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13900 | https://www.ashp.org/-/media/assets/policy-guidelines/docs/therapeutic-guidelines/therapeutic-guidelines-monitoring-vancomycin-ASHP-IDSA-PIDS.pdf | ASHP REPORT Address correspondence to Dr. Rybak (m.rybak@wayne.edu).
Keywords: nephrotoxicity, pharmacokinetics and pharmacodynamics, target attainment, vancomycin, vancomycin consensus guideline © American Society of Health-System Pharmacists 2020. All rights reserved. For permissions, please e-mail: journals.
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DOI 10.1093/ajhp/zxaa036 Michael J. Rybak, PharmD, MPH, PhD, FCCP, FIDP, FIDSA, Anti-Infective Research Laboratory, Department of Pharmacy Practice, Eugene Applebaum College of Pharmacy & Health Sciences, Wayne State University, Detroit, MI, School of Medicine, Wayne State University, Detroit, MI, and Detroit Receiving Hospital, Detroit, MI Jennifer Le, PharmD, MAS, FIDSA, FCCP, FCSHP, BCPS-AQ ID, Skaggs School of Pharmacy and Pharmaceutical Sciences, University of California San Diego, La Jolla, CA Thomas P. Lodise, PharmD, PhD, Albany College of Pharmacy and Health Sciences, Albany, NY, and Albany Medical Center Hospital, Albany, NY Donald P. Levine, MD, FACP, FIDSA, School of Medicine, Wayne State University, Detroit, MI, and Detroit Receiving Hospital, Detroit, MI John S. Bradley, MD, JSB, FIDSA, FAAP, FPIDS, Department of Pediatrics, Division of Infectious Diseases, University of California at San Diego, La Jolla, CA, and Rady Children’s Hospital San Diego, San Diego, CA Catherine Liu, MD, FIDSA, Division of Allergy and Infectious Diseases, University of Washington, Seattle, WA, and Vaccine and Infectious Disease Division, Fred Hutchinson Cancer Research Center, Seattle, WA Bruce A. Mueller, PharmD, FCCP, FASN, FNKF, University of Michigan College of Pharmacy, Ann Arbor, MI Manjunath P. Pai, PharmD, FCCP, University of Michigan College of Pharmacy, Ann Arbor, MI Annie Wong-Beringer, PharmD, FCCP, FIDSA, University of Southern California School of Pharmacy, Los Angeles, CA John C. Rotschafer, PharmD, FCCP, University of Minnesota College of Pharmacy, Minneapolis, MN Keith A. Rodvold, PharmD, FCCP, FIDSA, University of Illinois College of Pharmacy, Chicago, IL Holly D. Maples, PharmD, University of Arkansas for Medical Sciences College of Pharmacy & Arkansas Children’s Hospital, Little Rock, AR Benjamin M. Lomaestro, PharmD, Albany Medical Center Hospital, Albany, NY T he first consensus guideline for ther-apeutic monitoring of vancomycin in adult patients was published in 2009. A committee representing 3 organiza-tions (the American Society for Health-System Pharmacists [ASHP], Infectious Diseases Society of America [IDSA], and Society for Infectious Diseases Pharmacists [SIDP]) searched and re-viewed all relevant peer-reviewed data on vancomycin as it related to in vitro and in vivo pharmacokinetic and phar-macodynamic (PK/PD) characteristics, including information on clinical effi-cacy, toxicity, and vancomycin resistance in relation to serum drug concentration and monitoring. The data were summar-ized, and specific dosing and monitoring recommendations were made. The pri-mary recommendations consisted of eliminating routine monitoring of serum peak concentrations, emphasizing a ratio of area under the curve over 24 hours to minimum inhibitory concentration (AUC/MIC) of ≥400 as the primary PK/ PD predictor of vancomycin activity, and promoting serum trough concentrations of 15 to 20 mg/L as a surrogate marker for the optimal vancomycin AUC/MIC if the MIC was ≤1 mg/L in patients with normal renal function. The guideline also recommended, albeit with limited data support, that actual body weight be used to determine the vancomycin dosage and loading doses for severe infections in pa-tients who were seriously ill.1 Since those recommendations were generated, a number of publications have evaluated the impact of the 2009 guidelines on clinical efficacy and tox-icity in patients receiving vancomycin for the treatment of methicillin-resistant Staphylococcus aureus (MRSA) infec-tions. It should be noted, however, that when the recommendations were orig-inally published, there were important issues not addressed and gaps in know-ledge that could not be covered ade-quately because of insufficient data. In fact, adequate data were not avail-able to make recommendations in the original guideline for specific dosing and monitoring for pediatric patients outside of the neonatal age group; spe-cific recommendations for vancomycin dosage adjustment and monitoring in the morbidly obese patient popu-lation and patients with renal failure, including specific dialysis dosage ad-justments; recommendations for the use of prolonged or continuous in-fusion (CI) vancomycin therapy; and safety data on the use of dosages that exceed 3 g per day. In addition, there were minimal to no data on the safety and efficacy of targeted trough concen-trations of 15 to 20 mg/L.
This consensus revision evaluates the current scientific data and contro-versies associated with vancomycin dosing and serum concentration moni-toring for serious MRSA infections (in-cluding but not limited to bacteremia, sepsis, infective endocarditis, pneu-monia, osteomyelitis, and meningitis) and provides new recommendations based on recent available evidence. Due to a lack of data to guide appro-priate targets, the development of this guideline excluded evaluation of van-comycin for methicillin-susceptible S. aureus (MSSA) strains, coagulase-negative staphylococci, and other pathogens; thus, the extrapolation of Therapeutic monitoring of vancomycin for serious methicillin-resistant Staphylococcus aureus infections: A revised consensus guideline and review by the American Society of Health-System Pharmacists, the Infectious Diseases Society of America, the Pediatric Infectious Diseases Society, and the Society of Infectious Diseases Pharmacists Supplementary material is available with the full text of this article at AJHP online.
Am J Health-Syst Pharm. 2020; XX:XX-XX AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 1 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING guideline recommendations to these pathogens should be viewed with ex-treme caution. Furthermore, serious invasive MRSA infections exclude nonbacteremic skin and skin structure and urinary tract infections. Since this guideline focuses on optimization of vancomycin dosing and monitoring, recommendations on the appropriate-ness of vancomycin use, combination or alternative antibiotic therapy, and multiple medical interventions that may be necessary for successful treat-ment of invasive MRSA infections are beyond the scope of this guideline and will not be presented.
Methods These are the consensus state-ments and guideline of ASHP, IDSA, the Pediatric Infectious Diseases Society (PIDS), and SIDP. Guideline panel com-position consisted of physicians, phar-macists, and a clinical pharmacologist with expertise in clinical practice and/or research with vancomycin. Committee members were assigned key topics re-garding vancomycin dosing and moni-toring. A draft document addressing these specific areas was reviewed by all committee members and made avail-able for public comments for 30 days through ASHP, IDSA, PIDS, and SIDP. The committee then met to review and revise the document based on the sub-mitted comments, suggestions, and recommendations. After careful discus-sion and consideration, the document was revised and circulated among the committee and supporting organiza-tions prior to final approval and publi-cation. A search of PubMed and Embase was conducted using the following search terms: vancomycin, pharmacoki-netics, pharmacodynamics, efficacy, re-sistance, toxicity, obesity, and pediatrics. All relevant and available peer-reviewed studies in the English-language litera-ture published from 1958 through 2019 were considered. Studies were rated by their quality of evidence, and the subse-quent recommendations were graded using the classification schemata de-scribed in Table 1.
Potential limitations of this review included the fact that there are few published randomized clinical trials of vancomycin dosing and monitoring available in the literature. Most pub-lished studies evaluating vancomycin dosing, dosage adjustment, and moni-toring were retrospective PK or PD clinical assessments or retrospective observational studies in patients with MRSA infections.
PK/PD efficacy targets. To optimize the dosing of any antimicro-bial agent, a firm understanding of the drug’s exposure-effect and exposure-toxicity links are required. While a va-riety of PD indices for vancomycin have been suggested, an AUC/MIC ratio of ≥400 (with the MIC determined by broth microdilution [BMD]) is the current ac-cepted critical PK/PD index in light of our limited experience and studies evaluating AUC/MIC values of <400.1,3-7 In vitro and in vivo assessments of PK/ PD models applicable to human MRSA infection have found that bactericidal activity (ie, a 1- to 2-log reduction in bac-terial inoculum in the animal model) is achieved when the vancomycin AUC/ MICBMD ratio approximates or exceeds 400. Furthermore, in vitro data sug-gest that an AUC of <400 potentiates the emergence of MRSA resistance and vancomycin-intermediate S. aureus strains.8,9 There are also mounting clin-ical data, albeit mostly retrospective in nature, in support of this PK/PD target for vancomycin.10-18 A summary of these investigations and their findings can be found in eTable 1.10-17,19-23 Clinical PK/PD Data: Adults While an AUC/MICBMD ratio of ≥400 is currently considered the optimal PK/PD “efficacy” target, it is important to recognize that this target has been largely derived from retrospective, single-center, observational studies of patients with MRSA bloodstream infec-tions.11-17 It is also important to recog-nize that most of the landmark clinical studies that established the contem-porary PK/PD efficacy target relied on simple vancomycin clearance (CL) for-mulas based on daily vancomycin dose and estimated renal function to deter-mine AUC values.10,11,13 Current evalu-ation of these data demonstrates that Table 1. Grading System for Recommendations Based on Quality of Evidencea Category and Grade Definition Strength of recommendation A Good evidence to support a recommendation for or against use B Moderate evidence to support a recommendation for or against use C Poor evidence to support a recommendation Quality of evidence I Evidence from 1 or more properly randomized con-trolled trials II Evidence from 1 or more well-designed clinical trials, without randomization; from cohort or case-controlled analytic studies (preferably from more than 1 center); from multiple time-series; or from dramatic results from uncontrolled experiments III Evidence from opinions of respected authorities, based on clinical experience, descriptive studies, or reports of expert committees aAdapted from the Canadian Task Force on the Periodic Health Examination.2 2 AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING these CL formulas provide imprecise estimates of the AUC.24-26 This finding is not surprising, as there is consid-erable interpatient variability in van-comycin exposure profiles in clinical practice, and it is not possible to gen-erate valid estimates of exposure vari-ables in a given individual based on CL formulas that are derived from glomer-ular filtration rate estimation equations alone.10,11,13 In most cases, the formula-based approach will overestimate van-comycin CL by approximately 40% to 50%.16 While it has been cumbersome to estimate AUC in the clinical setting in the past, Neely and colleagues24 re-cently demonstrated that Bayesian soft-ware programs (refer to Therapeutic Monitoring section) can be used to gen-erate accurate and reliable estimates of the daily AUC values with trough-only PK sampling. However, the accuracy of AUC estimation is higher with peak and trough measurements compared to trough-only PK sampling.24 Using this validated Bayesian method to es-timate the daily AUC in a single-center, retrospective study of patients with MRSA bloodstream infections, Lodise and colleagues16 found that outcomes were maximized when day 1 and day 2 AUC/MICBMD ratios exceeded 521 and 650, respectively. Employing the same Bayesian approach to estimate daily AUC values, Casapao and col-leagues17 also noted that the risk of vancomycin treatment failure among patients with MRSA infective endo-carditis was greatest among those with an AUC/MICBMD ratio of ≤600 and that this exposure-failure relationship persisted after adjusting for factors such as intensive care unit (ICU) ad-mission, presence of heteroresistant vancomycin-intermediate S. aureus, and other comorbidities. In contrast to the studies by Lodise et al and Casapao et al, several small-scale, retrospec-tive clinical evaluations of vancomycin exposure-response reported lower Bayesian-derived thresholds for AUC/ MIC since the AUC was measured at steady-state conditions and indexed to the MIC, as determined by the Etest (bioMérieux USA, Hazelwood, MO) method, to arrive at an AUC/MICEtest value.12,14,15 The MICEtest value tends to be 1.5- to 2-fold higher than the MICBMD value; therefore, it is likely that the AUC threshold needed for response from these 3 studies,12,14,15 if calculated using the MICBMD, would align with the studies by Lodise et al16 and Casapao et al.17 In an effort to surmount the limitations associated with previous single-center, retrospective vancomy cin exposure-response clinical ana-lyses, a multicenter, observational prospective study was performed to evaluate the relationship between the prespecified day 2 AUC/MIC ratios (ie, AUC/MICEtest of ≥320 and AUC/MICBMD of ≥650) and outcomes in adult pa-tients (n = 265) with MRSA bacteremia. In the multivariate analyses, treatment failure rates were not significantly dif-ferent between the prespecified day 2 AUC/MIC groups. Post hoc global out-comes analyses suggested that patients in the 2 lowest AUC exposure quintiles (ie, those with an AUC of ≤515 mg·h/L) experienced the best global outcome (defined as absence of both treatment failure and acute kidney injury [AKI]). While global outcomes were similar in the 2 lowest AUC-exposure quin-tiles, only 20% of the study population (n = 54) had an AUC of ≤400 mg·h/L, and it is unclear if efficacy outcomes are maintained at an AUC less than this threshold of 400 mg·h/L.23 Notably, the higher AUC value cited above (515 mg·h/L) provides a new index that in-corporates both efficacy and AKI that is still within the recommended AUC range of 400 to 600 mg·h/L (assuming a MIC of 1 mg/L).
Collectively, recent studies highlight the importance of generating valid esti-mates of the AUC values through Bayesian modeling techniques when conducting vancomycin exposure-outcomes analyses in patients. Current vancomycin exposure-effectiveness data originated largely from studies of MRSA bacteremia, with some studies for pneumonia and infective en-docarditis and none for osteomyelitis and meningitis. Furthermore, outcomes data for a MIC of 2 mg/L are limited, sug-gesting the need for more studies to ascer-tain the optimal AUC/MIC target for this MIC value or consideration for the use of alternative antibiotics. The currently avail-able data also highlight the critical need for large-scale, multicenter, randomized, vancomycin dose–optimized clinical out-comes trials. As data from future prospec-tive, multicenter clinical studies emerge, it is important that clinicians recognize that our current understanding of the PK/PD target associated with maximal effect and toxicity is subject to change, and this may ultimately alter the current way we dose vancomycin to optimize effect and mini-mize toxicity.
Toxicodynamics: AKI A major concern with vancomycin use is the occurrence of AKI. While multiple definitions of vancomycin-associated AKI have been employed in the literature, most studies defined it as an increase in the serum creatinine (SCr) level of ≥0.5 mg/dL, or a 50% in-crease from baseline in consecutive daily readings, or a decrease in calcu-lated creatinine CL (CLcr) of 50% from baseline on 2 consecutive days in the absence of an alternative explanation.1 Recently, it has been proposed that a more sensitive threshold (ie, an increase in SCr of ≥0.3 mg/dL over a 48-hour pe-riod) may be considered as an indicator of vancomycin-associated AKI. This threshold was adopted from the Acute Kidney Injury Network (AKIN) and the Kidney Disease: Improving Global Outcomes (KDIGO) criteria.27-29 The incidence of vancomycin-associated AKI has varied across published studies. In a meta-analysis by van Hal and colleagues,29 the prevalence of vancomycin-associated AKI varied from 5% to 43%. Similarly, a recent meta-analysis of 13 studies by Sinha Ray et al30 reported that the relative risk of AKI with vancomycin was 2.45 (95% confidence interval, 1.69-3.55), with an attributable risk of 59%. Most episodes of AKI developed between 4 and 17 days after initiation of therapy. Many patients, especially those who are critically ill, do not fully recover renal AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 3 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING function after AKI,31 and even mild AKI can significantly decrease long-term survival rates, increase morbidity, pro-long hospitalizations, and escalate healthcare costs.22,32 With any drug, an understanding of its toxicodynamic profile is required for optimal dosing. Several studies, largely retrospective in nature, have attempted to quantify the relationship between vancomycin exposure and probability of AKI.33,34 Although data are limited, the collective literature suggests that the risk of AKI increases as a function of the trough concentration, espe-cially when maintained above 15 to 20 mg/L.29 Similarly, there are recent data to suggest that the risk of AKI in-creases along the vancomycin AUC continuum, especially when the daily AUC exceeds 650 to 1,300 mg·h/L.24,33-35 Furthermore, animal studies corrob-orate the finding that increased AUC rather than trough concentration is a strong predictor of AKI.36,37 Suzuki et al33 evaluated the mean vancomycin AUC in relation to AKI. Most patients who developed AKI had AUC values between 600 and 800 mg·h/L, compared with 400 to 600 mg·h/L in those without AKI (P = 0.014). Furthermore, Lodise and colleagues34 showed that the probability of AKI in-creased 2.5-fold among patients with AUC values above 1,300 mg·h/L com-pared with patients with lower values (30.8% vs 13.1%, P = 0.02). Although AUC values above 1,300 mg·h/L were associ-ated with a substantial increase in AKI, an AUC exposure-response relationship appeared to exist, and the probability of a nephrotoxic event increased as a func-tion of the daily AUC and patient’s body weight.38 A study by Zasowski et al21 also reported a similar relationship between Bayesian-estimated vancomycin AUC thresholds and AKI in 323 patients; AUC values of ≥1,218 mg·h/L for 0 to 48 hours, ≥677 for 0 to 24 hours, and ≥683 for 24 to 48 hours or troughs of ≥18.2 mg/L were associated with a 3- to 4-fold in-creased risk of nephrotoxicity. Similarly, the aforementioned multicenter, pro-spective study of patients with MRSA bloodstream infections found that AKI increased along the day 2 AUC con-tinuum in a stepwise manner and that patients with day 2 AUC values of ≥793 mg·h/L were at the greatest risk for AKI.23 Given the understanding about po-tential toxic concentrations, there are also data to suggest that AUC-guided vancomycin dosing may reduce the oc-currence of vancomycin-associated AKI. In a retrospective, quasi-experimental study of 1,280 hospitalized patients, Finch et al20 compared the incidence of nephrotoxicity in patients monitored by individualized AUC vs trough concen-tration. AUC-guided dosing was found to be independently associated with a significant decrease in AKI (odds ratio [OR], 0.52; 95% CI, 0.34-0.80; P = 0.003).20 Median Bayesian-estimated AUC was significantly lower with AUC-guided dosing vs trough monitoring (474 [SD, 360-611] mg·h/L vs. 705 [SD, 540-883] mg·h/L; P < 0.001). In the prospective study by Neely et al,22 252 patients were monitored via troughs of 10 to 20 mg/L in year 1 vs Bayesian-estimated AUC values of ≥400 mg·h/L in years 2 and 3 of the investigation. Nephrotoxicity oc-curred in 8% of subjects in year 1 and in 0% and 2% of subjects in years 2 and 3, respectively (P = 0.01). The median trough concentration and AUC values associated with AKI were 15.7 mg/L and 625 mg·h/L, as compared with values of 8.7 mg/L and 423 mg·h/L in subjects without AKI (P = 0.02).22 Collectively, the published clinical exposure-response analyses suggest that a daily AUC of ≥400 is the driver of effectiveness and that the risk of AKI is related to AUC and trough values. More importantly, these data provide the foundation for the current under-standing of the therapeutic window for vancomycin. When evaluating the toxicodynamics of vancomycin, it is im-portant to recognize other factors that may complicate or exacerbate the risk of AKI. Host-related factors associated with nephrotoxicity include increased weight, pre-existing renal dysfunc-tion, and critical illness. Concurrent administration of nephrotoxic agents such as aminoglycosides, loop di uretics, amphotericin B, intravenous (i.v.) contrast dye, and vasopressors has been shown to increase the risk of nephrotoxicity. Recently, piperacillin/ tazobactam and flucloxacillin have been reported to increase the risk for AKI in patients receiving vanco-mycin.39-44 It is unclear if the threshold for vancomycin-induced AKI varies according to these covariates, but clin-icians should be mindful of the poten-tial for additional risk when prescribing vancomycin to patients when these conditions are present.34,40-50 Based on the current best avail-able evidence, daily vancomycin AUC values (assuming a MIC of 1 mg/L) should be maintained between 400 and 600 mg·h/L to minimize the likelihood of nephrotoxicity and maximize effi-cacy for suspected or definitive serious invasive MRSA infections. Once culture results or the clinical presentation rule out invasive MRSA infection, the em-piric use of vancomycin at guideline-recommended exposures should be de-escalated, either by a decrease in vancomycin exposure or initiation of alternative antibiotics. Extrapolation of guideline recommendations to non-invasive MRSA and other pathogens should be viewed with extreme caution.
Therapeutic Monitoring Therapeutic monitoring has cen-tered on maintaining trough con-centrations between 15 and 20 mg/L for serious infections due to MRSA. Previous expert guidelines recom-mended monitoring trough concen-trations as a surrogate marker for the AUC/MIC ratio based on the historical difficulty in estimating the AUC in clin-ical practice.1,5 In the past, calculation of AUC in clinical practice involved col-lection of multiple vancomycin serum concentrations during the same dosing interval, with subsequent use of PK software that was not readily available at all institutions. As such, the guide-line viewed trough-directed dosing as a more practical alternative to AUC/MIC-guided dosing in clinical practice.
Although the recommendation to maintain trough values between 15 and 20 mg/L for serious infections 4 AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING due to MRSA has been well integrated into practice, the clinical benefits of maintaining higher vancomycin trough values have not been well docu-mented.38,51-55 From a PK/PD perspec-tive, it is not surprising that there are limited clinical data to support the range of 15 to 20 mg/L. Recent studies have demonstrated that trough values may not be an optimal surrogate for AUC values.26,56,57 While trough attain-ment ensures achievement of a min-imum cumulative exposure, a wide range of concentration-time profiles can result in an identical trough value. Patel et al26 reported a wide range of AUC values from several different dosing regimens yielding similar trough values. The therapeutic discordance between trough and AUC values is not surprising, as the AUC is the integrated quantity of cumulative drug exposure (ie, the serum drug concentration–time curve over a defined interval). In con-trast, the trough represents a single ex-posure point at the end of the dosing interval. In clinical practice, monitoring of trough concentrations will translate into achievement of one specific min-imum daily AUC value, whereas the 24-hour AUC (AUC24) largely represents the average concentration during that time period [AUC24 (mg·h/L) = average concentration (mg/L) x 24 (hours)]. For troughs of 15 to 20 mg/L, this typically equates to a daily AUC in excess of 400 mg·h/L. However, there is considerable variability in the upper range of AUC values associated with a given trough value. Although trough-only moni-toring is practical, the potential limita-tions surrounding the practice suggest that trough monitoring may be insuffi-cient to guide vancomycin dosing in all patients.
Although the AUC/MIC ratio is con-sidered the PK/PD driver of efficacy for vancomycin, clinicians trying to opti-mize vancomycin treatment for patients with serious MRSA infections may be best advised to use AUC-guided dosing and assume a MICBMD of 1 mg/L (unless it is known, through BMD, to be greater or less than 1 mg/L). The MIC value is of less importance for several reasons. First, the range of vancomycin MIC values among contemporary MRSA isolates is narrow, and the BMD MIC90 in most institutions is 1 mg/L or less.58-62 Second, measurement of MIC values is imprecise, with dilution of ±1 log2 and variation of 10% to 20% considered ac-ceptable; therefore, the variability of reported MIC values encountered in routine clinical practice is likely to re-flect measurement error.63 Third, there is a high degree of variability between commercially available MIC testing methods relative to the BMD method (see Vancomycin Susceptibility Testing section). Last, MIC results are typically not available within the first 72 hours of index culture collection, yet cur-rent data indicate that the vancomycin AUC/MIC ratio needs to be optimized early in the course of infection.
Daily AUC values (assuming a MICBMD of 1 mg/L) should be main-tained between 400 and 600 mg·h/L to maximize efficacy and minimize the likelihood of AKI. In the past, AUC monitoring required the collection of multiple concentrations over the same dosing interval. With these data, a clini-cian would calculate the AUC using the linear-trapezoid rule. This approach required precise collection of vanco-mycin concentrations, making it largely impractical outside of a research set-ting. However, this is no longer the case. It is now possible to accurately estimate the AUC with limited PK sampling. One such approach involves the use of Bayesian software programs to esti-mate the vancomycin AUC value with minimal PK sampling (ie, 1 or 2 van-comycin concentrations) and provide AUC-guided dosing recommendations in real time. An alternative approach involves use of 2 concentrations (peak and trough) and simple analytic PK equations to estimate AUC values.57,64 Bayesian-derived AUC moni-toring. Bayesian-guided dosing is based in part on Bayes’ Theorem, as it quantifies the sequential relationship between the estimated probability distribution of an individual patient’s PK parameter values (eg, volume [Vd] or CL) prior to administering the drug based on the way the drug behaved in a population of prior patients (the Bayesian prior) and the revised probability distribution of a specific patient’s PK parameter values using exact dosing and drug concentration data (the Bayesian conditional poste-rior). In short, Bayesian dose optimi-zation software uses a well-developed vancomycin population PK model as the Bayesian prior, together with the individual patient’s observed drug concentrations in the data file, to cal-culate a Bayesian posterior parameter value distribution for that patient. The dose optimization software then calculates the optimal dosing reg-imen based on the specific patient’s profile.65-67 An advantage of the Bayesian ap-proach is that vancomycin concentra-tions can be collected within the first 24 to 48 hours rather than at steady-state conditions (after the third or fourth dose), and this information can be used to inform subsequent dosing (adap-tive feedback control). As part of their output, Bayesian dosing programs pro-vide innovative treatment schemes, such as front-loading doses with subse-quent transition to a lower maintenance dosing regimen, to rapidly achieve target concentrations within the first 24 to 48 hours among critically ill patients. The Bayesian approach also provides the ability to integrate covariates, such as CLcr, in the structural PK models (the Bayesian prior density file) that account for the pathophysiological changes that readily occur in critically ill patients. Incorporation of covariates that account for these “dynamic” changes serves as a way to identify dosing schemes that optimize effect and predict future dosing in a patient who has an evolving PK profile.67 Bayesian dose-optimizing software programs are now readily available and can be used in real time to identify the optimal vancomycin dosage that readily achieves the AUC target (assuming a MICBMD of 1 mg/L).66,68 Bayesian programs offer numerous advantages over the traditional first-order equa-tion software programs. Using richly AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 5 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING sampled vancomycin PK data from 3 studies comprising 47 adults with varying renal function, Neely and col-leagues24 demonstrated that Bayesian software programs, embedded with a PK model based on richly sampled van-comycin data as the Bayesian prior, can be used to generate accurate and reli-able estimates of the daily AUC values with trough-only PK sampling. Of note, there was limited inclusion of special populations in this study, and it is un-clear if this trough-only Bayesian AUC estimation approach can be applied to obese patients, critically ill patients, pediatric patients, and patients with unstable renal function. A random-ized controlled study of 65 subjects by Al-Sulaiti et al69 showed that estimating AUC using both peak and trough con-centrations (vs trough-only estimates) may improve vancomycin-associated therapeutic cure. Until more data are available, it is preferred to estimate the Bayesian AUC using 2 vancomycin con-centrations (peak and trough).
First-order PK analytic equa-tions. Alternatively, the AUC can be accurately estimated based on the col-lection of 2 timed steady-state serum vancomycin concentrations and the use of first-order PK equations.57 The equations used to compute AUC from 2 samples are based in part on an original approach proposed by Begg, Barclay, and Duffull70 for aminoglycosides and modified by Pai and Rodvold.57 It is preferred that a near steady-state, postdistributional peak (1 to 2 hours after end of infusion) and trough con-centrations within the same dosing interval (if possible) are used when estimating the AUC with the equation-based methods.
The major advantage of this ap-proach is that it is simpler and re-lies on fewer assumptions than the Bayesian approach. The first-order PK equations used to estimate the AUC are also familiar to most clinicians, facilitating ease of use in practice. Once the AUC24 is estimated, the clini-cian simply revises the total daily dose to achieve the desired AUC24, as alter-ations of total daily dose will provide proportional changes in observed AUC24.
6,71-73 The major limitation of this approach is that it is not adaptive like the Bayesian approach, as it can only provide a snapshot of the AUC for the sampling period. As such, this AUC cal-culation will not be correct if a physio-logic change such as renal dysfunction occurs during or after the sampling period. Furthermore, it is extremely dif-ficult to estimate the vancomycin AUC24 with the equation-based method in pa-tients who receive multiple dosing re-gimens within a 24-hour period. If the vancomycin dosing interval is more frequent than once a day, the AUC24 will be a function of the number of identical doses administered during that interval (eg, AUC must be multiplied by 2 for a 12-hour dosing interval to calculate the true AUC24). It is also highly preferred that concentrations are collected near steady-state conditions.
Despite its drawbacks, this es-timate of AUC is a clear step above trough-only or peak-only concentra-tion interpretation and is familiar to most clinicians. Several large medical centers within the United States have already adopted this approach of ac-quiring 2 postdose serum concentra-tion estimates of the AUC to perform routine vancomycin dosing and moni-toring and have demonstrated a con-siderable improvement in safety over the current trough-only concentration monitoring method.37,64 PK sampling time. Timing of achievement of targeted AUC values (assuming a MICBMD of 1 mg/L) re-mains unclear. The early AUC/MIC target ratios derived in animal models were based on the AUC value from 0 to 24 hours.3,4 More recent clinical assess-ments that identified a link between AUC/MIC ratio and outcomes also as-sessed the AUC values achieved early in the course of therapy.1,3,5-7,10,13,20-22 The 2009 vancomycin guideline stated that the trough should be assessed prior to steady-state conditions (ie, prior to the fourth dose).1,5 In fact, steady-state conditions are difficult to determine in clinical practice, and the timing of the fourth dose is more dependent on the dosing interval (ie, 12 vs 24 hours) than steady-state conditions. Given the importance of early, appropriate therapy,74 targeted AUC exposures should be achieved early during the course of therapy, preferably within the first 24 to 48 hours. If monitoring is initiated after the first dose, the con-tribution of the loading dose to the actual AUC may vary depending on the magnitude of the loading dose vs maintenance doses. The decision to delay therapeutic monitoring beyond 48 hours should be based on severity of infection and clinical judgment.
Summary and recommendations: 1. In patients with suspected or definitive serious MRSA infections, an individu-alized target of the AUC/MICBMD ratio of 400 to 600 (assuming a vancomycin MICBMD of 1 mg/L) should be advo-cated to achieve clinical efficacy while improving patient safety (A-II). Doses of 15 to 20 mg/kg (based on actual body weight) administered every 8 to 12 hours as an intermittent infusion are recommended for most patients with normal renal function when as-suming a MICBMD of 1 mg/L (A-II). In patients with normal renal function, these doses may not achieve the thera-peutic AUC/MIC target when the MIC is 2 mg/L. 2. Given the narrow vancomycin AUC range for therapeutic effect and minimal AKI risk, the most accurate and optimal way to manage van-comycin dosing should be through AUC-guided dosing and monitoring (A-II). We recommend to accomplish this in one of two ways.
a. One approach relies on the collec-tion of 2 concentrations (obtained near steady-state, postdistributional peak concentration [Cmax] at 1 to 2 hours after infusion and trough concentration [Cmin] at the end of the dosing interval), preferably but not required during the same dosing interval (if possible) and utilizing first-order PK equations to estimate the AUC (A-II).
6 AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING b. The preferred approach to monitor AUC involves the use of Bayesian software programs, embedded with a PK model based on richly sampled vancomycin data as the Bayesian prior, to optimize the delivery of vancomycin based on the collection of 1 or 2 vancomycin concentrations, with at least 1 trough. It is preferred to obtain 2 PK samples (ie, at 1 to 2 hours post infusion and at end of the dosing interval) to estimate the AUC with the Bayesian approach (A-II). A trough concentration alone may be sufficient to estimate the AUC with the Bayesian approach in some patients, but more data are needed across different patient populations to confirm the viability of using trough-only data (B-II). 3. When transitioning to AUC/MIC monitoring, clinicians should con-servatively target AUC values for patients with suspected or docu-mented serious infections due to MRSA assuming a vancomycin MICBMD of 1 mg/L or less at most institutions. Given the importance of early, appropriate therapy, vanco-mycin targeted exposure should be achieved early during the course of therapy, preferably within the first 24 to 48 hours (A-II). As such, the use of Bayesian-derived AUC monitoring may be prudent in these cases since it does not require steady-state serum vancomycin concentrations to allow for early assessment of AUC target attainment. 4. Trough-only monitoring, with a target of 15 to 20 mg/L, is no longer recommended based on efficacy and nephrotoxicity data in patients with serious infections due to MRSA (A-II). There is insufficient evidence to provide recommendations on whether trough-only or AUC-guided vancomycin monitoring should be used among patients with noninva-sive MRSA or other infections. 5. Vancomycin monitoring is re-commended for patients receiving vancomycin for serious MRSA infections to achieve sustained targeted AUC values (assuming a MICBMD of 1 mg/L unless it is known to be greater or less than 1 mg/L by BMD). Independent of MRSA infection, vancomycin monitoring is also recommended for all pa-tients at high risk for nephrotoxicity (eg, critically ill patients receiving concurrent nephrotoxins), patients with unstable (ie, deteriorating or significantly improving) renal func-tion, and those receiving prolonged courses of therapy (more than 3 to 5 days). We suggest the frequency of monitoring be based on clinical judgment; frequent or daily moni-toring may be prudent for hemo-dynamically unstable patients (eg, those with end-stage renal disease), with once-weekly monitoring for hemodynamically stable patients (B-II).
Vancomycin Susceptibility Testing With the MIC being a component of the vancomycin AUC/MIC targeted surrogate for efficacy, it is important to be aware of local and national van-comycin susceptibility patterns for MRSA. Although in some centers there has been a steady increase in the av-erage vancomycin MIC over several decades, recent national and inter-national studies that have evaluated MRSA susceptibility to glycopeptides, lipopeptides, and beta-lactams have demonstrated that vancomycin MICs have remained constant over time, with a MIC of ≤1 mg/L demonstrated for more than 90% of isolates.58-62 A meta-analysis of 29,234 MRSA strains from 55 studies revealed that the MIC deter-minations performed by BMD, Etest, and automated systems were predomi-nately 1 mg/L and that there was no ev-idence of a MIC creep phenomenon.75 Furthermore, a global surveillance pro-gram reported that 95% of 57,319 MRSA isolates had MICs of 1 mg/L, with no signs of MIC creep over 20 years.76 While there does not seem to be a large number of organisms with a vanco-mycin MIC of ≥2 mg/L when reference methods are used, there is considerable variability in MIC results between the susceptibility testing methods.
The challenge is that, according to the Clinical Laboratory Standards Institute (CLSI), acceptable variability for MIC measurement methods is within ±1 doubling dilution (essential agree-ment), such that current susceptibility testing methods are unable, with high reproducibility, to distinguish MICs of 1 mg/L from MICs of 0.5 mg/L or 2 mg/L. Most institutions routinely perform MIC testing using automa ted systems: BD Phoenix (BD, Franklin Lakes, NJ), MicroScan WalkAway (Beckman Coulter, Brea, CA), or Vitek 2 (bioMérieux), and in some cases the Etest methodology (bioMérieux). In a study of 161 MRSA blood isolates, when using the essential agreement defini-tion of ±1 log2 dilution error, Vitek 2 and MicroScan WalkAway demonstrated a 96.3% agreement with BMD, whereas BD Phoenix demonstrated an 88.8% agreement.77 The Etest method had the lowest agreement with BMD, at 76.4% (results were consistently higher by 1 to 2 dilutions). The Etest will likely pro-duce a higher value (0.5 to 2 dilutions higher) than BMD. In another study, 92% of the strains were demonstrated to have a vancomycin MIC of 1 mg/L by BMD; corresponding figures were 70% for MicroScan WalkAway and Etest and 41% for Vitek 1.78 Rybak et al79 compared MicroScan WalkAway, Vitek 2, BD Phoenix, and Etest to BMD methods among 200 MRSA strains. In contrast to previous studies, these investigators used an absolute agreement definition of ±0 log2 dilution error to better charac-terize the precision. Using this def-inition, results with BD Phoenix and MicroScan WalkAway had the highest agreement with BMD (66.2% and 61.8%, respectively), followed by Vitek 2 (54.3%). As noted above, Etest tended to produce results that were 1 to 2 dilutions higher (agree-ment with BMD was 36.7%). However, when compared to BMD, Etest identi-fied a MIC of 2 mg/L 80% of the time. When compared to BMD, Micro Scan WalkAway (prompt method) AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 7 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING overcalled MIC values of 1 mg/L by 74.1%, and BD Phoenix and Vitek 2 undercalled MIC values of 2 mg/L by 76% and 20%, respectively.
The high variability of MIC results among the 4 systems compared to BMD clearly poses a challenge to the clinician making treatment decisions based on MIC and poses questions as to the most relevant MIC method.79 This variability between MIC values and testing methods routinely per-formed at most institutions further supports the use of AUC (assuming a MICBMD of 1 mg/L) to guide vanco-mycin empiric dosing. For nonserious infections, this variability may be in-consequential. In a critically ill patient infected by MRSA, who may require prompt achievement of the target AUC/MIC, it is imperative to verify the MIC by a standardized method (preferably BMD, as Etest may result in a higher MIC than BMD) as soon as possible to avoid a delay in effec-tive therapy. An AUC/MICBMD of 400 to 600 is approximately equivalent to an AUC/MICEtest of 200 to 400, re-flecting values that are 1 to 2 dilutions higher than those yielded by Etest. Furthermore, there are no data to sup-port decreasing the dose to achieve the targeted AUC/MIC of 400 to 600 if the MIC is less than 1 mg/L.
Summary and recommendations: 6. Based on current national vancomycin susceptibility surveillance data, under most circumstances of empiric dosing, the vancomycin MIC should be assumed to be 1 mg/L. When the MICBMD is >1 mg/L, the probability of achieving an AUC/MIC target of ≥400 is low with conventional dosing; higher doses may risk unnecessary toxicity, and the decision to change therapy should be based on clinical judgment. In addition, when the MICBMD is <1 mg/L, we do not recom-mend decreasing the dose to achieve the AUC/MIC target. It is important to note the limitations in automated sus-ceptibility testing methods, including the lack of precision and variability in MIC results depending on method used (B-II).
Continuous Infusion vs Intermittent Infusion Administration of vancomycin by continuous infusion (CI) has been evaluated as an alternative to intermit-tent infusion (II) with potential advan-tages of earlier target attainment, less variability in serum concentrations, ease of drug level monitoring (less de-pendence on sampling time or multiple concentrations to calculate AUC), and lower the potential risk of AKI.
Comparative studies. Published studies that compared intermittent to continuous administration primarily focused on 2 distinct populations, adult critically ill patients in the ICU with sus-pected or documented infections and those receiving outpatient antimicro-bial therapy (OPAT) for bone and joint infections.80-89 Most studies compared CI to II for the risk of AKI and attain-ment of target serum concentrations; only 4 studies included other outcome endpoints such as treatment failure and mortality.80,84,87,89 Measures of vanco-mycin drug exposure reported in clin-ical trials include trough and average steady-state concentrations and AUC24. One challenge when comparing clin-ical outcomes between CI and II is the lack of consistent reporting of exposure parameters between groups treated using the 2 dosing strategies. For CI, the most commonly reported drug ex-posure parameter was the steady-state concentration, while for II it was the trough concentration. For future inves-tigations it would be beneficial to report AUC and/or average steady-state con-centration for both CI and II groups to enable direct comparison of drug expo-sure between groups and correlate with efficacy and safety endpoints.
Critically ill patients. A total of 7 studies compared CI vs II of vanco-mycin in critically ill patients.81-87 Only one study, by Wysocki et al,80 evalu-ated both efficacy and safety in a pro-spective randomized trial comparing vancomycin CI (n = 61) and II (n = 58) in 119 patients. Most patients had pneumonia or bacteremia, mostly due to MRSA. Mean serum steady-state and trough concentrations attained were 24 mg/L and 15 mg/L, respectively, for both the CI and II groups. AUC24 values were comparable between the CI and II groups, with significantly less variability in the CI group (P = 0.026). Clinical failure rates were similar in the CI and II groups on day 10 (21% vs 26%) and at end of treatment (21% vs 19%), although the mean AUC24 was shown to be lower in the CI group than in the II group (596 [SD, 159] mg·h/L vs 685 [SD, 260] mg·h/L, P < 0.05). Nephrotoxicity occurred in 18% of pa-tients overall, with similar rates in the CI and II groups (16% vs 19%). However, dialysis was required more often in those who received CI vs II (6 of 10 pa-tients vs 3 of 11 patients). Risk factors for nephrotoxicity such as diabetes and concomitant diuretic, aminoglycoside, and iodine use were similar between groups. It is notable that the study only had 23% power to detect a difference in clinical outcomes between groups.1 Another study compared mor-tality among critically ill burn patients receiving CI (n = 90) or II (n = 81).84 Mortality rates in the hospital and on days 14 and 28 were numerically higher for those receiving CI, but the differ-ences did not reach statistical signifi-cance (10% vs 6.2%, 18.9% vs 11%, and 32% vs 21%, respectively). However, when mortality was compared by treat-ment indications, those who received CI for non–gram-positive sepsis had significantly higher mortality (70% vs 16.7%, P = 0.001); nearly half of this sub-group had gram-negative bacteremia or candidemia. It is possible that the dif-ference in outcome may be attributed to differences in the management of those infections and not directly related to van-comycin administration. Nephrotoxicity occurred numerically less frequently in the CI group than in the II group (per-centage of patients with increase in Scr of 0.5 mg/dL at end of therapy, 6.7% vs 14.8%). While higher mean vancomycin concentrations were noted in the CI 8 AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING group relative to the II group (20 [SD, 3.8] mg/L vs 14.8 [SD, 4.4] mg/L, P < 0.001), which would be expected when com-paring steady-state and trough concen-trations, AUC24 was not reported, thereby precluding comparison of drug exposure between the CI and II groups.
Five other studies compared serum drug concentrations achieved and the risk of nephrotoxicity between CI and II in critically ill patients.81-83,85,86 As ex-pected, the ranges of measured vanco-mycin concentrations from the studies were significantly higher in the CI groups than in the II groups (steady-state con-centrations of 20-25 mg/L vs troughs of 10-15 mg/L, respectively). Another study showed that a higher percentage of patients attained a vancomycin con-centration of >20 mg/L at least once during the treatment course with CI vs II administration (63.2% vs 44.9%, P = 0.065).82 One study reported lower mean AUC24 with CI vs II (529 [SD, 98] mg·h/L vs 612 [SD, 213] mg·h/L, P value not stated), and increased steady-state concentration compared with trough (25 ± 4 vs 17 ± 4.7 mg/L, respectively, P = 0.42) with CI vs. II.83 The discordance observed in the relationship of trough concentration and AUC24 underscores the importance of measuring AUC24 to compare relative drug exposure with CI vs II in future studies.
In general, the rate of nephrotoxicity was reported to be similar or numeri-cally lower with CI vs II administration (range, 4%-16% vs 11%-19%); the same trend but higher rates were reported in studies that applied the AKIN criteria for nephrotoxicity (26%-28% vs 35%-37%).81-83,85,86 In addition, Saugel et al85 noted significantly less frequent need for renal replacement therapy (RRT) during van-comycin treatment for patients in the CI group than for those in the II group (7% [7 of 94 patients] vs 23% [12 of 52 pa-tients] required RRT; P = 0.007). Of in-terest, in the largest retrospective study comparing CI and II, conducted in 1,430 ICU patients, Hanrahan et al86 reported a higher rate of nephrotoxicity in those re-ceiving CI vs II (25% [161 of 653 patients] vs 20% [77 of 390 patients]; P = 0.001); bivariate analysis indicated that every 1-mg/L increase in serum concentra-tion was associated with an 11% increase in the risk of nephrotoxicity, with lower odds in those receiving II. However, lo-gistic regression analysis indicated the contrary in that II was associated with an 8-fold higher odds of nephrotoxicity (95% confidence interval, 2.87-23.41). The lack of information provided on confounding variables such as receipt of concomitant nephrotoxins and relative AUCs between treatment groups pre-clude drawing a definitive conclusion regarding the safety of CI, especially in light of the disparate results of bivariate and logistic regression analyses.
Patients receiving OPAT. To date there have been 2 studies comparing the efficacy of vancomycin adminis-tration by CI vs II in patients whose therapy was initiated in the hospital and continued as OPAT. Duration of therapy ranged from 30 days to 14 weeks.87,89 Most patients were treated for bone and joint and skin structure–related infec-tions. In a small prospective study, rates of osteomyelitis cure, defined as re-maining asymptomatic 12 months after completion of therapy, did not differ significantly between groups (94% vs 78%, P = 0.3), but only 27 patients were evaluable.87 Another study retrospec-tively evaluated the efficacy of vanco-mycin in patients with MRSA infections; most had bone and joint and skin struc-ture–related infections, while 10% had bloodstream infections or endocar-ditis.89 Rates of clinical failure were sim-ilar in the CI and II groups (19% [25 of 133 patients] vs 25% [9 of 36], P = 0.41) after excluding 29% of study patients who had subtherapeutic serum van-comycin concentrations for more than 1 week. However, it is not clear how frequent serum concentrations were monitored, if in-hospital treatment du-ration before OPAT differed between groups, and whether treatment success rates differed by type of infection.
In studies that evaluated the safety of CI vancomycin as OPAT, treatment dura-tion ranged from 4 to 14 weeks, with a re-ported average mean steady-state serum concentration of 13 to 30 mg/L.87,88 In a retrospective matched cohort study of 80 patients, a trend towards less fre-quent occurrence of nephrotoxicity was observed in the CI group vs the II group (10% vs 25%, P = 0.139), and when neph-rotoxicity did occur it had a later onset in the CI group (P = 0.036).88 Patients were matched by age, comorbid condi-tions, gender, baseline Scr, and receipt of concurrent nephrotoxins; those who had an Scr of ≥1.5 mg/dL at baseline, developed nephrotoxicity as inpatients prior to OPAT, or experienced hypo-tension resulting in renal dysfunction were excluded. In another retrospective study,90 the same investigators identified a steady-state average concentration of 28 mg/L as the threshold breakpoint for the development of nephrotoxicity using CART (classification and regression tree) analysis: Nephrotoxicity occurred in 71.4% (5 of 7) and 11.6% (11 of 95) pa-tients with steady-state concentrations of ≥28 mg/L and <28 mg/L, respectively. In one prospective study of an elderly cohort (mean age, 70 years) receiving high-dose vancomycin therapy by CI, with targeting of a steady-state concen-tration of 30 to 40 mg/L for a median duration of 6 weeks, nephrotoxicity oc-curred in 32% of patients. Additionally, 4 patients in that study developed leukopenia.91 Dosing and other consider-ations for use of CI. Most published studies of critically ill patients receiving vancomycin CI employed a loading dose of 15 to 20 mg/kg followed by daily maintenance infusions at doses of 30 to 40 mg/kg (up to 60 mg/kg) to achieve a target steady-state concentration of 20 to 25 mg/L. By simply multiplying the steady-state concentration by 24, a target steady-state concentration of 20 to 25 mg/L would equate to an AUC24/ MIC of 480 to 600 (assuming a MIC of 1 mg/L). Of note, the PK/PD target for CI has not been validated. All of the PK/ PD data supporting an AUC24/MIC ratio of >400 as the best correlate for clinical outcomes were derived from patients who received II vancomycin dosing.
Rapid attainment of target serum concentrations has been cited as a po-tential advantage of CI over II when treating acute infections, particularly AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 9 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING in ICU patients early during the course of infection. In 2 comparative studies, target steady-state concentrations of 20 to 25 mg/L were achieved more rapidly with use of CI vs II: in a mean time of 36 (SD, 31) hours vs 51 (SD, 39) hours (P = 0.03) in one study and 16 (SD, 8) hours vs 50 (SD, 21) hours (P < 0.001) in the other.81,83 Importantly, less var-iability in the steady-state concentra-tion and fewer blood samples (a single steady-state concentration vs both peak and trough concentrations) are required to calculate AUC24 among pa-tients receiving CI vs II. Timing of blood sampling for trough determinations is critical during II, whereas steady-state concentration can be measured any time after steady state has been reached during CI. In addition, vanco-mycin administration by CI in patients receiving OPAT has the theoretical ad-vantage of a need for less frequent ac-cess to the i.v. catheter and thus less complications resulting from thrombus formation or infections. On the other hand, incompatibility of vancomycin with certain drugs (particularly at high concentrations), that are commonly administered in the critical care setting is a notable challenge of vancomycin CI.92,93 The use of proper concentration, alternative agents, independent lines, or multiple catheters may be warranted if vancomycin is to be administered by CI.
Summary and recommendations: 7. The pharmacokinetics of CI suggest that such regimens may be a reason-able alternative to conventional II dosing when the AUC target cannot be achieved (B-II). Based on cur-rently available data, a loading dose of 15 to 20 mg/kg, followed by daily maintenance CI of 30 to 40 mg/kg (up to 60 mg/kg) to achieve a target steady-state concentration of 20 to 25 mg/L may be considered for crit-ically ill patients (B-II). AUC24 can be simply calculated by multiplying the steady-state concentration (ie, the desired therapeutic range of 20 to 25 mg/L throughout the entire dosing interval) by a factor of 24. Attaining the desired drug exposure may be more readily accomplished given the ease of sampling time and dosage adjustment by changing the rate of infusion, which is a highly desirable feature in critically ill patients (B-II). 8. The risk of developing nephrotoxicity with CI appears to be similar or lower than that with intermittent dosing when targeting a steady-state concen-tration of 15 to 25 mg/L and a trough concentration of 10 to 20 mg/L (B-II). Definitive studies are needed to com-pare drug exposure based on meas-ured AUC24 and factors that predispose to development of nephrotoxicity, such as receipt of concomitant nephro-toxins, diuretics, and/or vasopressor therapy in patients receiving CI vs II of vancomycin. 9. Incompatibility of vancomycin with other drugs commonly coadministered in the ICU requires the use of independent lines or mul-tiple catheters when vancomycin is being considered for CI (A-III).
Loading Doses Loading doses of vancomycin have been evaluated in several studies during the past decade.94-109 Providing loading doses of 20 to 35 mg/kg based on ac-tual body weight rapidly achieves tar-geted ranges of serum vancomycin concentrations and decreases the risk of subtherapeutic concentrations during the first days of therapy. Loading doses are recommended in patients who are criti-cally ill or in the ICU,95-102 require dialysis or renal replacement therapy,102-106 or are receiving vancomycin CI therapy.94-98,105,108 While this approach is not currently sup-ported by evidence from large random-ized clinical trials, vancomycin loading doses can be considered in the treatment of serious MRSA infections. Vancomycin should be administered in a dilute solu-tion (eg, concentrations of no more than 5 mg/mL) and infused over a period of not less than 60 minutes or at a rate of 10 to 15 mg/min (≥1 hour per 1,000 mg) to minimize infusion-related adverse events. An infusion rate of 10 mg/min or less is associated with fewer infusion-related events. Loading doses of 25 to 35 mg/kg will require infusion times of at least 2 to 3 hours.99 After administra-tion of the loading dose, the initiation of the maintenance dose should occur at the next dosing interval (eg, an interval of every 6 hours indicates initiating the maintenance dose 6 hours after the start of the loading dose).
In most studies that have employed loading doses, vancomycin dosing was based on actual body weight. While this practice is commonplace, dosing by actual body weight assumes there is a linear relationship between key population PK parameters (ie, Vd and clearance) and the body size descriptor employed. While a wide variety of ac-tual weight–based estimates of Vd (for example, 0.4 to 1 L/kg) have been re-ported in the literature,6 mounting data suggest that it is not entirely ac-curate to describe vancomycin Vd as being proportional to body weight, particularly among obese patients (refer to Dosing in Obesity section). As noted in several recent articles dis-cussing vancomycin PK in obesity, as weight increases the coefficient used to calculate Vd decreases.48,110,111 At this point, dosing should be based on ac-tual body weight, with doses capped at 3,000 mg (refer to Dosing in Obesity section.112 More intensive therapeutic monitoring should also be performed in obese patients.
Summary and recommendations: 10. In order to achieve rapid attain-ment of targeted concentrations in critically ill patients with suspected or documented serious MRSA in-fections, a loading dose of 20 to 35 mg/kg can be considered for intermittent-infusion administration of vancomycin (B-II).1 11. Loading doses should be based on actual body weight and not exceed 3,000 mg (refer to Dosing in Obesity section). More intensive and early therapeutic monitoring should 10 AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING also be performed in obese patients (B-II).
Dosing in Obesity The original vancomycin dosing strategies predate our current defin-itions of obesity and understanding of drug pharmacokinetics in obesity. Obesity is defined as a body mass index (BMI) of ≥30 kg/m2 and is currently divided into 3 tiers: class I obesity (30.0-34.9 kg/m2), class II obesity (35.0-39.9 kg/m2), and class III, or morbid, obesity (≥40 kg/m2).113 The prevalence of obesity increased from approxi-mately 10% in the 1950s to 39.8% in 2015-2016, and the average US adult weighs approximately 83 kg, compared to the historical standard of 70 kg.114,115 This shift in the distribution of body size is relevant to the calculation of vancomycin doses based on patient body weight. Obesity may be associated with an increased risk of vancomycin-induced nephrotoxicity, in part due to supratherapeutic exposure resulting from maintenance doses calculated using actual body weight.45,116 The selection of vancomycin loading dose is dependent on the es-timated Vd. Pharmacokinetic studies have repeatedly demonstrated that the vancomycin Vd increases with actual body weight; however, this PK param-eter does not increase with actual body weight in a proportionate manner and is not reliably predictable in obese indi-viduals.111,117-121 Blouin and colleagues111 demonstrated a statistically significant difference in weight-indexed Vd be-tween obese and nonobese patients. Similarly, using data from 704 patients, Ducharme and colleagues118 found that mean weight-indexed vancomycin Vd decreased with increasing body size. The average weight-indexed Vd in a study by Bauer and colleagues119 was much lower in 24 morbidly obese pa-tients (0.32 L/kg) than in 24 patients of normal weight (0.68 L/kg, P < 0.001). Recent studies in obese adults corrob-orate these findings and suggest that lower Vd estimates of approximately 0.5 L/kg or weight-independent central tendency estimates approaching 75 L are observed in obese adults.112,120,121 The nonlinear relationship between vancomycin Vd and body weight can be resolved with piecewise functions of al-ternate weight descriptors, allometric scaling, use of lower mg/kg doses with increasing body size, or capping the dose at a threshold.118,122 The underlying rationale for a loading dose is rapid at-tainment of therapeutic concentra-tions. Therefore, using actual body weight–based loading doses of 20 to 25 mg/kg (doses lower than previously recommended), with consideration of capping doses at 3,000 mg, is the most practical strategy in obese patients with serious infections.112 For example, this strategy would result in calculated loading doses of 1,500 to 2,500 mg in patients weighing 80 to 99 kg, 2,000 to 3,000 mg in those weighing 100 to 119 kg, and 2,500 to 3,000 mg in pa-tients with a weight of ≥120 kg (doses rounded to the nearest 250 mg). The decision of whether or not to employ a loading dose, as well as the magnitude of this dose, should be driven by the severity of infection and the urgency to achieve a therapeutic concentration rather than by body size alone.
Empiric maintenance dosing of vancomycin is reliant on estimated CL. Vancomycin CL is predicted by kidney function, which is most commonly estimated as CLcr with the Cockcroft-Gault equation using patient age, sex, Scr, and body size.123 There is consider-able controversy regarding the optimal body size metric for this calculation in obese patients.124 The Cockcroft-Gault equation predates the global standard-ization of Scr measurement traceable to isotopic-dilution mass spectrometry (IDMS) standards advocated to reduce intralaboratory and interlaboratory measurement variability.124 A recent population PK study by Crass and col-leagues112 of obese patients (n = 346) with BMI values of 30.1 to 85.7 kg/m2 and body weights of 70 to 294 kg pro-vided an equation to estimate vanco-mycin CL based on age, sex, Scr (IDMS traceable), and allometrically scaled body weight. This model or similar approaches to estimating vancomycin CL, such as that defined by Rodvold and colleagues,125 can be used to esti-mate the total daily maintenance dose. The population model–estimated van-comycin CL multiplied by the target AUC estimates the initial daily main-tenance dose.112,120,122 For example, studies report an average vancomycin CL of approximately 6 L/h in obese patients that equates to achieving an AUC of approximately 500 mg·h/L with a daily dose of 3,000 mg. Empiric van-comycin maintenance dosages above 4,500 mg/day are not expected in obese adults, because vancomycin CL rarely exceeds 9 L/h.112,120,121 Population PK models of vanco-mycin cannot account for more than 50% of the interindividual variabili ty, which supports therapeutic drug monitoring (TDM) in this popula-tion.117,118,120,122 A reliable estimate of vancomycin Vd is necessary for AUC estimation when AUC is based solely on a trough concentration measure-ment.24,121,126,127 This bias is addressed and precision is improved by meas-urement of both a peak (collected at least 1 hour after the end of infu-sion) and a trough concentration to estimate AUC accurately in obese patients.126 Once a reliable PK es-timate of vancomycin elimination is determined by using these 2 con-centration measurements, subsequent vancomycin AUC estimation is achiev-able with trough-only measurements by Bayesian methods in physiologically stable patients.57 For critically ill obese patients with unstable physiology, ad-ditional work to design adaptive feed-back models to tailor doses is needed.
Summary and recommendations: 12. A vancomycin loading dose of 20 to 25 mg/kg using actual body weight, with a maximum dose of 3,000 mg, may be considered in obese adult patients with serious infections (B-II). Initial mainte-nance doses of vancomycin can be computed using a population PK AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 11 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING estimate of vancomycin clearance and the target AUC in obese patients. Empiric maintenance doses for most obese patients usually do not exceed 4,500 mg/day, depending on their renal function (B-II). Early and fre-quent monitoring of AUC exposure is recommended for dose adjustment, especially when empiric doses exceed 4,000 mg/day (A-II). Measurement of peak and trough concentrations is recommended to improve the accu-racy of vancomycin AUC estimation and maintenance dose optimi-zation in obese patients, aligning with recommendations 2 and 5 for nonobese adults.
Renal Disease and Renal Replacement Therapies Intermittent hemodialysis. Despite the common use of vanco-mycin in patients receiving hemodial-ysis, there are few published outcome studies that provide guidance on the optimal PK/PD targets in this popula-tion. Previously published drug dosing recommendations generally targeted a predialysis serum concentration, even though other PD targets may be more appropriate. Predialysis vancomycin concentration to MRSA MIC ratios of >18.6 have been associated with im-proved bacteremic patient outcomes, suggesting that serum concentration monitoring is essential throughout the course of therapy.128 Dosing to achieve predialysis vancomycin concentrations of 10 to 20 mg/L, as has been done clin-ically,129 results in mean AUC24 values ranging from 250 to 450 mg·h/L, with some values below the AUC/MIC goals recommended in other populations.130 Outcome studies validating the AUC24h goal of 400 to 600 mg·h/L used in other patient populations have not been con-ducted in the hemodialysis population. Nonetheless, the maintenance doses recommended in this section aim to reach this AUC24 target (ie, 400-600 mg·h/L), as recommended throughout this document.
Many dialysis-related factors affect the degree of vancomycin exposure in these patients. These considerations include the amount of time between vancomycin dose administration and the scheduled time of the next dialysis session,104 whether the dose is given during dialysis or after hemodialysis has ended, and the dialyzer’s per-meability if the dose is administered intradialytically.131 Dialysis frequency also plays a role in dosing decisions. For non–critically ill patients receiving hemodialysis, 2 or 3 days is the most common interdialytic period. Some critically ill patients with severe catab-olism and AKI may require more than thrice-weekly hemodialysis for optimal metabolic control, and their main-tenance vancomycin doses should be based on serum concentration monitoring.132 Vancomycin dosing in patients with acute or chronic kidney failure has transformed over time due to the changes in dialysis technology and techniques.133 Older (pre-1990s) hemodialyzers were not very perme-able to large molecules. Vancomycin (with a molecular weight of 1,450 Da) was not considered “dialyzable” be-cause it poorly crossed the hemodi-alysis membranes of the era. Indeed, even today’s vancomycin package in-sert, based on PK studies conducted in the 1980s, states that “vancomycin is poorly removed by dialysis.” 134 As hemodialysis membrane technology has improved, dialyzers have become far more permeable. Vancomycin is cleared substantially by contemporary high-permeability hemodialyzers135,136; consequently, vancomycin dosing strategies have changed substantially as well. For example, in spite of the package insert statement “In anuria, a dose of 1000 mg every 7 to 10 days has been recommended” and the statement that “vancomycin is poorly removed by dialysis,” 134 far more fre-quent doses are needed to maintain therapeutic serum concentrations in patients receiving hemodialysis. The extent of vancomycin removal by dial-ysis is dependent on the permeability of the hemodialyzer used131; conse-quently, investigators have developed and published a wide variety of vanco-mycin dosing protocols in an attempt to compensate for the increase in vanco-mycin dialytic CL caused by increases in dialyzer permeability.
An added complication of appro-priate vancomycin dosing in patients receiving hemodialysis is the prevailing practice of administering the drug during the final hours of the hemodialysis pro-cess, thus resulting in some of the infused drug being removed immediately by the hemodialyzer. This practice started back when low-permeability dialyzers were used and little vancomycin was elimin-ated by hemodialysis. The practice has persisted at most dialysis units because most dialysis units treat 3 shifts of patients per day, and holding a dialysis chair for 60 to 90 additional minutes while vanco-mycin infuses into a patient is not cost-ef-fective. Indeed, it is more cost-effective to infuse “extra” vancomycin during the hemodialysis session to compensate for intradialytic loss than it is to keep a dial-ysis unit open later to allow vancomycin infusions. Intradialytically infused vanco-mycin results in reduced delivery of drug to the patient, similar to a first-pass phe-nomenon. The extent of intradialytic drug removal is variable and depends on pa-tient and dialysis system factors, the most important of which is dialyzer membrane permeability.135,137-139 Approximately 20% to 40% of an intradialytically adminis-tered vancomycin dose is removed by the simultaneous hemodialysis, with the highly permeable dialyzers tending to the higher end of this range.137,140,141 Maintenance dosing strategies that do not provide a dose with every he-modialysis session (eg, a maintenance dose is given with every second or third hemodialysis session) have been studied,102,142,143 but none have been found to meet vancomycin exposure goals in the last day of the dosing in-terval without giving massive doses that result in very high peak concen-trations. Consequently, maintenance vancomycin doses are recommended to be administered with each hemo-dialysis session to ensure therapeutic serum concentrations throughout the dosing interval. In the typical 12 AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING thrice-weekly hemodialysis schedule, 25% larger doses are needed for the 3-day interdialytic period (eg, Friday to Monday) to maintain sufficient vanco-mycin exposure on the third day.130,144 Dosing that is weight based ap-pears to be superior to standard dosing schemes that do not account for patient size. Further, doses should be based on actual body weight rather than a calcu-lated body weight (see Dosing in Obesity section for considerations on how to dose morbidly obese patients). Because vancomycin is water soluble, vanco-mycin dosing in fluid overloaded patients should also be based on actual body weight at the time of dosing rather than on some calculated adjusted weight.102-105 Serum concentration monitoring is a valuable tool to guide vancomycin dosing in patients receiving dialysis, provided that serum concentrations are obtained and interpreted correctly. For example, blood sampling for assess-ment of vancomycin concentrations should not occur during or for at least 2 hours after a hemodialysis treatment. These samples will not be reflective of the true vancomycin body load because of the dialytic removal of vancomycin. Vancomycin serum concentrations will be low immediately following a dialysis treatment but will rebound substan-tially as drug redistributes from the tis-sues back to the blood over the next few hours.131,142,145 Dosing decisions based on serum concentrations obtained during or soon after hemodialysis will be inherently incorrect and could re-sult in administration of doses higher than necessary.145 Serum concentration monitoring performed with blood sam-ples obtained prior to the hemodialysis treatment is recommended to guide dosing, although other serum concen-tration monitoring techniques have been suggested.146 Dosing to achieve predialysis vanco-mycin concentrations of 10 to 20 mg/L, as has been conducted clinically,129 re-sults in mean AUC24 values ranging from 250 to 450 mg·h/L, often below the AUC/ MIC goals recommended in other popu-lations.130 Outcome studies validating the AUC target of 400 to 600 mg·h/L used in other patient populations have not been conducted in the hemodialysis popula-tion. While determination of AUC/MIC attainment is recommended, limited serum concentration monitoring is pos-sible in patients receiving hemodialysis in the outpatient setting for 2 reasons. The first reason is that frequent phlebotomy must be avoided in order to preserve future hemodialysis vascular access needs; the second is that it is imprac-tical to obtain blood samples aside from the predialysis sample that is obtained from the blood catheter inserted for use in the dialysis process. Patients leave the dialysis unit after hemodialysis and do not return until the next dialysis session days later. Consequently, since data are unavailable for an optimal AUC target in these patients, and no data are available to demonstrate efficacy below an AUC threshold value of 400, the goal should be to attain the AUC target of 400 to 600 mg·h/L used in other patient popula-tions. It is most practical to continue monitoring based on predialysis con-centrations and extrapolate these values to estimate AUC. Maintaining predialysis concentrations between 15 and 20 mg/L is likely to attain the AUC target of 400 to 600 mg·h/L in the previous 24 hours, with higher AUC/MIC values occurring on days prior.
Summary and recommendations: 13. The following tabulation outlines recommended vancomycin loading and maintenance doses for pa-tients receiving hemodialysis, with accounting for permeability of the dialyzer and whether the dose is ad-ministered intradialytically or after dialysis ends (B-II).
Timing and Dialyzer Permeability Vancomycin Dose, mg/kga After dialysis ends Low permeability Loading: 25 Maint.: 7.5b High permeability Loading: 25 Maint.: 10b Timing and Dialyzer Permeability Vancomycin Dose, mg/kga Intradialytic Low permeability Loading: 30 Maint.: 7.5-10b High permeability Loading: 35 Maint.: 10-15b aFrom references 104, 129, 130, 137, 138, 140, and 147.
bThrice-weekly dose administration. 14. Since efficacy data are unavailable for AUC values of <400 mg·h/L, monitoring based on predialysis serum concentrations and extrapo-lating these values to estimate AUC is most practical. Maintaining predialysis concentrations be-tween 15 and 20 mg/L is likely to achieve the AUC of 400 to 600 mg·h/L in the previous 24 hours (C-III). Predialysis serum con-centration monitoring should be performed not less than weekly and should drive subsequent dosing, as opposed to a strict weight-based recommendation, although these recommended doses provide a useful starting point until serum concentrations have been deter-mined (B-II).
Hybrid hemodialysis ther-apies. Contemporary renal replace-ment therapies used to treat kidney disease have expanded well beyond thrice-weekly, 3- to 4-hour hemodial-ysis sessions. In the outpatient setting, shorter, more frequent home hemodi-alysis treatments are used in a growing number of patients. In the inpatient set-ting, various types of “hybrid” hemodi-alysis therapies are employed. These hybrid treatments go by many names, including prolonged intermittent renal replacement therapy (PIRRT) and slow-low efficiency dialysis (SLED). Essentially these hybrid therapies use standard hemodialysis machines that run at slower blood and dialysate flow rates and for longer durations (usually 6 to 12 hours per day). Even hemodi-alysis itself differs in the inpatient and outpatient settings, as patients with AKI AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 13 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING are often hemodynamically unstable and lack sufficient vascular access for robust blood flow through the dialysis vascular access. All these hybrid dial-ysis therapies clear vancomycin to a different extent than standard intermit-tent hemodialysis.148,149 The timing of the vancomycin dose in relation to the hybrid hemodialysis session is essen-tial in determining a dosing regimen. If hybrid hemodialysis is started soon after the dose is administered, much of the dose will be removed, whereas the same vancomycin dose given after the dialysis session ends will yield a much larger AUC24 and much higher average serum concentrations. As is the case with any hemodialysis therapy, serum concentrations obtained during or within 2 hours from the end of hemo-dialysis will be artificially low because dialysis will have efficiently removed vancomycin from the blood, and van-comycin located in the tissues will not have had time to redistribute back into the bloodstream. Calculation of main-tenance doses based on an intra- or postdialytic vancomycin serum con-centration may result in doses that are too high. Caution is recommended in basing any maintenance dosing on these serum concentration values.
Little has been published on the patient outcomes achieved when van-comycin is used in patients receiving hybrid dialysis. Authors of one small case series of 27 courses of vancomycin given to patients receiving a hybrid he-modialysis therapy reported that pre-scribers have tried a wide variety of dosing schemes.150 By these authors’ criteria, 89% of the prescribed vanco-mycin doses in their institution were too low. Given the absence of outcome data in patients receiving these ther-apies, it seems prudent to use the same vancomycin AUC goal recommended throughout this document (400 to 600 mg·h/L assuming a MIC of 1 mg/L).
Summary and recommendations: 15. Loading doses of 20 to 25 mg/kg actual body weight should be used, recognizing that these hybrid dialysis therapies efficiently remove vanco-mycin (B-III). Initial doses should not be delayed to wait for a dialysis treat-ment to end. Maintenance doses of 15 mg/kg should be given after hybrid hemodialysis ends or during the final 60 to 90 minutes of dialysis, as is done with standard hemodialysis (B-III).130 Concentration monitoring should guide further maintenance doses.
Continuous renal replacement therapies. The use of continuous renal replacement therapy (CRRT) mo-dalities like continuous venovenous hemofiltration (CVVH), continuous venovenous hemodialysis (CVVHD), and continuous venovenous hemodiafiltration (CVVHDF) has grown in popularity in critically ill patients with AKI because of their superior ability to provide fluid and solute balance. Provided these therapies operate in an uninterrupted fashion, vancomycin CL is relatively constant over the dosing interval, although CL may decline as the hemodiafilter clogs over time.151 Vancomycin is removed by CRRT and its CL is related closely to the rate of ultrafiltrate/dialysate flow,105 with hemodiafilter type being of lesser importance, because contemporary hemodiafilters are all very permeable to the drug.
In patients on CRRT, serum concen-tration attainment goals often are not met with conventional dosing.84,152 Although outcomes studies specific to patients re-ceiving CRRT have not been conducted, it seems prudent to apply the same van-comycin AUC/MIC target (ie, 400-600) in these critically ill patients as is recom-mended throughout this document.
Summary and recommendations: 16. Loading doses of 20 to 25 mg/kg by actual body weight should be used in patients receiving CRRT at con-ventional, KDIGO-recommended effluent rates of 20 to 25 mL/kg/h (B-II).153 Initial maintenance dosing for CRRT with effluent rates of 20 to 25 mL/kg/h should be 7.5 to 10 mg/kg every 12 hours (B-II). Maintenance dose and dosing in-terval should be based on serum con-centration monitoring, which should be conducted within the first 24 hours to ensure AUC/MIC targets are met.154 In fluid overloaded patients, doses may be reduced as patients become euvolemic and drug Vd de-creases. The use of CI of vancomycin in patients receiving CRRT appears to be growing,84,105 and this method could be used in place of intermittent vancomycin dosing, especially when high CRRT ultrafiltrate/dialysate flow rates are employed (B-II).
Pediatric Patients In 2011, prior to the availability of alternative agents for MRSA in pediat-rics, vancomycin was recommended as the drug of choice for invasive MRSA infections in children.5 Although there are limited prospective, comparative data on the value of vancomycin thera-peutic monitoring in adults with respect to improving outcomes and decreasing toxicity, virtually no prospectively col-lected data on outcomes of MRSA infec-tion in newborns, infants and children exist. Further, for newborns (particu-larly premature infants) compared with older infants, immature renal elimina-tion mechanisms and a relative increase in Vd by body weight further complicate dosing guidelines during the first sev-eral weeks of life. Additional complexity for dosing strategies during early child-hood is based on a continual maturation of glomerular filtration, which is directly related to vancomycin CL. The glomer-ular filtration rate increases through the first years of life to rates in school-aged children that are greater than those in adults, with subsequent decline during the teen years to adult normal rates. Such a diversity of PK parameter values based on developmental pharmacology from neonates to adolescents pro-vides a challenge to develop general-ized vancomycin dosing. However, this has improved with the application of population-based PK models using al-lometric scaling and renal maturation 14 AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING covariates. In a population-based PK study by Colin and colleagues155 that evaluated vancomycin PK throughout the entire age continuum from infancy to geriatric years using pooled data from 14 studies, age, weight, and kidney func-tion were important factors in estimating clearance. Careful monitoring in the pediatric population is prudent, espe-cially with the evident dynamic changes in renal function in this population. As with adults, comorbidities and concur-rent medications can influence vanco-mycin tissue distribution, elimination, and toxicity.
Limitation of outcomes data. Recent retrospective studies on bac-teremic S. aureus infections (both MRSA and MSSA strains) in children treated with vancomycin suggest that trough concentrations of >15 mg/L were not associated with improved outcomes, yet an increase in AKI was observed.156-158 Furthermore, an-other retrospective pediatric study evaluating outcomes of MRSA bac-teremia as a function of an AUC/ MICBMD of ≥400 did not show im-proved outcomes.159 Similarly, van-comycin trough concentrations of <10 mg/L, as compared with concen-trations of >10 mg/L, were not associ-ated with increased 30-day mortality and recurrent bacteremia in chil-dren, although the lower concentra-tions were associated with prolonged bacteremia.160 In the absence of prospective, com-parative outcomes data in children re-garding unique AUC/MIC exposures necessary for clinical and microbio-logic success in treating serious MRSA infections in different neonatal and pediatric populations to validate the observations reported in adults (see Clinical PK/PD Data: Adults section), dosing in children should be designed to achieve an AUC of 400 mg·h/L and potentially up to 600 mg·h/L (assuming a MIC of 1 mg/L). This PD target range, specifically a range closer to an AUC/ MIC of 400 rather than 600, has been widely used by investigators to model pediatric dosing and therapeutic moni-toring. With inadequate PK studies and outcomes data to support the higher end of the AUC target range in pediatrics, it is prudent to aim for an AUC/MIC of 400 in pediatrics to limit the development of exposure-related AKI. Furthermore, in pediatrics, an AUC/MIC target of 400 is more readily achievable than it is in adults and cor-relates to trough concentrations of 7 to 10 mg/L rather than concentrations of 15 to 20 mg/L as are reported in adults. This wide variability in trough concentrations between these popula-tions with regard to achieving an AUC/ MIC of 400 corroborates the need for an AUC-guided approach to dosing and monitoring. It is possible that in otherwise healthy children with fewer comorbidities than are typically seen in adults, a lower target may yield out-comes equivalent to an AUC of 400 to 600 mg·hr/L. The decision to retain or increase AUC target exposure should be based on clinical judgment in the management of these patients.
With use of currently recommended vancomycin dosages of 45 to 60 mg/ kg/day, widespread treatment failures in children have not been reported in the literature, which may be reflective of a younger host with a more robust systemic and immunologic response to infection, a different management approach (surgical and antibiotic) to invasive MRSA infection, lack of associ-ated comorbidities, or publication bias. Prospective comparative clinical trials involving children with documented infections treated with different vanco-mycin dosages or exposures have not been published.
Empiric maintenance regimen. Published retrospective PK/PD data in children suggest that current van-comycin dosing of 45 to 60 mg/kg/day (in divided doses administered every 6 to 8 hours) may be insufficient to achieve currently recommended tar-gets for adults of an AUC of 400 to 600 mg·h/L (assuming a MIC of 1 mg/L).1 In fact, higher dosages, ranging from 60 to 80 mg/kg/day and given in di-vided doses every 6 hours, may be needed to achieve these targets for MRSA strains with a vancomycin MIC of 1 mg/L or less, presumably as a re-sult of greater vancomycin CL than is seen in adults.1,161-164 For children in-fected by MRSA pathogens with a MIC of >1 mg/L, it is unlikely that the target exposure can be reliably achieved with previously investigated dosages of van-comycin in children.
Le and colleagues164 utilized population-based PK modeling to an-alyze 1,660 vancomycin serum con-centrations obtained at 2 institutions from 2003 to 2011 among 702 children older than 3 months of age with varying comorbidities. They demonstrated that 4 important factors (age, weight, renal function as assessed by SCr, and MIC) contributed to vancomycin exposure. Monte Carlo simulations were created using population-based PK modeling with Bayesian estimation and MICs of clinical isolates as determined by Etest, with 85% of clinical isolates demon-strated to have a MICEtest of 1 mg/L or less. To achieve an AUC/MICEtest of ≥400 in 90% of subjects, a dosage of 80 mg/kg/ day was necessary, particularly in those less than 12 years of age with normal renal function. At a dosage of 80 mg/ kg/day, the median AUC and median trough concentration were 675 mg·h/L and 16 mg/L, respectively. As expected, subjects 12 years of age or older achieved similar exposure at lower dosages of 60 to 70 mg/kg/day. At a dosage of 60 to 70 mg/kg/day (divided doses admin-istered every 6 hours), an AUC of 400 mg·h/L correlated to a mean trough of 8 to 9 mg/L.164 The clinical applicability of this PK model for vancomyin CL estima-tion to determine AUC exposure was val-idated in a small study by Ploessl et al.165 Other studies corroborated Le and colleagues’ findings regarding the need to use higher dosages, ranging from 60 to 80 mg/kg/day, depending on age and renal function.162,164,166,167 Using the liter-ature for vancomyin CL published in or before 2000 and Bayesian estimation for one 25-kg base subject, Frymoyer et al163 evaluated the relationship be-tween AUC and trough concentrations, showing that a dosage of 60 mg/kg/day achieved trough concentrations of 7 to 10 mg/L and an AUC/MIC of ≥400 in AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 15 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING 90% of children for MRSA pathogens with a MIC of 1 mg/L. However, their finding may not be extrapolatable to the entire pediatric population given the variable ages and renal function. In a second study, these investigators dem-onstrated that a dosage of 60 mg/kg/day achieved AUC/MICBMD values between 386 and 583 (assuming a MICBMD of 1 mg/L) in children 2 to 12 years of age, indicating that some younger children may require higher doses to achieve target AUC/MICBMD.162 The probability of target attainment was not provided, and dosages above 60 mg/kg/day were not evaluated in this study.
Two retrospective studies that util-ized non-Bayesian methods evaluated trough concentration targets of 10 to 20 mg/L (a higher range than that used by Le et al164 and Frymoyer et al163, who also assessed AUC) in children 1 month to 18 years of age. An interesting finding of the study of Madigan et al166 was that a dosage of 60 mg/kg/day achieved the target trough concentration in only 17% of preschool-aged children 2 to 5 years old, which was the lowest attainment for any pediatric age group. Eiland and colleagues161 showed that dosages of 70 to 80 mg/kg/day were necessary to achieve trough concentrations of 10 to 20 mg/L. Another study, by Abdel et al,168 demonstrated that dosages higher than 60 mg/kg/day were necessary to achieve an AUC/MIC of ≥400 in children with cancer. The mean age in this study co-hort was 6 (SD, 2.5) years; it is possible that young age with greater CL may have been a contributing factor for the need for an increased dose, an observation uncovered in studies by Le et al164 and Madigan et al.166 As a drug that demonstrates renal elimination, vancomycin requires dosage adjustment in children with acute or chronic renal insufficiency. Le and col-leagues169 conducted a population-based PK analysis with Bayesian methods that evaluated 63 case-control pairs (matched by age and weight) with 319 vancomycin serum concentrations. The mean age of this study cohort was 13 (SD, 6) years. The investigators reported that a vancomycin dosage of 45 mg/kg/day (ie, 15 mg/kg every 8 hours) in renally impaired chil-dren achieved AUC exposure similar to that achieved with a dosage of 60 mg/kg/ day in children with normal renal func-tion. Notably, they showed that in 87% of children with initial renal impairment, vancomycin CL improved (with a lag in the recovery of renal function as assessed by SCr) within the first 5 days of therapy, indicating some degree of renal function recovery, a finding that provides support for ongoing vancomycin TDM. In ad-dition, vancomycin CL does not always correlate well with renal function (as as-sessed by creatinine CL) in children, par-ticularly in those who are acutely ill in the ICU setting and have varying degrees of renal dysfunction. Rapid return of renal function may occur over the first few days after ICU admission. As such, therapeutic monitoring of both serum concentrations and renal function should be conducted during vancomycin therapy.170,171 Loading doses. Loading doses of 25 to 30 mg/kg for critically ill adults have been suggested to achieve steady-state concentrations more quickly, but preliminary data on pediatric patients suggest that the benefit of a loading dose of 30 mg/kg is quickly lost if the maintenance dose is insufficient to provide adequate ongoing exposure.167 However, the concept of a loading dose accompanied by a daily maintenance dose sufficient to achieve the target exposure and initiated at a specified time after the loading dose should be investigated.
Minimizing AKI risk. Similar to the literature on adults, the literature in pediatrics suggests, in aggregate, that the risk of AKI increases as a func-tion of vancomycin exposure, espe-cially when the trough concentration exceeds 15 to 20 mg/L. In fact, Fiorito and colleagues158 reported in a recent meta-analysis of 10 pediatric studies that troughs of ≥15 mg/L increased the OR for AKI by 2.7-fold (95% confidence interval, 1.82-4.05) and that AKI was further correlated with a stay in the pe-diatric ICU. McKamy and colleagues172 published results of the first study that uncovered the association between vancomycin trough concentrations greater than 15 to 20 mg/L and AKI in pediatric patients. In addition, they showed that children who received con-current nephrotoxic drugs (particularly furosemide) and stayed in the pediatric ICU were also more likely to experi-ence AKI. Four studies published later corroborated these findings, indicating that the interplay of multiple factors in addition to vancomycin exposure contributed to AKI.173-176 Interestingly, Sinclair et al174 reported that a 5-mg/kg dose augmentation or each additional day of vancomycin use increased the risk of AKI. Knoderer and colleagues173 evaluated late-onset AKI (defined as occurring after 7 days of vancomycin therapy) and observed that young age (<1 year) was independently associated with late AKI.
One pediatric study evaluated the relationship of AKI with vancomycin AUC and trough concentrations, both derived by Bayesian estimation. Le and colleagues19 conducted a large population-based PK analysis using 1,576 serum concentrations collected from 680 pediatric subjects. A contin-uous exposure-response relationship was observed, with 10%, 33%, and 57% of patients who respectively achieved AUC values of ≥400, 800, and 1,000 mg·h/L experiencing AKI. Even after adjusting for ICU stay and concomi-tant use of nephrotoxic drugs, an AUC of ≥800 mg·h/L and trough concentra-tions of ≥15 mg/L were independently associated with a greater than 2.5-fold increased risk of AKI. The linkage of AUC to AKI, along with the strong cor-relation between AUC and trough con-centrations (Spearman’s coefficient, 0.963; P < 0.001), reinforces AUC as a plausible PK/PD parameter for thera-peutic monitoring that encompasses both therapeutic and toxic responses. Vancomycin exposure should be main-tained at an AUC of <800 mg·h/L to minimize AKI risk. As such, vanco-mycin dosages of ≥100 mg/kg/day should be avoided given that the pro-jected median AUC and trough values are 843 mg·h/L and 21 mg/L, respec-tively, at a dosage of 100 mg/kg/day.164 However, enhanced renal clearance of 16 AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING vancomycin may transiently occur in specific situations in children, in which case the dose of vancomycin may need to be higher than is usually prescribed to achieve an AUC of 400 mg·h/L, highlighting the need for therapeutic monitoring.
Therapeutic monitoring. Recent literature on vancomycin in pediatrics has focused primarily on PK analysis to support optimal dosing. Le and col-leagues177 conducted a population-based PK analysis in 138 pediatric subjects who were more than 3 months of age, evaluating 712 serum vanco-mycin concentrations (collected mostly after the third or fourth dose). They showed that both accuracy and preci-sion for estimating AUC24 (calculated by total daily dose over vancomycin CL, with the integration of Bayesian estima-tion) were improved using 2 concentra-tions (peak and trough), compared with trough-only monitoring. Furthermore, the 2-concentration approach im-proved the prediction of future AUC exposure in patients.177 Another study, by Stockmann et al,178 evaluated AUC‐ based vancomycin monitoring in 23 pediatric patients with cystic fibrosis. The researchers demonstrated that 2 concentrations calculated using a standard PK equation and a trough concentration calculated using a Bayesian population-based PK model produced similar AUC estimations. Despite the availability of limited studies on vancomycin monitoring in pediatrics, the findings appear con-gruent with adult data supporting AUC-guided therapeutic moni-toring that incorporates the Bayesian method, especially if only a single trough concentration is available. Furthermore, this AUC-guided moni-toring approach also appears prudent in order to predict toxicity in light of AKI data in pediatrics.
Overall, there are limited pediatric outcomes data to support the AUC target correlated with drug effectiveness in adults. Some of the differences found between adults and children treated for MRSA infections with vancomycin in-clude the complexity of vancomycin CL in the various pediatric age groups, and the differences in tissue site-of-infection drug exposure (eg, common occurrence of multifocal complicated osteomyelitis in children requiring therapeutic bone concentrations, with rare occurrence of MRSA endocarditis) suggest that fur-ther studies in children that incorporate prospective assessment of clinical out-comes and large sample size are needed to identify the optimal dosing strat-egies for MRSA infections in pediatrics. Until additional data are available, the AUC target used in adults (ie, from 400 up to 600 mg·h/L [assuming a MIC of 1 mg/L]) appears to be the most appro-priate initial target for vancomycin ex-posures in all pediatric age groups. For most children across the pediatric age groups, assuming a vancomycin MIC of 1 mg/L, published data suggest that a dosage of 60 to 80 mg/kg/day (given in divided doses every 6 hours) is required to achieve an AUC target of 400 to 600 mg·h/L.
Summary and recommendations: 17. Based on an AUC target of 400 mg·h/L (but potentially up to 600 mg·hr/L, assuming a vancomycin MIC of ≤1 mg/L for MRSA) from adult data, the initial recommended vancomycin dosage for children with normal renal function and suspected serious MRSA infections (including pneumonia, pyomyositis, multifocal osteomyelitis, complicated bacte-remia, and necrotizing fasciitis) is: • 60 to 80 mg/kg/day, in divided doses given every 6 hours, for chil-dren ages 3 months to less than 12 years or • 60 to 70 mg/kg/day, in divided doses given every 6 to 8 hours, for those ≥12 years old (A-II).
The maximum empiric daily dose is usually 3,600 mg in children with ad-equate renal function (C-III). Most children generally should not require more than 3,000 mg/day, and doses should be adjusted based on observed concentrations to achieve the AUC/MIC target. Early monitoring of observed concentrations is recommended when doses exceed 2,000 to 3,000 mg/day (A-III). Furthermore, close monitoring of observed concentrations and renal func-tion is prudent in patients with poor or augmented renal clearance, as resolu-tion of their renal function abnormal-ities may occur within the first 5 days of therapy. 18. AUC-guided therapeutic moni-toring for vancomycin, preferably with Bayesian estimation, is sug-gested for all pediatric age groups, based on developmental changes of vancomycin CL documented from the newborn to the adolescent. Based on current available data, the suggestion for AUC-guided monitoring in pediatrics aligns with the approach for adults, including the application of Bayesian esti-mation for 1 trough concentration or first-order PK equations with 2 concentrations (B-II). The Bayesian AUC-guided dosing strategy may be an optimal approach to individu-alize vancomycin therapy in pediat-rics since it can incorporate varying ages, weights, and renal function. Both serum concentrations of van-comycin and renal function should be monitored since vancomycin CL and CLcr are not always well correl-ated in pediatrics. Furthermore, ag-gressive dosing to maintain target AUC exposure and decrease the risk of potential AKI in treatment of MRSA infection necessitates drug monitoring. 19. Therapeutic monitoring may begin within 24 to 48 hours of vancomycin therapy for serious MRSA infections in children, as in adults (B-III). Any delay in therapeutic monitoring should be based on severity of infec-tion and clinical judgment. Dosing adjustment should be made for those with renal insufficiency, those with obesity (see Pediatric Obesity), and those receiving concurrent nephro-toxic drug therapy. Following the initial dose, dosing adjustment is AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 17 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING important for those with acute renal insufficiency, but subsequent adjust-ment (particularly within the first 5 days of therapy) may be necessary for those experiencing recovery of renal function. Sustained or subse-quent decreases in dosage may be needed, particularly for those with chronic renal insufficiency and those receiving concurrent nephrotoxic drug therapy (B-III). 20. Vancomycin exposure may be op-timally maintained below the thresh-olds for AUC of 800 mg·h/L and for trough concentrations of 15 mg/L to minimize AKI (B-II). The safety of vancomycin dosages above 80 mg/ kg/day has not been prospectively evaluated. Avoiding vancomycin dosages of ≥100 mg/kg/day is sug-gested since they are likely to surpass these thresholds (B-III). 21. Insufficient data exist on which to base a recommendation for a loading dose among the nonobese pediatric population. Loading doses from adult studies may be considered, but further studies are needed to elucidate the appropriate dose for the various pediatric popu-lations from the neonate to adoles-cent (C-III).
Pediatric obesity. Vancomycin is a large glycopeptide molecule that is hydrophilic, suggesting that distri-bution into tissues with high lipid con-centrations, such as adipose tissue, is decreased, as noted above for adults (see Dosing in Obesity section). When vancomycin dosing is based on total body weight (mg/kg) for both obese and nonobese children, serum con-centrations have been documented to be higher in obese children, assuming that renal CL is similar between the 2 populations.179 Moffett retrospectively compared vancomycin pharmacoki-netics in 24 obese children who were matched with 24 nonobese control children.180 Vancomycin dose admin-istration per child was slightly higher in the obese children, which resulted in increased trough concentrations. Similarly, 2 retrospective non-Bayesian studies by Heble et al181 and Miller et al182 documented higher vancomycin trough concentrations in overweight and obese children, as compared to normal-weight children, with dosing based on total body weight. No increase in AKI was noted in the overweight children.182 Collectively, non-Bayesian studies of obese children have evaluated main-tenance regimens ranging from 40 to 80 mg/kg/day (calculated using total body weight), with some instituting maximum doses of 1 to 2 g over 1 to 2 hours.180,181,183,184 As an alternative to total body weight, one study recom-mended the use of body surface area to dose vancomycin, which necessitates establishing a different dosing regimen and obtaining height measurements that may not always be readily avail-able in clinical practice.185 Body sur-face area is not typically used for dosing medications in children, except for chemotherapeutic agents.186 Using a Bayesian population-based PK analysis of 389 vancomycin serum concentrations collected from 87 pairs of obese and nonobese children (matched by age and baseline SCr), Le and colleagues187 showed that the Vd was strongly correlated with actual or total body weight and that CL correl-ated with allometric weight (ie, weight x 0.75) and body surface area. Using this PK model, Nguyen and colleagues164,166 concluded, using Monte Carlo simu-lations with Bayesian estimation, that vancomycin 60 mg/kg/day dosed by total body weight, as compared with other weight measures, resulted in the highest rate of achievement of the target AUC/MIC of ≥400 in obese children (ie, the target was achieved in 76% when vancomycin was given by total body weight, in 66% when given by adjusted body weight, and in 31% when given by allometric weight). Furthermore, when given dosages of vancomycin of 60 mg/kg/day by total body weight, fewer obese children of <12 vs ≥12 years of age achieved an AUC/MIC of ≥400 (70% and 84%, respectively), an age-based observation also identified in nonobese children.164,166 Interestingly, the use of a 20-mg/kg loading dose based on total body weight in obese children increased achievement of an AUC/MIC of ≥400, especially within the first 12 hours of therapy. In addition, 1 of every 5 obese children had an AUC of ≥ 800 mg·h/L, indicating that routine therapeutic and safety monitoring is prudent.188 Summary and recommendations: 22. Data suggest that obese children are likely to have vancomycin exposures that may be statistically greater than those in normal-weight children when doses are calculated on a mg/kg basis, but these differences are not known to be of sufficient clinical importance to suggest different mg/kg empiric vancomycin dosages in obese children at this time. Similar to nonobese chil-dren, obese children < 12 years old, compared with those ≥ 12 years, may require a higher mg/kg dose (B-II). 23. Therapeutic monitoring is likely to be of particular value in obese children, both for therapeutic response and the risk of AKI. The specific recom-mendations for therapeutic moni-toring in nonobese children may also apply for obese children (B-II). 24. A loading dose of 20 mg/kg by total body weight is recommended in obese children (A-III).
Neonates. Vancomycin ther-apeutic monitoring is important in neonates, based on developmental considerations of prominent increasing renal function that occurs over the first several weeks of life.189 Models to predict vancomycin dosing have var-iously incorporated weight-based dosing, chronologic age–based dosing, postmenstrual age–based dosing, SCr-based dosing (except for the first week of life, when transplacental maternal cre-atinine in the neonatal circulation ren-ders neonatal SCr values inaccurate in estimating renal function), or combin-ations of these strategies. Regardless of which model is used, therapeutic moni-toring in the neonate is essential due to 18 AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING the rapid maturation of renal function over the first weeks of life.
Mehrotra et al190 compared 4 models for predicting vancomycin serum con-centrations, based on their population PK model, using a standard weight-based dose, a postmenstrual age–based dose, a postmenstrual and postnatal age–based dose, and a SCr-based dose. Serum creatinine–based dosing pre-dicted trough concentrations with the smallest variability in both term and pre-term neonates. However, for those who wish to achieve a target exposure based on high trough concentrations within a narrow range of 15 to 20 mg/L, it should be noted that only 13% to 21% of neo-nates were within this range across the 4 dosing regimens. Marqués-Miñana et al191 also developed a population PK model and proposed dosing based on postmenstrual age. SCr-based rather than postmenstrual or postconceptional age–based dosing has been supported by Irikura et al192 and Capparelli et al.193 However, when evaluating published neonatal PK models, no consensus on an optimal dosing regimen was achieved by experts on neonatal vancomycin, Zhao et al reported.194 After evaluating the predictive performance of 6 models, Zhao et al concluded the importance of evaluating analytical techniques for SCr and vancomycin concentrations best explained the variability of predictions between the models. Zhao et al found that the Jaffé method overestimated SCr concentrations when compared to the enzymatic method and that for vanco-mycin concentrations, the fluorescence polarization immunoassay method and enzyme-multiplied immunoassay method assays showed different predic-tive performances as well.194 With the knowledge that AUC, as compared with trough concentrations, is a more achievable target in pediat-rics, Frymoyer and colleagues195 evalu-ated the association between AUC and trough concentrations in neonates. Using 1,702 vancomycin concentrations (measured by the homogenous particle-enhanced turbidimetric inhibition immunoassay) collected from 249 neo-nates up to 3 months of age, population PK analysis was conducted to create a model for vancomycin CL that was based on weight, postmenstrual age, and SCr (measured by a modified ki-netic Jaffé reaction). Monte Carlo simulations with Bayesian estimation demonstrated that trough concentra-tions ranging from 7 to 11 mg/L were highly predictive of an AUC24 of >400 mg·h/L in at least 90% of neonates. Dosages to achieve this PK/PD target ranged from 15 to 20 mg/kg every 8 to 12 hours, depending on postmenstrual age and SCr.194 Stockmann et al196 later sup-ported the predictive performance and generalizability of this model in their study of 243 neonates with 734 vanco-mycin concentrations. While a trough concentration of 11 mg/L predicted the attainment of an AUC of ≥400 mg·h/L in 93% of neonates, Stockmann and colleagues noted that a trough concen-tration alone did not precisely predict AUC and concluded that Bayesian ap-proaches to support vancomycin dosing decisions for neonates in the clinical setting are needed.196 Furthermore, Cies et al197 reported differences in vanco-mycin pharmacokinetics, in particular rapid vancomycin CL, in neonates with extracorporeal oxygenation life sup-port, reiterating the need for Bayesian-derived dosing decision support in this vulnerable population. Lastly, Leroux et al198 demonstrated the success of the clinical integration of a model-based vancomycin dosing calculator, devel-oped from a population PK study, that was successful in improving the rate of attainment of a serum concentration of 15 to 25 mg/L from 41% to 72% without any cases of AKI.
As an alternative to intermittent ad-ministration, CI of vancomycin has been evaluated in infants. In a multicenter, randomized controlled trial involving 111 infants less than 90 days of age, Gwee et al199 showed that the use of CI resulted in fewer dose adjustments and a lower mean daily dose than intermittent ad-ministration. The target trough concen-trations were 10 to 20 mg/L for II, and the steady-state concentrations were 15 to 25 mg/L for CI. The AUC and clinical out-comes, including nephrotoxicity, could not be evaluated rigorously in this study due to the small sample size. Overall, the clinical utility of CI in neonates re-quires further evaluation, as the most common pathogen causing late-onset sepsis requiring vancomycin therapy is Staphylococcus epidermidis, with limited cases of S. aureus sepsis. While the op-timal AUC/MIC target for S. epidermidis is not well studied, a lower target may be reasonable, but further data to support this recommendation are needed.
The incidence of vancomycin-associated AKI reported in neonates has been low, ranging from 1% to 9%.200 Nonetheless, a positive correla-tion between increasing vancomycin trough concentrations and AKI has been reported by Bhargava et al.201 Furthermore, in a large, retrospec-tive, multicenter, propensity score– matched cohort study of 533 neonates receiving vancomycin and gentamicin and 533 receiving gentamicin alone, Constance et al202 concluded that AKI was not associated with vancomycin alone but may occur in the presence of other recognized risk factors, in-cluding patent ductus arteriosus, concomitant nonsteroidal anti-in-flammatory drug use, 1 or more pos-itive blood cultures, low birth weight, and higher scores for severity of ill-ness and risk of mortality.
Summary and recommendations: 25. Doses recommended to achieve an AUC of 400 mg·h/L (assuming a MIC of 1 mg/L) in neonates and infants up to 3 months old range from 10 to 20 mg/kg every 8 to 48 hours, depending on postmenstrual age, weight, and SCr (A-II). AUC-guided therapeutic dosing and monitoring, preferably with Bayesian estimation, can best achieve the target vanco-mycin exposure likely to be required for a successful outcome of treat-ment for a MRSA infection for all neonates, regardless of gestational and chronologic age. The specific recommendations for AUC-guided therapeutic monitoring in children AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 19 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING Table 2. Primary Recommendations for Vancomycin Dosing and Therapeutic Drug Monitoring A. ADULTS AND PEDIATRIC PATIENTS 1. In patients with suspected or definitive serious MRSA infections, an individualized target AUC/MICBMD ratio of 400 to 600 (assuming a vancomycin MICBMD of 1 mg/L) should be advocated to achieve clinical efficacy while improving patient safety (A-II). 2. When transitioning to AUC/MIC monitoring, clinicians should conservatively target AUCs for patients with suspected or documented serious infections due to MRSA, assuming a vancomycin MICBMD of 1 mg/L or less at most institutions. Given the importance of early, appropriate therapy, vancomycin targeted exposure should be achieved early during the course of therapy, preferably within the first 24 to 48 hours (A-II). As such, the use of Bayesian-derived AUC monitoring may be prudent in these cases since it does not re-quire steady-state serum vancomycin concentrations to allow for early assessment of AUC target attainment.
3. Trough-only monitoring, with a target of 15 to 20 mg/L, is no longer recommended, based on efficacy and nephrotoxicity data in patients with serious infections due to MRSA (A-II). There is insufficient evidence to provide recommendations on whether trough-only or AUC-guided vancomycin monitoring should be used among patients with noninva-sive MRSA or other infections. 4. Vancomycin monitoring is recommended for patients receiving vancomycin for serious MRSA infections to achieve a sustained targeted AUC (assuming a MICBMD of 1 mg/L unless it is known to be greater or less than 1 mg/L by BMD). Independent of MRSA infection, vancomycin monitoring is also recommended for all patients at high risk for nephrotoxicity (eg, critically ill patients receiving concurrent nephrotoxins), patients with unstable (ie, deteriorating or significantly improving) renal function, and those receiving prolonged courses of therapy (more than 3 to 5 days). We suggest the frequency of monitoring be based on clinical judgment; frequent or daily monitoring may be prudent for hemodynamically unstable patients (eg, those with end-stage renal disease), with once-weekly monitoring for hemodynamically stable patients (B-II). 5. Based on current national vancomycin susceptibility surveillance data, under most circumstances for empiric dosing, the vancomycin MIC should be assumed to be 1 mg/L. When the MICBMD is greater than 1 mg/L, the probability of achieving an AUC/MIC target of ≥400 is low with conventional dosing; higher doses may risk unnecessary tox-icity, and the decision to change therapy should be based on clinical judgment. In addition, when the MICBMD is less than 1 mg/L, we do not recommend decreasing the dose to achieve the AUC/MIC target. It is important to note the limitations in automated susceptibility testing methods, including the lack of precision and variability in MIC results depending on the method used (B-II).
6. The pharmacokinetics of continuous infusion suggest that such regimens may be a reasonable alternative to conventional intermittent infusion dosing when the AUC target cannot be achieved (B-II). 7. Incompatibility of vancomycin with other drugs commonly coadministered in the ICU requires the use of independent lines or multiple catheters when vancomycin is being con-sidered for continuous infusion (A-III).
B. ADULTS 8. Given the narrow vancomycin AUC range for therapeutic effect and minimal associated risk of acute kidney injury (AKI), the most accurate and optimal way to manage vanco-mycin dosing should be through AUC-guided dosing and monitoring (A-II). We recommend to accomplish this in one of two ways: a. One approach relies on the collection of 2 concentrations (obtained near the steady-state, post-distributional peak concentration at 1 to 2 hours after infusion and trough at end of dosing interval), preferably but not required during the same dosing interval (if possible) and utilizing first-order pharmacokinetic (PK) equations to estimate the AUC (A-II). b. The preferred approach to monitor AUC involves the use of Bayesian software programs, embedded with a PK model based on richly sampled vancomycin data as the Bayesian prior, to optimize the delivery of vancomycin based on the collection of 1 or 2 vancomycin concentrations, with at least 1 trough. It is preferred to obtain 2 PK samples (ie, 1 to 2 hours post infusion and at the end of the dosing interval) to estimate the AUC with the Bayesian approach (A-II). A trough concentration alone may be sufficient to estimate the AUC with the Bayesian approach in some patients, but more data across different patient populations are needed to confirm the viability of using trough-only data (B-II). 9. Doses of 15 to 20 mg/kg (based on actual body weight) administered every 8 to 12 hours as an intermittent infusion are recommended for most patients with normal renal func-tion when assuming a MICBMD of 1 mg/L (A-II). In patients with normal renal function, these doses may not achieve therapeutic AUC/MIC targets when the MIC is 2 mg/L. Continued on next page 20 AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING 10. Continuous Infusion: Based on current available data, a loading dose of 15 to 20 mg/kg, followed by daily maintenance CI of 30 to 40 mg/kg (up to 60 mg/kg), to achieve a target steady-state concentration of 20 to 25 mg/L may be considered for critically ill patients (B-II). AUC24 can be simply calculated when multiplying the steady-state concen-tration (ie, desired therapeutic range of 20 to 25 mg/L throughout the entire dosing interval) by a factor of 24 (B-II). Attaining the desired drug exposure may be more readily accomplished, given the ease of sampling time and dosage adjustment, by changing the rate of infusion, which is a highly desirable feature in critically ill patients (B-II).
11. The risk of developing nephrotoxicity with CI appears to be similar or lower compared to intermittent dosing when targeting a steady-state concentration of 15 to 25 mg/L and a trough concentration of 10 to 20 mg/L, respectively (B-II). Definitive studies are needed to compare drug exposure based on measured AUC24 and factors that predispose to development of nephrotoxicity, such as receipt of concomitant nephrotoxins, diuretics, and/or vasopressor therapy in patients receiving continuous vs intermittent infusion of vancomycin. 12. In order to achieve rapid attainment of targeted concentrations in critically ill patients with suspected or documented serious MRSA infections, a loading dose of 20 to 35 mg/ kg can be considered for intermittent administration of vancomycin (B-II). Loading doses should be based on actual body weight and not exceed 3,000 mg. More intensive and early therapeutic monitoring should also be performed in obese patients (B-II).
13. Adult Obesity: A vancomycin loading dose of 20 to 25 mg/kg using actual body weight, with a maximum of 3,000 mg, may be considered in obese adult patients with serious infections (B-II). Empiric maintenance doses for most obese patients usually do not exceed 4,500 mg/day, depending on their renal function (B-II). Early and frequent moni-toring of AUC exposure is recommended for dose adjustment, especially when empiric doses exceed 4,000 mg/day (A-II).
14. Intermittent Hemodialysis: Since efficacy data are unavailable for an AUC of <400 mg · h/L, monitoring based on predialysis serum concentrations and extrapolating these values to estimate AUC is most practical. Maintaining predialysis concentrations between 15 and 20 mg/L is likely to achieve the AUC of 400 to 600 mg · h/L in the previous 24 hours (C-III). Predialysis serum concentration monitoring should be performed not less than weekly and should drive subsequent dosing rather than a strict weight-based recommendation, although these recommended doses provide a useful starting point until serum concentrations have been determined (B-II). 15. Hybrid Dialysis Therapies (eg, Slow-Low Efficiency Dialysis [SLED]): Loading doses of 20 to 25 mg/kg actual body weight should be used, recognizing that these hybrid dialysis therapies efficiently remove vancomycin (B-III). Initial doses should not be delayed to wait for a dialysis treatment to end. Maintenance doses of 15 mg/kg should be given after hybrid hemodialysis ends or during the final 60 to 90 minutes of dialysis, as is done with standard hemodialysis (B-III). Concentration monitoring should guide further maintenance doses. 16. Continuous Renal Replacement Therapy (CRRT): Loading doses of 20 to 25 mg/kg by actual body weight should be used in patients receiving CRRT at conventional, KDIGO-recommended effluent rates of 20 to 25 mL/kg/h (B-II). Initial maintenance dosing for CRRT with effluent rates of 20 to 25 mL/kg/h should be 7.5 to 10 mg/kg every 12 hours (B-II). Maintenance dose and dosing interval should be based on serum concentration monitoring, which should be conducted within the first 24 hours to ensure AUC/MIC targets are met. In fluid overloaded patients, doses may be reduced as patients become euvolemic and drug Vd decreases. The use of CI vancomycin in patients receiving CRRT ap-pears to be growing, and this method could be used in place of intermittent vancomycin dosing, especially when high CRRT ultrafiltrate/dialysate flow rates are employed (B-II). C. PEDIATRIC PATIENTS 17. Based on an AUC target of 400 mg · h/L (but potentially up to 600 mg · hr/L assuming a MIC of ≤1 mg/L) from adult data, the initial recommended vancomycin dosage for chil-dren with normal renal function and suspected serious MRSA infections is 60 to 80 mg/kg/day, divided every 6 to 8 hours, for children ages 3 months and older (A-II). Continued from previous page Table 2. Primary Recommendations for Vancomycin Dosing and Therapeutic Drug Monitoring B. ADULTS Continued on next page AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 21 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING 18. The maximum empiric daily dose is usually 3,600 mg/day in children with adequate renal function (C-III). Most children generally should not require more than 3,000 mg/day, and doses should be adjusted based on observed concentrations to achieve the AUC/MIC target. Early monitoring of observed concentrations is recommended when doses exceed 2,000 to 3,000 mg/day (A-III). Furthermore, close monitoring of observed concentrations and renal function is prudent in patients with poor or augmented renal clear-ance as resolution of their renal function may occur within the first 5 days of therapy. 19. AUC-guided therapeutic monitoring for vancomycin, preferably with Bayesian estimation, is suggested for all pediatric age groups, based on developmental changes of vanco-mycin CL documented from the newborn to the adolescent. Based on current available data, the suggestion for AUC-guided monitoring in pediatrics aligns with the approach for adults, including the application of Bayesian estimation for 1 trough concentration, or first-order PK equations with 2 concentrations (B-II). The Bayesian AUC-guided dosing strategy may be an optimal approach to individualize vancomycin therapy in pediatrics since it can incorporate varying ages, weights, and renal function. Both serum concentrations of vancomycin and renal function should be monitored since vancomycin CL and creatinine CL are not always well correlated in pediatrics. Furthermore, ag-gressive dosing to maintain target AUC exposure and decrease the risk of potential AKI in treatment of MRSA infection necessitates drug monitoring. 20. Therapeutic monitoring may begin within 24 to 48 hours of vancomycin therapy for serious MRSA infections in children, as in adults (B-III). Any delay in therapeutic monitoring should be based on severity of infection and clinical judgment. Dosing adjustment should be made for those with renal insufficiency, or those with obesity, or for those re-ceiving concurrent nephrotoxic drug therapy. Following the initial dose, dosing adjustment is important for those with acute renal insufficiency, but subsequent adjustment (par-ticularly within the first 5 days of therapy) may be necessary for those experiencing recovery of renal function. Sustained or subsequent decreases in dosage may be needed, particularly for those with chronic renal insufficiency and those receiving concurrent nephrotoxic drug therapy (B-III). 21. Vancomycin exposure may be optimally maintained below the thresholds for AUC of 800 mg · h/L and for trough concentrations of 15 mg/L to minimize AKI (B-II). The safety of vancomycin above 80 mg/kg/day has not been prospectively evaluated. Avoiding vancomycin dosages of ≥100 mg/kg/day is suggested since they are likely to surpass these thresholds (B-III).
22. Insufficient data exist on which to base a recommendation for a loading dose among the nonobese pediatric population. Loading doses from adult studies may be considered, but further studies are needed to elucidate the appropriate dose for the various pediatric populations, from neonates to adolescents (C-III).
23. Pediatric Obesity: Data suggest that obese children are likely to have vancomycin exposures that may be statistically greater than in normal-weight children when doses are calculated on a mg/kg basis, but these differences are not known to be of sufficient clinical importance to suggest different mg/kg empiric vancomycin dosages in obese chil-dren at this time. Similar to nonobese children, obese children < 12 years old, compared with those ≥ 12 years, may require higher mg/kg doses (B-II). 24. Pediatric Obesity: Therapeutic monitoring is likely to be of particular value in obese children, both for therapeutic response and to minimize the risk of AKI. The specific re-commendations for therapeutic monitoring in nonobese children may also apply for obese children (B-II). A loading dose of 20 mg/kg by total body weight is recommended in obese children (A-III). 25. Neonates: Dosages recommended to achieve an AUC of 400 mg · hr/L (assuming a MIC of 1 mg/L) in neonates and infants up to 3 months old range from 10 to 20 mg/kg every 8 to 48 hours depending on postmenstrual age, weight, and SCr (A-II). Abbreviations (not defined in body of table): AUC, area under the curve; BMD, broth micodilution; CL, clearance; ICU, intensive care unit; KDIGO, Kidney Disease: Improving Global Outcomes; MIC, minimum inhibitory concentration; MRSA, methicillin-resistant Staphylococcus aureus; SCr, serum creatinine; Vd, volume of distribution.
Continued from previous page Table 2. Primary Recommendations for Vancomycin Dosing and Therapeutic Drug Monitoring C. PEDIATRIC PATIENTS 22 AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING should also apply for neonates (see recommendation 18, A-III).
Conclusion To optimize vancomycin use for the treatment of serious infections caused by MRSA, we recommend targeting an AUC/MICBMD ratio of 400 to 600 (as-suming an MICBMD of 1 mg/L) for em-piric dosing in both adult and pediatric patients to maximize clinical efficacy and minimize AKI risk. Furthermore, the AUC should be therapeutically monitored using 1 or 2 postdose con-centrations (ie, a peak concentration measured after the early vancomycin tissue distribution phase and a trough level measured prior to the next dose), preferably integrating the Bayesian ap-proach. The primary recommendations are summarized in Table 2. The suc-cessful use of these guideline recom-mendations to positively impact patient outcomes may require multifaceted interventions, including educational meetings, guideline implementation, and dissemination of educational ma-terial on vancomycin dosing, moni-toring, and nephrotoxicity.203,204 While valuable literature pertaining to adults, children, and neonates has emerged since the last vancomycin guideline, future studies in all patient popula-tions are necessary to address existing gaps, including (1) efficacy data to sup-port vancomycin use in specific pa-tient populations (including neonates and pediatric patients and patients with renal disease and obesity) and for other types of infections, (2) efficacy data for specific pathogens, including coagulase-negative staphylococcus and Streptococcus species; (3) robust pediatric efficacy data for MRSA and other gram-positive pathogens causing different types of serious infections; (4) optimal loading and maintenance dosing regimens for patients with obe-sity and renal insufficiency; (5) efficacy benefit and the need for a dosing al-gorithm (specifically incorporating a loading dose followed by maintenance infusion); and (6) toxicodynamics of vancomycin CI in critically ill patients.
Disclosures Dr. Wong-Beringer received a grant from Merck & Co. and consulted for Rempex Pharmaceuticals, INSMED, Merck & Co., Nabriva Therapeutics, GlaxoSmithKline, Paratek Pharmaceuticals, Achaogen, Inc., Bayer HealthCare, and SIGA Technologies. Dr. Bradley served on a planning committee for the US Food and Drug Administration and European Medicines Agency at the American College of Clinical Pharmacy’s annual meeting; participated in the design of a a pediatric clinical trial; served on an executive committee for the United States Committee on Antimicrobial Susceptibility Testing; and consulted for Achaogen, Allergan, ContraFect, GlaxoSmithKline, Janssen, Melinta, Merck, Nabrive, Pfizer, Theravance, and Zavante. Dr. Liu received a research grant from Nohla Therapeutics and was a member of an Independent Efficacy Adjudication Committee with Theravance. Dr. Le received research awards from the Sternfels Prize for Drug Safety Innovation, Duke University, the National Institutes of Health (NIH) and National Institute of Child Health and Human Development, and JMI Laboratories; was an invited advisory board member for FDA, the Asian Pacific Health Foundation, and Infectious Diseases and Therapy journal. Dr. Levine served on the data safety monitoring board for Contrafect and served on an adjudication panel for Novartis. Dr. Lodise received grants from the Antibiotic Resistance Leadership Group (ARLG), Merck & Co, and Motif Bio PLC; served as a health outcomes project consultant for Paratek Pharmaceuticals, Allergan, Merck & Co., and Melinta Therapeutics; served on advi-sory boards for Paratek Pharmaceuticals, Motif Bio PLC, Achaogen, Nabriva, and Tetraphase; served as a consultant to Paratek Pharmaceuticals, ARLG, Allergan, Merck & Co., Melinta Therapeutics, Motif Bio PLC, Achaogen, Nabriva, and Tetraphase; and was a speaker for Melinta Therapeutics, Tetraphase, and Sunovion. Dr. Maples served as an international working group member for the European Cystic Fibrosis Society and North American Cystic Fibrosis Society, served on an advisory panel for the Centers for Disease Control and Prevention and Pew Charitable Trust, and was on a committee for the Arkansas Health Department. Dr. Mueller received research grants from Merck & Co. and Hope Pharmaceutical and served on an advisory board for NxStage and Baxter. Dr. Pai received a grant from Merck, Inc., served on an advisory board for Shinogi and Paratek Pharmaceuticals, and served on the meet the professor program for Merck. Dr. Rodvold re-ceived a grant from Theravance Biopharm, NIH, ARLG, and Allergan; consulted for BLC, Entasis, Merck, Paratek Pharmaceuticals, Shionogi, Tetraphase, and Wockhardt; was a speaker at the American Society for Microbiology and European Society for Clinical Microbiology and Infectious Diseases ASM/ESCMID conference; served on the 2015– 2019 Program Committee and was 2016–2018 Program Co-Chairperson for the American Microbiology Society; and was a member of the 2017–2019 Antimicrobial Resistance Committee for IDSA. Dr. Rybak received re-search grants from Bayer Pharmaceuticals, the NIH Research Project Grant (RO1) Program, Merck, Allergan, the Michigan Department of Health and Human Services, Accelerate Diagnostics, Inc., NIH, Contrafect, Motif Biosciences, and the Michigan Translational Research and Commercialization Program; and served as a grant review panel member for NIH. The other authors have declared no potential conflicts of interest.
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AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 29 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING Authors(Year Published) Population, Design, Infection Type(s) Method to Determine AUC24 Method to Determine MIC AUC/MIC Breakpoint/ Target Outcome(s) Measured Reference Moise-Broder et al (2004) Adults; retrospective; S. aureus lower respiratory infections (n = 107) Dose24h/clearance BMD ≥350BMD Bacterial eradication 10 Kullar et al (2011) Adults; retrospective; MRSA bacteremia (n = 320) Dose24h/clearance BMD/Etest ≥421BMD Composite failure (based on 30-day mortality and persistent signs & symptoms of infection, >7 days of bacteremia) 11 Holmes et al (2013) Adults; retrospective; MRSA bac-teremia (n = 182) Dose24h/clearance BMD/Etest >373BMD/271.5Etest 30-day all-cause mortality 13 Jung et al (2014) Adults; retrospective; MRSA bacteremia (n = 76) Dose24h/clearance BMD/Etest <430BMD/ 398.5Etest 30-day all-cause mortality 15 Brown et al (2012) Adults; retrospective; MRSA bacteremia (n = 50) Bayesian Etest ≥211 Attributable mortality 12 Gawronoski et al (2013) Adults; retrospective; MRSA bacteremia & osteomyelitis (n = 59) Bayesian Etest >292 Time to bacterial clearance 14 Lodise et al (2014) Adults; retrospective; MRSA bac-teremia (n = 123) Bayesian BMD/Etest 521BMD/303Etest Composite failure (based on 30-day mortality, >7 days of bacteremia, recurrence of bacteremia within 60 days of discontinuation of therapy) 16 Casapao et al (2015) Adults; retrospective; MRSA bacteremia & endocarditis (n = 139) Bayesian BMD >600 Composite failure (based on >7 days of bacteremia and/or 30-day attributable mortality) 17 Le et al (2015) Pediatric patients; retrospective; all pediatric infection types (n = 680) Bayesian NA ≥800 Nephrotoxicity 19 Finch et al (2017) Adults; retrospective, quasi-study design; all infection types except UTI, SSSI, meningitis; surgical prophylaxis (n = 1,300) AUC derived from multiple samples NA <400 Nephrotoxicity 20 eTable 1. Summary of Adult and Pediatric Studies With Outcome Assessment Continued on next page 30 AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 Downloaded from by ASHP user on 19 March 2020 ASHP REPORT GUIDELINE ON VANCOMYCIN MONITORING Authors(Year Published) Population, Design, Infection Type(s) Method to Determine AUC24 Method to Determine MIC AUC/MIC Breakpoint/ Target Outcome(s) Measured Reference Zasowski et al (2017) Adults; retrospective; pneumonia or bloodstream infection (n = 323) Bayesian NA >700 Nephrotoxicity 21 Neely et al (2018) Adults; prospective; all infection types (n = 252) Bayesian NA ≥400 Nephrotoxicity, resolution or improve-ment in signs & symptoms, relapse, and mortality 22 Lodise et al (2019) Adults; multicenter, observational, prospective; MRSA bacteremia Bayesian BMD/Etest No threshold identi-fied, but only 20% of study population had AUC/MICBMD ratio of <420 Composite failure (based on >7 days of bacteremia and/or 30-day mortality) 23 Abbreviations: AUC, area under the curve; BMD, broth microdilution; MRSA, methicillin-resistant Staphylococcus aureus; NA, not applicable; SSSI, skin and skin structure infection; UTI, urinary tract infection.
Continued from previous page eTable 1. Summary of Adult and Pediatric Studies With Outcome Assessment AM J HEALTH-SYST PHARM | VOLUME XX | NUMBER XX | XXXX XX, 2020 31 Downloaded from by ASHP user on 19 March 2020 |
13901 | https://wumbo.net/concepts/unit-circle-chart-degrees/ | Unit Circle Chart Degrees
Unit Circle Chart Degrees
The unit circle chart shows the position of the points on the unit circle that are formed by dividing the circle into equal parts. This chart shows the unit circle divided into 12 equal parts.
See the two alternate versions of this chart which measure angles in radians:
Unit Circle Chart τ (tau)
Unit Circle Chart π (pi)
Unit Circle Chart 4
Unit Circle Chart 8
Unit Circle Chart 12
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13902 | https://www.youtube.com/watch?v=GZCOd4w_9Nc | Vectors: Line of intersection & the angle between two planes
Joel Speranza Math
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Transcript:
Intro all right so we're going to jump through this pretty quickly we want to look at two planes so a plane and another plane here we're going to find the line of intersection so if the planes are parallel they're never going to intersect but if they're not parallel they are going to intersect so let's just put these two together like that and you can see that when two planes intersect they intersect along a line and it's our job to find the equation of that line now i'm giving the impression that they meet like that but they could meet sort of like like that like that they could meet on like all sorts of weird angles and stuff so there's multiple different ways that these planes could meet but they're always going to have a line of intersection if they're not parallel all right so we have two planes Converting to cartesian here plane one and plane two now converting these to cartesian form is our first step so that's going to be relatively easy it's just going to x plus y minus 3 z equals 6 and 2 x minus y plus z equals four so there are two cartesian forms now what i can do is subtract one from the other and make something cancel out it doesn't matter what cancels out now looking at these two equations it's going to be easier if i add them together because x plus two x will be three x y plus negative y be zero so that gets rid of my y and negative three z plus z is negative two z equals six plus four which is ten all right so now i have this neat little thing here now this all comes back to understanding what a plane is right a plane is a flat plane and we're going to figure out what we're going to let our x value be absolutely everything and then we're going to see what z and y values get spat out the other side so i'm going to let x equal lambda so that's my first step now if x equals lambda what does z equal well let's try it out 3 lambda minus 2 z equals 10 and then we just rearrange to make z the subject so that's 10 minus 3 lambda over negative 2 equals z so now i have this which is a parametric equation for z but i also have this which is a parametric equation for x now what about a parametric equation for y well what can i do here i've got this equation here with x and z in it and i now know what x and z are so using that equation i can sub in x and z and find out what y is in terms of lambda so let's do that Writing parametric equations we've got this equation here which is going to be 2 lambda minus y plus 10 minus 3 lambda over negative 2 equals 4 and then just rearrange that to make y the subject i'm just going to move that over there move that over there and probably multiply by negative one because i don't want negative y i want positive y all right so i've done a little bit of the heavy lifting there it's just something to keep in mind here there was a negative y here i put everything over two in this step which gives us negative y now when i simplify that i get 18 minus 7 lambda but then multiplying by the negative 1 reverses that that top okay so i am really finished here i've got an equation in terms of parametric form so i can say that x equals lambda y equals 7 lambda minus 18 over 2 and z equals 10 minus 3 lambda over 2. all right so now that i've got these parametric equations i can write them in cartesian form i can write it in vector form whatever it is what did we find the line of intersection between those two planes all right so i'm not going to go further but you can set that up in vector former in cartesian form however you want it to look so The angle between two planes second part of the video is finding the angle between two planes now here are my two planes i built these myself i put these together like this and i know we just did the line of intersection so we know what that is right we want to find the angle between them and that angle might be small like that it might be 90 degrees it might be that now we always talk about the acute angle there we don't talk about the larger angle so the angle between them is always going to be somewhere between 0 and 90. now why if i got a pen behind each ear because finding the angle between two planes is really hard but finding the angle between two normals is really really easy so there's my there's my planes that's the angle between them right there this is the angle between the two normals which is really really easy to find because we've got a normal there we've got a normal there we know how to find the angle between vectors now that is a 90 degree angle between the normal and there that is a 90 degree angle between the normal and there you can see we've created a quadrilateral between these so if those two are 90 if i know what that is that's going to be 180 minus that i'll just draw that little quadrilateral for you all right there it is in pictorial form if there's right angles there and the angle between the vectors is theta then the angle between the two planes would be 180 minus theta or potentially it could be just theta it really depends on how large that angle is there right because if that angle there is very very large these normals are going to meet up here at a small angle and you can see there is this large angle here but we always talk about the acute angle between two planes so we'd actually be looking for that one there which would be equal to that one up there totally worth building that thing Finding the angle so relatively straightforward here when we find the angle between two vectors it's the dot product on this side i've just put them in column vectors here so the dot product of this and this is equal to the magnitude of this times the magnitude of this cosine theta and that's the angle i'm looking for so calculate the dot product negative 2 calculate the magnitudes root 11 and root 6. now rearranging root 11 times root 6 is root 66 so that comes to the bottom here inverse cos we get an angle of 104.25 now if i were to get out my planes again 104.25 is a bit more than 90. so that's 90 so like the angle the planes actually probably look a bit like that so that's the angle between this is the angle i wanted to so the answer between for the angle between the two planes would be 180 minus that and that is job done we found a line of intersection between two planes and the angle between them |
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Sum of Square roots formula.
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I would like to know if there is formula to calculate sum of series of square roots $\sqrt{1} + \sqrt{2}+\dotsb+ \sqrt{n}$ like the one for the series $1 + 2 +\ldots+ n = \frac{n(n+1)}{2}$.
Thanks in advance.
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summation
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edited Apr 19, 2015 at 14:14
Chappers
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asked Apr 19, 2015 at 14:05
lk42392lk42392
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$\begingroup$ Real or integer square roots ? As a first approximation, $\frac23n^{3/2}$. For better, use the Euler-MacLaurin summation formula en.wikipedia.org/wiki/… $\endgroup$
user65203
– user65203
2015-04-19 14:16:05 +00:00
Commented Apr 19, 2015 at 14:16
$\begingroup$ Integer roots. Thanks @YvesDaoust $\endgroup$
lk42392
– lk42392
2015-04-19 18:31:45 +00:00
Commented Apr 19, 2015 at 18:31
$\begingroup$ If you want to sum only integers, that is, numbers of the form floor(n), then that needs to be stated clearly in the question. $\endgroup$
Dan Asimov
– Dan Asimov
2024-01-16 18:10:11 +00:00
Commented Jan 16, 2024 at 18:10
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For integer square roots, one should note that there are runs of equal values and increasing lengths
$$1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4\dots$$
For every integer $i$ there are $(i+1)^2-i^2=2i+1$ replicas, and by the Faulhaber formulas
$$\sum_{i=1}^m i(2i+1)=2\frac{2m^3+3m^2+m}6+\frac{m^2+m}2=\frac{4m^3+9m^2+5m}{6}.$$
When $n$ is a perfect square minus $1$, all runs are complete and the above formula applies, with $m=\sqrt{n+1}-1$.
Otherwise, the last run is incomplete and has $n-\left(\lfloor\sqrt n\rfloor\right)^2+1$ elements.
Hence, with $m=\lfloor\sqrt n\rfloor$,
$$S_n=\frac{4(m-1)^3+9(m-1)^2+5(m-1)}{6}+m\left(n-m^2+1\right)\ =m\left(n-\frac{2m^2+3m-5}6\right).$$
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edited Apr 19, 2015 at 19:37
answered Apr 19, 2015 at 19:31
user65203user65203
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The definition of harmonic numbers is $$H_p^{(-a)}=\sum_{i=1}^p i^a $$ When $a$ is not a positive integer, there is no closed form but, as Yves Daoust commented, there are quite nice expansions.
For example, if $n=\frac 12$ as in the post, you have $$H_p^{\left(-\frac{1}{2}\right)}=\frac{2 p^{3/2}}{3}+\frac{\sqrt{p}}{2}+\zeta \left(-\frac{1}{2}\right)+\frac{1}{24\sqrt p}+O\left(\left(\frac{1 }{p}\right)^{5/2}\right)$$ where $\zeta \left(-\frac{1}{2}\right)\approx -0.2078862250$.
For example, for $p=10$, the exact value is $\approx 22.46827819$ while the above approximation gives $\approx 22.46827983$. By itself, the first term already gives $21.0819$; the sum of first and second term gives $\approx 22.6629$. For $p=100$, the approximation leads to $12$ exact significant figures.
There are similar expansions for any value of the exponent $a$
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answered Apr 19, 2015 at 15:26
Claude LeiboviciClaude Leibovici
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$\begingroup$ Claude, do you know of reading material on this ? The Wikipedia article on this seems just to be sums of $1/n$ $\endgroup$
Will Jagy
– Will Jagy
2021-08-06 04:10:02 +00:00
Commented Aug 6, 2021 at 4:10
$\begingroup$ @WillJagy. Have a look at the excellent (as always !) answer from robjohn in math.stackexchange.com/questions/1583452/… . Cheers :-) $\endgroup$
Claude Leibovici
– Claude Leibovici
2021-08-06 05:51:57 +00:00
Commented Aug 6, 2021 at 5:51
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For an easier solution notice that $f(x) = \sqrt{x}$ is a monotone increasing function, hence for every $[k ,k+1]$ $$ \int_{k-1}^{k} \sqrt{x} dx<\sqrt{k}<\int_{k}^{k+1} \sqrt{x} dx $$ Now sum over k, you'll get a sharp approximation
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answered Apr 19, 2015 at 14:33
AlexAlex
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$\begingroup$ Could you elaborate, please ? I do not see how to sum over $k$. I have the feeling that I am in a loop. Thanks. $\endgroup$
Claude Leibovici
– Claude Leibovici
2015-04-19 14:55:29 +00:00
Commented Apr 19, 2015 at 14:55
3
$\begingroup$ I believe you meant $\sum\sqrt k$ in the middle... $\endgroup$
Simply Beautiful Art
– Simply Beautiful Art
2016-04-12 20:15:32 +00:00
Commented Apr 12, 2016 at 20:15
4
$\begingroup$ These bounds give you a per-term bound. If you sum all three "sides" to this inequality over $\sum_{k=1}^n$, you'll get that $$\int_0^n ! \sqrt{x} \, \mathrm{d}x < \sum_{k=1}^n \sqrt{k} < \int_1^{n+1} \sqrt{x} \, \mathrm{d}x,$$ as desired. $\endgroup$
Nicolás Kim
– Nicolás Kim
2017-10-05 16:36:13 +00:00
Commented Oct 5, 2017 at 16:36
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Refer to the docs. Simple as that!
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answered Apr 19, 2015 at 14:11
Aditya AgarwalAditya Agarwal
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For a better upper bound than Alex's answer,
$$\sum_{n=1}^xn^{1/2}\le\frac23\left(x+\frac12\right)^{3/2}$$
And if you want to improve upon that,
$$\sum_{n=1}^xn^{1/2}\approx\frac23\left(x+\frac12\right)^{3/2}\underbrace{-0.22474487139}_{\zeta(-1/2)}$$
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edited Sep 23, 2017 at 11:35
answered Apr 12, 2016 at 20:14
Simply Beautiful ArtSimply Beautiful Art
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$\begingroup$ This is very beautiful. Thank you! $\endgroup$
Dmitry Kamenetsky
– Dmitry Kamenetsky
2021-09-02 04:45:39 +00:00
Commented Sep 2, 2021 at 4:45
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13904 | https://www.math.uci.edu/~mathcircle/materials/2013.11.02MSMC.pdf | Problem 0: Warm-Up Problems 1) Find the quotient and remainder when: (a) 108 is divided by 3.
(b) 129 is divided by 7.
2) What is the remainder of: (a) 16 × 2 after division by 5.
(b) 12039102394 + 12340984921423 after division by 5.
3) Which of the following are true?
7 ≡1 (mod 3) 7 ≡3 (mod 3) 7 ≡147 (mod 3) 76 ≡−152 (mod 3) 4) If x ≡5 (mod 8) and y ≡6 (mod 8), what number is x + y congruent to modulo 8? What number is xy congruent to modulo 8.
5) A remainder class modulo n is the collection of integers which give the same remainder when divided by n. Dividing by n gives us n possible remainders, 0, 1, 2, . . . n −1 and so there are n remainder classes. For example, if n = 2 the two remainder classes are: 0 = {. . . , −9, −6, −3, 0, 3, 6, . . .} 1 = {. . . , −8, −5, −2, 1, 4, 7, . . .} 2 = {. . . , −7, −4, −1, 2, 5, 8, . . .} What are the remainder classes modulo 4?
Remainders when Dividing Given two integers a and n, we may divide a by n. If n does not divide a evenly, then there is a remainder r, which will be less than n. For example, 2500(r17) 25 62517 50 125 125 17 For any integer a and divisor n, there are two integers q and r called the quotient and remainder such that a = q × n + r and 0 ≤r < n So from our example above: 62517 = 2500 · 25 + 17.
Remainders when Dividing Given two integers a and n, we may divide a by n. If n does not divide a evenly, then there is a remainder r, which will be less than n. For example, 2500(r17) 25 62517 50 125 125 17 For any integer a and divisor n, there are two integers q and r called the quotient and remainder such that a = q × n + r and 0 ≤r < n So from our example above: 62517 = 2500 · 25 + 17.
Remainders when Dividing Given two integers a and n, we may divide a by n. If n does not divide a evenly, then there is a remainder r, which will be less than n. For example, 2500(r17) 25 62517 50 125 125 17 For any integer a and divisor n, there are two integers q and r called the quotient and remainder such that a = q × n + r and 0 ≤r < n So from our example above: 62517 = 2500 · 25 + 17.
Remainders when Dividing Given two integers a and n, we may divide a by n. If n does not divide a evenly, then there is a remainder r, which will be less than n. For example, 2500(r17) 25 62517 50 125 125 17 For any integer a and divisor n, there are two integers q and r called the quotient and remainder such that a = q × n + r and 0 ≤r < n So from our example above: 62517 = 2500 · 25 + 17.
Modular Arithmetic Cheat Sheet We say that a is congruent to b modulo n and write: a ≡b (mod n) if a and b have the same remainder when divided by n.
If a ≡c (mod n) and b ≡d (mod n) then, a + b ≡c + d (mod n) a −b ≡c −d (mod n) a × b ≡c × d (mod n) Modular Arithmetic Cheat Sheet We say that a is congruent to b modulo n and write: a ≡b (mod n) if a and b have the same remainder when divided by n.
If a ≡c (mod n) and b ≡d (mod n) then, a + b ≡c + d (mod n) a −b ≡c −d (mod n) a × b ≡c × d (mod n) Modular Arithmetic Cheat Sheet We say that a is congruent to b modulo n and write: a ≡b (mod n) if a and b have the same remainder when divided by n.
If a ≡c (mod n) and b ≡d (mod n) then, a + b ≡c + d (mod n) a −b ≡c −d (mod n) a × b ≡c × d (mod n) Modular Arithmetic Cheat Sheet We say that a is congruent to b modulo n and write: a ≡b (mod n) if a and b have the same remainder when divided by n.
If a ≡c (mod n) and b ≡d (mod n) then, a + b ≡c + d (mod n) a −b ≡c −d (mod n) a × b ≡c × d (mod n) Problem 1: A Clock Arithmetic 1) Marty the marathon runner has a unique method of resting after his marathons: he goes into hibernation. If Marty enters his cryo-static sleep chamber at 6 pm, and sets the timer for 76 hours, what time will he wake up?
2) On Monday morning, Sam broke his stubborn ways and decided to try green eggs and ham.
Much to his suprise, he loved them. He loved them so much, he decided he would eat green eggs and ham for the next 200 days. What day of the week will it be when Sam can stop eating green eggs and ham?
3) In the year 2340, Joanna, who is a spectacular snow-hoverboard rider, charters a flight to the moon to ride its killer slopes. She is going to try and break the world-record and complete a 7110◦rotation. If she is facing due west when she enters the air offthe jump, and she rotates counter-clockwise what direction should she be facing if she is to break the record?
Problem 2: A Modular Game This is a two player game between player A and player B.
Modular Addition Game: The game is played as follows: To start, player A picks a positive integer n Player B decides who will go first A running total T is set to 0 At each player’s turn, they pick a number m and add it to T The player loses if T is congruent to a previous total modulo n Here is an example play through: Alice and Bob play the Modular Addition Game. Alice chooses the modulus n = 5 and Bob decides to go first. The turns are as follows: Round # 1 2 3 4 5 Alice -88 -23 Bob 7 102 13 Running total T 7 -81 21 -2 11 On the last play, Bob loses, because he makes the running total 11 and 11 ≡21 (mod 5).
What number should Bob have picked instead?
Try the game for yourself!
Player A: Player B: Modulus n: Round # 1 2 3 4 5 6 7 8 9 10 Player A Player B Total T How many moves can a game have? What strategy should player A follow? What about player B?
Player A: Player B: Modulus n: Round # 1 2 3 4 5 6 7 8 9 10 Player A Player B Total T Player A: Player B: Modulus n: Round # 1 2 3 4 5 6 7 8 9 10 Player A Player B Total T Player A: Player B: Modulus n: Round # 1 2 3 4 5 6 7 8 9 10 Player A Player B Total T Problem 3: A Frightful Halloween Todd, Eileen, Jane, and Eric decide to go to a haunted house for Halloween. Todd, who is deathly scared of ghosts, asks Eileen, Jane, and Eric to go through the haunted house before him and count all the ghosts they see. Eileen, Jane, and Eric agree to this, but they decide to play a trick on Todd. They come up with this plan: Eileen will count the ghosts by 3’s, Jane will count the ghosts by 5’s, and Eric will count the ghosts by 19’s. After getting through the haunted house they come back to Todd: Eileen says: “I counted the ghosts by 3’s and there were 2 left over. ” Jane says: “I counted the ghosts by 5’s and there were 3 left over.” Eric says: “I counted the ghosts by 19’s and there were 7 left over.” What is the smallest number of ghosts Todd should be prepared for? What if Eileen, Jane, and Eric are really mischievous and say: “There are at least 500 ghosts and at most 1500 ghosts” (this is a big haunted house!). How many ghosts should Todd watch out for? (There is more than one number!) Problem 4: Even, Threeven, and Throdd Oh My!
In Miss Cardinality’s fascinating Math class, Jack and Danielle discuss the properties of even and odd numbers. “Jack, did you know that an even number times and odd number is even?
Or that an even number times an even number is even?” says Danielle. Jack responds, “No, I didn’t! But did you know that an even plus an odd is odd? Or that an even plus an even is even?” To which Danielle responds, “No, I didn’t realize that! How cool!” Terry, who fancies himself as an amateur philosopher, overhears this conversation and can’t help himself: “But how do you guys know that an even times an odd is even? Or that an even plus an odd is odd? If I give you a one-hundred digit even and a two-thousand digit odd could you prove to me that if we multiply them we will get an even number and if we add them, we will get an odd number? How can you say that you know 21031240128492141254184912840129021912484146 times 1129048219047821940759073518273589235728357 is odd??” Jack and Danielle, who are human, and hence smarter than calculators, decide they do not want to compute 21031240128492141254184912840129021912484146×1129048219047821940759073518273589235728357 Instead, they want to find a clever way to prove to Terry that they are indeed correct. Can you help them out? In addition, figure out (and prove!) the following rules: a) even + even = ?
b) even + odd = ?
c) odd + odd = ?
d) even × even = ?
e) even × odd = ?
f) odd × odd = ?
After disproving Terry’s doubts so handedly, Jack and Danielle are having so much fun, they decide to call a number threeven if it is divisbile by 3 throdd1 if it leaves a remainder of 1 when divided by 3 throdd2 if it leaves a remainder of 2 when divided by 3 Using these definitons, they figure out rules for adding and multiplying like above. Can you figure them out also?
a) threeven + throdd1 = ?
b) threeven + throdd2 = ?
c) throdd1 −throdd2 = ?
d) throdd2 −throdd1 =?
e) threeven × throdd1 = ?
f) throdd1 × throdd2 = ?
Problem 5: Rules Are Made for . . . Division?
Cole Sear claims he has a sixth-sense and can tell if a number is divisble by 3. “I see divisors of three” he claims. Johnathan is highly suspect of this and so challenges Cole to a divisbility-of-3-duel.
Johnathan: “84” Cole: “Yes” Johnathan, goes and checks and sees that 84 = 3 · 28. “Well that was an easy one, let’s try something harder” thinks Johnathan.
Johnathan: “189” Cole: “Yes” Again, Johnathan checks and sees that 189 = 3 · 63. “Ok, let me give him one that’s not divisible by 3” Johnathan says to himself.
Johnathan: “973” Cole: “No” “Wow! He actually is pretty good” thinks Johnathan.
Johnathan: “1539” Cole: “Yes” Johnathan: “2762” Cole: “No” Johnathan: “1011457” Cole: “No” Johnathan: “333699817245” Cole: “Yes” Johnathan checks and all these are correct too! How is Cole doing this?
There is a rule for divsion by 9 and 11 as well. Can you think of what the rule for 9 would be? What about the rule for division by 11?
Problem 6: A Calendar for the Ages The Gregorian calendar, which most nations use today, is based on a year containing 365 days. Every fourth year is a leap year (containing 366 days),and years divisible by 100 if and only if they are also divisible by 400. For example, 1900 was not a leap year, but 2000 was.
a) Show that the calendar repeats itself every 28 years that do not include a turn of the century which is not a leap year.
b) What day of the week was January 1, 1901? (January 1, 2013 was a Tuesday).
c) In what years of the 20th century does February have five Sundays?
d) What day of the week, if any, can never be February 29?
Problem 7: Bonus Some More Useful Facts About Modular Arithmetic Let p be a prime number, and n be an integer not divisible by p. If for some integers a, b, we have na ≡nb (mod p) then a ≡b (mod p). (In essence, we can “divide” by n).
Let p be a prime number and a an integer not divisible by p.
Then ap−1 ≡1 (mod p).
If p is a prime number then for any integer a, ap ≡a (mod p).
a ≡b (mod n) if and only if a −b is divisible by n if and only if a = kn + b for some integer k If p is a prime and p divides ab then p divides a or p divides b Use these facts, and some ingenuity to solve the following problems: 1) Find the remainder when 2100 is divided by 101.
2) Find the remainder when 3102 is divided by 101.
3) Prove that 3003000 −1 is divisible by 1001.
4) Prove that 7120 −1 is divisible by 143.
5) The sum of the numbers a, b, and c is divisible by 30. Prove that a5 + b5 + c5 is divisible by 30.
6) Let p be a prime number, and a, b be two integers. Prove that (a + b)p ≡ap + bp (mod p).
(Notice how much easier this is! (3a+2b)5 = 243a5+810a4b+1080a3b2+720a2b3+240ab4+32b5 but (3a + 2b)5 ≡3a + 2b (mod 5)) 7) Show that if a ≡b (mod 3) then 2 3(a2+ab+b2) can be written as the sum of three non-negative squares.
8) Let n be a natural number not divisible by 17. Prove that either n8 + 1 or n8 −1 is divisible by 17.
9) Let p be a prime not equal to 3. Prove that the number 111 . . . 1 (p ones) is not divisible by p.
10) Let p > 5 be a prime. Prove that the number 111 . . . 1 (p −1 ones) is divisible by p.
11) Let p be a prime. Prove that 1 · 2 · 3 · · · (p −1) ≡−1 (mod p). |
13905 | https://www.cs.cornell.edu/courses/cs2800/2014fa/lnotes/07_incl_excl.pdf | I n c l u s i o n -E x c l u s i o n C S 2 8 0 0 : D i s c r e t e S t r u c t u r e s , F a l l 2 0 1 4 S i d C h a u d h u r i e c a m p u s n e w s .
c o m t e l e g r a p h .
c o .
u k f t b i e .
c o m i a m l y o n s .
c o m P r o b a b i l i t y o f g e n e r a l c e l e b r a t i o n i n c l a s s ●B u s e s a r e d e l a y e d / c a n c e l l e d 1 0 % o f t h e t i m e d u r i n g t h e w i n t e r ●I o v e r s l e e p 2 0 % o f t h e t i m e A n a l o g o u s : a s e r i a l c i r c u i t A B Broken 20% of the time A n a l o g o u s : a s e r i a l c i r c u i t A B Broken 10% of the time W h a t ' s t h e p r o b a b i l i t y n o c u r r e n t i s f o w i n g ?
A B Broken 20% of the time Broken 10% of the time M o d e l i n g ●A: e v e n t t h a t b u s e s a r e d e l a y e d – ( o r f r s t c o m p o n e n t b r e a k s ) ●B: e v e n t t h a t I o v e r s l e e p – ( o r s e c o n d c o m p o n e n t b r e a k s ) ●Late = A ∪ B: e v e n t t h a t I a m l a t e – ( o r c u r r e n t i s b l o c k e d ) P r o b a b i l i t y o f a U n i o n ●K o l m o g o r o v ' s 3 r d A x i o m g u a r a n t e e s a s i m p l e f o r m u l a f o r t h e p r o b a b i l i t y o f t h e u n i o n o f m u t u a l l y e x c l u s i v e e v e n t s i n a p r o b a b i l i t y s p a c e P(E1 ∪ E2 ∪ E3 ∪ …) = P(E1) + P(E2) + P(E3) + … P r o b a b i l i t y o f a U n i o n ●K o l m o g o r o v ' s 3 r d A x i o m g u a r a n t e e s a s i m p l e f o r m u l a f o r t h e p r o b a b i l i t y o f t h e u n i o n o f m u t u a l l y e x c l u s i v e e v e n t s i n a p r o b a b i l i t y s p a c e P(E1 ∪ E2 ∪ E3 ∪ …) = P(E1) + P(E2) + P(E3) + … ●B u t w h a t i f t h e e v e n t s a r e n o t m u t u a l l y e x c l u s i v e ?
S A B A∩B S A B A∩B S A B A∩B C o u n t i n g E l e m e n t s |A ∪ B| C o u n t i n g E l e m e n t s |A ∪ B| = |A ∪ (B \ A)| C o u n t i n g E l e m e n t s |A ∪ B| = |A ∪ (B \ A)| = |A| + |B \ A| C o u n t i n g E l e m e n t s |A ∪ B| = |A ∪ (B \ A)| = |A| + |B \ A| = |A| + |B \ A| + |A ∩ B| – |A ∩ B| C o u n t i n g E l e m e n t s |A ∪ B| = |A ∪ (B \ A)| = |A| + |B \ A| = |A| + |B \ A| + |A ∩ B| – |A ∩ B| C o u n t i n g E l e m e n t s |A ∪ B| = |A ∪ (B \ A)| = |A| + |B \ A| = |A| + |B \ A| + |A ∩ B| – |A ∩ B| = |A| + |(B \ A) ∪ (A ∩ B)| – |A ∩ B| C o u n t i n g E l e m e n t s |A ∪ B| = |A ∪ (B \ A)| = |A| + |B \ A| = |A| + |B \ A| + |A ∩ B| – |A ∩ B| = |A| + |(B \ A) ∪ (A ∩ B)| – |A ∩ B| = |A| + |B| – |A ∩ B| C o u n t i n g E l e m e n t s |A ∪ B| = |A ∪ (B \ A)| = |A| + |B \ A| = |A| + |B \ A| + |A ∩ B| – |A ∩ B| = |A| + |(B \ A) ∪ (A ∩ B)| – |A ∩ B| = |A| + |B| – |A ∩ B| A s i m i l a r r e s u l t h o l d s f o r p r o b a b i l i t i e s T h e I n c l u s i o n -E x c l u s i o n P r i n c i p l e ( f o r t w o e v e n t s ) F o r t w o e v e n t s A, B i n a p r o b a b i l i t y s p a c e : P(A ∪ B) = P(A) + P(B) – P(A ∩ B) T h e I n c l u s i o n -E x c l u s i o n P r i n c i p l e ( f o r t w o e v e n t s ) F o r t w o e v e n t s A, B i n a p r o b a b i l i t y s p a c e : P(A ∪ B) = P(A) + P(B) – P(A ∩ B) D o n ' t u s e t h i s t o “ p r o v e ” K o l m o g o r o v ' s A x i o m s !
!
!
P r o o f : P(A ∪ B) = P(A ∪ (B \ A)) ( s e t t h e o r y ) T h e I n c l u s i o n -E x c l u s i o n P r i n c i p l e ( f o r t w o e v e n t s ) P r o o f : P(A ∪ B) = P(A ∪ (B \ A)) ( s e t t h e o r y ) = P(A) + P(B \ A) ( m u t . e x c l .
, s o A x i o m 3 ) T h e I n c l u s i o n -E x c l u s i o n P r i n c i p l e ( f o r t w o e v e n t s ) P r o o f : P(A ∪ B) = P(A ∪ (B \ A)) ( s e t t h e o r y ) = P(A) + P(B \ A) ( m u t . e x c l .
, s o A x i o m 3 ) = P(A) + P(B \ A) + P(A ∩ B) – P(A ∩ B) ( A d d i n g 0 = P(A ∩ B) – P(A ∩ B) ) T h e I n c l u s i o n -E x c l u s i o n P r i n c i p l e ( f o r t w o e v e n t s ) P r o o f : P(A ∪ B) = P(A ∪ (B \ A)) ( s e t t h e o r y ) = P(A) + P(B \ A) ( m u t . e x c l .
, s o A x i o m 3 ) = P(A) + P(B \ A) + P(A ∩ B) – P(A ∩ B) ( A d d i n g 0 = P(A ∩ B) – P(A ∩ B) ) = P(A) + P((B \ A) ∪ (A ∩ B)) – P(A ∩ B) ( m u t . e x c l .
, s o A x i o m 3 ) T h e I n c l u s i o n -E x c l u s i o n P r i n c i p l e ( f o r t w o e v e n t s ) P r o o f : P(A ∪ B) = P(A ∪ (B \ A)) ( s e t t h e o r y ) = P(A) + P(B \ A) ( m u t . e x c l .
, s o A x i o m 3 ) = P(A) + P(B \ A) + P(A ∩ B) – P(A ∩ B) ( A d d i n g 0 = P(A ∩ B) – P(A ∩ B) ) = P(A) + P((B \ A) ∪ (A ∩ B)) – P(A ∩ B) ( m u t . e x c l .
, s o A x i o m 3 ) = P(A) + P(B) – P(A ∩ B) ( s e t t h e o r y ) T h e I n c l u s i o n -E x c l u s i o n P r i n c i p l e ( f o r t w o e v e n t s ) W i l l I b e l a t e f o r c l a s s ?
●P(Late) = P(A ∪ B) W i l l I b e l a t e f o r c l a s s ?
●P(Late) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) W i l l I b e l a t e f o r c l a s s ?
●P(Late) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.1 + 0.2 – ???
W i l l I b e l a t e f o r c l a s s ?
●P(Late) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.1 + 0.2 – ???
L e t ' s m a k e t h e m o d e l i n g a s s u m p t i o n A a n d B a r e i n d e p e n d e n t : P(A ∩ B) = P(A) P(B) W i l l I b e l a t e f o r c l a s s ?
●P(Late) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = P(A) + P(B) – P(A) P(B) W i l l I b e l a t e f o r c l a s s ?
●P(Late) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = P(A) + P(B) – P(A) P(B) = 0.1 + 0.2 – 0.1 × 0.2 W i l l I b e l a t e f o r c l a s s ?
●P(Late) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = P(A) + P(B) – P(A) P(B) = 0.1 + 0.2 – 0.1 × 0.2 = 0.28 T h e I n c l u s i o n -E x c l u s i o n P r i n c i p l e ( f o r t h r e e e v e n t s ) F o r t h r e e e v e n t s A, B, C i n a p r o b a b i l i t y s p a c e : P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C) T h e I n c l u s i o n -E x c l u s i o n P r i n c i p l e F o r e v e n t s A1, A2, A3, … An i n a p r o b a b i l i t y s p a c e : P(∪i=1 n Ai)=∑i=1 n P(A i)−∑1≤i< j≤n P( Ai∩A j) +∑1≤i< j<k≤n P( Ai∩A j∩Ak)−...+(−1) n−1P(∩i=1 n Ai) T h e I n c l u s i o n -E x c l u s i o n P r i n c i p l e F o r e v e n t s A1, A2, A3, … An i n a p r o b a b i l i t y s p a c e : =∑k=1 n ((−1) k−1∑I⊆{1,2,... n} |I|=k P(∩i∈I Ai)) +∑1≤i< j<k≤n P( Ai∩A j∩Ak)−...+(−1) n−1P(∩i=1 n Ai) P(∪i=1 n Ai)=∑i=1 n P(A i)−∑1≤i< j≤n P( Ai∩A j) |
13906 | https://flexbooks.ck12.org/cbook/ck-12-middle-school-math-concepts-grade-8/section/9.15/primary/lesson/solving-linear-systems-by-substitution-msm8/ | Skip to content
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9.15
Solving Linear Systems by Substitution
Written by:Brenda Meery | Jen Kershaw, M.ed
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
Kathleen’s rectangular garden has a perimeter of 280 yards. The length is three times the width. If she wants to buy fencing panels to surround it, then Kathleen must determine the garden's dimensions.
How can Kathleen use a system of equations to calculate the length and width of her rectangular garden?
In this concept, you will learn to solve linear systems by substitution.
Substitution Method
There is an ordered pair @$\begin{align}(x,y)\end{align}@$ that makes both equations in the following system of linear equations true. This means that ‘@$\begin{align}x\end{align}@$’ in the first equation equals ‘@$\begin{align}x\end{align}@$’ in the second equation and ‘@$\begin{align}y\end{align}@$’ in the first equation equals ‘@$\begin{align}y\end{align}@$’ in the second equation.
@$$\begin{align}\begin{array}{rcl}
3y &=& x-2 \
y-x &=& 4
\end{array}\end{align}@$$
The solution to a system of two linear equations with two variables @$\begin{align}x\end{align}@$ and @$\begin{align}y\end{align}@$ is a value for @$\begin{align}x\end{align}@$ and a value for @$\begin{align}y\end{align}@$ that will make both equations true. One way to find these values for the variables is by using the substitution method. The substitution method involves solving one of the equations for one of the variables and replacing that variable in the other equation with its equivalent expression. The result will be an equation with one variable which can be solved using algebra.
Let’s solve the given system of linear equations using the substitution method.
@$$\begin{align}\begin{array}{rcl}
3y &=& x-2 \
y-x &=& 4
\end{array}\end{align}@$$
First, solve the second equation for the variable @$\begin{align}y\end{align}@$ by adding @$\begin{align}x\end{align}@$ to both sides of the equation.
@$$\begin{align}\begin{array}{rcl}
y-x &=& 4 \
y-x+x &=& 4+x
\end{array}\end{align}@$$
Next, simplify both sides of the equation.
@$$\begin{align}\begin{array}{rcl}
y-x+x &=& 4+x \
y &=& 4+x
\end{array}\end{align}@$$
Next, substitute this expression for @$\begin{align}y\end{align}@$ into the first equation.
@$$\begin{align}\begin{array}{rcl}
y &=& 4+x \
3y &=& x-2 \
3(4+x) &=& x-2
\end{array}\end{align}@$$
Next, do the multiplication on the left side of the equation to clear parentheses.
@$$\begin{align}\begin{array}{rcl}
3(4+x) &=& x-2 \
12+3x &=& x- 2
\end{array}\end{align}@$$
Next, subtract 12 from both sides of the equation and simplify.
@$$\begin{align}\begin{array}{rcl}
12+3x &=& x-2 \
12-12+3x &=& x-2-12 \
3x &=& x-14
\end{array}\end{align}@$$
Next, subtract @$\begin{align}x\end{align}@$ from both sides of the equation and simplify to isolate the variable on one side.
@$$\begin{align}\begin{array}{rcl}
3x &=& x-14 \
3x-x &=& x-x-14 \
2x &=& -14
\end{array}\end{align}@$$
Then, divide both sides of the equation by 2 to solve for the variable @$\begin{align}x.\end{align}@$
@$$\begin{align}\begin{array}{rcl}
2x &=& -14 \
\overset{1} {\frac{\cancel{2}x}{\cancel{2}}} &=& -\overset{7} {\frac{\cancel{14}}{\cancel{2}}} \
x &=& -7
\end{array}\end{align}@$$
Next, substitute the value for @$\begin{align}x\end{align}@$ into the expression for @$\begin{align}y\end{align}@$ and simplify the right side of the equation.
@$$\begin{align}\begin{array}{rcl}
y &=& 4+x \
y &=& 4+(-7) \
y &=& 4-7 \
y &=& -3
\end{array}\end{align}@$$
The answer is @$\begin{align}x=-7\end{align}@$ and @$\begin{align}y=-3\end{align}@$.
The solution for the system of linear equations is:
@$$\begin{align}\dbinom{x}{y} = \dbinom{\text{-}7}{\text{-}3}\end{align}@$$
Examples
Example 1
Earlier, you were given a problem about Kathleen and her rectangular garden. She needs to figure out the dimensions of the garden. How can Kathleen use a system of equations to help her figure this out?
First, write a system of linear equations to represent the information given in the story.
Let @$\begin{align}l\end{align}@$ represent length and let @$\begin{align}w\end{align}@$ represent the width.
The length of the garden is three times the width and the perimeter is 280 yards.
@$$\begin{align}\begin{array}{rcl}
l &=& 3w \
2l+2w &=& 280
\end{array}\end{align}@$$
Next, substitute the expression for @$\begin{align}l\end{align}@$ into the second equation.
@$$\begin{align}\begin{array}{rcl}
l &=& 3w \
2(3w)+2w &=& 280
\end{array}\end{align}@$$
Next, perform the multiplication on the left side of the equation to clear parentheses and simplify.
@$$\begin{align}\begin{array}{rcl}
2(3w)+2w &=& 280 \
6w+2w &=& 280 \
8w &=& 280
\end{array}\end{align}@$$
Next, divide both sides of the equation by 8 to solve for @$\begin{align}w\end{align}@$.
@$$\begin{align}\begin{array}{rcl}
8w &=& 280 \
\overset{1} {\frac{\cancel{8}w}{\cancel{8}}} &=& \overset{35} {\frac{\cancel{280}}{\cancel{8}}} \
w &=& 35
\end{array}\end{align}@$$
The answer is 35.
The width of the garden is 35 yards.
Next, substitute the value for @$\begin{align}w\end{align}@$ into the expression for the variable @$\begin{align}l\end{align}@$ and simplify the right side of the equation.
@$$\begin{align}\begin{array}{rcl}
l &=& 3w \
l &=& 3(35) \
l &=& 105
\end{array}\end{align}@$$
The answer is 105.
The length of the garden is 105 yards.
The solution for the system of linear equations is @$\begin{align}\dbinom{w}{l} = \dbinom{35}{105}\end{align}@$ which means the dimensions of Kathleen’s garden are 35 yards by 105 yards.
Example 2
Solve the following system of linear equations using the substitution method.
@$$\begin{align}
\begin{array}{rcl}
2x+3y &=& 6 \
x+y &=& 6
\end{array}\end{align}@$$
First, solve the second equation for the variable @$\begin{align}x\end{align}@$ by subtracting @$\begin{align}y\end{align}@$ from both sides of the equation.
@$$\begin{align}\begin{array}{rcl}
x+y &=& 6 \
x+y-y &=& 6-y
\end{array}\end{align}@$$
Next, simplify the left side of the equation.
@$$\begin{align}\begin{array}{rcl}
x+y-y &=& 6-y \
x&=& 6-y
\end{array}\end{align}@$$
Next, substitute this expression for @$\begin{align}x\end{align}@$ into the first equation.
@$$\begin{align}\begin{array}{rcl}
x &=& 6-y \
2x+3y &=& 6 \
2(6-y)+3y &=& 6
\end{array}\end{align}@$$
Next, do the multiplication on the left side of the equation to clear parentheses.
@$$\begin{align}\begin{array}{rcl}
2(6-y)+3y &=& 6 \
12-2y+3y &=& 6
\end{array}\end{align}@$$
Next, simplify the left side of the equation.
@$$\begin{align}\begin{array}{rcl}
12-2y+3y &=& 6 \
12+y &=& 6
\end{array}\end{align}@$$
Next, subtract 12 from both sides of the equation to solve for the variable @$\begin{align}y.\end{align}@$
@$$\begin{align}\begin{array}{rcl}
12+y &=& 6 \
12-12+y &=& 6-12
\end{array}\end{align}@$$
Next, simplify both sides of the equation.
@$$\begin{align}\begin{array}{rcl}
12-12+y &=& 6-12 \
y &=& -6
\end{array}\end{align}@$$
Next, substitute the value for @$\begin{align}y\end{align}@$ into the expression for @$\begin{align}x\end{align}@$ and simplify the right side of the equation.
@$$\begin{align}\begin{array}{rcl}
x &=& 6-y \
x &=& 6-(-6) \
x &=& 6+6 \
x &=& 12
\end{array}\end{align}@$$
The answer is @$\begin{align}x=12\end{align}@$ and @$\begin{align}y=-6\end{align}@$.
The solution for the system of linear equations is:
@$$\begin{align}\dbinom{x}{y} = \dbinom{12}{-6}\end{align}@$$
Example 3
Solve the following system of linear equations using the substitution method.
@$$\begin{align}\begin{array}{rcl}
2y &=& x+4 \
y &=& 3x
\end{array}\end{align}@$$
First, substitute the expression for @$\begin{align}y\end{align}@$ into the first equation.
@$$\begin{align}\begin{array}{rcl}
y &=& 3x \
2y &=& x+4 \
2(3x) &=& x+4
\end{array}\end{align}@$$
Next, perform the multiplication on the left side of the equation to clear parentheses.
@$$\begin{align}\begin{array}{rcl}
2(3x) &=& x+4 \
6x &=& x+4
\end{array}\end{align}@$$
Next, subtract @$\begin{align}x\end{align}@$ from both sides of the equation and simplify to isolate the variable on one side.
@$$\begin{align}\begin{array}{rcl}
6x &=& x+4 \
6x-x &=& x-x+4
\end{array}\end{align}@$$
Next, simplify both sides of the equation.
@$$\begin{align}\begin{array}{rcl}
6x-x &=& x-x+4 \
5x &=& 4
\end{array}\end{align}@$$
Next, divide both sides of the equation by 5 to solve for the variable @$\begin{align}x.\end{align}@$
@$$\begin{align}\begin{array}{rcl}
5x &=& 4 \
\overset{1} {\frac{\cancel{5}}{\cancel{5}}x} &=& \frac{4}{5} \
x &=& \frac{4}{5}
\end{array}\end{align}@$$
Next, substitute the value for @$\begin{align}x\end{align}@$ into the expression for @$\begin{align}y\end{align}@$ and simplify the right side of the equation.
@$$\begin{align}\begin{array}{rcl}
y &=& 3x \
y &=& 3 \left(\frac{4}{5} \right) \
y &=& \frac{12}{5} \
\end{array}\end{align}@$$
The answer is @$\begin{align}x = \frac{4}{5}\end{align}@$ and @$\begin{align}y=\frac{12}{5}\end{align}@$.
The solution for the system of linear equations is:
@$$\begin{align}\dbinom{x}{y} = \dbinom{\frac{4}{5}}{\frac{12}{5}}\end{align}@$$
Example 4
Solve the following system of linear equations using the substitution method.
@$$\begin{align}\begin{array}{rcl}
x+y &=& 4 \
2x-3y &=& 18
\end{array}\end{align}@$$
First, solve the first equation for the variable ‘@$\begin{align}x\end{align}@$’ by subtracting ‘@$\begin{align}y\end{align}@$’ from both sides of the equation.
@$$\begin{align}\begin{array}{rcl}
x+y &=& 4 \
x+y-y &=& 4-y
\end{array}\end{align}@$$
Next, simplify the left side of the equation.
@$$\begin{align}\begin{array}{rcl}
x+y-y &=& 4-y \
x &=& 4-y
\end{array}\end{align}@$$
Next, substitute this expression for ‘@$\begin{align}x\end{align}@$’ into the first equation.
@$$\begin{align}\begin{array}{rcl}
x &=& 4-y \
2x-3y &=& 18 \
2(4-y)-3y &=& 18
\end{array}\end{align}@$$
Next, do the multiplication on the left side of the equation to clear parentheses.
@$$\begin{align}\begin{array}{rcl}
2(4-y)-3y &=& 18 \
8-2y-3y &=& 18
\end{array}\end{align}@$$
Next, simplify the left side of the equation.
@$$\begin{align}\begin{array}{rcl}
8-2y-3y &=& 18 \
8-5y &=& 18
\end{array}\end{align}@$$
Next, subtract 8 from both sides of the equation to isolate the variable ‘@$\begin{align}y\end{align}@$’.
@$$\begin{align}\begin{array}{rcl}
8-5y &=& 18 \
8-8-5y &=& 18-8
\end{array}\end{align}@$$
Next, simplify both sides of the equation.
@$$\begin{align}\begin{array}{rcl}
8-8-5y &=& 18-8 \
-5y &=& 10
\end{array}\end{align}@$$
Next, divide both sides of the equation by -5 to solve for the variable ‘@$\begin{align}y\end{align}@$’.
@$$\begin{align}\begin{array}{rcl}
-5y &=& 10 \
\overset{1} {\frac{\cancel{-5}}{\cancel{-5}}y} &=& \overset{-2} {\frac{\cancel{10}}{\cancel{-5}}} \
y &=& -2
\end{array}\end{align}@$$
Next, substitute the value for ‘@$\begin{align}y\end{align}@$’ into the expression for ‘@$\begin{align}x\end{align}@$’ and simplify the right side of the equation.
@$$\begin{align}\begin{array}{rcl}
x &=& 4-y \
x &=& 4-(-2) \
x &=& 4+2 \
x &=& 6
\end{array}\end{align}@$$
The answer is @$\begin{align}x=6\end{align}@$ and @$\begin{align}y=-2\end{align}@$.
The solution for the system of linear equations is:
@$$\begin{align}\dbinom{x}{y} = \dbinom{6}{-2}\end{align}@$$
Example 5
Solve the following system of linear equations using the substitution method.
@$$\begin{align}\begin{array}{rcl}
3y &=& x-22 \
y &=& 4x
\end{array}\end{align}@$$
First, substitute the expression for ‘@$\begin{align}y\end{align}@$’ into the first equation.
@$$\begin{align}\begin{array}{rcl}
y &=& 4x \
3y &=& x-22 \
3(4x) &=& x-22
\end{array}\end{align}@$$
Next, perform the multiplication on the left side of the equation to clear parentheses.
@$$\begin{align}\begin{array}{rcl}
3(4x) &=& x-22 \
12x &=& x-22
\end{array}\end{align}@$$
Next, subtract ‘@$\begin{align}x\end{align}@$’ from both sides of the equation and simplify to isolate the variable on one side.
@$$\begin{align}\begin{array}{rcl}
12x &=& x-22 \
12x -x &=& x-x-22
\end{array}\end{align}@$$
Next, simplify both sides of the equation.
@$$\begin{align}\begin{array}{rcl}
12x-x &=& x-x-22 \
11x &=& -22
\end{array}\end{align}@$$
Next, divide both sides of the equation by 11 to solve for the variable ‘@$\begin{align}x\end{align}@$’.
@$$\begin{align}\begin{array}{rcl}
11x &=& -22 \
\overset{1} {\frac{\cancel{11}}{\cancel{11}}x} &=& \overset{-2} {\frac{\cancel{-22}}{\cancel{11}}} \
x &=& -2
\end{array}\end{align}@$$
Next, substitute the value for ‘@$\begin{align}x\end{align}@$’ into the expression for ‘@$\begin{align}y\end{align}@$’ and simplify the right side of the equation.
@$$\begin{align}\begin{array}{rcl}
y &=& 4x \
y &=& 4(-2) \
y &=& -8
\end{array}\end{align}@$$
The answer is @$\begin{align}x=-2\end{align}@$ and @$\begin{align}y=-8\end{align}@$.
The solution for the system of linear equations is:
@$$\begin{align}\dbinom{x}{y} = \dbinom{-2}{-8}\end{align}@$$
Review
Solve each linear system by substitution.
1. @$\begin{align}\begin{array}{rcl}
y &=& 8 \
2y+2x &=& 2
\end{array}
\end{align}@$
2. @$\begin{align}\begin{array}{rcl}
x &=& 16 \
2y+2x &=& 2
\end{array}\end{align}@$
3. @$\begin{align}\begin{array}{rcl}
2x &=& 8 \
2y+x &=& 10
\end{array}\end{align}@$
4. @$\begin{align}\begin{array}{rcl}
3y &=& 9 \
y+2x &=& 11
\end{array}\end{align}@$
5. @$\begin{align}\begin{array}{rcl}
y-8 &=& 8 \
2y+x &=& 20
\end{array}\end{align}@$
6. @$\begin{align}\begin{array}{rcl}
4y &=& 8 \
y-2x &=& 8
\end{array}\end{align}@$
7. @$\begin{align}\begin{array}{rcl}
x-2 &=& 4 \
4x+y &=& 12
\end{array}\end{align}@$
8. @$\begin{align}\begin{array}{rcl}
y &=& x+8 \
y+2x &=& 11
\end{array}\end{align}@$
9. @$\begin{align}\begin{array}{rcl}
2y+8 &=& 12 \
y+2x &=& 20
\end{array}\end{align}@$
10. @$\begin{align}\begin{array}{rcl}
y-3 &=& 6 \
3y+3x &=& 9
\end{array}\end{align}@$
11. @$\begin{align}\begin{array}{rcl}
4y-1 &=& 11 \
y-4x &=& 5
\end{array}\end{align}@$
12. @$\begin{align}\begin{array}{rcl}
2y-8 &=& 8 \
2y+2x &=& 2
\end{array}\end{align}@$
13. @$\begin{align}\begin{array}{rcl}
4x+y &=& -2 \
-2x-3y &=& 1
\end{array}\end{align}@$
14. @$\begin{align}\begin{array}{rcl}
y &=& 2x \
6x-y &=& 8
\end{array} \end{align}@$
15. @$\begin{align}\begin{array}{rcl}
x+4y &=& -6 \
2x+10y &=& -6
\end{array}\end{align}@$
Review (Answers)
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13907 | https://math.stackexchange.com/questions/1264571/how-to-find-fracd-tanhkxd-x | derivatives - How to find $ \frac{d (\tanh(kx))}{d x}=?$ - Mathematics Stack Exchange
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How to find d(tanh(k x))d x=?d(tanh(k x))d x=?
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Asked 10 years, 5 months ago
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I am tried to resolve the problem
d(tanh(k x))d x=?d(tanh(k x))d x=?
where k k is positive value.
I found one solution that is
d(tanh(k x))d x=k 2 cosh 2(k x)d(tanh(k x))d x=k 2 cosh 2(k x)
Is it right? If is not true, could you give me the true solution. Thanks
derivatives
partial-derivative
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edited May 3, 2015 at 13:47
Joe
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asked May 3, 2015 at 13:23
JohnJohn
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No, it is not right.
d(tanh(k x))d x=1 cosh 2(k x)⋅d(k x)d x=k cosh 2(k x).d(tanh(k x))d x=1 cosh 2(k x)⋅d(k x)d x=k cosh 2(k x).
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answered May 3, 2015 at 13:27
mathlovemathlove
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Sorry. I mistake the 2 in the formula John –John 2015-05-03 13:29:41 +00:00 Commented May 3, 2015 at 13:29
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Just the classical chain rule.
y=tanh(u(x))y=tanh(u(x))
y′=sech 2(u(x))d u(x)d x y′=sech 2(u(x))d u(x)d x
I do not know where the 2 2 comes from.
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answered May 3, 2015 at 13:28
Claude LeiboviciClaude Leibovici
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Sorry. I mistake the 2 in the formula John –John 2015-05-03 13:29:46 +00:00 Commented May 3, 2015 at 13:29
So, what is final solution without 2 John –John 2015-05-03 13:30:42 +00:00 Commented May 3, 2015 at 13:30
See mathlove's answer. With respect to mine, if u(x)=k x u(x)=k x, d u(x)d x=k d u(x)d x=k Claude Leibovici –Claude Leibovici 2015-05-03 13:33:04 +00:00 Commented May 3, 2015 at 13:33
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13908 | https://www.quora.com/What-are-the-next-two-numbers-of-this-sequence-1-2-5-14-41 | Something went wrong. Wait a moment and try again.
Reasoning Puzzles
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5
What are the next two numbers of this sequence 1, 2, 5, 14, 41, ___?
Rajdeep Das
Quality Analyst at Ekodus Private Limited
·
Author has 70 answers and 71.3K answer views
·
6y
To get the pattern check from right to left
143 -1 = 41
53–1= 14
23 -1= 5
13–1= 2
Therefore the next two numbers will be
413 - 1 = 122
1223–1 = 365
Related questions
What is the sequence of 2, 5, 14, 41?
Can you master the next number of this sequence: 1, 5, 14, 40?
What is the next number in this sequence: 26, -17, 6, 5, -2, 3, 2, 1?
What is the next number in 0, 1, 4, 13, 40, 121, __?
What would be the next number in this sequence: 5, 0, 6, 1, 7, 2, 8?
Arvinder Singh
I like solving algebraic problems.
·
Author has 117 answers and 322.5K answer views
·
6y
I will represent the difference between first two terms as a1 and between second and third term as a2and so on.
Here we have,
a1=2−1=1=30
a2=5−2=3=31
a3=14−5=9=32
a4=41−14=27=33
Guess what comes next?
41+34=122
Similarly,
122+35=365
∴ The next two terms are : 122,365 (Answer)
Please upvote if you liked the answer.
Mohammad Afzaal Butt
B.Sc in Mathematics & Physics, Islamia College Gujranwala (Graduated 1977)
·
Author has 24.6K answers and 22.9M answer views
·
6y
2=1+305=2+3114=5+3241=14+33
Next number
=41+34=41+81=122
Assistant
Bot
·
1y
To find the next two numbers in the sequence 1,2,5,14,41,___, we can look for a pattern.
Let's examine the differences between consecutive terms:
2−1=1
5−2=3
14−5=9
41−14=27
Now, let's look at the differences of these differences:
3−1=2
9−3=6
27−9=18
These differences are 2,6,18, which are 2×1,2×3,2×9 or 21,22×3,23×3.
If we notice a pattern, the second differences appear to be multiplied by 3 each time.
Continuing this pattern, the next difference would be 2×33=54.
Now let's calculate the next terms in the original s
To find the next two numbers in the sequence 1,2,5,14,41,___, we can look for a pattern.
Let's examine the differences between consecutive terms:
2−1=1
5−2=3
14−5=9
41−14=27
Now, let's look at the differences of these differences:
3−1=2
9−3=6
27−9=18
These differences are 2,6,18, which are 2×1,2×3,2×9 or 21,22×3,23×3.
If we notice a pattern, the second differences appear to be multiplied by 3 each time.
Continuing this pattern, the next difference would be 2×33=54.
Now let's calculate the next terms in the original sequence:
The next term after 41 will be 41+54=95.
The next difference after 54 would be 2×34=162.
Therefore, the term following 95 will be 95+162=257.
So, the next two numbers in the sequence are 95 and 257.
Related questions
What is the next number in the series 2, 12, 36, 80, 150?
What is the next number in this tricky sequence 1, 2, 6, 41, _?
What comes next: 2, 2, 3, 5, 6, 11, 12, 23, 24?
What is the next term to this series, 2, 3, 7, 16, 32, and 57?
What is the sequence numbers of 1/2, 1, 3/2, 2, 5/2?
Augustus Kerr
Former Application Programmer
·
Author has 131 answers and 52.6K answer views
·
2y
Originally Answered: What are the next two numbers in the sequence 1, 2, 4, 10, 13, 37, __, __?
·
Two sequences:
1, 4, 13, …
2, 10, 37, …
1: 1 + 3^1 = 4; 4 + 3^2 = 13; 13 + 3^3 = 40
2: 2 + 2^3 = 10; 10 + 3^3 = 37; 37 + 4^3 = 101
Or.
Initial 1, 2 are seed values
Step 1: 4 = 2 + 1 (+ 1)
Step 2: 10 = 4 + 2 + 1 (+ 2 + 1)
Step 3: 13 = 10 (+ 2 + 1)
Step 4: 37 = 13 + 10 + 4 + 2 + 1 (+ 4 + 2 + 1)
Step 5: 40 = 37 (+ 2 + 1)
Step 6: 128 = 40 + 37 + 13 + 10 + 4 + 2 + 1 (+ 10 + 4 + 2 + 1)
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Olav Surland Hansen
Studied Computer Programming & Mathematics at University of Oslo (Graduated 2002)
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6y
Element n is equal to element (n-1) + 3 to the power of (n-2)
For various n
n = 2 gives 1 + 3^(2–2) = 1 + 1 = 2
n = 3 gives 2 + 3^(3–2) = 2 + 3 = 5
.
.
.
n = 5 gives 14 + 3^3 = 14 + 27 = 41
n = 6 gives 41 + 3^4 = 41 + 81 = 122
Imad Zghaib
BA in Mathematics & Engineering, Free University of Brussels (ULB) (Graduated 1989)
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6y
U1=1
U2=3∗U1−1=3∗5−1=15−1=14
U3=3∗U2−1=3∗14−1=42−1=41
U4=3∗U3−1=3∗41−1=123−1=122
We have sequence of general pattern:
Un=3∗Un−1−1
And the number after 41 is 122 :
1, 2, 5, 14, 41, 122,….
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Richard Fettig
Studied Electrical and Electronics Engineering & Physics (Graduated 1982)
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6y
122 and 365. X3 -1. 1x3–1=2. 2x3–1=5. 5x3–1 =14. 14x3–1 =41. So the pattern is last nimber times 3 subtract 1. So 41x3–1 =122. And 122x3–1 is 365
Fidelis Ochili
Works at Fidmak Investment Limited (1999–present)
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6y
1+3^0=2
2+3^1 =5
5+3^2 =14
14+3^3=41
41+3^4=122
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Bruce Stevens
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Related
What are the next two terms of the sequence 1, 5, 14, 30, 55?
The simplest description, of this increasing sequence is, it's an increasing cubic sequence.
The general term rule, of this increasing cubic sequence, when n = 1, 2, 3, …, is:
t(n) = ( ((2n^3) + (3n^2) + n) / 6 ).
The next two terms, when n = 6, and 7, are:
When n = 6:
t(6) = ( ((2 (6^3)) + (3 (6^2)) + 6) / 6 ),
t(6) = ( ((2 216) + (3 36) + 6) / 6 ),
t(6) = ( (432 + 108 + 6) / 6 ),
t(6) = (546 / 6),
t(6) = 91.
When n = 7:
t(7) = ( ((2 (7^3)) + (3 (7^2)) + 7) / 6 ),
t(7) = ( ((2 343) + (3 49) + 7) / 6 ),
t(7) = ( (686 + 147 + 7) / 6 ),
t(7) = (840 / 6),
t(7) = 140.
P.S.
Google search: “cubic seq
The simplest description, of this increasing sequence is, it's an increasing cubic sequence.
The general term rule, of this increasing cubic sequence, when n = 1, 2, 3, …, is:
t(n) = ( ((2n^3) + (3n^2) + n) / 6 ).
The next two terms, when n = 6, and 7, are:
When n = 6:
t(6) = ( ((2 (6^3)) + (3 (6^2)) + 6) / 6 ),
t(6) = ( ((2 216) + (3 36) + 6) / 6 ),
t(6) = ( (432 + 108 + 6) / 6 ),
t(6) = (546 / 6),
t(6) = 91.
When n = 7:
t(7) = ( ((2 (7^3)) + (3 (7^2)) + 7) / 6 ),
t(7) = ( ((2 343) + (3 49) + 7) / 6 ),
t(7) = ( (686 + 147 + 7) / 6 ),
t(7) = (840 / 6),
t(7) = 140.
P.S.
Google search: “cubic sequence formula”
6a = 3rd difference,
(12a + 2b) = the first, 2nd difference,
(7a + 3b + c) = (term_2 — term_1),
(a + b + c + d) = term_1
Steve Smith
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Related
What is the next number in this sequence, 1, 2, 5, 14, 42, ___?
They look like a sequence called the Catalan numbers.
Catalan number - Wikipedia
if so, then 132 is the next term.
Tammy Sellards
Related
What is the next number in this sequence, 1, 2, 5, 14, 42, ___?
Chuck Coker
Homework doer, enjoys doing people's homework for them.
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Related
What is the next number in this sequence: 2, 3, 4, 5, 4, ___?
The next number is 3.
The sequence is the number of letters in Roman numerals, where x=20..24. The next number would be the number of letters in Roman numerals for 25.
XX (2), XXI (3), XXII (4), XXIII (5), XXIV (4), XXV (3)
The sequence could also be the the distances between x and 1 in a square spiral with positive integers and 1 at the center, where x=23..27. The distance from 28 to 1 is 3.
49 26--27--28--29--30--31 | | | 48 25 10--11--12--13 32 | | | | | 47 24 9 2---3 14 33 | | | | | | | 46 23 8 1 4 15 34 | | | |
The next number is 3.
The sequence is the number of letters in Roman numerals, where x=20..24. The next number would be the number of letters in Roman numerals for 25.
XX (2), XXI (3), XXII (4), XXIII (5), XXIV (4), XXV (3)
The sequence could also be the the distances between x and 1 in a square spiral with positive integers and 1 at the center, where x=23..27. The distance from 28 to 1 is 3.
49 26--27--28--29--30--31 | | | 48 25 10--11--12--13 32 | | | | | 47 24 9 2---3 14 33 | | | | | | | 46 23 8 1 4 15 34 | | | | | | 45 22 7---6---5 16 35 | | | | 44 21--20--19--18--17 36 | | 43--42--41--40--39--38--37
The sequence could also be the number of dots in Morse Code, where x=2..6. The next number is 7, with 3 dots.
Or the next number could be 7.
xGreatestPrimeFactor(x)×SmallestPrimeFactor(x)
1, 2, 3, 4, 5, 4, 7, …
Original question: “What is the next number in this sequence: 2, 3, 4, 5, 4, ___?”
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13909 | https://mathoverflow.net/questions/450534/the-function-hx-4-pi-t-n-2-int-mathbbrd-e-frac-x-y24t | reference request - The function $H(x) =(4\pi t)^{-n/2} \int_{\mathbb{R}^d} e^{\frac{-|x-y|^2}{4t}}(1+|y|)^m\,dy$ attains its maximum in $x=0.$ - MathOverflow
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The function H(x)=(4 π t)−n/2∫R d e−|x−y|2 4 t(1+|y|)m d y H(x)=(4 π t)−n/2∫R d e−|x−y|2 4 t(1+|y|)m d y attains its maximum in x=0.x=0.
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In this paper, Lemma 6, Pinsky proves that H(x)=(4 π t)−n/2∫R d e−|x−y|2 4 t(1+|y|)m d y H(x)=(4 π t)−n/2∫R d e−|x−y|2 4 t(1+|y|)m d y attains its maximum in x=0.x=0. I didn't understand the proof very well, but I was visually convinced that the result is true. From the figure below at a=0 a=0 the area under the graph appears to be maximum if x=0. My question is, if considering the function G(x)=(4 π t)−n/2∫R d e−|x−y|4 t(1+|y|)m d y G(x)=(4 π t)−n/2∫R d e−|x−y|4 t(1+|y|)m d y the same result is true? i.e. the square in |x−y||x−y| has some importance? the maximum is reached at x=0 x=0. In that case how to do it? Would the argument be similar to Pinsky's Lemma 6?
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edited Jul 11, 2023 at 0:22
Michael Hardy
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asked Jul 11, 2023 at 0:06
IlovemathIlovemath
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1 I think he simply used spherical coordinates and then proved that the inner integral is nonpositive for each r.Thomas Kojar –Thomas Kojar 2023-07-11 00:13:03 +00:00 Commented Jul 11, 2023 at 0:13
1 For the (3.2) integral you can try the conversion done here math.stackexchange.com/questions/3816246/…, of turning into the dot-product wrt to the standard vector e1. And then maybe the idea from here math.stackexchange.com/questions/4340255/… to deal with the x 1 x 1.Thomas Kojar –Thomas Kojar 2023-07-11 00:43:09 +00:00 Commented Jul 11, 2023 at 0:43
2 The claim also follows from the Riesz rearrangement inequality en.wikipedia.org/wiki/Riesz_rearrangement_inequality (set g(x)=e−|x|/4 t g(x)=e−|x|/4 t, h(x)=(1+|x|)−m h(x)=(1+|x|)−m, and f=δ x f=δ x (or an approximation thereof)).Terry Tao –Terry Tao 2023-07-11 04:02:55 +00:00 Commented Jul 11, 2023 at 4:02
@TerryTao Thank you, I was unaware of this wonderful result! I still have a lot to learn.Ilovemath –Ilovemath 2023-07-11 12:21:38 +00:00 Commented Jul 11, 2023 at 12:21
@TerryTao, I couldn't understand why the function itself will have a maximum at 0 0, due to the Riesz rearrangement inequality, won't the right side have a maximum at 0 0?Ilovemath –Ilovemath 2023-07-13 21:14:40 +00:00 Commented Jul 13, 2023 at 21:14
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Of course, we have to assume that t>0 t>0 -- otherwise, G(x)G(x) will be either undefined (if t=0 t=0) or infinite for all x x (if t<0 t<0). Also, in Pinsky's Lemma 6 it is assumed that m≤0 m≤0.
Let us prove the following statement, which contains both Pinsky's Lemma 6 and your conjecture (and much more):
Proposition 1: For x∈R d x∈R d, let
G(x):=∫R d d y g(|x−y|)f(|y|),(1)(1)G(x):=∫R d d y g(|x−y|)f(|y|),
where g g and f f are nonnegative functions defined on [0,∞)[0,∞) such that
c:=∫R d d y g(|y|)<∞(2)(2)c:=∫R d d y g(|y|)<∞
and g(|y|)g(|y|) is log concave in y y. Then (i) G(x)G(x) is nonincreasing in |x||x| if f f is nonincreasing and (ii) G(x)G(x) is nondecreasing in |x||x| if f f is nondecreasing.
Proof: Let us first prove part (i) of the conclusion of Proposition 1, where f f is nonincreasing. Without loss of generality (wlog), the function f f is right continuous. So, for all real t≥0 t≥0
f(t)=f(∞−)+∫(t,∞)(−d f(u))=f(∞−)+∫(0,∞)(−d f(u))f u(t),f(t)=f(∞−)+∫(t,∞)(−d f(u))=f(∞−)+∫(0,∞)(−d f(u))f u(t),
where f(∞−):=lim t→∞f(t)f(∞−):=lim t→∞f(t) and f u(t):=1(t<u)f u(t):=1(t<u). So, in view of (1)(1) and (2)(2),
G(x):=c f(∞−)+∫(0,∞)(−d f(u))G u(x),(3)(3)G(x):=c f(∞−)+∫(0,∞)(−d f(u))G u(x),
where
G u(x):=∫R d d y g(|x−y|)f u(|y|).G u(x):=∫R d d y g(|x−y|)f u(|y|).
Note that f u(|y|)f u(|y|) is log concave in y y and hence in (x,y)(x,y). Also, g(|x−y|)g(|x−y|) is log concave in (x,y)(x,y), since g(|y|)g(|y|) is log concave in y y. So, g(|x−y|)f u(|y|)g(|x−y|)f u(|y|) is log concave in (x,y)(x,y). So, by the Prékopa–Leindler inequality (cf. e.g. this), G u(x)G u(x) is log concave in x∈R d x∈R d. Also, the function G u G u is radial: G u(x)G u(x) depends on x x only through |x||x|. So, G u(x)G u(x) is nonincreasing in |x||x|.
Since f f is nonincreasing, the Lebesgue--Stieltjes measure (−d f(u))(−d f(u)) is nonnegative. So, in view of (3)(3), G(x)G(x) is nonincreasing in |x||x|, which proves part (i) of the conclusion of Proposition 1.
Let us now prove part (ii) of the conclusion of Proposition 1, where f f is nondecreasing. Replacing f f by min(M,f)min(M,f) for a real M>0 M>0, letting M→∞M→∞, and recalling the monotone convergence theorem, we see that wlog f≤M f≤M for some real M>0 M>0. Then the function M−f M−f is nonnegative and nonincreasing. So, by (2)(2) and part (i) of the conclusion of Proposition 1,
M c−G(x)=M c−∫R d d y g(|x−y|)f(|y|)=∫R d d y g(|x−y|)(M−f(|y|))M c−G(x)=M c−∫R d d y g(|x−y|)f(|y|)=∫R d d y g(|x−y|)(M−f(|y|))
is nonincreasing in |x||x|. So, G(x)G(x) is nondecreasing in |x||x|.
Proposition 1 is completely proved. □◻
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edited Jul 11, 2023 at 8:38
answered Jul 11, 2023 at 8:32
Iosif PinelisIosif Pinelis
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impressive! Thank you very much for the explanation, I did not know this result. Is it from a book? Or did you create?Ilovemath –Ilovemath 2023-07-11 12:11:20 +00:00 Commented Jul 11, 2023 at 12:11
@Ilovemath : There are many applications of the Prékopa–Leindler inequality. Here we have one of them. I have not seen this particular application elsewhere in this same form.Iosif Pinelis –Iosif Pinelis 2023-07-11 15:44:23 +00:00 Commented Jul 11, 2023 at 15:44
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0Weak convergence of f(x,e i t x)f(x,e i t x)
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13910 | https://www.weather.gov/aprfc/normaldepthcalc | Normal Depth Calculator
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One the most commonly used equations governing Open Channel Flow is known as the Mannings's Equation. It was introduced by the Irish Engineer Robert Manning in 1889 as an alternative to the Chezy Equation. The Mannings equation is an empirical equation that applies to uniform flow in open channels and is a function of the channel velocity, flow area and channel slope.
Where:
Q = Flow Rate, (ft 3/s)
v = Velocity, (ft/s)
A = Flow Area, (ft 2)
n = Manning's Roughness Coefficient
R = Hydraulic Radius, (ft)
S = Channel Slope, (ft/ft)
Under the assumption of uniform flow conditions the bottom slope is the same as the slope of the energy grade line and the water surface slope. The Manning’s n is a coefficient which represents the roughness or friction applied to the flow by the channel. Manning’s n-values are often selected from tables, but can be back calculated from field measurements. In many flow conditions the selection of a Manning’s roughness coefficient can greatly affect computational results.
Instructions: Select variable to solve, adjust slider bars, click on graph to modify the cross section. CSV cross section data can be loaded in the input box below. This online calculator is for demonstration and educational purposes only.
Solve For:
Slope: (ft/ft)
WSE: (ft)
Channel Manning n:
Flow: (ft 3/s)
Flow Area:2693.2 (ft 2)
Wetted Perimeter:349.1 (ft)
Max Depth:14.29 (ft)
Average Velocity:7.43 (ft/s)
Top Width:342 (ft)
Iterations:2499
Froude Number:0.47
Load CSV XS Data Below (station,elevation)
Update Plot
Select HECRAS Geometry:
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13911 | https://brainly.com/question/23758776 | [FREE] The relationship between the decay constant (λ) and the half-life (t_{1/2}) is given by the equation - brainly.com
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The relationship between the decay constant (λ) and the half-life (t 1/2) is given by the equation t 1/2=λ 0.693. Use the equation for N to derive this relationship.
Hint: When t=t 1/2, N 0N=0.5.
A sample of carbon-14 initially consists of 5×1 0 24 particles. Carbon-14 has a half-life of 5730 years.
a. What is the decay constant for carbon-14? (Answer in units of yr−1).
b. How many radioactive particles of the sample remain after 100 years?
c. What percentage of the radioactive particles remains after 500 years?
d. How many radioactive particles of the sample remain after 1000 years?
e. How much time will it take for 50% of the particles to decay?
f. How much time will it take for 99% of the particles to decay?
g. How many half-lives will it take for 99% of the sample to decay?
h. What is the initial decay rate of the sample? (Answer in decays/yr.)
i. After 200 years, what is the decay rate of the sample?
j. How long will it take for the decay rate to decrease to 1 0 15 decays/year?
k. How many half-lives have passed after the time you found in part (j)?
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Final answer:
The relationship between the decay constant (λ) and the half-life (t₁/₂) can be derived using the equation N = Noe^(-λt) and by knowing that when t = t(1/2), N/No = 0.5. The decay constant for carbon-14 is approximately 1.21 × 10^-4 y^-1. To find the number of radioactive particles that remain after a certain time, we can use the equation N = Noe^(-λt) and substitute the given values.
Explanation:
The relationship between the decay constant (λ) and the half-life (t₁/₂) is given by the equation t(1/2) = 0.693/λ. This relationship can be derived using the equation N = Noe^(-λt) and by knowing that when t = t(1/2), N/No = 0.5.
To find the decay constant (λ) for carbon-14, we can use the equation t₁/₂ = 0.693/λ. Rearranging this equation, we can solve for λ: λ = 0.693/t₁/₂. Substituting the given half-life of carbon-14 (5730 years), we get λ = 0.693/5730 y^-1, which is approximately 1.21 × 10^-4 y^-1.
To find the number of radioactive particles in the sample that remain after a certain time, we can use the equation N = Noe^(-λt) and substitute the given values. For example, after 100 years, we have N = (5 × 10^24)e^(-1.21 × 10^-4 100), which can be calculated to find the number of remaining particles.
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The decay constant for carbon-14 is approximately 1.21 × 10^-4 yr^-1. After 100 years, about 4.938 × 10^{24} particles remain, and 95.46% of the particles remain after 500 years. It takes about 38,000 years for 99% of the particles to decay, which corresponds to approximately 6.63 half-lives.
Explanation
The relationship between the decay constant (λ) and the half-life (t 1/2) can be derived from the equation for radioactive decay:
N=N 0e−λ t
Where:
N is the quantity remaining after time t
N 0 is the initial quantity
λ is the decay constant
e is the base of the natural logarithm
When t=t 1/2, it is known that N 0N=0.5. Substituting this into the decay equation gives:
0.5=e−λ t 1/2
Taking the natural logarithm (ln) of both sides:
ln(0.5)=−λ t 1/2
We know that ln(0.5)=−0.693, therefore:
−0.693=−λ t 1/2
Rearranging this gives the relationship:
λ=t 1/20.693
This shows how the decay constant is related to the half-life of the substance.
Now, for the carbon-14 sample:
a. To find the decay constant for carbon-14 with a half-life of 5730 years:
λ=5730 0.693≈1.21×1 0−4 yr−1
b. To find the number of radioactive particles remaining after 100 years:
Using N=N 0e−λ t:
N=(5×1 0 24)e−1.21×1 0−4×100≈5×1 0 24 e−0.0121≈4.938×1 0 24 particles
c. The percentage of particles that remain after 500 years:
First, find N:
N=(5×1 0 24)e−1.21×1 0−4×500≈5×1 0 24 e−0.0605≈4.773×1 0 24 particles
Percentage remaining:
N 0N=5×1 0 24 4.773×1 0 24×100≈95.46%
d. After 1000 years, using the same method:
N=(5×1 0 24)e−1.21×1 0−4×1000≈5×1 0 24 e−0.121≈4.443×1 0 24 particles
e. It takes 5730 years for 50% of the particles to decay since that is the definition of half-life.
f. For 99% decay, we can find the time t for which N 0N=0.01:
0.01=e−λ t Taking ln of both sides:
ln(0.01)=−λ t→t=λ−ln(0.01)≈1.21×1 0−4−(−4.605)≈38,000 years
g. Since the half-life is 5730 years, it takes approximately 6.63 half-lives for 99% decay, which is 39,400 years.
h. The initial decay rate can be found using:
R=λ N 0≈1.21×1 0−4×(5×1 0 24)≈6.05×1 0 20 decays/year
i. After 200 years, calculate N:
N=(5×1 0 24)e−1.21×1 0−4×200≈4.911×1 0 24 particles
Then find the new decay rate:
R=1.21×1 0−4⋅N≈1.21×1 0−4⋅4.911×1 0 24≈5.94×1 0 20 decays/year
j. To reach a decay rate of 1 0 15 decays/year, we can set up:
R=λ N≈1.21×1 0−4⋅(5×1 0 24 e−λ t)=1 0 15
Solving this equation gives:
t≈38,000 years
k. From part (j), dividing the total time by the half-life gives:
5730 38,000≈6.63 half-lives
Examples & Evidence
For example, if you start with a sample of carbon-14 and measure its quantity, after one half-life (5730 years), you would have half of the original amount. If you continue this process, after two half-lives (11,460 years), you would have a quarter of the original amount, showcasing the exponential decay of carbon-14 over time.
The calculations rely on the defined half-life of carbon-14, which is widely accepted in scientific literature as 5730 years, confirming that the derived decay constant and remaining particle calculations are accurate.
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Community Answer The rate of decay for a particular type of radioactive particle is relatively constant, and can be represented using the equation N (t) = Noe (-In 2) t ti Where t is time, N is the mass of the sample, and t1 is the half-life (time it takes for half of the initial sample to decay). The half-life of Carbon-14 is about 5730 years. How many years would it take a 1000 gram sample to decay to only 300 grams? Round your answer to the nearest tenth.
Community Answer The rate of decay for a particular type of radioactive particle is relatively constant, and can be represented using the equation Where t is time, N is the mass of the sample, and is the half-life (time it takes for half of the initial sample to decay). The half-life of Carbon-14 is about 5730 years. How many years would it take a 1000 gram sample to decay to only 400 grams
Community Answer Where t is time, N is the mass of the sample, and t12 is the half-life (time it takes for half of the initial sample to decay). The half-life of Carbon-14 is about 5730 years. How many years would it take a 1000 gram sample to decay to only 400 grams
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13912 | https://oeis.org/A365627 | A365627 - OEIS
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A365627
Number of parking functions of length n with cars parking at most 4 spots away from their preferred spot.
1
1, 1, 3, 16, 125, 1296, 15511, 212978, 3321091, 58196400, 1134161181, 24333706866, 569786870013, 14455456239756, 394940662364775, 11560567008386106, 360947377705705971, 11973823441677468648, 420576028975783973061, 15593290472977894193850
(list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
LINKS
Table of n, a(n) for n=0..19.
CROSSREFS
Column k=4 of A365623.
Sequence in context: A365626A000950A320254 A245012A000951A000272
Adjacent sequences: A365624A365625A365626 A365628A365629A365630
KEYWORD
nonn
AUTHOR
J. Carlos Martínez Mori, Sep 13 2023
STATUS
approved
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13913 | https://artofproblemsolving.com/wiki/index.php/Asymptote:_Basics?srsltid=AfmBOop7Eb2PLlEBUIsVa4XdFozFkuyhIxq5ejSlTABxX_axZnu93MIJ | Art of Problem Solving
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Asymptote: Basics
Asymptote (Vector Graphics Language)
Getting Started - Basics - Drawing - Labeling - Filling - Useful functions - Examples - Macros and Packages
Help - Reference - Advanced Asymptote - 3D Graphics - CSE5 Package - How to
Syntax
In the AoPS community, all Asymptote diagrams are declared with the [asy] tag and ended with the [/asy] tag. On the AoPS Wiki, Asymptote is declared like HTML tags, using and . Using the Asymptote package in LaTeX, asymptote is declared using \begin{asy} and \end{asy}.
Each command in Asymptote must be separated by a semicolon (;), similar to programming languages like C and Java. This convention tells Asymptote where each command ends. Commands do not necessarily need to be on separate lines, since semicolons are the only method by which Asymptote interprets new commands. However, putting the commands on separate lines is often useful to separate your commands to organize your code and improve readability for people looking to interpret your code. Whitespaces before and after commands are also not read by Asymptote, so any line can be indented as far as desired for clarity's sake.
To write comments (Lines that are not interpreted by Asymptote), start a line with two forward slashes (//. Multi-line comments can be declared with / and ended with /.
Applying what we've learned, we can now write clear, legible, Asymptote code:
/
This is a program
that draws a triangle
/
//Below, we draw the first segment
draw((0,0)--(50,50));
//And the next two segments, written without a line break between them
draw((50,0)--(50,50));draw((50,0)--(0,0));
Variables and Data Types
When coding in Asymptote, it's often helpful to store data in data types. For example, a real number like or a pair of numbers like . Each new variable that you create must be declared using a data type, with the command [datatype] [variable];. For example, if you wanted to declare the variable to have type integer, you can use the command
int n;
After it's declared, you can store a specific value in a variable using the = symbol, as in
n=3;
These two commands can be abbreviated by the single command int n=3;, and several integers can also be declared at once (int m,n,d;) or even many declarations and assignations at once: int a,b=2,c,d=5;.
The most commonly used data types in Asymptote are given in the following table:
Size and Unitsize
Asymptote is a primarily coordinate-based graphics language. Each point is a pair where is the -coordinate and is the -coordinate.
However, there are many ways to choose a Cartesian coordinate system for the plane; one must pick the placement of origin and the scale on each of the - and -axis. Asymptote will place your image in the center of your output page after it is drawn, so placement of origin is actually irrelevant. By default, the unit length in both the and directions is the PostScript bigpoint, which has length inches. Thus, if you do not change the scaling on the picture, the points and are exactly one inch apart when drawn in Asymptote. However, drawing in bigpoints is inconvenient if you wish to draw a figure that is exactly 3cm wide.
The function unitsize can be used to specify the unit length for your picture. This function takes up to 3 arguments: the picture you want to scale the axes for (if this isn't specified, it defaults to currentpicture, the picture you are drawing on), the unit length in the x direction, and the unit length in the y direction. If only one real argument is given, both the x and y unit sizes are set to this number. Thus the command
unitsize(72);
will tell Asymptote that from now on, your unit length is inch. Be careful when you are redefining your unit length - now that unitsize is set to , the points and are actually inches apart!
Asymptote has the built-in constants pt (1/72.27 inches), inch, cm, and mm for convenience when defining lengths, so the above command can also be stated
unitsize(1inch);
The other useful function is size, which specifies the exact width and height of the box that your picture (if unspecified as a first argument, this will again default to currentpicture) will be fit into. If only one number is given, both the width and the height will be set to this number. For example, the command
size(5cm,5cm);
or just
size(5cm);
will fit the diagram to a 5cm x 5cm box regardless of the specified unitsizes.
As an example, make an Asymptote document containing the following two lines:
unitsize(2inch);
draw(unitsquare);
and see what happens as you change the inch size to several other values.
Next: Drawing
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Prove that in an n(n+1) table filled with integers, we can always cross out some columns and make the sum of the integers in each row, even
The boxes of an n (n+1) table ( n rows and n+1 columns) are filled with integers. Prove that one can cross out zero or several columns ( not all of them ) so that after this operation, the sum of the numbers in each row is even.
From the book, Mathematical circles Russian experience.
P.S: Also, can someone help me complete my partial attempt at solving this using induction.
This is my attempt using induction: I first proved for the 1 2 table. Then successfully proved that whenever it is true for the n (n+1) case, it would also be true for the n (n+2) case. Now, I needed to prove that whenever it is true for the n (n+2) case, it would also be true for the (n+1) (n+2) case. But, this is where I failed. Can someone please help me with the proof for the last part ?
4 Answers 4
The key idea is to
reduce the numbers modulo 2, and then treat the columns as vectors in $\mathbb F_2^{n}$. Call these vectors $c_1,\dots, c_{n+1}$. Here $\mathbb F_2$ is the field of integers modulo 2.
Then, since there are $n+1$ columns,
these column vectors must be linearly dependent: there exists $a_1,\dots, a_{n+1}\in {0,1}$, not all zero, so that $a_1c_1+\cdots+a_{n+1}c_{n+1}$ is the zero vector (modulo 2).
Then we are done by
crossing out the columns for which $a_i=0$.
Edit: Here's another more elementary solution that's trickier to think of (in my opinion), but accessible to high-school students.
Note that all that matters is the parity of the numbers, we just have to keep track of odd/even-ness of the row-sums rather than the actual values. Now,
there are $n$ rows, so there are $2^n$ possibilities for the the parity of the row sums. However, there are $2^{n+1}-1$ ways to delete some (but not all) columns, which is certainly bigger than $2^n$.
This means
there are at least two different ways to delete the columns that yield the same pattern of parities for the row sums.
Now the trick is that
if you pick out the columns that are left undeleted in exactly one of these two ways, and delete the remaining columns, that gets us done!
Why? Well, let's explain this with an example.
Suppose in the first choice of columns, the columns $1,3,5,6$ get left undeleted, and in the second case, the columns $1,4,5,7,8$ stay undeleted. The assumption says adding the columns $1+3+5+6$ gives the same pattern of odds and evens in the row sums, as adding the columns $1+4+5+7+8$.
But that means,
if I add all of them together, ending up with 2 copies of column 1, two copies of column 5, and one copy each of columns $3,4,6,7,8$, we get even sums in each row! However, the columns that get added twice don't contribute to the parity at all, so adding columns $3,4,6,7$ and $8$ would give even sums in each row as well.
@Ankoganit's brilliant answer gives the proof on two different levels, the easier of which should be accessible with high school mathematics. I liked the proof quite a lot, so I wanted to try and rewrite it so that it can be understood with very little maths at all. Here's what I ended up with.
To reiterate the puzzle setup,
With all that, our task is to prove that in addition to the "all columns off" state, there is always another way to make every row total show "even".
Following the method in @Ankoganit's answer (or at least something that works in an analogous fashion), we can start by toggling all the columns off, and then, one by one, we'll try every possible combination of columns, and write down the exact pattern of odds and evens in the row sums.
For example, if the numbers and row sums look like this with all the columns A to E toggled on
we'd start systematically writing down what the row sums are when only a portion of the columns is toggled on:
and so on.
Then comes the first clever bit: no matter what the actual numbers in the grid are, we are always guaranteed to find a pair. That is, we can always find two column combinations that will produce the exact same pattern of odds and evens for the row sums. Why? Because of the grid shape, and something called the pigeonhole principle: There are more columns than there are rows, so there are also more column combinations (each column is either on or off) than there are possible distinct patterns of row sums (each row sum is either odd or even). If you try to match each column combination to a different row sum pattern, you will run out of row sum patterns first, so by necessity, some row sum pattern(s) must have more than one column combination that produces them.
So, we know we can always find a pair, which is when the second clever bit of Ankoganit's answer steps in: We can do addition to such a pair! If we toggle one of the combinations first, and then the second one, then each row sum will be either "even+even=even", or "odd+odd=even", and that's exactly what we wanted!
There's a final catch though: what if both the combinations in the pair included the same column? We cannot very well include the same column twice, now can we?
Turns out, this is not a problem at all. We just need to notice that adding a number changes parity in exactly same way that subtracting does, so if we need to toggle on a column that was already on, we can achieve the exact same effect by toggling that column back off.
That covers all the cases, so the proof is complete, and since we never used the grid size or the specific contents of the grid in the proof, it applies to all grids that fit the initial description. Just for fun and illustration purposes, we can then apply it to our example case:
and now that we finally found a pair, we can add A+E to B to achieve the desired "even-even-even-even" at "A+B+E":
Finally, to demonstrate the "other kind of addition", we can notice that this was not actually the first pair we found: at the start, we had already calculated A+B+C+D+E=even-even-even-odd, which is the same pattern as C+D, so we could have also done A+B+C+D+E + C+D = A+B+C+D+E-C-D = A+B+E = even-even-even-even to reach the same result.
A+B+C+D+E=even-even-even-odd
C+D
A+B+C+D+E + C+D = A+B+C+D+E-C-D = A+B+E = even-even-even-even
Hope this was worth your time, and if you find something that could be made even more clear, please do drop a comment; that would be much appreciated.
Maybe I do not understand the question. But in the 23 table
you cannot cross out any one or two columns and get even sums.
On the other hand the original table contains even sums in both rows.
"Several" means (for me as a non-English speaker) at least one. so this is a counterexample.
What is probably meant is: Prove that one can cross out 0 or more columns (but not all of them) so that after this operation, the sum of the numbers in each row is even.
At OP's special request here is a proof by induction:
Note: I first tried writing the following without using "modulo 2" but now think it is not worth the clutter.
Thus we rephrase given an n x (n+1) table of numbers mod 2 we can always find a nonempty set of columns that sum to the zero vector.
The assertion is easily confirmed for 1x2 tables, i.e. n=1.
Now assume it has been established for (n-1) x n. Given an n x (n+1) table T pick and remove a column c and set it aside. After also removing any row r we can apply the assumption to find a nonempty set S(c,r) of columns that does not include the c-th such that their sum as column vectors vanishes everywhere except possibly in row r. If the r-th row also vanishes the proof is complete. As this last statement holds for any r, to finish, we need only consider the case where for all r the sum of S(c,r) is $e_r$, the r-th standard basis vector. Having the entire standard basis at our disposal we needn't know any linear algebra to see that we can form any vector, in particular, we can express the c-th column in terms of the other columns. As we are modulo 2 all coefficients are either one or zero, i.e. "yes" or "no". Recall that we have so far avoided using column c. Adding it now yields the zero vector as sum of a set of columns that include column c hence is nonempty.
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13915 | https://ncert.nic.in/textbook/pdf/kemh108.pdf | vNatural numbers are the product of human spirit. – DEDEKIND v 8.1 Introduction In mathematics, the word, “sequence” is used in much the same way as it is in ordinary English. When we say that a collection of objects is listed in a sequence, we usually mean that the collection is ordered in such a way that it has an identified first member, second member, third member and so on. For example, population of human beings or bacteria at different times form a sequence. The amount of money deposited in a bank, over a number of years form a sequence.
Depreciated values of certain commodity occur in a sequence. Sequences have important applications in several spheres of human activities.
Sequences, following specific patterns are called progressions. In previous class, we have studied about arithmetic progression (A.P). In this Chapter, besides discussing more about A.P.; arithmetic mean, geometric mean, relationship between A.M.
and G.M., special series in forms of sum to n terms of consecutive natural numbers, sum to n terms of squares of natural numbers and sum to n terms of cubes of natural numbers will also be studied.
8.2 Sequences Let us consider the following examples: Assume that there is a generation gap of 30 years, we are asked to find the number of ancestors, i.e., parents, grandparents, great grandparents, etc. that a person might have over 300 years.
Here, the total number of generations = 300 10 30 = Fibonacci (1175-1250) Chapter SEQUENCES AND SERIES 8 Reprint 2025-26 136 MATHEMATICS The number of person’s ancestors for the first, second, third, …, tenth generations are 2, 4, 8, 16, 32, …, 1024. These numbers form what we call a sequence.
Consider the successive quotients that we obtain in the division of 10 by 3 at different steps of division. In this process we get 3,3.3,3.33,3.333, ... and so on. These quotients also form a sequence. The various numbers occurring in a sequence are called its terms. We denote the terms of a sequence by a1, a2, a3, …, an, …, etc., the subscripts denote the position of the term. The nth term is the number at the nth position of the sequence and is denoted by an. The nth term is also called the general term of the sequence.
Thus, the terms of the sequence of person’s ancestors mentioned above are: a1 = 2, a2 = 4, a3 = 8, …, a10 = 1024.
Similarly, in the example of successive quotients a1 = 3, a2 = 3.3, a3 = 3.33, …, a6 = 3.33333, etc.
A sequence containing finite number of terms is called a finite sequence. For example, sequence of ancestors is a finite sequence since it contains 10 terms (a fixed number).
A sequence is called infinite, if it is not a finite sequence. For example, the sequence of successive quotients mentioned above is an infinite sequence, infinite in the sense that it never ends.
Often, it is possible to express the rule, which yields the various terms of a sequence in terms of algebraic formula. Consider for instance, the sequence of even natural numbers 2, 4, 6, … Here a1 = 2 = 2 × 1 a2 = 4 = 2 × 2 a3 = 6 = 2 × 3 a4 = 8 = 2 × 4 .... .... ....
.... .... ....
.... .... ....
.... .... ....
a23 = 46 = 2 × 23, a24 = 48 = 2 × 24, and so on.
In fact, we see that the nth term of this sequence can be written as an = 2n, where n is a natural number. Similarly, in the sequence of odd natural numbers 1,3,5, …, the nth term is given by the formula, an = 2n – 1, where n is a natural number.
In some cases, an arrangement of numbers such as 1, 1, 2, 3, 5, 8,.. has no visible pattern, but the sequence is generated by the recurrence relation given by a1 = a2 = 1 a3 = a1 + a2 an = an – 2 + an – 1, n > 2 This sequence is called Fibonacci sequence.
Reprint 2025-26 SEQUENCES AND SERIES 137 In the sequence of primes 2,3,5,7,…, we find that there is no formula for the nth prime. Such sequence can only be described by verbal description.
In every sequence, we should not expect that its terms will necessarily be given by a specific formula. However, we expect a theoretical scheme or a rule for generating the terms a1, a2, a3,…,an,… in succession.
In view of the above, a sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it. Sometimes, we use the functional notation a(n) for an.
8.3 Series Let a1, a2, a3,…,an, be a given sequence. Then, the expression a1 + a2 + a3 +,…+ an + ...
is called the series associated with the given sequence .The series is finite or infinite according as the given sequence is finite or infinite. Series are often represented in compact form, called sigma notation, using the Greek letter ∑(sigma) as means of indicating the summation involved. Thus, the series a1 + a2 + a3 + ... + an is abbreviated as 1 n k k a = ∑ .
Remark When the series is used, it refers to the indicated sum not to the sum itself.
For example, 1 + 3 + 5 + 7 is a finite series with four terms. When we use the phrase “sum of a series,” we will mean the number that results from adding the terms, the sum of the series is 16.
We now consider some examples.
Example 1 Write the first three terms in each of the following sequences defined by the following: (i) an = 2n + 5, (ii) an = 3 4 n − .
Solution (i) Here an = 2n + 5 Substituting n = 1, 2, 3, we get a1 = 2(1) + 5 = 7, a2 = 9, a3 = 11 Therefore, the required terms are 7, 9 and 11.
(ii) Here an = 3 4 n − . Thus, 1 2 3 1 3 1 1 0 4 2 4 a , a ,a − = = − = − = Reprint 2025-26 138 MATHEMATICS Hence, the first three terms are 1 1 2 4 – , – and 0.
Example 2 What is the 20th term of the sequence defined by an = (n – 1) (2 – n) (3 + n) ?
Solution Putting n = 20 , we obtain a20 = (20 – 1) (2 – 20) (3 + 20) = 19 × (– 18) × (23) = – 7866.
Example 3 Let the sequence an be defined as follows: a1 = 1, an = an – 1 + 2 for n ≥ 2.
Find first five terms and write corresponding series.
Solution We have a1 = 1, a2 = a1 + 2 = 1 + 2 = 3, a3 = a2 + 2 = 3 + 2 = 5, a4 = a3 + 2 = 5 + 2 = 7, a5 = a4 + 2 = 7 + 2 = 9.
Hence, the first five terms of the sequence are 1,3,5,7 and 9. The corresponding series is 1 + 3 + 5 + 7 + 9 +...
EXERCISE 8.1 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are: 1. an = n (n + 2) 2.
an = 1 n n+ 3.
an = 2n 4. an = 2 3 6 n − 5. an = (–1)n–1 5n+1 6.
an 2 5 4 n n + = .
Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are: 7. an = 4n – 3; a17, a24 8.
an = 2 7 ; 2n n a 9.
an = (–1)n – 1n3; a9 10.
20 ( – 2) ; 3 n n n a a n = + .
Reprint 2025-26 SEQUENCES AND SERIES 139 Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 11.
a1 = 3, an = 3an – 1 + 2 for all n > 1 12.
a1 = – 1, an = 1 n a n −, n ≥ 2 13.
a1 = a2 = 2, an = an – 1–1, n > 2 14.
The Fibonacci sequence is defined by 1 = a1 = a2 and an = an – 1 + an – 2, n > 2.
Find 1 n n a a + , for n = 1, 2, 3, 4, 5 8.4 Geometric Progression (G . P.) Let us consider the following sequences: (i) 2,4,8,16,..., (ii) 1 1 1 1 9 27 81 243 – – , , , ... (iii) .
,.
,.
,...
01 0001 000001 In each of these sequences, how their terms progress? We note that each term, except the first progresses in a definite order.
In (i), we have a a a a a a a 1 2 1 3 2 4 3 2 2 2 2 = = = = , , , and so on.
In (ii), we observe, a a a a a a a 1 2 1 3 2 4 3 1 9 1 3 1 3 1 3 = = = = , , , and so on.
Similarly, state how do the terms in (iii) progress? It is observed that in each case, every term except the first term bears a constant ratio to the term immediately preceding it. In (i), this constant ratio is 2; in (ii), it is – 1 3 and in (iii), the constant ratio is 0.01.
Such sequences are called geometric sequence or geometric progression abbreviated as G .P.
A sequence a1, a2, a3, …, an, … is called geometric progression, if each term is non-zero and a a k k + 1 = r (constant), for k ≥ 1.
By letting a1 = a, we obtain a geometric progression, a, ar, ar2, ar3,…., where a is called the first term and r is called the common ratio of the G.P. Common ratio in geometric progression (i), (ii) and (iii) above are 2, – 1 3 and 0.01, respectively.
Reprint 2025-26 140 MATHEMATICS As in case of arithmetic progression, the problem of finding the nth term or sum of n terms of a geometric progression containing a large number of terms would be difficult without the use of the formulae which we shall develop in the next Section. We shall use the following notations with these formulae: a = the first term, r = the common ratio, l = the last term, n = the numbers of terms, Sn = the sum of first n terms.
8.4.1 General term of a G .P. Let us consider a G.P. with first non-zero term ‘a’ and common ratio ‘r’. Write a few terms of it. The second term is obtained by multiplying a by r, thus a2 = ar. Similarly, third term is obtained by multiplying a2 by r. Thus, a3 = a2r = ar2, and so on.
We write below these and few more terms.
1st term = a1 = a = ar1–1, 2nd term = a2 = ar = ar2–1, 3rd term = a3 = ar2 = ar3–1 4th term = a4 = ar3 = ar4–1, 5th term = a5 = ar4 = ar5–1 Do you see a pattern? What will be 16th term?
a16 = ar16–1 = ar15 Therefore, the pattern suggests that the nth term of a G.P. is given by an = arn–1.
Thus, a, G.P. can be written as a, ar, ar2, ar3, … arn – 1; a, ar, ar2,...,arn – 1... ;according as G.P. is finite or infinite, respectively.
The series a + ar + ar2 + ... + arn–1 or a + ar + ar2 + ... + arn–1 +...are called finite or infinite geometric series, respectively.
8.4.2. Sum to n terms of a G .P. Let the first term of a G.P. be a and the common ratio be r. Let us denote by Sn the sum to first n terms of G.P. Then Sn = a + ar + ar2 +...+ arn–1 ... (1) Case 1 If r = 1, we have Sn = a + a + a + ... + a (n terms) = na Case 2 If r ≠ 1, multiplying (1) by r, we have rSn = ar + ar2 + ar3 + ... + arn ... (2) Subtracting (2) from (1), we get (1 – r) Sn = a – arn = a(1 – rn) This gives or ( 1) S 1 n n a r r − = − Example 4 Find the 10th and nth terms of the G.P. 5, 25,125,… .
Solution Here a = 5 and r = 5. Thus, a10 = 5(5)10–1 = 5(5)9 = 510 and an = arn–1 = 5(5)n–1 = 5n .
Reprint 2025-26 SEQUENCES AND SERIES 141 Example 5 Which term of the G.P., 2,8,32, ... up to n terms is 131072?
Solution Let 131072 be the nth term of the given G.P. Here a = 2 and r = 4.
Therefore 131072 = an = 2(4)n – 1 or 65536 = 4n – 1 This gives 48 = 4n – 1.
So that n – 1 = 8, i.e., n = 9. Hence, 131072 is the 9th term of the G.P.
Example 6 In a G.P., the 3rd term is 24 and the 6th term is 192.Find the 10th term.
Solution Here, a ar 3 2 24 = = ... (1) and a ar 6 5 192 = = ... (2) Dividing (2) by (1), we get r = 2. Substituting r = 2 in (1), we get a = 6.
Hence a10 = 6 (2)9 = 3072.
Example 7 Find the sum of first n terms and the sum of first 5 terms of the geometric series 2 4 1 3 9 ...
+ + + Solution Here a = 1 and r = 2 3 . Therefore Sn = 2 1 3 (1 ) 2 1 1 3 n n a r r − − = − − = 2 3 1 3 n − In particular, 5 5 2 S 3 1 3 = − = 211 3 243 × = 211 81 .
Example 8 How many terms of the G.P. 3 3 3 2 4 , , ,... are needed to give the sum 3069 512 ?
Solution Let n be the number of terms needed. Given that a = 3, r = 1 2 and 3069 S 512 n = Since (1 ) S 1 n n a – r r = − Reprint 2025-26 142 MATHEMATICS Therefore 1 3(1 ) 3069 1 2 6 1 1 512 2 1 2 n n − = = − − or 3069 3072 = 1 1 2n − or 1 2n = 3069 1 3072 − 3 1 3072 1024 = = or 2n = 1024 = 210, which gives n = 10.
Example 9 The sum of first three terms of a G.P. is 13 12 and their product is – 1.
Find the common ratio and the terms.
Solution Let a r , a, ar be the first three terms of the G.P. Then a ar a r + + = 13 12 ... (1) and ( ) ( ) 1 a a ar – r = ... (2) From (2), we get a3 = – 1, i.e., a = – 1 (considering only real roots) Substituting a = –1 in (1), we have 1 13 1 12 – – –r r = or 12r2 + 25r + 12 = 0.
This is a quadratic in r, solving, we get 3 4 or 4 3 r – – = .
Thus, the three terms of G.P. are : 4 3 –3 3 4 –4 , 1, for = and , 1, for = 3 4 4 4 3 3 – r – r , Example10 Find the sum of the sequence 7, 77, 777, 7777, ... to n terms.
Solution This is not a G.P., however, we can relate it to a G.P. by writing the terms as Sn = 7 + 77 + 777 + 7777 + ... to n terms Reprint 2025-26 SEQUENCES AND SERIES 143 = 7 [9 99 999 9999 to term] 9 ...
n + + + + = 2 3 4 7 [(10 1) (10 1) (10 1) (10 1) terms] 9 ...n − + − + − + − + = 2 3 7 [(10 10 10 terms) (1+1+1+... terms)] 9 ...n – n + + + = 7 10(10 1) 7 10 (10 1) 9 10 1 9 9 n n n n − − − = − − .
Example 11 A person has 2 parents, 4 grandparents, 8 great grandparents, and so on.
Find the number of his ancestors during the ten generations preceding his own.
Solution Here a = 2, r = 2 and n = 10 Using the sum formula Sn = ( 1) 1 n a r r − − We have S10 = 2(210 – 1) = 2046 Hence, the number of ancestors preceding the person is 2046.
8.4.3 Geometric Mean (G .M.) The geometric mean of two positive numbers a and b is the number ab . Therefore, the geometric mean of 2 and 8 is 4. We observe that the three numbers 2,4,8 are consecutive terms of a G.P. This leads to a generalisation of the concept of geometric means of two numbers.
Given any two positive numbers a and b, we can insert as many numbers as we like between them to make the resulting sequence in a G.P.
Let G1, G2,…, Gn be n numbers between positive numbers a and b such that a,G1,G2,G3,…,Gn,b is a G.P. Thus, b being the (n + 2)th term,we have 1 n b ar + = , or 1 1 n b r a + = .
Hence 1 1 1 G n b ar a a + = = , 2 1 2 2 G n b ar a a + = = , 3 1 3 3 G n b ar a a + = = , 1 G n n n n b ar a a + = = Reprint 2025-26 144 MATHEMATICS Example12 Insert three numbers between 1 and 256 so that the resulting sequence is a G.P.
Solution Let G1, G2,G3 be three numbers between 1 and 256 such that 1, G1,G2,G3 ,256 is a G.P.
Therefore 256 = r4 giving r = ± 4 (Taking real roots only) For r = 4, we have G1 = ar = 4, G2 = ar2 = 16, G3 = ar3 = 64 Similarly, for r = – 4, numbers are – 4,16 and – 64.
Hence, we can insert 4, 16, 64 between 1 and 256 so that the resulting sequences are in G.P.
8.5 Relationship Between A.M. and G.M.
Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively.
Then A andG 2 a b ab + = = Thus, we have A – G = 2 a b ab + − = 2 2 a b ab + − = ( ) 2 0 2 a b − ≥ ... (1) From (1), we obtain the relationship A ≥ G.
Example 13 If A.M. and G.M. of two positive numbers a and b are 10 and 8, respectively, find the numbers.
Solution Given that A.M.
10 2 a b + = = ... (1) and G.M.
8 ab = = ... (2) From (1) and (2), we get a + b = 20 ... (3) ab = 64 ... (4) Putting the value of a and b from (3), (4) in the identity (a – b)2 = (a + b)2 – 4ab, we get (a – b)2 = 400 – 256 = 144 or a – b = ± 12 ... (5) Reprint 2025-26 SEQUENCES AND SERIES 145 Solving (3) and (5), we obtain a = 4, b = 16 or a = 16, b = 4 Thus, the numbers a and b are 4, 16 or 16, 4 respectively.
EXERCISE 8.2 1.
Find the 20th and nth terms of the G.P. 5 5 5 2 4 8 , , , ...
2.
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
3.
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.
4.
The 4th term of a G.P. is square of its second term, and the first term is – 3.
Determine its 7th term.
5.
Which term of the following sequences: (a) 2 2 2 4 is128 ?
, , ,...
(b) 3 3 3 3 is729 ?
, , ,...
(c) 1 1 1 1 is 3 9 27 19683 , , ,...
?
6.
For what values of x, the numbers 2 7 7 2 – , x, – are in G.P.?
Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 7.
0.15, 0.015, 0.0015, ... 20 terms.
8.
7 , 21 , 3 7 , ... n terms.
9.
1, – a, a2, – a3, ... n terms (if a ≠ – 1).
10.
x3, x5, x7, ... n terms (if x ≠ ± 1).
11.
Evaluate 11 1 (2 3 ) k k= + ∑ .
12.
The sum of first three terms of a G.P. is 39 10 and their product is 1. Find the common ratio and the terms.
13.
How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?
14.
The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G .P.
15.
Given a G.P. with a = 729 and 7th term 64, determine S7.
Reprint 2025-26 146 MATHEMATICS 16.
Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
17.
If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
18.
Find the sum to n terms of the sequence, 8, 88, 888, 8888… .
19.
Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1 2 .
20.
Show that the products of the corresponding terms of the sequences a, ar, ar2, …arn – 1 and A, AR, AR2, … ARn – 1 form a G.P, and find the common ratio.
21.
Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
22.
If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq – r br – pcP – q = 1.
23.
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.
24.
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is 1 n r .
25.
If a, b, c and d are in G.P. show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2 .
26.
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
27.
Find the value of n so that a b a b n n n n + + + + 1 1 may be the geometric mean between a and b.
28.
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio ( ) ( ) 3 2 2 : 3 2 2 + − .
29.
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A A G A G ( )( ) ± + − .
30.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour ?
Reprint 2025-26 SEQUENCES AND SERIES 147 31.
What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
32.
If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Miscellaneous Examples Example 14 If a, b, c, d and p are different real numbers such that (a2 + b2 + c2)p2 – 2(ab + bc + cd) p + (b2 + c2 + d2) ≤ 0, then show that a, b, c and d are in G.P.
Solution Given that (a2 + b2 + c2) p2 – 2 (ab + bc + cd) p + (b2 + c2 + d2) ≤ 0 ... (1) But L.H.S.
= (a2p2 – 2abp + b2) + (b2p2 – 2bcp + c2) + (c2p2 – 2cdp + d2), which gives (ap – b)2 + (bp – c)2 + (cp – d)2 ≥ 0 ... (2) Since the sum of squares of real numbers is non negative, therefore, from (1) and (2), we have, (ap – b)2 + (bp – c)2 + (cp – d)2 = 0 or ap – b = 0, bp – c = 0, cp – d = 0 This implies that b c d p a b c = = = Hence a, b, c and d are in G.P.
Miscellaneous Exercise On Chapter 8 1.
If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and 1 ( ) 120 n x f x = = ∑ , find the value of n.
2.
The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
3.
The first term of a G.P. is 1. The sum of the third term and fifth term is 90.
Find the common ratio of G.P.
4.
The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers. 5. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Reprint 2025-26 148 MATHEMATICS 6.
If a bx a bx b cx b cx c dx c dx x + − = + − = + − ≠ ( ), 0 then show that a, b, c and d are in G.P. 7.
Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P.
Prove that P2Rn = Sn. 8.
If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P. 9.
If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15.
10.
The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that ( ) ( ) 2 2 2 2 : a b m m – n : m – m – n = + .
11.
Find the sum of the following series up to n terms: (i) 5 + 55 +555 + … (ii) .6 +. 66 +. 666+… 12.
Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms.
13.
A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?
14.
Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?
15.
A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.
16.
A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.
Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.
17.
A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
18.
150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on.
It took 8 more days to finish the work. Find the number of days in which the work was completed.
Reprint 2025-26 SEQUENCES AND SERIES 149 Summary ®By a sequence, we mean an arrangement of number in definite order according to some rule. Also, we define a sequence as a function whose domain is the set of natural numbers or some subsets of the type {1, 2, 3, ....k}. A sequence containing a finite number of terms is called a finite sequence. A sequence is called infinite if it is not a finite sequence.
®Let a1, a2, a3, ... be the sequence, then the sum expressed as a1 + a2 + a3 + ...
is called series. A series is called finite series if it has got finite number of terms.
®A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout. This constant factor is called the common ratio. Usually, we denote the first term of a G.P. by a and its common ratio by r. The general or the nth term of G.P. is given by an= arn – 1.
The sum Sn of the first n terms of G.P. is given by ( ) ( ) – 1 1– S 1 1 1 – n n n a r a r = or , if r r – r ≠ ®The geometric mean (G.M.) of any two positive numbers a and b is given by ab i.e., the sequence a, G, b is G.P.
Historical Note Evidence is found that Babylonians, some 4000 years ago, knew of arithmetic and geometric sequences. According to Boethius (510), arithmetic and geometric sequences were known to early Greek writers. Among the Indian mathematician, Aryabhatta (476) was the first to give the formula for the sum of squares and cubes of natural numbers in his famous work Aryabhatiyam, written around 499. He also gave the formula for finding the sum to n terms of an arithmetic sequence starting with pth term. Noted Indian mathematicians Brahmgupta Reprint 2025-26 150 MATHEMATICS — v v v v v — (598), Mahavira (850) and Bhaskara (1114-1185) also considered the sum of squares and cubes. Another specific type of sequence having important applications in mathematics, called Fibonacci sequence, was discovered by Italian mathematician Leonardo Fibonacci (1170-1250). Seventeenth century witnessed the classification of series into specific forms. In 1671 James Gregory used the term infinite series in connection with infinite sequence. It was only through the rigorous development of algebraic and set theoretic tools that the concepts related to sequence and series could be formulated suitably.
Reprint 2025-26 |
13916 | https://www.youtube.com/watch?v=oia7fG3RpS8 | Place Value – Teaching Children about the Concept of Zero
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This video helps teachers explain to children the importance of 'zero', especially in relation to place value, and as a placeholder for numbers in our base 10 structure.
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in this video we are going to talk about the importance of zero place value is a key concept for children to get to grips with without an understanding of place value they will struggle to read and write numbers let alone add subtract multiply or divide a really important digit children need to learn about is zero explain that the word zero comes from a Latin word serum which means empty or blank children need to recognize zero as a label for an empty collection of things tell children that a collection of things is called a set a set might be something like all the people living in the world that's a big set an empty set is something with no things for example the set of people living in the world who are five meters tall the set of people living in the world who have six eyes the sets of people living in the world who have antlers explain that because there are no people in the world after over five metres tall then we say this set is empty and we use zero to symbolize it being empty emphasize to children that zero should always be referred to as zero and not nothing naught or none can children think of some more examples of their own talk about some empty sets the funnier the better children also need to recognize zero as a placeholder for numbers in our base-10 structure tell children that in place value the zero symbol is used as a placeholder to signify the absence of value in a particular position for example in the number twenty the ones position is empty so the zero stands in the empty place and holds the position explain that without the zero being in this place then 20 would be two so the role of zero as a placeholder is crucial repeat again that the zero is holding the place for the ones to show that the ones column is empty look at some different examples together and explain where the zero is standing and what place it is holding for example in the number 105 the tens place is empty so the zero is standing in its place and holding its position write the following number on the board say that someone might say this number as 58 can children say why this wouldn't be right ask children what the number does say and ask them to explain what the zero is doing explain that in the number 508 the zero is holding the tens position go through some more examples together and ask what the zero is doing in other words what place is it standing in and hold in position four to end tell children that a number doesn't usually begin with zero for example naught 375 would have the value of 375 as there are no thousands but explain that sometimes the zero is used especially when writing dates another thing to remember and to share with children is that zero isn't the lowest number their experience may be limited but explain that there are numbers below zero that are much smaller and these are called negative numbers you |
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From Wikipedia, the free encyclopedia
Presence of internal cells of the cervical canal outside the cervix
Not to be confused with Ectropion.
Medical condition
| Cervical ectropion |
| Other names | Cervical eversion |
| Cervical ectropion |
| Specialty | Gynecology |
Cervical ectropion is a condition in which the cells from the 'inside' of the cervical canal, known as glandular cells (or columnar epithelium), are present on the 'outside' of the vaginal portion of the cervix. The cells on the 'outside' of the cervix are typically squamous epithelial cells. Where the two cells meet is called the transformation zone, also known as the stratified squamous epithelium. Cervical ectropion can be grossly indistinguishable from early cervical cancer and must be evaluated by a physician to determine risks and prognosis. It may be found incidentally when a vaginal examination (or pap smear test) is done. The area may look red because the glandular cells are red. While many women are born with cervical ectropion, it can be caused by a number of reasons, such as:
Hormonal changes, meaning it can be common in young women
Using oral contraceptives
Pregnancy.
Signs and symptoms
[edit]
Cervical ectropion can be associated with excessive, non-purulent vaginal discharge due to the increased surface area of columnar epithelium containing mucus-secreting glands as well as intermenstrual bleeding (bleeding outside of regular menses). It may also give rise to post-coital bleeding, as fine blood vessels present within the columnar epithelium are easily traumatized.
Causes
[edit]
Cervical ectropion is a normal phenomenon, especially in the ovulatory phase in younger women, during pregnancy, and in women taking oral contraceptive, which increases the total estrogen level in the body. It also may be a congenital problem by the persistence of the squamocolumnar junction which is normally present prior to birth.
Mucopurulent cervicitis may increase the size of the cervical ectropion.
Mechanism
[edit]
The squamocolumnar junction, where the columnar secretory epithelium of the endocervical canal meets the stratified squamous covering of the ectocervix, is located at the external os before puberty. As estrogen levels rise during puberty, the cervical os opens, exposing the endocervical columnar epithelium onto the ectocervix. This area of columnar cells on the ectocervix forms an area that is red and raw in appearance called an ectropion (cervical erosion). It is then exposed to the acidic environment of the vagina and, through a process of squamous metaplasia, transforms into stratified squamous epithelium.
Treatment
[edit]
Usually no treatment is indicated for clinically asymptomatic cervical ectropions. Hormonal therapy may be indicated for symptomatic erosion. If it becomes troublesome to the patient, it can be treated by discontinuing oral contraceptives, cryotherapy treatment, or by using ablation treatment under local anesthetic. Ablation involves using a preheated probe (100 °C) to destroy 3–4 mm of the epithelium. In post-partum erosion, observation and re-examination are necessary for 3 months after labour.
References
[edit]
^ Aggarwal, Pearl; Ben Amor, Anissa (31 May 2023). "Cervical Ectropion". StatPearls. StatPearls Publishing.
^ "UpToDate".
^ Standring: Gray's Anatomy, 40th ed.
^ Bope: Conn's Current Therapy 2011, 1st Edition.
^ Standring: Gray's Anatomy, 40th ed.
External links
[edit]
| | |
--- |
| Classification | D ICD-10: N86 MeSH: D002579 DiseasesDB: 2288 |
| v t e Female reproductive system |
| Internal | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- | | Adnexa | | | | | | | | | | --- --- --- --- | | Ovaries | | | | --- | | Follicles | Corpus + Hemorrhagicum + Luteum + Albicans Thecae + Externa + Interna Follicular antrum + Follicular fluid Corona radiata Zona pellucida Membrana granulosa Perivitelline space | | Other | Germinal epithelium Tunica albuginea Cortex + Cumulus oophorus + Stroma Medulla Retia ovarii | | Oogenesis | Oogonium Oocytogenesis Oocyte Ootidogenesis Ootid Ovum | | | Fallopian tubes | Isthmi Ampullae Infundibula Fimbriae Ostia Intramural segments | | Ligaments | Ovarian ligaments Suspensory ligaments + Pampiniform plexus | | Wolffian vestiges | Gartner's ducts Epoophora + Vesicular appendages Paroophora | | | Uterus | | | | --- | | Regions | Body + Cavity + Fundus Isthmus Cervix + Canal + Internal os + External os + Supravaginal portion Horns | | Layers | Wall + Endometrium + Myometrium + Perimetrium Parametrium Epithelium | | Ligaments | Round ligaments Broad ligaments Cardinal ligaments Uterosacral ligaments Pubocervical ligaments | | General | Glands Urogenital diaphragm | | | Vestibular glands | Bartholin's glands Skene's glands | | Vagina | Canal + Rugae + Walls + Fornices Support structures Epithelium | |
| External | | | | | | | | | | --- --- --- --- | | Vulva | | | | --- | | Labia | Mons pubis Labia majora + Pudendal cleft + Labiocrural folds + Anterior commissure + Posterior commissure + Dartos muliebris + Posterior labial nerves + Anterior labial nerves + Perineal nerve Labia minora + Frenulum Posterior labial arteries Posterior labial veins Interlabial sulci | | Clitoris | Root + Crura + Bulbs + Suspensory ligament + Fundiform ligament Body + Corpora cavernosa + Trabeculae of corpora cavernosa + Infra-corporeal residual spongy part + Angle Commissure of bulbs Pars intermedia Glans + Corona + Frenulum Arteries + Dorsal arteries + Deep arteries + Arteries of bulbs + Internal pudendal artery Veins + Superficial dorsal veins + Deep dorsal veins + Veins of bulbs + Internal pudendal veins Nerves + Dorsal nerves + Pudendal nerve Clitoral hood Fascia Tunica albuginea Septum | | Vestibule | Fossa Vaginal orifice + Hymen Vestibular gland openings Urethra + External urethral orifice | | |
| Blood supply | Arteries + Ovarian arteries + Uterine artery + Arcuate artery + Vaginal artery + Spiral arteries Veins + Ovarian veins + Uterine vein + Uterine venous plexuses + Vaginal venous plexus |
| Other | G-spot Urethral sponge Perineal sponge Rectouterine pouch Vesicouterine pouch Uterotubal junctions |
Retrieved from "
Category:
Cervix
Hidden categories:
Articles with short description
Short description is different from Wikidata
Cervical ectropion
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13918 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10777755/ | An Unpleasant Souvenir: Whipworm as an Incidental Finding During a Screening Colonoscopy - PMC
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J Investig Med High Impact Case Rep
. 2024 Jan 9;12:23247096231224328. doi: 10.1177/23247096231224328
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An Unpleasant Souvenir: Whipworm as an Incidental Finding During a Screening Colonoscopy
Lefika Bathobakae
Lefika Bathobakae, MD, MPH
1 St. Joseph’s University Medical Center, Paterson, NJ, USA
Find articles by Lefika Bathobakae
1,✉, Tyler Wilkinson
Tyler Wilkinson, MSc
2 St. George’s University, Grenada, West Indies
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2, Saif Yasin
Saif Yasin, BS
2 St. George’s University, Grenada, West Indies
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2, Rammy Bashir
Rammy Bashir, MSc
2 St. George’s University, Grenada, West Indies
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2, Nargis Mateen
Nargis Mateen, MD
1 St. Joseph’s University Medical Center, Paterson, NJ, USA
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1, Ruhin Yuridullah
Ruhin Yuridullah, MD
1 St. Joseph’s University Medical Center, Paterson, NJ, USA
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1, Yana Cavanagh
Yana Cavanagh, MD
1 St. Joseph’s University Medical Center, Paterson, NJ, USA
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1, Walid Baddoura
Walid Baddoura, MD
1 St. Joseph’s University Medical Center, Paterson, NJ, USA
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1, Jin Suh
Jin Suh, MD, FACP, FIDSA
1 St. Joseph’s University Medical Center, Paterson, NJ, USA
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1
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1 St. Joseph’s University Medical Center, Paterson, NJ, USA
2 St. George’s University, Grenada, West Indies
✉
Lefika Bathobakae, MD, MPH, Internal Medicine, St. Joseph’s University Medical Center, 165 Barclay St., Paterson, NJ 07503, USA. Email: lbathoba@sgu.edu
Received 2023 Nov 23; Accepted 2023 Dec 15; Collection date 2024 Jan-Dec.
© 2024 American Federation for Medical Research
This article is distributed under the terms of the Creative Commons Attribution-NonCommercial 4.0 License ( which permits non-commercial use, reproduction and distribution of the work without further permission provided the original work is attributed as specified on the SAGE and Open Access page (
PMC Copyright notice
PMCID: PMC10777755 PMID: 38193443
Abstract
Trichuriasis is a neglected tropical disease caused by Trichuris trichiura that spreads through the ingestion of embryonated eggs in contaminated soil, water, or food. In nonendemic areas, T trichiura infestation is very rare and sporadic and is often diagnosed in immigrants from endemic countries such as the Philippines. Whipworms feed on human blood and also erode the colonic mucosa, thereby evoking an inflammatory response. In milder forms, trichuriasis can be asymptomatic and often an incidental diagnosis on screening colonoscopy. Heavily infested patients usually present with abdominal pain, nausea, vomiting, tenesmus, chronic diarrhea, iron deficiency anemia, or stunted growth. T trichiura worms can be removed with biopsy forceps during a colonoscopy; however, most patients require a course of albendazole, mebendazole, or ivermectin. We describe a unique case of T trichiura as an incidental finding during a screening colonoscopy. The whipworms were retrieved using biopsy forceps and the patient was treated with albendazole. At the time of the colonoscopy, the patient did not exhibit any specific symptoms related to the worm infestation.
Keywords:Trichuris trichiura, trichuriasis, whipworm, colonoscopy, colon, infectious colitis, albendazole, eosinophilia
Introduction
Trichuris trichiura is a common soil-transmitted helminth that causes trichuriasis in human hosts.1-3 It is one of more than 70 recognized species of Trichuris, with a long history of association with humans, as evidenced by the presence of its eggs in coprolites from archeological sites.2Trichuris trichiura is transmitted via ingestion of embryonated eggs in contaminated soil or food and can cause a long-term infection.2-4 Trichuriasis is endemic in tropical and subtropical countries as the causative agent thrives well in warm and humid climate.2,5 This parasitic infection is more prevalent in areas with poor sanitation and limited access to clean water2 such as sub-Saharan Africa, India, China, and the Caribbean.6 The severity of the symptoms has a positive correlation with the burden of the worms within the human gastrointestinal tract.6 Signs and symptoms range from no symptoms to profuse diarrhea, abdominal pain, hematochezia, tenesmus, anemia, and rectal prolapse in severe cases.1,3,4
Identification of T trichiura eggs in stool specimens is the primary diagnostic method especially in symptomatic patients.3 Colonoscopy with biopsy has also been employed in cases where there is a high index of suspicion for trichuriasis despite a negative stool test.3 Like in our case, T trichiura has also been found incidentally on screening colonoscopies.3 Treatment involves the use of antihelminthic medications such as albendazole, mebendazole, and ivermectin, with the duration of treatment depending on the severity of the infection.1,3 Endoscopic removal with biopsy forceps has also been reported in some cases.7 Herein, we describe a unique case of a geriatric Asian female with a history of peptic ulcer disease who was found to have T trichiura worm during a screening colonoscopy. The parasite was treated with a 3-day course of oral albendazole, and posttreatment stool ova and parasite test results were negative.
Case Report
A 71-year-old female with a medical history of asthma, hypertension, and peptic ulcer disease presented to our ambulatory center for screening colonoscopy and an esophagogastroduodenoscopy (EGD) due to a dilated pancreatic duct and pancreatic cyst recently seen on imaging. The patient was in her usual state of health until 2 years ago when she experienced dull pain on the left side of her abdomen. The abdominal pain was intermittent, nonradiating, and associated with watery diarrhea, and a 30-pound weight loss over a few months. She denied experiencing nausea, vomiting, melena, hematochezia, or tenesmus. The patient stated that she attributed the pain to gastric ulcers and the weight loss to stress. Since it was at the start of the COVID-19 pandemic, the patient was unable to see a doctor due to fear of exposing herself to the virus. She stayed in the Philippines until international travel resumed.
When she returned to the United States 2 years later, the patient visited her doctor for evaluation of abdominal pain, and the blood work was significant for elevated pancreatic enzymes. Abdominal ultrasound showed a 1.5 × 1.9 × 1.7 cm hypoechoic focus in the pancreatic tail concerning for a pancreatic cyst versus cystic mass. An abdominal magnetic resonance imaging (MRI) and magnetic resonance cholangiopancreatography (MRCP) without contrast showed distention of the pancreatic duct throughout the tail extending to a lesion in the body measuring 1.5 × 1.5 × 1.5 cm and worrisome for neoplasm. Also seen was the cystic lesion in the pancreatic tail of the pancreas measuring 1.2 × 1.4 × 1.1 cm concerning for a sequela of pancreatitis or an intraductal papillary mutinous neoplasm.
While undergoing further workup, the patient continued to experience occasional dull pain in the left lower abdomen, but without diarrhea. She was referred to an interventional gastroenterologist for an EGD with endoscopic ultrasound-guided fine-needle aspiration, which revealed pancreatic abnormalities consistent with atrophy of the entire pancreas. The pancreatic duct had a dilated endosonographic appearance in the entire pancreas. The pancreatic duct measured up to 6 mm in diameter. A 1.9 cm cystic lesion was observed in the pancreatic tail.
Given the history of colonoscopy with polypectomy 5 years ago, the patient also underwent a screening colonoscopy, which showed nonbleeding internal hemorrhoids and a single nonbleeding colonic angioectasia that was treated with argon plasma coagulation to prevent bleeding (Figure 1). Worms were found in the ascending colon and in the cecum (Figure 1) and were removed using cold forceps for histology. The examined portion of the ileum was normal. Histopathological examination revealed a colonic mucosa with eosinophilia and active colitis consistent with the patient’s parasitic infestation. The morphology of the parasite was consistent with T trichiura (whipworm). The patient was treated with oral albendazole 400 mg daily, for 3 days. Stool ova and parasite tests after the treatment were negative.
Figure 1.
Open in a new tab
Endoscopic image showing worms in the ascending colon (panels A and B), a single medium-sized localized angioectasia in the sigmoid colon (panel C), and nonbleeding internal hemorrhoids on retroflexion (panel D).
Molecular testing showed AccuCEA: 648 ng/mL, amylase: 485 U/L with no KNAS or GNAS mutations. On routine follow-up, the patient continued to complain of lower abdominal pain and weight loss. Repeat imaging of the abdomen demonstrated 4.7 × 8.0 × 4.8 cm pancreatic body and tail mass with extensive invasion of the surrounding vasculature. Two weeks later, the patient underwent a laparoscopic peritoneal washing, laparoscopic biopsy of the pancreatic mass, subclavian chemo port placement, and diagnostic laparoscopy. Cytology was positive for malignant cells consistent with pancreatic adenocarcinoma. The patient is currently undergoing neoadjuvant chemotherapy with FOLFIRINOX (folinic acid, fluorouracil, irinotecan hydrochloride, and oxaliplatin) and is estimated to complete 6 to 8 cycles before repeat imaging for restaging. She continues to have the lower abdominal pain which improves with gabapentin and is most likely due to the pancreatic cancer.
Discussion
Trichuriasis is a neglected parasitic infection caused by the whipworm T trichiura and is endemic in tropical regions due to a warm and humid climate.2,8 It is estimated that 500 million people are infected worldwide, with at least 1 billion people at risk of infection every year.9 Trichuriasis cases in nonendemic countries are sporadic and related to international travel to endemic countries.9 Our patient for example, had recently traveled to Asia where she spent at least 2 years. Trichuriasis is acquired through fecal-oral transmission when a human host consumes a water or food contaminated with T trichiura eggs.3,4 These eggs eventually hatch in the small intestines and migrate to the large intestines. Their presence activates an immune response that is responsible for the recruitment of eosinophils, lymphocytes, and plasma cells. The parasite usually resides in the terminal ileum and cecum; however, it is not uncommon for the entire colon and rectum to be involved. The adult worm lays eggs, which will ultimately be shed in the feces.2
T trichiura infestation is usually without symptoms, but a higher worm burden can lead to intestinal mucosal injury.8,10 Symptoms include loss of appetite, weight loss, abdominal pain and distension, bloody diarrhea, and tenesmus.2,7 Chen et al11 reported an interesting case of T trichiura as an endoscopic finding in a farmer with a 6-month history of abdominal pain and a positive fecal blood test. Complications of untreated or unresolved trichuriasis include iron deficiency anemia, severe dehydration, asthenia, rectal prolapse, and trichuris dysentery syndrome (TDS).2,6,7,12 Zanwar et al12 described a unique case of TDS in a pediatric patient mimicking inflammatory bowel disease (IBD). The patient presented with bloody stools and generalized abdominal pain for 2 months, and blood work was significant for severe microcytic anemia and eosinophilia. Diagnostic colonoscopy revealed numerous worms in the entire colon, and the infection was resolved with oral albendazole. Our patient had a short course of watery diarrhea while in the Philippines but none was reported while in the United States. Abdominal pain, which persisted even after treatment was most likely due to the pancreatic cancer. At the time of the colonoscopy, the patient did not exhibit any specific symptoms related to the worm infestation.
A comprehensive history and physical examination can lead to a timely diagnosis of trichuriasis or related parasitic infections. This investigation is augmented with blood work, an infectious stool workup, and in some cases, a colonoscopy. Eosinophilia on a complete blood count, albeit unspecific, typically raises a suspicion of a parasitic infection.3 Microcytic anemia may be seen in long-term infections as the parasite damages the intestinal mucosa and feeds on blood. T trichiura eggs can be observed on stool microscopic examinations as barrel-shaped ova.10-12 In endemic regions, the World Health Organization (WHO) recommends the use of the KATO-KATZ technique to quantify the burden of T trichiura eggs per gram of stool samples for a quicker and lower-cost diagnsosis.8,13 Stool ova and parasite test has a very low sensitivity (2.7%-6.4%), thus, symptomatic patients with negative stool analysis warrant a colonoscopy or sigmoidoscopy to confirm a diagnosis.1,3,11 On colonoscopy, the white worms can be seen attached to the mucosa in the ascending colon or cecum.1,2 Dangling worms with their anterior portions threaded into the colonic mucosa give the classic “coconut cake rectum” look.8 Tokmak et al10 reported a unique case of trichuriasis that was first seen on a computed tomography (CT) scan as an irregular and nodular marked thickening of the cecum and ascending colon. The diagnosis was confirmed on colonoscopy and histopathology. A repeat CT scan 6 months after treatment revealed a colonic wall with normal thickness.
Once a diagnosis of trichuriasis is confirmed, treatment modalities may range from endoscopic removal to medications, or both. In nonendemic areas, colonoscopy serves as an important diagnostic and therapeutic tool, allowing direct visualization and retrieval of worms attached to the mucosa at their anterior ends.1,14,15 Ona et al14 described a case series of 2 unrelated Bangladeshi women who were found to have whipworm infection on a diagnostic colonoscopy. The 2 patients presented with abdominal pain, anorexia, and iron deficiency anemia, and the worms were removed using biopsy forceps. The patients’ symptoms resolved with albendazole. Kim et al15 also described a case of long-term TDS in a young North Korean defector that was managed with endoscopic worm retrieval and albendazole. Similarly in our case, the whipworms were removed with biopsy forceps, and the patient was treated with albendazole to ensure complete eradication of the parasite.
In endemic regions, trichuriasis is treated with a 3-day course of albendazole 400 mg or mebendazole 500 mg, or mebendazole 100 mg twice daily.16 Mebendazole, given at 100 mg twice daily for 3 days, shows a 70% success rate, with a second course recommended if no cure is achieved within 3 to 4 weeks.6 Albendazole, though slightly less effective, offers better adherence as a single-dose regimen, potentially achieving cure rates of up to 80% when extended to 3 days.6 For severe infections resistant to standard treatment, therapeutic colonoscopy with pincers is effective, ensuring successful outcomes.5 Combination therapies, such as albendazole with ivermectin, have promising potential particularly in pediatric patients.6 Monitoring treatment effectiveness is very important due to the partial efficacy of anthelmintic drugs against T trichiura, necessitating close follow-up for appropriate intervention.16 In cases of severe anemia secondary to T trichiura infection, iron supplementation is recommended as a supportive measure.16 Ensuring patient compliance is essential, with home visits by health care professionals if needed, especially in challenging follow-up scenarios.6 Proper patient education on hygiene practices is critical to prevent reinfection, especially in endemic regions.3,16
Conclusion
Trichuriasis is a neglected parasitic infection caused by T trichiura worms and is endemic to tropical and subtropical regions. Trichuris trichiura infection spreads via the fecal-oral route as the embryonated eggs are ingested in a contaminated medium. Although, trichuriasis is largely asymptomatic, some patients may present with gastrointestinal symptoms and iron deficiency anemia. Trichuris trichiura infestation in the United States is extremely rare and sporadic, thus it is often an incidental finding on screening colonoscopy. Trichuriasis is treated endoscopically and with antihelminthic medications such as albendazole, mebendazole, or ivermectin.
Footnotes
Authors’ Contribution: LB and TW conceptualized the idea of this case report. SY, RB, NM, and RY wrote some parts of the case report and fully participated in the patient’s care. YC, WB, and JS fact-checked, edited, and proofread the final version of this case report.
The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.
Funding: The author(s) received no financial support for the research, authorship, and/or publication of this article.
Ethics Approval: Our institution does not require IRB approval/waiver for case reports.
Consent: The patient consented to the publication of this case report.
ORCID iD: Lefika Bathobakae
Data Availability Statement: Further enquiries can be directed to the corresponding author
References
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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
[hide]
1 Proofs
2 Generalizations
2.1 Weighted AM-GM Inequality
2.2 Mean Inequality Chain
2.3 Power Mean Inequality
3 Problems
3.1 Introductory
3.2 Intermediate
3.3 Olympiad
4 See Also
Proofs
Main article: Proofs of AM-GM
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
Main 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
Main article: 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.
Problems
Introductory
For nonnegative real numbers , demonstrate that if then . (Solution)
Find the maximum of for all positive . (Solution)
Intermediate
Find the minimum value of for .
(Source)
Olympiad
Let , , and be positive real numbers. Prove that
(Source)
See Also
Proofs of AM-GM
Mean Inequality Chain
Power Mean Inequality
Cauchy-Schwarz Inequality
Inequality
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Coordination Compounds · Chemistry · JEE Main
MCQ (Single Correct Answer)
Start Practice
MCQ (Single Correct Answer)
[Match the LIST-I with LIST-II
| LIST-I(Complex/ Species) | LIST-II(Shape & magnetic moment) |
--- |
| A. [Ni(CO)4] | I. Tetrahedral, 2.8 BM |
| B. [Ni(CN)4]2– | II. Square planar, 0 BM |
| C. [NiCl4]2– | III. Tetrahedral, 0 BM |
| D. [MnBr4]2– | IV. Tetrahedral, 5.9 BM |
Choose the correct answer from the options given below:
JEE Main 2025 (Online) 8th April Evening Shift](/past-years/jee/question/pmatch-the-blist-ib-with-blist-iibpstyletab-jee-main-chemistry-28sk9xcpp9g9jox3)[The number of species from the following that are involved in sp3d2 hybridization is :
[Co(NH3)6]3+, SF6, [CrF6]3−, [CoF6]3−, [Mn(CN)6]3−, and [MnCl6]3−
JEE Main 2025 (Online) 8th April Evening Shift](/past-years/jee/question/pthe-number-of-species-from-the-following-that-are-involve-jee-main-chemistry-g0pxn9xpribtyskg)Given below are two statements:
Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism.
Statement II: cis- and trans- platin are heteroleptic complexes of Pd.
In the light of the above statements, choose the correct answer from the options given below:
JEE Main 2025 (Online) 8th April Evening Shift[Match List - I with List - II.
| List - I (Complex) | List - II (Primary valency and Secondary valency) |
--- |
| (A) [Co(en)2Cl2]Cl | (I) 3, 6 |
| (B) [Pt(NH3)2Cl(NO2)] | (II) 3, 4 |
| (C) Hg [Co(SCN)4] | (III) 2, 6 |
| (D) [Mg (EDTA)]2− | (IV) 2, 4 |
Choose the correct answer from the options given below :
JEE Main 2025 (Online) 7th April Evening Shift](/past-years/jee/question/pmatch-blist-ib-with-blist-iibptable-bo-jee-main-chemistry-vp5biw77bxvo832s)[The number of unpaired electrons responsible for the paramagnetic nature of the following complex species are respectively :
[Fe(CN)6]3−, [FeF6]3−, [CoF6]3−, [Mn(CN)6]3−
JEE Main 2025 (Online) 7th April Evening Shift](/past-years/jee/question/pthe-number-of-unpaired-electrons-responsible-for-the-para-jee-main-chemistry-yv4qvz0qkthckirg)['X' is the number of acidic oxides among VO2, V2O3, CrO3, V2O5 and Mn2O7. The primary valency of cobalt in [Co(H2NCH2CH2NH2)3]2(SO4)3 is Y. The value of X + Y is _________.
JEE Main 2025 (Online) 7th April Evening Shift](/past-years/jee/question/px-is-the-number-of-acidic-oxides-among-vosub2sub-jee-main-chemistry-k4kzvxnlfdha7zpi)An octahedral complex having molecular composition $\mathrm{Co} \cdot 5 \mathrm{NH}_3 \cdot \mathrm{Cl}^2 . \mathrm{SO}_4$ has two isomers A and B. The solution of A gives a white precipitate with $\mathrm{AgNO}_3$ solution and the solution of B gives white precipitate with $\mathrm{BaCl}_2$ solution. The type of isomerism exhibited by the complex is,
JEE Main 2025 (Online) 7th April Morning Shift[' $X$ ' is the number of electrons in $t_{2 g}$ orbitals of the most stable complex ion among $\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_6\right]^{3+},\left[\mathrm{FeCl}_6\right]^{3-}, \quad\left[\mathrm{Fe}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$. The nature of oxide of vanadium of the type $\mathrm{V}_2 \mathrm{O}_{\mathrm{X}}$ is :
JEE Main 2025 (Online) 4th April Evening Shift](/past-years/jee/question/p-x--is-the-number-of-electrons-in-t2-g-orbitals-jee-main-chemistry-some-basic-concepts-of-chemistry-0x1ofgg5bcdzo3ez)[The correct order of $\left[\mathrm{FeF}_6\right]^{3-},\left[\mathrm{CoF}_6\right]^{3-},\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ complex species based on the number of unpaired electrons present is:
JEE Main 2025 (Online) 4th April Evening Shift](/past-years/jee/question/pthe-correct-order-of-leftmathrmfef6right3-l-jee-main-chemistry-some-basic-concepts-of-chemistry-aviajxzaisnfazne)Which one of the following complexes will have $\Delta_{\mathrm{o}}=0$ and $\mu=5.96$ B.M?
JEE Main 2025 (Online) 4th April Morning Shift[Number of stereoisomers possible for the complexes, $\left[\mathrm{CrCl}_3(\mathrm{py})_3\right]$ and $\left[\mathrm{CrCl}_2(\mathrm{ox})_2\right]^{3-}$ are respectively $(p y=$ pyridine,$o x=$ oxalate $)$
JEE Main 2025 (Online) 4th April Morning Shift](/past-years/jee/question/pnumber-of-stereoisomers-possible-for-the-complexes-lef-jee-main-chemistry-some-basic-concepts-of-chemistry-hqcgpcdxfbsmuyhi)[Identify the diamagnetic octahedral complex ions from below ;
A. $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$
B. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
C. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
D. $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{~F}_3\right]$
Choose the correct answer from the options given below:
JEE Main 2025 (Online) 3rd April Evening Shift](/past-years/jee/question/pidentify-the-diamagnetic-octahedral-complex-ions-from-bel-jee-main-chemistry-some-basic-concepts-of-chemistry-v99t5160gxh7ts9c)[The correct order of the complexes $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{3+}(\mathrm{A}),\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}(\mathrm{B}),\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}(\mathrm{C})$ and $\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right]^{2+}(\mathrm{D})$ in terms of wavelength of light absorbed is
JEE Main 2025 (Online) 3rd April Morning Shift](/past-years/jee/question/the-correct-order-of-the-complexes-leftmathrmcoleft-jee-main-chemistry-some-basic-concepts-of-chemistry-ml0hlzypmdlfcroc)[$$ \text { Match the LIST-I with LIST-II } $$
| LIST-I(Molecules/ion) | LIST-II(Hybridisation of central atom) |
--- |
| A. | $$\mathrm{PF}_5$$ | I | $$\mathrm{dsp}^2$$ |
| B | $$\mathrm{SF}_6$$ | II | $$\mathrm{sp}^3 \mathrm{~d}$$ |
| C | $$\mathrm{Ni}(\mathrm{CO})_4$$ | III | $$\mathrm{sp}^3 \mathrm{~d}^2$$ |
| D | $$\left[\mathrm{PtCl}_4\right]^{2-}$$ | IV | $$\mathrm{sp}^3$$ |
$$ \text { Choose the correct answer from the options given below: } $$
JEE Main 2025 (Online) 3rd April Morning Shift](/past-years/jee/question/-text--match-the-list-i-with-list-ii---table-idtab-jee-main-chemistry-some-basic-concepts-of-chemistry-zsa23bfggrg6ptor)[The type of hybridization and the magnetic property of $\left[\mathrm{MnCl}_6\right]^{3-}$ are,
JEE Main 2025 (Online) 2nd April Evening Shift](/past-years/jee/question/the-type-of-hybridization-and-the-magnetic-property-of-lef-jee-main-chemistry-some-basic-concepts-of-chemistry-aujowyfvhejocl6x)[The d-orbital electronic configuration of the complex among $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+},\left[\mathrm{CoF}_6\right]^{3-}$, $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ and $\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ that has the highest CFSE is :
JEE Main 2025 (Online) 2nd April Evening Shift](/past-years/jee/question/the-d-orbital-electronic-configuration-of-the-complex-among-jee-main-chemistry-some-basic-concepts-of-chemistry-icdbcirht3vnsfii)Given below are two statements :
Statement (I) : In octahedral complexes, when $\Delta_0<\mathrm{P}$ high spin complexes are formed. When $\Delta_0>P$ low spin complexes are formed.
Statement (II) : In tetrahedral complexes because of $\Delta_t < P$, low spin complexes are rarely formed.
In the light of the above statements, choose the most appropriate answer from the options given below :
JEE Main 2025 (Online) 2nd April Morning Shift[Identify the homoleptic complexes with odd number of $d$ electrons in the central metal :
(A) $\left[\mathrm{FeO}_4\right]^{2-}$
(B) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$
(C) $\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]^{2-}$
(D) $\left[\mathrm{CoCl}_4\right]^{2-}$
(E) $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{~F}_3\right]$
Choose the correct answer from the options given below :
JEE Main 2025 (Online) 29th January Evening Shift](/past-years/jee/question/pidentify-the-homoleptic-complexes-with-odd-number-of-d-jee-main-chemistry-v5colalqzpxsnqw8)[The calculated spin-only magnetic moments of $K_3[Fe(OH)_6]$ and $K_4[Fe(OH)_6]$ respectively are :
JEE Main 2025 (Online) 29th January Evening Shift](/past-years/jee/question/pthe-calculated-spin-only-magnetic-moments-of-k3feoh-jee-main-chemistry-svdpln7jidz1t5nz)[The correct increasing order of stability of the complexes based on $\Delta_0$ value is :
I. $[\text{Mn}(\text{CN})_6]^{3-}$
II. $[\text{Co}(\text{CN})_6]^{4-}$
III. $[\text{Fe}(\text{CN})_6]^{4-}$
IV. $[\text{Fe}(\text{CN})_6]^{3-}$
JEE Main 2025 (Online) 29th January Morning Shift](/past-years/jee/question/pthe-correct-increasing-order-of-stability-of-the-complexe-jee-main-chemistry-qpkydkklkvacewgq)[21
Match List - I with List - II.
| List - I (Complex) | List - II (Hybridisation & Magnetic characters) |
--- |
| (A) [MnBr4]2- | (I) d2sp3 & diamagnetic |
| (B) [FeF6]3- | (II) sp3d2 & paramagnetic |
| (C) [Co(C2O4)3]3- | (III) sp3 & diamagnetic |
| (D) [Ni(CO)4] | (IV) sp3 & paramagnetic |
Choose the correct answer from the options given below :
JEE Main 2025 (Online) 29th January Morning Shift](/past-years/jee/question/pmatch-list-i-with-list-iip-table-border1--jee-main-chemistry-iylt3iw023zguloj)[22
Match List - I with List - II.
| List - I (Complex) | List - II (Hybridisation of central metal ion) |
--- |
| (A) [CoF6]3- | (I) d2sp3 |
| (B) [NiCl4]2- | (II) sp3 |
| (C) [Co(NH3)6]3+ | (III) sp3d2 |
| (D) [Ni(CN)4]2- | (IV) dsp2 |
Choose the correct answer from the options given below:
JEE Main 2025 (Online) 28th January Evening Shift](/past-years/jee/question/pmatch-list-i-with-list-iibr-table-border1--jee-main-chemistry-n6raqbevdlswu4xx)23
The conditions and consequence that favours the $t_{2 \mathrm{~g}}{ }^3 \mathrm{e}_{\mathrm{g}}{ }^1$ configuration in a metal complex are :
JEE Main 2025 (Online) 24th January Evening Shift24
When Ethane-1,2-diamine is added progressively to an aqueous solution of Nickel (II) chloride, the sequence of colour change observed will be:
JEE Main 2025 (Online) 24th January Evening Shift25
One mole of the octahedral complex compound $\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3$ gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with excess of $\mathrm{AgNO}_3$ solution to yield two moles of $\mathrm{AgCl}_{(\mathrm{s})}$. The structure of the complex is:
JEE Main 2025 (Online) 24th January Morning Shift[26
Identify the coordination complexes in which the central metal ion has $\mathrm{d}^4$ configuration.
(A) $\left[\mathrm{FeO}_4\right]^{2-}$
(B) $\quad\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$
(C) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$
(D) (E) $\left[\mathrm{NiF}_6\right]^{2-}$
Choose the correct answer from the options given below :
JEE Main 2025 (Online) 23rd January Evening Shift](/past-years/jee/question/pidentify-the-coordination-complexes-in-which-the-central-jee-main-chemistry-some-basic-concepts-of-chemistry-rs7bnbqcn4z8jauq)27
$\mathrm{CrCl}_3 \cdot \mathrm{xNH}_3$ can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of $0.558^{\circ} \mathrm{C}$. Assuming $100 \%$ ionisation of this complex and coordination number of Cr is 6 , the complex will be (Given $\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
JEE Main 2025 (Online) 23rd January Morning Shift28
The d-electronic configuration of an octahedral Co (II) complex having magnetic moment of 3.95 BM is:
JEE Main 2025 (Online) 23rd January Morning Shift29
The complex that shows Facial - Meridional isomerism is :
JEE Main 2025 (Online) 23rd January Morning Shift30
The correct order of the following complexes in terms of their crystal field stabilization energies is :
JEE Main 2025 (Online) 22nd January Evening Shift[31
Identify the homoleptic complex(es) that is/are low spin.
(A) $\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]^{2-}$
(B) $\left[\mathrm{CoF}_6\right]^{3-}$
(C) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
(D) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
(E) $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
Choose the correct answer from the options given below :
JEE Main 2025 (Online) 22nd January Evening Shift](/past-years/jee/question/pidentify-the-homoleptic-complexes-that-isare-low-spin-jee-main-chemistry-some-basic-concepts-of-chemistry-muf9fzrobvtkvphe)[32
From the magnetic behaviour of $\left[\mathrm{NiCl}_4\right]^{2-}$ (paramagnetic) and $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ (diamagnetic), choose the correct geometry and oxidation state.
JEE Main 2025 (Online) 22nd January Morning Shift](/past-years/jee/question/pfrom-the-magnetic-behaviour-of-leftmathrmnicl4rig-jee-main-chemistry-some-basic-concepts-of-chemistry-j4wxrizntgrm8cps)33
In which of the following complexes the CFSE, $\Delta_o$ will be equal to zero?
JEE Main 2025 (Online) 22nd January Morning Shift[34
Match List I with List II
| LIST I | LIST II |
| A. | $$\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$$ | I. | $$sp^3$$ |
| B. | $$\left[\mathrm{Ni}(\mathrm{CO})_4\right]$$ | II. | $$sp^3d^2$$ |
| C. | $$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$$ | III. | $$dsp^2$$ |
| D. | $$\mathrm{Na}_3\left[\mathrm{CoF}_6\right]$$ | IV. | $$d^2sp^3$$ |
Choose the correct answer from the options given below:
JEE Main 2024 (Online) 9th April Evening Shift](/past-years/jee/question/pmatch-list-i-with-list-iip-pstyle-typetextcss-jee-main-chemistry-some-basic-concepts-of-chemistry-b0tmqertzxlku59m)35
The coordination environment of $$\mathrm{Ca}^{2+}$$ ion in its complex with $$\mathrm{EDTA}^{4-}$$ is :
JEE Main 2024 (Online) 9th April Evening Shift[36
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The total number of geometrical isomers shown by $$[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2]^{+}$$ complex ion is three.
Reason (R): $$[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2]^{+}$$ complex ion has an octahedral geometry.
In the light of the above statements, choose the most appropriate answer from the options given below :
JEE Main 2024 (Online) 9th April Morning Shift](/past-years/jee/question/pgiven-below-are-two-statements--one-is-labelled-as-asser-jee-main-chemistry-some-basic-concepts-of-chemistry-w9ooqmcmc2gzaggm)[37
Match List I with List II
| LIST I(Complex ion) | LIST II(Spin only magnetic moment in B.M.) |
| A. | $$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$$ | I. | 4.90 |
| B. | $$\left[\mathrm{NiCl}_4\right]^{2-}$$ | II. | 3.87 |
| C. | $$\left[\mathrm{CoF}_6\right]^{3-}$$ | III. | 0.0 |
| D. | $$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$$ | IV. | 2.83 |
Choose the correct answer from the options given below :
JEE Main 2024 (Online) 8th April Evening Shift](/past-years/jee/question/pmatch-list-i-with-list-iip-pstyle-typetextcss-jee-main-chemistry-some-basic-concepts-of-chemistry-gkcqwvg9vuiukcxi)[38
Number of Complexes with even number of electrons in $$\mathrm{t_{2 g}}$$ orbitals is -
$$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$$
JEE Main 2024 (Online) 8th April Morning Shift](/past-years/jee/question/pnumber-of-complexes-with-even-number-of-electrons-in-m-jee-main-chemistry-some-basic-concepts-of-chemistry-zikthgfzvzdwhuon)39
An octahedral complex with the formula $$\mathrm{CoCl}_3 \cdot \mathrm{nNH}_3$$ upon reaction with excess of $$\mathrm{AgNO}_3$$ solution gives 2 moles of $$\mathrm{AgCl}$$. Consider the oxidation state of $$\mathrm{Co}$$ in the complex is '$$x$$'. The value of "$$x+n$$" is __________.
JEE Main 2024 (Online) 8th April Morning Shift40
Given below are two statements:
Statement I: $$\mathrm{N}\left(\mathrm{CH}_3\right)_3$$ and $$\mathrm{P}\left(\mathrm{CH}_3\right)_3$$ can act as ligands to form transition metal complexes.
Statement II: As N and P are from same group, the nature of bonding of $$\mathrm{N}\left(\mathrm{CH}_3\right)_3$$ and $$\mathrm{P}\left(\mathrm{CH}_3\right)_3$$ is always same with transition metals.
In the light of the above statements, choose the most appropriate answer from the options given below:
JEE Main 2024 (Online) 8th April Morning Shift[41
Match List I with List II
| LIST I(Compound) | LIST II(Colour] |
| A. | $$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{xH_2O}$$ | I. | Violet |
| B. | $$\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-}$$ | II. | Blood Red |
| C. | $$[\mathrm{Fe}(\mathrm{SCN})]^{2+}$$ | III. | Prussian Blue |
| D. | $$\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4\cdot12 \mathrm{MoO}_3$$ | IV. | Yellow |
Choose the correct answer from the options given below:
JEE Main 2024 (Online) 8th April Morning Shift](/past-years/jee/question/pmatch-list-i-with-list-iip-pstyle-typetextcss-jee-main-chemistry-some-basic-concepts-of-chemistry-kjk3bxco8cnd6klx)[42
Given below are two statements :
Statement I : $$\mathrm{PF}_5$$ and $$\mathrm{BrF}_5$$ both exhibit $$\mathrm{sp}^3 \mathrm{~d}$$ hybridisation.
Statement II : Both $$\mathrm{SF}_6$$ and $$[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$$ exhibit $$\mathrm{sp}^3 \mathrm{~d}^2$$ hybridisation.
In the light of the above statements, choose the correct answer from the options given below :
JEE Main 2024 (Online) 6th April Evening Shift](/past-years/jee/question/pgiven-below-are-two-statements-p-pstatement-i-m-jee-main-chemistry-some-basic-concepts-of-chemistry-igiq5hmdhd7t1e04)[43
Match List I with List II.
| LIST ITetrahedral Complex | LIST IIElectronic configuration |
| A. | $$\mathrm{TiCl}_4$$ | I. | $$\mathrm{e}^2, \mathrm{t}_2^0$$ |
| B. | $$\left[\mathrm{FeO}_4\right]^{2-}$$ | II. | $$\mathrm{e^4, t_2^3}$$ |
| C. | $$\left[\mathrm{FeCl}_4\right]^{-}$$ | III. | $$\mathrm{e}^0, \mathrm{t}_2^0$$ |
| D. | $$\left[\mathrm{CoCl}_4\right]^{2-}$$ | IV. | $$\mathrm{e}^2, \mathrm{t}_2^3$$ |
Choose the correct answer from the options given below :
JEE Main 2024 (Online) 6th April Evening Shift](/past-years/jee/question/pmatch-list-i-with-list-iip-pstyle-typetextcss-jee-main-chemistry-some-basic-concepts-of-chemistry-lvwnpfbr8qk2a5vs)[44
The correct IUPAC name of $$[\mathrm{PtBr}_2(\mathrm{PMe}_3)_2]$$ is :
JEE Main 2024 (Online) 6th April Evening Shift](/past-years/jee/question/pthe-correct-iupac-name-of-mathrmptbr2mathrmpme-jee-main-chemistry-some-basic-concepts-of-chemistry-arpayevdsqrjuout)[45
Consider the following complexes
(A) $$\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right]^{2+}$$, (B) $$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$$, (C) $$ \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{3+} $$, (D) $$\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{2+}$$
The correct order of A, B, C and D in terms of wavenumber of light absorbed is :
JEE Main 2024 (Online) 6th April Morning Shift](/past-years/jee/question/pconsider-the-following-complexesp-pa-leftmath-jee-main-chemistry-some-basic-concepts-of-chemistry-tty33ypqg3rsghkh)46
Match List I with List II
| LIST I(Hybridization) | LIST II(Orientation in Shape) |
| A. | sp$$^3$$ | I. | Trigonal bipyramidal |
| B. | dsp$$^2$$ | II. | Octahedral |
| C. | sp$$^3$$d | III. | Tetrahedral |
| D. | sp$$^3$$d$$^2$$ | IV. | Square planar |
Choose the correct answer from the options given below:
JEE Main 2024 (Online) 6th April Morning Shift[47
The number of complexes from the following with no electrons in the $$t_2$$ orbital is ______.
$$\mathrm{TiCl}_4,\left[\mathrm{MnO}_4\right]^{-},\left[\mathrm{FeO}_4\right]^{2-},\left[\mathrm{FeCl}_4\right]^{-},\left[\mathrm{CoCl}_4\right]^{2-}$$
JEE Main 2024 (Online) 5th April Evening Shift](/past-years/jee/question/pthe-number-of-complexes-from-the-following-with-no-electr-jee-main-chemistry-some-basic-concepts-of-chemistry-gnpkdsckmgfimo6d)48
The metal atom present in the complex MABXL (where A, B, X and L are unidentate ligands and $$\mathrm{M}$$ is metal) involves $$\mathrm{sp}^3$$ hybridization. The number of geometrical isomers exhibited by the complex is :
JEE Main 2024 (Online) 5th April Evening Shift49
The correct order of ligands arranged in increasing field strength.
JEE Main 2024 (Online) 5th April Morning Shift[50
Which one of the following complexes will exhibit the least paramagnetic behaviour ? [Atomic number, $$\mathrm{Cr}=24, \mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27$$]
JEE Main 2024 (Online) 5th April Morning Shift](/past-years/jee/question/pwhich-one-of-the-following-complexes-will-exhibit-the-lea-jee-main-chemistry-some-basic-concepts-of-chemistry-02bxn6l0wlc1q1ms)[51
If an iron (III) complex with the formula $$\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_x(\mathrm{CN})_y\right]^-$$ has no electron in its $$e_g$$ orbital, then the value of $$x+y$$ is
JEE Main 2024 (Online) 4th April Evening Shift](/past-years/jee/question/pif-an-iron-iii-complex-with-the-formula-leftmathrm-jee-main-chemistry-some-basic-concepts-of-chemistry-mhnjikbk1ref5z4x)[52
The number of unpaired d-electrons in $$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$$ is ________.
JEE Main 2024 (Online) 4th April Evening Shift](/past-years/jee/question/pthe-number-of-unpaired-d-electrons-in-leftmathrmco-jee-main-chemistry-some-basic-concepts-of-chemistry-exyzdpxeznnvbdgf)53
A first row transition metal in its +2 oxidation state has a spin-only magnetic moment value of $$3.86 \mathrm{~BM}$$. The atomic number of the metal is
JEE Main 2024 (Online) 4th April Evening Shift[54
Number of complexes from the following with even number of unpaired "$$\mathrm{d}$$" electrons is ________ $$[\mathrm{V}(\mathrm{H}_2 \mathrm{O})_6]^{3+},[\mathrm{Cr}(\mathrm{H}_2 \mathrm{O})_6]^{2+},[\mathrm{Fe}(\mathrm{H}_2 \mathrm{O})_6]^{3+},[\mathrm{Ni}(\mathrm{H}_2 \mathrm{O})_6]^{3+},[\mathrm{Cu}(\mathrm{H}_2 \mathrm{O})_6]^{2+}$$ [Given atomic numbers: $$\mathrm{V}=23, \mathrm{Cr}=24, \mathrm{Fe}=26, \mathrm{Ni}=28 \mathrm{Cu}=29$$]
JEE Main 2024 (Online) 4th April Morning Shift](/past-years/jee/question/pnumber-of-complexes-from-the-following-with-even-number-o-jee-main-chemistry-some-basic-concepts-of-chemistry-yu3bmnbm52iy79uj)55
The correct sequence of ligands in the order of decreasing field strength is :
JEE Main 2024 (Online) 4th April Morning Shift[56
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{CoF}_6\right]^{3-}$ are respectively known as :
JEE Main 2024 (Online) 1st February Evening Shift](/past-years/jee/question/leftmathrmcoleftmathrmnh3right6right3-jee-main-chemistry-2vo8kpzghoyb10nw)57
Given below are two statements :Statement (I) : Dimethyl glyoxime forms a six-membered covalent chelate when treated with $\mathrm{NiCl}_2$ solution in presence of $\mathrm{NH}_4 \mathrm{OH}$. Statement (II) : Prussian blue precipitate contains iron both in $(+2)$ and $(+3)$ oxidation states. In the light of the above statements, choose the most appropriate answer from the options given below :
JEE Main 2024 (Online) 1st February Evening Shift58
Which of the following compounds show colour due to d-d transition?
JEE Main 2024 (Online) 1st February Evening Shift59
Which of the following complex is homoleptic?
JEE Main 2024 (Online) 1st February Morning Shift[60
Given below are two statements :Statement (I) : A solution of $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ is green in colour. Statement (II) : A solution of $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is colourless. In the light of the above statements, choose the most appropriate answer from the options given below :
JEE Main 2024 (Online) 1st February Morning Shift](/past-years/jee/question/given-below-are-two-statements-brbr-bstatement-i-jee-main-chemistry-aw2dehrmmh27no6a)61
Select the option with correct property -
JEE Main 2024 (Online) 31st January Evening Shift[62
Match List I with List II
| List - I(Complex ion) | List - II(Electronic Configuration) |
| (A) | $$\mathrm{[Cr(H_2O)_6]^{3+}}$$ | (I) | $$t_{2 g}{ }^2 e_g^0$$ |
| (B) | $$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$$ | (II) | $$t_{2 g}{ }^3 e_g{ }^0$$ |
| (C) | $$\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$$ | (III) | $$t_{2 g}{ }^3 e_g{ }^2$$ |
| (D) | $$\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$$ | (IV) | $$t_{2 g}{ }^6 e_g^2$$ |
Choose the correct answer from the options given below:
JEE Main 2024 (Online) 31st January Evening Shift](/past-years/jee/question/pmatch-list-i-with-list-iip-pstyle-typetextcss-jee-main-chemistry-some-basic-concepts-of-chemistry-ljq6dtzjsrulu7q8)[63
The correct statements from following are:
A. The strength of anionic ligands can be explained by crystal field theory.
B. Valence bond theory does not give a quantitative interpretation of kinetic stability of coordination compounds.
C. The hybridization involved in formation of $$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$$ complex is $$\mathrm{dsp}^2$$.
D. The number of possible isomer(s) of cis- $$\left[\mathrm{PtCl}_2(\mathrm{en})_2\right]^{2+}$$ is one
Choose the correct answer from the options given below:
JEE Main 2024 (Online) 31st January Morning Shift](/past-years/jee/question/pthe-correct-statements-from-following-arep-pa-the-jee-main-chemistry-some-basic-concepts-of-chemistry-bce22somu76pmn9t)64
The coordination geometry around the manganese in decacarbonyldimanganese $$(0)$$ is
JEE Main 2024 (Online) 30th January Evening Shift65
Choose the correct statements from the following :
(A) Ethane-1, 2-diamine is a chelating ligand.
(B) Metallic aluminium is produced by electrolysis of aluminium oxide in presence of cryolite.
(C) Cyanide ion is used as ligand for leaching of silver.
(D) Phosphine act as a ligand in Wilkinson catalyst.
(E) The stability constants of $$\mathrm{Ca}^{2+}$$ and $$\mathrm{Mg}^{2+}$$ are similar with EDTA complexes.
Choose the correct answer from the options given below :
JEE Main 2024 (Online) 30th January Morning Shift66
A reagent which gives brilliant red precipitate with Nickel ions in basic medium is
JEE Main 2024 (Online) 29th January Evening Shift67
Match List I with List II
| List - I(Substances) | List - II(Element Present) |
| (A) | Ziegler catalyst | (I) | Rhodium |
| (B) | Blood Pigment | (II) | Cobalt |
| (C) | Wilkinson catalyst | (III) | Iron |
| (D) | Vitamin $$\mathrm{B_{12}}$$ | (IV) | Titanium |
Choose the correct answer from the options given below:
JEE Main 2024 (Online) 29th January Morning Shift68
In which one of the following metal carbonyls, $$\mathrm{CO}$$ forms a bridge between metal atoms?
JEE Main 2024 (Online) 29th January Morning Shift69
Identity the incorrect pair from the following :
JEE Main 2024 (Online) 27th January Evening Shift70
Identify from the following species in which $$\mathrm{d}^2 \mathrm{sp}^3$$ hybridization is shown by central atom :
JEE Main 2024 (Online) 27th January Evening Shift[71
Consider the following complex ions
$$\begin{aligned} & \mathrm{P}=\left[\mathrm{FeF}_6\right]^{3-} \ & \mathrm{Q}=\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \ & \mathrm{R}=\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \end{aligned}$$
The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :
JEE Main 2024 (Online) 27th January Morning Shift](/past-years/jee/question/pconsider-the-following-complex-ionsp-pbeginalign-jee-main-chemistry-some-basic-concepts-of-chemistry-hjp3x05a6bgv3hxw)72
Yellow compound of lead chromate gets dissolved on treatment with hot $$\mathrm{NaOH}$$ solution. The product of lead formed is a :
JEE Main 2024 (Online) 27th January Morning Shift73
The complex with highest magnitude of crystal field splitting energy $\left(\Delta_{0}\right)$ is :
JEE Main 2023 (Online) 15th April Morning Shift[74
The total number of stereoisomers for the complex $$\left[\mathrm{Cr}(o x)_{2} \mathrm{ClBr}\right]^{3-}$$ (where $$o x=$$ oxalate) is :
JEE Main 2023 (Online) 13th April Evening Shift](/past-years/jee/question/pthe-total-number-of-stereoisomers-for-the-complex-left-jee-main-chemistry-some-basic-concepts-of-chemistry-l6urkcz7sd82u9ak)75
Which of the following complexes will exhibit maximum attraction to an applied magnetic field?
JEE Main 2023 (Online) 13th April Evening Shift[76
The mismatched combinations are
A. Chlorophyll - Co
B. Water hardness - EDTA
C. Photography $$-\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}$$
D. Wilkinson catalyst $$-\left[\left(\mathrm{Ph}_{3} \mathrm{P}\right)_{3} \mathrm{RhCl}\right]$$
E. Chelating ligand - D-Penicillamine
Choose the correct answer from the options given below :
JEE Main 2023 (Online) 13th April Morning Shift](/past-years/jee/question/pthe-mismatched-combinations-arep-pa-chlorophyll-c-jee-main-chemistry-some-basic-concepts-of-chemistry-lyvlru9anmcujjd7)[77
Match List I with List II
| LIST IComplex | LIST IICFSE ($$\Delta_0$$) |
| A. | $$\mathrm{[Cu(NH_3)_6]^{2+}}$$ | I. | $$-0.6$$ |
| B. | $$\mathrm{[Ti(H_2O)_6]^{3+}}$$ | II. | $$-2.0$$ |
| C. | $$\mathrm{[Fe(CN)_6]^{3-}}$$ | III. | $$-1.2$$ |
| D. | $$\mathrm{[NiF_6]^{4-}}$$ | IV. | $$-0.4$$ |
Choose the correct answer from the options given below:
JEE Main 2023 (Online) 12th April Morning Shift](/past-years/jee/question/pmatch-list-i-with-list-iip-pstyle-typetextcss-jee-main-chemistry-some-basic-concepts-of-chemistry-sqf7w91ppq0n2dfd)[78
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : $$\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$$ absorbs at lower wavelength of light with respect to $$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}$$
Reason R : It is because the wavelength of the light absorbed depends on the oxidation state of the metal ion.
In the light of the above statements, choose the correct answer from the options given below:
JEE Main 2023 (Online) 11th April Evening Shift](/past-years/jee/question/pgiven-below-are-two-statements-one-is-labelled-as-assert-jee-main-chemistry-some-basic-concepts-of-chemistry-7bdu3yrszp0kzkf2)[79
If $$\mathrm{Ni}^{2+}$$ is replaced by $$\mathrm{Pt}^{2+}$$ in the complex $$\left[\mathrm{NiCl}_{2} \mathrm{Br}_{2}\right]^{2-}$$, which of the following properties are expected to get changed ?
A. Geometry
B. Geometrical isomerism
C. Optical isomerism
D. Magnetic properties
JEE Main 2023 (Online) 11th April Evening Shift](/past-years/jee/question/pif-mathrmni2-is-replaced-by-mathrmpt2-jee-main-chemistry-some-basic-concepts-of-chemistry-k0esyyqgjmdjvfht)[80
Match List I with List II
| LIST IComplex | LIST IIColour |
| A. | $$Mg(N{H_4})P{O_4}$$ | I. | brown |
| B. | $${K_3}[Co{(N{O_2})_6}]$$ | II. | white |
| C. | $$MnO{(OH)_2}$$ | III. | yellow |
| D. | $$F{e_4}{[Fe{(CN)_6}]_3}$$ | IV. | blue |
Choose the correct answer from the options given below :
JEE Main 2023 (Online) 11th April Evening Shift](/past-years/jee/question/pmatch-list-i-with-list-iip-pstyle-typetextcss-jee-main-chemistry-some-basic-concepts-of-chemistry-heki8mxzglayw2ce)[81
The magnetic moment is measured in Bohr Magneton (BM).
Spin only magnetic moment of $$\mathrm{Fe}$$ in $$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$$ and $$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$$ complexes respectively is :
JEE Main 2023 (Online) 11th April Evening Shift](/past-years/jee/question/pthe-magnetic-moment-is-measured-in-bohr-magneton-bmp-jee-main-chemistry-some-basic-concepts-of-chemistry-qnmxotg1zzg71qlc)82
The complex that dissolves in water is :
JEE Main 2023 (Online) 11th April Morning Shift83
The set which does not have ambidentate ligand(s) is :
JEE Main 2023 (Online) 11th April Morning Shift84
Which of the following complex has a possibility to exist as meridional isomer?
JEE Main 2023 (Online) 11th April Morning Shift[85
The correct order of the number of unpaired electrons in the given complexes is
A. $$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$$
B. $$\left[\mathrm{Fe} \mathrm{F}_{6}\right]^{3-}$$
C. $$\left[\mathrm{CoF}_{6}\right]^{3-}$$
D. $$\left.[\mathrm{Cr} \text { (oxalate})_{3}\right]^{3-}$$
E. $$\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$$
Choose the correct answer from the options given below:
JEE Main 2023 (Online) 10th April Evening Shift](/past-years/jee/question/pthe-correct-order-of-the-number-of-unpaired-electrons-in-jee-main-chemistry-some-basic-concepts-of-chemistry-1mshh6ddxwqkgxjs)[86
Match List I with List II
| List - IComplex | List - IICrystal Field splitting energy ($$\Delta_0$$) |
| A. | $${[Ti{({H_2}O)_6}]^{2 + }}$$ | I. | $$-1.2$$ |
| B. | $${[V{({H_2}O)_6}]^{2 + }}$$ | II. | $$-0.6$$ |
| C. | $${[Mn{({H_2}O)_6}]^{3 + }}$$ | III. | 0 |
| D. | $${[Fe{({H_2}O)_6}]^{3 + }}$$ | IV. | $$-0.8$$ |
Choose the correct answer from the options given below:
JEE Main 2023 (Online) 10th April Evening Shift](/past-years/jee/question/pmatch-list-i-with-list-iip-pstyle-typetextcss-jee-main-chemistry-some-basic-concepts-of-chemistry-7gr0txqcoszlc3yh)87
The octahedral diamagnetic low spin complex among the following is :
JEE Main 2023 (Online) 10th April Morning Shift[88
Match List I with List II
| LIST ICoordination Complex | LIST IINumber of unpaired electrons |
| A. | $$\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}$$ | I. | 0 |
| B. | $$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$$ | II. | 3 |
| C. | $$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$$ | III. | 2 |
| D. | $$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$$ | IV. | 4 |
Choose the correct answer from the options given below:
JEE Main 2023 (Online) 8th April Evening Shift](/past-years/jee/question/pmatch-list-i-with-list-iip-pstyle-typetextcss-jee-main-chemistry-some-basic-concepts-of-chemistry-rnevlxzfgoofz5dz)89
Which of the following complex is octahedral, diamagnetic and the most stable?
JEE Main 2023 (Online) 8th April Morning Shift90
The correct order of spin only magnetic moments for the following complex ions is
JEE Main 2023 (Online) 8th April Morning Shift[91
The IUPAC name of $$\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]$$ is:-
JEE Main 2023 (Online) 6th April Evening Shift](/past-years/jee/question/pthe-iupac-name-of-mathrmk3leftmathrmcoleft-jee-main-chemistry-some-basic-concepts-of-chemistry-fa94ztcojlmmjh5o)[92
Given below are two statements, one is labelled as Assertion $$\mathbf{A}$$ and the other is labelled as Reason $$\mathbf{R}$$.
Assertion A: The spin only magnetic moment value for $$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$$ is $$1.74 \mathrm{BM}$$, whereas for $$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$$ is $$5.92 \mathrm{BM}$$.
Reason $$\mathbf{R}$$ : In both complexes, $$\mathrm{Fe}$$ is present in +3 oxidation state.
In the light of the above statements, choose the correct answer from the options given below:
JEE Main 2023 (Online) 6th April Morning Shift](/past-years/jee/question/pgiven-below-are-two-statements-one-is-labelled-as-assert-jee-main-chemistry-some-basic-concepts-of-chemistry-n6ethcwof6bymxxk)93
The complex cation which has two isomers is :
JEE Main 2023 (Online) 1st February Evening Shift94
Which of the following complex will show largest splitting of d-orbitals?
JEE Main 2023 (Online) 1st February Morning Shift95
Which of the following are the example of double salt?
A. $$\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$$
B. $$\mathrm{CuSO}_{4}\cdot 4 \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$$
C. $$\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O}$$
D. $$\mathrm{Fe}(\mathrm{CN})_{2}\cdot4 \mathrm{KCN}$$
Choose the correct answer :
JEE Main 2023 (Online) 1st February Morning Shift[96
A solution of $$\mathrm{FeCl_3}$$ when treated with $$\mathrm{K_4[Fe(CN)_6]}$$ gives a prussium blue precipitate due to the formation of :
JEE Main 2023 (Online) 1st February Morning Shift](/past-years/jee/question/pa-solution-of-mathrmfecl3-when-treated-with-ma-jee-main-chemistry-some-basic-concepts-of-chemistry-p3ij9c6iwpq00u27)97
Cobalt chloride when dissolved in water forms pink colored complex $$\underline{\mathrm{X}}$$ which has octahedral geometry. This solution on treating with conc $$\mathrm{HCl}$$ forms deep blue complex, $$\underline{\mathrm{Y}}$$ which has a $$\underline{\mathrm{Z}}$$ geometry. $$\mathrm{X}, \mathrm{Y}$$ and $$\mathrm{Z}$$, respectively, are
JEE Main 2023 (Online) 31st January Morning Shift[98
Match List I with List II:
| List I (Complexes) | List II (Hybridisation) |
--- |
| A. | $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ | I. | $\mathrm{sp}^{3}$ |
| B. | $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ | II. | dsp$^{2}$ |
| C. | $\left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ | III. | $\mathrm{sp}^{3}\mathrm{d}^{2}$ |
| D. | $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ | IV. | $\mathrm{d}^{2} \mathrm{sp}^{3}$ |
JEE Main 2023 (Online) 30th January Evening Shift](/past-years/jee/question/pmatch-list-bib-with-list-biibp-style-type-jee-main-chemistry-pymrgrnywjbuyoqd)[99
The $\mathrm{Cl}-\mathrm{Co}-\mathrm{Cl}$ bond angle values in a fac- $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]$ complex is/are :
JEE Main 2023 (Online) 30th January Evening Shift](/past-years/jee/question/the-mathrmcl-mathrmco-mathrmcl-bond-angle-values-jee-main-chemistry-fo24myhnarcpg5zm)100
To inhibit the growth of tumours, identify the compounds used from the following :
A. EDTA
B. Coordination Compounds of Pt
C. D - Penicillamine
D. Cis - Platin
Choose the correct answer from the option given below :
JEE Main 2023 (Online) 30th January Morning Shift101
Which of the following is correct order of ligand field strength?
JEE Main 2023 (Online) 30th January Morning Shift102
Correct order of spin only magnetic moment of the following complex ions is :
(Given At.no. Fe : 26, Co : 27)
JEE Main 2023 (Online) 29th January Evening Shift103
Chiral complex from the following is :
Here en = ethylene diamine
JEE Main 2023 (Online) 29th January Morning Shift[104
Match List I with List II
| List ICoordination entity | List IIWavelength of light absorbed in nm |
| A. | $$\mathrm{[CoCl(NH_3)_5]^{2+}}$$ | I. | 310 |
| B. | $$\mathrm{[Co(NH_3)_6]^{3+}}$$ | II. | 475 |
| C. | $$\mathrm{[Co(CN)_6]^{3-}}$$ | III. | 535 |
| D. | $$\mathrm{[Cu(H_2O)_4]^{2+}}$$ | IV. | 600 |
Choose the correct answer from the options given below :
JEE Main 2023 (Online) 25th January Evening Shift](/past-years/jee/question/pmatch-list-i-with-list-iip-pstyle-typetextcss-jee-main-chemistry-some-basic-concepts-of-chemistry-zoinbxkc7ryiltcb)[105
The hybridization and magnetic behaviour of cobalt ion in $$\mathrm{[Co(NH_3)_6]^{3+}}$$ complex, respectively is :
JEE Main 2023 (Online) 24th January Evening Shift](/past-years/jee/question/pthe-hybridization-and-magnetic-behaviour-of-cobalt-ion-in-jee-main-chemistry-some-basic-concepts-of-chemistry-cthvdzyxvabbeoql)106
Which of the following cannot be explained by crystal field theory?
JEE Main 2023 (Online) 24th January Evening Shift[107
The primary and secondary valencies of cobalt respectively in $$\mathrm{[Co(NH_3)_5Cl]Cl_2}$$ are :
JEE Main 2023 (Online) 24th January Morning Shift](/past-years/jee/question/pthe-primary-and-secondary-valencies-of-cobalt-respectivel-jee-main-chemistry-some-basic-concepts-of-chemistry-fi9hedm7fj2otxqn)108
Octahedral complexes of copper(II) undergo structural distortion (Jahn-Teller). Which one of the given copper (II) complexes will show the maximum structural distortion? (en - ethylenediamine; $$\mathrm{H}_{2} \mathrm{~N}_{-} \mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$$)
JEE Main 2022 (Online) 29th July Evening Shift[109
Match List I with List II
| List - I (Complex) | List - II (Hybridization) |
| (A) | $$Ni{(CO)_4}$$ | (I) | $$s{p^3}$$ |
| (B) | $${[Ni{(CN)_4}]^{2 - }}$$ | (II) | $$s{p^3}{d^2}$$ |
| (C) | $${[Co{(CN)_6}]^{3 - }}$$ | (III) | $${d^2}s{p^3}$$ |
| (D) | $${[Co{F_6}]^{3 - }}$$ | (IV) | $$ds{p^2}$$ |
Choose the correct answer from the options given below :
JEE Main 2022 (Online) 28th July Evening Shift](/past-years/jee/question/pmatch-list-i-with-list-iip-pstyle-typetextcss-jee-main-chemistry-some-basic-concepts-of-chemistry-nbykaxgim9unoomj)110
Low oxidation state of metals in their complexes are common when ligands :
JEE Main 2022 (Online) 27th July Evening Shift111
$$\mathrm{Fe}^{3+}$$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of :
JEE Main 2022 (Online) 27th July Evening Shift112
The metal complex that is diamagnetic is (Atomic number: $$\mathrm{Fe}, 26 ; \mathrm{Cu}, 29)$$
JEE Main 2022 (Online) 26th July Evening Shift[113
The correct order of energy of absorption for the following metal complexes is :
A : [Ni(en)3]2+ , B : [Ni(NH3)6]2+ , C : [Ni(H2O)6]2+
JEE Main 2022 (Online) 25th July Evening Shift](/past-years/jee/question/pthe-correct-order-of-energy-of-absorption-for-the-followi-jee-main-chemistry-some-basic-concepts-of-chemistry-0q9frd2ylbeevfgm)114
Correct formula of the compound which gives a white precipitate with BaCl2 solution, but not with AgNO3 solution, is :
JEE Main 2022 (Online) 30th June Morning Shift115
Given below are two statements.
$$\bullet$$ Statement I : In CuSO4 . 5H2O, Cu-O bonds are present.
$$\bullet$$ Statement II : In CuSO4 . 5H2O, ligands coordinating with Cu(II) ion are O-and S-based ligands.
In the light of the above statements, choose the correct answer from the options given below.
JEE Main 2022 (Online) 29th June Evening Shift[116
Given below are two statements :
Statement I : [Ni(CN)4]2$$-$$ is square planar and diamagnetic complex, with dsp2 hybridization for Ni but [Ni(CO)4] is tetrahedral, paramagnetic and with sp3-hybridication for Ni.
Statement II : [NiCl4]2$$-$$ and [Ni(CO)4] both have same d-electron configuration have same geometry and are paramagnetic.
In light the above statements, choose the correct answer from the options given below :
JEE Main 2022 (Online) 28th June Morning Shift](/past-years/jee/question/pgiven-below-are-two-statements-p-pstatement-i--ni-jee-main-chemistry-some-basic-concepts-of-chemistry-ro8srndulv0acywl)[117
Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers : Mn = 25; Fe = 26)
A. [FeF6]3$$-$$
B. [Fe(CN)6]3$$-$$
C. [MnCl6]3$$-$$ (high spin)
D. [Mn(CN)6]3$$-$$
Choose the correct answer from the options given below :
JEE Main 2022 (Online) 27th June Evening Shift](/past-years/jee/question/parrange-the-following-coordination-compounds-in-the-incre-jee-main-chemistry-some-basic-concepts-of-chemistry-o1ezvtkohd8ap2w2)118
Which of the following will have maximum stabilization due to crystal field?
JEE Main 2022 (Online) 27th June Morning Shift119
Which statement is not true with respect to nitrate ion test?
JEE Main 2022 (Online) 26th June Morning Shift120
Transition metal complex with highest value of crystal field splitting ($$\Delta$$0) will be :
JEE Main 2022 (Online) 24th June Evening Shift[121
Match List - I with List - II :
| List - I | List -II |
| (A) | $${[PtC{l_4}]^{2 - }}$$ | (I) | $$s{p^3}d$$ |
| (B) | $$Br{F_5}$$ | (II) | $${d^2}s{p^3}$$ |
| (C) | $$PC{l_5}$$ | (III) | $$ds{p^2}$$ |
| (D) | $${[Co{(N{H_3})_6}]^{3 + }}$$ | (IV) | $$s{p^3}{d^2}$$ |
Choose the most appropriate answer from the options given below :
JEE Main 2022 (Online) 24th June Morning Shift](/past-years/jee/question/pmatch-list-i-with-list-ii-p-pstyle-typetext-jee-main-chemistry-some-basic-concepts-of-chemistry-tklfvcctfs2lnxtk)122
The Crystal Field Stabilization Energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion (Mz+) are $$-$$0.8 $$\Delta$$0 and 3.87 BM, respectively. Identify (Mz+) :
JEE Main 2021 (Online) 1st September Evening Shift123
The potassium ferrocyanide solution gives a Prussian blue colour, when added to :
JEE Main 2021 (Online) 1st September Evening Shift[124
Spin only magnetic moment in BM of [Fe(CO)4(C2O4)]+ is :
JEE Main 2021 (Online) 31st August Evening Shift](/past-years/jee/question/spin-only-magnetic-moment-in-bm-of-fecosub4subcsub-jee-main-chemistry-some-basic-concepts-of-chemistry-ojmckq1tgrlpbwgh)125
The denticity of an organic ligand, biuret is :
JEE Main 2021 (Online) 31st August Morning Shift126
Indicate the complex/complex ion which did not show any geometrical isomerism :
JEE Main 2021 (Online) 26th August Evening Shift[127
Arrange the following Cobalt complexes in the order of increasing Crystal Field Stabilization Energy (CFSE) value.Complexes : $$\mathop {{{[Co{F_6}]}^{3 - }}}\limits_A ,\mathop {{{[Co{{({H_2}O)}_6}]}^{2 + }}}\limits_B ,\mathop {{{[Co{{(N{H_3})}_6}]}^{3 + }}}\limits_C and \mathop {{{[Co{{({en})}_3}]}^{3 + }}}\limits_D $$Choose the correct option :
JEE Main 2021 (Online) 26th August Evening Shift](/past-years/jee/question/arrange-the-following-cobalt-complexes-in-the-order-of-incre-jee-main-chemistry-some-basic-concepts-of-chemistry-ycqfwjnc5oxansfz)[128
Given below are two statements :Statement I : $${[Mn{(CN)_6}]^{3 - }}$$, $${[Fe{(CN)_6}]^{3 - }}$$ and $${[Co{({C_2}{O_4})_3}]^{3 - }}$$ are d2sp3 hybridised.Statement II : $${[MnCl)_6}{]^{3 - }}$$ and $${[Fe{F_6}]^{3 - }}$$ are paramagnetic and have 4 and 5 unpaired electrons, respectively.In the light of the above statements, choose the correct answer from the options given below :
JEE Main 2021 (Online) 27th July Evening Shift](/past-years/jee/question/given-below-are-two-statements-brbrstatement-i---jee-main-chemistry-some-basic-concepts-of-chemistry-m9mg39zstxr1ex0n)[129
The type of hybridisation and magnetic property of the complex [MnCl6]3$$-$$, respectively, are :
JEE Main 2021 (Online) 27th July Morning Shift](/past-years/jee/question/the-type-of-hybridisation-and-magnetic-property-of-the-compl-jee-main-chemistry-some-basic-concepts-of-chemistry-absaslf3qfrbjlvb)[130
The number of geometrical isomers found in the metal complexes [PtCl2(NH3)2], [Ni(CO)4], [Ru(H2O)3Cl3 and [CoCl2(NH3)4]+ respectively, are :
JEE Main 2021 (Online) 27th July Morning Shift](/past-years/jee/question/the-number-of-geometrical-isomers-found-in-the-metal-complex-jee-main-chemistry-some-basic-concepts-of-chemistry-cmzbbszeakf8suo2)131
Which one of the following metal complexes is most stable?
JEE Main 2021 (Online) 25th July Evening Shift132
Which one of the following species responds to an external magnetic field?
JEE Main 2021 (Online) 25th July Morning Shift133
Spin only magnetic moment of an octahedral complex of Fe2+ in the presence of a strong field ligand in BM is :
JEE Main 2021 (Online) 20th July Evening Shift134
Which one of the following species doesn't have a magnetic moment of 1.73 BM, (spin only value)?
JEE Main 2021 (Online) 20th July Evening Shift135
According to the valence bond theory the hybridization of central metal atom is dsp2 for which one of the following compounds?
JEE Main 2021 (Online) 20th July Morning Shift136
The correct order of intensity of colors of the compounds is :
JEE Main 2021 (Online) 20th July Morning Shift137
The secondary valency and the number of hydrogen bonded water molecule(s) in CuSO4 . 5H2O, respectively, are :
JEE Main 2021 (Online) 18th March Evening Shift[138
The correct structures of trans-[NiBr2(PPh3)2] and meridonial-[Co(NH3)3(NO2)3], respectively, are
JEE Main 2021 (Online) 18th March Morning Shift](/past-years/jee/question/the-correct-structures-of-trans-nibrsub2subpphsub3-jee-main-chemistry-some-basic-concepts-of-chemistry-579g9mqjlnzv6uex)[139
Match List - I with List - II :
| List - I | List - II |
--- |
| (a) | $$[Co{(N{H_3})_6}][Cr{(CN)_6}]$$ | (i) | Linkage isomerism |
| (b) | $$[Co{(N{H_3})_3}{(N{O_2})_3}]$$ | (ii) | Solvate isomerism |
| (c) | $$[Cr{({H_2}O)_6}C{l_3}$$ | (iii) | Co-ordination isomerism |
| (d) | $$cis - {[CrC{l_2}{(ox)_2}]^{3 - }}$$ | (iv) | Optical isomerism |
Choose the correct answer from the options given below :
JEE Main 2021 (Online) 17th March Evening Shift](/past-years/jee/question/match-list-i-with-list-ii-brbrtable-thead--jee-main-chemistry-some-basic-concepts-of-chemistry-3pioqc9ftyy64e3l)[140
In which of the following order the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment?(i) [FeF6]3$$-$$(ii) [Co(NH3)6]3+(iii) [NiCl4]2$$-$$(iv) [Cu(NH3)4]2+
JEE Main 2021 (Online) 25th February Evening Shift](/past-years/jee/question/in-which-of-the-following-order-the-given-complex-ions-are-a-jee-main-chemistry-some-basic-concepts-of-chemistry-ytute0mnii8eucow)[141
The hybridization and magnetic nature of $${[Mn{(CN)_6}]^{4 - }}$$ and $${[Fe{(CN)_6}]^{3 - }}$$, respectively are :
JEE Main 2021 (Online) 25th February Morning Shift](/past-years/jee/question/the-hybridization-and-magnetic-nature-of-mncn64-jee-main-chemistry-some-basic-concepts-of-chemistry-mziv2lytd3uidpk9)[142
The calculated magnetic moments (spin only value) for species $${[FeC{l_4}]^{2 - }}$$, $${[Co{({C_2}{O_4})_3}]^{3 - }}$$ and $$MnO_4^{2 - }$$ respectively are :
JEE Main 2021 (Online) 24th February Evening Shift](/past-years/jee/question/the-calculated-magnetic-moments-spin-only-value-for-specie-jee-main-chemistry-some-basic-concepts-of-chemistry-rjrxtp9wdatcrds0)143
For a d4 metal ion in an octahedral field, the correct electronic configuration is :
JEE Main 2020 (Online) 6th September Evening Slot144
The species that has a spin-only magnetic moment of 5.9 BM, is : (Td = tetrahedral)
JEE Main 2020 (Online) 6th September Morning Slot[145
Consider the complex ions, trans-[Co(en)2Cl2]+ (A) and cis-[Co(en)2Cl2]+ (B). The correct statement regarding them is :
JEE Main 2020 (Online) 5th September Evening Slot](/past-years/jee/question/consider-the-complex-ions-brtrans-coensub2subcl-jee-main-chemistry-coordination-compounds-m4bkfyrnp3ijfzii)146
The values of the crystal field stabilization energies for a high spin d6 metal ion in octahedral and tetrahedral fields, respectively, are :
JEE Main 2020 (Online) 5th September Morning Slot[147
The Crystal Field Stabilization Energy (CFSE) of [CoF3(H2O)3] ($$\Delta $$0 < P) is :
JEE Main 2020 (Online) 4th September Evening Slot](/past-years/jee/question/the-crystal-field-stabilization-energy-brcfse-of-cofs-jee-main-chemistry-coordination-compounds-2bpqqkawhzwjgcbf)148
The one that can exhibit highest paramagnetic behaviour among the following is : gly = glycinato; bpy = 2, 2'-bipyridine
JEE Main 2020 (Online) 4th September Evening Slot149
The molecule in which hybrid MOs involve only one d-orbital of the central atom is :
JEE Main 2020 (Online) 4th September Evening Slot[150
The number of isomers possible for [Pt(en)(NO2)2] is :
JEE Main 2020 (Online) 4th September Morning Slot](/past-years/jee/question/the-number-of-isomers-possible-for-brptennosub2su-jee-main-chemistry-coordination-compounds-44tjnlzzfe5gphbc)151
The pair in which both the species have the same magnetic moment (spin only) is :
JEE Main 2020 (Online) 4th September Morning Slot[152
The d-electron configuration of [Ru(en)3 ]Cl2 and [Fe(H2O)6]Cl2 , respectively are :
JEE Main 2020 (Online) 3rd September Evening Slot](/past-years/jee/question/the-d-electron-configuration-of-ruensub3sub-clsub-jee-main-chemistry-coordination-compounds-wzcm7u6wqq61zjm8)[153
Complex A has a composition of H12O6Cl3Cr. If the complex on treatment with conc.H2SO4 loses 13.5% of its original mass, the correct molecular formula of A is : [Given: atomic mass of Cr = 52 amu and Cl = 35 amu]
JEE Main 2020 (Online) 3rd September Evening Slot](/past-years/jee/question/complex-a-has-a-composition-of-hsub12subosub6subcl-jee-main-chemistry-coordination-compounds-8drbali4pdfjosmu)154
The complex that can show optical activity is :
JEE Main 2020 (Online) 3rd September Morning Slot[155
The electronic spectrum of [Ti(H2O)6]3+ shows a single broad peak with a maximum at 20,300 cm-1 . The crystal field stabilization energy (CFSE) of the complex ion, in kJ mol-1, is :
JEE Main 2020 (Online) 3rd September Morning Slot](/past-years/jee/question/the-electronic-spectrum-of-tihsub2subosub6sub-jee-main-chemistry-coordination-compounds-fgiw3hxbyo1inpcr)156
The one that is not expected to show isomerism is :
JEE Main 2020 (Online) 2nd September Evening Slot[157
Simplified absorption spectra of three complexes ((i), (ii) and (iii)) of Mn+ ion are provided below; their $$\lambda $$max values are marked as A, B and C respectively. The correct match between the complexes and their $$\lambda $$max values is (i) M(NCS)6 (ii) MF6 (iii) [M(NH3)6]n+
JEE Main 2020 (Online) 2nd September Evening Slot](/past-years/jee/question/simplified-absorption-spectra-of-three-complexes-i-ii-jee-main-chemistry-coordination-compounds-8de5wvva35vvwlcq)158
For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following statements: (I) both the complexes can be high spin. (II) Ni(II) complex can very rarely be low spin. (III) with strong field ligands, Mn(II) complexes can be low spin. (IV)aqueous solution of Mn(II) ions is yellow in colour. The correct statements are :
JEE Main 2020 (Online) 2nd September Morning Slot159
Consider that a d6 metal ion (M2+) forms a complex with aqua ligands, and the spin only magnetic moment of the complex is 4.90 BM. The geometry and the crystal field stabilization energy of the complex is
JEE Main 2020 (Online) 2nd September Morning Slot[160
The correct order of the spin-only magnetic moments of the following complexes is : (I) [Cr(H2O)6]Br2 (II) Na4[Fe(CN)6] (III) Na3[Fe(C2O4)3] ($$\Delta $$0 $$>$$ P) (IV) (Et4N)2[CoCl4]
JEE Main 2020 (Online) 9th January Evening Slot](/past-years/jee/question/the-correct-order-of-the-spin-only-magnetic-moments-of-the-f-jee-main-chemistry-coordination-compounds-48jwt4qxt3odivss)[161
The isomer(s) of [Co(NH3)4Cl2] that has/have a Cl–Co–Cl angle of 90°, is/are :
JEE Main 2020 (Online) 9th January Evening Slot](/past-years/jee/question/the-isomers-of-conhsub3subsub4subclsub2su-jee-main-chemistry-hydrocarbons-husfiiaokj588zly)[162
[Pd(F)(Cl)(Br)(I)]2– has n number of geometrical isomers. Then, the spin-only magnetic moment and crystal field stabilisation energy [CFSE] of [Fe(CN)6]n–6, respectively, are: [Note : Ignore the pairing energy]
JEE Main 2020 (Online) 9th January Morning Slot](/past-years/jee/question/pdfclbrisup2sup-has-n-number-of-geometrical-jee-main-chemistry-analytical-chemistry-zzgq5zlbvjduqwll)163
Complex X of composition Cr(H2O)6Cln has a spin only magnetic moment of 3.83 BM. It reacts with AgNO3 and shows geometrical isomerism. The IUPAC nomenclature of X is :
JEE Main 2020 (Online) 9th January Morning Slot[164
Among (a) – (d) the complexes that can display geometrical isomerism are :(a) [Pt(NH3)3Cl]+ (b) [Pt(NH3)Cl5]– (c) [Pt(NH3)2Cl(NO2)] (d) [Pt(NH3)4ClBr]2+
JEE Main 2020 (Online) 8th January Evening Slot](/past-years/jee/question/among-a--d-the-complexes-that-can-display-geometrical-i-jee-main-chemistry-analytical-chemistry-zvlx3osakwcgqqg9)[165
The correct order of the calculated spin-only magnetic moments of complexs (A) to (D) is:(A) Ni(CO)4 (B) [Ni(H2O)6]Cl2 (C) Na2[Ni(CN)4] (D) PdCl2(PPh3)2
JEE Main 2020 (Online) 8th January Evening Slot](/past-years/jee/question/the-correct-order-of-the-calculated-spin-only-magnetic-momen-jee-main-chemistry-analytical-chemistry-uydpeepilglftddl)166
The complex that can show fac- and mer-isomers is :
JEE Main 2020 (Online) 8th January Morning Slot167
The number of possible optical isomers for the complexes MA2B2 with sp3 and dsp2 hydridized metal atom. respectively, is : Note : A and B are unidentate netural and unidentate monoanionic ligands, respectively.
JEE Main 2020 (Online) 7th January Evening Slot[168
Among the statements(a)-(d) the incorrect ones are :(a) Octahedral CO(III) complexes with strong fields ligands have very high magnetic moments.(b) When $$\Delta $$0 < P, the d-electron configuration of Co(III) in an octahedral complex is $$t_{eg}^4e_g^2$$ (c) Wavelength of light absorbed by [Co(en)3]3+ is lower than that of [CoF6]3- (d) If the $$\Delta $$0 for an octahedral complex of CO(III) is 18,000 cm-1, the $$\Delta $$t for its tetrahedral complex with the same ligand be 16,000 cm-1
JEE Main 2020 (Online) 7th January Evening Slot](/past-years/jee/question/among-the-statementsa-d-the-incorrect-ones-are-br-a-jee-main-chemistry-analytical-chemistry-j2ec6smekkxlbv7g)[169
The theory that can completely/properly explain the nature of bonding in [Ni(Co)4] is :
JEE Main 2020 (Online) 7th January Morning Slot](/past-years/jee/question/the-theory-that-can-completelyproperly-explain-the-nature-o-jee-main-chemistry-analytical-chemistry-trqsxvkjzwe3yxc0)[170
The IUPAC name of complex [Pt(NH3)2Cl(NH2CH3)]Cl is:
JEE Main 2020 (Online) 7th January Morning Slot](/past-years/jee/question/the-iupac-name-of-complex-ptnhsub3subsub2subcl-jee-main-chemistry-analytical-chemistry-l9n1mnzpiztrrl4d)[171
The coordination numbers of Co and Al in [Co(Cl)(en)2]Cl and K3[Al(C2O4)3], respectively, are : (en = ethane-1, 2-diamine)
JEE Main 2019 (Online) 12th April Evening Slot](/past-years/jee/question/the-coordination-numbers-of-co-and-al-in-coclensub2-jee-main-chemistry-coordination-compounds-ooqm56bh6wvhauof)172
The compound used in the treatment of lead poisoning is :
JEE Main 2019 (Online) 12th April Evening Slot173
The complex ion that will lose its crystal field stabilization energy upon oxidation of its metal to +3 state is :
JEE Main 2019 (Online) 12th April Morning Slot174
Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale).
JEE Main 2019 (Online) 12th April Morning Slot[175
The crystal field stabilization energy (CFSE) of [Fe(H2O)6]Cl2 and K2[NiCl4] respectively, are :
JEE Main 2019 (Online) 10th April Evening Slot](/past-years/jee/question/the-crystal-field-stabilization-energy-cfse-of-fehsub2-jee-main-chemistry-coordination-compounds-uqkoypkgom3ch7dj)176
The species that can have a trans-isomer is : (en = ehane-1, 2-diamine, ox = oxalate)
JEE Main 2019 (Online) 10th April Morning Slot[177
Three complexes, [CoCl(NH3)5] 2+(I), [Co(NH3)5H2O]3+ (II) and [Co(NH3)6] 3+(III) absorb light in the visible region. The correct order of the wavelength of light absorbed by them is :
JEE Main 2019 (Online) 10th April Morning Slot](/past-years/jee/question/three-complexes-brcoclnhsub3subsub5sub-su-jee-main-chemistry-coordination-compounds-4vul43o83ggvacei)178
The maximum possible denticities of a ligand given below towards a common transition and inner-transition metal ion, respectively, are :
JEE Main 2019 (Online) 9th April Evening Slot179
The correct statements among I to III are :(I) Valence bond theory cannot explain the color exhibited by transition metal complexes.(II) Valence bond theory can predict quantitatively the magnetic properties of transtition metal complexes.(III) Valence bond theory cannot distinguish ligands as weak and strong field ones.
JEE Main 2019 (Online) 9th April Evening Slot[180
The degenerate orbitals of [Cr(H2O)6]3+ are :
JEE Main 2019 (Online) 9th April Morning Slot](/past-years/jee/question/the-degenerate-orbitals-of-crh2o63-are-jee-main-chemistry-some-basic-concepts-of-chemistry-b9zdpiqjrxbsb3di)181
The one that will show optical activity is :(en = ethane-1,2-diamine)
JEE Main 2019 (Online) 9th April Morning Slot182
The compound that inhibits the growth of tumors is :
JEE Main 2019 (Online) 8th April Evening Slot[183
The calculated spin-only magnetic moments (BM) of the anionic and cationic species of [Fe(H2O)6]2 and [Fe(CN)6], respectively, are :
JEE Main 2019 (Online) 8th April Evening Slot](/past-years/jee/question/the-calculated-spin-only-magnetic-moments-bm-of-the-anioni-jee-main-chemistry-some-basic-concepts-of-chemistry-xqgzabfhuqvm6lua)[184
The correct order of the spin-only magnetic moment of metal ions in the following low-spin complexes, [V(CN)6]4–,[Fe(CN)6]4–, [Ru(NH3)6]3+, and [Cr(NH3)6]2+ , is :
JEE Main 2019 (Online) 8th April Morning Slot](/past-years/jee/question/the-correct-order-of-the-spin-only-magnetic-moment-of-metal-jee-main-chemistry-some-basic-concepts-of-chemistry-ejm23ixokplawv9f)185
The following ligand is :
JEE Main 2019 (Online) 8th April Morning Slot186
The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 BM. The suitable ligand for this complex is -
JEE Main 2019 (Online) 12th January Evening Slot[187
The pair of metal ions that can give a spin only magnetic moment of 3.9 BM for the complex [M(H2O)6]Cl2, is -
JEE Main 2019 (Online) 12th January Morning Slot](/past-years/jee/question/the-pair-of-metal-ions-that-can-give-a-spin-only-magnetic-mo-jee-main-chemistry-some-basic-concepts-of-chemistry-wo3oqbqjdn2ff101)[188
The metal d-orbitals that are directly facing the ligands in K3[Co(CN)6] are -
JEE Main 2019 (Online) 12th January Morning Slot](/past-years/jee/question/the-metal-d-orbitals-that-are-directly-facing-the-ligands-in-jee-main-chemistry-some-basic-concepts-of-chemistry-eejk9ln6vp4cxwf4)189
Mn2(CO)10 is an organometallic compound due to the presence of -
JEE Main 2019 (Online) 12th January Morning Slot190
The number of bridging CO ligand(s) and Co-Co bond(s) in Co2(CO)8, respectively are :
JEE Main 2019 (Online) 11th January Evening Slot[191
The coordination number of Th in K4[Th(C2O4)4(OH2)2] is : (C2O$${_4^{2 - }}$$ = Oxalato)
JEE Main 2019 (Online) 11th January Evening Slot](/past-years/jee/question/the-coordination-number-of-th-in-k4thc2o44oh22-is-c-jee-main-chemistry-some-basic-concepts-of-chemistry-k82nwrsginv3i7uq)192
Match the metals (column I) with the coordination compound(s)/enzyme(s) (column II)
| (Column I) Metals | (Column II) Coordination compounds(s) enzyme(s) |
--- |
| (A) | Co | (i) | Wilkinson catalyst |
| (B) | Zn | (ii) | Chlorophyl |
| (C) | Rh | (iii) | Vitamin B12 |
| (D) | Mg | (iv) | Carbonic anhydrase |
JEE Main 2019 (Online) 11th January Morning Slot193
A reaction of cobalt (III) chloride and ethylenediamine in a 1 : 2 mole ratio generates two isomeric products A (violet coloured) and B (green coloured). A can show optical activity, but B is optically inactive. What type of isomers does A and B represcent?
JEE Main 2019 (Online) 10th January Evening Slot194
The difference in the number of unpaired electrons of a metal ion in its high spin and low-spin octahedral complexes is two. The metal ion is :
JEE Main 2019 (Online) 10th January Evening Slot[195
The total number of isomers for a square planar complex[M(F)(Cl)(SCN)(NO2)] is
JEE Main 2019 (Online) 10th January Morning Slot](/past-years/jee/question/the-total-number-of-isomers-for-a-square-planar-complex-mf-jee-main-chemistry-some-basic-concepts-of-chemistry-ge49h4fkvumyojdd)196
The complex that has highest crystal field splitting energy ($$\Delta $$), is :
JEE Main 2019 (Online) 9th January Evening Slot197
Homoleptic octahedral complexes of a metal ion 'M3+' with three monodenate ligands L1, L2 and L3 adsorb wavelenths in the region of green, blue and red respectively. The increasing order of the ligands strength is :
JEE Main 2019 (Online) 9th January Evening Slot[198
Two complexes [Cr(H2O)6]Cl3 (A) and [Cr(NH3)6]Cl3 (B) are violet and yellow coloured, respectively. The incorrect statement regarding them is :
JEE Main 2019 (Online) 9th January Morning Slot](/past-years/jee/question/consider-the-compound-a-crh2o6cl3-yellow-b-crnh36-jee-main-chemistry-some-basic-concepts-of-chemistry-q5nr1ywoyvf0jodj)199
In Wilkinson's catalyst, the hybridization of central metal ion and its shape are respectively :
JEE Main 2018 (Online) 16th April Morning Slot200
Which of the following complexes will shows geometrical isomerism ?
JEE Main 2018 (Online) 16th April Morning Slot[201
The oxidation states of Cr in [Cr(H2O)6]Cl3, [Cr(C6H6)2] and K2[Cr(CN)2(O)2(O2)(NH3)] respectively are :
JEE Main 2018 (Offline)](/past-years/jee/question/the-oxidation-states-of-cr-in-crh2o6cl3-crc6h62-and-2018-marks-4-fea2e7hjegrohbyr.htm)[202
Consider the following reaction and statements:[Co(NH3)4Br2]+ + Br- $$\to$$ [Co(NH3)3Br3] + NH3 (I) Two isomers are produced if the reactant complex ion is a cis-isomer (II) Two isomers are produced if the reactant complex ion is a trans-isomer (III) Only one isomer is produced if the reactant complex ion is a trans-isomer (IV) Only one isomer is produced if the reactant complex ion is a cis – isomer The correct statements are
JEE Main 2018 (Offline)](/past-years/jee/question/consider-the-following-reaction-and-statements-conh34br2-2018-marks-4-qxdyecfhkbmurv1o.htm)203
The correct order of spin-only magnetic moments among the following is : (Atomic number : Mn = 25, Co = 27, Ni = 28, Zn = 30)
JEE Main 2018 (Online) 15th April Evening Slot[204
The total number of possible isomers for square-planar [Pt(Cl)(NO2)(NO3)(SCN)]2- is :
JEE Main 2018 (Online) 15th April Evening Slot](/past-years/jee/question/the-total-number-of-possible-isomers-for-square-planar-ptc-jee-main-chemistry-some-basic-concepts-of-chemistry-41bdxqqekmg30xd9)205
The correct combination is :
JEE Main 2018 (Online) 15th April Morning Slot[206
[Co2(CO)8] displays :
JEE Main 2017 (Online) 9th April Morning Slot](/past-years/jee/question/co2co8-displays-jee-main-chemistry-some-basic-concepts-of-chemistry-dyxqrjc4ollqjbz0)207
On treatment of 100 mL of 0.1 M solution of CoCl3. 6H2O with excess AgNO3; 1.2 $$\times$$ 1022 ions are precipitated. The complex is :
JEE Main 2017 (Offline)208
Which of the following is an example of homoleptic complex?
JEE Main 2016 (Online) 10th April Morning Slot209
Which one of the following complexes will consume more equivalents of aqueous solution of Ag(NO3)?
JEE Main 2016 (Online) 9th April Morning Slot210
Identify the correct trend given below : (Atomic No.=Ti : 22, Cr : 24 and Mo : 42)
JEE Main 2016 (Online) 9th April Morning Slot[211
The pair having the same magnetic moment is : [At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]
JEE Main 2016 (Offline)](/past-years/jee/question/the-pair-having-the-same-magnetic-moment-is-at-no-cr-2016-marks-4-tfd6qu4vpk9f6elx.htm)212
Which one of the following complexes shows optical isomerism?(en = ethylenediamine)
JEE Main 2016 (Offline)213
Which of the following compounds is not colored yellow?
JEE Main 2015 (Offline)[214
The number of geometric isomers that can exist for square planar [Pt (Cl) (py) (NH3) (NH2OH)]+ is (py = pyridine) :
JEE Main 2015 (Offline)](/past-years/jee/question/the-number-of-geometric-isomers-that-can-exist-for-square-pl-2015-marks-4-sgydxdfyjdla5dsk.htm)215
The octahedral complex of a metal ion M3+ with four monodentate ligands L1, L2, L3 and L4 absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is :
JEE Main 2014 (Offline)216
The equation which is balanced and represents the correct product(s) is :
JEE Main 2014 (Offline)217
Which of the following complex species is not expected to exhibit optical isomerism?
JEE Main 2013 (Offline)218
Which among the following will be named as dibromidobis (ethylene diamine) chromium(III) bromide?
AIEEE 2012[219
Which of the following facts about the complex [Cr (NH3)6 ]Cl3 is wrong?
AIEEE 2011](/past-years/jee/question/which-of-the-following-facts-about-the-complex-cr-nh36-c-2011-marks-4-7tnfurrxh2mrzkad.htm)[220
The magnetic moment (spin only) of [NiCl4]2− is
AIEEE 2011](/past-years/jee/question/the-magnetic-moment-spin-only-of-nicl42-is-2011-marks-4-4ecfpwwypwonud2m.htm)221
A solution containing 2.675g of CoCl3. 6NH3 (molar mass = 267.5 g mol–1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78 g of AgCl (molar mass = 143.5 g mol–1). The formula of the complex is : (At. Mass of Ag = 108 u)
AIEEE 2010222
Which one of the following has an optical isomer ?(en = ethylenediamine)
AIEEE 2010223
Which of the following has an optical isomer ?
AIEEE 2009224
Which of the following pairs represents linkage isomers ?
AIEEE 2009225
In which of the following octahedral complexes of Co (at. no. 27), will the magnitude of $$\Delta _o$$ be the highest?
AIEEE 2008[226
The coordination number and the oxidation state of the element ‘E’ in the complex [E(en)2(C2O4)]NO2 (where (en) is ethylene diamine) are, respectively,
AIEEE 2008](/past-years/jee/question/the-coordination-number-and-the-oxidation-state-of-the-eleme-2008-marks-4-cn2wgmcrjcjg9d4e.htm)227
Which one of the following has a square planar geometry?
AIEEE 2007[228
The IUPAC name for the complex [Co(NO2)(NH3)5]Cl2 is
AIEEE 2006](/past-years/jee/question/the-iupac-name-for-the-complex-cono2nh35cl2-is-2006-marks-4-fzgiaiobadwc2yms.htm)229
How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral complex with a Ca2+ ion?
AIEEE 2006230
In Fe(CO)5, the Fe – C bond possesses :
AIEEE 2006[231
Nickel (Z = 28) combines with a uninegative monodentate ligand X– to form a paramagnetic complex [NiX4]2− . The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively
AIEEE 2006](/past-years/jee/question/nickel-z-28-combines-with-a-uninegative-monodentate-liga-2006-marks-4-qv8n6cjid5ghsz6b.htm)[232
The oxidation state of Cr in [Cr(NH3)4Cl2]+ is :
AIEEE 2005](/past-years/jee/question/the-oxidation-state-of-cr-in-crnh34cl2-is-2005-marks-4-mpwtq2oy9tnqz8ep.htm)[233
The IUPAC name of the coordination compound K3[Fe(CN)6] is
AIEEE 2005](/past-years/jee/question/the-iupac-name-of-the-coordination-compound-k3fecn6-is-2005-marks-4-o6qiyzxzuwmbqihq.htm)234
The value of the ‘spin only’ magnetic moment for one of the following configurations is 2.84 BM. The correct one is :
AIEEE 2005235
Which of the following compounds shows optical isomerism?
AIEEE 2005236
Which one of the following cyano complexes would exhibit the lowest value of paramagnetic behaviour? (At. No. Cr = 24, Mn = 25, Fe = 26, Co = 27)
AIEEE 2005237
The coordination number of central metal atom in a complex is determined by :
AIEEE 2004238
Among the properties (a) reducing (b) oxidising (c) complexing, the set of properties shown by CN– ion towards metal species is :
AIEEE 2004239
The correct order of magnetic moments (spin only values in B.M.) among is :(Atomic numbers: Mn = 25; Fe = 26, Co =27)
AIEEE 2004240
Coordination compound have great importance in biological systems. In this context which of the following statements is incorrect?
AIEEE 2004241
Which one the following has largest number of isomers? (R = alkyl group, en = ethylenediamine)
AIEEE 2004242
Which one of the following complexes in an outer orbital complex?
AIEEE 2004[243
Ammonia forms the complex ion [Cu(NH3)4]2+ with copper ions in alkaline solutions but not in acidic solutions. What is the reason for it?
AIEEE 2003](/past-years/jee/question/ammonia-forms-the-complex-ion-cunh342-with-copper-ions-2003-marks-4-spbn67dgocrmtgwi.htm)244
One mole of the complex compound Co(NH3)5Cl3, gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles of AgNO3 solution to yield two moles of AgCl (s). The structure of the complex is :
AIEEE 2003[245
In the coordination compound, K4[Ni(CN)4], the oxidation state of nickel is :
AIEEE 2003](/past-years/jee/question/-in-the-coordination-compound-k4nicn4-the-oxidation-st-2003-marks-4-fyrfarqlq6avks0j.htm)246
The most stable ion is :
AIEEE 2002247
CH3 - Mg - Br is an organometallic compound due to :
AIEEE 2002248
The type isomerism present in nitropentammine chromium $$\left( {{\rm I}{\rm I}{\rm I}} \right)$$ chloride is :
AIEEE 2002
Numerical
[1
The number of paramagnetic metal complex species among $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+},\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$, $\left[\mathrm{MnCl}_6\right]^{3-},\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{CoF}_6\right]^{3-},\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ and $\left[\mathrm{FeF}_6\right]^{3-}$ with same number of unpaired electrons is __________.
JEE Main 2025 (Online) 7th April Evening Shift](/past-years/jee/question/the-number-of-paramagnetic-metal-complex-species-among-lef-jee-main-chemistry-dqmzbx1zykwnjbg3)[2
The number of paramagnetic complexes among $\left[\mathrm{FeF}_6\right]^{3-},\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$, $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-},\left[\mathrm{MnCl}_6\right]^{3-}$, and $\left[\mathrm{CoF}_6\right]^{3-}$, which involved $\mathrm{d}^2 \mathrm{sp}^3$ hybridization is _________.
JEE Main 2025 (Online) 7th April Morning Shift](/past-years/jee/question/pthe-number-of-paramagnetic-complexes-among-leftmathrm-jee-main-chemistry-some-basic-concepts-of-chemistry-dyjdujjihszifwh0)3
A metal complex with a formula $\mathrm{MCl}_4 \cdot 3 \mathrm{NH}_3$ is involved in $\mathrm{sp}^3 \mathrm{~d}^2$ hybridisation. It upon reaction with excess of $\mathrm{AgNO}_3$ solution gives ' $x$ ' moles of AgCl . Consider ' $x$ ' is equal to the number of lone pairs of electron present in central atom of $\mathrm{BrF}_5$. Then the number of geometrical isomers exhibited by the complex is _________.
JEE Main 2025 (Online) 4th April Evening Shift4
The number of optical isomers exhibited by the iron complex $(\mathrm{A})$ obtained from the following reaction is___________.
$$ \mathrm{FeCl}_3+\mathrm{KOH}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightarrow \mathrm{~A} $$
JEE Main 2025 (Online) 3rd April Morning Shift[5
The spin-only magnetic moment value of $\mathrm{M}^{\mathrm{n}+}$ ion formed among $\mathrm{Ni}, \mathrm{Zn}, \mathrm{Mn}$ and Cu that has the least enthalpy of atomisation is_________ . (in nearest integer) Here n is equal to the number of diamagnetic complexes among $\mathrm{K}_2\left[\mathrm{NiCl}_4\right],\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2$, $\mathrm{K}_3\left[\mathrm{Mn}(\mathrm{CN})_6\right]$ and $\left[\mathrm{Cu}\left(\mathrm{PPh}_3\right)_3 \mathrm{I}\right]$
JEE Main 2025 (Online) 2nd April Evening Shift](/past-years/jee/question/pthe-spin-only-magnetic-moment-value-of-mathrmmmath-jee-main-chemistry-some-basic-concepts-of-chemistry-z5stgnhduvgwh8t4)[6
A transition metal (M) among $\mathrm{Mn}, \mathrm{Cr}, \mathrm{Co}$ and Fe has the highest standard electrode potential $\left(\mathrm{M}^{3+} / \mathrm{M}^{2+}\right)$. It forms a metal complex of the type $\left[\mathrm{M}(\mathrm{CN})_6\right]^{4-}$. The number of electrons present in the $\mathrm{e}_{\mathrm{g}}$ orbital of the complex is ___________.
JEE Main 2025 (Online) 2nd April Morning Shift](/past-years/jee/question/pa-transition-metal-m-among-mathrmmn-mathrmcr--jee-main-chemistry-some-basic-concepts-of-chemistry-gvulbi5v9big9sen)[7
Consider the following low-spin complexes
$$ \mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{NO}_2\right)_6\right], \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right], \mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right], \mathrm{Cu}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right] \text { and } \mathrm{Zn}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right] $$
The sum of the spin-only magnetic moment values of complexes having yellow colour is ________ B.M. (answer in nearest integer)
JEE Main 2025 (Online) 29th January Evening Shift](/past-years/jee/question/pconsider-the-following-low-spin-complexesp-p-math-jee-main-chemistry-4ft6rb47p8vcryed)8
The spin only magnetic moment ( $\mu$ ) value (B.M.) of the compound with strongest oxidising power among $\mathrm{Mn}_2 \mathrm{O}_3, \mathrm{TiO}$ and VO is ____________ B.M. (Nearest integer).
JEE Main 2025 (Online) 28th January Evening Shift[9
Total number of molecules/species from following which will be paramagnetic is __________. $\mathrm{O}_2, \mathrm{O}_2^{+}, \mathrm{O}_2^{-}, \mathrm{NO}, \mathrm{NO}_2, \mathrm{CO}, \mathrm{K}_2\left[\mathrm{NiCl}_4\right],\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3, \mathrm{~K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$
JEE Main 2025 (Online) 28th January Evening Shift](/past-years/jee/question/total-number-of-moleculesspecies-from-following-which-will-jee-main-chemistry-pmjdtbkhmw4gfelo)10
The complex of $\mathrm{Ni}^{2+}$ ion and dimethyl glyoxime contains __________ number of Hydrogen (H) atoms.
JEE Main 2025 (Online) 22nd January Evening Shift11
Consider the following test for a group-IV cation.
$$\mathrm{M}^{2+}+\mathrm{H}_2 \mathrm{S} \rightarrow \mathrm{A} \text { (Black precipitate)+ byproduct }$$
$$\mathrm{A}+\text { aqua regia } \rightarrow \mathrm{B}+\mathrm{NOCl}+\mathrm{S}+\mathrm{H}_2 \mathrm{O}$$
$$\mathrm{B}+\mathrm{KNO}_2+\mathrm{CH}_3 \mathrm{COOH} \rightarrow \mathrm{C}+\text { byproduct }$$
The spin-only magnetic moment value of the metal complex $$\mathrm{C}$$ is _________ $$\mathrm{BM}$$ (Nearest integer)
JEE Main 2024 (Online) 9th April Evening Shift12
Number of ambidentate ligands among the following is _________.
$$\mathrm{NO}_2^{-}, \mathrm{SCN}^{-}, \mathrm{C}_2 \mathrm{O}_4^{2-}, \mathrm{NH}_3, \mathrm{CN}^{-}, \mathrm{SO}_4^{2-}, \mathrm{H}_2 \mathrm{O} \text {. }$$
JEE Main 2024 (Online) 9th April Morning Shift[13
Total number of unpaired electrons in the complex ions $$[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$$ and $$[\mathrm{NiCl}_4]^{2-}$$ is ________.
JEE Main 2024 (Online) 8th April Evening Shift](/past-years/jee/question/ptotal-number-of-unpaired-electrons-in-the-complex-ions--jee-main-chemistry-some-basic-concepts-of-chemistry-bv13cllnvup4rvt6)14
The 'spin only' magnetic moment value of $$\mathrm{MO}_4{ }^{2-}$$ is ________ BM. (Where M is a metal having least metallic radii. among $$\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}$$ and $$\mathrm{Zn}$$ ).
(Given atomic number: $$\mathrm{Sc}=21, \mathrm{Ti}=22, \mathrm{~V}=23, \mathrm{Cr}=24, \mathrm{Mn}=25$$ and $$\mathrm{Zn}=30$$)
JEE Main 2024 (Online) 8th April Morning Shift15
The difference in the 'spin-only' magnetic moment values of $$\mathrm{KMnO}_4$$ and the manganese product formed during titration of $$\mathrm{KMnO}_4$$ against oxalic acid in acidic medium is ________ $$\mathrm{BM}$$. (nearest integer)
JEE Main 2024 (Online) 6th April Morning Shift16
The spin-only magnetic moment value of the ion among $$\mathrm{Ti}^{2+}, \mathrm{V}^{2+}, \mathrm{Co}^{3+}$$ and $$\mathrm{Cr}^{2+}$$, that acts as strong oxidising agent in aqueous solution is _________ BM (Near integer).
(Given atomic numbers : $$\mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Co}: 27$$)
JEE Main 2024 (Online) 5th April Morning Shift[17
The 'Spin only' Magnetic moment for $$\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$$ is _________ $$\times 10^{-1} \mathrm{~BM}$$. (given $$=$$ Atomic number of $$\mathrm{Ni}: 28$$)
JEE Main 2024 (Online) 31st January Morning Shift](/past-years/jee/question/pthe-spin-only-magnetic-moment-for-leftmathrmnil-jee-main-chemistry-some-basic-concepts-of-chemistry-og8pux2ssn6egjey)[18
Number of complexes which show optical isomerism among the following is ________.
$$\text { cis- }\left[\mathrm{Cr}(\mathrm{ox})_2 \mathrm{Cl}_2\right]^{3-},\left[\mathrm{Co}(\text {en})_3\right]^{3+}, \text { cis- }\left[\mathrm{Pt}(\text {en})_2 \mathrm{Cl}_2\right]^{2+}, \text { cis- }\left[\mathrm{Co}(\text {en})_2 \mathrm{Cl}_2\right]^{+}, \text {trans- }\left[\mathrm{Pt}(\text {en})_2 \mathrm{Cl}_2\right]^{2+}, \text { trans- }\left[\mathrm{Cr}(\mathrm{ox})_2 \mathrm{Cl}_2\right]^{3-}$$
JEE Main 2024 (Online) 30th January Evening Shift](/past-years/jee/question/pnumber-of-complexes-which-show-optical-isomerism-among-th-jee-main-chemistry-some-basic-concepts-of-chemistry-nvbggo2gmzzkv1ng)[19
The Spin only magnetic moment value of square planar complex $$\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}\left(\mathrm{NH}_2 \mathrm{CH}_3\right)\right] \mathrm{Cl}$$ is _________ B.M. (Nearest integer)
(Given atomic number for $$\mathrm{Pt}=78$$)
JEE Main 2024 (Online) 27th January Evening Shift](/past-years/jee/question/pthe-spin-only-magnetic-moment-value-of-square-planar-comp-jee-main-chemistry-some-basic-concepts-of-chemistry-clra8d5axyqu6gkv)20
The homoleptic and octahedral complex of $\mathrm{Co}^{2+}$ and $\mathrm{H}_{2} \mathrm{O}$ has ___________ unpaired electron(s) in the $t_{2\mathrm{g}}$ set of orbitals.
JEE Main 2023 (Online) 15th April Morning Shift[21
The ratio of spin-only magnetic moment values $$\mu_{\text {eff }}\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-} / \mu_{\text {eff }}\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$$ is _________.
JEE Main 2023 (Online) 11th April Morning Shift](/past-years/jee/question/pthe-ratio-of-spin-only-magnetic-moment-values-mutex-jee-main-chemistry-some-basic-concepts-of-chemistry-sqhfvqksmprxfz7c)22
For a metal ion, the calculated magnetic moment is $$4.90 ~\mathrm{BM}$$. This metal ion has ___________ number of unpaired electrons.
JEE Main 2023 (Online) 10th April Evening Shift23
In potassium ferrocyanide, there are ________ pairs of electrons in the $$t_{2g}$$ set of orbitals.
JEE Main 2023 (Online) 10th April Morning Shift[24
The observed magnetic moment of the complex $$\left.\left[\operatorname{Mn}(\underline{N} C S)_{6}\right)\right]^{x^{-}}$$ is $$6.06 ~\mathrm{BM}$$. The numerical value of $$x$$ is __________.
JEE Main 2023 (Online) 8th April Evening Shift](/past-years/jee/question/pthe-observed-magnetic-moment-of-the-complex-leftleft-jee-main-chemistry-some-basic-concepts-of-chemistry-kivg02qaghknjijn)[25
Number of ambidentate ligands in a representative metal complex $$\left[\mathrm{M}(\mathrm{en})(\mathrm{SCN})_{4}\right]$$ is ___________.
[en = ethylenediamine]
JEE Main 2023 (Online) 6th April Morning Shift](/past-years/jee/question/pnumber-of-ambidentate-ligands-in-a-representative-metal-c-jee-main-chemistry-some-basic-concepts-of-chemistry-waa9nlqm9s7mmsor)[26
The spin only magnetic moment of $$\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$$ complexes is _________ B.M. (Nearest integer)
(Given : Atomic no. of Mn is 25)
JEE Main 2023 (Online) 1st February Evening Shift](/past-years/jee/question/pthe-spin-only-magnetic-moment-of-leftmathrmmnleft-jee-main-chemistry-some-basic-concepts-of-chemistry-hysv9swlsk76upwr)[27
If the CFSE of $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ is $-96.0 \mathrm{~kJ} / \mathrm{mol}$, this complex will absorb maximum at wavelength ___________ $\mathrm{nm}$. (nearest integer) Assume Planck's constant (h) $=6.4 \times 10^{-34} \mathrm{Js}$, Speed of light $(\mathrm{c})=3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}$ and Avogadro's Constant $\left(\mathrm{N}_{\mathrm{A}}\right)=6 \times 10^{23} / \mathrm{mol}$
JEE Main 2023 (Online) 31st January Evening Shift](/past-years/jee/question/if-the-cfse-of-leftmathrmtileftmathrmh2-mathr-jee-main-chemistry-o4xjiubpr6pjkvi7)28
The sum of bridging carbonyls in $$\mathrm{W(CO)_6}$$ and $$\mathrm{Mn_2(CO)_{10}}$$ is ____________.
JEE Main 2023 (Online) 29th January Morning Shift[29
Total number of moles of AgCl precipitated on addition of excess of AgNO$$_3$$ to one mole each of the following complexes $$\mathrm{[Co(NH_3)_4Cl_2]Cl,[Ni(H_2O)_6]Cl_2,[Pt(NH_3)_2Cl_2]}$$ and $$\mathrm{[Pd(NH_3)_4]Cl_2}$$ is ___________.
JEE Main 2023 (Online) 25th January Evening Shift](/past-years/jee/question/ptotal-number-of-moles-of-agcl-precipitated-on-addition-of-jee-main-chemistry-some-basic-concepts-of-chemistry-tlka0giq9qx4fs5n)[30
The number of paramagnetic species from the following is _____________.
$$\mathrm{{[Ni{(CN)_4}]^{2 - }},[Ni{(CO)_4}],{[NiC{l_4}]^{2 - }}}$$
$$\mathrm{{[Fe{(CN)_6}]^{4 - }},{[Cu{(N{H_3})_4}]^{2 + }}}$$
$$\mathrm{{[Fe{(CN)_6}]^{3 - }}\,and\,{[Fe{({H_2}O)_6}]^{2 + }}}$$
JEE Main 2023 (Online) 25th January Morning Shift](/past-years/jee/question/pthe-number-of-paramagnetic-species-from-the-following-is-jee-main-chemistry-some-basic-concepts-of-chemistry-g8jan0fpokrlnn9h)[31
The d-electronic configuration of $$\mathrm{[CoCl_4]^{2-}}$$ in tetrahedral crystal field in $${e^mt_2^n}$$. Sum of "m" and "number of unpaired electrons" is ___________
JEE Main 2023 (Online) 24th January Morning Shift](/past-years/jee/question/pthe-d-electronic-configuration-of-mathrmcocl42--jee-main-chemistry-some-basic-concepts-of-chemistry-tzf1lsqqamz3ivny)[32
Sum of oxidation state (magnitude) and coordination number of cobalt in $$\mathrm{Na}\left[\mathrm{Co}(\mathrm{bpy}) \mathrm{Cl}_{4}\right]$$ is _________.
JEE Main 2022 (Online) 29th July Evening Shift](/past-years/jee/question/psum-of-oxidation-state-magnitude-and-coordination-numbe-jee-main-chemistry-some-basic-concepts-of-chemistry-nppzzknvvoixzv8o)[33
$$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$$ should be an inner orbital complex. Ignoring the pairing energy, the value of crystal field stabilization energy for this complex is $$(-)$$ ____________ $$\Delta_{0}$$. (Nearest integer)
JEE Main 2022 (Online) 29th July Morning Shift](/past-years/jee/question/pleftmathrmfemathrmcn6right3--should-jee-main-chemistry-some-basic-concepts-of-chemistry-punmer9n0ltebvhw)[34
Total number of relatively more stable isomer(s) possible for octahedral complex $$\left[\mathrm{Cu}(\mathrm{en})_{2}(\mathrm{SCN})_{2}\right]$$ will be _________.
JEE Main 2022 (Online) 28th July Morning Shift](/past-years/jee/question/ptotal-number-of-relatively-more-stable-isomers-possible-jee-main-chemistry-some-basic-concepts-of-chemistry-seqcezpftkkefkgq)35
The conductivity of a solution of complex with formula $$\mathrm{CoCl}_{3}\left(\mathrm{NH}_{3}\right)_{4}$$ corresponds to 1 : 1 electrolyte, then the primary valency of central metal ion is __________.
JEE Main 2022 (Online) 27th July Morning Shift[36
The difference between spin only magnetic moment values of $$\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\mathrm{Cl}_{2}$$ and $$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$$ is ___________.
JEE Main 2022 (Online) 26th July Morning Shift](/past-years/jee/question/pthe-difference-between-spin-only-magnetic-moment-values-o-jee-main-chemistry-some-basic-concepts-of-chemistry-b7vrychjogsoph9j)[37
Consider the following metal complexes :
$$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}$$
The spin-only magnetic moment value of the complex that absorbes light with shortest wavelength is _____________ B. M. (Nearest integer)
JEE Main 2022 (Online) 25th July Morning Shift](/past-years/jee/question/pconsider-the-following-metal-complexes-p-pleft-jee-main-chemistry-some-basic-concepts-of-chemistry-uywhnfn81m4n60dq)[38
In the following brown complex, the oxidation state of iron is +_____________.
$${[Fe{({H_2}O)_6}]^{2 + }} + NO \to \mathop {{{[Fe{{({H_2}O)}_5}(NO)]}^{2 + }}}\limits_{\text{Brown complex}} + {H_2}O$$
JEE Main 2022 (Online) 30th June Morning Shift](/past-years/jee/question/pin-the-following-brown-complex-the-oxidation-state-of-ir-jee-main-chemistry-some-basic-concepts-of-chemistry-miawiomppngeeevx)[39
Spin only magnetic moment ($$\mu$$s) of $${K_3}[Fe{(CN)_6}]$$ is ____________ B.M.
(Nearest integer)
JEE Main 2022 (Online) 30th June Morning Shift](/past-years/jee/question/pspin-only-magnetic-moment-musubssub-of-k3-jee-main-chemistry-some-basic-concepts-of-chemistry-mdu8dex2cltr0l3s)[40
$${[Fe{(CN)_6}]^{4 - }}$$
$${[Fe{(CN)_6}]^{3 - }}$$
$${[Ti{(CN)_6}]^{3 - }}$$
$${[Ni{(CN)_4}]^{2 - }}$$
$${[Co{(CN)_6}]^{3 - }}$$
Among the given complexes, number of paramagnetic complexes is ____________.
JEE Main 2022 (Online) 28th June Evening Shift](/past-years/jee/question/pfecn64-p-pfecn63--jee-main-chemistry-some-basic-concepts-of-chemistry-kneffy6bop1xwvqk)41
(a) CoCl3.4NH3, (b) CoCl3.5NH3, (c) CoCl3.6NH3 and (d) CoCl(NO3)2.5NH3.
Number of complex(es) which will exist in cis-trans form is/are _______________.
JEE Main 2022 (Online) 28th June Evening Shift[42
Number of complexes which will exhibit synergic bonding amongst, $$[Cr{(CO)_6}]$$, $$[Mn{(CO)_5}]$$ and $$[M{n_2}{(CO)_{10}}]$$ is ___________.
JEE Main 2022 (Online) 28th June Morning Shift](/past-years/jee/question/pnumber-of-complexes-which-will-exhibit-synergic-bonding-a-jee-main-chemistry-some-basic-concepts-of-chemistry-omwef9b6t12aaapl)43
Acidified potassium permanganate solution oxidises oxalic acid. The spin-only magnetic moment of the manganese product formed from the above reaction is ____________ B.M. (Nearest integer)
JEE Main 2022 (Online) 27th June Morning Shift[44
Reaction of [Co(H2O)6]2+ with excess ammonia and in the presence of oxygen results into a diamagnetic product. Number of electrons present in t2g-orbitals of the product is ___________.
JEE Main 2022 (Online) 26th June Evening Shift](/past-years/jee/question/preaction-of-cohsub2subosub6subsup2sup-jee-main-chemistry-some-basic-concepts-of-chemistry-ziynjbjsodchbin4)45
The spin-only magnetic moment value of an octahedral complex among CoCl3.4NH3, NiCl2.6H2O and PtCl4.2HCl, which upon reaction with excess of AgNO3 gives 2 moles of AgCl is ___________ B.M. (Nearest integer)
JEE Main 2022 (Online) 26th June Morning Shift[46
Amongst FeCl3.3H2O, K3[Fe(CN)6] and [Co(NH3)6]Cl3, the spin-only magnetic moment value of the inner-orbital complex that absorbs light at shortest wavelength is ____________ B.M. [nearest integer]
JEE Main 2022 (Online) 25th June Evening Shift](/past-years/jee/question/pamongst-feclsub3sub3hsub2subo-ksub3subf-jee-main-chemistry-some-basic-concepts-of-chemistry-5ha8cfuhcqyuahga)[47
If [Cu(H2O)4]2+ absorbs a light of wavelength 600 nm for d-d transition, then the value of octahedral crystal field splitting energy for [Cu(H2O)6]2+ will be ____________ $$\times$$ 10$$-$$21 J. [Nearest Integer]
(Given : h = 6.63 $$\times$$ 10$$-$$34 Js and c = 3.08 $$\times$$ 108 ms$$-$$1)
JEE Main 2022 (Online) 25th June Morning Shift](/past-years/jee/question/pif-cuhsub2subosub4subsup2sup-absorbs-jee-main-chemistry-some-basic-concepts-of-chemistry-vit8foxshdiqhljw)[48
In the cobalt-carbonyl complex : [Co2(CO)8], number of Co-Co bonds is "X" and terminal CO ligands is "Y". X + Y = ___________.
JEE Main 2022 (Online) 24th June Morning Shift](/past-years/jee/question/pin-the-cobalt-carbonyl-complex--cosub2subcosub-jee-main-chemistry-some-basic-concepts-of-chemistry-jkq1yapbptrilhky)[49
The sum of oxidation states of two silver ions in [Ag(NH3)2] [Ag(CN)2] complex is _____________.
JEE Main 2021 (Online) 1st September Evening Shift](/past-years/jee/question/the-sum-of-oxidation-states-of-two-silver-ions-in-agnhsub-jee-main-chemistry-some-basic-concepts-of-chemistry-lcmgtaxq1gohvxuv)[50
The number of optical isomers possible for [Cr(C2O4)3]3$$-$$ is ____________.
JEE Main 2021 (Online) 27th August Evening Shift](/past-years/jee/question/the-number-of-optical-isomers-possible-for-crcsub2sub-jee-main-chemistry-some-basic-concepts-of-chemistry-slqqdqupwqjqgllv)51
1 mol of an octahedral metal complex with formula MCl3 . 2L on reaction with excess of AgNO3 gives 1 mol of AgCl. The denticity of Ligand L is ____________. (Integer answer)
JEE Main 2021 (Online) 27th August Morning Shift[52
The overall stability constant of the complex ion [Cu(NH3)4]2+ is 2.1 $$\times$$ 1013. The overall dissociations constant is y $$\times$$ 10$$-$$14. Then y is __________. (Nearest integer)
JEE Main 2021 (Online) 26th August Evening Shift](/past-years/jee/question/the-overall-stability-constant-of-the-complex-ion-cunhsub-jee-main-chemistry-some-basic-concepts-of-chemistry-od9puecfxumnwqag)53
The ratio of number of water molecules in Mohr's salt and potash alum is ____________ $$\times$$ 10$$-$$1. (Integer answer)
JEE Main 2021 (Online) 26th August Morning Shift54
3 moles of metal complex with formula Co(en)2Cl3 gives 3 moles of silver chloride on treatment with excess of silver nitrate. The secondary valency of Co in the complex is ___________.(Round off to the nearest integer)
JEE Main 2021 (Online) 27th July Evening Shift55
The number of geometrical isomers possible in triamminetrinitrocobalt (III) is X and in trioxalatochromate (III) is Y. Then the value of X + Y is _______________.
JEE Main 2021 (Online) 27th July Morning Shift56
Three moles of AgCl get precipitated when one mole of an octahedral co-ordination compound with empirical formula CrCl3.3NH3.3H2O reacts with excess of silver nitrate. The number of chloride ions satisfying the secondary valency of the metal ion is ______________.
JEE Main 2021 (Online) 25th July Morning Shift[57
The total number of unpaired electrons present in [Co(NH3)6]Cl2 and [Co(NH3)6]Cl3 is :
JEE Main 2021 (Online) 22th July Evening Shift](/past-years/jee/question/the-total-number-of-unpaired-electrons-present-in-conhsub-jee-main-chemistry-some-basic-concepts-of-chemistry-xhgokvtqggrj3ml9)[58
An aqueous solution of NiCl2 was heated with excess sodium cyanide in presence of strong oxidizing agent to form [Ni(CN)6]2$$-$$. The total change in number of unpaired electrons on metal centre is _______________.
JEE Main 2021 (Online) 20th July Evening Shift](/past-years/jee/question/an-aqueous-solution-of-niclsub2sub-was-heated-with-exce-jee-main-chemistry-some-basic-concepts-of-chemistry-kpejd11lf1hsf0yx)[59
The spin-only magnetic moment value for the complex [Co(CN)6]4$$-$$ is __________ BM.[At. no. of Co = 27]
JEE Main 2021 (Online) 20th July Morning Shift](/past-years/jee/question/the-spin-only-magnetic-moment-value-for-the-complex-cocn-jee-main-chemistry-some-basic-concepts-of-chemistry-mxltpnnvibulkd1k)[60
The total number of unpaired electrons present in the complex K3[Cr(oxalate)3] is _____________.
JEE Main 2021 (Online) 18th March Morning Shift](/past-years/jee/question/the-total-number-of-unpaired-electrons-present-in-the-comple-jee-main-chemistry-some-basic-concepts-of-chemistry-k7sryuf0yicufq8k)61
On complete reaction of FeCl3 with oxalic acid in aqueous solution containing KOH, resulted in the formation of product A. The secondary valency of Fe in the product A is __________. (Round off to the Nearest Integer).
JEE Main 2021 (Online) 17th March Evening Shift[62
[Ti(H2O)6]3+ absorbs light of wavelength 498 nm during a d $$-$$ d transition. The octahedral splitting energy for the above complex is ____________ $$\times$$ 10$$-$$19 J. (Round off to the Nearest Integer). h = 6.626 $$\times$$ 10$$-$$34 Js; c = 3 $$\times$$ 108 ms$$-$$1
JEE Main 2021 (Online) 16th March Evening Shift](/past-years/jee/question/tihsub2subosub6subsup3sup-absorbs-light-jee-main-chemistry-some-basic-concepts-of-chemistry-wuj8kksosx0btekm)63
The equivalents of ethylene diamine required to replace the neutral ligands from the coordination sphere of the trans-complex of CoCl3 . 4NH3 is _________. (Round off to the Nearest Integer).
JEE Main 2021 (Online) 16th March Morning Shift[64
The number of stereoisomers possible for [Co(ox)2(Br)(NH3)]2$$-$$ is ___________. [ox = oxalate]
JEE Main 2021 (Online) 26th February Evening Shift](/past-years/jee/question/the-number-of-stereoisomers-possible-for-cooxsub2sub-jee-main-chemistry-some-basic-concepts-of-chemistry-vsvsy4aj8agqls4a)[65
Number of bridging CO ligands in [Mn2(CO)10] is __________.
JEE Main 2021 (Online) 26th February Morning Shift](/past-years/jee/question/number-of-bridging-co-ligands-in-mnsub2subcosub10-jee-main-chemistry-some-basic-concepts-of-chemistry-k3y53hiakyz5wi1e)66
The spin only magnetic moment of a divalent ion in aqueous solution (atomic number 29) is _________ BM.
JEE Main 2021 (Online) 25th February Evening Shift[67
Considering that $$\Delta $$0 > P, the magnetic moment (in BM) of [Ru(H2O)6]2+ would be _________.
JEE Main 2020 (Online) 5th September Evening Slot](/past-years/jee/question/considering-that-delta-sub0sub--p-the-magnetic-jee-main-chemistry-coordination-compounds-mfpwscq70radl8cw)68
The total number of coordination sitesin ethylenediaminetetraacetate (EDTA4–) is _____.
JEE Main 2020 (Online) 5th September Morning Slot[69
The oxidation states of iron atoms in compounds (A), (B) and (C), respectively, are x, y and z. The sum of x, y and z is ________. Na4[Fe(CN)5(NOS)] (A) Na4[FeO4] (B) [Fe2(CO)9] (C)
JEE Main 2020 (Online) 2nd September Morning Slot](/past-years/jee/question/the-oxidation-states-of-iron-atoms-in-compounds-a-b-and-jee-main-chemistry-coordination-compounds-tapitlxahllmydqk)70
Complexes (ML5) of metals Ni and Fe have ideal square pyramidal and trigonal bipyramidal grometries, respectively. The sum of the 90°, 120° and 180° L-M-L angles in the two complexes is ________.
JEE Main 2020 (Online) 8th January Evening Slot
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13921 | https://quizlet.com/616970377/3-cell-membrane-physiology-icfecf-flash-cards/ | Cell membrane Physiology; ICF/ECF Flashcards | Quizlet
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3. Cell membrane Physiology; ICF/ECF
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Protoplasm
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cell nucleus and cytoplasm
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at least partially embedded in the plasma membrane
Ion channels, proton pumps, G-protein coupled receptors
Choose an answer
1
Protoplasm
2
Properties of the bilayer
3
integral membrane proteins
4
Partitioning of cell
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Terms in this set (59)
Protoplasm
cell nucleus and cytoplasm
Nucleus
nuclear envelope, nucleolus, nucleoplasm and chromatin
Cytoplasm
Endoplasm containing organelles, cytoplasmic inclusions and cytosol (hyaloplasm) and Ectoplasm (Actin)
Cytoskeleton
microtubules, intermediate filaments and microfilaments
Partitioning of cell
1.cytoplasmic matrix (organells Inclisions)
Plasma membrane (separates cells from environment)
Nucleear envelope
important characteristic of the cell membrane
FLUIDITY
fluidity of the cell membrane
Helps in endocytosis, exocytosis, membrane growth.
New phospholipids keep adding to the membrane and this helps in membrane biogenesis.
Allows the cell membrane to readily reform when its continuity is disturbed.
4.Proteins can move freely within the membrane because of fluidity.
Cell Membrane
a selectively permeable, functional barrier that separates the interstitial fluid and the intracellular fluid
Functions of the Cell Membrane
To maintain the shape of the cell.
To control the passage of all substances in and out of the cell.
The cell membrane bears receptors that may be specific for particular hormones or enzymes.
Membrane proteins help to maintain the structural integrity of the cell.
5.high degree of specialization in some cells. Example: rod and cone cells present in the retina
Components of the Cell Membrane
phospholipids, integral proteins, paripheral proteins, carbohydrates, cholesterol
hydrophilic molecules
polar since can form bonds with water
hydrophobic molecules
nonpolar molecules that are not soluble in water
Properties of the bilayer
Asymmetric: The inner and outer layers have different phospholipids.
Non-uniform: has domains (e.g., lipid raft)
Concept of "fluid mosaic": Lipid bilayer allows free lateral movement of proteins and lipids.
5 types of molecules that pass through the lipid bilayer
Small, nonpolar molecules (O2, CO2) - can pass through the lipid bilayer
Small, polar molecules (H2O) - slower that small nonpolar but they can cross without the help of proteins
Large, nonpolar molecules (carbon rings) - These rings can pass through but it is also slow process
Large, polar molecules (ex: simple sugar - glucose) and 5. ions - charge of an ion, and the size and charge of large polar molecules, makes it too difficult to pass through the nonpolar region of the phospholipid membrane without help
cell membrane not permeable to:
Large uncharged polar molecules (e.g., glucose)
Ions ( e.g., Na⁺, K⁺)
Charged polar molecules (e.g., ATP, amino acids)
3 main factors that influence cell membrane fluidity
Temperature (When it's cold they are found closer together and when it's hot they move farther apart.)
Cholesterol - like a glue (holds the phospholipids together so that they don't separate too far, letting unwanted substances in, or compact too tightly, restricting movement across the membrane)
Saturated and unsaturated fatty acids (The double bonds create kinks in the chains, making it harder for the chains to pack tightly.)
Transition temperature (10°C-40°C)
Selective permeability
Transition temperature (10°C-40°C)
At low temperatures, the membrane changes to a gel-like solid state.
The membrane loses fluidity (becomes more rigid).
Cholesterol helps preserve membrane fluidity at low temperatures
Permeable to
Gases (CO2, CO, O2)
Small uncharged polar molecules (H2O, ethanol, urea)
Not permeable to
Large uncharged polar molecules (e.g., glucose)
Ions ( e.g., Na⁺, K⁺)
Charged polar molecules (e.g., ATP, amino acids)
membrane components
Phospholipids
Glycolipids
Cholesterol
Phospholipids
50% of membrane lipids
Amphipathic lipids with a phosphate head (polar) and 2 fatty-acid tails (nonpolar)
Form a double layer (lipid bilayer)
proportion of saturated versus unsaturated fatty acids determines membrane fluidity.
Glycolipids
approximately 2%
Lipid attached to carbohydrate, outward-facing
Play a role in cell-cell interactions
Determine ABO blood type
Participate in glycocalyx formation
Inflammatory response
Viral recognition of host cells
Cholesterol
Maintains membrane integrity without the need of a cell wall
Lipid with a sterol ring
Hydroxyl group interacts with water
Sterol ring embedded in the membrane
Decreases membrane fluidity
Increases the phase-transition range of temperature
Can also be packaged in lipoproteins for transport through the bloodstream
lipid rafts
lipids assembled in a defined patch in the cell membrane
planar lipid rafts
contain a family of proteins known as flotillins in addition to lipids and cholesterol. Flotillins - scaffolding proteins.
found in neurons and enriched in flotillin
Caveolar rafts
small flask-shaped invaginations of plasma membrane enriched with integral membrane proteins - caveolins.
Membrane protein functions
Transporters
Enzymes (catalysts)
receptors (for specific hormones or neurotransmitters)
Cell-surface identity markers
Cell-to-cell adhesion proteins
Attachments to the cytoskeleton
Broad Classification of Proteins
Integral membrane proteins
Peripheral proteins
Lipid-anchored proteins:
integral membrane proteins
at least partially embedded in the plasma membrane
Ion channels, proton pumps, G-protein coupled receptors
peripheral proteins
not embedded in the lipid bilayer;
Cell signaling and protein-protein interactions
The cytoskeletal proteins, ankyrin and spectrin, link to actin
lipid-anchored proteins
attached to a lipid
G-proteins
Carbohydrates that are present at the surface of the membrane
glycoproteins or glycolipids
glycocalyx
The carbohydrate layer on the external surface of cell membrane
cell coat
Functions of the glycocalyx
(a) Special adhesion molecules present in the layer enable the cell to adhere to specific types of cells, or to specific extracellular molecules.
(b) The layer contains antigens. Eg. Major histocompatibility antigens.
(c) Most molecules in the glycocalyx are negatively charged causing adjoining cells to repel one another. This force of repulsion maintains the space between cells.
Junctional Complexes (Unspecialised contacts)
Cell adhesion molecules (CAMs) or Intermediate link proteins.
Junctional Complexes (Specialised junctional structures)
Anchoring junctions, occluding tight junctions and communicating gap junctions
Hereditary spherocytosis
Autosomal-dominant inherited disease of the RBCs.
morphological change of the erythrocytes to spherocytes.
development of hemolytic anemia (due to membrane defect)
Dysfunctional membrane proteins interfere with the cell's ability to be flexible to travel from the arteries to the smaller capillaries.
difference in shape also makes the red blood cells more prone to rupture.
degraded in the spleen.
Hemolytic anemia
disorder in which red blood cells are destroyed faster than they can be made
Treatment of hereditary spherocytosis
iron folic acid supplementation, blood transfusion and total or partial splenectomy.
% of fluid in the body
60%
Intracellular fluid (ICF)
fluid within cells
contains large amounts of intracellular ions such as Mg, phosphates, and K.
Extracellular fluid (ECF)
fluid surrounds the cells;
contains Na, Cl, and bicarbonate ions
spaces of ECF
2 spaces:
Intravascular space - fluid that include blood (limited within the lumen of the heart and blood vessels)
Extravascular space — surrounds the tissues
Extravascular space can be
interstitial (space directly surrounding cells; the microenvironment of the cells)
transcellular (areas in the body where only small amounts of fluid can be accommodated; e.g., peritoneal cavity, joints, and the inside of the eyeball)
Dehydration
decreased volume of the extracellular fluid
Signs of dehydration in children
Sunken anterior fontanelle (Затонувший передний родничок)
Sunken eyes
Thirst
Reduced urine output (oliguria)
Lethargy (Вялость)
Skin pinch test - skin stays raised
Total body water volume
40 L
60% of body weight
intracellular fluid volume
25L
40% of body weight
extracellular fluid volume
15L
20% of body weight
intestitial fluid volume
12L
80% of ECF
plasma volume
3 L
20% of ECF
what does plasma contain that does not cross
plasma proteins (globulin)
concentrstion of Na K and Cl
Na: 145 mM(milimol) ECF
K: 120 ICF
CL: 110 ECF
why children more prone to severe dehydration
very hugh metabolic rate
prone to vomiting and diarrhea
GABA (gamma-aminobutyric acid)
A major inhibitory neurotransmitter
Which of the following is true about ion concentration difference between ICF and ECF?
Potassium concentration is higher in ECF.
Sodium and calcium are primarily intracellular ions.
Calcium concentration is higher in ICF.
Sodium concentration is higher in ECF.
Chloride concentration is higher in ICF.
Sodium concentration is higher in ECF.
Which channels are open at rest?
K+
What is an electrogenic pump?
A pump that creates an electrical potential across the membran
Which pump is electrogenic?
Na/K ATPase
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13922 | http://mathhints.com/pre-algebra/negative-numbers/ | Skip to content
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Negative Numbers and Absolute Value
| | |
--- |
| Negative Numbers on the Number Line | Multiplying and Dividing Negative Numbers |
| Absolute Value | Summary Table of Negative Number Operations |
| Adding and Subtracting Negative Numbers |
Negative Numbers on the Number Line
Negative numbers seem a little scary at first, but they really aren’t that bad. Let’s first re-introduce our number line:
Notice how the negative integers (the ones with the minus in front of them) are the same distance from zero ($ 0$) as the positive numbers — but they are to the left of the $ 0$. That’s all negative numbers are; they just go backward the same way that positive numbers go forward.
Absolute Value
The absolute value of a number (also called the modulus) is the distance from $ 0$, so it is always a positive number. It is written with two lines around the number, and it is simply the positive value of what’s inside the lines, whether the number is positive or negative.
It can get a little more complicated in algebra when we work with variables, or unknowns, but for now, here are examples to show how really simple the concept is:
$ \left| {-5} \right|=5\,\,\,\,\,\,\,\,\,\,\,\,\,\left| 5 \right|=5\,\,\,\,\,\,\,\,\,\,-\left| {-5} \right|=-5\,\,\,\,\,\,\,\,\,\,\,\left| 0 \right|=0$
Two numbers are called opposites if they are the same distance from $ 0$. For example, $ 2$ and $ -2$ are opposites.
Remember that numbers with a larger absolute value can actually be smaller when the numbers are negative – for example, $ -6<-5$, and, in the case of fractions, $ \displaystyle -\frac{3}{4}<-\frac{1}{2}$. So if we’re comparing negative numbers, it’s actually backwards compared to what we’re used to.
Let’s think of an example of how a negative might exist in real life. Let’s say your mom is having a little sister and the new baby is supposed to be born in $ 5$ months. In a weird way, your baby sister is $ =5$ months old. Next month, she will be $ -4$ months old, and so on. On the day she is born, she is $ 0$, and then she’ll start being positive months after that. Again, negative numbers are the same “distance” (distance is always positive) away from 0, but just in the opposite direction. If we wanted to know how long it is until she’s $ 9$ months old, we’d add the $ 5$ months before she’s $ 0$ to the $ 9$ months after she’s $ 0$ to get $ 14$ months. This is actually a subtraction problem, when you think about it: $ 9-\left( {-5} \right)=14$ – weird!. Can you guess that when we have two negatives together, it becomes a positive?
More advanced topics on Absolute Value are found in the Solving Absolute Value Equations and Inequalities section.
Adding and Subtracting Negative Numbers
Let’s talk about adding and subtracting negative numbers. For the number $ -2$, this means that we are two places to the left of the $ 0$ and four places to the left of the $ 2$. Remember that when we add, we count to the right, and when we subtract, or add a negative number, we count to the left. Thus, to add $ -2$ and $ 4$, we’d get $ 2$. There are some rules about adding and subtracting negative numbers that we’ll talk about shortly.
Let’s do some more addition and add some subtraction. As just mentioned, adding means moving to the right, and subtracting means moving to the left, as in the following graphic:
Remember that $ \boldsymbol {-3+5}$ is the same as $ \boldsymbol {5-3}$ and $ \boldsymbol {5+-6}$ is the same as $ \boldsymbol {5-6}$; it’s OK to memorize this! Also note that the sign of a number comes before it and many times there is an invisible plus sign before numbers, like when they are at the beginning.
Here are some rules for adding and subtracting negative numbers:
Adding two positive numbers yields a positive number. For example, $ 4+4=8$.
Adding two negative numbers yields a negative number. Add the two numbers and put negative in front of it. For example, $ -4+-4=-8$ . This is the same as $ -4-4=-8$.
Adding a negative number is the same thing as subtracting that number. For example, $ \displaystyle -4+-4\text{ }=-4-4=-8$.
When adding a positive number and negative number, subtract the absolute values of the two numbers (larger – smaller) and make the sign whichever has largestoriginal number without the signs (in other words, the largest absolute value). Some examples:
For $ -4+10$, since one is positive and the other one is negative, subtract $ 4$ from $ 10$ (the absolute values of the numbers), to get $ 6$. Since $ \left| {10} \right|>\left| {-4} \right|$ and the $ 10$ is positive, the answer is a positive $ 6$.
For $ -15+8$, subtract $ 8$ from $ 15$, to get $ 7$. Since $ \left| {-15} \right|>\left| 8 \right|$ and 15 has the – sign before it, add a “–” to the 7, to get –7.
If you have two minuses in a row, turn those into pluses. For example, if you have $ 4-\left( {-8} \right)=4- -8$, it’s the same as $ 4++\,8$, which is $ 4+8$, which is 12.
Don’t worry; once you do a lot of these, they will become second nature!
Multiplying and Dividing Negative Numbers
Now let’s talk about multiplying and dividing with negative numbers. This is actually easier, as there as fewer rules:
Multiplying or dividing two positive numbers results in a positive number. For example, $ 5\times 5=25$.
Multiplying or dividing two negative numbers results in a positive number. For example, $ \displaystyle -5\times -5=25$, and $ \displaystyle \frac{{-5}}{{-5}}=1$.
Multiplying or dividing a positive number with a negative number is always negative. For example, $ -5\times 5=-25$.
Summary Table of Negative Number Operations
Here is a table that sums all this up. Interesting how multiplication with negative numbers is easier than addition!
| | |
--- |
| Multiplication and Division | Addition and Subtraction |
| $ \begin{array}{c}\text{Positive }\times \,\,\text{Positive}=\text{Positive}\4\times 3=12\end{array}$ | $ \begin{array}{c}\text{Positive }+\,\,\text{Positive}=\text{Positive}\4+4=8\end{array}$ |
| $ \begin{array}{c}\text{Negative }\times \,\,\text{Negative}=\text{Positive}\-5\times -5=25\end{array}$ | $ \begin{array}{c}\text{Negative }+\,\,\text{Negative}=\text{Negative}\-4+-5=-9\end{array}$ |
| $ \begin{array}{c}\text{Negative }\times \,\,\text{Positive}=\text{Negative}\-4\times 4=-16\end{array}$ | $ \displaystyle \text{Negative }+\,\,\text{Positive }=\,\,\text{Depends}\,\,\text{:)}$ When you add a positive number and negative number, you’ll want to subtract the absolute values of the two numbers (larger – smaller) and make the sign the same sign of the largest absolute value of the numbers: $ -15+8=$ Subtract $ 8$ from $ 15$ to get $ 7$; since $ \left| {-15} \right|>\left| 8 \right|$ and $ 15$ has the – sign before it, you would add a “–” to the $ 7$ to get $ -7$. $ -8+15=$ Subtract $ 8$ from $ 15$ to get $ 7$; since $ \left| {15} \right|>\left| {-8} \right|$ and $ 15$ is positive, you get $ 7$. NOTE: Subtracting a negative number is the same thing as adding numbers. $ 5-\left( {-10} \right)=5- -10=5++10=5+10=15$ |
You really need to practice with negative numbers, because you will use them a lot all throughout your mathematical life; maybe not so much in your “real” life, but when you take math classes through Calculus in high school, you’ll still be using them. And, as always, be careful with fractions!
Learn these rules and practice, practice, practice!
On to Powers, Exponents, Radicals and Scientific Notation– you are ready!!
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13923 | https://www.mrbigler.com/Physics-1/Notes/06g_Tension.pdf | Tension Page: 320 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler Tension Unit: Forces in One Dimension NGSS Standards/MA Curriculum Frameworks (2016): N/A AP® Physics 1 Learning Objectives/Essential Knowledge (2024): 2.3.A, 2.3.A.1, 2.3.A.2, 2.3.A.3, 2.3.A.3.i, 2.3.A.3.ii, 2.3.A.3.iii, 2.3.A.3.iv, Mastery Objective(s): (Students will be able to…) • Set up and solve problems involving pulleys and ropes under tension. Success Criteria: • Expressions involving tension and acceleration are correct including the sign (direction). • Equations for all parts of the system are combined correctly algebraically. • Algebra is correct and rounding to an appropriate number of significant figures is reasonable. Language Objectives: • Explain how the sign of all of the forces in a pulley system relate to the direction that the system will move. Tier 2 Vocabulary: pulley, tension Labs, Activities & Demonstrations: • Atwood machine Notes: tension ( T F ,T ): the pulling force on a rope, string, chain, cable, etc. Tension is its own reaction force; tension always travels through the rope in both directions at once, and unless there are additional forces between one end of the rope and the other, the tension at every point along the rope is the same. The direction of tension is always along the rope. For example, in the following picture a blindfolded person pulls on a rope with a force of 100 N. The rope transmits the force to the scale, which transmits the force to the other rope and then to the wall. This causes a reaction force (also tension) of 100 N in the opposite direction. The scale attached to the rope measures 100 N, because that is the amount of force (tension) that is stretching the spring in the scale. Tension Page: 321 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler If we replace the brick wall with a person who is pulling with a force of 100 N, the blindfolded person has no idea whether the 100 N of resistance is coming from a brick wall or another person. Thus, the forces acting on the blindfolded person (and the scale) are the same. Of course, the scale doesn’t “know” either, so it still reads 100 N. A popular demonstration in physics classrooms is to set up the equivalent situation, using a scale with hanging weights on both sides: As you have undoubtedly realized, each rope pulls against the scale with a force of 5 N. The spring inside the scale pulls back with the same 5 N of force (in both directions), so the scale must read 5 N. Tension Page: 322 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler Pulleys pulley: a wheel used to change the direction of tension on a rope The tension remains the same in all parts of the rope. In the example at the right (with one pulley), it takes 100 N of force to lift a 100 N weight. The pulley changes the direction of the force, but the amount of force does not change. If the rope is pulled 10 cm, the weight is lifted by the same 10 cm. Up to this point, we have chosen a single direction (left/right or up/down) to be the positive direction. With pulleys, we usually define the positive and negative directions to follow the rope. In this example, we would most likely choose the positive direction to be the direction that the rope is pulled. Instead of saying that positive is upward for the weight and downward for the hook, we would usually say that the positive direction is counter-clockwise (), because that is the direction that the pulley will turn. Mechanical Advantage If we place a second pulley just above the weight that we want to lift, two things will happen when we pull on the rope: 1. As we pull on the rope, there is less rope between the two pulleys. This means the lower pulley will move upward. 2. The rope going around the lower pulley will be lifting the 100 N weight from both sides. This means each side will support half of the weight (50 N). Therefore, the tension in every part of the rope is 50 N, which means it takes half as much force to lift the weight. 3. The length of rope that is pulled is divided between the two sections that go around the lower pulley. This means that pulling the rope 20 cm will raise the weight half as much (10 cm). CP1 & honors (not AP®) Tension Page: 323 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler Notice that when the force is cut in half, the length of rope is doubled. The double pulley is effectively trading force for distance. Later, in the Energy, Work & Power unit starting on page 429, we will see that force times distance is work (change in energy). This means using half as much force but pulling the rope twice as much distance takes the same amount of energy to lift the weight. As you would expect, as we add more pulleys, the force needed is reduced and the distance increases. This reduction in force is called mechanical advantage. mechanical advantage: the ratio of the force applied by a machine divided by the force needed to operate it. The mechanical advantage of a pulley system is equal to the number of ropes supporting the hanging weight. It is therefore also equal to the number of pulleys. The mechanical advantage of the above system is 2:1 (or just 2). If we add a third pulley, we can see that there are now three sections of the rope that are lifting the 100 N weight. This means that each section is holding up ⅓ of the weight. This means that the tension in the rope is ⅓ of 100 N, or 33⅓ N, but we now need to pull three times as much rope to lift the weight the same distance. A two-pulley system has a mechanical advantage of 2, because it applies twice as much force to the weight as you need to apply to the rope. Similarly, a 3-pulley system has a mechanical advantage of 3, and so on. The mechanical advantage of any pulley system equals the number of ropes participating in the lifting. block and tackle: a system of two or more pulleys (which therefore has a mechanical advantage of 2 or more) that is used for lifting heavy loads. CP1 & honors (not AP®) Tension Page: 324 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler Atwood’s Machine Atwood’s machine is named for the English mathematician George Atwood. The machine is a device with a single pulley in which one weight, which is pulled down by gravity, is used to lift a second weight. Atwood invented the machine in 1784 to verify Isaac Newton’s equations of motion with constant acceleration. To illustrate how Atwood’s experiment works, consider the system to the right. To simplify the problem, we will assume that the rope and the pulley have negligible mass, and the pulley operates with negligible friction. Let us choose the positive direction as the direction that turns the pulley clockwise (). (We could have chosen either direction to be positive, but it makes intuitive sense to choose the direction that the system will move when we release the weights.) The force on the mass on the right is its weight, which is (10)( 10) 100N m = + = g . (We use a positive value for g because gravity is attempting to pull this weight in the positive direction.) The force on the mass on the left is (5)( 10) 50N m = − = − g . (We use a negative value for g because gravity is attempting to pull this weight in the positive direction.) The net force on the system is therefore 100 ( 50) 50N net = + − = = F F . The masses are connected by a rope, which means both masses will accelerate together. The total mass is 15 kg. Newton’s Second Law says: 2 m s 50 15 50 3.3 15 net m = + = = = + = a F a a F I.e., the system will accelerate at 2 m s 3.3 in the positive direction (clockwise). Atwood performed experiments with different masses and observed behavior that was consistent with both Newton’s second law, and with Newton’s equations of motion. Notice that the solution to finding acceleration in a problem involving Atwood’s machine is to simply find the net force, add up the total mass, and use net m = a F . Tension Page: 325 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler An important feature of Newton’s second law is that it can be applied to an entire system, or to any component of the system. For the Atwood’s machine pictured, we found that: Entire system: 2 m s 50N (5kg 10kg)(3.3 ) net m = + = + F a We can apply Newton’s second law to each block separately: 1, 1 2, 2 net net m m = = F a a F Because the blocks are connected via the same rope, the acceleration is the same for both blocks. This means that we can apply Newton’s second law to either of the blocks to determine the tension in the rope: Block #1: 1, ,1 1, 1 1 1 ) ) ] (5) ( 50) 16.6 66.6 N ( ( [ (5)(10) (3.3) net T g net T T T T m m m − = − = − − = = = − = F a g a F F F F F F F Block #2: (same calculation; yields the same result) We can do the same calculation for Block #2, with the same result for T F . (Remember that we chose the positive direction to be the direction that the system moves. This means the positive direction is up for block #1, but down for block #2.) 2, ,2 2, 2 2 2 ( ( ( [(10)(10) (3.3) 10 ) ) ] 10) 33.3 66 0 .6 N net g T net T T T T Q.E.D.
m m m − = = = = = = − − − F F F F F F F F a g a Tension Page: 326 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler Notice that the tension (66.6N) must be greater than the weight of the smaller block (50 N), and less than the weight of the larger block (100 N). (This should be obvious from the free-body diagrams.) Tension Page: 327 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler A variation of Atwood’s machine is to have one of the masses on a horizontal table (possibly on a cart to reduce friction). This means that the net force is only the action of gravity on the hanging mass. Consider the problem illustrated to the right. To simplify the problem, we will assume that the pulley has negligible mass, and that both the pulley and the cart are frictionless. The 10 kg for the mass on the left includes the mass of the cart. The forces on the two masses are: FBD for the cart: FBD for the hanging mass: 1,net T = F F 2,net g T = −F F F Gravity and the normal force cancel for the cart. The tensions cancel because they are equal (it’s the same rope) and are in opposite directions. This means that the only uncancelled force is the force of gravity on the 4 kg mass. This uncancelled force is the net force, which is N (4)(10) 40 net mg = = = F . The total mass is 10 + 4 = 14 kg. Now that we have the net force and the total mass, we can find the acceleration using Newton’s Second Law: 2 m s 40 14 40 2.86 14 net m = = = = a a a F Tension Page: 328 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler To find the tension, we can apply Newton’s second law to the cart: , (10)(2.86) 28.6N T T net cart cart F F a F m = = = Again, we can get the same result by applying Newton’s second law to the hanging mass: , (4)(2.86) (4)(10) 11.4 40 N 40 11.4 28.6 g T g T T T T net hang hang F F a F F F F F F m − = = = = − = − − = − Notice that the tension (28.6 N) must be less than the weight of the hanging block (40 N). (Again, this should be obvious from the free-body diagram for the hanging block.) Alternative Approach In most physics textbooks, the solution to Atwood’s machine problems is presented as a system of equations. The strategy is: • Draw a free-body diagram for each block. • Apply Newton’s 2nd Law to each block separately, giving 1 net F m a = for block 1 and 2 net g T a F F F m = − = , which becomes 2 2 net T a F m F m g = − = for block 2. • Set the two net F equations equal to each other, eliminate one of T F or a, and solve for the other. This is really just a different presentation of the same approach, but most students find it less intuitive. Tension Page: 329 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler Homework Problems 1. (M) For the pulley system shown at the right: a. (M) What is the mechanical advantage of the system? b. (M) How much force needs to be applied to the rope in order to lift the hanging weight? c. (M) If the 120 N weight is to be lifted 0.5 m, how far will the rope need to be pulled? CP1 & honors (not AP®) Tension Page: 330 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler 2. (M – AP® & honors; A – CP1) A block with a mass of m1 sitting on a frictionless horizontal table is connected to a hanging block of mass m2 by a string that passes over a pulley, as shown in the figure below. Assuming that friction, the mass of the string, and the mass of the pulley are negligible, derive expressions for the rate at which the blocks accelerate and the tension in the rope. (If you are not sure how to solve this problem, do #3 below and use the steps to guide your algebra.) Answer: = = + + 2 1 2 1 2 1 2 ; T a m m g m m m m g m F honors & AP® Tension Page: 331 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler 3. (S – AP® & honors; M – CP1) A block with a mass of 2.0 kg sitting on a frictionless horizontal table is connected to a hanging block of mass 6.0 kg by a string that passes over a pulley, as shown in the figure below. Assuming that friction, the mass of the string, and the mass of the pulley are negligible, at what rate do the blocks accelerate? What is the tension in the rope? (You must start with the equations in your Physics Reference Tables and show all of the steps of GUESS. You may not use the answer to question #2 above as a starting point unless you have already solved that problem.) Answer: 2 m s 7.5 ; 15N T a F = = 4. (M) Two masses, m1 = 3 kg and m2 = 8 kg, are connected by an ideal (massless) rope over an ideal pulley (massless and frictionless). What is the acceleration of the system? What is the tension in the rope? Answer: 2 m s 4.5 a = ; 43.6N T F = Tension Page: 332 Big Ideas Details Unit: Forces in One Dimension Use this space for summary and/or additional notes: Class Notes For Physics 1: Mechanics In Plain English Jeff Bigler 5. (S) A block with a mass of 3.0 kg sitting on a horizontal table is connected to a hanging block of mass 5.0 kg by a string that passes over a pulley, as shown in the figure below. The force of friction between the upper block and the table is 10 N. At what rate do the blocks accelerate? What is the tension in the rope? Answer: 2 m s 5 ; 25N T a F = = honors & AP® |
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13925 | https://www.analog.com/media/en/training-seminars/design-handbooks/Op-Amp-Applications/Section3.pdf | USING OP AMPS WITH DATA CONVERTERS H Op Amp History 1 Op Amp Basics 2 Specialty Amplifiers 3 Using Op Amps with Data Converters 1 Introduction 2 ADC/DAC Specifications 3 Driving ADC Inputs 4 Driving ADC/DAC Reference Inputs 5 Buffering DAC Outputs 4 Sensor Signal Conditioning 5 Analog Filters 6 Signal Amplifiers 7 Hardware and Housekeeping Techniques OP AMP APPLICATIONS USING OP AMPS WITH DATA CONVERTERS INTRODUCTION 3.1 CHAPTER 3: USING OP AMPS WITH DATA CONVERTERS Walt Kester, James Bryant, Paul Hendriks SECTION 3-1: INTRODUCTION Walt Kester This chapter of the book deals with data conversion and associated signal conditioning circuitry involving the use of op amps. Data conversion is a very broad topic, and this chapter will provide only enough background material so that the reader can make intelligent decisions regarding op amp selection. There is much more material on the subject available in the references (see References 1-5). Figure 3-1: Typical sampled data system showing potential amplifier applications Figure 3-1 above shows a generalized sampled data system and some possible applications of op amps. The analog input signal is first buffered and filtered before it is applied to the analog-to-digital converter (ADC). The buffer may or may not be required, depending upon the input structure of the ADC. For example, some ADCs (such as switched capacitor) generate transient currents at their inputs due to the internal conversion architecture, and these currents must be isolated from the signal source. A suitable buffer amplifier provides a low impedance drive and absorbs these currents. In some cases, an op amp is required to provide the appropriate gain and offset to match the signal to the input range of the ADC. ADC DAC fS DSP SAMPLED AND QUANTIZED WAVEFORM RECONSTRUCTED WAVEFORM LPF OR BPF LPF OR BPF VREF VREF ANALOG INPUT ANALOG OUTPUT 1 fS 1 fS OP AMP APPLICATIONS 3.2 Another key component in a sampled data system is the anti-aliasing filter which removes signals that fall outside the Nyquist bandwidth, fs/2. Normally this filter is a lowpass filter, but it can be a bandpass filter in certain undersampling applications. If the op amp buffer is required, it may be located before or after the filter, depending on system considerations. In fact, the filter itself may be an active one, in which case the buffering function can be performed by the actual output amplifier of the filter. More discussions regarding active filters can be found in Chapter 5 of this book. After the signal is buffered and filtered, it is applied to the ADC. The full scale input voltage range of the ADC is generally determined by a voltage reference, VREF. Some ADCs have this function on chip, while others require an external reference. If an external reference is required, its output may require buffering using an appropriate op amp. The reference input to the ADC may be connected to an internal switched capacitor network, causing transient currents to be generated at that node (similar to the analog input of such converters). Therefore some references may require a buffer to isolate these transient currents from the actual reference output. Other references may have internal buffers that are sufficient, and no additional buffering is needed in those cases. Figure 3-2: Data converter amplifier applications The output of the ADC is then processed digitally by an appropriate processor, shown in the diagram as a digital signal processor (DSP). DSPs are processors that are optimized to perform fast repetitive arithmetic, as required in digital filters or fast Fourier transform (FFT) algorithms. The DSP output then drives a digital-to-analog converter (DAC) which converts the digital signal back into an analog signal. The DAC analog output must be filtered to remove the unwanted image frequencies caused by the sampling process, and further buffering may be required to provide the proper signal amplitude and offset. The output filter is generally placed between the DAC and the buffer amplifier, but their positions can be reversed in certain applications. It is also possible to combine the filtering and buffering function if an active filter is used to condition the DAC output. Gain setting Level shifting Buffering ADC transients from signal source Buffering voltage reference outputs Buffering DAC outputs Active anti-aliasing filter before ADC Active anti-imaging filter after DAC USING OP AMPS WITH DATA CONVERTERS INTRODUCTION 3.3 Trends in Data Converters It is useful to examine a few general trends in data converters, to better understand any associated op amp requirements. Converter performance is first and foremost; and maintaining that performance in a system application is extremely important. In low frequency measurement applications (10Hz bandwidth signals or lower), sigma-delta ADCs with resolutions up to 24 bits are now quite common. These converters generally have automatic or factory calibration features to maintain required gain and offset accuracy. In higher frequency signal processing, ADCs must have wide dynamic range (low distortion and noise), high sampling frequencies, and generally excellent AC specifications. In addition to sheer performance, other characteristics such as low power, single supply operation, low cost, and small surface mount packages also drive the data conversion market. These requirements result in application problems because of reduced signal swings, increased sensitivity to noise, etc. In addition, many data converters are now produced on low-cost foundry CMOS processes which generally make on-chip amplifier design more difficult and therefore less likely to be incorporated on-chip. Figure 3-3: Some general trends in data converters As has been mentioned previously, the analog input to a CMOS ADC is usually connected directly to a switched-capacitor sample-and-hold (SHA), which generates transient currents that must be buffered from the signal source. On the other hand, data converters fabricated on Bi-CMOS or bipolar processes are more likely to have internal buffering, but generally have higher cost and power than their CMOS counterparts. It should be clear by now that selecting an appropriate op amp for a data converter application is highly dependent on the particular converter under consideration. Generalizations are difficult, but some meaningful guidelines can be followed. The most obvious requirement for a data converter buffer amplifier is that it not degrade the DC or AC performance of the converter. One might assume that a careful reading of the op amp datasheets would assist in the selection process: simply lay the data converter and the op amp datasheets side by side, and compare each critical performance specification. It is true that this method will provide some degree of success, however in order to perform an accurate comparison, the op amp must be specified under the exact operating conditions required by the data converter application. Such factors as gain, gain setting resistor values, source impedance, output load, input and output signal amplitude, input and output common-mode (CM) level, power supply voltage, etc., all affect op amp performance. Higher sampling rates, higher resolution, higher AC performance Single supply operation (e.g., +5V, +3V) Lower power Smaller input/output signal swings Maximize usage of low cost foundry CMOS processes Smaller packages Surface mount technology OP AMP APPLICATIONS 3.4 It is highly unlikely that even a well written op amp datasheet will provide an exact match to the operating conditions required in the data converter application. Extrapolation of specified performance to fit the exact operating conditions can give erroneous results. Also, the op amp may be subjected to transient currents from the data converter, and the corresponding effects on op amp performance are rarely found on datasheets. Converter datasheets themselves can be a good source for recommended op amps and other application circuits. However this information can become obsolete as newer op amps are introduced after the converter’s initial release. Analog Devices and other op amp manufacturers today have on-line websites featuring parametric search engines, which facilitate part selection (see Reference 1). For instance, the first search might be for minimum power supply voltage, e.g., +3V. The next search might be for bandwidth, and further searches on relevant specifications will narrow the selection of op amps even further. Figure 3-4 below summarizes the selection process. Figure 3-4: General amplifier selection criteria While not necessarily suitable for the final selection, this process can narrow the search to a manageable number of op amps whose individual datasheets can be retrieved, then reviewed in detail before final selection. From the discussion thus far, it should be obvious that in order to design a proper interface, an understanding of both op amps and data converters is required. References 2-6 provide background material on data converters. The next section of this chapter addresses key data converter performance specifications without going into the detailed operation of converters themselves. The remainder of the chapter shows a number of specific applications of op amps with various data converters. The amplifier should not degrade the performance of the ADC/DAC AC specifications are usually the most important z Noise z Bandwidth z Distortion Selection based on op amp data sheet specifications difficult due to varying conditions in actual application circuit with ADC/DAC: z Power supply voltage z Signal range (differential and common-mode) z Loading (static and dynamic) z Gain Parametric search engines may be useful ADC/DAC data sheets often recommend op amps (but may not include newly released products) USING OP AMPS WITH DATA CONVERTERS INTRODUCTION 3.5 REFERENCES: INTRODUCTION 1. ADI website, at 2. Walt Kester, Editor, Practical Analog Design Techniques, Analog Devices, 1995, ISBN: 0-916550-16-8. 3. Walt Kester, Editor, High Speed Design Techniques, Analog Devices, 1996, ISBN: 0-916550-17-6. 4. Chapters 3, 8, Walt Kester, Editor, Practical Design Techniques for Sensor Signal Conditioning, Analog Devices, 1999, ISBN: 0-916550-20-6. 5. Chapters 2, 3, 4, Walt Kester, Editor, Mixed-Signal and DSP Design Techniques, Analog Devices, 2000, ISBN: 0-916550-23-0. 6. Chapters 4, 5, Walt Kester, Editor, Linear Design Seminar, Analog Devices, 1995, ISBN: 0-916550-15-X. OP AMP APPLICATIONS 3.6 NOTES: USING OP AMPS WITH DATA CONVERTERS ADC/DAC SPECIFICATIONS 3.7 SECTION 3-2: ADC/DAC SPECIFICATIONS Walt Kester ADC and DAC Static Transfer Functions and DC Errors The most important thing to remember about both DACs and ADCs is that either the input or output is digital, and therefore the signal is quantized. That is, an N-bit word represents one of 2N possible states, and therefore an N-bit DAC (with a fixed reference) can have only 2N possible analog outputs, and an N-bit ADC can have only 2N possible digital outputs. The analog signals will generally be voltages or currents. The resolution of data converters may be expressed in several different ways: the weight of the Least Significant Bit (LSB), parts per million of full scale (ppm FS), millivolts (mV), etc. It is common that different devices (even from the same manufacturer) will be specified differently, so converter users must learn to translate between the different types of specifications if they are to compare devices successfully. The size of the least significant bit for various resolutions is shown in Figure 3-5 below. Figure 3-5: Quantization: the size of a least significant bit (LSB) As noted above (and obvious from this table), the LSB scaling for a given converter resolution can be expressed in various ways. While it is convenient to relate this to a full scale of 10V, as in the figure, other full scale levels can be easily extrapolated. Before we can consider op amp applications with data converters, it is necessary to consider the performance to be expected, and the specifications that are important when operating with data converters. The following sections will consider the definition of errors and specifications used for data converters. RESOLUTION N 2-bit 4-bit 6-bit 8-bit 10-bit 12-bit 14-bit 16-bit 18-bit 20-bit 22-bit 24-bit 2N 4 16 64 256 1,024 4,096 16,384 65,536 262,144 1,048,576 4,194,304 16,777,216 VOLTAGE (10V FS) 2.5 V 625 mV 156 mV 39.1 mV 9.77 mV (10 mV) 2.44 mV 610 µV 153 µV 38 µV 9.54 µV (10 µV) 2.38 µV 596 nV ppm FS 250,000 62,500 15,625 3,906 977 244 61 15 4 1 0.24 0.06 % FS 25 6.25 1.56 0.39 0.098 0.024 0.0061 0.0015 0.0004 0.0001 0.000024 0.000006 dB FS -12 -24 -36 -48 -60 -72 -84 -96 -108 -120 -132 -144 600nV is the Johnson Noise in a 10kHz BW of a 2.2kΩ Resistor @ 25°C Remember: 10-bits and 10V FS yields an LSB of 10mV, 1000ppm, or 0.1%.
All other values may be calculated by powers of 2. OP AMP APPLICATIONS 3.8 The first applications of data converters were in measurement and control, where the exact timing of the conversion was usually unimportant, and the data rate was slow. In such applications, the DC specifications of converters are important, but timing and AC specifications are not. Today many, if not most, converters are used in sampling and reconstruction systems where AC specifications are critical (and DC ones may not be). Figure 3-6: DAC transfer functions Figure 3-6 above shows the transfer characteristics for a 3-bit unipolar ideal and non-ideal DAC. In a DAC, both the input and output are quantized, and the graph consists of eight points— while it is reasonable to discuss a line through these points, it is critical to remember that the actual transfer characteristic is not a line, but a series of discrete points. Similarly, Figure 3-7 below shows the transfer characteristics for a 3-bit unipolar ideal and non-ideal ADC. Note that the input to an ADC is analog and is therefore not quantized, but its output is quantized. Figure 3-7: ADC transfer functions The ADC transfer characteristic therefore consists of eight horizontal steps (when considering the offset, gain and linearity of an ADC we consider the line joining the midpoints of these steps). DIGITAL INPUT ANALOG OUTPUT FS 000 001 010 011 100 101 110 111 DIGITAL INPUT FS 000 001 010 011 100 101 110 111 NON-MONOTONIC IDEAL NON-IDEAL ANALOG INPUT DIGITAL OUTPUT FS 000 001 010 011 100 101 110 111 QUANTIZATION UNCERTAINTY ANALOG INPUT FS 000 001 010 011 100 101 110 111 MISSING CODE IDEAL NON-IDEAL USING OP AMPS WITH DATA CONVERTERS ADC/DAC SPECIFICATIONS 3.9 The (ideal) ADC transitions take place at ½ LSB above zero, and thereafter every LSB, until 1½ LSB below analog full scale. Since the analog input to an ADC can take any value, but the digital output is quantized, there may be a difference of up to ½ LSB between the actual analog input and the exact value of the digital output. This is known as the quantization error or quantization uncertainty as shown in Figure 3-7. In AC (sampling) applications this quantization error gives rise to quantization noise which will be discussed shortly. The integral linearity error of a converter is analogous to the linearity error of an amplifier, and is defined as the maximum deviation of the actual transfer characteristic of the converter from a straight line. It is generally expressed as a percentage of full scale (but may be given in LSBs). There are two common ways of choosing the straight line: end point and best straight line. In the end point system, the deviation is measured from the straight line through the origin and the full scale point (after gain adjustment). This is the most useful integral linearity measurement for measurement and control applications of data converters (since error budgets depend on deviation from the ideal transfer characteristic, not from some arbitrary "best fit"), and is the one normally adopted by Analog Devices, Inc. The best straight line, however, does give a better prediction of distortion in AC applications, and also gives a lower value of "linearity error" on a datasheet. The best fit straight line is drawn through the transfer characteristic of the device using standard curve fitting techniques, and the maximum deviation is measured from this line. In general, the integral linearity error measured in this way is only 50% of the value measured by end point methods. This makes the method good for producing impressive datasheets, but it is less useful for error budget analysis. For AC applications, it is even better to specify distortion than DC linearity, so it is rarely necessary to use the best straight line method to define converter linearity. The other type of converter non-linearity is differential non-linearity (DNL). This relates to the linearity of the code transitions of the converter. In the ideal case, a change of 1 LSB in digital code corresponds to a change of exactly 1 LSB of analog signal. In a DAC, a change of 1 LSB in digital code produces exactly 1 LSB change of analog output, while in an ADC there should be exactly 1 LSB change of analog input to move from one digital transition to the next. Where the change in analog signal corresponding to 1 LSB digital change is more or less than 1 LSB, there is said to be a DNL error. The DNL error of a converter is normally defined as the maximum value of DNL to be found at any transition. If the DNL of a DAC is less than –1 LSB at any transition (Figure 3-6, again), the DAC is non-monotonic; i.e., its transfer characteristic contains one or more localized maxima or minima. A DNL greater than +1 LSB does not cause non-monotonicity, but is still undesirable. In many DAC applications (especially closed-loop systems where non-monotonicity can change negative feedback to positive feedback), it is critically important that DACs are monotonic. DAC monotonicity is often explicitly specified on datasheets, OP AMP APPLICATIONS 3.10 but if the DNL is guaranteed to be less than 1 LSB (i.e., |DNL| ≤ 1LSB) then the device must be monotonic, even without an explicit guarantee. ADCs can be non-monotonic, but a more common result of excess DNL in ADCs is missing codes (Figure 3-7, again). Missing codes (or non-monotonicity) in an ADC are as objectionable as non-monotonicity in a DAC. Again, they result from DNL > 1 LSB. Quantization Noise in Data Converters The only errors (DC or AC) associated with an ideal N-bit ADC are those related to the sampling and quantization processes. The maximum error an ideal ADC makes when digitizing a DC input signal is ±1/2LSB. Any AC signal applied to an ideal N-bit ADC will produce quantization noise whose rms value (measured over the Nyquist bandwidth, DC to fs/2) is approximately equal to the weight of the least significant bit (LSB), q, divided by √12. This assumes that the signal is at least a few LSBs in amplitude so that the ADC output always changes state. The quantization error signal from a linear ramp input is approximated as a sawtooth waveform with a peak-to-peak amplitude equal to q, and its rms value is therefore q/√12 (see Figure 3-8 below). Figure 3-8: Ideal N-bit ADC quantization noise It can be shown that the ratio of the rms value of a full scale sinewave to the rms value of the quantization noise (expressed in dB) is: SNR = 6.02N + 1.76dB, Eq. 3-1 where N is the number of bits in the ideal ADC. Note— this equation is only valid if the noise is measured over the entire Nyquist bandwidth from DC to fs/2. If the signal bandwidth, BW, is less than fs/2, then the SNR within the signal bandwidth BW is increased because the amount of quantization noise within the signal bandwidth is less. DIGITAL CODE OUTPUT ANALOG INPUT ERROR q = 1LSB SNR = 6.02N + 1.76dB + 10log FOR FS SINEWAVE fs 2•BW RMS ERROR = q/√12 USING OP AMPS WITH DATA CONVERTERS ADC/DAC SPECIFICATIONS 3.11 The correct expression for this condition is given by: SNR N dB fs BW = + + ⋅ 6 02 176 10 2 .
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log . Eq. 3-2 The above equation reflects the condition called oversampling, where the sampling frequency is higher than twice the signal bandwidth. The correction term is often called processing gain. Notice that for a given signal bandwidth, doubling the sampling frequency increases the SNR by 3dB. ADC Input-Referred Noise The internal circuits of an ADC produce a certain amount of wideband rms noise due to thermal and kT/C effects. This noise is present even for DC input signals, and accounts for the fact that the output of most wideband (or high resolution) ADCs is a distribution of codes, centered around the nominal DC input value, as is shown in Figure 3-9 below. Figure 3-9: Effect of ADC input-referred noise on "grounded input" histogram To measure its value, the input of the ADC is grounded, and a large number of output samples are collected and plotted as a histogram (sometimes referred to as a grounded-input histogram). Since the noise is approximately Gaussian, the standard deviation (σ) of the histogram is easily calculated, corresponding to the effective input rms noise. It is common practice to express this rms noise in terms of LSBs, although it can be expressed as an rms voltage. n n+1 n+2 n+3 n+4 n–1 n–2 n–3 n–4 NUMBER OF OCCURRENCES RMS NOISE (σ) P-P INPUT NOISE ≈6.6 × RMS NOISE OUTPUT CODE OP AMP APPLICATIONS 3.12 Calculating Op Amp Output Noise and Comparing it with ADC Input-Referred Noise In precision measurement applications utilizing 16- to 24-bit sigma-delta ADCs operating on low frequency (<20Hz, e.g.) signals, it is generally undesirable to use a drive amplifier in front of the ADC because of the increased noise due to the amplifier itself. If an op amp is required, however, the op amp output 1/f noise should be compared to the input-referred ADC noise. The 1/f noise is usually specified as a peak-to-peak value measured over the 0.1Hz to 10Hz bandwidth and referred to the op amp input (see Chapter 1 of this book). Op amps such as the OP177 and the AD707 (input voltage noise 350nV p-p) or the AD797 (input voltage noise 50nV p-p) are appropriate for high resolution measurement applications if required. Figure 3-10: Op amp noise model for a first-order circuit with resistive feedback The general model for calculating the referred-to-input (RTI) or referred-to-output (RTO) noise of an op amp is shown in Figure 3-10 above. This model shows all possible noise sources. The results using this model are relatively accurate, provided there is less than 1dB gain peaking in the closed loop frequency response. For higher frequency applications, 1/f noise can be neglected, because the dominant contributor is white noise. An example of a practical noise calculation is shown in Figure 3-11 (opposite). In this circuit, a wideband, low distortion amplifier (AD9632) drives a 12-bit, 25MSPS ADC (AD9225). The input voltage noise spectral density of the AD9632 (4.3nV/√Hz) dominates the op amp noise because of the low gain and the low values of the external feedback resistors. The noise at the output of the AD9632 is obtained by multiplying the input voltage noise spectral density by the noise gain of 2. To obtain the rms noise, the noise spectral density is multiplied by the equivalent noise bandwidth of 50MHz which is set by the single-pole lowpass filter placed between the op amp and the ADC input. CLOSED LOOP BW = fCL – + VN ∼ R2 R1 R3 IN– IN+ VOUT NOISE GAIN = 1 + R2 R1 NG = ∼ ∼ ∼ VN,R1 VN,R3 VN,R2 RTI NOISE = VN 2 + 4kTR3 + 4kTR1 R2 R1+R2 2 R2 R1+R2 2 + IN+ 2R32 + IN– 2 R1•R2 R1+R2 2 + 4kTR2 R1 R1+R2 2 R1 R1+R2 2 BW • RTO NOISE = NG • RTI NOISE 4kTR1 4kTR3 4kTR2 A B GAIN FROM "A" TO OUTPUT GAIN FROM "B" TO OUTPUT = – R2 R1 = BW = 1.57 fCL USING OP AMPS WITH DATA CONVERTERS ADC/DAC SPECIFICATIONS 3.13 Note that the closed-loop bandwidth of the AD9632 is 250MHz, and the input bandwidth of the AD9225 is 105MHz. With no filter, the output noise of the AD9632 would be integrated over the full 105MHz ADC input bandwidth. However, the sampling frequency of the ADC is 25MSPS, thereby implying that signals above 12.5MHz are not of interest, assuming Nyquist operation (as opposed to undersampling applications where the input signal can be greater than the Nyquist frequency, fs/2). The addition of this simple filter significantly reduces noise effects. Figure 3-11: Noise calculations for the AD9632 op amp driving the AD9225 12-bit, 25MSPS ADC The noise at the output of the lowpass filter is calculated as approximately 61µV rms which is less than half the effective input noise of the AD9225, 166µV rms. Without the filter, the noise from the op amp would be about 110µV rms (integrating over the full equivalent ADC input noise bandwidth of 1.57×105MHz = 165MHz). Figure 3-12: Proper positioning of the antialiasing filter will reduce the effects of op amp noise This serves to illustrate the general concept shown in Figure 3-12 above. In most high speed system applications a passive antialiasing filter (either lowpass for baseband sampling, or bandpass for undersampling) is required, and placing this filter between the op amp and the ADC will serve to reduce the noise due to the op amp. fFILTER AMP AMP LPF OR BPF LPF OR BPF ADC ADC fFILTER fs fs fCL fCL AMP NOISE INTEGRATED OVER fCL OR fADC, WHICHEVER IS LESS AMP NOISE INTEGRATED OVER FILTER BW, fFILTER fADC fADC IN GENERAL, fFILTER < << fADC < fCL fs 2 AD9225 12-bit ADC fs = 25MSPS Vni Vni AD9632 75Ω 103Ω 274Ω + -274Ω VREF VINA 50Ω 1.0kΩ 100pF 0.1uF Noise Bandwidth = 1.57 • 1 2π RC = 50MHz AD9632 OP AMP SPECIFICATIONS Input Voltage Noise = 4.3nV/√Hz Closed-Loop Bandwidth = 250MHz AD9225 ADC SPECIFICATIONS Effective Input Noise = 166µV rms Small Signal Input BW = 105MHz AD9632 Output Noise Spectral Density = 2 • 4.3nV/√Hz = 8.6nV/√Hz Vni = 8.6nV/√Hz • 50MHz = 61µV rms Vni = 8.6nV/√Hz • 50MHz = 61µV rms R = C = OP AMP APPLICATIONS 3.14 Quantifying and Measuring Converter Dynamic Performance There are various ways to characterize the AC performance of ADCs. In the early years of ADC technology (over 30 years ago) there was little standardization with respect to AC specifications, and measurement equipment and techniques were not well understood or available. Over nearly a 30 year period, manufacturers and customers have learned more about measuring the dynamic performance of converters, and the specifications shown in Figure 3-13 below represent the most popular ones used today. Figure 3-13: Quantifying ADC dynamic performance Practically all the specifications represent the converter’s performance in the frequency domain, and all are related to noise and distortion in one manner or another. Figure 3-14: Measuring ADC / DAC dynamic performance ADC outputs are analyzed using fast Fourier transform (FFT) techniques, and DAC outputs are analyzed using conventional analog spectrum analyzers, as shown in Figure 3-14 above. In the case of an ADC, the input signal is an analog sinewave, and in the case of a DAC, the input is a digital sinewave generated by a direct digital synthesis (DDS) system. SIGNAL GEN.
ADC BUFFER MEMORY M SAMPLES DSP M-POINT FFT DDS DAC ANALOG SPECTRUM ANALYZER fs 2 Resolution = fs / M f Signal-to-Noise-and-Distortion Ratio (SINAD, or S/N +D) Effective Number of Bits (ENOB) Signal-to-Noise Ratio (SNR) Analog Bandwidth (Full-Power, Small-Signal) Harmonic Distortion Worst Harmonic Total Harmonic Distortion (THD) Total Harmonic Distortion Plus Noise (THD + N) Spurious Free Dynamic Range (SFDR) Two-Tone Intermodulation Distortion Multi-tone Intermodulation Distortion USING OP AMPS WITH DATA CONVERTERS ADC/DAC SPECIFICATIONS 3.15 Signal-to-Noise-and-Distortion Ratio (SINAD), Signal-to-Noise Ratio (SNR), and Effective Number of Bits (ENOB) SINAD and SNR deserve careful attention (see Figure 3-15 below), because there is still some variation between ADC manufacturers as to their precise meaning. Signal-to-noise-and Distortion (SINAD, or S/N+D) is the ratio of the rms signal amplitude to the mean value of the root-sum-square (RSS) of all other spectral components, including harmonics, but excluding DC. SINAD is a good indication of the overall dynamic performance of an ADC as a function of input frequency, because it includes all components which make up noise (including thermal noise) and distortion. It is often plotted for various input amplitudes. SINAD is equal to THD+N if the bandwidth for the noise measurement is the same. Figure 3-15: SINAD, ENOB, and SNR The SINAD plot shows where the AC performance of the ADC degrades due to high-frequency distortion, and is usually plotted for frequencies well above the Nyquist frequency so that performance in undersampling applications can be evaluated. SINAD is often converted to effective-number-of-bits (ENOB) using the relationship for the theoretical SNR of an ideal N-bit ADC: SNR = 6.02N + 1.76dB. The equation is solved for N, and the value of SINAD is substituted for SNR: ENOB SINAD dB = −176 6 02 .
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. Eq. 3-3 Signal-to-noise ratio (SNR, or SNR-without-harmonics) is calculated the same as SINAD except that the signal harmonics are excluded from the calculation, leaving only the noise terms. In practice, it is only necessary to exclude the first 5 harmonics since they dominate. The SNR plot will degrade at high frequencies, but not as rapidly as SINAD because of the exclusion of the harmonic terms. SINAD (Signal-to-Noise-and-Distortion Ratio): z The ratio of the rms signal amplitude to the mean value of the root-sum-squares (RSS) of all other spectral components, including harmonics, but excluding DC.
ENOB (Effective Number of Bits): ENOB = SNR (Signal-to-Noise Ratio, or Signal-to-Noise Ratio Without Harmonics: z The ratio of the rms signal amplitude to the mean value of the root-sum-squares (RSS) of all other spectral components, excluding the first 5 harmonics and DC SINAD – 1.76dB 6.02 SINAD (Signal-to-Noise-and-Distortion Ratio): z The ratio of the rms signal amplitude to the mean value of the root-sum-squares (RSS) of all other spectral components, including harmonics, but excluding DC.
ENOB (Effective Number of Bits): ENOB = SNR (Signal-to-Noise Ratio, or Signal-to-Noise Ratio Without Harmonics: z The ratio of the rms signal amplitude to the mean value of the root-sum-squares (RSS) of all other spectral components, excluding the first 5 harmonics and DC SINAD – 1.76dB 6.02 SINAD – 1.76dB 6.02 OP AMP APPLICATIONS 3.16 Many current ADC datasheets somewhat loosely refer to SINAD as SNR, so the design engineer must be careful when interpreting these specifications. Figure 3-16: AD9220 12-bit, 10MSPS ADC SINAD and ENOB for various input signal levels A SINAD/ENOB plot for the AD9220 12-bit, 10MSPS ADC is shown in Figure 3-16. Analog Bandwidth The analog bandwidth of an ADC is that frequency at which the spectral output of the fundamental swept frequency (as determined by the FFT analysis) is reduced by 3dB. It may be specified for either a small signal bandwidth (SSBW), or a full scale signal (FPBW- full power bandwidth), so there can be a wide variation in specifications between manufacturers. Figure 3-17: ADC Gain (bandwidth) and ENOB versus frequency shows importance of ENOB specification Like an amplifier, the analog bandwidth specification of a converter does not imply that the ADC maintains good distortion performance up to its bandwidth frequency. In fact, 13.0 12.2 11.3 10.5 9.7 8.8 8.0 7.2 6.3 ENOB SINAD (dB) ANALOG INPUT FREQUENCY (MHz) ADC INPUT FREQUENCY (Hz) ENOB GAIN (FS INPUT) ENOB (FS INPUT) ENOB (-20dB INPUT) FPBW = 1MHz 10 100 1k 10k 100k 1M 10M GAIN USING OP AMPS WITH DATA CONVERTERS ADC/DAC SPECIFICATIONS 3.17 the SINAD (or ENOB) of most ADCs will begin to degrade considerably before the input frequency approaches the actual 3dB bandwidth frequency. Figure 3-17 (opposite) shows ENOB and full scale frequency response of an ADC with a FPBW of 1MHz, however, the ENOB begins to drop rapidly above 100kHz. Harmonic Distortion, Worst Harmonic, Total Harmonic Distortion (THD), Total Harmonic Distortion Plus Noise (THD + N) There are a number of ways to quantify the distortion of an ADC. An FFT analysis can be used to measure the amplitude of the various harmonics of a signal. The harmonics of the input signal can be distinguished from other distortion products by their location in the frequency spectrum. Figure 3-18 below shows a 7MHz input signal sampled at 20MSPS and the location of the first 9 harmonics. Aliased harmonics of fa fall at frequencies equal to |±Kfs±nfa|, where n is the order of the harmonic, and K = 0, 1, 2, 3,.... The second and third harmonics are generally the only ones specified on a datasheet because they tend to be the largest, although some datasheets may specify the value of the worst harmonic. Figure 3-18: Location of harmonic distortion products: Input signal = 7MHz, sampling rate = 20MSPS Harmonic distortion is normally specified in dBc (decibels below carrier), although in audio applications it may be specified as a percentage. Harmonic distortion is generally specified with an input signal near full scale (generally 0.5 to 1dB below full scale to avoid clipping), but it can be specified at any level. For signals much lower than full scale, other distortion products due to the DNL of the converter (not direct harmonics) may limit performance. Total harmonic distortion (THD) is the ratio of the rms value of the fundamental signal to the mean value of the root-sum-square of its harmonics (generally, only the first 5 are significant). THD of an ADC is also generally specified with the input signal close to full scale, although it can be specified at any level. RELATIVE AMPLITUDE FREQUENCY (MHz) fa 1 2 3 4 5 6 7 8 9 10 3 6 9 8 5 7 HARMONICS AT: |±Kfs±nfa| n = ORDER OF HARMONIC, K = 0, 1, 2, 3, . . .
2 4 = 7MHz fs = 20MSPS OP AMP APPLICATIONS 3.18 Total harmonic distortion plus noise (THD+ N) is the ratio of the rms value of the fundamental signal to the mean value of the root-sum-square of its harmonics plus all noise components (excluding DC). The bandwidth over which the noise is measured must be specified. In the case of an FFT, the bandwidth is DC to fs/2. (If the bandwidth of the measurement is DC to fs/2, THD+N is equal to SINAD. Spurious Free Dynamic Range (SFDR) Probably the most significant specification for an ADC used in a communications application is its spurious free dynamic range (SFDR). The SFDR specification is to ADCs what the third order intercept specification is to mixers and LNAs. Figure 3-19: Spurious free dynamic range (SFDR) SFDR of an ADC is defined as the ratio of the rms signal amplitude to the rms value of the peak spurious spectral content (measured over the entire first Nyquist zone, DC to fs/2). SFDR is generally plotted as a function of signal amplitude and may be expressed relative to the signal amplitude (dBc) or the ADC full scale (dBFS) as shown in Figure 3-19 above. For a signal near full scale, the peak spectral spur is generally determined by one of the first few harmonics of the fundamental. However, as the signal falls several dB below full scale, other spurs generally occur which are not direct harmonics of the input signal. This is because of the differential non-linearity of the ADC transfer function as discussed earlier. Therefore, SFDR considers all sources of distortion, regardless of their origin. FULL SCALE (FS) dB SFDR (dBc) fs 2 INPUT SIGNAL LEVEL (CARRIER) WORST SPUR LEVEL SFDR (dBFS) FREQUENCY USING OP AMPS WITH DATA CONVERTERS ADC/DAC SPECIFICATIONS 3.19 Two Tone Intermodulation Distortion (IMD) Two tone IMD is measured by applying two spectrally pure sinewaves to the ADC at frequencies f1 and f2, usually relatively close together. The amplitude of each tone is set slightly more than 6dB below full scale so that the ADC does not clip when the two tones add in-phase. Second and third-order product locations are shown in Figure 3-20 below. Notice that the second-order products fall at frequencies which can be removed by digital filters. However, the third-order products 2f2–f1 and 2f1–f2 are close to the original signals, and are almost impossible to filter. Unless otherwise specified, two-tone IMD refers to these third-order products. The value of the IMD product is expressed in dBc relative to the value of either of the two original tones, and not to their sum. Figure 3-20: Second and third-order intermodulation products for f1 = 5MHz, f2 = 6MHz Note, however, that if the two tones are close to fs/4, then the aliased third harmonics of the fundamentals can make the identification of the actual 2f2–f1 and 2f1–f2 products difficult. This is because the third harmonic of fs/4 is 3fs/4, and the alias occurs at fs – 3fs/4 = fs/4. Similarly, if the two tones are close to fs/3, the aliased second harmonics may interfere with the measurement. The same reasoning applies here; the second harmonic of fs/3 is 2 fs/3, and its alias occurs at fs – 2 fs/3 = fs/3. The concept of second and third-order intercept points is not valid for an ADC, because the distortion products don't vary predictably (as a function of signal amplitude). The ADC doesn't gradually begin to compress signals approaching full scale (there is no 1dB compression point); it acts as a hard limiter as soon as the signal exceeds the input range, producing extreme distortion due to clipping. Conversely, for signals much below full scale, the distortion floor remains relatively constant and is independent of signal level. Multi-tone SFDR is often measured in communications applications. The larger number of tones more closely simulates the wideband frequency spectrum of cellular telephone systems such as AMPS or GSM. High SFDR increases the receiver’s ability to capture small signals in the presence of large ones, and prevent the small signals from being masked by the intermodulation products of the larger ones. FREQUENCY: MHz 2 = SECOND ORDER IMD PRODUCTS 3 = THIRD ORDER IMD PRODUCTS NOTE: f1 = 5MHz, f2 = 6MHz f2 - f1 2f1 - f2 2f2 - f1 f1 f2 2f1 2f2 f2 + f1 2f1 + f2 3f1 2f2 + f1 3f2 2 3 3 2 3 3 1 4 5 6 7 10 11 12 15 16 17 18 OP AMP APPLICATIONS 3.20 REFERENCES: ADC/DAC SPECIFICATIONS 1. Walt Kester, Editor, Practical Analog Design Techniques, Analog Devices, 1995, ISBN: 0-916550-16-8. 2. Walt Kester, Editor, High Speed Design Techniques, Analog Devices, 1996, ISBN: 0-916550-17-6. 3. Chapters 3, 8, Walt Kester, Editor, Practical Design Techniques for Sensor Signal Conditioning, Analog Devices, 1999, ISBN: 0-916550-20-6. 4. Chapters 2, 3, 4, Walt Kester, Editor, Mixed-Signal and DSP Design Techniques, Analog Devices, 2000, ISBN: 0-916550-23-0. 5. Chapters 4, 5, Walt Kester, Editor, Linear Design Seminar, Analog Devices, 1995, ISBN: 0-916550-15-X. USING OP AMPS WITH DATA CONVERTERS DRIVING ADC INPUTS 3.21 SECTION 3-3: DRIVING ADC INPUTS Walt Kester, Paul Hendriks Introduction Op amps are often used as drivers for ADCs to provide the gain and level shifting required for the input signal to match the input range of the ADC. An op amp may be required because of the antialiasing filter impedance matching requirements. In some cases, the antialiasing filter may be an active filter and include op amps as part of the filter itself. Some ADCs also generate transient currents on their inputs due to the conversion process, and these must be isolated from the signal source with an op amp. This section examines these and other issues involved in driving high performance ADCs. To begin with, one shouldn't necessarily assume that a driver op amp is always required. Some converters have relatively benign inputs and are designed to interface directly to the signal source. There is practically no industry standardization regarding ADC input structures, and therefore each ADC must be carefully examined on its own merits before designing the input interface circuitry. In some applications, transformer drive may be preferable. Figure 3-21: General op amp requirements in ADC driver applications Assuming an op amp is required for one reason or another, the task of its selection is a critical one and not at all straightforward. Figure 3-21 above lists a few of the constraints and variables. The most important requirement is that the op amp should not significantly degrade the overall DC or AC performance of the ADC. At first glance, it would appear that a careful comparison of an op amp datasheet with the ADC datasheet would allow an appropriate choice. However, this is rarely the case. The problem is that the op amp performance specifications must be known for the exact circuit configuration used in the ADC driver circuit. Even a very complete datasheet is unlikely to provide all information required, due to the wide range of possible variables. Minimize degradation of ADC / DAC performance specifications Fast settling to ADC/DAC transient High bandwidth Low noise Low distortion Low power Note: Op amp performance must be measured under identical conditions as encountered in ADC / DAC application z Gain setting resistors z Input source impedance, output load impedance z Input / output signal voltage range z Input signal frequency z Input / output common-mode level z Power supply voltage (single or dual supply) z Transient loading OP AMP APPLICATIONS 3.22 Although the op amp and ADC datasheets should definitely be used as a guide in the selection process, it is unlikely that the overall performance of the op amp/ADC combination can be predicted accurately without actually prototyping the circuit, especially in high performance applications. Various tested application circuits are often recommended on either the op amp or the ADC datasheet, but these can become obsolete quickly as new op amps are released. In most cases, however, the ADC datasheet application section should be used as the primary source for tested interfaces. Op Amp Specifications Key to ADC Applications The two most popular applications for ADCs today are in either precision high-resolution measurements or in low distortion high speed systems. Precision measurement applications require ADCs of at least 16 bits of resolution, and sometimes up to 24 bits. Op amps used with these ADCs must be low noise and have excellent DC characteristics. In fact, high resolution measurement ADCs are often designed to interface directly with the transducer, eliminating the need for an op amp entirely. Figure 3-22: Key op amp specifications If op amps are required, it is generally relatively straightforward to select one based on well-understood DC specifications, as listed in Figure 3-22 above. It is much more difficult to provide a complete set of op amp AC specifications because they are highly dependent upon the application circuit. For example, Figure 3-23 (opposite) shows some key specifications taken from the table of specifications on the datasheet for the AD8057/AD8058 high speed, low distortion op amp (see Reference 1). Note that the specifications depend on the supply voltage, the signal level, output loading, etc. It should also be emphasized that it is customary to provide only typical AC specifications (as opposed to maximum and minimum values) for most op amps. In DC z Offset, offset drift z Input bias current z Open loop gain z Integral linearity z 1/f noise (voltage and current) AC (Highly application dependent!) z Wideband noise (voltage and current) z Small and Large Signal Bandwidth z Harmonic Distortion z Total Harmonic Distortion (THD) z Total Harmonic Distortion + Noise (THD + N) z Spurious Free Dynamic Range (SFDR) z Third Order Intermodulation Distortion z Third Order Intercept Point USING OP AMPS WITH DATA CONVERTERS DRIVING ADC INPUTS 3.23 addition, there are restrictions on the input and output CM signal ranges, which are especially important when operating on low voltage dual (or single) supplies. Figure 3-23: AD8057/AD8058 op amp key AC specifications, G = +1 Most op amp datasheets contain a section that provides supplemental performance data for various other conditions not explicitly specified in the primary specification tables. For instance, Figure 3-24 below shows the AD8057/AD8058 distortion as a function of frequency for G = +1 and VS = ±5V. Unless it is otherwise specified, the data represented by these curves should be considered typical (it is usually marked as such). Figure 3-24: AD8057/AD8058 op amp distortion versus frequency G = +1, RL = 150Ω, VS = ±5V Note however that the data in both Fig. 3-24 (and also the following Figure 3-25) is given for a DC load of 150Ω. This is a load presented to the op amp in the popular application of driving a source and load-terminated 75Ω cable. Distortion performance is generally better with lighter DC loads, such as 500Ω - 1000Ω (more typical of many ADC inputs), and this data may or may not be found on the datasheet. Input Common Mode Voltage Range Output Common Mode Voltage Range Input Voltage Noise Small Signal Bandwidth THD @ 5MHz, VO = 2V p-p, RL = 1kΩ THD @ 20MHz, VO = 2V p-p, RL = 1kΩ VS = ±5V VS = +5V –4.0V to +4.0V –4.0V to +4.0V 7nV/√Hz 325MHz – 85dBc – 62dBc +0.9V to +3.4V +0.9V to +4.1V 7nV/√Hz 300MHz – 75dBc – 54dBc OP AMP APPLICATIONS 3.24 On the other hand, Figure 3-25 below shows distortion as a function of output signal level for a frequencies of 5MHz and 20MHz. Figure 3-25: AD8057/AD8058 op amp distortion versus output voltage G = +1, RL = 150Ω, VS = ±5V Whether or not specifications such as those just described are complete enough to select an op amp for an ADC driver application depends upon the ability to match op amp specifications to the actually required ADC operating conditions. In many cases, these comparisons will at least narrow the op amp selection process. The following sections will examine a number of specific driver circuit examples using various types of ADCs, ranging from high resolution measurement to high-speed, low distortion applications. Driving High Resolution Sigma-Delta Measurement ADCs The AD77XX family of ADCs is optimized for high resolution (16–24 bits) low frequency transducer measurement applications. Details of operation can be found in Reference 2, and general characteristics of the family are listed in Figure 3-26 below. Figure 3-26: High resolution low frequency measurement ADCs Some members of this family, such as the AD7730, have a high impedance input buffer which isolates the analog inputs from switching transients generated in the front-end programmable gain amplifier (PGA) and the sigma-delta modulator. Therefore, no special THD @ THD @ Resolution: 16 - 24 bits Input signal bandwidth: <60Hz Effective sampling rate: <100Hz Generally Sigma-Delta architecture Designed to interface directly to sensors (< 1 kΩ) such as bridges with no external buffer amplifier (e.g., AD77XX - series) z On-chip PGA and high resolution ADC eliminates the need for external amplifier If buffer is used, it should be precision low noise (especially 1/f noise) z OP177 z AD707 z AD797 USING OP AMPS WITH DATA CONVERTERS DRIVING ADC INPUTS 3.25 precautions are required in driving the analog inputs. Other members of the AD77XX family, however, either do not have the input buffer, or if one is included on-chip, it can be switched either in or out under program control. Bypassing the buffer offers a slight improvement in noise performance. The equivalent input circuit of the AD77XX family without an input buffer is shown below in Figure 3-27. The input switch alternates between the 10pF sampling capacitor and ground. The 7kΩ internal resistance, RINT, is the on-resistance of the input multiplexer. The switching frequency is dependent on the frequency of the input clock and also the internal PGA gain. If the converter is working to an accuracy of 20-bits, the 10pF internal capacitor, CINT, must charge to 20-bit accuracy during the time the switch connects the capacitor to the input. This interval is one-half the period of the switching signal (it has a 50% duty cycle). The input RC time constant due to the 7kΩ resistor and the 10pF sampling capacitor is 70ns. If the charge is to achieve 20-bit accuracy, the capacitor must charge for at least 14 time constants, or 980ns. Any external resistance in series with the input will increase this time constant. Figure 3-27: Driving unbuffered AD77XX-series Σ∆ ADC inputs There are tables on the datasheets for the various AD77XX ADCs, which give the maximum allowable values of REXT in order to maintain a given level of accuracy. These tables should be consulted if the external source resistance is more than a few kΩ. Note that for instances where an external op amp buffer is found to be required with this type of converter, guidelines exist for best overall performance. This amplifier should be a precision low-noise bipolar-input type, such as the OP177, AD707, or the AD797. HIGH IMPEDANCE > 1GΩ SWITCHING FREQ DEPENDS ON fCLKIN AND GAIN CINT 10pF TYP REXT RINT 7kΩ ~ REXT Increases CINT Charge Time and May Result in Gain Error Charge Time Dependent on the Input Sampling Rate and Internal PGA Gain Setting Refer to Specific Data Sheet for Allowable Values of REXT to Maintain Desired Accuracy Some AD77XX-Series ADCs Have Internal Buffering Which Isolates Input from Switching Circuits AD77XX-Series (WITHOUT BUFFER) VSOURCE OP AMP APPLICATIONS 3.26 Op Amp Considerations for Multiplexed Data Acquisition Applications Multiplexing is a fundamental part of many data acquisition systems. Switches used in multiplexed data acquisition systems are generally CMOS-types shown in Figure 3-28 below. Utilizing P-Channel and N-Channel MOSFET switches in parallel minimizes the change of on-resistance (RON) as a function of signal voltage. On-resistance can vary from less than five to several hundred ohms depending upon the device. Variation in on-resistance as a function of signal level (often called RON-modulation) causes distortion if the multiplexer drives a load, therefore RON flatness is also an important specification. Because of the effects of non-zero RON and RON-modulation, multiplexer outputs should be isolated from the load with a suitable buffer op amp. A separate buffer is not required if the multiplexer drives a high input impedance, such as a PGA, SHA or ADC— but beware! Some SHAs and ADCs draw high frequency pulse current at their sampling rate and cannot tolerate being driven by an unbuffered multiplexer. Figure 3-28: Basic CMOS analog switch Key multiplexer specifications are switching time, on-resistance, on-resistance flatness, and off-channel isolation, and crosstalk. Multiplexer switching time ranges from less than 20ns to over 1µs, RON from less than 5Ω to several hundred ohms, and off-channel isolation from 50 to 90dB. A number of CMOS switches can be connected to form an analog multiplexer, as shown in Figure 3-29 (opposite). The number of input channels typically ranges from 4 to 16, and some multiplexers have internal channel-address decoding logic and registers, while with others, these functions must be performed externally. Unused multiplexer inputs must be grounded or severe loss of system accuracy may result. In applications requiring an op amp buffer, it should be noted that when the multiplexer changes channels it is possible to have a full scale step function into the op amp and the ADC which follows it. P-CH N-CH P-CH N-CH VIN VOUT +VS –VS –VS +VS OFF ON + – SIGNAL VOLTAGE RON NMOS PMOS CMOS USING OP AMPS WITH DATA CONVERTERS DRIVING ADC INPUTS 3.27 Op amp settling time must be fast enough so that conversion errors do not result. It is customary to specify the op amp settling time to 1 LSB, and the allowed time for this settling is generally the reciprocal of the sampling frequency. Figure 3-29: Typical multiplexed data acquisition system requires fast settling op amp buffer Driving Single-Supply Data Acquisition ADCs with Scaled Inputs The AD789X and AD76XX family of single supply SAR ADCs (as well as the AD974, AD976, and AD977) includes a thin film resistive attenuator and level shifter on the analog input to allow a variety of input range options, both bipolar and unipolar. Figure 3-30: Driving single-supply data acquisition ADCs with scaled inputs A simplified diagram of the input circuit of the AD7890-10 12-bit, 8-channel ADC is shown in Figure 3-30 above. This arrangement allows the converter to digitize a ±10V input while operating on a single +5V supply. ADC N-BITS ADDRESS REGISTER ADDRESS DECODER RON RON CHANNEL ADDRESS CHANNEL 1 CHANNEL M CLOCK OP AMP BUFFER CMOS SWITCHES fS fS SETTLING TIME TO 1 LSB < 1/fS CH 1 CH 2 +2.5V REFERENCE + _ ~ REFOUT/ REFIN VINX AGND RS 2kΩ R2 7.5kΩ R3 10kΩ 30kΩ +2.5V TO ADC REF CIRCUITS TO MUX, SHA, ETC.
±10V 0V TO +2.5V AD7890-10 12-BITS, 8-CHANNEL VS R1 R1, R2, R3 ARE RATIO-TRIMMED THIN FILM RESISTORS +5V OP AMP APPLICATIONS 3.28 Within the ADC, the R1/R2/R3 thin film network provides attenuation and level shifting to convert the ±10V input to a 0V to +2.5V signal that is digitized. This type of input requires no special drive circuitry, because R1 isolates the input from the actual converter circuitry that may generate transient currents due to the conversion process. Nevertheless, the external source resistance RS should be kept reasonably low, to prevent gain errors caused by the RS/R1 divider. Driving ADCs with Buffered Inputs Some ADCs have on-chip buffer amplifiers on their analog input to simplify the interface. This feature is most often found on ADCs designed on either bipolar or BiCMOS processes. Conversely, input amplifiers are rarely found on CMOS ADCs because of the inherent difficulty associated with amplifier design in CMOS. Figure 3-31: AD9042 ADC is designed to be driven directly from 50Ω source for best SFDR A typical input structure for an ADC with an input buffer is shown in Figure 3-31 above for the AD9042 12-bit, 41MSPS ADC. The effective input impedance is 250Ω, and an external 61.9Ω resistor in parallel with this internal 250Ω provides an effective input termination of 50Ω to the signal source. The circuit shows an AC coupled input. An internal reference voltage of 2.4V sets the input CM voltage of the AD9042. The input amplifier precedes the ADC sample-and-hold (SHA), and therefore isolates the input from any transients produced by the conversion process. The gain of the amplifier is set such that the input range of the ADC is 1Vp-p. In the case of a single-ended input structure, the input amplifier serves to convert the single-ended signal to a differential one, which allows fully differential circuit design techniques to be used throughout the remainder of the ADC. FROM 50Ω SOURCE RT 61.9Ω 250Ω 250Ω AD9042 INPUT = 1V p-p + -+2.4V REF.
50Ω VOFFSET TO SHA USING OP AMPS WITH DATA CONVERTERS DRIVING ADC INPUTS 3.29 Driving Buffered Differential Input ADCs Figure 3-32 below shows two possible input structures for an ADC with buffered differential inputs. The input CM voltage is set with an internal resistor divider network in Figure 3-32A (left), and by a voltage reference in Figure 3-32B (right). In single supply ADCs, the CM voltage is usually equal to one-half the power supply voltage, but some ADCs may use other values. Although the input buffers provide for a simplified interface, the fixed CM voltage may limit flexibility in some DC coupled applications. Figure 3-32: Simplified input circuit of typical buffered ADC with differential inputs It is worthwhile noting that differential ADC inputs offer several advantages over single ended ones. First, many signal sources in communications applications are differential, such as the output of a balanced mixer or an RF transformer. Thus an ADC that accepts differential inputs interfaces easily in such systems. Secondly, maintaining balanced differential transmission in the signal path and within the ADC itself often minimizes even-order distortion products as well as improving CM noise rejection. Third, (and somewhat more subtly), a differential ADC input swing of say, 2Vp-p requires only 1Vp-p from twin driving sources. On low voltage and single-supply systems, this lower absolute level of drive can often make a real difference in the dual amplifier driver distortion, due to practical headroom limitations. Given all of these points, it behooves the system engineer to operate a differential-capable ADC in the differential mode for best overall performance. This may be true even if a second amplifier need be added for the complementary drive signal, since dual op amps are only slightly more expensive than singles. GND AVDD VINB R1 R1 R2 R2 INPUT BUFFER SHA VINA INPUT BUFFER SHA VREF VINA VINB Input buffers typical on BiMOS and bipolar processes Difficult on CMOS Simplified input interface - no transient currents Fixed common-mode level may limit flexibility (A) (B) OP AMP APPLICATIONS 3.30 Driving CMOS ADCs with Switched Capacitor Inputs CMOS ADCs are quite popular because of their low power and low cost. The equivalent input circuit of a typical CMOS ADC using a differential sample-and-hold is shown in Figure 3-33 below. While the switches are shown in the track mode, note that they open/ close at the sampling frequency. The 16pF capacitors represent the effective capacitance of switches S1 and S2, plus the stray input capacitance. The CS capacitors (4pF) are the sampling capacitors, and the CH capacitors are the hold capacitors. Although the input circuit is completely differential, this ADC structure can be driven either single-ended or differentially. Optimum performance, however, is generally obtained using a differential transformer or differential op amp drive. Figure 3-33: Simplified input circuit for a typical switched capacitor CMOS sample-and-hold In the track mode, the differential input voltage is applied to the CS capacitors. When the circuit enters the hold mode, the voltage across the sampling capacitors is transferred to the CH hold capacitors and buffered by the amplifier A (the switches are controlled by the appropriate sampling clock phases). When the SHA returns to the track mode, the input source must charge or discharge the voltage stored on CS to a new input voltage. This action of charging and discharging CS, averaged over a period of time and for a given sampling frequency fs, makes the input impedance appear to have a benign resistive component. However, if this action is analyzed within a sampling period (1/fs), the input impedance is dynamic, and certain input drive source precautions should be observed. The resistive component to the input impedance can be computed by calculating the average charge that is drawn by CH from the input drive source. It can be shown that if CS is allowed to fully charge to the input voltage before switches S1 and S2 are opened that the average current into the input is the same as if there were a resistor equal to 1/(CSfS) connected between the inputs. Since CS is only a few picofarads, this resistive component is typically greater than several kΩ for an fS = 10MSPS. VINB + -SWITCHES SHOWN IN TRACK MODE A VINA CP 16pF CP 16pF S1 S2 S3 S4 S5 S7 S6 CS 4pF CS 4pF CH 4pF CH 4pF USING OP AMPS WITH DATA CONVERTERS DRIVING ADC INPUTS 3.31 Over a sampling period, the SHA's input impedance appears as a dynamic load. When the SHA returns to the track mode, the input source should ideally provide the charging current through the RON of switches S1 and S2 in an exponential manner. The requirement of exponential charging means that the source impedance should be both low and resistive up to and beyond the sampling frequency. The output impedance of an op amp can be modeled as a series inductor and resistor. When a capacitive load is switched onto the output of the op amp, the output will momentarily change due to its effective high frequency output impedance. As the output recovers, ringing may occur. To remedy this situation, a series resistor can be inserted between the op amp and the SHA input. The optimum value of this resistor is dependent on several factors including the sampling frequency and the op amp selected, but in most applications, a 25 to 100Ω resistor is optimum. Single Ended ADC Drive Circuits Although most CMOS ADC inputs are differential, they can be driven single-ended with some AC performance degradation. An important consideration in CMOS ADC applications are the input switching transients previously discussed. Figure 3-34: Single-ended input transient response of CMOS switched capacitor SHA (AD9225) For instance, the input switching transient on one of the inputs of the AD9225 12-bit, 25MSPS ADC is shown above in Figure 3-34. This data was taken driving the ADC with an equivalent 50Ω source impedance. During the sample-to-hold transition, the input signal is sampled when CS is disconnected from the source. Notice that during the hold-to-sample transition, CS is reconnected to the source for recharging. The transients consist of linear, nonlinear, and CM components at the sample rate. In addition to selecting an op amp with sufficient bandwidth and distortion performance, the output should settle to these transients during the sampling interval, 1/fs. The general Hold-to-Sample Mode Transition Sample-to-Hold Mode Transition Hold-to-Sample Mode Transition - CS Returned to Source for “recharging”. Transient Consists of Linear, Nonlinear, and Common-Mode Components at Sample Rate . Sample-to-Hold Mode Transition - Input Signal Sampled when CS is disconnected from Source. OP AMP APPLICATIONS 3.32 circuit shown below in Figure 3-35 is typical for this type of single-ended op amp ADC driver application. In this circuit, series resistor RS has a dual purpose. Typically chosen in the range of 25-100Ω, it limits the peak transient current from the driving op amp. Importantly, it also decouples the driver from the ADC input capacitance (and possible phase margin loss). Figure 3-35: Optimizing single-ended switched capacitor ADC input drive circuit Another feature of the circuit are the dual networks of RS and CF. Matching both the DC and AC the source impedance for the ADC's VINA and VINB inputs ensures symmetrical settling of CM transients, for optimum noise and distortion performance. At both inputs, the CF shunt capacitor acts as a charge reservoir and steers the CM transients to ground. In addition to the buffering of transients, RS and CF also form a lowpass filter for VIN, which limits the output noise of the drive amplifier into the ADC input VINA. The exact values for RS and CF are generally optimized within the circuit, and the recommended values given on the ADC datasheet. The ADC data sheet information should also be consulted for the recommended drive op amp for best performance. To enable best correlation of performance between environments, an ADC evaluation board should used (if available). This will ensure confidence when the ADC data sheet circuit performance is duplicated. ADI makes evaluation boards available for many of their ADC and DAC devices (plus of course, op amps), and general information on them is contained in Chapter 7 of this book. VINA VINB VREF AD922X + – RS RS CF CF 10µF 0.1µF VIN USING OP AMPS WITH DATA CONVERTERS DRIVING ADC INPUTS 3.33 Op Amp Gain Setting and Level Shifting in DC Coupled Applications In DC coupled applications, the drive amplifier must provide the required gain and offset voltage, to match the signal to the input voltage range of the ADC. Figure 3-36 below summarizes various op amp gain and level shifting options. The circuit of Figure 3-36A operates in the non-inverting mode, and uses a low impedance reference voltage, VREF, to offset the output. Gain and offset interact according to the equation: VOUT = [1 + (R2/R1)] • VIN – [(R2/R1) • VREF] Eq. 3-4 The circuit in Figure 3-36B operates in the inverting mode, and the signal gain is independent of the offset. The disadvantage of this circuit is that the addition of R3 increases the noise gain, and hence the sensitivity to the op amp input offset voltage and noise. The input/output equation is given by: VOUT = – (R2/R1) • VIN – (R2/R3) • VREF Eq. 3-5 Figure 3-36: Op amp gain and level shifting circuits The circuit in Figure 3-36C also operates in the inverting mode, and the offset voltage VREF is applied to the non-inverting input without noise gain penalty. This circuit is also attractive for single-supply applications (VREF > 0). The input/output equation is given by: VOUT = – (R2/R1) • VIN + [(R4/(R3+R4))(1 +(R2/R1)] • VREF Eq. 3-6 Note that the circuit of Fig. 3-36A is sensitive to the impedance of VREF, unlike the counterparts in B and C. This is due to the fact that the signal current flows into/from VREF, due to VIN operating the op amp over its CM range. In the other two circuits the CM voltages are fixed, and no signal current flows in VREF. VIN VREF VOUT R R • VIN = − 2 1 R R3 + R4 R R • VREF + + 4 1 2 1 NOISE GAIN R R = + 1 2 1 VOUT R R • VIN = − 2 1 R R3 + R4 R R • VREF + + 4 1 2 1 VOUT R R • VIN = − 2 1 VOUT R R • VIN = − 2 1 R R3 + R4 R R • VREF + + 4 1 2 1 R R3 + R4 R R • VREF + + 4 1 2 1 NOISE GAIN R R = + 1 2 1 NOISE GAIN R R = + 1 2 1 + + + VIN VIN VREF VREF ---R1 R2 R3 R1 R2 R4 R3 C B A R1 R2 NOISE GAIN R R = + 1 2 1 R2 R1 VREF − • VOUT • VIN = + 1 R2 R1 NOISE GAIN R R = + 1 2 1 R2 R1 VREF − • R2 R1 R2 R1 VREF − • VOUT • VIN = + 1 R2 R1 VOUT • VIN = + 1 R2 R1 R2 R1 NOISE GAIN R R R = + 1 2 1 3 || R2 R3 VREF − • VOUT = • VIN R2 R1 − NOISE GAIN R R R = + 1 2 1 3 || R2 R3 VREF − • VOUT = • VIN R2 R1 − R2 R3 VREF − • R2 R3 R2 R3 VREF − • VOUT = • VIN R2 R1 − OP AMP APPLICATIONS 3.34 A DC coupled single-ended op amp driver for the AD9225 12-bit, 25MSPS ADC is shown in Figure 3-37 below. This circuit interfaces a ±2V input signal to the single-supply ADC, and provides transient current isolation. The ADC input voltage range is 0 to +4V, and a dual supply op amp is required, since the ADC minimum input is 0V. The non-inverting input of the AD8057 is biased at +1V, which sets the output CM voltage at VINA to +2V for a bipolar input signal source. Note that the VINA and VINB source impedances are matched for better CM transient cancellation. The 100pF capacitors act as small charge reservoirs for the input transient currents, and also form lowpass noise filters with the 33Ω series resistors. Figure 3-37: DC coupled single-ended level shifter and driver for the AD9225 12-bit, 25MSPS CMOS ADC A similar level shifter and drive circuit is shown in Figure 3-38 below, operating on a single +5V supply. In this circuit the bipolar ±1V input signal is interfaced to the input of the ADC whose span is set for 2V about a +2.5V CM voltage. The AD8041 rail-to-rail output op amp is used. The +1.25V input CM voltage for the AD8041 is developed by a voltage divider from the external AD780 2.5V reference. Figure 3-38: DC coupled single-ended, single-supply ADC driver / level shifter using external reference Note that single-supply circuits of this type must observe op amp input and output CM voltage restrictions, to prevent clipping and excess distortion. 33Ω 52.3Ω +5V AD9225 VINA VINB + -AD8057 +5V 1kΩ 0.1µF 0.1µF 1kΩ 33Ω 1kΩ 1kΩ 1kΩ 10µF 0.1µF + -5V 100pF 100pF INPUT ±2V +1.0V +2.0V - /+2V +2.0V VREF 10µF +5V 52.3Ω INPUT + +5V AD922X VINA VINB + -AD8041 +5V 0.1µ 1kΩ 0.1µF 1kΩ 33Ω AD780 2.5V REF.
1kΩ 1kΩ 10µF 33Ω + 100pF 100pF ± 1V +2.5V - /+ 1V +1.25V +2.5V USING OP AMPS WITH DATA CONVERTERS DRIVING ADC INPUTS 3.35 Drivers for Differential Input ADCs Most high performance ADCs are now being designed with differential inputs. A fully differential ADC design offers the advantages of good CM rejection, reduction in second-order distortion products, and simplified DC trim algorithms. Although they can be driven single-ended as previously described, a fully differential driver usually optimizes overall performance. Figure 3-39: Differential input ADCs offer performance advantages Waveforms at the two inputs of the AD9225 12-bit, 25MSPS CMOS ADC are shown in Figure 3-40A, designated as VINA and VINB. The balanced source impedance is 50Ω, and the sampling frequency is set for 25MSPS. The diagram clearly shows the switching transients due to the internal ADC switched capacitor sample-and-hold. Figure 3-40B shows the difference between the two waveforms, VINA − VINB. Figure 3-40: Differential input transient response of CMOS switched capacitor SHA (AD9225) Note that the resulting differential charge transients are symmetrical about mid-scale, and that there is a distinct linear component to them. This shows the reduction in the CM transients, and also leads to better distortion performance than would be achievable with a single-ended input. High common-mode noise rejection Flexible input common-mode voltage levels Reduced input signal swings helps in low voltage, single-supply applications Reduced second-order distortion products Simplified DC trim algorithms because of internal matching Requires high performance differential driver VINA VINB VINA-VINB (A) (B) Differential charge transient is symmetrical around mid-scale and dominated by linear component Common-mode transients cancel with equal source impedance OP AMP APPLICATIONS 3.36 Transformer coupling into a differential input ADC provides excellent CM rejection and low distortion if performance to DC is not required. Figure 3-41 shows a typical circuit. The transformer is a Mini-Circuits RF transformer, model #T4-6T which has an impedance ratio of 4 (turns ratio of 2). The schematic assumes that the signal source has a 50Ω source impedance. The 1:4 impedance ratio requires the 200Ω secondary termination for optimum power transfer and low VSWR. The Mini-Circuits T4-6T has a 1dB bandwidth from 100kHz to 100MHz. The center tap of the transformer provides a convenient means of level shifting the input signal to the optimum CM voltage of the ADC. The AD922X CML (common-mode level) pin is used to provide the +2.5 CM voltage. Figure 3-41: Transformer coupling into AD922x ADC Transformers with other turns ratios may also be selected to optimize the performance for a given application. For example, a given input signal source or amplifier may realize an improvement in distortion performance at reduced output power levels and signal swings. Hence, selecting a transformer with a higher impedance ratio (i.e. Mini-Circuits #T16-6T with a 1:16 impedance ratio, turns ratio 1:4) effectively "steps up" the signal level thus reducing the driving requirements of the signal source. Note the 33Ω series resistors inserted between the transformer secondary and the ADC input. These values were specifically selected to optimize both the SFDR and the SNR performance of the ADC. They also provide isolation from transients at the ADC inputs. Transients currents are approximately equal on the VINA and VINB inputs, so they are isolated from the primary winding of the transformer by the transformer's CM rejection. Transformer coupling using a CM voltage of +2.5V provides the maximum SFDR when driving the AD922X-series. By driving the ADC differentially, even-order harmonics are reduced compared with the single-ended circuit. +5V AD922X VINA VINB 0.1µF +2.5V 33Ω CML 33Ω 200Ω 1:2 Turns Ratio RF TRANSFORMER: MINI-CIRCUITS T4-6T 2Vp-p 49.9Ω USING OP AMPS WITH DATA CONVERTERS DRIVING ADC INPUTS 3.37 Driving ADCs with Differential Amplifiers There are many applications where differential input ADCs cannot be driven with transformers because the frequency response must extend to DC. In these cases, op amps can be used to implement the differential drivers. Figure 3-42 shows how the dual AD8058 op amp can be connected to convert a single-ended bipolar signal to a differential one suitable for driving the AD922X family of ADCs. The input range of the ADC is set for a 2Vp-p signal on each input (4V span), and a CM voltage of +2V. The A1 amplifier is configured as a non-inverting op amp. The 1kΩ divider resistors level shift the +/-1V input signal to +1V +/–0.5V at the non-inverting input of A1. The output of A1 is therefore +2V +/–1V. Figure 3-42: Op amp single-ended to differential DC-coupled driver with level shifting. The A2 op amp inverts the input signal, and the 1kΩ divider resistors establish a +1V CM voltage on its non-inverting input. The output of A2 is therefore +2V –/+1V. This circuit provides good matching between the two op amps because they are duals on the same die and are both operated at the same noise gain of 2. However, the input voltage noise of the AD8058 is 20nV/√Hz, and this appears as 40nV/√Hz at the output of both A1 and A2 thereby introducing possible SNR degradation in some applications. In the circuit of Fig. 3-42, this is mitigated somewhat by the 100pF input capacitors which not only reduce the input noise but absorb some of the transient currents. It should be noted that because the input CM voltage of A1 can go as low as +0.5V, dual supplies must be used for the op amps. +5V V IN B V IN B V INA V INA + -AD8058 33Ω 33Ω +1.0V 10 µF 10 µF 0.1µF + 1 Ω k 1 Ω k VREF 100pF 100pF 1 Ω k 1 Ω k 1 Ω k 1 Ω k 1 Ω k 1 Ω k 1 Ω k 1 Ω k 1 Ω k 1 Ω k + -AD8058 +2.0V - /+ 1V +2.0V 1 Ω k 1 Ω k 33Ω 33Ω INPUT AD922X +1V +/- 0.5V Set for 4 Volt p-p Differential Input Span +5V -5V A1 A2 1/2 1/2 ±1V +2.0V +/- 1V OP AMP APPLICATIONS 3.38 A block diagram of the AD813X family of fully differential amplifiers optimized for ADC driving is shown in Figure 3-43 (see References 3-5). Figure 3-43A shows the details of the internal circuit, and Figure 3-43B shows the equivalent circuit. The gain is set by the external RF and RG resistors, and the CM voltage is set by the voltage on the VOCM pin. The internal CM feedback forces the VOUT+ and VOUT- outputs to be balanced, i.e., the signals at the two outputs are always equal in amplitude but 180° out of phase per the equation, VOCM = ( VOUT+ + VOUT- ) / 2 . Eq. 3-7 The circuit can be used with either a differential or a single-ended input, and the voltage gain is equal to the ratio of RF to RG. Figure 3-43: AD813X differential ADC driver functional diagram and equivalent circuit The AD8138 has a 3dB small-signal bandwidth of 320MHz (G = +1) and is designed to give low harmonic distortion as an ADC driver. The circuit provides excellent output gain and phase matching, and the balanced structure suppresses even-order harmonics. Figure 3-44 (opposite) shows the AD8138 driving the AD9203 10-bit, 40MSPS ADC (see Reference 6). This entire circuit operates on a single +3V supply. A 1Vp-p bipolar single-ended input signal produces a 1Vp-p differential signal at the output of the AD8138, centered around a CM voltage of +1.5V (mid-supply). Each of the differential inputs of the AD8138 swing between +0.625V and +0.875V, and each output swings between +1.25V and +1.75V. These voltages fall within the allowable input and output CM voltage range of the AD8138 operating on a single +3V supply. ~ RF RF RG RG VOUT– VOUT+ + – GAIN = RF RG VIN+ VIN– EQUIVALENT CIRCUIT: VOCM + – + – – + + – RF RF RG RG VIN+ VIN– VOUT+ VOUT– VOCM V+ V– (A) (B) USING OP AMPS WITH DATA CONVERTERS DRIVING ADC INPUTS 3.39 The circuit as shown operates on a 1Vp-p single-ended bipolar input signal, and the input span of the AD9203 ADC is set for 1Vp-p differential. If the signal input amplitude is increased to 2Vp-p, the span of the AD9203 must be set for 2Vp-p differential. Under these conditions, each of the AD8138 inputs must swing between +0.5V and +1V, and each of the outputs between +1V and +2V. Figure 3-44: AD8138 driving AD9203 10-bit, 40MSPS ADC As shown in Figure 3-45 below, increasing the amplitude in this manner offers a slight improvement in low frequency SINAD due to the improvement in low frequency SNR. Figure 3-45: SINAD and ENOB for AD9203 12-bit, 40MSPS ADC driven by AD8138 differential amplifier But at the same time however, a degradation occurs in the high frequency SINAD because of the larger distortion due to the larger signal swings. 9.8 9.5 9.2 8.8 8.5 8.2 7.9 7.5 ENOB INPUT FREQUENCY (MHz) SINAD (dBc) + – AD9203 AIN-AIN+ VIN ±0.5V 49.9Ω 499Ω 499Ω 499Ω 523Ω 10kΩ 10kΩ 49.9Ω 49.9Ω +1.5V VOCM AD8138 0.1µF 0.1µF 20pF 0.1µF 20pF +0.75V +/ - 0.125V +1.5V - / + 0.25V +1.5V +/ - 0.25V Set for 1V p-p Differential Input Span +3V OP AMP APPLICATIONS 3.40 Overvoltage Considerations The input structures of most high performance ADCs are sensitive to overvoltage conditions because of the small geometry devices used in the designs. Although ADC inputs generally have ESD protection diodes connected from the analog input to each supply rail, these diodes are not designed to handle the large currents that can be generated from typical op amp drivers. Two good rules of thumb are to (1) limit the analog input voltage to no more than 0.3V above or below the positive and negative supply voltages, respectively, and (2) limit the analog input current to 5mA maximum in overvoltage conditions. Figure 3-46: ADC input overvoltage protection circuits Several typical configurations for the drive amp/ADC interface are shown in Figure 3-46 above. In Figure 3-46A, the ADC requires no additional input protection because both the op amp and the ADC are driven directly from the same supply voltages. While the RS resistor is not required for overvoltage protection, it does serve to isolate the capacitive input of the ADC from the output of the op amp. Figure 3-46B shows a dual supply op amp driving a single supply ADC, with the +5V supply is shared between the two devices. The diode protects the input of the ADC in case the output of the op amp is driven below ground. A Schottky diode is used because of its low forward voltage drop and its low capacitance. The RS resistor is split into two equal resistors, and they are chosen to limit the ADC input fault current to 5mA maximum. Note that the RS resistor in conjunction with the ADC input capacitance forms a lowpass filter. If RS is made too large, the input bandwidth may be restricted. Figure 3-46C shows the condition where the op amp and the ADC are driven from separate supplies. Two Schottky diodes are required to protect the ADC under all power supply and signal conditions. As in 3-46A, the RS/2 resistors limit ADC fault current. RS/2 +5V +5V AMP -5V ADC RS/2 (C) Separate Supplies RS/2 +5V AMP -5V ADC RS/2 (B) Same +5V Supply AMP ADC +5V RS (A) Single or Dual, Same Supplies (Or –5V) USING OP AMPS WITH DATA CONVERTERS DRIVING ADC INPUTS 3.41 REFERENCES: DRIVING ADC INPUTS 1. Data sheet for AD8057/AD8058 Low Cost, High Performance Voltage Feedback, 325 MHz Amplifiers, 2. Chapter 8 of Walt Kester, Editor, Practical Design Techniques for Sensor Signal Conditioning, Analog Devices, 1999, ISBN: 0-916550-20-6. 3. Data sheet for AD8131 Low-Cost, High-Speed Differential Driver, 4. Data sheet for AD8132 Low-Cost, High-Speed Differential Amplifier, 5. Data sheet for AD8138 Low Distortion Differential ADC Driver, 6. Data Sheet for AD9203 10-Bit, 40 MSPS, 3V, 74mW A/D Converter, 7. Chapters 3-6, Walt Kester, Editor, Practical Analog Design Techniques, Analog Devices, 1995, ISBN: 0-916550-16-8. 8. Chapters 4, 5, Walt Kester, Editor, High Speed Design Techniques, Analog Devices, 1996, ISBN: 0-916550-17-6. 9. Chapters 3, 4, Walt Kester, Editor, Mixed-Signal and DSP Design Techniques, Analog Devices, 2000, ISBN: 0-916550-23-0. 10. Chapters 4, 5, Walt Kester, Editor, Linear Design Seminar, Analog Devices, 1995, ISBN: 0-916550-15-X. OP AMP APPLICATIONS 3.42 NOTES: USING OP AMPS WITH DATA CONVERTERS DRIVING ADC/DAC REFERENCE INPUTS 3.43 SECTION 4-4: DRIVING ADC/DAC REFERENCE INPUTS Walt Jung, Walt Kester It might seem odd to include a section on voltage references in a book devoted primarily to op amp applications, but the relevance will shortly become obvious. Unfortunately, there is little standardization with respect to ADC/DAC voltage references. Some ADCs and DACs have internal references, while others do not. In some cases, the DC accuracy of a converter with an internal reference can often be improved by overriding the internal reference with a more accurate and stable external one. Although the reference element itself can be either a bandgap, buried zener, or XFET™ (see Reference 1), practically all references have some type of output buffer op amp. The op amp isolates the reference element from the output and also provides drive capability. However, this op amp must obey the general laws relating to op amp stability, and that is what makes the topic of references relevant to the discussion. Figure 3-47 below summarizes voltage reference considerations. Figure 3-47: ADC / DAC voltage reference considerations Note that a reference input to an ADC or DAC is similar to the analog input of an ADC, in that the internal conversion process can inject transient currents at that pin. This requires adequate decoupling to stabilize the reference voltage. Adding such decoupling might introduce instability in some reference types, depending on the output op amp design. Of course, a reference data sheet may not show any details of the output op amp, which leaves the designer in somewhat of a dilemma concerning whether or not it will be stable and free from transient errors. Fortunately, some simple lab tests can exercise a reference circuit for transient errors, and also determine stability for capacitive loading. Figure 3-48 shows the transients associated with the reference input of a typical successive approximation ADC. The ADC reference voltage input must be stabilized with a sufficiently large decoupling capacitor, in order to prevent conversion errors. The value of the capacitor required as CB may range from below 1µF, to as high as 100µF. This capacitor must of course have a voltage rating greater than the reference voltage. Physically, it will be of minimum size when purchased in a surface mount style. Data converter accuracy determined by the reference, whether internal or external Bandgap, buried zener, XFET™ generally have on-chip output buffer op amp Transient loading can cause instability and errors External decoupling capacitors may cause oscillation Output may require external buffer to source and sink current Reference voltage noise may limit system resolution OP AMP APPLICATIONS 3.44 Note that in this case, a 1µF capacitor on the reference input is required to reduce the transients to an acceptable level. Note that the capacitor size can be electrically larger for further noise reduction— the tradeoff here is of course cost and PCB real estate. The AD780 will work with capacitors of up to 100µF. Figure 3-48: Successive approximation ADCs can present a transient load to the reference A well-designed voltage reference is stable with heavy capacitive decoupling. Unfortunately, some are not, as shown in Figure 3-49 below, where the addition of CL to the reference output (a 0.01µF capacitor) actually increases the amount of transient ringing. Such references are practically useless in data converter applications, because some amount of local decoupling is almost always required at the converter. Figure 3-49: Make sure reference is stable with large capacitive loads A suitable op amp buffer might be added between the reference and the data converter. But, there are many good references available (such as the AD780) which are stable with an output capacitor. This type of reference should be chosen for a data converter application, rather than incurring the further complication and expense of an op amp. AD780 VIN CB VREF SCOPE SCOPE START CONVERT START CONVERT CB = 0.01µF CB = 0.22µF CB = 1µF TOP TRACE VERTICAL SCALE: 5V/div.
ALL OTHER VERTICAL SCALES: 5mV/div.
HORIZONTAL SCALE: 1µs/div. SAR ADC REFERENCE UNDER TEST PULSE GENERATOR VIN CL RL 1mA to 2mA STEP SCOPE TOP TRACE: NO LOAD (CL = 0) 50mV/div.
BOTTOM TRACE: CL = 0.01µF 200mV/div.
BOTH TRACES: 5µs/div. USING OP AMPS WITH DATA CONVERTERS DRIVING ADC/DAC REFERENCE INPUTS 3.45 If very low noise levels are required from a reference, an additional low pass filter followed by a low noise op amp can be used to achieve the desired performance. The reference circuit of Figure 3-50 is one such example (see References 2 and 3). This circuit uses external filtering and a precision low-noise op-amp to provide both very low noise and high DC accuracy. Reference U1 is a 2.5, 3.0, 5, or 10V reference with a low noise buffered output. The output of U1 is applied to the R1-C1/C2 noise filter to produce a corner frequency of about 1.7 Hz. Electrolytic capacitors usually imply DC leakage errors, but the bootstrap connection of C1 causes its applied bias voltage to be only the relatively small drop across R2. This lowers the leakage current through R1 to acceptable levels. Since the filter attenuation is modest below a few Hertz, the reference noise still affects overall performance at low frequencies (i.e., <10 Hz). Figure 3-50: Low-noise op amp with filtering yields reference noise performance (1.5 to 5nV/√Hz @ 1kHz) A precision low noise unity-gain follower, such as the OP113, then buffers the output of the filter. With less than ±150µV of offset error and under 1µV/°C drift, the buffer amplifier’s DC performance will not seriously affect the accuracy/drift of most references. For example, an ADR292E for U1 will have a typical drift of 3ppm/°C, equivalent to 7.5µV/°C, higher than the buffer amplifier. Almost any op amp will have a current limit higher than a typical IC reference, so this circuit allows greater current output. It also removes any load related thermal errors that might occur when the reference IC is loaded directly. Even lower noise op-amps are available, for 5-10V use. The AD797 offers 1kHz noise performance less than 2nV/√Hz in this circuit, compared to about 5nV/√Hz for the OP113. With any buffer amplifier, Kelvin sensing can be used at the load point, a technique that eliminates I×R related output voltage errors. 0.1µF VIN VO GND R1 1kΩ 10kΩ R2 C1 100µF 25V + + C2 100µF 25V 100Ω 1.1kΩ + 100µF,25V 3.3Ω + 10µF 25V 2 6 4 3 2 7 4 6 100Ω U1 U2 + U1: AD586, AD587, REF01 REF02, REF05, REF10 U2: OP113, OP27 OP176, AD797 +15V +15V DIODES: 1N4148 OP AMP APPLICATIONS 3.46 REFERENCES: DRIVING ADC/DAC REFERENCE INPUTS 1. Chapter 2, Walt Kester, Editor, Practical Design Techniques for Power and Thermal Management, Analog Devices, 1998, ISBN: 0-916550-19-2. 2. Walt Jung, "Build an Ultra-Low-Noise Voltage Reference," Electronic Design Analog Applications Issue, June 24, 1993. 3. Walt Jung, "Getting the Most from IC Voltage References," Analog Dialogue, Vol. 28, No., 1994, pp. 13-21. USING OP AMPS WITH DATA CONVERTERS BUFFERING DAC OUTPUTS 3.47 SECTION 3-5: BUFFERING DAC OUTPUTS Walt Kester, Paul Hendriks General Considerations Another important op amp application is buffering DAC outputs. Modern IC DACs provide either voltage or current outputs. Figure 3-51 below shows three fundamental configurations, all with the objective of using an op amp for a buffered output voltage. Figure 3-51: Buffering DAC outputs with op amps Figure 3-51A shows a buffered voltage output DAC. In many cases, the DAC output can be used directly, without additional buffering. If an additional op amp buffer is needed, it is usually configured in a non-inverting mode, with gain determined by R1 and R2. There are two basic methods for dealing with a current output DAC. In Figure 3-51B, a voltage is simply developed across external load resistor, RL. An external op amp can be used to buffer and/or amplify this voltage if required. Many DACs supply fullscale currents of 20mA or more, thereby allowing reasonable voltages to be developed across fairly low value load resistors. For instance, fast settling video DACs typically supply nearly 30mA fullscale current, allowing 1V to be developed across a source and load terminated 75Ω coaxial cable (representing a DC load of 37.5Ω to the DAC output). A direct method to convert the output current into a voltage is shown in Figure 3-51C, This circuit is usually called a current-to-voltage converter, or I/V. In this circuit, the DAC output drives the inverting input of an op amp, with the output voltage developed across the R2 feedback resistor. In this approach the DAC output always operates at virtual ground (which may give a linearity improvement vis-à-vis Fig. 3-51B). The general selection process for an op amp used as a DAC buffer is similar to that of an ADC buffer. The same basic specifications such as DC accuracy, noise, settling time, bandwidth, distortion, etc., apply to DACs as well as ADCs, and the discussion will not be repeated here. Rather, some specific application examples will be shown. N + – VOUT N + – IOUT N + – IOUT (A) (B) (C) RL R2 R1 R2 R2 R1 VOUT 1 + R2 R1 IOUT•RL –IOUT•R2 1 + R2 R1 + OP AMP APPLICATIONS 3.48 Differential to Single-Ended Conversion Techniques A general model of a modern current output DAC is shown in Figure 3-52 below. This model is typical of the AD976X and AD977X TxDAC™ series (see Reference 1). Current output is more popular than voltage output, especially at audio frequencies and above. If the DAC is fabricated on a bipolar or BiCMOS process, it is likely that the output will sink current, and that the output impedance will be less than 500Ω (due to the internal R/2R resistive ladder network). On the other hand, a CMOS DAC is more likely to source output current and have a high output impedance, typically greater than 100kΩ. Figure 3-52: Model of high speed DAC output Another consideration is the output compliance voltage— the maximum voltage swing allowed at the output in order for the DAC to maintain its linearity. This voltage is typically 1V to 1.5V, but can vary depending upon the DAC. Best DAC linearity is generally achieved when driving a virtual ground, such as an op amp I/V converter. Modern current output DACs usually have differential outputs, to achieve high CM rejection and reduce the even-order distortion products. Fullscale output currents in the range of 2mA to 20mA are common. In most applications, it is desirable to convert the differential output of the DAC into a single-ended signal, suitable for driving a coax line. This can be readily achieved with an RF transformer, provided low frequency response is not required. Figure 3-53 (opposite) shows a typical example of this approach. The high impedance current output of the DAC is terminated differentially with 50Ω, which defines the source impedance to the transformer as 50Ω. ROUT ROUT IFS – I I IOUT IOUT RSET IFS 2 - 20mA typical Bipolar or BiCMOS DACs sink current, ROUT < 500Ω CMOS DACs source current, ROUT > 100kΩ Output compliance voltage < ±1V for best performance USING OP AMPS WITH DATA CONVERTERS BUFFERING DAC OUTPUTS 3.49 The resulting differential voltage drives the primary of a 1:1 RF transformer, to develop a single-ended voltage at the output of the secondary winding. The output of the 50Ω LC filter is matched with the 50Ω load resistor RL, and a final output voltage of 1Vp-p is developed. Figure 3-53: Differential transformer coupling The transformer not only serves to convert the differential output into a single-ended signal, but it also isolates the output of the DAC from the reactive load presented by the LC filter, thereby improving overall distortion performance. An op amp connected as a differential to single-ended converter can be used to obtain a single-ended output when frequency response to DC is required. In Figure 3-54 below the AD8055 op amp is used to achieve high bandwidth and low distortion (see Reference 2). The current output DAC drives balanced 25Ω resistive loads, thereby developing an out-of-phase voltage of 0 to +0.5V at each output. The AD8055 is configured for a gain of 2, to develop a final single-ended ground-referenced output voltage of 2Vp-p. Note that because the output signal swings above and below ground, a dual-supply op amp is required. Figure 3-54: Differential DC coupled output using a dual supply op amp The CFILTER capacitor forms a differential filter with the equivalent 50Ω differential output impedance. This filter reduces any slew induced distortion of the op amp, and the optimum cutoff frequency of the filter is determined empirically to give the best overall distortion performance. LC FILTER MINI-CIRCUITS T1-1T 1:1 RDIFF = 50Ω RLOAD = 50Ω VLOAD = ± 0.5V IOUT IOUT 0 TO 20mA 20 TO 0mA ± 10mA CMOS DAC IOUT IOUT 0 TO 20mA 20 TO 0mA CMOS DAC AD8055 + – +5V –5V 25Ω 25Ω 0V TO +0.5V +0.5V TO 0V CFILTER 500Ω 500Ω 1kΩ 1kΩ ± 1V f3dB = 1 2π • 50Ω• CFILTER OP AMP APPLICATIONS 3.50 A modified form of the Fig. 3-54 circuit can also be operated on a single supply, provided the CM voltage of the op amp is set to mid-supply (+2.5V). This is shown in Figure 3-55 below. The output voltage is 2Vp-p centered around a CM voltage of +2.5V. This CM voltage can be either developed from the +5V supply using a resistor divider, or directly from a +2.5V voltage reference. If the +5V supply is used as the CM voltage, it must be heavily decoupled to prevent supply noise from being amplified. Figure 3-55: Differential DC coupled output using a single-supply op amp Single-Ended Current-to-Voltage Conversion Single-ended current-to-voltage conversion is easily performed using a single op amp as an I/V converter, as shown in Figure 3-56 below. The 10mA fullscale DAC current from the AD768 (see Reference 3) develops a 0 to +2V output voltage across the 200Ω RF. Figure 3-56: Single-ended I/V op amp interface for precision 16-bit AD768 DAC Driving the virtual ground of the AD8055 op amp minimizes any distortion due to nonlinearity in the DAC output impedance. In fact, most high resolution DACs of this type are factory trimmed using an I/V converter. IOUT IOUT 0 TO 10mA AD768 16-BIT BiCMOS DAC AD8055 + – +5V RF = 200Ω –5V RDAC||CDAC CF 0 TO +2.0V For RDAC ≈ RF, make CF ≈ CIN RDAC (CDAC + CIN) RF For RDAC >> RF, make CF ≈ CDAC + CIN 2π RF fu fu = Op Amp Unity Gain-Bandwidth Product IOUT IOUT 0 TO 20mA 20 TO 0mA CMOS DAC AD8055 + – +5V 25Ω 25Ω 0V TO +0.5V +0.5V TO 0V CFILTER 500Ω 500Ω 2kΩ 1kΩ ± 1V f3dB = 1 2π • 50Ω• CFILTER +2.5V +5V 2kΩ +2.5V REF 1kΩ SEE TEXT +5V USING OP AMPS WITH DATA CONVERTERS BUFFERING DAC OUTPUTS 3.51 It should be recalled, however, that using the single-ended output of the DAC in this manner will cause degradation in the CM rejection and increased second-order distortion products, compared to a differential operating mode. The CF feedback capacitor should be optimized for best pulse response in the circuit. The equations given in the diagram should only be used as guidelines. A more detailed analysis of this circuit is given in Reference 6. Differential Current-to-Differential Voltage Conversion If a buffered differential voltage output is required from a current output DAC, the AD813X-series of differential amplifiers can be used as shown in Figure 3-57 below. Figure 3-57: Buffering high speed DACs using AD813X differential amplifier The DAC output current is first converted into a voltage that is developed across the 25Ω resistors. The voltage is amplified by a factor of 5 using the AD813X. This technique is used in lieu of a direct I/V conversion to prevent fast slewing DAC currents from overloading the amplifier and introducing distortion. Care must be taken so that the DAC output voltage is within its compliance rating. The VOCM input on the AD813X can be used to set a final output CM voltage within the range of the AD813X. If transmission lines are to be driven at the output, adding a pair of 75Ω resistors will allow this. IOUT IOUT 0 TO 20mA 0 TO +0.5V 20 TO 0mA +0.5 TO 0V CMOS DAC + – AD813X VOCM 2.49kΩ 2.49kΩ 5V p-p DIFFERENTIAL OUTPUT 25Ω 25Ω 499Ω 499Ω OP AMP APPLICATIONS 3.52 An Active Lowpass Filter for Audio DAC Figure 3-58 below shows an active lowpass filter which also serves as a current-to-voltage converter for the AD1853 sigma-delta audio DAC (see Reference 4). The filter is a 4-pole filter with a 3dB cutoff frequency of approximately 75kHz. Because of the high oversampling frequency (24.576MSPS when operating the DAC at a 48KSPS throughput rate), a simple filter is all that is required to remove aliased components above 12MHz). Figure 3-58: A 75kHz 4-pole gaussian active filter for buffering the output of AD1853 stereo DAC The diagram shows a single channel for the dual channel DAC output. U1A and U1B I/V stages form a 1-pole differential filter, while U2 forms a 2-pole multiple-feedback filter that also performs a differential-to-single-ended conversion. A final fourth passive pole is formed by the 604Ω resistor and the 2.2nF capacitor across the output. The OP275 op amp was chosen for operation at U1 and U2, and for its quality audio characteristics (see Reference 5). For further details of active filter designs, see Chapter 5 of this book. 4.12kΩ 4.12kΩ 330pF 330pF 2.74kΩ 2.74kΩ 2.74kΩ 2.74kΩ 2.94kΩ 2.94kΩ 220pF 220pF 604Ω 49.9kΩ 2.2nF +15V –15V +15V –15V 402Ω 402Ω – + + – – + ROUT+ ROUT– VREF +2.75V 0mA TO +1.5mA +1.5mA TO 0mA RIGHT CHANNEL OUTPUT U1A 1/2 OP275 U1B 1/2 OP275 U2A 1/2 OP275 NOTE: ONLY RIGHT CHANNEL SHOWN FROM AD1853 DAC 680pF 680pF 220pF USING OP AMPS WITH DATA CONVERTERS BUFFERING DAC OUTPUTS 3.53 REFERENCES: BUFFERING DAC OUTPUTS 1. Data sheet for AD9772A 14-Bit, 160 MSPS TxDAC+® with 2x Interpolation Filter, 2. Data sheet for AD8055/AD8056 Low Cost, 300 MHz Voltage Feedback Amplifiers, 3. Data sheet for AD768 16-Bit, 30 MSPS D/A Converter, 4. Data sheet for AD1853 Stereo, 24-Bit, 192 kHz, Multibit Σ∆ DAC, 5. Data sheet for OP275 Dual Bipolar/JFET, Audio Operational Amplifier, 6. Chapters 5, Walt Kester, Editor, Practical Design Techniques for Sensor Signal Conditioning, Analog Devices, 1999, ISBN: 0-916550-20-6. OP AMP APPLICATIONS 3.54 NOTES: |
13926 | https://arxiv.org/pdf/1409.7610 | Generalized Convergence Rates Results for Linear Inverse Problems in Hilbert Spaces
Roman Andreev ∗ Peter Elbau † Maarten V. de Hoop ‡
Lingyun Qiu § Otmar Scherzer ¶
September 19, 2018
Abstract
In recent years, a series of convergence rates conditions for regulariza-tion methods has been developed. Mainly, the motivations for developing novel conditions came from the desire to carry over convergence rates re-sults from the Hilbert space setting to generalized Tikhonov regularization in Banach spaces. For instance, variational source conditions have been developed and they were expected to be equivalent to standard source con-ditions for linear inverse problems in a Hilbert space setting (see Schuster et al ). We show that this expectation does not hold. However, in the standard Hilbert space setting these novel conditions are optimal, which we prove by using some deep results from Neubauer , and generalize existing convergence rates results. The key tool in our analysis is a novel source condition, which we put into relation to the existing source con-ditions from the literature. As a positive by-product, convergence rates results can be proven without spectral theory, which is the standard tech-nique for proving convergence rates for linear inverse problems in Hilbert spaces (see Groetsch ).
1 Introduction
In this paper we consider for some (not exactly known) data y ∈ R (L) the operator equation
Lu = y , (1)
∗
Johann Radon Institute for Computational and Applied Mathematics (RICAM), Al-tenbergerstrasse 69, A-4040 Linz, Austria ( roman.andreev@oeaw.ac.at )
†
Computational Science Center, University of Vienna, Oskar-Morgenstern Platz 1, A-1090 Vienna, Austria ( peter.elbau@univie.ac.at )
‡
Center for Computational and Applied Mathemematics, Purdue University, West Lafayette, IN 47907, USA ( mdehoop@purdue.edu )
§
Institute for Mathematics and its Applications, University of Minnesota, Minneapolis, MN 55455, USA ( qiu.lingyun@ima.umn.edu )
¶
Computational Science Center, University of Vienna and RICAM (otmar.scherzer@univie.ac.at )
1
arXiv:1409.7610v1 [math.FA] 26 Sep 2014
where L : U → V is a bounded linear operator between two real Hilbert spaces
U and V , and R(L) denotes its range. Given some approximate data yδ ∈ V with ‖y − yδ ‖ ≤ δ, the objective is to reconstruct the minimal norm solution u† ∈ U , that is the element fulfilling
Lu † = y and ‖u†‖ = inf {‖ u‖ : Lu = y} .
Such a minimal norm solution exists for every y ∈ R (L) and is uniquely defined, see for example [3, Theorem 2.5]. The method of choice for performing this task is Tikhonov regularization, that is to find for arbitrary α > 0 the regularized solution
uδα := argmin
u∈U
{‖Lu − yδ ‖2 + α‖u‖2} . (2) Standard results on Tikhonov regularization guarantee the existence and unique-ness of the minimizer uδα and that uδα converges to u† for an appropriate choice of α depending on δ as δ ↘ 0, see for instance [3, Theorem 5.1 and Theorem 5.2]. Convergence rates conditions, moreover, guarantee a certain convergence rate ‖uδα − u†‖ = O(f (δ)) as δ ↘ 0 if again α is chosen to depend in the right way on δ.Two kinds of such convergence rates conditions have been developed:
• source conditions and
• variational source conditions [6, 9, 5, 10]. The goal of this paper is to put the different source conditions into perspective, together with three novel variational source conditions, which are presented here for the first time. The main results on the relations in between the source conditions are summarized in a table form (cf. Figure 1). Aside from these particular conditions the novelties are to show that these conditions are in fact more general than the classical source conditions, and they are optimal in the sense that convergence rates of a certain order are only possible if these con-ditions are satisfied. The argumentation is based on a result from Neubauer . Moreover, as a side product, this clarifies some assertion from on the equivalence of standard and variational source conditions.
2 Relations of Source Conditions in the Stan-dard Setting
The key to obtain convergence rate results for the regularized solution uδα, de-fined in (2), of the problem (1) is to impose conditions on the minimal norm solution u†. In the literature, various kind of such source conditions have been introduced. 2Definition 1. Let U and V be real Hilbert spaces, L : U → V be a bounded linear operator, and y ∈ R (L). Moreover, let u† denote the minimum-norm solution of the operator equation (1). Then, we say that the problem fulfills
• the standard source condition, see , with the parameter ν ∈ (0 , 2] if
u† ∈ R (( L∗L) ν
2
); (3)
• the homogeneous variational inequality with the parameter ν ∈ (0 , 1] if there exists a constant β ≥ 0 such that 2〈u†, u 〉 ≤ β‖Lu ‖ν ‖u‖1−ν for every u ∈ U ; (4)
• the inhomogeneous variational inequality with the parameter μ ∈ (0 , 1], introduced in for the case μ = 1 and in for general μ ∈ (0 , 1] in the setting of non-linear problems, if there exist constants β ≥ 0 and γ ∈ [0 , 1) such that 2〈u†, u 〉 ≤ β‖Lu ‖μ + γ‖u‖2 for every u ∈ U ; (5)
• the symmetrized variational inequality with the parameter ν ∈ (0 , 2] if there exists a constant β ≥ 0 such that 2〈u†, u 〉 ≤ β‖L∗Lu ‖ ν
2
‖u‖1− ν
2
for every u ∈ U. (6)
Remark: Let ρ ∈ (0 , 2]. The family of variational source conditions, 2〈u†, u 〉 ≤ β‖(L∗L) ρ
2
u‖ νρ ‖u‖1− νρ , ν ≤ ρ,
puts the homogeneous variational inequality, the symmetrized variational in-equality, and the standard source condition under one umbrella, when we set
ρ = 1, ρ = 2, and ρ = ν (see the proof of Lemma 3 (ii) ), respectively. However, Proposition 11 and Proposition 10 show that all these variational source condi-tions with the same parameter ν and a parameter ρ > ν are equivalent to each other. Note that the inhomogeneous variational inequality is not homogeneous with respect to u ∈ U , as opposed to the other three source conditions. Let us first discuss the relation between the first three source conditions.
Lemma 2. Let U and V be real Hilbert spaces, L : U → V be a bounded linear operator, y ∈ R (L), and ν ∈ (0 , 1] . Then, we have that (i) the standard source condition for ν implies the homogeneous variational inequality with the same parameter ν,(ii) the homogeneous variational inequality with the parameter ν implies the inhomogeneous variational inequality with the parameter μ = 2ν
1+ ν
, and
3(iii) the inhomogeneous variational inequality with the parameter μ = 1 implies the standard source condition with the parameter ν = 1 .Proof: Let u† denote the minimum-norm solution of the operator equation (1).
(i) If the standard source condition is fulfilled for some ν ∈ (0 , 1], then there exists an element ω ∈ U with ( L∗L) ν
2
ω = u†. Using now the interpolation inequality
‖(L∗L)r u‖ ≤ ‖ (L∗L)q u‖ rq ‖u‖1− rq for all u ∈ U, 0 < r ≤ q, (7) see for example [3, Chapter 2.3], with r = ν
2
and q = 12 , it follows for every
u ∈ U that 2〈u†, u 〉 = 〈2ω, (L∗L) ν
2
u〉 ≤ 2‖ω‖‖ (L∗L) ν
2
u‖ ≤ 2‖ω‖‖ Lu ‖ν ‖u‖1−ν ,
which is of the form (4) with the parameter ν.
(ii) If u† fulfills the variational inequality (4) for some parameters ν ∈ (0 , 1] and β ≥ 0, then Young’s inequality implies for every u ∈ U that 2〈u†, u 〉 ≤ β‖Lu ‖ν ‖u‖1−ν ≤ 1 + ν
2 β 21+ ν ‖Lu ‖ 2ν
1+ ν
1 − ν
2 ‖u‖2,
so that the inhomogeneous variational inequality with the parameter μ =
2ν
1+ ν
is fulfilled.
(iii) If u† fulfills the inequality (5) for μ = 1 and some constants β ≥ 0 and
γ ∈ [0 , 1), then, by evaluating it at u = tv for arbitrary v ∈ U and t > 0, we find in the limit t ↘ 0 that 2〈u†, v 〉 ≤ β‖Lv ‖ = β‖(L∗L) 12 v‖ for every v ∈ U. (8) Now, it can be shown, see [9, Lemma 8.21], that if T : U → U is a bounded linear operator, then u† ∈ R (T ∗) if and only if there exists a constant C > 0 such that 〈u†, v 〉 ≤ C‖T v ‖ for all v ∈ U .Thus, with T = ( L∗L) 12 , we find that (8) is equivalent to u† ∈ R (( L∗L) 12 ).
Remark: That the standard source condition for a parameter ν ∈ (0 , 1] implies the inhomogeneous variational inequality with the parameter μ = 2ν
1+ ν
was already realized in . The case ν = 1 has been treated in more generality in [9, Table 3.1]. Thus, the homogeneous variational inequality and the inhomogeneous vari-ational inequality cover only the parameter range ν ∈ (0 , 1] compared to the standard source condition. However, the symmetrized variational inequality is an extension of the standard source condition in the full parameter range
ν ∈ (0 , 2], as the following lemma shows. 4Lemma 3. Let U and V be real Hilbert spaces, L : U → V be a bounded linear operator, and y ∈ R (L).Then, we have that (i) the standard source condition with a parameter ν ∈ (0 , 2] implies the sym-metrized variational inequality with the same parameter ν,(ii) the symmetrized variational inequality with the parameter ν = 2 is equiv-alent to the standard source condition with the parameter ν = 2 .Proof: Let u† denote the minimum-norm solution of the problem (1).
(i) From the inequality (6) with some parameters ν ∈ (0 , 1] and β ≥ 0, we obtain by applying the Cauchy–Schwarz inequality for every u ∈ U that 2〈u†, u 〉 ≤ β〈L∗Lu, u 〉 ν
2
‖u‖1−ν ≤ β‖L∗Lu ‖ ν
2
‖u‖1− ν
2
,
which is the symmetrized variational inequality with the parameter ν.
(ii) The symmetrized variational inequality with the parameter ν = 2 states that there exists a constant β ≥ 0 so that
〈u†, u 〉 ≤ β‖L∗Lu ‖ for all u ∈ U.
Now, as in the proof of Lemma 2 (iii) , this is equivalent to u† ∈ R (L∗L), see [9, Lemma 8.21].
The following two examples illustrate that the degree of ill-posedness of the operator L is a criterion for equivalency of the different source conditions. Finer results, establishing in particular the equivalence of the source conditions (4) and (5) and the corresponding convergence rates, see Proposition 8, will be derived in Theorem 12 below.
Example 4. Let U and V be real Hilbert spaces, ν ∈ (0 , 1] , L : U → V be a bounded linear operator so that (L∗L)ν/ 2 has closed range, and y ∈ R (L).Then, (i) the standard source condition with parameter ν,(ii) the homogeneous variational inequality with parameter ν, and (iii) the inhomogeneous variational inequality with parameter μ = 2ν
1+ ν
are equivalent. Proof: In view of Lemma 2, we only need to show that (iii) implies (i) . To that end recall that, if T : U → U is a bounded linear self-adjoint operator then its nullspace N (T ) is the orthogonal complement of the range R(T ), and
U = R(T ) ⊕ N (T ). Since the range of T = ( L∗L)ν/ 2 is closed by assumption, we have the orthogonal decomposition
U = R(( L∗L) ν
2
) ⊕ N (( L∗L) ν
2
). (9) 5Observe now that, if u ∈ N (( L∗L) ν
2
) then
‖Lu ‖2 = 〈Lu, Lu 〉 = 〈(L∗L)ν/ 2u, (L∗L)1−ν/ 2u〉 = 0 ,
so that Lu = 0. Therefore, if u† satisfies (5) with some constants β ≥ 0 and
γ ∈ [0 , 1), then 2〈u†, u 〉 ≤ γ‖u‖2 for every u ∈ N (( L∗L)ν/ 2).
Substituting u by tu in the above inequality with t > 0, we arrive at 2t〈u†, u 〉 ≤ t2γ‖u‖2 for every u ∈ N (( L∗L)ν/ 2), t > 0.
Dividing by t and letting t go to 0, this implies 〈u†, u 〉 = 0 whenever u ∈N (( L∗L)ν/ 2). By the orthogonality of the decomposition (9) we have u† ∈R(( L∗L)ν/ 2), which is (i) .
Remark: As in Example 4, one can also show that the standard source condition and the symmetrized variational inequality with the same parameter ν ∈ (0 , 2] are equivalent if ( L∗L) ν
2
has closed range.
Example 5. Let U be a real, separable Hilbert space with orthonormal basis
{ϕn}n∈N. We define the compact linear operator L : U → U by L(ϕn) = 2 −nϕn.(Note that its range is not closed, for the range of a compact operator is closed if and only if it is finite-dimensional.) Then, for the data
y = ∑
n≥1
2− 32 nϕn ∈ R (L),
the problem (1) fulfills the homogeneous variational inequality with the param-eter ν = 12 , but not the standard source condition with parameter ν = 12 . In particular, the two source conditions are not equivalent. However, the standard source condition is fulfilled for every parameter ν < 12 .Proof: The minimum-norm solution u† can be directly calculated to be
u† = L−1y = ∑
n≥1
2− n
2
ϕn. (10) Now, since L is self-adjoint by definition so that we have ( L∗L) 12 = L, we see that u† /∈ R (( L∗L) 14 ) because L− 12 u† = ∑
n≥1
ϕn /∈ U .However, we have for every ν < 12 that L−ν u† = ∑
n≥1
2n(ν− 12 )ϕn is in U ,and therefore u† is in the range of ( L∗L) ν
2
for every ν < 12 .For u ∈ U arbitrary we write u = ∑
n≥1
2 n
2
γnϕn with some γn ∈ R. Then
〈u†, u 〉 = ∑
n≥1
γn, ‖u‖2 = ∑
n≥1
2n |γn|2 , and ‖Lu ‖2 = ∑
n≥1
2−n |γn|2 .
6Now we can show that the homogeneous variational inequality with parameter
ν = 12 is fulfilled, more precisely, that we have
〈u†, u 〉 ≤ 2√2‖u‖ 12 ‖Lu ‖ 12 for every u ∈ U. (11) Indeed, set S := ∑
n≥1
|γn| and let N ∈ N be such that
12
S ≤ A := ∑
n≤N
|γn| and 12 S ≤ B := ∑
n≥N
|γn|.
Observe that, using the Cauchy–Schwarz inequality,
A2 ≤ ∑
k≤N
2k ∑
n≤N
2−n|γn|2 ≤ (2 N +1 − 1) ‖Lu ‖2,
and
B2 ≤ ∑
k≥N
2−k ∑
n≥N
2n|γn|2 ≤ 2−N +1 ‖u‖2.
Since we have by definition of S that 〈u†, u 〉 ≤ S and by the choice of N that
S ≤ 2√AB , the inequality (11) follows. This proof is largely from . The proof of Proposition 11 below is a more elaborate version of the same idea.
Remark: In the above proof we noted that
u† ∈ R (( L∗L) ρ
2
) for every ρ ∈ [0 , ν ). (12) This property is a general consequence of the variational source condition (4). This follows from Proposition 8 and [8, Corollary 2.4]. However, if u† satisfies (12), it need not satisfy (4): take u† as in (10) but with L(ϕn) := n−22−nϕn. Then (12) holds for ν = 1 /2. But 〈u†, ϕ n〉 = 2 −n/ 2
is not bounded in terms of ‖Lϕ n‖1/2‖ϕn‖1/2 = n−12−n/ 2 uniformly in n ≥ 1.
3 Rates Results without Spectral Theory
We briefly review the convergence rate results which follow from the introduced source conditions.
Definition 6. Let U and V be real Hilbert spaces, L : U → V be a bounded linear operator, and y ∈ R (L). Moreover, let u† denote the minimum-norm solution of the operator equation (1). Then, we say that the problem has
• a noise-free convergence rate of order σ if there exists a constant C > 0so that the regularized solution
uα = argmin
u∈U
(‖Lu − y‖2 + α‖u‖2) , (13) fulfills that
‖uα − u†‖ ≤ Cα σ for every α > 0,
7• a convergence rate of order ρ if there exists a constant C > 0 so that the regularized solutions
uα(˜ y) = argmin
u∈U
(‖Lu − ˜y‖2 + α‖u‖2) , α > 0, ˜y ∈ V, (14) fulfill for every δ > 0 the inequality sup
{
inf
α> 0
‖uα(˜ y) − u†‖ : ˜y ∈ V, ‖˜y − y‖ ≤ δ
}
≤ Cδ ρ. (15) The classical convergence results now state that if a problem (1) fulfils the standard source condition for some parameter ν ∈ (0 , 2], then it has a conver-gence rate of order ν
1+ ν
, see [4, Corollary 3.1.4]. For ν, μ ∈ (0 , 1] the same result can be obtained under the weaker source conditions (4) and (5), see [5, 10]. The simple proof is added here for completeness.
Lemma 7. Let L : U → V be a bounded linear operator between two real Hilbert spaces U and V , and y ∈ R (L). Moreover, let u† denote the minimum-norm solution of the problem (1) and assume that it fulfils the inhomogeneous varia-tional inequality (5) for some parameters μ ∈ (0 , 1] , β ≥ 0, and γ ∈ (0 , 1) .Then, for every choice of yδ ∈ V with ‖yδ − y‖ ≤ δ for some δ > 0 and every
α > 0, the corresponding regularized solution
uδα = argmin
u∈U
(‖Lu − yδ ‖2 + α‖u‖2)
satisfies
‖uδα − u†‖2 ≤ 21 − γδ2
α + β 22−μ (2 − μ)2(1 − γ) α μ
2−μ
. (16)
Proof: From the definition of the minimizer uδα, it follows that
‖Lu δα − yδ ‖2 + α‖uδα‖2 ≤ δ2 + α‖u†‖2.
This inequality together with the variation inequality (5) yields
‖Lu δα − yδ ‖2 + α‖uδα − u†‖2 ≤ δ2 + 2 α〈u†, u † − uδα〉≤ δ2 + αβ ‖L(uδα − u†)‖μ + αγ ‖uδα − u†‖2.
Now, observing that 12 ‖L(uδα − u†)‖2 − δ2 ≤ ‖ Lu δα − yδ ‖2,
which is a consequence of the triangle inequality and the fact that a ≤ b + c
implies a2 ≤ 2( b2 + c2), we further find that 12 ‖L(uδα − u†)‖2 + α(1 − γ)‖uδα − u†‖2 ≤ 2δ2 + αβ ‖L(uδα − u†)‖μ.
8Applying then Young’s inequality to the last term, we end up with 12 ‖L(uδα − u†)‖2 + α(1 − γ)‖uδα − u†‖2 ≤ 2δ2 + 2 − μ
2 (αβ ) 22−μ + μ
2 ‖L(uδα − u†)‖2,
which in particular implies (16).
Proposition 8. Assume that L : U → V is a bounded linear operator between two real Hilbert spaces U and V , and y ∈ R (L).Then, if the problem (1) fulfills the inhomogeneous variational inequality with the parameter μ = 2ν
1+ ν
for some ν ∈ (0 , 1] , it has (i) a noise-free convergence rate of order ν
2
and (ii) a convergence rate of order ν
1+ ν
.Proof: Let u† be the minimal-norm solution of (1).
(i) In the noise free case, Lemma 7 with δ = 0 and μ = 2ν
1+ ν
directly implies for the regularized solution uα defined by (13) the inequality
‖uα − u†‖ ≤ Cα μ
2(2 −μ)
= Cα ν
2
for all α > 0for some constant C > 0.
(ii) In the noisy case, Lemma 7 yields for arbitrary δ > 0 and data ˜ y ∈ V with
‖˜y − y‖ ≤ δ the inequality inf
α> 0
‖uα(˜ y) − u†‖2 ≤ ‖ uδ2−μ (˜ y) − u†‖2 ≤ Cδ μ = Cδ 2ν
1+ ν
(17) for some constant C > 0. Here, uα(˜ y) denotes the regularized solu-tion (14).
Remark: Because of Lemma 2, the homogeneous variational inequality with a parameter ν ∈ (0 , 1] therefore also implies a noise-free convergence rate of order ν
2
and a convergence rate of order ν
1+ ν
.
4 On converse results of Neubauer
In this section we go deeper into the results of Neubauer . In the Hilbert space setting, Neubauer characterized the minimum-norm solution for which the problem has a convergence rate of order νν+1 for some ν ∈ (0 , 2) in terms of its spectral tail. (Note that Neubauer writes 2 ν where we write ν.)
Definition 9. Let U and V be real Hilbert spaces, L : U → V be a bounded linear operator, and y ∈ R (L). We say that the minimum-norm solution u† of the problem (1) has spectral tail of order ν if there exists a constant C > 0 so that
‖E[0 ,λ ]u†‖2 ≤ C2λν for all λ ≥ 0, (18) where A 7 → EA denotes the (projection-valued) spectral measure of the opera-tor L∗L.9Proposition 10. Let L : U → V be a bounded linear operator between two real Hilbert spaces U and V , and y ∈ R (L).Then, for every ν ∈ (0 , 2) , it is equivalent for the problem (1) that (i) it has a noise-free convergence rate of order ν
2
,(ii) it has a convergence rate of order νν+1 , and (iii) its minimum-norm solution has a spectral tail of order ν.Proof: Neubauer showed in [8, Theorem 2.1] that the condition (i) is equivalent to (iii) , and proved in [8, Theorem 2.6] that (iii) is equivalent to the fact that there exists a constant C ≥ 0 so that sup
{
inf
α> 0
‖uα(˜ y) − u†‖ : ˜y ∈ V, ‖Q(˜ y − y)‖ ≤ δ
}
≤ Cδ νν+1 (19) for every δ ≥ 0, where Q denotes the orthogonal projection onto the range R(L)and the regularized solution uα(˜ y) is defined by (14). It therefore only remains to show that (19) is equivalent to a convergence rate of order νν+1 .It is clear that (19) implies such a convergence rate, since the supremum in the definition (15) of the convergence rate is taken over a smaller set than in (19). For the other direction, we define for arbitrary ˜ y ∈ V with ‖Q(˜ y − y)‖ ≤ δ,the element ˆ y := y + Q(˜ y − y). Then,
‖ˆy − y‖ = ‖Q(˜ y − y)‖ ≤ δ,
and the optimality conditions for the regularized solutions uα(˜ y) and uα(ˆ y) yield
uα(˜ y) − uα(ˆ y) = ( αI + L∗L)−1L∗(˜ y − ˆy).
Since now L∗ = L∗Q, we have L∗(˜ y − ˆy) = L∗Q(˜ y − ˆy) = 0 and therefore,
uα(˜ y) = uα(ˆ y). As ˜ y was arbitrary subject to ‖Q(˜ y − y)‖ ≤ δ, condition (15) with ρ = νν+1 implies (19).
Neubauer also gave a counterexample to show that the standard source condition with parameter ν ∈ (0 , 2), which implies the three equivalent condi-tions of Proposition 10, is not equivalent to them, see also Example 5. However, we will show in the following that the homogeneous variational inequality with parameter ν ∈ (0 , 1) and the inhomogeneous variational in-equality with parameter μ = 2ν
1+ ν
are indeed equivalent to the conditions of Proposition 10.
Proposition 11. Let L : U → V be a bounded linear operator between two real Hilbert spaces U and V , and let y ∈ R (L).Then for arbitrary ν ∈ (0 , 2) and ρ > ν the conditions that
10 (i) the maximum-norm solution u† of the problem (1) has a spectral tail of order ν and (ii) there exists a constant β ≥ 0 so that
2〈u†, u 〉 ≤ β‖(L∗L) ρ
2
u‖ νρ ‖u‖1− νρ for all u ∈ U (20)
are equivalent. Proof: We first show that (i) implies (ii) .Let A 7 → EA denote the (projection-valued) spectral measure of L∗L. For arbitrary u ∈ U , we define the signed measure A 7 → μu†,u (A) = 〈EAu†, u 〉 and set for λ ∈ [0 , ∞]
Aλ := |μu†,u |([0 , λ ]) and Bλ := |μu†,u |([ λ, ∞)) , (21) where |μu†,u | denotes the variation of the measure μu†,u .Let now Λ := inf {λ ≥ 0 : Aλ ≥ 12 A∞}. Then, since λ 7 → Aλ is right-continuous, there holds AΛ ≥ 12 A∞. Moreover, since Λ is minimal and λ 7 → Bλ
is left-continuous, it also follows that BΛ ≥ 12 A∞.We now estimate AΛ with the inequality (23) with T = L∗L and ρ = 0, which yields
AΛ = |μu†,u |([0 , Λ]) ≤ ‖ E[0 ,Λ] u†‖‖ u‖.
If the spectral tail of u† has order ν, we have a constant C > 0 so that
‖E[0 ,λ ]u†‖ ≤ Cλ ν
2
and thus
AΛ ≤ C‖u‖Λ ν
2
. (22) For BΛ, we also use the inequality (23) with T = L∗L, and get for arbitrary
ρ ∈ R, the upper bound
BΛ = |μu†,u |([Λ , ∞)) ≤ ‖ (L∗L) ρ
2
u‖
(∫
[Λ ,∞)
1
λρ dμu†,u † (λ)
) 12
.
Choosing now ρ > ν , we can estimate the integral with (24) (using that the measure μu†,u † satisfies μu†,u † ([0 , λ ]) = ‖E[0 ,λ ]u†‖2 ≤ C2λν ) and find
BΛ ≤ C
√
1 − νρ
‖(L∗L) ρ
2
u‖Λ ν−ρ
2
.
Therefore, recalling that Λ was chosen so that AΛ ≥ 12 A∞ and BΛ ≥ 12 A∞,we have 2〈u†, u 〉 ≤ 2A∞ ≤ 4A1− νρ
Λ
B νρ
Λ
≤ 4C
(1 − νρ ) ν
2ρ
‖u‖1− νρ ‖(L∗L) ρ
2
u‖ νρ ,
which is the condition (ii) .11 For the other direction, we remark that from the inequality (20) for some constant β ≥ 0, we find for every λ ≥ 0 that
‖E[0 ,λ ]u†‖2 = 〈E[0 ,λ ]u†, u †〉 ≤ β
2 ‖(L∗L) ρ
2
E[0 ,λ ]u†‖ νρ ‖E[0 ,λ ]u†‖1− νρ .
Now, since E is the spectral measure of L∗L, we have that ‖(L∗L) ρ
2
E[0 ,λ ]u†‖ ≤
λ ρ
2
‖E[0 ,λ ]u†‖, see for example [2, Chapter X.2.9, Corollary 9], and we therefore obtain that
‖E[0 ,λ ]u†‖2 ≤ β
2 λ ν
2
‖E[0 ,λ ]u†‖,
which concludes the proof.
Remark: In fact, it can be seen from this proof that condition (ii) in Propo-sition 11 also implies condition (i) in the case ρ = ν = 2, which corresponds to the result that the standard source condition for ν = 2 yields a noise-free convergence rate of order 1. Finally, we can summarize all the statements in an equivalence result be-tween the different source conditions and convergence rates.
Theorem 12. Let L : U → V be a bounded linear operator between two real Hilbert spaces U and V and y ∈ R (L).Then, for every ν ∈ (0 , 2) , it is equivalent for the problem (1) that (i) it fulfils the symmetrized variational inequality with parameter ν,(ii) it has a noise-free convergence rate of order ν
2
,(iii) it has a convergence rate of order νν+1 ,(iv) its minimum-norm solution u† has a spectral tail of order ν,and if ν ∈ (0 , 1) these are additionally equivalent to (v) the homogeneous variational inequality with parameter ν and (vi) the inhomogeneous variational inequality with parameter μ = 2ν
1+ ν
.Proof: We already know from Proposition 10 that (ii) , (iii) , and (iv) are equiv-alent conditions. Moreover, we know from Lemma 2 that (v) implies (vi) , and from Proposition 8 that (vi) implies (ii) and (iii) .Now, Proposition 11 with ρ = 1 shows that (iv) implies (v) , which proves the equivalence of all conditions but (i) .And finally, the equivalence of (i) and (iv) follows directly from Proposi-tion 11 with ρ = 2. We briefly comment on the case ν = 1. We have already seen in Lemma 2 that in this case the standard source condition, the homogeneous variational inequality and the inhomogeneous variational inequality (all with the parameter 1) are equivalent. Moreover, because of Proposition 8 they also imply all the conditions of Proposition 10. However, the converse is not true. 12 Example 13. Let U be a real, separable Hilbert space with orthonormal basis
{ϕn}n∈N. We define the compact linear operator L : U → U by Lϕ n = n− 12 ϕn.Then, for the data
y = ∑
n≥1
n− 32 ϕn ∈ R (L),
the minimum-norm solution u† of problem (1) has spectral tail of order 1, but the problem does not fulfill the inhomogeneous variational inequality with the parameter μ = 1 .Proof: We see that the minumum norm solution u† is explicitly given by
u† = ∑
n≥1
n−1ϕn
and thus has a spectral tail of order 1:
‖E[0 ,λ ]u†‖2 = ∑
n≥λ−2
|〈 u†, ϕ n〉| 2 = ∑
n≥λ−2
n−2 ≤ Cλ 2
for all λ ∈ [0 , ‖L‖] for some constant C > 0, where A 7 → EA again denotes the spectral measure of L∗L.However, for uN = N −1 ∑
n≤N
ϕn, we find that 2〈u†, u N 〉‖Lu N ‖ + ‖uN ‖2 = 2N −1HN
N −1H1/2
N
N −2 ∑
n≤N
1 ≥ H1/2
N
,
where HN := ∑
n≤N
n−1 denotes the N -th harmonic number. Because HN →∞ as N → ∞ , the inhomogeneous variational inequality with parameter μ = 1 cannot be satisfied.
The condition (4) seems to be the natural condition for convergence rates. It is a necessary and sufficient condition for the rate O(α ν
2
), while the standard range condition (3) leaves a small gap. Theorem 12 guarantees that the variational source conditions are optimal conditions for convergence rates.
Acknowledgment
This work has been supported by the Austrian Science Fund (FWF) within the national research network Geometry + Simulation (project S11704, Variational Methods for Imaging on Manifolds). This research was supported in part by National Science Foundation grant CMG DMS-1025318, and in part by the members of the Geo-Mathematical Imaging Group at Purdue University. 13 Standard source condition with parameter ν
for ν∈(0 ,1], see Lemma 2 (i)
Homogeneous variational inequality with parameter ν
for ν∈(0 ,1], see Lemma 2 (ii)
Inhomogeneous variational inequality with parameter 2νν+1
Convergences rates of order νν+1
for ν∈(0 ,1], see Proposition 8
for ν∈(0 ,2), see Proposition 10 and
Noise-free convergence rates of order ν
2
for ν∈(0 ,2), see Proposition 10 and
Spectral characterization of order ν
for ν∈(0 ,1), see Proposition 11 with ρ= 1
for ν= 1, see Lemma 2 (iii) ;or for ν∈(0 ,1] if R(L∗L) is closed, see Example 4
Symmetrized variational imequality with parameter ν
for ν∈(0 ,2), see Proposition 11 with ρ= 2
for ν∈(0 ,2], see Lemma 3 (i)
for ν= 2, see Lemma 3 (ii)
Figure 1: Relation between the different source conditions and the convergence rate results.
Conclusion
In this paper we have developed a series of novel variational source conditions as alternatives to classical source conditions to prove convergence rates results for Tikhonov regularization in an Hilbert space setting. In many cases the new source conditions provide optimal convergence rates, opposed to the standard source conditions. The interplay between various source conditions and conver-gence rates is developed in detail and summarized in Table 1. As a side product we could clarify an open question in . An open question is of course how these results can be generalized to non-linear ill–posed problems, to Banach spaces or general topological spaces, and to other regularization methods.
Appendix
Lemma 14. Let T : U → U be a self-adjoint, non-negative definite, bounded linear operator on a real Hilbert space U , u†, u ∈ U , and Λ ≥ 0. We denote with
A 7 → EA the projection-valued spectral measure of T and define for all v, w ∈ U
the signed measure A 7 → μv,w (A) = 〈EAv, w 〉.Then, we have for all 0 ≤ a ≤ b and every ρ ∈ R that
|μu†,u |([ a, b ]) ≤
(∫
[a,b ]
λ−ρdμu†,u † (λ)
) 12 (∫
[a,b ]
λρdμu,u (λ)
) 12
. (23) 14 Proof: Using the spectral representation theorem, see for instance [2, Chap-ter X.5.3, Corollary 4], we may assume that the operator T is a multiplication operator u 7 → mu , where m ≥ 0 is a bounded measurable function on a measure space (Ω , Σ, μ ) and U is the Lebesgue space U = L2(Ω; μ). In this case, the spectral measure of T is given by 〈EAu†, u 〉 = ∫
m−1(A)
u†udμ. Therefore, for every ρ ∈ R, we have the representation
∫
[a,b ]
λρd|μu†,u |(λ) =
∫
m−1([ a,b ])
mρ|u†| | u| dμ.
Thus we can estimate with the Cauchy–Schwarz inequality for arbitrary
ρ ∈ R:
|μu†,u |([ a, b ]) =
∫
m−1([ a,b ])
|u†| | u| dμ(λ)
≤
(∫
m−1([ a,b ])
m−ρ|u†|2dμ
) 12 (∫
m−1([ a,b ])
mρ|u|2dμ
) 12
=
(∫
[a,b ]
λ−ρdμu†,u † (λ)
) 12 (∫
[a,b ]
λρdμu,u (λ)
) 12
.
Lemma 15. Let μ be a non-negative finite Borel measure on R with compact support in [0 , ∞). Let 0 ≤ ν < ρ . Suppose that there exists a constant C > 0
such that μ([0 , λ )) ≤ Cλ ν for all λ ≥ 0. Then
∫
[Λ ,∞)
λ−ρdμ(λ) ≤ C ρρ − ν Λν−ρ for all Λ > 0. (24)
Proof: For λ ≥ 0 define I(λ) := μ([0 , λ )) and g(λ) := λ−ρ. Then we can write the above integral as a Stieltjes integral, and apply integration by parts,
∫
[Λ ,∞)
λ−ρdμ(λ) =
∫ ∞
Λ
g(λ)d I(λ)= g(λ)I(λ)|λ=∞
λ=Λ
−
∫ ∞
Λ
I(λ)d g(λ).
Because μ is finite and g(∞) = 0 it follows that g(λ)I(λ)|λ=∞
λ=Λ
≤ 0. There-fore, and taking into account that g is smooth, it follows that
∫
[Λ ,∞)
λ−ρdμ(λ) ≤ −
∫ ∞
Λ
I(λ)d g(λ) = −
∫ ∞
Λ
I(λ)g′(λ)d λ.
Then we use the assumption I(λ) ≤ Cλ ν and the monotonicity of g, to see that ∫
[Λ ,∞)
λ−ρdμ(λ) ≤ − C
∫ ∞
Λ
λν g′(λ)d λ = C ρρ − ν Λν−ρ.
Thus the assertion is proved.
15 References
J. Cheng and M. Yamamoto. One new strategy for a priori choice of regular-izing parameters in Tikhonov’s regularization, Inverse Problems , volume 16, L31–L38, 2000. N. Dunford and J.T. Schwartz. Linear Operators I,II . Wiley, New York, 1963. H. W. Engl, M. Hanke, and A. Neubauer. Regularization of inverse prob-lems , volume 375 of Mathematics and its Applications . Kluwer Academic Publishers Group, Dordrecht, 1996. C. W. Groetsch. The Theory of Tikhonov Regularization for Fredholm Equa-tions of the First Kind . Pitman, Boston, 1984. T. Hein and B. Hofmann. Approximate source conditions for nonlinear ill-posed problems – chances and limitations. Inverse Problems , 25:035003, 2009. B. Hofmann, B. Kaltenbacher, C. P¨ oschl, and O. Scherzer. A convergence rates result for Tikhonov regularization in Banach spaces with non-smooth operators. Inverse Probl. , 23(3):987–1010, 2007. B. Hofmann and M. Yamamoto. On the interplay of source conditions and variational inequalities for nonlinear ill-posed problems. Appl. Anal. ,89(11):1705–1727, 2010. A. Neubauer. On converse and saturation results for Tikhonov regularization of linear ill-posed problems. SIAM J. Numer. Anal. , 34:517–527, 1997. O. Scherzer, M. Grasmair, H. Grossauer, M. Haltmeier, and F. Lenzen. Vari-ational methods in imaging , volume 167 of Applied Mathematical Sciences .Springer, New York, 2009. T. Schuster, B. Kaltenbacher, B. Hofmann, and K. S. Kazimierski. Regu-larization methods in Banach spaces , volume 10 of Radon Series on Com-putational and Applied Mathematics . Walter de Gruyter GmbH & Co. KG, Berlin, 2012.
16 |
13927 | https://pmc.ncbi.nlm.nih.gov/articles/PMC2747339/ | Review of the safety and efficacy of imiglucerase treatment of Gaucher disease - PMC
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. 2009 Sep 15;3:407–417. doi: 10.2147/btt.2009.3497
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Review of the safety and efficacy of imiglucerase treatment of Gaucher disease
Deborah Elstein
Deborah Elstein
1 Gaucher Clinic, Shaare Zedek Medical Center, Jerusalem, Israel
Find articles by Deborah Elstein
1,✉, Ari Zimran
Ari Zimran
1 Gaucher Clinic, Shaare Zedek Medical Center, Jerusalem, Israel
Find articles by Ari Zimran
1
Author information
Article notes
Copyright and License information
1 Gaucher Clinic, Shaare Zedek Medical Center, Jerusalem, Israel
✉
Correspondence: Deborah Elstein, Gaucher Clinic, Shaare Zedek Medical Center, PO Box 3235, Jerusalem 91031, Israel, Tel +972-2-655-5093, Fax +972-2-651-7979, Email elstein@szmc.org.il
Issue date 2009; Collection date 2009.
© 2009 Elstein and Zimran, publisher and licensee Dove Medical Press Ltd.
This is an Open Access article which permits unrestricted noncommercial use, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC2747339 PMID: 19774208
Abstract
Most patients who suffer from symptomatic Gaucher disease will benefit from enzyme replacement therapy (ERT) with imiglucerase. The safety profile is excellent, only a small percentage of those exposed developing antibodies; similarly, very few patients require pre-medication for allergic reactions. Within 3 to 5 years of imiglucerase therapy, best documented at doses of 30 to 60 units/kg/infusion, hepatosplenomegaly can be expected to be reduced so that the liver volume will be maintained at 1 to 1.5 times normal (30% to 40% reduction from advent of ERT) and spleen volume to ≤ 2 to 8 times normal (50% to 60% reduction from advent of ERT). For anemic and thrombocytopenic patients, with 2 to 5 years of imiglucerase, hemoglobin levels are expected to be ≥ 11 g/dL for women and children and ≥ 12 g/dL for men; and platelet counts in patients with an intact spleen, depending on the baseline value, should approximately be doubled. Bone crises and bone pain but not irreversible skeletal damage will improve in most patients. Nonetheless, some features and some symptomatic patients apparently do not respond equally well and/or perhaps inadequately. The benefit for patients with the neuronopathic forms is primarily in improved visceral and hematological signs and symptoms. There are still several unresolved issues, the high per-unit cost being an important one, which have spurred the development of biosimilar enzymes as well as chaperone therapies currently in clinical trials.
Keywords: Gaucher disease, enzyme replacement therapy, imiglucerase, substrate reduction therapy, pharmacological chaperones, cost
Introduction: Gaucher disease is a rare recessive disorder
Gaucher disease, the most prevalent glycolipid storage disorder, is a result of the genetic defect in the lysosomal enzyme β-glucocerebrosidase, and consequent accumulation of substrate, glucocerebroside, in the monocyte-macrophage system.1 Clinical heterogeneity is attributable, to a large extent, to (~300) mutations within the glucocerebrosidase gene (located at chromosome 1q21);2 however, the importance of epigenetic and environmental influences are beginning to be appreciated.3 Three clinical forms have been delineated, based on absence (type 1) or presence (types 2 and 3) of neurological signs4 and categorization of this sort is useful when talking about management options and genetic counseling.
Type 1: non-neuronopathic form
Type 1, the non-neuronopathic form, is the most prevalent, with an ethnic predilection among Ashkenazi Jews that may be due to a selective advantage which has not yet been identified.5 Many patients are virtually asymptomatic and are diagnosed incidentally.6 Age of onset of symptoms and disease course are variable even among patients homozygous for the common N370S (1226G) mutation which is considered mild and protective of neurological involvement.7 Anemia and/or thrombocytopenia are common findings on presentation. In children, height retardation may be noted.8 Osteopenia and Erlenmeyer flask deformity of the distal femur are common;9 symptomatic skeletal features are less prevalent, including “bone crises”, osteonecrosis of joints, and pathological fractures. Various imaging modalities assess/quantify bone involvement,10,11 but pre-symptomatic prediction of onset and rate of progression of bone disease has not been achieved. Lung involvement, especially infiltrative disease, is an uncommon but serious complication usually in splenectomized patients with liver involvement. Although the N370S mutation was considered to confer protection from neurological involvement, patients with this mutation may be at increased risk for parkinsonism12,13 and/or may develop peripheral nerve abnormalities.14,15
Type 2: acute perinatal lethal neuronopathic form
Type 2 is a lethal neuronopathic form usually associated with compound heterozygosity for a severe mutation and a null mutation,16–18 characterized by hypertonic posturing, strabismus, trismus, and retroflexion of the head during the first 6 months of life and death following aspiration pneumonia and/or apnea/laryngospasm by 2 years.19 Massive hepatosplenomegaly and lung involvement are usually seen.16
Type 3: sub-acute neuronopathic form with three variants
Type 3 is the more heterogeneous neuronopathic form, presenting in childhood and characterized by the pathognomonic sign of horizontal supranuclear gaze palsy (HSGP).20 Most patients have at least one L444P (1448C) mutation. Sub-classification of this form is based on relative prominence of neurological versus visceral findings.21 Type 3a patients exhibit mild-to-moderate hepatosplenomegaly and slowly progressive neurologic deterioration; recurrent myoclonic seizures are common. Type 3b has aggressive visceral involvement but only HSGP; and type 3c is marked by homozygosity for the D409H (1342C) mutation, mild visceral disease, HSGP, and importantly, progressive and fatal calcifications of left heart (mitral and aortic) valves, aorta, and other arteries.22
Nonetheless, while it serves the purposes of this review to maintain demarcations between disease types to highlight the potential effects of therapeutic interventions, one may also view Gaucher phenotypes as a continuum of clinical manifestations that obscures rigid classifications.23
Surrogate markers to assess efficacy of interventions
Although amelioration of disease-specific symptoms and/or signs, eg, reduction in hepatosplenomegaly and improved blood counts, is a natural measure of efficacy of treatment, because of variability of response and/or the extended time-course to achieve a measureable response in disease-specific symptoms, chitotriosidase activity levels which are elevated hundreds-fold in Gaucher disease, have been used in the past decade as the preferred surrogate marker.24 But, because of genetic deficiency of chitotriosidase in ~6% of the population, CCL 18 (PARC) levels are additionally evaluated25 and research is ongoing to indentify novel biomarkers that would better predict bony complications. Some clinicians use these surrogate markers as a guide to initiation of treatment and/or dosage changes.
Enzyme replacement therapy (ERT): the early years
Nearly 50 years ago DeDuve suggested that lysosomal storage disease may be treatable by enzyme replacement.26 With the development of adequate purification techniques, glucocerebrosidase was derived from human placental tissue;27 subsequent deglycosylation to expose mannose residues28 targeted the product to mannose receptors on the macrophages.29 Commercialization resulted in the development of alglucerase (Ceredase®; Genzyme Inc., Cambridge, MA, USA), the placental derivative, which was tested in a seminal 9-month clinical trial of 12 patients with type 1 disease.30 Within the first 5 years of alglucerase, its safety and efficacy in improving hemoglobin levels and platelet counts, and in reducing splenic and hepatic enlargement were affirmed.
In 1994, the human recombinant form, imiglucerase (Cerezyme®; Genzyme Inc, Cambridge, MA, USA), was approved based on two clinical trials: the first comparing safety and efficacy of imiglucerase with alglucerase, at (high-dose) regimen of 60 units/kg body weight/2 weeks;31 the second compared frequency of administration of imiglucerase, every 2 weeks versus 3 times a week, at (low-dose) 15 units/kg body weight.32 Conclusions noted no significant differences between and among the groups. Imiglucerase gradually replaced alglucerase, which is only available today for a handful of patients who are unable to tolerate imiglucerase.
Assessment modalities to quantify change: recommendation to reduce use less commonly available resources and decrease invasive modalities for routine evaluations
At minimum, monitoring disease expression and effects of ERT includes serial evaluation of hematological parameters and reduction in hepatosplenomegaly. Although ultrasonography may be recommended for repeat assessments as the least invasive and as having no risk of radiation,33 especially in children,34 it is the least used because of issues of reliability and reproducibility and the importance of observer experience. Computed tomography (CT) was universally employed in the past,35 and today, magnetic resonance imaging (MRI),36 especially because of justifiable concerns about radiation with periodic assessments is preferred. Similarly, scintigraphy37 is rarely employed.
The most accurate but least available tool for the bones is quantitative chemical shift imaging (QCSI).38
There are several bone and skeletal assessment scores,39–42 in addition to actual bone densitometry42,43 and bone marker42 evaluations. Echocardiography44 is recommended for follow-up of pulmonary hypertension. There are also specific tools for quality of life45 assessment and cognitive function46 which are optimal for clinical trials rather than routine evaluations. In children, assessments are age-appropriate47 and in patients with type 3 disease, cognitive functioning48 is important as well.
Based on years of experience, the Gaucher community may/should re-consider the need for minutely accurate routine spleen and liver volumes (not in clinical trials) and rely on physical examination and ultrasonography combined with improved hematology and biomarkers for routine evaluations of patients.
The Gaucher registry and “therapeutic goals”: benchmarks for optimal management?
Since 1991, the Genzyme Corporation has sponsored an international database of global outcome assessments, the International Collaborative Gaucher Group (ICGG). What was originally a commitment of the manufacturer to regulatory agencies for post-marketing surveillance, has evolved into a powerful tool of information about patient care. Published reports with ICGG data highlight patient demographics, and safety and efficacy of imiglucerase,49 as well as enable regulatory adjustments such as addition of type 3 to drug indications and modification regarding imiglucerase in pregnancy (see below). Indirectly, the almost predictable efficacy of imiglucerase has resulted in a new tool for assessing outcome. Based on ICGG input, the concept of therapeutic goals was introduced.50 These benchmarks reflect past experience in treated patients registered in the ICGG, but imply that future competitors may have to improve upon these outcomes to compete/supersede imiglucerase as standard care.
“Therapeutic goals” for ERT
The findings/expectations51 regarding hepatosplenomegaly are to reduce and maintain liver volume 1 to 1.5 times normal (by 30% to 40% by years 3 to 5 of ERT) and reduce and maintain spleen volume ≤ 2 to 8 times normal (by 50% to 60% by years 2 to 5 of ERT). For achievement of the therapeutic goals for anemic and thrombocytopenic patients, the findings/expectations respectively for hemoglobin are ≥ 11 g/dL for women and children and ≥ 12 g/dL for men; and for platelet counts in patients with an intact spleen and depending on the baseline value, a 1.5 to 2.0 fold-increase or doubling of counts by years 2 to 5 of ERT but not necessarily normalization of counts.
More recently, Weinreb et al52 presented the accumulated experience of the ICGG registry in achieving the above four goals. After 4 years of ERT, the findings in 195 patients (who were chosen based on various criteria and of sufficient data points but who are claimed to be representative of the entire cohort of 4760 patients) in achieving said goals was 23.6% to 54.9%. There were two additional therapeutic goals, reduction in bone pain and bone crises, but it is still unclear whether ERT directly impacts these symptoms: most patients at baseline do not suffer from bone crises (episodes of bone pain that are localized and self-limiting); however, many patients suffer chronic bone pain, but achievement of this goal was the most modest (7.7%) of the six parameters evaluated.52 Nonetheless, other registry-based studies have shown significant improvements in bone parameters within 4 years of ERT.53 Children assessed in a separate study showed equally impressive improvements in the visceral response and linear growth.54
Therapeutic goals for hemoglobin improvement and liver volume reduction ie, the less Gaucher-specific parameters were more often met than for splenic volume and platelet counts which are more specifically characteristic of Gaucher disease.1
Finally, the impact of alglucerase/imiglucerase on the surrogate markers, chitotriosidase55 and CCL 18 (PARC)56 has lent credibility to their reliability to track interventions.
ERT for bone disease
Among patients from the seminal alglucerase trial radiological evidence of skeletal change was noted only 42 months after advent of therapy,57,58 but pathological damages (eg, osteonecrosis, bone infarcts, fractures) were not reversed. Importantly, improved MRI signals with ERT are not necessarily clinically relevant since no definitive correlation exists between MRI findings and incidence/severity of skeletal complications such as avascular necrosis.59
The most probable advantage for patients prone to bone disease (beyond not undergoing total60 or partial splenectomy),61 is early administration of ERT as preventative, especially in children.62 This premise is based on fewer cases of avascular necrosis among children born after ERT availability relative to children growing up before availability of ERT.62 The challenge is still how to identify children at risk during the pre-symptomatic stage of the disease. This is particularly difficult in patient homozygous for the N370S mutation.
Osteopenia is not uncommon in the general population but apparently progression to osteoporosis, ie, decreased bone mineral density (BMD) as generally estimated by dual-energy X-ray absorptiometry (DEXA), is characteristic of adults with Gaucher disease.63 Imiglucerase (60 units/kg/2 weeks) significantly improves BMD in adult patients with improvement of BMD at the lumbar spine64 and femoral neck65 after > 3 to 4 years. Moreover, biochemical markers for bone formation increased, markers for bone resorption decreased, with resolution of bone crises, decreased bone pain, and fewer skeletal complications.65
In summary, it cannot be said with absolute certainty that ERT will prevent or reverse skeletal complications.
Co-administration of imiglucerase and alendronate for bone involvement
Bisphosphonates are well-documented as effective in increasing BMD in at-risk adult populations.66 A clinical trial of co-administration of imiglucerase with oral alendronate (40 mg/day) in adult patients proved that combined therapy was significantly better for Gaucher-related osteopenia relative to imiglucerase alone, although skeletal lesions remained irreversible.67 However, this placebo-controlled prospective study unfortunately did not include an alendronate-only arm which may have been equally effective. This speculation is based on the authors’ experience with osteopenic/osteoporotic patients without other significant Gaucher-related signs/symptoms, who have demonstrated impressive improvement in BMD using bisphosphonates and calcium supplementation alone.
It may be posited therefore that while various facets of bone disease in patients with Gaucher disease are specific Gaucher signs and theoretically amenable to ERT, there are a myriad of epigenetic and environmental factors, comparable to those seen in healthy populations,68 that impact a global marker such as BMD, and hence ERT would not be expected to completely correct all deficits.
Another tenable hypothetical construct is that the large enzyme molecules are equally incapable of traversing the blood-brain barrier as the small blood vessels of bony matrix and/or Gaucher-cell-infiltrated marrow which may have the character of immortalized cells and/or dead tissue. Be that as it may, preventing bone involvement and reversing destructive processes of (type 1 and type 3) Gaucher disease, remains unresolved.69
Other uncommon sites of Gaucher involvement where the effect of enzyme replacement is imperfect: lung and brain
The earliest report of ERT in severely affected patients with non-neuronopathic disease and infiltrative lung involvement reported improved functioning,70 but not always of clinical significance.71 Children with type 272 or type 373 neuronopathic disease who are more prone to pulmonary complications, generally showed no improvement in pulmonary features with ERT.
Pulmonary hypertension (PH) has been noted in some patients on ERT. It has been suggested that there is a predisposition for PH in patients with type 1, especially in the presence of additional genetic factors and epigenetic modifiers.74 Others have suggested a causal relationship44 between PH and ERT, particularly among young splenectomized women, but this has been difficult to prove. Treatment withdrawal may be considered in these patients who evince primary-like PH and progressive increases in TI gradient (>30 mmHg) with ERT during routine echocardiographic monitoring. Alternatively, adding PH-specific therapy to ERT has also been beneficial.75
As implied, ERT cannot be expected to change the natural course of neurological progression in type 2 disease76 or type 3 disease77 regardless of dosage regimens.
Summary of the clinical efficacy of imiglucerase
In summary, and with 15 years of experience with imiglucerase, the important conclusion seems to be that at advent of therapy most patients suffer from several clinically relevant signs and symptoms of Gaucher disease, and that these gradually (2 to 5 years) improve with exposure to ERT so that near-normalization of many parameters is possible, although some parameters or perhaps some segment of the bell-curve of symptomatic patients, do not respond equally well or perhaps not even adequately. Those patients with neurological features will probably only see benefit in improved visceral and hematological signs and symptoms.
Safety of imiglucerase
Most of the side effects listed for alglucerase and imiglucerase during the clinical trials and afterwards, were infrequently observed, typically mild, and almost always transient in nature. This excellent safety has allowed home therapy in many countries,78,79 and the administration of ERT during pregnancy despite the original warning in the package insert.80,81 The development of anti-glucocerebrosidase antibodies has been reported to occur among 15% of patients;82 primarily non-neutralizing IgG antibodies.
Unresolved issues
There were some concerning issues that were raised by clinicians based on patient reports: some related to the prevalence and impact of side effects of ERT such as weight gain and diabetes and metabolic syndrome-like conditions;83 or evolution of neurological symptoms in type 1 patients;84 the incidence of new bony complications in treated patients;85,86 and severe allergic reactions87 requiring pre-medication for infusions and the adverse related to both infusions per se and the pre-medications.
Recently, it has been implied that there is an increased risk for cancer among patients with the N370S allele88 and that there is also decreased survival among patients with type 1 disease which is ascribed to cancers, as well as cardiovascular and cerebrovascular events,89 the latter heretofore not having figured prominently as Gaucher-specific signs, and unexplainable based on studies of lipid-cholesterol profiles in Gaucher disease.90 These potentially immunologically based events have raised theoretical concerns about the nature of glucocerebroside storage and its potentially beneficial rami-fications.91 Because of the putative association of ERT with increased rates of malignancy, immune mediated disorders, and altered metabolic pathways ERT although indicated for patients with moderate to severe disease, may not be justified in patients with very mild disease.91
The dosage controversy has never been adequately resolved.92,93 The first and possibly overriding concern was the cost, but beyond this was the question of whether higher doses actually translate into more effective therapy.94 Moreover, as the above concerns about the inimical aspects of ERT and particularly high-dose ERT are raised, and the question of indications for therapy re-evaluated, the concept of maintenance regimens and/or (intermittent or complete) withdrawal95,96 and not just dose adjustments97 after achieving near-normalization of hematological and visceral parameters with ERT should also be re-considered.
Important unresolved issues that spur the initiatives for new modalities: cost
Cost remains a critical constraint in providing imiglucerase to symptomatic patients, although it must be noted that the Genzyme Corporation has been generous with a worldwide programs for compassionate use. However, expensive therapy for a potentially non-lethal disease such as type 1 Gaucher disease has both ethical ramifications98 because of issues of scarce resources, as well as economic considerations because national health budgets99 that subsidize expensive therapies are prioritized by societal value systems. Cost per unit is high; the manufacturer recommended regimen is high-dose; asymptomatic patients are treated prophylactically; and there is a commitment to life-long treatment because maintenance regimens at lower doses and/or drug vacations are not sanctioned. Thus, a less expensive enzyme, potentially safe and effective as imiglucerase, would be an attractive option and this despite the above clinically-relevant qualifiers
Biosimilar enzymatic preparations
Both Shire Human Genetics Therapies (Cambridge, MA, USA) and Protalix Pharmaceuticals (Carmiel, Israel) are currently in clinical trials with respective infusible enzyme therapies and are preparing New Drug Administration submissions to the FDA. The Shire enzyme, velaglucerase alfa, has the advantages of being produced in a human cell line and having the normal human sequence100 (imiglucerase has a point mutation) while the Protalix enzyme uses a high-yield plant cell system that is easily up-scalable in disposable bioreactors and free from any exposure to mammalian tissues.101 Both are being tested at two doses and in treatment-naïve patients as well as in patients who had been exposed to imiglucerase; velaglucerase alfa also has a head-to-head comparison with imiglucerase at the high-dose regimen. Because being dependent on a single source for therapy is not desirable, even for rare diseases, comparable treatments should be encouraged. Should these biosimilars actualize their potential as equally safe and comparably effective as imiglucerase plus be less expensive, they would be well-positioned to capture some of the imiglucerase market share.
Other therapeutic modalities: substrate reduction (inhibition) therapy
In 1996, Radin suggested that enzyme therapy, a “spectacularly expensive mode of treatment should be replaceable with a suitable enzyme inhibitor that simply slows formation of the lipid, and matches the rate of synthesis with the rate of the defective, slowly working beta-glucosidase”.102 The iminosugar N-butyl deoxynojirimycin (miglustat), an inhibitor of glucosyltransferase, the first committed step in the biosynthesis of glycolipids, was a harbinger of this new class of oral substrate inhibitors for lysosomal storage diseases but also indirectly introduced pharmacological chaperoning as a therapeutic modality (see below). Because of its more problematic safety profile, miglustat (Zavesca®; Actelion Pharmaceuticals) was approved by the EMEA (2002) only for adult patients with mild to moderate disease unsuitable for standard enzyme therapy and by the FDA (2003) with a similar caveat, for adults for whom ERT is not a therapeutic option. In addition, miglustat therapy showed improved BMD as early as 6 months after advent of therapy.103
In a switch-over trial that evaluated miglustat in patients clinically stable on imiglucerase, tolerability of miglustat and imiglucerase, alone and in combination, pharmacokinetic profile, organ reduction, and chitotriosidase activity were assessed. Combination therapy did not show a clinically significant benefit.104
In retrospect, one might raise the question why combination therapy would be recommended for type 1 disease, but the lack of interference between modalities may be noteworthy. On the other hand, for type 3 disease, small molecule therapy also has the potential to cross the blood-brain barrier;105 but in the case of miglustat, it was not less expensive and the clinical trial failed to meet the primary end point and/or to make a substantial impact on neuronopathic features of the disease. The clinical trial recruited patients with type 3 disease receiving (high-dose) imiglucerase and added miglustat (200 mg, tid);106 but a single case report of an adult, after 2 years of combined miglustat (200 mg, tid) and imiglucerase (60 units/kg/2 weeks), showed fewer tonic-clonic seizures and speech improved.107
A ceramide analog as SRT
The Genzyme Corporation has begun clinical trials with substrate reduction therapy with GENZ 112638 a ceramide analog of the substrate (rather than the sugar moiety as in miglustat). To date, preliminary results of efficacy are very encouraging with improvement in key clinical parameters, including bone, in patients with type 1 disease. Unlike miglu-stat, this compound is probably incapable of traversing the blood–brain barrier. Phase 3 trials will be recruiting in the 2009 to 2010 quartiles.
Other therapeutic modalities: chaperone therapy
“Enzyme enhancement therapy” or “chaperone-mediated therapy” offers a novel therapeutic strategy to increase residual function of mutant proteins. Enhancement of the mutant enzyme is achieved by employing small molecules to properly traffic mis-folded and/or unstable mutant enzymes from the endoplasmic reticulum (ER)108 and prevent ER-associated degradation in proteasomes.109 This therapeutic approach is especially applicable to Gaucher disease because only a modest increase in residual glucocerebrosidase is sufficient to ameliorate the clinical phenotype.110 Small molecule chaperones should be able to cross the blood-brain barrier.
The first potent in vitro inhibitor of glucocerebrosidase that met criteria as a pharmacological chaperone was isofagomine tartrate. It proved to be active-site-specific and capable of increasing residual glucocerebrosidase activity in fibroblasts (from patients with Gaucher disease) with the N370S mutation.111 Recent reports of the clinical safety of isofagomine (commercial name: Plicera™) in a phase 2 trial by Amicus Therapeutics (Cranberry, NJ, USA) show good tolerability and safety. However, the patients were those on ERT who discontinued for a 4-week period only in order to participate in the current trial; this may be insufficient to extrapolate that clinical parameters were maintained, since there was no period of ERT wash-out, and hence no information can be derived from this trial about efficacy; this awaits the report of the phase 2 trial in patients naïve to treatment. Experience with patients who had received ERT for varying periods and then had withdrawn from ERT for varying periods, showed a very slow wash-out of the beneficial effects that had accrued from ERT.95
A second pharmacological chaperone is ambroxol hydrochloride (ExSAR Corporation, Princeton Junction, NJ, USA). A high-throughput screen of chemical compound libraries identified this non-carbohydrate-based inhibitory molecule capable of raising glucocerebrosidase activity in fibroblasts (from patients with Gaucher disease) with the N370S mutation and the more rare F213I mutation; comparable fibroblasts treated with ambroxol had decreased levels of glucocerebrosidase in the ER and increased levels in lysosomes.112,113 Ambroxol was originally developed as a mucolytic agent (Mucosolvan®; Boehringer Ingelheim GmbH, Ingelheim, Germany) to improve expectoration in conditions associated with viscid mucus114 but also promotion of prenatal lung maturation and prophylaxis against newborn respiratory distress syndrome.115 An investigator-initiated pilot study with ambroxol in patients with type 1 Gaucher disease was begun in 2009 at the Shaare Zedek Medical Center.
Diltiazem, an inhibitor of L-type Ca(2+) channels, at neutral pH as found in the ER was also shown to exhibit biochemical characteristics of a glucocerebrosidase pharmacological chaperone: in vitro it increased enzyme activity in normal cells, in N370S/N370S and other Gaucher mutant cells.116 This drug is currently administered to patients with type 3 disease in the UK as another investigator-initiated study.
The ethical dilemma of expensive non-curative drugs for rare diseases
In discussing the successes of imiglucerase, it would be remiss not to highlight the medical ethical dilemma of an expensive non-curative drug such as imiglucerase for a rare but not lethal disease such as Gaucher disease. It is obvious that this allocation precludes availability of ever-more limited financial resources for more common and more fatal disorders. In 1983, the American Congress by virtue of the legal imperative of the Orphan Drug Act (similar legislation was subsequently enacted in Europe and Japan) encouraged clinical research into the several thousand “orphan” disorders each of which affects fewer than 200,000 Americans, by offering tax breaks and exclusive marketing rights. The reality of this legislation was to give pharmaceutical companies brave enough to invest in “orphan drugs” that finally make it to the pharmacy shelf, a captive audience whose only drug of choice was often personally unaffordable. Although no society would elect to curtail treatment on the basis of cost in the case of abundance of resources, this is not the reality in any country in the world today. Thus, whereas one applauds the ultimately humanitarian outcome of underwriting medical research into diseases that had heretofore been neglected because few patients were involved, the phenomenon of a single “gold standard” treatment is no longer realistic nor desirable: per unit cost matters a lot. Moreover, in the case of type 1 Gaucher disease where patients have enjoyed imiglucerase therapy for 15 years, the concept of a plateauing of the clinical response is no longer dismissed: there are good responders, ie, those achieving near-normalization of the key clinical parameters within 2 to 5 years, and there are also less good responders who do not derive a complete clinical remission. There are those with brain, bone, and lung signs and symptoms that have emerged, whether before treatment or whether while on ERT, whose disease progresses in these organs and whose quality of life is considerably impaired. Indeed, there is the difficult question of whether ERT is ethically condonable in type 2 Gaucher disease.117 For this reason, the interest in new formulations and new modalities is a sign of the economics of our time, but also an appreciation of the fact that ERT is not the ultimately best resolution for all patients with the various forms of Gaucher disease.
Conclusion: hopes for the future
Clearly, all the above therapeutic modalities miss the point of providing a cure: although maintenance regimens have not been addressed, it is probable that symptomatic patients will continue to need specific therapy for Gaucher disease over the course of a lifetime. It is also manifestly clear that once brain, bone, and lung involvement become symptomatic, the probability of any of the above modalities to reverse the pathology is low indeed. Only gene therapy118 may prevent emergence of any symptoms and would be curative for all forms of the disease as well. Bone marrow transplantation (BMT) has been used in the past,119 albeit in limited cases and mostly in children with type 3 disease, in order to ameliorate the enzyme deficiency by providing enzyme competent cells from the donor marrow. Enzyme levels were restored and glucocerebroside levels in plasma were normalized one year after BMT in all the engrafted patients,120 but this modality has been abandoned in non-neuronopathic Gaucher disease because of its inherent risks for a non-life-threatening disorder. Studies of human CD34 cells were carried out to evaluate their potential use in a gene therapy approach to Gaucher disease and trials in humans were initiated.121,122 Unhappily progress has been erratic and new clinical trials in patients with Gaucher disease are not yet on the horizon. However, based on a new murine model of type 1 Gaucher disease and using low-risk conditioning regimens (non-myeloablative doses of busulfan) there have been encouraging data that only a low level of normal or gene-corrected cells with engraftment can induce a beneficial therapeutic outcome.123
Abbreviations
BMD
bone mineral density
BMT
bone marrow transplantation
CT
computed tomography
DEXA
dual-energy X-ray absorptiometry
EMEA
European Medicines Agency
ER
endoplasmic reticulum
ERT
enzyme replacement therapy
FDA
(US) Food and Drug Administration
HSGP
horizontal supranuclear gaze palsy
ICGG
International Collaborative Gaucher Group
MRI
magnetic resonance imaging
PH
pulmonary hypertension
QCSI
quantitative chemical shift imaging
SRT
substrate reduction therapy
TI
tricuspid insufficiency
tid
three times a day
Footnotes
Disclosures
DE receives consulting fees from Shire HGT. AZ receives consultancy fees from Shire Human Genetic Therapies; receives consultancy fees and has options in Protalix Therapeutics and sits on their Scientific Advisory Board; receives support from Genzyme Therapeutics for participation in the ICGG registry.
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Abstract
Introduction: Gaucher disease is a rare recessive disorder
Type 1: non-neuronopathic form
Type 2: acute perinatal lethal neuronopathic form
Type 3: sub-acute neuronopathic form with three variants
Surrogate markers to assess efficacy of interventions
Enzyme replacement therapy (ERT): the early years
Assessment modalities to quantify change: recommendation to reduce use less commonly available resources and decrease invasive modalities for routine evaluations
The Gaucher registry and “therapeutic goals”: benchmarks for optimal management?
“Therapeutic goals” for ERT
ERT for bone disease
Co-administration of imiglucerase and alendronate for bone involvement
Other uncommon sites of Gaucher involvement where the effect of enzyme replacement is imperfect: lung and brain
Summary of the clinical efficacy of imiglucerase
Safety of imiglucerase
Unresolved issues
Important unresolved issues that spur the initiatives for new modalities: cost
Biosimilar enzymatic preparations
Other therapeutic modalities: substrate reduction (inhibition) therapy
A ceramide analog as SRT
Other therapeutic modalities: chaperone therapy
The ethical dilemma of expensive non-curative drugs for rare diseases
Conclusion: hopes for the future
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13928 | https://brainly.com/question/2040809 | [FREE] Why does a hurricane weaken when it moves toward land? A. There is no low-pressure area on land. B. It’s - brainly.com
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Why does a hurricane weaken when it moves toward land?
A. There is no low-pressure area on land.
B. It’s not fed with moisture on land.
C. There is no high-pressure area on land.
D. Gravity pulls it back.
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The correct answer of the given question above would be the second option, option B. A hurricane weakens when it moves toward land because it is not fed with moisture on land. It gets its moisture from the waters of the ocean, and once it hits the land, it looses that strength. Hope this answer helps.
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A hurricane weakens when it moves toward land because it loses its moisture source and is affected by cooler temperatures that reduce its energy. The correct answer is option B: It's not fed with moisture on land. Without warm water and moisture, the storm no longer has the energy it needs to sustain itself.
Explanation
A hurricane weakens when it moves toward land mainly for two reasons: it loses its source of moisture and it is cut off from the warm water that fuels it.
Loss of Moisture: Hurricanes are formed over warm ocean waters where they absorb moisture from the sea surface. This moisture is critical for maintaining the storm's strength. When a hurricane moves onto land, it can no longer gather moisture from the ocean, leading to a loss of energy and strength.
Cooler Surface Temperatures: As hurricanes travel over land, they also encounter cooler land surfaces, which do not provide the necessary energy that warm ocean water does. Cooler temperatures reduce evaporation, which is important for the storm's heat and energy.
As a result of these factors, the hurricane begins to dissipate and weaken, often leading to reduced wind speeds and rainfall intensity. Although hurricanes can still bring significant weather conditions after making landfall, they generally become less powerful over time.
Understanding this process is vital for predicting the behavior of hurricanes and preparing for their impacts on coastal areas.
Examples & Evidence
An example of this phenomenon can be seen with Hurricane Katrina, which weakened significantly after making landfall in Louisiana. Initially, it was fueled by the warm waters of the Gulf of Mexico, but once it moved inland, it lost strength and began to dissipate. Another example is Hurricane Sandy, which also weakened as it approached the New Jersey coast due to a similar loss of moisture and energy.
Meteorological studies show that hurricanes draw energy from warm ocean water, and this energy source is significantly diminished when the storm moves over land. Data collected during past hurricanes consistently demonstrate that storms weaken shortly after making landfall, further confirming that the lack of moisture and heat from the ocean is key to their dissipation.
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13929 | https://citizendium.org/wiki/Acetone | Acetone - Citizendium
Acetone
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Acetone.
Acetone (also called propanone (IUPAC name), 2-propanone, propan-2-one, 2-oxo-propane, β{\displaystyle \beta }-ketopropane, dimethyl ketone or dimethyl ketal), is an aprotic colorless solvent widely used in organic chemistry reactions and the main ingredient in many fingernail polish removers. Because it has a medium-high dipole moment it can be used for organic reactions in which ionic reagants are used or in which ionic reaction intermediates are produced. Nucleophilic reactions proceed faster in acetone, compared to alcohols, because anions are less well solvated. Acetone is miscible with many compounds including water, alcohols, ethers and many other solvents.
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Contents
1 Physical properties
2 Household uses
3 Acid-base features
4 Carbon-carbon bond formation
5 Halogenation
6 Haloform reaction
7 Hazards
8 References
Physical properties
Acetone, chemical formula CH 3 C(=O)CH 3, is a planar, highly polar molecule with a large dipole (μ{\displaystyle \mu } = 2.88) and large dielectric constant (ϵ{\displaystyle \epsilon } = 20.7). The presence of a central carbonyl group, with only one carbon on each side, makes it the smallest ketone. Its high dielectric constant means that it can separate ionic charges fairly well. Acetone has a very high vapor pressure, is flammable and boils at 56.5 Celsius. Although acetone has two possible forms, the keto- and the enol-forms (see figure), the keto- form is greatly favored for acetone.
Household uses
Acetone is often the main ingredient, along with acetonitrile, in fingernail polish removers, although its use is being faded out due to environmental concerns. It is also used as a degreasing agent and glassware cleanser. Superglue and many epoxide mixtures use acetone as the solvent. Because water is an azeatrope when acetone evaporates, it can be used as a drying agent.
Acid-base features
The oxygen atom of acetone has two lone electron pairs that can be shared with Lewis acids, and thus is a Lewis base. Lewis acids such as AlCl 3 or BF 3 can form complexes with acetone in which the metal atom shares some of the oxygen atoms electron pairs. The resulting partial positive charge on the carbonyl oxygen atom is stabilized by a resonance structure in which the carbonyl double bond becomes a single bond and donates electrons to the oxygen. As a result, the carbonyl carbon atom becomes partially charged and becomes more electrophilic. Protonation of the carbonyl oxygen also enhances the electrophility in a similar manner.
Acetone can also function as a very weak acid, with a pK a of 19. When a base removes one proton from either of the non-carbonyl carbons of acetone, a carbanion is produced in which one of the carbon has a negative charge. This structure has a resonance form, an enolate ion, in which a double bond is formed between this negative carbon and the carbonyl carbon, the C-O bond becomes a single bond, and the negative charge is placed on the oxygen atom. This is the deprotonated enol form.
Carbon-carbon bond formation
Ketones can be used in aldol condensation reactions, although they are less reactive than aldehydes in this regard. Aldol condensation reactions occur when the carbanion carbon of an enolate reacts with the carbonyl carbon of a ketone or aldehyde, forming a new carbon-carbon bond. Acetone can react with itself, to produce the α{\displaystyle \alpha },β{\displaystyle \beta }-unsaturated ketone 4-methyl-3-penten-2-one, or with other carbonyl compounds.
Organometallic reagents, such a Grignard reagents or organolithium reagents, can also be used as nucleophiles to react with acetone. For example, R 3 C-MgBr, a Grignard reagent, can react with acetone to produce the tertiary alcohol R 3 C-C(OH)(CH 3)2. The formation of new carbon-carbon bonds in such a manner is critical to the synthesis of many complex organic molecules.
Halogenation
Ketones can be halogenated at one of the alpha carbon atoms when reacted with halogens in acidic conditions. Protonation of the carbonyl oxygen favors formation of the enolate (carbanion) form. Halogenation of the C=C bond then yields primarily the singly halogenated product. Thus, the bromination of acetone yields primary H 3 CC(=O)CH 2 Br, with lesser amounts of BrH 2 CC(=O)CH 2 Br and H 3 CC(=O)CHBr 2.
Haloform reaction
Acetone behaves very differently in the presence of halogens in basic solutions. Rather than becoming halogenated, methyl ketones are oxidized to a carbonylic acid in which the methyl group has been removed. The reaction of acetone with excess iodine (I 2) and sodium hydroxide (NaOH) yields sodium acetate, Na+:[H 3 CCO 2]-, and triiodomethane (iodoform), CHI 3, an antiseptic agent commonly used in hospitals.
Hazards
Both the liquid and vapor phases of acetone are extremely flammable, with a flash point of -20C. It is also a severe contact hazard. Excess exposure can lead to coughing or unconsciousness. Contact with eyes may lead to perminent damage.
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13930 | https://www.reddit.com/r/askmath/comments/1dnyuym/why_isnt_the_outcome_66_treated_as_two_separate/ | Why isn't the outcome (6,6) treated as two separate outcomes when you roll two dice? : r/askmath
Skip to main contentWhy isn't the outcome (6,6) treated as two separate outcomes when you roll two dice? : r/askmath
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Why isn't the outcome (6,6) treated as two separate outcomes when you roll two dice?
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13932 | https://s.wayne.edu/mbk/files/2020/05/MAT-7410-The-Class-Number-1.pdf | MAT 7410 (Advanced Algebra II) The Class Number Mohammad Behzad Kang May 1, 2020 Abstract This is an expository paper on the class number of a finite field extension of Q, intended as a survey of work done on the class group and class number of a number field. We first develop some background to gain an understanding of what the class number is, then proceed to build tools to discuss computations of the class number, and, finally, we discuss work primarily done by Gauss on binary quadratic forms and the form class group to learn about the foundations of the class number.
1 Understanding What the Class Number Is 1.1 Background Technology Definition 1.1.1. A number field (or algebraic number field) F is a finite field extension of Q. As such, F may be viewed as a finite-dimensional vector space over Q, with finite degree [F : Q] over Q.
• If F has degree 2 over Q, F is called a quadratic field. Examples of quadratic fields include Q( √ 7), Q( √ 8) = Q( √ 2), Q(ω) = Q(√−3), where ω is a primitive cube root of unity, and Q(√−5). Every quadratic field may be written in the form Q( √ d), where d ̸= 0, 1 is a square-free integer. If d < 0, Q( √ d) is called an imaginary quadratic field, and if d > 0, Q( √ d) is called a real quadratic field.
• If F has degree 3 over Q, F is called a cubic field. By the primitive element theorem, any cubic field may be written in the form Q(α) for some α ∈F such that the minimal polynomial of α over Q has degree 3. Examples of cubic fields include Q( 3 √ 2) and Q( 3 √ 5).
• If F is of the form Q(ζn), where ζn is a primitive n-th root of unity, F is called a cyclotomic field, and F has degree φ(n) over Q.
• Most generally, if f(x) is an irreducible polynomial of degree n over Q, then F = Q[x]/(f(x)) is a number field of degree n over Q. By the primitive element theorem, F may be written in the form Q(α) for some α ∈F such that the minimal polynomial of α has degree n over Q.
Our main focus will be on quadratic fields, but we will also include some results related to cubic and cyclotomic fields.
Definition 1.1.2. The ring of integers OK of a number field K is the subring of K consisting of algebraic integers in K. That is, OK is the set of elements α ∈K such that α is a root of a monic polynomial in Z[x].
As such, α ∈K will belong to OK if its minimal monic polynomial over Q is in Z[x]. These elements are also called the integral elements of K over Z, forming the integral closure of Z in K, which contains Z as a subring. OK may be viewed as a finitely-generated Z-module with an integral basis b1, b2, ..., bn ∈OK such that any element in OK can be written as a linear combination of basis elements with coefficients in Z.
When defining the class number of a number field, it will be necessary to think about certain ideals in this ring.
1 Proposition 1.1.3. The ring of integers of the quadratic field K = Q( √ d) is given by OK = ( Z[ √ d] d ≡2, 3 (mod 4) Z[ 1+ √ d 2 ] d ≡1 (mod 4) , where d ̸= 0, 1 is a square-free integer. (Theorem 3.4, ) We quickly note that it is not possible that d ≡0 (mod 4), since this would violate d being square-free.
Proving the above proposition amounts to first showing that OK contains Z[ 1+ √ d 2 ] if d ≡1 (mod 4) (OK contains Z, contains 1+ √ d 2 as it is a root of f(x) = x2 −x + 1−d 4 , and OK is a ring, so this follows), and contains Z[ √ d] if d ≡2, 3 (mod 4) (which holds since OK contains Z, contains √ d as it is a root of g(x) = x2 −d, and OK is a ring). We simply show from here that there are no more integral elements of K over Z, which uses no more technology than that of minimal polynomials and elementary modular arithmetic.
Remark 1.1.4. If ζn be a primitive n-th root of unity and K = Q(ζn), OK = Z[ζn] . One can also show that, if d is an integer that is not a perfect cube and K is the cubic field Q( 3 √ d), OK = Z[ 3 √ d].
Definition 1.1.5. A Dedekind domain R is an integral domain such that R is Noetherian (that is, if we have ideals I1 ⊆I2 ⊆· · ·Ik−1 ⊆Ik ⊆Ik+1 ⊆· · · of R, then there is some n ∈N such that In = In+1), R is integrally closed in its field of fractions S (that is, R equals the integral closure of R in S), and every nonzero prime ideal of R is maximal.
• One can show that, if R is a Dedekind domain, then R is a unique factorization domain if and only if R is a principal ideal domain, because any UFD S is a PID if and only if every nonzero prime ideal of S is maximal .
The setting of Dedekind domains is of great importance for us, because if K is a number field, then OK is a Dedekind domain (see Theorem 3.1.3 of ). Further, due to this, every ideal in OK uniquely factors as a product of prime ideals (due to Theorems 12.2 and 12.3 of ).
Definition 1.1.6. If K is a number field, and J ⊆K, J is a fractional ideal of OK if aJ = {aj : j ∈J} is an ideal of OK for some a ∈OK such that a ̸= 0. If aJ is a principal ideal of OK, then J is principal fractional ideal of OK. (, Definition 12.4) We will see that it is these ideals of OK that drive the definition of the class number of K.
If J is a (principal) fractional ideal of OK, some authors may simply write that J is a (principal) fractional ideal of K. Further, any ideal of OK is a fractional ideal.
1.2 The Class Number With all of the machinery in Section 1.1, we are now ready to introduce the notion of the class group and corresponding class number.
Definition 1.2.1. The class group (often called the ideal class group to distinguish from the form class group, which we discuss in Section 3) of a number field K (or of OK) is the quotient group ClK (or ClOK or Cl(K)) given by ClK = (fractional ideals of OK)/(principal fractional ideals of OK). The order of the class group is called the class number of K. (, Definition 12.9) It is important to note that the class number of a number field K is always finite. Class numbers are typically studied in the context of a number field, which is our primary focus. However, one may consider the class number of a general Dedekind domain, and I’m told by Andrew Salch that there are a relatively small subset of mathematicians such as Frank Okoh that also study the class number of non-Dedekind domains.
2 Example 1.2.2. Suppose that the class number of a number field K is 1. Then, every fractional ideal of OK is a principal fractional ideal of OK. Since any ideal of OK is a fractional ideal, it follows that every ideal of OK is a principal ideal of OK, and OK is a principal ideal domain. This implies OK is actually a unique factorization domain, since OK is a Dedekind domain. This leads to an important statement : a number field K has class number 1 ⇐ ⇒OK is a PID ⇐ ⇒OK is a UFD.
Continuing with this train of thought, it turns out that the larger the class number of a number field K, the greater the failure of OK to be a principal ideal domain, or, equivalently, a unique factorization domain. In this way, the importance of the class number of K lies in its ability to measure the failure of OK to be a unique factorization domain.
Computations of the class group and class number for a number field K are frequently done using algebraic tools such as the Minkowski bound, the trace and norm of elements α ∈K, and the discriminant of K. In the next section, we introduce these tools and use them to handle some of these computations.
2 Calculating the Class Number 2.1 Computational Tools A nice general reference for these computational tools is contained in Sections 3 and 4 of .
Proposition 2.1.1. If K is a number field with [K : Q] = n, there are exactly n embeddings of K into C.
Let’s convince ourselves of this. By the primitive element theorem, K = Q[x]/(f) = Q(γ) for some γ ∈K such that the minimal polynomial f of γ over Q has degree n. As an irreducible polynomial over C, this minimal polynomial has n distinct roots in C. Letting β1, β2, ..., βn be the roots of these roots, the n unique embeddings from K into C are given by σi : Q[x]/(f) , →C : x 7− →βi, where 1 ≤i ≤n. There are no more distinct embeddings than these n, since the image of x must be one of the βi.
We will continue to denote the n distinct embeddings of a number field of degree n over Q into C by σ1, σ2, ..., σn. If α ∈K, σ1(α), σ2(α), ..., σn(α) are called the conjugates of α. σ is real if σ(K) ⊂R and complex if σ(K) ⊂C.
If K is a number field with [K : Q] = n, and α, β ∈K, one can consider the multiplication map α : K − →K : β 7− →αβ. The characteristic polynomial of this map is given by Pα(x) = Πn i=1(x −σi(α)), and it has coefficients lying in Q. Recalling that the trace of a linear operator is the sum of its eigenvalues and the determinant of the operator is the product of the eigenvalues, this polynomial is precisely what gives rise to the following definition.
Definition 2.1.2. Let K be a number field, and let α ∈K. The norm of α is given by N(α) = NK/Q(α) = Πn i=1σi(α) ∈Q, and the trace of α is given by Tr(α) = TrK/Q(α) = Pn i=1 σi(α) ∈Q.
In particular, the trace is additive and the norm is multiplicative – that is, for α, β ∈K, Tr(α + β) = Tr(α) + Tr(β) and N(αβ) = N(α)N(β). The trace is also Q-linear.
Definition 2.1.3. Let K be a number field such that [K : Q] = n, with b1, b2, ..., bn an integral basis for OK. Then the discriminant of K is given by ∆K = σ1(b1) σ1(b2) · · · σ1(bn) σ2(b1) σ2(b2) · · · σ2(bn) · · · · · · · · · · · · · · · · · · σn(b1) σn(b2) · · · σn(bn) 2 = Tr(b1b1) Tr(b1b2) · · · Tr(b1bn) Tr(b2b1) Tr(b2b2) · · · Tr(b2bn) · · · · · · · · · · · · · · · · · · Tr(bnb1) Tr(bnb2) · · · Tr(bnbn) ∈Q 3 Example 2.1.4.
(a) Let K be a quadratic field Q( √ d), and consider α = a+b √ d ∈K. There are 2 distinct embeddings of K into C, given by σ1(α) = a + b √ d and σ2(α) = a −b √ d (and, note that, if K is imaginary, both embeddings are complex, while if K is real, neither are complex). This gives Tr(α) = (a + b √ d) + (a −b √ d) = 2a and N(a + b √ d) = (a + b √ d)(a −b √ d) = a2 −db2. If d ≡1 mod 4, an integral basis for OK is given by b1 = 1, b2 = 1+ √ d 2 , and we get ∆K = 1 1+ √ d 2 1 1− √ d 2 2 = d. If d ≡2, 3 (mod 4), an integral basis for OK is given by b1 = 1, b2 = √ d, and we get ∆K = 2 0 0 2d = 4d. Thus, we get a useful formula for the discriminant of the quadratic field Q( √ d), given by ∆K = ( 4d d ≡2, 3 (mod 4) d d ≡1 (mod 4) .
(b) Let K = Q( 3 √ d) = Q(θ) be a cubic field, where d is not a perfect cube, and consider α = a + bθ + cθ2 ∈K. There are 3 distinct embeddings of K into C, given by σ1(α) = α, σ2(α) = a + bζ3θ + cζ2 3θ2 and σ3(α) = a + bζ2 3θ + cζ3θ2, where ζ3 is a cube root of unity ⇒Tr(α) = σ1(α) + σ2(α) + σ3(α) = 3a + bθ(1 + ζ3 + ζ2 3) + cθ2(1 + ζ3 + ζ2 3) = 3a, and N(α) = σ1(α) · σ2(α) · σ3(α) = a3 + 2b2 + 4c3 + 6abc. An integral basis for OK is given by b1 = 1, b2 = 3 √ d and b3 = 3 √ d2. Thus, we get ∆K = 3 0 0 0 0 3d 0 3d 0 = −27d2.
(c) Let K = Q(ω), where ω is a primitive 5-th root of unity. Since [K : Q] = φ(5) = 4, there are 4 distinct embeddings of K into C, given by σ1 : ω 7− →ω, σ2 : ω 7− →ω2, σ3 : ω 7− →ω3, and σ4 : ω 7− →ω4. Let α = a+bω+cω2+dω3+eω4 ∈K. Then, Tr(α) = (a+bω+cω2+dω3+eω4)+(a+bω2+cω4+dω+eω3)+(a+ bω3+cω+dω4+eω2)+(a+bω4+cω3+dω2+eω1) = 4a+(b+c+d+e)(ω+ω2+ω3+ω4) = 4a−(b+c+d+e).
Calculating N(α) = σ1(α) · σ2(α) · σ3(α) · σ4(α) is algebraically much more complicated. An integral basis for OK is given by b1 = 1, b2 = ω, b3 = ω2, and b4 = ω3. Thus, we get ∆K = 1 ω ω2 ω3 1 ω2 ω4 ω 1 ω3 ω ω4 1 ω4 ω3 ω2 2 = 125. In fact, if K is the number field Q(ζn), where ζn is a primitive n-th root of unity (n ≥3), this agrees with the formula for the discriminant of K, given by ∆K = (−1)φ(n)/2 nφ(n) Πp|npφ(n)/(p−1) .
The discriminant of a number field K is necessary to use a primary tool for computing the class number of K, the Minkowski bound.
Definition 2.1.5. Let K be a number field of degree n over Q. Then, every ideal class in ClK contains an integral ideal in OK of norm less than or equal to Minkowski’s constant MK = p |∆K|( 4 π)r2 n!
nn , where r2 is half of the number of complex field embeddings of K into C. This is known as Minkowski’s bound. As a consequence, ClK is generated by prime ideals of norm at most MK.
The norm of an ideal I ⊆OK above refers to its absolute norm, N(I) = |OK/I|.
An important consequence of Minkowski’s bound (which we mentioned earlier in this paper) is that, if K is a number field, as the number of integral ideals in OK of a specified norm is finite, the class group of K must have finite order (see 5.3.6 of ).
We briefly note that the Hurwitz constant of a number field K serves as another upper bound to help compute the class number of K, and can be computed from an integral basis of K. However, the bound one achieves from Minkowski’s constant is more accurate. Ireland-Rosen’s bound is another less robust bound to assist with class number calculations.
Remark 2.1.6. Recall that if K = Q(α) is a number field, K contains Q and OK contains Z. For a prime ideal p of OK, p ∩Z is a prime ideal of Z. That is, p ∩Z = (p) for some prime number p, and we say that the prime p of OK lies over (p). More generally, prime ideals of an extension field lie over prime ideals of 4 the ground field. Thus, since prime ideals in Q are generated by prime numbers, to help determine prime ideals of OK, we look to determine prime ideal factorizations in K.
For a given (p), we can often find the primes p in OK lying over (p) by factoring the minimal polynomial f of α over Q into irreducible factors mod p. The ideal p =< p, h(α) > in OK lies over (p), then, if h is one of these irreducible factors. This correspondence between primes in OK lying over (p) and irreducible factors of f modulo p is one consequence of theory developed by Dedekind. Even when this does not suffice, we may still find primes p of OK lying over (p) by finding prime ideals of OK/pOK and pulling these back to OK .
Remark 2.1.7. If p is prime in OK, its norm N(P) is |OK/p|. The ideal pOK is the product of prime ideals with norm a power of p, and each power is determined by the ramification index of the corresponding prime ideal. Further, since every nonzero prime ideal of OK factors as a product of prime ideals of OK, one can show that the ideal p in OK is prime if it has norm p. Essentially, one can learn about the factorization of the ideal xOK in the ring of integers of a number field K by looking at the factorization of the norm of x as an element of Z.
A helpful consequence of this theory is that if a prime p ramifies in K (that is, (p) = pOK factors as a product of powers of prime ideals, and one of these powers is greater than 1), there is only one ideal of norm p in OK. Dedekind’s Theorem states that a rational prime number is ramified in K if and only if p|∆K. If a prime p splits as a product of two distinct prime ideals, then there are exactly 2 ideals of norm p in OK .
2.2 Examples Example 2.2.1.
(a) Let K = Q( √ d) be an imaginary quadratic field.
n = 2 and r2 = 2, so Minkowski’s bound gives MK = (2/π) p |4d| if d ≡2, 3 (mod 4) and MK = (2/π) p |d| if d ≡1 (mod 4). We now use this to compute the class number of K when d = −1, −2, −3 and −7. For d = −1 and d = −2, we have MK = 4/π ≈1.27 and MK = 4 √ 2/π ≈1.80, respectively. For d = −3 and d = −7, we have MK = 2 √ 3/π ≈1.10 and MK = 2 √ 7/π ≈1.68, respectively. In all four cases, the Minkowski bound is less than 2 ⇒every ideal class in ClK must contain the trivial ideal (1), as this is the only ideal of norm 1 in OK. Thus, there can only be one ideal class, and the class number of K is exactly 1. That is, ClK is trivial, and OK is a UFD.
(b) Now, let K = Q( √ d) be a real quadratic field. n = 2 and r2 = 0, so Minkowski’s bound gives MK = √ |4d| 2 if d ≡2, 3 (mod 4) and MK = √ |d| 2 if d ≡1 (mod 4). We now use this to compute the class number of K when d = 2, 3, 5 and 13. For d = 2 and d = 3, we have MK = √ 2 and MK = √ 3, respectively.
For d = 5 and d = 13, we have MK = √ 5/2 ≈1.12 and MK = √ 13/2 ≈1.80, respectively. Since the Minkowski bound is less than 2 for all four cases, we use the same argument as (a) to conclude that in every case, the class number of K is 1.
Example 2.2.2. Let K = Q(ζ5). Then n = 4, r2 = 2 and ∆K = 125, so Minkowski’s bound gives MK ≈1.7.
Thus, by the same arguments in (a), K has class number 1.
An important takeaway from the last two examples is, if we get MK < 2 for a number field K, we can conclude immediately that K has class number 1.
Example 2.2.3.
(a) Let K = Q(√−5). n = 2, r2 = 2, and ∆K = −20, so Minkowski’s bound gives MK = (2/π) √ 20 ≈2.85.
By Dedekind’s Theorem, 2 ramifies in K (which we can verify by observing that x2+5 factors as (x+1)2 mod 2, so that (2) = 2OK = (2, 1+√−5)2) ⇒there is only one ideal of norm 2 in OK, and the class number is at most 2. However, OK = Z[√−5] is not a unique factorization domain, since 6 = 2·3 = (1+√−5)(1−√−5), and the only units in OK are 1 and −1 (that is, none of those 4 factors are associates). Thus, the class number of K cannot be 1 ⇒the class number of K is 2.
(b) Let K = Q( √ 6). n = 2, r2 = 0, and ∆K = 24, so Minkowski’s bound gives MK = √ 6 ≈2.45. By Dedekind’s Theorem, 2 ramifies in K, and the rest of the argument proceeds as in (a), where OK = Z[ √ 6] 5 is not a unique factorization domain since −2 = (2 + √ 6)(2 − √ 6), and both factors are associates, because (2 + √ 6)/(2 − √ 6) = −5 −2 √ 6, which is a unit. Thus, the class number of K here is also 2.
Example 2.2.4.
(a) Let K = Q( √ 17). n = 2, r2 = 0 and ∆K = 17, so Minkowski’s bound gives MK = √ 17 2 ≈2.06. By Dedekind’s Theorem, 2 does not ramify in K – in fact, 2 splits as a product of two distinct prime ideals, and there are 2 ideals of norm 2 in OK. However, these ideals are principal, since −2 = [(3+ √ 17)/2][(3− √ 17)/2].
Thus, every ideal class contains a principal ideal, and the class number of K is 1.
(b) Let K = Q( √ 14). n = 2, r2 = 0 and ∆K = 56, so Minkowski’s bound gives MK = √ 14 ≈3.74. 3 remains prime, so there are no ideals of norm 3. 2 ramifies by Dedekind’s Theorem, so there is only one ideal of norm 2 in OK. But, this ideal is principal, since 2 = (4 + √ 14)(4 − √ 14). Thus, similar to (a), we get trivial class group, and the class number of K is 1.
Section 4.3.2 in is helpful in addressing the different cases considered when factoring the ideal (p) in OK for a quadratic field K. In the last example, case c2 was used for part (a), while cases a2 and b were used for part (b).
Example 2.2.5. Let K = Q(21/3). n = 3, r2 = 1 and ∆K = −108, so Minkowski’s bound gives MK ≈2.94.
Thus, every ideal in OK is equivalent to one whose norm is at most 2 – that is, we only need to check the prime ideals of norm at most 2 in OK, as each ideal class group contains such an ideal. The only ideal of norm 1 in OK is the full ring of integers. Consider a prime ideal p in OK of norm 2, lying over an ideal (p) of Z. In OK, we have the factorization 2OK = (2, 21/3)2 = (21/3)3. Thus, the ideal class group of OK is generated by (1) and (21/3), both principal. Therefore, the class number of K is 1.
Completing computations to find the class number of an imaginary quadratic field Q( √ d) partially addresses problems posed by Gauss in his work Disquisitiones Arithmeticae (1801). Gauss conjectured the following : • As d approaches −∞, the class number of Q( √ d) tends to ∞.
This was later proved by Heilbronn in 1934. In fact, it turns out that for a given class number, there are finitely many imaginary quadratic fields with that class number.
• For low class numbers, Gauss gave different lists of which imaginary quadratic fields have that class number, and conjectured this list to be complete.
For class numbers 1, 2 and 3, work of Baker, Stark, Heegner, and Oesterl´ e proved Gauss’ lists to indeed be complete. Watkins solved the problem for class number up to 100 in 2004. For class number 1, Gauss’ list was d = −1, −2, −3, −7, −11, −19, −43, −67, −163, now known as the Heegner numbers.
Providing such lists became known as the class number problem.
• There are infinitely many quadratic fields with class number 1.
This is still an open problem. A list of real quadratic fields K = Q( √ d) known to have class number 1 is complete up to d = 100, and contains that list. The article contains similar results for cubic and cyclotomic fields. Further, discusses a complicated formula, the class number formula, to compute the class number of a quadratic field along with related results.
Gauss formulated his work on the class number of quadratic fields in terms of binary quadratic forms, and defined a product for them. His work, largely included in Disquisitiones Arithmeticae, was clarified and successfully translated to have consequences related to ideal class groups of quadratic fields only later, after, for example, Kummer and Dedekind put in work to define ideals (see page 74 of , and the brief discussion on the next few pages as a supplement for what we study in Section 3). In fact, Gauss’ work with binary quadratic forms motivated Kummer and Kronecker, inspired by Gauss’ composition of classes of quadratic forms, to define number fields and ideal class groups in general. In the next section, we will take a closer look at the foundational work of Gauss and others on the class number in terms of the theory on these forms that they developed, and show its connection to the modern study of the class number of quadratic fields.
A general reference for what we discuss in Section 3 is (Sections 2A, 3A, 3D, and 7D, in particular).
6 3 Foundational Theory Using Quadratic Forms & Its Connection to the Class Group of Quadratic Fields 3.1 Relevant Theory of Quadratic Forms The theory of quadratic forms we discuss was first introduced and studied by Gauss, but built upon by others such as Lagrange, Dedekind and Dirichlet.
Definition 3.1.1. A binary quadratic form is a polynomial q(x, y) = ax2 + bxy + cy2 in two variables x and y, with coefficients a, b and c and discriminant ∆= b2 −4ac (and, it follows that ∆≡0 (mod 4) or ∆≡1 (mod 4)). When a, b, c ∈Z, q(x, y) is known as an integral binary quadratic form (often simply called a binary quadratic form, which is the convention we will use). We denote the space of binary quadratic forms of discriminant ∆by F∆.
If ∆̸= 0, q(x, y) is called nondegenerate; otherwise, q(x, y) is degenerate. q(x, y) is primitive if gcd(a, b, c), also known as the content of the form, is equal to 1.
A binary quadratic form is definite if ∆< 0, positive definite if ∆< 0 and a > 0, and indefinite if ∆> 0.
When ∆< 0, we may call q(x, y) an imaginary quadratic form, and if ∆> 0, q(x, y) is called a real quadratic form. If q(x, y) is a (positive) definite binary quadratic form, we say it is reduced if |b| ≤a ≤c and b ≥0 if a = c or a = |b|. If q(x, y) is an indefinite form, it is reduced if | √ ∆−2|c|| < b < √ ∆.
Two binary quadratic forms q(x, y) and q′(x, y) are equivalent, or properly equivalent (and, we may write q(x, y) ∼q′(x, y)) , if q′(x, y) = q(αx + βy, γx + δy) and the integer matrix A = α β γ δ has determinant 1 (i.e., A ∈SL2(Z)). q(x, y) and q′(x, y) are improperly equivalent if q′(x, y) = q(αx + βy, γx + δy) and A has determinant ±1 (as was considered by Lagrange). If q(x, y) and q′(x, y) are equivalent, they have the same discriminant ∆; however, there are inequivalent forms that have equal discriminants (a brief discussion of much of this is covered in ).
In their treatment of the theory of binary quadratic forms that leads to the ideal class group of quadratic fields, many authors restrict their attention to positive definite forms, for simplicity. Further, when working with binary quadratic forms, attention is often restricted to primitive forms, because every form is a multiple of a primitive one. Lastly, there is a notion of equivalence of binary quadratic forms under an action of GL2(Z), and since SL2(Z) is a subgroup of GL2(Z), this action restricts to the action of SL2(Z) on binary quadratic forms described above (see Section 1.2 of ).
The proper equivalence ∼defined in the above definition indeed defines an equivalence relation on the set of binary quadratic forms : (i) q(x, y) = q(1x + 0y, 0x + 1y) ⇒q(x, y) ∼q(x, y) , (ii) q(x, y) ∼q′(x, y) ⇒q′(x, y) = q(αx + βy, γx + δy) for some α, β, γ, δ ∈Z such that αδ −βγ = 1 ⇒ q(x, y) = q′(δx −βy, −γx + αy) ⇒q′(x, y) ∼q(x, y), and (iii) q(x, y) ∼q′(x, y), q′(x, y) ∼q′′(x, y) for binary quadratic forms q(x, y), q′(x, y), q′′(x, y) ⇒ q′(x, y) = q(α1x + β1y, γ1x + δ1y) and q′′(x, y) = q′(α2x + β2y, γ2x + δ2y) for some αi, γi, βi, δi ∈Z (i = 1, 2) such that α1δ1 −β1γ1 = α2δ2 −β2γ2 = 1 ⇒ q′′(x, y) = q((α1α2 + β1γ2)x + (α1β2 + β1γ2)y, (γ1α2 + δ1γ2)x + (γ1β2 + δ1δ2)y) ⇒q(x, y) ∼q′′(x, y).
We may form a factor set of the relation ∼consisting of the collection of proper equivalence classes of binary quadratic forms defined by ∼. In particular, as was considered by Gauss, we may study the set of proper equivalence classes of binary quadratic forms under ∼with a fixed discriminant, Cl(∆) = F∆/ ∼.
Brahmagupta’s composition law states that, given two numbers a2 + nb2 and c2 + nd2, we have (a2 + nb2)(c2 + nd2) = (ac −nbd)2 + n(ad + bc)2 = X2 + nY 2 (where X = ac −nbd, Y = ad + bc), i.e., that the set of numbers of the form a2 + nb2 are closed under multiplication. Gauss, restricting his attention to primitive binary quadratic forms, took inspiration from this in defining a composition q1(x1, y1) ◦q2(x2, y2) of two binary quadratic forms q1 and q2 of discriminant ∆to produce a form q3(x, y) with the same discriminant, where x, y are quadratic expressions in x1, y1, x2, y2. The especially remarkable property of this composition is that it gives Cl(∆) the structure of a finite abelian group. With this, we may now formally define the form class group (see Sections 4.2 and 4.3 of ).
7 Definition 3.1.2. The set of proper equivalence classes of binary quadratic forms with discriminant ∆ under composition of forms Cl(∆) = F∆/ ∼is called the form class group of discriminant ∆. The form class group is a finite abelian group, with order the class number |Cl(∆)| = h(∆) of discriminant ∆.
It is worth briefly commenting on the group structure of Cl(∆). Gauss provided a simple reduction algorithm that, given a binary quadratic form q(x, y), produced a properly equivalent reduced binary quadratic form q′(x, y) in a finite number of steps. Using Gauss’ algorithm, one can show that, if q(x, y) has negative discriminant, q(x, y) is properly equivalent to a unique reduced binary quadratic form (and, with this, we can construct a canonical representative of each equivalence class of Cl(∆)), and there is a cycle of reduced binary quadratic forms such that each element of a proper equivalence class of binary quadratic forms of positive discriminant is equivalent to one of the forms in this cycle . Further, the identity element of Cl(∆) is the principal class of discriminant ∆, the proper equivalence class of the principal form. If ∆≡0 (mod 4), the principal form is given by x2 −∆ 4 y2, and if ∆≡1 (mod 4), the principal form is given by x2 −xy + 1−∆ 4 (see Definition 2.7 of ). The inverse of the binary quadratic form q1(x, y) = ax2 + bxy + cy2 is given by q2(x, y) = ax2 −bxy + cy2. Finiteness of |Cl(∆)| = h(∆) was due to a result of Lagrange, although Gauss’ work sufficed to see this.
Example 3.1.3.
(i) There is only 1 proper equivalence class in Cl(−8), with the reduced binary quadratic form representative x2 + 2y2. Thus, h(−8) = 1.
(ii) Consider the form class group of discriminant −20. There are 2 distinct proper equivalence classes in Cl(−20) with reduced binary quadratic form representatives x2 + 5y2 and 2x2 + 2xy + 3y2. Thus, h(−20) = 2.
(iii) Consider the form class group of discriminant −56. There are 4 distinct proper equivalence classes in Cl(−56) with reduced binary quadratic form representatives q1 = x2 + 14y2, q2 = 2x2 + 7y2, q3 = 3x2 + 2xy + 5y2, and q4 = 3x2 −2xy + 5y2. Thus, h(−56) = 4. Further, using Gauss’ method for composing quadratic forms, one can verify the identities q3 ◦q3 ∼q4 ◦q4 ∼q2, and conclude that Cl(−56) is isomorphic to the cyclic group of order 4 (Example 4.4.6 and Exercise 4.15 of ).
In light of the above example, let us revisit the ideal class group of a quadratic field Q( √ d) from the first two sections of this paper for a moment. Q(√−8) = Q(√−2) has class number 1, as shown in example 2.2.1. Q(√−20) = Q(√−5) has class number 2, as shown in Example 2.2.3. One can also show that Q(√−56) = Q(√−14) has class group isomorphic to C4, and thus, has class number 4. It is no coincidence that, in all of these cases, the class number of K = Q( √ d) occurring as the order of the ideal class group of K agrees with the class number h(∆) of the form class group of discriminant ∆, the discriminant of K, as we will see in Section 3.2.
3.2 Connection to Class Groups of Quadratic Fields Definition 3.2.1. Let K be a number field. An element a of K is totally positive if σ(a) > 0 for any real embedding σ of K into R. A totally positive principal fractional ideal of OK, then, is an ideal in OK of the form (a) = aOK. The narrow class group of K is the quotient group C+ K = IK/P + K, where IK is the group of fractional ideals of OK and P + K is the group of totally positive principal fractional ideals of OK. |C+ K| is the narrow class number of K.
Let K be a quadratic number field Q( √ d). We note that if d < 0, there are no real embeddings σ of K into R, so every element of K is totally positive (that is, totally positive principal fractional ideals of OK coincide with principal fractional ideals of OK, and we see that C+ K is just the ideal class group of K), and if d > 0, we need σ1(λ), σ2(λ) > 0 in order for λ = a + b √ d ∈K to be totally positive, where σ1 : K − →R : a + b √ d 7− →a + b √ d and σ2 : K − →R : a + b √ d 7− →a −b √ d .
Theorem 3.2.2. Let K = Q( √ d) be a quadratic field with discriminant ∆. Then Cl(∆) is isomorphic to the narrow class group of K.
8 We note that, if d < 0, one can simply say that Cl(∆) is isomorphic to the ideal class group of K, following our note of the fact that the ideal class group of K is equal to the narrow class group of K in this case.
Thus, the narrow class number of K coincides with the class number of K if d < 0. However, if d > 0, the ideal class group of K may be half the size of the narrow class group of K (as is the case with Q( √ 3), which has class number 1 but narrow class number 2).
The above Theorem is at the heart of the connection between the form class group that we’ve discussed in this section and the ideal class group of a quadratic number field K discussed in Sections 1 and 2. It is significant in translating Gauss’ work on binary quadratic forms and his construction of the form class group of a specified discriminant to the more modern notions of the ideal class group and class number of a quadratic field K that measures the failure of OK to be a unique factorization domain.
Amazingly, Gauss’ work went far beyond defining a proper equivalence of binary quadratic forms to construct the form class group. He also built a looser notion of equivalence on forms to construct a genus of forms . This concept is out of the scope of this paper, but I’m certainly interested in continuing to read up on this idea.
References Class Number Problem. number problem.
List of Number Fields with Class Number One. of number fields with class number one.
Robert Ash. A Course in Algebraic Number Theory. Dover Publications, 2010.
Steven Charlton. Quadratic Forms and Quadratic Number Fields. spc/teaching/primes 17/handout2 quadratic field.pdf.
Brian Conrad.
Dedekind Domains.
math210b-dedekind-domains.pdf.
Keith Conrad. Factoring in Quadratic Fields. quadraticgrad.pdf.
David Cox. Primes of the Form x2 + ny2. Wiley-Interscience, 1989.
Daniel E. Flath. Introduction to Number Theory. Wiley, 1989.
Andrew Granville. Binary Quadratic Forms.
pdf.
Minhyong Kim. Math 3704 Lecture Notes.
Kimball Martin. Binary Quadratic Forms.
Kimball Martin. Prime Ideals.
Kimball Martin. Primes in Extensions.
James S. Milne. Algebraic number theory (v3.01), 2008. Available at www.jmilne.org/math/.
Nicholas Phat Nguyen. A Note on Cyclotomic Integers.
05390.pdf.
Corentin Perret-Gentil. The Correspondence Between Binary Quadratic Forms and Quadratic Fields.
Master’s thesis, 2012.
pdf.
Roy Zhao. The Class Number Formula for Quadratic Fields and Related Results, 2016.
math.ethz.ch/∼pink/Theses/2016-Junior-Paper-Roy-Zhao.pdf.
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Andrae John Banton
n!/(n - 5)! = 72 n!/(n - 3)!
(n - 3)!/(n - 5)! = (n - 3)(n - 4) = n^2 - 7n + 12 = 72 -->
n^2 - 7n - 60 = 0 --> (n + 5)(n - 12) = 0 --> n = 12 only since n>=5
11y
9
View all 4 replies
Vinod Dhale
nPr = n! / (n-r)!
nP5 = 72 nP3 ------------------ (I)
nP5 = n! / (n-5)! ------------- (II)
nP3 = n! / (n-3)! ------------- (III)
From (I), (II), & (III)
n! / (n-5)! = 72 n! / (n-3)!
After cancellation & re-arrangement of terms
(n-3)! = 72 (n-5)!
(n-3) (n-4) (n-5)! = 72 (n-5)!
(n-3) (n-4) = 72
n²-7n-60 = 0
After solving this equation,
(n-12) (n+5) = 0
So n=12 & n≠ -5
10y
7
Tanmoy Sengupta
12
10y
3
View all 3 replies
Arslan Afzal
N=-43.2
11y
2
View all 6 replies
Rithika Reethu
n = 12.
10y
Sindhukumari Parameswaran
nP5 = 72 nP3 ------------------ (I)
nP5 = n! / (n-5)! ------------- (II)
nP3 = n! / (n-3)! ------------- (III)
From (I), (II), & (III)
n! / (n-5)! = 72 n! / (n-3)!
After cancellation & re-arrangement of terms
(n-3)! = 72 (n-5)!
(n-3) (n-4) (n-5)! = 72 (n-5)!
(n-3) (n-4) = 72
n²-7n-60 = 0
After solving this equation,
(n-12) (n+5) = 0
So n=12 & n≠ -5
10y
P Siahaan
n!/(n-5)!=72 n!/(n-3)!
n!/(n-5)!=72 n!/(n-3)(n-4)(n-5)!
(n-3)(n-4)=72
(n-12)(n+5)=0
and n=12
10y
Piiyush Sharma
Parikshit Sharma
10y
Indranil Mukherjee
12, -5
10y
Anik Rahman
12
10y |
13934 | https://pmc.ncbi.nlm.nih.gov/articles/PMC3585029/ | Soft substrates normalize nuclear morphology and prevent nuclear rupture in fibroblasts from a laminopathy patient with compound heterozygous LMNA mutations - PMC
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. 2013 Jan 1;4(1):61–73. doi: 10.4161/nucl.23388
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Soft substrates normalize nuclear morphology and prevent nuclear rupture in fibroblasts from a laminopathy patient with compound heterozygous LMNA mutations
Chiara Tamiello
Chiara Tamiello
1 Department of Biomedical Engineering; Eindhoven University of Technology; Eindhoven, The Netherlands
Find articles by Chiara Tamiello
1,, Miriam AF Kamps
Miriam AF Kamps
2 Department of Molecular Cell Biology; Cardiovascular Research Institute Maastricht; Maastricht University Medical Centre; Maastricht, The Netherlands
3 Department of Dermatology; Maastricht University Medical Centre; GROW-School for Oncology and Developmental Biology; Maastricht, The Netherlands
Find articles by Miriam AF Kamps
2,3, Arthur van den Wijngaard
Arthur van den Wijngaard
4 Department of Clinical Genetics; Cardiovascular Research Institute Maastricht; Maastricht University Medical Centre; Maastricht, The Netherlands
Find articles by Arthur van den Wijngaard
4, Valerie L R M Verstraeten
Valerie L R M Verstraeten
3 Department of Dermatology; Maastricht University Medical Centre; GROW-School for Oncology and Developmental Biology; Maastricht, The Netherlands
Find articles by Valerie L R M Verstraeten
3, Frank PT Baaijens
Frank PT Baaijens
1 Department of Biomedical Engineering; Eindhoven University of Technology; Eindhoven, The Netherlands
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1, Jos LV Broers
Jos LV Broers
2 Department of Molecular Cell Biology; Cardiovascular Research Institute Maastricht; Maastricht University Medical Centre; Maastricht, The Netherlands
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2, Carlijn CV Bouten
Carlijn CV Bouten
1 Department of Biomedical Engineering; Eindhoven University of Technology; Eindhoven, The Netherlands
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1
Author information
Copyright and License information
1 Department of Biomedical Engineering; Eindhoven University of Technology; Eindhoven, The Netherlands
2 Department of Molecular Cell Biology; Cardiovascular Research Institute Maastricht; Maastricht University Medical Centre; Maastricht, The Netherlands
3 Department of Dermatology; Maastricht University Medical Centre; GROW-School for Oncology and Developmental Biology; Maastricht, The Netherlands
4 Department of Clinical Genetics; Cardiovascular Research Institute Maastricht; Maastricht University Medical Centre; Maastricht, The Netherlands
Correspondence to: Chiara Tamiello, Email: c.tamiello@tue.nl
Copyright © Landes Bioscience
This is an open-access article licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License. The article may be redistributed, reproduced, and reused for non-commercial purposes, provided the original source is properly cited.
PMC Copyright notice
PMCID: PMC3585029 PMID: 23324461
Abstract
Laminopathies, mainly caused by mutations in the LMNA gene, are a group of inherited diseases with a highly variable penetrance; i.e., the disease spectrum in persons with identical LMNA mutations range from symptom-free conditions to severe cardiomyopathy and progeria, leading to early death. LMNA mutations cause nuclear abnormalities and cellular fragility in response to cellular mechanical stress, but the genotype/phenotype correlations in these diseases remain unclear. Consequently, tools such as mutation analysis are not adequate for predicting the course of the disease.
Here, we employ growth substrate stiffness to probe nuclear fragility in cultured dermal fibroblasts from a laminopathy patient with compound progeroid syndrome. We show that culturing of these cells on substrates with stiffness higher than 10 kPa results in malformations and even rupture of the nuclei, while culture on a soft substrate (3 kPa) protects the nuclei from morphological alterations and ruptures. No malformations were seen in healthy control cells at any substrate stiffness. In addition, analysis of the actin cytoskeleton organization in this laminopathy cells demonstrates that the onset of nuclear abnormalities correlates to an increase in cytoskeletal tension.
Together, these data indicate that culturing of these LMNA mutated cells on substrates with a range of different stiffnesses can be used to probe the degree of nuclear fragility. This assay may be useful in predicting patient-specific phenotypic development and in investigations on the underlying mechanisms of nuclear and cellular fragility in laminopathies.
Keywords: lamina, laminopathies, substrate stiffness, nuclear shape alteration, nuclear rupture
Introduction
The structural continuity between the intracellular (nucleus and cytoskeleton) and the extracellular environment of adherent cells is crucial for cell fate.1,2 The coupling between nucleus and cytoskeleton via proteins embedded in the nuclear envelope and the connection between cytoskeletal filaments and the extracellular matrix (ECM) via focal adhesions, together are part of the mechanotransduction mechanism, i.e., the process of converting physical forces into biochemical signals and integrating these signals into cellular responses.3-5
Some of the structural connections between the nucleus and the cytoskeleton are altered by mutations in the LMNA gene, which encodes for lamina-type proteins, i.e., lamin A, lamin C and lamin AΔ10. Lamins are located just underneath the inner nuclear membrane of most differentiated somatic cells and form the nuclear lamina, a fibrillar network part of the nuclear envelope which plays a crucial role in the maintenance of nuclear shape and gives structural support to the nucleus.6,7 Consequent to disturbances in the structural connections with the cytoskeleton and in the nuclear lamina assembly, LMNA mutations lead to decreased cellular stiffness and increased mechanical weakness leading to increased sensitivity to mechanical stress.8,9 Abnormal nuclear morphology, compromised nuclear integrity and tendency to spontaneous nuclear disruption, even in the absence of external forces, are also reported for these cells.8-16
The family of genetic diseases associated with mutations in the LMNA gene is called laminopathies. Laminopathies are associated with a diverse array of tissue-specific degenerative disorders as well as syndromes with overlapping features. The most important pathologies included are: different types of striated muscle diseases, such as Limb-girdle muscular dystrophy, Emery-Dreifuss muscular dystrophy and dilated cardiomyopathy; abnormalities in adipose tissue development, including Familial Partial Lipodystrophy, type II (Dunningan syndrome) and type II diabetes; peripheral nerve diseases such as Charcot Marie-Tooth disease and systemic failure diseases such as Hutchinson Gilford Progeria Syndrome (premature aging). Most of the symptoms develop in the postnatal phase and may lead to early death.17 The molecular mechanisms giving rise to tissue-specific laminopathies are still largely unknown. The complexity of these diseases is further exemplified by the fact that identical genetic mutations can give rise either to a severe disease phenotype in one patient or no clinical symptoms at all in another person. These observations indicate that mutation analysis alone is not conclusive for diagnosis or prognosis of laminopathy development and consequent functional losses.
Here we propose to use cell culture on substrates with different stiffness to probe laminopathy cells from a progeroid syndrome patient with compound heterozygous mutations in the LMNA gene, consisting of p.T528M in combination with p.M540T.18 We hypothesize that soft substrates can protect nuclei of these laminopathy cells from morphological disturbances and structural weakness, as in this case lower forces are propagated to the weakened nucleus. We examined dermal fibroblasts from the laminopathy patient and healthy control dermal fibroblasts seeded on collagen-I coated polyacrylamide gels (PA gels) with stiffness varying over a physiologic range (3−80 kPa) and glass substrates as control. After 48 h from seeding, we analyzed nuclear shape and rupture, as well as actin cytoskeleton organization, which is the main determinant of cell shape, structure and cellular stiffness.19-21 Our results show that that only on soft substrates (3 kPa) the laminopathy cells tested respond similar to healthy control cells. Interestingly, we were able to probe the intracellular response of these cells by varying the stiffness of the extracellular environment. This suggests that modulation of substrate stiffness is an attractive tool to investigate mechanical functioning and fragility of genetically affected cells of individual patients as a phenotypic marker of the disease stage.
Results
We investigated the intracellular effect of increasing substrate stiffness on diseased dermal fibroblasts, isolated from a patient suffering from a progeroid syndrome due to compound heterozygous missense mutations (p.T528M and p.M540T) in the LMNA gene (LMNA mut) and we compared these findings with control human fibroblast cell line (NHDFα). For this purpose both cell types were seeded on collagen I coated polyacrylamide (PA) gels with stiffnesses ranging from 3 kPa to 80 kPa, as well as on collagen I coated glass substrates. Both cell types adhered and elongated when plated on the surface of the collagen I coated PA gels and glass substrates, except for the 3 kPa where fewer cells adhered and reduced cell spreading after attachment was observed after 48 h from seeding (Fig.1A). Fluorescent staining (phalloidin-TRITC) of the actin cytoskeleton at 48 h after seeding suggested that both cell types could equally sense the stiffness of the substrates as their actin cytoskeleton became more stretched and organized in bundles for substrates stiffer than 3 kPa (Fig.1B). Quantitative measurement of cell area and aspect ratio confirmed that soft substrates (3 kPa) with E = 3 kPa elicit significant lower cell spreading and elongation in both cell types (Fig.1C-D). However, no significant differences were observed on the 10, 20, 80 kPa PA gels and glass.
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Figure 1. Effect of substrate stiffness on cell morphology and actin cytoskeleton organization. (A)Representative bright field images of NHDFα and LMNA mut cells seeded on polyacrylamide gels with stiffness ranging from 3 kPa to 80 kPa taken 48 h after seeding. Fewer and less spread cells were present on 3 kPa polyacrylamide gels than on stiffer substrates for both cell types. Scale bar: 100 μm. (B) Actin organization in NHDFα and LMNA mut (phalloidin-TRITC, red) showed increased organization and tension on substrates stiffer than 3 kPa. Green color is given by lamin B1 staining. Scale bar: 20 μm. (C and D) Cell area and aspect ratio presented as box-and-whisker plots. The measurements of NHDFα and LMNA mut did not show significant difference, thus the values were considered as a group.
Nuclear shape of LMNA mut is abnormal on stiff substrates but preserved on soft substrates
Morphologically visible nuclear abnormalities are common in laminopathy cells.22 These abnormalities, seen as nuclear blebs, herniations and so-called honeycomb structures after immunostaining, seem to indicate the presence of weak spots at the nuclear membrane and/or nuclear interior. Here, we tested whether the extracellular substrate stiffness affects the frequency of these nuclear abnormalities in the LMNA mut cells. From the images of DAPI and lamin B1 immunolabeled nuclei, it became obvious that few abnormally shaped nuclei were seen in cells seeded at low stiffness after 48 h from seeding (between 3 and 4% of all nuclei) (Fig. S1). Representative images of normal and abnormally shaped nuclei are shown in Figure 2A. Further quantitative analysis of 600 nuclei per cell genotype showed that on soft substrates (3 kPa) both LMNA mut and NHDFα nuclei overall have a normal appearance (2.9 ± 0.4% NHDFα, 3.7 ± 0.4% LMNA mut, Fig.2B). However, while in the NHDFα control fibroblasts abnormally shaped nuclei were detected in about 3.0 ± 0.7% of the cells regardless of the substrate stiffness, a significant increase of abnormally shaped LMNA mut nuclei was observed on 10, 20, 80 kPa PA gels and glass substrates (respectively 8.2 ± 0.7%, 26.9 ± 5.0%, 44.7 ± 1.7%, 22.5 ± 2.4%) (Figs.2B and C). The fraction of misshapen nuclei in LMNA mut cells increased significantly on the 80 kPa gel (up to 44.7 ± 1.7% compared with 26.9 ± 5.0% on 20 kPa). A reason for this significant increase could be the higher cell density observed on the 80 kPa gels seeded with LMNAmut cells. As in a side experiment we observed increased nuclear aberrations with increased cell density, we therefore hypothesize that cell-cell contact played a role in the formation of nuclear abnormalities (Fig. S2). On the glass substrate results were similar to those on the 20 kPa PA gels (22.5 ± 2.4%). The findings on glass are in agreement with earlier studies, showing that 36% of all cells from this LMNA mut patient cultured on glass substrates had irregularly shaped nuclei with blebs, honeycomb figures, large and poorly defined protrusions.18
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Figure 2. Nuclear morphological abnormality regulation by substrate stiffness. (A) Immunofluorescent labeling of cell nuclei with DAPI (blue), lamin B1 (green) and overlay of the two in the most right panels allowed to distinguish between normally (upper row) and abnormally shaped nuclei (second and third row). In particular, the nucleus in the second row shows a protrusion and in the third row a bleb can be observed. Scale bars: 10 μm. (B) Frequency of abnormally shaped nuclei on increasing PA gel stiffness for LMNA mut and control NHDFα. Values represent means from at least 300 cells from two experiments. Bars represent SEM p < 0.05, p < 0.01 vs NHDFα on the same substrate stiffness (C) Statistical analyses of differences in frequency of misshapen nuclei for LMNA mut and NHDFα on the different substrate stiffness’s. , p < 0.05; no star, p > 0.05.
LMNA mut cells show a defective actin cytoskeleton on stiff substrates but not on soft substrates
In order to provide insight into the role of the actin cytoskeleton on the onset of nuclear abnormalities (protective mechanism of a soft environment on nuclear integrity), we investigated actin fiber organization using phalloidin-TRITC labeling. The actin cytoskeleton is known indeed to respond to substrate stiffness and affect cell shape and migration. Confocal microscopy of the phalloidin stained cells plated on 3 kPa PA gels showed a rounded morphology for both cell genotypes, with little polymerized actin formation that barely formed bundling of tensed fibers. In the perinuclear region there seemed to be no actin fibers, while we observed fibers running on top of the nucleus (actin cap) (Fig.3A). At 10 kPa and higher stiffnesses, cells demonstrated the typical well-spread and flattened morphology with development of bundles of tense stress fibers (Figs.1A and 3B). According to Khatau et al.,23LMNA mutant cells can lack the characteristic actin cap running above the nucleus. After analysis of confocal z-stacks of both cell genotypes in our study, we could not confirm a difference in actin cap presence. However, we did detect aberrations in actin cytoskeleton organization in about 5% of the LMNA mut cells plated on 10, 20, 80 kPa and glass, at 48 h after seeding but not on the 3 kPa. These aberrations included detachment of actin stress fibers in the perinuclear region with formation of a speckled pattern of actin which suggests actin depolymerisation in these areas (Fig.3C−E). Similar observations were already reported for cells cultured on glass coverslips.7,17,24,25
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Figure 3. Influence of substrate stiffness on cytoskeletal actin organization and aberrations. Confocal z-series taken from half height of the whole cell and relative orthogonal cross sections of NHDFα and LMNA mut immunocytochemical stained for F-actin in red (phalloidin-TRITC) and Lamin B1 in green at 48 h after seeding. (A) Representative fibroblast seeded on 3 kPa PA gels. It shows short and not tensed actin fibers, which are missing in the perinuclear region. An actin cap is running above the nucleus (white arrowhead). No differences could be noticed between LMNA mut and NHDFα. Thus no aberrations could be detected in the actin cytoskeleton of cells plated on soft substrates. (B)Control NHDFα on PA gel stiffer than 3 kPa, precisely on the 20 kPa PA gel. Actin stress fibers are tensed and well-structured also in the perinuclear region. The actin cap made of thick stress fibers run above the nucleus the (white arrowhead) nucleus. (C and D) Representative aberrations found in LMNA mut seeded on 20 (C)and 80 kPa (D) PA gels. Cells have a misshapen nucleus. Yellow arrowhead indicates the lack of actin fibers in the perinuclear region (D and E) and a speckled distribution of actin (E). The actin cap is running above the nucleus (white arrowhead). Scale bars: 20 μm. (E) Representative images of cells on three substrate stiffnesses. NHDFα (left panel) and LMNA mut (right panel) on 3kPa (F), 20 kPa (G) and 80 kPa (H) PA gels. Scale bar: 50 μm.
Disruptions of the actin-cytoskeleton and trypsinization partially normalize nuclear abnormalities in LMNA mut cells
To further prove the correlation between actin cytoskeletal tension and the onset of nuclear abnormalities, LMNA mut cells grown on collagen I coated glass bottom culture dishes were incubated for different period of times with cytochalasin D (cytoD), which inhibits actin dynamics and, consequently, causes disruption of the actin-cytoskeleton (Fig.4). Next, the drug was removed and LMNA mut cells were allowed to recover in normal growth medium for 1 h to overnight. Confocal microscopy on phalloidin-TRITC and DAPI stained LMNA mut cells showed that the short treatment (30 min) followed by 1 h recovery (short treatment + short recovery) disrupted the actin-cytoskeleton only mildly compared with untreated control LMNA mut (Fig.4A and B). Yet, there was no difference between the frequency of misshapen nuclei in this group and in the untreated control LMNA mut (18.4 ± 2.1% vs 22.3 ± 4.0%, Fig.4E). There was presumably not enough time for the nucleus to respond to the decrease in cytoskeletal tension or the degree of disruption did not allow any response. In contrast, a three hour treatment followed by an hour recovery (long treatment + short recovery) leads to serious disruption of the actin cytoskeleton and significantly less misshapen nuclei (11.8 ± 0.8%, Fig.4C). Upon three hours cytoD treatment followed by overnight recovery (long treatment + long recovery), the actin cytoskeleton completely recovered from the treatment and tensed stress fibers were visible (Fig.4D). The frequency of misshapen nuclei (19.1 ± 1.6%) was found to be comparable to that of untreated LMNA mut cells. Moreover we analyzed the changes in nuclear morphology due to cellular detachment of LMNA mut cells by trypsin, followed by re-adhesion to a glass substrate (Fig.5). At 30 and 60 min after seeding, nuclear folding due to trypsin treatment did not yet allow a reliable analysis of nuclear shape. At this stage of attachment the actin cytoskeleton was largely disorganized, seen as absence of tense stress fibers in these cells. Starting from 2 till 8 h after seeding the frequency of misshapen nuclei was significantly lower than that at 72 h (11.0 ± 2.0%, 13.3 ± 3.8%, 14.6 ± 2.3%, 15.6 ± 2.2%, 28.3 ± 3.5% respectively at 2, 4, 8, 24 and 72 h). At these time points actin reorganization did take place in the lower regions of the cell, making contact with the glass substrate, but stress fibers were absent at close distance to the nucleus. While after 24 h of attachment the actin organization was mainly restored, showing actin fibers in close contact with the nucleus, it took even longer (up to 72 h) until cells were fully stretched, and showing the regular percentage of abnormal nuclei (Fig.5A and B)
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Figure 4. Effects of transient cytoD treatment on LMNA mut nuclei. Representative confocal sections of LMNA mut seeded on collagen I coated glass substrates incubated with cytoD 1μM and recovered in normal growth medium. After fixation, cells were stained with DAPI (blue) to check for nuclear abnormalities and phalloidin-TRITC (red) to check for actin organization. (A)Untreated control LMNA mut. (B) Short treatment + short recovery: LMNA mut treated for 30 min with cytoD and recovered for 1 h. (C) Long treatment + short recovery: LMNA mut treated for 3 h with cytoD and recovered for 1 h. (D) Long treatment + long recovery LMNA mut treated for 3 h with cytoD and recovered overnight. Scale bar: 20 μm. (E) Frequency of misshapen nuclei in LMNA mut upon treatment with cytoD. At least 600 cells were assessed per each group., p < 0.05; no star, p > 0.05
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Figure 5. Alterations in nuclear shape and actin organization upon attachment of cells after trypsin treatment. (A) Representative confocal sections of LMNA mut seeded on collagen I coated glass substrates at 30 min, 1, 2, 4, 24 and 72 h after seeding. Cells were immunocytochemical stained for F-actin in red (phalloidin-TR) and lamin A/C in green. Inset at 4 h: 3D view (generated by ImageJ 3D-viewer, showing the position of the nucleus (green)at the upper region of the cell, with very few tense actin stress fibers (red) surrounding the nucleus). Scale bar: 10 μm. (B) Frequency of misshapen nuclei after seeding. ., p < 0.05; no star, p > 0.05. (C) Changes in nuclear bleb size upon attachment, visualized by immunofluorescence using the Jol2 lamin A/C antibody. Note the increase in size as well as the aberrant shape of the nuclear blebs. Note also that in most blebs a typical honeycomb structure of the lamina staining can be seen. Scale bar: 10 μm.
Strikingly, not only the number of cells with blebs, but also bleb size itself increased considerably with time, ranging from 1−25 µm 2 after 2 h (average 5.24 µm 2, n = 12) to 3−62 µm 2 (average 24.4 µm 2, n = 10, Figure 5C). This shows that nuclear morphology becomes partially normalized after trypsinization, which hydrolyzes the protein-protein bonds that attach cells to the extracellular matrix and consequently induces cell rounding along with reduction of cytoskeletal tension.
All together, these results suggest a direct correlation between the level of actin-cytoskeleton tension and the prominence of nuclear abnormalities.
Cellular compartmentalization in LMNA mut cells is not compromised on soft substrates
Given the increased presence of abnormally shaped nuclei in LMNAmut cells cultured on substrates stiffer than 3 kPa, we tested whether this was correlated with a loss of cellular compartmentalization. We chose promyeolocytic leukemia nuclear bodies (PML-NBs) as marker, as these assemblies of PML proteins are normally confined to the nuclear interior of non-proliferating cells (Fig.6A).26 Earlier studies by De Vos et al. and Houben et al. showed that frequent loss of PML-NBs from the nucleus to the cytoplasm can be found in laminopathy cells.14,27,28 In the current experiment, approximately 600 cells for each genotype, on each substrate, were screened manually for PML-NBs localization using fluorescent microscopy. We observed cytoplasmic PML-NBs (cytPML-NBs) in cases of abnormally shaped nuclei as well as for intact nuclei (Fig.6B and C). Therefore it is not possible to directly correlate abnormalities in the nuclear shape to the loss of cellular compartmentalization. Similarly to previous findings,14 4.4 ± 1.1% NHDFα control cells demonstrated cytPML-NBs, regardless of the substrate stiffness. On the 3kPa substrate, LMNA mut and NHDFα control cells showed no significant differences in the frequency of cells with cytPML-NBs (3.1 ± 0.5% LMNA mut and 2.0 ± 0.2% NHDFα). However, we did observe a gradual increase of LMNA mut cells with cytPML-NBs with increasing stiffness of the substrates between 3 and 20 kPa (from 3.1 ± 0.5% to 12.8 ± 1.2%), indicating increased frequency of cytPML-NBs in LMNA mut cells.
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Figure 6. PML-NBs localization as a marker for cellular compartmentalization. (A−C) Confocal sections representative of cell nuclei were immunolabeled with Lamin B1 (red), DAPI (blue) and PML-NBs (green) to investigate the localization of PML-NBs. Nuclei were counterstained with DAPI (blue). The most right panel shows the triple overlay. Scale bars: 10 μm. (A) Nuclei showing normal morphology and internal localization of PML-NBs. Cellular compartmentalization is intact. (B)Cytoplasmic localization of PML-NBs (cytPML-NBs) around a nucleus showing an abnormal morphology (white arrowhead). Loss of cellular compartmentalization is indicated by the exit of PML-NBs to the cytoplasm. (C) CytPML-NBs could be found also in normally shaped nuclei (white arrowhead) indicating that loss of cellular compartmentalization is not directly related to nuclear morphology abnormalities. (D) Relative frequency of NHDFα and LMNA mut showing cytPML-NBs. Values represent means from at least 600 cells from 2 experiments. Bars represent SEM p < 0.05, p < 0.01 vs NHDFα on the same substrate stiffness (E)Statistical analyses of differences in frequency of cytPML-NBs for LMNA mut and NHDFα on the different substrate stiffness’s. , p < 0.05; no star, p > 0.05.
Nuclear ruptures in LMNA mut cells increase with substrate stiffness, but are prevented on soft substrates
A recent study by De Vos et al.16 showed the occurrence of spontaneous nuclear ruptures in cells from laminopathy patients cultured on glass substrates. These ruptures never occurred in wild type cells (NHDFα) under the same culturing conditions. Based on these results and triggered by the finding of variable cytPML-NBs on different substrate stiffness, we hypothesized that mechanical cues provided by the extracellular environment might affect the frequency of nuclear rupture events. For this purpose we monitored living cells (about 20) for two hours at one or two minute intervals under a fluorescent microscope on 3, 10, 20, 80 kPa PA gels and glass substrate after 24−36 h from transfection with EYFP-nuclear localization signal (EYFP-NLS), which helped to check for nuclear integrity. Correct expression of EYFP-NLS was revealed by a constant intense intranuclear fluorescent signal. In NHDFα cells, as well as in LMNA mut cells on 3 kPa substrates we could not detect a nuclear rupture event in any cell examined. In contrast, for the stiffer substrates an increased frequency of nuclear rupture was detected in the LMNA mut cells, increasing from 20% (4/20) of LMNA mut with nuclear rupture on the 10 kPa substrate to 34.5% (10/29) on 80 kPa. The ruptures were visible as a sudden transient efflux of EYFP-NLS from the nucleus to the cytoplasm (Fig.7 and Video S1). This phenomenon, which lasts about 20 min and can occur repetitively in the same cell, was followed by restoration of EYFP-NLS signal in the nucleus and was not lethal for the cells. All together, these results confirm that soft substrates do not compromise the nuclear integrity of LMNA mut cells, while stiff environments do.
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Figure 7. Spontaneous ruptures of the nuclear membrane do not occur on soft substrates. (A) Montage of selected images from a time-lapse recording of LMNA mut cell cultured on 10 kPa PA gel and transfected with EYFP-NLS, sampled at 1 min intervals for 2 h (Video S1). Nuclear membrane rupture causes the decrease in intranuclear EYFP signal and increase in cytoplasmic EYFP signal (at 0:36, yellow arrowhead). Subsequently, the nuclear signal is gradually up taken by the nucleus and the rupture appears to be restored (at 0:51, orange arrowhead) (B) Evolution in time of EYFP-NLS (mean) intensity in the nucleus of the cell shown in (A).(C) Frequency of spontaneous nuclear membrane ruptures on the different substrates for LMNA mut. Error bars represent the square root of the number of recording.
Discussion
In this study we showed that, on soft substrates with stiffness of 3 kPa, abnormal nuclear morphology and nuclear ruptures in dermal fibroblasts from a laminopathy patient with compound heterozygosity for mutations in LMNA can be normalized. Normalization of nuclear shape at low substrate stiffness, i.e., in presence of low cytoskeletal tension, indicates that nuclear abnormalities correlate to the mechanical properties of the ECM, such as the collagen I-coated PA gels used in here. For the purpose of this study and in view of a future clinical application, we chose to investigate dermal fibroblasts because it is and easily accessible cell source to probe and investigate.29
A crucial finding is that the nuclei of LMNA mut cells used in this study do not develop an abnormal morphology when they are cultured on soft gels (3 kPa), while on stiffer substrates nuclei appear to have a misshapen shape (Fig.2) as reported also by Verstraeten et al.18 Abnormal nuclear phenotypes are normally found in cells with LMNA mutations,10,11,18,30 but the relevance of morphological abnormalities in the pathogenesis of laminopathies is not unravelled. Nuclear abnormalities are indeed not present in all diseased cells and there is no direct association between nuclear abnormalities and disease phenotype or severity.31 However, previous studies have been performed only on glass or stiff silicon substrates and did not consider the mechanotransduction feedback-loop, which influences the cellular response based on the ECM mechanical cues. Still, in order to establish correlations between genotype and phenotype repeated measures using cells from different patients or families of patients are needed, as the phenotypic variability in this family of diseases may lead to different responses of the nucleus to developing intracellular tension.
One reason for increased nuclear abnormalities, nuclear ruptures and loss of cellular compartmentalization might be that, on soft substrates, the nuclear membrane is exposed to reduced cytoskeletal forces, transduced from the ECM. This can be inferred by our results on the actin cytoskeleton organization and from the partial normalization of the nuclar abnormalities upon disruption of the actin cytoskeleton by cytoD and after cell trypsinization. When actin is not assembled into tensed stress fibers, it is likely that the force exerted on the nucleus is not enough to tear apart the nuclear membrane or to compress the nucleus, causing nuclear rupture at weak spots.
Furthermore, disorganization of the actin cytoskeleton in the perinuclear region of LMNA mut, was observed particularly on stiffer substrates and gave indication for an abnormal distribution of forces exerted to the nucleus, enhancing nuclear morphology disturbance. In contrast to Kathau et al.,23 we observed in the LMNA mut cells the presence of the nuclear shaping actin cap. Therefore, to explain these findings, we propose that a pulling mechanism in addition to a compressive pushing mechanism might play a role in altering nuclear morphology. We suspect that, on stiff substrates, the actin cap presses tightly against the nucleus and, in addition, organization of the stress fibers around the nucleus is abnormal, enhancing the probability of disturbance and rupture in the morphology of the genetically disorganized and weakened nucleus (Fig.8). This physical model can explain the observations on the different substrate stiffnesses. However the significant increase of misshapen nuclei on the 80 kPa PA gel is, in our opinion, due to an increase of cell-cell contact and increase in cell area and aspect ratio, which imply increase of cytoskeletal forces exerted on the nucleus. Reasons for the increase in cell area and cell aspect ratio might be found in changes in adhesive properties of the substrates. Hydrogels of increasing stiffnesses lead to increasing anchoring densities and thereby increase in cell spreading,32 while collagen absorption onto glass substrate could determine an anchoring density similar to that of the 20 kPa PA gel. While these in vitro studies cannot be directly interpolated to the in vivo situation our assay to measure nuclear weakness could well predict the development of a laminopathy phenotype in patients. A common denominator in (nearly) all laminopathies is the loss of specific tissues, seen as muscular dystrophies and/or lipodystrophies. For each of these laminopathies, its value will have to be proven.
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Figure 8. Proposed mechanism for actin cytoskeleton organization and effects on nuclear abnormalities on soft and stiff substrates. Schematic representation of the cross section of a cell seeded on substrates with different stiffness. Red represents the actin cytoskeleton while green the nucleus (dark green nuclear lamina). (A) When seeded on a soft substrate NHDFα and LMNA mut cells have the same response. They develop not tensed actin stress fibers which are not directly connected to the nucleus in the perinuclear region. Fibers run below and on top of the nucleus. Since low forces are exerted on the nucleus, the onset of nuclear abnormalities is prevented. (B) NHDFα cells seeded on a stiff substrate (stiffer than 3 kPa) develop tensed actin stress fibers which are well-organized in the whole cytoplasm. Stress fibers connect to the nucleus and form also the actin cap running on top of the nucleus. (C)LMNA mut cells on top of a stiff substrate develop tensed stress fibers. These stress fibers appear to be lacking in the perinuclear region and aggregates of actin are often visible. The pushing action of the actin cap (stress fibers running on top of the nucleus) together with the uneven distribution of pushing actin in the area around the nucleus lead to the development of nuclear abnormalities, further resulting in nuclear damage.
Taken together our data suggests that we were able to probe the response of the nucleus from the outside of the LMNAmut cells by using the mechanoresponsive pathways of the actin cytoskeleton. However, presently, we cannot rule out the involvement of microtubules, as they are known to be connected to actin via kinesin 1 and to the nuclear membrane via nesprin-4.3 Studies on nucleus and cytoskeletal elements co-transfected laminopathy cells could give insights on the precise mechanisms of nuclear rupture.
Substrate stiffness appears to modulate also nuclear integrity. Indeed, we detected that repetitive disruptions of the nuclear membrane, previously reported by De Vos et al.16 in cells from different laminopathy patients under standard culturing conditions, are prevented on soft substrates (3 kPa) but increasingly occurs on stiffer substrates. It is not clear how the cells can survive a repetitive disruption of the nuclear membrane, as mixing of cytoplasmic and nuclear components prevents appropriate nuclear localization of nuclear factors that can be crucial for several mechanisms (such as replication, transcription).
Also PML-NBs, often lost from the nucleus in laminopathy cells cultured on a glass substrate, were retained in the nucleus of LMNA mut seeded on soft substrate. However, since we observed cytoplasmic localization of PML-NBs without nuclear abnormalities (Fig.6), the presence of PML bodies in the cytoplasm is not indicative for disfunctions of the nuclear lamina. Moreover, in a parallel study a direct correlation between occurrence of cytoplasmic PML-NBs and nuclear rupture as seen with EYFP-NLS could not be established: while in some cases of nuclear rupture PML-NBs moved out of the nuclei, in other cases this did not happen. Conversely, leakage of PML-NB proteins into the cytoplasm or incomplete import of PML-NB proteins can cause cytoplasmic assembly of PML bodies without nuclear rupture.16
In conclusion, despite the fact that the data reported were from cells of only one laminopathy patient with rare compound mutation in the LMNA gene, our findings suggest that soft substrates could be used protect and possibly rescue cell from laminopathy patients with morphological disturbances and structural weakness. This study shows the value of using substrate stiffness based approach for improved diagnosis of genetically diseased cells in order to understand the interplay between genotype and phenotype. Elucidating the mechanotransduction pathways involved in the response of LMNA mutated cells to changes in the extracellular environment will also help to provide new insight into the genotype phenotype correlations.
Materials and methods
Cell cultures
Cells used in this study were primary skin fibroblasts. The laminopathies cells (LMNA mut) were obtained from a skin biopsy taken from a two-year-old male subject diagnosed with apparently typical Hutchinson Gilford Progeria Syndrome, which showed compound heterozygous mutations (LMNA^T528M/M540T).18 Informed consent was obtained from the parents of the proband for this study. Normal human dermal fibroblasts (NHDFα) obtained from the European Collection of Cell Cultures (Salisbury, United Kingdom) were used as a control. Details about culturing can be found in Supplementary Material.
Transfection for live-cell imaging
LMNA mut were transiently transfected with an EYFP-NLS construct33 (kind gift from Dr J. Goedhart, University of Amsterdam) using GeneJammer (Invitrogen, 204132) according to manufacturer's instructions at a GeneJammer/DNA ratio of 6:1 (μl per μg DNA). Transfection was performed 24 h after seeding of the cells and culture medium was changed 4 h after transfection to minimize cytotoxicity.
Coated polyacrylamide (PA) gels and glass substrates
Polyacrylamide (PA) gels coated with collagen I were used to create 2 dimensional substrates with controlled stiffness for NHDFα and LMNA mut. PA gel stiffness (expressed as elastic modulus, E) was controlled by modulating the bis-acrylamide crosslink concentration and was verified using an indentation test.34 The method used for the preparation was adapted from Pelham and Wang.35 Precursor mixtures of PA gels were made from acrylamide (40%, Sigma) and N, N', N'-methylene-bisacrylamide (bis-AA, 2%, Sigma-Aldrich) mixed with MilliQ water and Hepes 50mM. Final acrylamide concentrations were 5% or 10%, while bis-AA varied between 0.03% and 3%. Details of PA gels preparation can be found in Supplementary Material. Rat tail collagen I (BD biosciences) was covalently coupled the surfaces of the gels using the sulfo-SANPAH (Pierce Biotechnology) crosslinker in order to provide cellular attachment. The elastic modulus (E-modulus) of the PA gels was determined on gels prepared on coverslips (Menzel) with 25 μl of solution. Indentation was applied to the center of the gels with a spherical indenter (2mm-diameter) while measuring force and indentation depth. Afterwards a numerical model was iteratively fitted to these experimental data using a parameter estimation algorithm. The mechanical properties of the PA gels were quantified in three gels for each group in the same day when the cells were seeded. The indentation test revealed that differences between the batches and in time were not significant (data not shown). The gels prepared with 5% acrylamide/0.01% bisacrylamide, 5% acrylamide/0.05% bisacrylamide 5% acrylamide/0.3% bisacrylamide, 10% acrylamide/0.26% bisacrylamide had elastic moduli of 3.8 ± 0.9, 9.9 ± 3.7, 19.8 ± 3.6 and 81.7 ± 2.4 kPa respectively (mean ± SD), as shown in Table 1. This stiffness range (3–80 kPa) was created to mimic physiologically-relevant stiffness values similar to fat tissue (3 kPa), muscle (10–20 kPa) and collagenous bone (> 20 kPa).
Table 1. Composition and elastic modulus of the polyacrylamide gels used as substrates.
| | 3 kPa | 10 kPa | 20 kPa | 80 kPa |
:---: :---:
| Acrylamide | 5% | 5% | 5% | 10% |
| Bis-Acrylamide | 0.01% | 0.03% | 0.3% | 0.26% |
| E (kPa) | 3.8 ± 0.9 | 9.9 ± 3.7 | 19.8 ± 3.6 | 81.7 ± 2 |
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The glass substrates (coverslips No. 0, 13-mm diameter; Menzel), were sterilized in 70% ethanol and, subsequently, coated with adsorbed collagen I.
The covalent binding of collagen I coating to the substrates was examined by immunolabeling. The antibody used were mouse monoclonal antibody to collagen I (IGg1, diluted 1:100, Sigma-Aldrich) and, as secondary antibodies, goat anti-mouse IGg1 Alexa 488 (diluted 1:500, Molecular Probes).
Immunofluorescence labeling and imaging
At 48 h after seeding, NHDFα and LMNA mut grown onto PA gels of 3, 10, 20, 80 kPa and glass bottom culture dishes coated with collagen I were washed with PBS and fixed with 4% formaldehyde in PBS (Sigma-Aldrich) for 10 min at room temperature. Next, they were permeabilized with 0.1% Triton-X-100 (Merck) in PBS for 10 min and incubated with 2% bovine serum albumin (BSA) in PBS in order to block non-specific binding. Afterwards, they were incubated for two hours with primary antibodies in NET-gel. The following primary antibodies were used: mouse MoAb to PML proteins (IgG1, diluted 1:200, sc-966, Santa Cruz) and rabbit polyclonal to LaminB1 (IgG1 1:500, diluted, ab16048, AbCam). After washing with PBS (three times, 10 min), secondary antibodies in NET-gel were applied for 1 h. Goat anti-mouse IgG1 Alexa 488 (diluted 1:500, Molecular Probes) was used against PML-NBs antibody while goat anti- rabbit IgG Alexa 555 or Alexa 488 (diluted 1:500, Molecular Probes) were used against Lamin B1 antibody. For F-actin staining phalloidin-TRITC (1:200, Molecular Probes) was used. After two washing steps with PBS, cells were incubated for 5 min with DAPI (1:500, Molecular Probes) for nuclear counterstaining. Imaging for the immunofluorescence studies was performed by means of an inverted confocal microscope connected to an inverted Axiovert 200M (Zeiss LSM 510 META, Zeiss). A C-Apochromat water-immersion objective (63×, NA = 1.2) was used to minimize the effects of spherical aberration when focusing deep into PA gels, while for cells plated on glass a Plan-Apochromat oil immersion objective was used (63× , NA = 1.4). The laser scanning microscope was used in the dual parameter setup, according to the manufacturer's specification, using dual wavelength excitation: the Ar laser at 488nm (30 mW) and the HeNe laser at 543 nm (1 mW). Z-series were generated by collecting a stack consisting of optical sections using a step size of 0.3–0.45 μm in the z-direction while a minimum pinhole opening was used (1AU). Alternatively, a Leica SPE confocal microscope was used, mounted on a DMI 4000 inverted microscope. Excitation lines were 405 nm (DAPI), 488 nm (FITC) and 532 nm (Phalloidin).
CytochalasinD treatment
At 24 h after seeding, LMNA mut grown onto collagen I coated glass bottom culture dishes were transiently treated with cytochalasin D (cytoD, Sigma-Aldrich) 1 μM to inhibit actin filament dynamics. Successively the growth medium was refreshed with normal growth medium. Three different treatments were performed:
• Short treatment + short recovery = 30 min cytoD treatment and 1 h recovery in normal growth medium
• Long treatment + short recovery = 3 h cytoD treatment and 1 h recovery in normal growth medium
• Long treatment + short recovery = 3 h cytoD treatment and overnight recovery in normal growth medium.
Finally, LMNA mut cells were fixed and stained for actin organization and nuclear abnormalities.
Cell spreading assay
Cells were detached using a trypsin solution containing 0.125% trypsin (Invitrogen Life Technologies, Breda, the Netherlands), 0.02 M EDTA and 0.02% glucose in PBS. Duration of trypsin treatment was kept to a minimum (approximately three min at 37°C), Trypsin was inactivated by adding an excess of culture medium with serum, and cells were seeded onto glass coverslips. Cells were allowed to attach for a variable period of time (0.5 h, 1 h, 2hr, 4 h, 8 h, 24 h or 72 h.) under standard culture conditions and were fixed and processed for immunofluorescence as described above. As primary antibody the mouse monoclonal lamin A/C antibody Jol2 (IgG1, diluted 1:20; a kind gift from Dr C.J. Hutchison (Durham University) was used. The percentage of cells with abnormal nuclei (blebs) was estimated by counting 3 × 100 cells per time point.
Image analysis
Brightfield images of cell cultures were obtained with an Axio Observer Z1 (Zeiss). For cell area and aspect ratio measurements, NHDFα and LMNA mut were manually outlined in relative brightfield images. Using ImageJ (1.45) freeware software, cell area and cell aspect ratio were measured. Cell aspect ratio was calculated as the length of the long axis of the cell divided by the length of the short axis. At least 60 cells were analyzed on the substrate of each type
Abnormal nuclei or cytoplasmic PML bodies were scored manually on about 100 cells in three random locations on two samples per each stiffness (600 cells in total) for each cell genotype. Nuclei were scored as abnormally shaped when their appearance, after laminB1 staining and DAPI counterstaining, showed abnormalities such as blebs, large and poorly defined protrusions and invaginations. In scoring of PML bodies, a second non-specific channel was acquired (550LP) to avoid counting of autofluorescent foci. Mitotic cells, identified by the shape of the DAPI staining, were rejected from the analysis, as they also show cytoplamsic PML bodies.
Live cell imaging
In order to perform live cell-imaging, LMNA mut were seeded on PA gels of 3, 10, 20, 80 kPa and on glass bottom dishes and, after 24 h, were transfected with EYFP-NLS. After 24 to 36 h from transfection, LMNA mut were supplemented with pre-warmed phenol-free serum-containing culture medium (DMEM 31053, Invitrogen) complemented with 15 mM Hepes. Evaporation of the medium was prevented by covering it with an approximate 2 mm layer of mineral oil (Sigma) previously washed with culture medium. Time-lapse recording with an interval of one or two minutes were taken using an inverted automated microscope (Leica DMRBE, Manheim, Germany) equipped with a black and white CCD-camera (CA4742–95, Hamamatsu, Bridgewater, NJ, USA). Image acquisition was achieved using Openlab software (Improvision, Lexington, MA, USA). The microscope was equipped with a heated stage which temperature was set at 37ffC. This allowed imaging the sample while keeping it in optimal cell culture conditions. A 20× (N.A.-0.45) objective was used. For image processing and analysis of time-lapse videos, ImageJ (1.45) freeware was used. Briefly, for each time-lapse recording the analysis of the fluctuations of the fluorescence intensity in representative nuclear regions was performed. Values were normalized and then plotted as function of the time.
Statistical analysis
Data are expressed as mean ± SEM and mean ± SD for PA gels elastic modulus. Statistical analysis was performed using StatGraphics (Manugistics, Inc., Rockville, MD). The data were analyzed by unpaired t-test (allowing different SD), one-way ANOVA (followed by Tukey's multiple comparison test) or, in case of non-Gaussian distribution, the Mann-Whitney or Kruskal Wallis tests (the latter when comparing more than two groups, followed by Tukey's multiple comparison test). A p-value of 0.05 was considered statistically significant.
Supplementary Material
Additional material
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nucl-4-61-s01.pdf (9.9MB, pdf)
Additional material
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Acknowledgments
The authors are grateful to Cor Semeins, Emanuela Fioretta and Marloes Janssen-van den Broek for helping with protocol development, gel preparation and the many helpful discussions.
Disclosure of Potential Conflicts of Interest
The authors declare that they have no financial interests in relation to the submitted work.
Footnotes
Previously published online: www.landesbioscience.com/journals/nucleus/article/23388
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13935 | https://en.wikipedia.org/wiki/Pound_sign | Jump to content
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Contents
(Top)
1 Origin
2 Usage
3 Other English variants
4 Historic variants
4.1 Double bar style
4.2 Other
5 Currencies that use the pound sign
5.1 Former currencies
6 Use with computers
7 Other uses
8 See also
9 Notes
10 References
Pound sign
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From Wikipedia, the free encyclopedia
Currency sign
This article is about the currency symbols "£" and "₤". For the sign "#", see Number sign. For the sign "lb", see pound (mass).
| £ |
| Pound sign |
| In Unicode | U+00A3 £ POUND SIGN (£) |
| Currency |
| Currency | Pound |
| Graphical variants |
| ￡ |
| U+FFE1 ￡ FULLWIDTH POUND SIGN |
| Different from |
| Different from | U+20A4 ₤ LIRA SIGN U+0023 # NUMBER SIGN |
| Category |
The pound sign (£) is the symbol for the pound unit of sterling – the currency of the United Kingdom and its associated Crown Dependencies and British Overseas Territories and previously of Great Britain and of the Kingdom of England. The same symbol is used for other currencies called pound, such as the Egyptian and Syrian pounds. The sign may be drawn with one or two bars depending on personal preference, but the Bank of England has used the one-bar style exclusively on banknotes since 1975.
In the United States, "pound sign" refers to the symbol # (number sign). In Canada, "pound sign" can mean £ or #.
Origin
[edit]
The symbol derives from the upper case Latin letter L, representing libra pondo, the basic unit of weight in the Roman Empire, which in turn is derived from the Latin word libra, meaning scales or a balance. The pound became an English unit of weight and in England became defined as the tower pound (equivalent to 350 grams) of sterling silver. According to the Royal Mint Museum:
It is not known for certain when the horizontal line or lines, which indicate an abbreviation,[a] first came to be drawn through the L. However, there is in the Bank of England Museum a cheque dated 7 January 1661 with a clearly discernible £ sign. By the time the Bank was founded in 1694 the £ sign was in common use.
However, the simple letter L, in lower- or uppercase, was used to represent the pound in printed books and newspapers until well into the 19th century. In the blackletter type used until the seventeenth century, the letter L is rendered as .
Usage
[edit]
When used for sterling, the pound sign is placed before the numerals (e.g., £12,000) and separated from the following digits by no space or only a thin space. In the UK, the sign is used without any prefix. In Egypt and Lebanon, a disambiguating letter is added (E£ or £E and £L respectively). In international banking and foreign exchange operations, the symbol is rarely used: the ISO 4217 currency code (e.g., GBP, EGP, etc.) is preferred.[b]
Other English variants
[edit]
In Canadian English, the symbol £ is called the pound sign. The symbol # has several uses and is sometimes called the pound sign too, though it is most often known as the number sign. (Telephone instructions for equipment manufactured in the United States often call # the pound key.)
In American English, the term "pound sign" usually refers to the symbol # (number sign), and the corresponding telephone key is called the "pound key". (As in Canada, the # symbol has many other uses.)
Historic variants
[edit]
Double bar style
[edit]
Banknotes issued by the Bank of England since 1975 have used only the single bar style as a pound sign. The bank used both the two-bar style (₤) and the one-bar style (£) (and sometimes a figure without any symbol whatever) more or less equally from 1725 to 1971 intermittently and sometimes concurrently. In typography, the symbols are allographs – style choices – when used to represent the pound; consequently fonts use U+00A3 £ POUND SIGN (Unicode) code point irrespective of which style chosen, (not U+20A4 ₤ LIRA SIGN despite its similarity). It is a font design choice on how to draw the symbol at U+00A3. Although most computer fonts do so with one bar, the two-bar style is not rare, as may be seen in the illustration above.
Other
[edit]
In the eighteenth-century Caslon metal fonts, the pound sign was identical to an italic uppercase J, rotated 180 degrees.]
Currencies that use the pound sign
[edit]
Egypt: Egyptian pound
Falkland Islands: Falkland Islands pound
Gibraltar: Gibraltar pound
Guernsey: Guernsey pound
Isle of Man: Manx pound
Jersey: Jersey pound
Saint Helena: Saint Helena pound
South Sudan: South Sudanese pound
Sudan: Sudanese pound
Syria: Syrian pound
United Kingdom: Pound sterling
Former currencies
[edit]
American Colonies:
Connecticut pound
Delaware pound
Georgia pound
Maryland pound
New Hampshire pound
New Jersey pound
New York pound
North Carolina pound
Pennsylvania pound
Rhode Island pound
South Carolina pound
Virginia pound
Australia: Australian pound
The Bahamas: Bahamian pound
Bermuda: Bermudian pound
British Mandatory Palestine: Palestine pound
Canada: Canadian pound
New Brunswick pound
Nova Scotian pound
Prince Edward Island pound
Cyprus: Cypriot pound
Fiji: Fijian pound
The Gambia: Gambian pound
Ghana: Ghanaian pound
Ireland: Irish pound
Israel: Israeli pound
Malta: Maltese pound
Newfoundland: Newfoundland pound
New Zealand: New Zealand pound
Rhodesia: Rhodesian pound
South Africa: South African pound
Tonga: Tongan pound
Western Samoa: Western Samoan pound
Yemen : Yemeni dinar
Use with computers
[edit]
In the Unicode standard, the pound sign is encoded at U+00A3 £ POUND SIGN (£) Whether the glyph is drawn with one or two bars is a type designer's choice as explained above; the key point is that the code is constant irrespective of the presentation chosen.[c]
The encoding of the £ symbol in position xA3 (16310) was first standardised by ISO Latin-1 (an "extended ASCII") in 1985. Position xA3 was used by the Digital Equipment Corporation VT220 terminal, Mac OS Roman, Amstrad CPC, Amiga, and Acorn Archimedes.
Many early computers (limited to a 7-bit, 128-position character set) used a variant of ASCII with one of the less-frequently used characters replaced by the £. The UK national variant of ISO 646 was standardised as BS 4730 in 1985. This code was identical to ASCII except for two characters: x23 encoded £ instead of #, while x7E encoded ‾ (overline) instead of ~ (tilde). MS-DOS on the IBM PC originally used a proprietary 8-bit character set Code page 437 in which the £ symbol was encoded as x9C; adoption of the ISO/IEC 8859-1 ("ISO Latin-1") standard code xA3 only came later with Microsoft Windows. The Atari ST also used position x9C. The HP LaserJet used position xBA (ISO/IEC 8859-1: º) for the £ symbol, while most other printers used x9C. The BBC Ceefax system which dated from 1976 encoded the £ as x23. The Sinclair ZX80 and ZX81 characters sets used x0C (ASCII: form feed). The ZX Spectrum and the BBC Micro used x60 (ASCII: `, grave). The Commodore 64 used x5C (ASCII: ) while the Oric computers used x5F (ASCII: _). IBM's EBCDIC code page 037 uses xB1 for the £ while its code page 285 uses x5B. ICL's 1900-series mainframes used a six-bit (64-position character set) encoding for characters, loosely based on BS 4730, with the £ symbol represented as octal 23 (hex 13, dec 19).
Other uses
[edit]
The logo of the UK Independence Party, a British political party, is based on the pound sign, symbolising the party's opposition to adoption of the euro and to the European Union generally.
The pound sign was used as an uppercase letter (the lowercase being ⟨ſ⟩, long s) to signify the sound [ʒ] in the early 1993–1995 version of the Turkmen Latin alphabet.
See also
[edit]
Latin letter L with stroke Ł ł
Semuncia 𐆒
Category:Currency symbols
Notes
[edit]
^ See scribal abbreviations
^ Prior to ISO 4217, abbreviations such as "stg" or "STG" were traditionally used to disambiguate sterling from other currencies that used the symbol.
^ There is a separate code point, U+20A4 ₤ LIRA SIGN Unicode notes that the "lira sign" is not widely used and was added due to both it and the pound sign being available on HP printers.
References
[edit]
^ Thomas Snelling (1762). A View of the Silver Coin and Coinage of England from the Norman Conquest to the Present Time. T. Snelling. p. ii. Retrieved 19 September 2016.
^ "A brief history of the pound". The Dozenal Society of Great Britain. Archived from the original on 2020-11-12. Retrieved 2011-01-14.
^ "The Origins of £sd". The Royal Mint Museum. Archived from the original on 8 March 2020.
^ For example, Samuel Pepys (2 January 1660). "Diary of Samuel Pepys/1660/January". Archived from the original on 23 September 2019. Retrieved 23 September 2019. Then I went to Mr. Crew's and borrowed L10 of Mr. Andrewes for my own use, and so went to my office, where there was nothing to do.
^ Dowding, Geoffrey (1962). An introduction to the history of printing types; an illustrated summary of main stages in the development of type design from 1440 up to the present day: an aid to type face identification. Clerkenwell [London]: Wace. p. 5.
^ Hayes, Adam (22 April 2022). "Egyptian Pound (EGP) Definition". Investopedia. Archived from the original on 7 August 2022. Retrieved 24 August 2022.
^ "Alexandria City Center to undergo LE 370 million expansion". Daily News Egypt. 10 June 2008.
^ "Lebanon". CIA World Factbook 1990 - page 178. Central Intelligence Agency. 1 April 1990. Archived from the original on 2022-06-21. Retrieved 2022-06-21 – via en.wikisource.org.
^ Barber, Katherine, ed. (2004). The Canadian Oxford dictionary (2nd ed.). Toronto: Oxford University Press. ISBN 0-19-541816-6.
^ William Safire (1991-03-24). "On Language; Hit the Pound Sign". New York Times. Archived from the original on 2010-07-21. Retrieved 2011-05-21.
^ a b "Withdrawn banknotes". Bank of England. Archived from the original on 15 January 2019. Retrieved 13 September 2019. ("£1 1st Series Treasury Issue" to "£5 Series B")
^ "Current banknotes". Bank of England. Archived from the original on 4 December 2019. Retrieved 8 November 2019.
^ a b "History of the use of the single crossbar pound sign on Bank of England's banknotes". Bank of England. Archived from the original on 25 March 2022. Retrieved 13 April 2022.
^ Howes, Justin (2000). "Caslon's punches and matrices" (PDF). Matrix. 20: 1–7. Archived from the original (PDF) on 2024-07-13.
^ The Unicode Consortium (11 June 2015). "The Unicode Standard, Version 10.0 | Character Code Charts" (PDF). Archived (PDF) from the original on 2019-06-13. Retrieved 2018-01-23.
^ The Unicode Consortium (26 August 2015). "The Unicode Standard, Version 10.0 | Character Code Charts" (PDF). Archived (PDF) from the original on 2021-02-25. Retrieved 2018-01-23.
^ Allen, Julie D., ed. (August 2015) . The Unicode Standard - Version 8.0 - Core Specification - Chapter 22.1. Currency Symbols (PDF). Mountain View, CA, USA: Unicode, Inc. pp. 751–752. ISBN 978-1-936213-10-8. Archived (PDF) from the original on 2016-12-06. Retrieved 2016-12-06. [...] Currency Symbols: U+20A0–U+20CF [...] Lira Sign. A separate currency sign U+20A4 LIRA SIGN is encoded for compatibility with the HP Roman-8 character set, which is still widely implemented in printers. In general, U+00A3 POUND SIGN may be used for both the various currencies known as pound (or punt) and the currencies known as lira. [...]
^ "UK Independence Party". Archived from the original on 24 August 2000. Retrieved 17 April 2017.
^ Clement, Victoria (2008). "Emblems of independence: script choice in post-Soviet Turkmenistan in the 1990s". International Journal of the Sociology of Language (192): 171–185.
| v t e Currency symbols |
| Circulating | ؋ ฿ ₵ ₡ ¢ $ ₫ ֏ € ƒ F ₲ ₴ ₭ ₾ ₼ ₦ ₱ £ 元 ﷼ ៛ ₽ ރ ₹ रू 𞱱 රු ૱ ௹ ꠸ Rs ₪ ⃀ ৳ ₺ ₸ ₮ ₩ ¥ 円 |
| Obsolete and historical | ₳ Դր. ₢$ ₰ ₯ ₠ ₣ ₤ ℒ𝓈 ₶ ₥ ℳ Pts ℛℳ ₷ 𐆚 𐆖 𐆙 𐆗 𐆘 |
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13936 | https://www.quora.com/The-square-root-of-a-negative-number-is-undefined-Why-is-the-square-root-of-%E2%80%93x-not-necessarily-undefined-and-what-does-this-mean-about-the-domain-and-range-of-f-x | The square root of a negative number is undefined. Why is the square root of –x not necessarily undefined and what does this mean about the domain and range of f(x) =? - Quora
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The square root of a negative number is undefined. Why is the square root of –x not necessarily undefined and what does this mean about the domain and range of f(x) =?
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Ross Rheingans-Yoo
amateur hobbyist mathematician · Upvoted by
Vladimir Novakovski
, silver medals, IOI 2001 and IPhO 2001 and
Alon Amit
, Lover of math. Also, Ph.D. · Author has 906 answers and 2.7M answer views
·Updated 11y
Originally Answered: Why is the square root of a negative number undefined? ·
The square root of a negative is undefined over the Reals because there is no Real number which, squared, is negative; it's as simple as that.
There are alternate number systems (e.g. Complex numbers, Algebraic numbers) where there are numbers which, when squared, yield a negative result. Over those number systems, square roots are defined for some or all negative numbers.
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More answers below
Why is the square root of -1 not undefined?
Why is the square root of a negative number undefined?
What is the square root of a negative number squared?
Is square root of any number always positive?
Why does the square root of -1 not have a negative sign?
Drake Way
Mathematics Hobbyist · Author has 4.2K answers and 3.3M answer views
·7y
Correction.
The square root of a negative number is not in the real numbers. In higher level mathematics, the square root of negatives becomes important and has many real life applications.
But ignore that for now.
The square root of -x is not undefined because -x is not necessarily a negative number.
Example:
If f(x)=√−x f(x)=−x, let's plug in an example:
f(−25)=√−(−25)f(−25)=−(−25)
=√25=25
=5.=5.
So clearly there are values for which f (x) is defined.
Basically, when −x≥0,√(−x)−x≥0,(−x) is defined, which means the domain is x≤0 x≤0, and the range is f(x)≥0.f(x)≥0.
Upvote ·
Neal Schermerhorn
Lives in Massachusetts (1967–present) · Author has 7K answers and 1.8M answer views
·11mo
Originally Answered: Why is the square root of a negative number undefined? ·
It is, and it isn’t.
If you are working strictly among the real numbers, it is undefined. The reason is pretty simple—if I square 2, I get 4, and if I square -2, I also get 4. A negative times a negative equals a positive. A positive times a positive equals a positive.
The only way a product of two numbers can be negative is if there is one positive number and one negative number. And that’s not going to be a square because the values are not the same.
So this means sqrt(-4) is undefined among the reals.
However, if you extend to the complex numbers, sqrt(-4) is well-defined. It’s +/- 2i.
Upvote ·
Phil Scovis
I play guitar. And sing in the car. · Author has 6.9K answers and 13.3M answer views
·7y
“-x” is read “negative x”, however, this is really incorrect. The ‘-’ symbol does not mean “negative”. It means “the opposite of”.
The opposite of -9 is 9. The square root of the opposite of -9 is 3. So if x is -9, then
√−x=3−x=3.
I’m sure that someone will point out that the square root of a negative number is not undefined; it’s just not a real number.
f(x), on the other hand, is undefined, for the simple reason that you haven’t defined it. So I have no idea what its domain and range might be.
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More answers below
Shouldn't the square root of zero be undefined as 0/0 is undefined?
What is the square root of 1 + square root of 1 + square root of 1 infinitely?
Why do we call "square root" a "square root"?
Who discovered the square and square roots?
What is the square root of the square of x? Is it only x, or +-x?
Jeremy Rubin
MIT student
·11y
Originally Answered: Why is the square root of a negative number undefined? ·
The square root of a negative number is indeed well defined. It is, however, imaginary. Imaginary does not mean that it does not exist though, it simply means that it has a different set of properties in contrast to the numbers we normally see.
i=√−1,i 2=−1 i=−1,i 2=−1
Edit:
To clarify why you just need this one case to describe all negative numbers, consider some positive value a
√−a=√−1∗a=√−1∗√a=√a∗i−a=−1∗a=−1∗a=a∗i
Upvote ·
9 4
Mark Ross
My work as a programmer required a certain knack for math. · Author has 22.6K answers and 9.3M answer views
·7y
It’s undefined in the Real Numbers, yes, but it is perfectly well defined in the Complex Numbers by defining i as the square root of -1.
The square root of -x might be defined if x is a negative number, that would make -x a positive number.
I’m not sure what ANY of that has to do with the domain and range of f(x).
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Muhammad Farooq Malik
Works at Pakistan Railways
·8y
Originally Answered: Why is the square root of a negative number undefined? ·
How we define a square root ? e.g. we say square root of 4 is 2. why ?
Square root of 4 is 2, because 2 x 2 = 4
Now can we define a number which is multiplied by itself and results in a negative number ? The answer is No. there is no such number which is defined by itself and gives a result with minus sign. For example -2 x -2 is also +4.
As such we can say that the square root of a negative number is undefined. Better to say that square root of a negative number is not a real number.
Upvote ·
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Darren Lorent
Author has 5.1K answers and 594.8K answer views
·3y
If x is negative, -x is positive. Note also that square roots of negative numbers are perfectly defined, they are just not real. An imaginary number has the property that i² = -1. This gives the other definition i = +/- √ (-1). We can then define the complex numbers as numbers of the form a + bi where a and b are any real numbers
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Chandresh Jain
Retired Chief Engineer at UP Irrigation Dptt
·7y
f(x) = sqrt(-x) can be defined. It has real values when x belongs to (-infinity, 0]
domainof f(x) = (-infinity,0]
range of f( x) =[0,+infinity)
Upvote ·
Dhrupit Dave
Student of Life. · Author has 266 answers and 831.4K answer views
·10y
Related
Why is the square root of -1 not undefined?
Short answer No it is not undefined it's imaginary number.
Here's the proof.
A square root of a number a is any number x with x2=a, or equivalently a root of the polynomial x2−a. The fundamental theorem of algebra implies that every complex number a has a square root. In fact, for a≠0, a has precisely two square roots, which are additive inverses.
You can already see that this is a bit of a problem in the non-negative real numbers. For this case, we choose a√ to be the unique non-negative square root, which has a lot of nice properties. Viewed as a function of a, a√ is continuous, and is a mul
Continue Reading
Short answer No it is not undefined it's imaginary number.
Here's the proof.
A square root of a number a is any number x with x2=a, or equivalently a root of the polynomial x2−a. The fundamental theorem of algebra implies that every complex number a has a square root. In fact, for a≠0, a has precisely two square roots, which are additive inverses.
You can already see that this is a bit of a problem in the non-negative real numbers. For this case, we choose a√ to be the unique non-negative square root, which has a lot of nice properties. Viewed as a function of a, a√ is continuous, and is a multiplicative homomorphism (i.e. ab−−√=a√b√). These properties are nice enough that it makes sense to call this choice of a√ the square root of a, instead of a square root of a. Of course, this is already abusing terminology a bit, but don't worry too much about that.
It is perfectly reasonable to try to extend the square root function a√ for a any complex number, since we know that complex numbers have square roots. But unlike in the positive reals, there's no really nice way to choose what the square root of a should be. In particular, for −1−−−√, we can choose either −i or i, and since complex conjugation preserves all the algebraic properties of C we shouldn't expect a nice way to do so purely based on algebraic considerations (like we had for the nonnegative reals).
Let's ignore this for a minute. Any complex number z can be written in the form z=reiθ where 0≤r<∞ and 0≤θ<2π , the polar representation for complex numbers. For z≠0 the choice of r and θ is unique. We can define z√=r√eiθ/2, which is a square root of z. Indeed, if we do this, then −1−−−√=i. So what's the issue with this?
For one thing, you lose the homomorphism property. There are cases where ab−−√≠a√b√. This is generic for all extensions, of the square root function, too. There is no we could avoid it by choosing a different definition. You can look at this question to see why this must fail generically.
Furthermore, our choice is not continuous, since limy→0+x−iy−−−−−√=−x√ for x,y real. We could make it continuous, but at the cost of not defining z√ for θ=0. This is what's called a branch cut. But this is precisely the case when z is a positive real number! Some people do this, but it's bad practice and probably confusing. We could define a branch cut elsewhere, for instance, only allow −π<θ<π and use the same formula. Now the branch cut is on the negative real axis, which is better since our notation doesn't conflict with the notation for real numbers, but now −1−−−√ is undefined, and limy→0+−1+iy−−−−−−√=i while limy→0−−1+iy−−−−−−√=−i. If you don't care about continuity at the negative real axis, you can extend the definition to θ=π, which again gives you back −1−−−√=i. We could also induce a branch cut other places, for instance on the positive imaginary axis, to avoid both the problem of disagreeing with the real square root and of being undefined or discontinuous on some part of the real axis, but then your branch cut has to change under complex conjugation, which can also be a problem.
There are other ways to address the issue. The nicest is the theory of Riemann surfaces. The idea here is that you think of z√ as a function on a set that is larger than just the complex plane. The Riemannn surface for square root is essentially 2 copies of the complex plane, split at the branch cuts and glued to each other. Here is an image, taken from Wikipedia, for the Riemann surface for the square root function:
This approach is not without fault either, since you're now no longer talking strictly about the square root of a complex number. The points on the Riemann surface tell you which square root to pick, and there is one point for each square root. Since -1 is a complex number, not a point on the surface, −1−−−√ doesn't make sense. However, there are 2 points corresponding to -1, one of which has square root i and the other one −i.
You can also consider z√ to be a multivalued function, returning a pair of numbers which are both of the square roots of z. This works fine, except for when you want to do any sort of actual calculations. In particular, in this approach, −1−−−√={i,−i}. This approach does have the homomorphism property that ab−−√=a√∗b√, once you define what it means to multiply sets (namely, {a,−a}∗{b,−b}={ab,−ab}). This definition does not agree with our definition for positive reals, but it's not as bad as before since the square root of a real number is at least an element of the set of its square roots as a complex number.
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Upvote ·
9 1
Mark Ross
My work as a programmer required a certain knack for math. · Author has 22.6K answers and 9.3M answer views
·9y
Originally Answered: Why is the square root of a negative number undefined? ·
It isn’t undefined, exactly. Mathematicians, for their own (very good) reasons came up with something called the Imaginary numbers and a member in it called i that has the property that i squared = -1 Root expressions using i are called Complex roots. So the root of any negative number (-n) can be expressed as in^½.
These ‘imaginary’ numbers have actually been found to be useful in some engineering calculations and, I believe, in gps calculations.
Upvote ·
9 1
Amin Guermazi
Lives in Tunisia
·5y
Originally Answered: Why is the root of minus undefined? ·
Simply because anything times itself will give a positive (or zero) result.
Ex: -3² = -3 -3 = 9
Upvote ·
9 1
9 1
Related questions
Why is the square root of -1 not undefined?
Why is the square root of a negative number undefined?
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Why does the square root of -1 not have a negative sign?
Shouldn't the square root of zero be undefined as 0/0 is undefined?
What is the square root of 1 + square root of 1 + square root of 1 infinitely?
Why do we call "square root" a "square root"?
Who discovered the square and square roots?
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Related questions
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Shouldn't the square root of zero be undefined as 0/0 is undefined?
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13937 | http://www1.us.elsevierhealth.com/studentconsult/Kumar_Robbins/virtual_microscope_slides/?srsltid=AfmBOoqx6ShM4tSyCdHy5KSbaNOU6Z8hORTrLSAxsC0AtHAu7ejUonsO | STUDENT CONSULT: Robbins & Cotran Pathologic Basis of Disease, 8th edition: Virtual Microscope
Blood Vessels
Vein - organizing and recanalizing thrombus
The Heart
Heart - glycogen storage disease
Heart - myocardial hypertrophy
Aorta - aortic arteriosclerosis
Heart and coronary artery - arteriosclerosis with organizing thrombus
Heart - acute transmural infarct (5 days old) with rupture
Heart - healing transmural infarct
Heart - healed transmural infarct
Heart - acute rheumatic carditis
Heart with mitral valve - infective endocarditis
Aorta - cystic medionecrosis
Heart - acute myocardial infarct
Heart - normal
Aorta - normal
White Blood Cells, Lymph Nodes, Spleen and Thymus
Spleen - Gaucher
Lymph Node - metastatic carcinoma of the breast
Lymph Node - sarcoidosis
Lymph Node - normal
Spleen - normal
The Lung
Lung - chronic passive congestion
Lung - acute passive congestion and edema
Lung - oxygen toxicity
Lung - pseudomonal bronchopneumonia
Lung - viral pneumonia
Lung - abscesses
Lung - pulmonary tuberculosis with cavity
Lung - emphysema
Lung - interstitial pulmonary fibrosis
Lung, bronchus - asthma
Lung, bronchus - squamous cell carcinoma
Lung, bronchus - small cell undifferentiated carcinoma
Lung, bronchus - adenocarinoma
Lung - normal
The Kidney
Kidney - acute infarct
Kidney - systemic lupus erythematosus
Kidney - arteriolar nephrosclerosis
Kidney - malignant nephrosclerosis
Kidney - acute glomerulonephritis
Kidney - membranous glomerulopathy
Kidney - crescentic glomerulonephritis
Kidney - acute pyelonephritis with necrotizing papillitis
Kidney - Renal cell carcinoma
Kidney - diabetic glomerulosclerosis
Kidney - normal
The Gastrointestinal Tract
Small intestine, jejunum - organizing peritonitis
Stomach - chronic peptic ulcer
Colon - chronic ulcerative colitis
Colon - villous adenoma
Colon - adenocarcinoma
Colon - amebiasis
Appendix - fibrous obliteration of the tip
Appendix - acute appendicitis
Stomach, fundus - normal
Appendix - normal
Stomach, pylorus - normal
Colon - normal
Small Intestine - normal
Esophagus - normal
Liver and Biliary Tract
Liver - fatty change (steatosis)/ alcoholic fatty liver
Liver - chronic passive congestion with centrilobular necrosis
Liver - metastatic carcinoma of pancreas
Liver - miliary tuberculosis
Liver - viral hepatitis
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Liver - alcoholic (portal) cirrhosis
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Liver - biliary cirrhosis
Liver - primary carcinoma of the liver (hepatoma)
Gallbladder - acute and chronic cholecystitis
Liver - normal
Gallbladder - normal
The Pancreas
Pancreas - acute interstitial pancreatitis
Pancreas - chronic (alcoholic) pancreatitis
Pancreas - adenocarcinoma
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Pancreas - normal
The Lower Urinary Tract and Male Genital System
Skin, genitalia - chancre of the prepuce
Prostate - adenocarcinoma
Testis - seminoma
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The Female Genital Tract
Fallopian Tube - suppurative salpingitis
Uterus, cervix - chronic cervicitis
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Breast - fibrocystic change
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The Endocrine System
Thyroid - Graves disease - hypertrophy and hyperplasia
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Adrenal gland - nodular hyperplasia
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Joint - rheumatoid arthritis
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Skeletal muscle - progressive muscular dystrophy
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Skin - healing surgical incision with foreign-body granuloma
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The Central Nervous System
Brain - cerebral edema
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Brain - acute bacterial meningitis
Brain - Alzheimer disease
Brain - organizing cerebral infarct
Brain - old cerebral infarct
Brain, cerebellum - astrocytoma
Brain, cerebrum - glioblastoma multiforme
Brain, cerebellum - medulloblastoma
Brain, meninges, and cerebrum - meningioma
Brain - herpes simplex virus encephalitis
Brain, cerebrum - multiple sclerosis
Spinal cord - amyotrophic lateral sclerosis
Brain, cerebrum - leukodystrophy (myelin stain)
Brain - normal
Brain, cerebellum - normal
Head and Neck
Larynx - squamous cell
Larynx, right vocal cord - normal |
13938 | https://www.etymonline.com/word/prosecution | Advertisement
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Origin and history of prosecution
prosecution(n.)
1560s, "the carrying out or following up of anything" (also literal, "action of pursuing, a following after," but this is obsolete), from French prosecution (late 13c.) and directly from Late Latin prosecutionem (nominative prosecutio) "a following," noun of action from past-participle stem of prosequi "to follow after; chase, pursue; attack, assail" (see prosecute). The meaning "legal action, the institution and carrying out of a suit at law" is from 1630s. Hence, transferred, "the party by whom legal proceedings are initiated" (1891).
also from 1560s
Entries linking to prosecution
prosecute(v.)
early 15c., prosecuten, "to follow up, pursue with a view to carry out or obtain" (some course or action), from Latin prosecutus, past participle of prosequi "follow after, accompany; chase, pursue; attack, assail, abuse," from pro- "forward" (see pro-) + sequi "follow" (from PIE root sekw- (1) "to follow"). Meaning "bring to a court of law, seek to obtain by legal process" is recorded from 1570s. The Latin verb in Old French became prosequer, vulgarly porsuir, which passed to English as pursue.
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Trends of prosecution
adapted from books.google.com/ngrams/ with a 7-year moving average; ngrams are probably unreliable.
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defense
c. 1300, "action of guarding or shielding from attack or injury; act of defending by fighting; a fortified place of refuge," from Old French defense, from Latin defensus, past participle of defendere "ward off, protect" (see defend). It also arrived (without the final -e) from Ol
pursuance
"act of following or pursuing," 1590s, from French poursuiance "act of pursuing," from Old French poursuir "to chase, pursue, follow; continue, carry on" (see pursue). Pursuance is not now used except in the moral sense, and then generally in the sense of following out : as, pur
plea
legal proceedings over which the Crown did not claim exclusive jurisdiction (as distinct from pleas of the Crown "public prosecution...
people
This also is the sense in the legal phrase The People vs., in U.S. cases of prosecution under certain laws (1801)....
trail
c. 1300, "to hang down loosely and flow behind" (of a gown, sleeve, etc.), from Old French trailler "to tow; pick up the scent of a quarry," ultimately from Vulgar Latin tragulare "to drag," from Latin tragula "dragnet, javelin thrown by a strap," probably related to trahere "to
trial
mid-15c., "act or process of testing, a putting to proof by examination, experiment, etc.," from Anglo-French trial, noun formed from trier "to try" (see try (v.)). Sense of "examining and deciding of the issues between parties in a court of law" is first recorded 1570s; extended
aggregation
early 15c., aggregacioun, originally in medicine (Chauliac), "formation of a pustule," from Medieval Latin aggregationem (nominative aggregatio), noun of action from past-participle stem of Latin aggregare "collect, bring together," from ad "to" (see ad-) + gregare "to collect in
action
mid-14c., accioun, "cause or grounds for a lawsuit," from Anglo-French accioun, Old French accion, action (12c.) "action; lawsuit, case," from Latin actionem (nominative actio) "a putting in motion; a performing, a doing; public acts, official conduct; lawsuit, legal action" (sou
accumulation
late 15c., "that which is heaped up, an accumulated mass," from Latin accumulationem (nominative accumulatio) "a heaping up," noun of action from past-participle stem of accumulare "to heap up, amass," from ad "to," here perhaps emphatic (see ad-), + cumulare "heap up," from cumu
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13939 | https://www.youtube.com/watch?v=WZtO3oERges | Area of Trapezoid
Let's Do Math
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We know how to find the area of any triangle. We can use that knowledge to help us solve the area of any trapezoid.
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159 comments
how to find the area of a trapezoid which is also known as a trapezium in different parts of the world a trapezoid is a quadrilateral with one pair of parallel sides as shown by these little arrowheads these are all examples of trapezoids when I first thought about finding the area of one as a student I was confused how can we possibly work out the area of one of these things they can be so different what can we do with them you won't be surprised to know there's a trick to it I'm going to show you the way to understand it and then how to do it really fast we're given a length for each parallel side and the height what if I do this to get us started on the area now we have two triangles we know how to find the area of a triangle I'll call this one a it has a base of 7 and a height of 4 and this is B it has a base of 5 and the same height of 4 let's work out the area of triangle a base times height over 2 7 times 4 is 28 and divide that by 2 gives me 14 centimeters squared and for triangle B we have 5 times 4 that's 20 and half of that is an area of 10 centimeters squared and you guessed it we can just add the area of these two triangles to get the area of the trapezoid and that is 24 centimeters squared that's pretty neat it gives us a clear strategy but it's kind of a clunky job don't you think doing two triangles and then adding them together there's a formula we can use that lets us work in a more elegant way it's also much quicker I called these triangles a and B for a very good reason the formula is base a plus base B you do that part first multiplied by the height and divided by 2 this formula lets us find the area of both those triangles that we made inside the trapezoid in one calculation nice let's do the same trapezoid again using the formula this time so you can see it in action we already know the area of course so here's the formula for the area of a trapezoid base a is 7 plus base b is 5 times the height which is 4 divided by 2 we do the math inside the brackets first so 7 plus 5 that's 12 times the height of 4 over 2 12 times 4 is 48 and we divide that by 2 half of 48 is 24 the area is 24 centimeters squared this calculation was much more streamlined isn't it cool it's so much more efficient think of it as 2 triangles combined into one shape the trapezoid use the formula a plus b times the height over to take down some notes and do the practice build the math muscle if you can work out the area of this triangle using the formula multiplying base by height and then dividing by 2 to get the area then you can move over and solve the area of trapezoids 2 using this formula as we saw it's like finding the area of two triangles that make up one trapezoid at one shot we add the two parallel sides and multiply by the height and / - and there's the area you can do this have fun and build your math muscle |
13940 | https://en.wikipedia.org/wiki?title=Talk:Coprime | Talk:Coprime integers - Wikipedia
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1 Awkward wording (a.k.a., "Huh?!" )8 comments
2 Wiki Education Foundation-supported course assignment1 comment
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5 Cases for three integers2 comments
Talk:Coprime integers
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Awkward wording (a.k.a., "Huh?!" )
[edit")]
From the text:
"...As specific examples, 14 and 25 are coprime, being commonly divisible only by 1, while 14 and 21 are not coprime, because they are both divisible by 7..."
Huh?!?! If you're trying to say that "14 and 25 are coprime, sharing no factors in common between them except 1," then for God's sake, please say that! I puzzled over that text uncomprehendingly for a good two minutes, trying to decipher the odd phrase commonly divisible, which I recall seeing nowhere else; it was only when I stopped ruminating on the first example long enough to take in the second, counterexample, that understanding finally came. Good grief! The Grand Rascal (talk) 08:31, 28 September 2020 (UTC)[reply]
TheGrandRascal, I agree. "Commonly divisible" is not a common phrase for saying "having a common divisor". I have fixed it. D.Lazard (talk) 08:51, 28 September 2020 (UTC)[reply][a] By chance, I read that a few days ago and found it perfectly understandable. The "proposed" version is also ok. [b] commonly divisible seems to be used: see ( LMSchmitt09:33, 28 September 2020 (UTC)[reply]Indeed, that's because it's a completely standard English language construction -- there's nothing jargony or obscure at all about it. --JBL (talk) 11:32, 28 September 2020 (UTC)[reply]JayBeeEll, it is grammatically correct, but semantically confusing. In mathematics, one must always take care when words have a mathematical meaning that differ from their common (usual) meaning. In the case of "common divisor", "common" refers to "common to two integers". In "commonly divisible", "commonly" is not used in its common meaning, but is used in place of "simultaneously divisible". So the formulation is confusing. By the way Khan Academy is not a reliable source for attesting a common use of "commonly divisible" D.Lazard (talk) 14:35, 28 September 2020 (UTC)[reply]To be clear: this is just forum-y chatting, I am not trying to get anyone to change the article. The most common meaning of "common" is probably "widespread; public". The second-most common meaning of "common" is "shared; joint". Both of these can be turned into an adverb: "commonly" meaning "frequently; often; usually" and "commonly" meaning "in common; jointly". The phrase "greatest common divisor" uses the second meaning of the adjective "common", and the phrase "commonly divisible" uses the corresponding adverbial form. Anyone who understands the phrase "greatest common divisor" should, by applying standard rules of English, be able to convert it to the corresponding adverbial form, and vice-versa. P.S. I feel like it is strangely common (at least, this is not the first time) for you, a PhD mathematician who writes English well but not fluently, to lecture me, a PhD mathematician who writes English fluently, on basic points of English grammar and its use in mathematics. I promise that I will never attempt to lecture you on fine points of French usage or grammar! --JBL (talk) 12:51, 29 September 2020 (UTC)[reply]I find the phrase grammatically, semantically, syntactically, and pragmatically correct and transparent. Inasmuch as I am a native speaker of English, but am neither a mathematician nor even mathematically adroit, I think the article speaks to people like me.What I find humorously anti-mathematical is the phrase "relatively prime" being applied even numbers, and that is what brought me here (from a Numberphile youTube video, where the host asked 'What do you mean by "relatively prime"?'. This brought an only slightly more clarifying answer. I had to examine the matter at greater length. __ "Nobody understands mathematics, you just get used to it" --Von Neumann. Perhaps he could have added:"Nobody understands mathematicians, you just get used to them."JohndanR (talk) 00:26, 14 January 2025 (UTC)[reply]
Well. It may not be my recommended wording, but at least it's a lot clearer than it was! Thanks. 😊 The Grand Rascal (talk) 09:13, 28 September 2020 (UTC)[reply]
Wiki Education Foundation-supported course assignment
[edit]
This article is or was the subject of a Wiki Education Foundation-supported course assignment. Further details are available on the course page. Student editor(s): Nbecker1.
Above undated message substituted from Template:Dashboard.wikiedu.org assignment by PrimeBOT (talk) 18:27, 16 January 2022 (UTC)[reply]
Clarity in the lead
[edit]
The lead can be difficult for someone first coming to this concept, even though the concept is not complicated. I suggest that a sentence like the one Dan Harkless suggested be used, but with changes to avoid objections about 'coprime' not being a noon. Another clarification would be to expand on the example of 8 and 9 by naming the prime factors of each (2 and 3). 2600:6C67:1C00:5F7E:6DF1:CD52:7E5F:D359 (talk) 15:25, 9 July 2022 (UTC)[reply]
Maths
[edit]
What is co prime number 103.66.81.7 (talk) 17:04, 10 July 2022 (UTC)[reply]
The answer is in the article. D.Lazard (talk) 18:06, 10 July 2022 (UTC)[reply]
Cases for three integers
[edit]
When three different random integers are selected, there are five different cases of the three numbers having a common divisor or being coprime to one another:
All three integers have a common divisor greater than 1.
Example: 4, 6, 8; 4, 6, and 8 are all divisible by 2.
Exactly one pair of the three integers have a common divisor greater than 1 and the other integer is relatively prime to both the other two integers.
Example: 3, 4, 10; 4 and 10 are divisible by 2, 3 is relatively prime to both 4 and 10.
Exactly two pairs of the three integers have a common divisor greater than 1 and one pair of integers is relatively prime.
Example: 4, 6, 9; 4 and 6 are both divisible by 2, 6 and 9 are both divisible by 3, and 4 and 9 are relatively prime.
All three pairs of the three integers have a common divisor greater than 1 but the greatest common divisor of all three numbers is 1.
Example: 6, 10, 15; 6 and 10 are both divisible by 2, 6 and 15 are both divisible by 3, 10 and 15 are both divisible by 5, but the greatest common divisor of 6, 10, and 15 is 1.
No two integers have a common divisor greater than 1 (pairwise coprime).
Example: 3, 4, 5; 3, 4, and 5 are all relatively prime to one another.
What is the probability that each of the five cases will occur when choosing three random integers? Ar Colorado (talk) 16:02, 2 December 2022 (UTC)[reply]
This depends on the probability law that is chosen, and, except for the case 4 (setwise coprimality), it should be an exercise of probability theory to deduce the probability from the case of the probability of pairwise coprimality. In any case, per WP:VERIFIABILITY, a reliable source is needed for mentioning the result in this article, D.Lazard (talk) 16:39, 2 December 2022 (UTC)[reply]
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13941 | https://www.britannica.com/animal/toadfish | toadfish
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toadfish, any of about 80 species of bottom-living fishes constituting the family Batrachoididae and the order Batrachoidiformes. They are found chiefly in the New World and mostly in warm seas—occasionally in freshwater. Toadfishes are heavy-bodied fishes with broad, flattened heads and large mouths equipped with strong teeth. They grow to a maximum of about 40 cm (16 inches) and either are scaleless or have small scales. Most can produce audible grunting or croaking sounds. Toadfishes are carnivorous and sometimes bite when touched.
They are divided into three groups: true toadfishes, such as the oyster toadfish (Opsanus tau), a common resident of shallow coastal waters along eastern North America; venomous toadfishes (Thalassophryne and Daector), found in Central and South America and notable for inflicting painful wounds with the hollow, venom-injecting spines on their dorsal fins and gill covers; and midshipmen (Porichthys), shallow-water American fishes named for numerous (600–840) small, buttonlike light organs arranged in rows along the body. |
13942 | https://s28543.pcdn.co/wp-content/uploads/sites/39/2021/03/Necessary-vs.-Sufficient-lesson.pdf | Necessary vs. Sufficient: Under the Right Conditions - Reasoning Series | Academy 4 Social Change Necessary vs. Sufficient: Lesson Plan Topic A necessary condition is one that is needed for the other half of a conditional statement to be true. A sufficient condition is one that is enough to guarantee the truth of the other part of the statement, though there may be other conditions that could also affirm the statement to be true. Possible subjects/classes Time needed ● Logic/reasoning ● Philosophy 45-60 minutes Video link: Objective: What will students know/be able to do at the end of class? Students will be able to... ● Differentiate between necessary and sufficient conditions. ● Improve their ability to reason through complex conditional scenarios. Key Concepts & Vocabulary Conditional statement Materials Needed Worksheet, poster making materials and/or computers with presentation software (PowerPoint, Google Slides, etc.) Before you watch Quick write: Define the words “necessary” and “sufficient” as you use them in your everyday vocabulary. While you watch Define conditional statement, necessary condition, and sufficient condition. Necessary vs. Sufficient: Under the Right Conditions - Reasoning Series | Academy 4 Social Change After you watch/discussion questions 1. Why is it important to know the difference between necessary and sufficient? In what fields might it be particularly important? 2. How can you improve your ability to recognize necessary and sufficient conditions? 3. Why do you think necessary and sufficient conditions are confused so often? Activity Ideas ● Create a visual representation of the difference and relationship between necessity and sufficiency. This can be a poster, a PowerPoint presentation, a performance, or anything creative. Use images or videos to explain the information. Challenge yourself to use as few words as possible while still making the information easy to understand. ● Select a criminal law and research the statute that states what makes a person guilty of breaking that law. Identify the necessary and sufficient conditions of guilt and innocence in regard to the law you have selected. ● Individually complete the Worksheet. Then review answers in small groups or as a class. Sources/places to learn more 1. Wireless Philosophy. (2015, February 2). CRITICAL THINKING - Fundamentals: More About Necessary and Sufficient Conditions [HD] . Retrieved from . 2. If x, then y | sufficiency and necessity (Article) . (n.d.). Khan Academy. Retrieved, from w/a/logic-toolbox--article--if-x-then-y--sufficiency-and-necessity . 3. Brennan, A. (2017). Necessary and sufficient conditions. In E. N. Zalta (Ed.), The Stanford Encyclopedia of Philosophy (Summer 2017). Metaphysics Research Lab, Stanford University. . |
13943 | https://mrciardullo.weebly.com/uploads/2/6/8/4/26847231/08_-_critical_angle_notes_2017_key.pdf | Refraction – Snell’s Law – Critical Angles When a ray of light passes from a medium into a less dense medium (with a smaller index of refraction), the refracted ray bends away from the normal. At a particular angle of incidence, the angle of refraction is 90°, and the refracted ray points along the surface of the medium. A critical angle is the angle of incidence that causes the refracted ray to point along the surface of the interface ( 90 R °). Critical Angle Total Internal Reflection Total internal reflection is the effect that occurs when an angle of incidence exceeds the critical angle. When this happens, there is no refracted ray, and all of the incident light is reflected back into the medium it came from following the law of reflection. Example: Light travels from water to air. The angle of incidence which results in a refracted angle of 90° is called the critical angle. Any angle greater than the critical angle will result in total internal reflection. Formulas to remember: 1 2 2 1 2 1 2 1 sin sin n n v v Examples: 1. What is the critical angle for an air-Lucite interface if the index of refraction of Lucite is 1.51? 2. What is the critical angle for a water-Lucite interface if the index of refraction for water is 1.33 and of Lucite is 1.51? 3. The critical angle for a certain liquid-air interface is 51.2º. What is the index of refraction of the liquid? Critical Angle Problems: 1. What is the critical angle for an air-glass interface if the index of refraction of glass is 1.50? (41.8⁰) 2. What is the critical angle for a water-Lucite interface if the index of refraction of water is 1.33 and of Lucite is 1.51? (61.8⁰) 3. The critical angle for a certain liquid-air interface is 48.8⁰. What is the index of refraction of the liquid? (1.33) 4. The speed of light in a clear liquid is 0.75c. What is the critical angle of the liquid? (49⁰) 5. A laser beam strikes the top surface of a block of glass at an angle θ1 and the refracted beam undergoes total internal reflection at the left vertical surface of the block as shown in the diagram. The index of refraction of the glass is 1.52. a) If the block is surrounded by air, find the maximum value of θ1. b) If the block is surrounded by water (n=1.33), find the maximum value of θ1. 6. In the following diagram, the index of refraction for ice is 1.31 and the index of refraction for oil is 1.45. The index of this type of glass is 1.52. a) If the angle θ1 of the incident ray is 36°, find the angle θ4 of the emergent ray. b) If total internal reflection occurs at the interface between the ice and air, find the critical angle of the incident ray. |
13944 | https://quizlet.com/371481301/trigonometry-flash-cards/ | Trigonometry Flashcards | Quizlet
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Use the graph to find the value of y = sin theta for the value of theta
240º
Choose an answer
1
C. 1
2
B. 0.7
3
B. 0
4
A. -0.9
Don't know?
Terms in this set (90)
Determine the period of the function y=3sin(2x)
d. Pi
Determine the amplitude or period as requested.
Amplitude of 1/3cos2x
d. 1/3
Determine the amplitude or period as requested.
Amplitude of y=cos3x
b. 1
Determine the amplitude or period as requested.
Amplitude of y=3-4sin3x
A. 4
Determine the amplitude or period as requested.
Amplitude of y=4sin 1/3x
A. 4
Determine the amplitude or period as requested.
Amplitude of y=-4cos1/3x
D. 4
Determine the amplitude or period as requested.
Amplitude of y=4cosx
B. 4
Determine the amplitude or period as requested.
Amplitude of y=sin5x
D. 1
Determine the amplitude or period as requested.
Amplitude of y=-3cos1/4x
D. 3
Determine the period of the function y=-2sin(pi/2)x
D. 4
Determine the amplitude of the function y = 2 cos x from the graph shown below:
b.
2
The b in both of the graph types: affects the period (or wavelength) of the graph. What happens to the period as the value of b gets larger?
c.
period decreases
Determine the amplitude or period as requested.
Amplitude of y=sin3x
D. 1
Determine the amplitude or period as requested.
Period of y=4sin1/3x
D. 6pi
In the expression y=asin(x), what does the a stand for?
b.
amplitude
Determine the amplitude of the function y=-1/2cosx
C. 1/2
Determine the amplitude or period as requested.
Amplitude of y=5cosx
B. 5
Determine the amplitude or period as requested.
Amplitude of y=-5sinx
c. 5
Determine the amplitude or period as requested.
Amplitude of y=-2sinx
c. 2
Determine the amplitude or period as requested.
Amplitude of y=-1/5sinx
b. 1/5
Determine the amplitude or period as requested.
Amplitude of y= -2sinx
c. 2
In the expression y=asin(x), what does the a stand for?
b.
amplitude
Determine the amplitude or period as requested.
Period of y=-1/3sin2x
d. pi.
Determine the amplitude of the function y = 2 cos x from the graph shown below:
b. 2
Determine the period of the function y = -3cos(pi/2)x
d.
10
Determine the amplitude or period as requested.
Amplitude of y=-1/5sinx
b. 1/5
The b in both of the graph types: affects the period (or wavelength) of the graph. What happens to the period as the value of b gets larger?
c.
period decreases
Determine the amplitude or period as requested.
Period of y=-2sinx
a. 2pi.
Determine the amplitude or period as requested.
Amplitude of y=3cos1/2x
b. 3
Determine the amplitude or period as requested.
Period of y=4sin1/3x
d. 6pi
Write the equation for the sine function shown below.
a.y=-sin5theta
A particular sound wave can be graphed using the function y=-10sin4x. Find the amplitude and period of the function.
A.
A particular sound wave can be graphed using the function y=-3sinx. Find the amplitude and period of the function
A.
Find the period of the graph shown below.
B. 2/3pi
Sketch one cycle of y = sin 2theta
C.
Use the graph to find the value of y = sin theta for the value of theta
240º
A. -0.9
Sketch one cycle of y = 4 sin 4theta
C.
Find the amplitude of the sine curve shown below.
c.
4
Find the amplitude of the sine curve shown below.
a.
2
Use the graph to find the value of y = sin theta for the value of theta
1/4 pi radians
b. 0.7
See an expert-written answer! We have an expert-written solution to this problem!
A particular sound wave can be graphed using the function y=1sin4x Find the amplitude and period of the function.
a.
Use the graph to find the value of y = sin theta for the value of theta
1/3 pi radians
c.
0.9
Write the equation for the sine function shown below.
c. y=5sin2theta
Find the period of the graph shown below.
a. 4pi.
Use the graph to find the value of y = sin theta for the value of theta.
270º
d.
-1
Use the graph to find the value of y = sin theta for the value of theta
360º
b.
0
Find the period of the graph shown below.
A. pi
Use the graph to find the value of y = sin theta for the value of theta.
330º
a.
-0.5
Find the period of the graph shown below.
A. 1/2pi.
Sketch one cycle of y = 2 sin theta
C.
Use the graph to find the value of y = sin theta for the value of theta.
90º
d.
1
Find the period, range, and amplitude of the cosine function
y=3/2 cos t/2
A.
Write a cosine function for the graph.
D. y=-2cos3theta
Suppose 8-in waves pass every 3 s. Write an equation that models the height of a water molecule as it moves from crest to crest.
D. 4cos 2pi/3 theta
Use a graphing calculator to solve the equation -6cos pi/3 theta=4 in the interval from 0 to 2pi. Round to the nearest hundredth.
b.
2.20, 3.80
cos(theta + 2pi) = cos theta
True
Write a cosine function for the graph.
C. -5cos theta
Write a cosine function for the graph.
B. y=3cos 3 theta
Write a cosine function for the graph.
C. 2cos4 theta
Write an equation of the cosine function with amplitude pi/2 and period 3.
A.
How does A compare to B?
A. Sin pi/2
B. cos(-pi)
A. A>=B
How does A compare to B?
A. Cos pi
B. Sin 3pi/2
c.
A = B
Write an equation of the cosine function with amplitude 2 and period 4pi.
C. y=2cos(1/2x)
Write a cosine function for the graph.
a. y=4cos2 theta
How does A compare to B?
A. sin pi/2
B. cos 0
c.
A = B
cos theta= cos (-theta)
true
How does A compare to B?
A. cos(theta+2pi)
B. cos(-theta)
c.
A = B
Write the equation for the inverse of the function.
y= Arctan(x+pi/2)
b. y=tanx=pi/2
Find the value of Tan^-1 squareroot 3/ 3
a. pi/6
Write the equation for the inverse of the function.
y = sin x
b.
y = Arcsin x
What is the value of Arc tan 1?
c. pi/4
Find the value of Arc cos 0
D. pi/2
Find the value of Sin^-1 0
B. 0
cos^1 (x) = -Cos^-1 x for -1<= x <= 1
False
Graph y= sin^-1 (-1/2x) the interval -5<= x <= 5
C.
Graph y=sin^-1 (1/4x) on the interval -5<= x <= 5
D.
Evaulate sin(Arccos(-9/square root 145))
A. 8/ squareroot 145
What is the value of cos(Tan^-1 1)?
D. square root 2 /2
Sin(Sin^-1x) = x for -1<= x <= 1
True
Find the value of tan(sin^-1(1/2))
c. square root 3/ 3
Write the equation for the inverse of the function.
y= pi/4 + sinx
c. y= Arcsin(x - pi/4)
Which describes the behavior of the function f(x)= tan^-1(x)
C.
Find the value of Sin^-1(tan pi/4)
d. Pi/2
Write the equation for the inverse of the function.
y = cos 2x
D.
Which function has a range of (-pi/2,pi/2)?
B. y= Tan ^-1 x.
Solve 2 cos theta + 2 = 3 in the interval from 0 to 2 theta Round to the nearest hundredth.
c.
1.05, 5.24
Write a cosine function for the graph.
y= -4 cos 4 theta
Find the amplitude and period of y = -5 sin (5x).
b.
amplitude = 5, period = 2/5 pi
Graph the following fnction on the interval -5 <= x <= 5
y= Arc cos (1/3x)
D.
Sketch one cycle of y = sin 2 theta
C.
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13945 | https://latin-dictionary.net/search/latin/capio | Latin Definitions for: capio (Latin Search) - Latin Dictionary and Grammar Resources - Latdict
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Latin search results for: capio
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capio, capere, additional, forms
1
verb
voice: irregular
Definitions:
arrest/capture
captivate
grasp
occupy
put on
take bribe
take hold, seize
Age:In use throughout the ages/unknown
Area:All or none
Geography:All or none
Frequency:Very frequent, in all Elementry Latin books, top 1000+ words
Source:“Oxford Latin Dictionary”, 1982 (OLD)
capio, capere, cepi, captus
2
verb
conjugation: 3 rd conjugation
voice: transitive
Definitions:
arrest/capture
captivate
grasp
occupy
put on
take bribe
take hold, seize
Age:In use throughout the ages/unknown
Area:All or none
Geography:All or none
Frequency:Very frequent, in all Elementry Latin books, top 1000+ words
Source:“Oxford Latin Dictionary”, 1982 (OLD)
capio, capionis
3
noun
declension: 3 rd declension
gender: feminine
Definitions:
taking/seizing
[usus ~ => getting ownership by continued possession]
Age:In use throughout the ages/unknown
Area:All or none
Geography:All or none
Frequency:For Dictionary, in top 10,000 words
Source:“Oxford Latin Dictionary”, 1982 (OLD)
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13946 | https://www.unitconverters.net/typography/pica-computer-to-centimeter.htm | Convert Pica (computer) to Centimeter
Please provide values below to convert pica (computer) to centimeter [cm], or vice versa.
| | | | |
--- --- |
| From: | | pica (computer) | switch |
| To: | | centimeter |
| | | | |
Pica (computer) to Centimeter Conversion Table
| Pica (computer) | Centimeter [cm] |
--- |
| 0.01 pica (computer) | 0.0042333333 cm |
| 0.1 pica (computer) | 0.0423333333 cm |
| 1 pica (computer) | 0.4233333333 cm |
| 2 pica (computer) | 0.8466666667 cm |
| 3 pica (computer) | 1.27 cm |
| 5 pica (computer) | 2.1166666667 cm |
| 10 pica (computer) | 4.2333333333 cm |
| 20 pica (computer) | 8.4666666667 cm |
| 50 pica (computer) | 21.1666666667 cm |
| 100 pica (computer) | 42.3333333333 cm |
| 1000 pica (computer) | 423.3333333333 cm |
How to Convert Pica (computer) to Centimeter
1 pica (computer) = 0.4233333333 cm
1 cm = 2.3622047244 pica (computer)
Example: convert 15 pica (computer) to cm:
15 pica (computer) = 15 × 0.4233333333 cm = 6.35 cm
Convert Pica (computer) to Other Typography Units |
13947 | https://www.carlisleschools.org/common/pages/UserFile.aspx?fileId=12259604 | Name: Period: Date:
Math Lab: Trig Identities Hexagon
Using the hexagon below, you will create a memory trick to learn the reciprocal, product/quotient, Pythagorean,
and cofunction trig identities and how identities can be used to evaluate trig functions. An identity is an
equation that is true for all x-values.
Buildi ng the Trig Identities Hexagon :
1] Draw a hexagon and put a “1” in the center
2] Write tan on the farthest left vertex
3] Use the Quotient Identity for tangent going clockwise (see below)
4] Fill in the Reciprocal Identities on opposite vertices
Quotient Identities
Along the outside edges of the hexagon, any trig function equals the quotient of the functions on the next two
consecutive vertices in either direction. Write the identities below in two equivalent forms for each:
Clockwise Counterc lockwise
tan 𝑥 = sin 𝑥
cos 𝑥 cot 𝑥 = cos 𝑥 = sec 𝑥 =
sin 𝑥 = csc 𝑥 = sin 𝑥 = csc 𝑥 =
cos 𝑥 = sec 𝑥 = tan 𝑥 = cot 𝑥 =
Reciprocal Identities
The two trig functions on any diagonal are reciprocals of each other. Write the six identities below:
sin 𝑥 = 1
csc 𝑥 cos 𝑥 = tan 𝑥 =
csc 𝑥 = sec 𝑥 = cot 𝑥 =Product Identities
The hexagon also shows that a function BETWEEN any two functions is equal to them multiplied together. If
they are opposite each other, then the 1 is between them. Write the identities below:
sin 𝑥 = tan 𝑥 ⋅ cos 𝑥 cos 𝑥 =
csc 𝑥 = sec 𝑥 =
tan 𝑥 = cot 𝑥 =
tan 𝑥 ⋅ cot 𝑥 = 1
Cofunction Identities
The trig functions cosine, cotangent, and cosecant on the right of the hexagon are the cofunctions of sine,
tangent, and secant on the left, respectively. So, sine and cosine are cofunctions; hence the “co” in cosine.
Cofunction identities are created fro m 𝑥 and 𝜋
2
− 𝑥 being a pair of complementary angles . That means they can
be found going left to right or going right to left on the hexagon. Write the identities below:
sin 𝑥 = cos (𝜋
2 − 𝑥 ) tan 𝑥 = sec 𝑥 =
cos 𝑥 = cot 𝑥 = csc 𝑥 =Deriving the Pythagorean Identities:
A is the center of the Unit Circle, B is a point on the circle in the first quadrant, and C is a right angle.
1] State the ratio 𝐵𝐶
𝐴𝐵
in terms of 𝑥 . This is the length of side a.
2] State the ratio 𝐴𝐶
𝐴𝐵
in terms of 𝑥 . This i s the length of side b.
3] Substitute the expressions you found in #1 and #2 into the
Pythagorean Theorem to create the first Pythagorean Identity.
4] Does the Pythagorean Identity hold true in the other three quadrants? Why or why not?
5] Divide each term in the Pythagorean Identity in #3 by cos 2 𝑥 to derive another form of the identity.
6] Divide each term in the Pythagorean Identity in #3 by sin 2 𝑥 to derive another form of the identity.
Pythagorean Identities
For each shaded triangle, the upper left function squared
plus the upper right function squared equals the bottom
function squared. Write the identities below in three
equivalent forms for each :
sin 2 𝑥 + cos 2 𝑥 = 1
sin 2 𝑥 = tan 2 𝑥 = cot 2 𝑥 =
cos 2 𝑥 = sec 2 𝑥 = csc 2 𝑥 =Even and Odd Identities
Recall that functions are considered even if their graphs have y-axis symmetry or odd if their graphs have
rotational symmetry about the origin . Algebraically, 𝑓 (−𝑥 ) = 𝑓 (𝑥 ) holds true for all even functions and
𝑓 (−𝑥 ) = −𝑓 (𝑥 ) holds true for all odd functions. First sketch each parent graph and decide if the function is
even or odd. Then w rite the identities using the algebraic definitions .
even or odd?
sin (−𝑥 ) =
even or odd?
cos (−𝑥 ) =
even or odd?
tan (−𝑥 ) =
even or odd?
csc (−𝑥 ) =
even or odd?
sec (−𝑥 ) =
even or odd?
cot (−𝑥 ) = |
13948 | https://www.openintro.org/go/?id=stat_nonlinear_relationships&referrer=/book/os/index.php | 1
OpenIntro online supplement
This material is an online resource of OpenIntro Statistics , a textbook available for free in PDF at openintro.org and in paperback for under $10 at amazon.com. This doc-ument is licensed to you under a Creative Commons license, and you are welcome to share it with others. For additional details on the license this document is under, see www.openintro.org/rights.php. 2
Fitting models for nonlinear trends
Prerequisites: Sections 1.1-1.6, 3.1, 4.1-4.4, 7.1-7.4, and 8.1 from OpenIntro Statistics are the bare minimum. Figure 1 presents two examples of nonlinear relationships between two numerical vari-ables. We’ll introduce two techniques for fitting these two data sets: (1) transforming the response variable and (2) fitting nonlinear model using polynomial terms in multiple re-gression. While these two methods are very useful, there is no “one size fits all” modeling solution, and there are plenty of situations where these two methods will be insufficient for your needs. If you find that nonlinearity or challenges with residuals cannot be adequately addressed using these methods, consider turning to additional statistical methods. 120 30 40 50 050 100 150 200 250
−5 0510 15 20 25 −5 05
Figure 1: Two pairs of numerical variables where each relationship is non-linear. The residuals may also show other deviations that must be consid-ered when modeling these data, including non-normal or heteroskedastic
residuals ( heteroskedastic means non-constant variance ). The techniques introduced in this section may be useful when the first, second, or fourth conditions for a simpler linear model are violated: 1. the model residuals should be nearly normal, 2. the variability of the residuals is nearly constant, 3. the residuals are independent, and 4. each variable is linearly related to the outcome.
1 Transformations on the response
Consider the scatterplot in the left panel of Figure 1. Here, the response y (vertical) tends to be positive but grow quickly. Additionally, the residuals show non-constant variance, because they are more variable for larger values of x (horizontal) and y. These two char-acteristics of the untransformed data are a clue that a transformation may be useful. In Section 1.6 of OpenIntro Statistics, we learned about the power of transformations to make skewed data more symmetric. If we look at a histogram of the x and y variables in Figure 2, we can see that x shows a very slight right skew and y is strongly right skewed.
1See the Supplement page on openintro.org for recommended free books that may be useful, or post a question on the online Public Forums on openintro.org.
3x
10 20 30 40 50 02468
y
050 100 150 200 250 300 0510 15 20 25 30 35
Figure 2: Histograms for both the x and y variables from the left panel of Figure 1. This suggests that it may be useful to transform the y variable. Had x been strongly right skewed, then we should have also considered using a transformation on x.There are many possible transformations, but one of the most common is the natural log-transformation (sometimes written as ln ). We’ll take the natural log for the y values and call this new variable y?:
y? = log y, where “log” is the natural log Figure 3 shows y? plotted against x. The data now show a linear relationship, where outliers are limited and the variability is roughly constant. Such an outcome is ideal, though far from guaranteed. We may now readily fit a linear model to the transformed scatterplot: ˆy? = 1 .03 + 0 .08 xy? = 1 .03 + 0 .08 x + residuals
In the first equation above, the formula has been written in the form used by OpenIntro Statistics. The second line is a more general way to write this formula. This general form is important when we are transforming data since we often want to back-transform the data. Here, we back-transform by substituting log( y) for y? and then solve for y:
y? = 1 .03 + 0 .08 x + residuals
log( y) = 1 .03 + 0 .08 x + residuals y = e1.03+0 .08 x+residuals
In this way, we can now enter a value for x and get an estimate for what value we think y
will take. This fitted line is shown in Figure 4. The predicted value for y in this model should not be confused with the expected (or mean) value of y for a given value of x, though the result may be somewhat close. The footnote provides an explanation of the difference for the interested reader. 2
2Suppose we collected many observations where x= 35. This model suggests that the distribution of the corresponding yvalues would be skewed as a result of the relationship between the residuals and the outcome ybeing nonlinear. The model (roughly speaking) estimates the median for each value of x.Because the median is not the same as a mean in a skewed distribution, the model will not provide the expected value of y, though often times it will be close.
420 30 40 50 2345
x
y = log(y)
Figure 3: A plot of y? (the result of transforming y by taking the natural logarithm log y) against x. The relationship between y? and x appears to be linear. 20 30 40 50 050 100 150 200 250
x
y
Figure 4: A nonlinear curve through the data generated by fitting a model of the form log y = β0 + β1x + residuals , then solving for y.5
TIP: Interpreting coefficients from a model that used log y
If the outcome in a model was transformed using the natural logarithm and the model fits well, then y tends to grow (or decay) exponentially relative to x.
Caution: Transformations can be abused
There is a very large number of possible transformations. If we keep trying trans-formations until one “works”, we have not effectively modeled our data. Rather, we have performed a complicated form of data fishing where we mine the data until we see structure. This apparent structure may just be due to chance. Therefore, think carefully about transformations before applying them. You are once again armed with knowledge that is both powerful and dangerous. This very brief introduction to transformations should be useful for informal projects. For a more complete review of this topic, visit Chapter 8 of Practical Regression and ANOVA in R ,which can be found in the Free Books section on the Supplements page of openintro.org.
2 Fitting a polynomial curve
Let’s take a look at the second nonlinear relationship we saw in Figure 1, which appears again in Figure 5 with a poorly-fit straight line. Here we see what appears to be a nonlinear relationship but where the residuals would be approximately homoskedastic (constant variance) if we could reasonably model the curve of the line. This is a good signal that we want to fit a curve but not perform a transformation. We can do so by generating a
polynomial basis of x: x1 = x, x2 = x2, x3 = x3, and so on. In short, we will use the variables x1, x2, x3, ... in a multiple regression model instead of simply the original variable x. We should note that it is uncommon to use terms beyond x2 = x2 and very rarely beyond x3 = x3.We start by fitting a linear model to the data, where the best-fitting straight line is shown in Figure 5 and summarized as
y = 0 .8441 − 0.0964 x + residuals
Even without checking the residual plot, it is evident that this line does not fit the data well, though Table 6 shows that the linear term is statistically significant.
⊙ Exercise 1 Suppose you were providing feedback to someone on a project, and the colleague had fit the line to the data shown in Figure 5. Suppose also that your colleague believes this model is sufficient because the estimate for the slope is statistically significant. Explain why the model is inappropriate. One possible explanation is provided in the footnote. 3
As a next step, we’ll add another variable to the model: x2 = x2. This new variable is itself a transformation on the variable x. However, rather than substituting x2 for x, we’ll fit a multiple regression model including both variables from the polynomial basis:
y = β0 + β1x1 + β2x2 + residuals
3Regression models require certain conditions to be met. In particular, the residuals must be indepen-dent of each other. However, when we look at the residuals from the fit in Figure 5, there is a clear trend in the residuals not captured by the straight line, meaning the independence condition is violated and the model is inadequate.
6−5 0 5 10 15 20 25 −5 05
x
y
Figure 5: Scatterplot with the best-fitting straight line, which does not fit the data well. Estimate Std. Error t value Pr( >|t|)(Intercept) 0.8441 0.5799 1.46 0.1487 x1 -0.0964 0.0397 -2.43 0.0169 Table 6: Summary for a straight line fit to the data shown in Figure 5. The best-fitting model of this form is shown in Figure 7, and the summary for the model is shown in Table 8.
Example 2 Write out the best fitting quadratic model using Table 8. The model may be written as
y = 2 .4252 − 0.7769 x1 + 0 .0295 x2 + residuals
= 2 .4252 − 0.7769 x + 0 .0295 x2 + residuals
In this example a quadratic model is still insufficient, so we will try a cubic polynomial (also known as a third-order polynomial). We will try fitting a model based on a cubic polynomial:
y = β0 + β1x1 + β2x2 + β3x3 + residuals
= β0 + β1x + β2x2 + β3x3 + residuals
Such a model is summarized in Figure 9 and Table 10. Estimate Std. Error t value Pr( >|t|)(Intercept) 2.4252 0.5079 4.78 0.0000 x1 -0.7769 0.0956 -8.13 0.0000 x2 0.0295 0.0039 7.55 0.0000 Table 8: Summary for a quadratic fit to the data shown in Figure 7. 7−5 0 5 10 15 20 25 −5 05
x
y
Figure 7: Scatterplot with the best-fitting quadratic line, which fits better than a straight-line but still misses some data structures. For example, the model underestimates much of the data for the range -5 to 0 and it overestimates nearly all of the data between 5 and 10. −5 0 5 10 15 20 25 −5 05
x
y
x
residuals
−5 0510 15 20 25 −6 −3 036
Figure 9: Scatterplot with the best-fitting cubic line. The residual plot shows no apparent structure, which is a good sign the model is fitting well. 8Estimate Std. Error t value Pr( >|t|)(Intercept) 2.0187 0.4242 4.76 0.0000 x1 -1.4202 0.1236 -11.49 0.0000 x2 0.1187 0.0136 8.75 0.0000 x3 -0.0026 0.0004 -6.77 0.0000 Table 10: Summary for a cubic line fit to the data shown in Figure 9
⊙ Exercise 3 Write out the best fitting quadratic model using Table 10. The solution is in the footnote. 4
The initial prognosis from the residual plot is that the cubic model fits very well. However, a complete analysis would include checking the model diagnostics carefully, which is a topic discussed in Section 8.3 of OpenIntro Statistics.
TIP: Stick with lower-order polynomials
If you want to try out using a polynomial term in your model, consider x2 and perhaps x3 if the model is still not a good fit. If a cubic polynomial will not model your data well, then be very cautious about trying higher-order polynomials. Instead, consider learning about regression splines, kernel smoothing, or some other statistical technique. See the textbook Elements of Statistical Learning for more information on advanced modeling techniques.
Caution: Do not extrapolate with transformed models or models that use polynomial terms
Extrapolation is already treacherous for any model, but it can be much worse for transformed data or data that includes polynomial terms, as the model can deviate very rapidly from the typical values observed in the original data set. |
13949 | https://www.themathdoctors.org/arithmetic-series-backward/ | Typesetting math: 100%
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Arithmetic Series, Backward
Here is a recent question about arithmetic sequences and series (specifically, reversing the process to find the number of terms given the sum), that nicely illustrates a common type of interaction with a student: gathering information about both problem and student, then guiding them to use what they know, or giving new information as needed. We’ll see how formulas can be used in solving problems, but are not all we need.
When will the sum reach 200,000?
At the very end of March, Daniel wrote to us with this question:
I currently have 3260 units of inventory/stock. Each month the stock purchased increases by 10 units (so next month will be 3270 and the month after, 3280). Assuming no stock is sold, stolen or destroyed, how many months will it take to reach 200,000 and how many units of stock would have been purchased in that final month?
Please include formula, thank you in advance.
Sometimes we get real life problems that can look a lot like word problems; but Daniel gave his age as high school, and the situation is a little too precise for real life! So we can infer the general context, but not Daniel’s level of knowledge, or what he has been learning that he is expected to use. So we need to start by finding out about those things.
What formulas do you know?
Doctor Rick answered, giving a hint while asking for information:
Hi, Daniel, thanks for writing to the Math Doctors.
This problem is about arithmetic series. Rather than provide you with a formula (reinforcing the wrong idea that math is all about formulas), I want to know what you have learned recently, so that I can help you apply what you already know.
What are your thoughts about the problem? When I say “arithmetic series”, that’s a big hint assuming you have learned about them — and I imagine the problem comes from a unit on arithmetic series. Have you learned any formulas or done example problems? That could be a starting point for thinking about your problem. Please show me what you can do, even if that is only to list possibly relevant definitions or formulas.
If the name “arithmetic series” means nothing, then what have you been learning? There might be another way to approach the problem.
It is quite common for people in or out of school to expect that a formula will solve every problem; but most problems require a mix of formulas and thinking about how to use them. Formulas are tools in our toolbox; problem-solving ability involves choosing the right tool for each step of a project.
First attempt
Daniel replied, providing just what we need to make a start at helping: his state of knowledge, and a first attempt.
Hi Doctor Rick,
Thank you for your response.
I apologise for not showing my workings.
I am aware of 2 formulas which I understand:
an = a1 + (n – 1)d
sn = (n/2)(a1 + an)
I have attached a page showing my workings but it is here where I also hit a wall.
In summary, while I suppose that I can keep randomly guessing “n” until I chance upon the correct number, this method is only feasible for homework because during tests and exams, I won’t have the luxury of time on my side to muddle through it randomly.
I’m fairly certain that a formula does exist for this but I can’t seem to figure it out on my own.
Thank you again for your assistance.
Kind Regards
He has made a good start. He identified the two formulas (one for the nth term of an arithmetic sequence, and one for the sum of n terms). He sees that the first term (the first amount purchased) is 3260, and the common difference (the increase from any term to the next) is 10. For practice, he applied the first formula to find the amount of the 20th purchase; this amounts to the fact that the first was 3260, and it has been increased by 10, 19 times, for a total of 3260+19(10)=3450.
Then, the total amount purchased in these first 20 months is given by the second formula, which in effect multiplies the average of the first and last terms by the number of terms: 202(3260+3450)=67,100.
But if this is just the beginning of a trial-and-error search for the value of n needed to reach a total of 200,000, it will take some time. We might, for example, guess that since the goal is more than twice the amount at 20 weeks, maybe we should try n=40. Then we get
a40=3260+10(40−1)=3650S40=402(3260+3650)=138,200
That isn’t enough, so we’d keep trying. It wouldn’t take too much longer if we make good guesses (not too cautious, and not too wild); but surely there is a more direct way?
Making an equation
Doctor Rick responded, suggesting the next “tool” that will be needed, which hopefully Daniel is aware of from other lessons, and didn’t realize it could be used here:
Thanks for the work. Now I can see what you know and where you are having trouble, so I’m in a much better position to help you!
There is another formula you will probably be using, but it’s not a formula specific to this kind of problem; it’s the quadratic formula.
The idea is to use both the formulas you showed to write an equation that says the sum of the first “n” terms of the series is (at least) 200 000. Then, instead of guess-and-check, you just solve the equation for n.
Here’s a start. You’ve got the equation
Sn = (n/2)(a1 + an)
and an equation for an:
an = a1 + (n – 1)d
Use the second equation replace an in the first equation, and replace Sn, a1, and d with values given by the problem. Then you’ll have the equation in n. Give this a try.
This is a key idea of algebra: When you need to find out what value of a variable will make some condition true, you express the condition as an equation and solve for the variable. We want to find the value of n for which Sn=200,000, so we need to express Sn as an expression in n, set that equal to 200,000, and solve.
Daniel answered,
Thank you for your response Doctor Rick,
I looked up the quadratic formula and the slew of alphabets and roots has completely baffled me.
Looking at the second part, I tried to follow your instructions as per the following
an = a1 + (n – 1)d
Sn = a1 + (n – 1)d
200,000 = 3260 + (n – 1)10
200,000 -3260 = 10n – 10
196,740 – 10 = 10n
196,730 = 10n
19,673 = n
Somehow it looks as though I’ve made things worse…
I’m not looking to be spoon-fed by any means and I apologise if I’m coming across that way, but it I am finding it slightly difficult to understand.
Please can you show me where I’ve gone wrong?
Kind Regards
So he clearly has never seen the quadratic formula, and perhaps is not even ready to learn it; this strongly suggests that he is expected either to solve the quadratic equation we’ll be writing using another method (such as factoring, if possible), or else just to use trial and error (“guess and check”) as he started to do.
We’ll get back to that; but for now we can look at his work, which since it didn’t result in a quadratic equation, contains an error somewhere. And we see it quickly: On the second line of work, he used the formula for the nth term as if it gave the nth sum. So what he’s actually done here is to find the index of the term that is equal to 200,000, instead of the term that makes the total equal to 200,000. (Actually, he made a small mistake in his algebra; can you find it? We won’t mention it to him, because this is the wrong thing to be doing anyway.)
Correcting the equation
Doctor Rick replied, pointing out the main error and then starting the correct work:
OK, now I know that you are not familiar with the quadratic formula. You see how important it is for us to find out what a student knows, in order to provide the right kind of help. Perhaps you can tell me what kind of class you are in, and what you have learned so far in that class … assuming this is for a class.
There are other ways to solve a quadratic equation besides the quadratic formula, but we aren’t ready to think about that until we have written a quadratic equation. Let’s look at what you did so far:
an = a1 + (n – 1)d
Sn = a1 + (n – 1)d
200,000 = 3260 + (n – 1)10
Somehow you copied the same formula twice. The first formula tells us how to find the nth term of an arithmetic sequence, given the first term (a1) and the common difference (d). We can’t use the same formula to find Sn, the sum of the first n terms of the sequence. What you wrote earlier was correct:
Sn = (n/2)(a1 + an)
The trouble is that in order to find Sn using this formula, we need to know the first term a1 and the nth term an; and we don’t know an yet! However, we do have a formula for an. So … and this may be where I confused you … what we cando is to replace an, in the formula for Sn, with the formula for an. Here’s what I get when I do that:
Sn = (n/2)(a1 + (a1 + (n – 1)d))
This can be simplified to obtain a nicer formula for Sn in terms of n, a1, and d. (Some students may memorize that formula; I don’t — I reconstruct the formula from yours, when I need it, just as I am doing now.) Or you could plug in the numbers you know right now, and then simplify the result.
If you do that correctly, you will have a quadratic equation in n. This will be the time to talk about quadratic equations and how to solve them. So let’s see what equation you come up with, and I’d like you also to tell me what you know about quadratic equations.
Daniel never replied, so we don’t know whether this start was enough, or what other tools are in his toolbox. But we can continue from here.
Simplifying the equation
First, let’s simplify the formula:
Sn=n2(a1+(a1+(n−1)d))=n2(2a1+dn−d)=12(dn2+(2a1−d)n)
Now we need to use the numbers from the problem (which we could have used earlier, if not for the desire to see if this is a memorable formula): a1=3260,d=10,Sn=200,000. Our equation becomes
12(10n2+6510n)=200,000
We can distribute the fraction and get
5n2+3255n=200,000
Now we can subtract 200,000 from both sides, and divide by 5:
n2+651n−40,000=0
We could try to factor this, but (knowing about the discriminant) I know that will be a waste of time. As a result, I strongly suspect that Daniel was expected just to use trial and error.
Using the quadratic formula
But we can use the quadratic formula, which says that the solutions to ax2+bx+c=0 are
x=−b±b2−4ac−−−−−−−√2a
Taking a=1,b=651,c=−40,000, we get
n=−651±6512−4(−40,000)−−−−−−−−−−−−−−−√2=−651±583,801−−−−−−−√2=−651±764.06872=56.53
(taking only the positive solution).
Of course, n has to be an integer! What did the problem ask for? “How many months will it take to reach 200,000?” We now know that the first whole number of months after which we will exceed 200,000 units is 57; at 57 months, we will have exceeded 200,000 units. Or we could say that during the 57th month we will reach the goal.
Let’s check the answer. First, to check the solution to the equation, we can take n=56.53 and see that we get the right sum:
Sn=12(10n2+6510n)=12(10(56.53)2+6510(56.53))=199,983.3545
This is less than 200,000 due to rounding. Rounding up, we get
Sn=12(10(56.54)2+6510(56.54))=200,021.558
So we have the right solution.
But we still need to find the amount added in that 57th month:
an=a1+(n−1)d=3260+(57−1)10=3820
And now that we have that, we can confirm the sum:
Sn=n2(a1+an)=572(3260+3820)=201,780
which will be the first time over 200,000.
A trial of trial and error
Let’s do one more thing: How much work would it take by trial and error, as we now think Daniel needs to do it?
Our last attempt looked like this:
a40=3260+10(40−1)=3650S40=402(3260+3650)=138,200
Doing that repeatedly would take more work than necessary. Instead, we can use our combined formula,
Sn=5n2+3255n
Using that formula, our last guess (40) would be
S40=5(40)2+3255(40)=138,200
as before. That’s still too small; let’s double our guess again (I’m not looking at the answer we found as I do this!). Taking n=80, we get
S80=5(80)2+3255(80)=292,400
That’s too large; let’s back off halfway to 60:
S60=5(60)2+3255(60)=213,300
Still a little high, but not much; let’s decrease it to 55:
S55=5(55)2+3255(55)=194,150
We went too far; let’s split the difference and try 57:
S57=5(57)2+3255(57)=201,780
The only remaining possibility is 56, so we check that:
S56=5(56)2+3255(56)=197,960
So we’ve found our answer: 57 months.
For a student who hasn’t learned the quadratic formula, this is still doable, but I wouldn’t consider it appropriate.
1 thought on “Arithmetic Series, Backward”
Rick Peterson
Re-reading this discussion I had with Daniel makes me wonder again how he could have been expected to solve this problem if he had never seen the quadratic formula. It seems unlikely that he would have seen, for instance, the method of completing the square without also seeing the quadratic formula. Perhaps he changed schools and was put into a class for which he hadn’t had all the prerequisites. Still, I got to thinking … Could we do better than basic guess-and-check in this situation?
Actually, first I was thinking about the question of whether the combined formula is in fact memorable. Doctor Peterson simplified it to
S[n] = (1/2)(dn^2+(2a − d)n)
This can also be written as
S[n] = an + dn(n - 1)/2
which we can understand as follows: We start with a in each of the n terms. We leave the first term alone, add d to the second term, add 2d to the third term, and so on. Thus we have n copies of a, for a total of an, to which we add d(1 + 2 + 3 + … + (n-1)). In the parentheses we see the (n-1)th “triangular number”, whose formula is T[n] = n(n+1)/2, or T[n-1] = (n-1)n/2. The result is the formula above.
Now, let’s make use of to solve the problem by iteration. We have a = 3260, d = 10, so
S[n] = 3260n + 5n(n - 1)
and we want to solve the equation
3260n + 5n(n - 1) = 200,000
I’ll factor out n on the left and divide both sides by 5 to start with:
n(652 + (n - 1)) = 40,000
n(651 + n) = 40,000
What if we suppose that n is small compared to 651? Then we can approximate n by solving the equation
651n = 40,000
The solution is n = 61.444. Now, use this value for n in that second factor:
n(651 + 61.444) = 712.444n = 40,000
Solving this equation gives n = 56.1448. Once again using this value for n in the second factor, we get
707.1447n = 40,000
n = 56.5656
It looks like we’ve got our solution to the nearest integer, in just three iterations! That’s not too bad …
But I really don’t think Daniel would have been expected to come up with this solution method.
Reply
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13951 | https://brilliant.org/wiki/basic-trigonometric-functions/ | Basic Trigonometric Functions
Mei Li,
Pi Han Goh,
Omkar Kulkarni,
and
Andrew Ellinor
Krishna Chaitanya Porandla
Kenny Lau
Calvin Lin
Jimin Khim
Eli Ross
contributed
Contents
Basic Trigonometric Functions
Specific Values - Basic
Specific Values - Intermediate
Problem Solving
Right Triangle Trigonometry
Isosceles Right Triangle
30∘-60∘-90∘ Right Triangle
Right Angle Trigonometry
Basic Trigonometric Functions
The trigonometric functions relate the angles in a right triangle to the ratios of the sides. Given the following triangle:
the basic trigonometric functions are defined for 0<θ<2π as
sinθ=cb,cosθ=ca,tanθ=ab.
If we consider the angle θ and label the sides with respect to θ, then a is the length of the "adjacent" side, b is the length of the "opposite" side, and c is the length of the hypotenuse. Then the basic trigonometric functions can be expressed as follows:
sinθ=hypotenuseopposite,cosθ=hypotenuseadjacent,tanθ=adjacentopposite.
For a review of converting between degrees and radians, see Degrees and Radians. However, a more useful definition comes from the unit circle. If we consider a circle with a radius of 1 unit, centered at the origin, then the angle θ inside the circle describes a right triangle when we drop a perpendicular to the x-axis from the point of intersection with the circle.
Unit Circle
Notice that the right triangle so described has a hypotenuse equal to the radius of the circle, an adjacent side equal to the x-coordinate of the point (x,y), and an opposite side equal to the y-coordinate. This gives rise naturally to the following refined definitions:
sinθ=radiusy,cosθ=radiusx,tanθ=xy.
As shown in the above diagram, since the radius is 1 in the unit circle, this simplifies to x=cosθ and y=sinθ.
These definitions have the advantage of being compatible with the triangle definition above, as well as allowing the evaluation of angles corresponding to any real number.
There are certain values of these functions which are useful to remember. They are:
θsinθcosθtanθ0∘202406π=30∘2123314π=45∘222213π=60∘232132π=90∘2420∞
The reason for writing them in this way is to aid remembering these terms. For example, the numerator for sinθ is simply the square root of 0,1,2,3 or 4. To visualize these values in the unit circle, see specific values of trigonometric functions.
What values of θ satisfy
0≤θ<2π and cosθ=0?
A good approach for a problem like this is to imagine the unit circle diagram shown above. In this unit circle diagram, cosθ is the x-coordinate. Thus, the problem is asking for angle θ whose x-coordinate is equal to 0, which occurs at the two points where the unit circle intersects the y-axis: θ=2π and θ=23π. Since these angles satisfy the given conditions, the two possible angles are θ=2π and θ=23π. □
If θ is an angle such that cosθ=0, what are the possible values of sinθ?
Solution 1:
As we saw above, cosθ=0 corresponds to points on the unit circle whose x-coordinate is 0. Since these points occur at the points of intersection with the y-axis, the possible values of sinθ are the possible y-coordinates, which are 1 and −1. □
Solution 2:
From the first example, if cosθ=0 and 0≤θ<2π , then θ=2π or θ=23π. Note that for all other values of θ outside of this range, we have sinθ=sin(θ±2π), so we can add or subtract multiples of 2π until θ lies in this range. For θ=2π, we have sinθ=1 and for θ=23π, we have sinθ=−1. Therefore, the possible values of sinθ are 1 and −1. □
If θ is an angle in a right triangle such that sinθ=21, what is the value of cosθ?
Since θ is an angle in a right triangle, it must be the case that 0<θ<2π.
From the table above, we see that the value θ such that sin(θ)=21 is θ=6π=30∘. Also, since the y-coordinate of the points on the unit circle are increasing as θ goes from 0 to 2π, the value θ=6π is the unique value such that 0<θ<2π and sinθ=21. Then, we have cosθ=cos2π=23. □
In the table of specific values for trigonometric functions above, why is there no value for tan2π? What happens to tanθ as θ gets closer and closer to 2π?
From the above definition of tanθ, we have tanθ=cosθsinθ=xy, where (x,y) are the x- and y-coordinates of the point of angle θ on the unit circle. As θ moves towards 2π, cosθ (the x-coordinate) becomes smaller and smaller, whereas sinθ (the y-coordinate) becomes closer and closer to 1. Therefore, the numerator of tanθ=xy approaches 1 and the denominator approaches 0, implying tanθ approaches infinity. □
To learn about the other trigonometric functions, read Reciprocal Trigonometric Functions and Inverse Trigonometric Functions.
Specific Values - Basic
There are certain values of the basic trigonometric functions which are useful to remember. They are:
θsinθcosθtanθ0∘202406π=30∘2123314π=45∘222213π=60∘232132π=90∘2420±∞
The reason for writing them in this way is to aid remembering these terms. For example, the numerator for sinθ is simply the square root of 0, 1, 2, 3, 4.
Visualization in Unit Circle
We can also visualize the cosine and sine values for these values in the unit circle:
Since the cosine function corresponds to x values, the cosine function will be positive when x values are positive and will be negative when x values are negative. Similarly, since the sine function corresponds to y values, the sine function will be positive when y values are positive and will be negative when y values are negative. This gives us the following behavior in the four quadrants of the plane:
Then by using the few specific values from the first quadrant, we can figure out the specific values of the cosine and sine functions in all quadrants. Here is the visualization for all quadrants:
Image courtesy commons.wikimedia.org
What are the values of θ in the range 0≤θ<2π such that sinθ=cosθ?
From the unit circle visualization above, we see that θ=4πand θ=45π satisfy
sin(4π)sin(45π)=cos(4π)=22=cos(45π)=−22.
We also observe that the line y=x intersects the unit circles for only these two values of θ, so the values of θ satisfying the required conditions are θ=4π and θ=45π. □
What values of θ in the range 0≤θ<2π satisfy sinθ≥22?
By drawing the line y=22, we would like to find the values of θ such that the the y-value of the angle θ on the unit circle lies above this line (since sinθ corresponds to the y-coordinate of the unit circle). This holds for θ∈[4π,43π], so these are the values of θ satisfying sinθ≥22. □
What are the values of θ in the range 0≤θ<2π such that sinθ=−cosθ?
From the unit circle visualization above, we see that θ=43πand θ=47π satisfy
sin(43π)sin(47π)=−cos(43π)=22=−cos(47π)=−22.
We also observe that the line y=−x intersects the unit circle for only these two values of θ, so the values of θ satisfying the required conditions are θ=43π and θ=47π. □
Calculate
3tan230∘+csc245∘−tan45∘9cot260∘+sec245∘−4sin90∘.
The correct answer is: 0.5
Specific Values - Intermediate
In order to obtain further values, we will need to use some of the trigonometric formulas like sum and difference and product to sum. If you are unfamiliar with them, please skip this section for now and come back to it later.
Let us see an application of the sum and difference formulas.
Evaluate cos15∘.
Using the difference formula for cos, we have
cos15∘=cos(45∘−30∘)=cos45∘cos30∘+sin45∘sin30∘=22×23+22×21=42(3+1)=46+2. □
Let us see an application of the double angle formula.
Evaluate sin15∘.
We know that cos30∘=1−2sin215∘, so sin215∘=42−3. Since sin15∘ is positive, we take the positive square root and get that
sin15∘=22−3. □
1
2
3
4
The angle θ whose cosine is equal to its tangent is given by
sinθ=nsin18∘.
Find the value of n.
This problem is part of the set Trigonometry.
The correct answer is: 2
sec2(9π)+sec2(92π)+sec2(94π)=?
The correct answer is: 36
Problem Solving
How many integers x are there such that 0<x∘<360 and 2sin2x∘<sinx∘?
We have 2sin2x∘−sinx∘=sinx∘(2sinx∘−1)<0 or 0<sinx∘<21.
If we only consider the first quadrant, then y=sinx∘ is an increasing function. So
0=sin0∘<sin1∘<sin2∘<⋯<sin29∘<sin30∘=21,
which means there are 29 solutions.
Using the same argument, we have another 29 solutions in the second quadrant. This gives us a total of 29+29=58 solutions. □
If we know that tan63∘<2<tan64∘ and tan84∘<9<tan85∘, how many integers x are there such that 0∘<x∘<90∘ and 2<tanx∘<9?
Because y=tanx is an increasing function,
2<tan64∘<tan65∘<tan66∘<⋯<tan84∘<9.
This gives us 84−64+1=21 integer solutions of x. □
0
1
2
3
4
Infinitely many solutions
How many real numbers x in radian satisfy the equation
cosxtanx+sinxcotx=1?
The correct answer is: 0
cos[sin−1(11−7)+sec−1(7−11)]
Evaluate the expression above.
The correct answer is: 0.00
Right Triangle Trigonometry
There are certain types of right triangles whose ratios of side lengths are useful to know. These are also found in specific values of trigonometric functions.
Isosceles Right Triangle
In an isosceles right triangle, the angles are 45∘, 45∘, and 90∘. For such a triangle, the two shorter sides of the triangle are equal in length and the hypotenuse is 2 times the length of the shorter side:
We can also see this relationship from the definition of sinθ and cosθ and using the specific value of θ=45∘:
sin45∘cos45∘=sin4π=21=hypotenuseopposite=cos4π=21=hypotenuseadjacent.
30∘-60∘-90∘ Right Triangle
In this right triangle, the angles are 30∘,60∘, and 90∘.
If the side opposite the 30∘ angle has length a, then the the side opposite the 60∘ angle has length a3 and the hypotenuse has length 2a. We can also see this from the definition of sinθ and cosθ and using the specific value of θ=60∘:
sin60∘cos60∘=sin(3π)=23=hypotenuseopposite=cos(3π)=21=hypotenuseadjacent.
Cite as:
Basic Trigonometric Functions.
Brilliant.org.
Retrieved
03:18, September 28, 2025,
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13952 | https://selfhelp.courts.ca.gov/criminal-court/overview/trial | Criminal trial overview | California Courts | Self Help Guide
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A basic overview of the main steps in a criminal jury trial.
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Criminal trial
Criminal trials are open to the public. Defendants have a right to a jury trial, which means a jury listens to both sides present their case and then decides if the defendant is guilty or not. If the defendant wants, they can ask for a judge to decide instead.
Trials can last from a day to many months. In general, the more witnesses in a case the longer it can last. Trials also tend to be longer if there is more than 1 defendant in the case.
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Jury selection
A jury has 12 members of the public. To decide who is on the jury, both sides get to ask potential jurors questions to make sure they can be fair. The judge may ask questions as well. This is called voir dire.
Issues the judge needs to decide before either side presents its case
When the jury is not in the courtroom, either side can ask the judge to decide whether or how certain evidence can be used, if at all. For example, if the defendant has a prior conviction the defense may ask the judge to decide whether the prosecution can bring it up if the defendant testifies.
### Opening statements
Both sides start by giving an overview of what they plan to show at the trial. This is not an argument, it's where they describe their case and what evidence they'll have.
### Prosecution presents its case
The prosecution presents its witnesses and evidence. The defense can ask the witnesses questions as well (called cross-examination). The prosecution must prove the defendant committed each crime the person is charged with beyond a reasonable doubt.
### Defense presents it case
The defense does not have to present a case. If they do, they can have witnesses testify or show evidence that supports their side. The prosecutor can ask the witnesses questions.
If the defendant wants, they can also testify. They do not have to. They have the right to remain silent. The jury cannot use the fact that the defendant chose not to testify against them (assume they are guilty).
If the defense does present a case, the prosecution gets another chance to present witnesses to respond to what the defense witnesses said.
### Closing arguments
Both sides get a final chance to convince the jury to vote guilty or not guilty. The prosecutor argues that they have proven the defendant is guilty beyond a reasonable doubt and that jury should vote guilty. The defendant can argue why the prosecutor did not prove their case, and even that the defendant is innocent.
Jury makes a decision
Jury instructions
Typically, after closing arguments, the judge gives the jury instructions. These are the legal standards the jury uses to decide a case. They include a legal definition of the crime.
Jury deliberations and verdict
The jurors meet in private and discuss the case and vote guilty or not guilty. This is called deliberations.To reach a final decision (a verdict), the jury must all agree that a person is guilty or not guilty of each charge.
If they vote not guilty, the person is acquitted and can't be tried again. Not guilty does not mean innocent. It means the jury was not convinced beyond reasonable doubt the person was guilty.
If the jurors all vote guilty, then the judge decides the sentence.
If jurors cannot agree, the judge can declare a mistrial. This is sometimes called a hung jury. The prosecutor may or may not decide to try the case again.
What's next?
If the jury found the defendant guilty or the defendant pled guilty, the next step is sentencing. This may happen shortly after the trial or a week to months later.
### Go back to an overview Go back to all the steps in a criminal court case.
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Defining Physiology: Principles, Themes, Concepts. Volume 2
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Micelles are aggregates of bile salt that forms a polar outer shell and a hydrophobic inner core. Long chain fatty acids, cholesterol and other hydrophobic molecules dissolves in the core and are transported in the surface of the enterocytes for processing.
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13954 | https://www.ejinme.com/article/S0953-6205(20)30369-1/fulltext | A middle-aged woman with numbness and weakness of the extremities - European Journal of Internal Medicine
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Internal Medicine FlashcardsVolume 83p78-79 January 2021
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A middle-aged woman with numbness and weakness of the extremities
Keizo Tanitame, M.D
Keizo Tanitame, M.D0000-0002-5825-376X
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Corresponding Author. Tel: +81-82-254-1818
tntrad@gmail.com
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Department of Diagnostic Radiology, Hiroshima Prefectural Hospital, 1-5-54 Ujinakanda, Hiroshima, 734-8530, Japan
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Department of Diagnostic Radiology, Hiroshima Prefectural Hospital, 1-5-54 Ujinakanda, Hiroshima, 734-8530, Japan
Publication History:
Received June 25, 2020; Accepted September 20, 2020; Published online September 23, 2020
DOI: 10.1016/j.ejim.2020.09.015 External LinkAlso available on ScienceDirect External Link
Copyright: © 2020 European Federation of Internal Medicine. Published by Elsevier B.V. All rights reserved.
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1 Case description
A 45-year-old woman presented to the neurology department with numbness and weakness of extremities, which had progressed during 3 months. The patient had tingling sensation in the left neck. Although muscle atrophy was not observed, manual muscle test revealed grade 4 weakness in bilateral deltoid muscles, right upper and left lower limbs. The patient had normal deep tendon reflex, and Babinski reflex was negative bilaterally. Laboratory blood tests showed no abnormalities. Spinal fluid examination revealed a slightly increased protein level of 56 mg/dL (normal range: 10–45 mg/dL) and no white blood cells. Nerve conduction study revealed prolonged distal latencies, reduction in conduction velocities, and delayed F wave latencies of the median, ulnar and tibial nerves, which was suggestive of predominantly distal sensorimotor demyelinating polyneuropathy. The patient had no family history of Charcot-Marie-Tooth (CMT) disease, and the CMT type 1A-related PMP22 gene duplication was absent in the gene analysis. Cervical magnetic resonance neurography (MRN) was performed for the evaluation of peripheral neuropathies (Fig.1). What is your diagnosis for this patient?
Figure viewer
Fig. 1 MRN using coronal fat-saturated T2-weighted imaging revealing symmetrical enlargement and increased signal intensity of cervical nerve roots.
2 Discussion section
The patient was diagnosed with idiopathic chronic inflammatory demyelinating polyneuropathy (CIDP) according to the American Academy of Neurology criteria after excluding the causes of secondary CIDP, such as diabetes mellitus, POEMS (polyneuropathy, organomegaly, endocrinopathy, M protein, and skin changes), monoclonal gammopathy, and malignancy. The patient's symptoms remitted after corticosteroid therapy, but relapsed 3 months later. Corticosteroids and intravenous immunoglobulin were repeatedly administered, and the patient has experienced remissions and exacerbations of the neurological symptoms.
CIDP is considered an autoimmune disorder targeting the myelin of peripheral nerve that progresses over more than 8 weeks. This rare disease is often underrecognized due to its varied presentation and the limitations of clinical, serologic, and electrophysiologic diagnostic criteria. Despite these limitations, early diagnosis and treatment are highly important in preventing axonal degeneration and recovering neurological function. Corticosteroids, intravenous immunoglobulin, and plasma exchange are the most common first-line therapies for CIDP patients
1.
Dimachkie, MM ∙ Barohn, RJ
Chronic inflammatory demyelinating polyneuropathy
Curr Treat Options Neurol. 2013; 15:350-366
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. Greater than 90% of CIDP patients initially improved with immunosuppressive treatment, but the relapse rate was approximately 50%
2.
Barohn, RJ ∙ Kissel, JT ∙ Warmolts, JR ...
Chronic inflammatory demyelinating polyradiculoneuropathy. Clinical characteristics, course, and recommendations for diagnostic criteria
Arch Neurol. 1989; 46:878-884
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. MRI evaluation of the spinal nerve roots is useful for the diagnosis of CIDP. CIDP is one of the hypertrophic neuropathies; the repetitive demyelination and remyelination results in enlargement of peripheral nerves and roots
3.
Freitas, MR ∙ Nascimento, OJ ∙ Soares, CN ...
Chronic inflammatory demyelinating polyradiculoneuropathy: two cases with cervical spinal cord compression
Arq Neuropsiquiatr. 2005; 63:666-669
Crossref
PubMed
Google Scholar
. Moreover, increased T2 signal intensity of nerve roots is considered to reflect the active inflammation with edema in the endoneurium and perineurium.
Declaration of Competing Interest
None declared. I disclose no actual or potential conflict of interest including any financial, personal or other relationships with other people or organizations within last three years.
References
1.
Dimachkie, MM ∙ Barohn, RJ
Chronic inflammatory demyelinating polyneuropathy
Curr Treat Options Neurol. 2013; 15:350-366
Crossref
Scopus (47)
PubMed
Google Scholar
2.
Barohn, RJ ∙ Kissel, JT ∙ Warmolts, JR ...
Chronic inflammatory demyelinating polyradiculoneuropathy. Clinical characteristics, course, and recommendations for diagnostic criteria
Arch Neurol. 1989; 46:878-884
Crossref
Scopus (581)
PubMed
Google Scholar
3.
Freitas, MR ∙ Nascimento, OJ ∙ Soares, CN ...
Chronic inflammatory demyelinating polyradiculoneuropathy: two cases with cervical spinal cord compression
Arq Neuropsiquiatr. 2005; 63:666-669
Crossref
PubMed
Google Scholar
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Using a table of logarithms
Posted on by John
My favorite quote from Richard Feynman is his remark that “nearly everything is really interesting if you go into it deeply enough.” This post will look at something that seems utterly trivial—looking up numbers in a table—and show that there’s much more to it when you dig a little deeper.
More than just looking up numbers
Before calculators were common, function values would be looked up in a table. For example, here is a piece of a table of logarithms from Abramowitz and Stegun, affectionately known as A&S.
But you wouldn’t just “look up” logarithm values. If you needed to know the value of a logarithm at a point where it is explicitly tabulated, then yes, you’d simply look it up. If you wanted to know the log of 1.754, then there it is in the table. But what if, for example, you wanted to know the log of 1.7543?
Notice that function values are given to 15 significant figures but input values are only given to four significant figures. If you wanted 15 sig figs in your output, presumably you’d want to specify your input to 15 sig figs as well. Or maybe you only needed 10 figures of precision, in which case you could ignore the rightmost column of decimal places in the table, but you still can’t directly specify input values to 10 figures.
Lagrange interpolation
If you go to the bottom of the column of A&S in the image above, you see this:
What’s the meaning of the mysterious square bracket expression? It’s telling you that for the input values in the range of this column, i.e. between 1.750 and 1.800, the error using linear interpolation will be less than 4 × 10−8, and that if you want full precision, i.e. 15 sig figs, then you’ll need to use Lagrange interpolation with 5 points.
So going back to the example of wanting to know the value of log(1,7543), we could calculate it using
0.7 × log(1.754) + 0.3 × log(1.755)
and expect the error to be less than 4 × 10−8.
We can confirm this with a little Python code.
```
from math import log >>> exact = log(1.7543) >>> approx = 0.7log(1.754) + 0.3log(1.755) >>> exact - approx 3.411265947494968e-08
```
Python uses double precision arithmetic, which is accurate to between 15 and 16 figures—more on that here—and so the function calls above are essentially the same as the tabulated values.
Now suppose we want the value of x = 1.75430123456789. The hint in square brackets says we should use Lagrange interpolation at five points, centered at the nearest tabulated value to x. That is, we’ll use the values of log at 1.752, 1.753, 1.754, 1.755, and 1.756 to compute the value of log(x).
Here’s the Lagrange interpolation formula, given in A&S as equation 25.2.15.
We illustrate this with the following Python code.
def interpolate(fs, p, h): s = (p2 - 1)p(p-2)fs/24 s -= (p - 1)p(p2 - 4)fs/6 s += (p2 - 1)(p2 - 4)fs/4 s -= (p + 1)p(p2 - 4)fs/6 s += (p2 - 1)p(p + 2)fs/24 return s xs = np.linspace(1.752, 1.756, 5) fs = np.log(xs) h = 0.001 x = 1.75430123456789 p = (x - 1.754)/h print(interpolate(fs, p, h)) print(np.log(x))
This prints
0.5620706206909348 0.5620706206909349
confirming that the interpolated value is indeed accurate to 15 figures.
Lagrange interpolation takes a lot of work to carry out by hand, and so sometimes you might use other techniques, such as transforming your calculation into one for which a Taylor series approximation converges quickly. In any case, sophisticated use of numerical tables was not simply a matter of looking things up.
Contemporary applications
A book of numerical tables enables you to do calculations without a computer. More than that, understanding how to do calculations without a computer helps you program calculations with a computer. Computers have to evaluate functions somehow, and one way is interpolating tabulated values.
For example, you could think of a digital image as a numerical table, the values of some ideal analog image sampled at discrete points. The screenshots above are interpolated: the HTML specifies the width to be less than that of the original screenshots,. You’re not seeing the original image; you’re seeing a new image that your computer has created for you using interpolation.
Interpolation is a kind of compression. A&S would be 100 billion times larger if it tabulated functions at 15 figure inputs. Instead, it tabulated functions for 4 figure inputs and gives you a recipe (Lagrange interpolation) for evaluating the functions at 15 figure inputs if you desire. This is a very common pattern. An SVG image, for example, does not tell you pixel values, but gives you equations for calculating pixel values at whatever scale is needed.
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One thought on “Using a table of logarithms”
Andrew Gelman
John:
Your Feynman quote (which I’d heard as “No problem is too small or too trivial if we really do something about it”) reminds me of the saying in statistics that God is in every leaf of every tree, as briefly discussed here:
Comments are closed. |
13958 | http://www.surface-tension.de/ | Surface tension values of some common test liquids for surface energy analysis
| | | | |
--- --- |
| Name | CAS Ref.-No. | Surface tension @ 20 °C in mN/m | Temperature coefficient in mN/(m K) |
| 1,2-Dichloro ethane | 107-06-2 | 33.30 | -0.1428 |
| 1,2,3-Tribromo propane | 96-11-7 | 45.40 | -0.1267 |
| 1,3,5-Trimethylbenzene (Mesitylene) | 108-67-8 | 28.80 | -0.0897 |
| 1,4-Dioxane | 123-91-1 | 33.00 | -0.1391 |
| 1,5-Pentanediol | 111-29-5 | 43.30 | -0.1161 |
| 1-Chlorobutane | 109-69-3 | 23.10 | -0.1117 |
| 1-Decanol | 112-30-1 | 28.50 | -0.0732 |
| 1-nitro propane | 108-03-2 | 29.40 | -0.1023 |
| 1-Octanol | 111-87-5 | 27.60 | -0.0795 |
| Acetone (2-Propanone) | 67-64-1 | 25.20 | -0.1120 |
| Aniline 22°C (AN) | 62-53-3 | 43.40 | -0.1085 |
| 2-Aminoethanol | 141-43-5 | 48.89 | -0.1115 |
| Anthranilic acid ethylester 22°C | 87-25-2 | 39.30 | -0.0935 |
| Anthranilic acid methylester 25 °C | 134-20-3 | 43.71 | -0.1152 |
| Benzene | 71-43-2 | 28.88 | -0.1291 |
| Benzylalcohol | 100-51-6 | 39.00 | -0.0920 |
| Benzylbenzoate (BNBZ) | 120-51-4 | 45.95 | -0.1066 |
| Bromobenzene | 108-86-1 | 36.50 | -0.1160 |
| Bromoform | 75-25-2 | 41.50 | -0.1308 |
| Butyronitrile | 109-74-0 | 28.10 | -0.1037 |
| Carbon disulfid | 75-15-0 | 32.30 | -0.1484 |
| Quinoline | 91-22-5 | 43.12 | -0.1063 |
| Chloro benzene | 108-90-7 | 33.60 | -0.1191 |
| Chloroform | 67-66-3 | 27.50 | -0.1295 |
| Cyclohexane | 110-82-7 | 24.95 | -0.1211 |
| Cyclohexanol 25 °C | 108-93-0 | 34.40 | -0.0966 |
| Cyclopentanol | 96-41-3 | 32.70 | -0.1011 |
| p-Cymene | 99-87-6 | 28.10 | -0.0941 |
| Decalin | 493-01-6 | 31.50 | -0.1031 |
| Dichloromethane | 75-09-2 | 26.50 | -0.1284 |
| Diiodomethane (DI) | 75-11-6 | 50.80 | -0.1376 |
| 1,3-Diiodopropane 23 °C | 627-31-6 | 46.51 | -0.1195 |
| Diethylene glycol (DEG) | 111-46-6 | 44.80 | -0.0841 |
| Dipropylene glycol | 25265-71-8 | 33.90 | -0.1070 |
| Dipropylene glycol monomethylether | 34590-94-8 | 28.41 | -0.1088 |
| Dodecyl benzene | 123-01-3 | 30.70 | |
| Ethanol | 64-17-5 | 22.10 | -0.0832 |
| Ethylbenzene | 100-41-4 | 29.20 | -0.1094 |
| Ethylbromide | 74-96-4 | 24.20 | -0.1159 |
| Ethylene glycol (EG) | 107-21-1 | 47.70 | -0.0890 |
| Formamide (FA) | 75-12-7 | 58.20 | -0.0842 |
| Fumaric acid diethylester 22°C | 623-91-6 | 31.40 | -0.1039 |
| Furfural (2-Furaldehyde) | 98-01-1 | 41.90 | -0.1225 |
| Glycerol (GLY) | 56-81-5 | 64.00 | -0.0598 |
| Ethylene glycol monoethyl ether (Ethyl Cellosolve) | 110-80-5 | 28.60 | -0.0918 |
| Hexachlorobutadiene | 87-68-3 | 36.00 | -0.0994 |
| Iodobenzene | 591-50-4 | 39.70 | -0.1123 |
| Isoamylchloride | 107-84-6 | 23.50 | -0.1078 |
| Isobutylchloride | 513-36-0 | 21.90 | -0.1144 |
| Isopropanol | 67-63-0 | 23.00 | -0.0789 |
| Isopropylbenzene | 98-82-8 | 28.21 | -0.1054 |
| Isovaleronitrile | 625-28-5 | 26.00 | -0.0827 |
| m-Nitrotoluene | 99-08-1 | 41.40 | -0.1140 |
| Mercury | 7439-97-6 | 425.41 | -0.2049 |
| Methanol | 67-56-1 | 22.70 | -0.0773 |
| Methyl ethyl ketone (MEK) | 78-93-3 | 24.60 | -0.1199 |
| Methyl naphthalene | 90-12-0 | 38.60 | -0.1118 |
| N,N-dimethyl acetamide (DMA) | 127-19-5 | 36.70 | -0.1395 |
| N,N-dimethyl formamide (DMF) | 68-12-2 | 37.10 | -0.1400 |
| N-methyl-2-pyrrolidone | 872-50-4 | 40.79 | -0.1156 |
| n-Decane (DEC) | 124-18-5 | 23.83 | -0.0920 |
| n-Dodecane (DDEC) | 112-40-3 | 25.35 | -0.0884 |
| n-Heptane | 142-82-5 | 20.14 | -0.0980 |
| n-Hexadecane (HDEC) | 544-76-3 | 27.47 | -0.0854 |
| n-Hexane (HEX) | 110-54-3 | 18.43 | -0.1022 |
| n-Octane (OCT) | 111-65-9 | 21.62 | -0.0951 |
| n-Tetradecane (TDEC) | 629-59-4 | 26.56 | -0.0869 |
| n-Undecane | 1120-21-4 | 24.66 | -0.0901 |
| n-Butylbenzene | 104-51-8 | 29.23 | -0.1082 |
| n-Propylbenzene | 103-65-1 | 28.99 | -0.1071 |
| Nitroethane | 79-24-3 | 31.90 | -0.1255 |
| Nitrobenzene | 98-95-3 | 43.90 | -0.1177 |
| Nitromethane | 75-52-5 | 36.80 | -0.1678 |
| o-Nitrotoluene | 88-72-2 | 41.50 | -0.1229 |
| Perfluoroheptane | 335-57-9 | 12.85 | -0.0972 |
| Perfluorohexane | 355-42-0 | 11.91 | -0.0935 |
| Perfluorooctane | 307-34-6 | 14.00 | -0.0902 |
| Phenylisothiocyanate | 103-72-0 | 41.50 | -0.1172 |
| Phthalic acid diethylester 22°C | 84-66-2 | 37.00 | -0.1018 |
| Polyethylen glycol 200 (PEG) | 25322-68-3 | 43.50 | -0.1170 |
| Polydimethyl siloxane (Baysilone M5) | 9016-00-6 | 19.00 | -0.0365 |
| Propanol 25 °C | 71-23-8 | 23.70 | -0.0777 |
| Pyridine | 110-86-1 | 38.00 | -0.1372 |
| 3-Pyridylcarbinol 23°C (PYC) | 100-55-0 | 47.68 | -0.1259 |
| Pyrrol (PY) | 109-97-7 | 36.60 | -0.1100 |
| sym-Tetrabromoethane | 79-27-6 | 49.70 | -0.1528 |
| tert-Butylchloride | 507-20-0 | 19.60 | -0.1072 |
| sym-Tetrachloromethane | 56-23-5 | 26.95 | -0.1224 |
| Tetrahydrofuran (THF) | 109-99-9 | 26.40 | -0.1277 |
| Thiodiglycol ( 2,2'-Thiobisethanol) (TDG) | 111-48-8 | 54.00 | -0.0830 |
| Toluene | 108-88-3 | 28.40 | -0.1189 |
| Tricresylphosphate (TCP) | 1330-78-5 | 40.90 | -0.0887 |
| Water (WA) | 7732-18-5 | 72.80 | -0.1514 |
| o-Xylene | 95-47-6 | 30.10 | -0.1101 |
| m-Xylene | 108-38-3 | 28.90 | -0.1104 |
| a-Bromonaphthalene (BN) | 90-11-9 | 44.40 | -0.0979 |
| a-Chloronaphthalene | 90-13-1 | 41.80 | -0.1064 |
For surface tension components (e.g. dispersive and
polar, hydrogen bonding, acid-base contributions etc.),
corresponding (single) references
and data at elevated temperatures up to +180 °C, and
surfactant property
prediction
(CMC, surface tension@CMC), please send an email to the address below.
For any further questions regarding this table please
don't hesitate to contact us at: quest[at]surface-tension.de
See also table of
solid surface energy data (SFE) for common polymers
Last updated: 25 Feb 2017 - www.surface-tension.de |
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Rounding Numbers 4.NBT.A.3 Question & Answer Key Resources Lumos StepUp: TNReady Online Practice and Assessments - Grade 4 Mathematics
Lumos StepUp: TNReady Online Practice and Assessments - Grade 4 Mathematics Rounding Numbers
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John says 73,495 rounded to the nearest ten is 73,500. Jose says 73,495 rounded to the nearest hundred is 73,500. Who is correct? Choose the correct answer from the drop down menu.
Resource: Question
Question Type: Drop-Down - Equation Completion
Standard(s):4.NBT.A.3
Standard Description: Use place value understanding to round multi-digit whole numbers to any place. (Grade 4 expectations in this domain are limited to whole numbers less than or equal to 1,000,000.)
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Triangle Geometric Centers: Bisectors, Medians & Incenters
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\Perpendicular Bisector (example) Definition in my own words: A line that cuts a side of the triangle in half while making a right angle with the side. Geometric Properties: Every triangle has three perpendicular bisectors, the intersection of the perpendicular bisectors is called the circumcenter. It is the point that is equidistant from all 3 vertices. Real-life example explanation: Three friends are located at points A,B,and C. By drawing a triangle and using perpendicular bisectors they can find the point where they should meet so that all are traveling an equal distance. Real life example picture ______________________
MEDIANS OF TRIANGLE
Definition in my own words
:A median of a triangle is a line segment joining a vertex of the triangle to the midpoint of the opposite side. Every triangle has three medians, one from each vertex. The medians intersect at a point called the centroid, which is the center of mass of the triangle.
Geometric Properties: Intersection Point (Centroid): The medians of a triangle intersect at a single point called the centroid. The centroid divides each median into a ratio of 2:1, where the longer segment is between the vertex and the centroid. Balance Point: The centroid is also the center of mass of the triangle, meaning it is the balance point of the triangle. Centroid Coordinates: If the vertices of the triangle are at coordinates ( ?
1 , ?
1 ) (x 1 ,y 1 ), ( ?
2 , ?
2 ) (x 2 ,y 2 ), and ( ?
3 , ?
3 ) (x 3 ,y 3 ), the centroid's coordinates can be found using the formula
Real-life example explanation: Imagine three friends, Alice, Bob, and Charlie, each living at different points A, B, and C, forming a triangle. They decide to order a pizza and want to find a central meeting point for delivery, such that the pizza shop is equidistant from all three houses.
Real life example picture _____
_________________
Angle Bistector of Triangles
Definition in My Own Words: An angle bisector of a triangle is a line that divides one of the triangle's angles into two equal smaller angles. Each triangle has three angle bisectors, one from each vertex. These bisectors intersect at a single point called the incenter, which is the center of the triangle's inscribed circle (incircle).
Geometric Properties: Intersection Point (Incenter): The angle bisectors of a triangle intersect at a point called the incenter. Equidistance: The incenter is equidistant from all three sides of the triangle. Incircle: The incenter is the center of the largest circle that can fit inside the triangle, known as the incircle.
Imagine three friends, Alice, Bob, and Charlie, who live at points A, B, and C, forming a triangle. They want to build a circular pizza oven that is equidistant from all three sides of the triangular plot of land they own.
1.
Draw the Triangle:
Connect the points A, B, and C to form a triangle.
2.
Construct Angle Bisectors:
For each vertex of the triangle, draw a line that divides the angle into two equal parts.
3.
Find the Incenter:
The point where all three angle bisectors intersect is the incenter. This is the point equidistant from all three sides of the triangle.
When they place the pizza oven at the incenter, it will be equally accessible from each side of their triangular plot, ensuring that the oven is centrally located.
Medians of Triangle
Definition in My Own Words: A median of a triangle is a line segment that joins a vertex of the triangle to the midpoint of the opposite side. Each triangle has three medians, one from each vertex. These medians intersect at a single point called the centroid, which is the center of mass of the triangle.
Geometric Properties:
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●
Intersection Point (Centroid):
The medians of a triangle intersect at a point called the centroid.
●
Balance Point:
The centroid divides each median into a ratio of 2:1, with the longer segment being between the vertex and the centroid.
●
Centroid Coordinates:
If the vertices of the triangle are at coordinates (x1,y1)(x_1, y_1)(x1,y1), (x2,y2)(x_2, y_2)(x2,y2), and (x3,y3)(x_3, y_3)(x3,y3), the centroid's coordinates can be found using the formula
Real-Life Example Explanation:
Imagine three friends, Alice, Bob, and Charlie, who want to order a pizza to be delivered to a central location that is the center of mass of their triangular area. They live at points A, B, and C, forming a triangle.
1.
Draw the Triangle:
Connect the points A, B, and C to form a triangle.
2.
Construct the Medians:
For each vertex of the triangle, draw a line segment to the midpoint of the opposite side.
3.
Find the Centroid:
The point where all three medians intersect is the centroid. This point acts as the center of mass for the triangular area.
When they choose the centroid as the delivery location, the pizza delivery person will reach a balanced central point, making it equally convenient for all three friends to get their pizza.
Real-Life Example Picture:
Imagine a triangle formed by the locations of three friends' houses and the pizza delivery point being at the centroid.
In this diagram, G represents the centroid, the ideal location for the pizza delivery, ensuring a central, balanced point for Alice, Bob, and Charlie. By placing the pizza delivery point at the centroid, the friends ensure that the pizza is delivered to a central location that is equidistant from all sides of their triangular area. You can visualize this scenario by drawing it on a map or using a drawing tool to see how the medians intersect at the centroid.
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13961 | https://www.chegg.com/homework-help/questions-and-answers/reduction-reactions-oxygen-fad-1-2-o-2-2e-2h-h-2-o-e-083v-fad-2e-2h-fadh-2-e-022v-potentia-q168777765 | Solved Below are the reduction reactions for oxygen and | Chegg.com
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Below are the reduction reactions for oxygen and FAD.12O2+2e-+2H+→H2O,E@'=0.83VFAD+2e-+2H+→FADH2,E@'=-0.22Va) What is the potential (E°) for the oxidation of FADH2 by oxygen?b) What is the ΔG° for the oxidation of FADH2 by oxygen?c) If we assume that the pumping of protons in conjunction with the oxidation of FADH2 requires 120kJ, what percentage of
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Question: Below are the reduction reactions for oxygen and FAD.12O2+2e-+2H+→H2O,E@'=0.83VFAD+2e-+2H+→FADH2,E@'=-0.22Va) What is the potential (E°) for the oxidation of FADH2 by oxygen?b) What is the ΔG° for the oxidation of FADH2 by oxygen?c) If we assume that the pumping of protons in conjunction with the oxidation of FADH2 requires 120kJ, what percentage of
Below are the reduction reactions for oxygen and FAD. 1 2 O 2+2 e-+2 H+→H 2 O,E@'=0.8 3 V FAD+2 e-+2 H+→F A D H 2,E@'=-0.2 2 V a) What is the potential (E°) for the oxidation of F A D H 2 by oxygen? b) What is the Δ G° for the oxidation of F A D H 2 by oxygen? c) If we assume that the pumping of protons in conjunction with the oxidation of F A D H 2 requires 1 2 0 k J, what percentage of the energy from F A D H 2 oxidation is stored in the proton gradient (think about how many protons are pumped when F A D H 2 transfers electrons to oxygen)? Please write it out
There are 3 steps to solve this one.Solution Share Share Share done loading Copy link Step 1 a).Calculate the standard cell potential (E∘) for the oxidation ofFADH A 2by oxygen:
Given reduction potentia... View the full answer Step 2 UnlockStep 3 UnlockAnswer Unlock Previous questionNext question
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Geogebra graphing help
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Asked 9 years, 11 months ago
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I'm trying to draw a wave in geogebra, something like the one on a pepsi logo. The wave is a vertical wave and I have no idea how to draw it.
I tried drawing a sine graph but I want a 'single' wave, which is 'vertical'
Any ideas?
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edited Oct 8, 2015 at 4:04
Martin Sleziak
56.3k 20 20 gold badges 211 211 silver badges 391 391 bronze badges
asked Oct 3, 2015 at 4:31
user268664user268664
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I sorted out with making a vertical wave now. Thx! But how do I have a 'single' wave?user268664 –user268664 2015-10-03 06:22:31 +00:00 Commented Oct 3, 2015 at 6:22
Geogebra or algebraic geometry ;)ClassicStyle –ClassicStyle 2015-10-08 04:10:48 +00:00 Commented Oct 8, 2015 at 4:10
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To reduce the “multiple” sine wave to a “single” one, multiply with a boolean value such as x 2<π 2 x 2<π 2, for instance. Just type sin(x) (x^2 < pi^2) into the input bar, and then hit enter.
To transform a horizontal wave into a vertical one, use Rotate[...].
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answered Oct 3, 2015 at 10:59
LucianLucian
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A vertical wave is not a function of x x. You might try x=cos y x=cosy, which is just a normal wave, except vertically.
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answered Oct 3, 2015 at 5:27
davidlowryduda♦davidlowryduda
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Technically true and mathematically correct, but GeoGebra does not accept as input explicit functions of argument other than x, and implicit equations are only accepted if they are polynomial in x and/or y.Lucian –Lucian 2015-10-03 11:05:21 +00:00 Commented Oct 3, 2015 at 11:05
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You can just reflect sin(x)sin(x) in the line y=x y=x thus:
web.geogebra.org/?command=Reflect[sin(x),y=x]
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edited Nov 12, 2015 at 14:14
answered Nov 10, 2015 at 20:07
murklemurkle
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13963 | https://arxiv.org/html/2407.10196v1 | A3S: A General Active Clustering Method with Pairwise Constraints
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Why HTML?Report IssueBack to AbstractDownload PDF
Table of Contents
Abstract
1 Introduction
2 Methodology
2.1 Notation and Definition
2.2 Theoretical Analysis
2.3 Adaptive Active Aggregation and Splitting
2.3.1 Initialization via Adaptive Clustering
2.3.2 Active Aggregation and Splitting
2.3.3 Discussion.
3 Experiments
3.1 Experimental Setup
3.2 Comparison with SOTA Active Clustering Methods
3.3 A3S for Different Clustering Algorithms
3.4 Influence of Estimated Pairwise Probability Quality
3.5 Performance of A3S on Large datasets.
3.6 Case Study for A3S
4 Related Work
5 Conclusion
A Proofs
A.1 Proof of Theorem 2.5
Term (V) in Eq.(12).
Putting Together.
A.2 Derivation of Eq.(2)
A.3 Justification of Approximation
A.4 Fast Transitive Inference
B Implementation Details of A3S
B.1 Estimating Pairwise Probability
B.2 Calculating Aggregation Probability
B.3 Feature Extraction
B.4 Hyperparameter Setting
B.5 Computing Resources
C More Experiment Results
References
License: arXiv.org perpetual non-exclusive license
arXiv:2407.10196v1 [cs.LG] 14 Jul 2024
A3S: A General Active Clustering Method with Pairwise Constraints
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Xun Deng Junlong Liu Han Zhong Fuli Feng Chen Shen Xiangnan He Jieping Ye Zheng Wang
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Abstract
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Active clustering aims to boost the clustering performance by integrating human-annotated pairwise constraints through strategic querying. Conventional approaches with semi-supervised clustering schemes encounter high query costs when applied to large datasets with numerous classes. To address these limitations, we propose a novel A daptive A ctive A ggregation and S plitting (A3S) framework, falling within the cluster-adjustment scheme in active clustering. A3S features strategic active clustering adjustment on the initial cluster result, which is obtained by an adaptive clustering algorithm. In particular, our cluster adjustment is inspired by the quantitative analysis of Normalized mutual information gain under the information theory framework and can provably improve the clustering quality. The proposed A3S framework significantly elevates the performance and scalability of active clustering. In extensive experiments across diverse real-world datasets, A3S achieves desired results with significantly fewer human queries compared with existing methods.
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Machine Learning, ICML
1 Introduction
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In the realm of data science, clustering algorithms have emerged as a cornerstone technology within the domain of unsupervised learning(Khanum et al., 2015; Celebi & Aydin, 2016). By automatically grouping similar data objects based on inherent structures and patterns within datasets, clustering provides an efficacious means to condense and structure complex information and is widely applied in image classification(Caron et al., 2020), social network analysis(He et al., 2022), etc. However, conventional clustering techniques often rely on static parameter settings and one-off computations, rendering them less adaptable to strange or expanding data environments (See Section 2.3.3). This context highlights the growing importance of human-computer collaborative active clustering approaches.
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Active clustering refers to a paradigm that actively selects side information(Anand et al., 2014), in the form of pairwise constraints, to maximally improve the clustering performance. Extensive work(González-Almagro et al., 2023) has explored the combination of the strategic selection of pairwise constraints and semi-supervised clustering (SSC)(Basu et al., 2002), and attains much lower query complexity compared to its semi-supervised counterpart(Bilenko et al., 2004). Despite the practicality, current active clustering methods often suffer from high computational and query costs when the number of classes is large.
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SSC-based active clustering methods primarily assess the uncertainty of all pairwise constraints, and iteratively choose the most uncertain ones for expert queries. This manner, while systematic, faces notable challenges: it potentially relies on the assumption that a small set of initial pairwise constraints will rapidly cover most real classes. This assumption becomes increasingly unreliable in scenarios with large sample numbers N 𝑁 N italic_N and class numbers K 𝐾 K italic_K. Therefore, some active clustering methods(Van Craenendonck et al., 2017; Shi et al., 2020) shift from SSC to a cluster-adjustment scheme. This scheme involves over-clustering data into k 𝑘 k italic_k clusters via a specific clustering method (where k 𝑘 k italic_k is greater than K 𝐾 K italic_K), and subsequently aggregates the resulting small clusters into larger ones based on pairwise constraints. However, it requires a proper cluster number k 𝑘 k italic_k as an input parameter, which is hard to determine in real applications. Moreover, the human query may be misleading when the selected samples are outliers, as they do not represent the majority sample of a cluster.
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This work aims to overcome the drawbacks of existing cluster adjustment schemes. We first present a theoretical result that identifies conditions where aggregating two clusters does not reduce the normalized mutual information (NMI) between the resulting clustering and the real clustering. Here, NMI measures the overlap between two clustering results, with larger values indicating better performance (see Definition2.3). In addition, to guide active human queries, we quantify the impact of merging two clusters on the expected NMI value difference from the theoretical side. This characterization enables us to actively select cluster pairs that maximize the NMI gain.
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Building upon these theoretical results, we propose Adaptive Active Aggregation and Splitting (A3S), a generic cluster adjustment framework for active clustering. A3S operates in two stages: the adaptive initialization stage and the active aggregation and splitting stage. The first stage autonomously identifies an appropriate cluster number and produces initial clustering results using a designated clustering algorithm (e.g., K-means and hierarchical clustering). In the latter stage, A3S proactively identifies the cluster pair that is expected to enhance the NMI value mostly. Afterward, it assesses whether the majority of samples in a cluster are of the same class (i.e., cluster purity in Definition2.4), then merges pure clusters queried to be in the same class by oracles, and divides impure clusters into pure subclusters and outliers. This stage will be repeated a few times to ensure convergence, where no more aggregation and splitting will happen.
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Our contributions are summarized as follows:
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•We introduce A3S, a general active clustering algorithm that improves the quality of clustering by optimizing the NMI value. A3S strategically selects cluster pairs for aggregation, aiming to maximize the expected NMI improvement within a limited query cost. The quantification of NMI gain is guided by our information-theoretical analysis (Theorem 2.4). Additionally, A3S implements precise splitting on clusters that fail purity tests, ensuring the correction of outlier samples.
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•Regarding implementation superiority, A3S can reveal the underlying ground truth clustering structure with substantially fewer human queries compared to traditional methods, especially when prior dataset knowledge is unknown. In addition, A3S locally adjusts the cluster labels, offering a more efficient alternative to rerunning semi-supervised clustering algorithms, and remains feasible for large datasets.
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•In terms of practical performance, by conducting comprehensive evaluations on various real-world datasets, we demonstrate that A3S consistently surpasses all baseline models. Notably, baseline methods typically require more than 4000 queries to achieve high-quality clustering results for datasets containing thousands of samples. In contrast, A3S achieves this performance with only a few hundred queries.
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2 Methodology
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Figure 1: The workflow of A3S which consists of the adaptive clustering stage and active aggregation and splitting stage.
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2.1 Notation and Definition
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Definition 2.1(Active Clustering).
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We denote the true classes of N 𝑁 N italic_N samples X={x 1,⋯,x N}𝑋 subscript 𝑥 1⋯subscript 𝑥 𝑁 X={x_{1},\cdots,x_{N}}italic_X = { italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , ⋯ , italic_x start_POSTSUBSCRIPT italic_N end_POSTSUBSCRIPT } by Y={y 1,⋯,y N}𝑌 subscript 𝑦 1⋯subscript 𝑦 𝑁 Y={y_{1},\cdots,y_{N}}italic_Y = { italic_y start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , ⋯ , italic_y start_POSTSUBSCRIPT italic_N end_POSTSUBSCRIPT }, where y i∈{1,⋯,K}subscript 𝑦 𝑖 1⋯𝐾 y_{i}\in{1,\cdots,K}italic_y start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ∈ { 1 , ⋯ , italic_K } and K 𝐾 K italic_K is the number of classes. Let the ground truth clustering of X 𝑋 X italic_X as C={c 1,⋯,c K}𝐶 subscript 𝑐 1⋯subscript 𝑐 𝐾 C={c_{1},\cdots,c_{K}}italic_C = { italic_c start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , ⋯ , italic_c start_POSTSUBSCRIPT italic_K end_POSTSUBSCRIPT }, and the initial clustering result as Ω={w 1,⋯,w k}Ω subscript 𝑤 1⋯subscript 𝑤 𝑘\Omega={w_{1},\cdots,w_{k}}roman_Ω = { italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , ⋯ , italic_w start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT }, where X=∪i=1 K c i=∪i=1 k w i 𝑋 superscript subscript 𝑖 1 𝐾 subscript 𝑐 𝑖 superscript subscript 𝑖 1 𝑘 subscript 𝑤 𝑖 X=\cup_{i=1}^{K}c_{i}=\cup_{i=1}^{k}w_{i}italic_X = ∪ start_POSTSUBSCRIPT italic_i = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_K end_POSTSUPERSCRIPT italic_c start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT = ∪ start_POSTSUBSCRIPT italic_i = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_k end_POSTSUPERSCRIPT italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT and c i∩c j=∅,w i∩w j=∅,∀i,j formulae-sequence subscript 𝑐 𝑖 subscript 𝑐 𝑗 subscript 𝑤 𝑖 subscript 𝑤 𝑗 for-all 𝑖 𝑗 c_{i}\cap c_{j}=\emptyset,w_{i}\cap w_{j}=\emptyset,\forall i,j italic_c start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = ∅ , italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ∩ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = ∅ , ∀ italic_i , italic_j. Active Clustering strategically selects sample pairs (x i,x j)subscript 𝑥 𝑖 subscript 𝑥 𝑗(x_{i},x_{j})( italic_x start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ), and requires the oracle to judge if y i=y j subscript 𝑦 𝑖 subscript 𝑦 𝑗 y_{i}=y_{j}italic_y start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT = italic_y start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT (two samples are must-linked) or y i≠y j subscript 𝑦 𝑖 subscript 𝑦 𝑗 y_{i}\neq y_{j}italic_y start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ≠ italic_y start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT (two samples are cannot-linked). It then updates the clustering result Ω Ω\Omega roman_Ω with the queried pairwise constraints accordingly. Active Clustering aims to utilize the queried constraints to maximally reduce the difference between C 𝐶 C italic_C and Ω Ω\Omega roman_Ω, which is measured by the normalized mutual information (NMI)(Kvalseth, 1987; Vinh et al., 2009).
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Definition 2.2(Cluster Adjustment Scheme).
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We define a cluster adjustment scheme as a label update strategy employed by active clustering algorithms. Specifically, it entails the algorithm’s process of either locally aggregating small clusters into larger ones or splitting impure clusters into several subclusters, guided by pairwise constraints.
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Definition 2.3(NMI).
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NMI is a measure of how much common information two clustering results share. Given N 𝑁 N italic_N samples and their two clusterings Ω={w 1,⋯,w k}Ω subscript 𝑤 1⋯subscript 𝑤 𝑘\Omega={w_{1},\cdots,w_{k}}roman_Ω = { italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , ⋯ , italic_w start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT } and Ω′={w 1′,⋯,w k′}superscript Ω′subscript superscript 𝑤′1⋯subscript superscript 𝑤′𝑘\Omega^{\prime}={w^{\prime}{1},\cdots,w^{\prime}{k}}roman_Ω start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = { italic_w start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , ⋯ , italic_w start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT }, we define the NMI value as
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n=2𝕀(Ω;Ω′)ℍ(Ω)+ℍ(Ω′)=2𝕀(ζ;ζ′)ℍ(ζ)+ℍ(ζ′),𝑛 2 𝕀 Ω superscript Ω′ℍ Ω ℍ superscript Ω′2 𝕀 𝜁 superscript 𝜁′ℍ 𝜁 ℍ superscript 𝜁′\displaystyle n=\frac{2\mathbb{I}(\Omega;\Omega^{\prime})}{\mathbb{H}(\Omega)+% \mathbb{H}(\Omega^{\prime})}=\frac{2\mathbb{I}(\zeta;\zeta^{\prime})}{\mathbb{% H}(\zeta)+\mathbb{H}(\zeta^{\prime})},italic_n = divide start_ARG 2 blackboard_I ( roman_Ω ; roman_Ω start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) end_ARG start_ARG blackboard_H ( roman_Ω ) + blackboard_H ( roman_Ω start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) end_ARG = divide start_ARG 2 blackboard_I ( italic_ζ ; italic_ζ start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) end_ARG start_ARG blackboard_H ( italic_ζ ) + blackboard_H ( italic_ζ start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT ) end_ARG ,
where ζ=(|w 1|/N,⋯,|w k|/N)𝜁 subscript 𝑤 1 𝑁⋯subscript 𝑤 𝑘 𝑁\zeta=(|w_{1}|/N,\cdots,|w_{k}|/N)italic_ζ = ( | italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | / italic_N , ⋯ , | italic_w start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT | / italic_N ) and ζ′=(|w 1′|/N,⋯,|w k′′|/N)superscript 𝜁′subscript superscript 𝑤′1 𝑁⋯subscript superscript 𝑤′superscript 𝑘′𝑁\zeta^{\prime}=(|w^{\prime}{1}|/N,\cdots,|w^{\prime}{k^{\prime}}|/N)italic_ζ start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = ( | italic_w start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | / italic_N , ⋯ , | italic_w start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_k start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT end_POSTSUBSCRIPT | / italic_N ) are two distributions induced by Ω Ω\Omega roman_Ω and Ω′superscript Ω′\Omega^{\prime}roman_Ω start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT, respectively. Here, 𝕀(⋅;⋅)𝕀⋅⋅\mathbb{I}(\cdot;\cdot)blackboard_I ( ⋅ ; ⋅ ) denotes mutual information, and ℍ(⋅)ℍ⋅\mathbb{H}(\cdot)blackboard_H ( ⋅ ) denotes entropy.
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We also introduce the notion of clustering purity(González-Almagro et al., 2023).
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Definition 2.4(Purity).
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We define the dominant class of a cluster w i subscript 𝑤 𝑖 w_{i}italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT as argmax j|w i∩c j|subscript 𝑗 subscript 𝑤 𝑖 subscript 𝑐 𝑗\arg\max_{j}|w_{i}\cap c_{j}|roman_arg roman_max start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT |. We label the sample in a cluster that does not belong to its dominant class or the sample that is a single cluster itself as an outlier. Then, the purity of w i subscript 𝑤 𝑖 w_{i}italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT is max j|w i∩c j||w i|subscript 𝑗 subscript 𝑤 𝑖 subscript 𝑐 𝑗 subscript 𝑤 𝑖\max_{j}\frac{|w_{i}\cap c_{j}|}{|w_{i}|}roman_max start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT divide start_ARG | italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | end_ARG start_ARG | italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT | end_ARG, and the purity of the clustering result Ω Ω\Omega roman_Ω is quantified by ∑i max j|w i∩c j|N subscript 𝑖 subscript 𝑗 subscript 𝑤 𝑖 subscript 𝑐 𝑗 𝑁\frac{\sum_{i}\max_{j}|w_{i}\cap c_{j}|}{N}divide start_ARG ∑ start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT roman_max start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | end_ARG start_ARG italic_N end_ARG.
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2.2 Theoretical Analysis
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Active clustering that adopts the cluster-adjustment scheme does not rely on semi-supervised clustering for updating clustering results. Instead, it emphasizes the use of must-link and cannot-link constraints as indicators to merge or split clusters(Van Craenendonck et al., 2018). This greedy strategy may, however, jeopardize the quality of clustering outcomes for several reasons. For example, the queried samples could be outliers, or the purity of a cluster might not be sufficiently high. In light of this, we introduce a pivotal theorem that offers clear guidance for aggregation actions. This is achieved through an evaluation of the NMI.
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Theorem 2.5(Guarantee for Cluster Aggregation).
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Denote the clustering of N 𝑁 N italic_N samples as Ω Ω\Omega roman_Ω, the ground truth clustering as C 𝐶 C italic_C, and the NMI value of Ω Ω\Omega roman_Ω with respect to C 𝐶 C italic_C as n 1 subscript 𝑛 1 n_{1}italic_n start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT. For any two clusters in Ω Ω\Omega roman_Ω, say w 1 subscript 𝑤 1 w_{1}italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT and w 2 subscript 𝑤 2 w_{2}italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT, suppose they have a common dominant class c 1 subscript 𝑐 1 c_{1}italic_c start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT with purities t 1 subscript 𝑡 1 t_{1}italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT and t 2 subscript 𝑡 2 t_{2}italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT respectively, where t 1,t 2∈[p,1]subscript 𝑡 1 subscript 𝑡 2 𝑝 1 t_{1},t_{2}\in[p,1]italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∈ [ italic_p , 1 ]. By aggregating w 1 subscript 𝑤 1 w_{1}italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT and w 2 subscript 𝑤 2 w_{2}italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT into a new cluster w 1,2=w 1∪w 2 subscript 𝑤 1 2 subscript 𝑤 1 subscript 𝑤 2 w_{1,2}=w_{1}\cup w_{2}italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT = italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∪ italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT, we arrive at a new clustering Ω⋆={w 1,2,w 3,⋯,w k}superscript Ω⋆subscript 𝑤 1 2 subscript 𝑤 3⋯subscript 𝑤 𝑘\Omega^{\star}={w_{1,2},w_{3},\cdots,w_{k}}roman_Ω start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT = { italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT , ⋯ , italic_w start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT } with NMI value of n 2 subscript 𝑛 2 n_{2}italic_n start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT. This aggregation positively impacts clustering performance (i.e., n 2≥n 1 subscript 𝑛 2 subscript 𝑛 1 n_{2}\geq n_{1}italic_n start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ≥ italic_n start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT) if p≥0.7 𝑝 0.7 p\geq 0.7 italic_p ≥ 0.7 and n 1≥2⋅(1.0586−min{t 1,t 2})subscript 𝑛 1⋅2 1.0586 subscript 𝑡 1 subscript 𝑡 2 n_{1}\geq 2\cdot(1.0586-\min{t_{1},t_{2}})italic_n start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≥ 2 ⋅ ( 1.0586 - roman_min { italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } ).
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Theorem2.5 delineates the conditions for non-deteriorating cluster aggregation, and the detailed proof is in AppendixA.1. Specifically, merging two clusters can achieve provable benefits when their purity is at least 0.7 0.7 0.7 0.7 and preceding NMI exceeds 2⋅(1.0586−min{t 1,t 2})⋅2 1.0586 subscript 𝑡 1 subscript 𝑡 2 2\cdot(1.0586-\min{t_{1},t_{2}})2 ⋅ ( 1.0586 - roman_min { italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } ). This finding is important as it suggests that the purity requirement for cluster aggregation is relatively mild, allowing for the inclusion of a small number of outliers within each cluster without compromising the overall clustering performance.
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We take a step further by formulating the expected gain in NMI value when we decide to query a sample pair from a cluster pair. This involves aggregating clusters if the query result is “must-link” and keeping them separated if the result is “cannot-link”.
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Definition 2.6(Expected NMI Gain).
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Suppose the dominant class of clusters w i subscript 𝑤 𝑖 w_{i}italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT and w j subscript 𝑤 𝑗 w_{j}italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT is c m subscript 𝑐 𝑚 c_{m}italic_c start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT and c n subscript 𝑐 𝑛 c_{n}italic_c start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT respectively, which remain unknown before querying. Let n 1 subscript 𝑛 1 n_{1}italic_n start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT and n 2 subscript 𝑛 2 n_{2}italic_n start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT represent the NMI values of the clustering result before and after aggregating the two clusters. We denote ℙ(c m=c n)ℙ subscript 𝑐 𝑚 subscript 𝑐 𝑛\mathbb{P}(c_{m}=c_{n})blackboard_P ( italic_c start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT = italic_c start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) as the probability that the oracle observes a ‘must-link’ result. Then, we define the expected NMI gain from this query as follows:
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𝔼[ΔNMI∣w i,w j]=ℙ(c m=c n)⋅(n 2−n 1),𝔼 delimited-[]conditional Δ NMI subscript 𝑤 𝑖 subscript 𝑤 𝑗⋅ℙ subscript 𝑐 𝑚 subscript 𝑐 𝑛 subscript 𝑛 2 subscript 𝑛 1\displaystyle\mathbb{E}[\Delta\texttt{NMI}\mid w_{i},w_{j}]=\mathbb{P}(c_{m}=c% {n})\cdot(n{2}-n_{1}),blackboard_E [ roman_Δ NMI ∣ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ] = blackboard_P ( italic_c start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT = italic_c start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) ⋅ ( italic_n start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT - italic_n start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) ,(1)
In what follows, we present how to estimate ℙ(c m=c n)ℙ subscript 𝑐 𝑚 subscript 𝑐 𝑛\mathbb{P}(c_{m}=c_{n})blackboard_P ( italic_c start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT = italic_c start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) and (n 2−n 1)subscript 𝑛 2 subscript 𝑛 1(n_{2}-n_{1})( italic_n start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT - italic_n start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ).
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We use e st=1/0 subscript 𝑒 𝑠 𝑡 1 0 e_{st}=1/0 italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 / 0 to signify if y s subscript 𝑦 𝑠 y_{s}italic_y start_POSTSUBSCRIPT italic_s end_POSTSUBSCRIPT equals y t subscript 𝑦 𝑡 y_{t}italic_y start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT or not, and denote the posterior pairwise probability as ℙ(e st=1)ℙ subscript 𝑒 𝑠 𝑡 1\mathbb{P}(e_{st}=1)blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ). We estimate the pairwise probability following previous probability clustering methods(e.g., Liu et al., 2022), and discuss the estimation details in AppendixB.1. Then we can express the aggregation probability ℙ(c m=c n)ℙ subscript 𝑐 𝑚 subscript 𝑐 𝑛\mathbb{P}(c_{m}=c_{n})blackboard_P ( italic_c start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT = italic_c start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) as follows:
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ℙ(c m=c n)=∏s∈w i,t∈w j ℙ(e st=1)∏s∈w i,t∈w j ℙ(e st=1)+∏s∈w i,t∈w j ℙ(e st=0),ℙ subscript 𝑐 𝑚 subscript 𝑐 𝑛 subscript product formulae-sequence 𝑠 subscript 𝑤 𝑖 𝑡 subscript 𝑤 𝑗 ℙ subscript 𝑒 𝑠 𝑡 1 subscript product formulae-sequence 𝑠 subscript 𝑤 𝑖 𝑡 subscript 𝑤 𝑗 ℙ subscript 𝑒 𝑠 𝑡 1 subscript product formulae-sequence 𝑠 subscript 𝑤 𝑖 𝑡 subscript 𝑤 𝑗 ℙ subscript 𝑒 𝑠 𝑡 0\mathbb{P}(c_{m}=c_{n})=\frac{\prod\limits_{s\in w_{i},t\in w_{j}}\mathbb{P}(e% {st}=1)}{\prod\limits{s\in w_{i},t\in w_{j}}\mathbb{P}(e_{st}=1)+\prod% \limits_{s\in w_{i},t\in w_{j}}\mathbb{P}(e_{st}=0)},blackboard_P ( italic_c start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT = italic_c start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) = divide start_ARG ∏ start_POSTSUBSCRIPT italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) end_ARG start_ARG ∏ start_POSTSUBSCRIPT italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) + ∏ start_POSTSUBSCRIPT italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 0 ) end_ARG ,(2)
The detailed derivation of Eq. (2) is in AppendixA.2.
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Moving forward, we focus on formulating n 2−n 1 subscript 𝑛 2 subscript 𝑛 1 n_{2}-n_{1}italic_n start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT - italic_n start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT. In line with the notations used in Theorem2.5, we define Δh=ℍ(Ω)−ℍ(Ω∗)Δ ℎ ℍ Ω ℍ superscript Ω\Delta h=\mathbb{H}(\Omega)-\mathbb{H}(\Omega^{})roman_Δ italic_h = blackboard_H ( roman_Ω ) - blackboard_H ( roman_Ω start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ) and proceed with the following result:
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n 2=2𝕀(Ω⋆;C)ℍ(Ω⋆)+ℍ(C)≈2𝕀(Ω;C)ℍ(Ω)+ℍ(C)−Δh,subscript 𝑛 2 2 𝕀 superscript Ω⋆𝐶 ℍ superscript Ω⋆ℍ 𝐶 2 𝕀 Ω 𝐶 ℍ Ω ℍ 𝐶 Δ ℎ\displaystyle n_{2}=\frac{2\mathbb{I}(\Omega^{\star};C)}{\mathbb{H}(\Omega^{% \star})+\mathbb{H}(C)}\approx\frac{2\mathbb{I}(\Omega;C)}{\mathbb{H}(\Omega)+% \mathbb{H}(C)-\Delta h},italic_n start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = divide start_ARG 2 blackboard_I ( roman_Ω start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT ; italic_C ) end_ARG start_ARG blackboard_H ( roman_Ω start_POSTSUPERSCRIPT ⋆ end_POSTSUPERSCRIPT ) + blackboard_H ( italic_C ) end_ARG ≈ divide start_ARG 2 blackboard_I ( roman_Ω ; italic_C ) end_ARG start_ARG blackboard_H ( roman_Ω ) + blackboard_H ( italic_C ) - roman_Δ italic_h end_ARG ,
where we use the fact that 𝕀(Ω∗;C)≈𝕀(Ω,C)𝕀 superscript Ω 𝐶 𝕀 Ω 𝐶\mathbb{I}(\Omega^{};C)\approx\mathbb{I}(\Omega,C)blackboard_I ( roman_Ω start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ; italic_C ) ≈ blackboard_I ( roman_Ω , italic_C ) when the purity of w i subscript 𝑤 𝑖 w_{i}italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT and w j subscript 𝑤 𝑗 w_{j}italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT is sufficiently large. Moreover, when the sizes of clusters w i subscript 𝑤 𝑖 w_{i}italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT and w j subscript 𝑤 𝑗 w_{j}italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT are significantly smaller than the sample size N 𝑁 N italic_N, the direct calculation gives that Δh≪ℍ(Ω)+ℍ(C)much-less-than Δ ℎ ℍ Ω ℍ 𝐶\Delta h\ll\mathbb{H}(\Omega)+\mathbb{H}(C)roman_Δ italic_h ≪ blackboard_H ( roman_Ω ) + blackboard_H ( italic_C ) (refer to AppendixA.3 for verification). Hence, we have
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n 2−n 1≈2𝕀(Ω;C)Δh(ℍ(Ω)+ℍ(C))2∝Δh,subscript 𝑛 2 subscript 𝑛 1 2 𝕀 Ω 𝐶 Δ ℎ superscript ℍ Ω ℍ 𝐶 2 proportional-to Δ ℎ\displaystyle n_{2}-n_{1}\approx\frac{2\mathbb{I}(\Omega;C)\Delta h}{(\mathbb{% H}(\Omega)+\mathbb{H}(C))^{2}}\propto\Delta h,italic_n start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT - italic_n start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≈ divide start_ARG 2 blackboard_I ( roman_Ω ; italic_C ) roman_Δ italic_h end_ARG start_ARG ( blackboard_H ( roman_Ω ) + blackboard_H ( italic_C ) ) start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT end_ARG ∝ roman_Δ italic_h ,(3)
Combing Eq. (1), Eq. (2), and Eq. (3), we obtain our estimators of 𝔼[ΔNMI∣w i,w j]𝔼 delimited-[]conditional Δ NMI subscript 𝑤 𝑖 subscript 𝑤 𝑗\mathbb{E}[\Delta\texttt{NMI}\mid w_{i},w_{j}]blackboard_E [ roman_Δ NMI ∣ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ] as follows:
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𝔼[ΔNMI∣w i,w j]∝RHS of Eq.(2)⋅Δh.proportional-to 𝔼 delimited-[]conditional Δ NMI subscript 𝑤 𝑖 subscript 𝑤 𝑗⋅RHS of Eq.(2)Δ ℎ\mathbb{E}[\Delta\texttt{NMI}\mid w_{i},w_{j}]\propto\text{RHS of Eq.~{}\eqref% {prob:ori}}\cdot\Delta h.blackboard_E [ roman_Δ NMI ∣ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ] ∝ RHS of Eq. ( ) ⋅ roman_Δ italic_h .(4)
where RHS denotes the right-hand side.
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2.3 Adaptive Active Aggregation and Splitting
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Building on the analysis of cluster aggregation and expected query impact, we detail our Adaptive Active Aggregation and Splitting framework, which comprises two distinct stages. First, the Adaptive Clustering stage is introduced in Section2.3.1, where we describe the generation of initial clustering results. Second, in Section2.3.2, we discuss selecting pairwise constraints and updating clustering during the Active Aggregation and Splitting stage. The A3S workflow is illustrated in Figure1, and the steps are summarized in Algorithm1.
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Figure 2: Case study for A3S. In iteration 1, the cluster w 6 subscript 𝑤 6 w_{6}italic_w start_POSTSUBSCRIPT 6 end_POSTSUBSCRIPT does not pass the purity test and the oracle invests 12 queries to split it into two pure subclusters. In iteration 2, the query result for the two central samples is must-link and they are merged into one cluster.
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2.3.1 Initialization via Adaptive Clustering
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Current active clustering methods often necessitate manually setting initial cluster numbers, a challenging task when the number of classes is unknown. In response, we propose using adaptive clustering methods to determine an appropriate cluster number for initialization. An adaptive clustering method organizes data based on local density, revealing the dataset’s inherent structure. It handles noise by isolating each noisy sample into a separate cluster (i.e., outlier), avoiding unsuitable data grouping. This approach ensures a more natural and purer clustering outcome. Classic adaptive clustering methods include Probabilistic Clustering(Liu et al., 2022), density-based clustering(Zhang et al., 2021; Khan et al., 2018), among others. Additionally, the quality of adaptive clustering can be significantly enhanced in multi-view clustering scenarios(Liu et al., 2023). In this process, we aim to obtain a suitable reference for the number of clusters (typically larger than real class numbers), rather than the optimal clustering, hence do not require a precise hyper-parameter search. Once the adaptive cluster number is established, we can employ the desired clustering algorithm to produce the initial clustering result.
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2.3.2 Active Aggregation and Splitting
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Query Strategy. To ensure a high success rate in establishing ‘must-link’ connections among selected cluster pairs, We employ a two-step query strategy. The first step involves filtering out low-quality cluster pairs, focusing on those with aggregation probabilities at the top. In the second step, we utilize Eq. (4) to calculate the expected NMI gain for these cluster pairs. Then we choose the pair with the highest NMI gain. By Theorem2.5, it is necessary to evaluate the purity of the two chosen clusters and select one representative sample (i.e., it belongs to the dominant class of this cluster) from each cluster to form a sample pair. This pair will then be subjected to queries by oracles. To facilitate this process, we specially designed a purity test.
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Purity Test. For convenience, we employ the sphere structure to depict a cluster, where the centroid sample of a cluster is denoted as j 0 subscript 𝑗 0 j_{0}italic_j start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT. The other samples in the cluster are marked as j i,ρ subscript 𝑗 𝑖 𝜌 j_{i,\rho}italic_j start_POSTSUBSCRIPT italic_i , italic_ρ end_POSTSUBSCRIPT, indicating that a sphere centered on sample i 𝑖 i italic_i with a radius of d(i,j i,ρ)𝑑 𝑖 subscript 𝑗 𝑖 𝜌 d(i,j_{i,\rho})italic_d ( italic_i , italic_j start_POSTSUBSCRIPT italic_i , italic_ρ end_POSTSUBSCRIPT ) includes ρ 𝜌\rho italic_ρ percent of the samples in the cluster. Considering that outlier samples typically reside in the outer regions of a cluster, and samples within impure clusters tend to be more sparsely distributed, we bifurcate the task of purity testing into two consecutive judgments.
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First, we evaluate how densely the samples are concentrated within a cluster. This assessment is formalized as the density test for a cluster w 𝑤 w italic_w, expressed in Eq. (5), where τ 𝜏\tau italic_τ represents a pre-set threshold that determines the level of strictness in this density test, and 𝟏(⋅)1⋅\mathbf{1}(\cdot)bold_1 ( ⋅ ) denotes the indicator function.
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𝒟𝒯(w)=𝟏(∑i∈w,j∈w(i)ℙ(e ij=1)∑i∈w|w(i)|>τ),𝒟 𝒯 𝑤 1 subscript formulae-sequence 𝑖 𝑤 𝑗 𝑤 𝑖 ℙ subscript 𝑒 𝑖 𝑗 1 subscript 𝑖 𝑤 𝑤 𝑖 𝜏\displaystyle\mathcal{DT}(w)=\mathbf{1}\Big{(}\frac{\sum_{i\in w,j\in w(i)}% \mathbb{P}(e_{ij}=1)}{\sum_{i\in w}|w(i)|}>\tau\Big{)},caligraphic_D caligraphic_T ( italic_w ) = bold_1 ( divide start_ARG ∑ start_POSTSUBSCRIPT italic_i ∈ italic_w , italic_j ∈ italic_w ( italic_i ) end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT = 1 ) end_ARG start_ARG ∑ start_POSTSUBSCRIPT italic_i ∈ italic_w end_POSTSUBSCRIPT | italic_w ( italic_i ) | end_ARG > italic_τ ) ,(5)
w(i)={j∣j∈w,ℙ(e ij=1)<ℙ(e ij i,0.5=1)}𝑤 𝑖 conditional-set 𝑗 formulae-sequence 𝑗 𝑤 ℙ subscript 𝑒 𝑖 𝑗 1 ℙ subscript 𝑒 𝑖 subscript 𝑗 𝑖 0.5 1\displaystyle w(i)={j\mid j\in w,\mathbb{P}(e_{ij}=1)<\mathbb{P}(e_{ij_{i,% \text{0.5}}}=1)}italic_w ( italic_i ) = { italic_j ∣ italic_j ∈ italic_w , blackboard_P ( italic_e start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT = 1 ) < blackboard_P ( italic_e start_POSTSUBSCRIPT italic_i italic_j start_POSTSUBSCRIPT italic_i , 0.5 end_POSTSUBSCRIPT end_POSTSUBSCRIPT = 1 ) }
If a cluster fails in the density test, we select a sample pair as (j 0,j j 0,0.7)subscript 𝑗 0 subscript 𝑗 subscript 𝑗 0 0.7(j_{0},j_{j_{0},\mathrm{0.7}})( italic_j start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT , italic_j start_POSTSUBSCRIPT italic_j start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT , 0.7 end_POSTSUBSCRIPT ), and require oracles to judge 𝒫(w)=𝟏(y j 0=y j j 0,0.7)𝒫 𝑤 1 subscript 𝑦 subscript 𝑗 0 subscript 𝑦 subscript 𝑗 subscript 𝑗 0 0.7\mathcal{P}(w)=\mathbf{1}(y_{j_{0}}=y_{j_{j_{0},\mathrm{0.7}}})caligraphic_P ( italic_w ) = bold_1 ( italic_y start_POSTSUBSCRIPT italic_j start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT end_POSTSUBSCRIPT = italic_y start_POSTSUBSCRIPT italic_j start_POSTSUBSCRIPT italic_j start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT , 0.7 end_POSTSUBSCRIPT end_POSTSUBSCRIPT ). It estimates whether the purity of this cluster is higher than 0.7 (i.e., satisfying the conditions in Theorem2.5). Overall, the purity test is as follows:
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𝒫𝒯(w)=𝒟𝒯(w)if 𝒟𝒯(w)else 𝒫(w).𝒫 𝒯 𝑤 𝒟 𝒯 𝑤 if 𝒟 𝒯 𝑤 else 𝒫 𝑤\displaystyle\mathcal{PT}(w)=\mathcal{DT}(w)\quad\text{if}\quad\mathcal{DT}(w)% \quad\text{else}\quad\mathcal{P}(w).caligraphic_P caligraphic_T ( italic_w ) = caligraphic_D caligraphic_T ( italic_w ) if caligraphic_D caligraphic_T ( italic_w ) else caligraphic_P ( italic_w ) .(6)
Clustering Update. In response to different outcomes in the purity test, we adopt the following strategies to update the clustering: (i) The purity test yields a result of 1 for both clusters, and we require the oracle to query their central samples. Should the query yield a must-link result, we will merge the clusters; if not, we will retain them as separate clusters. (2) The purity test yields 0 for at least one cluster, indicating a need for refinement, we proceed to split it into multiple subclusters with enhanced purity. Given the proven effectiveness of sample-based active clustering in managing small-scale datasets, as noted by Basu et al. (2004), we apply this approach for the splitting task, and the detail is described in Algorithm2.
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Algorithm 1 Adaptive Active Aggregation and Splitting
Input: Data X 𝑋 X italic_X, query limit Q max subscript 𝑄 Q_{\max}italic_Q start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT, index q=0 𝑞 0 q=0 italic_q = 0, clustering algorithm 𝒜 𝒜\mathcal{A}caligraphic_A
Initialization: Using 𝒜 𝒜\mathcal{A}caligraphic_A to generate initial clustering Ω Ω\Omega roman_Ω
for iter in 1:L 2:1 subscript 𝐿 2 1:L_{2}1 : italic_L start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT do
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Get a batch of k 𝑘 k italic_k candidate cluster pairs whose aggregation probability ranks top-k 𝑘 k italic_k as 𝒞={(w i,w j)|w i,w j∈Ω}𝒞 conditional-set subscript 𝑤 𝑖 subscript 𝑤 𝑗 subscript 𝑤 𝑖 subscript 𝑤 𝑗 Ω\mathcal{C}={(w_{i},w_{j})|w_{i},w_{j}\in\Omega}caligraphic_C = { ( italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) | italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∈ roman_Ω }
Use Eq.(4) to measure cluster pairs in 𝒞 𝒞\mathcal{C}caligraphic_C and choose the top pair
if q<Q max 𝑞 subscript 𝑄 q<Q_{\max}italic_q < italic_Q start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT then
Implement Purity Test on w 1 subscript 𝑤 1 w_{1}italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT and w 2 subscript 𝑤 2 w_{2}italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT with Eq. (6)
if Both clusters pass the test then
Select their central samples as x 1 subscript 𝑥 1 x_{1}italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT and x 2 subscript 𝑥 2 x_{2}italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT
Require oracle to query (x 1,x 2)subscript 𝑥 1 subscript 𝑥 2(x_{1},x_{2})( italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT )
Aggregate w 1 subscript 𝑤 1 w_{1}italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT and w 2 subscript 𝑤 2 w_{2}italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT if (x 1,x 2)subscript 𝑥 1 subscript 𝑥 2(x_{1},x_{2})( italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) is must-linked
else
Split w 1 subscript 𝑤 1 w_{1}italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT or/and w 2 subscript 𝑤 2 w_{2}italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT with Algorithm2
end if
Update constraints set with Algorithm3 Update q 𝑞 q italic_q by adding the newly invested number of queries
else
Terminate and return the result
end if
end for
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Algorithm 2 Subcluster Partition
Input: Cluster w 𝑤 w italic_w; subcluster lists 𝒩={}𝒩\mathcal{N}={}caligraphic_N = { }
Sort samples in w 𝑤 w italic_w in ascending order by their distance to the centroid
for i in w 𝑤 w italic_w do
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Select one sample from each subcluster in 𝒩 𝒩\mathcal{N}caligraphic_N, and get S 𝑆 S italic_S
Query i 𝑖 i italic_i with sample in S 𝑆 S italic_S till a must-link is reached or all samples in S 𝑆 S italic_S have been queried
if i 𝑖 i italic_i is must-linked to j 𝑗 j italic_j in S 𝑆 S italic_S then
Move i 𝑖 i italic_i from w 𝑤 w italic_w to the subcluster 𝒩 u subscript 𝒩 𝑢\mathcal{N}{u}caligraphic_N start_POSTSUBSCRIPT italic_u end_POSTSUBSCRIPT, j∈𝒩 u 𝑗 subscript 𝒩 𝑢 j\in\mathcal{N}{u}italic_j ∈ caligraphic_N start_POSTSUBSCRIPT italic_u end_POSTSUBSCRIPT
else
Move i 𝑖 i italic_i from w 𝑤 w italic_w to an empty subcluster, and add the new subcluster 𝒩 i={i}subscript 𝒩 𝑖 𝑖\mathcal{N}_{i}={i}caligraphic_N start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT = { italic_i } to 𝒩 𝒩\mathcal{N}caligraphic_N
end if
end for
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Algorithm 3 Fast Transitive Inference
Input: State matrix S 𝑆 S italic_S, new constraints (s,t)𝑠 𝑡(s,t)( italic_s , italic_t ).
for i 𝑖 i italic_i in (s,t)𝑠 𝑡(s,t)( italic_s , italic_t ) do
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Get ℳℒ={j|S[i,j]=1}ℳ ℒ conditional-set 𝑗 𝑆 𝑖 𝑗 1\mathcal{ML}={j|S[i,j]=1}caligraphic_M caligraphic_L = { italic_j | italic_S [ italic_i , italic_j ] = 1 }
Get 𝒞ℒ={j|S[i,j]=−1}𝒞 ℒ conditional-set 𝑗 𝑆 𝑖 𝑗 1\mathcal{CL}={j|S[i,j]=-1}caligraphic_C caligraphic_L = { italic_j | italic_S [ italic_i , italic_j ] = - 1 }
Let S[p,q]=1 𝑆 𝑝 𝑞 1 S[p,q]=1 italic_S [ italic_p , italic_q ] = 1, for p,q∈ℳℒ 𝑝 𝑞 ℳ ℒ p,q\in\mathcal{ML}italic_p , italic_q ∈ caligraphic_M caligraphic_L
Let S[p,q]=−1 𝑆 𝑝 𝑞 1 S[p,q]=-1 italic_S [ italic_p , italic_q ] = - 1, for p∈ℳℒ,q∈𝒞ℒ formulae-sequence 𝑝 ℳ ℒ 𝑞 𝒞 ℒ p\in\mathcal{ML},q\in\mathcal{CL}italic_p ∈ caligraphic_M caligraphic_L , italic_q ∈ caligraphic_C caligraphic_L
end for
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Transitive Inference. The must-link and cannot-link constraints possess the transitivity property (e.g., (x 1,x 2),(x 2,x 3)subscript 𝑥 1 subscript 𝑥 2 subscript 𝑥 2 subscript 𝑥 3(x_{1},x_{2}),(x_{2},x_{3})( italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) , ( italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ) are must-linked, then (x 1,x 3)subscript 𝑥 1 subscript 𝑥 3(x_{1},x_{3})( italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ) is must-linked). To store the constraints, we define a state matrix as S={s ij}N×N,s ij∈{−1,0,1}formulae-sequence 𝑆 subscript subscript 𝑠 𝑖 𝑗 𝑁 𝑁 subscript 𝑠 𝑖 𝑗 1 0 1 S={s_{ij}}{N\times N},s{ij}\in{-1,0,1}italic_S = { italic_s start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT } start_POSTSUBSCRIPT italic_N × italic_N end_POSTSUBSCRIPT , italic_s start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT ∈ { - 1 , 0 , 1 }. Here, 1/-1 denotes must-link/cannot-link, and 0 indicates an unqueried state. To avoid unnecessary queries, we need to augment the constraints set each time a new constraint is added. We assert that this expansion is only relevant to the preceding constraints that share a common sample with the new constraint, and propose a Fast Transitive Inference (Algorithm3) method to update the constraints. The correctness of this assertion and algorithm is proved in AppendixA.4.
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2.3.3 Discussion.
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Complexity Analysis. The computational complexity of the A3S algorithm comprises three distinct components. During the pairwise probability estimation, the complexity is O(NM)𝑂 𝑁 𝑀 O(NM)italic_O ( italic_N italic_M ), where M 𝑀 M italic_M denotes the number of neighbors considered for each sample. This is due to the need to only account for the neighboring clusters in each query. In the adaptive clustering process, the complexity depends on the specific algorithm, which we mark as O(A)𝑂 𝐴 O(A)italic_O ( italic_A ). The complexity of the query strategy is less than O(k 2+kL 2)𝑂 superscript 𝑘 2 𝑘 subscript 𝐿 2 O(k^{2}+kL_{2})italic_O ( italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_k italic_L start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ), where k 𝑘 k italic_k is the initial cluster number and L 2 subscript 𝐿 2 L_{2}italic_L start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT denotes the iteration number for A3S. Because there are k 2 superscript 𝑘 2 k^{2}italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT cluster pairs at the beginning, and we only need to re-calculate k 𝑘 k italic_k cluster pairs in each iteration. Consequently, the computational complexity of A3S is at most O(NM+A+k 2+kL 2)𝑂 𝑁 𝑀 𝐴 superscript 𝑘 2 𝑘 subscript 𝐿 2 O(NM+A+k^{2}+kL_{2})italic_O ( italic_N italic_M + italic_A + italic_k start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT + italic_k italic_L start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ).
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Application of A3S. A3S excels in two key scenarios: First, in clustering tasks lacking prior data information (strange environment), like the number of classes or sample distribution, where traditional methods falter or require extensive human queries. Second, in ongoing real-world applications needing regular data aggregation (expanding environment), such as weekly updates. Here, A3S adeptly merges new with existing clusters, efficiently managing redundancy. It’s particularly suitable when each period’s data is already high-quality, naturally meeting purity constraints.
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3 Experiments
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We organize the experiments as follows: we explain the experimental setup in Section3.1; we compare A3S with state-of-the-art active clustering methods and present the detailed results in Section3.2; then we compare the performance of A3S when applied to different clustering algorithms in Section3.3; lastly, we explore the influence of components in A3S in Section3.4 to 3.5.
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3.1 Experimental Setup
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Datasets. We sampled six datasets from four real-world image sources for the experiments: Market-1501(Zheng et al., 2015), which comprises human body images from 1501 individuals. We use two subsets: MK20 (351 images from 20 people) and MK100 (1650 images from 100 people); Humbi(Yu et al., 2020), a large multiview image dataset focused on human expressions like faces, and we extracted a subset Humbi-Face containing 5600 face images from 100 different people; Handwritten(Dua et al., 2017), a collection containing 2000 samples of handwritten digits from ‘0’ to ‘9’. We use the Fourier coefficient features in the experiments. (4) MS1M(Guo et al., 2016), a substantial benchmark dataset commonly used in face recognition tasks, and we sampled two large subsets: MS1M-10k, MS1M-100k. The details of these datasets are shown in Table1. Regarding the commonly used benchmarks in previous constrained clustering methods (e.g., UCI datasets (Asuncion & Newman, 2007)), they are not appropriate for our problems due to the very small sample and class sizes. Consequently, we have not evaluated the performance on those benchmarks. We seek to demonstrate that our method is generally workable for different types of data/applications with a wide range of cluster numbers and sample numbers.
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Figure 3: Comparing the performance of A3S and baselines on four datasets in terms of query count. More queries are invested for baselines to illustrate their characteristics.
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Baselines. To validate the performance of our methods, a set of baselines and state-of-the-art algorithms are compared. Random(Basu et al., 2003) randomly selects pairwise constraints. FFQS(Basu et al., 2004) uses the farthest-first scheme to acquire diversified samples and pairwise constraints. NPU(Xiong et al., 2013) uses the classic entropy-based principle to select informative samples to construct pairwise constraints. We use PCKMeans(Basu et al., 2003) as the semi-supervised clustering algorithm for these three methods, because PCKMeans best suits the pairwise constraints manner, and is suitable for large data sets with sparse high-dimensional data(Cai et al., 2023). URASC(Xiong et al., 2016) aims to iteratively query pairwise constraints that can maximally reduce the uncertainty of spectral clustering. COBRA over-cluster a dataset with K-means, then iteratively selects the closest cluster pairs for querying.
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Table 1: Statistical information about six datasets. N 𝑁 N italic_N, K 𝐾 K italic_K, and D 𝐷 D italic_D represent the sample numbers, class numbers, and the dimension of features. b 𝑏 b italic_b indicates whether the sample quantities between different classes are balanced.
MK20 MK100 Handwritten Humbi-Face MS1M-10k MS1M-100k
N 𝑁 N italic_N 351 1650 2000 5600 10000 100000
K 𝐾 K italic_K 20 100 10 100 146 1469
D 𝐷 D italic_D 256 256 76 256 512 512
b 𝑏 b italic_b✗✗✓✓✗✗
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Implementation. For the baseline methods, we maintain the same hyperparameter settings as reported in their original papers to ensure fairness in the comparison. Note that these baseline methods require the real cluster number as input, which is provided in our experiments. In many clustering applications, however, this number is typically not known beforehand, thus they are at an advantage. For the initialization of A3S 1 1 1 The code is available at we use isotonic regression to learn pairwise probability and utilize Fast Probabilistic Clustering (FPC)(Liu et al., 2022) as the adaptive clustering method. Here, we choose FPC because it only requires the pairwise probability for automatic data grouping, which is easy to implement. COBRA cannot assign a proper initial cluster number itself, so we use the same k 𝑘 k italic_k as A3S for a fair comparison. More details are presented in AppendixB.
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Evaluation. As discussed in Vinh et al. (2009), NMI can exhibit bias towards fine-grained clustering. Therefore, in addition to NMI, we employ the Adjusted Rand Index (ARI)(Hubert & Arabie, 1985) to evaluate the performance. NMI and ARI fall within the range of (0,1] and [-1,1] respectively, with larger values indicating superior clustering performance. To further investigate whether A3S effectively resolves the category fission and recover the real clustering structure, we introduce two supplementary metrics: (1) the Fission Rate (Υ=k K Υ 𝑘 𝐾\Upsilon=\frac{k}{K}roman_Υ = divide start_ARG italic_k end_ARG start_ARG italic_K end_ARG), where K 𝐾 K italic_K is the real class number and k 𝑘 k italic_k is the resulting cluster number; and (2) the entropy ratio between resulting clusters (Ω Ω\Omega roman_Ω) and real class partitions (C 𝐶 C italic_C), r=ℍ(Ω)ℍ(C)𝑟 ℍ Ω ℍ 𝐶 r=\frac{\mathbb{H}(\Omega)}{\mathbb{H}(C)}italic_r = divide start_ARG blackboard_H ( roman_Ω ) end_ARG start_ARG blackboard_H ( italic_C ) end_ARG. When Υ Υ\Upsilon roman_Υ and r 𝑟 r italic_r approaches 1, we conclude that A3S has effectively mitigated the category fission problem, and successfully recovered the true structure of C 𝐶 C italic_C.
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3.2 Comparison with SOTA Active Clustering Methods
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We compare A3S and five baseline methods on four datasets with different numbers of queries in terms of both NMI and ARI. The results are shown in Figure3. Overall, A3S has higher NMI and ARI values than other methods on these data sets when setting the same number of queries, and A3S requires only a small amount of queries to improve the NMI and ARI values significantly. In addition, we observe that both A3S and COBRA (cluster-based methods) improve steadily with the increase of queries, while Random, FFQS, NPU and URASC (semi-supervised clustering based methods) show fluctuations on all data sets. This is because genuine supervisory information can sometimes be detrimental to clustering, as it may introduce violations (e.g., (x i,x j)subscript 𝑥 𝑖 subscript 𝑥 𝑗(x_{i},x_{j})( italic_x start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) is cannot-linked, but their similarity to x k subscript 𝑥 𝑘 x_{k}italic_x start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT is both very high). This problem also exists for the latest semi-supervised clustering methods such as PCSKM(Vouros & Vasilaki, 2021). However, it does not mean that this line of work is not applicable. One common advantage of them is that they can ultimately improve the NMI and ARI value to 1.0 if enough queries can be provided (typically less than N×log(N)𝑁 𝑁 N\times\log(N)italic_N × roman_log ( italic_N )). They are a good choice when the target is to reveal the cluster identity for all samples accurately, or only low-quality features are available and the NMI of the initial clustering is lower than 0.2.
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The detailed running result of A3S is in Table2. A3S significantly reduces the fission rate and the entropy ratio to almost 1.0 on all datasets, with a high clustering purity. This validates that A3S can effectively reveal the true clustering structure of data. It’s important to highlight that these results are obtained without any prior knowledge about the number of classes or class distribution. Additionally, A3S exhibits robustness to the choice of adaptive cluster number. As detailed in AppendixC, increasing the adaptive number does not significantly alter the number of queries needed to obtain the desired result.
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Next, we delineate the distinctions in the results yielded by A3S and COBRA. Although COBRA quickly improves the NMI and ARI values, it cannot further enhance the clustering even when more queries are invested (all must-link clusters have already been discovered), leading to a performance ceiling dictated by cluster purities. Recent methods like COBRAS (Van Craenendonck et al., 2018) and AQM+MEE (Deng et al., 2023b) also encounter similar issues. In contrast, A3S not only delivers high-quality clusters but also identifies a subset of outlier samples. This approach enables continued augmentation of the NMI and ARI values through strategic querying of these outliers in combination with existing clusters. For instance, A3S and COBRA reach NMI values of 0.93 and 0.90 on MK100 separately, but the overall clustering purity of A3S is 0.9752, far surpassing COBRA that is 0.8509. Further, by querying the outlier samples of A3S with their neighbor clusters until ‘must-link’ is observed and they are aggregated to the corresponding clusters, A3S can reach an NMI of 0.99 with less than 500 more queries.
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Table 2: Detailed results of A3S in Figure3: the total running time (seconds), initial and ending fission rate (Υ Υ\Upsilon roman_Υ) and entropy ratio (r 𝑟 r italic_r), and the final clustering purity.
Time Υ init subscript Υ init\Upsilon_{\mathrm{init}}roman_Υ start_POSTSUBSCRIPT roman_init end_POSTSUBSCRIPT Υ end subscript Υ end\Upsilon_{\mathrm{end}}roman_Υ start_POSTSUBSCRIPT roman_end end_POSTSUBSCRIPT r init subscript 𝑟 init r_{\mathrm{init}}italic_r start_POSTSUBSCRIPT roman_init end_POSTSUBSCRIPT r end subscript 𝑟 end r_{\mathrm{end}}italic_r start_POSTSUBSCRIPT roman_end end_POSTSUBSCRIPT purity
MK20 0.93 2.05 0.95 1.25 1.01 0.9886
MK100 5.58 2.83 0.94 1.30 1.08 0.9752
Handwritten 6.39 8.6 1.00 1.55 1.00 0.9165
Humbi-Face 37.62 3.48 1.01 1.19 1.0 0.9745
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3.3 A3S for Different Clustering Algorithms
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Figure 4: Performance of A3S on MK100 and Humbi-Face when utilizing different clustering algorithms to generate the initial clustering result.
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Table 3: The ARI value between the clustering results of different clustering algorithms. We use F, K, S, and A to represent Fast Probabilistic Clustering, K-means Clustering, Spectral Clustering, and Agglomerative Clustering.
MK100 Humbi-Face
F K S A F K S A
F 1.000 0.581 0.187 0.597 1.000 0.600 0.272 0.805
K 0.581 1.000 0.140 0.408 0.600 1.000 0.178 0.573
S 0.187 0.140 1.000 0.249 0.272 0.178 1.000 0.276
A 0.597 0.408 0.249 1.000 0.805 0.573 0.276 1.000
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Figure 5: Performance of A3S on Handwritten when 1, 2, or 4 views of the feature are used.
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To assess the compatibility of A3S on different clustering algorithms, we additionally use three classic clustering algorithms to generate the initial clustering (adaptive cluster number is provided by FPC): K-means clustering(Choo et al., 2020), Spectral clustering(Von Luxburg, 2007) and Agglomerative Clustering(Murtagh & Legendre, 2014). We test these versions of A3S on MK100 and Humbi-Face, and the results are shown in Figure4. Besides, we quantify the difference between these initial clustering results with their mutual ARI value in Table3. We have two observations: first, the initial clustering outcomes derived from different algorithms exhibit substantial variability (mutual ARI value is typically lower than 0.6); second, A3S demonstrates a consistent ability to enhance the clustering performance across various algorithms. The results validate that A3S is robust to the initial clustering results, and can be effectively applied to other clustering algorithms without modification. In contrast, previous active clustering methods are typically designed and applicable to a specific clustering algorithm like DBSCAN(Mai et al., 2013) and Spectral Clustering(Xiong et al., 2016).
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3.4 Influence of Estimated Pairwise Probability Quality
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Better pairwise probability can lead to improved clustering performance, but its impact on A3S is unexplored. Multi-view clustering(Yang & Wang, 2018) is the major approach in this domain, and we utilize Handwritten to investigate this aspect. The complete Handwritten dataset comprises four distinct feature types for each sample. Following the setup in Liu et al. (2022), we initially learn the pairwise probabilities using features from each view. Subsequently, pairwise probabilities are aggregated across V 𝑉 V italic_V different views employing the following formula(Liu et al., 2022): ℙ(e ij=1|d ij 1,⋯,d ij V)=∏m=1 V ℙ(e ij=1|d ij m)∏m=1 V ℙ(e ij=1|d ij m)+∏m=1 V ℙ(e ij=0|d ij m)ℙ subscript 𝑒 𝑖 𝑗 conditional 1 superscript subscript 𝑑 𝑖 𝑗 1⋯superscript subscript 𝑑 𝑖 𝑗 𝑉 superscript subscript product 𝑚 1 𝑉 ℙ subscript 𝑒 𝑖 𝑗 conditional 1 superscript subscript 𝑑 𝑖 𝑗 𝑚 superscript subscript product 𝑚 1 𝑉 ℙ subscript 𝑒 𝑖 𝑗 conditional 1 superscript subscript 𝑑 𝑖 𝑗 𝑚 superscript subscript product 𝑚 1 𝑉 ℙ subscript 𝑒 𝑖 𝑗 conditional 0 superscript subscript 𝑑 𝑖 𝑗 𝑚\mathbb{P}(e_{ij}=1|d_{ij}^{1},\cdots,d_{ij}^{V})=\frac{\prod_{m=1}^{V}\mathbb% {P}(e_{ij}=1|d_{ij}^{m})}{\prod_{m=1}^{V}\mathbb{P}(e_{ij}=1|d_{ij}^{m})+\prod% {m=1}^{V}\mathbb{P}(e{ij}=0|d_{ij}^{m})}blackboard_P ( italic_e start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT = 1 | italic_d start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 1 end_POSTSUPERSCRIPT , ⋯ , italic_d start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_V end_POSTSUPERSCRIPT ) = divide start_ARG ∏ start_POSTSUBSCRIPT italic_m = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_V end_POSTSUPERSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT = 1 | italic_d start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_m end_POSTSUPERSCRIPT ) end_ARG start_ARG ∏ start_POSTSUBSCRIPT italic_m = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_V end_POSTSUPERSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT = 1 | italic_d start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_m end_POSTSUPERSCRIPT ) + ∏ start_POSTSUBSCRIPT italic_m = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_V end_POSTSUPERSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT = 0 | italic_d start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_m end_POSTSUPERSCRIPT ) end_ARG, where d ij m superscript subscript 𝑑 𝑖 𝑗 𝑚 d_{ij}^{m}italic_d start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_m end_POSTSUPERSCRIPT is the pairwise distance in the m 𝑚 m italic_m-th view.
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In the investigation of A3S’s performance with varying views (one/two/four) as illustrated in Figure5, we observed that employing multi-view clustering slightly improves the initial clustering performance, but significantly boosts A3S’s overall effectiveness. This approach leads to quicker convergence and nearly perfect NMI and ARI values. The improvement is largely attributed to multi-view aggregated pairwise probabilities, which make outlier samples distinguishable from neighbor samples that are from different classes, thereby preventing them from being in one cluster during initial clustering. Empirically, In four-view scenarios, despite a higher initial fission rate of 10.1, the initial clustering purity improves from 0.916 to 0.989.
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Figure 6: Left: performance of A3S on MS1M-10k and MS1M-100k. Right: trade-off between initial clustering quality and initial clustering purity for COBRA on Handwritten.
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3.5 Performance of A3S on Large datasets.
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We further test A3S on two large datasets MS1M-10k and MS1M-100k, where the initial cluster number is determined by FPC for both A3S and COBRA. The result is in Figure6. We have two observations: (1) A3S can reach near-optimal ARI value (0.995) with only 3100 queries for MS1M-100k in 8581.86 seconds, which validates that it is scalable to large datasets. As far as we know, this is the largest dataset ever used in active clustering research. (2) COBRA is sensitive to the initial cluster number and can waste too many queries to get a bad clustering result.
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3.6 Case Study for A3S
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We present a case study in Figure2 to demonstrate A3S’s mechanism using a two-class dataset. Initially, data points are clustered into seven groups, including two outlier clusters. In the first iteration, clusters w 6 subscript 𝑤 6 w_{6}italic_w start_POSTSUBSCRIPT 6 end_POSTSUBSCRIPT and w 7 subscript 𝑤 7 w_{7}italic_w start_POSTSUBSCRIPT 7 end_POSTSUBSCRIPT are selected for their largest expected impact on the NMI value. During the purity test, w 6 subscript 𝑤 6 w_{6}italic_w start_POSTSUBSCRIPT 6 end_POSTSUBSCRIPT fails the density test, leading to a subsequent query between its central sample (j 0 subscript 𝑗 0 j_{0}italic_j start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT) and the margin sample (j j 0,0.7 subscript 𝑗 subscript 𝑗 0 0.7 j_{j_{0},0.7}italic_j start_POSTSUBSCRIPT italic_j start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT , 0.7 end_POSTSUBSCRIPT) of the sphere that is centered at j 0 subscript 𝑗 0 j_{0}italic_j start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT and contains 70% samples of w 6 subscript 𝑤 6 w_{6}italic_w start_POSTSUBSCRIPT 6 end_POSTSUBSCRIPT. The cannot-link result indicates w 6 subscript 𝑤 6 w_{6}italic_w start_POSTSUBSCRIPT 6 end_POSTSUBSCRIPT fails the purity test, so it is split into two sub-clusters using Algorithm2, which costs 12 queries. In the second iteration, clusters w 9 subscript 𝑤 9 w_{9}italic_w start_POSTSUBSCRIPT 9 end_POSTSUBSCRIPT and w 7 subscript 𝑤 7 w_{7}italic_w start_POSTSUBSCRIPT 7 end_POSTSUBSCRIPT are selected, both passing the purity test. Their central samples are queried, and they are merged into w 10 subscript 𝑤 10 w_{10}italic_w start_POSTSUBSCRIPT 10 end_POSTSUBSCRIPT following the must-link result.
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A3S addresses outlier detection through purity tests and subcluster partitioning. However, the purity test can discover most but not all low-quality clusters whose purity is lower than 0.7. As a remedy, we could resort to multi-view features which can significantly improve the clustering purity and reduce the number of outliers, as shown in Section3.4. We remark that when multi-view data is available, the purity test and subcluster partition of A3S can be omitted, streamlining the procedure. In contrast, COBRA attempts to mitigate outlier issues by opting for a larger cluster number to enhance initial clustering purity, which is inefficient as depicted in the right image of Figure6 (the clustering purity increases very slowly).
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4 Related Work
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Query Strategy in Active Clustering. Recent Active clustering methods have embraced a trend of incorporating sample uncertainty into their query strategies. These methods frequently utilize entropy to quantify uncertainty(Abin, 2016; Xiong et al., 2016; Shi et al., 2020). A common task involves estimating the probability of a sample belonging to different clusters or neighborhoods(Xiong et al., 2013). Additionally, alternative criteria such as maximum expected error reduction(Wang & Davidson, 2010) and maximum expected clustering change(Biswas & Jacobs, 2014) have been proposed to assess the stability of clustering results when perturbing the similarity values between two samples.
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Constraints in Semi-supervised Clustering. When using the must-link and cannot-link constraints to perform SSC, two aspects are usually taken into consideration: the transitive inference of constraints and the combination of constraints to specific clustering algorithms. A few studies(Lutz et al., 2021) address transitive inference with graph-based techniques instead of a brute-force manner. In addition, recent studies that optimize clustering results with constraints in SSC have explored various approaches, Vouros & Vasilaki (2021) explores Kmeans clustering for high dimensional data; Yang et al. (2022) attempts to correctly infer the number of clusters for hierarchical clustering; Ren et al. (2018) tries to utilize prior knowledge to determine a proper cluster number for density-based clustering; Chen & Zhong (2022) develops a graph-based SSC that is robust to noise, but is not suitable for large datasets. However, there remains a relatively unexplored potential in integrating these modern SSC approaches with active clustering, which presents a promising avenue for future research.
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5 Conclusion
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This paper studies the cluster-adjustment scheme in active clustering, offering theoretical guidance for non-deteriorating cluster aggregation and quantifying the impact of human queries and aggregation operations. We then propose A3S, a general framework that does not rely on the dataset prior. Through extensive testing, A3S demonstrates its effectiveness on diverse real-world datasets with varying class numbers and distributions. We will explore the application of A3S on more complex multi-view datasets and gigantic datasets at the million level in the future.
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Acknowledgements
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This work is supported by the National Key Research and Development Program of China (2022YFB3104701) and Alibaba Group through Alibaba Research Intern Program. We appreciate the reviewers for their insightful feedback and advice, these constructive criticism and recommendations have been invaluable in helping us improve the quality of this work.
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Impact Statement
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This paper presents a work that aims to advance the field of Active Clustering. There are some potential societal consequences of our work, none of which we feel must be specifically highlighted here.
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Appendix A Proofs
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A.1 Proof of Theorem 2.5
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Proof of Theorem 2.5.
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Let p 𝑝 p italic_p and q 𝑞 q italic_q denote the sizes of w 1 subscript 𝑤 1 w_{1}italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT and w 2 subscript 𝑤 2 w_{2}italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT, respectively. We further assume that the class index of these outliers is i j∈{1,2,⋯,K}subscript 𝑖 𝑗 1 2⋯𝐾 i_{j}\in{1,2,\cdots,K}italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∈ { 1 , 2 , ⋯ , italic_K }, where j∈{1,2,⋯,(1−t 1)p+(1−t 2)q}𝑗 1 2⋯1 subscript 𝑡 1 𝑝 1 subscript 𝑡 2 𝑞 j\in{1,2,\cdots,(1-t_{1})p+(1-t_{2})q}italic_j ∈ { 1 , 2 , ⋯ , ( 1 - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_p + ( 1 - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_q }. For ease of presentation, for any class index i∈{1,2,⋯,K}𝑖 1 2⋯𝐾 i\in{1,2,\cdots,K}italic_i ∈ { 1 , 2 , ⋯ , italic_K }, we use s i subscript 𝑠 𝑖 s_{i}italic_s start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT to denote the class size, i.e., |c i|=s i subscript 𝑐 𝑖 subscript 𝑠 𝑖|c_{i}|=s_{i}| italic_c start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT | = italic_s start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT. Without loss generality, we assume that q≥p 𝑞 𝑝 q\geq p italic_q ≥ italic_p throughout this proof.
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By the definition of mutual information, we have
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𝕀(Ω∗;C)𝕀 superscript Ω 𝐶\displaystyle\mathbb{I}(\Omega^{};C)blackboard_I ( roman_Ω start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ; italic_C )=𝕀(Ω;C)+∑τ=1 K ℙ(w 1,2∩c τ)logℙ(w 1,2∩c τ)ℙ(w 1,2)ℙ(c τ)−∑τ=1 K ℙ(w 1∩c τ)logℙ(w 1∩c τ)ℙ(w 1)ℙ(c τ)absent 𝕀 Ω 𝐶 superscript subscript 𝜏 1 𝐾 ℙ subscript 𝑤 1 2 subscript 𝑐 𝜏 ℙ subscript 𝑤 1 2 subscript 𝑐 𝜏 ℙ subscript 𝑤 1 2 ℙ subscript 𝑐 𝜏 superscript subscript 𝜏 1 𝐾 ℙ subscript 𝑤 1 subscript 𝑐 𝜏 ℙ subscript 𝑤 1 subscript 𝑐 𝜏 ℙ subscript 𝑤 1 ℙ subscript 𝑐 𝜏\displaystyle=\mathbb{I}(\Omega;C)+\sum_{\tau=1}^{K}\mathbb{P}(w_{1,2}\cap c_{% \tau})\log\frac{\mathbb{P}(w_{1,2}\cap c_{\tau})}{\mathbb{P}(w_{1,2})\mathbb{P% }(c_{\tau})}-\sum_{\tau=1}^{K}\mathbb{P}(w_{1}\cap c_{\tau})\log\frac{\mathbb{% P}(w_{1}\cap c_{\tau})}{\mathbb{P}(w_{1})\mathbb{P}(c_{\tau})}= blackboard_I ( roman_Ω ; italic_C ) + ∑ start_POSTSUBSCRIPT italic_τ = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_K end_POSTSUPERSCRIPT blackboard_P ( italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) roman_log divide start_ARG blackboard_P ( italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG start_ARG blackboard_P ( italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ) blackboard_P ( italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG - ∑ start_POSTSUBSCRIPT italic_τ = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_K end_POSTSUPERSCRIPT blackboard_P ( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) roman_log divide start_ARG blackboard_P ( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG start_ARG blackboard_P ( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) blackboard_P ( italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG
−∑τ=1 K ℙ(w 2∩c τ)logℙ(w 2∩c τ)ℙ(w 2)ℙ(c τ)superscript subscript 𝜏 1 𝐾 ℙ subscript 𝑤 2 subscript 𝑐 𝜏 ℙ subscript 𝑤 2 subscript 𝑐 𝜏 ℙ subscript 𝑤 2 ℙ subscript 𝑐 𝜏\displaystyle\qquad-\sum_{\tau=1}^{K}\mathbb{P}(w_{2}\cap c_{\tau})\log\frac{% \mathbb{P}(w_{2}\cap c_{\tau})}{\mathbb{P}(w_{2})\mathbb{P}(c_{\tau})}- ∑ start_POSTSUBSCRIPT italic_τ = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_K end_POSTSUPERSCRIPT blackboard_P ( italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) roman_log divide start_ARG blackboard_P ( italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG start_ARG blackboard_P ( italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) blackboard_P ( italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG
=𝕀(Ω;C)+∑τ=1 K|w 1,2∩c τ|NlogN⋅|w 1,2∩c τ||w 1,2|⋅|c τ|⏟(I)−∑τ=1 K|w 1∩c τ|NlogN⋅|w 1∩c τ||w 1|⋅|c τ|⏟(II)absent 𝕀 Ω 𝐶 subscript⏟superscript subscript 𝜏 1 𝐾 subscript 𝑤 1 2 subscript 𝑐 𝜏 𝑁⋅𝑁 subscript 𝑤 1 2 subscript 𝑐 𝜏⋅subscript 𝑤 1 2 subscript 𝑐 𝜏 I subscript⏟superscript subscript 𝜏 1 𝐾 subscript 𝑤 1 subscript 𝑐 𝜏 𝑁⋅𝑁 subscript 𝑤 1 subscript 𝑐 𝜏⋅subscript 𝑤 1 subscript 𝑐 𝜏 II\displaystyle=\mathbb{I}(\Omega;C)+\underbrace{\sum_{\tau=1}^{K}\frac{|w_{1,2}% \cap c_{\tau}|}{N}\log\frac{N\cdot|w_{1,2}\cap c_{\tau}|}{|w_{1,2}|\cdot|c_{% \tau}|}}{\displaystyle\mathrm{(I)}}-\underbrace{\sum{\tau=1}^{K}\frac{|w_{1}% \cap c_{\tau}|}{N}\log\frac{N\cdot|w_{1}\cap c_{\tau}|}{|w_{1}|\cdot|c_{\tau}|% }}{\displaystyle\mathrm{(II)}}= blackboard_I ( roman_Ω ; italic_C ) + under⏟ start_ARG ∑ start_POSTSUBSCRIPT italic_τ = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_K end_POSTSUPERSCRIPT divide start_ARG | italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ⋅ | italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG start_ARG | italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT | ⋅ | italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG end_ARG start_POSTSUBSCRIPT ( roman_I ) end_POSTSUBSCRIPT - under⏟ start_ARG ∑ start_POSTSUBSCRIPT italic_τ = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_K end_POSTSUPERSCRIPT divide start_ARG | italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ⋅ | italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG start_ARG | italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | ⋅ | italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG end_ARG start_POSTSUBSCRIPT ( roman_II ) end_POSTSUBSCRIPT
−∑τ=1 K|w 2∩c τ|NlogN⋅|w 2∩c τ||w 2|⋅|c τ|⏟(III).subscript⏟superscript subscript 𝜏 1 𝐾 subscript 𝑤 2 subscript 𝑐 𝜏 𝑁⋅𝑁 subscript 𝑤 2 subscript 𝑐 𝜏⋅subscript 𝑤 2 subscript 𝑐 𝜏 III\displaystyle\qquad-\underbrace{\sum{\tau=1}^{K}\frac{|w_{2}\cap c_{\tau}|}{N% }\log\frac{N\cdot|w_{2}\cap c_{\tau}|}{|w_{2}|\cdot|c_{\tau}|}}_{\displaystyle% \mathrm{(III)}}.- under⏟ start_ARG ∑ start_POSTSUBSCRIPT italic_τ = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_K end_POSTSUPERSCRIPT divide start_ARG | italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ⋅ | italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG start_ARG | italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT | ⋅ | italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG end_ARG start_POSTSUBSCRIPT ( roman_III ) end_POSTSUBSCRIPT .(7)
Then we bound these three terms respectively. For Term (I), we have
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(I)I\displaystyle\mathrm{(I)}( roman_I )=|w 1,2∩c 1|NlogN⋅|w 1,2∩c 1||w 1,2|⋅|c 1|+∑τ=2 K|w 1,2∩c τ|NlogN⋅|w 1,2∩c τ||w 1,2|⋅|c τ|absent subscript 𝑤 1 2 subscript 𝑐 1 𝑁⋅𝑁 subscript 𝑤 1 2 subscript 𝑐 1⋅subscript 𝑤 1 2 subscript 𝑐 1 superscript subscript 𝜏 2 𝐾 subscript 𝑤 1 2 subscript 𝑐 𝜏 𝑁⋅𝑁 subscript 𝑤 1 2 subscript 𝑐 𝜏⋅subscript 𝑤 1 2 subscript 𝑐 𝜏\displaystyle=\frac{|w_{1,2}\cap c_{1}|}{N}\log\frac{N\cdot|w_{1,2}\cap c_{1}|% }{|w_{1,2}|\cdot|c_{1}|}+\sum_{\tau=2}^{K}\frac{|w_{1,2}\cap c_{\tau}|}{N}\log% \frac{N\cdot|w_{1,2}\cap c_{\tau}|}{|w_{1,2}|\cdot|c_{\tau}|}= divide start_ARG | italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ⋅ | italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | end_ARG start_ARG | italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT | ⋅ | italic_c start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | end_ARG + ∑ start_POSTSUBSCRIPT italic_τ = 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_K end_POSTSUPERSCRIPT divide start_ARG | italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ⋅ | italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG start_ARG | italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT | ⋅ | italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG
=t 1p+t 2q NlogN(t 1p+t 2q)(p+q)s 1+∑j=1(1−t 1)p+(1−t 2)q 1 NlogN⋅|w 1,2∩c i j|(p+q)s i j.absent subscript 𝑡 1 𝑝 subscript 𝑡 2 𝑞 𝑁 𝑁 subscript 𝑡 1 𝑝 subscript 𝑡 2 𝑞 𝑝 𝑞 subscript 𝑠 1 superscript subscript 𝑗 1 1 subscript 𝑡 1 𝑝 1 subscript 𝑡 2 𝑞 1 𝑁⋅𝑁 subscript 𝑤 1 2 subscript 𝑐 subscript 𝑖 𝑗 𝑝 𝑞 subscript 𝑠 subscript 𝑖 𝑗\displaystyle=\frac{t_{1}p+t_{2}q}{N}\log\frac{N(t_{1}p+t_{2}q)}{(p+q)s_{1}}+% \sum_{j=1}^{(1-t_{1})p+(1-t_{2})q}\frac{1}{N}\log\frac{N\cdot|w_{1,2}\cap c_{i% {j}}|}{(p+q)s{i_{j}}}.= divide start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_p + italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_q end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ( italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_p + italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_q ) end_ARG start_ARG ( italic_p + italic_q ) italic_s start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_ARG + ∑ start_POSTSUBSCRIPT italic_j = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ( 1 - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_p + ( 1 - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_q end_POSTSUPERSCRIPT divide start_ARG 1 end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ⋅ | italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT | end_ARG start_ARG ( italic_p + italic_q ) italic_s start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT end_ARG .(8)
Similarly, we have
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(II)II\displaystyle\mathrm{(II)}( roman_II )=|w 1∩c 1|NlogN⋅|w 1∩c 1||w 1|⋅|c 1|+∑τ=2 K|w 1∩c τ|NlogN⋅|w 1∩c τ||w 1|⋅|c τ|absent subscript 𝑤 1 subscript 𝑐 1 𝑁⋅𝑁 subscript 𝑤 1 subscript 𝑐 1⋅subscript 𝑤 1 subscript 𝑐 1 superscript subscript 𝜏 2 𝐾 subscript 𝑤 1 subscript 𝑐 𝜏 𝑁⋅𝑁 subscript 𝑤 1 subscript 𝑐 𝜏⋅subscript 𝑤 1 subscript 𝑐 𝜏\displaystyle=\frac{|w_{1}\cap c_{1}|}{N}\log\frac{N\cdot|w_{1}\cap c_{1}|}{|w% {1}|\cdot|c{1}|}+\sum_{\tau=2}^{K}\frac{|w_{1}\cap c_{\tau}|}{N}\log\frac{N% \cdot|w_{1}\cap c_{\tau}|}{|w_{1}|\cdot|c_{\tau}|}= divide start_ARG | italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ⋅ | italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | end_ARG start_ARG | italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | ⋅ | italic_c start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | end_ARG + ∑ start_POSTSUBSCRIPT italic_τ = 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_K end_POSTSUPERSCRIPT divide start_ARG | italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ⋅ | italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG start_ARG | italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | ⋅ | italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG
=t 1p NlogNt 1 s 1+∑j=1(1−t 1)p 1 NlogN⋅|w 1∩c i j|ps i j,absent subscript 𝑡 1 𝑝 𝑁 𝑁 subscript 𝑡 1 subscript 𝑠 1 superscript subscript 𝑗 1 1 subscript 𝑡 1 𝑝 1 𝑁⋅𝑁 subscript 𝑤 1 subscript 𝑐 subscript 𝑖 𝑗 𝑝 subscript 𝑠 subscript 𝑖 𝑗\displaystyle=\frac{t_{1}p}{N}\log\frac{Nt_{1}}{s_{1}}+\sum_{j=1}^{(1-t_{1})p}% \frac{1}{N}\log\frac{N\cdot|w_{1}\cap c_{i_{j}}|}{ps_{i_{j}}},= divide start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_p end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_ARG start_ARG italic_s start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_ARG + ∑ start_POSTSUBSCRIPT italic_j = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ( 1 - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_p end_POSTSUPERSCRIPT divide start_ARG 1 end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ⋅ | italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT | end_ARG start_ARG italic_p italic_s start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT end_ARG ,(9)
and
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(III)III\displaystyle\mathrm{(III)}( roman_III )=|w 2∩c 1|NlogN⋅|w 2∩c 1||w 2|⋅|c 1|+∑τ=2 K|w 2∩c τ|NlogN⋅|w 2∩c τ||w 2|⋅|c τ|absent subscript 𝑤 2 subscript 𝑐 1 𝑁⋅𝑁 subscript 𝑤 2 subscript 𝑐 1⋅subscript 𝑤 2 subscript 𝑐 1 superscript subscript 𝜏 2 𝐾 subscript 𝑤 2 subscript 𝑐 𝜏 𝑁⋅𝑁 subscript 𝑤 2 subscript 𝑐 𝜏⋅subscript 𝑤 2 subscript 𝑐 𝜏\displaystyle=\frac{|w_{2}\cap c_{1}|}{N}\log\frac{N\cdot|w_{2}\cap c_{1}|}{|w% {2}|\cdot|c{1}|}+\sum_{\tau=2}^{K}\frac{|w_{2}\cap c_{\tau}|}{N}\log\frac{N% \cdot|w_{2}\cap c_{\tau}|}{|w_{2}|\cdot|c_{\tau}|}= divide start_ARG | italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ⋅ | italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | end_ARG start_ARG | italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT | ⋅ | italic_c start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT | end_ARG + ∑ start_POSTSUBSCRIPT italic_τ = 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_K end_POSTSUPERSCRIPT divide start_ARG | italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ⋅ | italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG start_ARG | italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT | ⋅ | italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT | end_ARG
=t 2q NlogNt 2 s 2+∑j=1(1−t 2)q 1 NlogN⋅|w 2∩c i j|qs i j.absent subscript 𝑡 2 𝑞 𝑁 𝑁 subscript 𝑡 2 subscript 𝑠 2 superscript subscript 𝑗 1 1 subscript 𝑡 2 𝑞 1 𝑁⋅𝑁 subscript 𝑤 2 subscript 𝑐 subscript 𝑖 𝑗 𝑞 subscript 𝑠 subscript 𝑖 𝑗\displaystyle=\frac{t_{2}q}{N}\log\frac{Nt_{2}}{s_{2}}+\sum_{j=1}^{(1-t_{2})q}% \frac{1}{N}\log\frac{N\cdot|w_{2}\cap c_{i_{j}}|}{qs_{i_{j}}}.= divide start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_q end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT end_ARG start_ARG italic_s start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT end_ARG + ∑ start_POSTSUBSCRIPT italic_j = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ( 1 - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_q end_POSTSUPERSCRIPT divide start_ARG 1 end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N ⋅ | italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT | end_ARG start_ARG italic_q italic_s start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT end_ARG .(10)
Plugging Eq.(A.1), Eq.(A.1), and Eq.(A.1) into Eq.(A.1), together with the fact that
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|w 1,2∩c i j|subscript 𝑤 1 2 subscript 𝑐 subscript 𝑖 𝑗\displaystyle|w_{1,2}\cap c_{i_{j}}|| italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT |≥max{|w 1∩c i j|,|w 2∩c i j|},∀j∈{1,2,⋯,(1−t 1)p+(1−t 2)q},formulae-sequence absent subscript 𝑤 1 subscript 𝑐 subscript 𝑖 𝑗 subscript 𝑤 2 subscript 𝑐 subscript 𝑖 𝑗 for-all 𝑗 1 2⋯1 subscript 𝑡 1 𝑝 1 subscript 𝑡 2 𝑞\displaystyle\geq\max{|w_{1}\cap c_{i_{j}}|,|w_{2}\cap c_{i_{j}}|},\quad% \forall j\in{1,2,\cdots,(1-t_{1})p+(1-t_{2})q},≥ roman_max { | italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT | , | italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_i start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT | } , ∀ italic_j ∈ { 1 , 2 , ⋯ , ( 1 - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_p + ( 1 - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_q } ,
we obtain that
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𝕀(Ω∗;C)𝕀 superscript Ω 𝐶\displaystyle\mathbb{I}(\Omega^{};C)blackboard_I ( roman_Ω start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ; italic_C )≥𝕀(Ω;C)+t 1p Nlogt 1p+t 2q t 1(p+q)+t 2q Nlogt 1p+t 2q t 2(p+q)⏟(IV)absent 𝕀 Ω 𝐶 subscript⏟subscript 𝑡 1 𝑝 𝑁 subscript 𝑡 1 𝑝 subscript 𝑡 2 𝑞 subscript 𝑡 1 𝑝 𝑞 subscript 𝑡 2 𝑞 𝑁 subscript 𝑡 1 𝑝 subscript 𝑡 2 𝑞 subscript 𝑡 2 𝑝 𝑞 IV\displaystyle\geq\mathbb{I}(\Omega;C)+\underbrace{\frac{t_{1}p}{N}\log\frac{t_% {1}p+t_{2}q}{t_{1}(p+q)}+\frac{t_{2}q}{N}\log\frac{t_{1}p+t_{2}q}{t_{2}(p+q)}}% {\displaystyle\mathrm{(IV)}}≥ blackboard_I ( roman_Ω ; italic_C ) + under⏟ start_ARG divide start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_p end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_p + italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_q end_ARG start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ( italic_p + italic_q ) end_ARG + divide start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_q end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_p + italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_q end_ARG start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( italic_p + italic_q ) end_ARG end_ARG start_POSTSUBSCRIPT ( roman_IV ) end_POSTSUBSCRIPT(11)
−((1−t 1)p+(1−t 2)q Nlog(p+q)−(1−t 1)p Nlogp−(1−t 2)q Nlogq)⏟(V).subscript⏟1 subscript 𝑡 1 𝑝 1 subscript 𝑡 2 𝑞 𝑁 𝑝 𝑞 1 subscript 𝑡 1 𝑝 𝑁 𝑝 1 subscript 𝑡 2 𝑞 𝑁 𝑞 V\displaystyle\qquad-\underbrace{\Big{(}\frac{(1-t{1})p+(1-t_{2})q}{N}\log(p+q% )-\frac{(1-t_{1})p}{N}\log p-\frac{(1-t_{2})q}{N}\log q\Big{)}}_{\displaystyle% \mathrm{(V)}}.- under⏟ start_ARG ( divide start_ARG ( 1 - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_p + ( 1 - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_q end_ARG start_ARG italic_N end_ARG roman_log ( italic_p + italic_q ) - divide start_ARG ( 1 - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_p end_ARG start_ARG italic_N end_ARG roman_log italic_p - divide start_ARG ( 1 - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_q end_ARG start_ARG italic_N end_ARG roman_log italic_q ) end_ARG start_POSTSUBSCRIPT ( roman_V ) end_POSTSUBSCRIPT .(12)
For Term (IV) in Eq.(11), by the Taylor expansion
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log(1+x)=∑u=1∞(−1)u−1 u⋅x u,1 𝑥 superscript subscript 𝑢 1⋅superscript 1 𝑢 1 𝑢 superscript 𝑥 𝑢\displaystyle\log(1+x)=\sum_{u=1}^{\infty}\frac{(-1)^{u-1}}{u}\cdot x^{u},roman_log ( 1 + italic_x ) = ∑ start_POSTSUBSCRIPT italic_u = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG ( - 1 ) start_POSTSUPERSCRIPT italic_u - 1 end_POSTSUPERSCRIPT end_ARG start_ARG italic_u end_ARG ⋅ italic_x start_POSTSUPERSCRIPT italic_u end_POSTSUPERSCRIPT ,
we have
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(IV)IV\displaystyle\mathrm{(IV)}( roman_IV )=t 1p N∑u=1∞(−1)u−1 u⋅((t 2−t 1)q t 1(p+q))u+t 2q N∑u=1∞(−1)u−1 u⋅[(t 1−t 2)p t 2(p+q)]u absent subscript 𝑡 1 𝑝 𝑁 superscript subscript 𝑢 1⋅superscript 1 𝑢 1 𝑢 superscript subscript 𝑡 2 subscript 𝑡 1 𝑞 subscript 𝑡 1 𝑝 𝑞 𝑢 subscript 𝑡 2 𝑞 𝑁 superscript subscript 𝑢 1⋅superscript 1 𝑢 1 𝑢 superscript delimited-[]subscript 𝑡 1 subscript 𝑡 2 𝑝 subscript 𝑡 2 𝑝 𝑞 𝑢\displaystyle=\frac{t_{1}p}{N}\sum_{u=1}^{\infty}\frac{(-1)^{u-1}}{u}\cdot\Big% {(}\frac{(t_{2}-t_{1})q}{t_{1}(p+q)}\Big{)}^{u}+\frac{t_{2}q}{N}\sum_{u=1}^{% \infty}\frac{(-1)^{u-1}}{u}\cdot\Big{[}\frac{(t_{1}-t_{2})p}{t_{2}(p+q)}\Big{]% }^{u}= divide start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_p end_ARG start_ARG italic_N end_ARG ∑ start_POSTSUBSCRIPT italic_u = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG ( - 1 ) start_POSTSUPERSCRIPT italic_u - 1 end_POSTSUPERSCRIPT end_ARG start_ARG italic_u end_ARG ⋅ ( divide start_ARG ( italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_q end_ARG start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ( italic_p + italic_q ) end_ARG ) start_POSTSUPERSCRIPT italic_u end_POSTSUPERSCRIPT + divide start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_q end_ARG start_ARG italic_N end_ARG ∑ start_POSTSUBSCRIPT italic_u = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG ( - 1 ) start_POSTSUPERSCRIPT italic_u - 1 end_POSTSUPERSCRIPT end_ARG start_ARG italic_u end_ARG ⋅ [ divide start_ARG ( italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_p end_ARG start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( italic_p + italic_q ) end_ARG ] start_POSTSUPERSCRIPT italic_u end_POSTSUPERSCRIPT
=∑v=1∞1 2v−1⋅[t 1p N⋅((t 2−t 1)q t 1(p+q))2v−1+t 2q N((t 1−t 2)p t 2(p+q))2v−1]absent superscript subscript 𝑣 1⋅1 2 𝑣 1 delimited-[]⋅subscript 𝑡 1 𝑝 𝑁 superscript subscript 𝑡 2 subscript 𝑡 1 𝑞 subscript 𝑡 1 𝑝 𝑞 2 𝑣 1 subscript 𝑡 2 𝑞 𝑁 superscript subscript 𝑡 1 subscript 𝑡 2 𝑝 subscript 𝑡 2 𝑝 𝑞 2 𝑣 1\displaystyle=\sum_{v=1}^{\infty}\frac{1}{2v-1}\cdot\Big{[}\frac{t_{1}p}{N}% \cdot\Big{(}\frac{(t_{2}-t_{1})q}{t_{1}(p+q)}\Big{)}^{2v-1}+\frac{t_{2}q}{N}% \Big{(}\frac{(t_{1}-t_{2})p}{t_{2}(p+q)}\Big{)}^{2v-1}\Big{]}= ∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG 1 end_ARG start_ARG 2 italic_v - 1 end_ARG ⋅ [ divide start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_p end_ARG start_ARG italic_N end_ARG ⋅ ( divide start_ARG ( italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_q end_ARG start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ( italic_p + italic_q ) end_ARG ) start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT + divide start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_q end_ARG start_ARG italic_N end_ARG ( divide start_ARG ( italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_p end_ARG start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( italic_p + italic_q ) end_ARG ) start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT ]
−∑v=1∞1 2v⋅[t 1p N⋅((t 2−t 1)q t 1(p+q))2v+t 2q N((t 1−t 2)p t 2(p+q))2v].superscript subscript 𝑣 1⋅1 2 𝑣 delimited-[]⋅subscript 𝑡 1 𝑝 𝑁 superscript subscript 𝑡 2 subscript 𝑡 1 𝑞 subscript 𝑡 1 𝑝 𝑞 2 𝑣 subscript 𝑡 2 𝑞 𝑁 superscript subscript 𝑡 1 subscript 𝑡 2 𝑝 subscript 𝑡 2 𝑝 𝑞 2 𝑣\displaystyle\qquad-\sum_{v=1}^{\infty}\frac{1}{2v}\cdot\Big{[}\frac{t_{1}p}{N% }\cdot\Big{(}\frac{(t_{2}-t_{1})q}{t_{1}(p+q)}\Big{)}^{2v}+\frac{t_{2}q}{N}% \Big{(}\frac{(t_{1}-t_{2})p}{t_{2}(p+q)}\Big{)}^{2v}\Big{]}.- ∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG 1 end_ARG start_ARG 2 italic_v end_ARG ⋅ [ divide start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_p end_ARG start_ARG italic_N end_ARG ⋅ ( divide start_ARG ( italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_q end_ARG start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ( italic_p + italic_q ) end_ARG ) start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT + divide start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_q end_ARG start_ARG italic_N end_ARG ( divide start_ARG ( italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_p end_ARG start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( italic_p + italic_q ) end_ARG ) start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT ] .(13)
For ease of presentation, we denote m=q/p≥1 𝑚 𝑞 𝑝 1 m=q/p\geq 1 italic_m = italic_q / italic_p ≥ 1. Then for any v≥1 𝑣 1 v\geq 1 italic_v ≥ 1, we have
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∑v=1∞1 2v⋅[t 1p N⋅((t 2−t 1)q t 1(p+q))2v+t 2q N((t 1−t 2)p t 2(p+q))2v]superscript subscript 𝑣 1⋅1 2 𝑣 delimited-[]⋅subscript 𝑡 1 𝑝 𝑁 superscript subscript 𝑡 2 subscript 𝑡 1 𝑞 subscript 𝑡 1 𝑝 𝑞 2 𝑣 subscript 𝑡 2 𝑞 𝑁 superscript subscript 𝑡 1 subscript 𝑡 2 𝑝 subscript 𝑡 2 𝑝 𝑞 2 𝑣\displaystyle\sum_{v=1}^{\infty}\frac{1}{2v}\cdot\Big{[}\frac{t_{1}p}{N}\cdot% \Big{(}\frac{(t_{2}-t_{1})q}{t_{1}(p+q)}\Big{)}^{2v}+\frac{t_{2}q}{N}\Big{(}% \frac{(t_{1}-t_{2})p}{t_{2}(p+q)}\Big{)}^{2v}\Big{]}∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG 1 end_ARG start_ARG 2 italic_v end_ARG ⋅ [ divide start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_p end_ARG start_ARG italic_N end_ARG ⋅ ( divide start_ARG ( italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_q end_ARG start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ( italic_p + italic_q ) end_ARG ) start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT + divide start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_q end_ARG start_ARG italic_N end_ARG ( divide start_ARG ( italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_p end_ARG start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( italic_p + italic_q ) end_ARG ) start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT ]
=∑v=1∞pq(t 1−t 2)2v 2vN(p+q)2v⋅[q 2v−1 t 1 2v−1+p 2v−1 t 2 2v−1]absent superscript subscript 𝑣 1⋅𝑝 𝑞 superscript subscript 𝑡 1 subscript 𝑡 2 2 𝑣 2 𝑣 𝑁 superscript 𝑝 𝑞 2 𝑣 delimited-[]superscript 𝑞 2 𝑣 1 superscript subscript 𝑡 1 2 𝑣 1 superscript 𝑝 2 𝑣 1 superscript subscript 𝑡 2 2 𝑣 1\displaystyle\qquad=\sum_{v=1}^{\infty}\frac{pq(t_{1}-t_{2})^{2v}}{2vN(p+q)^{2% v}}\cdot\Big{[}\frac{q^{2v-1}}{t_{1}^{2v-1}}+\frac{p^{2v-1}}{t_{2}^{2v-1}}\Big% {]}= ∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG italic_p italic_q ( italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG start_ARG 2 italic_v italic_N ( italic_p + italic_q ) start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG ⋅ [ divide start_ARG italic_q start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT end_ARG start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT end_ARG + divide start_ARG italic_p start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT end_ARG start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT end_ARG ]
≤∑v=1∞mp 2(3/10)2v 2vNp 2v(1+m)2v⋅(1+m 2v−1)⋅(10p 7)2v−1 absent superscript subscript 𝑣 1⋅𝑚 superscript 𝑝 2 superscript 3 10 2 𝑣 2 𝑣 𝑁 superscript 𝑝 2 𝑣 superscript 1 𝑚 2 𝑣 1 superscript 𝑚 2 𝑣 1 superscript 10 𝑝 7 2 𝑣 1\displaystyle\qquad\leq\sum_{v=1}^{\infty}\frac{mp^{2}(3/10)^{2v}}{2vNp^{2v}(1% +m)^{2v}}\cdot(1+m^{2v-1})\cdot\Big{(}\frac{10p}{7}\Big{)}^{2v-1}≤ ∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG italic_m italic_p start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ( 3 / 10 ) start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG start_ARG 2 italic_v italic_N italic_p start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT ( 1 + italic_m ) start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG ⋅ ( 1 + italic_m start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT ) ⋅ ( divide start_ARG 10 italic_p end_ARG start_ARG 7 end_ARG ) start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT
≤3p 20N∑v=1∞1 v⋅(3 7)2v−1≤0.0716p N,absent 3 𝑝 20 𝑁 superscript subscript 𝑣 1⋅1 𝑣 superscript 3 7 2 𝑣 1 0.0716 𝑝 𝑁\displaystyle\qquad\leq\frac{3p}{20N}\sum_{v=1}^{\infty}\frac{1}{v}\cdot\Big{(% }\frac{3}{7}\Big{)}^{2v-1}\leq\frac{0.0716p}{N},≤ divide start_ARG 3 italic_p end_ARG start_ARG 20 italic_N end_ARG ∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG 1 end_ARG start_ARG italic_v end_ARG ⋅ ( divide start_ARG 3 end_ARG start_ARG 7 end_ARG ) start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT ≤ divide start_ARG 0.0716 italic_p end_ARG start_ARG italic_N end_ARG ,(14)
where the first inequality uses m=q/p 𝑚 𝑞 𝑝 m=q/p italic_m = italic_q / italic_p and the assumption that t 1,t 2∈[0.7,1]subscript 𝑡 1 subscript 𝑡 2 0.7 1 t_{1},t_{2}\in[0.7,1]italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∈ [ 0.7 , 1 ], the second inequality follows the fact that m(1+m 2v−1)≤(1+m)2v 𝑚 1 superscript 𝑚 2 𝑣 1 superscript 1 𝑚 2 𝑣 m(1+m^{2v-1})\leq(1+m)^{2v}italic_m ( 1 + italic_m start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT ) ≤ ( 1 + italic_m ) start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT, and the last inequality follows that
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∑v=1∞1 v⋅(3 7)2v−1≤3 7+∑v=2∞1 2⋅(3 7)2v−1=267 560 superscript subscript 𝑣 1⋅1 𝑣 superscript 3 7 2 𝑣 1 3 7 superscript subscript 𝑣 2⋅1 2 superscript 3 7 2 𝑣 1 267 560\displaystyle\sum_{v=1}^{\infty}\frac{1}{v}\cdot\Big{(}\frac{3}{7}\Big{)}^{2v-% 1}\leq\frac{3}{7}+\sum_{v=2}^{\infty}\frac{1}{2}\cdot\Big{(}\frac{3}{7}\Big{)}% ^{2v-1}=\frac{267}{560}∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG 1 end_ARG start_ARG italic_v end_ARG ⋅ ( divide start_ARG 3 end_ARG start_ARG 7 end_ARG ) start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT ≤ divide start_ARG 3 end_ARG start_ARG 7 end_ARG + ∑ start_POSTSUBSCRIPT italic_v = 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG 1 end_ARG start_ARG 2 end_ARG ⋅ ( divide start_ARG 3 end_ARG start_ARG 7 end_ARG ) start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT = divide start_ARG 267 end_ARG start_ARG 560 end_ARG
and simple calculations.
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On the other hand, for any v≥1 𝑣 1 v\geq 1 italic_v ≥ 1, we have
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∑v=1∞1 2v−1⋅[t 1p N⋅((t 2−t 1)q t 1(p+q))2v−1+t 2q N((t 1−t 2)p t 2(p+q))2v−1]superscript subscript 𝑣 1⋅1 2 𝑣 1 delimited-[]⋅subscript 𝑡 1 𝑝 𝑁 superscript subscript 𝑡 2 subscript 𝑡 1 𝑞 subscript 𝑡 1 𝑝 𝑞 2 𝑣 1 subscript 𝑡 2 𝑞 𝑁 superscript subscript 𝑡 1 subscript 𝑡 2 𝑝 subscript 𝑡 2 𝑝 𝑞 2 𝑣 1\displaystyle\sum_{v=1}^{\infty}\frac{1}{2v-1}\cdot\Big{[}\frac{t_{1}p}{N}% \cdot\Big{(}\frac{(t_{2}-t_{1})q}{t_{1}(p+q)}\Big{)}^{2v-1}+\frac{t_{2}q}{N}% \Big{(}\frac{(t_{1}-t_{2})p}{t_{2}(p+q)}\Big{)}^{2v-1}\Big{]}∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG 1 end_ARG start_ARG 2 italic_v - 1 end_ARG ⋅ [ divide start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT italic_p end_ARG start_ARG italic_N end_ARG ⋅ ( divide start_ARG ( italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_q end_ARG start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ( italic_p + italic_q ) end_ARG ) start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT + divide start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT italic_q end_ARG start_ARG italic_N end_ARG ( divide start_ARG ( italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_p end_ARG start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ( italic_p + italic_q ) end_ARG ) start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT ]
=∑v=1∞pq(t 1−t 2)2v−1(2v−1)N(p+q)2v−1⋅[p 2v−2 t 2 2v−2−q 2v−2 t 1 2v−2]absent superscript subscript 𝑣 1⋅𝑝 𝑞 superscript subscript 𝑡 1 subscript 𝑡 2 2 𝑣 1 2 𝑣 1 𝑁 superscript 𝑝 𝑞 2 𝑣 1 delimited-[]superscript 𝑝 2 𝑣 2 superscript subscript 𝑡 2 2 𝑣 2 superscript 𝑞 2 𝑣 2 superscript subscript 𝑡 1 2 𝑣 2\displaystyle\qquad=\sum_{v=1}^{\infty}\frac{pq(t_{1}-t_{2})^{2v-1}}{(2v-1)N(p% +q)^{2v-1}}\cdot\Big{[}\frac{p^{2v-2}}{t_{2}^{2v-2}}-\frac{q^{2v-2}}{t_{1}^{2v% -2}}\Big{]}= ∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG italic_p italic_q ( italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT end_ARG start_ARG ( 2 italic_v - 1 ) italic_N ( italic_p + italic_q ) start_POSTSUPERSCRIPT 2 italic_v - 1 end_POSTSUPERSCRIPT end_ARG ⋅ [ divide start_ARG italic_p start_POSTSUPERSCRIPT 2 italic_v - 2 end_POSTSUPERSCRIPT end_ARG start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 italic_v - 2 end_POSTSUPERSCRIPT end_ARG - divide start_ARG italic_q start_POSTSUPERSCRIPT 2 italic_v - 2 end_POSTSUPERSCRIPT end_ARG start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 italic_v - 2 end_POSTSUPERSCRIPT end_ARG ]
=∑v=1∞pq(t 1−t 2)2v+1(2v+1)N(p+q)2v+1⋅[p 2v t 2 2v−q 2v t 1 2v].absent superscript subscript 𝑣 1⋅𝑝 𝑞 superscript subscript 𝑡 1 subscript 𝑡 2 2 𝑣 1 2 𝑣 1 𝑁 superscript 𝑝 𝑞 2 𝑣 1 delimited-[]superscript 𝑝 2 𝑣 superscript subscript 𝑡 2 2 𝑣 superscript 𝑞 2 𝑣 superscript subscript 𝑡 1 2 𝑣\displaystyle\qquad=\sum_{v=1}^{\infty}\frac{pq(t_{1}-t_{2})^{2v+1}}{(2v+1)N(p% +q)^{2v+1}}\cdot\Big{[}\frac{p^{2v}}{t_{2}^{2v}}-\frac{q^{2v}}{t_{1}^{2v}}\Big% {]}.= ∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG italic_p italic_q ( italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 2 italic_v + 1 end_POSTSUPERSCRIPT end_ARG start_ARG ( 2 italic_v + 1 ) italic_N ( italic_p + italic_q ) start_POSTSUPERSCRIPT 2 italic_v + 1 end_POSTSUPERSCRIPT end_ARG ⋅ [ divide start_ARG italic_p start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG - divide start_ARG italic_q start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG ] .(15)
Furthermore, we have
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∑v=1∞pq(t 1−t 2)2v+1(2v+1)(p+q)2v+1⋅[p 2v t 2 2v−q 2v t 1 2v]superscript subscript 𝑣 1⋅𝑝 𝑞 superscript subscript 𝑡 1 subscript 𝑡 2 2 𝑣 1 2 𝑣 1 superscript 𝑝 𝑞 2 𝑣 1 delimited-[]superscript 𝑝 2 𝑣 superscript subscript 𝑡 2 2 𝑣 superscript 𝑞 2 𝑣 superscript subscript 𝑡 1 2 𝑣\displaystyle\sum_{v=1}^{\infty}\frac{pq(t_{1}-t_{2})^{2v+1}}{(2v+1)(p+q)^{2v+% 1}}\cdot\Big{[}\frac{p^{2v}}{t_{2}^{2v}}-\frac{q^{2v}}{t_{1}^{2v}}\Big{]}∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG italic_p italic_q ( italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT 2 italic_v + 1 end_POSTSUPERSCRIPT end_ARG start_ARG ( 2 italic_v + 1 ) ( italic_p + italic_q ) start_POSTSUPERSCRIPT 2 italic_v + 1 end_POSTSUPERSCRIPT end_ARG ⋅ [ divide start_ARG italic_p start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG - divide start_ARG italic_q start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG ]≥−∑v=1∞pq|t 1−t 2|2v+1(2v+1)N(p+q)2v+1⋅[p 2v t 2 2v+q 2v t 1 2v]absent superscript subscript 𝑣 1⋅𝑝 𝑞 superscript subscript 𝑡 1 subscript 𝑡 2 2 𝑣 1 2 𝑣 1 𝑁 superscript 𝑝 𝑞 2 𝑣 1 delimited-[]superscript 𝑝 2 𝑣 superscript subscript 𝑡 2 2 𝑣 superscript 𝑞 2 𝑣 superscript subscript 𝑡 1 2 𝑣\displaystyle\geq-\sum_{v=1}^{\infty}\frac{pq|t_{1}-t_{2}|^{2v+1}}{(2v+1)N(p+q% )^{2v+1}}\cdot\Big{[}\frac{p^{2v}}{t_{2}^{2v}}+\frac{q^{2v}}{t_{1}^{2v}}\Big{]}≥ - ∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG italic_p italic_q | italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT | start_POSTSUPERSCRIPT 2 italic_v + 1 end_POSTSUPERSCRIPT end_ARG start_ARG ( 2 italic_v + 1 ) italic_N ( italic_p + italic_q ) start_POSTSUPERSCRIPT 2 italic_v + 1 end_POSTSUPERSCRIPT end_ARG ⋅ [ divide start_ARG italic_p start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG start_ARG italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG + divide start_ARG italic_q start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT end_ARG ]
≥−p N∑v=1∞1 2v+1⋅(3 10)2v+1 absent 𝑝 𝑁 superscript subscript 𝑣 1⋅1 2 𝑣 1 superscript 3 10 2 𝑣 1\displaystyle\geq-\frac{p}{N}\sum_{v=1}^{\infty}\frac{1}{2v+1}\cdot\Big{(}% \frac{3}{10}\Big{)}^{2v+1}≥ - divide start_ARG italic_p end_ARG start_ARG italic_N end_ARG ∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG 1 end_ARG start_ARG 2 italic_v + 1 end_ARG ⋅ ( divide start_ARG 3 end_ARG start_ARG 10 end_ARG ) start_POSTSUPERSCRIPT 2 italic_v + 1 end_POSTSUPERSCRIPT
≥−0.0096p N,absent 0.0096 𝑝 𝑁\displaystyle\geq-\frac{0.0096p}{N},≥ - divide start_ARG 0.0096 italic_p end_ARG start_ARG italic_N end_ARG ,(16)
where the second inequality uses the facts that (p+q)2v+1≥q 2v+1+qp 2v superscript 𝑝 𝑞 2 𝑣 1 superscript 𝑞 2 𝑣 1 𝑞 superscript 𝑝 2 𝑣(p+q)^{2v+1}\geq q^{2v+1}+qp^{2v}( italic_p + italic_q ) start_POSTSUPERSCRIPT 2 italic_v + 1 end_POSTSUPERSCRIPT ≥ italic_q start_POSTSUPERSCRIPT 2 italic_v + 1 end_POSTSUPERSCRIPT + italic_q italic_p start_POSTSUPERSCRIPT 2 italic_v end_POSTSUPERSCRIPT and t 1,t 2∈[0.7,1]subscript 𝑡 1 subscript 𝑡 2 0.7 1 t_{1},t_{2}\in[0.7,1]italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∈ [ 0.7 , 1 ], and the last inequality follows that
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∑v=1∞1 2v+1(3 10)2v+1≤9 1000+1 5∑v=2∞(3 10)2v+1=9 1000+243 455000<0.0096.superscript subscript 𝑣 1 1 2 𝑣 1 superscript 3 10 2 𝑣 1 9 1000 1 5 superscript subscript 𝑣 2 superscript 3 10 2 𝑣 1 9 1000 243 455000 0.0096\displaystyle\sum_{v=1}^{\infty}\frac{1}{2v+1}\Big{(}\frac{3}{10}\Big{)}^{2v+1% }\leq\frac{9}{1000}+\frac{1}{5}\sum_{v=2}^{\infty}\Big{(}\frac{3}{10}\Big{)}^{% 2v+1}=\frac{9}{1000}+\frac{243}{455000}<0.0096.∑ start_POSTSUBSCRIPT italic_v = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT divide start_ARG 1 end_ARG start_ARG 2 italic_v + 1 end_ARG ( divide start_ARG 3 end_ARG start_ARG 10 end_ARG ) start_POSTSUPERSCRIPT 2 italic_v + 1 end_POSTSUPERSCRIPT ≤ divide start_ARG 9 end_ARG start_ARG 1000 end_ARG + divide start_ARG 1 end_ARG start_ARG 5 end_ARG ∑ start_POSTSUBSCRIPT italic_v = 2 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ∞ end_POSTSUPERSCRIPT ( divide start_ARG 3 end_ARG start_ARG 10 end_ARG ) start_POSTSUPERSCRIPT 2 italic_v + 1 end_POSTSUPERSCRIPT = divide start_ARG 9 end_ARG start_ARG 1000 end_ARG + divide start_ARG 243 end_ARG start_ARG 455000 end_ARG < 0.0096 .
Combining Eq.(A.1), Eq.(A.1), Eq.(A.1), and Eq.(A.1), we obtain that
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(IV)≥−0.0812p N.IV 0.0812 𝑝 𝑁\displaystyle\mathrm{(IV)}\geq-\frac{0.0812p}{N}.( roman_IV ) ≥ - divide start_ARG 0.0812 italic_p end_ARG start_ARG italic_N end_ARG .(17)
Term (V) in Eq.(12).
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For Term (V) in Eq.(12), we have
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(V)V\displaystyle\mathrm{(V)}( roman_V )=(1−t 1)p+(1−t 2)q Nlog(p+q)−(1−t 1)p Nlogp−(1−t 2)q Nlogq absent 1 subscript 𝑡 1 𝑝 1 subscript 𝑡 2 𝑞 𝑁 𝑝 𝑞 1 subscript 𝑡 1 𝑝 𝑁 𝑝 1 subscript 𝑡 2 𝑞 𝑁 𝑞\displaystyle=\frac{(1-t_{1})p+(1-t_{2})q}{N}\log(p+q)-\frac{(1-t_{1})p}{N}% \log p-\frac{(1-t_{2})q}{N}\log q= divide start_ARG ( 1 - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_p + ( 1 - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_q end_ARG start_ARG italic_N end_ARG roman_log ( italic_p + italic_q ) - divide start_ARG ( 1 - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_p end_ARG start_ARG italic_N end_ARG roman_log italic_p - divide start_ARG ( 1 - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_q end_ARG start_ARG italic_N end_ARG roman_log italic_q
=(1−t 1)p Nlogp+q p+(1−t 2)q Nlogp+q q.absent 1 subscript 𝑡 1 𝑝 𝑁 𝑝 𝑞 𝑝 1 subscript 𝑡 2 𝑞 𝑁 𝑝 𝑞 𝑞\displaystyle=\frac{(1-t_{1})p}{N}\log\frac{p+q}{p}+\frac{(1-t_{2})q}{N}\log% \frac{p+q}{q}.= divide start_ARG ( 1 - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_p end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_p end_ARG + divide start_ARG ( 1 - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_q end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_q end_ARG .(18)
Furthermore, we have
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(1−t 1)p Nlogp+q p+(1−t 2)q Nlogp+q q 1 subscript 𝑡 1 𝑝 𝑁 𝑝 𝑞 𝑝 1 subscript 𝑡 2 𝑞 𝑁 𝑝 𝑞 𝑞\displaystyle\frac{(1-t_{1})p}{N}\log\frac{p+q}{p}+\frac{(1-t_{2})q}{N}\log% \frac{p+q}{q}divide start_ARG ( 1 - italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) italic_p end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_p end_ARG + divide start_ARG ( 1 - italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) italic_q end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_q end_ARG(19)
≤(1−min{t 1,t 2})p Nlogp+q p+(1−min{t 1,t 2})q Nlogp+q q absent 1 subscript 𝑡 1 subscript 𝑡 2 𝑝 𝑁 𝑝 𝑞 𝑝 1 subscript 𝑡 1 subscript 𝑡 2 𝑞 𝑁 𝑝 𝑞 𝑞\displaystyle\qquad\leq\frac{(1-\min{t_{1},t_{2}})p}{N}\log\frac{p+q}{p}+% \frac{(1-\min{t_{1},t_{2}})q}{N}\log\frac{p+q}{q}≤ divide start_ARG ( 1 - roman_min { italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } ) italic_p end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_p end_ARG + divide start_ARG ( 1 - roman_min { italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } ) italic_q end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_q end_ARG
=(1−min{t 1,t 2})N⋅[plogp+q p+qlogp+q q].absent⋅1 subscript 𝑡 1 subscript 𝑡 2 𝑁 delimited-[]𝑝 𝑝 𝑞 𝑝 𝑞 𝑝 𝑞 𝑞\displaystyle\qquad=\frac{(1-\min{t_{1},t_{2}})}{N}\cdot\Big{[}p\log\frac{p+% q}{p}+q\log\frac{p+q}{q}\Big{]}.= divide start_ARG ( 1 - roman_min { italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } ) end_ARG start_ARG italic_N end_ARG ⋅ [ italic_p roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_p end_ARG + italic_q roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_q end_ARG ] .(20)
Let
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Δh=1 N⋅[plogp+q p+qlogp+q q].Δ ℎ⋅1 𝑁 delimited-[]𝑝 𝑝 𝑞 𝑝 𝑞 𝑝 𝑞 𝑞\displaystyle\Delta h=\frac{1}{N}\cdot\Big{[}p\log\frac{p+q}{p}+q\log\frac{p+q% }{q}\Big{]}.roman_Δ italic_h = divide start_ARG 1 end_ARG start_ARG italic_N end_ARG ⋅ [ italic_p roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_p end_ARG + italic_q roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_q end_ARG ] .(21)
Combining Eq.(A.1), Eq.(19), and Eq.(21), we obtain
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(V)≤(1−min{t 1,t 2})⋅Δh.V⋅1 subscript 𝑡 1 subscript 𝑡 2 Δ ℎ\displaystyle\mathrm{(V)}\leq(1-\min{t_{1},t_{2}})\cdot\Delta h.( roman_V ) ≤ ( 1 - roman_min { italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } ) ⋅ roman_Δ italic_h .(22)
Putting Together.
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When t 1,t 2≥0.7 subscript 𝑡 1 subscript 𝑡 2 0.7 t_{1},t_{2}\geq 0.7 italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ≥ 0.7, plugging Eq.(17) and Eq.(22) into Eq.(11) and Eq.(12), we have
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𝕀(Ω∗;C)≥𝕀(Ω;C)−0.0812p N−(1−min{t 1,t 2})⋅Δh.𝕀 superscript Ω 𝐶 𝕀 Ω 𝐶 0.0812 𝑝 𝑁⋅1 subscript 𝑡 1 subscript 𝑡 2 Δ ℎ\displaystyle\mathbb{I}(\Omega^{};C)\geq\mathbb{I}(\Omega;C)-\frac{0.0812p}{N% }-(1-\min{t_{1},t_{2}})\cdot\Delta h.blackboard_I ( roman_Ω start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ; italic_C ) ≥ blackboard_I ( roman_Ω ; italic_C ) - divide start_ARG 0.0812 italic_p end_ARG start_ARG italic_N end_ARG - ( 1 - roman_min { italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } ) ⋅ roman_Δ italic_h .(23)
Recall that the Δh Δ ℎ\Delta h roman_Δ italic_h defined in Eq.(21) takes the form
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Δh Δ ℎ\displaystyle\Delta h roman_Δ italic_h=1 N⋅[plogp+q p+qlogp+q q]absent⋅1 𝑁 delimited-[]𝑝 𝑝 𝑞 𝑝 𝑞 𝑝 𝑞 𝑞\displaystyle=\frac{1}{N}\cdot\Big{[}p\log\frac{p+q}{p}+q\log\frac{p+q}{q}\Big% {]}= divide start_ARG 1 end_ARG start_ARG italic_N end_ARG ⋅ [ italic_p roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_p end_ARG + italic_q roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_q end_ARG ]
=p N⋅(log(1+m)+mlog(1+1 m))absent⋅𝑝 𝑁 1 𝑚 𝑚 1 1 𝑚\displaystyle=\frac{p}{N}\cdot\Big{(}\log(1+m)+m\log\big{(}1+\frac{1}{m}\big{)% }\Big{)}= divide start_ARG italic_p end_ARG start_ARG italic_N end_ARG ⋅ ( roman_log ( 1 + italic_m ) + italic_m roman_log ( 1 + divide start_ARG 1 end_ARG start_ARG italic_m end_ARG ) )
≥2log2⋅p N,absent 2⋅2 𝑝 𝑁\displaystyle\geq 2\log 2\cdot\frac{p}{N},≥ 2 roman_log 2 ⋅ divide start_ARG italic_p end_ARG start_ARG italic_N end_ARG ,(24)
where the second equality uses m=q/p 𝑚 𝑞 𝑝 m=q/p italic_m = italic_q / italic_p, the last inequality uses m≥1 𝑚 1 m\geq 1 italic_m ≥ 1. Putting Eq.(23) and Eq.(A.1) together, we have
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𝕀(Ω∗;C)≥𝕀(Ω;C)−(1.0586−min{t 1,t 2})⋅Δh.𝕀 superscript Ω 𝐶 𝕀 Ω 𝐶⋅1.0586 subscript 𝑡 1 subscript 𝑡 2 Δ ℎ\displaystyle\mathbb{I}(\Omega^{};C)\geq\mathbb{I}(\Omega;C)-(1.0586-\min{t_% {1},t_{2}})\cdot\Delta h.blackboard_I ( roman_Ω start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ; italic_C ) ≥ blackboard_I ( roman_Ω ; italic_C ) - ( 1.0586 - roman_min { italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } ) ⋅ roman_Δ italic_h .(25)
Then, we calculate the entropy after fusion.
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ℍ(Ω∗)ℍ superscript Ω\displaystyle\mathbb{H}(\Omega^{})blackboard_H ( roman_Ω start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT )=ℍ(Ω)−p+q Nlogp+q N+p Nlogp N+q Nlogq N absent ℍ Ω 𝑝 𝑞 𝑁 𝑝 𝑞 𝑁 𝑝 𝑁 𝑝 𝑁 𝑞 𝑁 𝑞 𝑁\displaystyle=\mathbb{H}(\Omega)-\frac{p+q}{N}\log\frac{p+q}{N}+\frac{p}{N}% \log\frac{p}{N}+\frac{q}{N}\log\frac{q}{N}= blackboard_H ( roman_Ω ) - divide start_ARG italic_p + italic_q end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_p + italic_q end_ARG start_ARG italic_N end_ARG + divide start_ARG italic_p end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_p end_ARG start_ARG italic_N end_ARG + divide start_ARG italic_q end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_q end_ARG start_ARG italic_N end_ARG
=ℍ(Ω)−p+q Nlog(p+q)+p Nlogp+q Nlogq absent ℍ Ω 𝑝 𝑞 𝑁 𝑝 𝑞 𝑝 𝑁 𝑝 𝑞 𝑁 𝑞\displaystyle=\mathbb{H}(\Omega)-\frac{p+q}{N}\log(p+q)+\frac{p}{N}\log p+% \frac{q}{N}\log q= blackboard_H ( roman_Ω ) - divide start_ARG italic_p + italic_q end_ARG start_ARG italic_N end_ARG roman_log ( italic_p + italic_q ) + divide start_ARG italic_p end_ARG start_ARG italic_N end_ARG roman_log italic_p + divide start_ARG italic_q end_ARG start_ARG italic_N end_ARG roman_log italic_q
=ℍ(Ω)−Δh absent ℍ Ω Δ ℎ\displaystyle=\mathbb{H}(\Omega)-\Delta h= blackboard_H ( roman_Ω ) - roman_Δ italic_h(26)
Recall that
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n 1=2𝕀(Ω;C)ℍ(Ω)+ℍ(C),n 2=2𝕀(Ω∗;C)ℍ(Ω∗)+ℍ(C).formulae-sequence subscript 𝑛 1 2 𝕀 Ω 𝐶 ℍ Ω ℍ 𝐶 subscript 𝑛 2 2 𝕀 superscript Ω 𝐶 ℍ superscript Ω ℍ 𝐶\displaystyle n_{1}=\frac{2\mathbb{I}(\Omega;C)}{\mathbb{H}(\Omega)+\mathbb{H}% (C)},\quad n_{2}=\frac{2\mathbb{I}(\Omega^{};C)}{\mathbb{H}(\Omega^{})+% \mathbb{H}(C)}.italic_n start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT = divide start_ARG 2 blackboard_I ( roman_Ω ; italic_C ) end_ARG start_ARG blackboard_H ( roman_Ω ) + blackboard_H ( italic_C ) end_ARG , italic_n start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT = divide start_ARG 2 blackboard_I ( roman_Ω start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ; italic_C ) end_ARG start_ARG blackboard_H ( roman_Ω start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ) + blackboard_H ( italic_C ) end_ARG .
By Eq.(25) and Eq.(A.1), we know that the sufficient condition of n 2≥n 1 subscript 𝑛 2 subscript 𝑛 1 n_{2}\geq n_{1}italic_n start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ≥ italic_n start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT is
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2[𝕀(Ω;C)−(1.0586−min{t 1,t 2})⋅Δh]ℍ(Ω)+ℍ(C)−Δh≥2𝕀(Ω;C)ℍ(Ω)+ℍ(C),2 delimited-[]𝕀 Ω 𝐶⋅1.0586 subscript 𝑡 1 subscript 𝑡 2 Δ ℎ ℍ Ω ℍ 𝐶 Δ ℎ 2 𝕀 Ω 𝐶 ℍ Ω ℍ 𝐶\displaystyle\frac{2[\mathbb{I}(\Omega;C)-(1.0586-\min{t_{1},t_{2}})\cdot% \Delta h]}{\mathbb{H}(\Omega)+\mathbb{H}(C)-\Delta h}\geq\frac{2\mathbb{I}(% \Omega;C)}{\mathbb{H}(\Omega)+\mathbb{H}(C)},divide start_ARG 2 [ blackboard_I ( roman_Ω ; italic_C ) - ( 1.0586 - roman_min { italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } ) ⋅ roman_Δ italic_h ] end_ARG start_ARG blackboard_H ( roman_Ω ) + blackboard_H ( italic_C ) - roman_Δ italic_h end_ARG ≥ divide start_ARG 2 blackboard_I ( roman_Ω ; italic_C ) end_ARG start_ARG blackboard_H ( roman_Ω ) + blackboard_H ( italic_C ) end_ARG ,
which is equivalent to
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n 1=2𝕀(Ω;C)ℍ(Ω)+ℍ(C)≥2⋅(1.0586−min{t 1,t 2}),subscript 𝑛 1 2 𝕀 Ω 𝐶 ℍ Ω ℍ 𝐶⋅2 1.0586 subscript 𝑡 1 subscript 𝑡 2\displaystyle n_{1}=\frac{2\mathbb{I}(\Omega;C)}{\mathbb{H}(\Omega)+\mathbb{H}% (C)}\geq 2\cdot(1.0586-\min{t_{1},t_{2}}),italic_n start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT = divide start_ARG 2 blackboard_I ( roman_Ω ; italic_C ) end_ARG start_ARG blackboard_H ( roman_Ω ) + blackboard_H ( italic_C ) end_ARG ≥ 2 ⋅ ( 1.0586 - roman_min { italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } ) ,
which concludes the proof of Theorem2.5.
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∎
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A.2 Derivation of Eq.(2)
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Derivation of Eq.(2).
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We consider the queried result of “must-link” (c m=c n subscript 𝑐 𝑚 subscript 𝑐 𝑛 c_{m}=c_{n}italic_c start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT = italic_c start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) between w i subscript 𝑤 𝑖 w_{i}italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT and w j subscript 𝑤 𝑗 w_{j}italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT as a conditional event, which is expressed by {∀s∈w i,∀t∈w j,e st=1∣∀(s,t)∈w iorw j,e st=1}conditional-set formulae-sequence for-all 𝑠 subscript 𝑤 𝑖 formulae-sequence for-all 𝑡 subscript 𝑤 𝑗 subscript 𝑒 𝑠 𝑡 1 formulae-sequence for-all 𝑠 𝑡 subscript 𝑤 𝑖 or subscript 𝑤 𝑗 subscript 𝑒 𝑠 𝑡 1{\forall s\in w_{i},\forall t\in w_{j},e_{st}=1\mid\forall(s,t)\in w_{i}\text% { or }w_{j},e_{st}=1}{ ∀ italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , ∀ italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ∣ ∀ ( italic_s , italic_t ) ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT or italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 }. Here the condition implies that samples within a cluster are assigned to the same class. Further, we follow the setup in probabilistic clustering(Lu & Leen, 2004; Liu et al., 2022) that models the distribution of clustering based on the pairwise probability. Then, the joint probability density of a clustering π=[z 1,z 2,⋯,z m\pi=[z_{1},z_{2},\cdots,z_{m}italic_π = [ italic_z start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_z start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , ⋯ , italic_z start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT] for m 𝑚 m italic_m samples is expressed as ℙ(π)=1 α∏s,t∈[1,2,⋯,m]ℙ(e st=1)I(z s=z t)×ℙ(e st=0)I(z s≠z t)ℙ 𝜋 1 𝛼 subscript product 𝑠 𝑡 1 2⋯𝑚 ℙ superscript subscript 𝑒 𝑠 𝑡 1 𝐼 subscript 𝑧 𝑠 subscript 𝑧 𝑡 ℙ superscript subscript 𝑒 𝑠 𝑡 0 𝐼 subscript 𝑧 𝑠 subscript 𝑧 𝑡\mathbb{P}(\pi)=\frac{1}{\alpha}\prod_{s,t\in[1,2,\cdots,m]}\mathbb{P}(e_{st}=% 1)^{I(z_{s}=z_{t})}\times\mathbb{P}(e_{st}=0)^{I(z_{s}\neq z_{t})}blackboard_P ( italic_π ) = divide start_ARG 1 end_ARG start_ARG italic_α end_ARG ∏ start_POSTSUBSCRIPT italic_s , italic_t ∈ [ 1 , 2 , ⋯ , italic_m ] end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) start_POSTSUPERSCRIPT italic_I ( italic_z start_POSTSUBSCRIPT italic_s end_POSTSUBSCRIPT = italic_z start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT ) end_POSTSUPERSCRIPT × blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 0 ) start_POSTSUPERSCRIPT italic_I ( italic_z start_POSTSUBSCRIPT italic_s end_POSTSUBSCRIPT ≠ italic_z start_POSTSUBSCRIPT italic_t end_POSTSUBSCRIPT ) end_POSTSUPERSCRIPT, where I(⋅)𝐼⋅I(\cdot)italic_I ( ⋅ ) is the indicator function, and α 𝛼\alpha italic_α is the normalization factor.
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Under the condition of ∀(s,t)∈w iorw j,e st=1 formulae-sequence for-all 𝑠 𝑡 subscript 𝑤 𝑖 or subscript 𝑤 𝑗 subscript 𝑒 𝑠 𝑡 1\forall(s,t)\in w_{i}\text{ or }w_{j},e_{st}=1∀ ( italic_s , italic_t ) ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT or italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1, the two events {∀s∈w i,∀t∈w j,e st=1}formulae-sequence for-all 𝑠 subscript 𝑤 𝑖 formulae-sequence for-all 𝑡 subscript 𝑤 𝑗 subscript 𝑒 𝑠 𝑡 1{\forall s\in w_{i},\forall t\in w_{j},e_{st}=1}{ ∀ italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , ∀ italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 } and {∀s∈w i,∀t∈w j,e st=0}formulae-sequence for-all 𝑠 subscript 𝑤 𝑖 formulae-sequence for-all 𝑡 subscript 𝑤 𝑗 subscript 𝑒 𝑠 𝑡 0{\forall s\in w_{i},\forall t\in w_{j},e_{st}=0}{ ∀ italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , ∀ italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 0 } are mutually exclusive. And we denote them as y(w i)=y(w j)𝑦 subscript 𝑤 𝑖 𝑦 subscript 𝑤 𝑗 y(w_{i})=y(w_{j})italic_y ( italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ) = italic_y ( italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) and y(w i)≠y(w j)𝑦 subscript 𝑤 𝑖 𝑦 subscript 𝑤 𝑗 y(w_{i})\neq y(w_{j})italic_y ( italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ) ≠ italic_y ( italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) for simplicity. Therefore, by the formula of conditional probability, we can obtain:
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ℙ(c m=c n)ℙ subscript 𝑐 𝑚 subscript 𝑐 𝑛\displaystyle\mathbb{P}(c_{m}=c_{n})blackboard_P ( italic_c start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT = italic_c start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT )=ℙ(∀s∈w i,∀t∈w j,e st=1∣∀(s,t)∈w i or w j,e st=1)\displaystyle=\mathbb{P}(\forall s\in w_{i},\forall t\in w_{j},e_{st}=1\mid% \forall(s,t)\in w_{i}\text{ or }w_{j},e_{st}=1)= blackboard_P ( ∀ italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , ∀ italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ∣ ∀ ( italic_s , italic_t ) ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT or italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 )
=ℙ(y(w i)=y(w j),∀(s,t)∈w iorw j,e st=1)ℙ(∀(s,t)∈w i or w j,e st=1))\displaystyle=\frac{\mathbb{P}(y(w_{i})=y(w_{j}),\forall(s,t)\in w_{i}\text{ % or }w_{j},e_{st}=1)}{\mathbb{P}(\forall(s,t)\in w_{i}\text{ or }w_{j},e_{st}=1% ))}= divide start_ARG blackboard_P ( italic_y ( italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ) = italic_y ( italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) , ∀ ( italic_s , italic_t ) ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT or italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) end_ARG start_ARG blackboard_P ( ∀ ( italic_s , italic_t ) ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT or italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) ) end_ARG
=ℙ(y(w i)=y(w j),∀(s,t)∈w i or w j,e st=1))ℙ(y(w i)=y(w j),∀(s,t)∈w i or w j,e st=1))+ℙ(y(w i)≠y(w j),∀(s,t)∈w i or w j,e st=1))\displaystyle=\frac{\mathbb{P}(y(w_{i})=y(w_{j}),\forall(s,t)\in w_{i}\text{ % or }w_{j},e_{st}=1))}{\mathbb{P}(y(w_{i})=y(w_{j}),\forall(s,t)\in w_{i}\text{% or }w_{j},e_{st}=1))+\mathbb{P}(y(w_{i})\neq y(w_{j}),\forall(s,t)\in w_{i}% \text{ or }w_{j},e_{st}=1))}= divide start_ARG blackboard_P ( italic_y ( italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ) = italic_y ( italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) , ∀ ( italic_s , italic_t ) ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT or italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) ) end_ARG start_ARG blackboard_P ( italic_y ( italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ) = italic_y ( italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) , ∀ ( italic_s , italic_t ) ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT or italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) ) + blackboard_P ( italic_y ( italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ) ≠ italic_y ( italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) , ∀ ( italic_s , italic_t ) ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT or italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) ) end_ARG
=1 A∏s∈w i,t∈w j ℙ(e st=1)∏s,t∈w i,s,t∈w j ℙ(e st=1)1 A[∏s∈w i,t∈w j ℙ(e st=1)+∏s∈w i,t∈w j ℙ(e st=0)]∏s,t∈w i,s,t∈w j ℙ(e st=1)absent 1 𝐴 subscript product formulae-sequence 𝑠 subscript 𝑤 𝑖 𝑡 subscript 𝑤 𝑗 ℙ subscript 𝑒 𝑠 𝑡 1 subscript product formulae-sequence 𝑠 𝑡 subscript 𝑤 𝑖 𝑠 𝑡 subscript 𝑤 𝑗 ℙ subscript 𝑒 𝑠 𝑡 1 1 𝐴 delimited-[]subscript product formulae-sequence 𝑠 subscript 𝑤 𝑖 𝑡 subscript 𝑤 𝑗 ℙ subscript 𝑒 𝑠 𝑡 1 subscript product formulae-sequence 𝑠 subscript 𝑤 𝑖 𝑡 subscript 𝑤 𝑗 ℙ subscript 𝑒 𝑠 𝑡 0 subscript product formulae-sequence 𝑠 𝑡 subscript 𝑤 𝑖 𝑠 𝑡 subscript 𝑤 𝑗 ℙ subscript 𝑒 𝑠 𝑡 1\displaystyle=\frac{\frac{1}{A}\prod_{s\in w_{i},t\in w_{j}}\mathbb{P}(e_{st}=% 1)\prod_{s,t\in w_{i},s,t\in w_{j}}\mathbb{P}(e_{st}=1)}{\frac{1}{A}[\prod_{s% \in w_{i},t\in w_{j}}\mathbb{P}(e_{st}=1)+\prod_{s\in w_{i},t\in w_{j}}\mathbb% {P}(e_{st}=0)]\prod_{s,t\in w_{i},s,t\in w_{j}}\mathbb{P}(e_{st}=1)}= divide start_ARG divide start_ARG 1 end_ARG start_ARG italic_A end_ARG ∏ start_POSTSUBSCRIPT italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) ∏ start_POSTSUBSCRIPT italic_s , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_s , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) end_ARG start_ARG divide start_ARG 1 end_ARG start_ARG italic_A end_ARG [ ∏ start_POSTSUBSCRIPT italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) + ∏ start_POSTSUBSCRIPT italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 0 ) ] ∏ start_POSTSUBSCRIPT italic_s , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_s , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) end_ARG
=∏s∈w i,t∈w j ℙ(e st=1)∏s∈w i,t∈w j ℙ(e st=1)+∏s∈w i,t∈w j ℙ(e st=0),absent subscript product formulae-sequence 𝑠 subscript 𝑤 𝑖 𝑡 subscript 𝑤 𝑗 ℙ subscript 𝑒 𝑠 𝑡 1 subscript product formulae-sequence 𝑠 subscript 𝑤 𝑖 𝑡 subscript 𝑤 𝑗 ℙ subscript 𝑒 𝑠 𝑡 1 subscript product formulae-sequence 𝑠 subscript 𝑤 𝑖 𝑡 subscript 𝑤 𝑗 ℙ subscript 𝑒 𝑠 𝑡 0\displaystyle=\frac{\prod_{s\in w_{i},t\in w_{j}}\mathbb{P}(e_{st}=1)}{\prod_{% s\in w_{i},t\in w_{j}}\mathbb{P}(e_{st}=1)+\prod_{s\in w_{i},t\in w_{j}}% \mathbb{P}(e_{st}=0)},= divide start_ARG ∏ start_POSTSUBSCRIPT italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) end_ARG start_ARG ∏ start_POSTSUBSCRIPT italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) + ∏ start_POSTSUBSCRIPT italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_t ∈ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 0 ) end_ARG ,
where A 𝐴 A italic_A is the normalization factor of the joint probability density for samples in {w i,w j}subscript 𝑤 𝑖 subscript 𝑤 𝑗{w_{i},w_{j}}{ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT }. ∎
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Remark on the condition. The assignment of classes for samples is subject to clustering. For instance, to categorize a group of sheep, various attributes like gender, age, health, and weight can be the basis for clustering. In our scenario to measure the likelihood of “must-link”, what we indeed care about is the probability that the two clusters should be merged during the clustering, rather than the actual classes of these samples (this is the mission of oracles). After all, each class assignment corresponds to a specific meaning in real applications.
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Remark on the aggregation probability. Previous studies often evaluate the likelihood of c m=c n subscript 𝑐 𝑚 subscript 𝑐 𝑛 c_{m}=c_{n}italic_c start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT = italic_c start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT with heuristic strategies. Common strategies include using the distance between the central samples of two clusters or the closest distance between the clusters themselves to determine which cluster pairs to query. However, these methods that focus on a single sample from a cluster, may become less effective for clusters with irregular shapes, and Eq. (2) provides a better estimation that takes into account the influence of all samples.
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Remark on the query style. A3S requires oracles to provide pairwise comparison results for each selected sample pair, which demands minimal domain-specific knowledge and is easier to implement (Xiong et al., 2016), especially when the number of classes is large. In contrast, traditional active learning necessitates that the oracle assigns specific labels to the selected samples or annotates them according to predefined rules (Deng et al., 2023a), causing much heavier costs during the annotation process.
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A.3 Justification of Approximation
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Regarding the Approximation of Entropy. If (p+q)≪N much-less-than 𝑝 𝑞 𝑁(p+q)\ll N( italic_p + italic_q ) ≪ italic_N, then
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Δh Δ ℎ\displaystyle\Delta h roman_Δ italic_h=p+q Nlog(p+q)−p Nlogp−q Nlogq absent 𝑝 𝑞 𝑁 𝑝 𝑞 𝑝 𝑁 𝑝 𝑞 𝑁 𝑞\displaystyle=\frac{p+q}{N}\log(p+q)-\frac{p}{N}\log p-\frac{q}{N}\log q= divide start_ARG italic_p + italic_q end_ARG start_ARG italic_N end_ARG roman_log ( italic_p + italic_q ) - divide start_ARG italic_p end_ARG start_ARG italic_N end_ARG roman_log italic_p - divide start_ARG italic_q end_ARG start_ARG italic_N end_ARG roman_log italic_q
<p+q Nlog(p+q)absent 𝑝 𝑞 𝑁 𝑝 𝑞\displaystyle<\frac{p+q}{N}\log(p+q)< divide start_ARG italic_p + italic_q end_ARG start_ARG italic_N end_ARG roman_log ( italic_p + italic_q )
<p+q NlogN p+q.absent 𝑝 𝑞 𝑁 𝑁 𝑝 𝑞\displaystyle<\frac{p+q}{N}\log\frac{N}{p+q}.< divide start_ARG italic_p + italic_q end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N end_ARG start_ARG italic_p + italic_q end_ARG .
Note that ℍ(Ω)=∑s s NlogN s ℍ Ω subscript 𝑠 𝑠 𝑁 𝑁 𝑠\mathbb{H}(\Omega)=\sum_{s}\frac{s}{N}\log\frac{N}{s}blackboard_H ( roman_Ω ) = ∑ start_POSTSUBSCRIPT italic_s end_POSTSUBSCRIPT divide start_ARG italic_s end_ARG start_ARG italic_N end_ARG roman_log divide start_ARG italic_N end_ARG start_ARG italic_s end_ARG, where s 𝑠 s italic_s is the cluster size like p 𝑝 p italic_p and q 𝑞 q italic_q. Together with p+q≪N much-less-than 𝑝 𝑞 𝑁 p+q\ll N italic_p + italic_q ≪ italic_N, we have
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Δh≪ℍ(Ω)<ℍ(Ω)+ℍ(C).much-less-than Δ ℎ ℍ Ω ℍ Ω ℍ 𝐶\displaystyle\Delta h\ll\mathbb{H}(\Omega)<\mathbb{H}(\Omega)+\mathbb{H}(C).roman_Δ italic_h ≪ blackboard_H ( roman_Ω ) < blackboard_H ( roman_Ω ) + blackboard_H ( italic_C ) .
Regarding the Mutual Information. If the purity of w 1 subscript 𝑤 1 w_{1}italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT and w 2 subscript 𝑤 2 w_{2}italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT is 1, and they belong to the same class c τ subscript 𝑐 𝜏 c_{\tau}italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT, then we have
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ℙ(w 1∩c τ)=ℙ(w 1),ℙ(w 2∩c τ)=ℙ(w 2),ℙ(w 1,2∩c τ)=ℙ(w 1,2)=ℙ(w 1)+ℙ(w 2).formulae-sequence ℙ subscript 𝑤 1 subscript 𝑐 𝜏 ℙ subscript 𝑤 1 formulae-sequence ℙ subscript 𝑤 2 subscript 𝑐 𝜏 ℙ subscript 𝑤 2 ℙ subscript 𝑤 1 2 subscript 𝑐 𝜏 ℙ subscript 𝑤 1 2 ℙ subscript 𝑤 1 ℙ subscript 𝑤 2\displaystyle\mathbb{P}(w_{1}\cap c_{\tau})=\mathbb{P}(w_{1}),\quad\mathbb{P}(% w_{2}\cap c_{\tau})=\mathbb{P}(w_{2}),\quad\mathbb{P}(w_{1,2}\cap c_{\tau})=% \mathbb{P}(w_{1,2})=\mathbb{P}(w_{1})+\mathbb{P}(w_{2}).blackboard_P ( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) = blackboard_P ( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) , blackboard_P ( italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) = blackboard_P ( italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) , blackboard_P ( italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) = blackboard_P ( italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ) = blackboard_P ( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) + blackboard_P ( italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) .
Hence, and we have 𝕀(Ω∗;C)=𝕀(Ω;C)𝕀 superscript Ω 𝐶 𝕀 Ω 𝐶\mathbb{I}(\Omega^{};C)=\mathbb{I}(\Omega;C)blackboard_I ( roman_Ω start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ; italic_C ) = blackboard_I ( roman_Ω ; italic_C ). This is because
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𝕀(Ω∗;C)𝕀 superscript Ω 𝐶\displaystyle\mathbb{I}(\Omega^{};C)blackboard_I ( roman_Ω start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ; italic_C )=𝕀(Ω;C)+ℙ(w 1,2∩c τ)logℙ(w 1,2∩c τ)ℙ(w 1,2)ℙ(c τ)absent 𝕀 Ω 𝐶 ℙ subscript 𝑤 1 2 subscript 𝑐 𝜏 ℙ subscript 𝑤 1 2 subscript 𝑐 𝜏 ℙ subscript 𝑤 1 2 ℙ subscript 𝑐 𝜏\displaystyle=\mathbb{I}(\Omega;C)+\mathbb{P}(w_{1,2}\cap c_{\tau})\log\frac{% \mathbb{P}(w_{1,2}\cap c_{\tau})}{\mathbb{P}(w_{1,2})\mathbb{P}(c_{\tau})}= blackboard_I ( roman_Ω ; italic_C ) + blackboard_P ( italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) roman_log divide start_ARG blackboard_P ( italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG start_ARG blackboard_P ( italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ) blackboard_P ( italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG
−ℙ(w 1∩c τ)logℙ(w 1∩c τ)ℙ(w 1)ℙ(c τ)−ℙ(w 2∩c τ)logℙ(w 2∩c τ)ℙ(w 2)ℙ(c τ)ℙ subscript 𝑤 1 subscript 𝑐 𝜏 ℙ subscript 𝑤 1 subscript 𝑐 𝜏 ℙ subscript 𝑤 1 ℙ subscript 𝑐 𝜏 ℙ subscript 𝑤 2 subscript 𝑐 𝜏 ℙ subscript 𝑤 2 subscript 𝑐 𝜏 ℙ subscript 𝑤 2 ℙ subscript 𝑐 𝜏\displaystyle\qquad-\mathbb{P}(w_{1}\cap c_{\tau})\log\frac{\mathbb{P}(w_{1}% \cap c_{\tau})}{\mathbb{P}(w_{1})\mathbb{P}(c_{\tau})}-\mathbb{P}(w_{2}\cap c_% {\tau})\log\frac{\mathbb{P}(w_{2}\cap c_{\tau})}{\mathbb{P}(w_{2})\mathbb{P}(c% {\tau})}- blackboard_P ( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) roman_log divide start_ARG blackboard_P ( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG start_ARG blackboard_P ( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) blackboard_P ( italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG - blackboard_P ( italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) roman_log divide start_ARG blackboard_P ( italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ∩ italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG start_ARG blackboard_P ( italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) blackboard_P ( italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG
=𝕀(Ω;C)+ℙ(w 1,2)log1 ℙ(c τ)−ℙ(w 1)log1 ℙ(c τ)−ℙ(w 2)log1 ℙ(c τ)absent 𝕀 Ω 𝐶 ℙ subscript 𝑤 1 2 1 ℙ subscript 𝑐 𝜏 ℙ subscript 𝑤 1 1 ℙ subscript 𝑐 𝜏 ℙ subscript 𝑤 2 1 ℙ subscript 𝑐 𝜏\displaystyle=\mathbb{I}(\Omega;C)+\mathbb{P}(w{1,2})\log\frac{1}{\mathbb{P}(% c_{\tau})}-\mathbb{P}(w_{1})\log\frac{1}{\mathbb{P}(c_{\tau})}-\mathbb{P}(w_{2% })\log\frac{1}{\mathbb{P}(c_{\tau})}= blackboard_I ( roman_Ω ; italic_C ) + blackboard_P ( italic_w start_POSTSUBSCRIPT 1 , 2 end_POSTSUBSCRIPT ) roman_log divide start_ARG 1 end_ARG start_ARG blackboard_P ( italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG - blackboard_P ( italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ) roman_log divide start_ARG 1 end_ARG start_ARG blackboard_P ( italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG - blackboard_P ( italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) roman_log divide start_ARG 1 end_ARG start_ARG blackboard_P ( italic_c start_POSTSUBSCRIPT italic_τ end_POSTSUBSCRIPT ) end_ARG
=𝕀(Ω;C).absent 𝕀 Ω 𝐶\displaystyle=\mathbb{I}(\Omega;C).= blackboard_I ( roman_Ω ; italic_C ) .
When the purity is less than 1.0, considering Eq.(25), and fact that Δh≪min{ℍ(Ω),ℍ(C)}much-less-than Δ ℎ ℍ Ω ℍ 𝐶\Delta h\ll\min{\mathbb{H}(\Omega),\mathbb{H}(C)}roman_Δ italic_h ≪ roman_min { blackboard_H ( roman_Ω ) , blackboard_H ( italic_C ) }, we have
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|𝕀(Ω∗;C)−𝕀(Ω;C)|<(1.0586−min{t 1,t 2})⋅Δh≪min{ℍ(Ω),ℍ(C)}.𝕀 superscript Ω 𝐶 𝕀 Ω 𝐶⋅1.0586 subscript 𝑡 1 subscript 𝑡 2 Δ ℎ much-less-than ℍ Ω ℍ 𝐶\displaystyle|\mathbb{I}(\Omega^{};C)-\mathbb{I}(\Omega;C)|<(1.0586-\min{t_{% 1},t_{2}})\cdot\Delta h\ll\min{\mathbb{H}(\Omega),\mathbb{H}(C)}.| blackboard_I ( roman_Ω start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ; italic_C ) - blackboard_I ( roman_Ω ; italic_C ) | < ( 1.0586 - roman_min { italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } ) ⋅ roman_Δ italic_h ≪ roman_min { blackboard_H ( roman_Ω ) , blackboard_H ( italic_C ) } .(27)
A.4 Fast Transitive Inference
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FTI involves expanding the set of constraints based on the information within the original set. For example, if (x 1,x 2)subscript 𝑥 1 subscript 𝑥 2(x_{1},x_{2})( italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ) and (x 2,x 3)subscript 𝑥 2 subscript 𝑥 3(x_{2},x_{3})( italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ) are must-link constraints, and (x 1,x 4)subscript 𝑥 1 subscript 𝑥 4(x_{1},x_{4})( italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT ) is a cannot-link constraint, it implies that (x 1,x 3)subscript 𝑥 1 subscript 𝑥 3(x_{1},x_{3})( italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT ) must be a must-link constraint, while (x 2,x 4)subscript 𝑥 2 subscript 𝑥 4(x_{2},x_{4})( italic_x start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT ) and (x 3,x 4)subscript 𝑥 3 subscript 𝑥 4(x_{3},x_{4})( italic_x start_POSTSUBSCRIPT 3 end_POSTSUBSCRIPT , italic_x start_POSTSUBSCRIPT 4 end_POSTSUBSCRIPT ) must be cannot-link constraints.
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To facilitate the process, we present an efficient method, Fast Transitive Inference (FTI), which is designed to discover the transitive closure for the constraint set in A3S. The implementation is shown in Algorithm3.
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The performance of FTI is guaranteed by TheoremA.1.
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Theorem A.1(Completeness of FTI).
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By executing the FTI algorithm every time a new human query is made, we can always get the latest transitive closure.
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Proof of Theorem A.1.
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Suppose the must-link sample sets with samples i 𝑖 i italic_i and j 𝑗 j italic_j are denoted as G i subscript 𝐺 𝑖 G_{i}italic_G start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT and G j subscript 𝐺 𝑗 G_{j}italic_G start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT, respectively. And the sample sets that are cannot-link with i 𝑖 i italic_i and j 𝑗 j italic_j are denoted as g i subscript 𝑔 𝑖 g_{i}italic_g start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT and g j subscript 𝑔 𝑗 g_{j}italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. When the constraints between i 𝑖 i italic_i and j 𝑗 j italic_j are queried, only the constraints of sample pairs within {G i,G j,g i,g j}subscript 𝐺 𝑖 subscript 𝐺 𝑗 subscript 𝑔 𝑖 subscript 𝑔 𝑗{G_{i},G_{j},g_{i},g_{j}}{ italic_G start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_G start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_g start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT } may change, as no must-link constraints are built between them and the rest samples. We discuss the two cases where i 𝑖 i italic_i and j 𝑗 j italic_j are must-linked or cannot-linked:
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1(i,j)𝑖 𝑗(i,j)( italic_i , italic_j ) is must-linked. First, FTI updates the constraints for sample pairs related to i 𝑖 i italic_i, then G i′=G i∪{j}superscript subscript 𝐺 𝑖′subscript 𝐺 𝑖 𝑗 G_{i}^{\prime}=G_{i}\cup{j}italic_G start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_G start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ∪ { italic_j }, G j′=G j∪G i superscript subscript 𝐺 𝑗′subscript 𝐺 𝑗 subscript 𝐺 𝑖 G_{j}^{\prime}=G_{j}\cup G_{i}italic_G start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_G start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∪ italic_G start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT, and g j′=g j∪g i superscript subscript 𝑔 𝑗′subscript 𝑔 𝑗 subscript 𝑔 𝑖 g_{j}^{\prime}=g_{j}\cup g_{i}italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∪ italic_g start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT. The constraints between i 𝑖 i italic_i and G j,g j subscript 𝐺 𝑗 subscript 𝑔 𝑗 G_{j},g_{j}italic_G start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT have not yet been updated. Then, FTI updates the constraints for sample pairs related to j 𝑗 j italic_j, then we have G i′=G i∪G j superscript subscript 𝐺 𝑖′subscript 𝐺 𝑖 subscript 𝐺 𝑗 G_{i}^{\prime}=G_{i}\cup G_{j}italic_G start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_G start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ∪ italic_G start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT and g i′=g i∪g j superscript subscript 𝑔 𝑖′subscript 𝑔 𝑖 subscript 𝑔 𝑗 g_{i}^{\prime}=g_{i}\cup g_{j}italic_g start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_g start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ∪ italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. And this means all ml constraints between G i subscript 𝐺 𝑖 G_{i}italic_G start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT and G j subscript 𝐺 𝑗 G_{j}italic_G start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT, and cl constraints between {G i,G j}subscript 𝐺 𝑖 subscript 𝐺 𝑗{G_{i},G_{j}}{ italic_G start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_G start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT } and {g i,g j}subscript 𝑔 𝑖 subscript 𝑔 𝑗{g_{i},g_{j}}{ italic_g start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT } are updated and stored in the state matrix S 𝑆 S italic_S.
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2(i,j)𝑖 𝑗(i,j)( italic_i , italic_j ) is cannot-linked. First, FTI updates the constraints for sample pairs related to i 𝑖 i italic_i, then g i′=g i∪{j}superscript subscript 𝑔 𝑖′subscript 𝑔 𝑖 𝑗 g_{i}^{\prime}=g_{i}\cup{j}italic_g start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_g start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ∪ { italic_j } and g j′=g j∪G i superscript subscript 𝑔 𝑗′subscript 𝑔 𝑗 subscript 𝐺 𝑖 g_{j}^{\prime}=g_{j}\cup G_{i}italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∪ italic_G start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT. Then FTI updates the constraints for sample pairs related to j 𝑗 j italic_j, then we have g i′=g i∪G j superscript subscript 𝑔 𝑖′subscript 𝑔 𝑖 subscript 𝐺 𝑗 g_{i}^{\prime}=g_{i}\cup G_{j}italic_g start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT start_POSTSUPERSCRIPT ′ end_POSTSUPERSCRIPT = italic_g start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ∪ italic_G start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. And this means all cl constraints between G i subscript 𝐺 𝑖 G_{i}italic_G start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT and G j subscript 𝐺 𝑗 G_{j}italic_G start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT are updated and stored in S 𝑆 S italic_S.
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Combining these two scenarios, we finish the proof of Theorem A.1 ∎
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Appendix B Implementation Details of A3S
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B.1 Estimating Pairwise Probability
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Following the setup in (Liu et al., 2022), we use isotonic regression to learn a regressor that estimates the pairwise posterior probability ℙ(e ij=1|d ij)ℙ subscript 𝑒 𝑖 𝑗 conditional 1 subscript 𝑑 𝑖 𝑗\mathbb{P}(e_{ij}=1|d_{ij})blackboard_P ( italic_e start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT = 1 | italic_d start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT ), where d ij subscript 𝑑 𝑖 𝑗 d_{ij}italic_d start_POSTSUBSCRIPT italic_i italic_j end_POSTSUBSCRIPT is the Euclidean distance between samples i 𝑖 i italic_i and j 𝑗 j italic_j. The estimation encompasses three steps:
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(1)Since ground truth labels are unavailable, we employ K-means clustering to generate pseudo labels for the samples. Alternatively, Fast Probabilistic Clustering (FPC) can also be used for this purpose, where similarity values between samples serve as a rudimentary approximation of pairwise probabilities. In practical applications, cosine similarity is particularly well-suited for FPC.
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(2)Next, we generate the training data for isotonic regression. We utilize the k-nearest neighbors of each sample to form sample pairs. The independent variable of isotonic regression is the Euclidean distance between two samples in the sample pair. Each pair is labeled as 0 or 1, indicating whether the two samples in each pair share the same pseudo label. The label serves as the dependent variable of isotonic regression.
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(3)Finally, we conduct isotonic regression on the gathered data, learning a function that maps the Euclidean distance between two samples to the pairwise probability, i.e., ℙ(e st=1|d st)ℙ subscript 𝑒 𝑠 𝑡 conditional 1 subscript 𝑑 𝑠 𝑡\mathbb{P}(e_{st}=1|d_{st})blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 | italic_d start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT ).
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Liu et al. (2022) also proposed Graph-context-aware refinement to enhance the quality of the posterior probability, but it is not an essential component of A3S. Therefore, we did not include it in our experiments. However, incorporating them would further enhance the performance of A3S, as they can improve the quality of the estimated merging probability between cluster pairs.
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B.2 Calculating Aggregation Probability
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We derive the aggregation probability based on the condition that the purity of two clusters are 1.0. In practical applications, the assumption that both clusters have a purity of 1.0 might not strictly hold. Specifically, the pairwise probability between major samples and outlier samples from w i subscript 𝑤 𝑖 w_{i}italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT and w j subscript 𝑤 𝑗 w_{j}italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT will degrade the aggregation probability (i.e., making the result biased towards 0). To deal with this issue, we propose a variant of ℙ(c m=c n)ℙ subscript 𝑐 𝑚 subscript 𝑐 𝑛\mathbb{P}(c_{m}=c_{n})blackboard_P ( italic_c start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT = italic_c start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ), denoted as ℙ knn(c m=c n)subscript ℙ knn subscript 𝑐 𝑚 subscript 𝑐 𝑛\mathbb{P}{\mathrm{knn}}(c{m}=c_{n})blackboard_P start_POSTSUBSCRIPT roman_knn end_POSTSUBSCRIPT ( italic_c start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT = italic_c start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ). This variant considers only the sample pairs within the k-nearest neighbors. Assume that |w i|≤|w j|subscript 𝑤 𝑖 subscript 𝑤 𝑗|w_{i}|\leq|w_{j}|| italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT | ≤ | italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT |, the formulation is as follows:
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ℙ knn(c m=c n)=∏s∈w it∈knn w j(s)ℙ(e st=1)∏s∈w it∈knn w j(s)ℙ(e st=1)+∏s∈w it∈knn w j(s)ℙ(e st=0),subscript ℙ knn subscript 𝑐 𝑚 subscript 𝑐 𝑛 subscript product 𝑠 subscript 𝑤 𝑖 𝑡 subscript knn subscript 𝑤 𝑗 𝑠 ℙ subscript 𝑒 𝑠 𝑡 1 subscript product 𝑠 subscript 𝑤 𝑖 𝑡 subscript knn subscript 𝑤 𝑗 𝑠 ℙ subscript 𝑒 𝑠 𝑡 1 subscript product 𝑠 subscript 𝑤 𝑖 𝑡 subscript knn subscript 𝑤 𝑗 𝑠 ℙ subscript 𝑒 𝑠 𝑡 0\mathbb{P}{\mathrm{knn}}(c{m}=c_{n})=\frac{\prod\limits_{s\in w_{i}t\in% \mathrm{knn}{w{j}}(s)}\mathbb{P}(e_{st}=1)}{\prod\limits_{s\in w_{i}t\in% \mathrm{knn}{w{j}}(s)}\mathbb{P}(e_{st}=1)+\prod\limits_{s\in w_{i}t\in% \mathrm{knn}{w{j}}(s)}\mathbb{P}(e_{st}=0)},blackboard_P start_POSTSUBSCRIPT roman_knn end_POSTSUBSCRIPT ( italic_c start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT = italic_c start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) = divide start_ARG ∏ start_POSTSUBSCRIPT italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT italic_t ∈ roman_knn start_POSTSUBSCRIPT italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT ( italic_s ) end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) end_ARG start_ARG ∏ start_POSTSUBSCRIPT italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT italic_t ∈ roman_knn start_POSTSUBSCRIPT italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT ( italic_s ) end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 1 ) + ∏ start_POSTSUBSCRIPT italic_s ∈ italic_w start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT italic_t ∈ roman_knn start_POSTSUBSCRIPT italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT ( italic_s ) end_POSTSUBSCRIPT blackboard_P ( italic_e start_POSTSUBSCRIPT italic_s italic_t end_POSTSUBSCRIPT = 0 ) end_ARG ,(28)
where knn w j(s)subscript knn subscript 𝑤 𝑗 𝑠\mathrm{knn}{w{j}}(s)roman_knn start_POSTSUBSCRIPT italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT end_POSTSUBSCRIPT ( italic_s ) denotes the nearest neighbors of s 𝑠 s italic_s in w j subscript 𝑤 𝑗 w_{j}italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. Empirical results show that A3S is not sensitive to the number of neighbors, and we consider 4 neighbors for each sample in our experiments.
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B.3 Feature Extraction
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The facial characteristics in the Humbi-Face dataset are extracted using a face recognition model. Similarly, the MK20 and MK100 datasets utilize a person re-identification model for body feature extraction (Liu et al., 2022). In the case of MS1M-10k and MS1M-100k, facial features are extracted using the Arcface model(Deng et al., 2019). The Handwritten dataset encompasses four types of features: average pixel features, Fourier coefficient features, Zernike moments features, and Karhunen-Loève coefficient features.
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B.4 Hyperparameter Setting
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The implementation of A3S involves two hyperparameters: the threshold for density test and the number of neighbors considered in Fast Probabilistic Clustering. We report our choice of these two parameters in Table4. In particular, the hyper-parameter τ 𝜏\tau italic_τ is used to filter out the clusters with low density, and it is set to be slightly lower than the average density among all clusters. For the selection of τ 𝜏\tau italic_τ, we first compute density using the formula in Eq. (5) for all clusters, then calculate the mean value as d 𝑑 d italic_d, and set τ 𝜏\tau italic_τ as d−0.1 𝑑 0.1 d-0.1 italic_d - 0.1. The final results of A3S are not sensitive to this value, and perturbing it to d 𝑑 d italic_d or d−0.05 𝑑 0.05 d-0.05 italic_d - 0.05 has a negligible impact on the final clustering result.
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Table 4: Hyperparameter setting of A3S.
dataset MK20 MK100 Handwritten Humbi-Face MS1M-10k MS1M-100k
τ 𝜏\tau italic_τ 0.5 0.8 0.8 0.5 0.5 0.5
neighbors 50 50 50 50 50 50
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B.5 Computing Resources
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We utilize a [GeForce RTX 3090 Ti] for feature extraction using DNN models. For the implementation of baseline methods and A3S, we perform the experiments on a machine equipped with an Intel(R) Xeon(R) Platinum 8163 CPU @ 2.50GHz.
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Appendix C More Experiment Results
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In our ablation study, we examine the impact of varying the adaptive cluster number on A3S. We define this number as r⋅k⋅𝑟 𝑘 r\cdot k italic_r ⋅ italic_k, where k 𝑘 k italic_k is the adaptive number determined by FPC and r 𝑟 r italic_r is a ratio factor set to values in the set 1, 1.5, 2. Utilizing the K-means algorithm, we generate initial clustering with the cluster number r⋅k⋅𝑟 𝑘 r\cdot k italic_r ⋅ italic_k, and present the corresponding results of A3S in Figure7. It’s observed that the performance of the initial clustering is relatively sensitive to the adaptive cluster number. However, the selection of this adaptive number has a minimal effect on the number of queries required to achieve the desired clustering outcome. This demonstrates A3S’s robustness to variations in the adaptive cluster number.
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Figure 7: The NMI and ARI performance of A3S when the adaptive cluster number is set as r⋅k⋅𝑟 𝑘 r\cdot k italic_r ⋅ italic_k, where k 𝑘 k italic_k is the adaptive number generated by FPC and r 𝑟 r italic_r is the ratio factor.
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13964 | https://www.youtube.com/watch?v=IgbUBgrTqxY | Learning Relative Speed:4 cases|Same Direction, Different Direction, Different Time|Formula,Examples
GMATWhiz
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Posted: 20 Jun 2022
Relative Speed questions are among the trickiest time and distance questions that can come in the test. However, if you understand the concept of GAP = Relative Speed, you can solve any relative speed question in under 2 mins easily, even some questions in under a minute!!
This video tutorial covers a basic introduction to the concept of all the variations covered in Relative Speed problems: Different Direction(Towards each other, Away from each other), Same Direction, Traveling at different time. All 4 cases have been illustrated in the video with the help of some useful examples and formula(s) required to solve the questions.
Watch this video to understand the concept and get a grasp on the basics of the same. You can check out the practice questions along with solutions to solidify the concept learnt here:
To learn more concepts tested on #GMAT Quant, you can register for a free trial of our course here:
If you want us to cover more topics in these shots and also intend to receive mail notifications, register here -
TIMESTAMPS
00:00 – Overview
00:42 – What are the different scenarios of Relative Speed ?
01:56 – Explanation of the Scenario 1 with the help of example
06:11 - Explanation of the Scenario 2 with the help of example
09:17 - Explanation of the Scenario 3 with the help of example
11:16 - Explanation of the Scenario 4 with the help of example
gmatwhiz #gmat #gmatprep #gmatquant #WednedaywithWhizzes #gmat700 #timedistance #relativespeed
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Transcript:
Overview [Music] hello everyone uh today we are going to cover a new concept today which is a part of um a surveillance video series that we have started recently uh today we are going to cover relative speed which is one of the topics which i've seen a lot of students struggle with so we thought why not um cover this topic if you have been watching our videos uh i would request you to kindly like all these videos and subscribe to our channel too so let's get started now uh first of all we need to understand what are the different kind of uh questions can come What are the different scenarios of Relative Speed ? in relative speed what are the different scenarios so in relative speed there are basically three kind of questions the first question could be a question in which two objects are moving from opposite direction and there are two different variations of it which you are going to talk about the second case could be when two objects are moving in the same direction and finally there could be a variation in which two objects are moving either in opposite or same direction but they might be doing it at different time or the third case is the slightly tricky one so i'll be focusing a lot on this case in this video but obviously i'll be covering the first two also to a certain extent the way we are going to undersolve any relative speed question is based on just one formula okay and that formula is gap is equal to relative speed times time taken okay so this is basically a variation of distance is equal to speed multiplied by time only the only difference is that we are going to focus on something called the gap distance which i'm going to explain a bit once we start solving looking at each and every case and relative speed is also something that you need to understand uh when two objects are moving in opposite in the same direction what are how do we consider relative speed okay so we'll understand all these cases with the help Explanation of the Scenario 1 with the help of example of examples and the first example is this one like there are two people jane and joe and they are 500 kilometers away and joe is traveling in this direction at a speed of 60 kilometers per hour and jane is traveling in this direction at 40 kilometers per hour usually the question would ask you um you know when will they meet at what time will they meet or how after how much time will they meet so notice in this question they're telling that if you're traveling at 10 a.m then at what time will they meet now try to visualize it what what is going what is happening here joe and jane they both start at 10 a.m and they start moving towards each other so after a certain point of time they'll meet at uh they'll meet somewhere they could be right in the middle or somewhere obviously since the speed is different they won't need in the middle but they would meet somewhere right and i need to find out that time now understand this that uh we don't know the time here but they are both traveling for the same time right they both started at 10 a.m so let's assume let's say they meet at 2 p.m hypothetically okay take hypothetical scenario so both of them from 10 am to 2 pm the both of them must have traveled for four hours right correct because they both started at the same time they are meeting at the same time now we need we don't know this time okay this is just a hypothetical scenario that i was talking about so first of all i'm explaining it to you conceptually and then we'll focus on the gap part so since we don't know the time let's assume the time to be t okay let's assume that after t time they meet okay so in t time joe will cover how much distance distance covered by joe is going to be the speed of joe let's write it as s1 for the moment and the time t okay distance covered by jain would be how much let's write it as speed as 2 and again the time t and together they are covering the total distance right the total distance of 500 kilometers so i can write the total distance of 500 kilometers must be the distance of joe that's this distance of jain the total distance is 500 which is equal to s1 times t plus s2 times t now notice an interesting thing here if i take t common i'll get the speed as s1 plus s2 times t okay now this s1 plus s2 is basically known as the relative speed okay so when two objects are moving in opposite direction okay this is what you need to remember when they are moving in opposite direction the relative speed is basically the sum of the speeds which is s1 plus s2 and it is logical also right they are coming towards each other so they would if they're moving towards each other they will meet quickly think of it in this way so to meet quickly the speed must be more so we need to add it so this is one way of you of you know remembering it whether you need to add or subtract okay this distance is known as the gap distance okay why the gap distance because this is the distance that they are covering this is a 500 kilometer that both of them will cover this is the gap that they need to cover so this is known as the gap and this is known as the time so when you get a question like this you don't have to actually write all these steps you can simply write gap is equal to 500 the speed is the sum of the speed which is 60 plus 40 and we need to find out the time which is t so this is going to be t 500 by 100 so they meet after 5 hours so if they are starting at 10 they'll meet at 3 pm okay you just add 5 hours to it and you get your answer but again the timing is not important the way we are solving it is important what you need to remember is if you're moving in opposite direction the speed is the sum of the speeds so the total formula becomes i'm writing it here on top gap is equal to s1 plus s2 times t when they are moving in the opposite direction important thing to note here is that they started at the same time okay that is why the time taken by both of them was taken as t this is one important fact that we are going to use later on Explanation of the Scenario 2 with the help of example now there could be another variation of this also okay the other variation could be that they might start they might have a gap of 500 but they are moving in opposite directions uh they're moving away from each other okay like this so this question basically states that they were 500 km apart they started moving away from each other simultaneously in the opposite direction at the same speed 60 40 after how much time will the distance between them would be 1100 kilometer so notice again the distance was previously 500 now after traveling in opposite direction they became 1100. understand this that this is also a question where where the two bodies are moving in uh opposite direction right uh the previous question also was in the same way in the previous question they were moving towards each other they were covering this gap of 500 but in this case they are moving in the opposite direction but they are moving now like this so think of it in this way this guy basically covers here and they're moving in this direction and this person is now moving in this direction but the point is that they are still moving in opposite directions now here we are still going to use the formula of gap gap is equal to s1 plus s2 times t but the mistake that people make in this question is regarding the gap now what should be the gap should the gap be 1100 should i take it as 100 should i take my gap as 500 because that is the gap that was initially present or should i take the gap as the difference 1100 minus 500 so there are three options 1100 that's the first option 500 is our second option and 1100 minus 500 is a third option which is 600. always remember gaap is the is basically think of it this way is the net net distance that they are covering okay net distance that they're covering after they start okay so in the previous question the net distance that they were covering was 500 after they started together covered 500 this was the gap they had to cover to meet okay in this question they are not covering this 500 okay so 500 would not be your answer they're also not covering 1100 so 1100 would also be not your answer they are basically covering this distance this plus this okay if you want to find out that distance then you would have to write it as 1100 minus 500 so see the question becomes a one-line question where you just need to solve like this and get the answer again i'm not solving it the answer obviously t is six so they take six hours uh so that's easy to find but i hope the the methodology is clear okay they are starting at the same time the gap was 500 but now they are moving away from each other so we need to focus on the difference of the distance because that is something that they're covering during that time okay so that is a gap that they are creating i hope it's perfectly clear now these were the two cases where bodies are moving in opposite direction Explanation of the Scenario 3 with the help of example you could also have a case where bodies are moving in the same direction something like this they would have a distance of 500 in between them and they would start moving towards them moving the same direction and the question would ask you after how much time would they meet okay now notice here that in this case the and and this one i'm going to do directly the formula would remain the same okay the only difference in the formula is that when they move in the same direction the relative speed becomes s 1 minus s 2 where obviously s1 is greater than s2 okay and gap is the net gap that i need to cover now notice jane and jody are both covering this distance so this is not the net distance that they're covering what does joe need to do joe has a higher speed of 60 and joe needs to meet jane so obviously since the speed is high joe would be able to meet jane some place but joe has to cover this net distance of 500 km the initial gap this is the initial gap that joe has to cover to meet jane the rest of the distance the same so the net gap that the zone needs to cover is 500 so in this case we'll just write it as 560 minus 40 and find out the value of t which is 500 by 20 which is 25 hours so joe will take 25 hours after 25 hours joe will meet jane that's it again focus on the fact that we are talking about the net distance that joe needs to cover you needs to cover the net distance of 500 important point again here is that the boat started moving simultaneously at the same time it's important okay so i hope the three cases are clear in the first case they started moving in the same direction towards each other in the second case they were moving away from each other so we had to cover the net distance and the third case was that they were moving in the same direction and one had to meet the other okay now the last is an interesting case in Explanation of the Scenario 4 with the help of example which you'll notice that uh jane and joe they are standing 500 kilometer apart okay joe is moving towards east okay at a speed of 60 kilometer per hour but joe started at 10 am and jane starts moving towards west so in this direction at 1 pm now notice the timing is different the question is still asking us when at what time will they meet now in this case if you use the graph formula and if you write 500 is equal to 60 plus 40 by the way uh jane's speed is missing here jn speed is 40 kmph let's assume this okay times t if you do this this is wrong this is wrong because they are not moving in the same as they are not moving at the same time remember when we did the opposite question when they're moving in opposite direction we assume that they are starting at the same time so they are taking the same time t to meet this is not happening here so what should we do in such cases remember that the first thing that you need to do is bring them at the same time okay so you need to ask yourself okay what is the same time who starts later one uh jane starts later at 1 pm right so find out where is joe at 1 pm we can find that out very easily from 10 am to 1 pm joe is travelling for 3 hours right if joe is traveling for 3 hours the distance that joe would cover is speed into time speed is 60 times 3 which is 180 kilometers so that means joe has already covered 180 kilometers and joe is here at 1pm now they are both at the same time 1 pm 1 pm the distance between them is 500 minus 180 which is 320 so all i need to figure out now is the time from 1 pm what they will take to meet okay so now the question becomes this okay the 180 is already covered and now i'm focusing at my time 1pm now the question is easy the gap between them is 320 the relative speed is 60 plus 40 a 40 again is missing but it has to be it will be given in the question stem and t is the time time from 1 pm just keep that in mind so 320 up 100 so that would be point two hours okay so if i have to find out the actual time we'll say three hours twelve minutes and note we are starting from one pm okay so 1 pm you add 3 hours to it you'll get 4 pm and then from the end there 4 4 12 basically is what you can say they'll meet okay so i hope this is clear this is the last case which is the most difficult case where people end up making a mistake but as you can see it's a one line question all you need to do first is make sure that you make the time same bring them at the same time once you bring them at the same time after that it is easy to solve you just use the ga formula and get the answer so i hope all these four parts are clear uh i will be sharing a few questions uh uh in three questions will be shared along with three solutions if this is clear try out those three questions the link you you'll be able to find the link on the top right corner in a few days okay |
13965 | http://www.newtonchineseschool.org/news/Algebra2_2020-%E9%9F%A9%E6%80%80%E4%B8%AD.pdf | Course Name: AoPS Algebra 2 Textbook: “Intermediate Algebra”, by Richard Rusczyk and Mathew Crawford, the Art of Problem Solving series, ISBN-13: 978-1-934124-04-8 Course Introduction: Algebra 2 covers complex numbers, quadratic equations, conics, polynomials, functions, logarithms, cleaver factorizations and substitutions, systems of equations, sequences and series, symmetric sums, advanced factoring methods, classical inequalities, functional equations, and more. This course goes beyond what you would find in a typical honors Algebra II curriculum, as it covers topics found in honors Algebra II and Precalculus classes, as well as many topics no found in most other curricula. This is a one-year course and is divided into two parts to be taught in two semesters, Algebra 2A and Algebra 2B. Algebra 2A covers Chapter 1 – Chapter 10, including complex numbers, conics, polynomial division, polynomial roots, factoring multivariable polynomials, sequences and series. Algebra 2B covers Chapter 11 – Chapter 20, including identities, induction, inequalities, exponents and logarithms, radicals, special classes of functions, and piecewise defined functions. This course is intended for high-performing students and will focus on problem solving skills. Students will learn via practicing solving problems in the classroom. Who Should Take Algebra 2A: Students should have a mastery of basic algebra up through and including quadratic equations before taking this course. Typically this class follows our “Algebra 1B” class. Students who have completed typical Algebra 1 may be ready for this class. An entrance exam should be taken to evaluate student’s readiness. Who Should Take Algebra 2B: Students should have a mastery of basic algebra up through and including polynomials, sequences and series, advanced factoring before taking this course. Typically this class follows our “Algebra 2A” class. Requirements for Students Registering for Algebra 2A or Algebra 2B In general, students in grade 8 – 10 are eligible to register for either one of Algebra 2 courses. All students interested in take any of these courses should pass the corresponding evaluation test. The test must be taken and submitted before the specified deadline posted by the school web site. The tests are posted on the school web site as well. Please check the class schedule at the school web site. Teaching Plan Algebra 2A Lesson 1 Review of Algebra I and Chapter 1 Lesson 2 Chapter 2: Functions Review Lesson 3 Chapter 3: Complex Numbers Lesson 4 Chapter 4: Quadratics (I) Lesson 5 Chapter 4: Quadratics (II) Lesson 6 Chapter 5: Conics (I) Lesson 7 Chapter 5: Conics (II) Lesson 8 Review of Midterm Exam Lesson 9 Chapter 6: Polynomial Division Lesson 10 Chapter 7: Polynomial Roots Part I (I) Lesson 11 Chapter 7: Polynomial Roots Part I (II) Lesson 12 Chapter 8: Polynomial Roots Part II (I) Lesson 13 Chapter 8: Polynomial Roots Part II (II) Lesson 14 Chapter 9: Factoring Multivariable Polynomials Lesson 15 Chapter 10: Sequences and Series Lesson 16 Review of Final Exam Algebra 2B Lesson 1 Chapter 11: Identities, Manipulations, and Induction (I) Lesson 2 Chapter 11: Identities, Manipulations, and Induction (II) Lesson 3 Chapter 12: Inequalities (I) Lesson 4 Chapter 12: Inequalities (II) Lesson 5 Chapter 13: Exponents and Logarithms (I) Lesson 6 Chapter 13: Exponents and Logarithms (II) Lesson 7 Chapter 14: Radicals Lesson 8 Review of Midterm Exam Lesson 9 Chapter 15: Special Classes of Functions (I) Lesson 10 Chapter 15: Special Classes of Functions (II) Lesson 11 Chapter 16: Piece-wise Defined Functions Lesson 12 Chapter 17: More Sequences and Series (I) Lesson 13 Chapter 17: More Sequences and Series (II) Lesson 14 Chapter 18: More Inequalities Lesson 15 Chapter 19: Functional Equations Lesson 16 Review of Final Exam About The Teacher: Huaizhong Han, received BS/MS from University of Science and Technology of China and PhD from University of Massachusetts - Amherst. He volunteers as the coach of Wayland Middle Schools MathCounts Team (2016 - 2020). In 2020, Wayland Middle School MC team was ranked 4th in State Contest and 2nd in MetroWest Chapter Contest. In 2018, he received the “Best Coach" award and Wayland Middle School received the “Most Improved School" award. |
13966 | https://www.youtube.com/watch?v=SsHUKGQkMQA | David Eppstein: Non-Euclidean Erdős--Anning Theorems
Computational Geometry
732 subscribers
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Posted: 2 Apr 2025
April 1, 2025
The Erdős--Anning theorem states that every point set in the Euclidean plane with integer distances must be either collinear or finite. More strongly, for any (non-degenerate) triangle of diameter~δ, at most O(δ2) points can have integer distances from all three triangle vertices. We prove the same results for any strictly convex distance function on the plane, and analogous results for every two-dimensional complete Riemannian manifold of bounded genus and for geodesic distance on the boundary of every three-dimensional Euclidean convex set. As a consequence, we resolve a 1983 question of Richard Guy on the equilateral dimension of Riemannian manifolds. Our proofs are based on the properties of additively weighted Voronoi diagrams of these distances.
To appear at SoCG 2025.
Transcript:
The title of the talk is non-ucuk uklidian erd shanning theorems and I will explain that um and but basically it's about points with all the distances between them being integers and so I thought I would start with some examples. So in one dimension it's super easy to find points at integer distances. you just use all of the integer points on the real line and their distances are integers from each other. I'm going to be talking about spaces that are not flat. And so it's you can't actually make a one-dimensional space that's not flat. They're all topologically equivalent if if they're geodessic spaces. But you could think of any curve and the distances along that curve. um and place points at unit distances along that curve and you would have exactly the same distance structure and this is what for instance the Romans did long ago by placing milestones along their roads. So this is kind of the canonical example of an infinite set of points at integer distances from each other. Now if you want to go to twodimensional point sets with integer distances from each other again it's easy if you just want a finite number of points if for instance uh any three integer distances as long as they satisfy the triangle inequality will be enough to define an integer triangle three points at integer distances from each other. And you know since ancient times people have studied right triangles at integer distances from each other. Now we don't need the triangle to be a right triangle in order to have integer distances. But if if we have many uh integer right triangles we can put them all together into one picture. um scale the triangles so that all of them have a unit length side and then place the the um other side along the x- axis. Place the unit side along the y- axis. Place them all together like this. All of these distances to this uh top point will be rational because you scaled them by a rational amount. All of the distances between two points on the axis will be rational because they are the differences between two rational numbers. And so we have infinitely many points with rational distances that are not all on one line. There's one of them that's not on the line. And you could put another copy of this uh top point uh reflected across the axis uh below the line. And now you have two points that are not on a line and everything else on a line. So not quite everything all on a line, but close. Now it should be not not much of a surprise that Leonard Oiler was interested in Can you see these? Do I need to move this off screen? Would that be better? Let's move this down here. We do see your cursor. You can see my cursor but not my Zoom windows. Okay. I'm what I'm seeing is my um That's better. Perhaps you're only sharing Zoom was covering part of my my slides. Um Leonard Oiler was interested in extensions of the Pythagorian theorem as were other mathematicians before him. Fibonacci for instance. Um here's a famous open problem coming from the work of Oiler. We don't know if it's possible to make a cuboid with right angles, um integer side lengths, integer face diagonals, and an integer body diagonal. you can make [Music] um six of these seven quantities be integers but we don't know how to make all seven of them integers. So that's a famous open problem. Another observation of oiler that's also closely related to this question of what point sets can have integer rational distances is uh this construction that um there exist infinitely many points at rational distances from each other on a unit circle. And one way to do that is to again use a 345 right triangle scaled so that the hypotenuse is one and repeatedly reflect it across one of its sides to another point on the circle to another point on the circle to another point on the circle. Keep going around because this angle is irrational. It's not a rational multiple of pi. You can keep going around and you'll never repeat the same point twice. But all of these points on the circle and the center of the circle have rational distances from each other. And so this gives you an infinite set of rational points of points with rational distances to each other. And actually no three of these points are collinear. So it's very far from being all on a line, but somehow it's all on a circle. It's it's still very constrained. And in fact, this question of can we get um points with rational distances that are not all on a line and not all on a circle is still quite open. So there exist sets of seven points in the plane. Here's one of them from these references where all of the distances are integers or rational numbers. We do not know if there are more than seven points in general position with no three on a line and no four on a circle that have integer or rational distances. Seven is the most we know. So this is it's a hard problem. It's an open problem. It's not a problem that I'm going to address, but it's it's related and it's important. Now maybe something much different from that is possible. Erdos um Erdos and Ulam speculated. I wouldn't call it a conjecture because um as far as I know what Ulam was trying to come up with in this proposal was what is the craziest thing that could happen? Can we manage to prove that that crazy thing cannot happen? And the craziest thing he could think of in terms of rational distances in the plane, very far from all being on a on a circle, very far from all being on a line, is maybe there's a dense subset of the entire plane where all of the points in that subset have rational distances from each other. It seems unlikely. It would disprove some major conjectures in mathematics unrelated to discrete geometry. But we don't know how to prove that it's not true except maybe addressing it by these these other bigger conjectures. So that's the Erdos Ulam problem and it's also unknown. What we know is that we can find infinitely many points on a circle plus one. We can find infinitely many points on a line plus two. And there are some sporadic finite examples. But those examples of infinitely many points on a line plus one, infinitely many points on a circle plus one have rational distances. They don't have integer distances. And that actually is an important difference. It doesn't sound like much rational integer. If you have a set of finitely many points with rational distances, then you can just scale them to get the distances to be integers. Find a largest common denominator of all the distances. Scale them by that scale factor. The distances turn into integers. But that gives you arbitrarily large subsets of the circle, for instance, with integer distances. But it does not give you infinite sets of points with integer distances because the more points you add to your set, the bigger the scale factor you need. And so there's not one common point set with all the distances integers uh infinitely many points. And so that is the subject of the airing theorem. Is it possible to get infinitely many points with integer distances that are not all on a line? The answer is no. Erish and Anning, Anning and Erish proved in 1945 that if you have a set of points in the uklidian plane with integer distances, then either it's all on a line, the set we started with, the the integers on the real line, or it's finite. Those are the only things that can happen. I've looked at their proof. It's not that enlightening. But in the same year in the same journal, Erish by himself published a onepage paper proving the same thing. Uh proving a little more strongly that if you have a set of points in the plane with diameter D and they're not all on a line, then there are only O of D^2 points. That has since been improved to sublinear in D. But Erish gave a fiveline proof. It's it's a very short simple proof. Um he wrote that an analogous theorem holds in higher dimensions. He didn't specify any details. And so that is the proof that I'm going to start with and this is the theorem that I'm going to be discussing. uh the question is that that I'm addressing in this work is can we get similar results in other metric spaces because the uklidian metric space it's nice it's um we can prove lots of things about it but um maybe we can generalize this to other metric spaces I started out looking at the hyperbolic plane for instance but there are many other less uniform metric spaces for which similar results might plausibly be true and for which in fact I can prove that they are true. But in order to get there, I want to go through Erdish's fiveline proof and then go through a different proof of the same theorem that uh is more flexible, allows me to generalize it to other metric spaces, but I will start out with the uklidian metric space. So this is the main idea not of my paper but of Erdish's paper. If you're given two points that are at integer distance D from each other, A and B, and you want have some third point C that is supposed to be at integer distances from both A and [Music] B, then C has to be on one of these curves. the distance from C to A and the distance from C to B, those two distances cannot be very different from each other. They have to be within D of each other by the triangle inequality. And so you can ask what is the difference between those two distances. It might be as much as in this example I've drawn here, the distance between A and B is three. So, um, C could be or or I guess I've labeled it X in this figure could be as much as three units farther from B than A. And if it is, it's on this ray vertically extending in the other direction up from from B from A. The two distances could differ by two. And then you get this branch of a hyperola. The same hyperola is over here but giving a different branch for the points that have uh difference too but in the other direction closer to B than to A. And all of these curves for each possible integer distance difference between the two distances from X to A and X to B. For each of those differences we get a hyperola a branch of a hyperola. the the two opposite rays that you can see here, you can think of them as being a degenerate hyperola. So if we have two points, then we know that that any third point with integer distances has to be somewhere in this system of hyperolas. If we have three points A, B, and C that are not all on a line, then we get these two systems of hyperolas. one for A and B, say another system of hyperolas for B and C, where the number of hyperolas in this system, including the degenerate one, is the number of possible differences between the two distances, which is at most d + one. And any point that is going to have integer distance to A, integer distance to B, and integer distance to C has to be on the blue system of hyperolas. And it has to be on the red system of hyperolas. And it would have to be on a third system of hyperolas between A and C. But I haven't drawn that one. And so for instance you can see this is this point over here has distance three to c four to a five to b because it's a reflection of b. This forms a a pythagorian rectangle. Um so these four points do form an integer distance set but it's a it's a point formed by the crossing of one of these red hyperolas and one of these blue hyperbolas. Now, it turns out that no two of these hyperolas can coincide the the actual hyperolas, the non- degenerate ones have different axes. Um, we have this horizontal axis for the red hyperolas and this blue reflection axis for the um blue hyperolas. You have to be a little bit careful of the u degenerate hyperolas but it turns out they can't coincide either. So for instance we do get a line here as another degenerate case of a hyperola. We get two rays on a parallel line for B and C but they don't coincide. They're they're uh offset from each other. So with a little bit of case analysis that Erish didn't do. He just said they're distinct. You can see that all of these hyperolas are distinct. And if you have two hyperolas, even the degenerate ones, two hyperolas can cross at most four times, you can have at most four crossings for any two hyperolas because they're quadratic algebraic curves. And um curves of degree d cross o of d^2 times. Here the degree is two. There's there's four crossings. Two squared So that's essentially the proof. We have a system of d +1 by d +1 hyperolas. Uh each pair of those hyperolas cross four times. Those crossings are the only additional points that you can have. And so your point set has to have at most four d +1^ squar points in it. Now I want to generalize this to other metrics. For instance, we might want to think about the L1 or L infinity metrics. I'm not going to be able to generalize to those metrics because in those metrics there do exist infinite sets of points with integer distances from each other the the integer grid. And so if I want to try to generalize this ashanding theorem to other metrics, I need to be able to distinguish these metrics for which the theorem is just not true from the other metrics that I'm trying to prove it for. But there's another more fundamental obstacle to any kind of generalization, which is this idea of using the degree of an algebraic curve. Erdish uses this fact that hyperbolas cross a constant number of times. That is a basic fact about algebraic curves. But the algebraic curves that come up here are very specific to uklidian distance. If we pick a different distance, we're going to get different curves. And there's no intrinsic reason to believe that they're always going to be algebraic curves. And so I would like to look for a different proof, a related proof, but a different proof where I can use topological ideas rather than algebraic ideas. And so that's what I want to go through in the next few slides. Another proof of the same theorem for points in the uklidian plane but one that is based on topological and combinatorial geometry ideas rather than algebraic geometry which is the basis of Erdisha's theorem and so instead of looking at the differences of the distances to two points this hyperola defined by the difference in distances from our unknown point x to a and to b. I want to look at the differences of the distances to three points simultaneously. The distance from a to x + 1 in this case is equal to the distance from b to x is equal to the distance from c to x + 2. and the plus one 0 + two are specific to this example. But for any three points that I would choose to have integer distances to each other and then any fourth point that is variable but at integer distances from all three. These distances from the four fourth point to the other three are going to differ from each other by integers because every distance in this imagined scenario is an integer but by integers that are relatively small. They're bounded by the distances between these three given points by the triangle inequality. And if I have three points and I want to look at a fourth point that has some sort of equidistant property adjusted with these plus ones and plus twos from those three points. Now we can use computational geometry. We can use boronoi diagrams because this fourth point in order for these three distances to be equal, this fourth point has to lie at the vertex of a voronoi diagram at a point where these three voronoi cells meet. This voro diagram is the minimization diagram of these three functions. And it's just an additively weighted vorono diagram. We take the distances we add we weight them by adding a small integer to each distance. We take the minimization diagram of those distances. We subdivide the plane into cells within which one of these three distances is the minimum and the integer point that we're looking for is on the boundary of all three of these voro cells. More importantly, it belongs to all of these cells. I'm going to think of a voro cell as being a closed set. So the voroy cell of C is the system of all points in the plane for which C is one of the minima in this minimization diagram. [Music] Um this yellow cell contains its boundary. It's a system of points in the plane for which a is one of the minima. And so this green point belongs to the intersection of all three of those closed cells. And so we need to understand what is the structure of additively weighted vori diagrams. I've written hyperbolic boundaries. By the way, these hyperolas that bisect pairs of cells in the additively weighted vorai diagram for uklitian distance are exactly the same hyperolas that I had in Erdish's proof. In fact, I drew this figure reusing some of those same hyperolas. So, what properties do additively weighted borai cells have? They're star-shaped. Uh if some site, this blue site here is one of the closest sites to this yellow point, a point in the Borai cell of this blue point and you move towards this blue point on the closest curve connecting these two points in a straight line. the distance from the yellow point to the blue point as you move reduces by exactly the amount that you've moved one to one. If you move in any other direction, the distance reduces more slowly or it might even increase. Um, and that actually is a property that is not true of L1 or L infinity. in L1 or L infinity there actually might be a continuous family of of directions in which you can move and reduce the distance um one one but regardless um and this is actually still true of L1 or infinity if you move in a straight line segment towards the closest site it remains a closest site it its distance decreases one to one the distance to all the other sites decreases at most one probably um more slowly and so this one remains closest and so you have this property that if this point is in the foreign cell everything on that line segment is in the boron cell and that's the definition of a star-shaped set one complication is that boray cells might not be two-dimensional if I have a site Why? One of the generators of the Vorni diagram. But that site, it has whatever additively weighted distance it has to itself, but it might have a closest neighbor that is not itself. It might be the case that it is tied with um this blue point for both of them being equal closest neighbors to this red site. And when that happens, what you get is a Vorno cell for the red site that forms a ray extending away from the blue site. All of these points on this ray are equidistant from the red and the blue site. And then we might have a third uh yellow site Z that's equidistant from the red site and the blue site and from itself. And so its vori cell is also a ray extending out in the same direction. And so we can get this nested system of voronoi cells or colinear sites. And when we get a system of of nested vorin cells like this, we actually get that um the number of triple intersection points of all of these triples of voroi cells can be infinite. And this is exactly what happens. This shouldn't be a surprise. This is what happens when you compute the additively weighted boroid diagram of um the integer points on the real line embedded in the uklidian plane with the waiting system set up so that all of these uh points are tied as equally close. you you simply um make the weight of each of these integers be it's the negation of that integer and you'll get a diagram like this. So this kind of degeneracy is something we have to worry about something we have to watch out for. Non- degenerate cells in an additively weighted boron diagram turn out to be interior or disjoint. There's an argument about if they were not if you had some point P that belonged to the interior of one cell and also belonged to another cell then we could find this nonstraight path that would have to be a shortest path but that's an impossibility for the uklidian um distance. So three non- degenerate cells have to be interior or disjoint and then it turns out that they can have at most two of these triple intersection points. In a standard Borney diagram for an unweighted distance every triple of cells has one intersection point. In the weighted case they can have two. Um two is possible. You can get something that looks like this. But if you tried to form three cells with three of these triple [Music] junctions, the fact that these cells are star-shaped means that you could draw segments from each of these junctions to each of the sites. Those segments would all be disjoint from each other. And that would give you a straight line drawing of the graph K33, the complete bipartite graph with three vertices on each side. And that graph is not planer. So it does not have a drawing like this and therefore these three junctions cannot exist. So any three degenerate cells have at most two triple intersections. Um degenerate triples of cells as long as they're non-olinear we don't get that nesting effect that I showed you earlier. And so it turns out there's at most one triple intersection. In the case where there's one degenerate cell inside another cell um its boronoi cell, the yellow voro cell forms a ray that gets stopped at the boundary between the blue and the red cells. And so this point where the ray stops is the one triple intersection of these three vori cells. If you have two degenerate cells and they're not colinear, then their rays diverge and you get no triple intersections. So this gives us a proof of the Erd Shanning theorem that if you have a system of non-olinear points with integer distances, then there must be O of D^2 of these points, the same bound that Erish proved. Because if you start with any non-linear triple, if the points are nonlinear, certainly there's a non-olinear triple within them. Let D be the max distance between the the points in that triple. Use all of the possible waitings where one weight is zero and the other two are at most D. And that will give you a VOI diagram with at most two triple points. There are O of D squared weightings and every point in your initial point set has to come from one of these diagrams with the weights given by the distances to the difference in distances between the farthest site to to the given point and the other sites. And that'll give you a system of three weights where the weight to the farthest site is zero and the other two are less than or equal to D. So it will be one of these four diagrams you've looked at. O of D squed borai diagrams O of one points per borai diagram O of D squared points in your given system of points. So that's a different proof of the Eric Shanning theorem. It maybe can be generalized more. So in order to generalize that theorem to other metric spaces in if we're using Erdishes the proof we need to understand the algebraic properties of these bis sector curves these curves that are equidistant from two points. For the new proof you can tell it's a new proof because it needs different properties to be able to generalize to other metric spaces. The properties we need are that the metric space is a geodessic space. That means that the distance between any two points is the length of the shortest curve between those two points. Uh the length of a curve can be defined as a limit of sums of distances of sequences of points along the curve. Um using the notion of a shortest curve, we can define what it means in any of these metric spaces for points to be colinear. A system of points is colinear if uh any triple of them belong to a shortest curve. It turns out that what we need is um not quite the fact that there's a unique shortest curve between any two points. That's true in uklidian space. It's not true of some other spaces. But we don't want to allow these shortest curves to merge or split with each other at interior points. that property is going to be important um in the detailed analysis of the voronoi diagrams that we need to make this proof generalize and in the uklidian play case we're we're working with uh the topology of the plane we used that topology in the fact that the graph k33 is non-planer um but we can generalize a little bit in that property we don't need to be working on a surface that is spherical or planer, but we do need to have bounded genus. And the reason for that, at least for this proof, the reason for that is that um if you have bounded genus, then there's a graph k3 something 2g + 3 that cannot be drawn on that surface without crossings. uh but the size of this graph that cannot be drawn without crossings is linear in the genus. So just as a quick example, this is a drawing of K36 drawn on a Taurus where you have to imagine that each of these hexagon edges wraps around from the top to the bottom, from the upper left to the lower right, from the upper right to the lower left. Um, and so this this hexagon when you glue pairs of opposite edges in that way gives you a topological Taurus. And on that Taurus we have three these three copies of this red point all get glued to one point. One red point and one blue point, one yellow point. The voronoid diagram of those three sites, the red site, the blue site, and the yellow site is the diagram that I'm showing here with six triple intersection points. So you can realize K36 on a hexagonal Taurus. You can realize it as the graph of Vorai sites and Vorni junctions on a flat hexagonal Taurus. And so trying to argue we don't have K33 on this topological space would would not work because we do have K33. We have K36. But we cannot draw K37 on this surface. There's an argument involving the oiler characteristic that that shows that it would not be possible. So now we can start looking for metric spaces that have these properties and we can prove using exactly the same proof. I don't need to go through the proof again because it's the same that all of these metric spaces obey an Erdish anning theorem. All of these metric spaces have the property that every integer distance point set is either finite or colinear in this sense of colinearity. So to start with, I've been talking about L1 and L infinity. Let's look more generally at convex distance functions. These are defined from any centrally symmetric convex body uh like this rounded rectangle where you define the distance from some point to another point to be the scale factor that you would need for the body centered at one of these points to touch the other point. And that's a symmetric definition. If you scaled the same body centered at the other point by the same scale amount, it would touch So this central rounded rectangle barely touches this point. These two points are at distance one. Scale it by a factor of two. These two points are at distance two, distance three, and so on. And if you start with a body like this one where it's allowed to have sharp corners, it's not necessarily smooth, but it's strictly convex. That means that there's no line segment in its boundary. Then for these uh bodies, for these distance functions, the shortest curves are just the uklidian line segment. And that's all we need to make this proof work. The boronite cells are star-shaped. they can be degenerate, but when they're non- degenerate, they're interior disjoint. Uh the non- degenerate cells have at most two triple intersections because we're still dealing with the topology of the plane. The degenerate cases look exactly the same and the proof works exactly in the same way and gives you the same bound on the number of points. For any system of points with integer distances, either they're all on a line or there are O of D squ of them. If you have a convex distance function that is not strictly convex, if it has a line segment in its boundary, then we can construct systems of infinitely many points at integer distances that don't look like they're all on a line and are not. So this is kind of an if and only if condition. If you're looking at a convex distance function, it obeys the air to shanning theorem if and only if it's a strictly convex distance function. We can apply the same theorem to the hyperbolic plane. The shortest curves are unique. They're hyperbolic line segments. They have all the same properties. We can apply it to surfaces of convex sets in 3D. Um and now you might think that this is trivial. If if you have a convex set with bounded diameter then um clearly there can be at most a bounded number of integer points on it regardless. But this applies even when to surfaces in 3D that are uh unbounded. For instance, imagine this cone extending upwards infinitely far. This cone is um locally approximately uklitian except at this one cone point. And the shortest curves on on a cone, there might be more than one of them. If you have a point over here and a point uh diametrically opposite, there could be two equal shortest curves, but those shortest curves pass straight through each point in its locally uklidian geometry. and therefore they cannot merge or split which is all that we need to prove all of the properties that we need at this one point that's not locally uklidian um it's actually not possible for any shortest curve to pass through that point. You can originate a a shortest curve from this point to any other point in the plane in the cone or vice versa. But a shortest curve between two points not at this apex does not pass through the apex. And so these surface metrics of convex sets in 3D have all the properties that we need. And any finite any set of points with integer distances must either be finite or lie on a single geodessic on this surface. We can extend it to complete remmanion metrics which maybe you might be less familiar with. Roughly what I mean is the surface geometry of smooth surfaces. You can think of them as embedded in 3D but they don't need to be. Um you they're not required to be convex. They can have negative curvature. Um you can have surfaces that are Taurus or higher genus. Um but in order to make our proof work because we want to forbid one of these graphs K3N we need to restrict our attention to bounded genus surfaces. Um we need the remmanion metric to be complete. Technically this is defined as saying that all Koshy sequences converge. What this means for instance is that in this cone surface if we took away this apex point if we punctured this surface by removing that one point everywhere else it would have a nice re remmanian geometry but it wouldn't be a complete remanian geometry because you can find a Koshy sequence of points that converges to this cone apex and any Koshy sequence in a complete Romanian manifold is required to have a limit point. And so to make it into a Romanian manifold, you would need to add this point. But the geometry is not locally uklidian here. And so you cannot make this punctured cone into a complete remmanian manifold. But for a complete remanion metric of bounded genus that the same theorem applies with basically the same proof. Here's another example showing that that completeness is a necessary property. This you can think of as being a cone with an infinite angle at the cone point point instead of an angle less than 2 pi. Or um another way of describing the same space is it's what you get using polar coordinates if you don't reduce the angle mod 2 pi. And you think of two different angles that differ by a multiple of 2 pi as giving you different points. In this geometry, we can actually find an infinite set of points at unit distance from each other that are not all on a line. And so this tells us that we need this this requirement that we look at a complete remmanian man metric. If we even allow one point of incompleteness, we can get a Romanian metric where the air shanning theorem does not hold. Okay. Now, this this idea of an of a unit distance set that it's a it's a stronger notion, right? It's it's we're not requiring all the distances to be merely integers. They have to be all equal to each other. And that leads to an idea that I want to conclude with the idea of um it's called the equilateral dimension of space. The equilateral dimension is just how many points can you find at unit distance from each other. In the plane, you can find three points at unit distance from each other. The three vertices of an equilateral triangle. In three dimensions, you can find four points, the four vertices of a regular tetrahedrin. And so the equilateral dimension of a space is defined as the maximum number of points at unit distance from each other. And so different distance functions are going to have different equilateral dimensions. The L infinity metric turns out to have equilateral dimension 2 to the D. Uh the points at a of a cube or of a hyper cube are all at equal distances from each other. You can scale that to be distance one. The L1 metric has equilateral dimension at least 2D because the points of a of a regular ocahedron or a cross polytope all have distance two from each other and you can scale that so they all have distance one. This is actually an open problem. We don't know whether that is the equilateral dimension of this metric or whether there might be other sets of points that are all at unit distance from each other and more of them but it's at least 2D. So in this paper describing equilateral dimension attributing most of it to Robert Kushner uh Richard Guy asks about the equilateral dimension of remanion manifolds the same manifolds that we've seen here and he asked you know how many points at unit distance can you find on these things if it's a topological sphere the answer is at most four you can there are four points at equal distance on a an actual geometric sphere uh four points of of a regular tetrahedrin. You can scale that to have distance one. Uh you cannot find five because unit length geodessics cannot cross each other in a ramanian metric. If two curves of unit distance crossed each other, you could shortcut the curves near the cross geodessics. So the equilateral dimension of a topological sphere is four the dimension of the surface plus two and and guy asked is that true in higher dimensions. Now if you look at bremanian manifolds of unbounded genus, we can form Romanian manifolds that have unbounded equilateral dimension essentially by taking this graph of K3 join to K4, join to K5, join to K6, join to K7, etc., and thickening all of those vertices and edges into little spherical surfaces connected by thin tubes. And if you do it in this in carefully enough, you can find a point on each of these spherical surfaces. Adjust the lengths of these tubes so that all of the points in each of these um graphs are at unit distance from each other. And so you have three points at unit distance, four points, five points, six points, seven points at unit distance, unbounded numbers of points at unit distance from each other. But then if you have this system of spheres and tubes all connected to each other, you can imagine embedding it in three-dimensional space and then filling in the gaps remaining in that space so that you get a three-dimensional Romanian manifold where the material you fill it in. Think of it as being like like honey or like glue like something that is very slow to move through. So that the distance that you would travel to move through these interial spaces between the tubes is much longer in the intrinsic metric of of the three-dimensional remmanian manifold that you get. And so that all of the shortest paths are the ones that stay on the tubes. And so you can embed this example in a three-dimensional Romanian manifold that has just the familiar topology of three-dimensional space or you can embed any single one of these examples in a single remmanian sphere of three dimensions. And so this answers guy's question in the negative that topological spheres of dimension D with a remmanian metric do not have bounded um equilateral dimension. that the equilateral dimension of a sphere can be arbitrarily large and the equilateral dimension of a Romanian manifold that has the same topology as ordinary three-dimensional space can um that can give you a single manifold where there are arbitrarily large uh integer distance unit distance point sets. Okay, so that's more or less all I wanted to say. The the main conclusion I want to draw from this is that an elegant proof, Erdish's proof certainly is elegant. He published it in five lines. It's not the same thing as a flexible proof. And by making a proof that is a little bit longer because there's more details to unpack with respect to the Voronai diagrams and their properties and their degenerate special cases. we get a more flexible proof that applies to a a huge uh continuous family of of infinitely many metric spaces. Now in the in the paper I don't have in these slides but in the in the paper itself you can improve the d squared bound on the number of points that you get in a non-olinear integer distance set to O of D for a strictly convex distance function and O of D to the 4/3 for a convex surface. that still doesn't match the sublinear bound that we get for ukodian spaces based on algebraic methods. But the other direction that I'd like to extend this further is three dimensions or higher. And in particular uh a question that I don't know is is the aerodynaming theorem true for three-dimensional hyperbolic space. Is it the case that any uh integer distance set of points in this specific space is either finite or colinear? I don't know. I would like to know. Okay. That's it. Um, the rest of the slides are just credits and references. So, I will |
13967 | https://www.youtube.com/watch?v=dsJS53hr8Q8 | Complex Fractions - Conjugate Method
Kelly Bow
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Posted: 3 Oct 2015
This video shows how to reduce a fraction with complex numbers by using the conjugate method. Rember i^2 = -1
Transcript:
okay now we're going to just review uh how to simplify fractions when they have complex numbers in them so remember we can't have irrational numbers in a denominator we need to have real numbers uh rational numbers there so again we're going to use the conjugate method and the conjugate method would say for that binomial you're going to multiply that denominator so the original was 4 + 2 I the conjugate is 4 subtract 2 I whatever you do to the denominator you've got to do to the numerator and then you're probably going to use the foil method you can use the distributive method or whichever method you're most comfortable with but you need to use a reliable method for multiplying all right so here we go so remember the foil method says the firsts the outers the inside numbers and the two last numbers you're going to do that with both the numerator and the denominator okay so this is what we did let's start with the numerator 7 4 is 28 7 -2 I is -4 I 8 4 is 4 8 I 4 is 32 I and then 8 I -2 I is -6 I I left off that negative sign GL I cut that -6 I remember um excuse me I S remember I SAR is equivalent to1 so when you multiply negative a negative that will end up being a positive but now let's go to the denominator the first 4 4 is 16 4 -2 I is 8 I 2 I pos4 is positive 8 I that's why we multiply by the conjugate we get the I's out of the denominator and then the two last 2 I -2 I is- 4 I SAR and again uh we'll have to simplify the I SAR in a moment and that's what I'm going to do now okay so we have we simplify these two things -14 I plus 32 I is postive 18 I and I 2ar is -1 again i^ 2 is1 so that brings us to this gives us6 16 28 subtract 16 is 12 and this becomes 16 + 4 because subtract subtract a negative that would be subtract ne4 gives you add positive4 now notice these all have something in common a two can be factored out so what I did is I just rewrote each number number factors of two and then notice the twos can cancel 6 + 9 I all over 10 let's see there we go that would be your final response that's it just be careful uh when you're doing the foil method but it's actually easier than than you'd think |
13968 | https://stewartcalculus.com/data/ESSENTIAL%20CALCULUS%202e/upfiles/instructor/ess_ax_1008.pdf | 1–3 ■Find the length of the given curve.
1.
, 2.
, 3.
, 4.
, , , 5.
, , , ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 6–8 ■Reparametrize the curve with respect to arc length mea-sured from the point where in the direction of increasing .
6.
7.
8.
, ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 9–14 ■ (a) Find the unit tangent and unit normal vectors and .
(b) Use Formula 9 to find the curvature.
9.
10.
11.
12.
13.
14.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ rt t 2, 2t 33, t rt s2t, e t, et rt 1 3t 3, t 2, 2t rt s2 cos t, sin t, sin t rt 6t, 3s2t 2, 2t 3 rt sin 4t, 3t, cos 4t Nt Tt 0 t 2 rt cos3t i sin3t j cos 2t k rt 1 2t i 3 t j 5t k rt e t sin t i e t cos t j t t 0 0 t 2 z t y t sin t x t cos t 0 t 1 z t 2 y t 2 x 2t 0 t 1 rt 6t i 3s2t 2 j 2t 3 k 0 t 2 rt e t, e t sin t, e t cos t a t b rt 2t, 3 sin t, 3 cot t 15–19 ■Use Theorem 10 to find the curvature.
15.
16.
17.
18.
19.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 20. Find the curvature of at the point (0, 1, 1).
21–23 ■Use Formula 11 to find the curvature.
21.
22.
23.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 24–25 ■Use the formula where the dots indicate derivatives with respect to (see Exer-24.
, 25.
, ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ y t cos t x t sin t y t 2 x t 3 t x y y x x 2 y 2 32 y ln x y sin x y sx rt s2t, e t, et rt sin t i cos t j sin t k rt t 2 2i t 2 4tj 2t k rt 2t 3 i 3t 2 j 6t k rt 1 ti 1 tj 3t 2 k rt i t j t 2 k SECTION 10.8 ARC LENGTH AND CURVATURE ■ 1 Click here for answers.
A Click here for solutions.
S ARC LENGTH AND CURVATURE 10.8 cise 32 in the text) to find the curvature of the parametric curve.
Copyright © 2013, Cengage Learning. All rights reserved.
2 ■ SECTION 10.8 ARC LENGTH AND CURVATURE Click here for solutions.
S Click here for exercises.
E ANSWERS 10.8 1.
√ 13 (b −a) 2.
√ 3 e2π −1 3. 8 4.
√ 3 + √ 2 2 ln √ 2 + √ 3 5. ln π2 4 + 2 + π 2 + π 4 π2 4 + 2 −ln √ 2 6. r (t (s)) = 1 √ 2s + 1 sin ln 1 √ 2s + 1 i + cos ln 1 √ 2s + 1 j 7. r (t (s)) = 1 + 2 √ 30s i + 3 + 1 √ 30s j − 5 √ 30s k 8. r (t (s)) = cos3 1 2 cos−1 1 −4 5s i + sin3 1 2 cos−1 1 −4 5s j + 1 −4 5s k 9. (a) 1 5 ⟨4 cos 4t, 3, −4 sin 4t⟩, ⟨−sin 4t, 0, −cos 4t⟩ (b) 16 25 10. (a) 1 1 + t2 1, √ 2t, t2 , 1 √ 2 (1 + t2) −2t, √ 2 1 −t2 , 2t (b) 1 3 √ 2 (1 + t2)2 11. (a) 1 √ 2 − √ 2 sin t, cos t, cos t , 1 √ 2 − √ 2 cos t, −sin t, −sin t (b) 1 √ 2 12. (a) 1 t2 + 2 t2, 2t, 2 , 1 t2 + 2 2t, 2 −t2, −2t (b) 2 (t2 + 2)2 13. (a) 1 e2t + 1 √ 2e2t, e2t, −1 , 1 e2t + 1 1 −e2t, √ 2et, √ 2et (b) √ 2e2t (e2t + 1)2 14. (a) 1 2t2 + 1 2t, 2t2, 1 , 1 2t2 + 1 1 −2t2, 2t, −2t (b) 2 (2t2 + 1)2 15.
2 (4t2 + 1)3/2 16.
3 (1 + 18t2)3/2 17.
√ 1 + 4t2 + t4 6 (t4 + t2 + 1)3/2 18.
√ 6 2 (2t2 −4t + 5)3/2 19.
√ 2 (1 + cos2 t)3/2 20.
√ 2 4 21.
2 (4x + 1)3/2 22.
|sin x| (1 + cos2 x)3/2 23.
|x| (x2 + 1)3/2 24.
6 |t| (9t2 + 4)3/2 25.
2 + t2 (1 + t2)3/2 Copyright © 2013, Cengage Learning. All rights reserved.
SECTION 10.8 ARC LENGTH AND CURVATURE ■ 3 Click here for exercises.
E SOLUTIONS 10.8 1. r′ (t) = ⟨2, 3 cos t, −3 sin t⟩, |r′ (t)| = 4 + 9 cos2 t + 9 sin2 t = √ 13, L = b a √ 13 dt = √ 13 (b −a) 2. r′ (t) = et, et (sin t + cos t) , et (cos t −sin t) , |r′ (t)| = et 1 + 2 sin2 t + 2 cos2 t = √ 3et, L = 2π 0 √ 3et dt = √ 3 e2π −1 3. r′ (t) = 6, 6 √ 2t, 6t2 , |r′ (t)| = 6 √ 1 + 2t2 + t4 = 6 1 + t2 , L = 1 0 6 1 + t2 dt = 1 3 · 6 t + t3 1 0 = 24 3 = 8 4. r′ (t) = ⟨2, 2t, 2t⟩, |r′ (t)| = 2 √ 1 + 2t2 L = 1 0 √ 1 + 2t2 dt = b a √ 2 sec3 θ dθ = √ 2 2 [ln |sec θ + tan θ| + tan θ sec θ]b a = √ 2 2 ln √ 1 + 2t2 + √ 2 t + √ 2 t √ 1 + 2t2 1 0 (or use Formula 21) = √ 2 2 ln √ 3 + √ 2 + √ 2 √ 3 = √ 3 + √ 2 2 ln √ 2 + √ 3 5. r′ (t) = ⟨cos t −t sin t, sin t + t cos t, 1⟩, |r′ (t)| = cos2 t + t2 sin2 t + sin2 t + t cos2 t + 1 = √ t2 + 2 L = π/2 0 √ t2 + 2 dt = b a sec3 θ dθ = ln √ t2 + 2 + t + 1 2t √ t2 + 2 π/2 0 (or use Formula 21) = ln π2 4 + 2 + π 2 + π 4 π2 4 + 2 −ln √ 2 6. r′ (t) = et (cos t + sin t) i + et (cos t −sin t) j, ds/dt = |r′ (t)| = et (cos t + sin t)2 + (cos t −sin t)2 = et 2 cos2 t + 2 sin2 t = √ 2et s (t) = t 0 |r′ (u)| du = t 0 √ 2eu du = √ 2 et −1 ⇒ 1 √ 2s + 1 = et ⇒ t (s) = ln 1 √ 2s + 1 . Therefore, r (t (s)) = 1 √ 2s + 1 sin ln 1 √ 2s + 1 i + cos ln 1 √ 2s + 1 j 7. r′ (t) = 2 i + j −5 k, ds/dt = |r′ (t)| = √4 + 1 + 25 = √ 30 and s (t) = t 0 |r′ (u)| du = t 0 √ 30 du = √ 30 t ⇒ t (s) = 1 √ 30s. Therefore, r (t (s)) = 1 + 2 √ 30s i + 3 + 1 √ 30s j − 5 √ 30s k.
8. |r′ (t)| = (−3 cos2 t sin t)2 + 3 sin2 t cos t 2 + (−2 sin 2t)2 = 9 sin2 t cos2 t cos2 t + sin2 t + 4 sin2 2t = 9 4 (2 sin t cos t)2 + 4 sin2 2t = 9 4 + 4 sin2 2t = 5 2 sin 2t Then for 0 ≤t ≤π 2 , s (t) = t 0 |r′ (u)| du = 5 2 t 0 sin 2u du = −5 4 [cos 2u]t 0 = 5 4 (1 −cos 2t) Therefore, t (s) = 1 2 cos−1 1 −4 5s and r (t (s)) = cos3 1 2 cos−1 1 −4 5s i + sin3 1 2 cos−1 1 −4 5s j + cos 2 · 1 2 cos−1 1 −4 5s k = cos3 1 2 cos−1 1 −4 5s i + sin3 1 2 cos−1 1 −4 5s j + 1 −4 5s k 9. (a) T (t) = r′ (t) |r′ (t)| = 1 √16 + 9 ⟨4 cos 4t, 3, −4 sin 4t⟩ = 1 5 ⟨4 cos 4t, 3, −4 sin 4t⟩ N (t) = T′ (t) |T′ (t)| = 5 16 · 5 ⟨−16 sin 4t, 0, −16 cos 4t⟩ = ⟨−sin 4t, 0, −cos 4t⟩ (b) κ (t) = |T′ (t)| |r′ (t)| = 16 5 · 5 = 16 25 10. (a) T (t) = r′ (t) |r′ (t)| = 1 6 (1 + t2) 6, 6 √ 2t, 6t2 = 1 1 + t2 1, √ 2t, t2 N (t) = T′ (t) |T′ (t)| = 1 |T′ (t)| − 2t (1 + t2)2 1, √ 2t, t2 + 1 1 + t2 0, √ 2, 2t = 1 + t2 √ 2 1 (1 + t2)2 −2t, √ 2 1 −t2 , 2t = 1 √ 2 (1 + t2) −2t, √ 2 1 −t2 , 2t (b) κ (t) = |T′ (t)| |r′ (t)| = √ 2 1 + t2 · 1 6 (1 + t2) = 1 3 √ 2 (1 + t2)2 Copyright © 2013, Cengage Learning. All rights reserved.
4 ■ SECTION 10.8 ARC LENGTH AND CURVATURE 11. (a) T (t) = r′ (t) |r′ (t)| = 1 2 sin2 t + 2 cos2 t − √ 2 sin t, cos t, cos t = 1 √ 2 − √ 2 sin t, cos t, cos t N (t) = T′ (t) |T′ (t)| = 1 2 cos2 t + 2 sin2 t − √ 2 cos t, −sin t, −sin t = 1 √ 2 − √ 2 cos t, −sin t, −sin t (b) κ (t) = |T′ (t)| |r′ (t)| = 1 √ 2 12. (a) r′ (t) = t2, 2t, 2 ⇒ |r′ (t)| = √ t4 + 4t2 + 4 = (t2 + 2)2 = t2 + 2. Then T (t) = r′ (t) |r′ (t)| = 1 t2 + 2 t2, 2t, 2 .
T′ (t) = −2t (t2 + 2)2 t2, 2t, 2 + 1 t2 + 2 ⟨2t, 2, 0⟩ = 1 (t2 + 2)2 −2t3, −4t2, −4t + 1 (t2 + 2)2 2t3 + 4t, 2t2 + 4, 0 = 1 (t2 + 2)2 4t, 4 −2t2, −4t |T′ (t)| = 1 (t2 + 2)2 16t2 + (16 −16t2 + 4t4) + 16t2 = 1 (t2 + 2)2 √ 4t4 + 16t2 + 16 = 1 (t2 + 2)2 4 (t2 + 2)2 = 2 t2 + 2 (t2 + 2)2 = 2 t2 + 2 Thus N (t) = T′ (t) |T′ (t)| = 1/ t2 + 2 2 2/ (t2 + 2) 4t, 4 −2t2, −4t = 1 t2 + 2 2t, 2 −t2, −2t (b) κ (t) = |T′ (t)| |r′ (t)| = 2/ t2 + 2 t2 + 2 = 2 (t2 + 2)2 13. (a) T (t) = 1 √ 2 + e2t + e−2t √ 2, et, −e−t = 1 et + et √ 2, et, −e−t = 1 e2t + 1 √ 2e2t, e2t, −1 T′ (t) = −2e2t (e2t + 1)2 √ 2et, e2t, −1 + 1 e2t + 1 √ 2et, 2e2t, 0 = 1 (e2t + 1)2 −2 √ 2e3t + √ 2e3t + √ 2et, −2e4t + 2e4t + 2e2t, 2e2t = 1 (e2t + 1)2 √ 2 et −e3t , 2e2t, 2e2t |T′ (t)| = 1 (e2t + 1)2 2 (e2t + e6t −2e4t) + 8e4t = √ 2et (e2t + 1)2 e2t + 1 = √ 2et e2t + 1 N (t) = 1 (e2t + 1)2 e2t + 1 √ 2et √ 2 et −e3t , 2e2t, 2e2t = 1 √ 2 (e2t + 1) et √ 2 et −e3t , 2e2t, 2e2t = 1 e2t + 1 1 −e2t, √ 2et, √ 2et (b) κ (t) = √ 2et (e2t + 1) · 1 et + e−t = √ 2e2t (e2t + 1)2 14. (a) T (t) = 1 √ 4t2 + 4t4 + 1 2t, 2t2, 1 = 1 2t2 + 1 2t, 2t2, 1 T′ (t) = − 2t2 + 1 −2 (4t) 2t, 2t2, 1 + 2t2 + 1 −1 ⟨2, 4t, 0⟩ = 1 (2t2 + 1)2 −8t2 + 4t2 + 2, −8t3 + 8t3 + 4, −4t = 1 (2t2 + 1)2 1 −2t2, 2t, −2t |T′ (t)| = 1 (2t2 + 1)2 √ 1 −4t2 + 4t4 + 8t2 = 2 2t2 + 1 N (t) = 2 (2t2 + 1)2 · 2t2 + 1 2 1 −2t2, 2t, −2t = 1 2t2 + 1 1 −2t2, 2t, −2t (b) κ (t) = 2 2t2 + 1 · 1 2t2 + 1 = 2 (2t2 + 1)2 15. r′ (t) = j −2t k, r′′ (t) = −2 k, |r′ (t)|3 = 4t2 + 1 3/2, |r′ (t) × r′′ (t)| = |−2i| = 2, κ (t) = |r′ (t) × r′′ (t)| |r′ (t)|3 = 2 (4t2 + 1)3/2 16. r′ (t) = ⟨1, −1, 6t⟩, r′′ (t) = ⟨0, 0, 6⟩, |r′ (t)|3 = √ 2 + 36t23 = 2 1 + 18t2 3/2, |r′ (t) × r′′ (t)| = |⟨−6, −6, 0⟩| = 6 √ 2, κ (t) = |r′ (t) × r′′ (t)| |r′ (t)|3 = 6 √ 2 [2 (1 + 18t2)]3/2 = 3 (1 + 18t2)3/2 #3) (by Theorem 10.7.5 Copyright © 2013, Cengage Learning. All rights reserved.
SECTION 10.8 ARC LENGTH AND CURVATURE ■ 5 17. r′ (t) = 6t2, −6t, 6 , r′′ (t) = ⟨12t, −6, 0⟩, |r′ (t)|3 = 63 t4 + t2 + 1 3/2, |r′ (t) × r′′ (t)| = 36 1, 2t, t2
= 36 √ 1 + 4t2 + t4, κ (t) = |r′ (t) × r′′ (t)| |r′ (t)|3 = 36 √ 1 + 4t2 + t4 63 (t4 + t2 + 1)3/2 = √ 1 + 4t2 + t4 6 (t4 + t2 + 1)3/2 18. r′ (t) = ⟨2t, 2t −4, 2⟩, r′′ (t) = ⟨2, 2, 0⟩, |r′ (t)|3 = 4t2 + 4t2 −16t + 16 + 4 3/2 = 8 2t2 −4t + 5 3/2, |r′ (t) × r′′ (t)| = 4 |⟨−1, 1, 2⟩| = 4 √ 6, κ (t) = |r′ (t) × r′′ (t)| |r′ (t)|3 = 4 √ 6 8 (2t2 −4t + 5)3/2 = √ 6 2 (2t2 −4t + 5)3/2 19. r′ (t) = ⟨cos t, −sin t, cos t⟩, r′′ (t) = ⟨−sin t, −cos t, −sin t⟩, |r′ (t)|3 = √ cos2 t + 1 3, |r′ (t) × r′′ (t)| = |⟨1, 0, −1⟩| = √ 2, κ (t) = |r′ (t) × r′′ (t)| |r′ (t)|3 = √ 2 (1 + cos2 t)3/2 20. r′ (t) = √ 2, et, −e−t . The point (0, 1, 1) corresponds to t = 0, and r′ (0) = √ 2, 1, −1 ⇒ |r′ (0)| = √ 2 2 + 12 + (−1)2 = 2.
r′′ (t) = 0, et, e−t ⇒ r′′ (0) = ⟨0, 1, 1⟩.
r′ (0) × r′′ (0) = 2, − √ 2, √ 2 , |r′ (0) × r′′ (0)| = 22 + − √ 2 2 + √ 2 2 = √ 8 = 2 √ 2 Then κ (0) = |r′ (0) × r′′ (0)| |r′ (0)|3 = 2 √ 2 23 = √ 2 4 .
21. y′ = 1 2√x, y′′ = − 1 4 (x)3/2 , κ (x) = |y′′ (x)| 1 + (y′ (x))2 3/2 = 1 4 |x3/2| 1 [1 + 1/(4x)]3/2 = 2 (4x + 1)3/2 22. y′ = cos x, y′′ = −sin x, κ (x) = |y′′ (x)| 1 + (y′ (x))2 3/2 = |sin x| (1 + cos2 x)3/2 23. y′ = 1 x, y′′ = −1 x2 , κ (x) = |y′′ (x)| 1 + (y′ (x))2 3/2 = −1 x2 1 (1 + 1/x2)3/2 = 1 x2 x23/2 (x2 + 1)3/2 = |x| (x2 + 1)3/2 24. κ (t)= | ˙ x¨ y −¨ x ˙ y| ( ˙ x2 + ˙ y2)3/2 = 3t2 (2) −(6t) (2t) (9t4 + 4t2)3/2 = 6t2 (t2)3/2 (9t2 + 4)3/2 = 6t2 |t|3 (9t2 + 4)3/2 = 6 |t| (9t2 + 4)3/2 25. κ (t) = | ˙ x¨ y −¨ x ˙ y| ( ˙ x2 + ˙ y2)3/2 = (sin t + t cos t) (−2 sin t −t cos t) −(2 cos t −t sin t) (cos t −t sin t) ( ˙ x2 + ˙ y2)3/2 = −2 sin2 t −3t sin t cos t −t2 cos2 t − 2 cos2 t −3t cos t sin t + t2 sin2 t ( ˙ x2 + ˙ y2)3/2 = − sin2 t + cos2 t 2 + t2 sin2 t + 2t cos t sin t + cos2 t + t2 cos2 t −2t sin t cos t + t2 sin t 3/2 = 2 + t2 (1 + t2)3/2 Copyright © 2013, Cengage Learning. All rights reserved. |
13969 | https://www.youtube.com/watch?v=-9HvzM-Mkvo | Finding Area by Coordinate Method
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Posted: 11 May 2020
Transcript:
Intro yeah it's just the coordinates what we're gonna end up doing now so go ahead and copy down the coordinates so you have them off of here because we're going to do area by coordinate you're just another thing in easting so we're gonna figure out area here on the coordinate method we've already done it by triangle method and by the double Meridian distance method or DMD method so once everybody has coordinates written down let me know okay okay move on good to go genes okay everybody can see that tried zooming in enough so it was readable hopefully it readable all right so give you this nice Point Numbers little spreadsheet here so the first thing we're going to do is populate the point numbers of the control points or the the Traverse points that we have so we know that we have two three and four okay we have to go back on top of ourselves so we have to put in the number two again so we're going around our Traverse two three four and then back to two okay so we have to put in two again there all right so then we can just populate the northing and easting all the way down to the sides so our northing was 100 at number two the easting was 1,000 at two at number three it's 450 two point eight six and 1153 point three five point four one eighty nine point eight one and point the easting for four fourteen forty two point six two and then for number two we just go ahead and write the same thing again okay as Population easy as that so we didn't have to do that changing the latitude or to change the departure we don't really have to do that so we've cut all the stuff already from the DMD so now we have to populate this information over here the dental area okay so what we're gonna do is we start with the point that comes after the first point and we multiply up okay so we times there we take and multiply these two numbers this is 1,000 these two numbers and we put that in the negative column okay so whatever that number comes out to be I need someone to shout out what it is I'm not 450 to 860 point zero zero okay and then we do it again go to the preceding one part of that not preceding one the next point number in the line we go we have times the easting or the normal thing of four times the easting of three or I put it in the same column and that's what twenty-seven are two seven three eight two three point seven zero so you're on 180 9.81 times eleven fifty three point three five two eighteen that was that number again nine seventeen point three six okay so we're going to do it again right here so we go from the northing of two to the easting of four and what does that come out to be point zero zero all right so now in order to get the double area and the pluses column we have to take the northing of the first times the easting of the second point okay Norling in the first times the easting of the second Oh copying it here that's eleven or one one five three three five point zero zero right all right so then we're at dude again so we're gonna take the northing of three times the northing of four are the easting of four sorry the easting of for another thing of three times the easting of four and whoever has that number go ahead and shout it okay and then the last one is the northing of four times the easting of two is that one eight nine eight one zero point zero zero okay all right so now we populated the double areas so now we're going to take and sum these up down each side and give that to me and then we'll sum up the negative column and you can shout that one also so you guys you can already see this is so much simpler than the Dieppe right area by DM D because you can use the straight coordinate anybody have the sum of the pluses nine five eight four four nine point eight nine I want the sum of the minuses okay so you probably already can guess what we're going to do with this we're going to take and add these together so we're gonna have nine five eight four four nine point eight nine plus a negative eat one six zero three nine point three six why did I call this a negative because it's in the negative column right so that's why this is a negative so what do we get there one for two for one 0.53 okay and we take and we'll take that and divide it by 2 so 1 4 2 4 1 zero point 5 3 divided by 2 what does that equal okay and that's square feet okay so this this one here was dental area this is now single area okay so now what's the a courage then so we take the seven one two zero five point two seven and divide by 43 560 and it's going to give us what one point six three acres okay pretty simple right that that process is pretty quick lot quicker than the DMD method way quicker okay like within the triangle method to the app coordinates pretty simple to come up with an area really fast now the only time that this doesn't work though well is if you have a curve associated with it but anything you have straight lines segments you can find an area pretty easily just by this method so alright everybody good there let's let's move on to the four-sided figure that we had from last couple days see if this one I've had to zoom in or out doesn't look like it all right so go ahead and copy down if you want to you can start populating that sheet right away with your northings and these things starting at 0.5 there you go little easier see no so we already solved the area on this I just hope you all can see that so once you have this L thought that northings and eastings I'll copy down we'll move on and actually solve this one okay all right let's populate this baby if you haven't done it already so we're going to go to point five six seven eight five okay that's the numbers from the country are the points that were in our Traverse so the northing at five one hundred point one zero zero point zero zero and the e sting is two thousand point zero zero northern at six four nine nine two four the e sting is two zero eight five point zero three more than s seven four nine nine point two four easting is two four five 3.76 eats Norling 100 the easting at eight two six two four point three six and then we go back to from bur five again we add that one in all right so that's my first step I can start either way I can do I can do the pluses or minuses so these ones you guys want to do first okay we can do that so then populate the pluses column first so whatever so we're gonna add it here this is how we figure out the pluses column zero eight five zero three point zero zero now here's one thing we know about this it doesn't really matter if I'm populating the double area with the northing in the plus column or the minus column as long as I'm consistent okay so I could when I did this multiplication here I could Multiplication have put it over here so as long as every time if I go down like this I'm always putting in the same column okay so doesn't really matter which column you decide to put it in just make sure you're consistent with the numbers you're putting in there okay so if I go six to seven northern is six to the easting of seven what's that number zero one five okay next one 1 million three hundred ten thousand one hundred eighty five point four nine okay we have one more to be pretty easy one so it looks like two hundred thousand I could unlock that alright so the next step we populated the plusses column like I said just as long as you're consistent so if I wanted to put these over here it doesn't really matter because we drop the negative once we get to this square footage anyway so but the next step would be northing of six times the easting of five what do we get there okay then the northing of seven times the easting of six you may have that number higher than 30 okay well northern have eat to the east Sina 7.00 that wasn't easy when I could do that on my head yeah next one okay so we got the rest of them populated so I had from five building a five to easting of eight and that gave me two six two four three six point zero zero okay so some Double Area loads up someone give me the pluses when they get it two one four three nine nine four two nine four three seven oh three okay the negatives some of the negatives right 3/8 okay let's take in put these together the negative two five four seven two two two point three eight what do we get for an answer there four eighty one two five okay so that's the double area right so we want to make that divided by two and that should give us 198 to forty point six three square feet yep sorry square feet okay and we want to turn this in a hunch so we take four or 198 to forty point six three divided by forty three five sixty and that's going to equal four point five five acres so if we look at it with the triangle formula a slide this in there real quick we came up with the same thing four point five five acres now we're off by about one square foot if you look at them so about one square foot that we're off as all that is so everybody gather that way way easier to use the coordinate system yes with the double meridian check ourself as we are going but this way deserve a are not really there's not really a way to check yourself as you go so yeah that's the nice thing about the double Meridian there's a couple different ways you can check yourself yes you do that the result is northern if you're taking the point the point that your say you're sitting on it then you take the point you're going to it's the northern times the easting yeah and then when you flip it over when you want to get that the next difference so it doesn't matter if you're populating the plus or minus just wanting to stick the same you're gonna take the the the new point time's the OL the previous point essentially so six you know so you're always taking fight a the previous northing times the next east thing and then when you get done populating one of these columns it doesn't matter which one you populate then you'll take the northing times the preceding easting the plus and minus it doesn't really matter which column you populated and just as long as you're consistent like I said so if I wanted to do these I could put it in this column fill on it too as long as every time I multiply down I'm consistent with whatever column I stick it in to stay in that column everyone that we go down on if I'm going up multiplying up like this I can pick a column just as long as everyone that I multiply up like this I stay in that column because we're taking one minus the other you're gonna get the same answer whether this numbers here or and that numbers there or vice versa it doesn't really matter you're still gonna get the same result here so still a double area we drop the sign here if it's a negative this come becomes a negative you're dropping the sign anyway because at that point you're turning into the square feet so there we understand that okay all right so we got one more so we can really hit home so we're gonna get that big one that we worked on so hopefully everybody can read those more things in these things so you go ahead popular them right here and the columns right away if you would like start at night and end at night and remember that when you go to the house yes yes yes three nice two guys don't you syllabus a movie well just just trendy yeah will do for sketches a lot yeah for instance that sir locate the property lines get you the house exactly in that way the years all right so let's move on then and will populate this one so we will nine ten Closing Traverse eleven twelve thirteen and back to nine boys we're going to wrap around to that last point because we're closing our Traverse if you can think of it that way so one hundred three thousand four seven three six nine and I'm just populating these right from that sketch that I just had up there okay all right so whenever some wants to spit out some some double areas for me Double Traverse that'd be fantastic so go we'll go from the northing and nine to the easting of ten okay that again hey next one and then the last one pretty easy one okay we populated the plus column we could have put those in the negative column don't really matter so all right so we're gonna go from the northing of ten to the easting of nine we're just gonna do that all the way down anyway have numbers for me [Music] okay so we got out of the negatives everybody following yet or ahead of me that's fine you very done this quite often so give me the sum of the pluses some of the negatives six-one all right slowed subtract those five eight seven seven three six four point zero seven are zero six sorry minus five five seven five four two point six one what do we get therefore the double area okay so I take that and three zero one nine six one point four five divided by two and that equals that's square feet and let's turn that into an area bye-bye 43 560 and that equals three point four seven acres okay right gather how we went about that and if I look back or off by three square feet from the total for the triangle method if we look triangle method we came up one fifty nine eight three and in this method we came up with one fifty nine eight one if we were owned up swear off by two square feet not terrible it's rounding in the math as all that is so both are both are acceptable if you have a difference when you're doing your homework on these if you're off by a couple square feet no big deal so there you guys just completed three problems from area by coordinates which is a lot simpler way of doing it so if a male asked me I'll leave this up here I can leave it up here for a little bit if you want to copy or do anything like that stop this |
13970 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4873908/ | Tumour assessment and staging: United Kingdom National Multidisciplinary Guidelines - PMC
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J Laryngol Otol
. 2016 May;130(Suppl 2):S53–S58. doi: 10.1017/S002221511600044X
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Tumour assessment and staging: United Kingdom National Multidisciplinary Guidelines
N Roland
N Roland
1 Department of ENT – Head & Neck Surgery, University Hospital Aintree, Liverpool, UK
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1,✉, G Porter
G Porter
2 Department of ENT – Head & Neck Surgery, University Hospitals Bristol, East Grinstead, UK
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2, B Fish
B Fish
3 Department of Otolaryngology – Head & Neck Surgery, Cambridge University Teaching Hospitals Trust, East Grinstead, UK
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3, Z Makura
Z Makura
4 Head & Neck Unit, Queen Victoria Hospital NHS Foundation Trust, East Grinstead, UK
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4
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1 Department of ENT – Head & Neck Surgery, University Hospital Aintree, Liverpool, UK
2 Department of ENT – Head & Neck Surgery, University Hospitals Bristol, East Grinstead, UK
3 Department of Otolaryngology – Head & Neck Surgery, Cambridge University Teaching Hospitals Trust, East Grinstead, UK
4 Head & Neck Unit, Queen Victoria Hospital NHS Foundation Trust, East Grinstead, UK
✉
Address for correspondence: Nick Roland, Department of ENT – Head & Neck Surgery, University Hospital, Aintree, Liverpool, UK E-mail: DrNJRoland@aol.com
© JLO (1984) Limited 2016
This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence ( which permits unrestricted re-use, distribution, and reproduction in any medium, provided the original work is properly cited.
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PMCID: PMC4873908 PMID: 27841118
Abstract
In general, the first decision to be made in a patient with a confirmed head and neck cancer is whether or not to treat the patient before deciding what form of management strategy is appropriate. There is no more important an aspect of head and neck cancer care than the initial evaluation of the patient and the patient's tumour. The practice requires specific expertise and judgement. The current tumour–node–metastasis system relies on morphology of the tumour (anatomical site and extent of disease) but the final decision on treatment hinges on a full assessment of the patient including physiological age and general condition. The aim of this paper is primarily to describe why and how we appraise a patient and their tumour. It addresses the general principles applicable to the topic of evaluation, classification and staging. In addition, the limitations and pitfalls of this process are described.
Recommendations
• All patients with head and neck cancer (HNC) should undergo tumour classification and staging prior to treatment. (R)
• Pre-therapeutic clinical staging of HNCs should be based on at least a C2 factor (evidence obtained by special diagnostic means, e.g. radiographic imaging (e.g. computed tomography, magnetic resonance imaging or ultrasound scan), endoscopy, biopsy and cytology). (R)
• Imaging to evaluate the primary site should be performed prior to biopsy to avoid the effect of upstaging from the oedema caused by biopsy trauma. (G)
• Panendoscopy is only recommended for symptomatic patients or patients with primary tumours known to have a significant risk of a second (synchronous) primary tumour. (G)
Introduction
There are many aspects affecting the outcome of patients with malignant head and neck tumours. These may relate to the tumour (e.g. the anatomical site and extent of the disease), the host (age, general condition and any concurrent disease) and management (treatment options, expertise available and patient preference).
Staging of head and neck cancer (HNC) is a system designed to express the relative severity, or extent, of the disease. The objectives are illustrated in Table 1.
Table I.
Objectives of staging
1.
To aid the clinician in the planning of treatment
2.
To give some indication of prognosis
3.
To assist in evaluation of the results of treatment
4.
To facilitate the exchange of information between treatment centres
5.
To contribute to the continuing investigation of human cancer
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The nature of staging has meant that the data to support the concept have been largely drawn from retrospective and observational studies. Much of the systems development has been through the opinion of expert panels using these data.
Both the International Union against Cancer (UICC) and the American Joint Committee on Cancer (AJCC) published rules on classification and staging which correspond in their 7th editions (2009) and have approval of all national tumour–node–metastasis (TNM) committees.1,2
Sites in the head and neck region
The TNM classification applies only to carcinomas and melanomas in the following sites: lip and oral cavity, pharynx (oropharynx, nasopharynx and hypopharynx), larynx, maxillary sinus, nasal cavity and ethmoid sinus, mucosal malignant melanoma, major salivary glands and thyroid gland. Each site is described having rules for classification, anatomical sites and subsites where appropriate, the clinical TNM (cTNM) classification, the pathological TNM (pTNM) classification, G histopathological grading, stage grouping and a summary. The main aspects are described here, but specific details can be found in the most recent UICC and AJCC TNM booklets.1,2
General rules
The TNM system for describing the anatomical extent of the disease is based on three components (Tables II–IV):
Table II.
An overview of the tnm staging terminology
T – Primary tumour
TX Primary tumour cannot be assessed
T0 No evidence of primary tumour
Tis Carcinoma in situ
T1, T2, T3, T4 Increasing size and/or local extent of the primary tumour
N – Regional lymph nodes
NX Regional lymph nodes cannot be assessed
N0 No evidence of regional lymph node metastases
N1, N2, N3 Increasing involvement of regional lymph nodes
M – Distant metastasis
M0 No distant metastasis
M1 Distant metastasis
The previously included MX category is now considered to be inappropriate.
The category M1 may be further specified according to the following notation:
Pulmonary PUL Bone marrow MAR
Osseous OSS Pleura PLE
Hepatic HEP Peritoneum PER
Brain BRA Adrenals ADR
Lymph nodes LYM Skin SKI
Other OTH
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Table III.
Histopathological grading system for squamous cell carcinoma
GX Grade of differentiation cannot be assessed
G1 Well differentiated
G2 Moderately differentiated
G3 Poorly differentiated
G4 Undifferentiated
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G=Histopathological grading
Table IV.
Optional descriptors used for histopathological reporting in squamous cell carcinoma
| Optional descriptors |
| Pn – Perineural invasion | |
| PnX | Perineural invasion cannot be assessed |
| Pn0 | No perineural invasion |
| Pn1 | Perineural invasion |
| L – Lymphatic invasion | |
| LX | Lymphatic invasion cannot be assessed |
| L0 | No lymphatic invasion |
| L1 | Lymphatic invasion |
| V – Venous invasion | |
| VX | Venous invasion cannot be assessed |
| V0 | No venous invasion |
| V1 | Microscopic venous invasion |
| V2 | Macroscopic venous invasion |
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T – Extent of the primary tumour
N – Absence or presence and extent of regional lymph node metastases
M – Absence or presence of distant metastases
All cases should be confirmed microscopically. Two classifications should be documented for each site, namely: cTNM (clinical (pre-treatment) classification) and pTNM (post-surgical histopathological classification). The clinical stage is essential to select and evaluate therapy, while the pathological stage provides the most precise data to estimate prognosis and calculate end results. It should be remembered that if there is doubt concerning the correct T, N or M category to which a particular case should be allotted, then the lower (i.e. less advanced) category should be chosen. After assigning the cTNM and pTNM categories, the patient should then be classified in a Stage Group. Once established, this must remain unchanged in the medical records.1,2
See site-specific chapters for each detailed tumour classification.
Histopathological grading
The histological grading of squamous cell carcinoma represents estimation by the pathologist of the expected biologic behaviour of the neoplasm. Although it is subject to inter- and intra-observer errors, it has been suggested such information in conjunction with other characteristics of the primary tumour is useful in the rational approach to therapy.3 The grade can be applied to all head and neck sites except thyroid.
Additional descriptors
Designation is now applicable when sentinel lymph node biopsy is attempted using the suffix (sn) after N stage. Optional descriptors for perineural invasion (Pn), lymphatic invasion (L) and venous invasion (V) may be used.
The absence or presence of residual tumour after treatment may be described by the symbol R. A recurrent tumour, when classified after a disease-free interval is identified by the prefix ‘r’. The prefix ‘a’ indicates that classification is first determined at autopsy. The suffix ‘m’ is used to indicate the presence of multiple primary tumours at a single site. In cases where multimodality treatment is used, the cTNM or pTNM is identified by a ‘y’ prefix which categorises the extent of tumour actually present at the time of that examination.
The C-factor, or certainty factor, reflects the validity of classification according to the diagnostic methods employed (C1–C5). C1 would be evidence from standard diagnostic means whereas C5 is evidence from autopsy. Generally speaking, pre-therapeutic clinical staging of HNCs is equivalent to C1, C2 and C3, whilst pathological classification is equivalent to C4.1,2
Related classifications
The World Health Organization (WHO) has developed a series aimed at classification of tumours. The WHO International Classification of Diseases for Oncology (ICD-O) is a coding system for neoplasms by topography and morphology and for indicating behaviour (e.g. malignant and benign).4 This coded nomenclature is identical in the morphology field for neoplasms to the Systemised Nomenclature of Medicine.5 It is recommended that the WHO classification of tumours is used for classification and definition of tumour types and that the ICD-O code is used for storage and retrieval of data.
Stage grouping
After TNM, classification of tumours should be assigned a stage grouping between 0 or I and IV (Tables V). The grouping adopted is designed to ensure, as far as possible, that each group is more or less homogeneous in respect of survival and that the survival rates for each cancer stage are distinctive. Carcinoma in situ is categorised as stage 0; cases with distant metastasis as stage IV. The exceptions to this grouping are for carcinoma of the nasopharynx, carcinoma of the thyroid (Tables VI and VII) and mucosal melanoma.1,2
Table V.
Stage grouping for head and neck cancers excluding nasopharynx, thyroid and mucosal melanoma
Stage 0 Tis N0 M0
Stage I T1 N0 M0
Stage II T2 N0 M0
Stage III T1, T2, T3 N1 M0
T3 N0 M0
Stage IVA T1, T2, T3 N2 M0
T4a N0, N1, N2 M0
Stage IVB Any T N3 M0
T4b Any N M0
Stage IVC Any T Any N M1
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Table VI.
Stage grouping for carcinoma of the nasopharynx
Stage 0 Tis N0 M0
Stage I T1 N0 M0
Stage II T1 N1 M0
T2 N0,N1 M0
Stage III T1,T2 N2 M0
T3 N0, N1, N2 M0
Stage IVA T4 N0, N1, N2 M0
Stage IVB Any T N3 M0
Stage IVC Any T Any N M1
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Table VII.
Stage grouping for thyroid carcinoma
Papillary or follicular under 45 years
Stage I Any T Any N M0
Stage II Any T Any N M1
Papillary or follicular 45 years and older
Stage I T1a, T1b N0 M0
Stage II T2 N0 M0
Stage III T3 N0 M0
T1, T2, T3 N1a M0
Stage IVA T1, T2, T3 N1b M0
Stage IVB T4a N0, N1 M0
Stage IVC T4b Any N M0
Any T Any N M1
Medullary
Stage I T1a, T1b N0 M0
Stage II T2, T3 N0 M0
Stage III T1, T2, T3 N1a M0
Stage IVA T1, T2, T3 N1b M0
Stage IVB T4a Any N M0
Stage IVC T4b Any N M0
Any T Any N M1
Anaplastic (all cases are stage IV)
Stage IVA T4a Any N M0
Stage IVB T4b Any N M0
Stage IVC Any T Any N M1
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Separate stage groupings are recommended for papillary and follicular, medullary and undifferentiated carcinomas
Methods of assessment
The aim is to define in each patient all of the factors relevant to the natural history and outcome of the relevant disease, thereby enabling a patient with cancer to be grouped with other similar cases. The sex and age of the patient, the duration and severity of symptoms and signs and the presence and severity of concurrent disease should all be documented.
Computed tomography (CT) and magnetic resonance imaging are now established as the mainstay investigations in the pre-operative work-up of patients with HNC, to delineate the extent and size of the primary tumour, to determine the presence (particularly when risk of occult nodes is >20 per cent), number and position of cervical lymph nodes, to search for an occult primary and to locate a synchronous primary or distant metastases (particularly the chest). Appropriate screening for synchronous tumours and distant metastases is particularly important in advanced tumours. Several studies have suggested that a CT scan should be obtained in preference to a plain chest X-ray as this may miss significant lung pathology.6 There is a growing body of evidence that points to the value of 18F fluoro-deoxyglucose-positron emission tomography/CT in the management of HNC patients and predicting patient-related outcomes. It is invaluable in the detection of the unknown primary and useful in the confirmation of residual or recurrent disease, but is not routinely used in initial staging assessment.7
Endoscopy and biopsy should be performed by a senior surgeon and in all cases by the head and neck surgeon responsible for any future procedure. This should include for each tumour a description, diagrammatic representation and preferably also photographic documentation. Routine panendoscopy (oesophagoscopy and bronchoscopy) is contentious. Proponents point out that these procedures require very little time, and may be performed easily during planned, direct laryngoscopy. A large meta-analysis found a small advantage to panendoscopy in detection of second primary tumours during analysis of multiple prospective studies.8 Opponents point out that the appropriate use of symptom directed investigations in addition to routine chest radiography have a similar detection rate compared with screening endoscopy and avoid unnecessary risk and expense in asymptomatic patients.9 McGarey et al.10 concluded that while rigid oesophagoscopy is safe, the utility is low for cancer staging and for detection of non-malignant oesophageal disease. Review of the literature and analysis of a large national cancer dataset indicate that the incidence of synchronous oesophageal malignant neoplasms in patients with head and neck squamous cell carcinoma is low and has been decreasing during the past three decades.10 Thus, screening oesophagoscopy should be limited to patients with head and neck squamous cell carcinoma who are at high risk for synchronous oesophageal malignant neoplasms.
There is a natural desire to confer a stage on the tumour at presentation in the clinic and certainly after endoscopy. This should be avoided. It is better to rely on descriptive text to avoid changing the stage as more information becomes available. The clinical (pre-treatment) classification (cTNM) based on examination, imaging, endoscopy and biopsy should be clearly documented in the case-file only when all of the above information is collated. The UICC book should be available in every theatre and clinic to assist in applying the correct stage.
Regional lymph nodes
The status of the regional lymph nodes in HNC is of such prognostic importance that they must be assessed for each patient and tumour. Lymph nodes are described as ipsilateral, bilateral, contralateral or midline; they may be single or multiple and are measured by size, number and anatomical location (Table VIII). Midline nodes are considered ipsilateral nodes except in the thyroid. Direct extension of the primary tumour into lymph nodes is classified as lymph node metastasis.1,2
Table VIII.
N STAGING FOR REGIONAL LYMPH NODES
NX Regional lymph nodes cannot be assessed
N0 No regional lymph node metastasis
N1 Metastasis in a single ipsilateral lymph node. 3 cm or less in greatest dimension
N2 N2a Metastasis in a single ipsilateral lymph node, more than 3 cm but not more than 6 cm in greatest dimension
N2b Metastasis in multiple ipsilateral lymph nodes, none more than 6 cm in greatest dimension
N2c Metastasis in bilateral or contralateral lymph nodes, none more than 6 cm in greatest dimension
N3 Metastasis in a lymph node more than 6 cm in greatest dimension
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Imaging for node detection and delineation is recommended in the following settings: the neck is being scanned as part of the evaluation of the primary tumour; there is a high chance of occult disease (e.g. supraglottic primary); to assess the extent of nodal disease; to define any deep nodal fixation; or if clinical assessment is difficult because of a short, fat or previously irradiated neck.
Lymph nodes are subdivided into specific anatomic sites and grouped into seven levels for ease of description. The pattern of lymphatic drainage varies for different anatomic sites. However, the location of the lymph node metastases has prognostic significance. Survival is significantly worse when metastases involve lymph nodes beyond the first echelon of lymphatic drainage.11 It is particularly poor for lymph nodes in the lower regions of the neck, i.e. levels IV and V (supraclavicular area).
International Union Against Cancer and AJCC recommend that each N-staging category be recorded to show, in addition to the established parameters, whether the nodes involved are located in the upper (U) or lower (L) regions of the neck, depending on their location above or below the lower border of the thyroid cartilage.1,2
The definitions of the N categories for all head and neck sites are the same (Table VIII) except thyroid (Table IX) and nasopharynx (Table X). The natural history and response to treatment of cervical nodal metastases from nasopharynx are different, in terms of their impact on prognosis, so they justify a different N classification. Regional lymph node metastases from well-differentiated thyroid cancer do not significantly affect the ultimate prognosis and therefore also justify a unique system.
Table IX.
N Staging for thyroid carcinoma
NX Regional lymph nodes cannot be assessed
N0 No regional lymph node metastasis
N1 Regional lymph node metastasis
N1a Metastasis in level VI (pre-tracheal, pre-laryngeal, paralaryngeal) nodes
N1b Metastasis in other unilateral, bilateral or contralateral cervical or retropharyngeal or superior mediastinal lymph node(s)
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Table X.
N Staging for nasopharynx
NX Regional lymph nodes cannot be assessed
N0 No regional lymph node metastasis
N1 Unilateral cervical, unilateral or bilateral retropharyngeal lymph node(s), 6 cm or less in greatest dimension, above supraclavicular fossa
N2 Bilateral cervical lymph node(s), 6 cm or less in greatest dimension, above supraclavicular fossa
N3 Metastasis in lymph node >6 cm and/or (in the supraclavicular fossa:
(N3a)Greater than 6 cm in dimension
(N3b)In the supraclavicular fossa
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Note: Midline nodes are considered ipsilateral nodes, and the supraclavicular triangle is defined by the lines joining the following three points – the superior margin of the clavicle at its sternal and acromial ends, and the point where the line of the neck meets the shoulder.
Pathological classification (pTNM)
The pT, pN and pM categories correspond to the T, N and M categories, respectively. The extent of the tumour in terms of the location and level of the lymph node should be documented. In addition, the number of nodes that contain tumour and the presence or absence of extracapsular spread of the tumour should be recorded. Histological examination of a selective neck dissection including central compartment specimen usually includes six or more lymph nodes; a radical or modified radical neck dissection specimen includes 10 or more lymph nodes.1,2
The current TNM system relies on morphology of the tumour (anatomical site and extent of disease) with little or no attention given to patient factors. However, the literature does suggest that symptom severity12 and comorbidity13 have a significant impact on outcomes. It is therefore recommended that these data be recorded. Definitions of TNM categories may be altered or expanded for clinical or research purposes as long as the basic definitions are recorded and not changed. Despite the obvious value of staging, both in the management of individual patients and for the grouping of patients in trials and reports of treatment, it does have its limitations. The most insidious of these is that attempts to increase the accuracy of staging leads to greater complexity, and hence paradoxically to more errors and an increased likelihood of non-compliance by the person responsible for staging. Advances in methods of collecting and recording data will hopefully reduce these errors. Changes in the TNM classification should and will only occur, based on the appropriate collection, presentation and analysis of data, in the forum of the UICC and AJCC.3,4 The principles, practice and limitations of the current staging system are well documented in many major texts.14–16 Changes between editions tend to be conservative and commentaries regarding HNC reflect this.17 It is seven years since the 7th edition of the UICC and AJCC staging manuals and the updated version is eagerly awaited. The early indications are that changes will be only subtle and few.
Key points
•
Staging of head and neck cancer is a system designed to express the relative severity, or extent, of the disease. It is meant to facilitate an estimation of prognosis and provide useful information for treatment decisions. Classification by anatomical extent of head and neck cancer as determined clinically and histopathologically is the TNM System
•
Radiological investigations to evaluate the primary site should be performed prior to biopsy to avoid the effect of upstaging from the oedema caused by biopsy trauma
•
The sex and age of the patient, the duration and severity of symptoms and signs, and the presence and severity of inter-current disease should all be documented
•
Assessment by endoscopy and biopsy should be performed by a senior surgeon and in all cases by the Head & Neck surgeon responsible for any future procedure
•
The clinical (pre-treatment) classification (cTNM) based on examination, imaging, endoscopy and biopsy should be clearly documented in the case-file only when all the information is collated
•
Individual TNM classifications should be assembled into four groups – stage groups (stages I–IV), each with similar survival outcomes
•
The UICC book should be available in every theatre, MDT meeting and clinic to assist in applying the correct stage.
References
1.Sobin LH, Wittekind CH, Gospodarowicz M. TNM Classification of Malignant Tumours, UICC, 7th edn. New York: Wiley-Liss, 2009 [Google Scholar]
2.Edge SB, Byrd DR, Compton CC, Fritz AG, Greene FL, Trotti A, eds. The AJCC Cancer Staging Manual, 7th edn. New York: Springer, 2009 [Google Scholar]
3.Roland NJ, Caslin AW, Nash J, Stell PM. The value of grading squamous cell carcinoma of the head and neck. Head Neck 1992;14:224–9 [DOI] [PubMed] [Google Scholar]
4.Fritz A, Percy C, Jack A, Shanmugaratnam K, Sobin L, Parkin DM et al. , eds. WHO International Classification of Diseases for Oncology ICD-O, 3rd edn. Geneva: WHO, 2000 [Google Scholar]
5.SNOMED International: The Systemised Nomenclature of Human and Veterinary Medicine. Northfield, Ill: College of American Pathologists, SNOMED CT. The Systematized Nomenclature of Medicine Clinical Terms. Copenhagen Denmark. The International Health Terminology Standards Development Organisation, (accessed 27 April 2016) [Google Scholar]
6.Ghosh SK, Roland NJ, Kumar A, Tandon S, Lancaster JL, Jackson SR et al. Detection of synchronous lung tumors in patients presenting with squamous cell carcinoma of the head and neck Head Neck 2009;31:1563–70 [DOI] [PubMed] [Google Scholar]
7.Sheikhbahaei S1, Marcus C1, Subramaniam RM. 18F FDG PET/CT and head and neck cancer: patient management and outcomes. PET Clin 2015;10:125–45 [DOI] [PubMed] [Google Scholar]
8.Haughey BH, Gates GA, Arfken CL, Harvey J. Meta analysis of second malignant tumours in head and neck cancer: the case for an endoscopic screening protocol. Ann Otol Rhinol Laryngol. 1992;101:105–12 [DOI] [PubMed] [Google Scholar]
9.Davidson J, Witterick I, Gilbert R, Brown D, Birt D, Freeman J et al. The role of panendoscopy in the management of mucosal head and neck malignancy. Head Neck 1999;22:1–6 [DOI] [PubMed] [Google Scholar]
10.McGarey PO Jr, O'Rourke AK, Owen SR, Shonka DC Jr, Reibel JF, Levine PA et al. Rigid esophagoscopy for head and neck cancer staging and the incidence of synchronous esophageal malignant neoplasms. JAMA Otolaryngol Head Neck Surg 2016;142:40–5 [DOI] [PubMed] [Google Scholar]
11.Jones AS, Roland NJ, Field JK, Phillips D. The level of cervical lymph node metastases: their prognostic relevance and relationship with head and neck squamous carcinoma primary sites. Clin Otolaryngol 1994;19:63–9 [DOI] [PubMed] [Google Scholar]
12.Pugliano FA, Picirillo JF, Zequeria MR, Fredrickson JM, Perez CA, Simpson JR et al. Symptoms as an index of biologic behaviour in head and neck cancer. Otolaryngol Head Neck Surg 1999;120:380–6 [DOI] [PubMed] [Google Scholar]
13.Picirillo JF. Importance of comorbidity in head and neck cancer. Laryngoscope 2000;110:593–602 [DOI] [PubMed] [Google Scholar]
14.Roland NJ. Staging of Head and Neck Cancer. Scott-Brown's otolaryngology, Head and Neck Surgery, Volume 2. 7th edn. London: Hodder Arnold, 2008;2359–71 [Google Scholar]
15.Roland NJ. Assessment of Head and Neck Cancer. Stell & Maran's Head and Neck Surgery, 5th edn. London: Hodder Arnold, 2010 [Google Scholar]
16.Gospodarowicz M, O'Sullivan B, Sobin L. Prognostic Factors in Cancer, 4th edn. New York: Wiley-Liss, 2009 [Google Scholar]
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Articles from The Journal of Laryngology and Otology are provided here courtesy of Cambridge University Press
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Abstract
Introduction
Sites in the head and neck region
General rules
Histopathological grading
Additional descriptors
Related classifications
Stage grouping
Methods of assessment
Regional lymph nodes
Pathological classification (pTNM)
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13971 | https://www.youtube.com/watch?v=pLqgMULp0Ew | How to Calculate the Vector Cross Product
Math and Science
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Posted: 24 Feb 2015
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In this lesson, the student will learn what the cross product is between two vectors. We will learn why cross products are used in calculations, and how to find the cross product of 3-D vectors.
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Transcript:
Introduction Hello, welcome to this lesson in engineering mechanics. Here we're going to cover the cross productduct and we're going to talk about the cross productduct in terms of math in terms of of the mathematical definition. We're not going to talk about moments in specific here, but I'm kind of giving you a little bit of a preview. Basically, the moment calculation that we've been doing in two dimensions is kind of a simplified version because it's in two dimensions, but in reality, it's a cross productduct of two vectors. And when we get to more complicated problems in three dimensions, it's difficult to do this two-dimensional calculation. Uh so what we do is resort to cross productduct in three dimensions. So you're going to need a couple of tricks in your tool bag in order to be able to calculate vectors uh vector cross productds. Now I know that everybody watching this has actually been exposed to cross productds before. You've probably been exposed to it in a physics course. You've probably been exposed to it in a calculus course. But just in case you're joining me now for the first time and you haven't taken those classes with me, I want to do this section and do it justice on the cross product. I'm going to warn you a little bit. It's going to get uh to be a slog uh to get through it in the middle. Uh but I I think sometimes when I teach these classes, it's worth five extra minutes to go through or maybe 15 extra minutes to go through some details. But when we get to the end of this lesson, I guarantee you that you will be more confident in your understanding of the cross productd than when you started. uh un unless you just use these things all the time. Maybe you already have that insight, but I bet most people watching it will be more comfortable at the end. The Cross Product So, just get through it with me, I guess, is what I'm trying to say. I promise there's a point to it. All right, so the cross productduct of of two vectors, uh we've already talked about the dot productduct a long time ago and the complement to that uh or the alternative to that sometimes we end up using the cross productduct. So, we say vector C is equal to the crossroduct of vector A time vector B. Now you should know since you've taken classes before that this is not a straight multiplication this x it indicates a cross productduct and since vectors have magnitude and direction then when we operate and cross a into b then the resulting vector that we get also has a magnitude and a direction. All right so that's a little different right off the bat than the dot productduct. Remember we dot two vectors together we get a number back a scaler. We just get four org -3 back for a dot productduct. But for a cross productduct, you're always going to get a vector back. Okay, a cross b will give us the vector c. Now the magnitude of this cross productduct or the magnitude of the result here um is something that we can use occasionally here. Um the magnitude notice I haven't put any bar on the top of it. So you can say the magnitude of the vector C is equal to the magnitude of A times the magnitude of B right um and times the sign of the angle between these guys here. This is a direct analog or or at least a direct comparison to what we did in the dot productduct. Remember in the dot productduct we said the magnitude of the dot productduct can be uh of a dob is a b uh times or magnitude of a time magnitude of b times the coine of the angle between them. Now this should look somewhat similar to you if you remember the dot The Direction product was defined as uh magnitude of a time magnitude of b time cosine of the angle between them and this is real similar magnitude of a time magnitude of b times the sign of the angle between them. But when you do this calculation in terms of a cross productduct, what you're getting back is the magnitude of the of the vector that's the c the uh the cross of those two guys here. So the the magnitude of this vector that's the cross productduct here is given by this calculation. But since it's a vector, it also has a direction. So the direction of this crossroduct is given by the right-h hand rule, which we've already given you a preview of. Because in some of those moment problems, I was kind of trying to to tell you, hey, if you curl your fingers and your your thumb comes out in the direction of the moment. So, it should come as no surprise the moment is going to end up being the crossroduct of two vectors, uh, it's going to be the cross productduct of the force and the moment arm basically. Uh, and we, uh, have a direction associated to that which comes about from the right-hand rule. So, to draw a picture of that and to make sure you understand this, it's easy to do that here. So you got to use your imagination a little bit because I'm I'm drawing on a two-dimensional board and really I'm trying to represent something that is is threedimensional in nature. So uh this has an angle theta between them. So if you can imagine I know it's kind of hard but imagine these two vectors being in the same plane like on a sheet of paper. Then I think when I draw this third one it'll be a little bit clear. The crossroduct of a cross b is going to be coming out of the plane here. And I'll draw it really high. Something like this. Vector C is A cross with B. So you got to imagine a sheet of paper lying here. I know I can't draw it on a flat board like this. In the plane of this paper is vector A and also vector B. I know I had to draw it up like that, but that's cuz I'm drawing it on a board, but basically vector A and vector B are in the same plane together. If I then cross A into B, I know it's drawn like this, but you got to realize it's in the same plane. cross A into B. Then the vector C, the direction of it is given by the right- hand rule. A cross B. You always cross the first vector into the second. So, I'm going A cross B like this. Remember, they're in the same plane. Vector C goes straight up out of there. Okay. So, a couple notes. I'm not going to write them down, but I'll I'll make sure you understand them. If theta becomes 90°, sine of 90 is one. And the cross productduct is going to be a maximum because you'll have magnitude of a time magnitude of b. This will be one if you have 90 degrees sine of 90. So the magnitude of the cross productduct is a maximum when you have 90°. This should come as no surprise since I've already hinted to you that the uh moment that we've been calculating uh is is basically a crossroduct. If you notice the whole time we've been looking for when does the line of action cross with the moment arm at 90°, right? Because that's when the the moment is the maximum. So 90° is important. That all comes back to this cross productduct thing here because what we end up having here 90° when you have a 90° angle here the cross productduct is a maximum. Okay. If you end up with an angle between these two vectors of 0 or 180 that means that these two vectors are going to line up with one another and the cross productduct will be zero. So in order to have a cross productduct you need to have some angle between the two vectors. Now remember this is in the like a plane here. So what I'm saying is when I have A and B at 90° to one another. I know it's kind of hard to see at 90° then vector C will be a maximum the cross productduct will extend out of the plane by the right-h hand rule at a maximum. But if these two vectors are lined up on top of one another at 0 degrees so A and B are kind of collinear then the cross productduct is zero because s of 0 degrees is zero. Same for 180. That would mean the vectors flipped over and they're both lined up in that direction. The third note I want to say is the direction of the cross productduct. It always follows the right-h hand rule. Um, but you may not realize it yet, but the direction of the cross productduct is always in a direction perpendicular to the plane that contains your original two vectors a and b. So in this case, we have a and b. If you're doing cross productducts, you're always going to be crossing them with the right-hand rule. And so the cross productduct is always going to come out of the page. Another way to look at that is if I have vectors on the board here and I cross them, the moment or the or the cross products coming always coming out of the plane that contains those two original vectors always comes out of that plane perpendicular to it. Um, this one warrants writing down. I'll tell you this right now. If you have a cross b that is not the same as vector b crossed with a it's not the same. Um, it it is not the same because the cross productduct has a direction. If I cross A into B, remember it's in the plane here, A into B, I'm going to get a vector pointed up like this. But if I cross B into A, that means I need to cut my fingers through vector B first. If they were the same vectors here, then I would be cutting oppositely and the cross productduct would go down. So remember, a vector is magnitude and direction. So if I cross A cross B, I'm going to end up with a cross productduct going one way. But if I flip my fingers to curl and go the opposite way, B cross A, the cross productduct is going the exact opposite way. So those two things are not the same. Even though the magnitudes might be the same, the directions aren't the same. And so uh that's just uh that's something that's that you always have to consider. Now that's different than the dot product. The dot product different didn't matter which direction you dotted. And that's why I'm pointing it out pointing it out here. When you have a cross productduct, you can't flip them because it's not straight up multiplication like you learned in third grade. It's a little more complicated than that. All right. So now we want to learn about something that I think um gives students problems Crossing Unit Vectors sometimes, but it's very easy to understand. I want to teach you how to cross. So I'll say crossing unit vectors because crossing unit vectors is the key to understanding how the cross productduct really works. And so what I want to do is let me draw uh a little three-dimensional situation here. So let me go ahead and draw let me think of how to do this. I'll do it right here. Here we have uh this guy here. And we have the x direction. We have the y direction, z direction, just like this. Now, because this is x, y, and z, you should know, and I'll I'll use black here. You should know that this guy here, there's actually a unit vector, which is just a vector pointing along the x direction with a length of one. We call that unit vector i hat, right? It's pointing along the x direction. And then we have a same unit vector or same length of the unit vector pointing in the y direction. We call this jhat. And then we have the same length of the unit vector pointing in the z direction. Need to make that a little taller like this. And we call that k hat. So i jk. We've been using this all the time. Okay. But what I want to show you is what's going to happen if you try to cross these vectors into each other in different ways. I mean don't forget these unit vectors. They're not magical. They're just vectors that point 90° from one another and they have a length of one. Other than that, they're just regular vectors. You can cross unit vectors just like you can cross any vector because they are vectors. They just have a length of one. So, let me draw a little table here and see if I can make this a little bit clearer for you. Okay. So, if I'm going to have I can have I and I can have J and I can have K and I can let me go ahead and maybe move this down just a little bit. Let's do it over here. I can have I, I can have J, and I can have K like this, right? And I can cross it with uh I, J. H. Let me give myself a little more space. J, or K. Okay, so basically there's a bunch of different combinations here I can cross these guys with. So, let me go ahead and do something like this. All I'm doing is drawing a table here. Basically, I can take vector I, I can cross it with itself. I can take vector I, I can cross it with vector J. I can take vector I, and I can cross it with vector K. What's going to happen in each of these cases? Well, you can just use the right-hand rule. I don't want you to calculate anything with math. I just want you to use your fingers. That's all you want to do. So what's going to happen if I take I and I cross it with itself into I? Well, we already said before that if I have uh two vectors that are lined up on top of one another, the angle between them zero sine of zero is zero. So anytime I have a vector crossed with itself, I just get zero because there's no way to cross it. I mean, it's just two vectors sitting on top of one another. You can't really cross it because the angle between them is zero. there's no angle that's there between them that you can actually sweep one into the other and so you're not going to get a cross productd. Okay? And the same thing happens when you try to cross J into J. And the same thing happens when you try to cross K into K because those vectors are just sitting on top of one another and you don't get a cross productd when that happens because the sign of zero is zero. But I can cross I into J without any problems. And to do that, you need to take your fingers and first cut them from I and sweep them into J. My thumb is now pointing up, which is in the K direction. So whenever I cross I into J, I get a new vector called K. Notice that K is perpendicular to the plane that contains I and J. That's what we always said is true with cross productds. The cross productduct answer that you get is always going to be perpendicular to the two vectors that you start with, to the plane that contains them. All right. Now, I can also take I and I can cross it into K. Right? I know it sounds a little weird, but if I take my fingers, if I cut through I and then cut through K, my fingers are going to be going, if you kind of visualize it, this way, they're going to be going opposite of what the way this way is pointing. So, they'll be pointing in the negative J direction. Don't forget this is the positive J direction. The negative J direction just goes this way. The negative K direction goes down. The negative X direction goes that way. So, if I sweep I into K, then you can visualize that I'm going to be pointing in this opposite direction, which is negative J. Now, we'll speed things up a little bit. If I'm going to go J crossed into I, J crossed into I, cut into J first, then into I, my fingers are pointing down, which is negative K direction. And then if I'm crossing J into K, J into K, I cut into J first, then into K. And you can see that I'm pointing in the I direction. which is like that. And then K crossed into I. K crossed into I cut through K then into I. I'm pointing in the J direction. And then if I cross K into J, K into J is going to be kind of use your imagination a little bit. K into J, that's going to be the negative I direction. Now, it's a little hard to visualize it completely because this is a a threedimensional drawing on a two-dimensional board. But whenever you you got to do a little mental gymnastics to kind of do that, but you really should get used to that because whenever you're doing these problems, they're all going to be drawn on a piece of paper. So, you have to kind of get the mental gymnastics of crossing them in, you know, with your fingers and which direction they're pointing. So, notice there's some symmetry here. You have zeros along the diagonal. You have K's here, but this one's negative, of course. You have J's here, but one of them's negative. And you have I's here, and this one's negative here. So, I have a negative here, a negative here, and a negative here. I'm just checking everything here to make sure I haven't lied to you. So, this is a chart. Now, you don't need to really keep this chart around. I'm just doing it to kind of gives you practice. The reason I put it here is to mostly show you how to do it. Um, you could write this down and use it as a reference, but really, it's just easier to draw the unit um coordinate system and just cross with your fingers. If you're trying to go J into K, then just do that. I'm trying to go K into into J, go the opposite way and so on. Okay. Um, what I want to do next is I want to show you how to cross two vectors in in the kind of the long way. Remember, they're just vectors. They just have a magnitude and an associated direction. Let me think. Maybe I can maybe I can do this on the other board. Let me swi switch this through like this. Okay. What if I had two vectors? I wanted to Crossing Two Vectors cross A and cross it with B. Now, these aren't unit vectors anymore. Let's say vector A is 1 I + 2 J + 3 K. And that's a vector and I'm going to cross it in with another vector that I'm going to write right here, which is 4 I + 5 J + 6 K like this. So this is vector A. That's this guy. I'll just write it down here. This is vector A. And this is vector B. Notice I have 1 2 3 4 5 6 just to make it a little bit easier. But these numbers could be anything. This is an The Hard Way arbitrary vector in threedimensional space. This is another arbitrary vector in threedimensional space. Now I'm going to kind of uh give you a little preview here before we do this. There is a hard way to do this. I'm going to show you that first because I I think it's important for you to know where it comes from and that the the cross productduct is not magical. I'll show you the rules to to to calculate it, but then when we get through that, I can kind of show you a much easier way that's just much easier to remember. It's not that it's it's really the same math. It's just easier to remember. And the cross productduct, the reason I'm spending time on it is because you use it in mechanics, but you also use it in electricity and magnetism and you use it in thermodynamics and you use it in physics and you use it everywhere in pure math. It's one of those things like multiplication that you just use everywhere in engineering. So, I really want you to get a good foundation in that. Now, if we're going to cross these two things together, and we are, um, notice you have three terms here and you have three terms here. So, if this were not a cross productd, if it was just multiplication, how would you do it? You would take the first thing here and you would multiply by this, and then with this, and then with this, and add them together, and then you would move your finger here, and then you would multiply this times this, then times this, then times this, and you'd add all those together. Then you would move your finger here and you would multiply this times this and then this and then this and you would add them all together. When you have large tromials like that with three terms and you're multiplying them. I mean I know you you can dump it into a computer and get the answer. But all that's happening is you're taking each term and multiplying by the other terms and you're taking the next next term multiplying by these terms and then you're taking this term multiplying by these terms and then you add all everything together and you simplify and that's how you multiply things in algebra really. Okay. Now the similar thing is going to happen in this cross productd. Okay? Because what we're going to have is it'll be this crossed with this. Now the numbers out in front just simply get multiplied together. So 1 4 will be four. But the vectors here you can't just multiply them. They have to be crossed. Okay? Because they're a vector. So when they're crossed like this it's I crossed with I. I crossed with I. Now I'm not going to do this cross productd. Now what I want to do is write all the terms out and then I want to simplify everything. You know from this discussion that I cross I is zero. There's nothing there. Uh because you can't when you cross a vector with itself there's just there's there's uh no angle between them. So it drops to zero. But for the purpose of this I want you to put your finger here and cross it with this 1 4 is 4. I cross I we're going to write down here. Then I'm going to add to it. Keep my finger here. I'm going to do this guy. 1 5 is five. And then I cross J. I cross J. Okay, I'm not going to do that yet. I'm going to write it down. I'm going to say plus I'm going to move my I'm going to keep my finger here. It'll be this one crossed with this one. 1 6 is 6 I cross K. Okay, you should sort of start to see a pattern here. Now, we're going to move to the next line and I'll move my finger from here to here. And again, I'll start over like this. So this crossed with this 2 4 is 8. And then I have J crossed I. J crossed I. And then I'm going to have this 2 5 is 10. J cross J. You should already know that J cross J is zero, but we're just going to write it down for now. And then here 2 6 is 12. And then I have J cross K from this to this. And then I go down to the next line. And then I move my finger. See, I was doing this one, then this one. So now I do this one. 3 4 is 12. And then I have K cross with I, right? And then I have this guy here. 3 5 is 15. K cross J like this. And then I have 3 6 is 18 K cross K, which you already should know is zero. And I've said it enough times that basically you know now that I cross I is just going to go to zero. J cross J is going to go to zero and K cross K also goes to zero because anytime you cross a vector with itself you get zero because there's no angle between them. All right. So now what we need to do is simplify this guy. That's really all that remains to be done. So let me just double check that I haven't made any mistakes here. We have five I cross J 6 I cross K 8 J cross I 12 J cross K 12 K cross I 15 K cross J okay that's right so then the next line 4 0 is 0 we don't have to write anything there five what is I cross J now I can just look at my table I cross J is K but if I don't have this table I just draw XYZ I cross J just cross I into J that gives you a K compon component. So what you're going to have here is okay the five comes from the multiplication the vector direction comes from the cross productduct of I and J. Then I have uh six coming from here and I cross K. What is I cross K? Well I when I cross with K is negative J. You can check I cross K is negative J. That's exactly what we have. So I'll just put the negative J here. I could put the negative here. I'm going to do that in in a sec second step. I just want to keep things clear. Then I have 8 J cross I and J cross I you can cut into J then into I is going to give you a negative K. Uh that was this one. Now we have the 12 J cross K. J cross K you can cut it into there is going to give you a positive I. And then over here you have a 12 K cross I. Uh K cross I is going to give you this guy which is going to be a positive J. And then we have 15 K cross J. K cross J gives me negative I. K cross J gives me negative I. So you can do it with the right hand rule like that. All right. So now we can simplify things here because now what you then you start you have to simplify these things. It's just like adding algebra right when you add things up in algebra you know 2x + 3x it's 5x because you're attaching it to the x's right 2 j + 2 j 4 j. So because these are unit vectors it doesn't really change that you're still trying to group terms. So for the i components you have an 12 i here and a -5 i here because of this negative sign. So what you end up with is -3 I right there. And then for the J components we have these guys. We have 12 and a -6. So that'll give us a positive 6 J. And for the K components we have the five and the8. So again that's -3 in the K direction. This is the crossroduct of A. Whoops. Crossed with B. So notice there's an I component, a J component, and a K component. It's pointing obliquely in threedimensional space someplace. The original two vectors were pointing obliquely in threedimensional space someplace. You can have negative uh components here, positive components, but you should always get a three-dimensional representation for a cross productduct in general. Now, this works bulletproof every single time, but it's just a little cumbersome because you have to write vector A and write vector B down and then you have to cross essentially cross everything with every other thing and then you have to simplify the cross productducts and then you have to add everything up. It's bulletproof, but it's just a little bit um ugly. It's just not very nice to do. Okay. So, what you would typically find in your book is in general, you'll find a formula for the cross productd as follows. And it'll be written like this. A cross with B. A cross with B. And it's going to look ugly. I'm going to warn you right now. This is how you write it in general. a Y B Z minus A Z B Y in the I direction. That's the I component of this answer. And then you have minus ax BZ ax B Z minus A Z BX in the J direction. And then you have plus some k component here which is ax b y oops ax b y minus a y bx k direction. So this is what you'll typically find in a book when you learn about the cross productd. They'll teach you what it is and then they'll put this thing in front of you and normally you'll kind of like throw up in your mouth a little bit because it just looks ugly. Let me make sure I wrote it down right. A y bzz minus a z b y in the i direction minus ax bz minus a z bx in the j direction plus ax b yus a y bx in the k direction. So what does all this stuff mean? See this is vector a and this is vector b. Now this is the x component a subx. This is the y component a suby. This is the z component a subz. Okay. Then you have b subx, b suby, b subz. So for each vector a and b, you have a x and a y and a z component. So all you do is you just plug it in here. The y component of a times the z component of b. Get that multiplication. Then you subtract it from this product. The z component of a minus the b component of y. So this is going to give you a number. And then you're going to uh that's going to be the x component of this guy. And then you do a similar calculation um for that guy there. So what what I want to do just really quickly and I'll move over here is to show you that this is actually true. So for our particular set of vectors here um ax is one um a y is 2 and a z is 3. And over here we have four five six. So b x is 4, b y is 5, b z is 6. So we have uh xyz for a and xyz for b. So if I wanted to use this canned formula for the cross productduct, the way I would do it is I would say a cross b is equal to the following. So a y which is 2. Okay. Uh bzz bz would be six. So 2 6 is 12. So what I would have is 12 minus and then I have uh z a z which is 3 minus bz which is y. So these guys here uh this would be 15. That's in the i direction because it's i direction here. Then I have a minus sign here I have to keep track of. Okay. Ax minus bz. ax or ax bz. So 1 6 gives me 6. And then a z minus bx or a z bx sorry 3 4 is 12 and that's in the j direction. And then plus here we have ax b y ax b y 1 5 is 5 and then a y bx a y bx 2 4 is 8 in the k direction. So what this is going to be 12 - 15 is -3 i. Okay, actually let me let me just make it a little bit clearer. I'll switch colors. -3 I 6 - 12 is -6. But then I have a negative out here, so it's positive 6 J. And then 5 - 8 is -3 K. So -3 I + 6 J - 3 K. -3 I + 6 J minus 3 K. So all of that showing that if I just bust out and do everything manually, I get the same exact thing that you you typically get in your textbook. All right, believe me, I know it is tough to follow all of this and to care, okay? But cross productds are going to be a part of your everyday life. You're going to be doing them all the time in mechanics and other classes. So let's take just a few minutes to review where we've been. All right. What we said is there's this RightHand Rule thing called a cross productd. The magnitude of it is a b sin theta. The direction is purely given by the right- hand rule. Okay. Then we said that the right-h hand rule works by curling your fingers and looking where your thumb points. So for any unit vector, we can cross any combination and we can figure out where it's going to point with the cross productd. Then I tried to lock it down with an example. Arbitrary vector a, arbitrary vector b. Cross everything with everything else and simplify and you get an answer. And then this is what you're given in your book. This is where it comes from. All of this stuff, if you mathematically simplify down, you can derive this. This thing really is coming from everything we've done above. But I want to simplify things even further for you because even though this is useful, this uh relation here, this equation is very hard to remember. Uh you're not going to remember it at least unless you really try hard. And even if you do remember it, um, you can easily Cross Product as a Determinate make a mistake. Did you notice how I was like ax times b y and so on and then crossing it the other way? It's very difficult to do. So here we're going to learn about some of the the most uh useful things I've ever learned in my life. And this one's called the crossroduct as a determinant. All right, as a determinant. And it's very easy to write down and it's very easy to implement and allow you to to calculate cross products till you're blue in the face. Very very simply here's how you do it. A cross B can be written as the determinant of a matrix. So what I'll write it is I'll say like this. The determinant of a matrix and the matrix is formed by putting I, J and K in the top. And then you have the x component, the y component, and the z component of a. And the x component of b, the y component of b, and the z component of b there. Now, at some point, I know this doesn't look any simpler yet, believe me, but it will. Um, at some point in your past, you probably studied matrices, and you should have learned how to calculate the determinant of a matrix, specifically how to calculate the determinant of a 2x2 matrix. And if you remember that or if you go back and review it from what I've taught you, the determinant of a 2x2 matrix is a crisscross pattern. It's you multiply the diagonals and you go the other way and multiply the diagonals. So if you don't even know what a determinant of a matrix is and have no idea and you haven't studied it for 15 years, probably should take a second and go review what a determinant of a 2x2 matrix is and what a determinant of a 3x3 matrix is here. I'm going to give you uh a practice problem or two here to nail it down. But you need to realize that the crossroduct of two vectors is written as the determinant of a 3x3 matrix where we write it with the components of the vectors like this. So for instance, if I have vector A which is 2 I + J + K and I have vector B which is I + 2 J + 3 K. I want to cross these guys together. Of course, I can use the formula on the previous board or I can bust it out manually, but instead I'm going to choose to write it down as a cross productduct. So, I'm going to say that vector A cross with vector B is the determinant of the following matrix. The first line of the matrix is always just the unit vectors. That's what makes this so easy, right? Then the next line is your A vector and it's just the numbers, the X and the Y and the Z component. That's all this represents. So you put two, this is a one and this is a one. You put numbers there. It's very simple. And then for vector B, this is going to be one from there, two from there, and three from there. So basically what you do is you write down the unit vectors. And then if you're going A cross B, the first vector here, the vector A get the components get written in the second line. And vector B, the components get written in the third line. Now all you need to do is learn how to calculate the determinant of this matrix. and then you got the answer. Okay, so let me show you how to do that. It's really simple. What you do is first of all you kind of draw a line through this column and a line through this column. Okay? So they intersect at the I location. So the first component you're going to calculate is going to be the I component of the answer. So you cross through this. You pretend it's not even here. And you cross through this. You pretend it's not even here. If I had two pieces of paper, I would cover this up and I would cover this up. So all that's left that I can actually see are these four numbers. Now I need to take the determinant of these four numbers. And you should remember from matrices matrices a long time ago that you multiply these diagonal elements and then you subtract the multiplication of these. So 1 3 is 3 minus that's from this direction minus 1 2 is 2. Okay, you've actually calculated the i component of the answer already. Okay, you didn't have to memorize anything other than you need to know how to calculate this determinant, but you don't have to memorize a Z B Y minus 2, you know, all that stuff. Basically, you just cover this, cover this. That's the I component is this multiplied minus this multiplied. Okay, then the next thing you have to remember is very important. You must put a minus sign for your second term. And then you're going to cross through the top and you're going to cross through this. Now, where they intersect is the J location. So, you're finding the J component. You must remember to put the negative sign out in front. So if I've covered this up and I've covered this up, the only numbers left are these numbers and these numbers. You need to pretend that those numbers are a nice 2x2 matrix here. So you're going to cover these up. You're going to take the determinant as if these were here. So 2 3 is 6 - 1 1 is 1. You're basically pretending this isn't there and you're treating this as a 2x2 matrix. 6 - 1. Boom. you are done with the answer to the second component here. Then you can go back to plus you're you're going to cover up this column and this guy which intersects at the K location. So you're finding the K component. Cover this up. Cover this up. The only thing revealed is this. 2 2 is 4 - 1 1 is 1. All right. So then what you have is a crossed with b is 3 - 2 is 1. So 1 I you can just write it as I 6 - 1 is 5 but there's a negative there so it's -5 J 4 - 1 is 3 so you get + 3 K this is the crossroduct of these two vectors all right now let me ask you a question even though I had to teach you how to do this was this easier or was this easier or was all of this easier where you had to plug everything in manually and all that I mean ultimately this is doing the exact same calculation. The multiplications here and the subtractions what you're doing here is exactly what you're doing when you cross out this and this and you do a crisscross multiplication to to get that determinant. It's basically uh you learn this kind of stuff in linear algebra but basically the determinant of matrix like that it that's how you calculate it. You criss-cross things and you find the determinant of the submatrices inside which are 2x2 matrices. The only thing you need to remember is that when you're finding the middle term, you have to stick a negative out in front in front of the J. Now, I'm going to do one more real quick and then we're going to call it a day. Uh, so I'm going to draw a little line here. If I have a is -2 I - 2 J + 3 K and B is I minus J minus 4 K. Okay, you could of course put everything into that enormous formula on the other side. Maybe not have any any way to remember it, but I'm going to encourage you to write it as A crossed with B is the determinant of the following matrix. The first line is always I, J, and K. Always. It never changes. Then you're crossing A into B. So you have -2 right under here, -2 here, and three here. And then for B, you have one. Then you have negative 1. and you have -4. And now you just have to go through the mechanics of writing down the determinant here. So what you're going to have is first I go down I cross out the first column and the first uh row here. So I'm working on the i component because that's where they intersect. The intersection of these two lines here that I'm crossing out. The only numbers re that are revealed or left over are these. So -2 4 gives me positive 8 minus this gives me -3. Now, don't get screwed up here. You have to have the minus when you're doing the crisscross multiplication, but the actual thing you multiply also gives you negative. Don't do too many things at once. Write down the minus negative so that you don't mess anything up. Then cross out this guy. Cross out this guy. So, I'm working on the J component, but I just need to remember to put a negative J here. So, I'll cross this out like this. The only numbers left are these. So -2 4 gives me positive 8 minus. Then I have 3 pos1 gives me a positive 3. Then I have a plus. I'm going to be working finally on the k component because I'll be crossing this out with this. This is what's left over. -2 -1 gives me pos2 minus and then I have so I have this minus this -2 this gives me -2. Again I need to keep the double negative here so I don't screw myself up. So then at the end of the day 8 - a -3 is 8 + 3 which is going to be 11 in the i direction. And then here I have 8 - 3 is five but this is a negative out here. So -5 in the j direction. And then here 2 plus or 2 -2 is 2 + 2 is four which is + 4 in the k direction. This is the crossroduct here. All right. So, I've done two examples to show you how to cross these things with um matrix methods. Basically, you just you focus on the first column. You cross this out. That's the i component. And then you do a criss-cross multiplication with a subtraction in between. Basically, find the determinant of this little submatrix here. Then you move on to the middle one. Make sure and put your negative sign. Crisscross these elements minus these. And then you work on the last one. Crisscross multiplication of these. And that's that's what you do. Now, it's really honestly up to you. You're going to be doing a lot of cross productds in this class and in lots of future classes. So, if you want to, you can just write it out every single time. You can just put it boom, put everything out, cross everything, figure out the directions of everything, simplify everything. Or if you prefer to memorize this really, this calculation is doing exactly what we're doing with determinants on the other side. You just have to label everything, put everything into the right spot, do the calculations. This stuff here that we're doing in this using this equation is exactly what you're doing here. It's just this is a visual way of representing the crisscross. Notice you're multiplying and you're subtracting here. You've multiplied things together like these guys and you're subtracting it. So really it's exactly the same thing. There's no difference. But this is just a visual way to represent what to do. I'm going on and on about it because to find the crossroduct of these vectors in this way with a determinant has honestly been one of the most useful things I've ever learned because you will be doing a test somewhere or you will be doing some other class way down the road and you'll have to cross two vectors and you won't remember you won't remember that giant equation even if you remember it now but then you'll remember you'll say wait a minute I can write that as a determinant I'll put I jk on the top a in the middle B on the bottom row and then you just have to Remember to go through and crisscross the submatrices and boom, you're done. So, I've talked enough. It's a very long lesson. Thanks for sticking with me. I want I could have just showed you this at the end. I was very tempted, but I wanted to show you that doing it the long way and kind of the evolution of the ways in which you can calculate the cross productd. There's nothing magical about any of that stuff. I just wanted to show that to you and finally culminate in what I really want you to do, which is this way. I'm going to be doing it this way as we go through the rest of the course when I need to calculate cross productducts. So I encourage you to do the same. Follow me on to the next lesson. We'll be talking about how to use this cross productduct to find the moment of vectors in three dimensions. |
13972 | https://simple.wikipedia.org/wiki/Cone | Published Time: 2005-07-06T20:41:29Z
Cone - Simple English Wikipedia, the free encyclopedia
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From Simple English Wikipedia, the free encyclopedia
A cone, marked with its height (h) and radius (r)
In common speaking and geometry, a cone is a solid object that one gets when one rotates a right triangle around one of its two short sides, the cone's axis. The disk made by the other short side is called the base, and the point of the axis which is not on the base is the cone's apex or vertex. An object that is shaped like a cone is conical.
In more technical terms, a cone is formed by a set of line segments, half-lines, or lines connecting a common point, the apex, to all of the points on a base that is in a plane that does not contain the apex. Depending on the author, the base may be restricted to be a circle, any one-dimensional quadratic form in the plane, any closed one-dimensional figure, or any of the above plus all the enclosed points. If the enclosed points are included in the base, the cone is a solid object; otherwise it is a two-dimensional object in three-dimensional space. In the case of a solid object, the boundary formed by these lines or partial lines is called the lateral surface; if the lateral surface is unbounded, it is a conical surface.
In the case of line segments, the cone does not extend beyond the base, while in the case of half-lines, it extends infinitely far. In the case of lines, the cone extends infinitely far in both directions from the apex, in which case it is sometimes called a double cone. Either half of a double cone on one side of the apex is called a nappe. The volume{\displaystyle } of a cone is one third of the product of the area of the base A B{\displaystyle A_{B}} and the height h{\displaystyle h}
V=1 3 A B h.{\displaystyle V={\frac {1}{3}}A_{B}h.}
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↑Alexander, Daniel C.; Koeberlein, Geralyn M. (2014-01-01). Elementary Geometry for College Students. Cengage Learning. ISBN9781285965901.
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13973 | https://www.youtube.com/watch?v=5jHaiQRpteY | Q10 of SMO Open 2024: Maximizing Area: Calculus OR AM-GM Inequality
Chan Lye Lee
5340 subscribers
7 likes
Description
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Posted: 7 Jun 2024
In this video, we solve a challenging problem from the Singapore Mathematical Olympiad (SMO) 2024 Open Section, Round 1. We determine the maximum area of a rectangle inscribed in the region bounded by the parabola ( y = 60 - x^2 ) and the x-axis.
Methods Covered:
1. Derivative Method: Using calculus to find the critical points and determine the maximum area.
2. AM-GM Inequality Method: Applying the Arithmetic Mean-Geometric Mean Inequality to solve the problem geometrically.
Both approaches lead us to the same solution, demonstrating the versatility of mathematical techniques.
Key Highlights:
- Step-by-step explanation of each method.
- Detailed derivation of the maximum area.
- Application of optimization principles.
This video is perfect for math enthusiasts and students preparing for competitive exams. Dive in to enhance your problem-solving skills and understand multiple methods to tackle similar problems.
Feel free to pause, rewind, and review each step as needed. Happy learning!
3 comments
Transcript: |
13974 | https://www.illinoisrealtors.org/blog/designated-agency-duties-4-good-reasons-for-an-exclusive-buyer-brokerage-agreement/ | Designated Agency Duties & 4 Good Reasons for an Exclusive Buyer Brokerage Agreement - Illinois REALTORS
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Designated Agency Duties & 4 Good Reasons for an Exclusive Buyer Brokerage Agreement
Writen by Betsy Urbance |
Published: June 22, 2022
_Elizabeth A. (Betsy) Urbance
General Counsel and Vice President of Legal Services_
Presumption of Designated Agency in Illinois
As every real estate licensee should know, the Illinois Real Estate License Act (RELA) contains specific statutory duties that real estate licensees owe to their clients. These duties are similar to common law fiduciary duties, but are found in Article 15, Section 15-15 of RELA. Hopefully, this concept is familiar and one of the first things newer brokers learn about.
Under RELA, even though the client’s brokerage relationship (sometimes set forth in an exclusive brokerage contract) is with the sponsoring brokerage company, the designated managing broker (DMB) for the company is responsible for appointing designated agents for the company’s clients. The designated agents are the “legal” agents and owe all of the statutory duties to their respective clients. The designated agent is presumed to be the legal agent for the client with whom they are working and must at least provide written (physical or electronic) disclosure to this effect.
Duties under RELA
The designated agent’s statutory duties include:
Performing according to the terms of the agency agreement whether it is written or oral (more on this later)
Promoting the client’s best interests by:
Seeking a transaction that meets the terms of the agreement or is otherwise acceptable to the client.
Presenting all offers to and from the client unless otherwise directed by the client.
Disclosing material facts about the transaction that the designated agent actually knows about and the information is not confidential to someone else to whom the agent owes duties (Note confidential information does not include information regarding the physical condition of the real estate).
Accounting for all money and property received from the client or for the client’s benefit.
Obeying the client’s lawful direction.
Promoting the client’s best interests over the interests of others including the agent’s own interests.
Exercising reasonable skill and care.
Keeping the client’s confidential information confidential (even within the agent’s own office because someone else within the office could be the legal or designated agent for the opposing parties in transactions); and
Complying with RELA and other applicable laws.
Brokerage Relationships & the MLS
A sponsoring brokerage company and its designated agents are free to enter non-exclusive, open, relationships with clients, or they might enter exclusive ones. Open or non-exclusive relationships won’t provide details setting forth the expectations between the company, the agent and the client. However, in a non-exclusive relationship the client must provide written consent for the broker to advertise the property for sale or lease and the broker must provide written disclosure to the client as to the identity of their designated agent(s).
On the other hand, under RELA, if the brokerage relationship is exclusive, there must be a written agreement between the company and the client. That written agreement must refer to the minimum services the exclusive broker will owe to the client, where the designated agent is required to have some involvement once contract negotiations begin in a purchase or lease transaction. This is a consumer protection measure to assure the client gets the professional assistance contracted for to the exclusion of any other brokers. Facts will vary as to what actions suffice for minimum services and it might well depend upon how much “hand holding” the client wants. If a client “waives” minimum services, then the agreement has become non-exclusive by definition under RELA.
Many brokerage companies have an office policy to take exclusive listing agreements with seller clients and they enter those listings (hard earned assets of their companies) into their Multiple Listing Service (MLS), thus exposing the listing to many potential buyer clients represented by other participating brokerage companies. These prospective buyers might come from any cooperating brokerage company, large or small, but where the co-op companies also participate in the MLS. The MLS offers a very diverse and competitive market exposing listings to a large pool of prospective buyers, along with an efficient mechanism whereby listing brokers offer to share a portion of their compensation with buyers’ brokers. The MLS has been tested and proven over time to be an efficient and very competitive marketplace resulting in countless successful transactions.
If Buyer Brokers are Paid Through MLS Offers of Compensation, Why Consider Exclusive Buyer Brokerage Agreements?
Four Good Reasons
1) The exclusive brokerage agreement will be in writing. This document contractually binds the buyer client to the brokerage company and their designated agent, and vice versa. The exclusive designated agent always owes their statutory duties to the buyer client, and while the same duties are owed in a non-exclusive relationship, such as serving the client’s best interests over the agent’s own interests; in a non-exclusive relationship, neither the brokerage company nor the designated agent has protection under RELA or the REALTOR® Code of Ethics. In other words, another company’s agent is completely free to interfere with a non-exclusive relationship. While “poaching” another agent’s buyer client is not an ideal business practice, if there is no exclusive relationship, there is no legal or ethical protection available to the non-exclusive agent, who is working hard and providing valuable services to their client.
2) The exclusive brokerage agreement provides a good written outline from which to frame discussion and education for the buyer client about the agent’s duties to the buyer client, the services they buyer agent will provide, the company’s duties to the buyer client and what the buyer should do in order to assist their exclusive agent in finding the property that best meets the buyer’s needs. See Illinois REALTORS® sample form Exclusive Buyer Brokerage Agreement.
This agreement refers to the buyer checklist which helps with consistency in treatment of all buyer clients.
The agreement helps the agent discuss and define the buyer’s needs and wants.
The agreement defines the market in which the designated agent will search, which can include the agent’s MLS as well as any other sources upon which the parties mutually agree.
The buyer agrees to provide relevant financial information.
The buyer agrees to be available for showings of properties that meet the buyer’s specifications.
The buyer agrees to pay the company a negotiated rate for the full array of services provided by their designated agent (this is sometimes qualified with language saying that buyer’s broker will first seek to be paid from the listing broker through the MLS offer of compensation).
It is very important for the designated agent to spend time explaining the exclusive agreement to buyer clients and to explain how the agreement can protect everyone involved.
3) The exclusive brokerage agreement can protect the company’s compensation or fee for services provided. The sponsoring brokerage company makes independent business decisions as to what amount it needs to be paid in order to cover expenses, pay their independent sponsored licensees according to their contracts and to make a reasonable profit. The MLS offers of compensation won’t always meet that rate. The non-exclusive agent is duty-bound to show all of the suitable properties in the defined market area, regardless of the amount of cooperating compensation offered in the MLS by the listing company. Remember the designated agent’s duty to serve your client’s best interest over the agent’s own interest? IF the non-exclusive designated agent purposely avoids showing their buyer client suitable properties due to a lower commission rate being offered, there is a very good argument that the designated agent is breaching their duty to their buyer client.
On the other hand, if the buyer has signed an exclusive brokerage agreement containing a provision requiring payment to the brokerage company in a pre-determined amount the buyer will know they are contractually obligated to pay the agreed fee for services provided. The interests of the buyer client and their broker are now more in line with each other.
In addition, having an exclusive buyer brokerage agreement, while not a guarantee the exclusive agent is the procuring cause in an MLS compensation dispute, it is a helpful factor. It can also provide another possible avenue for payment in the event the buyer’s broker is not the procuring cause in a successful transaction. How many buyer brokers have been called upon to help a buyer after the buyer has already gone almost all the way down the transaction road providing many hours of services, but wants another buyer’s broker to come in as “the closer.”
4) Having the exclusive written contract can serve as legal protection should the sponsoring brokerage company elect to seek enforcement. There is no requirement for a sponsoring broker to pursue enforcement of their contracts. Many brokers elect not to seek enforcement for diplomatic and other business reasons. However, sometimes circumstances are such that the sponsoring broker wants to pursue remedies provided for in the exclusive brokerage agreement. A final point to consider is that where a buyer has negotiated and agreed to be bound to an exclusive brokerage relationship, it makes said relationship more “real” or committed to their company and their designated agent. The buyer has “skin in the game.”
About the writer: Elizabeth A. (Betsy) Urbance, General Counsel and Vice President of Legal Services has served the association’s members as General Counsel since 2018 and prior to that she was Legal Hotline Attorney since 1994. Urbance is a 1984 graduate of Western Illinois University and received her law degree from the University of Missouri School of Law in 1987. She is licensed in both Illinois and Missouri.
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13975 | https://puzzling.stackexchange.com/questions/128377/are-these-colored-sets-closed-under-multiplication | Skip to main content
Are these colored sets closed under multiplication?
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A set is called closed under multiplication if the product of any two (not necessarily distinct) elements of the set is always a member of the set. For example, the integers are closed under multiplication.
Suppose that every real number is colored either green or blue such that the product of any three (not necessarily distinct) green numbers is a green number and the product of any three (not necessarily distinct) blue numbers is a blue number. Let G be the set of green numbers and B be the set of blue numbers.
Assume that you know nothing about the coloring other than it satisfies the above conditions.
Question 1: Is it necessarily true that at least one of the sets is closed under multiplication?
Question 2: Is it necessarily true that both sets are closed under multiplication?
Question 3: Is it possible that both sets are closed under multiplication?
Attribution: Mostly momath.org, partly me
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edited Sep 16, 2024 at 7:36
Will.Octagon.Gibson
asked Sep 15, 2024 at 8:58
Will.Octagon.GibsonWill.Octagon.Gibson
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Does your definition of closedness apply to two identical numbers?
– Nautilus
Commented
Sep 16, 2024 at 7:08
Yes. For example consider the 2-element set S={a,b}. Set S is closed under multiplication if and only if a2, ab and b2 are all elements of S. I have updated my question.
– Will.Octagon.Gibson
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Sep 16, 2024 at 7:32
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Question 1: Is it necessarily true that at least one of the sets is closed under multiplication?
Yes. Otherwise, you'd have g1g2=b3 and b1b2=g3 for some greens gi and some blues bi. But then g1g2g3=b1b2b3, a contradiction.
Question 2: Is it necessarily true that both sets are closed under multiplication?
No. The greens can be the negatives, which are not closed under multiplication.
Question 3: Is it possible that both sets are closed under multiplication?
Yes. The set of blues can be {0}.
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edited Sep 16, 2024 at 3:19
Will.Octagon.Gibson
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answered Sep 15, 2024 at 9:29
msh210msh210
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For Question 3, do you think it is possible that both sets are infinite AND closed under multiplication?
– Will.Octagon.Gibson
Commented
Sep 16, 2024 at 5:27
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@Will.Octagon.Gibson Take the set of real numbers with magnitude less than 1, and the other set is its complement (in the reals), then both sets are closed under multiplication. Modulo little variations as to whether 0 and/or 1 are in the first set vs in the second, for a total of 4 variants on the same idea, I'd even think that is the only possible solution as for partitioning the real numbers in two infinite sets: though I cannot think of a proof...
– user93836
Commented
Sep 16, 2024 at 10:35
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@MatthieuM.: If neither set is closed under multiplication (ie assume it's not true that at least one is closed), then there must be elements that satisfy g1g2=b3 and vice versa because that's what the definition of not being closed means. That doesn't immediately help, because as you say that's a 2-product and we only have a rule for a 3-product - but to be able to apply that rule, just multiply the two equations together (making sure all the gs and all the bs are on the same side).
– psmears
Commented
Sep 16, 2024 at 13:15
5
@MatthieuM.: As an alternative proof: observe that the number 1 must be either green or blue. Assuming it's blue: for any b2, b3, it's true that 1×b2×b3 is blue (since that's a 3-product). But that's just b2b3 - so in this case the blues are closed under multiplication. If instead 1 is green, by a similar argument the greens must be closed in that case. So whichever colour includes 1 is closed under multiplication.
– psmears
Commented
Sep 16, 2024 at 13:18
1
@psmears: Rereading the problem, I notice my understanding was flawed. There's talk about product of 2 numbers (closed under multiplication) and product of 3 numbers (g1g2g3 is green & b1b2b3 is blue), and I muddled the both of them, thinking that "closed under multiplication" referred to product of 3 numbers. My mistake.
– Matthieu M.
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Sep 16, 2024 at 13:36
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Problem statement and headline result
This is a partial answer to the tougher question asked in comments on msh210's answer:
(3bis) How can the real numbers be partitioned in two sets that are both closed under multiplication and infinite?
Per Julio Di Egidio (comment on msh210's answer):
Take the set of real numbers with magnitude less than 1, and the other set is its complement (in the reals), then both sets are closed under multiplication. Modulo little variations as to whether 0 and/or 1 are in the first set vs in the second, for a total of 4 variants on the same idea, I'd even think that is the only possible solution as for partitioning the real numbers in two infinite sets: though I cannot think of a proof...
I prove in the following that for any substantially different solution, at least one of the sets is dense in R (that is: for any real x, at any tolerance ϵ> 0, there exists an element of the set within distance ϵ of x). This does not disprove the existence of such solutions, but it puts considerable constraints upon them; at the very least it hints that one should not be able to find them "by hand".
Let us call A and B the sets that form such a partition (the "green" and "blue" sets in the original). (added per psmears' comment)
Proof
Lemma 1: A and B are stable by the operation "take the opposite".
Proof: assume without loss of generality that −1∈A. Assume there exist r∈A such that −r∈B. Then (−1)×r is not in A, which is contradictory. Hence no such r (or −r) exists, hence A (or B) is stable by the operation "take the opposite".
This allows us to consider only non-negative numbers.
Lemma 2: let α>0 be an irrational positive number. Let x>0 be a non-negative real number. Then, for all ϵ>0, there exists m,n integers such that 0<(n−mα)<ϵ.
Proof: This is a weak version of Dirichlet's approximation theorem. The only difficulty is that we want n−mα to be non-negative.
Define xk,−nk as the decimal and integer parts of −kα (i.e., −kα=−nk+xk with nk∈N and 0≤xk<1). Because α is irrational, the set xk is dense in [0,1( (well-known result - see for instance this question on math.se). In particular, there exists k such that xk<ϵ. Then, m=k,n=nk satisfy the desired condition.
Definition: Let 0<p<1<q. We say p and q are "co-exponential" if logq(p)∈Q (or, equivalently by definition of the logarithm, if ∃r∈Q,pr=q).
Lemma 3: assume there exists 0<p<1<q with p,q∈A not co-exponential. Then A is dense in R.
Proof:
First, notice that since p,q∈A, then pmqn∈A for any integers m,n.
Define α:=logq(p), which is irrational by hypothesis. Then pmqn=qn−mα.
Let x>0; we want to find m,n such that pmqn is "close to" x. That is equivalent to finding m,n such that n−mα is "close to" logq(x). (A rigorous proof requires a bit more complexity, but it works because the logq function is continuously differentiable on any non-negative real.) Define f(m,n):=n−mα.
Let ϵ>0 be our tolerance. From lemma 2, we can find m′,n′ such that 0<f(m′,n′)<ϵ. Define k as the quotient of the Euclidean division of logq(x) by f(m′,n′). Then kf(m′,n′)=f(km′,kn′)≤logq(x)<(k+1)f(m′,n′)<f(km′,kn′)+ϵ. Therefore m=km′,n=kn′ satisfies the condition.
Theorem: Either ]0,1[⊂A and ]1,∞[⊂B, or ]0,1[⊂B and ]1,∞[⊂A, or at least one of A and B is dense is R.
First, we consider the edge cases, and show they come down either to Julio Di Egidio's solution or a trivial solution (where either set is finite):
if A∩]0,1[ is empty, then
]0,1[⊂B
Furthermore
if B∩]1,∞[ is empty, then ]1,∞[⊂A (case 2 of the theorem).
Otherwise, let q∈B∩]1,∞[. For any x>0, there exists N such that qN>x; then, qN is in B and x/qN is in ]0,1[ hence also in B, thus x is in B. Therefore ]1,∞[⊂B. Combined with the previous point, B is dense in R (case 3 of the theorem).
if B∩]0,1[ is empty, the same reasoning as above applies, reversing the roles of A and B (yielding cases 1 and 3 respectively).
Having dealt with the edge cases, let us assume there exists pA∈A with 0<pA<1 and pB∈B with 0<pB<1.
Consider the set Sp:={pr,r∈Q}. That set is countable (because rationals are countable). As a consequence, SpA∪SpB is countable. Since ]1,∞[ is uncountable, it contains elements that are not in SpA∪SpB. Let q be one such element. Then neither pA and q nor pB and q are co-exponential.
If q∈A, then by lemma 3 applied to pA,q, A is dense in R. Otherwise, q∈B, and by the same lemma applied to pB,q, B is dense in R.
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edited Sep 17, 2024 at 9:52
answered Sep 16, 2024 at 15:51
UJMUJM
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There are a lot of these other solutions! First, here is a recipe to colour R so that the sets are closed under addition: Suppose we pick any Q-basis of R, call it S. For each element r∈S, we can then make the binary choice on colouring r blue and −r green or vice versa. Having done all these choices, we can extend the colouring to all positive rational linear combinations (and colour 0 any which way.) This clearly gives a partition into infinite sets closed under addition. This colouring can then be made multiplicative by applying ...
– Tim Seifert
Commented
Sep 17, 2024 at 9:22
... the exponential function and extending to negatives symmetrically. Note that the "nice" solutions correspond to the choice of picking the colours among r and −r (in the first comment) according to the sign. I haven't thought this through, but I suspect that the existence of these other exotic colourings could well be independent of ZF.
– Tim Seifert
Commented
Sep 17, 2024 at 9:23
What are A and B? They're not defined anywhere in the question or here (I assume they're any "blue" and "green" sets?). I'm assuming "take the opposite" means "negate"?
– psmears
Commented
Sep 17, 2024 at 9:48
@TimSeifert Nicely done. Now to prove those are the only solutions... (Also, why did I not think to work with additive stability after lemma 1? It’s much simpler to reason this way!)
– UJM
Commented
Sep 17, 2024 at 9:49
Oh hang on, my first idea doesn't work at all! These subsets are far from forming a partition :/ But here is an idea that does work and gives just as many colourings: Fix any linear surjection ϕ:R→Q. Then the preimages ϕ−1(Q<0) and ϕ−1(Q≥0) do form a partition and are closed under addition. The rest can be done the same way. Here, it is actually clear that the usual coloring is not of this form, since for every colouring of this type, one subset contains an infinite-dimensional Q-vector space ...
– Tim Seifert
Commented
Sep 17, 2024 at 10:30
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If 1 is blue, the blue set is closed, so Q1 is true.
So if no blue numbers are infinite, their absolute values can't exceed 1 because otherwise they can all grow farther and farther from 0 indefinitely. If we paint all the numbers from −1 to 1 blue and the rest green, then both sets will be closed, making Q3 true.
There are a few simple ways to create closed (blue) sets:
- All consist of the numbers x,x2,x3,...: Then three numbers from the other green set can give a product equal to a member of the closed set and create a contradiction.
- Numbers from 0 to 1: Then two negative numbers over −1 and a positive one above 1 can yield a blue product, also leading to a contradiction.
- Numbers from −1 to 1: Makes both sets closed.
- All non-negative numbers: Makes only the blue set closed.
Only the last option works for Q2.
Additional observation/stuff that can be useful to expand the answer:
Take all powers of a positive number. Any positive number can be written as a power of it, which turns closedness under multiplication into closedness of exponents under addition.
If x is green, then so is (√x), because the blue set is closed. Then it goes like x1/2n. By extension, (x1/2n)3+2k numbers are all green. (x1/2n)2 also works, so all that remains is (x1/2n)6k, but if (x1/2n)4k∗(x1/2n)k∗(x1/2n)k work, so do these. That is, all integer-exponent powers of x1/2n are green.
b2=b
b3=b
g3=g
b1/3=b
g1/3=g
g1/2=−g1/2=g
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edited Sep 17, 2024 at 12:29
answered Sep 17, 2024 at 7:47
NautilusNautilus
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Not sure what makes my answer wrong here. Genuinely confused.
– Nautilus
Commented
Sep 17, 2024 at 8:49
2
It's not wrong, but it's not good either. 1) You don't add anything that other answers did not add before. 2) You don't explain very well your reasonning 3) The sentences are confusing and hard to follow (like the last paragraph of the first spoiler block: "a few simple ways", but you start with a contradiction ?!? What are you trying to show there ?). Did not downvote, just telling why I think this answer is not well received
– Alois Christen
Commented
Sep 17, 2024 at 9:30
1- I see plenty of answers that get upvoted despite not adding anything, so it's annoying that these double standards come my way. 2 & 3- English isn't my first language, and not everyone is equally talented with languages. And what do you mean by "hard to follow" anyway? 90% of y'all are smarter than me, so it's not like anyone will have a problem understanding simpler ideas than theirs.
– Nautilus
Commented
Sep 17, 2024 at 9:39
The top answer didn't delve into the reasoning either. Why the heck should it ONLY apply to me? People are just looking for excuses to downvote me unless my answer is completely perfect and airtight.
– Nautilus
Commented
Sep 17, 2024 at 9:41
4
Your answer currently only has 1 vote which happens to be a downvote. A single vote, either upvote or downvote, doesn’t really mean very much.
– Will.Octagon.Gibson
Commented
Sep 17, 2024 at 10:20
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13976 | https://www.youtube.com/watch?v=yJSme00hQZw | IMO 2013 - P2: The great combinatorics problem with colors, points, and lines
Shefs of Problem Solving
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Posted: 2 Apr 2025
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Transcript:
Hello fellow problem solvers. So today we're going to be doing a problem from the 2039 IMO problem number two. This is one of those problems that you just find beautiful and it's nice at the very minimum. And at the very minimum I would invite you to try this out for a minimum of half an hour. Ideally 90 minutes to two hours not more than 4 and 1/2 hours. If you'd like to go along with us give this a go for the next 30 minutes. And now let's begin. And at the end of this, I'll also tell you a funny story about a friend of mine was actually at the Simo. So let's begin with this problem. We have let's understand what's happening here. We have that 4,37 points. Okay, 2013 are red, 2014 are blue. They're in a Colombian configuration. That's where the admiral was. If and only if we have that none of them are on the same line. That's basically what a Colombian configuration is. And then we put some lines. Okay. So we have some of these points and a configuration of lines is good. If the points in every region formed by these lines are what's it called? that every region doesn't have a two points of the same color. Yeah. So no, every region doesn't have two points which are of different colors. That's what a good configuration is and we need to find a case such that this is we can always if we have K lines we can always make a good arrangement or Colombian configuration. So what is the first thing you would do with this sort of problem? For me, this is true when I did it. This is going to be true now when I'm doing it again. We when you don't know what's happening like 4,000 looking at 4,27 points 2014, 2013. Those are big numbers. There are lots of possibilities. It's too difficult for me to comprehend like all the possible configurations of 4,000ish points. So, forget about 4,000. Let's say, okay, let's make let's make this n and n + one and work our way up until 2013. So, we're going to try to see like what happens if we have one red and two blue points, two red, three blue points, then three and four. And this is what I'll invite you to do for the next 10 minutes. Of course, please read the problem. And now, let me clear up the board and go through all of these cases. And now that the board has been cleared up, let's look at one and two. So say we have one blue point and two red ones. Okay, so red point here, a red point here, and a blue point just not on the line. Well, this is not a blue marker. But we'll make let's see how my blue marker looks today. It's decent. Let's use the blue one. Okay. Well, if we have this, I can just like draw a line here. I can make it as close as to these two red points such that the blue will be cut off. So there's no issue there really. Let's look at two free. I mean K is one here, right? No matter what the configuration is, they're going to form a triangle. And then we can do this. Now let's look at what happens if we have three red points. One, two, three red and two blues. So now it actually matters where the blue ones are. Say I have a blue one here and a blue one here. Okay, now these are points of a convex pentagon and I should be able to do this in two. In fact, I need at least two lines because why do I need two lines? Well, can we prove this? I think we could because this is a convex pentagon and we need huh if the point these lines need to cut through the these neighboring these h the lines I think need to cut through like every red point connected by a blue point needs to have a line cut through it otherwise they're not separated. Now this is a good way to think about in general this idea of what is necessary like you can give us a say lower bound because mind you this sort of this thing with K you need to show two things one that the given K we're going to find always works no matter the configuration but also you'll need to show that there's a configuration in which you can't do K minus one right you need to show both those things like If K if the answer was say 10, you need to show that 10 always works. We can always use use 10 lines. And also you need to show that there's a configuration where you can't do it with nine lines. You can't separate them out. So this seems okay. We have two lines. Now what can we are two enough? I invite you to pause for three minutes and think about whether or not two lines are enough. And the answer is well you can see like looking at different cases I can just no matter the if instead of the blue one being here the blue one one was say somewhere here I mean this would literally be the same situation but say it was like somewhere inside say here so it's not collinear I could like I need two lines I've shown in the previous example so I can just create like this little thing like a tube that collect sects these two blue points and leaves out the rest. Right? So that's an idea. And now for three and four, I invite you pause for another 10 minutes and see if you have an idea because and try to solve it the problem for three and four because right now we're getting kals for this configuration of two and three points. And now the thing with three and four, we're going to have let's see some we're going to have four red points. And now we're going to have three blue ones. Okay. Now the tube argument doesn't really work here. I mean in this configuration we could tube these two and that would give us two lines and another one here and it would give us a third line but I'm curious like we could do it in free. I'm now curious whether the solution is K is equal to the smaller one of these. Like for evens, it seems like it has to be right. For even if the smaller one is even, it seems like we'll just be able tooop tube them up and be done with it. But our trouble is that the smaller one is odd. So we can't just like tube them up. So, we have to figure out a different approach. So, let's see what what do we have? Okay, if this one's on the outside, we could also finish up. What if they're all like inside here? We'll also need to figure out like what this inside outside thing means. I mean, then I can actually do them with two lines. Let's say there was a red point inside here as well. And maybe one on this side. Maybe instead of the red point being here, it's hereish. Maybe it's between these two. Can I chop them off? I mean, I I still should be able to just like making them these two very small. I can get two of them. And then the third one I can just get out like this. What am I using here? That's the question. What am I using here for free? When I have three and four points, what is something that is important to me? Is there some structure we can look at here? This is where I would invite you, yes you, to pause for another five, 10 minutes. See if there's any sort of formalization you can give it. Like if there's one on the sides, on the edges, so to speak, we can get one line and put the other ones in tubes. How would you formalize these on the edges? Really think about that and actually try to push this problem forward because it's a beautiful problem. I would really like you to try and solve it yourself. I've given you a lot of cool ideas here. Now the next sort of bit is okay well let's call let's use a convex hull. Let's which is let's make a thing a engon such that inside it are all the points in this configuration. So because we have four and three let's say there's two points inside. Now what do we have in which case are we done? Pause for a few minutes ask yourself that. And the answer is if one of the points on the convex engon on the convex hull is blue right I'm saying three blue and four red then I can cut this one out and then the other two blue wherever there are I can just put them in a tube right so with that I will be done now the only question is what If all the points in this convex engon gone are in fact red and say there's not say there's three points in the say like this triangle covers all the points and then we have four other points inside. What do we do in this case? Now we have all of these are there's a red one, a red one, a red one and say a red one. What would we do in this situation? Well, for me, the idea is I mean, wait a second. If I cut out these two red ones, I can cut them out, one line, then I'm left with two reds and three blues. And if I have two of one and three of the other, I can use two tubes. So, I can then sort of finish inductively. and be done. And now here this in mind is there a way to perhaps generalize this to a bigger case. How are we going to now solve the problem for our 2013 and 2014 points? What is K? And also mind you, wait a sec. Wait a second. I'm getting a bit too ahead of myself. We're now only now showing some we're like going towards the idea that if I have n blue points and I have n + one red points that k is equal to n. Right? That's what I'm showing. That's sort of what I'm like trying to say that I can complete them with k but I need with n but I need to show that there's a configuration which I can't complete them with n minus one. So we still need to pause at this three or four and try to construct from configuration where we need absolutely need at least three of these lines. And here's where I invite you to pause again. If you haven't come this far, pause for another 10 minutes. Try to push the problem forward. Maybe even 15 minutes. 20 if you have the time. Please pause. Pause. Yes. Now let's get back to the problem. So what do we have? We are going to try to see how to construct this such that we need K to be equal three for the case uh three blue and four red. So now we had this idea of a pentagon here where it was like sort of necessary five. Now what if we had a seven gone? How many do I have? One, two, three, four, five, six, and say seven. And now if I just alternate their colors like this, then I've sort of forced myself to have how many lines? Well, every black every line goes through two segment goes through at least two segments, right? And if I take a blue point, if I connect it to these two neighboring red points, there needs to be a line to separate this point from this one. This these two points, there needs to be a line through this segment and a line through this segment. So that's one. To separate this blue one from these two points, there needs to be a line from through from through both of these segments. And that's another line. separate these two, we need another line here. And that's a third line. So in essence, we need at least three lines in this case. And now is the time for you. Your turn. Get to 2013. Like figure out how you'll get to 2013, what you want to prove. And pause and try to do that for 20 minutes. If you can, you'll feel very good about yourself that you've been able to solve an IMO problem number two in combinatorries and it's a beautiful problem and let me clear up the board now and now the board is cleared up. Let's see what would we do here. We would sort of try to prove the claim inductively saying that if we have end points of one color and n + one of another color in a colombian configuration meaning you know if we are colinear then we need at least k is equal to n lines to cut them up and rather on the induction side we can just prove sufficient like that if we have n lines That's enough to separate them out. And the way we'd do this is we'd say, okay, let's draw the convex hull around these two n plus1 points. So here's that convex hull. And now we say one of two things is true. We're going to prove this by induction. We prove this for n equals 1, two, and three. Now for the convex hole we say there's one of two cases. So this is number of red points and this is the number of blue points. Now if n is even if n is even we can pair up the blue points and put them into those tubes and then we'll have every tube consists of two lines. We can always do this such it separates out from the any two points because what we'd say is take two points doesn't matter which one but they're blue and take the red point that is closest to it let that be on a distance the two these two points then draw a line that's parallel to the one connecting these two blue points and that is of a distance d over two from both of them that's just that's a way that you can prove proof and it would be nice to actually prove these things like just by saying this you've actually proved it. If there was a red point inside it would be of a shorter distance than d contradiction, right? So if n is even we're immediately done. So for all intents and purposes assume then n is odd. Now we're looking at n is odd case. And what happens there? Well, we say if n is on the if n is odd, then n plus one is even. Now, if one of the points on a convex hull is blue, case one like actually case two subcase one. If one of these points is blue, then we say, okay, we're going to draw one line to separate this one out from all the other points. And we can do that. We can get it as close as we can get this line such that the distance from this point to the line is the shortest possible distance among any other points to this line and that the other points are on this side something like that's sort of the argument I mean we can keep moving the line up and up and up until we reach this point then okay so now with that in mind we have this point separated and now by induction we have actually not by induction we've separated out this point with one line and then we have n minus one blue points now how many now what can we do with them well we can pair them pair them up into these tubes when we pair them up into tubes because there's we can pair them up 2 by two of them and we'll use n minus one lines. So in total we're using k is equal to n minus one + n + 1 is equal to n lines. Now if all the points on the convex hull are red, there's at least three points on the convex hull. So there's two neighboring points which are red. And we'll say we take a line that is that separates out these two points from the rest of them. And now what do we have on the other points? We have n minus one red points and n blue ones. And by deductive hypothesis, we know that we can separate these out using k= n minus one lines. So the other ones are going to be separated with n minus one lines and the one other line we're going to have k is equal to n lines is enough. And we prove this now by induction. Tada. And this is showing that k is less than or equal to 2013 because we've shown if we have 2013 that is enough. Now can we show that there's a configuration where we have or we need 20 13 lines and we say yes regular it's easier if you say it's a regular or no put them on a circle even better I mean if it's a regular engon it's on a circle but regular 2017 engon label the points A1 A2 all the way till A427. And then we say, let's color the odd ones. I think if I color the odd Yeah, let's color the odd ones red. And now we have that. We need to separate out the points a 2 k minus one and a 2k with a line. for k is equal to for k 0 uh not zero from k is equal to 1 through k is 20 13 2013 I need to separate out so wait a second I need to separate out a1 with a2 oh wait a second I need to separate out these two and a2k k and a2k + one for k is equal to 1 through 2013. Now that means that there needs to be a what's it called? A line going through each all of these what's it called? How how do we put this? There isn't if a line goes through one of these segments, one of these uh 4,026 segments, then it's going through the circle as well. Then it's going through the arc A. If it goes through the line segment ai ai + one then it's going through the arc the shorter arc ai ai + one on the circle and given every line can go through most two points on a circle is that I'm cur I think yeah that's self-evident but I'm thinking how do I prove that now we have This every line goes for at least what's it called? One of these. So every line can go for at most two segments of these 20 of these 4,026. So we are going to need at least 4,026 over two lines to make them separate out these points which is equal to 2013. So K 2013 K needs to be greater than or equal to 2013 less than or equal 2013 and therefore K is equal to 2013 is our answer. And this finishes up the problem. Now I absolutely love this problem. It's one of my favorite problems from the IMO. I wish I had it. Now for the funny story. So, a friend of mine from a from our team went to the IMO and he got um he got sick during the first day and he needed to finish early. He needed to finish both days early. And what happened was he either he misread or it was badly or the first problem was badly translated. So he skipped over the first problem at the IMO, worked on the second problem, worked on it for like 90 minutes and or so and then submitted his paper and he worked on 90 minutes both days and he was a point or two away from bronze medal. But he solved the second problem. And it just goes to show like when you're going to these competitions, you should also be aware that you don't get food poisoning or something like that because that can derail you. Like imagine that in onethird of the time he was two points away from bronze. That's why we call him the machine or the boss. And this finishes up our the story and this beautiful little problem. And as always, thanks for problem solving. |
13977 | https://en.wiktionary.org/wiki/ducitur | ducitur - Wiktionary, the free dictionary
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13978 | https://www.doubtnut.com/qna/11480951 | EAN of cobalt is 36 in [Co(NH3)2O2(en)br] Thus O2 is .
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What is effective atomic number (EAN) ?
Calculate EAN of cobalt (Z=27) in [CO(NH3)6]+3 and of zinc (z=30) in [Zn(NH3)4]SO4.
(i) Write the IUPAC name of the complex : [Ce(NH3)2Cl2(en)Cl] .
(ii) Write the formula of the complexes : pentamminenitrito-o-cobalt (III) .
Knowledge Check
EAN of cobalt is 36 in [Co(NH3)2O2(en)Cl] thus ,O2 is
EAN of cobalt is 36 in [Co(NH3)2O2(en)Cl] thus ,O2 is
Oxidation state of cobalt in [Co(NH3)4(H2O)Cl]SO4 is
Name the following complexes according to the IUPAC system of nomenclature
(i) Co(NH3)4(H2O)Br2
(ii) Na[Au(CN)2]
(iii) Na3[Fe(C2O4)3]
(iv) [Co(en)3[Cl3
The EAN of cobalt in the complex ion [Co(en)2Cl2]+ is
Oxidation number of cobalt in [Co(NH3)6]Cl2Br is -
Which of the following will show optical isomers?
(I) cis-[Co(NH3)2(en)2]3+
(II) trans-[IrCl2(C2O4)2]3−
(III) [Rh(en)3]3+
(IV) cis-[IR(H2O)3Cl3]
Which of the following will show optical isomers?
(I) cis-[Co(NH3)2(en)2]3+
(II) trans-[IrCl2(C2O4)2]3−
(III) [Rh(en)3]3+
(IV) cis-[IR(H2O)3Cl3]
CENGAGE CHEMISTRY-COORDINATION COMPOUNDS-Exercises Single Correct (Isomerism)
Which has maximum EAN of the underbold atoms? (Cr=24,Co=27,Fe=26,Ni=...
Give EAN value of Mg in [Mg(EDTA)]^(2-) .
EAN of cobalt is 36 in [Co(NH(3))(2)O(2)(en)br] Thus O(2) is .
EAN of Fe in [Fe(C(2)O(4)(3)]^(3-) is .
The EAN of Fe atom in .
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The number of geometrical and optical isomers of [Cr(NH(3))(3)(NO(3)...
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1 Xronos Tutorial
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MAC1140 Introduction and Syllabus
2 Goals of this Section
This is the introduction to the overall course and it contains the syllabus as well as grade information.
2.1 Goals of this Course
This course has several concurrent but different goals.
2.2 Expect Differences
What makes this course different from previous courses?
2.3 Virtues of CBT
Contextual Based Learning (CBT) has many virtues, knowing why we are learning how we are will help your studying and learning process.
2.4 Methods to Prepare
Suggestions on Studying and Learning
2.5 General Syllabus
This is the syllabus for the course with everything but grading and the calendar
2.6 The Point of Grades
This is the grading rubric for the course, including the assignments, how many points things are worth, and how many points are needed for each letter grade.
2.7 Pre-Requisites
This section covers the skills that a MAC1140 student is expected to be fluent in.
Mathematical Modeling
3 Goals of this Section
This section is on learning to use mathematics to model real-life situations.
3.1 Terminology To Know
These are important terms and notations for this section.
3.2 What is mathematical reasoning?
This section aims to introduce the idea of mathematical reasoning and give an example of how it is used.
3.3 Logical Deduction
This section analyzes the previous example in detail to develop a three phase deductive process to develop a mathematical model.
3.4 Types of Information
This section aims to explore and explain different types of information.
3.5 Is this actually math?
This section aims to show how mathematical reasoning is different than ‘typical reasoning’, as well as showing how what we are doing is mathematical.
3.6 Math as a Language
This section contains important points about the analogy of mathematics as a language.
3.7 Numeric Model Walkthrough
This is a detailed numeric model example and walkthrough.
3.8 Embrace Laziness!
This section aims to show the virtues, and techniques, in generalizing numeric models into ‘generalized’ models.
3.9 Variables and Their Roles
This section explains types and interactions between variables.
3.10 Generalized Model Walkthrough
This is an example of a detailed generalized model walkthrough
Variables, Functions, Graphing, and Universal Properties
4 Goals of this Section
This section is on functions, their roles, their graphs, and we introduce the Library of Functions
4.1 Terminology To Know
These are important terms and notations for this section.
4.2 Relationship vs. Equations
In this section we discuss a very subtle but profoundly important difference between a relationship between information, and an equation with information.
4.3 Relationship vs. Functions
In this section we discuss what makes a relation into a function.
4.5 Functions Require Context
In this section we demonstrate that a relation requires context to be considered a function.
4.8 Domain, Codomain, and Range
In this section we cover Domain, Codomain and Range.
4.10 Set Notation
In this section we cover how to actual write sets and specifically domains, codomains, and ranges.
4.12 Function Notation
This section covers function notation, why and how it is written.
4.14 f(x) Notation
This section covers notation.
4.16 Function Composition
We cover the idea of function composition and it’s effects on domains and ranges.
5 Graphing Introduction
This section introduces graphing and gives an example of how we intuitively use it.
5.1 Terminology To Know
These are important terms and notations for this section.
5.2 French History and Dinosaurs!
This section introduces the origin an application of graphing.
5.3 Graphing To Relate Variables
This section describes how we will use graphing in this course; as a tool to visually depict a relation between variables.
5.4 Graphs Aren’t Precise
This section describes how accuracy and precision are different things, and how that relates to graphs.
5.5 Using Graphs
This section covers what graphs should be used for, despite being imprecise.
5.6 Vertical Line Test
This section describes the vertical line test and why it works.
6 Library Of Functions
This is an introduction and list of the so-called “library of functions”.
6.1 Parent Functions
This section provides the specific parent functions you should know.
7 Universal Properties
This section introduces the idea of studying universal properties to avoid memorizing vast amounts of information.
7.1 Terminology To Know
These are important terms and notations for this section.
7.2 Geometric Vs Analytic Viewpoints
We discuss what Geometric and Analytic views of mathematics are and the different roles they play in learning and practicing mathematics.
7.3 Geometric Perspective
We discuss the geometric perspective and what its role is in learning and practicing mathematics.
7.4 Analytic Viewpoint
We discuss the analytic view of mathematics such as when and where it is most useful or appropriate.
7.5 Intro: Rigid Translations
An introduction to the ideas of rigid translations.
7.6 Rigid Translations: Geometrics
This section describes the geometric perspective of Rigid Translations.
7.8 Rigid Translations: Analytics
This section describes the analytic perspective of what makes a Rigid Translation.
7.10 Transforms: Geometric
This section describes the geometric interpretation of what makes a transformation
7.11 Transforms: Analytic
This section describes the analytic interpretation of what makes a transformation and how to use the function notation to perform (or read) a transformation quickly and easily.
7.13 Transform And Translates
This covers doing transformations and translations at the same time. In particular we discuss how to determine what order to do the translations/transformations in.
7.15 Points of Interest on Graphs - Zeros
This section describes types of points of interest (PoI) in general and covers zeros of functions as one such type.
7.16 Points of Interest on Graphs - Extrema
This section describes extrema of a function as points of interest (PoI) on a graph.
7.18 Points of Interest on Graphs - Discontinuities
This section describes discontinuities of a function as points of interest (PoI) on a graph.
7.20 Algebra with Functions
This section describes how to perform the familiar operations from algebra (eg add, subtract, multiply, and divide) on functions instead of numbers or variables.
7.22 Equals Signs are Magic!
This section describes the very special and often overlooked virtues of the ‘equals sign’. It also includes when and why you should “set something equal to zero” which is often overused or used incorrectly.
7.23 One and Zero; the Most Useful of Numbers
This section describes the very special and often overlooked virtue of the numbers Zero and One.
7.24 Inverse Functions
This section introduces the geometric viewpoint of invertability.
7.25 Horizontal Line Test
This section discusses the Horizontal Line Test
7.27 Inverse Function - Analytic View
This section introduces the analytic viewpoint of invertability, as well as one-to-one functions.
Exploration of Functions
8 Polynomial Functions
This section is an exploration of polynomial functions, their uses and their mechanics.
8.1 Terminology To Know
These are important terms and notations for this section.
8.2 What is (and isn’t) a Polynomial?
We know an awful lot about polynomials, but it relies on the very specific structure of a polynomial, and thus it is paramount that one can correctly recognize what is, and isn’t, a polynomial to use these tools.
8.4 Fundamental Theorem of Algebra
This section covers one of the most important results in the last couple centuries in algebra; the so-called “Fundamental Theorem of Algebra.”
8.6 An Interjection into Polynomial History!
This section is a quick foray into math history, and the history of polynomials!
8.7 Exponents and Extrema: An Example
This section contains a demonstration of how odd versus even powers can effect extrema.
8.9 Exponents and Extrema 2: Local Extrema
This section contains information on how exponents effect local extrema
8.11 Curvature and Graphing
This section shows and explains graphical examples of function curvature.
8.12 Factoring: Introduction
Some information on factoring before we delve into the specifics.
8.13 Factoring: Round One!
First dive into factoring polynomials. This section covers factoring quadratics with leading coefficient of by factoring the coefficients.
8.15 Factoring; Grouping Method
Factor higher polynomials by grouping terms
8.17 Factoring; AC Method
How to factor when the leading coefficient isn’t one.
8.19 Factoring; Special Forms
Factor polynomials quickly when they are in special forms
8.21 Completing the Square
This section introduces the technique of completing the square.
8.23 Polynomial Long Division
In this section we explore how to factor a polynomial out of another polynomial using polynomial long division
8.24 Polynomial Synthetic Division
Factor one polynomial by another polynomial using polynomial synthetic division
8.26 Rational Root Theorem
Find factors via rational root theorem
8.28 Complex Numbers
Intro to complex numbers and conjugates
8.29 Simplifying Complex Numbers
Simplifying complex numbers
8.32 Quadratic Formula
Exploring the usefulness and (mostly) non-usefulness of the quadratic formula
8.33 Comprehensive Factoring Quiz
A Comprehensive Factoring Practice Quiz.
9 Radical Functions
This section is an exploration of radical functions, their uses and their mechanics.
9.1 Terminology To Know
These are important terms and notations for this section.
9.2 Why Radicals?
This section introduces radicals and some common uses for them.
9.3 Simplifying Numeric Radicals
This section introduces radicals and some common uses for them.
9.5 Types of Radicands
This section introduces two types of radicands with variables and covers how to simplify them... or not.
9.7 Type 2 Radicals
This section discusses how to handle type two radicals.
9.9 Type 1 Radicals
This section discusses how to handle type one radicals.
9.11 Square Root: the Inverse Function
This section views the square root function as an inverse function of a monomial. This is used to explain the dreaded symbol and when to use (and not use) absolute values.
9.13 Solving Unsimplified Radicals
This section shows techniques to solve an equality that has a radical that can’t be simplified into a non radical form. This has potential drawbacks which is also covered in this section.
10 Exponential Functions: Goals
This section is an exploration of exponential functions, their uses and their mechanics.
10.1 Terminology To Know
These are important terms and notations for this section.
10.2 A Review of Exponential Functions
This section reviews the basics of exponential functions and how to compute numeric exponentials.
10.4 Properties of Exponentials
This section gives the properties of exponential expressions. Most of these should be familiar, although we go into slightly more details as to how and why these properties hold in some cases.
10.7 Properties of the Exponential Function
This section gives the properties of exponential functions. There is a subtlety between the function and the expression form which will be explored, as well as common errors made with exponential functions.
10.9 Exponential Growth and Decay
This section discusses the two main modeling uses of exponentials; exponential growth, and exponential decay.
11 Logarithmic Functions
This section is an exploration of logarithmic functions, their uses and their mechanics.
11.1 Terminology To Know
These are important terms and notations for this section.
11.2 Introduction and Notation of Logarithms
This section is a quick introduction to logarithms and notation (and ways to avoid the notation).
11.4 Properties Of Logs
This is one of the most vital sections for logarithms. We cover primary and secondary properties of logs, which are pivotal in future math classes as these properties are often exploited in otherwise difficult mechanical situations.
11.6 Common Mistakes Of Logs
This is one of the most vital sections for logarithms. We cover primary and secondary properties of logs, which are pivotal in future math classes as these properties are often exploited in otherwise difficult mechanical situations.
11.7 Change of Base formula
This is one of the most vital sections for logarithms. We cover primary and secondary properties of logs, which are pivotal in future math classes as these properties are often exploited in otherwise difficult mechanical situations.
11.8 Examples of Logs
This is a demonstration of several examples of using log rules to handle logs mechanically.
12 Piecewise Functions
This section is an exploration of the piece-wise function; specifically how and why they are used and their mechanics.
12.1 Piecewise Functions: The Geometric View
This section discusses the geometric view of piecewise functions.
12.2 Piecewise Functions: The Analytic View
This section discusses the analytic view of piecewise functions.
12.4 Piecewise Functions: Computation
This section discusses how to compute values using a piecewise function
13 Absolute Value Functions
This section is an exploration of the absolute value function; specifically how and why they are used and their mechanics.
13.1 Absolute Value: Geometric View
This discusses Absolute Value as a geometric idea, in terms of lengths and distances.
13.2 Absolute Value: Analytic View
This discusses the absolute value analytically, ie how to manipulate absolute values algebraically.
13.4 Absolute Value: Solving Equalities
This section is on how to solve absolute value equalities.
14 Rational Functions
This section is an exploration of rational functions; specifically those functions that are made by taking a ratio (ie fraction) of polynomials.
14.1 Terminology To Know
These are important terms and notations for this section.
14.2 What is a rational function?
We discuss what makes a rational function, and why they are useful.
14.4 Domain of rational functions
We discuss one of the most important aspects of rational functions; the domain restrictions.
14.6 Vertical Asymptotes
We discuss the circumstances that generate vertical asymptotes in rational functions.
14.8 Holes in Domains of Rational Functions
We discuss the circumstances that generate holes in the domain of rational functions rather than vertical asymptotes.
14.10 Horizontal Asymptotes
We discuss the circumstances that generate horizontal asymptotes and what they mean.
mac1140nowell
Precalculus Algebra MAC1140
Polynomial Functions
Simplifying Complex Numbers
Simplifying complex numbers
There are a surprising number of consequences to the fact that , and one of these is how far one can simplify a complex number. Indeed, it is always possible to put any complex number into the form , where and are real numbers. This is not always obvious, however there is a set technique to accomplish that task. As usual we start with demonstrating the technique in a general case, and then give a concrete example for one to use in comparison.
If we want to simplify an expression, it is always important to keep in mind what we mean when we say ’simplify’. Typically in the case of complex numbers, we aim to never have a complex number in the denominator of any term. To accomplish this, we will first make an observation that may seem to be a non sequitur, but will prove to be pivotal.
Lets see what happens if we multiply by it’s complex conjugate; . We get:
We end up getting , a real number! This will allow us to simplify the complex nature out of a denominator. We demonstrate how in the following example.
Simply the complex number .
Applying the observation from the previous explanation; we multiply the top and bottom (multiplying by one cleverly) of our fraction by the conjugate of the bottom to get:
Notice that the result, is vastly easier to deal with than .
As we saw above, any (purely) numeric expression or term that is a complex number, can always be reduced using this technique to the form where and are some real numbers. Because of this, we say that the form is the “standard form” of a complex number.
1 : Simplify the following complex expression into standard form.
Remember to multiply the top and bottom of the fraction by the conjugate of the bottom; i.e. and foil out. This should get you a real denominator.
Also notice that the “” is already provided, so you don’t need to type into your answer.
2 : Simplify the following complex expression into standard form.
Remember to multiply the top and bottom of the fraction by the conjugate of the bottom; i.e. and foil out. This should get you a real denominator.
Also notice that the “” is already provided, so you don’t need to type into your answer.
3 : Simplify the following complex expression into standard form.
Remember to multiply the top and bottom of the fraction by the conjugate of the bottom; i.e. and foil out. This should get you a real denominator.
Also notice that the “” is already provided, so you don’t need to type into your answer.
4 : Simplify the following complex expression into standard form.
Remember to multiply the top and bottom of the fraction by the conjugate of the bottom; i.e. and foil out. This should get you a real denominator.
Also notice that the “” is already provided, so you don’t need to type into your answer.
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abs∣[?]∣\left|\blue{[?]}\right|∣[?]∣
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paren([?])\left(\blue{[?]}\right)([?])
floor⌊[?]⌋\lfloor \blue{[?]} \rfloor⌊[?]⌋
factorial[?]!\blue{[?]}![?]!
exp[?][?]{\blue{[?]}}^{\blue{[?]}}[?][?]
sub[?][?]{\blue{[?]}}{\blue{[?]}}[?][?]
frac[?][?]\dfrac{\blue{[?]}}{\blue{[?]}}[?][?]
int∫[?]d[?]\displaystyle\int{\blue{[?]}}d\blue{[?]}∫[?]d[?]
defi∫[?][?][?]d[?]\displaystyle\int{\blue{[?]}}^{\blue{[?]}}\blue{[?]}d\blue{[?]}∫[?][?][?]d[?]
deriv d d[?][?]\displaystyle\frac{d}{d\blue{[?]}}\blue{[?]}d[?]d[?]
sum∑[?][?][?]\displaystyle\sum_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}[?]∑[?][?]
prod∏[?][?][?]\displaystyle\prod_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}[?]∏[?][?]
root[?][?]\sqrt[\blue{[?]}]{\blue{[?]}}[?][?]
vec⟨[?]⟩\left\langle \blue{[?]} \right\rangle⟨[?]⟩
mat([?])\left(\begin{matrix} \blue{[?]} \end{matrix}\right)([?])
⋅\cdot⋅
infinity∞\infty∞
arcsin arcsin([?])\arcsin\left(\blue{[?]}\right)arcsin([?])
arccos arccos([?])\arccos\left(\blue{[?]}\right)arccos([?])
arctan arctan([?])\arctan\left(\blue{[?]}\right)arctan([?])
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cos cos([?])\cos\left(\blue{[?]}\right)cos([?])
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sec sec([?])\sec\left(\blue{[?]}\right)sec([?])
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cot cot([?])\cot\left(\blue{[?]}\right)cot([?])
log log([?])\log\left(\blue{[?]}\right)lo g([?])
ln ln([?])\ln\left(\blue{[?]}\right)ln([?])
alpha α\alpha α
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eta η\eta η
theta θ\theta θ
iota ι\iota ι
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lambda λ\lambda λ
mu μ\mu μ
nu ν\nu ν
xi ξ\xi ξ
omicron ο\omicron ο
pi π\pi π
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psi ψ\psi ψ
omega ω\omega ω
Gamma Γ\Gamma Γ
Delta Δ\Delta Δ
Theta Θ\Theta Θ
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13980 | https://e-jmt.org/journal/view.php?doi=10.12671/jkfs.2014.27.3.185 | J Musculoskelet Trauma : Journal of Musculoskeletal Trauma
Articles
Department of Orthopedic Surgery, Keimyung University School of Medicine, Daegu, Korea.
Copyright © 2014 The Korean Fracture Society. All rights reserved.
This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Abstract
Financial support: None.
Conflict of interest: None.
REFERENCES
Values are presented as mean±standard deviation or number.
Statistically significant.
Statistically significant.
Figure & Data
REFERENCES
Citations
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| Associated Factors of Radial Nerve Palsy Combined with Humerus Shaft Fracture J Korean Fract Soc. 2014;27(3):185-190. Published online July 31, 2014 DOI: |
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Demographic Data
Values are presented as mean±standard deviation or number.
Statistically significant.
Characteristics of Fractures
Statistically significant.
Values are presented as mean±standard deviation or number.
Statistically significant.
Values are presented as mean±standard deviation or number.
Statistically significant.
Statistically significant.
Statistically significant.
Table 1
Table 2
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13981 | https://openstax.org/books/college-physics-2e/pages/14-section-summary | Ch. 14 Section Summary - College Physics 2e | OpenStax
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College Physics 2e
Section Summary
College Physics 2eSection Summary
Contents Contents
Highlights
Table of contents
Preface
1 Introduction: The Nature of Science and Physics
2 Kinematics
3 Two-Dimensional Kinematics
4 Dynamics: Force and Newton's Laws of Motion
5 Further Applications of Newton's Laws: Friction, Drag, and Elasticity
6 Uniform Circular Motion and Gravitation
7 Work, Energy, and Energy Resources
8 Linear Momentum and Collisions
9 Statics and Torque
10 Rotational Motion and Angular Momentum
11 Fluid Statics
12 Fluid Dynamics and Its Biological and Medical Applications
13 Temperature, Kinetic Theory, and the Gas Laws
14 Heat and Heat Transfer Methods
Introduction to Heat and Heat Transfer Methods
14.1 Heat
14.2 Temperature Change and Heat Capacity
14.3 Phase Change and Latent Heat
14.4 Heat Transfer Methods
14.5 Conduction
14.6 Convection
14.7 Radiation
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
15 Thermodynamics
16 Oscillatory Motion and Waves
17 Physics of Hearing
18 Electric Charge and Electric Field
19 Electric Potential and Electric Field
20 Electric Current, Resistance, and Ohm's Law
21 Circuits and DC Instruments
22 Magnetism
23 Electromagnetic Induction, AC Circuits, and Electrical Technologies
24 Electromagnetic Waves
25 Geometric Optics
26 Vision and Optical Instruments
27 Wave Optics
28 Special Relativity
29 Quantum Physics
30 Atomic Physics
31 Radioactivity and Nuclear Physics
32 Medical Applications of Nuclear Physics
33 Particle Physics
34 Frontiers of Physics
A | Atomic Masses
B | Selected Radioactive Isotopes
C | Useful Information
D | Glossary of Key Symbols and Notation
Answer Key
Index
Search for key terms or text.
Close
14.1 Heat ---------
Heat and work are the two distinct methods of energy transfer.
Heat is energy transferred solely due to a temperature difference.
Any energy unit can be used for heat transfer, and the most common are kilocalorie (kcal) and joule (J).
Kilocalorie is defined to be the energy needed to change the temperature of 1.00 kg of water between 14.5 º C 14.5ºC 14.5ºC and 15.5 º C 15.5ºC 15.5ºC.
The mechanical equivalent of this heat transfer is 1.00 kcal=4186 J.1.00 kcal=4186 J.1.00 kcal=4186 J.
14.2 Temperature Change and Heat Capacity -----------------------------------------
The transfer of heat Q Q Q that leads to a change Δ T Δ T Δ T in the temperature of a body with mass m m m is Q=mc Δ T Q=mc Δ T Q=mc Δ T, where c c c is the specific heat of the material. This relationship can also be considered as the definition of specific heat.
14.3 Phase Change and Latent Heat ---------------------------------
Most substances can exist either in solid, liquid, and gas forms, which are referred to as “phases.”
Phase changes occur at fixed temperatures for a given substance at a given pressure, and these temperatures are called boiling and freezing (or melting) points.
During phase changes, heat absorbed or released is given by: Q=mL,Q=mL,Q=mL,
where L L L is the latent heat coefficient.
14.4 Heat Transfer Methods --------------------------
Heat is transferred by three different methods: conduction, convection, and radiation.
14.5 Conduction ---------------
Heat conduction is the transfer of heat between two objects in direct contact with each other.
The rate of heat transfer Q/t Q/t Q/t (energy per unit time) is proportional to the temperature difference T 2−T 1 T 2−T 1 T 2−T 1 and the contact area A A A and inversely proportional to the distance d d d between the objects: Q t=kA(T 2−T 1)d.Q t=kA T 2−T 1 d.Q t=kA T 2−T 1 d.
14.6 Convection ---------------
Convection is heat transfer by the macroscopic movement of mass. Convection can be natural or forced and generally transfers thermal energy faster than conduction. Table 14.4 gives wind-chill factors, indicating that moving air has the same chilling effect of much colder stationary air. Convection that occurs along with a phase change can transfer energy from cold regions to warm ones.
14.7 Radiation --------------
Radiation is the rate of heat transfer through the emission or absorption of electromagnetic waves.
The rate of heat transfer depends on the surface area and the fourth power of the absolute temperature: Q t=σ e A T 4,Q t=σ e A T 4,Q t=σ e A T 4,
where σ=5.67×10−8 J/s⋅m 2⋅K 4 σ=5.67×10−8 J/s⋅m 2⋅K 4 σ=5.67×10−8 J/s⋅m 2⋅K 4 is the Stefan-Boltzmann constant and e e e is the emissivity of the body. For a black body, e=1 e=1 e=1 whereas a shiny white or perfect reflector has e=0 e=0 e=0, with real objects having values of e e e between 1 and 0. The net rate of heat transfer by radiation is
Q net t=σ e A(T 4 2−T 4 1)Q net t=σ e A T 2 4−T 1 4 Q net t=σ e A T 2 4−T 1 4
where T 1 T 1 T 1 is the temperature of an object surrounded by an environment with uniform temperature T 2 T 2 T 2 and e e e is the emissivity of the object.
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Book title: College Physics 2e
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13982 | https://medium.com/nissenyeh/%E5%93%B2%E5%AD%B8%E6%98%AF%E4%BB%80%E9%BA%BC-%E5%93%B2%E5%AD%B8%E7%B3%BB%E5%9C%A8%E5%81%9A%E4%BB%80%E9%BA%BC-21c462f6d5cb | 哲學是什麼？哲學系在做什麼？ - nissenlab - Medium
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尼桑的學習筆記
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哲學是什麼？哲學系在做什麼？
一個哲學系學生的喃喃自語
Nissen Yeh
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Sep 11, 2018
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身為一個哲學系的學生，我一直都想談談我覺得「哲學系」或「哲學」對我的意義是什麼。就我的觀察，哲學在大眾的眼裡通常是兩種情況——要麼被過度低估，要麼就是被過度高估。一派人覺得念哲學沒有用處，讀哲學會餓死；另一派人覺得哲學很有用，可以訓練邏輯思考能力。
恩….我必須說，兩種說法我都同意。一來，哲學確實很沒用，市面上哲學偶爾有被過度吹捧的現象。二來，哲學確實也很有用，哲學訓練確實可以幫助學生更好地面對許多問題。
所以我要怎樣才能同時同意兩個矛盾的立場？這留到文末再進行描述。
首先，我想先從一些常見的現象開始。
我想哲學系的學生多多少少都會被問到「哲學系在幹嘛？」、「學哲學有什麼用？」而我最常聽到的答案是「哲學系會帶領學生更深入思考許多問題，並且很注重鍛鍊學生的邏輯、分析能力。」
此時，雖然自己身為哲學系的學生，但內心還是會有許多吐槽飄過去，像是「難道其他科系就不訓練學生邏輯和分析能力了嗎？」、「說的好像除了哲學系之外，其他領域的學生都不會訓練大腦。」
然後，好奇的人可能會繼續追問「所以思考那些沒有答案的問題有什麼用啊？」此時認真的哲學系同學就會說「哲學重要的是思考過程，不是那些答案。」
然後我內心就會難過地在內心想「唉…感覺就是因為哲學太沒用了，所以只好強調過程。」「就像比賽輸了很丟臉，所以只好說結果不重要，重要的是過程」然後內心默默為哲學系難過3秒。
好吧，其實以上說法都是沒有錯。確實，在哲學系當中是更有機會訓練到邏輯分析的能力。也確實，在哲學的思考中，過程也比結果重要。只是說每當有人問我「哲學系在幹嘛？」我自己並不會想這樣回答，一來，邏輯分析能力應該是每個領域的理想教育模式下要培養的能力，非哲學系的專利，二來，過程重於結果是有理由的，這並不是理所當然的特性。
所以，我會怎麼介紹哲學或哲學系呢？
如果對方是真的好奇（不是隨口找話題問問，而內心沒有真的很想知道），我就會直接跟他們講哲學的歷史——因為我認為只要解哲學的歷史，基本就會知道現在的大學哲學在幹嘛，還有為什麼哲學有那些奇怪的特性了。
哲學的起源——從繁榮到哲學之死
對現在人來說，一提到「哲學」可能就是直覺想到艱深、邏輯、抽象、困難…等形容詞。但回溯到哲學的原始含意，哲學的概念其實再單純不過。
從字根來看（我知道哲學系的同學肯定聽過100遍了，但我還是要這樣切入），哲學（philosophy）的原始意涵是「愛智慧」，是兩個古希臘字「愛」(philia) 與「智慧」(sophia)的集合，因此在一開始——「任何關於對智慧、知識的追求，其實都可以算是哲學的一部分」。從這個定義出發，即使是「晚餐吃什麼比較不會胖」之類問題，廣義而言也可以是一種哲學問題。
認真回顧哲學史，就會看到古代哲學家所思考的許多問題，放在現代而言根本不太像是哲學問題。比如說畢達格拉斯（Pythagoras）研究幾何學、畢氏定理，亞里斯多德（Aristotle）主張重的物體會比輕的物體更快落地。這些問題，放在現今都會被認為是數學、物理範圍的議題。但在那個時代，這些各類知識都被泛泛的稱作為哲學。因為只要是對智慧的追求，其實都能算是一種「愛智慧」的表現。
在古代，哲學的守備範圍包含了世間萬物的知識
這個現象維持非常久，直至17、18世紀，科學也還未從哲學中完全脫離出來，在當時，「哲學研究」跟我們以為的哲學研究很不同，它是包含了數學、光學和各種科學。比如說知名哲學家笛卡爾有一本書叫作《哲學原理》（Principia philosophiae），書名雖然有哲學兩字，但內容跟哲學並沒有關係，基本都是光學研究。
另一個可能更熟悉的例子——牛頓（Sir Isaac Newton），他最偉大的著作叫做《自然哲學的數學原理》(Philosophiæ Naturalis Principia Mathematica)，除了書名中「自然哲學」，書中除了連「物理」這個詞彙都沒有。在當時，數學、物理、生物….這些對學科最簡略的分類方式的都還未出現，而是通通統稱為自然哲學。
牛頓的《自然哲學的數學原理》
必須往後到19、20世紀，現代所熟悉的學科，才逐漸明確地被劃分出來。
但隨著細分領域的興起，哲學探討的守備範圍也逐漸被瓜分。
許多原本屬於哲學的問題，開始被其他領域瓜分出去：
「人類從哪裡來，要往哪裡去？」可以由生物演化學家接手，「宇宙的源頭是什麼？」可以由物理學家接手，「意識與靈魂是什麼？」可以由心理學家接手。
「人們一直以來就在尋問ㄧ大堆的問題：我們怎麼理解我們身居其中的世界？宇宙怎麼行為？實在的本質是甚麼？所有一切從何而來？宇宙需要一個創造者嗎？我們中的大多數人都不會花太多的時間去操心這些問題，但幾乎我們所有人都會在某些時候為它們發愁。
在傳統上，這些問題是哲學問題，然而哲學死了。哲學已經跟不上科學尤其是物理學的現代發展。在我們探索知識的進程中，科學家已成為發現的火炬手。」
——霍金(Stephen William Hawking)・《大設計》(The Grand Design)
霍金的《大設計》
霍金在大設計宣稱「哲學已死」，引起了許多人的批評。先不論他的觀點是否過於偏激，原本大包大攬的哲學領域，到了當代確實只遺留了那些未被其他領域接手、無法由科學方法解決的問題。
舉例來說，「什麼是道德？」對於這個問題，演化學家可能會從演化的視角探討人類道德的概念的是如何慢慢發展出來的。但愛智慧（狡猾）的哲學家永遠可以追問下去，他們可能會問「為什麼這樣就可稱為道德呢？」、「道德的準確定義是什麼？」。一直問，一直問，問題終究會停在ㄧ個點上讓科學家再也無法回答，而只能透過純粹邏輯辯證探討。而這些純粹智性的領域——就是現代意義哲學發揮的空間。
因此，如果有人要我用一句話描述大學哲學系在教什麼？我會很直白的說，我覺得大學哲學就是——「那些被撿剩問題的集合，然後再加上哲學史。」
回答問題時間
了解了哲學的歷史，就可以開始回答一些常見的問題
為什麼很多人說學哲學可以幫助人鍛鍊邏輯、分析能力？
這背後並沒有什麼特別高尚的理由，其實任何領域的學習，像是寫程式、念醫學、寫歌曲都可以（也應該要）鍛鍊學生的邏輯和分析能力。只是哲學特別的地方在於，比起實際的知識，哲學思考更仰賴純粹的智性推理。並且，由於哲學問題缺乏標準答案。因此思考過程的精緻和準確程度，就更是成為這個思考行為是否有價值的重要判准。
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以科學理工領域來說，在課堂上，許多問題，像是水的沸點、純金的密度….都有正確答案，因此不管學生到底是用死背或其實根本不懂原理，但凡他只要有能力把正確答案寫上去，就可以獲得分數。但哲學課堂中，由於問題很少有標準答案，因此唯一能幫助學生獲取分數的方式，就只有他主張的支持理由究竟是個好理由還是爛理由。
因此為什麼在哲學系上更能訓練邏輯、分析能力？因為哲學問題許多沒有標準答案，所以在哲學課堂上，學生就只能在分析和論述能力下更多的功夫。
為什麼說哲學重要的是過程而不是結果
理由也是一樣的，因為哲學的探討範圍，已經限縮到基本只能用智性探討，而沒有客觀的標準答案。當一個問題，可以藉由科學方法持續進行拆解，那它基本就還不能算是哲學問題，而會是心理學、物理學….等科學問題。
真正的哲學議題上，基本就只能用智性和邏輯思辨來進行分析。而至於科學結論本身，雖然有可能變成分析過程中所使用的例子或分析的對象，但就僅僅是作為手段而不是知識追求的終點。
既然哲學問題更多依賴邏輯思辨的問題，因而沒有客觀的標準答案，那重要的就自然不會是結果，而是推導出這樣結果的過程，這就是「哲學重要的是過程而不是結果」的緣由。
研究沒有標準答案的哲學有什麼好處
雖然在前面我把哲學形容的像是廚餘ㄧ樣，認為它是知識界中的剩餘物。這一方面反映了我的真實想法，但這不等於我否認了哲學的價值。哲學的沒有答案或許讓人覺得很空虛，但反過來說，這世界上許多重要的問題正是沒有標準答案的問題。
從近的看，許多生活的難題就是沒有標準答案的問題。比如說「畢業要念研究所還是去工作」，對於這類問題，在不同人或不同的前提前就會有100種答案。對於一個家中經濟困難的人來說，可能適合去工作。對於一個有錢有閒愛學習的人而言，可能適合去念研究所。而哲學就是這類問題的典型，在研究哲學時，學生時常需要像這樣大量拆解各種前提和假設，並藉此尋求一個較好的解答。
從遠的看，更多至關人生重要的議題也是沒有標準答案，比如說「人生的意義」、「怎樣算是幸福的人生」，這些哲學的老生常談也是完全沒有任何人或是客觀證據可以提供標準答案，只能透過自己的分析和判斷去進行決策。
當然，對於以上問題，即使不讀哲學系也可以很好的進行思考。我只是想說，沒有標準答案可能反而是世間常態。而正是在這樣模糊的世界，對於哲學問題的思考，可能有助於對我們尋找方向提供一點幫助。
如果重念哲學系，會給自己的建議
以下是如果我重新念哲學系，我會給自己幾點提醒（說得好像我已經畢業了）
1・念哲學系最重要的不是學哲學知識，而是去學背後的思考方式
除非要走學術，不然課堂上那些具體的哲學知識對未來的就業跟生活其實沒什麼幫助。至少於我而言，光知道笛卡爾的二元論、柏拉圖理型世界在講什麼，除了聊天時可以裝一下，在現實生活中很少有什麼實際運用這些知識的機會。因此，除非能真正學會思考問題的方法，不然閱讀《理想國》跟閱讀市面上《一年炒股100萬》之類的書是其實沒什麼區別的（後者可能更有用）。
在哲學系四年裡中，如果最後只學到一堆具體的「哲學知識」（知道某某哲學家講了什麼理論），但最後卻無法把這些背後的思考模式抽象出來，並且應用在未來生活中，那不如去讀一些擁有更實用知識的學科，例如醫學、資工…等。
2・要培養其他技能
除了學習背後的思考方式，課餘一定要花更多時間培養其他的技能。所謂邏輯、分析能力終究是一種隱性能力，我相信其他科系的畢業生也沒有人會在求職履歷上註記自己沒有邏輯能力。因此，不管再如何的會分析問題，如果缺乏真正的具體實踐或創造，那所謂上思想的優勢根本就只是一種嘴砲。
再來，在學習新技能上，即使哲學的訓練對於思考、分析問題有些許幫助，這也不等於可以少下苦工。即使對於思考問題有點小優勢，但人家本科生好歹也是花了四年培養的技能，不太可能輕易的被一個外行人隨便超越。因此我這裡所謂的「學習其他技能」，絕對不是那種「假日來去上個才藝班吧」那種等級的學習。我覺得如果不希望在鄰近畢業之際遇見求職窘境（這是自我告白嗚嗚），那在就學時也應該把自己真正當成那個領域的本科生，和他們用相同的標準投入學習和付出。
哲學到底是被高估？還是被低估？
回到開頭我提到的那個論述：
哲學在大眾的眼裡通常是兩種情況 — — 要麼被過度低估，要麼就是被過度高估。一派人覺得念哲學沒有用處，讀哲學會餓死；另一派人覺得哲學很有用，可以訓練邏輯思考能力。
我說，我同時同意這兩個觀點——這表面上看起來很矛盾，但其實我這對兩句話當中「哲學」解讀是不一樣的。
對於「哲學沒有用處」，這裡的「哲學」我的解讀是「哲學知識」（這邊不討論走學術或是在一些特別離奇的情境）。具體的哲學知識確實沒有用處，與其知道笛卡爾、柏拉圖的在說什麼，不如去研究哪種食物的卡路里比較低。
對於「哲學很有用處」，這裡的「哲學」我的解讀是「哲學思考」。雖然哲學知識本身沒有用處，但正是由於知識本身沒用，因此思考過程中對於推理的訓練就很有幫助的，它於我而言像是一種「腦力訓練」，可以幫助我在其他問題上用更有趣或是清晰的角度進行思考和決策。
一些總結
・在歷史當中，哲學是一個逐漸被瓜分的領域，哲學的定義一直隨著情境和歷史不同而變動。因此古代哲學、大學哲學跟一般人口中的哲學，指涉的對象可能完全不同。
・「哲學知識」跟「哲學背後的思考模式」是兩個不同的概念。比起學習、背誦具體的哲學知識，最重要的其實是學習後者。如果搞混了這兩者的區別，可能會造成許多精力上的浪費。
・我覺得哲學要麼被過度高估，要麼就是被過度低估：
(哲學被過度高估)很多人看到哲學經典或理論就很崇拜，覺得這些知識肯定很厲害。但我認為「哲學知識」反而是哲學中最無用的部分，它不僅複雜難懂而且離生活很遠。因此如果好不容易讀完《理想國》，卻僅僅只學到當中的故事細節（說真的知道什麼叫「洞穴諭」又能怎麼樣），而沒有產生自己的思考和觀點，那其實不如不要讀。
(哲學被過度低估)雖然比起更加實用的醫學、電腦科學，哲學知識在現實生活中沒有用處，但這些知識的價值其實是體現於當人思考該問題的過程。比起其他知識領域，哲學的學習更能有效的讓學習者進行思維上的訓練，而這些被訓練出的思考能力，絕對能夠是面對其他難題的強大武器。
總之，作為一個實用主義者，我覺得如果花了心思去學了哲學卻無法因其受益（像是培養出適合自己價值觀、獨特的視角或是擁有更好的人生）。那即使知道了那些複雜艱澀的哲學知識、名詞又怎麼樣呢？還不如去研究那些有標準答案且更靠近生活的實用知識。畢竟，學習應該是為了讓自己的生活變得更好，而不是變得複雜。
最後，感謝你的閱讀，希望本篇有幫你從另ㄧ個層面更認識哲學。對於我的觀點，我敢說很多人未必會同意，但還是希望我可以提供你一些啟發。：）
＊2021.2.23更新：作者（aka 我本人XD）畢業了，想知道對於這五年的反思，可以參考《我的哲學，我的世界觀》 裡面記錄了我大學五年，思考的三件事「這是個怎樣的世界？」、「怎麼跟社會相處？」、「怎麼獲得幸福？」。
我的臉書：葉妮姍（Nissen Yeh） -------------------- ### 不時分享一些學習心得在臉書，歡迎認識交流>.< www.facebook.com
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莊凱翔
Jan 14, 2019
謝謝你的論述，這是我看過最清晰的哲學解讀。
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褚襄烈 Stephen Chu Marriage educator
Dec 8, 2021
謝謝 Nissen!簡單扼要很清楚，很喜歡!關於打賞之事，太哲學了。我願意花5美金美買這篇文章放到我的網站(因為很清楚。不過，通常1、2元也該夠了，畢竟只是一篇文章，好吧，很好的文章，但是只有書本的1/20以下內容，還可以一賣再賣啊。不加水、料仍香甜!)
但我想，在Midium提供的選擇上，以一杯咖啡的吸引(唉，不誠實啊。既以美金算咖啡，就該以美國咖啡計價啊，不是嗎？麥當勞一杯99分，好的也就3元。看看，讀了您裝後，邏輯推理進步了吧!)，卻只能選「每月」5到100美金，大吃一驚啊!網上偶遇，竟一生…more
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定風客
Apr 19, 2024
Inspiring!
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13983 | https://math.mit.edu/research/highschool/primes/materials/2014/Lou-Murin.pdf | ON THE STRONGLY REGULAR GRAPH OF PARAMETERS (99, 14, 1, 2) SUZY LOU AND MAX MURIN Abstract. In an attempt to find a strongly regular graph of parameters (99, 14, 1, 2) or to disprove its existence, we studied its possible substructure and constructions.
1. Introduction Throughout the paper, the character ∼will denote adjacency; G will denote the graph with the parameters under question, assuming it exists; and V will denote the vertex set of G.
Definition 1.1. A strongly regular graph with parameters (n, k, λ, µ) is a k-regular graph on n vertices such that a pair of vertices has λ neighbors in common if they are adjacent, and µ neighbors in common otherwise. There are many parameter sets for which it can be proven that no strongly regular graph exists, but for many other parameter sets, neither existence nor existence of a corresponding strongly regular graph has been shown. One of these parameter sets is (99, 14, 1, 2).
Though these graphs are easy to define, it is not yet well understood for which parameter sets there exist at least one corresponding graph. Because of this lack of understanding, it is desirable to find out whether or not a parameter set such as (99, 14, 1, 2) might correspond to a graph. In addition, it is unknown whether there is a Moore graph, a graph with diameter k and girth 2k+1, with 57 vertices and girth 5.
If this graph exists, it would be strongly regular and would complete the classification of Moore graphs. Though this particular problem is not related to Moore graphs, these two facts contribute to the interest in strongly regular graphs.
In studying this graph, we created several unsuccessful attempts at construction.
We also found certain properties, such as bounds for the chromatic number and the size of a maximal independent set, possible substructures, and possible orders of automorphisms. In Section 2 we will first examine the ways the strongly regular graph of parameters (9, 4, 1, 2) could potentially be a substructure. In Section 3 we will discuss attempted constructions with Fano-planes, followed by a discussion in Section 4 of maximal independent sets and a discussion of a triangle decomposition in Section 5. Section 6 will contain a discussion of possible orders of automorphisms in the graph. Section 7 will discuss the relationship of G with rotational block designs, and finally Section 8 will discuss the structures that arise from an automorphism of order 7.
2. The srg(9, 4, 1, 2) as a substructure Let H be the unique strongly regular graph of parameters (9, 4, 1, 2).
Key words and phrases. Strongly regular graphs.
The project was supported by the PRIMES-USA program of MIT.
1 2 S. LOU AND M. MURIN Theorem 2.1. If G contains H minus an edge as a subgraph, then it contains H as an induced subgraph.
Proof. Suppose, for the sake of contradiction, that there H minus an edge was an induced subgraph in G.
According a lemma of Wilbrink and Brouwer, the following equation holds for an induced subgraph of a strongly regular graph with parameters (n, k, λ, µ), such that the induced subgraph has N vertices, of degree d1, . . . , dN, and M edges: (n −N) −(kN −2M) + λM + µ N 2 −M − N X i=1 di 2 = x0 + N X j=3 j −1 2 xj, where xj denotes the number of vertices outside the subgraph adjacent to exactly j vertices in the induced subgraph.
One may verify that no vertex outside of the subgraph is adjacent to more than 3 vertices in this particular subgraph; otherwise, at least one of the parameters is violated. Therefore, applying the lemma above, x0 + x3=5.
The induced subgraph is illustrated in Fig. 1.
Figure 1. An illustration of the induced subgraph, which we shall prove does not exist in G.
For convenience, the induced subgraph on the vertices labeled Xa, Xb, Ya, Yb, Za, Zb will be referred to as the "prism." Consider the bold edges. Each of these edges must form a triangle with another vertex. Keeping the third and fourth parameters in mind, we find that the two vertices that form a triangle with these edges are not adjacent to any vertex of the triangular prism and do not coincide.
Vertices X and Z share vertex Y as a common neighbor and have one more common neighbor. Again by examining the third and fourth parameters we find that this other ON THE STRONGLY REGULAR GRAPH OF PARAMETERS (99, 14, 1, 2) 3 common neighbor is not adjacent to any vertex of the triangular prism and does not coincide with the vertices previously mentioned.
This results in the subgraph shown in Fig. 2: Figure 2. A more detailed subgraph that must exist in G if the in-duced subgraph shown in Fig. 1 exists.
Let S be the set of the 60 vertices such that they are the neighbors of a vertex in the prism, that are themselves not in the prism.
By the fourth parameter, X and Z are each adjacent to 2 vertices in S. Similarly, vertex Y is adjacent to no such vertices. We find that X and Z are hence adjacent to 9 vertices that are not in S, and for both X and Z, 2 of these 9 neighbors are already drawn in the diagram. Similarly, Y has 10 neighbors that do not belong to S, and two of them are drawn in the diagram.
Given this information, we can directly compute x0: it is 5. Thus, x3 = 0.
Figure 3. All the vertices that have 2 or more neighbors in S ∩{1, 2, 3} Consider a vertex v belonging to this set of five vertices. By the fourth parameter, exactly twelve of its neighbors belong to S. We also know that v shares two neighbors with each of X, Y , Z. How is this possible? That would seem to make 18 neighbors 4 S. LOU AND M. MURIN of v, so we must be overcounting. At least 4 neighbors of v must either be adjacent to two of X, Y , Z, or simultaneously be adjacent to one of X, Y , and Z and belong to S. (Recall it is impossible for a vertex belonging to S to be adjacent to more than one of X, Y , and Z.) That is to say, each of these five are adjacent to four of the green vertices in the subgraph shown in Fig. 3, where H −4, H −5, H −6, H −7 are the vertices that simultaneously are adjacent to one of X, Y , and Z and belong to S.
Choosing 4 vertices from 7 is the same as not choosing 3 from the 7, and we note that because of the fourth parameter, at least 2 members of the set {H −1, H − 2, H −6, H −7} must not be chosen, and similarly, at least two members of the set {H −1, H −3, H −4, H −5} must not be chosen. That means that H −1 can never be chosen. Then H −1 has no neighbors from the vertices constituting x0.
However, since H −1 does not belong to S, we can count its neighbors: 2 are shown in the previous diagram; by the fourth parameter, 8 of its neighbors belong to S. Its remaining neighbors must be adjacent to one or more of X, Y , and Z, but not to any vertices of the triangular prism. But it already shares two neighbors with Y , and for each of vertex X and vertex Z it needs only 1 more common neighbor. Then H −1 has only 12 neighbors, contradiction.
□ 3. Labelings with Fano planes Definition 3.1. A Fano plane is a set of seven 3-element subsets, called lines, of {1, . . . , 7} such that every pair of lines share exactly one element. The elements are also called points.
There are 30 distinct Fano planes, which can be grouped into two disjoint sets of 15 Fano planes, such that two Fano-planes in the same set have the following property: the two Fano planes share exactly one 3-element set, one Fano plane can be obtained from the other by cyclically permuting the three elements of this shared 3-element set. In addition, if we cyclically permute three elements of a single set of a Fano plane, the result is a Fano plane in the same set of 15 Fano planes as the initial Fano plane.
Suppose one of these disjoint sets of 15 Fano planes is {F0, F1, . . . , F14}. Suppose that F0, F2n−1, and F2n share a line for n ∈1, . . . , 7. Then we can label G as follows: Call a central vertex F0, and its 14 neighbors F0 . . . F14, such that F0, F2n−1, and F2n form a triangle.
For the other 84 vertices in the graph, suppose a vertex is the common neighbor of Fi and Fj. If Fk is the Fano-plane such that Fi, Fj, Fk all share a line, and e is the shared line, then label the vertex as (Fk, e).
One attempt at construction was to create rules for adjacency among the vertices (Fk, e). However, none of the rules that were tried worked. A few rules that were the most noteworthy were the following.
1. Consider neighbors of Fx of the form (F, l). Connect two of them if the line the Fano-plane portion of their labellings share is in F0. This rule is equivalent to the impossible construction with srg(9, 4, 1, 2).
2. Connect (Fx, lm) and (Fy, ln) if lm and ln are disjoint.
3. Consider Fx and Fy that form a triangle with F0. Connect a neighbor of Fx of form (F, l) and a neighbor of Fy of form (F, l) if the Fano-plane part of their label is the same.
ON THE STRONGLY REGULAR GRAPH OF PARAMETERS (99, 14, 1, 2) 5 4. The following is not a rule, but rather a set of conditions for a rule.
Let e be the line Fk shares with F0. In that case, Fk can be obtained from F0 by cyclically permuting e. Consider the vertex e ∩l, which we will call P. Define the root of a vertex (Fk, l) to be the point such that P occupies the position it previously occupied before the rotation.
The root has the following property: Consider any one of the four lines of Fk that do not contain P, and consider two points of that line, a and b. Then a, b, and the root do not form a line in F0. It is the only point with this property.
Consider (Fk, l); suppose its root is P and that it is connected to Fa and Fb. Then consider all the lines that contain P but are not included in F0. There are 12 such lines. For each line lm that is one of the twelve, consider the three vertices such that lm is included in their label. Choose one of them whose root is part of l and connect (Fk, l) to it.
The reason these conditions cannot be met is the condition that (Fk, l) shares exactly one vertex with Fa and with Fb: there is no way to meet this condition.
4. Independent sets and 2-block designs of parameters (22, 4, 2).
Theorem 4.1. An independent set in G of size 9 cannot be maximal.
Proof. Suppose there were such a maximal independent set, I. Let xi be the number of vertices in V \ I adjacent to exactly i vertices in I, and let yi denote the set of vertices with i neighbors in I. For i ≥6, xi = 0. This is because a vertex vj with j neighbors in I needs 2(9 −j) + j = 18 −j common neighbors with the vertices of I, and has 14 −j neighbors in V \ I. But if vj had a neighbor with 6 or more neighbors in I, then it would have to have at least 19 −j common neighbors with the vertices of I. Thus, if there were a vertex with 6 or more neighbors in I, it would have no neighbors in V \ I, which is clearly impossible.
Lemma 4.2. Related to the above observation, for all vertices vj in V \ I, if a, b, c, and d are the number of neighbors of vj with, respectively 2, 3, 4, and 5 neighbors in I, a + 2b + 3c + 4d = 4. This also means that a vertex in V \ I can never serve as a common neighbor for a vertex in y5 and a vertex with 2 or more neighbors in I.
Proof. A vertex vj with j neighbors in I needs 2(9−j)+j = 18−j common neighbors with the vertices of I. The actual number of common neighbors is 2a+3b+4c+5d+ (14 −j −a −b −c −d). Setting this equal to 18 −j yields the above.
□ We have x1 + x2 + x3 + x4 + x5 = 90 x1 + 2x2 + 3x3 + 4x4 + 5x5 = 9 ∗14 = 126 x2 + 3x3 + 6x4 + 10x5 = 2 9 2 = 72.
In addition, we have x5 ≤3. Suppose, on the contrary, we had 4 vertices A, B, C, D such that each had 5 neighbors in I. If A and B shared only one common neighbor, no vertex could have more than 4 neighbors in I without sharing 3 neighbors with A or B. Thus, they must share 2 common neighbors. Then C and D must both be adjacent to the vertex of I adjacent to neither A nor B, as well as 2 vertices adjacent to A only and 2 vertices adjacent to B only. Then C and D share 3 common neighbors, contradiction. Note that if x5 = 3, then any pair of vertices with 5 neighbors in I 6 S. LOU AND M. MURIN share two common neighbors in I. In this case, 6 vertices in I have 2 neighbors in y5, and 3 vertices in I have 1 neighbor in y5.
The only solutions to the above equations are: (78, 0, 0, 12, 0), (77, 0, 6, 4, 3), (77, 2, 0, 10, 1), (76, 4, 0, 8, 2), (77, 1, 3, 7, 2), (79, 3, 3, 5, 3), (75, 6, 0, 6, 3).
However, none of these solutions work. In the first solution, (78, 0, 0, 12, 0), consider the 12 vertices with 4 neighbors in I. For every time a vertex in I is adjacent to such a vertex, it gains a common neighbor with a different vertex in I 3 times. In total, it must gain 2 ∗8 = 16 common neighbors in I, but 16 is not divisible by 3, so this is impossible.
Now note that the number of edges from y2 ∪y3 to y4 ∪y5 is at most |y2 ∪y3|, because each vertex of y2 ∪y3 has at most one neighbor in y4 ∪y5. On the other hand, it is also at least |y4|−|y5|, because at most |y5| members of y4 have a neighbor in y5, and the rest must have a neighbor in y2 ∪y3. For the third, fourth, and fifth solutions, this causes an immediate contradiction.
In the second solution, (77, 0, 6, 4, 3), Lemma 4.2 implies that the 3 vertices in y5 each have 2 neighbors in y3, and none in y4. Thus every vertex in y4 ∪y5 must have a neighbor in y2 ∪y3, but this means there are at least 7 edges between y2 ∪y3 and y4 ∪y5, while on the other hand there are at most 6; contradiction.
In the sixth solution, (79, 3, 3, 5, 3), consider the 5 vertices in y4. As mentioned before, 6 vertices in I have 2 neighbors in y5, and 3 vertices in I have 1 neighbor in y5. Each vertex in y4 must have at least 2 neighbors in I that have 1 neighbor in y5, or else it shares at least 3 neighbors with a vertex in y5. Thus, each serves as a common neighbor either 1 or 3 times for the 3 vertices in I that have 1 neighbor in y5. These 3 vertices share a common neighbor a total of 6 times, so we see that each of the 5 vertices in y4 has exactly 2 neighbors out of these 3 vertices in I. Then none of them is adjacent to a vertex in y5. Then the only way for Lemma 4.2 to be fulfilled with respect to the vertices in y5 is for a vertex to be adjacent to 2 vertices in y3, or 1 vertex in y3 and 2 vertices in y2. Then 2 vertices in y5 share a common neighbor in y3 or y2, as well as 2 common neighbors in I, contradiction.
In the seventh solution, (75, 6, 0, 6, 3), as above, the 6 vertices in y4 must each be adjacent to exactly 2 vertices in I that have 1 neighbor in y5. Thus, no edges exist between y4 and y5. Then to fulfill the lemma, each vertex in y5 has 4 neighbors in y2. Then 2 vertices in y5 share at least 2 neighbors in y2 as well as 2 neighbors in I, contradiction.
Thus, all possibilities lead to a contradiction.
□ Theorem 4.3. The largest independent set of a strongly regular graph of parameters (99, 14, 1, 2) has size at most 22. If it has size 22 then every vertex not belonging to the independent set has exactly 4 neighbors in the independent set.
Proof. Let I be a maximal independent set with n vertices. The set S of all vertices with at least one neighbor in I has size 99 −n. The number of edges between S and I is 14n.
Suppose S = {s1, . . . , s99−n}. Let F(i) be the number of neighbors si has in I.
Therefore, P99−n i=1 F(i) = 14n. In addition, because 2 nonadjacent vertices share 2 neighbors, P99−n i=1 F(i) 2 = 2 n 2 = n2 −n. Therefore, P99−n i=1 F(i)2 = 2n2 + 12n.
By the RMS-AM inequality, qP99−n i=1 F(i)2 99−n = q 2n2+12n 99−n ≥ P99−n i=1 F(i) 99−n = 14n 99−n. After some algebra, this turns into −n2 −5n + 594 ≥0, so −27 ≤n ≤22. Thus, the ON THE STRONGLY REGULAR GRAPH OF PARAMETERS (99, 14, 1, 2) 7 maximum size of I is 22, as desired. Equality holds when F(i) is equal across all values of i; thus, for all i, F(i) = 14∗22 77 = 4.
□ The 77 vertices outside the independent set all have 4 neighbors in S.
5. Miscellaneous.
Assuming existence, G has a unique triangle decomposition. Consider a triangle and all triangles adjacent to it. This is illustrated in Fig. 4.
Figure 4. An illustration of a triangle in G and all adjacent triangles.
Since there are 99∗14 2 = 693 edges in the graph, there are 693 3 = 231 disjoint triangles in the graph. We can consider a graph T on 231 vertices such that each triangle in G is a vertex of T and two vertices in T are adjacent iffthe corresponding triangles in G share a vertex. The graph T is 18-regular, and because G has diameter 2, T has diameter 3. In the above diagram, a central vertex in T is called v.
Lemma 5.1. The chromatic number of G is between 5 and 11.
Proof. The lower bound of 5 is a direct consequence of the fact that the maximum size of an independent set in G is 22.
By the second parameter, there is a perfect matching between the vertices Xi and Yi. Similarly, there is a perfect matching between the vertices Yi and Zi, and Xi and Zi. Then these edges determine disjoint cycles of total length 36, each cycle of length divisible by 3.
As a result, the induced subgraph on the 36 vertices is 3-regular.
By Brook’s Theorem, these vertices can be 3-colored. Assign them an arbitrary 3-coloring.
Let the set V ′ = V \ {X1, . . . , X12, Y1, . . . , Y12, Z1, . . . , Z12}.
Now, consider the vertices of V ′. By the fourth parameter, each one has two neighbors of the form Xi, two neighbors of the form Yi, and two neighbors of the form Zi. Thus, the induced 8 S. LOU AND M. MURIN subgraph on these vertices is 8-regular. Again by Brook’s Theorem, these vertices can be 8-colored. Assign them an arbitrary 8-coloring with colors that have not been used before.
Now consider X, Y , and Z, which have not been assigned colors. Assign them 3 distinct colors from the 8-coloring on the vertices in V ′.
Thus, the graph is 11-colorable.
□ Let us shift our attention again to T, so that a vertex refers to a triangle, and a G-vertex refers to a vertex in the traditional sense. Consider the set of vertices distance 2 from v. Now define the following: α : The number of such vertices connected to exactly one G-vertex distance 1 from v.
α −vertex : A vertex with the above property.
β : The number of such vertices connected to exactly two G-vertices distance 1 from v.
β −vertex : A vertex with the above property.
γ : The number of such vertices connected to exactly three G-vertices distance 1 from v.
γ −vertex : A vertex with the above property.
One may easily verify the following equations: α + β = 180 β + 3γ = 36 α −3γ = 144.
Also, the number of vertices distance 3 from v is 32 −γ = 20 + β 3. The value γ is an integer between 0 and 12. Note that γ cannot be 11: as we noted before, the perfect matchings between the vertices of form Xi, Yi; Yi, Zi, and Xi, Zi fall into disjoint cycles of lengths divisible by 3 and summing to 36. But if γ = 11, we have 11 3-cycles and three loose edges that do not fall into cycles; contradiction.
If γ = 12 for all vertices in T, then this reduces to the impossible labeling with srg(9, 4, 1, 2). This is apparent after relabeling the previous diagram as in Fig. 5.
Figure 5. A relabeling of the diagram in Fig. 4 that demonstrates the connection between γ = 12 and the first structure discussed.
ON THE STRONGLY REGULAR GRAPH OF PARAMETERS (99, 14, 1, 2) 9 6. Possible orders of automorphisms of G Theorem 6.1. srg(99, 14, 1, 2) has no automorphisms of p > 14, where p is prime.
Proof. An automorphism of the graph G of order p must have at least one orbit of order p, since p is prime. However, not every point of the graph can be in such an orbit, since p ∤99. Since the graph is connected, at least one point in an orbit must connect to a point P not in an orbit. However, by applying the automorphism, we can see that the P connects to every point in the orbit, so deg P = 14 ≥p > 14.
This is a contradiction, so no automorphisms of order p exist.
□ Theorem 6.2. srg(99, 14, 1, 2) has no automorphisms of order 13.
Proof. Assume such an automorphism π exists. Then, as before, we must have at least one automorphism of order 13, and points not in any orbits. By connectedness, there must exist a point P which connects to an orbit; thus, P must connect to every point in that orbit. P must also connect to exactly one more point, which cannot be in an orbit.
By the property λ = 1, P must share a common neighbor with each of these points. Thus, two points A and B in the orbit must be connected to each other. Since π has order 13, and A and B are in the same orbit, there exists a positive integer n < 13 such that πn(A) = B. Since π is an automorphism, πn is also an automorphism. Thus, B connects to πn(B). However, P and B are connected, and have two common neighbors: A and πn(B). This is a contradiction, so π cannot exist.
□ Theorem 6.3. srg(99, 14, 1, 2) has no automorphism of order 11.
Proof. Assume that some such automorphism π exists. Let n be the number of orbits of size 11 of π. Then, 1 ≤n ≤9. First, examine the case that n < 9. In this case, orbits of size 1 exist. By connectedness, there must be a point P that connects to an orbit. Each of the points in the orbit must connect to another neighbor of P. Since P has 3 neighbors outside of the orbit and 11 inside it, there must be two points in the orbit that connect. As before, contradiction. Thus, n = 9.
Let us label the orbits A1 through A9. Then, let us define a matrix M by Mij being the number of points in Ai that any point in Aj connects to. Note that Mij = Mji, so M is symmetric.
If any two points in the same orbit Ai, P and Q, are connected, then there exists a j such that P = πj(Q). Then, since π is an automorphism, πj(P) ∼πj(Q) = P.
πj(P) is also in Ai. If the same process is repeated with P and πj(P), we get back Q. Therefore, for any point in Ai, all of its neighbors in Ai can be paired, and thus Mii is even for all i.
Let us consider the eigenvalues of M. The eigenvalues of the adjacency matrix of G are 14, 3, and −4.
Any eigenvector of M corresponds to an eigenvector of G, the correspondence being to set every point in Ai to the corresponding value in the eigenvector of M. By the definition of M, the eigenvalue must also be equal: therefore, the eigenvalues of M must be in the set {14, 3, −4}. Since the adjacency matrix of G has the eigenvalue 14 with multiplicity 1, M can have this eigenvalue with multiplicity at most 1; the vector ⟨1, 1, 1, 1, 1, 1, 1, 1, 1⟩has this eigenvalue, so the multiplicity of the eigenvalue 14 is exactly 1. Next, note that the sum of the eigenvalues is equal to the trace, and that the diagonal of M must contain only positive even integers. Thus, the sum of the eigenvalues must be an even positive 10 S. LOU AND M. MURIN integer that is a sum of 14 and eight values from the set {3, −4}. The only such even positive sums are 38, 24, and 10.
Let us now consider the matrix M 2. Since Mij counts the number of ways to get from one specific point in Ai to any point in Aj by a path of length exactly 1, (M 2)ij counts the number of ways to get from any specific point in Ai to some point in Aj along a path of length exactly 2. In other words, (M 2)ij is the number of common neighbors one specific point of Ai has with all of the points of Aj. First, consider the case that i = j. Any point has degree 14, so it has 14 common neighbors with itself. Then, any point of the orbit Ai is connected to Mii points on the same orbit by the definition of M. For each one it is connected to, it has 1 common neighbor; otherwise, it has 2. Thus, (M 2)ii = 14 + Mii + 2(10 −Mii) = 34 −Mii. If i ̸= j, then (M 2)ij = Mij + 2(11 −Mij) = 22 −Mij, similarly. Therefore, (M 2 + M)ij = 34 if i = j and 22 otherwise.
By definition, (M 2)ii = 9 X j=1 M 2 ij = 34 −Mii.
This implies that every value in M must be at most 5.
Since the degree of any vertex is 14, the sum of the values of any row must be equal to 14. By trying every possibility, it can be shown that there are only seven possible rows of M that fulfill these two equations, up to permutation: 0, 4, 3, 2, 1, 1, 1, 1, 1 0, 4, 2, 2, 2, 2, 1, 1, 0 0, 3, 3, 3, 2, 1, 1, 1, 0 0, 3, 3, 2, 2, 2, 2, 0, 0 2, 4, 2, 2, 1, 1, 1, 1, 0 2, 3, 3, 2, 2, 1, 1, 0, 0 4, 2, 2, 1, 1, 1, 1, 1, 1 The first value in each listing must lie on the diagonal.
Note that the value 5 occurs in no row. Therefore, it is never possible to have a value of 5 in M. Thus, the sum of the eigenvalues can never be 38, so it must be either 24 or 10.
Since we know every possible row, we can now try every possibility. No such matrix exists. Therefore, there is no automorphism of order 11.
□ 7. Block designs of the parameters (22, 4, 2).
Definition 7.1. A block design of parameters (22, 4, 2) is comprised of a set S of 22 values (for convenience let them be the integers from 0 to 21), called treatments, and a set B of 77 4-subsets of S, called blocks, such that every value k in S is in exactly 14 members of B; and every pair of distinct values in S is in exactly 2 members of B.
Let G be some srg(99, 14, 1, 2) that has an independent set S of size 22. Then, let G\S be B. Let the graph G′ be G with every edge between two points of B removed; G′ is bipartite with parts S and B. As noted earlier, every vertex in B must have degree 4 in G′. Let B′ = {{k ∈S | k ∼b} | b ∈B}. Since the srg parameter µ is 2, every pair of members of S must have two common neighbors. Thus, (S, B′) is a ON THE STRONGLY REGULAR GRAPH OF PARAMETERS (99, 14, 1, 2) 11 (22, 4, 2)-block design. Note, however, that not every block design corresponds with a potential G′: some block designs have repeated blocks, or blocks that share three elements: this would lead to two points in B having four or three common neighbors, respectively.
Every possible graph that has an independent set of size 22 can thus be associated with a block design. One notable family of block designs is the family of cyclic, or 1-rotational, block designs, which have the property that if any block {a1, a2, a3, a4} is in the design, then the block {a1 + 1, a2 + 1, a3 + 1, a4 + 1} is also in the design (addition modulo 22). Some members of this family have repeated blocks or blocks that share three members; these do not form viable graphs.
Every block in a cyclic block design is part of a family of blocks produced by adding one to each element. For a block b, let f(b) be the block generated by adding 1 to every element of b modulo 22. Clearly, f 22(b) = b. Thus, the order of b under f must divide 22. If the order was 1, then for some k in b, k + 1, k + 2, k + 3, and k + 4 would also have to be in b, contradiction. A similar contradiction arises for an order of 2. Thus, the orders of all blocks must be either 11 or 22. If the order is 11, then the block must be equal to {k1, k1 + 11, k2, k2 + 11}; if it is not of this form, it must have order 22.
Another family of block designs is the family of 2-rotational block designs, with the property that if any block {a1, a2, a3, a4} is in the design, then the block {a1 + 2, a2 + 2, a3 + 2, a4 + 2} is also in the design. Thus, the family of 2-rotational block designs is a superset of the family of cyclic block designs. Defining the function g(b) as f(f(b)), with f as before, then every block in such a design must have g11(b) = 1.
As before, no block can have an order of 1, so every block in such a block design has order 11.
An attractive potential construction of G from a 2-rotational block design is to require that if two points in G′ labeled by blocks b1 and b2 are connected, then so are g(b1) and g(b2). This, however, would mean that g would be an automorphism of G with order 11.
As shown above, this is impossible, so the most attractive construction does not work. It might still be possible to create an srg(99, 14, 1, 2) from a 2-rotational block design through some method.
8. Automorphisms of order 7.
Assume that some automorphism π of G of order 7 existed. Then, since 7 does not divide 99, there must be at least one orbit of size 1. Since the graph is connected, at least one orbit of size 1 must connect to an orbit of size 7. Let us call the point in this orbit P, and let us call the orbit of size 7 to which P connects A. Since P is connected to a point in A, it must have exactly one common neighbor with that point. No element of A can connect to any other orbits of size 1, because then P would have seven common neighbors with that orbit. Thus, the common neighbor must be in an orbit of size 7; call that orbit B. It is connected to A and P.
Therefore, P connects to two orbits of size 7, A and B; P has degree 14, so it does not connect to any other orbits of size 1. Thus, any other orbits of size 1 must be connected to two orbits of size 7 similarly. Thus, if there are k orbits of size 1, there must be at least 2k orbits of size 7. This requires 15k points, so k ≤99/15 < 7. The number of orbits of size 7 is (99 −k)/7, which must be an integer, so k must be 1 modulo 7. Therefore, k must equal 1, and thus P is the only orbit of size 1.
12 S. LOU AND M. MURIN Let us label the members of A and B as A1 through A7 and B1 through B7 respec-tively, so that π(Ai) = Ai+1. We know that every point in A must connect to a point in B, so WLOG let Ai connect to Bi. No Ai can also connect to any Bj for i ̸= j, since that would cause Ai and P to have two common neighbors, even though they are connected to each other. Similarly, Ai is not connected to Aj for any j.
Since Ai and Bj are not connected for i ̸= j, they must share two common neigh-bors, one of which must be P.
Let this common neighbor be called Qi,j.
Note that Ai+1 and Bj+1 have the common neighbor Qi+1,j+1, but also Ai+1 = π(Ai) and Bj+1 = π(Bj), so Qi+1,j+1 = π(Qi,j).
Let us define Cα β as Q−α+β,α+β for α ∈{1, 2, 3}. Then, Cα β+k = Q−α+β+k,α+β+k = πk(Q−α+β,α+β) = πk(Cα β ). Also, Ai ∼Cα i+α and Cα i ∼Bi+α. Define Dα β as Qα+β,−α+β for α ∈{1, 2, 3}. As before, Dα β+k = πk(Dα β). Also, Bi ∼Cα i+α and Cα i ∼Ai+α.
Now, note that any Ai and Aj for unequal i and j must share a common neighbor; let this be Ri,j = Rj,i. Let us similarly define the neighbor of Bi and Bj as R′ i,j. As before, Ri+k,j+k = πk(Ri,j). Then, we can define Eα β as R−α+β,α+β for α ∈{1, 2, 3}.
Therefore, as before, Eα ( β + k) = πk(Eα β ). Similarly define F α β as R′ −α+β,α+β. Once again, F α ( β + k) = πk(F α β ).
Let us now define a 15 by 15 matrix M. The columns and rows correspond to P, A, B, C1, C2, C3, D1, D2, D3, E1, E2, E3, F 1, F 2, and F 3 in that order, and Mij is defined as the number of times any point in the ith orbit is adjacent to some point in the jth orbit.
Lemma 8.1. Either Mii = 0 or Mii = 2.
Proof. If there is an edge within Ci, then clearly there is a 7-cycle within Ci. If there are 2 7-cycles, then it is easy to verify that the parameters of the graph are violated.
□ Further note that no point in P, A, or B is adjacent to any other point in its own orbit, so M11 = M22 = M33 = 0. Thus, the trace is at most 12 × 2 = 24, and as before, must be even. The eigenvalues of M must be a subset of the eigenvalues of G, as shown in the proof of Theorem 6.3, so M has one eigenvalue equal to 14, the rest being either 3 or −4. If a of the eigenvalues are −4, then the trace is equal to 14 + 42 −7a, which must be divisible by 7. Thus, the trace is either 0 or 14.
9. Acknowledgements We are grateful to MIT PRIMES-USA for providing us with this project, and to Dr. Peter Csikvari for suggesting the problem and mentoring us.
References H. A. Wilbrink and A. E. Brouwer, A (57, 14, 1) strongly regular graph does not exist, Indaga-tiones Mathematicae (Proceedings) 86 (1983), 117–21. |
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In alcoholic fermentation, pyruvate changes to alcohol and carbon dioxide. NAD+ also forms from NADH, allowing glycolysis to continue making ATP. This type of fermentation is carried out by yeasts and some bacteria. It is used to make bread, wine, and biofuels.
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Glycolysis Inputs
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Alcoholic Fermentation
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Fermentation
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Citric Acid Cycle output
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Terms in this set (43)
Glycolysis Inputs
glucose, 2 ATP, 2 NAD+, 4 ADP + P
GLYCOLYSIS OUTPUTS
2 pyruvates, 4 ATP, 2 NADH + H+, 2 H2O
GLYCOLYSIS location
cytoplasm
Citric Acid Cycle input
2 acetyl CoA, 2 oxaloacetate, 2 ADP + P, 6 NAD+, 2 FAD
Citric Acid Cycle output
4 CO2, 2 ATP, 6 NADH + H+, 2 FADH2
Citric acid cycle location
mitochondrial matrix
ETC INPUT
10 NADH + H+, 2 FADH2
ETC OUTPUT
10 NAD+, 2 FAD, H2O, 34 ATP
FERMENTATION INPUT
glucose, 2 ATP
FERMENTATION OUTPUT
2 lactate or 2 alcohol, 2 CO2, 4 ATP
Anabolic
A process in which large molecules are built from small molecules
Catabolic
A process in which large molecules are broken down for energy
Fermentation
Process by which cells release energy in the absence of oxygen
Endergonic
absorbs energy
Exergonic
releases energy
What is a coupled reaction?
a pair of reactions, one exergonic and one endergonic, that are linked together such that the energy produced by the exergonic reaction provides the energy needed to drive the endergonic reaction example is the formation of ATP
Why is NAD important?
NAD is a molecule that promotes cellular function for the creation of energy, and plays a key role in mitochondrial function. In cellular respiration, NAD+ transfers electrons from one molecule to another. When NAD+ accepts an electron, it becomes reduced, and becomes NADH. NADH carries electrons all the way to the ETC, where it will then donate the hydrogen ions.
What does reduced mean?
Reducing refers to any process in which electrons are added to an atom or ion, as by removing oxygen or adding hydrogen.
What does oxidized mean?
Oxidized when electrons are chemically combined with oxygen
What is oxidative Phosphorylation?
The metabolic pathway in which cells use enzymes to oxidize nutrients, thereby releasing the chemical energy stored within in order to produce ATP, occurs in the mitochondria, needs to happen in order to conduct cellular respiration.
Inputs/Outputs of Glycolysis
Input to glycolysis: Glucose + 2 ATP, 2 NAD+, 4 ADP + Pi
Output of Glycolysis: 2 pyruvate, 4 ATP, NADPH, 2 H2O
Inputs/Outputs of Pyruvate
Input to Pyruvate oxidation: 2 Pyruvate, 2 CoA, 2 NAD+
Output of Pyruvate oxidation: 2 Acetyl CoA, 2 CO2, 2 NADH
Input/Outputs of the Krebs Cycle
Input to Citric Acid Cycle( Krebs Cycle): 2 Acetyl CoA, 2 Oxaloacetate, 2 ADP + Pi, 6 NAD+, 2 FAD
Output of Citric Acid Cycle: 2 CoA, 4 CO2, 2 ATP, 6 NADH, 2 FADH2
Input/Output of ETC
Input to Electron Transport chain/ Chemiosmosis: 10 NADH, 2 FADH2, 6 O2, 34 ADP + Pi,
Output of Electron Transport Chain/Chemiosmosis: 10 NAD+, 2 FAD, H2O, 34 ATP
Input/Output of Fermentation
Input to Fermentation: 2 pyruvate, 2 NADH
Output of Fermentation: 2 ATP, 2 NAD+, 2 ethanol and 2 CO2 or Lactate
Review glycolysis-Where does each process occur and what are the energy carriers?
Glycolysis - occurs in cytoplasm; splits glucose into two molecules of the three-carbon molecule pyruvate; transfers energy to produce two molecules of ATP and loads high energy electrons into electron carriers (NADH).
Review Krebs Cycle- Where does each process occur and what are the energy carriers?
he Krebs Cycle harnesses the energy which remains in pyruvate after glycolysis. The Krebs Cycle removes energy from citric acid in small steps, storing it in diverse energy carrier molecules: ATP, NADH and FADH2. The Krebs Cycle produces two molecules of CO2 per Acetyl-CoA, completing the breakdown of glucose. The Krebs cycle takes place inside the mitochondria-- the mitochondrial matrix. The Krebs cycle produces the CO2 that you breathe out. This stage produces most of the energy ( 34 ATP molecules, compared to only 2 ATP for glycolysis and 2 ATP for Krebs cycle).
Review ETC-Where does each process occur and what are the energy carriers?
A series of proteins in which the high-energy electrons from the Krebs cycle are used to convert ADP into ATP. The electron transport chain takes place in the inner membranes of the mitochondrion.
True or False: In the first step of the Krebs Cycle, acetyl CoA is used to drive the reaction to citric acid.
True
How is the proton gradient created?
At the inner mitochondrial membrane, a high energy electron is passed along an ETC. The energy released pumps hydrogen out of the matrix space. The gradient created by this drives hydrogen back through the membrane, through ATP synthase.
Why is ATP synthase important, where is it located?
ATP synthase is important to power all cellular processes as it produces ATP, which captures energy and is used as fuel for everything in the cell. During electron transport, participating protein complexes push protons from the matrix out to the intermembrane space. This creates a concentration gradient of protons where ATP synthase then can produce ATP out of hydrogen ions. ATP synthase is located in the inner mitochondrial membrane.
What does 2, 2 and 32 refer to?
The net production of ATP: 2 in glycolysis, 2 in Krebs, and 32 in ETC
What is fermentation, where does it occur and why does it occur?
Fermentation is a respiration pathway where glucose molecules are broken down without the use of oxygen. It occurs in the cytoplasm and because there is not enough oxygen to break down organic molecules so it produces a lot less ATP (2) and ethanol.
What is the role of oxygen?
Oxygen is the final electron acceptor of the electron transport chain in the final step of cellular respiration. Oxygen combines with electrons and hydrogen ions to produce water.
What is substrate-level phosphorylation?
Substrate-level phosphorylation: is a metabolism reaction that results in the production of ATP by the transfer of a phosphate group from a substrate directly to ADP. Transferring from a higher energy (whether phosphate group attached or not) into a lower energy product-- occurs both in glycolysis and the citric acid cycle
What is Electron transport?
Electron transport is the final stage of aerobic respiration. In this stage, energy from NADH and FADH2 is transferred to ATP. During electron transport, energy is used to pump hydrogen ions across the mitochondrial inner membrane, from the matrix into the intermembrane space.
What is Chemiosmosis?
Chemiosmosis is the movement of ions across a semipermeable membrane bound structure, down their electrochemical gradient. An example of this would be the formation of adenosine triphosphate (ATP) by the movement of hydrogen ions (H+) across a membrane during cellular respiration or photosynthesis.
What is the role of oxygen in cellular respiration?
Oxygen is the final electron acceptor of the ETC
True or False: For glycolysis 2ATP in 4 ATP made for a net of 2 ATP.
TRUE
What is the net gain for aerobic cellular respiration?
36 ATP
Lactic Acid Fermentation
In lactic acid fermentation, pyruvate (also known as pyruvic acid) from glycolysis changes to lactic acid. In the process, NAD+ forms from NADH. NAD+, in turn, lets glycolysis continue. This results in additional molecules of ATP. This type of fermentation is carried out by the bacteria in yogurt. It is also used by your own muscle cells when you work them hard and fast.
Alcoholic Fermentation
In alcoholic fermentation, pyruvate changes to alcohol and carbon dioxide. NAD+ also forms from NADH, allowing glycolysis to continue making ATP. This type of fermentation is carried out by yeasts and some bacteria. It is used to make bread, wine, and biofuels.
Know the difference between oxidative phosphorylation and substrate level phosphorylation.
The main difference between substrate level phosphorylation and oxidative phosphorylation is that substrate level phosphorylation is a direct phosphorylation of ADP with a phosphate group by using the energy obtained from a coupled reaction whereas oxidative phosphorylation is the production of ATP from the oxidized NADH and FADH .
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13986 | https://unepdhi.org/wp-content/uploads/sites/2/2022/03/Freshwater_Strategic_Priorities.pdf | Freshwater Strategic Priorities 2022–2025 to implement UNEP’s Medium-Term Strategy Produced by the Inter-Divisional Water Group, a water community of practice across UNEP, for implementation of UNEP’s Medium-Term Strategy.
March 2022 summary.
The protection, management and restoration of freshwater ecosystems is fundamental to combating the triple planetary crises of biodiversity loss, pollution and climate change. As water is under threat, despite it being central to life, there is a compelling need for freshwater action. To meet the water-related Sustainable Development Goals (SDGs), UNEP seeks to facilitate measurable and substantive progress on freshwater issues at global, regional and national levels. UNEP supports countries to promote the management, protection and restoration of the world’s freshwater ecosystems, while increasing resilience to natural disasters and conflict.
UNEP collects and transforms data into actionable information and decision support tools. It helps countries understand and determine the state of freshwater bodies, supports the implementation of plans and policies around water and develops and disseminates decision support tools that promote progress. 01 02 Why does the United Nations Enviroment Programme (UNEP) have a freshwater strategic document?
Water and water-related ecosystems play a fundamental role in the health of the environment, providing services to people and communities and combating the impacts of climate change and all economic activities. Working on the environmental aspects of monitoring, managing and protecting water resources and freshwater ecosystems has been an integral part of UNEP’s mandate since its inception and has been embodied in various freshwater strategies in the past (UNEP 2017).
Embedded in its Medium-Term Strategy, UNEP recognizes three interlinked planetary crises for urgent understanding, prioritization and action: • Climate change • Nature and biodiversity loss • Pollution and waste The impacts of all three of these inter-related planetary crises are directly, and in some cases disproportionately, felt on freshwater bodies, which are essential for the lives, livelihoods and health of people, economies and the planet. At the same time, when functional, including protected, restored and well managed, freshwater bodies are strong allies in combating all three crises. Freshwater bodies can help protect and restore biodiversity, mitigate pollution through water filtration and purification and contribute to climate stability by providing both mitigation and adaptation benefits.
03 How does this freshwater strategic document support UNEP’s Medium-Term Strategy 2022-2025? UNEP’s Medium-Term Strategy for 2022-25 (MTS) (UNEP 2021), adopted by the UN Environment Assembly and serving as the guidance for UNEP’s work, recognizes the three planetary crises and thus principal areas of action to address them (thematic sub-programmes). These three thematic sub-programmes are underpinned by two foundational sub-programmes and facilitated by two enabling sub-programmes (Figure 1). Multilateral Environmental Agreements Thematic subprogrammes Climate Action Nature Action Chemicals and Pollution Action Enabling subprogrammes Foundational subprogrammes Strategic objectives Digital transformations Climate stability Living in harmony with nature Pollution-free planet Finance and economic transformations Environmental governance Science-policy Climate change Biodiversity loss Pollution A planetary and human crisis caused by unsustainable patterns of consumption and production What the science says Towards the Sustainable Development Goals For people, prosperity and equity Figure 1 The seven sub-programmes of UNEP’s Medium-Term Strategy 2022-25 04 For an external audience, the aim is to: • create awareness among partners and stakeholders of the state of freshwater ecosystems and the importance of protecting and restoring them.
• provide information on the work of UNEP related to freshwater. • inspire and promote collaboration with other organizations for the benefit of freshwater, other ecosystems, people and economies. Within UNEP, the document is intended to provide a common framework for the various groups and units that each contribute to UNEP’s freshwater-related work.
The compelling need for freshwater action Despite some progress in certain areas, the world is not on track to meet the water-related SDGs and their targets (UN-Water 2021) nor is it on track to deliver long-term sustainability by 2050. Urgent action and strengthened international cooperation are needed (UNEP 2019) to reverse these negative trends and restore planetary and human health. For the three thematic sub-programmes of the MTS, the following drivers and trends illustrate the urgent need for freshwater action.
Freshwater and climate action Climate change significantly impacts water, and healthy freshwater ecosystems are needed to help adapt to climate change effects. This relationship is fundamental to sustainable development because: • The impacts of climate change are experienced by most societies through seasonal and inter-annual changes in rainfall, and snowfall patterns, as well as shortage or excess of water felt through droughts and floods. Pollution through sediment and storm run-off, forest fires and saltwater intrusion into groundwater are further impacts. • Over the last 20 years, 90 per cent of major disasters were caused by floods, droughts and other water-related events. With more frequent droughts, people in water-scarce areas will increasingly depend on groundwater because of its buffer capacity and resilience to climate variability.
• Climate change challenges the sustainability of water resources, which are already under severe pressure in many regions of the world. More than two billion people currently live under water stress, a factor likely to increase due to climate change.
05 • The increased frequency and severity of extreme climate events also impacts water quality and management. During periods of drought, less dilution and flushing-out of water pollution take place; and during floods nutrients, surface-bound contaminants and solid waste are flushed into rivers and lakes. Overflowing of sewage systems, which can contaminate and compromise the quality of otherwise suitable water sources, is also common. Rising average temperatures of surface waters affects water chemistry and may trigger more frequent incidents of harmful algal and bacterial blooms, as impressively illustrated in recent cyanobacteria blooms.
• Changes in access to water and rain-fed grazing lands are often a factor behind community and regional conflicts.
• Healthy freshwater ecosystems and areas adjacent to them, including lakes, wetlands, rivers and their floodplains, can serve as natural sponges and buffers to flooding and storm surges. Mangroves, which exist at the confluence of marine and freshwater ecosystems, are essential defences against extreme weather events and a cradle of biodiversity.
• Wetlands, especially peatlands, act as massive carbon sinks and thus exacerbate climate change if not properly managed. When managed, protected and restored to functionality, wetlands help to store carbon.
• Sustainable water and water ecosystem management are thus an essential part of the solution to both mitigating and adapting to climate change. But beyond that, they are also an essential element to meet other SDGs essential to climate action, including SDGs 1 (on poverty eradication) and 2 (on food security) and an underlying factor in creating stable and prosperous societies.
06 Freshwater and nature action Maintaining a healthy relationship between water, nature and people is more important than ever because: • Healthy resilient ecosystems, such as wetlands and forests, and access to abundant and clean water go hand in hand. The value provided by ecosystems has demonstrated that their benefits far exceed conservation costs. The 2011 economic value of ecosystem services has been globally estimated at $125 trillion. In the same year, global GDP was estimated at US$75 trillion.
• Freshwater ecosystems are particularly biodiverse, supporting 6 per cent of all known species and 55 per cent of fish species depending on them for their survival. Yet they are also going extinct more rapidly than terrestrial or marine species, with around one third of all freshwater biodiversity facing extinction due to invasive species, pollution, habitat loss and over-harvesting.
• Joined up management of surface and groundwater resources is essential. Groundwater makes an important contribution to river flow and is heavily used for irrigation; groundwater-dependent ecosystems supply almost half of all drinking water. In arid and semi-arid regions, groundwater is often the only reliable water resource and plays an integral role in supporting a healthy environment.
• The relationship between healthy nature and reliable water has always been central to development, but many communities around the world are highly vulnerable. Regrettably, the COVID-19 pandemic is affecting the fragile social and economic progress of many communities around the globe and exacerbating existing vulnerabilities due to declining water quality and reduced water supplies.
• Fiscal stimulus packages, as invoked by many governments in the wake of COVID-19, offer opportunities to restore nature and reduce threats to both water quality and availability. It is not an opportunity UNEP can let slip by.
• Water supply and safe sanitation are fundamental to human health and hygiene, but UNEP needs to reach beyond that. Deficiencies in a country’s water resources and wastewater management systems adversely affect ecosystem diversity and the ability to protect populations against the negative effects of pandemics/epidemics. Inclusive decision-making 07 is needed to improve access to and management of water resources and address water pollution, wastewater treatment, water use and services, water resources management, freshwater ecosystem health and enhance coordination between sectors and stakeholders at all levels. Freshwater and chemicals and pollution action Water pollution is intricately linked to human and ecosystem health, with effects on food security, because: • Around one-third of all rivers in Latin America, Africa and Asia suffer from severe pathogenic pollution; severe organic pollution is found in around one-seventh of all rivers; and severe and moderate salinity pollution in around one-tenth of all rivers. Together, such widespread pollution risks the health of people, the freshwater fishing industry (threatening food security and livelihoods) and the use of river water for irrigation, industry and other purposes. For several years, the overall ecological status in European rivers has not improved (European Environment Agency 2018) despite the EU Water Framework Directive.
• Solid waste, including macro and micro plastic pollution, is a major concern in freshwater bodies. Lakes and rivers are recipients and conduits for large amounts of plastic pollution entering coastal and marine habitats. UNEP recently launched a first guidance on freshwater monitoring for plastics and response strategies, which are currently being piloted in selected freshwater bodies (including lakes and rivers) around the world. • Many countries encounter massive pollution of freshwater systems from industrial effluent that includes heavy metals and other contaminants, rendering freshwater bodies unfit for their intended use.
• Like surface water, groundwater is threatened by both anthropogenic and natural contaminants, such as saline intrusion, nutrients, pesticides and other chemical substances. But despite the importance of groundwater for people and ecosystems, information and data on groundwater quality are sparse, with often less information available in countries of the Global South. Groundwater is an invisible resource that remains out of sight and out of mind for most people (World Water Quality Alliance 2021).
08 • Anthropogenic nutrient sources contribute more than 70 per cent to river nutrient loading (UNEP 2020) curbing global nutrient cycles requires paradigm shifts in food and waste systems. Most of the increase in river nutrient loading has been in Asia, but eutrophication and harmful algal blooms are now spreading in many river basins, including in Europe. Climate change is another key driver contributing to the risk.
• Waterborne diseases due to viral or bacterial contamination are already a significant cause of death and disability worldwide, exacerbated by new and emerging pollutants. Emerging pollutants are not easily removed by current wastewater treatment technologies and are of increasing concern. They include certain veterinary and human pharmaceuticals, pesticides, antimicrobial agents, flame retardants, detergent metabolites, microplastics and microfibres. Endocrine-disrupting chemicals are of particular concern as they are now widely distributed through the freshwater system on all continents. Their long-term impacts on human health include foetal underdevelopment, child neurodevelopment and male infertility.
• Human illnesses and deaths due to antibiotic- and antimicrobial-resistant infections are increasing rapidly and are projected to become a main cause of deaths worldwide by 2050. Antibiotics reach the aquatic environment from a wide range of sources, including treated and untreated human waste, agriculture, animal husbandry and aquaculture. Antibiotic-resistant bacteria are now found in both source water and treated drinking water worldwide.
• There is a need to encourage drastic reductions in pesticide and herbicide use, as well as antimicrobial applications in production that adversely affect soil and water quality.
• On the other hand, and after a piloting and scoping effort carried out by the European Commission across Europe, viral residual and DNA signals detectable in wastewater effluent will receive substantial attention in serving as a sentinel system for early warning of pathogen risks, including COVID-19.
09 UNEP’s mandate and comparative advantage on fresh water UNEP is the leading environmental authority in the United Nations system and uses its expertise to strengthen environmental standards and practices while helping implement environmental obligations at the country, regional and global levels. UNEP’s mission is to provide leadership and encourage partnership in caring for the environment by inspiring, informing and enabling nations and peoples to improve their quality of life without compromising that of future generations. Following the Rio+20 Summit in 2012, the UN General Assembly gave UNEP a stronger mandate and role in promoting a strong science-policy interface, providing capacity-building to countries as well as access to technology and strengthening its regional presence to assist countries in the implementation of their national environmental policies.
In short, and as stated in the MTS, UNEP’s overall mandate is to coordinate global responses to environmental challenges and related emerging issues, within and outside the UN, while keeping watch over the state of the world environment and linking science to policymaking.
10 For the freshwater domain, this means: • Collect data, analyse and report on the global status of ambient water quality, water resources management and the environmental health of freshwater ecosystems.
• Provide policy advice and harmonization and support institutional reform to strengthen national and regional cooperation to improve climate resilient water resources management, to prevent water pollution and to protect and restore freshwater ecosystems. • Build capacity through training and tools, facilitate access to data and information to enhance evidence-based assessment, action planning and decision-making processes, as well as policy and strategy formulation for improved management of freshwater resources.
• Support piloting of innovative management practices, including nature-based solutions, and implementation of national and regional strategic action plans that contribute to climate, nature and pollution actions.
• Utilize UNEP’s convening power to bring together Member States, agencies and actors across all of society through partnerships, networks and other mechanisms, fostering collaboration on assessment and implementation of water-related aspects of the global environmental agenda. • Scale up impact through data by mobilizing local actors and knowledge, including digital transformation, to foster local agenda achievements based on social engagement. • Provide freshwater-related inputs to strengthen UNEP’s communications and awareness-raising campaigns.
11 In the framework of the SDGs, UNEP has been entrusted with the co-custodianship of three water indicators (SDG 6.3.2, 6.5.1 and 6.6.1)1 related to the following freshwater targets, forming an important part of UNEP’s mandate on fresh water: • SDG 6.3: By 2030, improve water quality by reducing pollution, eliminating dumping and minimizing release of hazardous chemicals and materials, halving the proportion of untreated wastewater and substantially increasing recycling and safe reuse globally • SDG 6.5: By 2030, implement integrated water resources management at all levels, including through transboundary cooperation as appropriate • SDG 6.6: By 2020, protect and restore water-related ecosystems, including mountains, forests, wetlands, rivers, aquifers and lakes.
In addition, UNEP’s freshwater activities contribute to reaching and monitoring several other water-related SDG targets, for example: • SDG 2.4 on agricultural production systems that help maintain ecosystems and strengthen capacity for adaptation to climate change • SDG 6.4 on increasing water-use efficiency across all sectors and securing sustainable withdrawals of fresh water • SDG 11.5 on reducing the impacts of water-related disasters • SDG 12.4 on sound management of chemicals and waste • SDG 13.1 on strengthening resilience and adaptive capacity to climate-related hazards and natural disasters in all countries • SDG 14.1 on reducing marine pollution • SDG 14.2 on ecosystem-based marine and coastal management, conservation and restoration • SDG 15.1 on conservation, restoration and sustainable use of terrestrial and inland freshwater ecosystems and their services.
SDG indicator 6.3.2: Proportion of bodies of water with good ambient water quality.
SDG indicator 6.5.1: Degree of implementation of integrated water resources management.
SDG indicator 6.6.1: Change in the extent of water-related ecosystems over time.
1 12 UNEP is well-positioned to fulfil its mandate on fresh water because: • has a long history and experience of monitoring the status of and changes to the environment, including in areas relating to fresh water.
• has the convening power to bring data providers, Earth Observation and modelling experts – as well as stakeholders and actors across society – to the table for policy dialogue, science-based assessments and environmental agreements. It convenes global experts and user communities of practice such as the World Water Quality Alliance and coordinates global environment monitoring and capacity development at scale, including the SDGs.
• administers or provides the secretariat for multilateral environmental agreements related to fresh water, including Convention on Biological Diversity, the Basel Convention on the Control of Transboundary Movements of Hazardous Wastes and their Disposal and the Stockholm Convention on Persistent Organic Pollutants.
• enhances its in-house capacity and expertise by partnering with a wide range of organizations and collaborating centres which increase UNEP’s capacity and depth of water-related expertise, innovation and technologies.
• is engaged, as an environmental authority, in several global actions and campaigns that relate to freshwater, including the UN Water Action Decade, the UN Global Acceleration Framework for SDG 6, the UN Decade on Ecosystem Restoration and the Clean Seas campaign.
13 Strategic priorities on fresh water Within the three thematic sub-programmes of the MTS, UNEP’s convening power and its mandate and work on fresh water underpin three tiers of work. Those tiers illustrate the different aspects of UNEP’s mandate, spanning the entire spectrum from being a global policy-setting and normative body for the environment to demonstrating best practice in implementation on the ground (Figure 2): • Data, monitoring, and assessment: Establishing a solid scientific basis for diagnosis and assessment of the freshwater environment through global, national and local data collection, modelling and remote observation, citizen science, analysis and presentation.
• Transforming data into actionable information and decision support tools: Using the scientific basis to create normative guidelines globally and nationally and to engage in countries and at scale to build partnership to co-design and disseminate tools and methodologies for addressing water-related issues in regions and countries.
• Support to action: Supporting countries in implementing technology innovation, including social process, and promoting and demonstrating nature-based solutions for water-related priority issues.
Figure 2 UNEP’s core areas of work with fresh water Multilateral Environmental Agreements Thematic Dimensions Water and Climate Action Water and Nature Action Water and Chemicals and Pollution Action Tiers of Work SDG Indicators Data, Monitoring, and Assessment Actionable Info and Decision Support Tools Support to Action SDG 6.5 SDG 6.6 SDG 6.3 Capacity-building | Awareness Raising | Policy Support 14 Some of UNEP’s programmes or initiatives are confined to a particular thematic sub-programme, and others cut across and address two or all of them. The following list outlines UNEP initiatives within the three tiers of work: Establishing Science: Data, monitoring, and assessment • Like all custodian agencies, UNEP supports national monitoring and coordinates global reporting to the UN High-Level Political Forum (HLPF) on Sustainable Development regarding progress on three freshwater-related SDG indicators: Indicator 6.3.2 on proportion of bodies of water with good ambient water quality; 6.5.1 on Integrated Water Resources Management (IWRM) implementation; and, together with the Ramsar Secretariat, 6.6.1 on changes in the extent of water-related ecosystems over time.
• Since the start of the SDG period, UNEP has been a core member of the UN-Water coordinated Integrated Monitoring Initiative for SDG 6. More than 185 countries have reported to the UN via UNEP. While processes for the above three indicators are coordinated, they have separate monitoring methodologies that each require dedicated specialist attention.
• To track, monitor and improve the health of freshwater ecosystems UNEP, together with international partners, has developed and run the Freshwater Ecosystems Explorer with high resolution Earth Observation data over the past 20 years.
• Since 1978, UNEP has run the Global Environment Monitoring System for Freshwater (GEMS/Water) Programme, mandated to support Member States in monitoring and assessing their water quality and reporting their data to UNEP’s global database on water quality. The database currently comprises over 15 Mio data on freshwater quality; GEMS also provides related capacity development programmes reflecting regional and national demand.
• Mandated by UNEA, UNEP developed a World Water Quality Assessment, in response to which the World Water Quality Alliance (WWQA) convened an open community of practice with several workstreams and currently over 150 partners. The central aim is to combine the best available data sources, including modelling and Earth Observation, and serve the normative as well as the action value chain by tailored data services.
15 Using science: Transforming data into actionable information and decision support tools UNEP builds on the above work to determine freshwater status and issues, identify and prioritize actions and develop and disseminate decision support tools that promote progress. Examples include: • Working with countries to develop action plans (or similar) that aim to address identified barriers.
• Capacity development initiatives, such as workshops, training and mass open online courses and webinars, together with consortia and platforms developed with the aim of providing a match-making service to institutions and individuals.
• Sensitization and knowledge products, such as data/information platforms, targeted publications, vital graphics and videos.
• Leading on social engagement with actors of society; so far this has resulted in the launch of Local Water Forums that address specific water issues in over 30 municipalities globally.
• Providing a comprehensive overview of all water-related tools, platforms and access to information and data on UNEP’s World Environment Situation Room.
16 Supporting countries: Support to action Support to action on climate, ecosystems and pollution issues. This is typically achieved through a combination of larger initiatives/programmes, as well as by working with countries and other partners using funding from a variety of sources. Climate, nature and chemical/pollution issues are often collectively combined and addressed in individual projects.
Examples of significant initiatives: • Support to water and climate action – Climate Technology Centre & Network; collaborating with the Adaptation Unit on elaborating and implementing Global Environment Facility (GEF)/Green Climate Fund (GCF)-funded projects; support to the GEF portfolio, climate adaptation support in countries around the world; support to Adaptation Fund special programmes; IWRM support programme; Water and Climate Coalition; support to innovation efforts in the form of e.g. digital twinning of the hydrological cycle in a changing climate aimed at high-resolution short-term forecasting services with the European Commission and science partners • Support to water and nature action – Water Action Decade and the Decade on Ecosystem Restoration • Support to water and chemical pollution action – Global Partnership on Marine Litter; Global Wastewater Initiative, World Water Quality Alliance and the Clean Seas campaign • Support to a holistic source-to-sea continuum approach to address both ecosystems and pollution flow, recognizing the linkages between land, fresh water, coast and sea – Global Programme of Action for the Protection of the Marine Environment from Land-based Activities; Action Platform for Source-to-Sea Management (S2S Platform); Ocean Action 2030 Coalition of the High-Level Ocean Panel • Implementing agency for the Global Environment Facility and Global Climate Fund, piloting nature-based solutions for water-related climate adaptation. More information about UNEP’s projects on nature-based solutions for water-related climate adaptation can be found here.
Many more examples and details of UNEP’s activities in the field of water can be found online at UNEP’s Water webpage.
17 Strategic partnerships UNEP executes most of its activities and programmes through partnerships with other organizations and will continue fostering partnerships and integrated solutions. Within the UN system, UNEP works through – and supports – UN-Water, the inter-agency coordination mechanism which counts more than 30 UN organizations, Funds and programmes as its members. UNEP collaborates with other UN agencies on environmentally related freshwater topics of priority concern.
Over the years UNEP has built on its expertise and experience to develop a well-respected, integrated approach to global environmental management. UNEP treats environmental issues as multi-dimensional, requiring a range of perspectives and expertise, and has taken on the challenge of ensuring that global environmental issues are addressed through this integrated approach.
The world’s environmental challenges, and specifically in fresh water, can only be addressed through productive partnerships that manage, mitigate and ultimately leverage a wide range of perspectives. UNEP’s approach to partnering – and which follows an all-of-society scope, including youth – ensures that governments have access to the right knowledge and expertise to solve their environmental challenges. This partnering strategy will drive the integrated approach externally, through collaboration with external agencies, organizations and communities, governments, civil society and increased engagement with the private sector.
The private sector ranges from multi-nationals, small and medium enterprises, to small-scale local operators and the informal sector. Collectively the private sector is a major water user, consumer and polluter worldwide. At the same time, in most countries, the private sector is the major employer, educator, innovator and income generator, both for governments in the form of corporate taxes as well as for individuals. Partnering with the private sector is essential to address many of the world’s freshwater- related problems, to lift people and countries out of poverty, and to achieve sustainable development.
18 In creating and strengthening partnerships, UNEP will: • Build on existing partnerships, such as those within the UN system which serve as Members of UN-Water – and with Non-Government organizations (NGOs), private sector and academic organizations which collaborate as Partners of UN-Water – and seek to develop new ones in the United Nations system, and among other global, regional or national organizations, research/ academia, NGOs, civil society and the private sector. Certain initiatives, referred to earlier, are already using partnerships and demonstrating the added value to deliver on UNEP’s mandate. This should be further encouraged.
• Implement strategic projects that empower governments, regional bodies and the private sector to successfully manage, conserve and protect freshwater resources. UNEP has a track record of delivering strategic projects at the policy and implementation level. Strategic and demonstrative projects are differentiated from pilot projects in that they tangibly show what to do, how to do it and then can be replicated at various scales.
• Draw on its experience and position as a global convenor of numerous conventions, networks and assessments to support country implementation of global goals on the environmental dimensions of fresh water, including providing support for monitoring, analysis and reporting at the global level on the UNEP-led water-related targets of the SDGs.
• Continue and strengthen collaboration with dedicated partner organizations and collaboration centres whose capacity, expertise and resources on fresh water can complement those of UNEP.
• Use the freshwater agenda to build trust and promote peace and cooperation among users, especially in conflict situations.
• Consider several forms of engagement and partnerships with the private sector, taking advantage of the multi-faceted roles that private sector actors can play (financing partners, knowledge and technology providers, dissemination agents, advocacy promoters, etc.), and acknowledging that it may require amendments to legal and institutional arrangements to ensure stable and fair conditions for market players.
19 Cross-cutting aspects Gender in water management In many parts of the world, women and girls typically suffer disproportionately from poor water availability and management of water, yet they are often essential for good water management (Trivedi 2018). UNEP ensures all its projects reflect the different needs of women and men and seeks to promote a strong gender perspective in environmental policies at the national, regional and international levels. UNEP is also focusing gender mainstreaming on internal policies and processes, and gender disaggregation in its collection of data and information.
Water and conflict Environmental degradation and disasters can be a factor leading to displacement, forced migration, local or regional conflicts, and with 90 per cent of natural disasters being water-related, there is scope for better water management as a way of preventing conflicts. UNEP contributes to a wide range of environmental issues related to disasters and conflicts, focusing on the impacts of natural disasters on water quality, urban areas and damage to natural and man-made infrastructure, including freshwater habitats and ecosystem services. Providing water security for people may go hand in hand with nature-based solutions, e.g. through recharge of aquifers.
Funding sources underpinning UNEP’s freshwater agenda UNEP receives core funds from Member States through its Environment Fund, project resources and in-kind contributions from bilateral donors, global funds and through partnerships such as those with UN-Water, centres of excellence, communities of practice and private sector collaborations. But scaling up ambitions to help countries accelerate their efforts to meet the water-related SDG targets requires additional funding. Additional cash and in-kind contributions and developing new resource mobilization strategies and business models, notably for the powerful partnerships, would help further UNEP’s freshwater agenda.
20 • European Environment Agency (2018). • Trivedi, A (2018). Women Are the Secret Weapon for Better Water Management. World Resources Institute. • UN-Water (2021). Summary Progress Update 2021 – SDG 6 – water and sanitation for all. Geneva, Switzerland. • United Nations Environment Programme (UNEP) (2017). UN Environment’s Freshwater Strategy 2017-2021. • United Nations Environment Programme (UNEP) (2019). Global Environment Outlook – GEO-6: Healthy Planet, Healthy People. Nairobi. • United Nations Environment Programme (UNEP) (2020). Monitoring Plastics in Rivers and Lakes: Guidelines for the Harmonization of Methodologies. Nairobi. report/monitoring-plastics-rivers-and-lakes-guidelines-harmonization-methodologies. • United Nations Environment Programme (UNEP) (2021). For people and planet: the UNEP strategy for 2022–2025. References • World Water Quality Alliance (2021). Assessing Groundwater Quality: A Global Perspective: Importance, Methods and Potential Data Sources. A report by the Friends of Groundwater in the World Water Quality Alliance. Information Document Annex for display at the 5th Session of the United Nations Environment Assembly, Nairobi 2021. United Nations Avenue, Gigiri P O Box 30552, 00100 Nairobi, Kenya Tel +254 720 200200 communication@unep.org www.unep.org |
13987 | https://ocw.mit.edu/courses/ids-338j-multidisciplinary-system-design-optimization-spring-2010/05a3b2acfee9c6903984904149aeda21_MITESD_77S10_lec07.pdf | 1 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Multidisciplinary System Design Optimization (MSDO) Numerical Optimization I Lecture 7 Karen Willcox 2 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Today’s Topics • Existence & Uniqueness of an Optimum Solution • Karush-Kuhn-Tucker Conditions • Convex Spaces • Unconstrained Problems 3 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Disclaimer!
• This is not a classic optimization class ...
• The aim is not to teach you the details of optimization algorithms, but rather to expose you to different methods.
• We will utilize optimization techniques – the goal is to understand enough to be able to utilize them wisely.
• If you plan to use optimization extensively in your research, you should take an optimization class, e.g. 15.093 4 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Learning Objectives After the next two lectures, you should: • be familiar with what gradient-based (and some gradient-free) optimization techniques are available • understand the basics of how each technique works • be able to choose which optimization technique is appropriate for your problem • understand what to look for when the algorithm terminates • understand why the algorithm might fail 5 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics How to Choose an Algorithm?
• Number of design variables • Type of design variables (real/integer, continuous/discrete) • Linear/nonlinear • Continuous/discontinuous objective behavior • Equality/inequality constraints • Discontinuous feasible spaces • Initial solution feasible/infeasible • Availability of gradient information • Simulation code (forward problem) runtime 6 Accelerator shape optimization (SLAC): Next generation accelerators have complex cavities that require shape optimization for improved performance and reduced cost.
Gradient-Based Optimization Applications • Gradient-based methods can be used to solve large-scale, highly complex engineering problems • Many recent advances in nonlinear optimization, e.g. in PDE-constrained optimization, exploiting structure of the problem, adjoints, preconditioning, specialized solvers, parallel computing, etc.
• We will discuss some basic methods – not state-of-the-art Earthquake inverse modeling (CMU Quake Project, 2003): Inversion of surface observations for 17 million elastic parameters (right: target; left: inversion result). Optimization problem solved in 24 hours on 2048 processors of an HP AlphaServer system.
Examples from O. Ghattas (UT Austin) and collaborators.
Courtesy of Omar Ghattas. Used with permission.
7 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Standard Problem Definition 1 2 min ( ) s.t. ( ) 0 1 ,.., ( ) 0 1 ,.., 1 ,.., j k u i i i J g j m h k m x x x i n x x x • For now, we consider a single objective function, J(x).
• There are n design variables, and a total of m constraints (m=m1+m2).
• The bounds are known as side constraints.
8 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Linear vs. Nonlinear The objective function is a linear function of the design variables if each design variable appears only to the first power with constant coefficients multiplying it.
1 2 3 ( ) 2 3.4 J x x x x is linear in x=[x1 x2 x3]T 1 2 2 3 ( ) 2 3.4 J x x x x x is nonlinear in x 1 2 3 ( ) cos( ) 2 3.4 J x x x x is nonlinear in x 9 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Linear vs. Nonlinear A constraint is a linear function of the design variables if each design variable appears only to the first power with constant coefficients multiplying it.
1 2 3 6 2 10 x x x is linear in x is nonlinear in x is nonlinear in x 2 1 2 3 6 2 10 x x x 1 2 3 6 sin( ) 2 10 x x x 10 Iterative Optimization Procedures Many optimization algorithms are iterative: where q=iteration number S=vector search directiona =scalar distance and the initial solution x0 is given The algorithm determines the search direction S according to some criteria.
Gradient-based algorithms use gradient information to decide where to move. Gradient-free algorithms use sampling and/or heuristics.
1 q q q q x x S 11 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Iterative Optimization Procedures MATLAB® demo 12 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Gradient Vector Consider a function J(x), x=[x1,x2,...,xn] The gradient of J(x) at a point x0 is a vector of length n: 0 1 0 0 2 0 ( ) ( ) ( ) ( ) n J x J x J J x x x x x Each element in the vector is evaluated at the point x0.
13 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Hessian Matrix Consider a function J(x), x=[x1,x2,...,xn] The second derivative of J(x) at a point x0 is a matrix of size nn: Each element in the matrix is evaluated at the point x0.
2 2 2 2 1 1 2 1 2 1 2 0 2 0 2 2 2 1 ( ) ( ) n n n J J J x x x x x J x x J J J x x x H x x 14 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Gradients & Hessian Example 2 3 1 1 2 3 2 3 ( ) 3 6 J x x x x x x x 15 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Taylor Series Consider scalar case: 0 0 2 0 0 0 2 2 1 ( ) ( ) ( ) ( ) 2 z z df d f f z f z z z z z dz dz When function depends on a vector: 0 0 0 0 0 ( ) ( ) ( ) ( ) 1( ) ( )( ) 2 T T J J J 0 x x x x x x x H x x x 1n n1 1n n1 nn The gradient vector and Hessian matrix can be approximated using finite differences if they are not available analytically or using adjoints (L8).
16 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Existence & Uniqueness of an Optimum Solution • Usually cannot guarantee that global optimum is found – multiple solutions may exist – numerical ill-conditioning start from several initial solutions • Can determine mathematically if have relative minimum • Under certain conditions can guarantee global optimum (special class of optimization problems or with global optimization methods) • It is very important to interrogate the “optimum” solution 17 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Existence & Uniqueness: Unconstrained Problems • Unconstrained problems: at minimum, gradient must vanish ||J(x)|| = 0 – x is a stationary point of J – necessary but not sufficient – here A,B,C,D all have J=0 A B C D J(x) x • Calculus: at minimum, second derivative > 0 • Vector case: at minimum, H(x) >0 (positive definite) 18 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics SPD Matrices • Positive definiteness: yTHy > 0 for all y0 • Consider the ith eigenmode of H: Hvi = l ivi • If H is a symmetric matrix, then the eigenvectors of H form an orthogonal set: vi Tvj = d ij • Any vector y can be thought of as a linear combination of eigenvectors: y = aivi • Then: • Therefore, if the eigenvalues of H are all positive, H is SPD.
2 , T T T i i j j i i j j j i i i j i j i a a a a a y Hy v H v v v 19 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Existence & Uniqueness: Unconstrained Problems Necessary and sufficient conditions for a minimum (unconstrained problem): 1. Gradient must vanish 2. Hessian must be positive definite J(x) x The minimum is only guaranteed to be a global optimum if H(x)>0 for all values of x (e.g. simple parabola).
local minimum at x J(x)=0 H(x)>0 20 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Existence & Uniqueness: Constrained Problems At optimum: – at least one constraint on design is active – J does not have to be zero In order to improve design: – move in direction that decreases objective – move in direction that does not violate constraints Usable direction = any direction that reduces objective ST J(x) 0 Feasible direction = a direction in which a small move will not violate constraints ST gi(x) 0 (for all active constraints i) Note that these conditions may be relaxed for certain algorithms.
A B C D J(x) x constraint optimum 21 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Lagrangian Functions • A constrained minimization can be written as an unconstrained minimization by defining the Lagrangian function: • L(x,l ) is called the Lagrangian function. •l i is the jth Lagrange multiplier • It can be shown that a stationary point x of L(x,l ) is a stationary point of the original constrained minimization problem.
1 2 1 1 1 ( , ) ( ) ( ) ( ) m m j j m k k j k L J g h x x x x 22 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Lagrangian Example 2 2 1 2 1 2 min ( ) 3 s.t. 1 J x x x x x 23 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Karush-Kuhn-Tucker (KKT) Conditions If x is optimum, these conditions are satisfied: 1. x is feasible 2. l j gj(x) = 0, j =1,..,m1 and l j 0 3. The Kuhn-Tucker conditions are necessary and sufficient if the design space is convex.
sign in ed unrestrict 0 0 ) ( ) ( ) ( 1 2 1 1 1 1 k m j m k k k m m j j j x h x g x J 24 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics KKT Conditions: Interpretation Condition 1: the optimal design satisfies the constraints Condition 2: if a constraint is not precisely satisfied, then the corresponding Lagrange multiplier is zero – the jth Lagrange multiplier represents the sensitivity of the objective function to the jth constraint – can be thought of as representing the “tightness” of the constraint – if l j is large, then constraint j is important for this solution Condition 3: the gradient of the Lagrangian vanishes at the optimum 25 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Convex Sets Consider a set, and imagine drawing a line connecting any two points in the set.
If every point along that line is inside the set, then the set is convex.
If any point along that line is outside the set, then the set is non-convex.
The line connecting points x1 and x2 is given by w = q x1 + (1-q )x2, 0q 1 26 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Convex Functions Informal definition: a convex function is one that will hold water, while a concave function will not hold water...
convex concave neither A function f(x) bounding a convex set is convex if: f(x) x x1 f(x1) x2 f(x2) 1 2 1 [ (1 ] ( ) (1 ) f f f x x x x 27 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Convex Spaces Pick any two points in the feasible region. If all points on the line connecting these points lie in the feasible region, then the constraint surfaces are convex.
If the objective function is convex, then it has only one optimum (the global one) and the Hessian matrix is positive definite for all possible designs.
If the objective function and all constraint surfaces are convex, then the design space is convex, and the KKT conditions are sufficient to guarantee that x is a global optimum.
In general, for engineering problems, the design space is not convex ...
28 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Convergence Criteria Algorithm has converged when ...
no change in the objective function is obtained OR the maximum number of iterations is reached Once the “optimal” solution has been obtained, the KKT conditions should be checked.
Types of Optimization Algorithms • Useful resource: Prof. Steven Johnson’s open-source library for nonlinear optimization • Global optimization • Local derivative-free optimization • Local gradient-based optimization • Heuristic methods 29 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Most methods have some convergence analysis and/or proofs.
Local Derivative-Free Optimization: Nelder-Mead Simplex Figures from • A simplex is a special polytope of N + 1 vertices in N dimensions – e.g., line segment on a line, triangle in 2D, tetrahedron in 3D • Form an initial simplex around the initial guess x0 • Repeat the following general steps: – Compute the function value at each vertex of the simplex – Order the vertices according to function value, and discard the worst one – Generate a new point by “reflection” – If the new point is acceptable, generate a new simplex. Expand or contract simplex size according to quality of new point.
• Converges to a local optimum when the objective function varies smoothly and is unimodal, but can converge to a non-stationary point in some cases • “fminsearch” in MATLAB Courtesy of Saša Singer. Used with permission.
Global Derivative-Free Optimization: DIRECT • DIRECT: DIviding RECTangles algorithm for global optimization (Jones et al., 1993) • Initialize by dividing domain into hyper-rectangles • Repeat – Identify potentially optimal hyper-rectangles – Divide potentially optimal hyper-rectangles – Sample at centers of new hyper-rectangles • Balances local and global search – Global convergence to the optimum – May take a large, exhaustive searchRepresentation of dividing rectangles algorithm for global optimization.
Image by MIT OpenCourseWare.
32 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Gradient-Based Optimization Process Calculate J(xq) Calculate Sq Perform 1-D search xq = xq-1 + a Sq x0, q=0 Converged?
Done yes no q=q+1 33 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Unconstrained Problems: Gradient-Based Solution Methods • First-Order Methods – use gradient information to calculate S – steepest descent method – conjugate gradient method – quasi-Newton methods • Second-Order Methods – use gradients and Hessian to calculate S – Newton method • Methods to calculate gradients in Lecture 8.
• Often, a constrained problem can be cast as an unconstrained problems and these techniques used.
34 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Steepest Descent Algorithm: choose x0, set x=x0 repeat until converged: S = -J(x) choose a to minimize J(x+a S) x = x + a S Sq = -J(xq-1) • doesn’t use any information from previous iterations • converges slowlya is chosen with a 1-D search (interpolation or Golden section) -J(x) is the direction of max decrease of J at x 35 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Conjugate Gradient S1 = -J(x0) Sq = -J(xq-1) + b qSq-1 • search directions are now conjugate • directions Sj and Sk are conjugate if SjT H Sk = 0 (also called H-orthogonal) • makes use of information from previous iterations without having to store a matrix 2 1 2 2 ( ) ( ) q q q J J x x 36 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Geometric Interpretation Figures from “Optimal Design in Multidisciplinary Systems,” AIAA Professional Development Short Course Notes, September 2002.
Adapted from: "Optimal Design in Multidisciplinary System." AIAA Professional Development Short Course Notes. September 2002.
Figures from "Op timal Design in Mu lt i d is ci pli na ry Sy st e ms " 37 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Newton’s Method Taylor series: 0 0 0 1 ( ) ( ) ( ) ( ) 2 T T J J J x x x x x H x x 0 x x x where differentiate: at optimum J(x)=0 0 0 ( ) ( ) ( ) J J x x H x x 0 0 ( ) ( ) 0 J x H x x 1 0 0 ( ) ( ) J x H x x 38 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Newton’s Method 1 0 0 ( ) ( ) J S H x x • if J(x) is quadratic, method gives exact solution in one iteration • if J(x) not quadratic, perform Taylor series about new point and repeat until converged • a very efficient technique if started near the solution • H is not usually available analytically, and finite difference is too expensive (nn matrix) • H can be singular if J is linear in a design variable 39 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Quasi Newton Sq = -Aq J(xq-1) • Also known as variable metric methods • Objective and gradient information is used to create an approximation to the inverse of the Hessian • A approaches H-1 during optimization of quadratic functions • Convergence is similar to second-order methods (strictly 1st order) • Initially: A=I, so S1 is steepest descent direction then: Aq+1 = Aq + Dq where D is a symmetric update matrix Dq = fn(xq-xq-1, J(xq)- J(xq-1), Aq) • Various methods to determine D e.g. Davidon-Fletcher-Powell (DFP) Broydon-Fletcher-Goldfarb-Shanno (BFGS) 40 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics One-Dimensional Search (Choosing a ) • Polynomial interpolation – pick several values for a – fit polynomials to J(a ) – efficient, but need to be careful with implementation • Golden section search – easy to implement, but inefficient • The one-dimensional search is one of the more challenging aspects of implementing a gradient-based optimization algorithm 41 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Polynomial Interpolation Example 1 1 2 2 J Sa J dJ/da 0 10 -5 1 6 -5 2 8 -5 1 2 3 4 ( J c c c c T 1 2 1 2 x x dJ J J J d x x S 42 © Massachusetts Institute of Technology - Prof. de Weck and Prof. Willcox Engineering Systems Division and Dept. of Aeronautics and Astronautics Lecture Summary • Gradient vector and Hessian matrix • Existence and uniqueness • Optimality conditions • Convex spaces • Unconstrained Methods The next lecture will focus on gradient-based techniques for nonlinear constrained optimization. We will consider SQP and penalty methods. These are the methods most commonly used for engineering applications.
MIT OpenCourseWare ESD.77 / 16.888 Multidisciplinary System Design Optimization Spring 2010 For information about citing these materials or our Terms of Use, visit: |
13988 | https://www.ncbi.nlm.nih.gov/books/NBK559240/ | Group A Streptococcal Infections - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Group A Streptococcal Infections
Ryan Newberger; Caitlyn M. Hollingshead.
Author Information and Affiliations
Authors
Ryan Newberger 1; Caitlyn M. Hollingshead 2.
Affiliations
1 University of Illinois College of Medicine at Peoria
2 The University of Toledo
Last Update: January 14, 2025.
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Continuing Education Activity
Group A Streptococcus (GAS) refers to the bacterial species "Streptococcus pyogenes," which is a gram-positive bacterium commonly found in nature and uniquely adapted to humans. GAS can potentially trigger epidemic waves, often driven by the emergence of new genotypes. GAS is associated with a wide range of infections, including pharyngitis, scarlet fever, impetigo, cellulitis, and erysipelas.Additionally, GAS infections can result in more severe conditions known as invasive GAS infections, such as streptococcal toxic shock syndrome and necrotizing fasciitis, which are increasingly occurring worldwide.
GAS is the leading bacterial cause of acute pharyngitis, although most cases are caused by self-limiting viral infections. Although antibiotics are often prescribed for acute pharyngitis when GAS is diagnosed, they are often overprescribed, even for viral cases. Beyond acute infections, GAS can trigger immune-mediated complications such as acute rheumatic fever, poststreptococcal glomerulonephritis, and sequelae of immune-mediated processes, such as rheumatic heart disease.Diagnosis involves throat swab cultures or rapid antigen detection tests. Antibiotics, particularly beta-lactams, are the treatment of choice but must be prescribed judiciously to avoid resistance. Accurate and timely recognition is critical due to GAS's high morbidity and mortality potential.This activity provides an overview of the clinical spectrum of diseases caused by GAS bacteria, including its epidemiology, clinical evaluation, and treatment. This activity also emphasizes the importance of collaboration and highlights the critical role of the interprofessional healthcare team in diagnosing and managing GAS infections to enhance patient outcomes.
Objectives:
Identify the clinical manifestations of group A streptococcal infections, including pharyngitis, scarlet fever, impetigo, cellulitis, and invasive forms such as necrotizing fasciitis and streptococcal toxic shock syndrome.
Implement evidence-based diagnostic methods, such as rapid antigen detection tests and throat cultures, to accurately diagnose group A streptococcal infections such as pharyngitis.
Select appropriate antibiotic therapy, such as beta-lactams, for confirmed group A streptococcal infections while considering patient factors, including allergies and local antibiotic resistance patterns.
Collaborate with the interprofessional healthcare team, including infectious disease specialists, pharmacists, and laboratory personnel, to optimize the diagnosis and treatment of group A streptococcal infections.
Access free multiple choice questions on this topic.
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Introduction
Group A Streptococcus(GAS) refers to the bacterial species "Streptococcus pyogenes,"which is a gram-positive bacterium that grows in pairs and chains. GAS is ubiquitously found in nature and is uniquely adapted to humans. GAS is responsible for a wide range of infections affecting the upper respiratory tract and skin, ranging from mild and superficial to severe and invasive forms (iGAS) (seeImage.Streptococcus pyogenes Bacterium).
GAS infections are increasing globally, with high morbidity and mortality rates.These infections can affect various areas of the body and can be categorized based on their location and depth. Examples include pharyngitis, scarlet fever, and impetigo (superficial keratin layer); cellulitis (subcutaneous tissue); erysipelas (superficial epidermis); and more severe conditions such as streptococcal toxic shock syndrome (STSS), myositis and myonecrosis (muscle), and necrotizing fasciitis (NF; fascia).In addition to causing infections, GAS can lead to immune-mediated sequelae, including acute rheumatic fever (ARF), post-streptococcal glomerulonephritis (PSGN), and complications from immune-mediated processes, such as rheumatic heart disease (RHD).
Although most cases of pharyngitis are viral, GAS is the leading bacterial cause of acute pharyngitis. GAS accounts for 5% to 15% of sore throat visits in adults and 20% to 30% in children presenting with pharyngitis.Prompt diagnosis and treatment are essential to prevent ARF and other sequelae.Established criteria, diagnostic methods, and guidelines for managing GAS pharyngitis are available that aim to ensure timely diagnosis and reduce the risk of suppurative complications, such as peritonsillar abscess, iGAS infections, nonimmune-mediated sequelae, and further transmission.
Despite the availability of guidelines for diagnosing GAS pharyngitis, antibiotics are often overprescribed, leading to unnecessary exposure and contributing to antibiotic resistance. Antibiotic resistance has been reported in various antibiotics, including penicillin, which is first-line therapy.Overprescription is driven by several factors, such as poor adherence to guidelines, challenges in accurately diagnosing GAS pharyngitis, misdiagnosing GAS carriage as an active infection, and pressure from patients and clinicians to prescribe antibiotics.Clinicians must exercise caution and prescribe antibiotics only when necessary to minimize antibiotic pressure and reduce the risk of resistance development.
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Etiology
GAS is a gram-positive, nonspore-forming, catalase- and oxidase-negative bacterium that grows in pairs and chains.When cultured on blood agar between 35 °C and 37 °C in a 10% carbon dioxide environment, GAS produces smooth, moist, grayish-white colonies with clear margins measuring over 0.5 mm. These colonies are surrounded by a zone of complete hemolysis (beta-hemolysis)(see Image. Comparisons of Hemolytic Activity of Groups A, B, G, and F Streptococci).
GAS is ubiquitous in nature and uniquely adapted to humans, with mucous membranes and skin serving as its only known reservoirs. GAS commonly causes a wide spectrum of infections in the upper respiratory tract and skin, ranging from mild and superficial to severe iGAS.iGAS infections generally occur in normally sterile sites,such as the bloodstream, cerebrospinal fluid, or pleura. Globally, the incidence of GAS and iGAS infections is rising, contributing to significant morbidity and mortality.
The Lancefield classification categorizes streptococci into serologic groups, labeled A to O, based on the reactions of antisera with carbohydrate antigens on the streptococcal cell wall. At least 20 serological groups have been identified, such as groups A, B, and C, and GAS belongs to Lancefield group A. Other streptococci from different Lancefield groups can cause similar syndromes to GAS. Notably, group B Streptococcus (GBS;S agalactiae) colonizes the human gastrointestinal and genital mucosa and can lead to conditions such as puerperal sepsis, neonatal infections, pneumonia, bacteremia, and meningitis.
GAS has also been subdivided based on serotypes of M- and T antigens expressed on their surface.Traditional serotyping methods to detect T- and M antigens have largely been replaced by sequence typing of the N-terminal part of the M-protein gene (emm), which is now primarily used for genotyping GAS, mainly for epidemiological purposes.Whole genome sequencing (WGS) is increasingly utilized to identify epidemic strains of GAS. Over 220 emm types have been classified based on the gene sequence of the M-protein.The streptococcal M-protein, a key virulence factor used for epidemiological typing, also serves as a potential vaccine antigen.
Numerous virulence determinants in GAS have been identified, aiding in adhesion, colonization, evasion of the innate immune system, invasion, and dissemination within the host.Key virulence factors include the M-protein, hyaluronic acid, streptokinase, and deoxyribonuclease (DNase)-B. Among its notable toxins are the pyrogenic toxins (also known as scarlatina or erythrogenic toxins), which are responsible for the rash in scarlet fever. These toxins also stimulate mononuclear cells to produce tumor necrosis factor (TNF)-alpha and interleukins (IL-1 and IL-6), thereby contributing to fever and shock in patients with STSS.
Certain bacterial superantigenic exotoxins, associated with syndromes such as STSS, trigger an atypical polyclonal activation of lymphocytes. This leads to rapid onset shock and multiorgan failure, contributing to high mortality. The primary superantigenic exotoxins identified are toxic shock syndrome toxin-1 (TSST-1) and enterotoxins.
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Epidemiology
GAS exclusively infects humans.GAS spreads through respiratory secretions, fomites, and direct contact with infected skin (eg, in cases of impetigo). Although GAS infections can occur at any age, children, older adults, and immunocompromised individuals are at higher risk of becoming infected.The incubation period for GAS is 2 to 5 days, during which patients are infectious and capable of transmitting the bacteria. Environmental factors and crowded settings—such as schools, households, and nursing homes—significantly increase transmission risk.Notably, GAS can also cause disease in young, healthy individuals, with one study reporting its occurrence in 25% of people without identifiable risk factors.
The epidemiology of GAS infections varies based on the type of infection. GAS can exist in the pharynx either as an asymptomatic carrier state or as a pathogen causing pharyngitis. Approximately 5% to 15% of individuals in the general population are estimated to be GAS carriers within populations.GAS pharyngitis is most common in children aged between 5 and 15 and represents the leading bacterial cause of acute pharyngitis in this group.GAS accounts for 5% to 15% of sore throat visits in adults.GAS pharyngitis usually occurs most frequently in the winter and early spring.Pharyngitis caused by GAS typically results from person-to-person transmission through oropharyngeal secretions and respiratory droplets from infected individuals.
Severe illness and iGAS infections exhibit a bimodal distribution, occurring most commonly in individuals aged 2 or younger and 50 or older.Risk factors associated with increased mortality include advanced age, male sex, nursing home residency, chronic underlying conditions, immunosuppression, recent surgery, septic shock, necrotizing fasciitis, concurrent viral infections, isolated bacteremia, and infection with emm type 1 or 3.The global prevalence of severe GAS infections is estimated at 18.1 million cases, with 1.78 million new cases and 616 million cases of GAS pharyngitis occurring annually.Globally, severe GAS infections account for approximately 500,000 deaths annually, with RHD and iGAS infections being the leading contributors to these deaths.The burden of iGAS is particularly high, with around 663,000 new cases and 163,000 deaths reported each year.
Skin and soft tissue are the most common infection sites, with 32% of patients experiencing cellulitis and 8% developing necrotizing fasciitis. Skin and soft tissue infections (SSTIs) can vary from mild to invasive, with the latter associated with high mortality. Impetigo, a superficial infection typically caused by either Staphylococcus aureus or GAS, primarily affects young children and infants, especially those of preschool age. SSTIs are highly contagious, with a global prevalence estimated at 11.2%, higher in children (12.3%) compared to adults (4.9%).
Cellulitis and erysipelas are skin infections caused by bacteria, including S aureus and GAS, and approximately 10% of the cases are caused by GAS alone.Population-based surveillance data from Europe report an incidence rate of about 3 cases per 100,000 individuals per year, with no specific peak age or demographic for infection.Population-based data in the United States estimate the rate of necrotizing fasciitis at 2.5 cases per 1,000,000 person-years.
GAS is responsible for a wide range of infections and has historically been associated with increased morbidity and high mortality, particularly before the advent of antibiotics. Notably, puerperal sepsis and scarlet fever were major causes of death, with case fatality rates (CFR) reaching approximately 30% in the 19th century.Infections related to iGAS continue to have high morbidity and mortality, with case fatality rates ranging from 10% to 30%.Mortality for severe invasive infections, such as STSS and necrotizing fasciitis, has been reported in various studies to range from 14% to 64%, with some studies citing rates as high as 80%.
A study in the United States examined iGAS data from 2005 to 2012 and reported case-fatality rates for septic shock, STSS, and necrotizing fasciitis at 45%, 38%, and 29%, respectively.The number of iGAS infections began rising in the United States in the late 1980s. The incidence remained stable from the mid-1990s until 2012, but it started to increase thereafter, with an average annual rate of 3.8 per 100,000 individuals from 2005 to 2012. Between 2014 and 2016, the incidence further rose from 4.8 to 5.8 per 100,000 individuals.
Epidemiological surveillance is crucial for tracking epidemics, particularly given the increasing incidence and burden of GAS infections, especially iGAS, worldwide. WGS has an important role in this effort.Since 2000, the dominant emm types in Europe and North America have been emm1 and emm3, with emm1 being the most prevalent type associated with invasive infections in high-income countries. The 7 emm types responsible for 50% to 70% of iGAS infections include emm1,emm28,emm89,emm3,emm12,emm4, and emm6,which are collectively referred to as M1 global.
A new emm1 sublineage, coined M1 UK, was identified in 2008 in the United Kingdom, which exhibited an increased expression of the scarlet fever toxin and streptococcal pyrogenic exotoxin A (speA), leading to a rise in cases of scarlet fever and invasive infections in the United Kingdom between 2014 to 2018. As a result, it became the dominant type in the country.Following the COVID-19 pandemic, 3 emerging M1 UK clades rapidly expanded across the United Kingdom, resulting in severe outcomes in children. All globally sequenced M1 UK isolates (speA) can be traced back to the United Kingdom, where they caused an epidemic and have since spread across Europe and internationally.Although declining immunity may contribute to streptococcal outbreaks, the genetic characteristics of M1 UK suggest a fitness advantage in pathogenicity and a remarkable ability to endure population bottlenecks. M1 UK is now the dominant strain in England. In addition, 2 other lineages—M113SNPs and M123SNPs—were also identified.
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Pathophysiology
GAS infections result from a complex interplay between host and bacterial factors that facilitate the establishment of infection. GAS utilizes various virulence factors, including toxins and other substances, to evade the host immune system and infect humans. The hyaluronic acid capsule of GAS acts as a camouflage mechanism by resembling human hyaluronic acid, allowing the bacterium to evade immune detection. The surface-associated protein (S-protein) protects the bacterium from phagocytic destruction. Additionally, GAS produces proteases that degrade host immune signaling molecules and extracellular DNases to neutralize host immune defenses.
Various surface substances, such as lipoteichoic acid and F-protein, enable GAS to adhere to host cells and facilitate colonization.Cytolytic toxins,including streptolysins and hyaluronidase, contribute to tissue destruction, allowing GAS to invade the host. Additionally, GAS has several factors, such as its capsule, G-protein, C5a peptidase, and M-protein, to evade host immune defenses. Among these, M-protein is particularly significant for GAS virulence, as it inhibits the phagocytosis of host immune cells.
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History and Physical
Obtaining a thorough history of the present illness and past medical history is crucial for patients presenting with symptoms of infection. GAS infections can present in diverse ways, influenced by factors such as the infection's location, the toxins produced, the invasiveness of the GAS strain, the patient's immune status, and whether the infection is superficial or deep.Clinicians must remain vigilant, as even mild-appearing infections can rapidly progress to severe conditions. For this reason, GAS should always be included in the differential diagnosis.
For instance, patients presenting with seemingly uncomplicated superficial SSTIs might have necrotizing fasciitis—a condition that demands immediate and aggressive treatment. To avoid missing such critical diagnoses, clinicians should perform thorough physical examinations and ask targeted questions. The severity of the condition may be more apparent in cases of GAS bacteremia or TSST-1. In these scenarios, clinicians should include GAS in the differential diagnosis, initiate appropriate antibiotic therapy, and obtain cultures. Each type of GAS infection presents distinct signs and symptoms, underscoring the need for clinicians to maintain a high level of suspicion to ensure accurate diagnosis and effective treatment.
The most common symptoms of GAS pharyngitis are the sudden onset of fever and sore throat. Patients may also report headaches, nausea, vomiting, and abdominal pain. When obtaining a patient history, it is common to identify exposure to close contact with a GAS infection, particularly among school-age children or individuals in communal living environments, such as nursing homes, where prevalence is higher due to close living conditions.
Upon examination, physical findings in GAS pharyngitis commonly present with generalized inflammation of the tonsils and pharynx, variable tonsillar exudates, a red and swollen uvula, and palatal petechiae. Tender cervical lymphadenopathy is frequently noted during palpation. Symptoms such as conjunctivitis, cough, coryza, or diarrhea are uncommon and, when present, are more suggestive of a viral etiology.Physical examination of the posterior oropharynx alone is insufficient to differentiate GAS from other causes of acute pharyngitis, such as viral pharyngitis, which is the most common cause. The Centor Criteria, designed to diagnose GAS pharyngitis based on clinical findings, are unreliable when used alone and must be supported by microbiological testing for an accurate diagnosis.
A thorough skin examination is essential, as the presence of a fine maculopapular erythematous rash along with a "strawberry tongue" strongly indicates scarlet fever. While scarlet fever is commonly associated with GAS streptococcal pharyngitis, it can also develop in cases of iGAS infections.Impetigo typically occurs in school-aged children and is characterized by distinct cutaneous lesions described as discrete to confluent "honey-crusted" areas, most commonly on the face and extremities.Usually,vital sign abnormalities are not associated with the clinical presentation, and the patient will have no additional physical examination findings apart from the characteristic lesions.
iGAS infections include necrotizing fasciitis, TSST-1, and any other infection affecting sterile body compartments. These infections are severe and require prompt treatment, as patients can deteriorate rapidly. Necrotizing fasciitis is a critical, life-threatening soft tissue infection that involves the skin, subcutaneous tissue, and fascia. This can resemble cellulitis, making it easy to misdiagnose. A key indicator of necrotizing fasciitis is pain that is disproportionate to the findings on physical examination, which is a common feature in these cases.Clinicians must maintain a high level of suspicion, as misdiagnosis or delays in treatment can lead to poor outcomes. Prompt intervention with aggressive surgical debridement and antibiotic therapy is critical.
STSS is a severe and life-threatening condition caused by GAS. This condition results from an iGAS infection where bacterial enterotoxins are released, leading to severe systemic symptoms. STSS often develops following a primary GAS infection, particularly in deep wound infections. Clinically, patients present with symptoms of severe sepsis, including tachycardia, hypotension, poor tissue perfusion, and signs of end-organ dysfunction.
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Evaluation
For individuals with GAS pharyngitis, throat swab cultures are the gold standard for identification. GAS grows easily on sheep blood agar and is catalase- and oxidase-negative. Confirmatory identification is achieved through Lancefield grouping and, more recently, techniques such as matrix-assisted laser desorption ionization time-of-flight (MALDI-TOF).However, culture results typically take 24 hours or longer, which delays critical information needed for treatment, isolation, and epidemiological purposes.
A rapid antigen detection test (RADT) can be used to diagnose GAS pharyngitis. The RADT has a sensitivity of approximately 85% in children, although variability exists, while its specificity is stable at 95%.Due to the high specificity of the RADT in children, antibiotic therapy is recommended without the need for further throat culture to distinguish infection from carriage. If the RADT result is negative, treatment decisions depend on national guidelines.
In the United States, however, in adults, due to the low general rate of GAS pharyngitis and corresponding low pretest probability, a negative RADT does not require confirmatory culture. A negative rapid swab result confers an extremely low post-test probability of GAS infection.For other sites of infection, whether superficial or deep (such as skin, blood, wound, or lung), performing a Gram stain and cultures are the appropriate diagnostic methods. Polymerase chain reaction (PCR) can also be used to identify specific GAS strains, particularly in complicated cases.
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Treatment / Management
Recognizing GAS infections promptly and accurately can be challenging due to the broad differential diagnoses associated with its clinical manifestations. GAS can be associated with increased morbidity and ranks among the top 10 infectious causes of mortality. GAS should always be considered in the differential diagnosis when evaluating patients, as it can lead to poor outcomes. Culture results are crucial for tailoring antibiotic therapy, and ensuring the initial empirical regimen covers GAS is essential.
Beta-lactam antibiotics are consistently effective against GAS and remain the preferred treatment for both noninvasive and iGAS infections. While reports of antibiotic resistance to penicillins and increased minimal inhibitory concentrations (MICs) to penicillin and cephalosporins have emerged—primarily due to mutations in the peptidoglycan synthetic enzyme pbp2x gene—resistance rates remain low. However, penicillin continues to be the gold standard for GAS treatment. For patients with penicillin allergies, macrolides (eg, erythromycin) and lincosamides (eg, clindamycin) are notable alternatives. However, resistance to these antibiotics has increased in the past decade, with variable prevalence of resistant GAS strains worldwide.
Reports from China indicate macrolide resistance rates as high as 90%, with resistance in some European countries ranging from 20% to 40% and lincosamide resistance reaching up to 19%; in other parts of Europe, these rates can be as low as 2%.This resistance is attributed to multiple factors, including macrolide resistance mechanisms involving the MLSb phenotype.Alternative antibiotics for penicillin-allergic patients can be considered if resistance to first-line agents is present. The choice of therapy should be guided by the location and severity of the infection, local antibiotic resistance patterns associated with GAS, and the patient's allergy profile.
More specifically, an oral antibiotic regimen is typically recommended for 10 days when treating GAS pharyngitis. Recommended regimens include penicillin V or amoxicillin for 10 days by mouth. An alternative treatment for GAS pharyngitis is a single intramuscular dose of penicillin G benzathine, particularly for patients who are unlikely to complete the full course of oral antibiotics.For penicillin-allergic patients, alternatives such as macrolides or clindamycin can be used, but local resistance patterns should be considered when selecting treatment.
Broad-spectrum antibiotics should be initiated for severe GAS infections, such as necrotizing fasciitis and STSS, to ensure coverage while awaiting culture results.For severe infections such as TSST-1 and necrotizing fasciitis, clindamycin is often added to the antibiotic regimen (eg, penicillin), as it can inhibit superantigen production, which enhances the phagocytosis of S pyogenes by reducing M-protein production.In addition to antibiotics, supportive measures, such as intravenous fluid administration and blood pressure management with vasopressors, should be implemented for these severe or systemic infections.
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Differential Diagnosis
Many bacterial and viral pathogens can cause symptoms similar to those caused by GAS. Creating a differential diagnosis for each syndrome is crucial, as GAS can present in various forms, including pharyngitis, impetigo, cellulitis, necrotizing fasciitis, bacteremia, septic shock, and pneumonia. Therefore, it is crucial to consider GAS in the differential diagnosis of these symptoms to ensure appropriate antibiotic treatment and accurate testing for diagnosis.
Noninfectious conditions can closely resemble GAS infections, making a thorough physical examination essential. For instance, cellulitis can present similarly to venous stasis changes, with findings also overlapping those of acute deep vein thrombosis.
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Prognosis
Generally, simple, noninvasive GAS infections have low morbidity and mortality rates, with patients experiencing good outcomes. However, more invasive and severe GAS infections carry significant morbidity and mortality. iGAS infections caused by emm1 are associated with more severe conditions, such as TSST-1, and a higher risk of ICU admission compared to other iGAS strains.
TSST-1 has a mortality rate of 5% to 10%, particularly in patients at the extremes of age or those with preexisting conditions. High mortality rates are reported in patients with necrotizing fasciitis. When necrotizing fasciitis occurs alongside STSS, the mortality rate can reach up to 60%.
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Complications
GAS is associated with significant morbidity and mortality. In addition to causing infections, GAS can trigger immune-mediated sequelae, including ARF and PSGN, as well as direct consequences such as RHD. The valvular damage caused by ARF can be permanent and, in severe cases, may necessitate surgical intervention, including valve repair or replacement.
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Deterrence and Patient Education
Isolation and prevention of exposure are crucial in limiting the spread of viral and bacterial infections, including GAS. This is particularly important for conditions such as impetigo and GAS pharyngitis, which are highly contagious through saliva, droplets, and skin contact. These infections are common among school-age children who have frequent close contact with peers. Therefore, isolating individuals with GAS infections is crucial by having them stay home from school, daycare, or work until they are no longer considered infectious. After 24 hours of antibiotic treatment, approximately 80% of patients are considered noninfectious.
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Enhancing Healthcare Team Outcomes
Diagnosing and managing GAS infections requires an interprofessional healthcare team, including infectious disease specialists, microbiologists, nursing staff, and infectious disease pharmacists. One of the most crucial ways the healthcare team can improve patient outcomes is by promptly recognizing the signs of GAS infection, particularly in severe cases such as iGAS infections, necrotizing fasciitis, and TSST-1. Early detection and appropriate management are essential in improving patient outcomes.
Healthcare professionals must be able to recognize signs of sepsis and shock, carefully evaluating physical examination findings, systemic signs of infection, and potential pathological foci. Timely recognition and intervention can significantly improve clinical outcomes, as severe GAS infections demand prompt administration of antibiotics and additional treatments, such as surgery for necrotizing fasciitis or drainage of pleural effusion in GAS pneumonia. Effective management requires coordination among infectious disease specialists, clinicians, nursing staff, and pharmacists to ensure optimal care for these infections.
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Figure
Streptococcus pyogenes Bacteria. This illustration depicts a photomicrograph of a specimen highlighting chain-linked S pyogenes bacteria. Public Health Image Library, Public Domain, via Centers for Disease Control and Prevention(more...)
Figure
Comparisons of Hemolytic Activity of Groups A, B, G, and F Streptococci. Group B Streptococcus appears as small, colorless colonies and causes beta-hemolysis. Centers for Disease Control and Prevention
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Disclosure:Caitlyn Hollingshead declares no relevant financial relationships with ineligible companies.
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13989 | https://www.ijapm.org/vol10/448-A1012.pdf | Frequently-Used Properties of the Floor Function Xingbo Wang1, 2 1 Department of Mechatronic Engineering, Foshan University, Foshan, China. 2 State Key Laboratory of Information Security, Institute of Information Engineering, Chinese Academy of Sciences, Beijing 100093, China. Corresponding author. Tel.: +86075782988845; email: xbwang@fosu.edu.cn, 153668@qq.com Manuscript submitted March 15, 2020; accepted June 8, 2020. doi: 10.17706/ijapm.2020.10.4.135-142 Abstract: The paper collects 42 frequently-used properties of the floor function, including 35 ones from other literatures and 7 newly added-and-proved ones. The collected properties cover basic inequalities, basic identities, conditional inequalities, conditional equalities and practical formulas. The paper is helpful for scholars of mathematics and computer science and technology in reading and writing scientific works, reasoning and designing algorithms. Key words: Computational science, floor function, formula, mathematical reasoning. 1. Introduction The floor function, which is also called the greatest integer function (see in ), is a function that takes an integer value. For arbitrary real number x, the floor function of x, denoted by ⌊𝑥⌋, is defined by an inequality of 𝑥−1 < ⌊𝑥⌋≤𝑥 or equivalently ⌊𝑥⌋≤𝑥< ⌊𝑥⌋+ 1. The floor function frequently occurs in many aspects of mathematics and computer science. However, as I stated in article , except the Graham's book , it is hard to find another book or a literature that introduces in general the know-of of the function although people can find something via the Internet, e.g., the wikipedia . Since Graham's book was first published 30 year’s ago and its following-up editions made few modifications on the part of the floor function, it is necessary to sort out the properties of the function as a reference for researchers. In 2017 and 2019, I proved respectively several formulas for the function and made brief summaries on the frequently-used properties by my work together with certain formulas collected from previous literatures, as seen in and . In the past two years, I proved several new results and thus I put them together with the 2019 summary to form this literature. 2. Definition and Notation The floor function of real number x is denoted by symbol ⌊𝑥⌋ that satisfies ⌊𝑥⌋≤𝑥< ⌊𝑥⌋+ 1; the fraction part of x is denoted by symbol { } x that satisfies𝑥= ⌊𝑥⌋+ {𝑥}; the ceiling function of x is denoted by symbol ⌈𝑥⌉that fits𝑥≤⌈𝑥⌉< 𝑥+ 1. In this whole article, 𝐴⇒𝐵means conclusion B can be derived from condition A; 𝐴⇔𝐵means B holds if and only if A holds. Symbol Z means the integer set,𝑥∈𝑍means x is an integer and 𝑥∉𝑍indicates x is not an integer. 3. Frequently Used Properties of the Floor Function The following properties of the floor functions are sorted by basic inequalities, conditional inequalities and International Journal of Applied Physics and Mathematics 135 Volume 10, Number 4, October 2020 basic equalities. 3.1. Basic Inequalities In the following inequalities, x and y are real numbers by default. (P1) + + + + 1 x y x y x y (P2) − − − − − + 1 1 x y x y x y x y (P3) , + + + + 2 2 x y x y x y (P4) + + + + + + ( ) ( ) m n x m n y mx my nx ny with m and n being positive integers (P5) + − + + + ( 1) nx ny n x y x y with n being a positive integer (P6) , xy x y with , 0 x y . (P7) y y x x with 1 x and 0 y . (P8) n x nx ; = { } 1 n x nx n x , where n is a positive integer. (P9) + − 1 1 q q p p for arbitrary positive integers p and q; 3.2. Conditional Inequalities In the following inequalities, x and y are real numbers, and n is an integer. (P10) x n x n , n x n x (P11) x n y x n y (P12) x y x y (P13) , x y x y 3.3. Basic Equalities In the following equalities, x and y are real numbers, m and n are integers. (P14) , + = + n x n x . (P15) = x x m m with 1 m . (P16) − − = − − , 1, x x x x x Z Z (P17) , ⌊𝑛𝑥⌋= ⌊𝑥⌋+ ⌊𝑥+ 1 𝑛⌋+. . . + ⌊𝑥+ 𝑛−1 𝑛⌋ with n>0, particularly, ⌊𝑥⌋+ ⌊𝑥+ 1 2⌋= ⌊2𝑥⌋ and ⌊ 𝑥 2⌋+ ⌊ 𝑥+1 2 ⌋= ⌊𝑥⌋. (P18) ⌊𝑥⌋= ⌊ 𝑥 𝑛⌋+ ⌊ 1+𝑥 𝑛⌋+. . . + ⌊ 𝑛−1+𝑥 𝑛 ⌋, particularly,⌊ 𝑥 2⌋+ ⌊ 𝑥+1 2 ⌋= ⌊𝑥⌋ (P19) − = + 1 1 n n m m with 1 m . (P20) , = x x with 0 x (P21) = log log b b x x with 0 x (P22) + = + log 1 log ( 1) b b m m with 1 m . (P23) ⌊ ⌊𝑎 𝑏⌋ 𝑐⌋= ⌊ 𝑎 𝑏𝑐⌋ for an arbitrary integer a and positive integers b and c. International Journal of Applied Physics and Mathematics 136 Volume 10, Number 4, October 2020 (P24) , ⌊ 𝑚+1 𝑛⌋= { ⌊ 𝑚 𝑛⌋ , 𝑛∤𝑚+ 1 ⌊ 𝑚 𝑛⌋+ 1, 𝑛|𝑚+ 1 (P25) = 1 1 n x x (P26) + + = + = + = + 1 4 1 4 2 4 3 n n n n n (P27) , It needs + 2 log 1 N binary bits to express decimal integer N in its binary expression. A positive integer n with base b has + log 1 b n digits. (P28) Let N be an integer; then − 2 0 N N . (P29) Let m and p be positive integers; then number of p’s multiples from 1 to m is calculated by m p . (P30) Let , m n and p be positive integers such that 1 p m n ; then number of p’s multiples from m to n is calculated by 𝜈(𝑚, 𝑛, 𝑝) = { ⌊𝑛 𝑝⌋−⌊𝑚 𝑝⌋, 𝑝∤𝑚 ⌊𝑛 𝑝⌋−⌊𝑚 𝑝⌋+ 1, 𝑝|𝑚 (P31) Arbitrary positive integer i yields − 1 2 2 i i i Arbitrary positive even integer e yields = 2 2 e e And arbitrary positive old integer o yields = − 2 1 2 o o (P32) Let and x be positive real numbers; then it holds − + 1 ( 1) x x x Particularly, if is a positive integer, say =n, then it yields + − ( 1) 1 n x nx n x (P33) For arbitrary positive real numbers ,x and y with x y , it holds − + − ( ) 0 x y y x (P34) . For arbitrary odd integer 7 n , it holds − + 2 1 1 log 2 n n (P35) For positive integer k and real number 0 x , it holds − − − 2 2 1 2 ,0 log 0 2 , log 2 k k k k x x x x k x International Journal of Applied Physics and Mathematics 137 Volume 10, Number 4, October 2020 4. Some New Results Here lists some newly found and proved equalities and inequalities. (P36) min( , ) x x x for positive numbers and x . Particularly, − 1 x x x When 0 1. Proof. See the following three steps: 1). x x by definition; 2). x x x x by (P13); 3). = − = − − −= − ( { }) { }) { }) 1 1 x x x x x x x x . (P37) For an arbitrary positive integer k and an arbitrary odd integer 1 N , it holds − = 1 2 2 k k N N Proof. Consider the case 1 k = and 2 1 N s = + with integer 0 s . It yields − = + = = = 1 1 ( ) ( ) 2 2 2 N N s s s Now assume 1 k and = + 2k N s r with 0 s being an integer and − 0 2 1 k r being odd; without loss of generality, let 2 1 r t = + with integer 1 0 2 1 k t − − ; then −= + 1 2 2 k N s t and − − = + = 1 1 2 2 k k N t s s . Meanwhile, it knows + = + = 2 1 2 2 k k N t s s Because = + − 0 2 1 2 1 k r t . Remark on (P37). The condition that N is odd is mandatory because this property does not hold for an even integer N. A simple counterexample is − 4 4 1 2 2 . Actually, when 1 k = taking 2 N s = yields = 2 N s while − = + − = − 1 1 1 2 2 N s s . When 1 k taking 2 2 k N s t = + with 0 s and 1 0 2 1 k t − − being integers leads to − = + = 1 2 2 k k N t s s and − − − = + = + − 1 1 2 1 1 2 2 2 2 k k k k N t t s s , which results in by (P2) − − = = 0, 0 1 1, 0 2 2 k k t N N t (P38) For an arbitrary positive integer k and even integer 2 N , it holds + = 1 2 2 k k N N Proof. Consider the case 1 k = and 2 N s = with integer 0 s . Then + = = = + = 1 1 ( ) ( ) 2 2 2 N N s s s Now assume 1 k and + = + = + + 1 2 2 2 1 k k N s r s t with integers 0 s and − − 1 0 2 1 k t ; then International Journal of Applied Physics and Mathematics 138 Volume 10, Number 4, October 2020 + = + = 1 2 2 k k N r s s Because = + − 0 2 1 2 1 k r t . Meanwhile by = + 2 2 k N s t with − − 1 0 2 1 k t , it holds − = + = 1 2 2 k k N t s s Thus it knows + = 1 2 2 k k N N when N is even. Remark on (P38). The condition that N is even is mandatory because this property does not hold for an odd integer N. A simple counterexample is + 3 1 3 2 2 . Readers can confirm the general cases by referring to the Remark on (P37). (P39) For arbitrary positive real numbers x and y satisfying 𝑥≥𝑦, it holds − = + 0 1 x y y x y Proof. First is to prove the necessity as following reasoning: − = + − − = − + − = 0 { } { } 0 { } { } 0 x y x x y y x y x y Since {𝑥} and {𝑦} are positive real numbers satisfying 0 ≤{𝑥} < 1 and 0 ≤{𝑦} < 1 , it knows −1 < {𝑥} −{𝑦} < 1 ⇒−1 ≤⌊{𝑥} −{𝑦}⌋≤0 and thus − + − = + { } { } 0 1 x y x y y x y (P40) For positive integer and 0 1 x , it holds −+ = − 2 log (2 1 ) 1 x and + = 2 log (2 ) x Proof. By property of T3 tree, − 2 1 is on level −2 of T3 (see in ). That is −= − − − = − 2 2 2 log (2 1) 1 log (2 1) 1 Since − −+ = 2 2 2 log (2 1) log (2 1 ) log (2 ) x , by (P13) it holds −= − −+ −+ 2 2 2 1 log (2 1) log (2 1 ) log (2 1 ) x x Likewise, + 2 1 is on level −1 of T3, namely + −= − + = 2 2 log (2 1) 1 1 log (2 1) Since + + 2 2 2 log 2 log (2 ) log (2 1) x , it knows + + = 2 2 log (2 ) log (2 1) x (P41). For integers 0 n and 0 , it holds −− = −− 1 1 2 2 n n Proof. By (P16), the proof considers two cases: +1 n is divisible by 2 and it is not. For the case +1 n is International Journal of Applied Physics and Mathematics 139 Volume 10, Number 4, October 2020 divisible by 2 , let + = 1 2 n s with 1 s being an integer; then + −− + = = − 1 1 1 0mod2 2 2 n n n s s And this time = − = − = − 1 2 1 1 2 2 n n s s s Consequently, it holds −− = −− 1 1 2 2 n n For the case +1 n is not divisible by 2 , let + = + 1 2 n s r ; then integer 1 s and −− − 0 2 1 0 1 2 2 r r Accordingly, it holds + = + − = + − + = = 1 2 ,0 2 1 2 1 1 2 2 n s r r n s r n n s And consequently −− = −− 1 1 2 2 n n (P42). Given positive integers k and N; let 𝑚= ⌊𝑙𝑜𝑔2 𝑁⌋−𝑘> 0. Then there must be an odd one among 1 2 , ,..., 2 2 k k N N + + and 2k m N + if 0 2k N is even. Similarly, there must be an even one among 1 2 , ,...
2 2 k k N N + + ,and 2k m N + if 0 2k N is odd. Proof. Consider the case 2k N is even. Suppose 2 2k N s = with 1 being a positive integer and 1 s being an odd integer; then 2 2 ,0 2 1 2 k k k k N N r s r r + = + = + − This follows 2k N s + = Which is odd. Now check the bound of . Since 2k N s + , it knows 2 2 2 log log k N N s N k N k + − − For the case 2k N is odd, suppose 2 1 2k N s = + with 1 being a positive integer and 𝑠≥1 being an International Journal of Applied Physics and Mathematics 140 Volume 10, Number 4, October 2020 odd integer; 2 2 2 ,0 2 1 2 k k k k k N N r s r r + = + = + + − and for a positive integer x, it holds 2 2 2 2 k x k x k x N r s − + + + = + Since 1 1 2 2 2 1 2 k k k k r + + + − , it knows 2 0 2 k k x r + + = when 1 x . Consequently, if 1 1 x − then 2 2 x k x N s − + = is even. Considering 2 2 2 2 2 k k k k k N N r s r + + = + = + + , it follows 2 log N k − . 5. Motivation of This Paper In writing a paper related with mathematics, computer science, physics and so on, mathematical reasoning plays a major role in the whole procedure. During a reasoning procedure, some minor evidences such as one or more formulas are often required to keep the reasoning correct. Since our primary middle school we have remembered tens of identities, inequalities, theorems and axioms in our minds. The things we have remembered do help us to write an excellent paper of science and technology. However, it is not so fortunate for researchers who research the number theory, the graph theory and the related subjects because they often have to face the floor function, which frequently occurs in the reasoning but does not have many citable formulas. As I mentioned in , the mathematical reasoning or modeling involved with the floor function always requires quite a lot of special skills related with inequalities together with discrete mathematics and it is of quite individuality because the function is defined with an inequality, + 1 x x x , and applied in occasions with integers and other discrete traits. For this reason, I have paid attention to the function and collected its formulas in my teaching and researching work. In 2017, I collected 31 properties and published them in . Among the 31 properties, I proved 5 ones. They are (P12), (P13), (P28), (P30), and (P31). In 2018, I collected 35 properties among which I proved 4 new ones and published them in . In this paper, I collected 42 properties among which the newly added 7 ones are proved in previous subsection. I am sure these collected properties are helpful for certain people to cite. At least I myself frequently look over them when I was writing a paper related with the issue. The motivation that I say so many words here is to show something on how to agitate interest in scientific research. An old thing like the floor function might contain a lot of new work and any new work cannot be lack of some old things. I hope the background and history of this paper are educationally meaningful. Conflict of Interest The author declares that there is no conflict of interests regarding the publication of this article. Author Contributions Professor Xingbo Wang contributes the whole research work of the paper. Acknowledgment The research is supported by the Open Project Program of the State Key Lab of CAD&CG (Grant No. A2002) and by Foshan University and Foshan Bureau of Science and Technology under project that constructs Guangdong Engineering Center of Information Security for Intelligent Manufacturing System. International Journal of Applied Physics and Mathematics 141 Volume 10, Number 4, October 2020 References Rosen, K. H. (2011). Elementary Number Theory and Its Applications (6th ed.). New York: Addison-Wesley. Wang, X. (2012). A mean-value formula for the floor function on integers. Mathproblems Journal, 2(4), 136-143. Graham, R. L., Knuth, D. E., & Patashnik, O. (1994). Integer Functions. Ch.3 in Concrete Mathematics: A Foundation for Computer Science (2nd ed.). Reading, MA: Addison-Wesley. Wikipedia. (2020). Floor and ceiling functions. Retrieved from _functions Wang, X. (2017). Brief summary of frequently-used properties of the floor function. IOSR Journal of Mathematics, 13(5), 46-48. Wang, X. (2019). Brief summary of frequently-used properties of the floor function: Updated 2018. IOSR Journal of Mathematics, 15(1), 30–33. Pan, C. D., & Pan, C. B. (2013). Elementary Number Theory (3rd ed.). Press of Peking University. Kuang, J. (2010). Applied Inequalities. Shandong Science and Technology Press. Liu, P. J. (2010). The Collection of Difficult Problems of Elementary Number Theory. Press of Harbin University. Wang, X. (2017). Some more new properties of consecutive odd numbers. Journal of Mathematical Research, 9(5), 47-59. Wang, X. (2017). Strategy for algorithm design in factoring RSA numbers. IOSR Journal of Computer Engineering (IOSR-JCE), 19(3), 1-7. Wang, X. (2018). Some new inequalities with proofs and comments on applications. Journal of Mathematics Research, 11(3), 15-19. Wang, X. (2018). Some inequalities on T3 tree. Advances in Pure Mathematics, 8(8), 711-719. Wang, X. (2018). T3 tree and its traits in understanding integers. Advances in Pure Mathematics, 8(5), 494-507. Wang, X. (2018). Difference property of an integer function. International Journal of Mathematics Trends and Technology, 55(3), 23-29. Copyright © 2020 by the authors. This is an open access article distributed under the Creative Commons Attribution License which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited (CC BY 4.0). Xingbo Wang was born in Hubei, China. He got his master and doctor’s degrees at National University of Defense Technology of China and had been a staff in charge of researching and developing CAD/CAM/NC technologies in the university. Since 2010, he has been a professor in Foshan University with research interests in computer application and information security. He is now the chief of Guangdong engineering center of information security for intelligent manufacturing system. Prof. Wang was in charge of more than 40 projects including projects from the National Science Foundation Committee, published 8 books and over 90 papers related with mathematics, computer science and mechatronics engineering, and invented 30 more patents in the related fields. Author’s formal photo International Journal of Applied Physics and Mathematics 142 Volume 10, Number 4, October 2020 |
13990 | https://www.wfb-bremen.de/en/page/bremen-invest/bremerhaven-capital-of-fish-fingers | The capital of fish fingers
Food and beverage
Bremerhaven exports fish fingers around the world
"This product enjoys great popularity", says Frosta Marketing Director Hinnerk Ehlers.
© WFB/Focke Strangmann
The first fish fingers were sold in Germany in 1959, made in Bremerhaven, of course. The seaport is still regarded as the capital of these crispy treats – no great surprise since Iglo and Frosta produce 2.7 billion of them here every year, exporting to around 20 countries. And the appetite for these breaded fish sticks remains unbroken.
German consumers love fish fingers
Fish fingers are a favourite with German consumers, especially children. On average, people in Germany eat 24 fish fingers a year. And the popularity of these tasty treats shows no sign of waning. Few people are aware that they are mainly made in Bremerhaven by Frozen Fish International, a subsidiary of Iglo, and by Frosta. The two competitors’ factories are even on opposite sides of the same street. The companies export to around 20 countries, as fish fingers are in high demand outside of Germany too. “Our products are very popular,” says Frosta’s marketing director, Hinnerk Ehlers. Fish fingers have been one of Frosta’s biggest sellers since the Bremerhaven-based frozen food manufacturer started making them again in 2014 after a lengthy break. The company also produces them for other brands.
Historical insight into the production of fish fingers at Iglo.
© Iglo GmbH
2.7 billion fish fingers are exported around the world from Bremerhaven every year
Iglo is very pleased with demand: the company reported double-digit growth in sales in 2017. Iglo is the market leader, producing 1.9 billion fish fingers every year in Bremerhaven. They are sold under a variety of brands outside of Germany, such as BirdsEye in the UK and Ireland, and Findus in Italy. Frosta produces 800 million a year. This makes the seaport the capital of fish fingers, or as Hinnerk Ehlers puts it a little more cautiously: “Bremerhaven is one of the largest production sites in the world.” According to Bremeninvest, the city on the Weser estuary is one of Germany’s leading fish processing sites, with a market share of over 50 per cent. And it’s not just fish fingers that are made here – Frozen Fish International’s site is one of the world’s largest makers of frozen fish products.
One of the most popular convenience foods ever
According to the German Frozen Food Institute, fish fingers have been – and remain to this day – one of Germany’s most popular convenience foods. The secret behind their continuing popularity is simple – they are both healthy and easy to prepare. “The trend towards convenience is continuing,” says Wolfgang Adlwarth of German market research company GfK. The fish finger is one of the products that is benefiting from this.
The idea originated in the UK
Almost 60 years ago, in 1959 to be precise, the first fish fingers were produced in Bremerhaven on an industrial scale by Solo Feinfrost GmbH, a predecessor of Iglo. Only a year later, the product appeared in its first commercial on TV. According to Iglo spokesperson Alfred Jansen, fish fingers were invented in the UK, where strips of cod were covered in breadcrumbs and fried. Iglo’s UK-based sister brand BirdsEye launched the first mass-produced frozen food products on the UK market in 1955. “The company developed the product specifically for children in order to get them to eat more fish,” Jansen quotes from the company’s history books.
Fish fingers have been popular for a long time. This is also shown by this historical photograph of Iglo from the 1960s.
© Iglo GmbH
Bremerhaven was once Europe’s largest fishing port
It is no coincidence that the idea made its way from the UK to Bremerhaven, which at the time was Europe’s largest fishing port. The fish fingers were branded Iglo (Dutch for igloo) in 1963 when Solo Feinfrost was renamed Iglo Feinfrost. Frosta, a competitor of Iglo, moved into fish finger production in the mid-1960s and started selling them under its own brand name ten years later. According to the German Frozen Food Institute, they are made mainly from Alaska pollock, Atlantic pollock or hake.
Visitors can see how fish fingers are made
For three years now, people have been able to visit Frosta in Bremerhaven and see how fish fingers are made – with no advance notice required. All along the production line, large glass panels have been installed that offer views of what goes on in the factory. “We want people to be able to see how we work,” says Ehlers. This is unusual in a food industry that normally prefers its production methods to remain behind closed doors. Tourist coaches now regularly stop off for a brief visit. The plant is in constant operation over three shifts on five days of the week, and for half a day on Saturdays. The factory is closed on Sundays.
At the beginning of each fish finger is a 7.5 kilo deep-frozen block. A saw cuts the frozen fish block into small blanks.
© WFB/Focke Strangmann
The size of a fish finger is precisely specified
The starting point of every fish finger, whether made by Frosta or Frozen Fish International, is a 7.5kg block of frozen fish. This consists of layers of fish fillets that have been processed at sea. The size of the blocks determines the size of the fish fingers: they are always 9cm long, 2.6cm wide and 1.1cm thick. This ensures that there is no waste.
Everyone has their own batter recipe
On Mondays, Frosta produces its own brand. Works manager Frank Hoogestraat takes visitors through the factory. A machine removes the cardboard covering from the block and a worker checks that no bits of it are still stuck to the fish. An automated saw splits the block into the small shapes which are then transported on a conveyor belt to be covered in batter. “The batter is made from flour, water, salt and spices,” says Hoogestraat. Every customer and every country has its own recipe. For the German and Austrian markets, the percentage of fish has to be at least 65 percent; in France and Eastern Europe it can be as low as 52 percent.
Production today at Frosta: Employee Christoph Roes checks the fish fingers, which are always 9 centimetres long, 2.6 centimetres wide and 1.1 centimetres high.
© WFB/Focke Strangmann
The fish inside the finger remains frozen throughout
The fish fingers then move on to be breaded and pass through an 11 metre long deep-fat fryer for around 30 seconds. “The core remains frozen throughout,” Hoogestraat explains. The frying process merely ensures that the coating sticks to the finger. In the next step, the fingers are flash frozen. “We cool them down to minus 18 degrees,” he adds. All along the production line, staff in white coats check that the machines are working properly. In the final step, a machine packs 15 fish fingers into a folding carton. They are now ready for shipping.
Frosta stopped making fish fingers for ten years
Frosta stopped making fish fingers for ten years as the business was unable to source enough MSC-certified fish. It only restarted production once a sufficient supply could be guaranteed, says Ehlers. Here is another astounding figure – Frosta produces 142,500 fish fingers every hour. The fish fingers made by Iglo are now also made from 100 percent MSC-certified fish. Fish finger fans will be glad to hear that production will continue for the foreseeable future.
Further information on the history of Frosta:
Press contact:
Friederike Ahlers, Public Relations, Frosta Tiefkühlkost GmbH, +49 40 85414086, friederike.ahlers@frosta.de
Alfred Jansen, Head of Corporate, Brand and Sustainability Communication, Iglo GmbH, +49 40 180249202, Alfred.Jansen@iglo.com
For the German version of this article, please visit BIS Bremerhaven.
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13991 | https://ocw.mit.edu/courses/6-1200j-mathematics-for-computer-science-spring-2024/mit6_1200j_s24_lec19.pdf | 6.1200J/18.062J Mathematics for Computer Science Thursday 25th April, 2024 Massachusetts Institute of Technology, Spring 2024 Z. Abel, B. Chapman, E. Demaine revised Sunday 28th April, 2024 Lecture 19: Conditional Probability 1 Probability Rules In the last lecture, we learned how to compute probabilities using the Tree Method. In this lecture, we will see how some of our tools for reasoning about sizes of sets carry over naturally to the world of probability, and we will learn how to express mathematically statements like “if the prize is behind door A, what is the probability that Monty opens door B?” Recall: Defnition 1. We defne the probability of an event A as X Pr[A] := Pr[ω] ω∈A An immediate consequence: Proposition 1 (Sum Rule). If A and B are disjoint events, then Pr[A ∪ B] = Pr[A] + Pr[B] Corollary 2 (Complement Rule). Pr[A ¯] = 1 − Pr[A] Corollary 3 (Diference Rule). Pr[A \ B] = Pr[A] − Pr[A ∩ B] Corollary 4 (Principle of Inclusion-Exclusion). Pr[A ∪ B] = Pr[A] + Pr[B] − Pr[A ∩ B] Corollary 5 (Union Bound). Pr[A ∪ B] ≤ Pr[A] + Pr[B] Corollary 6 (Monotonicity Rule). If A ⊆ B are events, then Pr[A] ≤ Pr[B] 2 Lecture 19: Conditional Probability The Sum Rule and Union Bound generalize: Proposition 7 (Sum Rule). If Ai are (pairwise) disjoint events, then " # [ X Pr Ai = Pr[Ai] i∈N i∈N Proposition 8 (Union Bound). " # [ X Pr Ai ≤ Pr[Ai] i∈N i∈N PIE generalizes to fnitely many events in the same way as for counting: Proposition 9 (Principle of Inclusion-Exclusion). If I is a fnite index set, then " # " # [ X \ Pr Ai = − (−1)|J| Pr Aj i∈I ∅⊂J⊆I j∈J 2 Conditional Probability In the last lecture, we studied the Monty Hall problem. In the analysis of this problem, it was useful to make statements such as “if the car is behind door 1, the contestant chooses door 1 with probability 1/3.” How do we express this mathematically in the theory of probability? Defnition 2. For two events A, B, the conditional probability of A given B is Pr[A ∩ B] Pr[A | B] = . Pr[B] This expression can be rewritten to obtain the “product rule” for joint probabilities: Corollary 10 (Product Rule). Pr[A ∩ B] = Pr[A | B] Pr[B] This can be extended to multiple events: Pr[A ∩ B ∩ C] = Pr[A | B ∩ C] Pr[B ∩ C] = Pr[A | B ∩ C] Pr[B | C] Pr[C]. The product rule is the justifcation for the “tree method” of computing probabilities from the last lecture: the numbers on the edges of the tree are the terms in the product. This means in particular that the numbers on the edges of the tree (except at the highest level) are conditional probabilities! Another extension of the product rule that is useful is Pr[A ∩ B | C] = Pr[A | B ∩ C] Pr[B | C]. This can be obtained by dividing both sides of the previous product rule by Pr[C]. 3 Lecture 19: Conditional Probability 3 Example 1: tournament Suppose Ash and Gary have a series of battles against each other, and the frst to win two battles wins the series. The probabilities of victory have the following behavior: 1. The frst battle is a toss-up: 1/2 probability of either trainer winning. 2. If a trainer has won the previous battle, he has a 2/3 chance of winning the next one. 3. There are no draws. Let A be the event that Ash wins the series, and B the event that he wins the frst battle. What is Pr[A | B]? To compute this, let’s use the Tree Method. First Second Third ω Pr[ω] A? B? W WW 1/3 × × 2/3 1/3 W 1/2 W W LW 1/18 × × L 1/3 W LL 1/9 × L 2/3 W LW W 1/9 × W LW L 1/18 1/3 2/3 L 1/2 L 1/3 LL 1/3 L 2/3 Pr[A ∩ B] Pr[A | B] = Pr[B] 1/3 + 1/18 = 1/2 7 = . 9 So far, this is just mechanical computation. 4 Lecture 19: Conditional Probability 4 Bayes’ rule What about Pr[B | A]? This can be calculated in exactly same manner to get 7/9. But what does it mean? Such a conditional probability is expressing an inference: what is the chance that Ash won the frst battle, given that we later observe that he won the whole series? It is often the case that we have a “model” that makes it easy to compute “forward” con ditional probabilities Pr[A | B], but we would really like to know the “backward” probability Pr[B | A] in order to infer something that we do not have direct observational access to. In general Pr[B | A] ̸= Pr[A | B], but they are related by a simple formula called Bayes’ rule Pr[A | B] Pr[B] Pr[B | A] = . Pr[A] This can be derived from the product rule, but this form is so useful that it’s worth com mitting to memory. An especially useful consequence of Bayes’ rule is the expression for the ratio of the conditional probabilities of two events B, C given A: Pr[B | A] Pr[A | B] Pr[B] = . Pr[C | A] Pr[A | C] Pr[C] 5 Example 2: Biased and fair coins We will now explore an application of Bayes’ rule. First, let’s name the terms appearing in it: Pr[A | B] · Pr[B] Pr[B | A] = . Pr[A] We refer to Pr[A | B] as the likelihood (of A given B), and Pr[B] as the prior probability of B. The left-hand side Pr[B | A] is the posterior probability of B. Suppose I have a biased coin (which always comes up heads when I fip it), and a fair coin (which comes up heads half the time, and tails half the time). Suppose I pick a coin with uniform probability, and fip it, observing heads. What is the chance that the coin I picked was fair? Mathematically, let H denote the event of seeing heads, F denote the event of picking a fair coin, and B the event of picking a biased coin. Then we have Pr[F | H] Pr[H | F ] Pr[F ] = Pr[B | H] Pr[H | B] Pr[B] 1/2 · 1/2 = 1 · 1/2 = 1/2. Thus, the chance that the coin was fair is 1/3, and the chance that it was biased is 2/3. 5 Lecture 19: Conditional Probability Now observe that if the prior probability changes, this changes as well. For instance, if Pr[F ] = 1 − ϵ, we see that we get a ratio of 1/2 · (1 − ϵ) . 1 · ϵ This ratio goes to ∞ as ϵ → 0, so Pr[F | H] → 1 as ϵ → 0. 6 Example 3: COVID testing In the next few examples, we’re going to see some examples of counterintuitive behavior arising from Bayes’ rule, where intuitive reasoning underestimates the efect of the prior. Suppose 10% of the MIT community has COVID, and everybody is required to take a COVID test. The tests have a false positive rate of 0.3, and a false negative rate of 0.1. If I test positive, what’s the chance I have COVID? • Events: H I am healthy, S I am sick, + I test positive, − I test negative. • Probabilities: Pr[H] = 0.9, Pr[+ | H] = 0.3, Pr[−| S] = 0.1. From these we deduce the complements: Pr[S] = 0.1, Pr[−| H] = 0.7, Pr[+ | S] = 0.9. • Conditional probability: We want to calculate Pr[S | +]. Use Bayes’ rule, again for the odds: Pr[S | +] Pr[+ | S] Pr[S] = Pr[H | +] Pr[+ | H] Pr[H] 0.9 · 0.1 = 0.3 · 0.9 1 = . 3 So I have 1/4 chance of being sick and a 3/4 chance of being healthy! Even though the test looks pretty good on paper, the base rate (the prior probability that I’m sick) is the dominant efect here: I should still think I’m more likely than not healthy, even if I get a positive test. Of course, in the real world, we don’t test everybody, so this is not realistic. We only test people with symptoms, so in reality, we care about Pr[S | +, has symptoms]. This is a much higher quantity in general! 6 Lecture 19: Conditional Probability 7 Example 4: Simpson’s paradox An analysis of 1973 UC Berkeley graduate admissions data revealed the following paradoxical facts: the admissions rate was higher for men than for women for the university as a whole (that is, the fraction of men applicants who were admitted was higher than the fraction of women applicants who were admitted). However, for each department, the admissions rate for men was lower than it was for women. How could this be? Was the data wrong? It turns out that these facts are both consistent with each other. Suppose there are only two departments: EE and CS. Defne the events A for a student being admitted, M/F for the student being male or female, EE/CS for the student applying to the EE or CS departments. The statement about university-wide admissions rates is Pr[A | M] > Pr[A | F ]. The statement about per-department admissions rates is Pr[A | F ∩ CS] ≥ Pr[A | M ∩ CS] Pr[A | F ∩ EE] ≥ Pr[A | M ∩ EE]. How do we square these with each other? Pr[A ∩ F ] Pr[A | F ] = Pr[F ] Pr[A ∩ F ∩ CS] + Pr[A ∩ F ∩ EE] = Pr[F ] Pr[A | F ∩ CS] Pr[CS | F ] Pr[F ] + Pr[A | F ∩ EE] Pr[EE | F ] Pr[F ] = Pr[F ] = Pr[A | F ∩ CS] Pr[CS | F ] + Pr[A | F ∩ EE] Pr[EE | F ] Pr[A | M] = Pr[A | M ∩ CS] Pr[CS | M] + Pr[A | M ∩ EE] Pr[EE | M]. So to make the overall rates favor men, we can adjust the values of Pr[CS | M] and Pr[CS | F ] accordingly. Here’s the intuition: suppose both CS and EE both mildly favor women, but CS is much more popular with women, and is also much harder to get into (for everyone). Then the value of Pr[A | F ] will be much smaller than Pr[A | M], simply because of the “base rate” of students applying to diferent departments, not because of a gender diference in conditional acceptance probabilities. As an extreme example, suppose 100 men and 100 women apply. 99 women and 1 man apply to CS. 99 men and 1 woman apply to EE. CS is super snobby and accepts 1 applicant (a woman). EE has no standards and rejects 1 applicant (a man). Now admissions rates are: • > 0 for women and 0 for men in CS, • 1 for women and < 1 for men in EE, • 2% for women and 98% for men overall. 7 Lecture 19: Conditional Probability We could interpret this as saying that the gender bias in admissions is not caused by a direct preference for men over women at the level of individual application readers, but rather by other aspects of the system that determine how likely students are to apply to each department. 8 Example 5: O. J. Simpson O. J. Simpson was a retired football player who was accused, and later acquitted, of the murder of his wife, Nicole. Question: Was O. J.’s history of abuse towards his wife was admissible into evidence? Prosecution: Abusers are 10× more likely than randos to be murderers. Therefore, abuse is likely precursor to murder, and should be taken into account. Defense: Probability of abusive husband murdering wife is ∼ 1/2500. Therefore, abuse history has negligible probative value. It would, however, bias the jury against Simpson, so should be barred. Who is right? Both are attempting to reason about conditional probability, specifcally the conditional probability that a husband murders his wife, given that he abuses her. Let’s make precise some events. • Let A be the event [Husband abuses Wife]. • Let G be the event [Husband murders Wife]. • Let M be the event [Wife is murdered]. Pr[G | A] Prosecution argued that is high, so knowing A dramatically increases the pos ¯ Pr[G | A] terior probability of G. Defense argued that Pr[G | A] is low, so knowing A cannot dramatically increase the posterior probability of G. Both neglected the fact that Nicole was murdered; the relevant probability is Pr[G | A ∩ M], which as it turns out, was around 80%: 80% of abusive widowers with murdered wives are the murderers. Probability and conditional probability are used and misused all the time, and even experts make (very public) mistakes. If in doubt, make everything precise and fall back on the fundamentals! MIT OpenCourseWare 6.1200J Mathematics for Computer Science Spring 2024 For information about citing these materials or our Terms of Use, visit: |
13992 | https://people.tamu.edu/~j-epstein/Math147/14a_M147_Notes_Ch5_2.pdf | Page 1 | © 2014 by Janice L. Epstein 5.2 Monotonicity and Concavity 5.2: Monotonicity and Concavity A function f defined on an interval I is called (strictly) increasing on I if ( ) ( ) 1 2 f x f x < whenever 1 2 x x < in I and is called (strictly) decreasing on I if ( ) ( ) 1 2 f x f x < whenever 1 2 x x < in I A function that is always increasing or always decreasing is called monotonic. First Derivative Test for Monotonicity Suppose f is continuous on [a,b] and differentiable on (a,b) If ( ) 0 f x ¢ > for all ( , ) x a b Î , then f is increasing on [a,b] If ( ) 0 f x ¢ < for all ( , ) x a b Î , then f is decreasing on [a,b] Second Derivative Test for Concavity Suppose f is twice differentiable on an open interval I If ( ) 0 f x ¢¢ > for all x I Î , then f is concave up on I If ( ) 0 f x ¢¢ < for all x I Î , then f is concave down on I A critical number of a function f is a number c in the domain of f such that either ( ) 0 f c ¢ = or ( ) f c ¢ does not exist. An inflection point of a function f is the point where a function changes concavity. Page 2 | © 2014 by Janice L. Epstein 5.2 Monotonicity and Concavity Where is the function increasing? decreasing? Where does the function have a local maximum? local minimum? Where is the function concave up? concave down? Where are the critical numbers and inflection points? -2 -1 1 2 -3 -2 -1 1 2 3 f Page 3 | © 2014 by Janice L. Epstein 5.2 Monotonicity and Concavity Example: The graph of the derivative of f is shown. (a) Where is the function increasing or decreasing? (b) Where might the function have a local maximum or minimum? (c) Where is the function concave up or concave down? (d) Where are the inflection points? (e) If (0) 0 f = , sketch a possible graph of f. -5 -4 -3 -2 -1 1 2 3 4 5 -4 -3 -2 -1 1 2 3 4 f’ Page 4 | © 2014 by Janice L. Epstein 5.2 Monotonicity and Concavity Example: Sketch a graph of f satisfying the following conditions: ( ) ( ) ( ) 0 on ,1 and ( ) 0 on 1, f x f x ¢ ¢ > -¥ < ¥ ( ) ( ) ( ) 0 on , 2 and 2, f x ¢¢ > -¥ -¥ ( ) ( ) 0 on 2,2 f x ¢¢ < - lim ( ) 2 and lim ( ) 0 x x f x f x -¥ ¥ =-= Example: Determine where each function is increasing, decreasing, concave up, and concave down. (a) ( ) 1/3 3 1 y x = - (b) 2 2 3 y x -= + |
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Spurious relationship
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From Wikipedia, the free encyclopedia
Apparent, but false, correlation between causally-independent variables
In statistics, a spurious relationship or spurious correlation is a mathematical relationship in which two or more events or variables are associated but not causally related, due to either coincidence or the presence of a certain third, unseen factor (referred to as a "common response variable", "confounding factor", or "lurking variable").
Examples
[edit]
An example of a spurious relationship can be found in the time-series literature, where a spurious regression is one that provides misleading statistical evidence of a linear relationship between independent non-stationary variables. In fact, the non-stationarity may be due to the presence of a unit root in both variables. In particular, any two nominal economic variables are likely to be correlated with each other, even when neither has a causal effect on the other, because each equals a real variable times the price level, and the common presence of the price level in the two data series imparts correlation to them. (See also spurious correlation of ratios.)
Another example of a spurious relationship can be seen by examining a city's ice cream sales. The sales might be highest when the rate of drownings in city swimming pools is highest. To allege that ice cream sales cause drowning, or vice versa, would be to imply a spurious relationship between the two. In reality, a heat wave may have caused both. The heat wave is an example of a hidden or unseen variable, also known as a confounding variable.
Another commonly noted example is a series of Dutch statistics showing a positive correlation between the number of storks nesting in a series of springs and the number of human babies born at that time. Of course there was no causal connection; they were correlated with each other only because of two independent coincidences. During the Pagan era, which can be traced back at least to medieval times more than 600 years ago, it was common for couples to wed during the annual summer solstice, because summer was associated with fertility. At the same time, storks would commence their annual migration, flying all the way from Europe to Africa. The birds would then return the following spring — exactly nine months later.
In rare cases, a spurious relationship can occur between two completely unrelated variables without any confounding variable, as was the case between the success of the Washington Commanders professional football team in a specific game before each presidential election and the success of the incumbent President's political party in said election. For 16 consecutive elections between 1940 and 2000, the Redskins Rule correctly matched whether the incumbent President's political party would retain or lose the Presidency. The rule eventually failed shortly after Elias Sports Bureau discovered the correlation in 2000; in 2004, 2012 and 2016, the results of the Commanders' game and the election did not match. In a similar spurious relationship involving the National Football League, in the 1970s, Leonard Koppett noted a correlation between the direction of the stock market and the winning conference of that year's Super Bowl, the Super Bowl indicator; the relationship maintained itself for most of the 20th century before reverting to more random behavior in the 21st.
Hypothesis testing
[edit]
Often one tests a null hypothesis of no correlation between two variables, and chooses in advance to reject the hypothesis if the correlation computed from a data sample would have occurred in less than (say) 5% of data samples if the null hypothesis were true. While a true null hypothesis will be accepted 95% of the time, the other 5% of the times having a true null of no correlation a zero correlation will be wrongly rejected, causing acceptance of a correlation which is spurious (an event known as Type I error). Here the spurious correlation in the sample resulted from random selection of a sample that did not reflect the true properties of the underlying population.
Detecting spurious relationships
[edit]
The term "spurious relationship" is commonly used in statistics and in particular in experimental research techniques, both of which attempt to understand and predict direct causal relationships (X → Y). A non-causal correlation can be spuriously created by an antecedent which causes both (W → X and W → Y). Mediating variables, (X → M → Y), if undetected, estimate a total effect rather than direct effect without adjustment for the mediating variable M. Because of this, experimentally identified correlations do not represent causal relationships unless spurious relationships can be ruled out.
Experiments
[edit]
In experiments, spurious relationships can often be identified by controlling for other factors, including those that have been theoretically identified as possible confounding factors. For example, consider a researcher trying to determine whether a new drug kills bacteria; when the researcher applies the drug to a bacterial culture, the bacteria die. But to help in ruling out the presence of a confounding variable, another culture is subjected to conditions that are as nearly identical as possible to those facing the first-mentioned culture, but the second culture is not subjected to the drug. If there is an unseen confounding factor in those conditions, this control culture will die as well, so that no conclusion of efficacy of the drug can be drawn from the results of the first culture. On the other hand, if the control culture does not die, then the researcher cannot reject the hypothesis that the drug is efficacious.
Non-experimental statistical analyses
[edit]
Disciplines whose data are mostly non-experimental, such as economics, usually employ observational data to establish causal relationships. The body of statistical techniques used in economics is called econometrics. The main statistical method in econometrics is multivariable regression analysis. Typically a linear relationship such as
is hypothesized, in which is the dependent variable (hypothesized to be the caused variable), for j = 1, ..., k is the jth independent variable (hypothesized to be a causative variable), and is the error term (containing the combined effects of all other causative variables, which must be uncorrelated with the included independent variables). If there is reason to believe that none of the s is caused by y, then estimates of the coefficients are obtained. If the null hypothesis that is rejected, then the alternative hypothesis that and equivalently that causes y cannot be rejected. On the other hand, if the null hypothesis that cannot be rejected, then equivalently the hypothesis of no causal effect of on y cannot be rejected. Here the notion of causality is one of contributory causality: If the true value , then a change in will result in a change in y unless some other causative variable(s), either included in the regression or implicit in the error term, change in such a way as to exactly offset its effect; thus a change in is not sufficient to change y. Likewise, a change in is not necessary to change y, because a change in y could be caused by something implicit in the error term (or by some other causative explanatory variable included in the model).
Regression analysis controls for other relevant variables by including them as regressors (explanatory variables). This helps to avoid mistaken inference of causality due to the presence of a third, underlying, variable that influences both the potentially causative variable and the potentially caused variable: its effect on the potentially caused variable is captured by directly including it in the regression, so that effect will not be picked up as a spurious effect of the potentially causative variable of interest. In addition, the use of multivariate regression helps to avoid wrongly inferring that an indirect effect of, say x1 (e.g., x1 → x2 → y) is a direct effect (x1 → y).
Just as an experimenter must be careful to employ an experimental design that controls for every confounding factor, so also must the user of multiple regression be careful to control for all confounding factors by including them among the regressors. If a confounding factor is omitted from the regression, its effect is captured in the error term by default, and if the resulting error term is correlated with one (or more) of the included regressors, then the estimated regression may be biased or inconsistent (see omitted variable bias).
In addition to regression analysis, the data can be examined to determine if Granger causality exists. The presence of Granger causality indicates both that x precedes y, and that x contains unique information about y.
Other relationships
[edit]
There are several other relationships defined in statistical analysis as follows.
Direct relationship
Mediating relationship
Moderating relationship
See also
[edit]
Causality
Correlation does not imply causation
Illusory correlation
Model specification
Omitted-variable bias
Post hoc fallacy
Statistical model validation
One in ten rule
Literature
[edit]
David A. Freedman (1983) A Note on Screening Regression Equations, The American Statistician, 37:2, 152-155, DOI: 10.1080/00031305.1983.10482729
Footnotes
[edit]
^ Burns, William C., "Spurious Correlations", 1997.
^ Pearl, Judea. "UCLA 81st Faculty Research Lecture Series". singapore.cs.ucla.edu. Retrieved 2019-11-10.
^ Yule, G. Udny (1926-01-01). "Why do we Sometimes get Nonsense-Correlations between Time-Series? A Study in Sampling and the Nature of Time-Series". Journal of the Royal Statistical Society. 89 (1): 1–63. doi:10.2307/2341482. JSTOR 2341482. S2CID 126346450.
^ Granger, Clive W. J.; Ghysels, Eric; Swanson, Norman R.; Watson, Mark W. (2001). Essays in Econometrics: Collected Papers of Clive W. J. Granger. Cambridge University Press. ISBN 978-0521796491.
^ Sapsford, Roger; Jupp, Victor, eds. (2006). Data Collection and Analysis. Sage. ISBN 0-7619-4362-5.
^ Hofheimer, Bill (October 30, 2012). "'Redskins Rule': MNF's Hirdt on intersection of football & politics". ESPN. Retrieved October 16, 2016.
^ Manker, Rob (November 7, 2012). "Redskins Rule: Barack Obama's victory over Mitt Romney tackles presidential predictor for its first loss". Chicago Tribune. Retrieved November 8, 2012.
^ Pohl, Robert S. (2013). Urban Legends & Historic Lore of Washington. The History Press. pp. 78–80. ISBN 978-1625846648.[permanent dead link]
^ Don Peppers. "Big Data. Super Bowl. Small Minds". Retrieved December 31, 2015.
References
[edit]
Gumbel, E.J. (1926), "Spurious correlation and its significance to physiology", Journal of the American Statistical Association, 21 (154): 179–194, doi:10.1080/01621459.1926.10502169
Banerjee, A.; Dolado, J.; Galbraith, J. W.; Hendry, D. F. (1993). Co-Integration, Error-Correction, and the Econometric Analysis of Non-Stationary Data. Oxford University Press. pp. 70–81. ISBN 0-19-828810-7.
Pearl, Judea (2000). Causality: Models, Reasoning and Inference. Cambridge University Press. ISBN 0521773628.
External links
[edit]
a website listing examples of spurious correlations
| v t e Common fallacies (list) | |
--- |
| Formal | | | | --- | | In propositional logic | Affirming a disjunct Affirming the consequent Denying the antecedent Argument from fallacy Masked man Mathematical fallacy | | In quantificational logic | Existential Illicit conversion Proof by example Quantifier shift | | Syllogistic fallacy | Affirmative conclusion from a negative premise Negative conclusion from affirmative premises Exclusive premises Existential Necessity Four terms Illicit major Illicit minor Undistributed middle | |
| Informal | | | | --- | | Equivocation | Equivocation False equivalence False attribution Moral equivalence Quoting out of context Loki's Wager No true Scotsman Reification + Map–territory relation | | Question-begging | Circular reasoning / Begging the question Loaded language + Leading question Compound question / Loaded question / Complex question No true Scotsman | | Correlative-based | False dilemma + Perfect solution Denying the correlative Suppressed correlative | | Illicit transference | Composition Division Ecological | | Secundum quid | Accident Converse accident | | Faulty generalization | Anecdotal evidence Sampling bias + Cherry picking + McNamara Base rate / Conjunction Double counting False analogy Slothful induction Overwhelming exception | | Ambiguity | Accent False precision Moving the goalposts Quoting out of context Slippery slope Sorites paradox Syntactic ambiguity | | Questionable cause | Animistic + Furtive Correlation implies causation + Cum hoc + Post hoc Gambler's + Inverse Regression Single cause Slippery slope Texas sharpshooter | | Appeals | Law/Legality Stone / Proof by assertion | | | --- | | Consequences | Argumentum ad baculum Wishful thinking | | Emotion | Children Fear Flattery Novelty Pity Ridicule In-group favoritism Invented here / Not invented here Island mentality Loyalty Parade of horribles Spite Stirring symbols Wisdom of repugnance | | | Genetic fallacy | | | | --- | | Ad hominem | Appeal to motive Association + Reductio ad Hitlerum + Reductio ad Stalinum Bulverism Poisoning the well Tone Tu quoque Whataboutism | | Authority + Accomplishment + Ipse dixit + Poverty / Wealth Etymology Nature Tradition / Novelty + Chronological snobbery | | | | Other fallacies of relevance | | | | --- | | Arguments | Ad nauseam + Sealioning Argument from anecdote Argument from silence Argument to moderation Argumentum ad populum | | Cliché The Four Great Errors I'm entitled to my opinion Ignoratio elenchi Invincible ignorance Moralistic / Naturalistic Motte-and-bailey fallacy Psychologist's fallacy Rationalization Red herring + Two wrongs make a right Special pleading Straw man | | | |
| | |
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Multi-Electron Atoms
10: Multi-electron Atoms
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Sat, 03 May 2025 22:38:46 GMT
Penetration and Shielding
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Delmar Larsen
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[ "article:topic", "electron shielding", "penetration", "showtoc:no", "license:ccbyncsa", "licenseversion:30", "author@Sidra Ayub", "author@Alan Chu" ]
[ "article:topic", "electron shielding", "penetration", "showtoc:no", "license:ccbyncsa", "licenseversion:30", "author@Sidra Ayub", "author@Alan Chu" ]
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Contents
1. Home
2. Bookshelves
3. Physical & Theoretical Chemistry
4. Supplemental Modules (Physical and Theoretical Chemistry)
5. Quantum Mechanics
6. 10: Multi-electron Atoms
7. Multi-Electron Atoms
8. Penetration and Shielding
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Supplemental Modules (Physical and Theoretical Chemistry)
Fundamentals
Atomic Theory
Physical Properties of Matter
Acids and Bases
Kinetics
Equilibria
Thermodynamics
Statistical Mechanics
Quantum Mechanics
1: Waves and Particles
2: Fundamentals of Quantum Mechanics
3: The Tools of Quantum Mechanics
4: The 7 Postulates of Quantum Mechanics
5: Particle in Boxes
6: One Dimensional Harmonic Oscillator
7: Angular Momentum
9: The Hydrogen Atom
10: Multi-electron Atoms
11: Angular Momenta Coupling
11: Molecules
12: Electron Scattering
12: Stationary Perturbation Theory
13: Fine and Hyperfine Structure
14: Systems of Identical Particles
15: Time-dependent Quantum Dynamics
16: Molecular Spectroscopy
17: Quantum Calculations
Chapter 14. Nuclear Magnetic Resonance
Zeeman Effect
Gen Chem Quantum Theory
Chemical Bonding
Electronic Structure of Atoms and Molecules
Spectroscopy
Nuclear Chemistry
Group Theory
Penetration and Shielding
Last updated May 3, 2025
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Wave Function of Multi-electron Atoms
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Page ID 1704
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Introduction
2. Orbital Penetration
3. Shielding
1. Example 1: Fluorine, Neon, and Sodium
1. Solution
Radial Distribution Graphs
Periodic Trends Due to Penetration and Shielding
Questions
Solutions
References
Penetration and shielding are two underlying principles in determining the physical and chemical properties of elements. We can predict basic properties of elements by using shielding and penetration characteristics to assess basic trends.
Introduction
Electrons are negatively charged and are pulled pretty close to each other by their attraction to the positive charge of a nucleus. The electrons are attracted to the nucleus at the same time as electrons repel each other. The balance between attractive and repulsive forces results in shielding. The orbital (n) and subshell (m l) define how close an electron can approach the nucleus. The ability of an electron to get close to the nucleus is penetration.
Coulomb's Law (an analogy with classical physics) can be used to describe the attraction and repulsion between atomic particles:
(1)F=kq 1q 1 r 2
The force that an electron feels is dependent on the distance from the nearest charge (i.e., an electron, usually with bigger atoms and on the outer shells) and the amount of charge. More distance between the charges will result in less force, and more charge will have more force of attraction or repulsion.
In the simplest case, every electron in an atom would feel the same amount of "pull" from the nucleus. For example, in Li, all three electrons might "feel" the +3 charge from the nucleus. However, this is not the case when observing atomic behavior. When considering the core electrons (or the electrons closest to the nucleus), the nuclear charge "felt" by the electrons (Effective Nuclear Charge(Z eff)) is close to the actual nuclear charge. As you proceed from the core electrons to the outer valence electrons, Z eff falls significantly. This is because of shielding, or simply the electrons closest to the nucleus decrease the amount of nuclear charge affecting the outer electrons. Shielding is caused by the combination of partial neutralization of nuclear charge by core electrons, and by electron-electron repulsion.
The amount of charge felt by an electron depends on its distance from the nucleus. The closer an electron comes to the nucleus, or the more it penetrates, the stronger its attraction to the nucleus. C ore electrons penetrate more and feel more of the nucleus than the other electrons.
(2)F electron−nucleus=kZe 2 r 2
with
Z is the charge of the nucleus (i.e., number of protons)
e is the charge of an electron or proton
r is the radius, or distance between the proton and the electron
Penetration and shielding result in an Effective force(F eff) that holds the outer electrons to the atom, akin to Equation 2, but with Z eff substituted for Z:
(3)F eff=kZ effe 2 r 2
Orbital Penetration
Penetration describes the proximity to which an electron can approach to the nucleus. In a multi-electron system, electron penetration is defined by an electron's relative electron density (probability density) near the nucleus of an atom. Electrons in different orbitals have different wavefunctions and therefore different radial distributions and probabilities (defined by quantum numbers n and m l around the nucleus). In other words, penetration depends on the shell (n) and subshell (m l). For example, we see that since a 2s electron has more electron density near the nucleus than a 2p electron, it is penetrating the nucleus of the atom more than the 2p electron. The penetration power of an electron, in a multi-electron atom, is dependent on the values of both the shell and subshell.
Within the same shell value (n), the penetrating power of an electron follows this trend in subshells (m l):
s>p>d>f
And for different values of shell (n) and subshell (l), penetrating power of an electron follows this trend:
1s>2s>2p>3s>3p>4s>3d>4p>5s>4d>5p>6s>4f....
and the energy of an electron for each shell and subshell goes as follows...
1s<2s<2p<3s<3p<4s<3d<4p....
The electron probability density for s-orbitals is highest in the center of the orbital, or at the nucleus. If we imagine a dartboard that represents the circular shape of the s-orbital and if the darts landed in correlation to the probability to where and electron would be found, the greatest dart density would be at the 50 points region but most of the darts would be at the 30 point region. When considering the 1s-orbital, the spherical shell of 53 pm is represented by the 30 point ring.
Electrons which experience greater penetration experience stronger attraction to the nucleus, less shielding, and therefore experience a larger Effective Nuclear Charge(Z eff), but shield other electrons more effectively.
Shielding
An atom (assuming its atomic number is greater than 2) has core electrons that are extremely attracted to the nucleus in the middle of the atom. However the number of protons in the nucleus are never equal to the number of core electrons (relatively) adjacent to the nucleus. The number of protons increase by one across the periodic table, but the number of core electrons change by periods. The first period has no core electrons, the second has 2, the third has 10, and etc. This number is not equal to the number of protons. So that means that the core electrons feel a stronger pull towards the nucleus than any other electron within the system. The valence electrons are farther out from the nucleus, so they experience a smaller force of attraction.
Shielding refers to the core electrons repelling the outer rings and thus lowering the 1:1 ratio. Hence, the nucleus has "less grip" on the outer electrons and are shielded from them. Electrons that have greater penetration can get closer to the nucleus and effectively block out the charge from electrons that have less proximity. For example, Z eff is calculated by subtracting the magnitude of shielding from the total nuclear charge. The value of Z eff will provide information on how much of a charge an electron actually experiences.
Figure 1: This image shows how inner electrons can shield outer electrons from the nuclear charge. (CC BY-SA 3.0; from Wikipedia).
Because the order of electron penetration from greatest to least is s, p, d, f; the order of the amount of shielding done is also in the order s, p, d, f.
Since the 2s electron has more density near the nucleus of an atom than a 2p electron, it is said to shield the 2p electron from the full effective charge of the nucleus. Therefore the 2p electron feels a lesser effect of the positively charged nucleus of the atom due to the shielding ability of the electrons closer to the nucleus than itself, (i.e. 2s electron).
These electrons that are shielded from the full charge of the nucleus are said to experience an effective nuclear charge ( Z eff)of the nucleus, which is some degree less than the full nuclear charge an electron would feel in a hydrogen atom or hydrogenlike ions. The effective nuclear charge of an atom is given by the equation:
(4)Z eff=Z−S
where.
Z is the atomic number (number of protons in nucleus) and
S is the shielding constant
We can see from this equation that the effective nuclear charge of an atom increases as the number of protons in an atom increases. Therefore as we go from left to right on the periodic table the effective nuclear charge of an atom increases in strength and holds the outer electrons closer and tighter to the nucleus. This phenomena can explain the decrease in atomic radii we see as we go across the periodic table as electrons are held closer to the nucleus due to increase in number of protons and increase in effective nuclear charge.
Example 1: Fluorine, Neon, and Sodium
What is the effective attraction Z eff experienced by the valence electrons in the three isoelectronic species: the fluorine anion, the neutral neon atom, and sodium cation?
Solution
Each species has 10 electrons, and the number of nonvalence electrons is 2 (10 total electrons - 8 valence) but the effective nuclear charge varies because each has a different atomic numberA.
The charge Z of the nucleus of a fluorine atom is 9, but the valence electrons are screened appreciably by the core electrons (four electrons from the 1s and 2s orbitals) and partially by the 7 electrons in the 2p orbitals.
Diagram of a fluorine atom showing the extent of effective nuclear charge. (CC BY-SA- 3.0; Wikipedia).
Z eff(F−)=9−2=7+
Z eff(Ne)=10−2=8+
Z eff(Na+)=11−2=9+
So the sodium cation has the greatest effective nuclear charge, and thus the smallest radius.
Radial Distribution Graphs
A radial distribution function graph describes the distribution of orbitals with the effects of shielding (Figure 2). The small peak of the 2s orbital shows that the electrons in the 2s orbital are closest to the nucleus. Therefore, it is the electrons in the 2p orbital of Be that are being shielded from the nucleus, by the electrons in the 2s orbital.
Figure 2: Orbital Penetration. A comparison of the radial probability distribution of the 2 s and 2 p orbitals for various states of the hydrogen atom shows that the 2 s orbital penetrates inside the 1 s orbital more than the 2 p orbital does. Consequently, when an electron is in the small inner lobe of the 2 s orbital, it experiences a relatively large value of Z eff, which causes the energy of the 2 s orbital to be lower than the energy of the 2 p orbital.
The following is the radial distribution of the 1s and 2s orbitals. Notice the 1s orbital is shifted to the right, while the 2s orbital has a node.
Periodic Trends Due to Penetration and Shielding
Effective Nuclear Charge (Z eff): The effective nuclear charge increases from left to right and increases from top to bottom on the periodic table.
Atomic Radius: The atomic radius decreases from left to right, and increases from top to bottom.
Ionization Energies: The ionization energies increase from left to right, and decrease from top to bottom.
Electronegativity: The electronegativity of the elements is highest near flourine. In general, it increases from left to right and decreases from top to bottom.
Questions
Which orbital is more effective in shielding? 1s or 2p?
True/False: The greater the penetration of an orbital, the greater the shielding capability of that orbital.
Find the Z eff of
Mg
C
F
Ca
Which of these have the smallest electron affinity? B, C, N, O, or F.
Which atom has a stronger effective nuclear charge and why? (assuming S is the same in both cases) Li, or N
Why does the Hydrogen electron experiences the full charge of the nucleus without any shielding?
Which atom has a smaller radii? Be or F?
Which electron has higher energy level? 2s or 2p? and why?
Why do the orbitals of a hydrogen atom increase energy as follows: 1s<2s=2p<3s=3p=3d<4s=4p=4d<....
Which electrons shields better in an atom? 2s or 2p? 3p or 3d?
Why can we relate classical physics to quantum mechanics when it comes to subatomic activity?
What is penetration?
Solutions
1s
T
a. 12-2 =10
b. 6-4=2
c. 9-7=2
d. 20-2=18
-
nitrogen atom has a stronger effective nuclear charge than lithium due to its greater number of protons in the nucleus holding the electrons tighter.
Hydrogen atom has only one electron total, therefore there are no other, lower energy (more penetrating), electrons available to help shield this electron from the nucleus.
Fluorine has a smaller radii than Beryllium due to its greater number of protons providing a greater effective nuclear charge on the outer electrons and therefore pulling them in tighter and providing a smaller atomic radii.
2p has higher energy level because the negatively charged electron experiences less of an effective nuclear charge than the 2s electron.
because a Hydrogen atom has only one electron, that experiences no shielding from other electrons and therefore its energy level only depends on its distance away from the nucleus, which is dependent on it value of (n).
2s shields the atom better than 2p because the s orbitals is much closer and surrounds the nucleus more than the p orbitals, which extend farther out. 3p shields better than 3d, because p orbitals are closer to the nucleus than the 3d orbitals.
Classical physics and quantum mechanics both can deal with subatomic activity such as electron interactions, orbital location, size, and shape, and distances to find forces of attractions.
Penetration is how well the outer electrons are shielded from the nucleus by the core electrons. The outer electrons therefore experience less of an attraction to the nucleus.
References
Petrucci, Ralph H., William S. Harwood, F. Geoffrey Herring, and Jeffry D. Madura. General Chemistry: Principles and Modern Applications, Ninth Edition. Pearson Education Inc. Upper Saddle River, New Jersey: 2007.
Raymond Chang. Physical Chemistry for Biological Sciences. Sausalito, California: University Science Books, 2005
R. S. Mulliken, Electronic Structures of Molecules and Valence. II General Considerations, Physical Review, vol. 41, pp. 49-71 (1932)
Anastopoulos, Charis (2008). Particle Or Wave: The Evolution of the Concept of Matter in Modern Physics. Princeton University Press. pp. 236–237. ISBN0691135126.
Penetration and Shielding is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Sidra Ayub & Alan Chu.
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13995 | https://reference.wolfram.com/language/ref/Laplacian.html | Wolfram Language & System Documentation Center
Laplacian
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Vector Analysis
Laplacian[f,{x1,…,xn}]
gives the Laplacian .
Laplacian[f,{x1,…,xn},chart]
gives the Laplacian in the given coordinates chart.
Details
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Laplacian ✖ Laplacian
Updated in 14.1
Laplacian[f,{x1,…,xn}]
✖
Laplacian[f,{x1,…,xn}]
gives the Laplacian .
Laplacian[f,{x1,…,xn},chart]
✖
Laplacian[f,{x1,…,xn},chart]
gives the Laplacian in the given coordinates chart.
Details
Laplacian is also known as Laplace–Beltrami operator. When applied to vector fields, it is also known as vector Laplacian.
Laplacian[f,x] can be input as f. The character ∇ can be typed as del or (/language/ref/character/Del.html). The list of variables x and the 2 are entered as a subscript and superscript, respectively.
An empty template can be entered as del2, and moves the cursor from the subscript to the main body.
All quantities that do not explicitly depend on the variables given are taken to have zero partial derivative.
Laplacian[f,{x1,x2,…}] yields a result with the same dimensions as f.
In Laplacian[f,{x1,…,xn},chart], if f is an array, it must have dimensions {n,…,n}. The components of f are interpreted as being in the orthonormal basis associated to chart.
For coordinate charts on Euclidean space, Laplacian[f,{x1,…,xn},chart] can be computed by transforming f to Cartesian coordinates, computing the ordinary Laplacian and transforming back to chart. »
A property of Laplacian is that if chart is defined with metric g, expressed in the orthonormal basis, then Laplacian[g,{x1,…,xn},chart] gives zero. »
Coordinate charts in the third argument of Laplacian can be specified as triples {coordsys,metric,dim} in the same way as in the first argument of CoordinateChartData. The short form in which dim is omitted may be used.
Laplacian[f,VectorSymbol[…]] computes the Laplacian with respect to the vector symbol. »
Laplacian works with SparseArray and structured array objects.
Examples
open all
close all
Basic Examples (4)Summary of the most common use cases
The Laplacian in three-dimensional Cartesian coordinates:
In:=1
✖
Out=1
The Laplacian in three-dimensional cylindrical coordinates:
In:=1
✖
Out=1
The Laplacian in two-dimensional polar coordinates:
In:=1
✖
Out=1
Use del to enter ∇, for the list of subscripted variables, and to enter the 2:
In:=1
✖
Out=1
Use del2 to enter the template , fill in the variables, press , and fill in the function:
In:=2
✖
Out=2
Scope (6)Survey of the scope of standard use cases
Laplacian applies to arrays of arbitrary rank:
In:=1
✖
Out=1
In:=2
✖
Out=2
In a curvilinear coordinate system, a vector with constant components may have a nonzero Laplacian:
In:=1
✖
Out=1
A Laplacian specifying metric, coordinate system, and parameters:
In:=1
✖
Out=1
Laplacian works on curved spaces:
In:=1
✖
Out=1
The Laplacian of the coordinate vector is SymbolicZerosArray[{n}]:
In:=1
✖
In:=2
✖
Out=2
The Laplacian of the squared norm is expressed in terms of SymbolicIdentityArray[{n}]:
In:=3
✖
Out=3
Use TensorExpand to simplify to the expected result, namely twice the dimension:
In:=4
✖
Out=4
The Laplacian of the squared norm in n dimensions:
In:=1
✖
Out=1
Activate the sum to get the simple result:
In:=2
✖
Out=2
Applications (3)Sample problems that can be solved with this function
Poisson's equation in spherical coordinates:
In:=1
✖
Out=1
Solve for a radially symmetric charge distribution :
In:=2
✖
Out=2
The Laplacian on the unit sphere:
In:=1
✖
Out=1
The spherical harmonics are eigenfunctions of this operator with eigenvalue :
In:=2
✖
Out=2
In:=3
✖
Out=3
The generalization of the Coulomb potential—the electric potential of a point charge—to n dimensions is:
In:=1
✖
Out=1
Since the charge density is only nonzero at the origin, the Laplacian must be equal to zero everywhere else:
In:=2
✖
Out=2
Simplify the result:
In:=3
✖
Out=3
Activating the result in specific dimensions and combining denominators shows the result is zero:
In:=4
✖
Out=4
This result can also be obtained in each dimension using spherical coordinates:
In:=5
✖
Out=5
Properties & Relations (8)Properties of the function, and connections to other functions
Laplacian preserves the shape of an array:
In:=1
✖
Out=1
In:=2
✖
Out=2
The Laplacian is equal to the divergence of the gradient:
In:=1
✖
Out=1
In:=2
✖
Out=2
Since Grad uses an orthonormal basis, the Laplacian of a scalar equals the trace of the double gradient:
In:=1
✖
Out=1
In:=2
✖
Out=2
For higher-rank arrays, this is the contraction of the last two indices of the double gradient:
In:=3
✖
Out=3
In:=4
✖
Out=4
In:=5
✖
Out=5
Compute Laplacian in a Euclidean coordinate chart c by transforming to and then back from Cartesian coordinates:
In:=1
✖
Out=1
In:=2
✖
Out=2
In:=3
✖
Out=3
The result is the same as directly computing Laplacian[f,{x1,…,xn},c]:
In:=4
✖
Out=4
The Laplacian of an array equals the Laplacian of its components only in Cartesian coordinates:
In:=1
✖
Out=1
In:=2
✖
Out=2
If chart is defined with metric g, expressed in the orthonormal basis, Laplacian[g,{x1,…,xn},chart] is zero:
In:=1
✖
Out=1
For a vector field in three-dimensional flat space, the Laplacian is equal to :
In:=1
✖
Out=1
In:=2
✖
Out=2
In:=3
✖
Out=3
In:=4
✖
Out=4
In a flat space of dimension , the Laplacian of a vector field equals . For :
In:=5
✖
In:=6
✖
In:=7
✖
In:=8
✖
Out=8
Laplacian preserves the symmetry structure of SymmetrizedArray objects:
In:=1
✖
Out=1
The Laplacian has the same symmetry as the input:
In:=2
✖
Out=2
Interactive Examples (1)Examples with interactive outputs
View expressions for the Laplacian of a scalar function in different coordinate systems:
In:=1
✖
In:=2
✖
Out=2
Wolfram Research (2012), Laplacian, Wolfram Language function, (updated 2024).
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Wolfram Research (2012), Laplacian, Wolfram Language function, (updated 2024).
Text
Wolfram Research (2012), Laplacian, Wolfram Language function, (updated 2024).
✖
Wolfram Research (2012), Laplacian, Wolfram Language function, (updated 2024).
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Wolfram Language. 2012. "Laplacian." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2024.
✖
Wolfram Language. 2012. "Laplacian." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2024.
APA
Wolfram Language. (2012). Laplacian. Wolfram Language & System Documentation Center. Retrieved from
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Wolfram Language. (2012). Laplacian. Wolfram Language & System Documentation Center. Retrieved from
BibTeX
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@misc{reference.wolfram_2025_laplacian, author="Wolfram Research", title="{Laplacian}", year="2024", howpublished="\url{ note=[Accessed: 25-August-2025]}
BibLaTeX
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13996 | https://www.chegg.com/homework-help/questions-and-answers/density-mercury-20-degrees-celsius-13545-kg-m-3-50-degrees-celsius-13472-kg-m-3-use-linear-q15618239 | Your solution’s ready to go!
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Question: The density of mercury at 20 degrees Celsius is 13545 kg/m^3, while 50 degrees Celsius it is 13472 kg/m^3. Use linear interpolation to find the density at normal body temperature (37 degrees Celsius).
The density of mercury at 20 degrees Celsius is 13545 kg/m^3,
while 50 degrees Celsius it is 13472 kg/m^3. Use linear
interpolation to find the density at normal body temperature (37
degrees Celsius).
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Write down the formula for linear interpolation, which is , where is the density at 37 degrees Celsius, and are the densities at 20 and 50 degrees Celsius, respectively, and and are 20 and 50 degrees Celsius, respectively.
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What does loss of roots refer to in the context of an equation
The misinterpretation of the role of the unknown (x) that appears to the left and right of the equal sign often leads to root loss, which is a common error when solving quadratic equations. This error arises due to a need for more understanding regarding the proper interpretation of the unknown values within the equation. It is crucial to approach the problem with a clear understanding of the roles and relationships of the variables involved to avoid such mistakes and achieve accurate results.
For example, try to solve the equation:
[x(2x-5) = x]
Solving equations can be tricky, and it’s common to make mistakes, especially if you’re new to it. It’s important to remember that simplifying both sides of the equation by removing x might seem like the right approach, but it can lead to confusion and errors. In this case, someone could be tempted to do the following simplification:
[\cancel{x}(2x-5) = \cancel{x}]
This simplification leads to obtaining a first-degree equation in (x) that has a unique solution.
[2x-5=1] [x= \frac{6}{2} = 3]
This error generates an incorrect solution and leads to the loss of a second solution.
In more general terms, the variable (x) should always be retained and never cancelled when solving equations to avoid the loss of solutions.
Correct method
It is essential to follow the correct procedure when dealing with second-degree equations. First, the equation should be returned to its standard form, simplifying it to the form (ax^2+bx+c).
From: [x(2x-5) = x]
We obtain:
\begin{align} 2x^2-5x-x &= 0 \[0.5em] 2x^2-6x &= 0 \[0.5em] 2x(x-3) &= 0 \end{align}
The equation is reduced to an incomplete quadratic equation of the second degree with the coefficient (c) being null. The equations admits two distinct solutions, (x_1 = 0) and (x_2 = 3).
The solution to the equation is: [x_1 = 0 \quad x_2 = 3]
The correct procedure demonstrates that there are two solutions, and neither is equal to the value found in the case of the wrong procedure.
To safeguard against losing any solutions when solving equations, it’s wise to adopt these key strategies:
Rearrange the equation so that one side equals zero, and then factor the resulting expression. This technique allows you to identify all possible solutions by applying the zero-product principle: if the product of factors is zero, then at least one of those factors must be zero.
Steer clear of dividing both sides of the equation by any term that includes the variable. If such a division seems unavoidable, be sure to examine the scenario where that variable expression equals zero as a separate case.
When dealing with even roots (like square roots), always remember to consider both the positive and negative outcomes. Neglecting one can lead to missing valid solutions.
Crucially, always double-check your potential solutions by plugging them back into the original equation. This step is particularly important after performing operations that could potentially add extraneous solutions or remove valid ones.
Continuing on this topic
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879Logarithmic Equations
598Trigonometric Equations: Part 1
368Homogeneous Trigonometric Equations
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13998 | https://www.collinsdictionary.com/us/dictionary/english/partial | PARTIAL definition in American English | Collins English Dictionary
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Definition of 'partial'
COBUILD frequency band
partial
(p ɑ rʃ ə l)
1.adjective
You use partial to refer to something that is not complete or whole.
He managed to reach a partial agreement with both republics.
...a partial ban on the use of cars in the city.
2.adjective [v-link ADJ to n/-ing]
If you are partial to something, you like it.
He is also partial to golf, music and chocolate.
Mollie confesses she is rather partial to pink.
3.adjective [v-link ADJ]
Someone who is partial supports a particular person or thing, for example, in a competition or dispute, instead of being completely fair.
I might be accused of being partial.
More Synonyms of partial
Collins COBUILD Advanced Learner’s Dictionary. Copyright © HarperCollins Publishers
American English pronunciation
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COBUILD frequency band
partial in American English
(ˈpɑrʃəl)
adjective Origin: ME parcial< MFr partial< LL partialis< L pars,part 1
1.
favoring one person, faction, etc. more than another; biased; prejudiced
2.
of, being, or affecting only a part; not complete or total
noun
3.partial tone
4.
a partial artificialdenture
Idioms:
partial to
Webster’s New World College Dictionary, 5th Digital Edition. Copyright © 2025 HarperCollins Publishers.
Derived forms
ˈpartially
adverb
COBUILD frequency band
partial in American English
(ˈpɑːrʃəl)
adjective
1.
being such in part only; not total or general; incomplete
partial blindness
a partial payment of a debt
2.
biased or prejudiced in favor of a person, group, side, etc., over another, as in a controversy
a partial witness
3.
pertaining to or affecting a part
4.
being a part; component; constituent
5.Botany
secondary or subordinate
a partial umbel
See partial to
noun
7.Bridge
a contract to make less than the number of tricks required for game; part-score
Acoustics&Music Seepartial tone
SYNONYMS 1. unfinished, imperfect, limited. 2. one-sided, unfair, unjust.ANTONYMS 1, 3. complete. 2. unbiased, fair.
Most material © 2005, 1997, 1991 by Penguin Random House LLC. Modified entries © 2019 by Penguin Random House LLC and HarperCollins Publishers Ltd
Derived forms
partially
adverb
partialness
noun
Word origin
[1375–1425; late ME parcial biased, particular ‹ MF ‹ LL partiālis pertaining to a part, equiv. to L parti- (s. of pars) part + -ālis-al 1]
COBUILD frequency band
partial in British English
(ˈpɑːʃəl)
adjective
1.
relating to only a part; not general or complete
a partial explanation
2.
biased
a partial judge
3.(postpositive; foll by to)
having a particular liking (for)
botany
a.
constituting part of a larger structure
a partial umbel
b.
used for only part of the life cycle of a plant
a partial habitat
c.
(of a parasite) not exclusively parasitic
mathematics
designating or relating to an operation in which only one of a set of independent variables is considered at a time
noun
Also called: partial tone music, acoustics
any of the component tones of a single musical sound, including both those that belong to the harmonic series of the sound and those that do not
mathematics
a partial derivative
▶USAGE Partially and partly are to some extent interchangeable, but partly should be used when referring to a part or parts of something: the building is partly (not partially) of stone, while partially is preferred for the meaning to some extent: his mother is partially (not partly) sighted
Collins English Dictionary. Copyright © HarperCollins Publishers
Derived forms
partially (ˈpartially)
adverb
partialness (ˈpartialness)
noun
Word origin
C15: from Old French parcial, from Late Latin partiālis incomplete, from Latin pars part
COBUILD frequency band
partial in Insurance
(p ɑ rʃ ə l)
adjective
(Insurance: Underwriting)
A partial loss is a situation in which property covered by insurance is damaged but not completely destroyed, so that the insurer does not have to pay the full amount.
A partial loss under an insurance policy does not completely destroy or render worthless the insured property.
A partial loss is a loss involving less than all of the values insured or calling on the policy to pay less than its maximum amounts.
A partial loss is a situation in which property covered by insurance is damaged but not completely destroyed, so that the insurer does not have to pay the full amount.
Collins COBUILD Key Words for Insurance. Copyright © HarperCollins Publishers
Examples of 'partial' in a sentence
partial
These examples have been automatically selected and may contain sensitive content that does not reflect the opinions or policies of Collins, or its parent company HarperCollins.
We welcome feedback: report an example sentence to the Collins team. Read more…
This is a partial fix for the first error.
The Guardian (2020)
Officials are counting on a partial reopening of stadiums.
The Guardian (2020)
The country is currently in partial economic and social lockdown as it battles a second wave of infections.
The Guardian (2020)
This surgeseems to be increasing due to a partial collapse in other areas of critical health care.
The Guardian (2020)
Lyme disease can cause a range of debilitatingsymptoms from fatigue and joint pain to heart problems and partial paralysis.
The Guardian (2020)
The result is death or partial paralysis.
Holford, Patrick The Family Nutrition Workbook (1988)
Options could include a partial flotation.
Times, Sunday Times (2008)
The partial recovery followed a 15 per centslump on Wednesday.
Times, Sunday Times (2011)
These microscopic partial views sharpenobservation for the most minute details, having been dissectedalmost beyond recognition.
Royal Society Notes and Records (2022)
Last year, a pair of biologists suggested a partial answer.
New Scientist (Earth) (2022)
Related word partners
partial
[x]
partial recovery
partial reimbursement
partial remission
partial removal of
partial restoration
partial return
partial shade
partial shutdown
partial solution
partial success
partial tear
partial truth
partial victory
partial withdrawal
Show more...
Trends of
partial
View usage over:
Source: Google Books Ngram Viewer
In other languages
partial
British English: partial /ˈpɑːʃəl/ ADJECTIVE
You use partial to refer to something that is true or exists to some extent, but is not complete or total.
The event was only a partial success.
American English: partial /ˈpɑrʃəl/
Arabic: جُزْئِيٌ
Brazilian Portuguese: parcial
Chinese: 部分的
Croatian: djelomičan
Czech: částečný
Danish: delvis
Dutch: gedeeltelijk
European Spanish: parcial
Finnish: osittainen
French: partiel
German: teilweise
Greek: μερικός
Italian: parziale
Japanese: 部分的な
Korean: 부분적인
Norwegian: delvis
Polish: częściowy
European Portuguese: parcial
Romanian: parțial
Russian: частичный
Spanish: parcial
Swedish: partiell
Thai: ซึ่งเป็นบางส่วน
Turkish: kısmi
Ukrainian: частковий
Vietnamese: một phần
Translate your text for free
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partial
Parti Québécois
parti-colored
parti-coloured
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partial ban
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All ENGLISH words that begin with 'P'
Related terms of
partial
partial ban
partial sum
partial to
partial list
partial score
View more related words
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