterize this parabola, we get
x=4−2t
y=3+6t−4t2
And we can verify that these parametric equations, when plotted, will produce the same graph, and it’s done by eliminating the parameter. So basically solving for t and substituting it into either equation until it’s gone. So we’ll take this parametric equation:
x=4−2t
And solve for t.
[math]t = [/math]
A2A
To parametrize something in mathematics means to write (usually a non-bijective function) in terms of what is called a parameter. So here’s an example of that:
This right here is the equation for a parabola:
y=−x2+5x−1
And when we parameterize this parabola, we get
x=4−2t
y=3+6t−4t2
And we can verify that these parametric equations, when plotted, will produce the same graph, and it’s done by eliminating the parameter. So basically solving for t and substituting it into either equation until it’s gone. So we’ll take this parametric equation:
x=4−2t
And solve for t.
t=12(4−x)
When we substitute this value of t back into the explicit solution for y, then we see that we end up with the same graph.
There are two reasons that come to mind as to why parametric equations are so useful.
First, they are useful for making functions out of non-functions. Let me ask you a question, is the following a function?
x2+y2=r2(1)
The correct answer here would be no, because it does not pass the “vertical line test”. Now, are the following equations functions?
x=cost(2)
y=sint(2)
They both are functions, because they pass the “vertical line test”.
Yet we see that, when plotted, both (1) and (2) produce the same graph (Discounting direction of motion for those who know their stuff). So we turned an equation into a set of two different functions.
Another reason why parametrics are so helpful is because they bring to the surface a potential underlying variable. Take the following example that I got from Parametric Equations:
Suppose that for every degree Fahrenheit above 70, a convenience store sells s bottles of sunscreen and s2 pints of ice cream. This implies that this algebraic relationship for predicting the amount of sunscreen or ice cream sold would be in order:
s=s2
But this would be incorrect, because this is only holds true for temperatures at or above 70 degrees Fahrenheit. So then it would be more correct to say the following:
s=t−70
s2=(t−70)2
Where t is temperature in Fahrenheit. This would give a much more accurate representation of what’s actually going on.
Howard Landman
BA Math (honors, UC Berkeley), top 100 Putnam, 2 published papers
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John George
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Let’s take a simple example: the unit circle.
A non-parameterized equation for the unit circle is x2+y2=1. This tells us that the circle consists of all the points that are distance 1 from the point (0,0).
A parameterized description of the same circle might be x=cos(θ),y=sin(θ). This says that the circle consists of points whose x and y coordinates are both functions of a single parameter θ.
One advantage of the parameterized description is that it makes it easy to see that the circle is a 1-dimensional curve. Changing the parameter, θ, moves you along the curve.
Let’s take a simple example: the unit circle.
A non-parameterized equation for the unit circle is x2+y2=1. This tells us that the circle consists of all the points that are distance 1 from the point (0,0).
A parameterized description of the same circle might be x=cos(θ),y=sin(θ). This says that the circle consists of points whose x and y coordinates are both functions of a single parameter θ.
One advantage of the parameterized description is that it makes it easy to see that the circle is a 1-dimensional curve. Changing the parameter, θ, moves you along the curve.
Parameterizing also may make it a little easier to compute slopes (tangent lines) of a curve at a given point (x,y). (For the special case of the unit circle, this is easy, it’s just -x/y, but for arbitrary curves it can be more painful.) Just compute the derivatives of x and y with respect to θ, then divide to get dy/dx=dy/dθdx/dθ. For the circle, that would be:
dydθ=cos(θ)=x
dxdθ=−sin(θ)=−y
dydx=−xy
To get the same result directly from the circle equation, you either need to rewrite y explicitly as a function of x and then take the derivative of the resulting function, and then worry about which square root to take:
y=±√1−x2=(1−x2)12
dydx=12(1−x2)−12(−2x)
dydx=(y2)−12(−x)=−x|y|
or (cleaner) do implicit differentiation:
ddx(x2+y2)=ddx1=0
2x+2ydydx=0
2ydydx=−2x
dydx=−2x2y=−xy
which is not horribly hard but maybe a little non-obvious.
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Wes Hansen
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Parametrization in Calculus I is just preparation for paramaterization in Vector or Multi-Variable Calculus. In Vector Calculus you start with an interval, most often thought of as an interval of time but in general it is your parameter interval, i.e. it contains all values which your parameter can possibly take. You then define a parametrized curve on this interval by assigning to each element of the interval, i. e. each unit of time, a vector.
For example, in physics we often wish to encode information about a physical system in a mathematical model; we can do this with a parametrized curve o
Parametrization in Calculus I is just preparation for paramaterization in Vector or Multi-Variable Calculus. In Vector Calculus you start with an interval, most often thought of as an interval of time but in general it is your parameter interval, i.e. it contains all values which your parameter can possibly take. You then define a parametrized curve on this interval by assigning to each element of the interval, i. e. each unit of time, a vector.
For example, in physics we often wish to encode information about a physical system in a mathematical model; we can do this with a parametrized curve on a time interval - time is our parameter. The dynamical systems which interest us have freedom of movement to various degrees, they can move up/down, left/right, or forward/backward, for example; each type of freedom of movement is called a “degree of freedom” and each “degree of freedom” can be described as a function of time - our parameter. So now, to model our system we construct what’s called a position vector (this terminology comes from Classical Mechanics - Newtonian/Maxwellian physics), which is basically an n-tuple, where n corresponds to the number of “degrees of freedom” our system has, and each element of our n-tuple is described as a function of our time interval, i.e. parametrized. If each function in our position vector has a derivative, then we can define a velocity vector for our system. And if each function in our velocity vector has a derivative, then we can define an acceleration vector for our system. This enables us to encode a great deal of information about our system in our model and we can use the tools of Calculus and Vector Algebra to investigate it. The parametrization you deal with in Calculus I is simply this same exact phenomena but constrained to two-dimensional vectors, i. e. ordered pairs or coordinates in Cartesian two-space.
So, in symbolism, your position vector is:
X(t)=(x1(t),x2(t),...,xn(t))
where each xi(t) is a function describing the movement of one “degree of freedom” of your system. Your system can be anything, from a bug crawling on a flat surface - two-dimensional (Calc I), to an entire ecological system, n-dimensional (Vector Calc). Of course then, your velocity vector is simply:
X′(t)=(x′1(t),x′2(t),...,x′n(t))
and your acceleration vector is simply:
X′′(t)=(x′′1(t),x′′2(t),...,x′′n(t))
I hope this provides a bit of clarity for you! If you’re studying Calculus, you might appreciate Paul Dawkins’ Calculus Notes (click on Downloads). He also has notes for College Algebra and Differential Equations.
Bob Chamberlain
Eagle Scout, Caltech grad, 53 yrs at JPL (ret), agnostic
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Author has 389 answers and 409.3K answer views
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6y
Related
In plain English, what does it mean to “parameterize” a model?
Q: In plain English, what does it mean to “parameterize” a model?
Mathematical models express the relationships between the things you have chosen to model. The relationships and the things are represented by numerical values. Those numerical values can be in the model either as fixed numbers (constants) or as “parameters” that you can vary when you “run” the model. Expressing the values that interest you as variables, rather than as constants, is called parameterizing the model.
For example, a model of the distance traveled by a cannon ball in a vacuum may include the mass of the ball, its init
Q: In plain English, what does it mean to “parameterize” a model?
Mathematical models express the relationships between the things you have chosen to model. The relationships and the things are represented by numerical values. Those numerical values can be in the model either as fixed numbers (constants) or as “parameters” that you can vary when you “run” the model. Expressing the values that interest you as variables, rather than as constants, is called parameterizing the model.
For example, a model of the distance traveled by a cannon ball in a vacuum may include the mass of the ball, its initial height, speed, and launch angle. It would also include the acceleration of gravity. If all of these were represented as numbers, the model could compute an estimate of the distance. But if, say, the acceleration of gravity were to be parameterized, the model could be used on the moons of Jupiter just as well as on our own moon.
If the ball’s ballistic coefficient and the characteristics of the wind were included in the model, the assumption of a vacuum could be dropped. If they are parameterized, the cannon ball could be an artillery shell and the model could be applied at various places on Earth. In this case, it might be well to parameterize the coordinates of the artillery piece as well so the model can account for the Coriolis force. Including the radius and rotational speed of the planet would allow application on other planets.
The more complicated the input to the model — that is, the more the model is parameterized — the more important it is to have good default values for the parameters so that the users of the model can vary only a few of them without having to conduct their own investigation into appropriate values to use for the parameters in which they are not interested.
Of course, adding additional parameters to a model generally means that not only the applicability, but the fidelity, precision, and accuracy of the model is improved. Whether any of that happens depends upon the skill and diligence of the modeler.
Mathematical modeling is the art of science.
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Tim Farage
Professor, Mathematics and Computer Science, Retired
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Author has 4.8K answers and 17.6M answer views
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What is the use of factorials in calculus or calculus-like problems (if any)?
You’d be surprised about how often factorials are used in calculus, as well as in other areas of mathematics.
For example, you all know about the sine function. In its simplest form, the sine of an angle in a right triangle is the ratio of the opposite side of the angle divided by the hypotenuse. You can actually draw a right triangle with any given angle, measure the side opposite of the anle and divide by the hypotenuse, and you will have computed the sine of the angle.
All is well. But this is an approximation. Is there a way to find the sine of an angle to any degree of accuracy as you’d lik
You’d be surprised about how often factorials are used in calculus, as well as in other areas of mathematics.
For example, you all know about the sine function. In its simplest form, the sine of an angle in a right triangle is the ratio of the opposite side of the angle divided by the hypotenuse. You can actually draw a right triangle with any given angle, measure the side opposite of the anle and divide by the hypotenuse, and you will have computed the sine of the angle.
All is well. But this is an approximation. Is there a way to find the sine of an angle to any degree of accuracy as you’d like?
Yup. It’s called the Taylor Series for sin(x). Here it is:
sin(x)=x−x33!+x55!−x77!+...
(Here, x must be in radians, not degrees).
This formula is ridiculous. What do factorials have to do with the sine function? You’ll need some Calculus II to find out.
Let’s go in the opposite direction. Calculating n! for large values of n can quickly cause your calculator to complain, and turn off just to spite you.
But look at the formula for approximating n!:
n!≈√2πn×(ne)n
What the heck?! How did π get into the picture? Once again, you’ll need Calculus II to answer that. Sorry.
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Tim Farage
Professor, Mathematics and Computer Science, Retired
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Author has 4.8K answers and 17.6M answer views
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In a simple way, what is a bijection in mathematics?
A bijection in mathematics is a relationship between two sets. It is also called a one-to-one correspondence.
The good news is that it is very easy to understand. There is a bijection between two sets if the elements from both sets can be paired off (without leaving out any elements).
Here is a picture of a bijection between two sets:
And there you have it.
A less abstract example would be if four opposite-sex married couples went out to dinner together. Then the set of four husbands could be paired with the set of four wives.
Bijections can be confusing when dealing with infinite sets.
For instance
A bijection in mathematics is a relationship between two sets. It is also called a one-to-one correspondence.
The good news is that it is very easy to understand. There is a bijection between two sets if the elements from both sets can be paired off (without leaving out any elements).
Here is a picture of a bijection between two sets:
And there you have it.
A less abstract example would be if four opposite-sex married couples went out to dinner together. Then the set of four husbands could be paired with the set of four wives.
Bijections can be confusing when dealing with infinite sets.
For instance, consider the set of Natural Numbers, N = { 0, 1, 2, 3, 4, … }.
And now consider the multiples of ten, T = { 0, 10, 20, 30, 40, … }.
Clearly T is a proper subset of N because every multiple of 10 is a Natural Number. Yet a whole bunch Natural Numbers that are not multiples of ten.
And yet there is a simple bijection between them:
0 <-> 0
1 <-> 10
2 <-> 20
3 <-> 30
4 <-> 40
etc.
Thus, these sets have the same size (technically called the same cardinality).
Are there infinite sets that are not the same size? Yup, but you didn’t ask about that.
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If you put your dirty clothes into a washing machine and turn it on, it will give them as clean clothes. Washing machine performs a function. It takes input as dirty clothes and produces output as clean clothes.
Washing machine can only process textile materials like clothes, kerchiefs, sheets etc., It can't take input as chicken soup or alluvial soil. Thus the domain of washing machine function is the textile materials. It produces only clean textiles. Thus the range of the washing machine is the clean textiles.
If you put your friend's dirty clothes into the washing machine, it will give frien
If you put your dirty clothes into a washing machine and turn it on, it will give them as clean clothes. Washing machine performs a function. It takes input as dirty clothes and produces output as clean clothes.
Washing machine can only process textile materials like clothes, kerchiefs, sheets etc., It can't take input as chicken soup or alluvial soil. Thus the domain of washing machine function is the textile materials. It produces only clean textiles. Thus the range of the washing machine is the clean textiles.
If you put your friend's dirty clothes into the washing machine, it will give friend's clean clothes and not your's.
Similarly, f(x) is a function in which x can be anything in its domain. It will produce exactly one same output in its range for each x.
For example, if f(x) is the washing machine function,
f(dirty clothes) = clean clothes
Now, see a mathematical function.
f(x) = x+3
Which means if x is 6, f(x) will always be none other than 9.
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Consider a line segment connecting two points (x1, y1) and (x2, y2). Now push x2 closer to x1 (and take y1 with it), so close you can’t see the separation. What’s the slope now? Imagine pushing x2 a million times closer to x1. Now, what is the slope?
In order to understand the details of this, you will need to k
Start with Zeno’s paradox. You can do it yourself. Just walk half way across a room, then walk half that distance, then half of that. Will you ever cross the room? The Greeks couldn’t handle this paradox, but Newton and Leibniz finally did, by explaining the concept of the infinitesimal.
Consider a line segment connecting two points (x1, y1) and (x2, y2). Now push x2 closer to x1 (and take y1 with it), so close you can’t see the separation. What’s the slope now? Imagine pushing x2 a million times closer to x1. Now, what is the slope?
In order to understand the details of this, you will need to know functions, but that’s still algebra.
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I earn money from home
Gaming, watching videos, and doing simple online tasks has earned me over $5,000 for this.
Whenever I tell people that I make money just being online from my couch, they usually laugh or roll their eyes and say something like:
“Come on, get a real job”
But the truth is, you can earn a decent side income without being a streamer or expert. If you know your way around puzzle games, casual apps, and even just watching videos, platforms like Freecash make it ridiculously easy to turn free time into real cash.
Why Freecash beats most other side hustles
I’ve tried other side gigs like reselling, freelancing, even dropshipping…and they were always the same story:
Time-consuming (20+ hours a week)
High effort for low return
Not flexible around my actual life
Freecash, on the other hand, flips that script. I can earn $300–$350 a week by spending around 30 minutes to an hour a day. Usually while relaxing, watching YouTube, or chatting on Discord.
The best part? It’s not just about gaming.
Sure, I like games. But I’m not grinding for hours on end. A lot of my earnings come from watching ads, signing up for free trials, and doing simple tasks while watching videos or binging Netflix.
Here’s the kind of stuff I’ve made money from on Freecash:
Puzzle games: Things like Bubble Pop and Woody Sort keep me engaged during shows, and I rack up daily bonuses effortlessly.
Strategy & tycoon games: Like Raid: Shadow Legends or Bank Tycoon. I even bought a $5 in-game pack once and got rewarded with $30!
Watching videos: Yes, seriously—you can earn by watching short clips, ads, or promotional content. Easy to do when you’re already on YouTube.
Surveys: I’ll knock out a few while watching sports. Takes zero brainpower and can add $10–$20 a day if you’re consistent.
Product sign-ups: I got paid to try services like Chime and Amazon Prime. No catch—I still use some of them today.
My typical routine
I spend a chunk of my downtime either watching Twitch, YouTube, or rewatching old shows, and Freecash fits right into that. It doesn’t replace a full-time job (unless you really grind), but it absolutely pays for gas, groceries, and the occasional DoorDash splurge.
I average $50 a day on Freecash, and that’s without stressing myself out.
Getting started is stupid easy:
Sign up to Freecash with Google, email, or Facebook (I used Google)
Grab a $5 welcome bonus instantly
Pick what you like: play, answer, watch, or sign up
Cash out easily (crypto, PayPal, gift cards, etc.)
Bonus tip: micro-purchases in some games give 5x or 6x the value back—I once got $30 for a $5 Monopoly GO pack.
I’ve earned over $1,000 in a month just stacking bonuses, climbing leaderboards, and doing tasks while watching content or playing casually. No selling, no shipping, no Zoom calls.
If you’re someone who:
Works a 9–5 and wants something light on the side
Is in between jobs
Spends time online watching or gaming anyway
Then honestly, Freecash might be your best low-effort option. It’s flexible, fun, and actually rewarding.
👉 Try Freecash out. Worst case? You made a few extra bucks. Best case? You just found your favorite new side hustle.
Andrew Winkler
PhD in Mathematics, Courant Institute of Mathematical Sciences, NYU (Graduated 1987)
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Author has 9.3K answers and 10.4M answer views
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2y
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What is an intutive explanation or meaning of limits in calculus? What does limits mean in simple words?
You can think of limits as a process of elimination.
If x is not the limit of a sequence, some distance around it excludes infinitely many elements of the sequence. And if the sequence does have a limit, in fact it can be chosen to exclude all but a finite number.
So if a limit exists, it is the unique point which does not get excluded. This is one way of setting to rest the qualm that a sequence “never arrives”. It doesn’t have to. It just has to uniquely describe it as the thing that doesn’t get excluded.
A better way to think of it is in terms of continuity. L is the limit of f at x if the fun
You can think of limits as a process of elimination.
If x is not the limit of a sequence, some distance around it excludes infinitely many elements of the sequence. And if the sequence does have a limit, in fact it can be chosen to exclude all but a finite number.
So if a limit exists, it is the unique point which does not get excluded. This is one way of setting to rest the qualm that a sequence “never arrives”. It doesn’t have to. It just has to uniquely describe it as the thing that doesn’t get excluded.
A better way to think of it is in terms of continuity. L is the limit of f at x if the function defined to be L at x but agreeing with f elsewhere is continuous.
The easiest way to think about continuity, in turn, is in terms of tolerances. You want f(a). But maybe you can’t exactly obtain a. And maybe if the cost is small, you don’t care. Maybe any y near f(x) is good enough. Can you impose a tolerance on x so that even if x isn’t a, just being within that tolerance of a will guarantee that f(x) is within your error cost budget? If so, for any positive cost, f is continuous at a.
The cost allowed is usually represented by epsilon, and the tolerance is represented by delta.
To make it concrete think of f as the direction your tires point as a function of the angle of the steering wheel. You don’t care exactly where your tires are pointing . You do care a lot that you can get as close to you need to by getting the steering wheel close enough to the ideal position.
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
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Author has 6.8K answers and 52.3M answer views
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2y
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What is pure mathematics in simple terms? What does pure mean in that phrase?
I like this clever saying:
APPLIED MATHEMATICS is a “PROBLEM" looking for a “SOLUTION”.
PURE MATHEMATICS is a “SOLUTION" looking for a “PROBLEM we can apply it to”.
The idea can be explained in very elementary terms:
Suppose some problem ends up in having to solve a basic quadratic equation.
All that matters at this point is finding the 2 solutions and deciding which one applies to the problem. This is APPLIED MATHS.
But PURE MATHS is considering all types of quadratic equations and the different types of solutions that can occur (i.e. 2 rational solutions, 1 rational solution, 2 irrational so
I like this clever saying:
APPLIED MATHEMATICS is a “PROBLEM" looking for a “SOLUTION”.
PURE MATHEMATICS is a “SOLUTION" looking for a “PROBLEM we can apply it to”.
The idea can be explained in very elementary terms:
Suppose some problem ends up in having to solve a basic quadratic equation.
All that matters at this point is finding the 2 solutions and deciding which one applies to the problem. This is APPLIED MATHS.
But PURE MATHS is considering all types of quadratic equations and the different types of solutions that can occur (i.e. 2 rational solutions, 1 rational solution, 2 irrational solutions, 2 complex solutions) totally FOR ITS OWN SAKE and because it is interesting!
Woody Weaver
Author has 662 answers and 385.4K answer views
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1y
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What is bijection in mathematics explained simply?
A bijection between two things is a pairing up of objects so that they are evenly matched with nothing left over.
Suppose a goatherd and a shepherd are trying to fairly trade animals, but really can’t count beyond three or so. Sheep are better than goats, so they want to exchange 2 for 1. They have a bunch of sheep and goats penned up, how do they make the trade? If it were one of us, it would be easy to say “you have 15 sheep, let me count out 30 goats, and we are done.” They can’t count that high. But what they can do is take two goats from one pen, walk them over to the sheep, swap two goats
A bijection between two things is a pairing up of objects so that they are evenly matched with nothing left over.
Suppose a goatherd and a shepherd are trying to fairly trade animals, but really can’t count beyond three or so. Sheep are better than goats, so they want to exchange 2 for 1. They have a bunch of sheep and goats penned up, how do they make the trade? If it were one of us, it would be easy to say “you have 15 sheep, let me count out 30 goats, and we are done.” They can’t count that high. But what they can do is take two goats from one pen, walk them over to the sheep, swap two goats for one sheep, and walk the sheep back to the other pen. They do it until one side runs out, and they’ve done a fair trade without having to count past two.
What they did was a bijection between pairs of goats and sheep. One pair of goats to one sheep, until one side is exhausted (and if we exhaust both at the same time, it’s a true bijection.) It’s a mapping that shows two things have the same counts — 30 goats (15 pairs) is the same as 15 sheep.
That concept becomes more powerful when the two groups are large. For example, we have the set of all integers, and the set of all even integers. You and I both know they are infinite in size. Which one is the “bigger” infinity? That’s where the bijection comes into play. We can associate an even integer 2n with the integer n [think of ordered pairs (2n, n) ] — that association pairs everything up: obviously every even integer is of the form 2n, so it covers everything, and the pairing is unique, the integer 2n is paired with n and nothing else. So like with the sheep and goats, somehow they have the same “size”. So that means the integers is the same “size” as the even integers — a set is the same size as a proper subset of itself — and that’s a pretty good definition of integers, actually.
What will probably sound even stranger is that the set of decimal expansion of numbers between [0,1] is bigger than the integers — they are both infinite, but the former is “more infinite”? That is, suppose you could write down a pairing (n, r) where the n’s are the integers and the r’s are the real numbers. So any real number has the form 0.d1d2d3…. You’ve written down a pairing of the (n,r), but I’m going to claim you missed an r — consider the number generated by taking the di such that di is 7, unless the i’th digit in di is 7, in which case you pick 3. So my r=0.777377777373777… or whatever. I’m going to claim it’s not in your list. If so, then it’s (n,r) for some n — but look at the nth digit of r. What can it be? If the nth digit of r is 7, I was supposed to pick 3, so that’s not the case. But if the nth digit of r isn’t 7, it was supposed to be 7! That’s a contradiction, when you look at the nth place, something is wrong, so my special r can’t be on your list in the first place! Thus there is no bijection between integers and reals, the reals are truely “bigger”!
It’s a really rich concept. You could look at the wiki page,
[Bijection - Wikipedia
One-to-one correspondence "One-to-one correspondence" redirects here. For one-to-one function, see injective function . A bijective function, f : X → Y , where set X is {1, 2, 3, 4} and set Y is {A, B, C, D}. For example, f (1) = D. In mathematics , a bijection , bijective function , or one-to-one correspondence is a function between two sets such that each element of the second set (the codomain ) is the image of exactly one element of the first set (the domain ). Equivalently, a bijection is a relation between two sets such that each element of either set is paired with exactly one element of the other set. A function is bijective if it is invertible ; that is, a function f : X → Y {\displaystyle f:X\to Y} is bijective if and only if there is a function g : Y → X , {\displaystyle g:Y\to X,} the inverse of f , such that each of the two ways for composing the two functions produces an identity function : g ( f ( x ) ) = x {\displaystyle g(f(x))=x} for each x {\displaystyle x} in X {\displaystyle X} and f ( g ( y ) ) = y {\displaystyle f(g(y))=y} for each y {\displaystyle y} in Y . {\displaystyle Y.} For example, the multiplication by two defines a bijection from the integers to the even numbers , which has the division by two as its inverse function. A function is bijective if and only if it is both injective (or one-to-one )—meaning that each element in the codomain is mapped from at most one element of the domain—and surjective (or onto )—meaning that each element of the codomain is mapped from at least one element of the domain. The term one-to-one correspondence must not be confused with one-to-one function , which means injective but not necessarily surjective. The elementary operation of counting establishes a bijection from some finite set to the first natural numbers (1, 2, 3, ...) , up to the number of elements in the counted set. It results that two finite sets have the same number of elements if and only if there exists a bijection between them. More generally, two sets are said to have the same cardinal number if there exists a bijection between them. A bijective function from a set to itself is also called a permutation , [ 1 ] and the set of all permutations of a set forms its symmetric group . Some bijections with further properties have received specific names, which include automorphisms , isomorphisms , homeomorphisms , diffeomorphisms , permutation groups , and most geometric transformations . Galois correspondences are bijections between sets of mathematical objects of apparently very different nature. For a binary relation pairing elements of set X with elements of set Y to be a bijection, four properties must hold: each element of X must be paired with at least one element of Y , no element of X may be paired with more than one element of Y , each element of Y must be paired with at least one element of X , and no element of Y may be paired with more than one element of X . Satisfying properties (1) and (2) means that a pair]( "en.wikipedia.org")
That picture there, mapping (1,D), (2,B), (3,C), and (4,A) is the heart of what a bijection is. And it generalizes, quite powerfully.
Cyrus Liu
Systems and Program Languages Researcher
·
4y
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In plain English, what does it mean to “parameterize” a model?
Assuming you have a model to calculate the area of a rectangle, all you need to do is input the length and the width your rectangle, this black box system would output the result for you. Now, let’s parameterize it, why and how? Well what if I want to calculate the area of a triangle, circle, pentagon? So we need to parameterize the model with a parameter shape, each time I ask this black box, besides the relevant length, I also tell it what kind of the shape I’m asking for the answer. So parameterize a model usually means you can have wider input, which the parameter is an abstract variable
Assuming you have a model to calculate the area of a rectangle, all you need to do is input the length and the width your rectangle, this black box system would output the result for you. Now, let’s parameterize it, why and how? Well what if I want to calculate the area of a triangle, circle, pentagon? So we need to parameterize the model with a parameter shape, each time I ask this black box, besides the relevant length, I also tell it what kind of the shape I’m asking for the answer. So parameterize a model usually means you can have wider input, which the parameter is an abstract variable, each time you ask answer, this parameter can be instantiated to different concrete objects. Like shape is the variable, when you actually need to calculate the area of a shape, it can be a rectangle, triangle, pentagon etc..
Related questions
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ϵ
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4994 | https://goldbook.iupac.org/terms/view/A00208 | IUPAC - aldehydes (A00208)
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Version 5.0.0 (12318 Terms)
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aldehydes
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Compounds RC(=O)H, in which a carbonyl group is bonded to one hydrogen atom and to one R group.
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PAC, 1995, 67, 1307. (Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995)) on page 1312 [Terms] [Paper]
See also:
PAC, 1990, 62, 2167. (Glossary of atmospheric chemistry terms (Recommendations 1990)) on page 2173 [Terms] [Paper]
Citation: 'aldehydes' in IUPAC Compendium of Chemical Terminology, 5th ed. International Union of Pure and Applied Chemistry; 2025. Online version 5.0.0, 2025. RISBibTexEndNote
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SAT Math Section: Distance Rate – SAT Practice Questions | Video Post
Molly Kiefer
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Looking for a little help with the SAT Math section?
In this video, Magoosh’s SAT expert Chris explains the Distance-Rate Formula!
You don’t have to be scared of these question types anymore, once you’ve got the Distance-Rate Formula in your pocket!
Watch the embedded video below, or scroll down for a full video transcript. 🙂
What Will I See in the “SAT Math Section: Distance Rate | SAT Practice Questions” Video?
In this 8-minute video, Magoosh’s SAT expert Chris will take you, step-by-step, through three SAT practice questions, explaining how you can identify whether a question calls for the Distance-Rate Formula and how to use the formula to solve even the hardest Distance-Rate problems!
If you liked this video, hit that Like button–or better yet, send it to a friend! Let’s all go to college 🎓🙌
“SAT Math Section: Distance Rate | SAT Practice Questions” Full Transcript
Hi, this is Chris, the SAT expert at Magoosh.
I’ve had over 15 years’ experience helping students ace the SAT, and today, I’m going to talk about the distance rate formula.Why do you have to know this?
Well, there’s a certain question type, a word problem about moving vehicles going in different directions, that freaks a lot of people out.
But this formula can make things a lot easier.
So here’s a distance problem, we have Stephen cycling at a constant rate of 10 miles per hour, and Gertrude cycling at a constant rate of 15 miles per hour.
So we know people are moving at a certain rate for a certain amount of time, and that’s how we know we’re in the distance formula world.
So we’re gonna come back to this question in a second, and we’re gonna go straight to that distance formula.
We have distance = rate x time.
Now, we’re gonna put this to work with us, with that very problem we just saw, but I’m gonna show you an easy way to remember this.
See how D stands for distance, Rate stands for R at the beginning, and Time starts with T, so there it is?
And we’re gonna make that, and simplify that into D = RT, that’s gonna be your magic formula.
You always wanna keep that in mind, but of course, know what those things stand for.
Okay, so let’s take a look back here at Stephen and Gertrude.
And not only am I going to give you that question, but I’m gonna give you some answer choices.
You can try this on your own if you want, but let’s do it together now.
So Stephen cycles at a constant rate of 10 miles an hour for two hours, and then stops, so let’s use our formula.
W know the rate, that’s the speed, so that’s 10 miles an hour for Stephen, let’s get that right there.
And then he goes for how long, for two hours.
So we know that if he’s going at this rate ,10 times 2, we have a distance of 20, and that’s essentially how far Stephen has gone.
Now, if we look at Gertrude, we can do the same math, but she’s going faster, and the question is, how long will it take her?
So we actually have the distance she needs to go, which is 20, and her speed, which is 15.
And now, we suddenly don’t know T, that’s what the question is asking for.
But we can plug this information back into that original equation, and that’s why it’s so handy.
Essentially, we use it twice, once for Stephen, we figured out how far he had gone.
And then we plugged that 20 down into the equation here, with Gertrude, and that gave us 20.
And then that equals 15, being the rate, T is the time.
We solve for T, next step, here, T is 20 over 15.
That is equal to 4 over 3, so what is 4 over 3, it’s the same as 1 and one-third, make that 4 a little bit better there.
1 and one-third, what’s one-third of an hour, 20 minutes, and therefore, 1 hour and 20 minutes it is.
Now, you don’t even have to figure out that one-third being 20 minutes.
If you look at the answer choices, you have 45 minutes ,which is less than an hour.
And you look at your, the number we got, four-thirds, that’s greater than 1 but less than 2.
And the only answer that would work is only answer choice C, so quick way of doing things, as well.
But the focus, of course, here is the distance-rate formula, let’s do some more.
This one’s gonna be a little bit harder, scarier, it’s the two trains, the dreaded scenario.
They are setting off from different cities located 300 miles from each other.
And they’re headed directed towards each other, but don’t worry, on different tracks.
The first train moves at a constant rate of 40 miles an hour, and the second moves at a constant rate of 60 miles an hour.
So what’s the key here, if objects are heading towards each other, we always combine their rate.
Again, if they’re headed directly at each other, you combine their speeds, and so 40 plus 60, that’s gonna be our rate, that’s 100.
And so we have the fact that they are D, 300 miles from each other.
And that they are going at 100 miles an hour, cuz we have to combine the rates, and we have to figure out the time.
And we get 300 = 100T, and then therefore, we can get T = 3.
And so therefore, we can see that in 3 hours, they’ve gone how far?
Well, in three hours, again, you’re combining their speeds together.
So this train over here goes 120 miles, which is 40 times 3, this train over here goes 180 miles, so all in all, they’ve traveled 300 miles.
Now, it would seem like the answer is c, 1 o’clock.
But notice that we’re not waiting for the two trains to get to the other cities, we’re asking when they meet each other.
And so they’re going to meet each other at this halfway point in time.
So in three hours, they’ve covered the total 300 miles, but half of 3 hours, which is 1.5 hours, they’ve actually connected.
That’s the point they intersect, so it’s actually half of 3, which gives us an hour and a half.
If they started at 10, that means that 11:30, they are going to pass each other, and that’s what makes it tricky.
Now, this is definitely a harder question, but the point is that we can still use distance is equal to rate times time.
We can still use that formula when we know that we’re combining rates, things coming together.
But now we’re gonna look at a different problem, the last problem here, which is a tricky one.
Here, we’re going to go back to Stephen and Gertrude, but we’re gonna put them in cars, instead of bicycles, but here, one is chasing the other one.
And so unlike the trains, when you’re combining, because they’re heading towards each other, here they’re in catch-up mode, or at least Gertrude is.
So Stephen, let’s read the problem here, leaves Farm Town, driving at a rate of 40 miles an hour, so where is he in one hour?
Well, he’s 40 hours from Gertrude, who starts driving at that point at 50 miles an hour, so when is she actually gonna reach Stephen?
Well, assuming again that they’re both continuing to drive, after one hour, he is 40 miles an hour, we know that.
But she is, what, 50 miles an hour, which is 10 miles an hour faster, let me get that 0 there, 10 miles an hour faster than Stephen.
So you can think of it this way, for every hour, she catches up.
So in one hour, she catches up 10 miles to him, two hours, she catches up 20 miles to him.
We know that she’s 40 miles behind, so how long will it take her to catch up to him?
Well, she’s gaining 10 miles an hour, that means, and now we can kinda use our distance-rate formula here.
She’s 40 miles away, and she’s catching up at a rate of 10 miles an hour.
We can just say, oh, okay, therefore T is equal to 4 hours, that’s the total number of time it will take her to catch.
So we can still use distance and rate, but we have to think of it differently, in terms of subtracting.
We’re subtracting, not adding together, which is the tricky part, but there’s our answer.
And there you have three different setups where we have distance, rate, and time, and we have need to find an answer, relying on our nifty little formula.
So there we are, you don’t have to be scared anymore of these question types, because you’ve got the distance-rate formula in your back pocket.
If you like this video, then click on the link in the description below.
That will take you to sat.Magoosh.com, where you can join thousands of other students who are prepping for the SAT.
If you want more helpful tips and strategies, then check out the videos on the left, and I will see you next time.
Want More SAT Study Tips?
Take a look at some of our other useful study tips to help you prepare for the SAT Math section:
Tips to CRUSH the SAT Math Section | Video Post
SAT Math Practice: How to Study Smarter and Score Higher
5 SAT Math Tricks Every Student Should Know
Happy studying! 🙂
Author
Molly Kiefer
Molly is one of Magoosh’s Content Creators. She designs Magoosh’s graphic assets, manages our YouTube channels and podcasts, and contributes to the Magoosh High School Blog.
Since 2014, Molly has tutored high school and college students preparing for the SAT, GRE, and LSAT. She began her tutoring journey while in undergrad, helping her fellow students master math, computer programming, Spanish, English, and Philosophy.
Molly graduated from Lewis & Clark College with a B.A. in Philosophy, and she continues to study ethics to this day. An artist at heart, Molly loves blogging, making art, taking long walks and serving as personal agent to her cat, who is more popular on Instagram than she is.
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It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25!
"As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang
Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025!
Below is a list of the different WOOT programs and the competitions they cover.
[list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads.
[]MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads.
[]CodeWOOT: USACO, IOI
[]PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam
[]ChemWOOT: USNCO, IChO
For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++.
Keep in mind some other important dates coming up fast!
[list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions!
[]The AMC 10A/12A will be held on November 5th
[]The AMC 10B/12B will be held on November 13th.
[]USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October.
[]Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology.
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.
Introductory: Grades 5-10
Prealgebra 1 Self-Paced
Prealgebra 1
Friday, Sep 5 - Jan 16
Monday, Sep 8 - Jan 12
Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT)
Sunday, Sep 21 - Jan 25
Thursday, Sep 25 - Jan 29
Wednesday, Oct 22 - Feb 25
Tuesday, Nov 4 - Mar 10
Friday, Dec 12 - Apr 10
Prealgebra 2 Self-Paced
Prealgebra 2
Tuesday, Sep 9 - Jan 13
Thursday, Sep 25 - Jan 29
Sunday, Oct 19 - Feb 22
Monday, Oct 27 - Mar 2
Wednesday, Nov 12 - Mar 18
Introduction to Algebra A Self-Paced
Introduction to Algebra A
Friday, Sep 5 - Jan 16
Thursday, Sep 11 - Jan 15
Sunday, Sep 28 - Feb 1
Monday, Oct 6 - Feb 9
Tuesday, Oct 21 - Feb 24
Sunday, Nov 9 - Mar 15
Friday, Dec 5 - Apr 3
Introduction to Counting & Probability Self-Paced
Introduction to Counting & Probability
Wednesday, Sep 3 - Nov 19
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Friday, Oct 3 - Jan 16
Sunday, Oct 19 - Jan 25
Tuesday, Nov 4 - Feb 10
Sunday, Dec 7 - Mar 8
Introduction to Number Theory
Friday, Sep 12 - Dec 12
Sunday, Oct 26 - Feb 1
Monday, Dec 1 - Mar 2
Introduction to Algebra B Self-Paced
Introduction to Algebra B
Sunday, Sep 7 - Jan 11
Thursday, Sep 11 - Jan 15
Wednesday, Sep 24 - Jan 28
Sunday, Oct 26 - Mar 1
Tuesday, Nov 4 - Mar 10
Monday, Dec 1 - Mar 30
Introduction to Geometry
Sunday, Sep 7 - Mar 8
Thursday, Sep 11 - Mar 12
Wednesday, Sep 24 - Mar 25
Sunday, Oct 26 - Apr 26
Monday, Nov 3 - May 4
Friday, Dec 5 - May 29
Paradoxes and Infinity
Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET)
Intermediate: Grades 8-12
Intermediate Algebra
Sunday, Sep 28 - Mar 29
Wednesday, Oct 8 - Mar 8
Sunday, Nov 16 - May 17
Thursday, Dec 11 - Jun 4
Intermediate Counting & Probability
Sunday, Sep 28 - Feb 15
Tuesday, Nov 4 - Mar 24
Intermediate Number Theory
Wednesday, Sep 24 - Dec 17
Precalculus
Tuesday, Sep 9 - Feb 24
Sunday, Sep 21 - Mar 8
Monday, Oct 20 - Apr 6
Sunday, Dec 14 - May 31
Advanced: Grades 9-12
Calculus
Sunday, Sep 7 - Mar 15
Wednesday, Sep 24 - Apr 1
Friday, Nov 14 - May 22
Contest Preparation: Grades 6-12
MATHCOUNTS/AMC 8 Basics
Wednesday, Sep 3 - Nov 19
Tuesday, Sep 16 - Dec 9
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Oct 6 - Jan 12
Thursday, Oct 16 - Jan 22
Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!)
MATHCOUNTS/AMC 8 Advanced
Thursday, Sep 4 - Nov 20
Friday, Sep 12 - Dec 12
Monday, Sep 15 - Dec 8
Sunday, Oct 5 - Jan 11
Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!)
Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!)
AMC 10 Problem Series
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!)
Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 10 Final Fives
Sunday, Sep 7 - Sep 28
Tuesday, Sep 9 - Sep 30
Monday, Sep 22 - Oct 13
Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, Oct 8 - Oct 29
Thursday, Oct 9 - Oct 30
AMC 12 Problem Series
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 12 Final Fives
Thursday, Sep 4 - Sep 25
Sunday, Sep 28 - Oct 19
Tuesday, Oct 7 - Oct 28
AIME Problem Series A
Thursday, Oct 23 - Jan 29
AIME Problem Series B
Tuesday, Sep 2 - Nov 18
F=ma Problem Series
Tuesday, Sep 16 - Dec 9
Friday, Oct 17 - Jan 30
WOOT Programs
Visit the pages linked for full schedule details for each of these programs!
MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT
Programming
Introduction to Programming with Python
Sunday, Sep 7 - Nov 23
Tuesday, Dec 2 - Mar 3
Intermediate Programming with Python
Friday, Oct 3 - Jan 16
USACO Bronze Problem Series
Wednesday, Sep 3 - Dec 3
Thursday, Oct 30 - Feb 5
Tuesday, Dec 2 - Mar 3
Physics
Introduction to Physics
Tuesday, Sep 2 - Nov 18
Sunday, Oct 5 - Jan 11
Wednesday, Dec 10 - Mar 11
Physics 1: Mechanics
Sunday, Sep 21 - Mar 22
Sunday, Oct 26 - Apr 26
0 replies
jwelsh
Sep 2, 2025
0 replies
J
H
9 Pythagorean Triples
ZMB038 146
Please put some of the ones you know, and try not to troll/start flame wars! Thank you :D
146 replies
ZMB038
May 19, 2025
melloncandy
an hour ago
J
H
Percent & Saline Mix
Kushagra2012 2
How many liters of a % saline solution must be added to liters of a % saline solution to produce a % saline solution? %? %. Any pattern?
2 replies
Kushagra2012
Saturday at 9:32 PM
evt917
2 hours ago
J
H
Two Liar Problems
matharama 11
1
Frederick never tells the truth on Tuesdays, Thursdays, and Saturdays. He always tells the truth on Sundays, Mondays, Wednesdays, and Fridays.
One day Matthew had this conversation with Frederick:
"What day is it today?" asked Matthew.
"Saturday," Frederick replied.
"And what day will it be tomorrow?"
"Wednesday."
On which day did this conversation take place?
First to answer correctly goes on the leaderboard!
Leaderboard
2
The Liar Paradox
Important Note!
There is no correct solution. Therefore, there is no leaderboard!
A man say that he is lying. Is what he says true or false?
11 replies
matharama
Sep 26, 2025
Wax_Wax
2 hours ago
J
H
Mathcounts prep as I qualed into my school MATHCOUNTS team
HoneyHap 13
Hey guys,
With somewhat great difficulty, I was able to make it into the school MATHCOUNTS team. Now, my next step is to strengthen my rookie problem-solving skills and level it up. I am thinking because I don’t I have a lot of time, to take Po Shen Loh’s self paced course “Competition Booster Pack: Module 2+4+5” - , so I can be ready to perform well on the Mathcounts chapter round. Do you guys recommend taking this course with not that much time left? I am saying as I am in 8th grade and have never qualed for MATHCOUNTS team until now. So, I am thinking I might be way behind than the rest of my MATHCOUNTS peers in PA.
Also, since I live in Pennsylvania, I want to know how competitive is PA in Mathcounts so I can see where I am at and how much practice I need to put in. If you guys have any helpful info, do let me know.
13 replies
HoneyHap
Yesterday at 9:41 PM
quasar1
3 hours ago
J
High School Math Grades 9-12, Ages 13-18
Grades 9-12, Ages 13-18
3
V New Topic k Locked
High School Math Grades 9-12, Ages 13-18
Grades 9-12, Ages 13-18
3
V New Topic k Locked
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2 users
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k a September Highlights and 2025 AoPS Online Class Information
jwelsh 0
It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25!
"As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang
Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025!
Below is a list of the different WOOT programs and the competitions they cover.
[list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads.
[]MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads.
[]CodeWOOT: USACO, IOI
[]PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam
[]ChemWOOT: USNCO, IChO
For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++.
Keep in mind some other important dates coming up fast!
[list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions!
[]The AMC 10A/12A will be held on November 5th
[]The AMC 10B/12B will be held on November 13th.
[]USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October.
[]Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology.
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.
Introductory: Grades 5-10
Prealgebra 1 Self-Paced
Prealgebra 1
Friday, Sep 5 - Jan 16
Monday, Sep 8 - Jan 12
Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT)
Sunday, Sep 21 - Jan 25
Thursday, Sep 25 - Jan 29
Wednesday, Oct 22 - Feb 25
Tuesday, Nov 4 - Mar 10
Friday, Dec 12 - Apr 10
Prealgebra 2 Self-Paced
Prealgebra 2
Tuesday, Sep 9 - Jan 13
Thursday, Sep 25 - Jan 29
Sunday, Oct 19 - Feb 22
Monday, Oct 27 - Mar 2
Wednesday, Nov 12 - Mar 18
Introduction to Algebra A Self-Paced
Introduction to Algebra A
Friday, Sep 5 - Jan 16
Thursday, Sep 11 - Jan 15
Sunday, Sep 28 - Feb 1
Monday, Oct 6 - Feb 9
Tuesday, Oct 21 - Feb 24
Sunday, Nov 9 - Mar 15
Friday, Dec 5 - Apr 3
Introduction to Counting & Probability Self-Paced
Introduction to Counting & Probability
Wednesday, Sep 3 - Nov 19
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Friday, Oct 3 - Jan 16
Sunday, Oct 19 - Jan 25
Tuesday, Nov 4 - Feb 10
Sunday, Dec 7 - Mar 8
Introduction to Number Theory
Friday, Sep 12 - Dec 12
Sunday, Oct 26 - Feb 1
Monday, Dec 1 - Mar 2
Introduction to Algebra B Self-Paced
Introduction to Algebra B
Sunday, Sep 7 - Jan 11
Thursday, Sep 11 - Jan 15
Wednesday, Sep 24 - Jan 28
Sunday, Oct 26 - Mar 1
Tuesday, Nov 4 - Mar 10
Monday, Dec 1 - Mar 30
Introduction to Geometry
Sunday, Sep 7 - Mar 8
Thursday, Sep 11 - Mar 12
Wednesday, Sep 24 - Mar 25
Sunday, Oct 26 - Apr 26
Monday, Nov 3 - May 4
Friday, Dec 5 - May 29
Paradoxes and Infinity
Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET)
Intermediate: Grades 8-12
Intermediate Algebra
Sunday, Sep 28 - Mar 29
Wednesday, Oct 8 - Mar 8
Sunday, Nov 16 - May 17
Thursday, Dec 11 - Jun 4
Intermediate Counting & Probability
Sunday, Sep 28 - Feb 15
Tuesday, Nov 4 - Mar 24
Intermediate Number Theory
Wednesday, Sep 24 - Dec 17
Precalculus
Tuesday, Sep 9 - Feb 24
Sunday, Sep 21 - Mar 8
Monday, Oct 20 - Apr 6
Sunday, Dec 14 - May 31
Advanced: Grades 9-12
Calculus
Sunday, Sep 7 - Mar 15
Wednesday, Sep 24 - Apr 1
Friday, Nov 14 - May 22
Contest Preparation: Grades 6-12
MATHCOUNTS/AMC 8 Basics
Wednesday, Sep 3 - Nov 19
Tuesday, Sep 16 - Dec 9
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Oct 6 - Jan 12
Thursday, Oct 16 - Jan 22
Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!)
MATHCOUNTS/AMC 8 Advanced
Thursday, Sep 4 - Nov 20
Friday, Sep 12 - Dec 12
Monday, Sep 15 - Dec 8
Sunday, Oct 5 - Jan 11
Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!)
Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!)
AMC 10 Problem Series
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!)
Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 10 Final Fives
Sunday, Sep 7 - Sep 28
Tuesday, Sep 9 - Sep 30
Monday, Sep 22 - Oct 13
Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, Oct 8 - Oct 29
Thursday, Oct 9 - Oct 30
AMC 12 Problem Series
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 12 Final Fives
Thursday, Sep 4 - Sep 25
Sunday, Sep 28 - Oct 19
Tuesday, Oct 7 - Oct 28
AIME Problem Series A
Thursday, Oct 23 - Jan 29
AIME Problem Series B
Tuesday, Sep 2 - Nov 18
F=ma Problem Series
Tuesday, Sep 16 - Dec 9
Friday, Oct 17 - Jan 30
WOOT Programs
Visit the pages linked for full schedule details for each of these programs!
MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT
Programming
Introduction to Programming with Python
Sunday, Sep 7 - Nov 23
Tuesday, Dec 2 - Mar 3
Intermediate Programming with Python
Friday, Oct 3 - Jan 16
USACO Bronze Problem Series
Wednesday, Sep 3 - Dec 3
Thursday, Oct 30 - Feb 5
Tuesday, Dec 2 - Mar 3
Physics
Introduction to Physics
Tuesday, Sep 2 - Nov 18
Sunday, Oct 5 - Jan 11
Wednesday, Dec 10 - Mar 11
Physics 1: Mechanics
Sunday, Sep 21 - Mar 22
Sunday, Oct 26 - Apr 26
0 replies
jwelsh
Sep 2, 2025
0 replies
J
H
IOQM 2025 Region 04 and 05 P16
Lunatic_Lunar7986 1
Find the number of ordered pairs , where and are positive integers such that and is a perfect square.
1 reply
Lunatic_Lunar7986
an hour ago
Lunatic_Lunar7986
22 minutes ago
J
H
IOQM 2025 Region 04 and 05 P12
Lunatic_Lunar7986 1
In a convex quadrilateral , the lengths of the diagonals are 12 and 16 and the line segments joining the midpoints of the opposite sides are of equal length. What is the maximum possible area of the quadrilateral .
1 reply
1 viewing
Lunatic_Lunar7986
an hour ago
Lunatic_Lunar7986
22 minutes ago
J
H
IOQM 2025 Region 04 and 05 P14
Lunatic_Lunar7986 1
The side of a square is 1 and it is also a chord of circle . The side does not intersect . The length of the tangent , drawn from to at the point is 2. If is the diameter of , then calculate .
1 reply
Lunatic_Lunar7986
an hour ago
Lunatic_Lunar7986
24 minutes ago
J
H
IOQM 2025 Region 04 and 05 P9
Lunatic_Lunar7986 2
Find the largest positive integer for which the inequality holds.
2 replies
1 viewing
Lunatic_Lunar7986
2 hours ago
Lunatic_Lunar7986
25 minutes ago
J
Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3
V New Topic k Locked
Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3
V New Topic k Locked
t 35,028 topics
w
4 users
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Topic
First Poster
Last Poster
H
k a September Highlights and 2025 AoPS Online Class Information
jwelsh 0
It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25!
"As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang
Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025!
Below is a list of the different WOOT programs and the competitions they cover.
[list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads.
[]MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads.
[]CodeWOOT: USACO, IOI
[]PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam
[]ChemWOOT: USNCO, IChO
For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++.
Keep in mind some other important dates coming up fast!
[list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions!
[]The AMC 10A/12A will be held on November 5th
[]The AMC 10B/12B will be held on November 13th.
[]USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October.
[]Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology.
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.
Introductory: Grades 5-10
Prealgebra 1 Self-Paced
Prealgebra 1
Friday, Sep 5 - Jan 16
Monday, Sep 8 - Jan 12
Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT)
Sunday, Sep 21 - Jan 25
Thursday, Sep 25 - Jan 29
Wednesday, Oct 22 - Feb 25
Tuesday, Nov 4 - Mar 10
Friday, Dec 12 - Apr 10
Prealgebra 2 Self-Paced
Prealgebra 2
Tuesday, Sep 9 - Jan 13
Thursday, Sep 25 - Jan 29
Sunday, Oct 19 - Feb 22
Monday, Oct 27 - Mar 2
Wednesday, Nov 12 - Mar 18
Introduction to Algebra A Self-Paced
Introduction to Algebra A
Friday, Sep 5 - Jan 16
Thursday, Sep 11 - Jan 15
Sunday, Sep 28 - Feb 1
Monday, Oct 6 - Feb 9
Tuesday, Oct 21 - Feb 24
Sunday, Nov 9 - Mar 15
Friday, Dec 5 - Apr 3
Introduction to Counting & Probability Self-Paced
Introduction to Counting & Probability
Wednesday, Sep 3 - Nov 19
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Friday, Oct 3 - Jan 16
Sunday, Oct 19 - Jan 25
Tuesday, Nov 4 - Feb 10
Sunday, Dec 7 - Mar 8
Introduction to Number Theory
Friday, Sep 12 - Dec 12
Sunday, Oct 26 - Feb 1
Monday, Dec 1 - Mar 2
Introduction to Algebra B Self-Paced
Introduction to Algebra B
Sunday, Sep 7 - Jan 11
Thursday, Sep 11 - Jan 15
Wednesday, Sep 24 - Jan 28
Sunday, Oct 26 - Mar 1
Tuesday, Nov 4 - Mar 10
Monday, Dec 1 - Mar 30
Introduction to Geometry
Sunday, Sep 7 - Mar 8
Thursday, Sep 11 - Mar 12
Wednesday, Sep 24 - Mar 25
Sunday, Oct 26 - Apr 26
Monday, Nov 3 - May 4
Friday, Dec 5 - May 29
Paradoxes and Infinity
Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET)
Intermediate: Grades 8-12
Intermediate Algebra
Sunday, Sep 28 - Mar 29
Wednesday, Oct 8 - Mar 8
Sunday, Nov 16 - May 17
Thursday, Dec 11 - Jun 4
Intermediate Counting & Probability
Sunday, Sep 28 - Feb 15
Tuesday, Nov 4 - Mar 24
Intermediate Number Theory
Wednesday, Sep 24 - Dec 17
Precalculus
Tuesday, Sep 9 - Feb 24
Sunday, Sep 21 - Mar 8
Monday, Oct 20 - Apr 6
Sunday, Dec 14 - May 31
Advanced: Grades 9-12
Calculus
Sunday, Sep 7 - Mar 15
Wednesday, Sep 24 - Apr 1
Friday, Nov 14 - May 22
Contest Preparation: Grades 6-12
MATHCOUNTS/AMC 8 Basics
Wednesday, Sep 3 - Nov 19
Tuesday, Sep 16 - Dec 9
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Oct 6 - Jan 12
Thursday, Oct 16 - Jan 22
Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!)
MATHCOUNTS/AMC 8 Advanced
Thursday, Sep 4 - Nov 20
Friday, Sep 12 - Dec 12
Monday, Sep 15 - Dec 8
Sunday, Oct 5 - Jan 11
Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!)
Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!)
AMC 10 Problem Series
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!)
Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 10 Final Fives
Sunday, Sep 7 - Sep 28
Tuesday, Sep 9 - Sep 30
Monday, Sep 22 - Oct 13
Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, Oct 8 - Oct 29
Thursday, Oct 9 - Oct 30
AMC 12 Problem Series
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 12 Final Fives
Thursday, Sep 4 - Sep 25
Sunday, Sep 28 - Oct 19
Tuesday, Oct 7 - Oct 28
AIME Problem Series A
Thursday, Oct 23 - Jan 29
AIME Problem Series B
Tuesday, Sep 2 - Nov 18
F=ma Problem Series
Tuesday, Sep 16 - Dec 9
Friday, Oct 17 - Jan 30
WOOT Programs
Visit the pages linked for full schedule details for each of these programs!
MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT
Programming
Introduction to Programming with Python
Sunday, Sep 7 - Nov 23
Tuesday, Dec 2 - Mar 3
Intermediate Programming with Python
Friday, Oct 3 - Jan 16
USACO Bronze Problem Series
Wednesday, Sep 3 - Dec 3
Thursday, Oct 30 - Feb 5
Tuesday, Dec 2 - Mar 3
Physics
Introduction to Physics
Tuesday, Sep 2 - Nov 18
Sunday, Oct 5 - Jan 11
Wednesday, Dec 10 - Mar 11
Physics 1: Mechanics
Sunday, Sep 21 - Mar 22
Sunday, Oct 26 - Apr 26
0 replies
jwelsh
Sep 2, 2025
0 replies
J
H
Help!! Why did MAA change the weighting
Jessepinkman10 4
I feel like I have a decent shot at making JMO before they released that new change. I was mocking 135-141 with usually 1 silly on amc10 and 9 on aime but now i honestly think im cooked bc i need like an 11 on aime and idk if i can do that in 5 months. Does anyone have any tips for me? 10th grade btw :(
4 replies
Jessepinkman10
Sep 26, 2025
Jessepinkman10
41 minutes ago
J
H
AIME Qualification
reeva28 7
Im currently a sophmore and I want to get better at competitive math. I know its a little late but I want to qualify for the AIME. Would anyone have any tips or recommendations or resources for what I should do to qualify?
7 replies
reeva28
5 hours ago
melloncandy
an hour ago
J
H
Computational Difficulty Measurement
imbadatmath1233 4
We all now the infamous mohs which detects the difficulty of olympiad style problems(invented by evan chen). I think we should have one for computational problems as well! I will call it cohs(computational olympiad hardness scale) and the difficulty rating is much like what we see on mohs(I was thinking about raising the scale from 0-100 but idk it was just a thought). For example 2024 amc10a p23 might be an example of a 0-5 cmohs problem whereas aime 15's might be good 20-30 cmohs. Let me know your thoughts and also should we expand it to 100?
4 replies
imbadatmath1233
Yesterday at 3:34 AM
NamelyOrange
2 hours ago
J
H
Nice substitution
Owjebra 65
Source: 2020 AMC 10A #14
Real numbers and satisfy and . What is the value of
65 replies
Owjebra
Jan 31, 2020
clod
2 hours ago
J
High School Olympiads Regional, national, and international math olympiads
Regional, national, and international math olympiads
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k a September Highlights and 2025 AoPS Online Class Information
jwelsh 0
It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25!
"As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang
Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025!
Below is a list of the different WOOT programs and the competitions they cover.
[list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads.
[]MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads.
[]CodeWOOT: USACO, IOI
[]PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam
[]ChemWOOT: USNCO, IChO
For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++.
Keep in mind some other important dates coming up fast!
[list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions!
[]The AMC 10A/12A will be held on November 5th
[]The AMC 10B/12B will be held on November 13th.
[]USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October.
[]Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology.
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.
Introductory: Grades 5-10
Prealgebra 1 Self-Paced
Prealgebra 1
Friday, Sep 5 - Jan 16
Monday, Sep 8 - Jan 12
Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT)
Sunday, Sep 21 - Jan 25
Thursday, Sep 25 - Jan 29
Wednesday, Oct 22 - Feb 25
Tuesday, Nov 4 - Mar 10
Friday, Dec 12 - Apr 10
Prealgebra 2 Self-Paced
Prealgebra 2
Tuesday, Sep 9 - Jan 13
Thursday, Sep 25 - Jan 29
Sunday, Oct 19 - Feb 22
Monday, Oct 27 - Mar 2
Wednesday, Nov 12 - Mar 18
Introduction to Algebra A Self-Paced
Introduction to Algebra A
Friday, Sep 5 - Jan 16
Thursday, Sep 11 - Jan 15
Sunday, Sep 28 - Feb 1
Monday, Oct 6 - Feb 9
Tuesday, Oct 21 - Feb 24
Sunday, Nov 9 - Mar 15
Friday, Dec 5 - Apr 3
Introduction to Counting & Probability Self-Paced
Introduction to Counting & Probability
Wednesday, Sep 3 - Nov 19
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Friday, Oct 3 - Jan 16
Sunday, Oct 19 - Jan 25
Tuesday, Nov 4 - Feb 10
Sunday, Dec 7 - Mar 8
Introduction to Number Theory
Friday, Sep 12 - Dec 12
Sunday, Oct 26 - Feb 1
Monday, Dec 1 - Mar 2
Introduction to Algebra B Self-Paced
Introduction to Algebra B
Sunday, Sep 7 - Jan 11
Thursday, Sep 11 - Jan 15
Wednesday, Sep 24 - Jan 28
Sunday, Oct 26 - Mar 1
Tuesday, Nov 4 - Mar 10
Monday, Dec 1 - Mar 30
Introduction to Geometry
Sunday, Sep 7 - Mar 8
Thursday, Sep 11 - Mar 12
Wednesday, Sep 24 - Mar 25
Sunday, Oct 26 - Apr 26
Monday, Nov 3 - May 4
Friday, Dec 5 - May 29
Paradoxes and Infinity
Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET)
Intermediate: Grades 8-12
Intermediate Algebra
Sunday, Sep 28 - Mar 29
Wednesday, Oct 8 - Mar 8
Sunday, Nov 16 - May 17
Thursday, Dec 11 - Jun 4
Intermediate Counting & Probability
Sunday, Sep 28 - Feb 15
Tuesday, Nov 4 - Mar 24
Intermediate Number Theory
Wednesday, Sep 24 - Dec 17
Precalculus
Tuesday, Sep 9 - Feb 24
Sunday, Sep 21 - Mar 8
Monday, Oct 20 - Apr 6
Sunday, Dec 14 - May 31
Advanced: Grades 9-12
Calculus
Sunday, Sep 7 - Mar 15
Wednesday, Sep 24 - Apr 1
Friday, Nov 14 - May 22
Contest Preparation: Grades 6-12
MATHCOUNTS/AMC 8 Basics
Wednesday, Sep 3 - Nov 19
Tuesday, Sep 16 - Dec 9
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Oct 6 - Jan 12
Thursday, Oct 16 - Jan 22
Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!)
MATHCOUNTS/AMC 8 Advanced
Thursday, Sep 4 - Nov 20
Friday, Sep 12 - Dec 12
Monday, Sep 15 - Dec 8
Sunday, Oct 5 - Jan 11
Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!)
Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!)
AMC 10 Problem Series
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!)
Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 10 Final Fives
Sunday, Sep 7 - Sep 28
Tuesday, Sep 9 - Sep 30
Monday, Sep 22 - Oct 13
Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, Oct 8 - Oct 29
Thursday, Oct 9 - Oct 30
AMC 12 Problem Series
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 12 Final Fives
Thursday, Sep 4 - Sep 25
Sunday, Sep 28 - Oct 19
Tuesday, Oct 7 - Oct 28
AIME Problem Series A
Thursday, Oct 23 - Jan 29
AIME Problem Series B
Tuesday, Sep 2 - Nov 18
F=ma Problem Series
Tuesday, Sep 16 - Dec 9
Friday, Oct 17 - Jan 30
WOOT Programs
Visit the pages linked for full schedule details for each of these programs!
MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT
Programming
Introduction to Programming with Python
Sunday, Sep 7 - Nov 23
Tuesday, Dec 2 - Mar 3
Intermediate Programming with Python
Friday, Oct 3 - Jan 16
USACO Bronze Problem Series
Wednesday, Sep 3 - Dec 3
Thursday, Oct 30 - Feb 5
Tuesday, Dec 2 - Mar 3
Physics
Introduction to Physics
Tuesday, Sep 2 - Nov 18
Sunday, Oct 5 - Jan 11
Wednesday, Dec 10 - Mar 11
Physics 1: Mechanics
Sunday, Sep 21 - Mar 22
Sunday, Oct 26 - Apr 26
0 replies
jwelsh
Sep 2, 2025
0 replies
J
H
Cyclic quadrilateral geometry in the style of V. Thebault
darij grinberg 93
Source: IMO Shortlist 2004 geometry problem G8
Given a cyclic quadrilateral , let be the midpoint of the side , and let be a point on the circumcircle of triangle . Assume that the point is different from the point and satisfies . Prove that the points ,, are collinear, where and .
Proposed by Dusan Dukic, Serbia and Montenegro
93 replies
darij grinberg
May 27, 2005
b1zmark
6 minutes ago
J
H
Olympiad problem - I can't solve it pls help
kjhgyuio 7
Source: smo 2016
It is given that x and y are positive integers such that x>y and
√x + √y=√2000
How many different possible values can x take?
7 replies
kjhgyuio
Apr 1, 2025
kjhgyuio
18 minutes ago
J
H
Mod3 complete system of residue
MathMaxGreat 1
Source: 2020 Winter NSMO
Prove that there exists infinite many positive integer pairs such that all contains a complete system of residue and .
1 reply
MathMaxGreat
3 hours ago
soryn
20 minutes ago
J
H
ab+bc+ca=k^2 inequality
jokehim 5
Source: my problem
Let then When does equality hold?
5 replies
1 viewing
jokehim
Saturday at 7:35 AM
lbh_qys
36 minutes ago
J
College Math Topics in undergraduate and graduate studies
Topics in undergraduate and graduate studies
3
V New Topic k Locked
College Math Topics in undergraduate and graduate studies
Topics in undergraduate and graduate studies
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k a September Highlights and 2025 AoPS Online Class Information
jwelsh 0
It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25!
"As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang
Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025!
Below is a list of the different WOOT programs and the competitions they cover.
[list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads.
[]MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads.
[]CodeWOOT: USACO, IOI
[]PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam
[]ChemWOOT: USNCO, IChO
For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++.
Keep in mind some other important dates coming up fast!
[list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions!
[]The AMC 10A/12A will be held on November 5th
[]The AMC 10B/12B will be held on November 13th.
[]USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October.
[]Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology.
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.
Introductory: Grades 5-10
Prealgebra 1 Self-Paced
Prealgebra 1
Friday, Sep 5 - Jan 16
Monday, Sep 8 - Jan 12
Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT)
Sunday, Sep 21 - Jan 25
Thursday, Sep 25 - Jan 29
Wednesday, Oct 22 - Feb 25
Tuesday, Nov 4 - Mar 10
Friday, Dec 12 - Apr 10
Prealgebra 2 Self-Paced
Prealgebra 2
Tuesday, Sep 9 - Jan 13
Thursday, Sep 25 - Jan 29
Sunday, Oct 19 - Feb 22
Monday, Oct 27 - Mar 2
Wednesday, Nov 12 - Mar 18
Introduction to Algebra A Self-Paced
Introduction to Algebra A
Friday, Sep 5 - Jan 16
Thursday, Sep 11 - Jan 15
Sunday, Sep 28 - Feb 1
Monday, Oct 6 - Feb 9
Tuesday, Oct 21 - Feb 24
Sunday, Nov 9 - Mar 15
Friday, Dec 5 - Apr 3
Introduction to Counting & Probability Self-Paced
Introduction to Counting & Probability
Wednesday, Sep 3 - Nov 19
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Friday, Oct 3 - Jan 16
Sunday, Oct 19 - Jan 25
Tuesday, Nov 4 - Feb 10
Sunday, Dec 7 - Mar 8
Introduction to Number Theory
Friday, Sep 12 - Dec 12
Sunday, Oct 26 - Feb 1
Monday, Dec 1 - Mar 2
Introduction to Algebra B Self-Paced
Introduction to Algebra B
Sunday, Sep 7 - Jan 11
Thursday, Sep 11 - Jan 15
Wednesday, Sep 24 - Jan 28
Sunday, Oct 26 - Mar 1
Tuesday, Nov 4 - Mar 10
Monday, Dec 1 - Mar 30
Introduction to Geometry
Sunday, Sep 7 - Mar 8
Thursday, Sep 11 - Mar 12
Wednesday, Sep 24 - Mar 25
Sunday, Oct 26 - Apr 26
Monday, Nov 3 - May 4
Friday, Dec 5 - May 29
Paradoxes and Infinity
Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET)
Intermediate: Grades 8-12
Intermediate Algebra
Sunday, Sep 28 - Mar 29
Wednesday, Oct 8 - Mar 8
Sunday, Nov 16 - May 17
Thursday, Dec 11 - Jun 4
Intermediate Counting & Probability
Sunday, Sep 28 - Feb 15
Tuesday, Nov 4 - Mar 24
Intermediate Number Theory
Wednesday, Sep 24 - Dec 17
Precalculus
Tuesday, Sep 9 - Feb 24
Sunday, Sep 21 - Mar 8
Monday, Oct 20 - Apr 6
Sunday, Dec 14 - May 31
Advanced: Grades 9-12
Calculus
Sunday, Sep 7 - Mar 15
Wednesday, Sep 24 - Apr 1
Friday, Nov 14 - May 22
Contest Preparation: Grades 6-12
MATHCOUNTS/AMC 8 Basics
Wednesday, Sep 3 - Nov 19
Tuesday, Sep 16 - Dec 9
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Oct 6 - Jan 12
Thursday, Oct 16 - Jan 22
Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!)
MATHCOUNTS/AMC 8 Advanced
Thursday, Sep 4 - Nov 20
Friday, Sep 12 - Dec 12
Monday, Sep 15 - Dec 8
Sunday, Oct 5 - Jan 11
Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!)
Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!)
AMC 10 Problem Series
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!)
Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 10 Final Fives
Sunday, Sep 7 - Sep 28
Tuesday, Sep 9 - Sep 30
Monday, Sep 22 - Oct 13
Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, Oct 8 - Oct 29
Thursday, Oct 9 - Oct 30
AMC 12 Problem Series
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 12 Final Fives
Thursday, Sep 4 - Sep 25
Sunday, Sep 28 - Oct 19
Tuesday, Oct 7 - Oct 28
AIME Problem Series A
Thursday, Oct 23 - Jan 29
AIME Problem Series B
Tuesday, Sep 2 - Nov 18
F=ma Problem Series
Tuesday, Sep 16 - Dec 9
Friday, Oct 17 - Jan 30
WOOT Programs
Visit the pages linked for full schedule details for each of these programs!
MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT
Programming
Introduction to Programming with Python
Sunday, Sep 7 - Nov 23
Tuesday, Dec 2 - Mar 3
Intermediate Programming with Python
Friday, Oct 3 - Jan 16
USACO Bronze Problem Series
Wednesday, Sep 3 - Dec 3
Thursday, Oct 30 - Feb 5
Tuesday, Dec 2 - Mar 3
Physics
Introduction to Physics
Tuesday, Sep 2 - Nov 18
Sunday, Oct 5 - Jan 11
Wednesday, Dec 10 - Mar 11
Physics 1: Mechanics
Sunday, Sep 21 - Mar 22
Sunday, Oct 26 - Apr 26
0 replies
jwelsh
Sep 2, 2025
0 replies
J
H
Fourier Series
EthanWYX2009 1
Source: 2025 Spring NSTE(2)-3
Let be real numbers. Define . Prove that:
Proposed by Site Mu
1 reply
EthanWYX2009
Aug 2, 2025
yofro
4 hours ago
J
H
Question on Goldbach Conjecture Verification
MDB001 5
Hello,
I was looking at the AoPS page on the Goldbach Conjecture, and something came to mind. When people say the conjecture has been verified up to 10^{18}, what exactly is the verification method?
Is it done by generating all primes up to that bound and checking their sums to cover all even numbers in the interval? Or is it done the other way around, starting with an even number and directly decomposing it into two primes?
If the method is the first, then the second seems more intriguing. Because if one could start from an even number, especially a large one, say 100 digits or more, and directly retrieve its two primes, what would that imply for the problem itself?
Thank you for clarifying. I would be very curious about your perspectives.
5 replies
MDB001
Yesterday at 9:23 AM
Tintarn
Yesterday at 10:18 PM
J
H
Group theory
frr12 1
We consider the set of square matrices with real coefficients of the form
Show that the group is isomorphic to the multiplicative group .
First of all I proved that is a group but I have no idea how can I suggest a isomophic for this goup.
1 reply
frr12
Yesterday at 8:14 PM
Etkan
Yesterday at 8:30 PM
J
H
Fourier series
We2592 4
Q) Expand the function
in a Fourier series of period .
please help for integration hint(upper lower limit and the )
4 replies
1 viewing
We2592
Sep 26, 2025
Mathzeus1024
Yesterday at 3:23 PM
J
Site Support Tech support and questions about AoPS classes and materials
Tech support and questions about AoPS classes and materials
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Site Support Tech support and questions about AoPS classes and materials
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k a September Highlights and 2025 AoPS Online Class Information
jwelsh 0
It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25!
"As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang
Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025!
Below is a list of the different WOOT programs and the competitions they cover.
[list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads.
[]MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads.
[]CodeWOOT: USACO, IOI
[]PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam
[]ChemWOOT: USNCO, IChO
For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++.
Keep in mind some other important dates coming up fast!
[list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions!
[]The AMC 10A/12A will be held on November 5th
[]The AMC 10B/12B will be held on November 13th.
[]USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October.
[]Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology.
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.
Introductory: Grades 5-10
Prealgebra 1 Self-Paced
Prealgebra 1
Friday, Sep 5 - Jan 16
Monday, Sep 8 - Jan 12
Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT)
Sunday, Sep 21 - Jan 25
Thursday, Sep 25 - Jan 29
Wednesday, Oct 22 - Feb 25
Tuesday, Nov 4 - Mar 10
Friday, Dec 12 - Apr 10
Prealgebra 2 Self-Paced
Prealgebra 2
Tuesday, Sep 9 - Jan 13
Thursday, Sep 25 - Jan 29
Sunday, Oct 19 - Feb 22
Monday, Oct 27 - Mar 2
Wednesday, Nov 12 - Mar 18
Introduction to Algebra A Self-Paced
Introduction to Algebra A
Friday, Sep 5 - Jan 16
Thursday, Sep 11 - Jan 15
Sunday, Sep 28 - Feb 1
Monday, Oct 6 - Feb 9
Tuesday, Oct 21 - Feb 24
Sunday, Nov 9 - Mar 15
Friday, Dec 5 - Apr 3
Introduction to Counting & Probability Self-Paced
Introduction to Counting & Probability
Wednesday, Sep 3 - Nov 19
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Friday, Oct 3 - Jan 16
Sunday, Oct 19 - Jan 25
Tuesday, Nov 4 - Feb 10
Sunday, Dec 7 - Mar 8
Introduction to Number Theory
Friday, Sep 12 - Dec 12
Sunday, Oct 26 - Feb 1
Monday, Dec 1 - Mar 2
Introduction to Algebra B Self-Paced
Introduction to Algebra B
Sunday, Sep 7 - Jan 11
Thursday, Sep 11 - Jan 15
Wednesday, Sep 24 - Jan 28
Sunday, Oct 26 - Mar 1
Tuesday, Nov 4 - Mar 10
Monday, Dec 1 - Mar 30
Introduction to Geometry
Sunday, Sep 7 - Mar 8
Thursday, Sep 11 - Mar 12
Wednesday, Sep 24 - Mar 25
Sunday, Oct 26 - Apr 26
Monday, Nov 3 - May 4
Friday, Dec 5 - May 29
Paradoxes and Infinity
Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET)
Intermediate: Grades 8-12
Intermediate Algebra
Sunday, Sep 28 - Mar 29
Wednesday, Oct 8 - Mar 8
Sunday, Nov 16 - May 17
Thursday, Dec 11 - Jun 4
Intermediate Counting & Probability
Sunday, Sep 28 - Feb 15
Tuesday, Nov 4 - Mar 24
Intermediate Number Theory
Wednesday, Sep 24 - Dec 17
Precalculus
Tuesday, Sep 9 - Feb 24
Sunday, Sep 21 - Mar 8
Monday, Oct 20 - Apr 6
Sunday, Dec 14 - May 31
Advanced: Grades 9-12
Calculus
Sunday, Sep 7 - Mar 15
Wednesday, Sep 24 - Apr 1
Friday, Nov 14 - May 22
Contest Preparation: Grades 6-12
MATHCOUNTS/AMC 8 Basics
Wednesday, Sep 3 - Nov 19
Tuesday, Sep 16 - Dec 9
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Oct 6 - Jan 12
Thursday, Oct 16 - Jan 22
Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!)
MATHCOUNTS/AMC 8 Advanced
Thursday, Sep 4 - Nov 20
Friday, Sep 12 - Dec 12
Monday, Sep 15 - Dec 8
Sunday, Oct 5 - Jan 11
Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!)
Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!)
AMC 10 Problem Series
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!)
Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 10 Final Fives
Sunday, Sep 7 - Sep 28
Tuesday, Sep 9 - Sep 30
Monday, Sep 22 - Oct 13
Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, Oct 8 - Oct 29
Thursday, Oct 9 - Oct 30
AMC 12 Problem Series
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)
AMC 12 Final Fives
Thursday, Sep 4 - Sep 25
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jwelsh
Sep 2, 2025
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Emoji rule
O33ochan 18
Why does every post have a rule where there can only be 5 emojis? Can't it have more? I don't know why this rule exists, but if Idid, I would understand why this rule is not meant to be annoying. Thank you!
18 replies
O33ochan
Saturday at 3:13 PM
doglover37
21 minutes ago
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Topics turning white
RollingPanda4616 12
Summary of the problem: When rapidly clicking down topics, some of the topics turn white.
Page URL: (found it here but might work elsewhere)
Steps to reproduce:
Go onto Site Support.
View the top ~10 threads, turning them gray.
Click the first topic "Suggestion Form".
Go down the list of topics as fast as you can, using the things on the left. Click "Read me first/ How to...", then immediately click "Community Safety", and so on.
Some topics will turn white randomly.
Expected behavior: The topics will stay gray, unless someone else posts.
Frequency: ~25%
Operating system(s): macOS Sequoia 15.4
Browser(s), including version: Google Chrome 140
Additional information: It only happens on some threads. Refreshing fixes it, but it is still reproduceable.
12 replies
RollingPanda4616
Sep 20, 2025
Staragon
29 minutes ago
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Is this a Glitch?
ZuzabKoit 5
While I was looking at a topic it suddenly started scrolling. I was doing nothing, but I had to close the tab or hit the return to top button for it to stop. Even when I tried scrolling, it overrided my mouse and kept scrolling. Is my account getting hacked or is this a glitch?
5 replies
ZuzabKoit
Sep 26, 2025
KangarooPrecise
3 hours ago
J
H
[RESOLVED] 8char limit wrongfully imposed
jkim0656 12
Hi SS,
I noticed that if you write a message and surround it in brackets [], it says that the message is shorter than eight characters even when it is definitely longer.
Chrome Browser
HP laptop
I did notice that if you write at least 8 characters after the text in [], it lets the message through.
12 replies
jkim0656
Yesterday at 1:05 AM
SpeedCuber7
Yesterday at 1:47 PM
J
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4997 | https://id.scribd.com/doc/231655878/Vieta-Jumping | Vieta Jumping
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Vieta Jumping
The document discusses the mathematical technique of Vieta jumping, which involves using a known root of an equation to find other roots. It provides four examples showing how Vieta jumping and Vietas formulas can be used to solve problems involving quadratic, quartic, and other polynomial equations. In each example, the author finds a solution, assumes it is not the only solution, and then uses properties of the roots to derive contradictions and prove their initial assumptions were incorrect.
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4998 | https://pmc.ncbi.nlm.nih.gov/articles/PMC7339241/ | Giardia lamblia: Laboratory Maintenance, Lifecycle Induction, and Infection of Murine Models - PMC
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Curr Protoc Microbiol
. Author manuscript; available in PMC: 2021 Jun 1.
Published in final edited form as: Curr Protoc Microbiol. 2020 Jun;57(1):e102. doi: 10.1002/cpmc.102
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Giardia lamblia: Laboratory Maintenance, Lifecycle Induction, and Infection of Murine Models
Marc Y Fink
Marc Y Fink
1 Department of Biology, Georgetown University, Washington, District of Columbia, USA
2 Current Address: Center for Vascular and Inflammatory Diseases, Department of Microbiology and Immunology, University of Maryland School of Medicine, Baltimore, Maryland, USA
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1,2, Danielle Shapiro
Danielle Shapiro
1 Department of Biology, Georgetown University, Washington, District of Columbia, USA
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1, Steven M Singer
Steven M Singer
1 Department of Biology, Georgetown University, Washington, District of Columbia, USA
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1 Department of Biology, Georgetown University, Washington, District of Columbia, USA
2 Current Address: Center for Vascular and Inflammatory Diseases, Department of Microbiology and Immunology, University of Maryland School of Medicine, Baltimore, Maryland, USA
3
Corresponding author Address: Department of Biology, Georgetown University, Reiss Science Building, Room 406, 37 th and O Streets NW, Washington, DC 20057., Phone Number: (202) 687-9884, steven.singer@georgetown.edu
PMC Copyright notice
PMCID: PMC7339241 NIHMSID: NIHMS1600189 PMID: 32515871
The publisher's version of this article is available at Curr Protoc Microbiol
Abstract
Giardia lamblia is a protozoan parasite that is found ubiquitously throughout the world and is a major contributor to diarrheal disease. Giardia exhibits a biphasic lifestyle existing as either a dormant cyst or a vegetative trophozoite. Infections are typically initiated through the consumption of cyst-contaminated water or food. Giardia was first axenized in the 1970s and can be readily maintained in a laboratory setting. Additionally, Giardia is one of the few protozoans that can be induced to complete its complete lifecycle using laboratory methods. In this article, we outline protocols to maintain Giardia and induce passage through its lifecycle. We also provide protocols for infecting and quantifying parasites in an animal infection model.
BASIC PROTOCOL 1: IN VITRO MAINTENANCE AND GROWTH OF GIARDIA TROPHOZOITES
BASIC PROTOCOL 2: IN VITRO ENCYSTATION OF GIARDIA CYSTS
BASIC PROTOCOL 3: IN VIVO INFECTIONS USING GIARDIA TROPHOZOITES
Keywords:Giardia lamblia, Giardiasis, murine infection, passage and maintenance, cryopreservation, animal model, protozoan parasite
INTRODUCTION
Human infection with Giardia lamblia (giardiasis) is regarded as one of the most common diarrheal diseases in the world with an estimated annual incidence of 280 million cases. Symptoms are characteristic of gastrointestinal distress; however, most cases are asymptomatic. Several studies indicate that subclinical Giardia infection is associated with childhood malnutrition. Paradoxically, another study suggests that the presence of this parasite may be protective against severe pediatric diarrheal disease. Adding to this complexity, long-term post-infectious syndromes in humans have been reported despite the eradication of this parasite.
Giardia lamblia (syn. G. duodenalis, G. intestinalis) is a flagellated protozoan microorganism that is found ubiquitously throughout the world. The first reported discovery of Giardia is attributed to Anton van Leeuwenhoek, the pioneer of microscopy when he examined his stool for microorganisms and was able to accurately describe this parasite’s features (Dobell, 1920). This article will focus on the growth and maintenance of Giardia lamblia (hereafter called Giardia) as this organism can be readily cultivated in a laboratory setting. This protocol has been optimized for the growth of G. lamblia, the only Giardia species that infects humans. G. muris is the natural rodent pathogen, but does not infect humans and is not readily cultivatable in axenic culture. Additionally, this article will not cover the isolation of Giardia from human clinical samples and will solely focus on culturing Giardia and infecting mice within a laboratory setting.
Giardia is considered to be a parasite since its replication and growth require the colonization of the human or mammalian intestinal tract (Adam, 2001). However, Giardia is not an intracellular pathogen and does not require the use of other cell types for expansion in vitro or in vivo. Instead, these cells adhere to surfaces and actively replicate via asexual binary fission if conditions are optimal. Thus, Giardia is readily capable of being cultivated and maintained indefinitely in a controlled laboratory setting as a monoculture using the methods described below.
The Giardia lifecycle has two distinct phases: a vegetative trophozoite and an infective cyst that is resistant to harsh environmental conditions. Cysts are found in most natural sources of water, such as rivers or lakes, however, cysts are more likely to be found in areas that are contaminated with animal fecal matter. Once cysts are ingested by a host, the excystation process begins after reaching the stomach. The acidity of the stomach triggers the breakdown of the Giardia cyst wall. In the small intestine, each cyst releases two excyzoites, which then generate four trophozoites following cell division. These trophozoites proceed to colonize the host’s intestinal tract, particularly the duodenum of the upper small intestine where bile is available to support growth and replication. Eventually, trophozoites initiate the encystation process and migrate towards the lower intestine where they are shed from infected hosts to the outside environment as infective cysts (Adam, 2001).
Giardia is divided into 8 genetic groups, termed assemblages A-H. Each assemblage is generally associated with particular host-specificity and most assemblages do not infect humans. The most relevant assemblages to human health are assemblages A and B since these two are the primary assemblages found in human giardiasis (Mayrhofer, Andrews, Ey, & Chilton, 1995). However, a small number of reports have described human infections with Giardia assemblages other than A and B (Lasek-Nesselquist, Welch, & Sogin, 2010; Zahedi, Field, & Ryan, 2017). Within these two human-infective assemblages: the WB strain (Assemblage A, ATCC #50803), the GS strain (Assemblage B, ATCC #50581), and H3 strain (Assemblage B, Waterbourne, Inc.) are three of the most commons strains used for in vitro and in vivo l aboratory experiments. Giardia cultures are typically grown using the trophozoite form of this parasite as most strains of Giardia will actively divide in modified TYI-S-33 media (Keister, 1983); however, the growth rate of trophozoites varies between the strain of Giardia being cultivated. Doubling times range from 8–12 hours depending mainly on the strain. Trophozoites are primarily grown in borosilicate glass tubes, but cell culture flasks (T-25, T-75) can also be used if a large number of trophozoites are needed (e.g., infection of mice). During growth, the bulk of trophozoites will adhere to the interior surface of the culture vessel and will also swim in the supernatant of the media. Mechanical or enzymatic means are not required to detach adherent trophozoites as a simple chilling of the culture vessel will trigger the detachment of cells. Giardia cyst formation can also be induced from trophozoite cultures using media that mimic the biochemistry of the small intestine and contains higher concentrations of bile or higher pH levels. Both forms of Giardia must be cultivated at 37°C, but do not require additional humidity or 5% CO 2 for growth. Strict anaerobic growth conditions are not necessary for cultivation since Giardia is considered a microaerophile and can tolerate small amounts of O 2; however, if large amounts of O 2 remain within the culture vessel, they will generate toxic compounds leading to a collapse of the Giardia culture. Trophozoites and media-generated cysts can be used in vitro to examine general Giardia biology or can be used in cell co-cultures with other mammalian cells to examine host cell responses.
CAUTION
Giardia sp. is a Biosafety Level 2 (BSL-2) pathogen. Giardia lamblia is capable of human infection and laboratory infections have been reported. Follow all appropriate guidelines and regulations for the use and handling of Giardia. Guidelines for biosafety and practice can be found in the latest edition of Biosafety in Microbiological and Biomedical Laboratories (Current edition: 5 th Edition) and refer to section VIII-C: Parasitic Agents. See UNIT 1A.1 and other pertinent resources (APPENDIX 1B) for more information.
NOTE
All solutions and equipment coming into contact with living cells must be sterile and proper aseptic technique must be maintained.
NOTE
All culture incubations should be performed in a 37°C incubator unless otherwise specified. Humidity or 5% CO 2 is not necessary for the growth of this parasite.
STRATEGIC PLANNING
BASIC PROTOCOL 1: IN VITRO MAINTENANCE AND GROWTH OF GIARDIA TROPHOZOITES
Giardia is an intestinal parasite that possesses a biphasic lifestyle comprised of a dormant cyst or vegetative trophozoite. The latter form is responsible for the colonization of the intestinal tract and directly contacts host cells. Accordingly, most cell culture models use Giardia trophozoites of strain WB, whereas murine infection models use either trophozoites of strain GS or cysts of strain H3. As noted above, WB is a member of assemblage A, and both GS and H3 are members of assemblage B. The preference for using WB trophozoites for in vitro work is mainly historical, although it is easier to perform transfection experiments using WB as plasmids replicate episomally in this strain, but not in GS (Singer, Yee, and Nash, 1998). However, WB trophozoites do not readily establish infections in adult mice. The preference for using GS trophozoites for infection in mice is mainly historical, while the use of H3 cysts is facilitated by the ability to purchase these commercially. Giardia is not an intracellular pathogen and does not require the use of other cell types for expansion in vitro or in vivo. Trophozoites are simple to cultivate within a laboratory setting and will expand in a short time. As Giardia is microaerophilic, trophozoites will not tolerate large amounts of oxygen and air-tight culture vessels that are devoid of additional air space must be used for culturing.
Materials (Unless specified, catalog numbers refer to Fisher Scientific, although other vendors are equally valid)
Giardia Media (Complete modified TYI-S33 medium) (see recipe)
Giardia freezing medium (see recipe)
70% Ethanol
Phosphate buffered saline (PBS) without calcium or magnesium
Frozen stock of G. lamblia GS clone H7 (ATCC 50581) or Luciferase-expressing trophozoites (e.g., Barash et al., 2017)
37°C Cell Culture Incubator
37°C Water or Bead Bath
Beckman Coulter Allegra X-30 refrigerated benchtop centrifuge (or equivalent) with appropriate 15- and 50-mL centrifuge tube adaptors
Inverted microscope equipped with phase-contrast optics
Hemocytometer
15 mL conical tube (Cat. #: 05–539-12) or equivalent
Fisherbrand disposable borosilicate glass tubes with threaded end 100mm (16 mL volume) (Cat. #: 14–959-35A)
Fisherbrand screw caps for disposable glass tubes (Cat. #: 14–957-82A)
Test-tube Slant Rack (e.g. Cat. #: 14–810N)
25-cm 2 (T-25) cell-culture flasks with plug-seal caps
A bucket filled with ice
Sterile cryovials
Freezing container capable of 1°C/min cooling rate (e.g. Cat. #: 07–210-00).
−150°C freezer or liquid nitrogen tank
Thawing Trophozoites
Work in a biological safety cabinet (BSC) while wearing a protective laboratory coat and disposable examination gloves.
Warm complete modified TYI-S33 medium (Giardia media) to 37°C in water or bead bath.
Clean BSC working area with soap and water and then sterilize with 70% ethanol. After warming, sterilize the Giardia media container by wiping with 70% ethanol before placing it inside the BSC.
Transfer 10 ml of warm media to a sterile 15 ml centrifuge tube.
Remove a frozen cryovial of G. lamblia from −80°C freezer or liquid nitrogen tank and quickly place it into a 37°C water bath.
After the freezing vial has mostly (but not completely) thawed, spray the cryovial with 70% ethanol to avoid transferring microbial contaminants from the water bath and place it into a sterile and clean BSC.
Using a sterile pipette, transfer thawed Giardia cells from the cryovial into the 15 mL conical tube containing Giardia media. Close the tube and mix by inversion several times.
Centrifuge the tube at 600 × g for 10 minutes at 4°C and aspirate supernatant to remove the freezing medium containing DMSO.
Gently resuspend the pellet in 15 mL of warm Giardia media and transfer contents into a sterile borosilicate glass culture tube until liquid fills the entire vessel, leaving ~5 mm of air space. Close securely to prevent gas exchange. Be sure not to overfill the container with media. Any spillover must be cleaned with 70% ethanol to avoid microbial growth on the outside of the tube.
Alternatively, the Giardia suspension can be transferred into a T-25 flask rather than a culture tube. The flask will require ~55 ml of media.
When adding suspension into either a culture tube or flask, the media must fill the entire culture container to reduce the amount of oxygen within the vessel. As Giardia is a microaerophilic organism, large amounts of space (oxygen) within the container will lead to media oxidation and parasite death.
Label the tube with strain, initials, and the date and then place the tube in a 37°C incubator to grow overnight. Place tubes on a slanted rack 10–15˚ from horizontal so that any air bubbles maintain minimal surface area contact with the media.
The next day, warm Giardia media to 37°C in water or bead bath, sterilize BSC and Giardia media container and place media into BSC.
Check the tube for cell viability and growth using an inverted phase-contrast microscope.
Live Giardia trophozoites will adhere to the bottom of the tube and also may be swimming or “spinning” in the supernatant (despite flowing with the movement of the liquid). Under higher magnification, adherent cells can also be seen beating their flagella. Dead cells will be floating with the movement of liquid and will not show any movement.
Sterilize the Giardia culture tube with 70% ethanol and place it into BSC.
Aspirate media from tube to remove any residual DMSO and replace it with sterile, warm Giardia media. Be sure to use a secondary trap containing bleach to kill any trophozoites in the aspirate.
Incubate tube in 37°C incubator. Check the tube for cell viability, growth, and contamination using an inverted phase-contrast microscope daily until ready for passage or use. This is generally when trophozoites reach 70–80% confluence.
Trophozoites of different strains exhibit different growth rates. The assemblage A stain WB may double in ~ 8 hours, while the assemblage B strain GS doubles roughly every 12 hours.
Passaging and Expanding Trophozoites
Work in biological safety cabinet (BSC) while wearing protective laboratory coat and disposable examination gloves.
Warm complete modified TYI-S33 medium (Giardia media) to 37°C in water or bead bath.
Clean BSC working area with soap and then sterilize with 70% ethanol. After warming, sterilize the Giardia media container with 70% ethanol and then place it inside.
Remove a G. lamblia culture tube from 37°C incubator and examine for cell viability, growth, and contamination.
Place the tube in the ice bucket and cover completely with ice.
Let the tube chill for 15–30 minutes as trophozoites will start to detach as media cools.
Sterilize the outside of the culture tube with 70% ethanol and place it into BSC.
Transfer a small volume of detached cells to a new sterile glass tube and completely fill the tube with fresh Giardia media.
A 1:100 dilution will generally result in a new culture being ready for passage or use after 48–72 hours. This will vary based on the strain being used.
Label tube with strain, initials, and date and place tube in 37°C incubator to grow.
If expanding trophozoites to culture a greater number of trophozoites then transfer cells into a T-25 cell culture flask rather than a glass tube in step 23.
Harvesting Trophozoites
Work in BSC while wearing a protective laboratory coat and disposable examination gloves.
Clean BSC working area with soap and then sterilize with 70% ethanol.
Remove a G. lamblia culture tube or flask from the 37°C incubator and examine for cell viability, growth, and contamination.
Confluent cultures in flasks or tubes should have ~10 6 trophozoites/ml. Lower yields could mean that the reagents used for media production are not optimal. Testing specific lots of casein and serum, in particular, may be necessary. Low yields might also reflect insufficient time on ice to fully detach trophozoites, use of old oxidized media (generally obvious from the appearance of hexagonal crystals of cystine in the cultures), or trophozoite death due to incubating cultures too long before harvest.
Place the tube or flask in an ice bucket, cover completely, and let cells detach for 30–45 minutes. Flasks require more time than tubes to chill.
After cells have detached, sterilize the outside of the tube or flask with 70% ethanol and, in a BSC, transfer all media to 15 mL conical tubes. Be sure to properly equalize and balance tubes.
If using a T-25 cell culture flask, then upscale to 50 mL conical tubes (any sterile polypropylene tube should be fine) and adjust accordingly.
Centrifuge tubes at 600 × g for 10 minutes at 4°C to pellet trophozoites.
Aspirate supernatants, combine pellets (if necessary), and wash trophozoites with 14 mL of sterile PBS. If using cells for in vitro encystation then do not resuspend pellet into PBS and instead proceed to Basic Protocol 2.
For cells resuspended in a 50 mL conical tube, add >25 mL of PBS. After centrifugation, the pellet will still be somewhat “loose” within a 50 mL conical tube, so be sure not to suck up the pellet during aspiration.
Centrifuge trophozoites again at the same speed, temperature, and time. Wash pellets for at least 4 times.
If using trophozoites for in vitro co-culture with epithelial cells or immune cells, then be sure to wash trophozoites as described. Residual yeast extract from Giardia media is highly stimulatory and can induce cytokine release from cells on its own. Saving aliquots of the supernatant from the final wash provides good control for residual media contamination.
Count cells using a hemocytometer before the final centrifugation step. Resuspend cells in sterile PBS at the desired concentration. Infection of mice proceeds on Basic Protocol 3.
Alternatively, cells can also be resuspended in mammalian cell culture medium in preparation for co-culture experiments; however, media requirements for co-cultures will vary.
Freezing Trophozoites
Work in BSC while wearing a protective laboratory coat and disposable examination gloves.
Clean BSC working area with soap and then sterilize with 70% ethanol.
Prepare Giardia freezing media A and B (see recipes). Keep the freezing media chilled by placing media in an ice bucket.
A single T-25 flask will be able to generate enough cells to freeze 5 cryovials that each has a total volume of 1 mL.
The final freezing medium is essentially TYI-S33 with an extra 10% FBS and 10% DMSO. Freezing media should remain ice-chilled since combination with DMSO will generate an exothermic reaction that results in the increased media temperature and may potentially harm the cells during the freezing process. Pure DMSO should not be ice-chilled as it will solidify at 0˚C.
Remove a G. lamblia culture flask from the 37°C incubator and examine for cell viability, growth, and contamination. Trophozoites should be frozen when confluency reaches 70–80%.
Place the entire flask into an ice bucket to chill and let the flask chill for 45–60 minutes.
Label cryovials with strain, initials, and date of freezing and chill on ice.
After the flask has sufficiently chilled, sterilize the outside with 70% ethanol and place it into BSC.
Transfer all the media within the tube into 1–2 50 ml conical tubes.
Centrifuge tubes at 600 × g for 10 minutes at 4°C to pellet trophozoites. Be sure to properly balance tubes in a centrifuge.
Remove supernatant from pelleted cells and resuspend the pelleted cells in 2.5 ml Freezing medium A (without DMSO).
Add 2.5 ml Freezing Medium B and mix gently.
DMSO destabilizes the membranes of the trophozoites and vigorous pipetting can kill cells.
Aliquot ~1 mL of cells into each pre-cooled cryovial and transfer vials into a freezing container capable of cooling a 1°C/min within a −80°C freezer.
After cells have frozen, take a single cryovial and thaw trophozoites (step 1) to check for viability and contamination. If cells are viable and uncontaminated then move cryovials to a −150°C freezer or liquid nitrogen tank for long term storage and cryopreservation.
BASIC PROTOCOL 2: IN VITRO ENCYSTATION OF GIARDIA CYSTS
The two protocols listed in this article aim to induce encystation through a combination of increased pH adjustment increased bile concentration and the addition of lactic acid salts. Gillin, Boucher, Rossi, and Reiner (1989) were the first to establish a formalized protocol to generate a robust number of viable cysts in vitro by expanding on the observation that primary bile salts and a slightly basic pH were able to induce encystation (Gillin, Reiner, & Boucher, 1988; Gillin et al., 1987). In this protocol, trophozoites are first starved of bile in a pre-encystation medium and then introduced to encystation medium #1 that contains porcine bile, lactic acid, and a more basic pH. The second protocol (encystation protocol #2) developed by Sun, McCaffery, Reiner, and Gillin (2003) incorporated excess bovine bile and is devoid of a formal pre-encystation medium other than Giardia media. This protocol requires less time to generate viable cysts and is more efficient for strain GS (Davids & Gillin, 2011). The initiation of cyst formation can be identified by the production of encystation specific vesicles (ESVs) 6–12 hours after the addition of either encystation media (Fig. 2). Alternately, encystation progress can be monitored using flow cytometry for DNA content. Trophozoites in G1 have 4N DNA content, in G2 they have 8N, and cysts should have 16N (Bernander, Palm & Svard, 2001). Additionally, viable cysts should form into typical cyst morphology, be resistant to water treatment, and also return into trophozoites after excystation. The quality of cysts derived from either protocol should be tested to determine the most appropriate protocol for a particular investigation. Neither protocol, as far as we are aware, can produce cysts capable of infecting animals. Cysts isolated from feces or purchased from a commercial supplier (i.e. H3 cysts from Waterbourne, Inc.) should be used for animal infections initiated by cysts.
Figure 2.
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Brightfield image of Giardia strain GS cultured in encystation medium #2 for 72 hours. A trophozoite is indicated by the red arrow, while green arrows indicate encysting cells with encystation vesicles.
Materials
Pre-Encystation Medium (see recipe)
Encystation Medium #1 (see recipe)
Encystation Medium #2 (see recipe)
Excystation Stage 1 Solution (see recipe)
Excystation Stage 2 Solution (see recipe)
Giardia Media (Complete modified TYI-S33 medium) (see recipe)
Deionized, distilled water (ddH 2 O)
37°C Cell Culture Incubator
37°C Water or Bead Bath
70% Ethanol
A bucket filled with ice
Logarithmic Giardia trophozoite culture in 12 mL glass culture tube
Beckman Coulter Allegra X-30 refrigerated benchtop centrifuge (or equivalent) with appropriate 15- and 50-mL centrifuge tube adaptors
Inverted microscope equipped with phase-contrast optics
Hemocytometer
15 mL conical tube
1.5 mL microcentrifuge tubes (e.g., Cat. #: 05–408-129)
Fisherbrand disposable borosilicate glass tubes with threaded end 100mm (12mL volume) (Cat. #: 14–959-35AA)
Fisherbrand screw caps for disposable glass tubes (Cat. #: 14–957-82A)
Encystation Protocol #1 (Gillin et al., 1989)
Prepare pre-encystation media and warm to 37°C in bead or water bath.
Work in BSC while wearing a protective laboratory coat and disposable examination gloves.
Clean BSC working area with soap and then sterilize with 70% ethanol.
Remove a G. lamblia culture tube from the 37°C incubator and examine for cell viability, growth, and contamination.
Prepare cells as described in Basic Protocol 1, Harvesting Trophozoites.
Pellet trophozoites and resuspend cells at a density of 5000 trophozoites/mL in pre-encystation media into a new clean, sterile glass culture tube or T-25 flask.
Culture cells for 3 days at 37°C.
After 3 days, prepare encystation medium #1 and warm it to 37°C.
Aspirate media and unadhered cells from the Giardia cultures in pre-encystation media.
Fill tubes or flasks with warmed encystation medium #1 until full to initiate the encystation process.
Culture for an additional 42–48 hours at 37°C.
Cyst production and recovery may vary with media and strain.
Check cultures for cyst formation and contamination.
Chill encysting cultures on ice for 10–20 minutes if using tubes and 30–45’ if using flasks.
Transfer media and detached cells to 15 mL or 50 mL conical tubes and centrifuge at 600 × g for 10 minutes at 4°C.
Resuspend pellets in 12 mL of sterile ddH2O to lyse trophozoites and place cells on ice for 10–20 minutes.
Cysts are water-resistant and will not lyse.
Spin as before and repeat water-treatment at least two more times.
After the last wash, count cysts with a hemocytometer and store at 4°C until use. In vitro cysts are not as stable as natural cysts and will lose viability quickly.
Encystation Protocol #2 (Sun et al., 2003)
Prepare encystation medium #2 and warm to 37°C in bead or water bath.
Work in BSC while wearing a protective laboratory coat and disposable examination gloves.
Clean BSC working area with soap and then sterilize with 70% ethanol.
Remove a G. lamblia culture from the 37°C incubator and examine for cell viability, growth, and contamination.
Aspirate Giardia media within a culture tube or flask and refill with warm encystation medium #2 to initiate the encystation process.
Culture tubes for 40–48 hours in 37°C incubator.
Cyst production and recovery may vary with media and strain. Using strain GS we obtain ~10% encystation rates with this protocol. (See Fig. 2).
Check cultures for cyst formation and contamination.
harvest and water-treat encysting cell cultures as described before (Encystation Protocol 1, steps 13–16).
Count cysts with a hemocytometer and store at 4°C until use.
Excystation Protocol (Boucher & Gillin, 1990)
Warm excystation stage 1 solution to 37°C in bead or water bath.
Clean BSC working area with soap and then sterilize with 70% ethanol.
Place 2×10 6 type 1 cysts in a 1.5 mL microcentrifuge tube.
Type 1 cysts appear smooth and have an oval shape with clearly defined intracellular organelles.
Spin for 5 minutes at 4°C at 135 × g.
Discard the supernatant and resuspend pellet with 1.5 mL of stage 1 solution and briefly vortex to mix.
Incubate cysts at 37°C for 30 minutes.
Prepare excystation stage 2 solution and warm in 37°C bead or water bath along with Giardia media.
Alternatively, Tyrode’s salt solution (without trypsin) can be pre-warmed and can be mixed with trypsin right before use.
Briefly vortex incubating cysts.
Sediment cysts for 5 minutes at room temperature at 135 × g
Discard the supernatant and resuspend cells in 1.5 mL of stage 2 solution. Briefly vortex to mix cysts.
Incubate tubes at 37°C for 1 hour. Vortex gently every 15 minutes.
Sediment cysts again as described before.
Discard the supernatant and resuspend cells in 1.5 mL of fresh (made day of) Giardia media. Briefly vortex to mix cysts.
Incubate tubes at 37°C for 1 hour with periodic agitation.
Concentrate excyzoites at 135 × g for 5 min.
Aspirate supernatant and resuspend in Giardia media.
Count excyzoite numbers using a hemocytometer.
Excyzoites will appear motile.
BASIC PROTOCOL 3: IN VIVO INFECTIONS USING GIARDIA TROPHOZOITES
Much of what we know about the host-pathogen interaction between Giardia and the mammalian immune system is through the use of animal models. Giardia can infect mice in a laboratory setting; however, the intestinal microbiota can render mice resistant to colonization and must be accounted for during infection experiments. Mice purchased from Jackson Labs are typically susceptible to infection while those from Taconic Farms have frequently been resistant to infection. Treatment with antibiotic cocktails can make resistant mice susceptible to colonization (Singer & Nash, 2000). When analyzing transgenic lines of mice, including knockout strains, comparisons with littermate controls are essential to limit the impact of differing microbiotas on infection outcomes. The trophozoite form of Giardia is primarily used for the initiation of infection, however, cysts (purchased from Waterborne, Inc.) have also been used. It is unclear if in vitro generated cysts are infectious. Most infection studies have examined parasite burden at different times post-infection by direct counting of duodenal contents using a hemocytometer as described below. Recently, the use of in vivo imaging has been reported using strains of Giardia expressing luciferase enzymes (Barash et al., 2017). This can reduce the number of animals needed as euthanasia is not required to determine parasite burdens. In mice infected with trophozoites of strain GS, parasite burdens typically peak between 6–7 days post-infection with most parasites eliminated by day 14.
NOTE: Protocols using live animals must be reviewed and approved by the Institutional Animal Care and Use Committee (IACUC) and must adhere to government regulations regarding the use and care of animals. Additionally, all lab personnel must adhere to institutional training guidelines and rules when using the In Vivo Imaging System for live imaging.
Materials
Giardia trophozoites prepared in Basic Protocol 1 or Giardia cysts purchased from Waterbourne, Inc.
5 – 10-week-old C57BL/6J mice (Jackson Laboratory, Stock no. 000664)
Phosphate buffered saline (PBS) without calcium or magnesium
70% Ethanol
Isoflurane (Baxter Healthcare 1001936060)
Sterile D-luciferin solution (15 mg/mL) (PerkinElmer Part Number #122799)
In Vivo Imaging System (Perkin Elmer, IVIS)
Inverted microscope equipped with phase-contrast optics
Hemocytometer
20 G x 1”, 2.25 mm Animal Feeding Needle
1-mL syringe
Dissection Scissors
Dissection Forceps
35 mm Petri Dish
Murine Infections with Giardia
Work in BSC while wearing a protective laboratory coat and disposable examination gloves.
Clear workspace with disinfectant and move mouse cage into a BSC and set aside.
Resuspend trophozoites to a target concentration of 1 million trophozoites in 0.1 mL using sterile PBS.
Infectious dosage can vary by investigator but should be standardized between experiments.
Attach an oral gavage needle to a 1 mL syringe and load with an appropriate volume of trophozoites. Clear any trapped bubbles within syringes before administration.
Restrain mouse with proper scruffing technique and gently insert the feeding needle into the esophagus. If any resistance is felt, remove the needle, and try again as resistance may indicate improper placement in the trachea rather than the esophagus. Slowly administer the infectious dose and wait 2–3 seconds before removing the needle.
Release the newly infected mouse into a new cage to not mix up mice. Repeat step 5 until all designated mice are infected.
In vivo live imaging of Trophozoites (Barash et al., 2017)
After the desired number of days, shave the ventral abdomen of each mouse in preparation for live imaging.
For optimal detection of luciferin-expressing trophozoites, a luciferin kinetic curve should be performed in any new animal model before the collection of reportable data.
Sedate mice using isoflurane (1.5 to 3%) in an induction chamber.
Inject D-luciferin (15 mg/mL) intraperitoneally with each mouse receiving 0.15 mg luciferin/g of body weight and track time of injection.
The total injection volume is dependent on the weight of mice, but should not exceed 200 uL. Injections should be staggered so that every mouse reaches maximum luciferin signal during imaging.
Transfer sedated mice imager chamber and position mice on their dorsal surface (abdomen facing upward).
Maintain chamber anesthesia with 1.5 to 2% isoflurane and 100% oxygen during imaging.
Collect images with 2-minute exposures constantly over a time of 10 minutes.
The bioluminescent signal will become increasingly stronger over time as luciferin substrate disseminates. The maximum luciferase signal typically appears between 10–20 minutes after the injection of luciferin. It is important to compare signals obtained at the same time after luciferin injection.
The final image collection should use 2- or 5-minute exposures that are dependent on luciferase signal strength.
Detection of photons should be analyzed within the region of interest (abdomen or intestine) to quantify bioluminescence using image analysis computer programs.
Bioluminescence should be quantified as total flux (photons/second) and should be normalized to the background level of uninfected (non-luciferase expressing) mice.
Quantification of Trophozoites within the murine small intestine
After the desired number of days, euthanize infected mice using guidelines approved by your Institutional Animal Care and Use Committee. Spray mouse abdomen with 70% Ethanol to disinfect the dissection area.
Make a slight midline incision and retract the skin to expose peritoneum.
Cut peritoneum to expose the peritoneal cavity.
Make 2 individual cuts to excise a 3 cm segment of the upper small intestine (duodenum). One cut should be immediately distal to the ligament of Treitz (~2–3 cm below the stomach) and the second should be 3 cm from the initial cut. The ligament of Treitz is the point where the common bile duct enters the small intestine. Giardia require bile for growth and are typically more abundant just below this point.
Transfer the excised intestinal segment to a petri dish filled with 3 mL of sterile PBS.
The volume of PBS and the length of the intestinal segment being used for parasites should be equivalent to give a unit of trophozoites/cm of intestine.
Chill segments for 20–30 minutes on ice
Use curved scissors to finely mince the intestinal segment and gently swirl the dish to dissociate tissue that has aggregated.
Count trophozoites using a hemocytometer. It may take practice to count accurately since the trophozoites swim across the slide and attach to both the slide and the coverslip. Focusing up and down is required for accurate counting.
Pipette directly into the solution as trophozoites will detach from tissue after mincing. Avoid tissue clumps when pipetting as the tissue will clog pipette tip or give inaccurate cell counts.
REAGENTS AND SOLUTIONS
Use deionized, distilled water in all recipes, and protocol steps. If denoted otherwise, when media is not in use, store at 4°C. Bovine serum must be heat-inactivated by placing at 56°C for 30 minutes. Peptone digest, yeast extract, and bovine serum should be lot tested. All recipes can be down- or up-scaled according to need.
6.5% Bile Solution
6.5 g Bovine and Ovine Bile (Sigma Aldrich, cat. no. B8381)
100 mL Water
Filter sterilize with a 0.45 μM filter unit
Store at 4°C for up to 2 months.
2.28% Ferric Ammonium Citrate
2.28 g Ammonium iron (III) Citrate (Sigma Aldrich, cat. no. F5879)
100 mL Water
Filter sterilize with a 0.45 μM filter unit.
Store at 4°C for up to 2 months.
20X Lactic Acid
0.55 g L-lactic acid hemicalcium salt (Sigma-Aldrich, cat. no. L2000)
50 mL Water
Filter sterilize with a 0.45 μM filter unit
Must be made fresh and used immediately on the day of preparation.
40X Porcine Bile
1 g Porcine Bile (Sigma Aldrich, cat. no. B8631)
100 mL Water
Filter sterilize with a 0.45 μM filter unit
Must be made fresh and used immediately on the day of preparation.
Giardia Media (Complete Modified TYI-S-33 Medium)
Solution 1:
10 g Peptone from milk solids (Sigma-Aldrich, cat. no. P6838)
5 g D-glucose
5 g Bacto Yeast Extract (Gibco, cat. no. 212750)
1 g Sodium Chloride
1.05 g Sodium Bicarbonate
400 mL Water
4 mL Sterile 6.5% Bile Solution
1.5 mL Sterile 2.28% Ferric Ammonium Citrate
Alternatively, Sodium Bicarbonate can be substituted for 0.3 g of 3% KH 2 PO 4 and 0.65 g K 2 HPO 4•3H 2 0.
As media components for Giardia media solution 1 solubilize and combine, media should turn from a somewhat opaque solution into a clear solution.
Solution 2:
1 g L-cysteine Hydrochloride (Sigma Aldrich, cat. no. C1276)
0.05 g Ascorbic Acid
30 mL Water
Adjust pH of solution 2 to 7.0–7.2 with 10N Sodium Hydroxide
Add solution 2 to solution 1 and mix for at least 20–30 minutes.
Add 50 mL Adult Bovine Serum and 5 mL Antibiotic-Antimycotic (Gibco, cat. no. 15240–062)
Sterile filter with 0.45 μm filter and aliquot in BSC (if needed).
Store up to 1–2 weeks at 4°C. Over time the cysteine in the media will oxidize to cystine and precipitate. Media should be discarded at this time and fresh media remade.
Encystation Medium #1
This recipe follows the Pre-encystation Medium with the following modification:
Adjust pH of solution 2 to 7.8 with 10M Sodium Hydroxide instead of 7.0–7.2
Solution 1:
10 g Peptone from milk solids (Sigma-Aldrich, cat. no. P6838)
5 g D-glucose
5 g Bacto Yeast Extract (Gibco, cat. no. 212750)
1 g Sodium Chloride
1.05 g Sodium Bicarbonate
400 mL Water
1.5 mL Sterile 2.28% Ferric Ammonium Citrate
Alternatively, Sodium Bicarbonate can be substituted for 0.3 g of 3% KH 2 PO 4 and 0.65 g K 2 HPO 4•3H 2 0.
As media components for Giardia media solution 1 solubilize and combine, media should turn from a somewhat opaque solution into a clear solution.
Solution 2:
1 g L-cysteine Hydrochloride (Sigma Aldrich, cat. no. C1276)
0.05 g Ascorbic Acid
30 mL Water
Adjust pH of solution 2 to 7.8 with 10N Sodium Hydroxide
Add solution 2 to solution 1 and mix for at least 20–30 minutes
Add 50 mL Adult Bovine Serum and 5 mL Antibiotic-Antimycotic (Gibco, cat. no. 15240–062)
Sterile filter with 0.45 μm filter
Aseptically add 25 mL of sterile 40X Porcine bile
Aseptically 50 mL of sterile 20X lactic acid
This media must be made fresh and used immediately on the day of preparation.
Encystation Medium #2
This recipe follows the Pre-encystation Medium with the following modifications:
Add 6.5g Bovine and Ovine Bile to solution 1
Replace 1 g L-cysteine HCL in solution 2 with 1 g of L-cysteine (non HCL)
Adjust pH of solution 2 to 7.8 with 10N Sodium Hydroxide instead of pH 7.0–7.2
Solution 1:
10 g Peptone from milk solids (Sigma-Aldrich, cat. no. P6838)
5 g D-glucose
5 g Bacto Yeast Extract (Gibco, cat. no. 212750)
1 g Sodium Chloride
1.05 g Sodium Bicarbonate
6.5 g Bovine and Ovine Bile (Sigma Aldrich, cat. no. B8381)
400 mL Water
1.5 mL Sterile 2.28% Ferric Ammonium Citrate
Alternatively, Sodium Bicarbonate can be substituted for 0.3 g of 3% KH 2 PO 4 and 0.65 g K 2 HPO 4•3H 2 0.
As media components for Giardia media solution 1 solubilize and combine, media should turn from a somewhat opaque solution into a clear solution.
Solution 2:
1 g L-cysteine (non HCL) (Sigma Aldrich, cat. no. C7352)
0.05 g Ascorbic Acid
30 mL Water
Adjust pH of solution 2 to 7.8 with 10M Sodium Hydroxide
Add solution 2 to solution 1 and mix for at least 20–30 minutes
Add 50 mL Adult Bovine Serum and 5 mL Antibiotic-Antimycotic (Gibco, cat. no. 15240–062)
Sterile filter with 0.45 μm filter
Aseptically add 50 mL of sterile 20X lactic acid
This media must be made fresh and used immediately on the day of preparation.
Excystation Stage 1 Solution
68 mg L-Cysteine (non HCL) (Sigma Aldrich, cat. no. C7352)
68 mg L-Glutathione Reduced (Sigma Aldrich, cat. no. G4251)
52 mg Sodium Bicarbonate
7 mL 1X Hank’s Balanced Salt Solution
18 mL Water
Adjust pH to 2.0 and sterilize with 0.45 μm filter
This media must be made fresh and used immediately on the day of preparation.
Excystation Stage 2 Solution
100 mg Trypsin Type II-S
10 mL 1X Tyrode’s Salt Solution, pH = 8
Mix and sterilize with 0.45 μm filter
This media must be made fresh and used immediately on day of preparation.
Pre-encystation Medium (Bile Starvation Medium)
This recipe follows the Giardia Media with the following modification:
Pre-encystation medium is not supplemented with 6.5% bile
Solution 1:
10 g Peptone from milk solids (Sigma-Aldrich, cat. no. P6838)
5 g D-glucose
5 g Bacto Yeast Extract (Gibco, cat. no. 212750)
1 g Sodium Chloride
1.05 g Sodium Bicarbonate
400 mL Water
1.5 mL Sterile 2.28% Ferric Ammonium Citrate
Alternatively, Sodium Bicarbonate can be substituted for 0.3 g of 3% KH 2 PO 4 and 0.65 g K 2 HPO 4•3H 2 0.
As media components for Giardia media solution 1 solubilize and combine, media should turn from a somewhat opaque solution into a clear solution.
Solution 2:
1 g L-cysteine Hydrochloride (Sigma Aldrich, cat. no. C1276)
0.05 g Ascorbic Acid
30 mL Water
Adjust pH of solution 2 to 7.0–7.2 with 10N Sodium Hydroxide
Add 50 mL Adult Bovine Serum and 5 mL Antibiotic-Antimycotic (Gibco, cat. no. 15240–062)
Add solution 2 to solution 1 and mix for at least 20–30 minutes
Sterile filter with 0.45 μm filter
This media must be made fresh and used immediately on the day of preparation.
Trophozoite Freezing Medium A
4.5 mL Giardia Media
0.5 mL Sterile Fetal Bovine Serum (FBS)
Trophozoite Freezing Medium B
4.0 mL Giardia Media
0.5 mL Sterile Fetal Bovine Serum (FBS)
0.5 mL DMSO
COMMENTARY
Background Information
Giardia lamblia is a flagellated protozoan that is found ubiquitously throughout the world. Infection with Giardia in humans (giardiasis) is regarded as one of the most common causes of diarrheal disease, with an estimated 280 million cases reported worldwide each year (Lane & Lloyd, 2002). Transmission of this parasite occurs via a fecal-oral route and is initiated through the ingestion of parasite contaminated water or food (Adam, 2001). The Giardia lifecycle has two distinct phases: a vegetative trophozoite that replicates in the lumen of the small intestine and an infective cyst that is resistant to harsh environmental conditions.
Monocultures of Giardia that were free from bacteria and yeast (axenic cultures) were first generated from the intestines of various animals (Fortess & Meyer, 1976; Meyer, 1970). Axenic Giardia cultures were initially grown in complex HSP-1 medium (Meyer, 1970, 1976) and then transitioned to Diamond’s TP-S-1 medium (Diamond, Harlow, & Cunnick, 1978; Visvesvara, 1980). A modified recipe of TYI-S-33 medium then became the preferred medium for Giardia cultivation when it was modified with L-cysteine (Gillin & Diamond, 1981a, 1981b) and bile (Keister, 1983), as these reagents were able to better replicate conditions within the upper small intestine, Giardia’s environmental niche. Keister’s recipe has withstood the test of time and is still the basis for cultivating Giardia trophozoites in a laboratory setting.
Giardia is unique in that it one of the few protozoans that can be induced to differentiate into cysts using in vitro methods that replicate the host’s intestinal tract (Lauwaet & Gillin, 2011). Each section of the intestine is important as the absorption of specific nutrients varies by intestinal location, thus, leading to differences in chemical composition between sections. Trophozoites that are swimming and migrating towards the colon with the host’s intestinal fluid can respond to this chemical change and are thought to encyst based on the changing availability of nutrients like bile and fatty acids. Accordingly, the majority of cysts in vivo are located distally within the gastrointestinal tract in the ileum, cecum, and colon (Gillin et al., 1987). As such, the two encystation protocols listed in this article induce encystation by a combination of pH adjustment, increased bile concentration leading to altered fatty acid availability, and the addition of lactic acid salts.
Giardia cysts are robust structures that are temperature resistant and can survive in water at 4°C for up to three months. Infected individuals can shed up to 10 billion cysts a day, and ingestion of as few as 10 cysts can initiate an infection (Rendtorff, 1954). Encystation specific vesicles (ESVs) are easily identifiable structures that form in trophozoites upon encystation induction after 8–24 hours in encystation media. ESVs appear as protuberances with Nomarski differential interference contrast (DIC) optics (Reiner, Douglas, & Gillin, 1989) or as dark spots under phase contrast (Davids & Gillin, 2011). However, viable cysts should also be resistant to water-treatment and also capable of excystation into trophozoites. Ultimately, the efficiency of each protocol will most likely vary between individual laboratories (discussed in Critical Parameters and Troubleshooting section).
Giardia cysts can be classified into two forms: type 1 and type 2. Type 1 cysts are considered to be the best quality for excystation. This type of cyst is oval and will appear bright under phase-contrast microscopy. The cyst wall and intracellular organelles are also clearly defined and easily identified. Additionally, cytosolic structures can be visualized inside the cyst and will be trypan-blue negative (white) (Davids & Gillin, 2011). Type 2 cysts are not as structured or defined as type 1 cysts. The cell body or cytoplasm does not fill up the entire cytosolic space and appears detached from the cyst wall and is trypan-blue positive (blue) (Davids & Gillin, 2011).
Most reports involving in vitro encystation protocols have been optimized for strain WB clone C6 (ATCC # 50803) and not strain GS (ATCC #50581). This may be problematic for in vivo studies as Giardia infection studies initiated by strain WB trophozoites are unable to properly colonize mice without prior antibiotic treatment to eliminate intestinal microbiota (Solaymani-Mohammadi & Singer, 2011). It is currently unknown if in vitro generated cysts of WB or GS strains are infective and truly replicate a meaningful biological host response. Animal infections initiated with Giardia cysts have used the H3 strain (Bartelt et al., 2017; Bartelt et al., 2013); however, these cysts have been purchased from Waterbourne, Inc as purified cysts that have been isolated from infected gerbils. As cysts are the natural infectious form of Giardia, it will be important to understand if host immune responses differ between the initiation of animal infections using cysts or trophozoites.
The Public Health Impact of Giardia
Developing countries that have poor water sanitation processes have a prevalence of giardiasis of 20% or greater (Roxstrom-Lindquist, Palm, Reiner, Ringqvist, & Svard, 2006). In contrast, the United States, which has adequate water sanitation, has an estimated annual prevalence of <1% (Scallan et al., 2011). In areas endemic for giardiasis, Giardia was found to be the 4 th most common pathogen found in stools of children younger than 1 year of age and the 2 nd most common pathogen among children between 1 and 2 years of age (Platts-Mills et al., 2015).
Symptoms of giardiasis typically appear 6 – 15 days after infection (Farthing, 1997) and are characterized by diarrhea, intestinal cramps, nausea, intestinal malabsorption, and reduction in brush-border disaccharidases. Long-term manifestations have also been linked to Giardia, including irritable bowel and chronic fatigue syndrome. These can develop in giardiasis patients years after this parasite has been eliminated (Hanevik et al., 2017; Litleskare et al., 2018). Sub-clinical infections also appear to be very common in endemic areas which makes treatment difficult to initiate. Giardiasis is usually treatable with metronidazole or albendazole yet each has varying efficacies (Ansell et al., 2015; Leitsch, Schlosser, Burgess, & Duchene, 2012; Solaymani-Mohammadi, Genkinger, Loffredo, & Singer, 2010) and vaccines are not currently available for human use (Singer, Fink, & Angelova, 2019).
Unlike other intestinal infections, intestinal inflammation is typically not evident during giardiasis (Oberhuber et al., 1996). The Global Enteric Multicenter Study (GEMS) has reported that children infected with Giardia have a reduce incidence of severe pediatric diarrheal disease, suggesting the presence of this parasite may be protective through unknown mechanisms (Kotloff et al., 2013; Muhsen, Cohen, & Levine, 2014; Muhsen & Levine, 2012). On the other hand, The Etiology, Risk Factors, and Interactions of Enteric Infections and Malnutrition and the Consequences for Child Health and Development (MAL-ED) project reported that Giardia detection was among the top five contributors to stunting (low height-for-age z-scores) in children worldwide (Rogawski et al., 2018). Additionally, this study found that cumulative parasite burdens were associated with childhood stunting, yet symptomatic diarrhea was not (Rogawski et al., 2018). The mechanisms by which repeated enteric infections lead to malnutrition and deficits in child development remain to be defined.
Giardia is detected using three diagnostic tests that require samples from suspected giardiasis patients. Traditional microscopy is based on the identification of microbes that appear to have the morphology of Giardia (Koehler, Jex, Haydon, Stevens, & Gasser, 2014). Light microscopy without the use of differential staining is unable to reveal this parasite to a species level (Thompson & Monis, 2004) and staining protocols may involve iodine (Zajac, Leib, & Burkholder, 1992), iron-hematoxylin (Garcia, 2009), Giemsa (Ament, 1972), or trichrome (Addiss, Juranek, & Spencer, 1991; Thornton, West, Dupont, & Pickering, 1983). Immunological-based detection methods such as fluorescence assays or enzyme-linked immunosorbent assay (ELISA) that detect Giardia antigens (i.e. cyst wall proteins) can also be used to detected Giardia and provide a greater level of sensitivity and specificity when compared to microscopic techniques. Many commercialized detection tests for humans and animals are based on this method; however, protocols are not standardized and detection results may vary due to a lack of cross-reactivity between Giardia strains, quality of the sample, and subjectivity of sample results during analysis (Geurden & Olson, 2011; Koehler et al., 2014). These two detection methods are unable to discriminate Giardia based on a species and assemblage level which may become problematic during the detection of microbes within a large cohort as patients may be infected with mixed Giardia assemblages (Almeida, Pozio, & Cacciò, 2010). Polymerase chain reaction (PCR)-based techniques are the primary methods used to detect Giardia as it rapidly provides the greatest level of sensitivity and specificity. Additionally, these methods are also able to quantify the parasite burden of an infected individual; however, PCR techniques arguably require the greatest technical skill and are the most cost-prohibitive for detection (Geurden & Olson, 2011).
Critical Parameters and Troubleshooting
Giardia media is normally simple to make and should be fine to use for 7–10 days after creation as long as it is stored at 4˚C. Most difficulty in preparing Giardia media is typically due to manufacturer lot variability, clogging of vacuum filters due to precipitant formation, or the lack of recognizing expired media. Bovine serum and peptone vary by manufacturer and will not support trophozoite growth if a lot isn’t suitable for Giardia culture. Yeast extract also suffers from lot variation, although typically is more consistent. Reagents should first be batch tested to identify a usable lot for culturing and then purchased in bulk for long-term storage as peptone and frozen serum both have long shelf lives. Sterile filter vacuums can easily be clogged due to precipitant formation or from a lack of mixing. Polyethylene sulfonate filters are preferred in our lab. Also, our lab has modified Keister’s original recipe (Keister, 1983) by replacing the KH 2 PO 4 and K 2 HPO 4•3H 2 O with sodium bicarbonate as a pH buffer, due to phosphates propensity to precipitate from solution and consequently clog vacuum filters. Trophozoite cultures do not seem affected by this switch and remain viable. Additionally, it is critical to mix Giardia media for at least 20 minutes or until the solution turns clear as this will ensure that all components have been solubilized. The expiration of Giardia media is easily identified when white precipitate settles at the bottom of the bottle. This is due to the oxidation of cysteine and appears as hexagonal crystal under a microscope. At this point, the media is no good and needs to be thrown away. Keeping the media in a tightly-sealed bottle with little air prolongs the shelf-life. Giardia media is capable of being frozen at −20°C for extended periods, however, some batches of peptone will precipitate after freezing and thawing.
Giardia is considered a microaerophile and can tolerate small amounts of O 2, however, large amounts of O 2 (large open spaces) within the culture vessel will lead to a decrease in the viability of Giardia trophozoites. Thus, when growing Giardia in a culture vessel it is critical to leave little to no air within the vessel. Cell culture flasks with vented caps should never be used to culture Giardia and only plug-seal caps are suitable. Growing tubes at a slight angle also reduces the surface-air interface.
If Giardia is being cultivated in preparation for animal infections, then it is important to consider which strain will be used for infection. Giardia infection studies indicate that strain WB trophozoites are unable to properly colonize mice without prior antibiotic treatment (Solaymani-Mohammadi & Singer, 2011), while infections initiated by strain GS trophozoites or H3 cysts do not require prior antibiotic treatment. Prior treatment with antibiotics should be carefully considered, however, as depletion of the intestinal microbiota will alter the host’s immune response (Keselman, Li, Maloney, & Singer, 2016). The interactions between Giardia and the intestinal microbiome are critical and have been summarized in Fink and Singer (2017).
Understanding Results
Live Giardia trophozoites will adhere to the bottom of the tube and also may be swimming or “spinning” in the supernatant (despite flowing with the movement of the liquid). Cells should appear to be teardrop in shape and not round. Under higher magnification, adherent cells can also be seen beating their flagella. Dead trophozoites will be floating with the movement of liquid and will not show any other movement. The rate of growth for Giardia trophozoites is dependent on the strain. Assemblage A (WB strain) of Giardia double every 6–8 hours, whereas assemblage B (GS or H3 strain), require 10–12 hours per doubling. All strains typically reach ~10 6 trophozoites/mL when confluent.
Cyst formation should begin with the addition of encystation medium; however, variability within an individual laboratory with regard to the efficiency of the encystation protocol used should be expected. Encysting trophozoites that produce encystation specific vesicles (ESVs) are good indicators that encystation machinery is active. Encysting cells will morph from teardrop shapes into ovals or circles that have a defined cyst wall.
Quantification of trophozoites from infected animals is highly variable and is dependent on infectious dose, mouse strain, and animal vendor. If parasites are viable and healthy, the peak burden of Giardia trophozoite will occur around day 6 or 7 in Giardia-infected C57BL/6J mice and the majority of parasites will be eliminated by day 14. In immunocompromised mice (e.g., SCID), Giardia-infections are unable to be controlled and parasite colonization will be extremely high throughout the infection. It is recommended that any mice used for Giardia infections are purchased from Jackson laboratories (Singer & Nash, 2000) and littermate controls should be used when appropriate.
Time Considerations
Adult bovine serum, peptone digest of casein, and yeast extract should all be lot tested first to identify a batch that is suitable for Giardia culture. The creation of all the media listed here should take approximately 30–60 minutes to complete. The 6.5% bile and 2.28% ferric ammonium citrate solutions for Giardia media and 20X lactic acid and 40X porcine bile solutions for encystation medium should be created first as all of these solutions are supplements. Thawing of Giardia trophozoites will take 5–10 minutes and the entire thawing procedure will take approximately 30–40 minutes. Generally, passaging and expanding trophozoite cultures can be accomplished within 45–60 minutes when using flask cultures, or within a shorter time of 15–20 minutes when using glass tubes. After chilling trophozoites, the freezing procedure should take approximately 20–40 minutes to complete. Cells being harvested to be used for either animal infections or cell culture assays are required to go through multiple washes to remove residual media; thus, harvesting of trophozoites will take between 1–2 hours depending on the number of washes conducted. All encystation procedures are simply just an exchange of media and should not take longer than 15 minutes to complete. The infection of mice is dependent on animal handling skills and the number of mice being infected; however, the infection of 10 mice should take 30–45 minutes. Quantification of parasites from infected mice is also dependent on the number of mice being analyzed with the imager and intestinal segments being assessed for parasites. After chilling of tissue, the completion of parasite counts for 1 mouse should take approximately 5–15 minutes. Time consideration for in vivo live imaging will vary due to experimental and imager setup, however, 2 or 3 mice may be able to be imaged concurrently. For a more accurate estimation of time, a luciferin kinetic curve should be performed in any new animal model.
Figure 1:
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An outline of this article covering the typical workflow when culturing Giardia in a laboratory setting.
Acknowledgements
The authors are supported by the National Institutes of Health grants AI-094492 and AI-109591 to SMS.
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4999 | https://cs.uwaterloo.ca/journals/JIS/VOL20/Yilmaz/yilmaz7.pdf | 23 11 Article 17.4.6 Journal of Integer Sequences, Vol. 20 (2017), 2 3 6 1 47 On The Pfaffians and Determinants of Some Skew-Centrosymmetric Matrices Fatih Yılmaz Polatlı Arts and Sciences Faculty Gazi University 06900 Ankara Turkey fatihyilmaz@gazi.edu.tr Tomohiro Sogabe Graduate School of Engineering Nagoya University Nagoya, 464-8603 Japan sogabe@na.cse.nagoya-u.ac.jp Emrullah Kırklar Polatlı Arts and Sciences Faculty Gazi University 06900 Ankara Turkey e.kirklar@gazi.edu.tr Abstract This paper shows that the Pfaffians and determinants of some skew centrosymmet-ric matrices can be computed by a paired two-term recurrence relation, or a general number sequence of second order. As a result, the complexities of the formulas are of order n. Furthermore, the formulas have no divisions at all, i.e., they fall into the class of breakdown-free algorithms.
1 1 Introduction The determinant is one of the basic parameters in matrix theory. The determinant of a square matrix A = (ai,j) ∈Cn×n is defined as det(A) = X σ∈Sn sgn(σ) n Y i=1 ai,σ(i), where the symbol Sn denotes the group of permutations of sets with n elements and the symbol sgn(σ) denotes the signature of σ ∈Sn.
The Pfaffian of a skew-symmetric matrix A = (ai,j) ∈C2k×2k is defined by Pf(A) = 1 2kk!
X σ∈S2k sgn(σ) k Y i=1 aσ(2i−1),σ(2i), (1) and is closely related to the determinant. In fact, Cayley’s theorem states that the square of the Pfaffian of a matrix is equal to the determinant of the matrix, i.e., det(A) = Pf(A)2.
Matrix A is called a centrosymmetric matrix if A = JAJ−1, where J is the anti-diagonal matrix whose anti-diagonal elements are one with all others being zero. If A = −JAJ−1, the matrix is said to be skew-centrosymmetric. Skew-centrosymmetric matrices arise in many fields of science including numerical solutions of certain differential equations, digital signal processing, information theory, statistics, linear systems theory, and some Markov processes [1, 2, 3, 4, 5, 6].
In general, the complexities of the Pfaffian and the determinant are of the order O(n3).
This paper describes efficient computational formulas for the Pfaffians and determinants of special matrices for which the complexities of the formulas are of the order O(n). The formulas have no divisions at all, i.e., the formulas fall into the class of breakdown-free algorithms.
2 Pfaffians of skew-centrosymmetric matrices Definition 1. An = (ai,j) and Bn = (bi,j) denote n-by-n matrices with the following ele-ments: ai,j = a, if j = i + 1; −a, if i = j + 1; 0, otherwise, bi,j = ( (−1)i+1b, if i + j = n + 1; 0, otherwise, 2 where 1 ≤i, j ≤n.
Definition 2. Fn and Gn denote 2-by-2 block matrices of the following form: Fn = Ak Bk (−1)kBk Ak , Gn = Ak −Bk (−1)k+1Bk Ak , where n = 2k.
For example, if n = 10, it follows from Definition 2 that F10 = A5 B5 (−1)5B5 A5 = 0 a 0 0 0 0 0 0 0 b −a 0 a 0 0 0 0 0 −b 0 0 −a 0 a 0 0 0 b 0 0 0 0 −a 0 a 0 −b 0 0 0 0 0 0 −a 0 b 0 0 0 0 0 0 0 0 −b 0 a 0 0 0 0 0 0 b 0 −a 0 a 0 0 0 0 −b 0 0 0 −a 0 a 0 0 b 0 0 0 0 0 −a 0 a −b 0 0 0 0 0 0 0 −a 0 .
We now describe algorithms for computing the Pfaffians of Fn and Gn.
Theorem 3. Let {fn} and {gn} be the recursively defined sequences below: fn = bgn−1 + a2fn−2 for f1 = b, gn = −bfn−1 + a2gn−2 for g1 = −b.
Then, for n = 2k, we obtain fk = Pf(Fn) and gk = Pf(Gn), where f−1 = 0, f0 = 1 and g−1 = 0, g0 = 1.
Proof. The proof is done by induction on k. For k = 1, F2 = A1 B1 −B1 A1 = 0 b −b 0 and G2 = A1 −B1 B1 A1 = 0 −b b 0 .
The definition of the Pfaffian in (1) clearly indicates that Pf(F2) = b and Pf(G2) = −b.
Thus, f1 = b = Pf(F2), g1 = −b = Pf(G2).
3 Let us assume that the recurrence relations hold for all t ≤k. Then we show that they hold for k = t + 1.
F2t+2 = At+1 Bt+1 (−1)t+1Bt+1 At+1 = 0 a 0 · · · 0 b −a 0 0 At −Bt .
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(−1)t+1Bt At 0 0 a −b 0 · · · 0 −a 0 .
(2) From the expansion formula along with 2t + 2 column of (2), it follows that Pf(F2t+2) = bPf(G2t) + aPf(M2t) = bgt + aPf(M2t), (3) where M2t = 0 a 0 · · · · · · 0 −a 0 a 0 · · · 0 0 a .
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0 At−1 Bt−1 .
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(−1)t−1Bt−1 At−1 0 0 .
(4) From the expansion formula along with the first row of (4), it follows that Pf(M2t) = aPf(F2t−2) = aft−1.
(5) From (3) and (5), we have ft+1 = bgt + a2ft−1.
The recurrence relation for gt+1 can be obtained similarly.
Corollary 4. fn = (−1)n−1bfn−1 + a2fn−2 with f−1 = 0 and f1 = 1.
Corollary 4 shows that the computational costs of Pf(Fn) and det(Fn)(= Pf(Fn)2) are of the order O(n). Furthermore, the recurrences in Corollary 4 have no divisions. Thus, no breakdown occurs during the computation.
3 Determinant of the skew-centrosymmetric matrix In this section, we consider the determinant of the matrix Fn with n = 2k. It is well known from that the determinant of the 2-by-2 block matrix holds A B C D = det(AD −CB) 4 if AC = CA. Applying the above formula to Fn in Definition 2, the determinant of matrix Fn equals that of Tk := A2 k −(−1)kB2 k. Thus, we have |Fn| = |Tk| = det −a2 + b2 0 a2 0 −2a2 + b2 0 ...
a2 0 ...
...
a2 ...
...
−2a2 + b2 0 a2 0 −a2 + b2 k×k .
The matrix Tk belongs to the set of k-tridiagonal matrices. Sogabe and El-Mikkawy considered a fast block diagonalization of k-tridiagonal matrices using permutation matrices.
Exploiting the block diagonalization method, we can rearrange the matrix Tk as below.
(i) We consider the case where k is odd. Let us define the following matrices: H k−1 2 = (hi,j) = −2a2 + b2, if i = j; a2, if i = j + 1 or j = i + 1; 0, otherwise and K k+1 2 = (ki,j) = −a2 + b2, if i = j = 1 or i = j = k+1 2 ; −2a2 + b2, if i = j = 2 . . . k−1 2 ; a2, if i = j + 1 or j = i + 1; 0, otherwise.
Then, P TTkP = H k−1 2 0 0 K k+1 2 !
, where the permutation matrix P is determined by using the method in . Obviously, det(P TTkP) = det Tk = det Fn = det(H k−1 2 ) det(K k+1 2 ).
(ii) We consider the case where k is even. Let us define N k 2 = (ni,j) = −a2 + b2, if i = j = k 2; −2a2 + b2, if i = j = 1 . . . k 2 −1; a2, if i = j + 1 or j = i + 1; 0, otherwise 5 and Q k 2 = (qi,j) = −a2 + b2, if i = j = 1; −2a2 + b2, if i = j = 2 . . . k 2; a2, if i = j + 1 or j = i + 1; 0, otherwise.
Then, P TTkP = N k 2 0 0 Q k 2 !
.
Obviously, det(P TTkP) = det Tk = det Fn = det(N k 2 ) det(Q k 2 ).
It can be seen that det(N k 2 ) = det(Q k 2 ).
El-Mikkawy obtained two-term recurrence relation for the determinants of tridiagonal matrices, i.e., vi = d1 a1 0 . . .
0 b2 d2 a2 ...
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0 b3 d3 ...
0 .
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...
...
...
ai−1 0 . . .
0 bi di , where vi = divi−1 −biai−1vi−2 for v0 = 1 and v−1 = 0. Using this relation and the Laplace expansion, we obtain the result. If k is even, then det(N k 2 ) = det(Q k 2 ) = (−a2 + b2)w k 2 −1 −a4w k 2 −2.
If k is odd, then det(K k+1 2 ) = −a2 + b22 w k−3 2 −2a4(−a2 + b2)w k−5 2 + a8w k−7 2 , det(H k−1 2 ) = w k−1 2 , where wi = −2a2 + b2 a2 . . .
0 a2 −2a2 + b2 ...
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...
...
a2 0 . . .
a2 −2a2 + b2 .
Here wi = (−2a2 + b2)wi−1 −a4wi−2 for w0 = 1 and w−1 = 0 .
Consequently, for n = 2k, we obtain 6 (i) If k is odd, det Fn = det Tk = w k−1 2 −a2 + b22 w k−3 2 −2a4(−a2 + b2)w k−5 2 + a8w k−7 2 .
(ii) If k is even, det Fn = det Tk = (−a2 + b2)w k 2 −1 −a4w k 2 −2 2 .
4 Examples Some examples of the Pfaffian and the determinant of the matrix Fn (n = 2k) are shown in the following tables. Here Fn, Pn, and Jn are the nth Fibonacci, Pell, and Jacobsthal numbers, respectively.
a = i, b = 1 a = i, b = 2 a = i √ 2, b = 1 k Pf(F2k) Pf(F2k) Pf(F2k) 1 F2 = 1 P2 = 2 J2 = 1 2 −F3 = −2 −P3 = −5 −J3 = −3 3 −F4 = −3 −P4 = −12 −J4 = −5 4 F5 = 5 P5 = 29 J5 = 11 5 F6 = 8 P6 = 70 J6 = 21 6 −F7 = −13 −P7 = −169 −J7 = −43 7 −F8 = −21 −P8 = −408 −J8 = −85 8 F9 = 34 P9 = 985 J9 = 171 .
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≡0, 1 (mod 4) Fk+1 Pk+1 Jk+1 ≡2, 3 (mod 4) −Fk+1 −Pk+1 −Jk+1 Examples of the Pfaffians 7 a = i, b = 1 a = i, b = 2 a = i √ 2, b = 1 k det (F2k) det (F2k) det (F2k) 1 F 2 2 P 2 2 J2 2 2 F 2 3 P 2 3 J2 3 3 F 2 4 P 2 4 J2 4 4 F 2 5 P 2 5 J2 5 5 F 2 6 P 2 6 J2 6 6 F 2 7 P 2 7 J2 7 7 F 2 8 P 2 8 J2 8 8 F 2 9 P 2 9 J2 9 .
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t F 2 t+1 P 2 t+1 J2 t+1 Examples of the determinants 5 Acknowledgment This work has been supported in part by JSPS KAKENHI (Grant No. 26286088).
The authors sincerely appreciate the referee’s comments that enhanced the quality of the manuscript.
References A. L. Andrew, Centrosymmetric matrices, SIAM Rev. 40 (1988), 697–698.
A. L. Andrew, Eigenvectors of certain matrices, Lin. Alg. Appl. 7 (1973), 151–162.
F. Zhang, Matrix Theory: Basic Results and Techniques, Springer, 1999.
L. Datta and S. Morgera, On the reducibility of centrosymmetric matrices- applications in engineering problems, Circ. Syst. Signal. Pr. 8 (1989), 71–96.
M. El-Mikkawy and F. Atlan, On solving centrosymmetric linear systems, Appl. Math.
4 (2013), 21–32.
J. Weaver, Centrosymmetric (cross-symmetric) matrices, their basic properties, eigenval-ues, and eigenvectors, Amer. Math. Monthly 92 (1985), 711–717.
R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer, 1999.
8 T. Sogabe and M. El-Mikkawy, Fast block diagonalization of k-tridiagonal matrices, Appl.
Math. Comput. 218 (2011), 2740–2743.
M. El-Mikkawy, A note on a three-term recurrence for a tridiagonal matrix, Appl. Math.
Comput. 139 (2003), 503–511.
2010 Mathematics Subject Classification: Primary 15A15; Secondary 15A23.
Keywords: Pfaffian, determinant, skew-centrosymmetric matrix.
(Concerned with sequences A000045, A000129, A001045.) Received June 20 2016; revised versions received July 8 2016; January 18 2017; February 2 2017. Published in Journal of Integer Sequences, February 11 2017.
Return to Journal of Integer Sequences home page.
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