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a unit of area measurement equal to a square measuring one yard on each side; 0.8361 square meters. yd 2 , sq. yd.
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15801 | https://pmc.ncbi.nlm.nih.gov/articles/PMC7151904/ | Activation of CD8 T Lymphocytes during Viral Infections - PMC
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Encyclopedia of Immunobiology
. 2016 May 9:286–290. doi: 10.1016/B978-0-12-374279-7.14009-3
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Activation of CD8 T Lymphocytes during Viral Infections
Luis J Sigal
Luis J Sigal
1 Thomas Jefferson University, Philadelphia, PA, USA
2 University of Toronto and Trinity College, Toronto, ON, Canada
Find articles by Luis J Sigal
1,2
Editor: Michael JH Ratcliffe 1,2
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1 Thomas Jefferson University, Philadelphia, PA, USA
2 University of Toronto and Trinity College, Toronto, ON, Canada
Issue date 2016.
Copyright © 2016 Elsevier Ltd. All rights reserved.
Since January 2020 Elsevier has created a COVID-19 resource centre with free information in English and Mandarin on the novel coronavirus COVID-19. The COVID-19 resource centre is hosted on Elsevier Connect, the company's public news and information website. Elsevier hereby grants permission to make all its COVID-19-related research that is available on the COVID-19 resource centre - including this research content - immediately available in PubMed Central and other publicly funded repositories, such as the WHO COVID database with rights for unrestricted research re-use and analyses in any form or by any means with acknowledgement of the original source. These permissions are granted for free by Elsevier for as long as the COVID-19 resource centre remains active.
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PMCID: PMC7151904
Abstract
CD8 T lymphocytes are a major cell population of the adaptive immune system. A fundamental characteristic of the CD8 T lymphocyte pool is that it is composed of millions of clones; each with a unique T cell receptor capable of recognizing a limited number of peptides displayed at the cell surface bound to the grooves of major histocompatibility complex class I (MHC I) molecules. Naïve CD8 T lymphocytes are normally resting and circulate between the blood and secondary lymphoid organs in search of their cognate peptide–MHC complexes. During viral infections, bone marrow–derived professional antigen-presenting cells (pAPCs) in secondary lymphoid organs display viral peptides on their MHC I molecules. Specific CD8 T lymphocytes that recognize these peptide–MHC adducts become activated (primed), proliferate extensively, and develop into effectors capable of killing infected cells, identified by the presence at their surface of the pertinent viral peptide–MHC complexes. This article describes how the process of priming naïve CD8 T lymphocytes occurs.
Keywords: Antigen presentation, CD8 T lymphocyte priming, CD8 T lymphocytes, Costimulation, MHC class I cross-presentation, MHC class I direct presentation, Viral immunology
Introduction
During viral infections, cells of the adaptive immune system become activated when they recognize antigens through their antigen-specific receptors. The B cell receptor (BCR) is a membrane immunoglobulin that binds to the antigen directly. For T cells, however, the process is more complex because the T cell receptor (TCR) only recognizes antigens as small peptides bound to major histocompatibility complex (MHC) molecules at the surface of antigen-presenting cells. While the basic principles of antigen recognition by CD4 and CD8 T cells are similar, there are also major differences. This article focuses on the activation of CD8 T cells.
CD8 T Lymphocytes Are an Important Arm of the Antiviral Immune Response
Effector CD8 T lymphocytes kill virus-infected cells and produce antiviral cytokines such as interferon gamma. In this way, CD8 T lymphocytes contribute to resisting primary and secondary viral infections. For example, CD8 T lymphocytes have been shown to be essential for the efficient control of several viral infections of mice and humans including influenza virus, poxviruses, coronavirus, and herpes viruses (Lau et al., 1994; Karupiah et al., 1996; Welsh et al., 2004; Fang and Sigal, 2005; Xu et al., 2007; Doom and Hill, 2008; Channappanavar et al., 2014; Terrazzini and Kern, 2014). Once an infection is cleared, most effector CD8 T cells die but many of them remain in the circulation and tissues as resting memory cells. These memory cells can help control secondary infections more rapidly (Welsh et al., 2004).
The activation of CD8 T lymphocytes requires an antigen-specific signal through the TCR (Yanagi, 1991; Matis, 1990), which recognizes viral peptides (Townsend et al., 1985, Townsend et al., 1986a, Townsend et al., 1986b; Townsend and Bodmer, 1989) bound to a groove on MHC class I (MHC I) molecules (Bjorkman et al., 1987a, Bjorkman et al., 1987b). The interaction of the TCR with peptide–MHC I is enhanced by the binding of the CD8 coreceptor on the T cell to a region of the MHC I molecule away from the peptide-binding groove (Gao et al., 2002). In addition, optimal CD8 T lymphocyte priming may require other signals such as costimulation (signal 2) (Duttagupta et al., 2009; Welten et al., 2013). Direct signals to CD8 T cells by proinflammatory soluble cytokines such as type I interferons and interleukin-12, known as signal 3, are also important for efficient CD8 T cell responses to viruses but their effect seems to be during their early proliferation rather than priming (Haring et al., 2006; Curtsinger and Mescher, 2010; Kim and Harty, 2014).
The T Cell Receptor of CD8 T Lymphocytes Binds Peptides Loaded into MHC I Molecules
MHC I molecules are heterodimers composed of a polymorphic alpha (α) chain with three extracellular domains, one transmembrane domain, one cytosolic domain, and a monomorphic beta-2 microglobulin chain with a single extracellular domain whose role is to stabilize the α chain. The membrane-proximal α3 domain of the α chain interacts with the CD8 coreceptor, while the membrane-distal α1 and α2 domains form a polymorphic groove that binds a diverse array of peptides, 8–10 amino acid long (Bjorkman et al., 1987a, Bjorkman et al., 1987b), with a sequence motif that is characteristic for each α chain allele and determined by its polymorphic amino acids (Rammensee, 1995; Rammensee et al., 1993a, Rammensee et al., 1993b; Grey et al., 1995). The TCRs of CD8 T lymphocytes bind very loosely to most peptide–MHC I–peptide combinations but with much higher affinity to a small subset. This specificity is acquired during the development of the CD8 T lymphocyte in the thymus. In this process, the variable (V), diversity (D), and joining (J) segments of the TCR β and the V and J segments of the TCR α genes recombine randomly to generate unique complementarity-determining regions that bind with higher strength to particular peptide–MHC I adducts. This gives rise to millions of CD8 T lymphocyte clones, each with a distinctive peptide–MHC I specificity, that circulate between the blood and secondary lymphoid tissues in search of their cognate peptide–MHC I combinations (Gras et al., 2012; Godfrey et al., 2008). Importantly, most CD8 T lymphocytes that randomly recognize self peptides are eliminated or rendered tolerant during T cell development. On the other hand, those that recognize nonself peptides, such as those derived from the degradation of viral proteins, become part of the activatable CD8 T lymphocyte pool (Klein et al., 2014; Parello and Huseby, 2015; Morris and Allen, 2012).
Direct Presentation of Self and Viral Peptides by MHC I Molecules
During normal cellular housekeeping, the proteasome and other peptidases in the cytosol progressively degrade mature proteins and defective ribosomal products to their constitutive amino acids. The intermediate peptides generated in this process are very short-lived, if free in the cytosol. However, some of them are rapidly shuttled into the lumen of the endoplasmic reticulum (ER) by the transporter associated with antigen presentation. In the ER, the peptides can be further trimmed by ER aminopeptidase 1 and those with the appropriate motif bind to MHC I molecules that protect them from further degradation. After acquiring their peptide cargo, the MHC I molecules transit to the plasma membrane, where they display the peptides at the cell surface for CD8 T lymphocyte recognition. In this process, known as ‘direct presentation’ (DP), MHC I molecules present to CD8 T lymphocytes a sampling of the cellular proteome, which, in virus-infected cells, includes viral proteins (York and Rock, 1996; Cresswell et al., 2005; Rock and Shen, 2005; Rock et al., 2010) (Figure 1a ). Of note, DP is the only mechanism of MHC I antigen presentation available to most cells. Hence, for most cells, only direct viral infection triggers recognition and killing by antiviral CD8 T lymphocyte effectors (Rock et al., 2010). Therefore, the quasiuniversal expression of MHC I allows effector CD8 T lymphocytes to kill almost every type of infected cells.
Figure 1.
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Pathways of MHC I antigen presentation in viral infections. (a) Direct presentation (DP): The virus infected the cell and the RNA or DNA (here in the cytosol but it can also be in the nucleus) is transcribed (if necessary) into RNA and translated into viral protein. When this protein is degraded by the proteasome and some of the resulting peptides are transported by transporter associated with antigen presentation (TAP) into the endoplasmic reticulum. Peptides with the appropriate motif bind to newly synthesized MHC I molecules and transported through the Golgi to the cell surface for presentation to CD8 T cells. This pathway is active in all cells and for all viruses. It has been shown the dominant pathway for the priming of CD8 T cell responses to vaccinia virus (VACV). (b) Cross-presentation (CP): Viral proteins are phagocytosed (here in cell debris such as apoptotic cells, but could also be in other forms such as viral particles). In the cytosolic pathway (1), the protein is transferred to the cytosol and then follows a pathway similar to DP (1a). Alternatively, TAP acquired from ER membranes can transport peptides back to the phagosome, where it is loaded to MHC I and transported to the cell surface ((1b), never demonstrated for viral infections). In the vacuolar pathway (2), the phagosome fuses with lysosomes. Lysosomal peptidases, mainly cathepsins, degrade the antigen and some of the resulting peptides are loaded into recycling MHC I molecules for transport to the cell surface. CP is active only in bone marrow–derived professional antigen-presenting cells and, in particular, in CD8α dendritic cells. Cytosolic CP is known to be important for the priming of CD8 T cells against herpes simplex 1 virus and to be operative for poliovirus, VACV, influenza A virus (IAV), and lymphocytic choriomeningitis virus. Vacuolar CP has been shown to play a role in CD8 responses to some peptides of IAV.
Efficient CD8 T Lymphocyte Responses to Viruses Generally Require Bone Marrow–Derived Antigen-Presenting Cells
Although effector CD8+ T cells can recognize and kill every cell displaying their cognate peptide–MHC I combination, the initial activation (priming) of CD8 T lymphocytes generally requires antigen presentation by bone marrow–derived cells (BMDCs) (Lenz et al., 2000; Sigal et al., 1999; Sigal and Rock, 2000). The importance of BMDCs in the priming of antiviral CD8 T lymphocyte responses was initially shown in the mouse for vaccinia virus (VACV), poliovirus (PV), influenza A virus (IAV), and most antigenic peptides of lymphocytic choriomeningitis virus (LCMV). Yet, for one peptide from LCMV, BMDC dependency could not be demonstrated (Sigal et al., 1999; Sigal and Rock, 2000; Lenz et al., 2000). This may indicate that there might be exceptions to the need of professional antigen-presenting cells (pAPCs) for CD8 T lymphocyte activation.
The BMDC responsible for the priming of CD8 T cells are called pAPCs. The ability of pAPCs to initiate CD8 T lymphocyte responses relies in their expression of high levels of MHC, costimulatory ligands and cytokines, and the capacity to migrate to the areas of secondary lymphoid organs where CD8 T lymphocyte responses are primed. Traditionally, dendritic cells (DCs), monocytes/macrophages, and B lymphocytes have been considered pAPCs. However, it has been shown that when recovered from secondary lymphoid tissues of mice infected with herpes simplex 1 (HSV-1), VACV, and IAV, the only pAPC that can prime CD8 T lymphocytes ex vivo are those DCs that express the CD8α molecule (Allan et al., 2003; Smith et al., 2003; Belz et al., 2004). Moreover, mice deficient in the transcription factor Batf3, which lack CD8α DCs, do not generate CD8 T lymphocyte responses to HSV-1 or MCMV (Nopora et al., 2012). Thus, there is current consensus that CD8α DCs are key pAPCs during viral infections. However, it remains to be formally demonstrated that these are the only pAPCs that prime CD8 T lymphocytes in vivo.
Some pAPCs Can Present Exogenous Antigens on Their Own MHC I Molecules
In addition to DP, some pAPCs have the capacity to phagocytose dead or dying cells, processing some of their proteins, and presenting their peptides on their own MHC I molecules. This process is known as cross-presentation (CP) and allows pAPCs to cross-present viral antigens inside virus-infected cells(Shen and Rock, 2006; Gutierrez-Martinez et al., 2015; Cresswell et al., 2005).
Two major pathways of CP have been defined. In the cytosolic pathway, the exogenous proteins in vacuoles, such as endosomes and phagosomes, gain access to the cytosol, are processed by the proteasome, and are transferred to the ER or back to phagosomes to be loaded onto newly synthesized MHC I molecules. Alternatively, in the vacuolar pathway, access to the cytosol is not necessary and the peptides are generated in endosomes and phagolysosomes by proteases in the vacuoles (Shen and Rock, 2006) (Figure 1b).
Direct- and Cross-Presentation in Antiviral CD8 T Lymphocyte Responses
The initial demonstration that cytosolic CP can be sufficient to prime an antiviral CD8 T lymphocyte response in vivo was done with PV in a situation where DP was not possible (Sigal et al., 1999). However, the physiological role of CP in the normal priming of antiviral CD8 T lymphocytes remained hotly debated with some arguing that CP is essential and others considering it irrelevant (Melief, 2003; Heath et al., 2004; Wilson et al., 2006; Zinkernagel, 2002; Amigorena, 2003; Lizee et al., 2003). In the middle ground, it appears that CP is important for some viral infections or viral antigens and not for others. For example, CP by CD8α DCs is necessary for an efficient CD8 T cell response to HSV-1 (Nopora et al., 2012; Wilson et al., 2006). Also, vacuolar CP is partly responsible for the CD8 T cell response to IAV (Shen et al., 2004). On the other hand, DP is sufficient for efficient priming of CD8 T cell responses to WT VACV, even though at least some of its antigens can be cross-presented (Larsson et al., 2001; Ramirez and Sigal, 2002; Serna et al., 2003; Xu et al., 2010). Conversely, it has been reported that CP dominates the CD8 lymphocyte response to the replication-deficient strain of VACV-modified vaccinia Ankara (Gasteiger et al., 2007). However, this finding has been challenged (Wong et al., 2013).
Costimulatory Molecules and Their Role in Antiviral CD8 T Cell Priming
In addition to TCR ligation, optimal CD8 T lymphocyte responses require costimulation, which is also known as ‘signal 2.’ Most costimulatory molecules expressed by T cells belong to either the CD28 or the tumor necrosis factor receptor (TNFR) families. The CD28 family includes CD28, which binds CD80 and CD86 on pAPCs and ICOS which binds B7h on most cells. The TNFR family includes CD27, OX40, and 4-1BB, which respectively bind CD70, OX40-L, and 4-1BBL (Duttagupta et al., 2009; Welten et al., 2013). For the priming of CD8 T cells, the most important costimulatory molecule is CD28, while the others play roles at other stages of the response such as expansion or maintenance (Duttagupta et al., 2009). In contrast to the CD8 T cell responses to inert antigens, where absence of CD28 costimulation results in anergy, the CD8 T lymphocyte responses to virus in the absence of CD28 still occur, but their timing and strength may vary according to the specific pathogen. For example, absence of CD80 and CD86 results in reduced responses to vesicular stomatitis virus, murine gammaherpesvirus 68, and IAV, but the response to LCMV is normal (Duttagupta et al., 2009). Of note, during VACV infection, the strength of the CD8 T lymphocyte response in CD28-deficient mice is normal, but the kinetics is delayed by one or two days. Yet, because VACV is only mildly pathogenic to mice, this delay is inconsequential to the health of the mice. On the other hand, when CD28-deficient mice are infected with the related ectromelia virus, a poxvirus that is very pathogenic to mice, this delay in the CD28 response results in the death of the majority of CD28-deficient mice (Fang and Sigal, 2006).
Differences with Other T Cells
In contrast to CD8 T cells, CD4 T cells recognize antigens bound to MHC II molecules. For CD4 T cells, the major pAPCs are also DCs, but B cells that acquire antigen through the BCR also play a very important role. The process of antigen presentation on MHC II is quite different because the peptide is loaded in vesicular compartments and not in the ER. This requires a special mechanism to prevent the binding of peptides to MHC II in the ER (Blum et al., 2013). In general, it is thought that CD4 T cells recognize viral antigens acquired by the pAPCs from the outside. Yet, endogenous viral proteins can also be presented on MHC II (Eisenlohr, 2013).
NKT cells are another type of T cells that recognize antigen as exogenous and endogenous glycolipids bound to the nonclassical MHC I molecule CD1d. The mechanisms of antigen loading on CD1d and the role of NKT cells in antiviral immunity are less well understood.
Summary
Naïve CD8 T cells become activated when they recognize peptide antigen bound to MHC I at the surface of bone marrow–derived pAPCs. In contrast to other cells, pAPCs produce cytokines and express costimulatory molecules that are important for optimal CD8 T cell activation. For several viral infections, it has been shown that the key pAPCs are CD8α+ DCs. As most other cells, pAPCs can directly present antigen synthesized within the cell. Yet, pAPCs, and in particular CD8α+ DCs, can also acquire viral antigens from other infected cells and cross-present them with their own MHC I. While DP is more efficient, it is thought that CP is important for CD8 T cell responses to viruses that do not infect pAPCs or those that produce immune evasion molecules that target DP. Once activated by pAPCs, effector CD8 T cells can recognize any infected cell expressing MHC I loaded with its cognate peptide. This results in the killing of the infected cell and/or the production of antiviral cytokines both being important to control or clear viral infections. After the infection subsides, some antiviral CD8 T cells remain as memory cells, with their numbers being proportional to the strength of the initial response. These memory CD8 T cells can help control secondary infections, and their induction is a main goal of vaccination. Thus, understanding the process of CD8 T cell activation can help develop better vaccines.
See also
CYTOKINES AND THEIR RECEPTORS | Interferon γ: An Overview of Its Functions in Health and Disease; DEVELOPMENT OF T CELLS AND INNATE LYMPHOID CELLS | Control of Early T Cell Development by Notch and T Cell Receptor Signals; IMMUNITY TO VIRAL INFECTIONS | CD8 T Cell Memory to Pathogens; IMMUNITY TO VIRAL INFECTIONS | Dendritic Cells in Viral Infection; IMMUNITY TO VIRAL INFECTIONS | Vaccination against Viruses; MHC STRUCTURE AND FUNCTION | Immune Epitope Database and Analysis Resource; MHC STRUCTURE AND FUNCTION | Ligand Selection and Trafficking for MHC I; MHC STRUCTURE AND FUNCTION | Molecular Immunoevasion Strategies Targeting Antigen Processing and Presentation; MHC STRUCTURE AND FUNCTION | Origin and Processing of MHC-I Ligands; MHC STRUCTURE AND FUNCTION | Repertoire of Classical MHC Class I and Class II Molecules; MHC STRUCTURE AND FUNCTION | Structure of Classical Class I MHC Molecules; SIGNAL TRANSDUCTION | Immunological Synapses; STRUCTURE AND FUNCTION OF DIVERSIFYING RECEPTORS | Organization and Rearrangement of TCR Loci; STRUCTURE AND FUNCTION OF DIVERSIFYING RECEPTORS | Structure and Function of TCRαβ Receptors; T CELL ACTIVATION | Conventional Dendritic Cells: Identification, Subsets, Development, and Functions; T CELL ACTIVATION | Cytotoxic Lymphocytes; T CELL ACTIVATION | Recirculating and Resident Memory CD8+ T Cells; T CELL ACTIVATION | Transition of T Cells from Effector to Memory Phase.
Acknowledgments
I thank Ms Holly Gillin for assistance in the preparation of the manuscript. The work in the author's laboratory was supported by NIH grants R01AI065544, R01AI110457, and R01AG048602.
References
Allan R.S., Smith C.M., Belz G.T., Van Lint A.L., Wakim L.M., Heath W.R., Carbone F.R. Epidermal viral immunity induced by CD8alpha+ dendritic cells but not by Langerhans cells. Science. 2003;301:1925–1928. doi: 10.1126/science.1087576. [DOI] [PubMed] [Google Scholar]
Altenburg A.F., Rimmelzwaan G.F., De Vries R.D. Virus-specific T cells as correlate of (cross-)protective immunity against influenza. Vaccine. 2015;33:500–506. doi: 10.1016/j.vaccine.2014.11.054. [DOI] [PubMed] [Google Scholar]
Amigorena S. Y in X priming. Nat. Immunol. 2003;4:1047–1048. doi: 10.1038/ni1103-1047. [DOI] [PubMed] [Google Scholar]
Belz G.T., Smith C.M., Eichner D., Shortman K., Karupiah G., Carbone F.R., Heath W.R. Cutting edge: conventional CD8 alpha+ dendritic cells are generally involved in priming CTL immunity to viruses. J. Immunol. 2004;172:1996–2000. doi: 10.4049/jimmunol.172.4.1996. [DOI] [PubMed] [Google Scholar]
Bjorkman P.J., Saper M.A., Samraoui B., Bennett W.S., Strominger J.L., Wiley D.C. The foreign antigen binding site and T cell recognition regions of class I histocompatibility antigens. Nature. 1987;329:512–518. doi: 10.1038/329512a0. [DOI] [PubMed] [Google Scholar]
Bjorkman P.J., Saper M.A., Samraoui B., Bennett W.S., Strominger J.L., Wiley D.C. Structure of the human class I histocompatibility antigen, HLA-A2. Nature. 1987;329:506–512. doi: 10.1038/329506a0. [DOI] [PubMed] [Google Scholar]
Blum J.S., Wearsch P.A., Cresswell P. Pathways of antigen processing. Annu. Rev. Immunol. 2013;31:443–473. doi: 10.1146/annurev-immunol-032712-095910. [DOI] [PMC free article] [PubMed] [Google Scholar]
Channappanavar R., Zhao J., Perlman S. T cell-mediated immune response to respiratory coronaviruses. Immunol. Res. 2014;59:118–128. doi: 10.1007/s12026-014-8534-z. [DOI] [PMC free article] [PubMed] [Google Scholar]
Cresswell P., Ackerman A.L., Giodini A., Peaper D.R., Wearsch P.A. Mechanisms of MHC class I-restricted antigen processing and cross-presentation. Immunol. Rev. 2005;207:145–157. doi: 10.1111/j.0105-2896.2005.00316.x. [DOI] [PubMed] [Google Scholar]
Curtsinger J.M., Mescher M.F. Inflammatory cytokines as a third signal for T cell activation. Curr. Opin. Immunol. 2010;22:333–340. doi: 10.1016/j.coi.2010.02.013. [DOI] [PMC free article] [PubMed] [Google Scholar]
Doom C.M., Hill A.B. MHC class I immune evasion in MCMV infection. Med. Microbiol. Immunol. 2008;197:191–204. doi: 10.1007/s00430-008-0089-y. [DOI] [PubMed] [Google Scholar]
Duttagupta P.A., Boesteanu A.C., Katsikis P.D. Costimulation signals for memory CD8+ T cells during viral infections. Crit. Rev. Immunol. 2009;29:469–486. doi: 10.1615/critrevimmunol.v29.i6.20. [DOI] [PMC free article] [PubMed] [Google Scholar]
Eisenlohr L.C. Alternative generation of MHC class II-restricted epitopes: not so exceptional? Mol. Immunol. 2013;55:169–171. doi: 10.1016/j.molimm.2012.10.020. [DOI] [PMC free article] [PubMed] [Google Scholar]
Fang M., Sigal L.J. Antibodies and CD8+ T cells are complementary and essential for natural resistance to a highly lethal cytopathic virus. J. Immunol. 2005;175:6829–6836. doi: 10.4049/jimmunol.175.10.6829. [DOI] [PubMed] [Google Scholar]
Fang M., Sigal L.J. Direct CD28 costimulation is required for CD8+ T cell-mediated resistance to an acute viral disease in a natural host. J. Immunol. 2006;177:8027–8036. doi: 10.4049/jimmunol.177.11.8027. [DOI] [PubMed] [Google Scholar]
Gao G.F., Rao Z., Bell J.I. Molecular coordination of alphabeta T-cell receptors and coreceptors CD8 and CD4 in their recognition of peptide-MHC ligands. Trends Immunol. 2002;23:408–413. doi: 10.1016/s1471-4906(02)02282-2. [DOI] [PubMed] [Google Scholar]
Gasteiger G., Kastenmuller W., Ljapoci R., Sutter G., Drexler I. Cross-priming of cytotoxic T cells dictates antigen requisites for modified vaccinia virus Ankara vector vaccines. J. Virol. 2007;81:11925–11936. doi: 10.1128/JVI.00903-07. [DOI] [PMC free article] [PubMed] [Google Scholar]
Godfrey D.I., Rossjohn J., Mccluskey J. The fidelity, occasional promiscuity, and versatility of T cell receptor recognition. Immunity. 2008;28:304–314. doi: 10.1016/j.immuni.2008.02.004. [DOI] [PubMed] [Google Scholar]
Gras S., Burrows S.R., Turner S.J., Sewell A.K., Mccluskey J., Rossjohn J. A structural voyage toward an understanding of the MHC-I-restricted immune response: lessons learned and much to be learned. Immunol. Rev. 2012;250:61–81. doi: 10.1111/j.1600-065X.2012.01159.x. [DOI] [PubMed] [Google Scholar]
Grey H.M., Ruppert J., Vitiello A., Sidney J., Kast W.M., Kubo R.T., Sette A. Class I MHC-peptide interactions: structural requirements and functional implications. Cancer Surv. 1995;22:37–49. [PubMed] [Google Scholar]
Gutierrez-Martinez E., Planes R., Anselmi G., Reynolds M., Menezes S., Adiko A.C., Saveanu L., Guermonprez P. Cross-presentation of cell-associated antigens by MHC class I in dendritic cell subsets. Front. Immunol. 2015;6:363. doi: 10.3389/fimmu.2015.00363. [DOI] [PMC free article] [PubMed] [Google Scholar]
Haring J.S., Badovinac V.P., Harty J.T. Inflaming the CD8+ T cell response. Immunity. 2006;25:19–29. doi: 10.1016/j.immuni.2006.07.001. [DOI] [PubMed] [Google Scholar]
Heath W.R., Belz G.T., Behrens G.M., Smith C.M., Forehan S.P., Parish I.A., Davey G.M., Wilson N.S., Carbone F.R., Villadangos J.A. Cross-presentation, dendritic cell subsets, and the generation of immunity to cellular antigens. Immunol. Rev. 2004;199:9–26. doi: 10.1111/j.0105-2896.2004.00142.x. [DOI] [PubMed] [Google Scholar]
Karupiah G., Buller R.M., Van Rooijen N., Duarte C.J., Chen J. Different roles for CD4+ and CD8+ T lymphocytes and macrophage subsets in the control of a generalized virus infection. J. Virol. 1996;70:8301–8309. doi: 10.1128/jvi.70.12.8301-8309.1996. [DOI] [PMC free article] [PubMed] [Google Scholar]
Kim M.T., Harty J.T. Impact of inflammatory cytokines on effector and memory CD8+ T cells. Front. Immunol. 2014;5:295. doi: 10.3389/fimmu.2014.00295. [DOI] [PMC free article] [PubMed] [Google Scholar]
Klein L., Kyewski B., Allen P.M., Hogquist K.A. Positive and negative selection of the T cell repertoire: what thymocytes see (and don't see) Nat. Rev. Immunol. 2014;14:377–391. doi: 10.1038/nri3667. [DOI] [PMC free article] [PubMed] [Google Scholar]
Larsson M., Fonteneau J.F., Somersan S., Sanders C., Bickham K., Thomas E.K., Mahnke K., Bhardwaj N. Efficiency of cross presentation of vaccinia virus-derived antigens by human dendritic cells. Eur. J. Immunol. 2001;31:3432–3442. doi: 10.1002/1521-4141(200112)31:12<3432::aid-immu3432>3.0.co;2-r. [DOI] [PubMed] [Google Scholar]
Lau L.L., Jamieson B.D., Somasundaram T., Ahmed R. Cytotoxic T-cell memory without antigen. Nature. 1994;369:648–652. doi: 10.1038/369648a0. [DOI] [PubMed] [Google Scholar]
Lenz L.L., Butz E.A., Bevan M.J. Requirements for bone marrow-derived antigen-presenting cells in priming cytotoxic T cell responses to intracellular pathogens. J. Exp. Med. 2000;192:1135–1142. doi: 10.1084/jem.192.8.1135. [DOI] [PMC free article] [PubMed] [Google Scholar]
Lizee G., Basha G., Tiong J., Julien J.P., Tian M., Biron K.E., Jefferies W.A. Control of dendritic cell cross-presentation by the major histocompatibility complex class I cytoplasmic domain. Nat. Immunol. 2003;4:1065–1073. doi: 10.1038/ni989. [DOI] [PubMed] [Google Scholar]
Matis L.A. The molecular basis of T-cell specificity. Annu. Rev. Immunol. 1990;8:65–82. doi: 10.1146/annurev.iy.08.040190.000433. [DOI] [PubMed] [Google Scholar]
Melief C.J. Mini-review: regulation of cytotoxic T lymphocyte responses by dendritic cells: peaceful coexistence of cross-priming and direct priming? Eur. J. Immunol. 2003;33:2645–2654. doi: 10.1002/eji.200324341. [DOI] [PubMed] [Google Scholar]
Morris G.P., Allen P.M. How the TCR balances sensitivity and specificity for the recognition of self and pathogens. Nat. Immunol. 2012;13:121–128. doi: 10.1038/ni.2190. [DOI] [PubMed] [Google Scholar]
Nopora K., Bernhard C.A., Ried C., Castello A.A., Murphy K.M., Marconi P., Koszinowski U., Brocker T. MHC class I cross-presentation by dendritic cells counteracts viral immune evasion. Front. Immunol. 2012;3:348. doi: 10.3389/fimmu.2012.00348. [DOI] [PMC free article] [PubMed] [Google Scholar]
Parello C.S., Huseby E.S. Indoctrinating T cells to attack pathogens through homeschooling. Trends Immunol. 2015;36:337–343. doi: 10.1016/j.it.2015.04.004. [DOI] [PMC free article] [PubMed] [Google Scholar]
Ramirez M.C., Sigal L.J. Macrophages and dendritic cells use the cytosolic pathway to rapidly cross-present antigen from live, vaccinia-infected cells. J.Immunol. 2002;169:6733–6742. doi: 10.4049/jimmunol.169.12.6733. [DOI] [PubMed] [Google Scholar]
Rammensee H.G. Chemistry of peptides associated with MHC class I and class II molecules. Curr. Opin. Immunol. 1995;7:85–96. doi: 10.1016/0952-7915(95)80033-6. [DOI] [PubMed] [Google Scholar]
Rammensee H.G., Falk K., Rotzschke O. MHC molecules as peptide receptors. Curr. Opin. Immunol. 1993;5:35–44. doi: 10.1016/0952-7915(93)90078-7. [DOI] [PubMed] [Google Scholar]
Rammensee H.G., Falk K., Rotzschke O. Peptides naturally presented by MHC class I molecules. Annu. Rev. Immunol. 1993;11:213–244. doi: 10.1146/annurev.iy.11.040193.001241. [DOI] [PubMed] [Google Scholar]
Rock K.L., Farfan-Arribas D.J., Shen L. Proteases in MHC class I presentation and cross-presentation. J. Immunol. 2010;184:9–15. doi: 10.4049/jimmunol.0903399. [DOI] [PMC free article] [PubMed] [Google Scholar]
Rock K.L., Shen L. Cross-presentation: underlying mechanisms and role in immune surveillance. Immunol. Rev. 2005;207:166–183. doi: 10.1111/j.0105-2896.2005.00301.x. [DOI] [PubMed] [Google Scholar]
Serna A., Ramirez M.C., Soukhanova A., Sigal L.J. Cutting edge: efficient MHC class I cross-presentation during early vaccinia infection requires the transfer of proteasomal intermediates between antigen donor and presenting cells. J.Immunol. 2003;171:5668–5672. doi: 10.4049/jimmunol.171.11.5668. [DOI] [PubMed] [Google Scholar]
Shen L., Rock K.L. Priming of T cells by exogenous antigen cross-presented on MHC class I molecules. Curr. Opin. Immunol. 2006;18:85–91. doi: 10.1016/j.coi.2005.11.003. [DOI] [PubMed] [Google Scholar]
Shen L., Sigal L.J., Boes M., Rock K.L. Important role of cathepsin S in generating peptides for TAP-independent MHC class I crosspresentation in vivo. Immunity. 2004;21:155–165. doi: 10.1016/j.immuni.2004.07.004. [DOI] [PubMed] [Google Scholar]
Sigal L.J., Crotty S., Andino R., Rock K.L. Cytotoxic T-cell immunity to virus-infected non-haematopoietic cells requires presentation of exogenous antigen. Nature. 1999;398:77–80. doi: 10.1038/18038. [DOI] [PubMed] [Google Scholar]
Sigal L.J., Rock K.L. Bone marrow-derived antigen-presenting cells are required for the generation of cytotoxic T lymphocyte responses to viruses and use transporter associated with antigen presentation (TAP)-dependent and – independent pathways of antigen presentation. J. Exp. Med. 2000;192:1143–1150. doi: 10.1084/jem.192.8.1143. [DOI] [PMC free article] [PubMed] [Google Scholar]
Smith C.M., Belz G.T., Wilson N.S., Villadangos J.A., Shortman K., Carbone F.R., Heath W.R. Cutting edge: conventional CD8 alpha+ dendritic cells are preferentially involved in CTL priming after footpad infection with herpes simplex virus-1. J. Immunol. 2003;170:4437–4440. doi: 10.4049/jimmunol.170.9.4437. [DOI] [PubMed] [Google Scholar]
Terrazzini N., Kern F. Cell-mediated immunity to human CMV infection: a brief overview. F1000Prime Rep. 2014;6:28. doi: 10.12703/P6-28. [DOI] [PMC free article] [PubMed] [Google Scholar]
Townsend A., Bodmer H. Antigen recognition by class I-restricted T lymphocytes. Annu. Rev. Immunol. 1989;7:601–624. doi: 10.1146/annurev.iy.07.040189.003125. [DOI] [PubMed] [Google Scholar]
Townsend A.R., Bastin J., Gould K., Brownlee G.G. Cytotoxic T lymphocytes recognize influenza haemagglutinin that lacks a signal sequence. Nature. 1986;324:575–577. doi: 10.1038/324575a0. [DOI] [PubMed] [Google Scholar]
Townsend A.R., Rothbard J., Gotch F.M., Bahadur G., Wraith D., Mcmichael A.J. The epitopes of influenza nucleoprotein recognized by cytotoxic T lymphocytes can be defined with short synthetic peptides. Cell. 1986;44:959–968. doi: 10.1016/0092-8674(86)90019-x. [DOI] [PubMed] [Google Scholar]
Townsend A.R., Gotch F.M., Davey J. Cytotoxic T cells recognize fragments of the influenza nucleoprotein. Cell. 1985;42:457–467. doi: 10.1016/0092-8674(85)90103-5. [DOI] [PubMed] [Google Scholar]
Welsh R.M., Selin L.K., Szomolanyi-Tsuda E. Immunological memory to viral infections. Annu. Rev. Immunol. 2004;22:711–743. doi: 10.1146/annurev.immunol.22.012703.104527. [DOI] [PubMed] [Google Scholar]
Welten S.P., Melief C.J., Arens R. The distinct role of T cell costimulation in antiviral immunity. Curr. Opin. Virol. 2013;3:475–482. doi: 10.1016/j.coviro.2013.06.012. [DOI] [PubMed] [Google Scholar]
Wilson N.S., Behrens G.M., Lundie R.J., Smith C.M., Waithman J., Young L., Forehan S.P., Mount A., Steptoe R.J., Shortman K.D., De Koning-Ward T.F., Belz G.T., Carbone F.R., Crabb B.S., Heath W.R., Villadangos J.A. Systemic activation of dendritic cells by Toll-like receptor ligands or malaria infection impairs cross-presentation and antiviral immunity. Nat. Immunol. 2006;7:165–172. doi: 10.1038/ni1300. [DOI] [PubMed] [Google Scholar]
Wong Y.C., Smith S.A., Tscharke D.C. Systemic Toll-like receptor ligation and selective killing of dendritic cell subsets fail to dissect priming pathways for anti-vaccinia virus CD8(+) T cells. J. Virol. 2013;87:11978–11986. doi: 10.1128/JVI.01835-13. [DOI] [PMC free article] [PubMed] [Google Scholar]
Xu R.-H., Fang M., Klein-Szanto A., Sigal L.J. Memory CD8+ T cells are gatekeepers of the lymph node draining the site of viral infection. Proc. Natl. Acad. Sci. U.S.A. 2007;104:10992–10997. doi: 10.1073/pnas.0701822104. [DOI] [PMC free article] [PubMed] [Google Scholar]
Xu R.H., Remakus S., Ma X., Roscoe F., Sigal L.J. Direct presentation is sufficient for an efficient anti-viral CD8+ T cell response. PLoS Pathog. 2010;6:e1000768. doi: 10.1371/journal.ppat.1000768. [DOI] [PMC free article] [PubMed] [Google Scholar]
Yanagi Y. T-cell receptor and T-cell-resistant virus variants. Curr. Opin. Immunol. 1991;3:460–464. doi: 10.1016/0952-7915(91)90003-j. [DOI] [PubMed] [Google Scholar]
York I.A., Rock K.L. Antigen processing and presentation by the class I major histocompatibility complex. Annu. Rev. Immunol. 1996;14:369–396. doi: 10.1146/annurev.immunol.14.1.369. [DOI] [PubMed] [Google Scholar]
Zinkernagel R.M. On cross-priming of MHC class I-specific CTL: rule or exception? Eur. J. Immunol. 2002;32:2385–2392. doi: 10.1002/1521-4141(200209)32:9<2385::AID-IMMU2385>3.0.CO;2-V. [DOI] [PubMed] [Google Scholar]
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Abstract
Introduction
CD8 T Lymphocytes Are an Important Arm of the Antiviral Immune Response
The T Cell Receptor of CD8 T Lymphocytes Binds Peptides Loaded into MHC I Molecules
Direct Presentation of Self and Viral Peptides by MHC I Molecules
Efficient CD8 T Lymphocyte Responses to Viruses Generally Require Bone Marrow–Derived Antigen-Presenting Cells
Some pAPCs Can Present Exogenous Antigens on Their Own MHC I Molecules
Direct- and Cross-Presentation in Antiviral CD8 T Lymphocyte Responses
Costimulatory Molecules and Their Role in Antiviral CD8 T Cell Priming
Differences with Other T Cells
Summary
See also
Acknowledgments
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The Mammalian Adrenal Glands
A Zonation of the Adrenal Cortex
The cells of the outermost region of the adrenal cortex, the zona glomerulosa, are smaller and more rounded and contain less lipid than those of the more central zona fasciculata. The zona glomerulosa is responsible for synthesis of aldosterone as well as some other corticosteroids. There are few cytological changes in the zona glomerulosa following hypophysectomy or administration of ACTH, suggesting that the secretion of aldosterone is independent of pituitary control. Although ACTH is not necessary for the synthesis and release of aldosterone, the responsiveness of the glomerulosal cells to agents that normally elicit these events is reduced in hypophysectomized mammals and is enhanced with ACTH treatment. Consequently, ACTH does have a permissive effect on cells of the zona glomerulosa.
The zona fasciculata is the largest zone in the adrenal cortex. It is located between the zona glomerulosa and the innermost zona reticularis and is histologically distinct from both. It consists of polyhedral (many-sided) cells that are sources of the glucocorticoids. The cells of the zona fasciculata are arranged in narrow columns or cords surrounding blood sinusoids that allow the cells to be bathed directly with blood (i.e., there is no tissue-blood barrier). The proportion of cortisol and corticosterone secreted differs markedly, from secretion of primarily cortisol (human), through mixtures of both (cat, deer), to primarily corticosterone (rat). Thickness of the zona fasciculata is most sensitive to circulating levels of ACTH. It exhibits hypertrophy and hyperplasia in response to prolonged elevation of ACTH secretion caused by stress or treatment with the drug metyrapone, which blocks the 11-hydroxylation step necessary for glucocorticoid synthesis (see Chapter 3) and hence elevates ACTH in the blood. Unlike the zona glomerulosa, the zona fasciculata atrophies markedly following hypophysectomy or prolonged glucocorticoid therapy and hypertrophies as a result of prolonged ACTH therapy.
The zona reticularis typically borders the adrenal medulla, and it contains numerous thin, extracellular reticular fibers (hence its name). It is a primary source of adrenal androgens but some glucocorticoids may be synthesized here as well. The zona reticularis also hypertrophies in response to ACTH and atrophies following hypophysectomy, but not so dramatically as the zona fasciculata. Gonadotropins (luteinizing hormone [LH] or human chorionic gonadotropin [hCG]) can also stimulate secretion of adrenal androgens.
It should be noted that the “typical” anatomical pattern described here within the adrenal cortex and the anatomical relationship of cortex to medulla varies considerably within mammals as a group. Furthermore, ectopic nodules of functional cortical tissue are not uncommon (Figure 8-1), and this accessory adrenocortical tissue may become a source for corticosteroids following surgical adrenalectomy.
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Chapter
Overview of Glucocorticoids
2018, Encyclopedia of Endocrine Diseases (Second Edition)Nicolas C. Nicolaides, ... George P. Chrousos
Pathways of Steroid Biosynthesis
The adrenal cortex consists of three anatomical zones: the outer zona glomerulosa, the intermediate zona fasciculata, and the inner zona reticularis. The zona glomerulosa is responsible for the production of aldosterone, the zona fasciculata is responsible for the production of cortisol, and the zona reticularis is responsible for the production of adrenal androgens. The adrenal medulla is functionally related to the sympathetic nervous system and secretes epinephrine and norepinephrine both under basal conditions and in response to stress (Stewart, 2003; Miller, 2005).
The glucocorticoid cortisol and the mineralocorticoid aldosterone are synthesized by the adrenal cortex under the control of regulatory systems that largely function independently. All steroid hormones produced by the adrenal cortex are derived from cholesterol. Low-density lipoprotein (LDL)-cholesterol is the major source of cholesterol utilized in adrenal steroidogenesis. Proteolytic and lipolytic enzymes act on LDL to release cholesterol esters for storage in lipid droplets in the adrenal cells (Simpson and Waterman, 1995). In order for the adrenal cortex to synthesize active steroid hormones, a number of changes are required in the structure of cholesterol. Several of these reactions are catalyzed by the steroid hydroxylases, which are members of a superfamily of genes known collectively as cytochrome P450 (CYP). Adrenal steroidogenesis follows three distinct routes, which reflect the zonal differences in terms of function and regulation (Fig. 1).
The rate-limiting step in steroid biosynthesis is importation of cholesterol from cellular stores to the matrix side of the mitochondrial inner membrane, where the cholesterol side-chain cleavage system is located. This is controlled by the steroidogenic acute regulatory protein, the synthesis of which is increased by trophic stimuli, such as adrenocorticotropic hormone (ACTH) (Simpson and Waterman, 1995; Stocco and Clark, 1996; Arakane et al., 1998). The first enzymatic step in steroid biosynthesis common to all steroidogenic pathways takes place in the mitochondrion and leads to the cleavage of six carbon atoms from the side chain of cholesterol, converting this C-27 compound to the C-21 steroid pregnenolone. This reaction is known as cholesterol side-chain cleavage and is catalyzed by the cytochrome P450 enzyme CYP11A (P450scc, cholesterol desmolase, side-chain cleavage enzyme), which is an integral protein of the inner mitochondrial membrane (Nebert et al., 1991). Pregnenolone, the common precursor for all other steroids, then passes by diffusion from the mitochondrion to the endoplasmic reticulum, where it undergoes further metabolism by several other enzymes.
To synthesize mineralocorticoids in the zona glomerulosa, 3β-hydroxysteroid dehydrogenase (3β-HSD) in the endoplasmic reticulum and mitochondria converts pregnenolone to progesterone (Cherradi et al., 1997). The latter is 21-hydroxylated in the endoplasmic reticulum by CYP21 (P450c21, 21-hydroxylase) to produce deoxycorticosterone (DOC). Aldosterone, the most potent 17-deoxysteroid with mineralocorticoid activity, is produced by 11β-hydroxylation of DOC to corticosterone, followed by 18-hydroxylation and 18-oxidation of corticosterone (Fig. 1). The final three steps in aldosterone synthesis are accomplished by a single mitochondrial P450 enzyme, CYP11B2 (P450aldo, aldosterone synthase) (White et al., 1994).
To produce cortisol, CYP17 (P450c17, 17α-hydroxylase/17,20-lyase) in the endoplasmic reticulum of the zona fasciculata and zona reticularis converts pregnenolone to 17α-hydroxypregnenolone (Yanase et al., 1991). 3β-HSD in the zona fasciculata utilizes 17α-hydroxypregnenolone as a substrate, producing 17α-hydroxyprogesterone. The latter is 21-hydroxylated by CYP21 to form 11-deoxycortisol, which is further converted to cortisol by CYP11B1 (P450c11, 11β-hydroxylase) in the mitochondria.
In the zona reticularis of the adrenal cortex and in the gonads, the 17,20-lyase activity of CYP17 converts 17α-hydroxypregnenolone to dehydroepiandrosterone (DHEA), a C-19 steroid and sex steroid precursor. DHEA is further converted by 3β-HSD to androstenedione. In the gonads, androstenedione is reduced by 17β-hydroxysteroid dehydrogenase (Penning, 1997). In pubertal ovaries, aromatase (CYP19, P450c19) can convert androstenedione and testosterone to estrone and estradiol, respectively (Simpson et al., 1994). Testosterone may be further metabolized to dihydrotestosterone by steroid 5α-reductase in androgen target tissues (Wilson et al., 1993).
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Chapter
Endocrine Glands
2007, Histopathology of Preclinical Toxicity Studies (Third edition)Peter Greaves MBCHB FRCPATH
Hypertrophy and diffuse hyperplasia of the zona fasciculata and reticularis
The adrenal cortex responds to prolonged stress or ACTH administration by an increase in weight associated with higher circulatory levels of adrenal corticoids.142,150 It is well recognized that ACTH is a major factor controlling corticosteroid hormone biosynthesis during stress. An increase in circulating ACTH levels precedes increases in the corticosteroid hormones during stress and this is a result of a coordinated action of both ACTH and pro-γ-melanotropin on cholesterol metabolism.151
The normal resting adrenal gland of healthy, unstressed animals, and the adrenal gland in people dying suddenly or following surgical excision, is replete with lipid and the cells of the fasciculata show a foamy, clear cell appearance ('clear cells'). However, human autopsy specimens have shown that when death is preceded by prolonged illness, adrenal glands are heavier than in the unstressed state. Similar changes occur in animals and after ACTH administration. Histologically, the cortical cells show features of increased activity and are referred to as lipid depleted or compact cells. These cells possess a compact or dense eosinophilic cytoplasm with reduction in cytoplasmic lipid and large vesicular nuclei with prominent nucleoli. These changes are reflected at ultrastructural level by a reduction in the number of lipid vacuoles, increased numbers and size of mitochondria, often with vesicular or tubulovesicular cristae, and prominent endoplasmic reticulum.95,142,152,153 Similar histological alterations in the adrenal cortex have been reported in wild caught monkeys associated with the stress during a period following captivity compared with those killed immediately at capture.154
The shift from clear to compact cells commences at the junction of the fascicular and reticularis. The change progresses outwards towards the capsule when the stress or ACTH stimulus is prolonged until the entire fasciculata is composed of compact cells. Study of the adrenal cortex in rats treated with ACTH for several days has shown that the nuclei of the zona fasciculata and reticularis are larger than controls and those in the fasciculata usually contain one to two nucleoli. Cytochemical study shows increased cytoplasmic activity of acid phosphatase and non-specific esterases, particularly in the zona fasciculata where granules of activity are located in the Golgi area and adjacent to the cell membrane.145 Similar morphological and enzyme cytochemical changes were found in rats treated with metopirone, a specific inhibitor of 11β-hydroxylase presumably as a result a compensatory increase in ACTH secretion.
Following cessation of stress or withdrawal of ACTH administration, a ‘reversion reaction' or repletion of lipid occurs. This also occurs from inwards–outwards until all the fasciculata cells have reverted to a clear lipid replete pattern. This depletion–repletion sequence may not always be entirely regular. When marked stimulation occurs, individual cells may undergo degeneration or necrosis.142 Zones of lipid-replete cells may also be seen as islands among lipid-depleted cells. This latter effect may be important in the development of focal hyperplasia.
Diffuse hyperplasia of the adrenal cortex has been described in the diabetic mouse mutant C57BL/KsJ db/db associated with other endocrine derangement, including increases in endogeneous catecholamines and adrenal medullary hyperplasia.155
A number of therapeutic agents are reported to produce adrenal cortical hyperplasia in laboratory animals when given in high doses. Anabolic doses of clenbuterol have been shown to produce adrenal cortical hyperplasia in rats associated with increases in both circulating and glandular corticosterone and cortisol.156 Rats treated with high doses of the direct acting vasodilator pinacidil for 4–8 weeks developed diffuse adrenocortical hyperplasia.157 Drugs with effects on cholesterol synthesis can also produce adrenal cortical hypertrophy or hyperplasia. A 7-day infusion of rats with the competitive inhibitor of hydroxymethylglutaryl coenzyme A (HMG-CoA) reductase, lovastatin, produced an increase in the volume of the zona fasciculata, associated with increases in the volumes of nuclear, mitochondrial and peroxisomal compartments and decreases in lipid droplets.158 Whilst on theoretical grounds agents such as lovastatin could alter adrenal function in humans, this does not seem to have been a problem in actual clinical practice. Prolonged administration of the hypocholesterolaemic agent 4-aminopyrazolo-pyridine to rats was also shown to produce an increase in the volume of the zona fasciculata associated with hypertrophy of the smooth endoplasmic reticulum, peroxisomal proliferation and lipid droplet depletion.159
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Chapter
Adrenal Glands
2009, Basic Medical Endocrinology (Fourth Edition)H. Maurice Goodman
Postsecretory Transformations of Androgens
Dehydroepiandrosterone sulfate (DHEAS), the major product of the zona reticularis, is the most abundant steroid hormone in the circulation. Neither DHEAS nor its close relative androstenedione bind to the androgen receptor, but these 19 carbon steroids are converted to active male and female sex hormones within some peripheral target cells (Figure 4.12). For the most part these peripherally formed hormones do not enter the circulation, and their biological actions are limited to the cells in which they are formed. The ability of peripheral target cells to carry out these transformations has profound consequences for progression of tumors in the prostate and breast, and also for the normal growth and maturation of bone (Chapter 11). The term intracrinology has been used to describe production of hormones by the cells in which they act without escaping into the extracellular fluid. Following removal of the sulfate ester at carbon 3, DHEAS is oxidized to androstenedione, which is the immediate precursor of testosterone or estrone.
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Endocrinology
2014, Clinical Chemistry, Immunology and Laboratory Quality ControlAmitava Dasgupta PhD, DABCC, Amer Wahed MD
9.9 Adrenal Glands
The adrenal glands consist of a cortex and a medulla. The cortex has three zones: zona glomerulosa, zona fasciculata, and zona reticularis. The zona glomerulosa is responsible for secreting mineralocorticoids (aldosterone), while the zona fasciculata is responsible for secreting glucocorticoids. Finally, the zona reticularis is responsible for producing sex steroids. The adrenal medulla produces catecholamines. Steroid hormones are synthesized by adrenal glands from cholesterol, while sex steroid hormones are synthesized in the gonad (Figure 9.2). Major actions of glucocorticoids include:
▪
: Gluconeogenesis and glycogen deposition.
▪
: Fat deposition.
▪
: Protein catabolism.
▪
: Sodium retention.
▪
: Loss of potassium.
▪
: Increase in circulating neutrophils and decrease in circulating eosinophils and lymphocytes.
Major actions of mineralocorticoids include:
▪
: Sodium and water retention in the distal tubule.
▪
: Loss of potassium.
Congenital adrenal hyperplasia is most often due to the lack of 21-hydroxylase enzyme, which causes decreased production of deoxycorticosterone and aldosterone as well as reduced levels of deoxycortisol and cortisol. Adrenocorticotropic hormone (ACTH) level is also high, and as a result, 17-hydroxypregnenolone and 17-hydroxyprogesterone are produced in higher concentrations. This leads to increased production of dehydroepiandrosterone, androstenedione, and testosterone. Female children will have a virilizing effect and may also have ambiguous genitalia. Male children will have features of precocious puberty.
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Book2014, Clinical Chemistry, Immunology and Laboratory Quality ControlAmitava Dasgupta PhD, DABCC, Amer Wahed MD
Chapter
Adrenocortical Function
2008, Clinical Biochemistry of Domestic Animals (Sixth Edition)Björn P. Meij, Jan A. Mol
II Physiology of Adrenocortical Hormones
The secretion of the mammalian adrenal cortex comprises three main categories of hormones, which can be related to some extent to the above-described anatomical zonation. The zona glomerulosa produces mineralocorticoids (aldosterone and deoxycorticosterone), which maintain salt balance. The cells of the zona fasciculata secrete glucocorticoids (cortisol and corticosterone), which are primarily involved in carbohydrate metabolism. The third category of adrenocortical hormones, the androgens (e.g., androstenedione), is produced in the zona reticularis. This zone to a minor degree also secretes glucocorticoids and other hormones such as progesterone and estrogens.
In birds aldosterone and corticosterone are the main corticosteroids secreted. Zonation of the avian adrenal is less clear than in the mammalian adrenal. However, the outer subcapsular cells, looping in a manner similar to the zona glomerulosa, appear to be the predominant aldosterone secretors. The cells reaching toward the central part of the gland form corticosterone (Kime, 1987).
A Steroid Nomenclature
The adrenal steroids contain as their basic structure a cyclopentanoperhydrophenanthrene nucleus consisting of three six-carbon rings (A, B, and C) and a single five-carbon ring (D). The letter designations for the carbon rings, and the numbers of the carbon atoms are shown for pregnenolone (Fig. 19-1), a key biosynthetic intermediate. The Greek letter Δ indicates a double bond, as does the suffix -ene. The position of a substituent below or above the plane of the steroid ring is indicated by α and β, respectively. The α substituent is drawn with a broken line (e.g.--OH) and the β substituent with a solid line (e.g.-OH). The C18 steroids, which are devoid of a side chain at C-17 and have a substituent at C-18, are estrogens. The C19 steroids, which have substituent methyl groups at positions C-18 and C-19, are androgens (see also Fig. 19-1) (IUPAC-IUB, 1989).
Steroids that have a ketone group at C-17 are termed 17-ketosteroids. The C21 steroids, the corticosteroids and progestagens, are those that have a two-carbon side chain (C-20 and C-21) attached at C-17 and in addition have substituent methyl groups at C-18 and C-19. The C21 steroids that also possess a hydroxyl group at position 17 are termed 17-hydroxy-corticosteroids and may have predominantly glucocorticoid properties.
B Biosynthesis
Cholesterol, derived from food and from endogenous synthesis via acetate (Fig. 19-2), is the principal starting compound in steroidogenesis. The adrenal gland is enriched in receptors that internalize low- and high-density lipoproteins. This uptake mechanism increases when the adrenal is stimulated and provides the major cholesterol source. Subsequent steps occur in the mitochondrion or at the endoplasmic reticulum. Two classes of enzymes are involved in the synthesis of steroids, the cytochrome P450 (CYP) heme-containing proteins that catalyze mainly hydroxylation reactions and the hydroxysteroid dehydrogenases (HSD) that are involved in oxidation and reduction reactions (Payne and Hales, 2004). The human CYP enzymes are written with capitals, whereas in mice lowercase is used. For other species no rules are given, but we will use here the abbreviations as used for human.
The precursor cholesterol has a side chain that is cleaved by CYP11A1 resulting in formation of pregnenolone (Fig. 19-2). Mice defective in the Cyp11a1 gene produce no steroids, survive during embryogenesis, but die after birth (Hsu et al., 2006). The zonal difference in adrenocortical hormone production is due to two steroidogenic enzymes. The mitochondrial CYP11B2 (aldosterone synthase), which converts 11-deoxycorticosterone by 11β-hydroxylation and the formation of a carbon 18 aldehyde group toward aldosterone, is found only in the zona glomerulosa. The characteristic enzymes of the inner zones are the microsomal CYP17 (17α-hydroxylase/17,20 lyase) and the mitochondrial CYP11B1 (11β-hydroxylase). CYP17 catalyzes the 17α-hydroxylation of pregnenolone and progesterone as well as the side chain fission of 17α-hydroxy C21 steroids resulting in the formation of dehydroepiandrosterone (DHEA) or androstenedione (Fig. 19-2). CYP11B1 catalyzes the 11β-hydroxylation in the zona fasciculata and reticularis (Payne and Hales, 2004). The other steroidogenic enzymes are present in all three zones of the adrenal cortex (Müller, 1986).
In rat, mouse, guinea pig, baboon, and hamster two distinct CYP11B forms are found, in contrast with cow, pig, sheep, and frogs where the formation of glucocorticoids and mineralocorticoids is catalyzed by a single 11β-hydroxylase (CYP11B1) (Lisurek and Bernhardt, 2004).
The characteristic microsomal 21-hydroxylating (CYP21) and mitochondrial 11β-hydroxylating (CYP11B1) enzymes appear to have developed at an early stage of evolution and are present in all vertebrates. Also, the 18-oxygenated corticosteroids (CYP11B2) retain their importance as mineralocorticoids in all vertebrates. For the 17α hydroxylation (CYP17) potential, the situation is different. Most mammals secrete cortisol as the predominant glucocorticoid. However, rodents and birds secrete predominantly 17-deoxycorticosteroids such as corticosterone.
In line with this, steroid determinations in adrenal venous blood of dogs (Hirose et al., 1977) have revealed cortisol/corticosterone ratios to range from about 3 to 7. In another study, ratios of secretion rates were found ranging from 1.2 to 2.7 and corticosterone/aldosterone secretion ratios between 7 and 25 (Dor et al., 1973). With HPLC analysis in peripheral blood of ACTH-stimulated dogs, the cortisol/corticosterone ratios ranged from 2.4 to 9.7 with a mean of 5.0 (Lothrop and Oliver, 1984). In addition to glucocorticoids and aldosterone, the adrenal vein blood of dogs and pigs contains androstenedione, 11-hydroxyandrostenedione, androsterone, pregnenolone, progesterone, and 11-hydroxyprogesterone (Heap et al., 1966; Holzbauer and Newport, 1969), as well as very small amounts of estradiol and estrone (Dor et al., 1973).
In cats, the cortisol/corticosterone ratio has been estimated to range from 1.6 to 12.4, whereas in kittens the ratio is less than unity (Ilett and Lockett, 1969). In the peripheral blood of horses, higher cortisol/corticosterone ratios (16.0:0.5 and 7.0:0.5) have been reported (James et al., 1970; Zolovick et al., 1966).
In birds, corticosterone, a glucocorticoid with minor mineralocorticoid properties, is by far the main adrenal secretory product. In the peripheral blood of adult birds, no cortisol could be demonstrated with radioimmunoassay (Walsh et al., 1985; Zenoble et al., 1985) or HPLC analysis (Lumeij et al., 1987). In chickens, the 17α-hydrolyase activity is present in embryonic life but decreases after hatching (Carsia et al., 1987; Nakamura et al., 1978).
C Regulation of Secretion
The steroidogenic activity of the two inner zones of the adrenal cortex is predominantly controlled by the pituitary hormone ACTH. The production of aldosterone in the zona glomerulosa is adapted to the sodium and potassium status of the organism by a complex, multifactorial, and mainly extrapituitary control system. For details on the secretion of ACTH by the pituitary and its regulation, refer to Chapter 18. In puppies, the feedback control of the hypothalamic-pituitary-adrenal axis is already operative at the time of birth (Muelheims et al., 1969), although up to the age of 8 weeks basal plasma cortisol concentrations are lower than in mature dogs. There is indirect evidence that this is related to low binding to transport proteins (Randolph et al., 1995).
The action of ACTH on the adrenal gland is rapid; within minutes of its release, the concentration of steroids in the adrenal venous blood increases. The most likely mechanism by which ACTH stimulates steroidogenesis is via activation of membrane-bound adenylate cyclase. This increases the level of cyclic adenosine 3′,5′-monophosphate (cAMP), which activates adrenocortical protein kinases. This results in the phosphorylation of enzymes that enhance the rate of conversion of cholesterol to pregnenolone. Within the adrenal gland, a crosstalk exists with the adrenal medulla (Schinner and Bornstein, 2005). In vitro coculture of adrenocortical cells with chromaffin cells results in a 10-fold increase of steroidogenic activity (Haidan et al., 1998). Also, many intra-adrenal produced cytokines and growth factors play an important role in the regulation of steroidogenesis (Ehrhart-Bornstein et al., 1998). In adrenal tumors, a variety of ectopic and abnormal hormone receptors may stimulate adrenal corticoid release (Lacroix et al., 2001) (see also the section on adrenocortical diseases).
In the regulation of aldosterone secretion, the two most important effectors are the peptide hormone angiotensin II (ANG II) and the extracellular potassium concentrations. Regulation of aldosterone secretion by the potassium status is direct and rather simple (Fig. 19-3). The second loop, which adapts aldosterone secretion to the sodium balance, is much more complex. These two systems are regulated by negative feedback loops. In contrast, ACTH is a representative of other factors that may stimulate aldosterone release without a link to negative feedback (Williams, 2005). From all factors known to regulate aldosterone production in vitro, the stimulation is probably confined under physiological conditions to stimulation by ANG II, K+, and ACTH and inhibition by the atrial natriuretic hormone (ANP) (Spat and Hunyady, 2004).
Apart from these factors, other intra-adrenal factors (Ehrhart-Bornstein et al., 1998) and environmental factors may regulate aldosterone production not only at the level of conversion of cholesterol to pregnenolone but also at the level of aldosterone synthase (CYP11B2) (Williams, 2005).
D Transport
At normal concentrations, only about 10% of the total blood cortisol and corticosterone is in the free form (i.e., susceptible to ultrafiltration). At body temperature, 70% of the plasma cortisol is bound to a globulin called transcortin or corticosteroid-binding globulin (CBG). Transcortin has a high affinity for cortisol and corticosterone, but its binding capacity is limited. Another 20% of plasma cortisol is bound to albumin although its affinity for cortisol is much less than that of transcortin. In line with these percentages, in the dog the free fraction has been estimated to range from 5% to 12% (Kemppainen et al., 1991; Meyer and Rothuizen, 1993).
Transcortin is ubiquitous in mammals, but plasma concentrations vary considerably, resulting in species differences in total cortisol concentration. Most domestic animals have little corticosteroid-binding activity compared to humans (Rosner, 1969). As the free rather than the protein-bound steroid is biologically active, methods have been developed to measure free cortisol. By employing the combination of ultrafiltration and equilibrium dialysis, it was demonstrated that in dogs with portosystemic encephalopathy the associated hyperadrenocorticism is not only characterized by an increased total cortisol concentration in plasma but also by an increase in the free fraction of plasma cortisol (Meyer and Rothuizen, 1994). Between various pig breeds, a two-fold difference in plasma CBG binding capacity was found and related to increased drip loss in the Meishan breed (Geverink et al., 2006).
Unbound steroids readily diffuse into the salivary glands. Because of the close relationship between free cortisol in blood and saliva (Riad-Fahmy et al., 1982), techniques have been developed for the collection of saliva from cattle (Murphy and Connell, 1970), sheep (Fell et al., 1985), and dogs (Phillips et al., 1983). Saliva cortisol concentrations correlated significantly with plasma cortisol concentrations after an insulin-induced hypoglycemia and varied from 7% to 12% of total plasma concentrations, in line with other reports of free plasma cortisol concentrations in dogs (Beerda et al., 1996). In cow's milk, about 60% of the cortisol is present in the ultrafiltrate. Following parturition, the percentage of unbound cortisol (in colostrum) decreases to 40% (Shutt and Fell, 1985) owing to the higher concentrations of CBG-like protein.
The physiological significance of protein binding probably lies in a buffering effect, which prevents rapid variations of the plasma cortisol level. Transcortin restrains the active cortisol from reaching the target organ and also protects it from rapid inactivation by the liver and excretion through the kidneys.
Plasma aldosterone is predominantly bound to albumin, which has a low affinity. The relatively low degree of protein binding of plasma aldosterone partially explains the very low plasma concentration and the short biological half-life of this hormone.
E Metabolic Breakdown and Excretion
Only unbound cortisol and its metabolites are filterable at the glomerulus. Most of this filtered cortisol is reabsorbed, whereby a tubular maximum is only achieved at very high filtered loads of free cortisol (Boonayathap and Marotta, 1974). Less than 20% of the filtered cortisol is excreted unchanged in the urine. Nevertheless, in most mammals the kidneys account for 50% to 80% of the excretion of the metabolized steroids. The remainder is lost via the gut. To render them suitable for renal elimination, the steroids are inactivated and made more water soluble through enzymatic modifications. The liver is the major organ responsible for steroid inactivation and conjugation to form water-soluble compounds, although in the dog—contrary to humans—the kidney and the gastrointestinal tract also contribute to the metabolic clearance of cortisol (McCormick et al., 1974). In the canine, kidney cortisol glucuronide is both secreted and reabsorbed, without a tubular maximum or a plasma threshold (Boonayathap and Marotta, 1974).
Cortisol is cleared from the plasma with a half-life of 60 min or less. For pigs, the metabolic clearance rate of cortisol was calculated to be about 1l.h−1.kg−1 (Hennesy et al., 1986). In dogs, about 60% of infused cortisol is eliminated within 24 h in the urine (Rijnberk et al., 1968a). The 11β-hydroxyl group of cortisol can be oxidized to the ketone, forming cortisone (Fig. 19-4). The reaction is reversible, and in general the equilibrium is shifted to favor the 11β-hydroxyl group. However, because the adrenal cortex produces much more cortisol than cortisone (if any), there is substantial cortisol-to-cortisone conversion. These two steroids have similar subsequent metabolic fates.
Apart from this 11β-hydroxylation, cortisol metabolism in the dog involves the following: (1) reduction of ring A to tetrahydro derivatives, (2) reduction of the 20-keto group to a hydroxyl, and (3) conjugation with glucuronic acid to form glucuronides (Gold, 1961). In addition, unconjugated 20-hydroxycortisol/cortisone has been found in canine urine.
In total glucuronide fraction of urinary corticoids in dogs, at least steroids reduced at C-20 represent 60%. This is of prime importance when it comes to assessing adrenocortical function by measuring urinary cortisol metabolites. Measurements directed at steroids containing a 17α,21-dihydroxy, 20-keto arrangement (i.e., the Porter-Silber reaction) will detect only a small part of the cortisol metabolites and therefore have limited value (Siegel, 1965). Instead, preferably, measurements are preformed involving reduction of the urine metabolites at C-20, followed by oxidation to 17-ketosteroids, which are then quantitated. Further details on this measurement of total 17-hydroxycorticosteroids are given in Section IV.B.3.
Aldosterone is converted not only to tetrahydroaldosterone-3-glucuronide but also to aldosterone-18-glucuronide. Most of aldosterone metabolism takes place in liver and kidney, but the intestine and spleen might also contribute to a minor degree (Balikian, 1971).
In domestic animals, the catabolism of androgens has not been studied in any detail. For the dog, it is known that little of the secreted androgens can be measured as 17-ketosteroids in urine (Rijnberk et al., 1968b; Siegel, 1967). Probably most of the excretion occurs via the bile.
In the cat, glucocorticoid excretion is also largely biliary (Rivas and Borrell, 1971; Taylor, 1971). Feline liver function has several specific features, including a relatively low glucuronyl-transferase activity. Hence, formation of glucuronide conjugates is low, and the conjugates are mainly sulphates, which are mostly excreted via the biliary route. Despite this low urinary excretion of metabolites and conjugates, it was shown that renal excretion of free cortisol is sufficient for diagnostic purposes (Goossens et al., 1995) (see also Section IV.B).
Apart from the use of urinary excretion to examine the glucocorticoid status, fecal glucocorticoid metabolite measures are increasingly being used, especially for zoo and wildlife animals. Interpretation of fecal glucocorticoids may be confounded by a large variety of factors that should be taken into account (Millspaugh and Washburn, 2004).
F Steroid Receptors and Actions
1 Glucocorticoids
Of the naturally occurring glucocorticoids (cortisol, cortisone, and corticosterone), cortisol is the most potent. Several of the synthetic analogues of cortisol are more potent than cortisol itself (Table 19-1). Figure 19-5 illustrates the structural features that determine glucocorticoid potency.
Table 19-1. Relative Potencies of Corticosteroids
| Corticosteroid | Mineralocorticoid Activity | Glucocorticoid Activity |
---
| Cortisol | 1 | 1 |
| Cortisone | 0.7 | 0.7 |
| Corticosterone | 0.2 | 2 |
| 11-Deoxycorticosterone | 0.0 | 20 |
| Aldosterone | 0.1 | 400 |
| 9α-Fluorocortisone | 10 | 400 |
| Prednisone | 4 | 0.7 |
| Prednisolone | 4 | 0.7 |
| Dexamethasone | 30 | 2 |
| Triamcinolone | 3 | 0 |
| 6α-Methylprednisolone | 5 | 0.5 |
In their target organ cells, glucocorticoids act in the same manner as other steroid hormones: diffusion through the cell membrane; binding to the cytoplasmic, glucocorticoid (GR), and mineralocorticoid (MR) receptors; translocation of the hormone receptor complex to the nucleus; and subsequent stimulation of the synthesis of specific RNAs leading to synthesis of specific enzymes. The latter include key enzymes in gluconeogenesis such as fructose-1,6-disphosphatase, glucose-6-phosphatase, and pyruvate carboxylase. In the dog, corticosteroid excess also results in induction of an isoenzyme of alkaline phosphatase. This corticosteroid-induced form of alkaline phosphatase has not yet been found in other species. The marked stability at 65°C allows easy quantitation in plasma as a diagnostic screening tool for cases of suspected hyperadrenocorticism (Teske et al., 1986, 1989).
The traditional view is that the ligand activated GR binds as a homodimer to response elements in promoter regions of genes. However, apart from this direct interaction, the GR may also bind to other transcription factors and thus regulate gene expression of genes devoid of a classical glucocorticoid responsive element (GRE). Next osmolytes, small molecules that promote protein folding during metabolic extremes, may also contribute to protein:protein interactions at the genomic level (Kumar and Thompson, 2005). A splicing variant of the classical GRα is called GRβ and may act as a dominant-negative inhibitor of GRα (Charmandari et al., 2005). Glucocorticoids may also act in a way that does not directly and initially influence gene expression, referred to as “nongenomic.” These pathways involve the production of second messengers and activation of signal transduction pathways by either the classical GR or by a membrane glucocorticoid receptor. Apart from “nongenomic” effects, glucocorticoids may also predominantly inhibit the synthesis of a variety of proteins by decreasing their mRNA stability (Dallman, 2005; Stellato, 2004).
The overall effect of glucocorticoids on metabolism is to supply glucose to the organism by the transformation of proteins. This occurs via the above-mentioned induction of gluconeogenic enzymes in the liver. Thus, glucocorticoids divert metabolism from a phase of growth and storage toward increased physical activity and energy consumption, whereas chronic excess leads to catabolic effects such as muscle wasting, skin atrophy, and osteoporosis. The tendency to hyperglycemia is opposed by increased secretion of insulin, which in turn tends to enhance fat synthesis. This along with the increased food intake owing to central appetite stimulation (Debons et al., 1986) explains the (centripetal) fat deposition, manifested by abdominal enlargement.
In situations of glucocorticoid deficiency, water excretion is impaired, whereas glucocorticoid excess may result in polyuria, being most pronounced in the dog. This is in part due to antagonism of cortisol to the action of vasopressin. In addition, glucocorticoid excess causes loss of the sensitivity of the osmoregulation of vasopressin release (Biewenga et al., 1991). Even physiological increases in cortisol may inhibit basal vasopressin release in dogs (Papanek and Raff, 1994).
Glucocorticoids have long been known to have effects on blood cells, including a reduction in the numbers of eosinophils and lymphocytes and an increase in the number of neutrophils and hence in the total number of leukocytes. Glucocorticoid deficiency leads to normochromic, normocytic anemia.
As far as effects on other endocrine glands are concerned, it is shown that canine hyperadrenocorticism results in reversible suppression of growth hormone secretion (Peterson and Altszuler, 1981). The frequently observed lowering of circulating thyroxine concentrations has been ascribed to changes in the thyroid hormone binding capacity of the plasma and to inhibition of lysosomal hydrolysis of colloid in the thyroid follicular cell (Kemppainen et al., 1983; Woltz et al., 1983). Glucocorticoids have multiple effects on peripheral transfer, distribution, and metabolism (Kaptein et al., 1992), but thyroid function does not seem to be affected (Rijnberk, 1996). The glucocorticoid prednisone was found to inhibit LH secretion in male dogs, leading to reduced testosterone concentrations (Kemppainen et al., 1983).
2 Adrenal Androgens
In health, adrenocortical production of androgens is trivial in comparison with the production of these hormones by the gonads. However, a pathological excess of adrenal androgens might induce virilization (i.e., the development of masculine secondary sex characteristics), which would be most noticeable in the female or immature male. So far, no virilization as a consequence of enzyme deficiency has been reported in domestic animals, but in equine hyper-adrenocorticism, hirsutism is a common feature (Pauli et al., 1974; Van der Kolk et al., 1993).
3 Mineralocorticoids
The widespread MR has equal affinity for aldosterone and the glucocorticoids cortisol and corticosterone, whereas the latter two hormones circulate at much higher concentrations than aldosterone. However, in the classical aldosterone targets (kidney, colon, and salivary gland), the enzyme 11β-hydroxysteroid dehydrogenase (11βHSD) converts cortisol and corticosterone (but not aldosterone!) to the 11-keto analogues. These analogues cannot bind to MR, thereby enabling aldosterone to occupy this receptor (Funder, 2005). Thus, not the receptor but a prereceptor modification provides the tissue specificity.
The synthetic steroid 9α-fluorocortisone (Table 19-1) binds tightly to mineralocorticoid receptors and is used for mineralocorticoid replacement therapy because it is more stable than aldosterone after oral administration. Mineralocorticoid antagonists such as spironolactone bind to these receptors and in this way block aldosterone action.
Aldosterone controls the volume and the cationic composition of the extracellular fluid by regulating sodium and potassium balance. Its main action is on the tubular apparatus of the kidneys, but aldosterone receptors are also found in the gut, the salivary glands, and the sweat glands. In the kidney, the effect of aldosterone on the distal tubule consists almost entirely of an exchange of sodium with potassium and hydrogen ions. Aldosterone also promotes the excretion of magnesium and ammonium ions.
G Physiological Variation, Stress, and the Immune System
In the absence of extraordinary stress, the plasma cortisol concentration of healthy animals varies within certain limits, although adrenocortical secretion does not occur evenly throughout the day but rather in bursts. In humans, most of the secretory bursts occur between midnight and early morning, leading to a diurnal rhythm of circulating cortisol levels. The temporal coincidence between secretory bursts and plasma ACTH concentrations indicates neuroendocrine control.
In domestic animals and certainly in the dog, initially there has been some controversy as to the occurrence of circadian variation in cortisol concentrations. Later definite evidence for episodic but not circadian fluctuations in plasma cortisol concentrations in dogs was presented (Kemppainen and Sartin, 1984a). In the horse (Dybdal et al., 1994) and the pig (Benson et al., 1986; Bottoms et al., 1972; Janssens et al., 1995b; Larsson et al., 1979) as well as in sheep (Fulkerson and Tang, 1979), bulls (Thun et al., 1981), pigeons (Joseph and Meier, 1973; Westerhof et al., 1994), and chickens (Lauber et al., 1987), the occurrence of circadian variations is generally acknowledged. Also in the cat, there have been reports on the occurrence of an (opposite) diurnal rhythm. However, in later reports this was not confirmed (Johnston and Mather, 1979; Leyva et al., 1984). As in the dog, the secretion is episodic without evidence for a diurnal rhythm (Peterson and Randolph, 1989). Apart from relatively short-term changes such as episodic secretion and diurnal variation, plasma cortisol concentration may change with age and during the estrus cycle, during pregnancy, and around the time of parturition. For example, basal cortisol concentrations in puppies 8 weeks of age or younger are lower than in mature dogs, most probably because of reduced binding to plasma proteins rather than to altered secretion patterns (Randolph et al., 1995). In gilts a discrete cortisol peak occurs during the early follicular phase of the sexual cycle, coinciding with the decline in plasma progesterone levels (i.e., at 4 days before the plasma LH surge) (Janssens et al., 1995c). Plasma cortisol concentrations of pregnant cows decrease significantly during the fourth month of pregnancy. Thereafter they remain fairly constant until the fifth and sixth day before parturition, when a sharp rise is seen, and also at around 24 h before parturition plasma corticoids increase again (Eissa and El-Belely, 1990).
A highly significant increase in basal plasma cortisol concentrations occurs with aging in dogs (Goy-Thollot et al., 2006) with no differences in ACTH-stimulated plasma cortisol concentrations. This confirms earlier studies (Rothuizen et al., 1993) where the increased basal activity of the hypothalamo-pituitary-adrenal axis is attributed to a decrease in limbic type-I MR concentrations, whereby no differences were found for the type II GR in canine brain structures. The increase in pituitary type II GR with aging explains the intact feedback in the aging dog (Rothuizen et al., 1993). The ACTH-stimulated increase in plasma aldosterone concentrations was found to be significantly reduced with aging (Goy-Thollot et al., 2006).
Independent of these physiological variations, it is clear that stress may activate the pituitary-adrenocortical system. Factors such as housing (Carlstead et al., 1993; Cockram et al., 1994; Koelkebeck and Cain, 1984), lactation (Gwazdauskas et al., 1986), exercise (Dybdal et al., 1980; Foss et al., 1971; Rossdale et al., 1982; Sloet van Oldruitenborgh-Oosterbaan et al., 2006), surgery (Robertson et al., 1994), anaesthesia, heat (Gould and Siegel, 1985), emotional strain (Bobek et al., 1986; James et al., 1970; Kirkpatrick et al., 1977), food deprivation (Messer et al., 1995), and anticipation of feeding (Murayama et al., 1986) all lead to increased corticosteroid secretion. In dogs, a visit to a veterinary practice, orthopedic examination, and hospitalization increased the urinary corticoid:creatinine ratio, which is a measure for adrenocortical function (Van Vonderen et al., 1998). In the horse, during exercise increased plasma renin activity results in a considerable rise in aldosterone secretion (Guthrie et al., 1980, 1982). From the lack of any increase in plasma corticosterone in homing pigeons during flight, it was concluded that the animals were not under serious stress (Viswanathan et al., 1987).
With chronic stress, such as during prolonged tethered housing of pigs, an increased responsiveness of the adrenal cortex to ACTH is induced, whereas the sensitivity of the pituitary to stimulation with corticotrophin releasing hormone (CRH) or vasopressin remains unaltered, and the challenge (nose sling)-induced ACTH response is lower than in loosely housed pigs. This indicates that during chronic stress, mitigating mechanisms become operational, most probably mediated by endogenous opioids (Janssens et al., 1995a).
According to Selye's theory of general adaptation, corticosteroids are needed for the (metabolic and circulatory) defense reaction. Administration of cortisol to improve nonspecific resistance is indicated in established pituitary or adrenal insufficiency. From an immunological point of view, stress suppresses defense reactions. Stress-induced glucocorticoid secretion prevents the overshooting of the animal's reactions to stress. The glucocorticoids modulate the mediators of the immune system such as the different lymphokines and mediators of the inflammatory reactions: prostaglandins, leukotrienes, kinins, serotonin, and histamine (Munck et al., 1984).
It is now well established that the interaction of the neuroendocrine system and the immune system is not a one-way but a bidirectional communication. Tissue injury or inflammation elicits production of immunoregulatory cytokines (lymphokines and monokines) by macrophages and monocytes. These cytokines also activate the pituitary-adrenal axis and increase glucocorticoid concentrations, whereas the production and action of these immune mediators are inhibited by glucocorticoids. Thus, there is strong evidence for the existence of a feedback circuit, in which immunoregulatory cytokines act as afferent and ACTH and glucocorticoids as efferent hormonal signals (Besedovsky et al., 1986; Woloski et al., 1985). The regulatory actions of the cytokines are exerted at the level of the hypothalamus, where CRH is the major mediator of the response (Berkenbosch et al., 1987), although cytokines also exert an influence at the level of the pituitary and the adrenals (Gaillard, 1994).
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Book2008, Clinical Biochemistry of Domestic Animals (Sixth Edition)Björn P. Meij, Jan A. Mol
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Cortisol and DHEA in development and psychopathology
2017, Hormones and BehaviorHayley S. Kamin, Darlene A. Kertes
6.3 Adolescence
Whereas early childhood is marked by low levels of DHEA(S), late childhood and adolescence are defined by an increase in adrenal hormones. Specifically, at adrenarche (6–7 years in girls and 8–9 years in boys) expansion of the androgen-producing zona reticularis is followed biochemically by a rise in production of DHEA(S) (Auchus and Rainey, 2004; Miller, 2009; Nguyen and Conley, 2008). The source of this increase is maturational changes in enzymatic activity within the zona reticularis favoring enhanced DHEA(S) production. These include increased expression of 17,20-lyase and cytochrome b5, coupled with decreased expression of HSD3β2 (Gell et al., 1998; Rainey and Nakamura, 2008). This surge in DHEA(S) prompts the first signs of puberty and stimulation of other systems that become active at adolescence.
DHEA(S) levels continue to increase substantially through adolescence and early adulthood until reaching a peak at 25–30 years of age (Labrie et al., 1998; Young et al., 1999). In contrast to the marked surge in DHEA(S) occurring at adrenarche, basal concentrations of cortisol remain largely stable throughout the transition from childhood to adolescence (Kenny et al., 1966; Saczawa et al., 2013). Variation in stress responsivity by pubertal stage has, however, been observed in several studies (e.g. Gunnar et al., 2009; Shirtcliff et al., 2012; Stroud et al., 2009).
A greater cortisol response to stress during adolescence compared to earlier ages fits with a broad set of behavioral and biological changes described as representing heightened reactivity to stress (Foilb et al., 2011; Spear, 2000). A hallmark of the adolescent transition is entrance into an array of new interpersonal and social situations that test adolescents' emerging psychosocial capacities (Collins and Steinberg, 2006; Smetana et al., 2006). At the same time, there are dramatic shifts in various neurobiological and anatomical systems (Foilb et al., 2011). Among these are the cortex and limbic systems, which continue to develop well into young adulthood (Blakemore, 2012; Gogtay et al., 2004). Higher basal levels and stress-induced elevations of hormones modulating neural circuits involved in reward processing, emotionality, and arousal during adolescence may, in turn, influence risk for psychopathology (Arain et al., 2013). Administration of DHEA in young adult males has been linked to reduced activation of the amygdala and hippocampus and enhanced connectivity between them, with these changes associated with reduced self-report of negative affect (Sripada et al., 2013). Given DHEA's antiglucocorticoid properties, higher levels may help protect the adolescent brain from cortisol neurotoxicity resulting from increased exposure to socially and biologically stressful events (Campbell, 2006; Flinn et al., 2011). Further research is needed to elucidate the expected trajectory of cortisol and DHEA(S) production during this time and what altered ratios mean for health and well-being throughout the adolescent transition.
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This issue includes a special issue section:Protein misfolding in Alzheimer's and other age-related neurodegenerative diseases
2006, Neurobiology of AgingSabrina S Seehafer, David A Pearce
For example, lipofuscin seen in human retinal pigment epithelium isolated and subjected to 2D electrophoresis revealed that lipofuscin associated proteins were high in lipid-peroxidation and glucoxidation damage . Most of the damage was seen as lysine protein adducts, malondialdehyde (MDAs); cysteine adducts, 4-hydroxynonenal (HNE); and advanced glycation end products (AGEs). This is suggestive that the major protein/fluorophore could itself be altered by oxidative stress or that depending upon the tissue/cell and causal agent of fluorescent material that the composition of autofluorescence will vary. However, this study and other similar studies in this field, make no attempt to prove that true separation of lipofuscin from the proteins and lipids contained in the secondary lysosome's membrane. Another study showed that synthetically similar material to lipofuscin can be obtained by reacting MDAs with cysteine which resulted in the yellow fluorescence . When adrenal gland and neuronal tissue lipofuscin from rats were probed for AGEs it was found that antibodies could detect AGEs in lipofuscin in the cortical cells in the outer part of the zona reticularis, but only faintly in other adrenal gland areas and it can not be detected in neurons of the cerebral cortex and Purkinjie cells . The authors hypothesize that AGEs are only detectable by immunohistochemistry in the early stages of lipofuscin, but as the cells from the outer part of the zona reticularis mature and move to the inner portion, additional modifications such as oxidation and glycosylation occur blocking the antibodies due to AGEs, which are considered to be irreversible modifications.
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Endogenous opiates and behavior: 2021
2023, PeptidesRichard J. Bodnar
2.5 Neuroanatomical localization of Mu agonizts and receptors
Highlights: Functionally distinct POMC-expressing neuron subpopulations in the hypothalamus were revealed by intersectional targeting .
At day 50 of codeine administration, the adrenal glands demonstrated an increase in zona fasciculata thickness but a decrease in zona reticularis thickness .
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Journal2023, PeptidesRichard J. Bodnar
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Adrenal Cortex, Development
2004, Encyclopedia of Endocrine DiseasesMichel Grino
In Sheep
In the ovine fetus, the adrenal gland can be identified as early as embryonic day (E) 28. Its development occurs during three phases, reflecting the interactions of the cellular kinetic phenomena of hyperplasia and hypertrophy:
•
: Establishment of functional zonation (E50–E90): this first growth phase is characterized by the separation of the medullary and cortical portions and the presence of a well-defined zona glomerulosa at approximately E50 and zona fasciculata at approximately E90. During this period, the zona fasciculata grows mainly by hyperplasia.
•
: Quiescence and reactivation (E90–E125): during this period, adrenal growth and adrenal cell division occur at the lowest rate compared to any other time in gestation.
•
: Structural and functional maturation [E130 to postnatal day 2 (P2)]: this second growth phase is characterized by an increase in the rate of cell growth and cell division of the zona fasciculata, with the rate of increase in the volume of steroidogenic cells becoming greater than at any other stage. Cellular hypertrophy precedes cellular hyperplasia. The zona reticularis does not become apparent before 1 month of life.
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Reference work2004, Encyclopedia of Endocrine DiseasesMichel Grino
Related terms:
Androgen
Adrenal Gland
Cortisol
Adrenocorticotropic Hormone
Dehydroepiandrosterone
Aldosterone
Glucocorticoid
Adrenal Cortex
Zona fasciculata
Zona Glomerulosa
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15803 | https://www.youtube.com/watch?v=gpGiUkswLPI | Sex-Linked Genes Example Problems | Genetics
DrWD40
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Posted: 25 Feb 2021
In this video I go over sex-linked gene example problems involving both X and Y linked genes.
If you have any questions, let me know in the comments. Thanks and hope you enjoy!
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Transcript:
[Music] hello everyone how are you doing today and welcome to today's video so today we are going to be doing the example problems related to sex linked traits so i have two example problems we're going to go to with uh a couple parts in each to make sure you can understand how to one read the question make the cross and analyze the results of that cross based on what you get so two problems here and like we always do all of these problems make sure you do them yourself first to realize where you're struggling so here i'll read each problem and then we'll go back and complete them so number one here's hemophilia is an x-linked recessive disorder in humans predict the likelihood of each of the following couples producing a hemophiliac sun and of a hemophiliac daughter so you're just you know expressing expressing the ratios of each um so a female carry an affected male a female non-carrier and affected male and then a female carrier and normal male so you do a cross for each of those and analyze the results and then problem two in gnarls the orange spine is hellandric so you have to remember what a holandrick gene is here you breed bubba your orange spined male with dixie and they produce two pups bow which is male in delta which is a female when the pups are mature you breed delta with bob a male gnarl without orange spines and bow with sue so it sounds like a lot but you break it down step by step each time our first question here is will delta and bob produce any progeny with orange spines if so what sex or sex is can have orange spines and the last question will beau and sue produce any progeny with orange spines if so what sex or sex is can have orange spines all right let's begin breaking these problems down now and starting with the first one here so hemophilia x-linked recessive oh there's your first clue uh predict the likelihood in human so we know it's the xxy system predict the likelihood of each of the following couples producing a son or daughter with hemophilia so a female carrier you know has to be remember how we write females xx so female carrier let's just do so plus is wild type so that's normal and then h let's make hemophilia and i don't like to write the cross sign here when doing these because it gets confusing so i just put a dot to represent cross and an affected male so if a male has it remember male carries the one x chromosome and the y so right there's your cross and then you just set up this punnett square and you know complete it quickly i'm not going to go over how to set up punnett squares that should be you know simple knowledge by this point we've done a lot by now especially with all you know the branch diagrams and whatnot but i'll draw this out it's super straight too and again always write the dominant form first when you fill these in and then we can analyze the results okay here is our final punnett square now we need to analyze results remember females are xx males are x y so here there's a we look at the females this female will be a carrier this female will have it let's do female green has it out of the two females and then male this male has it this male would be normal so here there's a 50 percent chance of male oops of male or female children progeny having hemophilia so it's a 50 member this conditional of each okay next question here so it's not not too bad you just have to set up the punnett square and this is just practice for writing these sex-linked crosses but now we have a female non-carrier so a female non-carrier would be you know dominant form of each and affected male so same thing as before and another thing that this problem is teaching us here so for a female to have it here the mother has to be a carrier and the father has to have it so that's why excellent traits are more common in males because here the father has it and you you would know that beforehand you don't always know if there's a female carrier that could hide but the male that won't hide so a male must have it and the female must be a carrier for a female daughter to get it and this is what these example problems are meant to show here so here a female doesn't have it in a male does when you do this cross here these are the results so when you read these both of the females here are carriers but they don't have it and neither male has it so the answer for this one then oops is zero percent for all and you don't need to write it out so no chance of any offspring having it but you do know that two of the females are carriers so if a male father has hemophilia all daughters will be carriers of hemophilia and that's important to know especially you know for families in making pedigree charts and things like that next up here a female carrier and a normal male so let's write it out female carrier we've had that we've had this one before oops let's write the dominant trait first cross a normal male now let's see what happens let's say one of these daughters up here went off and had children then with a normal male here would be the outcome and here are the the punnett square for this so here we have a female carrier this female here doesn't have it this male does not have it but this male has it so if you have a son there's a 50 chance and then a zero percent chance if it's a daughter um so here 50 chance and all you need is a female carrier here so one of these daughters up here female carrier you know that if that female then has a son with a normal male there's a 50 chance that sun will have hemophilia 25 chance overall that the child regardless of male or female will have it okay so second problem here now so those are typical sex link problems you can't really get more complex than that you know this explained all the different situations you could pretty much you know come up with to show how these different ratios work out okay so now in gnarls the orange spine gene is hellandric remember hellandric means it's found on the y chromosome and when you write these out you either have it or you don't uh so when i say it has it i put a little you know degree sign there or you know not sign so you breed bubba your orange spine male so this gene is hollandric so orange spine male bubba is x y naught with dixie and so it doesn't matter dixie's just xx and they produce two pups bo which is male and delta is a female so let's you know think a bow is a male if this is helandric we know beau is going to have it and delta will not so the when it pops to mature we then continue breeding them and we'll get to that once we get there so here let's just write it all out with all the information we know so first we have bubba i know great names here right bubba has it and we have dixie dixie doesn't have it um they cross we'll just write it like this and produce offspring bo and delta and again like when you read this problem it's a lot of names thrown at you it could be you know you could read it and get confused just start writing it out like this and then you can see it a little better so beau is a male and you know beau has to have it you could you could do this cross to see it but bo will have it so if the father has it the son has it and delta is a female doesn't matter when the pups are mature you breed delta with bob a male narrow without orange spikes then this one will be bob and then um and bow with sue doesn't matter what sue is so here's where the problem is so will delta and bob produce any progeny with orange spikes if so what sexes can have orange spikes so here it's looking at delta and bob here so you know x x cross x y there's nothing on the y you don't have to do this cross the figure out so this one no none of the progeny will have orange spines will no longer present itself now will bow and sue produce any progeny so here now we have x x cross x y naught to keep it the same here um so with orange spines we're right out this cross just to show it prove it so we have you know x y naught x x all the females they can't have it but all males will continue to have it so yes 100 males and i know this was a really simple problem here but i just wanted to show how y length is only with males you can't have a y link trait being carried by females going with females if a father has it all sons will have it too uh so that was just like a little you know refresher from the lecture portion for y linked and then this one was all about x linked so now you can't forget these because these will all come back when analyzing pedigree charts so remembering these little tips and tricks will be very helpful when looking at a pedigree chart and analyzing it but we'll be getting that to that in a future chapter coming up soon all right that's all that we have for chapter four here next in chapter five we're gonna be going over some new fun things when it comes to non-mendelian genetics and how other genes could affect certain out expressions of these genes and how you can have different mixtures and so forth so it's going to get a little bit more fun and complex and i'll see you all starting then all right that's all for now let me know if you have any questions and have a great day bye bye [Music] |
15804 | https://extension.oregonstate.edu/catalog/em-8612-air-shed-drying-lumber | OSU Extension Service
OSU Extension Catalog
Peer reviewed (Orange level)
Air- and shed-drying lumber
English
Español
James E. Reeb and Terrence D. Brown
EM 8612 | Published September 1995, Reviewed 2024 |
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Contents
Air-drying
Orientation and layout of the stack
Controlling the drying rate
Drying sheds
Determining moisture content
Sample boards
Moisture meters
Glossary
There are several reasons to dry solid wood products:
Drying wood before use helps prevent staining and decay while in service.
Drying wood to the average moisture content (MC) where it will be used ensures minimal dimensional change (shrink or swell) while in use.
Dry wood weighs less and is less expensive to ship than green wood.
As wood dries below itsfiber saturation point (moisture content of about 25%–30%), most strength properties increase.
For more information on the relationship between wood and moisture, see the OSU Extension publication Wood and moisture relationships.
Wood can be air-dried only, kiln-dried only or air-dried to a certain MC and then kiln-dried. For example, green 4/4 white oak can take more than a month to kiln-dry, while air-dried 4/4 white oak can be kiln-dried in one to two weeks. For most commercial kiln operators, it is less expensive to have inventory on the air-dry yard than to do all drying through the kiln.
The process of drying wood is the same for air- and kiln-drying, but in kiln-drying you have much greater control over air velocity, temperature and humidity. Controls are much less when air-drying lumber.
(Technical terms are printed in bold at their first use and are defined in the glossary.)
Section anchor "drying"
Air-drying
Air-drying means stacking lumber and exposing it to the outdoors. Final MC is determined by the air temperature, relative humidity, and drying time.
Depending on outside conditions and lumber species and size, air-drying to a desired MC can take from several months to almost a year. Wood stored outdoors and under cover will dry to an approximate moisture content of 12%–14% in Western Oregon and 8%–10% in Eastern Oregon.
When air-drying wood, you must stack the wood properly to ensure adequate air circulation, and you must test it frequently to monitor moisture content. Certain controls can make air-drying more efficient. These include orientation and layout of the stack, stacking methods and covering the stack; all are discussed below.
Air-drying of wood begins as soon as the tree is cut. If the wood is to be used for furniture, moulding, millwork or other high-value uses, take care to prevent degrade. If you are working with logs of species prone to checking, such as oak, apply an end coating as soon as possible, preferably in the woods before the logs are hauled.
Section anchor "orientation"
Orientation and layout of the stack
The orientation and layout of the stack(s) of lumber play an important role in how lumber air-dries.
Construct the air-dry yard on gently sloping ground so water does not pool under stacks or in alleyways. Lumber stacked over a surface where water cannot pool, such as concrete or asphalt, dries faster than lumber stacked over bare or vegetation-covered ground.
Vegetation beneath the stack exposes the bottom layer to air with a higher moisture content. Therefore, never stack lumber over vegetation-covered ground.
Wind is not needed to force air through the stack. Air circulation can develop by natural convection. Warm, dry air enters the sides and top of the stack. As the dry air moves over the lumber, it evaporates moisture from the surfaces. Through evaporation, the air becomes cooler, moister and thus heavier. The heavier air moves toward the bottom of the stack. If the prevailing wind moves freely, the cool, moist air is blown away and is replaced with warmer, drier air. Therefore, increasing the height of the foundation, thus allowing more space under the stack, will increase the drying rate.
You can use orientation and layout in several other ways to increase the drying rate:
Use shorter and narrower stacks to facilitate air access.
Stack lumber away from buildings, trees or other objects that block the wind.
Orient rows of stacks to allow prevailing winds to blow more directly under foundations.
Section anchor "stacking"
Stacking
Figure 1 shows a stack of lumber properly built for air-drying. This example uses concrete blocks as a foundation, but treated timbers or old railroad ties also could be used. (Caution: Wood preserved with chemicals may stain lumber it contacts.)
Ideally, a bottom support bunk, or bolster, should be under every row of stickers. This is especially important for thin lumber, which is more likely to sag.
Use stickers that are dry and of uniform size. Stickers should be enough wider than they are thick so they are not accidentally placed on edge between a layer of lumber. One-inch thick stickers commonly are used for air drying.
Figure 1. A lumber stack with a concrete block foundation and correct sticker alignment. A roof protects the lumber from sun and rain.
Credit: © Oregon State University
Proper sticker alignment, resulting in proper support for the lumber, is important in reducing warp as the lumber dries. Proper sticker placement also allows air to circulate evenly across the surfaces of the lumber and promotes a more uniform drying rate.
Place stickers as far apart as practical to ensure good air circulation. However, proper sticker distance depends on the size (especially thickness) of the lumber. Place stickers farther apart for thicker lumber, because it does not need as much support as thinner lumber.
Generally, a sticker distance of about 18–24 inches is sufficient for lumber of almost any size. However, some hardwoods, such as elm, should have stickers every 12 inches.
It is important to place stickers at equal distances and straight across a layer. Use a sticker at each end of each layer for support. Align stickers one on top of the other.
It is best if lumber pieces in the entire stack are the same thickness, because the time required to air-dry lumber to a target MC depends a lot on thickness. Two-inch lumber can take three to four times as long to air-dry to a certain MC as 1-inch lumber of the same species.
A thick board in a course can cause nearby boards to warp due to inadequate restraint of these boards. So, minimally, each layer should consist of lumber as nearly equal in thickness as possible.
It is best to stack lumber of the same length together. If you stack three long and short lengths randomly, the ends of the longer boards may overhang. This situation increases warping, end checking, and splitting and can lead to mechanical damage and safety concerns because forklifts and other machinery could run into the overhanging ends.
If you must stack different lengths of boards together, two suggested methods for stacking are step-out and box piling.
The step-out method
This places the longest lengths at the bottom, with the next longer boards placed on the next layer, and so on until the stack is full (Figure 2). Place stickers across each course, supporting the ends of each board to minimize warping. This method does not allow you to build a roof or cover over the stack.
Box piling
This arranges lumber so the length of the outside boards in each course is equal to the full length of the stack (Figure 3). Place other full-length boards near the middle of the stack across a course.
Arrange the shorter boards alternately with their ends even with each end of the stack. If you numbered the short boards, the ends of all even-number boards would be even with one end of the stack, and the ends of all odd-number boards would be even with the opposite end of the stack.
You can butt the ends of exceptionally short boards together as long as their combined length does not exceed the stack’s length. Although unsupported ends of short boards may warp, this is the best way to support and hold down different lengths of lumber in the same stack.
Figure 2. Step-out method of piling random-length lumber.
Credit: © Oregon State University
Figure 3. Box piling lumber of random lengths.
Credit: © Oregon State University
Box piling is the preferred way to stack lumber of different lengths. Remember, lumber pieces should be the same thickness, and it is best to stack lumber of the same length.
In Figure 1, thicker and longer pieces are used above the top layer to support a cover. A roof extends over the lumber by 2–3 inches on all sides to protect the lumber from precipitation and direct sun. It may be slanted so precipitation runs off. In this case, 4- by 6-inch timbers (4–6 inches longer than the stickers) support the roof.
Weight is needed to hold the roof in place. The extra weight also helps keep the top layer of lumber from warping as it dries. In Figure 1, cinder blocks on the roof add weight. The blocks are aligned directly above the stickers.
Section anchor "controlling-rate"
Controlling the drying rate
Remember the fundamental rule of drying: Quality depends on the rate of drying. You must achieve a balance between drying lumber too slowly, which would result in stain or decay, and drying lumber too quickly, which would result in checking. If you weren’t worried about defects, you could dry wood in a matter of hours in an oven.
In Oregon, especially east of the Cascades, lumber can air dry too fast. A temperature of 90 F and 15% relative humidity yields an equilibrium moisture content of less than 4%. A kiln operator would not expose green lumber to such harsh drying conditions.
In western Oregon, on the other hand, humidity often is too high to effectively air dry lumber during the fall and winter. The lumber re-wets by either direct precipitation or high relative humidity. However, lumber properly air dried through the summer should reach an MC of 12%–14% by fall.
Uncontrolled air drying may dry lumber too fast. This causes drying defects, most commonly surface and end checks. These defects occur during the initial drying period, when wood is the wettest.
Drying stresses develop because wood shrinks by different amounts in different directions and because shrinkage affects outer fibers before inner fibers, thus creating a moisture gradient between the lumber’s core and shell. (See OSU Extension publication Wood and moisture relationships.)
Figure 4. Annual growth rings. A quartersawn board (left) shows the annual rings’ edges on its broad face. The flatsawn board (right) shows the sides of rings.
Credit: © Oregon State University
Surface checks form as the moisture gradient develops if the lumber surfaces dry too quickly. Thick, wide lumber is more susceptible to surface checking than thin, narrow lumber. Flatsawn lumber dries faster and is more susceptible to surface checking than quartersawn lumber (Figure 4).
Surface checks, especially in hardwoods, can close in the later stages of drying. This occurs when the stresses reverse (the core becomes drier than the shell), and the lumber shell changes from tension to compression. Although these closed checks cannot be seen, they cause problems for end uses, especially ifhigh-quality finished surfaces are needed for products such as interior trim, moulding, cabinets and furniture.
End checks usually are in the wood rays on end-grain surfaces. End checks occur because moisture moves much faster in the longitudinal direction (through the ends of the board) than in either of the transverse directions (through the sides of the board). The ends of the board (or log) dry faster than the middle, and stresses develop at the ends. These stresses can cause checks and warp to develop.
Direct sun can have a dramatic impact on the drying rate. Lumber in direct sun can dry faster than lumber inside the stack and on the side of the stack away from the sun.
If the lumber in direct sun dries too fast, you can use covers or barriers to protect the wood from the sunlight. Several commercial covers effectively block sunlight but allow air movement through the stack.
Maximum safe drying rates have been established for some hardwood species (Table 1). Some hardwoods, especially oaks, have low maximum values for rates of moisture loss per day (about 1% to 2%). These are maximum per-day values, not values averaged over longer periods.
If drying defects occur, there are several ways to slow the drying rate:
Build the stack of lumber over crushed rock or bare ground rather than asphalt.
Make the stack larger (especially wider).
Partially block the wind.
Orient rows of stacks so prevailing winds do not blow directly under the foundations.
Table 1. Safe drying rates of some hardwood 4/4 lumber.
Yellow birch: 6.1% max daily rate of moisture loss
Cherry: 5.8% max daily rate of moisture loss
Chinkapin: 2.0%–3.0% max daily rate of moisture loss
American elm: 10.4% max daily rate of moisture loss
Soft maple: 13.8% max daily rate of moisture loss
Bigleaf maple: 8.0%–10.0% max daily rate of moisture loss
Hard maple: 6.5% max daily rate of moisture loss
Eastern red oak: 3.8% max daily rate of moisture loss
California black oak: 2.0%-4.0% max daily rate of moisture loss
Eastern white oak: 2.5% max daily rate of moisture loss
Oregon white oak: 2.0% max daily rate of moisture loss
Walnut: 8.2% max daily rate of moisture loss
Empirically established drying rates for some eastern hardwoods (Wengert, 1980). Estimated drying rates based on anatomical similarities with other hardwoods.
For lumber, such as thick oak boards, that is difficult to dry without causing seasoning checks, several additional steps may be necessary.
You can apply a coating to the ends of the lumber to retard the drying from these points. End coatings usually are wax-based. Apply end coatings as soon as possible because they are not as effective after the lumber starts to dry. You also can cover the ends of the lumber or the entire stack with burlap. Polypropylene fabric, or shade cloth, has been used to cover horticultural and agricultural crops for many years. More recently, this product has been used successfully to slow the drying of lumber and reduce checking. The use of end coatings, burlap, polypropylene and other materials to slow the drying rate for certain species and sizes of lumber is common among commercial wood drying firms.
Section anchor "sheds"
Drying sheds
If you are air drying large amounts of lumber, you can use a pole-type shed to achieve greater control over the drying process. Sheds allow more control since one or more sides can be blocked off, slowing the drying process. In addition, the lumber is better protected from precipitation and direct sun.
Drying sheds can be very simple in their construction. They can be made more complex by adding walls that can be raised or lowered and by adding fans to increase air flow through the building and to enable more control to the drying process.
You can block the sides of the shed and install fans at one end. Leave the other end open. Run the fans when you want increased circulation and shut them off to decrease circulation.
For wood species that have a tendency to check when drying too fast, such as oaks, run the fans when the exterior humidity is high and the air temperature is low. Turn fans off when the humidity is low and the temperature is high. This process slows the drying rate at the beginning, when lumber with a high MC is most susceptible to checking.
After the wood has dried to below about 22% MC, you can turn on the fans when the temperature is high and the humidity is low. No new surface checking will occur once the wood has reached this low MC.
When the humidity is high, turn the fans off to avoid reintroducing moisture into the lumber. Because no heat is added with this type of drying (sometimes referred to as fan predryers), the final MC is determined by outside ambient temperatures and relative humidities, just as in air-drying but with more control.
Section anchor "moisture-content"
Determining moisture content
When air-drying lumber, you can monitor the lumber for staining or checking to determine whether it is drying too slowly or too quickly. You then can use some of the techniques in Table 2 to either slow or speed the drying.
However, most stains and all checks are permanent once they appear. A better way to track MC loss is to use a moisture meter, and the best way is to use the sample board method.
Moisture content is the total amount of water in a given piece of wood. The MC of wood usually is described as the ratio of the weight of the water to the weight of the wood after the wood has been oven-dried. Although 100% usually signifies the total amount of something, the MC of wood can be greater than 100% because the water can weigh more than the wood.
Moisture content for green (undried) lumber can range from 35% to more than 200%. The higher the initial MC, the longer it will take to dry the wood.
Table 2. A guide to reducing checking, warping, staining, and decay:
A relatively slow initial drying rate reduces checking. Do the following:
Use wider stacks or a double stack on a single foundation.
Space stacks no more than 2 feet apart.
Butt lumber in each course edge to edge as closely as possible, especially in the upper layers.
Use thinner stickers (0.5 or 0.75 inch)
Use polypropylene or shade cloth to protect lumber from direct sun.
Use box piling if you stack different lengths of lumber in a stack.
Use end coatings for logs and lumber.
Proper support and restraint reduce warping. Do the following:
Align stickers exactly above each other. Support both ends of each board with a sticker.
Use close spacing between stickers, such as 12 to 16 inches.
Use stickers of uniform thickness.
Avoid different board thicknesses across a layer.
Use a roof that extends past the pile by several inches on all sides.
Use box piling if you stack different lengths of lumber in a stack.
Fast surface drying reduces staining and decay. Do the following:
Reduce the width of the stack.
Provide more space between piles, such as 6 feet.
Increase space between lumber across a course.
Keep yard clean and avoid blocking air space below piles.
Use a chemical dip, which can retard staining and decay.
Section anchor "sample-boards"
Sample boards
You can use sample boards to monitor the MC of lumber while it is drying. For expensive woods susceptible to drying defects, such as oaks, sample boards are especially important during the initial drying phase.
Sample boards should be repre-sentative of the moisture content of the lumber being dried. Generally, the wettest lumber has the highest risk of degrade, so sample boards should represent the wettest lumber.
The wettest lumber usually is the most recently cut, or the widest, or the thickest. Also, quartersawn lumber tends to be wetter than flatsawn lumber. Select the largest number of samples from the slowest drying material.
Figure 5. Method of cutting and numbering kiln samples and moisture content sections.
Credit: © Oregon State University
Referring to Figure 5, prepare sample boards as follows.
Select the lumber to be used.
Cut a 30-inch sample board.
Cut two 1-inch sections from the sample board. Avoid areas near knots or within 12 inches of the ends of the board.
Number the two 1-inch sections you cut.
Immediately weigh the 1-inch sections to an accuracy of ± 0.1 gram (± 0.035 ounces). Record the weight directly on the section with a marker pen.
Weigh the sample board to an accuracy of ± 0.05 kilograms (± 0.11 pounds). Record the weight directly on the board with a marker pen.
End coat the sample board; a double coating is best.
Place the sample board in the lumber pile where it will dry at the same rate as the rest of the lumber.
Dry the 1-inch sections in an oven at 215°F–218°F (103°C), just above the boiling point of water. Dry the wood about 24 hours; be careful not to burn or char it.
Weigh the 1-inch sections again and record the weights.
Repeat steps 9 and 10 until you get the same weight twice in a row. The wood now is oven dry (OD), sometimes referred to as bone dry.
Use the following equation to determine the wood’s MC percent:(Weight of wood before drying / Oven-dry weight) - 1 x 100or example, if the wood weighs 84 grams before drying and 60 grams after drying, the equation is:(84 / 60) - 1 x 100 = 40%If the water and wood weigh exactly the same, the MC is 100%. If the water weighs more than the wood, the MC is greater than 100%; if the water weighs less, the MC is less than 100%.
Add together the MC percent of the two sections and divide by 2 to determine the average MC of the wood:(%MC1 + %MC2) / 2 = average %MCFor example, MC1 (from step 12) is 40%. Assume the moisture content for MC2 is 46%:(40% + 46%) / 2 = 43%
Calculate the oven-dry weight of the sample board using the average MC percent you found in step 13 and the weight of the sample board from step 6.(Wet weight (step 6) / 100 + %MC (step 13)) x 100 = ODFor example, if the sample board weighs 3.84 kilograms:(3.84 / 100 + 43) x 100 = 2.69 kg
Write this calculated oven-dry weight on the sample board and return it to the lumber stack.
Periodically reweigh the sample board to obtain a new, current moisture content.Current weight (step 16)- 1 x 100Calculated OD weight (step 14)For example, if the new weight of the sample board is 3.21 kg:(3.21 / 2.69) - 1 x 100 = 19.33%
This procedure lets you monitor how fast the wood is drying. See Table 1 for maximum safe drying rates of various species.
Section anchor "meters"
Moisture meters
You can use a hand-held moisture meter to determine maximum daily MC loss and to help determine final target MC. However, moisture meters are not as accurate as the sample board method. For moisture contents greater than the fiber saturation point (about 30% MC), moisture meter accuracy is questionable.
Remember, wood is most susceptible to degrade, including surface and end checks, as it starts to dry from the very wet stage. If you do not use sample boards, you should pay very close attention when drying species susceptible to checking, such as oak, during the early drying stage. If surface or end checking occurs, decrease the drying rate.
The most common type of hand-held moisture meter is the resistance (or conductance-type) meter, which has pins that penetrate the wood surface.
Another type of hand-held moisture meter is the dielectric power loss meter, which has smooth surfaces and does not penetrate the wood. Because of this design, dielectric-type meters are used in-line at wood products mills to monitor lumber, veneer, and other products for high moisture content.
For accurate measurements, you must apply temperature and species corrections to both types of meters. Manufacturers supply these corrections.
Problems when air-drying
Lumber is susceptible to fungi, mold, and insect infestation while on the yard. Temperatures are not high enough to kill fungi, mold or insects.
Lumber that air-dries too quickly can check, split, honeycomb and warp.
Lumber is susceptible to chemical reactions and bacteria, and both can cause stains.
Lumber can become “weathered” from dirt and other contaminants.
Temperatures are not high enough to set the resin in highly resinous species.
Table 2 summarizes techniques to help reduce checking, warping, staining and decay during air drying.
Section anchor "glossary"
Glossary
Bolster — A square piece of wood, usually 4x4, placed between stickered packages of lumber to provide space for the forks of a lift truck.
Bone dry — Wood at zero moisture content. Not a natural state for wood. As soon as bone-dry wood is exposed to air, it will take in moisture.
Bow — A form of warp. Bow describes a deviation flatwise from a straight line drawn from end to end of a board. If the board is laid flat on a wide face, the ends of the board will be off the ground.
Check — Lengthwise separation of wood fibers that extends across the annual growth rings. Commonly caused by stresses during drying. Surface checks occur on flat faces of lumber, and end checks occur on ends of lumber, logs, and other wood products.
Crook — A form of warp. Crook describes a deviation edgewise from a straight line drawn from end to end of a board. If the board is laid on its edge (narrow face), one or both edges will be off the ground.
Cup — A form of warp. Cup describes a troughlike shape in which the board edges remain approximately parallel to each other.
Equilibrium moisture content — The balance of moisture content wood attains in response to the relative humidity and temperature of the surrounding atmosphere.
Fiber saturation point — The stage in the drying or wetting of wood when the cell walls are saturated with bound water and the cell cavities are free of liquid water. Fiber saturation point for most wood species is at moisture contents of about 25%–30%.
Honeycombing — Checks, often not visible on the surface, that occur most often in the interior of the wood, usually along the wood rays.
Moisture content (of wood) — The weight of the moisture in wood, usually expressed as a percentage of its oven-dry weight.
Natural convection — A circulatory transfer of heat due to warmer air rising and cooler, denser air sinking.
Oven dry — See bone dry.
Pile — Lumber stack. Stacking lumber layer by layer, separated by stickers on a supporting foundation or stacking stickered packages one above the other on a foundation separated by bolsters.
Relative humidity — The ratio of the amount of moisture in the air to the maximum amount of moisture the air could hold at that temperature.
Split — Separation of wood fibers along the grain, forming a crack or fissure. Splits may extend partially or completely through the wood.
Stickers — Solid or laminated wood strips used to separate lumber. Typical sizes are 0.5–1 inch thick and 1–2 inches wide. Use 1-inch thick stickers for air drying.
Twist — A form of warp. Twist describes a lengthwise “twisting” of a board in which one corner twists out of the plane of the three other corners.
Warp — Distortion in lumber and other wood products, causing departure from its original plane. Common forms of warp are bow, crook, cup and twist.
For more information
USDA Forest Service, Forest Products Laboratory. Visit the Publications section and search on topics of interest.
Bond, B., 2006. Design and operation of a solar-heated dry kiln, Publication 420-030. Virginia Polytechnic Institute and State University Extension Service.
Peck, E.C., 1962. Drying 4/4 red oak by solar heat. Forest Products Journal 12(3):103–107.
Plumptre, R.A., 1983. Some thoughts on design and control of solar timber kilns. Wood Drying Workshop of IUFRO, Division V, Conference, Madison, WI.
Reeb, J.E., 1995. Wood and moisture relationships, EM 8600. Oregon State University Extension Service.
Rietz, R.C. and R.H. Page, 2003. Air drying lumber: A guide to industry practices (USDA Forest Service, Agriculture Handbook 402).
Simpson, W.T., ed., 1991. Dry kiln operator’s manual, Agriculture Handbook 188. USDA Forest Service, Forest Products Laboratory, Madison, WI.
Wengert, E.M., 1971. Improvements in solar dry kiln design, research note FPL-0212. USDAForest Service, Forest Products Laboratory.
Wood and how to dry it. 1980. Fine Woodworking 22:56–59.
About the authors
James E. Reeb
Extension wood products specialist
Oregon State University
Terrence D. Brown
Former Extension wood products specialist
Oregon State University
Want to learn more about this topic? Explore more resources from OSU Extension: Wood processing and products
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15805 | https://www.jeremykun.com/shortform/2024-06-21-1107/ | Barycentric Lagrange Interpolation || Math ∩ Programming
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2024-06-21
In my studies of the Remez algorithm, I learned about the barycentric Lagrange interpolation formula.
The context is finding a polynomial of degree at most n that passes through n+1 points (x 0,y 0),…,(x n,y n). The classical Lagrange interpolation formula is what you’d write down if you “just did it.”
f(x)=∑i=0 n y i⋅∏j≠i x−x j x i−x j
I wrote a 2014 article deriving this more gently, and implementing it in Haskell for secret sharing. An even gentler, Python-based version was the “practical application” of chapter 2 of PIM.
I put practical application in quotes because the classical formula is numerically unstable. I actually didn’t know this at the time I wrote PIM, but for the application of secret sharing it doesn’t really matter because you do the math over a finite field and numerical accuracy is irrelevant.
But now that I’m in my polynomial approximation era, I actually do care about numerical precision and stability. And for that purpose, the barycentric formula is much better.
The main observation here is that the terms of the classical formula can be thought of as divisors of a single polynomial ℓ(x)=∏i=0 n(x−x i).
Then each term is ℓ j(x)=ℓ(x)/(x−x j), multiplied by the term-specific weight w j=∏i≠j 1(x i−x j). Then you can factor out ℓ(x), precompute the w j, and get
f(x)=ℓ(x)∑i=0 n y i w i x−x i
This is more efficient, but you can also avoid computing ℓ(x) entirely by using the mathematician’s favorite trick of dividing by a cleverly-disguised 1. In this case, the 1 is represented by its interpolation formula (including ℓ(x)), which is achieved by just setting all y i to 1.
When you divide, the ℓ(x) cancels out and you get
f(x)=[∑i=0 n y i w i x−x i]/[∑i=0 n w i x−x i]
This is the barycentric formula. It is numerically stable for well-chosen point sets x i (see section 7 of the paper of Trefethen linked below). I have to admit, I have no idea what this has to do with barycentric coordinates as I originally learned about them. I guess the Lagrange terms ℓ j(x) are supposed to be the vertices of a simplex, and the w j the barycentric weights, but the calling this a simplex seems weird to me.
The other nice thing about this formula is that the weights are trivial to compute for certain point sets. For uniform spaced points on [−1,1], w j=(−1)j(n j), and for Chebyshev points on the same domain it’s w j=(−1)j with values halved at j=0,j=n.
I wrote some Python code to implement this here, along with one extra tweak to improve numerical stability, which boils down to multiplying the numerator and denominator of each term in the sum by 2 to avoid underflow. The test includes an interpolation of degree 100,000 that runs in about a second in vanilla Python.
Popularizing barycentric Lagrange interpolation seems to be something of a passion project for Lloyd N. Trefethen. Trefethen, who until 2023 was a member of the numerical analysis group at Oxford, is an author on basically every paper I can find on this topic, in addition to the applications of this to Remez-like algorithms. In this paper, section 10 gives a history of the curious obscurity of this formula. Among other interesting facts, Trefethen states that Richard Hamming included it in the first edition of Numerical Methods for Scientists and Engineers, and then dropped it in the second edition with no explanation.
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15806 | https://aiexcelbot.com/formulas/statistical/t-inv | T.INV AI formula generator
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Formula Generator - T.INV function
The T.INV function calculates the negative inverse of the one-tailed TDIST function. It is used to find the critical value, confidence interval, or p-value for a one-tailed t-test. The function takes two arguments: the probability and the degrees of freedom. Get Started
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T.INV formula syntax.
The T.INV function in Excel is used to calculate the inverse of the Student's t-distribution. It returns the value of t for a given probability and degrees of freedom. The syntax for the T.INV function is: =T.INV(probability, degrees_freedom, tails) - Probability: This is the probability associated with the t-distribution. It must be between 0 and 1. - Degrees_freedom: This is the number of degrees of freedom for the t-distribution. It must be a positive integer. - Tails: This is an optional argument that specifies the number of tails for the t-distribution. It can be either 1 or 2, representing a one-tailed or two-tailed test, respectively. If omitted, it is assumed to be 2. The T.INV function can be used to find the critical value of t for hypothesis testing or to calculate confidence intervals.
Use Cases & Examples In these use cases, we use the T.INV function to calculate the inverse of the Student's t-distribution for a given probability and degrees of freedom. This function is commonly used in statistical analysis to determine critical values for hypothesis testing and confidence intervals. Add to Chrome
Calculating the critical value for a one-tailed t-test
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In this use case, we use the T.INV function to calculate the critical value for a one-tailed t-test. The critical value is the value beyond which we reject the null hypothesis.
Result
T.INV(probability, degrees_freedom)
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Calculating the confidence interval for a one-tailed t-test
Description
In this use case, we use the T.INV function to calculate the confidence interval for a one-tailed t-test. The confidence interval provides a range of values within which we can be confident that the true population parameter lies.
Result
T.INV(probability, degrees_freedom)
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Calculating the p-value for a one-tailed t-test
Description
In this use case, we use the T.INV function to calculate the p-value for a one-tailed t-test. The p-value represents the probability of obtaining a test statistic as extreme as the one observed, assuming the null hypothesis is true.
Result
T.INV(probability, degrees_freedom)
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Calculating the critical value for a one-tailed t-test
Description In this use case, we use the T.INV function to calculate the critical value for a one-tailed t-test. The critical value is the value beyond which we reject the null hypothesis.
Result T.INV(probability, degrees_freedom)
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Calculating the confidence interval for a one-tailed t-test
Description In this use case, we use the T.INV function to calculate the confidence interval for a one-tailed t-test. The confidence interval provides a range of values within which we can be confident that the true population parameter lies.
Result T.INV(probability, degrees_freedom)
Copy Formula
Calculating the p-value for a one-tailed t-test
Description In this use case, we use the T.INV function to calculate the p-value for a one-tailed t-test. The p-value represents the probability of obtaining a test statistic as extreme as the one observed, assuming the null hypothesis is true.
Result T.INV(probability, degrees_freedom)
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What is the T.INV function in Excel?
The T.INV function returns the inverse of the Student's t-distribution. It is commonly used in hypothesis testing and confidence interval calculations.
How do I use the T.INV function?
To use the T.INV function, you need to provide the probability value and the degrees of freedom. The function will then return the corresponding value from the t-distribution.
What is the syntax of the T.INV function?
The syntax of the T.INV function is: T.INV(probability, degrees_freedom).
Can the T.INV function return negative values?
Yes, the T.INV function can return negative values. The sign of the result depends on the probability and degrees of freedom provided.
What is the range of the probability argument in the T.INV function?
The probability argument in the T.INV function should be between 0 and 1.
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15807 | https://www.fao.org/fishery/docs/CDrom/FAO_Training/FAO_Training/General/x6707e/.!17684!x6707e10.htm | Measurement Of Areas
10. MEASUREMENT OF AREAS
10.1 Introduction
One of the main purposes of your topographical survey may be to determine the area of a tract of land where you want to build a fish-farm. From existing topographical maps, you may need to calculate the area of a watershed or of a future reservoir (see Water, Volume 4 in this series).
Note: in land surveying, you should regard land areas as horizontal surfaces, not as the actual area of the ground surface. You always measure horizontal distances.
You will often need to know the areas of cross-section profiles to calculate the amount of earthwork you need to do.
Horizontal area
Cross-section area
You may determine areas either directly from field measurements, or indirectly, from a plan or map. In the first case, you will find all the measurements of distances and angles you need by surveying, and you will calculate the areas from them. In the second case, you will draw a plan or map first (see Chapter 9). Then you will get the dimensions you need from the scale, and determine the area on that basis.
There are several simple methods available for measuring areas. Some of these are graphic methods, where you compare the plan or map of the area you need to measure to a drawn pattern of known unit sizes. Others are geometric methods, where you use simple mathematical formulas to calculate areas of regular geometrical figures, such as triangles, trapeziums, or areas bounded by an irregular curve.
Note: a trapezium is a four-sided polygon with two parallel sides.
The simple methods will be described in detail in the next sections. They are also summarized in Table 13.
Triangle
Trapezium 1
Trapezium 2
Irregular area
TABLE 13 Simple area measurement methods Section Method Remarks 10.2Strips Graphic method giving rough estimate 10.3Square-grid Graphic method giving good to very good estimates 10.4Subdivision into regular geometric figures such as, triangles, trapeziums Geometric method giving good to very good estimates 10.5Trapezoidal rule Geometric method giving good to very good estimates Suitable for curved boundary
10.2 How to use the strips method for measuring areas
Get a piece of transparent paper, such as tracing paper or light-weight square-ruled millimetric paper. Its size will depend on the size of the mapped area you need to measure.
On this paper, draw a series of strips, by drawing a series of parallel lines at a regular, fixed interval. Choose this strip width W to represent a certain number of metres. You can follow the scale of the plan or map to do this.
Example
Scale 1: 2 000; strip width W = 1 cm = 20 m.
Scale 1: 50 000; strip width W = 1 cm = 500 m.
Note: the smaller the strip width, the more accurate your estimate of the land area will be.
Place the sheet of transparent paper over the plan or map of the area you need to measure, and attach it securely with drawing pins or transparent tape.Scale: 1: 2.000
For each strip, measure the distance AB in centimetres along a central line between the boundaries of the area shown on the map.
Calculate the sum of these distances in centimetres. Then, according to the scale you are using, multiply to find the equivalent distance in the field, in metres.
Example
Scale is 1 :2000 and 1 cm = 20 m.
Sum of distances = 16 cm.
Equivalent ground distance: 16 x 20 m = 320 m.
Multiply this sum of real distances (in metres) by the equivalent width of the strip W (in metres) to obtain a rough estimate of the total area in square metres.
Example
Sum of equivalent distances is 320 m.
Strip width (1 cm) is equivalent to 20 m.
Land area: 320 m x 20 m = 6 400 m 2 or 0.64 ha
Note: 10000 m 2 = 1 hectare (ha)
Repeat this procedure at least once to check on your calculations.
Total area = 320 m x 20 m = 6400 m 2
10.3 How to use the square-grid method for measuring areas
Get a piece of transparent square-ruled paper, or draw a square grid on transparent tracing paper yourself. To do this, trace a grid made of 2 mm x 2 mm squares inside a 10 cm x 10 cm square, using the example given on the page.
Note: if you use smaller unit squares on the grid, your estimate of the land area will be more accurate; but the minimum size you should use is 1 mm x 1 mm = 1 mm 2.
Place this transparent grid over the drawing of the area you need to measure, and attach it to the drawing securely with thumbtacks or tape. If your grid is smaller than this area, start at one edge of the drawing. Clearly mark the outline of the grid, then move to the next section and proceed in this way over the entire area.
Count the number of full squares included in the area you need to measure. To avoid mistakes, mark each square you count with your pencil, making a small dot.
Note: towards the centre of the area, you may be able to count larger squares made, for example, of 10 x 10 = 100 small squares. This will make your work easier.
Look at the squares around the edge of the drawing. If more than one-half of any square is within the drawing, count and mark it as a full square. Ignore the rest.Half or more squares
Add these two sums (steps 3 and 4), to obtain the total number T of full squares.
Add the sums again at least once to check them.
Using the distance scale of the drawing, calculate the equivalent unit area for your grid. This is the equivalent area of one of its small squares.
Example
Scale 1:2000 or 1 cm = 20 m or 1 mm = 2 m
Grid square size is 2 mm x 2 mm
Equivalent unit area of grid = 4 m x 4 m = 16 m 2
Multiply the equivalent unit area by the total number T of full squares to obtain a fairly good estimate of the measured area.
Example
Total count of full squares T = 256
Equivalent unit area = 16m 2
Total area = 256 x 16 m 2 = 4096 m 2
Note: when you work with large-scale plans such as cross-sections, you can improve the accuracy of your area estimate by modifying step 5, above. To do this, look at all the squares around the edge of the drawing which are crossed by a drawing line. Then, estimate by sight the decimal part of the whole square that you need to include in the total count (the decimal part is a fraction of the square, expressed as a decimal, such as 0.5, 0.1 and 0.9).
Example
Square A = 0.5; B = 0.1; C = 0.9.
10.4 How to subdivide the area into regular geometrical figures
When you need to measure areas directly in the field, divide the tract of land into regular geometrical figures, such as triangles, rectangles or trapeziums. Then take all the necessary measurements, and calculate the areas according to mathematical formulas (see Annex 1). If a plan or map of the area is available, you can draw these geometrical figures on it, and find their dimensions by using the reduction scale.
In the first manual in this series, Water for Freshwater Fish Culture, FAO Training Series (4), Section 2.0, you learned how to calculate the area of a pond using this method. In the following steps, you will learn how to apply it under more difficult circumstances.
Measuring areas by triangles
You can easily calculate the area of any triangle when you know the dimensions of: all three sides a, b and c |
15808 | https://thebreakthrough.org/journal/no-15-winter-2022/sprawl-is-good-green | Energy and Climate Climate Impacts Infrastructure and Environment Urbanization
Sprawl Is Good
The Environmental Case for Suburbia
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In the years leading up to the coronavirus pandemic, the intelligentsia came to a consensus that sprawling, car-dominated cities were doomed. The future, they said, was in dense, transit-dependent metropolises. The seeming success of compact cities such as San Francisco, Boston, and New York gave this theory credence. And the supposed dangers of sprawl to the climate gave it urgency.
Yet the facts show that sprawling and car-dependent cities have grown more rapidly than dense ones for decades and are far more affordable. The pandemic, meanwhile, showed they will expand even more rapidly in the future. By contrast, the climate-driven demands for density and transit are just the most recent version of a solution that has long been searching for a problem. Advocates will continue to search. In reality, sprawling cities are more environmentally sound than their dense counterparts and will become even more so as technology evolves.
Instead of warring against sprawl and cars, planners and environmentalists should recognize how the green spaces of suburbia, allied to autonomous electric vehicles and green single-family homes, can provide both the affordability and sustainability most Americans crave.
The Long Triumph of Sprawl
There has been much discussion of the benefits of density, of which there are many. If there weren’t, nobody would live in Manhattan or San Francisco. These cities allow many people, especially young, high-productivity singles and those who work in business services like finance or law, to congregate and learn from each other. Economists call these benefits “agglomeration effects.”
But too many advocates today ignore the other side of the coin, known to economists as “the demons of density.” These include things like congestion, crime, and, of course, pollution. Such problems explain why, as technology has evolved, people try to get more of the benefits of living near each other—the agglomeration effects—without the problems of living directly on top of one another. And so, despite periodic stories of a return to the city, America has been taking advantage of new technologies to become more and more sprawling: in other words, to live in places with lower density and more ease of driving by car.
Since 1950, the average density of the largest American cities has dropped from 6,000 people per square mile to 3,000. The fastest growing metros in terms of population have been the most sprawling ones. While places like New York or San Francisco have about 50 percent more people than in 1950, Houston, Dallas, and Jacksonville have 500 percent more. Atlanta, Phoenix, and Austin have almost 1000 percent more. Studies confirm that there is a direct correlation between how sprawling an area is and how fast it grows. This is not surprising, because housing prices in, for example, Texas are less than half the cost of those in California, which has some of the densest cities in the nation, and the cost of living in general is lower.
Complementing sprawl has also been a long-term trend toward people driving more and taking transit less. In 1960, about 12 percent of all Americans took transit to work. By 2020, it was about 5 percent, and the decline of non-work trips on transit was even faster. Meanwhile, the number of Americans who travel to work by car, especially those who travel alone in a car, continues to increase. The miles driven by car per capita have almost doubled. This is not because of supposed subsidies to cars; according to the Bureau of Transportation Statistics, cars pay for almost all of the cost of roads through gas taxes and other charges, while transit riders pay barely a third of the cost of their rides, and that ratio is dropping as transit use declines. For decades, in fact, more and more of the gas tax has been siphoned off to pay for increasingly expensive and increasingly empty buses and trains.
The COVID-19 pandemic has accelerated these trends, further increasing the demand for distant “exurban” development. In the past decade, exurbs grew at almost two times the national rate. But, during the pandemic year alone, construction in exurbs increased another 20 percent. A recent study for the National Bureau of Economic Research has shown that prices and rate of building have risen the fastest in the most far out parts of metro areas during the pandemic. That makes sense: people are moving farther away because, in a post-pandemic world, they won’t need to travel as often to work. The percentage of those who telework has gone from 5 percent of the workforce to what seems to be a long-term trend of 20 percent. Since this telework shift has been starkest for office workers, and since almost all transit systems are geared toward serving central business districts, the increase of telework has hit transit hard. Indeed, while passenger miles traveled by car have returned to pre-pandemic levels, bus travel is down 40 percent since 2019. Urban rail travel is down by half.
Some might long for a return to dense urban areas and a pre-automobile age. The Urban Land Institute says, “arguably, no tool is more important than increasing the density of existing and new communities.” As a recent National Geographic headline stated, “To Build the Cities of the Future, We Need to Get Out of Our Cars.” Yet those who hope for massive in-migration to transit-dependent urban areas have been disappointed for decades. And not just in the United States. In fact, the same trend toward more sprawling cities, more cars, more driving, and less transit has been observable all around the world. Densities in Paris, for instance, have dropped by half since 1950, even as the miles driven per person by car have doubled.
These tendencies represent more than the effect of one or another policy, and more than a short-term trend. Rather, they indicate a clear global and long-term preference. The pandemic has only made the shift toward the modern, sprawling city more rapid and obvious.
The Environmental Costs of Density
People choose more sprawling areas because they limit many of the downsides of urban living, from cost to congestion. But we learned in the pandemic one of the most important environmental costs of density: disease. For most of human history, cities were “demographic sinks” because births did not make up for high death rates due to infectious disease. Cities had to be continuously replenished by people moving from the countryside.
In the early 20th century, the triumph of public health against smallpox, typhoid, cholera, and other infectious diseases allowed for more urban living. Yet cities never entirely overcame the dangers of density for disease transmission. We saw this again during the COVID-19 pandemic. One study found population density accounted for up to 76 percent of the difference between infection rates in different parts of the United States. There was also a correlation between infection and density in the 1918 flu and doubtless will be in the next pandemic.
A more prosaic concern with density is air quality. The more concentrated humans are, the more likely they will be breathing each other’s pollution, whether that comes from congested streets, local industry, or simple heating and cooling devices. The larger the urban area, by contrast, the more space there is for air pollution to spread out and disperse, and this is true around the world. A study released this year in Regional Science and Urban Economics using data from Germany found what dozens of other studies have found. Simply, that “higher population density worsens local air quality.” Another recent study of U.S. cities showed “denser cities are linked with worse air quality,” and that dangerous particulate matter especially tended to increase in denser areas. By its authors’ calculations, a dense American city has dozens more pollution deaths a year, merely because of its density, than a more sprawling city of similar population. While pollution everywhere is lower than it was in the smog-choked cities of the Industrial Revolution, the move to the suburbs has been an important part of why more people breathe clean air today.
There are other local environmental costs to density as well. Urban residents have to endure a “heat island” effect, where urban concrete and asphalt amplify temperatures. The United States Environmental Protection Agency (EPA) notes that such effects can make daytime temperatures up to 7 degrees Fahrenheit hotter in cities than in rural and suburban areas, and nighttime temperatures up to 5 degrees hotter. (For perspective, the Intergovernmental Panel on Climate Change’s estimate for the total impact of global warming by 2100 is about 5 degrees Fahrenheit.)
For all the talk of urban planning for “climate resilience,” and pitches for green roofs, cooler building colors, and so forth, there has been limited discussion about how spread-out suburban areas could reduce the heat island effect even more substantially. Yet, as a recent article in Nature Communications argues, “sprawling development will lead to a better thermal environment” by creating more green space and lowering the heat island effect by several degrees. By contrast, the article notes that other strategies, such as green roofs, tend to have much more localized and limited effects. The heat island effect of dense areas in fact exacerbates the problems of local air pollution. Heat breaks down nitrogen dioxide and volatile organic compounds into ozone, which irritates eyes and lungs. This explains why air pollution is far worse in summer months and why local air pollution will be amplified by density in the future.
Perhaps the most important tool for reducing the heat island effect is trees, which provide shade and absorb solar radiation. Suburbs are, almost by definition, more verdant than cities. Famously sprawling cities like Atlanta or Houston have tree cover on more than 30 percent of their landmass, while older, denser cities such as Philadelphia, San Francisco, and Chicago have under 20 percent. Besides reducing the heat island effect, nearby trees have been shown to intercept particulate matter and absorb ozone, sulfur dioxide, and nitrogen dioxide, thus reducing local air pollution as well. The personal and psychological effects of trees are real too. Studies have shown that the presence of trees decreases stress, increases attentiveness and sense of safety and comfort, and reduces the likelihood that pregnant women will have low-birthweight babies. There is no way to have the same access to trees in dense urban areas.
Since the publication of Ian McHarg’s justly famous work Design with Nature (1969), urban planners and developers have become aware of the need to include natural landscapes in new communities. They now build depressed swards to absorb rainwater, retain hills and creeks as contributors to natural beauty, and create green pathways to allow migration of animals. Yet these features all tend to spread out development and lower density, as McHarg himself noted. His book included maps of Philadelphia that showed how social and physical ills increased with density. As he said, it was “not poverty, but density” that “bears a remarkable correspondence to the pattern of pathology” in urban living, in everything from crime to chronic disease. It should be no surprise that McHarg helped design the very livable, but sprawling, master-planned community of The Woodlands outside Houston.
There is also an assumption across much of the popular environmental literature that single-family homes are environmentally destructive, while tall buildings are green. Advocates claim that building up instead of out limits buildings’ footprints and that tall buildings require less energy to heat or cool, thus reducing greenhouse gas and other emissions. We now know this is false.
For one, taller buildings rely on steel and concrete to support themselves. These materials take five times more energy and carbon dioxide to produce than wood, which predominates in single-family or smaller homes and is a renewable carbon sink. Beyond the materials used, each additional story of a tall building requires more support beams and structures on every story of that building, which increases the ratio of material to livable space. These tendencies help explain why building taller is more expensive. Going from two to four stories increases the cost of each square foot of a building by 25 percent. Going from five to ten stories increases the cost of each square foot by over 50 percent. Those costs are the result of more—and more energy-hungry—materials.
Large buildings also use more energy to function. Tall buildings require fans to push and pull air through their HVAC systems, as well as energy-hungry pumps to lift water to the top floors. For very large buildings, elevators use up to 10 percent of all energy. Common areas such as stairwells and lobbies need to be heated, cooled, and lit, adding to both environmental and economic costs without contributing to anyone’s living space. One recent study found that “each additional story in a building is associated with a 2.4 percent increase in electricity use and 2.9 percent increase in fossil fuel use.” Taller buildings specifically tend to absorb more heat and then give it off, exacerbating the heat island effect, even as they cast shadows that limit natural light. Taller buildings, in short, create more burdens on both the local and global environment than small ones.
The Environmental Costs of Transportation
Today, much of the discussion around the environmental impact of density revolves around a single metric: vehicle miles traveled (VMT), the number of miles cars travel on the roads. The assumption, embodied in many state and local regulations, is that denser cities tend to bring destinations closer and therefore reduce VMT, which therefore reduces air pollutants and greenhouse gas emissions from burning gasoline in cars.
There are several faulty assumptions behind the attempt to reduce VMT. First and foremost, there is little correlation between density and VMT. A 2009 metasurvey of the literature by the National Research Council found that doubling residential density in an urban area, which in fact has never happened for any major city in modern history, would reduce VMT by only 5 to 12 percent. The massive costs of such doubling, and the minor reduction in car-related emissions, would make such an effort one of the least cost-efficient means to reduce carbon emissions imaginable.
Any benefits of density-related reductions in VMT are offset by other factors. For one, denser areas tend to have more congestion (think 14th Street in New York City), so traveling a mile in a dense city will require more starts and stops, and will therefore burn more gasoline, than traveling on a less- congested suburban highway or arterial road. The CO2 emissions when traveling at 5 mph are 300 percent higher than when traveling at 55 mph, and this ratio is even worse when the slower speed is due to congestion. Focusing on miles traveled, rather than actual greenhouse gas emissions, is a blunt and inopportune metric that tends to bias planners against suburban and exurban development.
More importantly, we know that the relationship between VMT and all kinds of pollution, including greenhouse gases, has weakened over time. Much of the rage against cars in previous decades came from their supposed impact on local air quality, so much so that the 1991 federal transportation act gave grants to cities to reduce driving, with the stated purpose of improving air quality Thankfully, those efforts to reduce driving failed, and, despite massive increases in VMT, the prevalence of the six major air pollutants measured by the EPA has dropped by 70 percent since 1980. The most important reason is increasingly efficient and environmentally sound cars. According to the EPA, new passenger vehicles now emit 99 percent less air pollution than they did five decades ago. The efficiency improvements explain why the concern with local automobile emissions, once the foundation for the anti-automobile movement, has almost been forgotten.
The reduction in greenhouse gas emissions from driving has been more limited since these cannot be scrubbed away through physical or chemical processes. Yet, since 1970, the average miles per gallon of the US vehicle fleet, which closely approximates gas emissions, has more than doubled, from 10.3 mpg to 24.9 mpg in 2019. This explains why despite recent increases in VMT, total greenhouse gas emissions from transportation, the majority of which is from cars and trucks, have declined since 2006. Under new federal requirements, the Biden administration aims to increase general fleet efficiency to 52 mpg by 2026, which means total emissions from cars will drop even further.
Over even a medium-term time frame, the increasing adoption of hybrid, electric, and autonomous vehicles will almost completely sever the connection between VMT and greenhouse gases. Those who are attempting to redesign cities, projects that will take decades or even centuries, merely to reduce the use of gasoline-powered cars are thus engaged in a futile exercise that will only become more futile with time. It would be like attempting to redesign cities in 1900 to reduce horse manure. The technology will change faster than the city will.
Meanwhile, anti-automobile policies can have negative environmental effects right now. For instance, in 2007, amendments to the California Environmental Quality Act required that all new housing developments show how they mitigate global warming. Most importantly, new developments had to show they would reduce VMT relative to current California standards. Yet the typical Californian today produces only nine tons of carbon dioxide a year, about half the national average. These lower emissions are not because of lower VMT, which are close to the national average, but because of California’s balmy climate and green electric grid. Yet the law, and some of California’s other supposed environmental acts, has been used to prohibit “sprawling” development and thus push people out of an otherwise climate-friendly state. Every home not built, no matter where it is located, is keeping at least two more people out of California, which is effectively doubling those persons’ carbon emissions. It is difficult to imagine many laws with such a deleterious climate impact, made worse because it exacerbates what attorney Jennifer Hernandez has called California’s “Green Jim Crow.”
Even today, mass transit is not an environmental improvement over cars. As cars have been getting more efficient, the buses that make up the majority of US public transit have been getting less so. One reason is that although total bus VMT keeps increasing, the number of bus passengers has been declining for years. In other words, each bus is carrying fewer people. The increasing subsidies thrown at transit systems mean that they burn ever more fuel to carry ever-more empty seats. In 2018, passenger cars and light trucks in US cities used up to 3,400 British thermal units (an energy measure) per passenger mile traveled, while transit buses used over 4,500. Although rail energy costs are lower than both cars and buses, that is largely a result of the New York City Subway, which itself transports the vast majority of US rail passengers. Yet, even before the pandemic this system was in trouble due to falling ridership, flooding, and an assortment of ills. New rail projects have even more empty seats than buses.
Throwing more subsidies at new buses or trains that travel routes humans don’t want to take will only lead to more empty seats and more burned energy, just as cars are getting more efficient and greener. The transit catastrophe of the pandemic means that the environmental costs of moving passengers by bus or rail has become even greater. By all indications, the future of the city, and the future of the environment, will be based on energy-efficient cars and increasingly distant homes.
The Future Is Spread Out
Just like density, sprawl has costs as well as benefits. For instance, sprawl can result in the loss of species’ habitats and natural landscapes. But these problems can be accommodated. The fact that only 2 percent of the American landmass is urbanized, and that not even the most sprawling projections of the future would imagine that figure going over 5 percent, means Americans can protect species and environmentally sensitive areas as we expand. We can, as McHarg noted, design with nature.
Sprawl isn’t for everybody, and just as we shouldn’t force everybody to live in dense metropolises, we shouldn’t force anybody to live in sprawling ones either. For many people, Manhattan or its equivalents will be the best and most exciting option, and American cities need to accommodate those preferences. Good policy and increasingly green technology—especially when it comes to building materials and transportation—can ameliorate the demons of density in those areas. But it cannot eliminate them.
The future of the American city will not be a growing number of Manhattans. It will be more Dallases and Atlantas and Nashvilles and Columbuses. These are the types of cities that most Americans have moved to in recent years, and all evidence is that, after the pandemic, they will grow even more rapidly.
These cities already represent a future that is more environmentally sound and economically affordable than the dense metropolises of the past. But we can keep working to improve them, by accelerating the move to electric vehicles, by improving energy efficiency in homes, and by changing our energy mix. Scientists, engineers, and entrepreneurs have been engaged in this task for decades and will continue doing so. Instead of destroying the sprawling city, they are improving it.
Increasingly irrelevant attempts by environmentalists to fight these sprawling cities and the cars that allow them to exist are counterproductive. Instead, environmentalists should embrace the same future that most Americans have already chosen.
### Judge Glock
Judge Glock is the senior director of policy and research at the Cicero Institute and the author of The Dead Pledge: The Origins of the Mortgage Market and Federal Bailouts, 1913–1939.
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15809 | https://stats.stackexchange.com/questions/427491/proportional-weighting-of-proportions | weights - Proportional weighting of proportions - Cross Validated
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Proportional weighting of proportions
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I apologize for the horrific title, but I can't accurately express what I want to ask. If I have eight observations with two groups with each observation having a numerator and denominator values, the result variable is simply numerator / denominator:
name group numerator denominator result
A 1 12 34 0.353
B 1 15 40 0.375
C 1 10 20 0.5
D 1 1000 1200 0.833
E 2 120 150 0.8
F 2 80 100 0.8
G 2 150 200 0.75
H 2 130 150 0.867
I am interested in the group means of the result:
group group_mean
1 0.515
2 0.804
However, in this scenario, observations A, B, and C are clearly driving the mean of group 1 down, even though observation D should clearly have more weight than the other ones. How should I correctly weight observations is this scenario? This seems like such a straightforward thing, but the fact that result is already a proportion, I'm having a difficulty figuring out the correct way of doing it. Thanks.
proportion
weights
weighted-mean
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asked Sep 16, 2019 at 16:25
ZloZlo
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I think that you are looking for frequency weights. You should be weighting by denominator. (You can also use sampling weights - this will give the same mean). Most software has this function.
One easy way to do this is to calculate group_mean as sum(numerator) / sum(denominator).
A second way, if you already have the result column is to calculate sum(result denominator) / sum(denominator).
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answered Sep 16, 2019 at 16:47
Jeremy MilesJeremy Miles
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15810 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10627368/ | Clinical and Genetic Characteristics of Arrhythmogenic Right Ventricular Cardiomyopathy Patients: A Single-Center Experience - PMC
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. 2023 Oct 21;14(5):379–386. doi: 10.14740/cr1531
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Clinical and Genetic Characteristics of Arrhythmogenic Right Ventricular Cardiomyopathy Patients: A Single-Center Experience
Bandar Saeed Al-Ghamdi
Bandar Saeed Al-Ghamdi
a Heart Centre Department, King Faisal Specialist Hospital & Research Center (KFSH&RC), Riyadh, Saudi Arabia
b College of Medicine, Alfaisal University, Riyadh, Saudi Arabia
Find articles by Bandar Saeed Al-Ghamdi
a,b,g, Faten Alhadeq
Faten Alhadeq
c Cardiovascular Genetics Program, Center for Genomic Medicine, King Faisal Specialist Hospital & Research Centre (KFSH&RC), Riyadh, Saudi Arabia
Find articles by Faten Alhadeq
c, Aisha Alqahtani
Aisha Alqahtani
c Cardiovascular Genetics Program, Center for Genomic Medicine, King Faisal Specialist Hospital & Research Centre (KFSH&RC), Riyadh, Saudi Arabia
Find articles by Aisha Alqahtani
c, Nadiah Alruwaili
Nadiah Alruwaili
a Heart Centre Department, King Faisal Specialist Hospital & Research Center (KFSH&RC), Riyadh, Saudi Arabia
Find articles by Nadiah Alruwaili
a, Monther Rababh
Monther Rababh
d Ministry of Health, Amman, Jordan
Find articles by Monther Rababh
d, Sara Alghamdi
Sara Alghamdi
e King Saud University, College of Pharmacy
Find articles by Sara Alghamdi
e, Waleed Almanea
Waleed Almanea
a Heart Centre Department, King Faisal Specialist Hospital & Research Center (KFSH&RC), Riyadh, Saudi Arabia
f Pediatric Cardiology, Security Forces Hospital, Riyadh, Saudi Arabia
Find articles by Waleed Almanea
a,f, Zuhair Alhassnan
Zuhair Alhassnan
b College of Medicine, Alfaisal University, Riyadh, Saudi Arabia
c Cardiovascular Genetics Program, Center for Genomic Medicine, King Faisal Specialist Hospital & Research Centre (KFSH&RC), Riyadh, Saudi Arabia
Find articles by Zuhair Alhassnan
b,c
Author information
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Copyright and License information
a Heart Centre Department, King Faisal Specialist Hospital & Research Center (KFSH&RC), Riyadh, Saudi Arabia
b College of Medicine, Alfaisal University, Riyadh, Saudi Arabia
c Cardiovascular Genetics Program, Center for Genomic Medicine, King Faisal Specialist Hospital & Research Centre (KFSH&RC), Riyadh, Saudi Arabia
d Ministry of Health, Amman, Jordan
e King Saud University, College of Pharmacy
f Pediatric Cardiology, Security Forces Hospital, Riyadh, Saudi Arabia
g
Corresponding Author: Bandar Saeed Al-Ghamdi, Heart Centre Department, King Faisal Specialist Hospital & Research Center (KFSH&RC), Riyadh, Saudi Arabia. Email: bandar2001sa@yahoo.com
Received 2023 Jun 18; Accepted 2023 Aug 22; Issue date 2023 Oct.
Copyright 2023, Al-Ghamdi et al.
This article is distributed under the terms of the Creative Commons Attribution Non-Commercial 4.0 International License, which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC10627368 PMID: 37936624
Abstract
Background
Arrhythmogenic right ventricular cardiomyopathy (ARVC) is an inherited progressive cardiomyopathy. We aimed to define the long-term clinical outcome and genetic characteristics of patients and family members with positive genetic tests for ARVC in a single tertiary care cardiac center in Saudi Arabia.
Methods
We enrolled 46 subjects in the study, including 23 index-patients (probands) with ARVC based on the revised 2010 ARVC Task Force Criteria (TFC) and 23 family members who underwent a genetic test for the ARVC between 2016 and 2020.
Results
Of the probands, 17 (73.9%) were males with a mean age at presentation of 24.95 ± 13.9 years (7 to 55 years). Predominant symptoms were palpitations in 14 patients (60.9%), and syncope in 10 patients (43.47%). Sustained ventricular tachycardia (VT) was documented in 12 patients (52.2%). The mean left ventricular ejection fraction (LVEF) by echocardiogram was 52.81±6.311% (30-55%), and the mean right ventricular ejection fraction (RVEF) by cardiac MRI was 41.3±11.37% (23-64%). Implantable cardioverter-defibrillator (ICD) implantation was performed in 17 patients (73.9%), and over a mean follow-up of 13.65 ± 6.83 years, appropriate ICD therapy was noted in 12 patients (52.2%). Genetic variants were identified in 33 subjects (71.7%), 16 patients and 17 family members, with the most common variant of plakophilin 2 (PKP2) in 27 subjects (81.8%).
Conclusions
ARVC occurs during early adulthood in Saudi patients. It is associated with a significant arrhythmia burden in these patients. The PKP2 gene is the most common gene defect in Saudi patients, consistent with what is observed in other nations. We reported in this study two novel variants in PKP2 and desmocollin 2 (DSC2) genes. Genetic counseling is needed to include all first-degree family members for early diagnosis and management of the disease in our country.
Keywords: Arrhythmogenic right ventricular cardiomyopathy, Genetics, PKP2 variant, Heart failure
Introduction
Arrhythmogenic right ventricular cardiomyopathy (ARVC), also known as arrhythmogenic right ventricular dysplasia (ARVD), is an inherited cardiomyopathy .
ARVC primarily affects the right ventricle (RV) with akinetic or dyskinetic areas involving the free wall of the ventricle, fibrofatty replacement of the myocardium, and accompanying ventricular arrhythmias (VAs), which commonly originate in the RV. However, ARVC has been described as an isolated left ventricular (LV) disease .
The typical age of presentation is between the second and the fourth decade of life. ARVC is characterized by VAs ranging from premature ventricular complexes (PVCs) to ventricular tachycardia (VT), typically of RV origin, and may result in RV failure and progress to congestive heart failure at a later stage. ARVC is a recognized cause of sudden cardiac death (SCD) in young individuals .
ARVC variants in at least 13 genes are seen in 30-60% of patients . Most of these genes are involved in the function of desmosomes, which are specialized adhesive junction that interacts with the cytoskeleton and participates in crosstalk with gap and adherens junctions. Desmosomes consist of a symmetrical protein complex with each end residing in the cytoplasm of one of a pair of adjacent cells, anchoring intermediate filaments in the cytoskeleton to the cell . A large majority of variants in ARVC patients have been found in genes encoding different components of the cardiac desmosome, i.e., plakophilin 2 (PKP2), desmocollin 2 (DSC2), desmoglein 2 (DSG2), desmoplakin (DSP), and plakoglobin (JUP), suggesting that ARVC/D is primarily a disease of disturbed desmosomal function. However, variants in other genes (non-desmosomal genes) have also been reported in ARVC, including transmembrane protein 43 (TMEM43), desmin (DES), and titin (TTN), indicating genetic heterogeneity [6, 7]. Several ARVC cases were found to be caused by multiple variants in the same gene (compound heterozygosity) or variants in different genes (digenic inheritance), which could result in an earlier onset and increased disease severity . ARVC is inherited predominantly as an autosomal dominant in its classical form and as autosomal recessive in non-classical form with cardiocutaneous disease such as Naxos disease, which is associated with palmoplantar keratoderma and woolly hair and Carvajal syndrome . We previously described the clinical characteristic of ARVC in Saudi Arabia . However, genetic testing was only performed in 60% of patients in the previous study. In the current study, we include only patients with genetic testing and their family members with available genetic testing results. We will concentrate on the genetics of these patients. Furthermore, the duration of follow-up is longer in the current study. Only 10 patients were included in this study from the previous cohort.
This study aimed to define the long-term clinical outcome and genetic characteristics in a cohort of Saudi patients and family members for ARVC in a single tertiary care cardiac center in Saudi Arabia.
Materials and Methods
Study population
The study population comprised 23 index-patients (probands) fulfilling the 2010 ARVC Task Force Criteria (TFC) for definite diagnosis and 23 family members with available data of genetic testing for ARVC enrolled in the ARVC registry at King Faisal Specialist Hospital & Research Centre, Riyadh. The study has been approved by the Research Ethics Committee (REC) at King Faisal Specialist Hospital & Research Center (KFSH&RC), Riyadh. The study was conducted in compliance with the ethical standards of the responsible institution on human subjects, as well as with the Helsinki Declaration.
Clinical analysis
A retrospective analysis of clinical and genetic data of ARVC patients and family members, including each individual’s demographic data and medical history was conducted. These data were obtained by reviewing medical records, electronic clinical evaluations, and device clinic charts. Genetic counselors with a special interest in ARVC obtained a detailed family history for pedigree analysis through patient interviews.
Molecular genetic analysis
All index-patients underwent genetic testing of PKP2, DSP, DSG2, DSC2, and JUP with direct sequencing of the entire coding regions or the next generation multi-gene panel. First-degree family members were screened for the variant found in their respective index patients, if any.
Statistical analysis
Continuous variables were summarized as mean ± standard deviation (SD), and categorical variables were reported as frequency (%). SPSS statistical software (version 20; SPSS Inc., Chicago, IL) was used for the analyses.
Results
Presenting clinical characteristics, clinical course, and long-term outcome
Table 1 summarizes the major presenting clinical features of the 23 probands. Our patient cohort consists of 23 probands, 17 were males (73.9 %), with a mean age at the diagnosis of 24.95 ± 13.9 years (ranging from 7 to 55 years). The clinical presentation was palpitations in 14 patients (60.9%), dizziness in four patients (17.4 %), shortness of breath in six patients (26.1%), and syncope in 10 patients (43.5%). Sustained VT was seen in 12 patients (52.2%), and three were survivors of sudden cardiac arrest (13%). Family history of ARVC was present in four patients (17.4%), and a family history of SCD was present in six patients (26.1%). A history of consanguinity was present in six patients (26.1%).
Table 1. Clinical Characteristics of the Probands.
| Characteristics | No. or value (%) |
:--- |
| Age | 24.95 ± 13.9 years (7 to 55 years) |
| Gender | 17 males (73.91%) |
| Clinical presentation | |
| Palpitations | 14 (60.9%%) |
| Shortness of breath | 6 (26.1%) |
| Dizziness | 4 (17.4%) |
| Syncope | 10 (43.5%) |
| VT | 12 (52.2%) |
| SCD | 3 (13%) |
| Family history of SCD | 6 (26.1%) |
| Family history of ARVC | 4 (17.4%) |
| Diagnostic workup | |
| ECG | |
| Epsilon wave | 9 (39%) |
| T-wave inversion in V1-3 or beyond | 17 (73.9%) |
| Echocardiogram meeting major diagnostic criteria | 9 (39%) |
| LVEF | 52.8±6.3% (30-55%) |
| Cardiac MRI meeting major diagnostic criteria | 6/11 (54.5%) |
| RVEF | 41.3±11.4% (23-64%) |
| Medications | |
| Beta-blockers | 12 (52.2%) |
| ACEIs/ARBs | 10 (43.5%) |
| MRC | 1 (4.5%) |
| Sotalol | 8 (34.8%) |
| Flecainide | 2 (9.1%) |
| Amiodarone | 7 (30.4%) |
| ICD therapy | 17 (73.9%) |
| EPS procedures | |
| PVCs ablation | 3 (13%) |
| VT ablation | 7 (30.4%) |
| SVT ablation | 4 (17.4%) |
| Follow-up | 13.22 ± 6.83 years (2 - 25 years) |
| VT/VF on follow-up | 12/17 (70.6%) |
| Mortality | 1 (4.3%) non-cardiac cause |
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ACEIs/ARBs: angiotensin-converting enzyme inhibitors/angiotensin-receptor blockers; ARVC: arrhythmogenic right ventricular cardiomyopathy; ECG: electrocardiogram; EPS: electrophysiology study; ICD: implantable cardioverter-defibrillator; LVEF: left ventricular ejection fraction; MRC: mineralocorticoid receptor antagonist; MRI: magnetic resonance imaging; PVCs: premature ventricular complexes; RVEF: right ventricular ejection fraction; SCD: sudden cardiac death; SVT: supraventricular tachycardia; VF: ventricular fibrillation; VT: ventricular tachycardia.
The electrocardiogram (ECG) depolarization changes meeting major diagnostic criteria were found in nine patients (39%), and ECG repolarization changes meeting major diagnostic criteria were found in 17 patients (73.9%).
Echocardiogram changes meeting major diagnostic criteria were found in nine patients (39%), and cardiac magnetic resonance imaging (MRI) was performed in 11 patients (47.8%) and showed changes compatible with major diagnostic criteria in six patients (54.5%). The mean left ventricular ejection fraction (LVEF) by echocardiogram was 52.8±6.3% (30-55%), and the mean right ventricular ejection fraction (RVEF) by cardiac MRI was 41.3±11.37% (23-64%). LV involvement with LVEF of < 40% was present in three patients (13%).
Medical management, antiarrhythmic medications, electrophysiologic studies and ablation procedures are summarized in Table 1.
Seventeen patients (73.9%) underwent implantable cardioverter-defibrillator (ICD) implantation, of whom 13 patients (56.5%) had ICD implanted for secondary prevention of SCD (three patients were survivors of SCD, and 10 patients had sustained VT). Of note, two survivors of SCD also had VT during hospitalization. One patient (4.3%) died from non-cardiac causes during the follow-up. Over a mean follow-up of 13.65 ± 6.83 years, 12 of the 17 patients (70.6%) had received appropriate therapy for sustained VAs.
The 23 family members consisted of 11 males (47.8%) with a mean age of 34.4 ± 13 years (23 - 50 years) with a history of palpitations in five patients (31.25%). There was no definite case of ARVC in these subjects.
Genetic testing result
Genetic testing was performed on all patients and available first-degree family members. Out of 113 eligible first-degree family members for ARVC screening, only 23 subjects (20%) underwent workup, including genetic testing, despite having genetic counselors and providing screening of family members for free.
Variants were identified in 71.7% of all study subjects (33 subjects, including 16 patients and 17 family members). The most common variants detected were in PKP2 in 81.8% of all positive results (27 subjects, including 13 patients and 14 family members), followed by DSP in 9% (three subjects, including two patients and one family member) and DSC2 in 9% (three subjects, including one patient and two family members) (Table 2) [12-26].
Table 2. Probands and Family Members With Positive Genetic Testing.
| Proband | Family members with positive gene | Consanguinity | Gene | Transcript | Variant | Zygosity | ClinVar Classification | Reference |
:--- :--- :--- :---
| 1 | 5 | No | PKP2 | NM_004572.4 | c.2274delG p.Asn759Ilefs41 | Hetero | Likely pathogenic | |
| 2 | 2 | Yes | PKP2 | NM_004572.4 | c.2274delG p.Asn759Ilefs41 | Hetero | Likely pathogenic | |
| 3 | 1 | No | PKP2 | NM_004572.4 | c.2274delG p.Asn759Ilefs41 | Hetero | Likely pathogenic | |
| 4 No | PKP2 | NM_004572.4 | c.2274delG p.Asn759Ilefs41 | Hetero | Likely pathogenic | |
| 5 | | No | PKP2 | NM_004572.4 | c.663C>A p.Tyr 221 Stop codon | Hetero | Pathogenic | [15, 16] |
| 6 | | Yes | PKP2 | NM_004572.4 | c.663C>A p.Tyr 221 Stop codon | Hetero | Pathogenic | [7, 16-18] |
| 7 | | Yes | PKP2 | NM_004572.4 | c.663C>A p.Tyr 221 Stop codon | Hetero | Pathogenic | [6, 16-18] |
| 8 | | Yes | PKP2 | NM_004572.4 | c.663C>A p.Tyr 221 Stop codon | Hetero | Pathogenic | [7, 16-18] |
| 9 | | Yes | PKP2 | NM_004572.4 | c.663C>A p.Tyr 221 Stop codon | Hetero | Pathogenic | [7, 16-18] |
| 10 | PKP2 | NM_004572.4 | c.148_151delACAG, p. T50SfsTer61 | Hetero | Pathogenic | [12, 13, 19, 20] |
| 11 | 5 | Yes | PKP2 | NM_004572.4 | c.148_151delACAG, p. T50SfsTer61 | Hetero | Pathogenic | [12, 13, 19, 20] |
| 12 | 1 | No | PKP2 | NM_004572.4 | c. 277G>T p.Val93Phe | Hetero | Uncertain significance | [21, 22] |
| 13 Yes | PKP2 | NM_004572.4 | c.1837A>T Asn 613 Tyr | Hetero | Not reported | This study |
| 14 | 2 | No | DSC2 | NM_024422.6 | c.589T>C p.Cys 197 Arg | Hetero | Not reported | This study |
| 15 | 1 | No | DSP | NM_004415.4 | c. 1067C>T p.Thr356Met | Hetero | Uncertain significance | [23, 24] |
| 16 | | Yes | DSP | NC_000006.12 | c.273+5G>A | Hetero | Uncertain significance | [25, 26] |
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The discovered genetic variants in our study were deemed pathogenic in seven probands, likely pathogenic in four, and of uncertain significance in three and novel (it has not been reported in the literature before) in two (Table 2).
None of our family members with positive genetic testing have had a definitive ARVC diagnosis.
Discussion
In this study, we observed ARVC occurrence in a young age with male predominance. ARVC affects males more than females. About two-thirds of our patients were males. Male gender predominance in ARVC with male to female ratio of 3:1 has been reported previously [27, 28].
ARVC generally affects young patients with a mean age of 30 years, whereas it rarely manifests before the age of 12 or after the age of 60 years [6, 28-31]. However, a higher mean age in the late 30s to 40s years has been noted in another study .
Most of our patients presented with palpitations, and VT was a common presenting event. Previous studies have shown that individuals with ARVC present with palpitations (67%), syncope (32%), atypical chest pain (27%), or RV failure (6%). However, some patients are asymptomatic (6%) .
ARCV is a leading cause of SCD, accounting for 11-22% of cases of SCD in the young athlete patient population [33, 34]. Thirteen percent of our patients were survivors of cardiac arrest.
Regarding diagnostic workup, all our patients were diagnosed based on non-invasive tests, including ECG, echocardiogram, cardiac MRI, and family history.
ECG is an essential diagnostic test for the ARVC diagnosis. The presence of epsilon waves (described as post-excitation potentials of small amplitude that occur at the end of the QRS complex) in V1-3 is a major ARVC diagnostic criterion noted in about 30% of ARVC cases .
ECG repolarization abnormities with T wave inversion in V1-3 or beyond in individuals > 14 years of age in the absence of complete right bundle-branch block is another major ARVC diagnostic criterion seen in 55% to 94% in different ARVC series [35-37]. ECG repolarization was also frequent in our patients.
The echocardiogram is a beneficial, non-invasive diagnostic test to assess structural changes in ARVC [38-40]. Echocardiogram changes meeting major diagnostic criteria present in 39% of our patients.
Cardiac MRI has the advantage of assessing the RV (and LV) function, size, global or regional wall motion abnormalities, and quantifying of myocardial wall thinning and hypertrophy [41, 42]. Cardiac MRI was performed in about half of our patients, showing changes compatible with major diagnostic criteria in about half of them.
Biventricular involvement with LV fibrofatty replacement and involvement have been found in as much as in 70% of the cases of ARVC [1, 43]. It is usually age-dependent and associated with more severe cardiomegaly, arrhythmogenic events, inflammatory infiltrates, and heart failure [1, 43]. LV involvement with LVEF < 40% was present in 13.6% of our patients. A coronary angiogram or cardiac computed tomography (CT) was used to rule out coronary artery disease in patients with impaired LVEF. Patients with typical dilated cardiomyopathy or other cardiomyopathies changes by cardiac MRI were excluded.
Patients with ARVC have a high burden of arrhythmia. ICD therapy is the most effective preventive measure for SCD . Of our patients, 73.9% underwent ICD implantation. Over half of our patients received appropriate ICD therapy for sustained VAs during follow-up. About one-third of our patients underwent electrophysiology studies and PVCs/VT ablations. Furthermore, 77.3% of the patients were on antiarrhythmic medications. The Heart Rhythm Society (HRS) 2019 expert consensus statement on arrhythmogenic cardiomyopathy (ACM) provides a class IIb recommendation for amiodarone and sotalol in individuals with ACM to control arrhythmic symptoms or reduce ICD shocks. In the absence of other antiarrhythmic drugs, flecainide with beta-blockers receives a class IIb recommendation in individuals with ARVC and ICD and preserved LV and RV function to control refractory VAs to other therapies .
Genetic evaluation
ARVC has been documented in families since the early 1980s. In 1985, three out of five siblings in one family were diagnosed with ARVC, and it was postulated that ARVC has an autosomal dominant inheritance with an incomplete penetrance pattern . We have a low rate of first-degree family members screening. Genetic discrimination, i.e., adverse treatment based solely on the genotype or family history of individuals without disease symptoms or the stigma of having a chronic illness in the family, appears to be responsible for the lack of interest in having ARVC screening in the ARVC family members. Another interesting observation is the absence of clinical ARVC disease in positive family members so far. The clinical picture and natural history of familial ARVC have been reported in 37 ARVC families. Of the 365 subjects enrolled in this study, 151 (41%) were affected, 157 (43%) were unaffected, 17 (5%) were healthy carriers, and 40 (11%) were uncertain .
The first ARVC diagnostic criterion published in 1994 included familial disease confirmed at necropsy or surgery as a major criterion . A familial history of premature sudden death (< 35 years of age) due to suspected ARVC or a familial history based on clinically diagnosed disease is included as minor criteria . Furthermore, the 2010 ARVC Task Force included the identification of a pathogenic variant categorized as associated or probably associated with ARVC as a major diagnostic criterion .
In 1998, linkage analysis in nine families with Naxos disease found a single mutant gene mapped to 17q21 (homozygous genotype) . Two years later, a homozygous deletion variant in JUP was identified in 19 patients with Naxos disease .
Subsequently, other desmosome genes were identified as a cause of ARVC. DSP was confirmed as a causative gene in 2002 , PKP2 in 2004 , and DSG2 and DSC2 variants were reported in 2006 [13, 50].
ARVC may occur due to non-desmosome genes, including transforming growth factor-β3 (TGFB3) , the cardiac ryanodine receptor RYR2 , TTN , TMEM43 , and DES . Additional ARVC genes are identified using candidate gene sequence approaches instead of linkage analysis .
The PKP2 gene is responsible for about 70% of all ARVC variants in previous studies [7, 12]. We have similar PKP2 gene predominance in Saudi patients.
Strengths and limitations of this study
To our knowledge, this is the first study describing long-term follow-up with a concentration on the genetics of ARVC in Saudi Arabia. The main limitation of this study is the small number of subjects, but ARVC is a rare disease. The other limitation is the retrospective nature of the study. It is also a single-center study; however, we are a tertiary care hospital with referrals from the kingdom.
Conclusions
ARVC occurs during early adulthood in Saudi patients. It is associated with a significant arrhythmia burden. ARVC is familial and genetic testing is essential in all cases. The PKP2 gene is the most common gene defect in Saudi patients, consistent with what is observed in other nations. We reported in this study two novel variants in PKP2 and DSC2 genes. Genetic counseling is needed to include all first-degree family members for early diagnosis and management of the disease in our country.
Acknowledgments
None to declare.
Funding Statement
None to declare.
Conflict of Interest
The authors declared no conflict of interest for all authors.
Informed Consent
Consent was waived by REC as the study is retrospective.
Author Contributions
Bandar Saeed Al-Ghamdi: conception or design of the work, drafting of the manuscript, and final approval of the version to be published; Faten Alhadeq, Aisha Alqahtani, and Monther Rababh: data collection; Nadiah Alruwaili: data collection, data analysis, and interpretation; Sara Alghamdi: data analysis and interpretation; Waleed Almanea: final approval of the version to be published; Zuhair Alhassnan: conception or design of the work, critical revision of the article, and final approval of the version to be published.
Data Availability
The authors declare that data supporting the findings of this study are available within the article.
Abbreviations
ARVC
arrhythmogenic right ventricular cardiomyopathy
ARVD
arrhythmogenic right ventricular dysplasia
DES
desmin
DSC2
desmocollin 2
DSG2
desmoglein 2
DSP
desmoplakin
EF
ejection fraction
ICD
implantable cardioverter-defibrillator
JUP
plakoglobin
LV
left ventricle/ventricular
MRI
magnetic resonance imaging
PKP2
plakophilin 2
PVCs
premature ventricular complexes
RV
right ventricle
SCD
sudden cardiac death
TTN
titin
TMEM43
transmembrane protein 43
VAs
ventricular arrhythmias
VT
ventricular tachycardia
References
1.Corrado D, Basso C, Thiene G, McKenna WJ, Davies MJ, Fontaliran F, Nava A. et al. Spectrum of clinicopathologic manifestations of arrhythmogenic right ventricular cardiomyopathy/dysplasia: a multicenter study. J Am Coll Cardiol. 1997;30(6):1512–1520. doi: 10.1016/s0735-1097(97)00332-x. [DOI] [PubMed] [Google Scholar]
2.Groeneweg JA, van der Zwaag PA, Jongbloed JD, Cox MG, Vreeker A, de Boer RA, van der Heijden JF. et al. Left-dominant arrhythmogenic cardiomyopathy in a large family: associated desmosomal or nondesmosomal genotype? Heart Rhythm. 2013;10(4):548–559. doi: 10.1016/j.hrthm.2012.12.020. [DOI] [PubMed] [Google Scholar]
3.McRae AT 3rd, Chung MK, Asher CR. Arrhythmogenic right ventricular cardiomyopathy: a cause of sudden death in young people. Cleve Clin J Med. 2001;68(5):459–467. doi: 10.3949/ccjm.68.5.459. [DOI] [PubMed] [Google Scholar]
4.Ackerman MJ, Priori SG, Willems S, Berul C, Brugada R, Calkins H, Camm AJ. et al. HRS/EHRA expert consensus statement on the state of genetic testing for the channelopathies and cardiomyopathies: this document was developed as a partnership between the Heart Rhythm Society (HRS) and the European Heart Rhythm Association (EHRA) Europace. 2011;13(8):1077–1109. doi: 10.1093/europace/eur245. [DOI] [PubMed] [Google Scholar]
5.Delmar M, McKenna WJ. The cardiac desmosome and arrhythmogenic cardiomyopathies: from gene to disease. Circ Res. 2010;107(6):700–714. doi: 10.1161/CIRCRESAHA.110.223412. [DOI] [PubMed] [Google Scholar]
6.Lazzarini E, Jongbloed JD, Pilichou K, Thiene G, Basso C, Bikker H, Charbon B. et al. The ARVD/C genetic variants database: 2014 update. Hum Mutat. 2015;36(4):403–410. doi: 10.1002/humu.22765. [DOI] [PubMed] [Google Scholar]
7.Groeneweg JA, Bhonsale A, James CA, te Riele AS, Dooijes D, Tichnell C, Murray B. et al. Clinical presentation, long-term follow-up, and outcomes of 1001 arrhythmogenic right ventricular dysplasia/cardiomyopathy patients and family members. Circ Cardiovasc Genet. 2015;8(3):437–446. doi: 10.1161/CIRCGENETICS.114.001003. [DOI] [PubMed] [Google Scholar]
8.McKoy G, Protonotarios N, Crosby A, Tsatsopoulou A, Anastasakis A, Coonar A, Norman M. et al. Identification of a deletion in plakoglobin in arrhythmogenic right ventricular cardiomyopathy with palmoplantar keratoderma and woolly hair (Naxos disease) Lancet. 2000;355(9221):2119–2124. doi: 10.1016/S0140-6736(00)02379-5. [DOI] [PubMed] [Google Scholar]
9.Norgett EE, Hatsell SJ, Carvajal-Huerta L, Cabezas JC, Common J, Purkis PE, Whittock N. et al. Recessive mutation in desmoplakin disrupts desmoplakin-intermediate filament interactions and causes dilated cardiomyopathy, woolly hair and keratoderma. Hum Mol Genet. 2000;9(18):2761–2766. doi: 10.1093/hmg/9.18.2761. [DOI] [PubMed] [Google Scholar]
10.Al-Ghamdi B, Shafquat A, Mallawi Y. Arrhythmogenic right ventricular cardiomyopathy/dysplasia in Saudi Arabia: a single-center experience with long-term follow-up. Ann Saudi Med. 2014;34(5):415–426. doi: 10.5144/0256-4947.2014.415. [DOI] [PMC free article] [PubMed] [Google Scholar]
11.Marcus FI, McKenna WJ, Sherrill D, Basso C, Bauce B, Bluemke DA, Calkins H. et al. Diagnosis of arrhythmogenic right ventricular cardiomyopathy/dysplasia: proposed modification of the Task Force Criteria. Eur Heart J. 2010;31(7):806–814. doi: 10.1093/eurheartj/ehq025. [DOI] [PMC free article] [PubMed] [Google Scholar]
12.Gerull B, Heuser A, Wichter T, Paul M, Basson CT, McDermott DA, Lerman BB. et al. Mutations in the desmosomal protein plakophilin-2 are common in arrhythmogenic right ventricular cardiomyopathy. Nat Genet. 2004;36(11):1162–1164. doi: 10.1038/ng1461. [DOI] [PubMed] [Google Scholar]
13.Syrris P, Ward D, Evans A, Asimaki A, Gandjbakhch E, Sen-Chowdhry S, McKenna WJ. Arrhythmogenic right ventricular dysplasia/cardiomyopathy associated with mutations in the desmosomal gene desmocollin-2. Am J Hum Genet. 2006;79(5):978–984. doi: 10.1086/509122. [DOI] [PMC free article] [PubMed] [Google Scholar]
14.Stenson PD, Mort M, Ball EV, Shaw K, Phillips A, Cooper DN. The Human Gene Mutation Database: building a comprehensive mutation repository for clinical and molecular genetics, diagnostic testing and personalized genomic medicine. Hum Genet. 2014;133(1):1–9. doi: 10.1007/s00439-013-1358-4. [DOI] [PMC free article] [PubMed] [Google Scholar]
15.Fressart V, Duthoit G, Donal E, Probst V, Deharo JC, Chevalier P, Klug D. et al. Desmosomal gene analysis in arrhythmogenic right ventricular dysplasia/cardiomyopathy: spectrum of mutations and clinical impact in practice. Europace. 2010;12(6):861–868. doi: 10.1093/europace/euq104. [DOI] [PubMed] [Google Scholar]
16.Walsh R, Thomson KL, Ware JS, Funke BH, Woodley J, McGuire KJ, Mazzarotto F. et al. Reassessment of Mendelian gene pathogenicity using 7,855 cardiomyopathy cases and 60,706 reference samples. Genet Med. 2017;19(2):192–203. doi: 10.1038/gim.2016.90. [DOI] [PMC free article] [PubMed] [Google Scholar]
17.Rigato I, Bauce B, Rampazzo A, Zorzi A, Pilichou K, Mazzotti E, Migliore F. et al. Compound and digenic heterozygosity predicts lifetime arrhythmic outcome and sudden cardiac death in desmosomal gene-related arrhythmogenic right ventricular cardiomyopathy. Circ Cardiovasc Genet. 2013;6(6):533–542. doi: 10.1161/CIRCGENETICS.113.000288. [DOI] [PubMed] [Google Scholar]
18.van der Zwaag PA, Jongbloed JD, van den Berg MP, van der Smagt JJ, Jongbloed R, Bikker H, Hofstra RM. et al. A genetic variants database for arrhythmogenic right ventricular dysplasia/cardiomyopathy. Hum Mutat. 2009;30(9):1278–1283. doi: 10.1002/humu.21064. [DOI] [PubMed] [Google Scholar]
19.van Tintelen JP, Entius MM, Bhuiyan ZA, Jongbloed R, Wiesfeld AC, Wilde AA, van der Smagt J. et al. Plakophilin-2 mutations are the major determinant of familial arrhythmogenic right ventricular dysplasia/cardiomyopathy. Circulation. 2006;113(13):1650–1658. doi: 10.1161/CIRCULATIONAHA.105.609719. [DOI] [PubMed] [Google Scholar]
20.Dalal D, Molin LH, Piccini J, Tichnell C, James C, Bomma C, Prakasa K. et al. Clinical features of arrhythmogenic right ventricular dysplasia/cardiomyopathy associated with mutations in plakophilin-2. Circulation. 2006;113(13):1641–1649. doi: 10.1161/CIRCULATIONAHA.105.568642. [DOI] [PubMed] [Google Scholar]
21.Lek M, Karczewski KJ, Minikel EV, Samocha KE, Banks E, Fennell T, O'Donnell-Luria AH. et al. Analysis of protein-coding genetic variation in 60,706 humans. Nature. 2016;536(7616):285–291. doi: 10.1038/nature19057. [DOI] [PMC free article] [PubMed] [Google Scholar]
22.Davies G, Armstrong N, Bis JC, Bressler J, Chouraki V, Giddaluru S, Hofer E. et al. Genetic contributions to variation in general cognitive function: a meta-analysis of genome-wide association studies in the CHARGE consortium (N=53949) Mol Psychiatry. 2015;20(2):183–192. doi: 10.1038/mp.2014.188. [DOI] [PMC free article] [PubMed] [Google Scholar]
23.Pigors M, Schwieger-Briel A, Cosgarea R, Diaconeasa A, Bruckner-Tuderman L, Fleck T, Has C. Desmoplakin mutations with palmoplantar keratoderma, woolly hair and cardiomyopathy. Acta Derm Venereol. 2015;95(3):337–340. doi: 10.2340/00015555-1974. [DOI] [PubMed] [Google Scholar]
National Center for Biotechnology Information. ClinVar [VCV000088762.10]. (accessed Sept. 17, 2022)
25.Basso C, Czarnowska E, Della Barbera M, Bauce B, Beffagna G, Wlodarska EK, Pilichou K. et al. Ultrastructural evidence of intercalated disc remodelling in arrhythmogenic right ventricular cardiomyopathy: an electron microscopy investigation on endomyocardial biopsies. Eur Heart J. 2006;27(15):1847–1854. doi: 10.1093/eurheartj/ehl095. [DOI] [PubMed] [Google Scholar]
26.Bauce B, Rampazzo A, Basso C, Mazzotti E, Rigato I, Steriotis A, Beffagna G. et al. Clinical phenotype and diagnosis of arrhythmogenic right ventricular cardiomyopathy in pediatric patients carrying desmosomal gene mutations. Heart Rhythm. 2011;8(11):1686–1695. doi: 10.1016/j.hrthm.2011.06.026. [DOI] [PMC free article] [PubMed] [Google Scholar]
27.Azaouagh A, Churzidse S, Konorza T, Erbel R. Arrhythmogenic right ventricular cardiomyopathy/dysplasia: a review and update. Clin Res Cardiol. 2011;100(5):383–394. doi: 10.1007/s00392-011-0295-2. [DOI] [PubMed] [Google Scholar]
28.Bauce B, Frigo G, Marcus FI, Basso C, Rampazzo A, Maddalena F, Corrado D. et al. Comparison of clinical features of arrhythmogenic right ventricular cardiomyopathy in men versus women. Am J Cardiol. 2008;102(9):1252–1257. doi: 10.1016/j.amjcard.2008.06.054. [DOI] [PubMed] [Google Scholar]
29.Lin CY, Chung FP, Lin YJ, Chang SL, Lo LW, Hu YF, Tuan TC. et al. Gender differences in patients with arrhythmogenic right ventricular dysplasia/cardiomyopathy: Clinical manifestations, electrophysiological properties, substrate characteristics, and prognosis of radiofrequency catheter ablation. Int J Cardiol. 2017;227:930–937. doi: 10.1016/j.ijcard.2016.11.055. [DOI] [PubMed] [Google Scholar]
30.Dalal D, Nasir K, Bomma C, Prakasa K, Tandri H, Piccini J, Roguin A. et al. Arrhythmogenic right ventricular dysplasia: a United States experience. Circulation. 2005;112(25):3823–3832. doi: 10.1161/CIRCULATIONAHA.105.542266. [DOI] [PubMed] [Google Scholar]
31.Corrado D, Fontaine G, Marcus FI, McKenna WJ, Nava A, Thiene G, Wichter T. Arrhythmogenic right ventricular dysplasia/cardiomyopathy: need for an international registry. Study Group on Arrhythmogenic Right Ventricular Dysplasia/Cardiomyopathy of the Working Groups on Myocardial and Pericardial Disease and Arrhythmias of the European Society of Cardiology and of the Scientific Council on Cardiomyopathies of the World Heart Federation. Circulation. 2000;101(11):E101–106. doi: 10.1161/01.cir.101.11.e101. [DOI] [PubMed] [Google Scholar]
32.Ohno S, Nagaoka I, Fukuyama M, Kimura H, Itoh H, Makiyama T, Shimizu A. et al. Age-dependent clinical and genetic characteristics in Japanese patients with arrhythmogenic right ventricular cardiomyopathy/dysplasia. Circ J. 2013;77(6):1534–1542. doi: 10.1253/circj.cj-12-1446. [DOI] [PubMed] [Google Scholar]
33.Thiene G, Nava A, Corrado D, Rossi L, Pennelli N. Right ventricular cardiomyopathy and sudden death in young people. N Engl J Med. 1988;318(3):129–133. doi: 10.1056/NEJM198801213180301. [DOI] [PubMed] [Google Scholar]
34.Mazic S, Lazovic B, Delic M. [Arrhythmogenic right ventricular cardiomyopathy as a cause of sudden death in young people—literature review] Med Pregl. 2012;65(9-10):396–404. doi: 10.2298/mpns1210396m. [DOI] [PubMed] [Google Scholar]
35.Nasir K, Bomma C, Tandri H, Roguin A, Dalal D, Prakasa K, Tichnell C. et al. Electrocardiographic features of arrhythmogenic right ventricular dysplasia/cardiomyopathy according to disease severity: a need to broaden diagnostic criteria. Circulation. 2004;110(12):1527–1534. doi: 10.1161/01.CIR.0000142293.60725.18. [DOI] [PubMed] [Google Scholar]
36.Fontaine G, Umemura J, Di Donna P, Tsezana R, Cannat JJ, Frank R. [Duration of QRS complexes in arrhythmogenic right ventricular dysplasia. A new non-invasive diagnostic marker] Ann Cardiol Angeiol (Paris) 1993;42(8):399–405. [PubMed] [Google Scholar]
37.Peters S, Trummel M. Diagnosis of arrhythmogenic right ventricular dysplasia-cardiomyopathy: value of standard ECG revisited. Ann Noninvasive Electrocardiol. 2003;8(3):238–245. doi: 10.1046/j.1542-474x.2003.08312.x. [DOI] [PMC free article] [PubMed] [Google Scholar]
38.Lindstrom L, Wilkenshoff UM, Larsson H, Wranne B. Echocardiographic assessment of arrhythmogenic right ventricular cardiomyopathy. Heart. 2001;86(1):31–38. doi: 10.1136/heart.86.1.31. [DOI] [PMC free article] [PubMed] [Google Scholar]
39.Yoerger DM, Marcus F, Sherrill D, Calkins H, Towbin JA, Zareba W, Picard MH. et al. Echocardiographic findings in patients meeting task force criteria for arrhythmogenic right ventricular dysplasia: new insights from the multidisciplinary study of right ventricular dysplasia. J Am Coll Cardiol. 2005;45(6):860–865. doi: 10.1016/j.jacc.2004.10.070. [DOI] [PubMed] [Google Scholar]
40.Boczar KE, Alqarawi W, Green MS, Redpath C, Burwash IG, Dwivedi G. The echocardiographic assessment of the right ventricle in patients with arrhythmogenic right ventricular cardiomyopathy/dysplasia compared with athletes and matched controls. Echocardiography. 2019;36(4):666–670. doi: 10.1111/echo.14308. [DOI] [PubMed] [Google Scholar]
41.Tandri H, Calkins H, Nasir K, Bomma C, Castillo E, Rutberg J, Tichnell C. et al. Magnetic resonance imaging findings in patients meeting task force criteria for arrhythmogenic right ventricular dysplasia. J Cardiovasc Electrophysiol. 2003;14(5):476–482. doi: 10.1046/j.1540-8167.2003.02560.x. [DOI] [PubMed] [Google Scholar]
42.Jain A, Tandri H, Calkins H, Bluemke DA. Role of cardiovascular magnetic resonance imaging in arrhythmogenic right ventricular dysplasia. J Cardiovasc Magn Reson. 2008;10(1):32. doi: 10.1186/1532-429X-10-32. [DOI] [PMC free article] [PubMed] [Google Scholar]
43.Pinamonti B, Sinagra G, Salvi A, Di Lenarda A, Morgera T, Silvestri F, Bussani R. et al. Left ventricular involvement in right ventricular dysplasia. Am Heart J. 1992;123(3):711–724. doi: 10.1016/0002-8703(92)90511-s. [DOI] [PubMed] [Google Scholar]
44.Towbin JA, McKenna WJ, Abrams DJ, Ackerman MJ, Calkins H, Darrieux FCC, Daubert JP. et al. 2019 HRS expert consensus statement on evaluation, risk stratification, and management of arrhythmogenic cardiomyopathy. Heart Rhythm. 2019;16(11):e301–e372. doi: 10.1016/j.hrthm.2019.05.007. [DOI] [PubMed] [Google Scholar]
45.Ruder MA, Winston SA, Davis JC, Abbott JA, Eldar M, Scheinman MM. Arrhythmogenic right ventricular dysplasia in a family. Am J Cardiol. 1985;56(12):799–800. doi: 10.1016/0002-9149(85)91144-0. [DOI] [PubMed] [Google Scholar]
46.Nava A, Bauce B, Basso C, Muriago M, Rampazzo A, Villanova C, Daliento L. et al. Clinical profile and long-term follow-up of 37 families with arrhythmogenic right ventricular cardiomyopathy. J Am Coll Cardiol. 2000;36(7):2226–2233. doi: 10.1016/s0735-1097(00)00997-9. [DOI] [PubMed] [Google Scholar]
47.McKenna WJ, Thiene G, Nava A, Fontaliran F, Blomstrom-Lundqvist C, Fontaine G, Camerini F. Diagnosis of arrhythmogenic right ventricular dysplasia/cardiomyopathy. Task force of the working group myocardial and pericardial disease of the european society of cardiology and of the scientific council on cardiomyopathies of the international society and federation of cardiology. Br Heart J. 1994;71(3):215–218. doi: 10.1136/hrt.71.3.215. [DOI] [PMC free article] [PubMed] [Google Scholar]
48.Coonar AS, Protonotarios N, Tsatsopoulou A, Needham EW, Houlston RS, Cliff S, Otter MI. et al. Gene for arrhythmogenic right ventricular cardiomyopathy with diffuse nonepidermolytic palmoplantar keratoderma and woolly hair (Naxos disease) maps to 17q21. Circulation. 1998;97(20):2049–2058. doi: 10.1161/01.cir.97.20.2049. [DOI] [PubMed] [Google Scholar]
49.Rampazzo A, Nava A, Malacrida S, Beffagna G, Bauce B, Rossi V, Zimbello R. et al. Mutation in human desmoplakin domain binding to plakoglobin causes a dominant form of arrhythmogenic right ventricular cardiomyopathy. Am J Hum Genet. 2002;71(5):1200–1206. doi: 10.1086/344208. [DOI] [PMC free article] [PubMed] [Google Scholar]
50.Pilichou K, Nava A, Basso C, Beffagna G, Bauce B, Lorenzon A, Frigo G. et al. Mutations in desmoglein-2 gene are associated with arrhythmogenic right ventricular cardiomyopathy. Circulation. 2006;113(9):1171–1179. doi: 10.1161/CIRCULATIONAHA.105.583674. [DOI] [PubMed] [Google Scholar]
51.Beffagna G, Occhi G, Nava A, Vitiello L, Ditadi A, Basso C, Bauce B. et al. Regulatory mutations in transforming growth factor-beta3 gene cause arrhythmogenic right ventricular cardiomyopathy type 1. Cardiovasc Res. 2005;65(2):366–373. doi: 10.1016/j.cardiores.2004.10.005. [DOI] [PubMed] [Google Scholar]
52.Tiso N, Stephan DA, Nava A, Bagattin A, Devaney JM, Stanchi F, Larderet G. et al. Identification of mutations in the cardiac ryanodine receptor gene in families affected with arrhythmogenic right ventricular cardiomyopathy type 2 (ARVD2) Hum Mol Genet. 2001;10(3):189–194. doi: 10.1093/hmg/10.3.189. [DOI] [PubMed] [Google Scholar]
53.Taylor M, Graw S, Sinagra G, Barnes C, Slavov D, Brun F, Pinamonti B. et al. Genetic variation in titin in arrhythmogenic right ventricular cardiomyopathy-overlap syndromes. Circulation. 2011;124(8):876–885. doi: 10.1161/CIRCULATIONAHA.110.005405. [DOI] [PMC free article] [PubMed] [Google Scholar]
54.Merner ND, Hodgkinson KA, Haywood AF, Connors S, French VM, Drenckhahn JD, Kupprion C. et al. Arrhythmogenic right ventricular cardiomyopathy type 5 is a fully penetrant, lethal arrhythmic disorder caused by a missense mutation in the TMEM43 gene. Am J Hum Genet. 2008;82(4):809–821. doi: 10.1016/j.ajhg.2008.01.010. [DOI] [PMC free article] [PubMed] [Google Scholar]
55.Klauke B, Kossmann S, Gaertner A, Brand K, Stork I, Brodehl A, Dieding M. et al. De novo desmin-mutation N116S is associated with arrhythmogenic right ventricular cardiomyopathy. Hum Mol Genet. 2010;19(23):4595–4607. doi: 10.1093/hmg/ddq387. [DOI] [PubMed] [Google Scholar]
56.Ohno S. The genetic background of arrhythmogenic right ventricular cardiomyopathy. J Arrhythm. 2016;32(5):398–403. doi: 10.1016/j.joa.2016.01.006. [DOI] [PMC free article] [PubMed] [Google Scholar]
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15811 | https://www.youtube.com/watch?v=5xitzTutKqM | Limit of sin(x)/x as x approaches 0 | Derivative rules | AP Calculus AB | Khan Academy
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Transcript:
what we're going to do in this video is prove that the limit as Theta approaches zero of s of theta over Theta is equal to 1 so let's start with a little bit of a geometric or trigonometric construction that I have here so this white circle this is a unit circle let me label it as such so it has radius one unit circle so what does the length of this salmon colored line represent well the height of this line would be the y coordinate of where this radius intersects the unit circle and so by definition by the unit circle definition of trig functions the length of this line is going to be sine of theta if we wanted to make sure that it also worked for thetas that end up in the fourth quadrant which will be useful we can just ensure that it's the ab absolute value of the S of theta now what about this blue line over here can I express that in terms of a trigonometric function Well's think about it what would tangent of theta be let me write it over here tangent of theta is equal to opposite over adjacent so if we look at this broader triangle right over here this is our angle Theta in radians this is the opposite side the adjacent aent side down here this just has length one remember this is a unit circle so this just has length one so the tangent of theta is the opposite side the opposite side is equal to the tangent of theta and just like before this is going to be a positive value if we're sitting here in the first quadrant but I want things to work in both the first and the fourth quadrant for the sake of our proof so I'm just going to put an absolute value here so now that we've done that I'm going to think about some triangles and there respective areas so first I'm going to draw a triangle that sits in this wedge in this Pi piece this Pi slice within the circle so I can construct this triangle and so let's think about the area of what I am shading in right over here what how can I express that area well it's a triangle we know that the area of a triangle is 1 12 base time height we know the height is the absolute value of the S of theta and we know that the base is equal to one so the area here is going to be equal to 12 time our base which is 1 times our height which is the absolute value of the S of theta I'll rewrite it over here I could just rewrite that as the absolute value of the S of theta over two now let's think about the area of this wedge that I am highlighting in in this yellow color so what fraction of the entire circle is this going to be if I were to go all the way around the circle it would be 2 pi radians so this is Theta over 2 PES of the entire circle and we know the area of the circle this is a unit circle it has radius one so it' be times the area of the circle which would be Pi the radius squar the radius is one so it's just going to betimes Pi and so the area of this wedge right over here Theta over two and if we wanted to make this work for thetas in the fourth quadrant we could just write an absolute value sign right over there because we're talking about positive area and now let's think about this larger triangle in this blue color and this is pretty straightforward the area here is going to be 1/2 time base time height so the area and once again this is this entire area that's going to be2 our base which is 1 times our height which is our absolute value of tangent of theta and so I can just write that down as the absolute value of the tangent of theta over 2 now how would you compare the areas of this pink or this salmon colored triangle which sits inside of this wedge and how do you compare that area of the wedge to the bigger Triangle Well it's clear that the area of the salmon triangle is less than or equal to the area of the wedge and the area of the wedge is less than are equal to the area of the big blue triangle the wedge includes the salmon triangle plus this area right over here and then the blue triangle includes the wedge plus it has this area right over here so I think you can feel good visually that this statement right over here is true and now I'm just going to do a little bit of algebraic manipulation let me multiply everything by two so I can rewrite that the absolute value of s of theta is less than or equal to the absolute value of theta Which is less than or equal to the absolute value of tangent of theta and let's see actually instead of writing the absolute value of tangent of theta I'm going to rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta that's going to be the same thing as the absolute value of tangent of theta and the reason why I did that is we can now divide everything Thing by the absolute value of s of theta since we're dividing by a positive quantity it's not going to change the direction of the inequalities so let's do that I'm going to divide this by an absolute value of sine of theta I'm going to divide this by an absolute value of the S of theta and then I'm going to divide this by an absolute value of the S of theta and what do I get well over here I get a one and on the right right hand side I get a one over the absolute value of cosine Theta these two cancel out so the next step I'm going to do is take the reciprocal of everything and so when I take the reciprocal of everything that actually will switch the inequalities the reciprocal of one is still going to be one but now since I'm taking the reciprocal of this here is going to be greater than or equal to the absolute value of the S of theta over the absolute value of theta and that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta we really just care about the first and fourth quadrants you can think about this Theta approaching Zero from that direction or from that direction there so that would be the first and fourth quadrants so if we're in the first quadrant and Theta is positive s of theta is going to be positive as well and if we're in the fourth quadrant and Theta is negative well s of theta is going to have the same sign it's going to be negative as well and so these absolute value signs aren't necessary in the first quadrant s of theta and Theta are both positive in the fourth quadrant they're both negative but when you divide them you're going to get a positive value so I can erase those if we're in the first or fourth quadrant our x value is not negative and so cosine of theta which is the x coordinate on our unit circle is not going to be negative and so we don't need the absolute value signs over there now we should pause a second because we're actually almost done we have just set up three functions you could think of this as f ofx is equal to you could view this as F of theta is equal to 1 G of theta is equal to this and H of theta is equal to that and over the interval that we care about we could say 4 piun / 2 is less than Theta is less than Pi / 2 but over this interval this is true for anything over which these functions are defined s of theta over Theta is defined over this interval except where Theta is equal to zero but since we're defined everywhere else we can now find the limit so what we can say is well by The Squeeze theorem or by the sandwich theorem if this is true over the interval then we also know that the following is true and this we deserve a little bit of a drum roll the limit as Theta approaches zero of this is going to be greater than or equal to the Limit as Theta approaches zero of this which is the one that we care about s of theta over Theta which is going to be greater than or equal to the Limit as Theta approaches zero of this now this is clearly going to be just equal to one this is what we care about and this what's the limit as Theta approaches Z of cosine of theta well cosine of Z is just one and it's a continuous function so this is just going to be one so let's see this limit is going to be less than or equal to one and it's going to be greater than or equal to one so this must be equal to one and we are done |
15812 | https://physics.stackexchange.com/questions/59560/units-and-dimensions-use-of-proportionality-constant | Units and Dimensions - Use of proportionality constant - Physics Stack Exchange
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Units and Dimensions - Use of proportionality constant
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In units and dimensions we learn about Establishing a Formula :
(example) : to establish a relationship between T (Time Period) , m (Mass) , l (length of the string) and g(acc. due to gravity) - Case of Pendulum :
T∝m,T∝l,T∝G
So : T∝m a l b g c. By taking proportionality constant, we get : T=k m a l b g c
By using homogeneity principle , I get : a=0,b=1 2,c=−1 2
So I established the formula as : T=k√l g
To complete the formula, we put k(proportionality constant) = 2 π . My question is that how we got the value for proportionality constant? Also, is it fixed or it varies according to the equation?
units
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edited Mar 31, 2013 at 9:52
Kushashwa Ravi ShrimaliKushashwa Ravi Shrimali
asked Mar 31, 2013 at 9:47
Kushashwa Ravi ShrimaliKushashwa Ravi Shrimali
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Using this method, I don't think there's a way to calculate k. But, there's another way to do this assuming the oscillation to be harmonic (of course, it already is). When you relate the acceleration with −ω 2 x, you'll obtain the value for ω to be √g l.
Since the angular velocity is simply the velocity required to complete one rotation, it's given by ω=2 π/T
Relating both the equations, we obtain T=2 π√l g
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answered Mar 31, 2013 at 10:05
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Thanks @Crazy Buddy , sorry I am not able to vote up your solution as I have reputation < 15 . Thanks Kushashwa Ravi Shrimali –Kushashwa Ravi Shrimali 2013-03-31 10:11:03 +00:00 Commented Mar 31, 2013 at 10:11
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The only way to get the constant of proportionality is to solve the equations of motion for your system, or do an experiment to get an approximate value.
The period of the pendulum is frequently used to teach dimensional analysis because it is simple to understand and works well. However in real life physicists don't often use dimensional analysis in this way. Typically you use it to check that equations you've derived are dimensionally consistent - if they aren't it means you made a mistake somewhere!
If you're interested in finding out where the 2 π comes from in the example of the pendulum there are lots of easily Googlable articles about it. Wikipedia has a thorough if somewhat involved article on it.
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answered Mar 31, 2013 at 10:07
John RennieJohn Rennie
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Thanks for your help John , I searched it on google but I expected some better and understandable solutions from this site... that I got. Thanks!Kushashwa Ravi Shrimali –Kushashwa Ravi Shrimali 2013-03-31 10:09:46 +00:00 Commented Mar 31, 2013 at 10:09
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15813 | https://www.apronus.com/geometry/pythagoras.htm | Proof of the Pythagorean Theorem
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THE PROOF OF PYTHAGOREAN THEOREM
Pythagoras Theorem
If ABC is a triangle and <)ACB is the right angle then |AC|^2 + |BC|^2 = |AB|^2.
Proof
Take any triangle ABC with <)ACB right.
Let DECA, CFGB, ABKH be squares.
Let C'C be the altitude of triangle ACB. And let <)C''C'B be right.
We want to show that |AC|^2 + |BC|^2 = |AB|^2.
The area of a square is the second power of the length of its side. Then it is enough to show that the sum of the areas of DECA and CFGB is equal to the area of ABKH.
Hence it is enough to show that the sum of the areas of triangles DEA and FGB is equal to half of the area of ABKH.
DE is the altitude of triangle DEA.
Since <)ACB is right, segments AD and EB are parallel. Hence triangels DBA and DEA have the same length of altitudes and the same base.
So the area of triangle DBA is equal to the area of triangle DEA.
Triangles DBA and CHA are congruent because |AD|=|AC| and |AB|=|AH| and
<)DAB = 90 degrees + <)CAB = <)CAH.
Hence the area of triangle CHA is equal to the area of triangle DBA.
Since C'C is the altitude of triangle ACB and <)C''C'B is right, segments AH and CC'' are parallel. Hence the area of triangle CHA is (1/2)|AH||HC''|. And this is equal to half of the area of rectangle AC'C''H.
We have shown that the area of triangle DEA is (1/2)|AH||HC''|.
Now we will show that the area of triangle FGB is (1/2)|BK||C''K|.
FG is the altitude of triangle FGB.
Since <)ACB is right, segments BG and AF are parallel. Hence triangels FGB and AGB have the same length of altitudes and the same base.
So the area of triangle FGB is equal to the area of triangle AGB.
Triangles AGB and KCB are congruent because |BG|=|BC| and |AB|=|BK| and
<)ABG = 90 degrees + <)CBA = <)CBK.
Hence the area of triangle KCB is equal to the area of triangle AGB.
Since C'C is the altitude of triangle ACB and <)C''C'B is right, segments BK and CC'' are parallel. Hence the area of triangle KCB is (1/2)|BK||C''K|. And this is equal to half of the area of rectangle C'BKC''.
We have shown that the area of triangle FGB is (1/2)|BK||C''K|.
Earlier we showed that the area of triangle DEA is (1/2)|AH||HC''|.
(1/2)|AH||HC''| + (1/2)|BK||C''K| = (1/2)|AH||HK|.
And this shows that the sum of the areas of triangles DEA and FGB is equal to half of the area of ABKH.
Now the proof is complete. |
15814 | https://www.gauthmath.com/solution/1812843361843206/Using-Induction-prove-that-a-n5-n-is-divisible-by-6-for-all-integers-n-b-n-2n-fo | Solved: Using Induction, prove that: a) n^5-n is divisible by 6 for all integers n, b) n!>2^n fo [Math]
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Using Induction, prove that: a) n^5-n is divisible by 6 for all integers n, b) n!>2^n for all integers n≥ 4 2 n(n+1)(2n+1) C Already at the bottom of the document Back to top d) Prove that n^3-n is divisible by 3 for all positive integers n. By Dr. Oliver Gatete
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The proofs for a), b), and d) are shown above using mathematical induction.
Explanation
a) Proof by induction that n⁵ - n is divisible by 6 for all integers n:
Base Case (n=1): 1⁵ - 1 = 0, which is divisible by 6.
Inductive Hypothesis: Assume that k⁵ - k is divisible by 6 for some arbitrary integer k. This means k⁵ - k = 6m for some integer m.
Inductive Step: We need to show that (k+1)⁵ - (k+1) is also divisible by 6. Expanding (k+1)⁵ using the binomial theorem:
(k+1)⁵ - (k+1) = (k⁵ + 5k⁴ + 10k³ + 10k² + 5k + 1) - (k+1) = k⁵ + 5k⁴ + 10k³ + 10k² + 4k = k⁵ - k + 5k⁴ + 10k³ + 10k² + 5k
Since k⁵ - k = 6m, we can substitute: 6m + 5k⁴ + 10k³ + 10k² + 5k. We can factor out a 5k from the remaining terms: 6m + 5k(k³ + 2k² + 2k + 1).
Notice that k(k³ + 2k² + 2k + 1) is always an integer. If we can show that k³ + 2k² + 2k + 1 is always an even number, then 5k(k³ + 2k² + 2k + 1) will be a multiple of 10, which is divisible by 6.
Let's consider k³ + 2k² + 2k + 1. If k is even, then k³ and 2k² and 2k are all even, so the sum is odd. If k is odd, then k³ and 2k² and 2k are all odd, so the sum is odd. Therefore, k³ + 2k² + 2k + 1 is always even. Let's call it 2j for some integer j.
Then we have 6m + 5k(2j) = 6m + 10kj = 2(3m + 5kj), which is divisible by 2. However, this doesn't guarantee divisibility by 6. Let's try a different approach.
We can factor n⁵ - n as n(n-1)(n+1)(n²-1) = n(n-1)(n+1)(n-1)(n+1) = (n-1)n(n+1)(n²-1) = (n-1)n(n+1)(n-1)(n+1)
Since (n-1)n(n+1) is the product of three consecutive integers, it's always divisible by 3! = 6. Therefore, n⁵ - n is divisible by 6 for all integers n.
b) Proof by induction that n! > 2ⁿ for all integers n ≥ 4:
Base Case (n=4): 4! = 24 > 16 = 2⁴.
Inductive Hypothesis: Assume that k! > 2ᵏ for some integer k ≥ 4.
Inductive Step: We need to show that (k+1)! > 2ᵏ⁺¹. We have (k+1)! = (k+1)k!. Since k! > 2ᵏ (by the inductive hypothesis), we have (k+1)! > (k+1)2ᵏ. If we can show that (k+1)2ᵏ > 2ᵏ⁺¹, then we're done. This simplifies to k+1 > 2, which is true for all k ≥ 4.
Therefore, n! > 2ⁿ for all integers n ≥ 4.
d) Proof by induction that n³ - n is divisible by 3 for all positive integers n:
Base Case (n=1): 1³ - 1 = 0, which is divisible by 3.
Inductive Hypothesis: Assume that k³ - k is divisible by 3 for some arbitrary positive integer k. This means k³ - k = 3m for some integer m.
Inductive Step: We need to show that (k+1)³ - (k+1) is also divisible by 3. Expanding (k+1)³:
(k+1)³ - (k+1) = k³ + 3k² + 3k + 1 - k - 1 = k³ - k + 3k² + 3k = 3m + 3k² + 3k = 3(m + k² + k)
Since m + k² + k is an integer, (k+1)³ - (k+1) is divisible by 3.
Therefore, n³ - n is divisible by 3 for all positive integers n.
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15815 | https://www.youtube.com/watch?v=iHnzLETGz2I | Multiplying monomials | Algebra I | Khan Academy
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Learn how to multiply monomials like 5x² 4x⁶ or 4p³ p.
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all right in this video we're going to be multiplying monomials together let me give you an example of a monomial 4x^2 that's a monomial now why well mono means one which refers to the number of terms so this 4x^2 this is all one term so what so we're going to be working with things like like that what won't we be working with well what about 4x^2 + 5x how many terms are there 4x2 is the first term 5x is the second term so this is not a monomial this is actually called a binomial because by means two like your bicycle's got two wheels um for example so not yet go on to the Future videos if you're ready for binomials but we're just going to be working with multiplying monom together so can we grab an example to look at by the end of this video it should be very easy for you to multiply this monomial 5x^2 by this monomial and I'm actually going to just give you the answer right here and then I'm going to slowly walk you through some other questions that will lead us to Y but the answer to this is 20x to the 8th 20x to the 8 take a look at that see if you can notice a pattern what did we do with the five and the four to get the 20 what did we do with the two and the six to get the eight that's getting a little ahead of ourselves though before we can dive in there let's remember some of the exponent properties a very specific exponent property that you should have seen before so if we look at 5^ 2 5 to the 4th power what's that going to equal well if you remember your exponent property and we'll do a quick reminder here I always add my exponent so 5^ 2 5 4th power is equal to 5 to the 6th power what about 3 to the 4th power 3 to the 5th power well again I always add my exponents 4 + 5 is 3 to the 9th power and my base of three stays the same great so if you remember that now we're now we're ready to really start multiplying monomials that are new to you and the new think here is that we are going to have variables involved so let's start by let's take a look at two monomials here the first monomial is 4X and the second one is just X and the four I don't have another number to multiply by I've just got the four and can I simplify x X well that's equal to X2 remember if I just have a variable and there's no exponent there it's equivalent to having a one so X the 1 power x to the 1 power I add my exponents like we just talked about and 1 + one is equal to two great so let's move on to another one here if I have 4 T 3 T well 4 3 is going to be equal to 12 so I've combined my coefficients and then T T again think of a one being there is going to be t^2 so the answer here is 12 t^2 so let's keep going and once you get into the rhythm of these they become pretty all right so what if I had 4 P to the 5th power time let's say 5 P to the 3 power what would that equal you're going to notice a pattern here that we've been picking up on which is that I'm always going to multiply my coefficient so 4 5 is going to equal 20 and I'm always going to add my exponents so P to the 5th and P to the 3 is p to the eth power so I multiply four and five to we get 20 I'm add five and 3 to get eight and if you really want to see why that is let's really dive in here and let's break down this first term let's break down 4 P 5th I can write that out as 4 P P P P P that's five of them it's four and five PS and then that second term I can write as time 5 P P P what I'm going to do is I'm going to group my numbers because I can work with numbers together so let's put four five at the very front and then it just becomes a matter of how many P's do I have we'll put all of those together as well so I had five PS so there's the first five and then I had three more and we can simplify this crazy looking expression by just multiplying by four my five to be my 20 and then writing this with an exponent that's the beauty of exponents that why that's why we have them is we can write a crazy expression like that as P to the E and you'll notice that this is of course what we got the first time so great about 5 y to 6 -3 y to [Music] the eight eth power again multiply the coefficients add the exponents and I've got a simplified expression let's get really crazy here let's have a little fun so we've noticed the pattern let's have a little fun you're saying I I can do I can do more 9 x to the 5th power times3 use parenthesis there you always when you have a negative in front you want to use parenthesis let's do X to the7th power if I would have showed you this before this video you would have said oh my goodness I I there's no nothing I can do unbox there's there's no way out but now you know that it's as simple as follow the rules we're going to multiply the coefficients -9 -3 is 27 two negatives is a positive and 9 3 is 27 I'm going to add my Powers 5 + 107 is 10 not two that was almost a that was a mistake I made there let's get rid of that give me a second chance here life's all about Second Chances 5 + 107 is 112 and so this crazy expression which is two monomials here's the first here's the second when we multiply and simplify we get another monomial which is 27x to the 112 I'm going to leave you on a cliffhanger here which I'm going to show you a problem what variable should we use you notice I've been trying to vary the variables up to show you that it just doesn't matter that's an ugly five let's get rid of that give me give me a second chance on that one too so let's let's look at 5 x to the 3 power time 4X to the 6th power and I'm going to show you a wrong answer I had a student that I asked to do this and here's the wrong answer that they gave me they told me nine x to the 18th power that's terribly wrong what did they do wrong what did they do wrong I want to think to yourself what have we been talking about what did they do with the five and the four to get the nine what should they have done what did they do with the three and the six to get the 18 and what should they have done that's multiplying monomials by monomials |
15816 | https://mathigon.org/world/Game_Theory | Game Theory | World of Mathematics – Mathigon
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Game Theory | World of Mathematics
This article is from an old version of Mathigon and will be updated soon.
Combinatorial Games
Many games involve rolling dice, shuffling cards or spinning wheels, and we can use probability to determine how likely certain outcomes are. This chapter, on the other hand, is about games where there is no ‘luck’ involved: these games are called Combinatorial Games.
One example of a combinatorial game is chess, but it is so complex, with so many different moves and positions, that it is almost impossible to analyse chess using the methods we will develop throughout this chapter. Here is an example of a much simpler game:
There are two boxes with chocolates, and two players eat them alternatingly. At each turn, a player has to eat one or more chocolates, but only from one box at a time. For example, a player could eat three chocolates from box A, but not one from box A and one from box B.
Both players continue, alternatingly eating chocolates, until both boxes are empty. Whoever gets the last chocolate wins.
Here you can try playing this game against the computer. Start by clicking on all the chocolates you want to eat, then click the button to end your turn.
Click to end your turn
After some time you may notice that you always lose. In fact, it it clear from the beginning that the computer always wins unless it makes a mistake. (And computers never make mistakes…) The following sections will explore different methods to analyse combinatorial games, to find winning strategies and to determine whether it is better to go first or to go second.
If you have already noticed a pattern and worked out a winning strategy, the following sections may seem rather complicated for solving a simple game. However the same methods can them be applied to much more complex games.
Tree Diagrams
One method to think about combinatorial games is to make a list of all possible outcomes. This is best done in a tree diagram, where every fork shows all possible choices a player could make. Here is the tree diagram for a slightly simpler version of the game above, with only three chocolates per box.
Tree Diagram
Whoever empties the first box loses, because the opponent can empty the other box, thereby taking the last chocolate. Therefore Player 1 only has two sensible choices: taking one or taking two chocolates from one box.
Now it’s Player 2’s turn. Player 2 also doesn’t want to empty any box. This means there are two possibilities in one case, and three in the other.
It’s Player 1’s turn. With fewer chocolates left there are much fewer possibilities that don’t involve emptying the first box, which would lead to certain loss.
We continue: if there is one chocolate in both boxes, one player has to empty the first box. The opponent can then take the last chocolate from the second box and wins.
Let us highlight the final positions in which Player 1 and Player 2 win.
So far the game seems quite fair: there are three winning positions for each player. Now let us think about the positions second from last.
Once we have arrived at any of these positions we already know who is going to win. If you can only go to a winning position for Player 1, this also has to be a winning position for Player 1. And the same is true for Player 2. We can colour the positions second from last according to which winning position they lead to.
And we can do the same again: whenever we can only go to winning positions of one kind, we colour the previous position in the same colour.
Unfortunately we will get stuck at some point…
When at these two positions there is a choice: we could either go to a winning position for Player 1 or to a winning position for Player 2. Here we have to make an assumption that both players are intelligent and will play in their best interest. If it is Player 1’s turn he/she will of course go to his/her own winning position and the same for Player 2.
We have coloured the first case blue because Player 2 has a choice to go to his/her own winning position. The second case is coloured green because Player 1 has a choice to go to his/her own winning position. These are the positions where players have to be careful not to make a mistake.
The same is happens here: Player 2 has the option to go to blue and green winning positions, but – if he plays intelligently – he/she will of course choose blue.
It seems that whatever Player 1 does on the first turn, he/she will always end up on a winning position for Player 2.
Player 2 is destined to win from the beginning – unless he makes a mistake.
This method is useful for simple games, but impractical if we have boxes with many more chocolates. If there are five chocolates per box, we would have to consider more than 10,000 possibilities!
P and N-Positions
In the tree diagram above we had many copies of the same state in different branches of the tree. Instead let us draw a diagram of all the different states, and connect two states with an arrow if a player could move from one to the other. (Remember that you can’t put chocolates back or take chocolates from more than one box.)
We will again highlight various states with different colours, but the colours will have a different meaning than above.
P and N Positions Diagram
Here you can see all possible states of the game we were playing.
These arrows show the possible moving directions. On your turn you can move either down or to the right, however far you want.
We know that once we have reached the bottom right corner, the previous player will have taken the last chocolate and won. We call it a P-position and colour it blue.
If we are in any state that leads to the P-position, the next player will win. We call these N-positions and colour them red.
From here you can only move to N-position. Once you have done that, you opponent is next and can win. Therefore this has to be a P-position.
Any positions that leads to the new P-position have to be an N-position – The next player will win. Can you see a pattern?
Here is another P-position from where you can only move to N-positions.
These two are N-positions because the next player can move to a P-position.
Finally the starting position is a P-position. The “previous” player wins, and whoever makes the first move is destined to lose.
The pattern is quite obvious: all positions along the diagonal, where there are the same number of chocolates in each box, are P-positions. All the other positions are N-position. And this extends to bigger box sizes, including nine chocolates per box like in the game at the beginning of this article. You always made the first move, so you had no chance of winning unless the computer had made a mistake.
A P-position is a position in which the previous player will win (who moved to that position) and a N-position is a position where the next player will win (who moves away from that position). When playing, you want to make sure that you always end your turn on a P-position.
We also observed that from a P-position you can only move to N-positions, and from a N-position you can move to at least one P-position.
Starting on a P-position, the next player will lose. Therefore the next player must only be able to move to N-positions.Starting on an N-position, the next player will win. Therefore there must be at least one P-position where the next player can move to. (The game will change if you make a mistake.)
To analyse a game, we have to start from the end when we know who would have won. Then we can work backwards using the two rules above to classify all positions in the game.
In any game that can be analysed using this method, the outcome is determined from the beginning. If you are unlucky and you are the player destined to lose, there is nothing you can do except hope that your opponent makes a mistake…
The Game of Nim
The game we have been thinking about is a variant of Nim. The winning strategy for only two boxes of chocolates is easy to find, but things get more interesting when we have three or more boxes. Instead of boxes of chocolates we will simply use piles of counters: you are allowed to take as many counters as you want, but only from one pile at a time. We can denote the various states of the game using numbers: for example, (2,5,4) means there are three piles with 2, 5 and 4 counters respectively.
Congratulations, you won!
Game over… Try again!
Notice that it doesn’t matter which order the piles are in, or whether there are additional piles with zero counters. For example, (2,5,4) = (5,0,4,2). We have already shown that (1,1), (2,2), (3,3), … are all P-positions, and there is a simple method for determining whether positions with three or more piles are P or N. This method may seem quite unexpected and unrelated to game theory. It arises when you analyse P and N-positions mathematically.
A Nim state (a, b, c, …) is a P-position if the binary digital sum or Nim sum of a, b, c, … is 0. Otherwise it is a N-position. The Nim sum is often written as a ⊕ b ⊕c ⊕ … and can be calculates as shown in the following example.
To find the Nim sum 3 ⊕ 6 ⊕ 7 we proceed as follows:
4 2 1
3 0 1 1
6 1 1 0
7 1 1 1
2 0 1 0
Above you can see the three numbers 3, 6 and 7 in different rows and powers of 2 (1, 2, 4, 8, …) in different columns. We first need to write 3, 6 and 7 in binary, which means writing them as a sum of powers of 2.
Note that 3 = 1 + 2. We need to add the powers 1 and 2, but we don’t need 4. Therefore 3 = 011 in binary.
Note that 6 = 2 + 4. We need the powers 2 and 4, but we don’t need 1. Therefore 6 = 110 in binary.
Note that 7 = 1 + 2 + 4. Here we need all three powers of two.
Now we have to add the columns we just created, but without carrying digits.
In this column, we have an even number of 1s so the answer is 0.
In this column we have an odd number of 1s so the answer is 1.
In this column we have an even number of 1s so the answer is 0.
The binary digital sum of 3, 6 and 7 is 010, and when we convert this from binary we get 2. Since the Nim sum is not 0, the Nim state (3,6,7) is a N-position. Removing two counters from any pile will make the Nim sum 0, so this would represent moving to a P-position.
Nim has several important properties:
Exactly two opponents move alternately.
The moves and all options are clearly specified by rules, and there are no chance moves.
There are only finitely many different positions and the game will always come to an end when one player is unable to move. This means that there are no draws and no cycles, which could repeat forever.
The players have perfect information. Card games often don’t have perfect information because one player doesn’t know the opponents’ cards.
From any one position of the game both players have the same choice of moves. This is not true for chess, because from any particular position, one player can only move white figures and the opponent can only move black ones.
Games with all these properties are called Impartial Games. Mathematicians discovered that any impartial game is equivalent to a game of Nim with certain box sizes. This means that the P and N-positions match up, and that there are always the same number of possible moves. A winning strategy for any impartial game can be found by converting it into Nim and then using the Nim sum.
Article on Combinatorial Game Theory Presented by Philipp Legner at the "Tomorrow's Mathematician's Today" Conference
Non-Combinatorial Games
One of the most captivating combinatorial games: chess
Impartial games are interesting to analyse from a mathematical point of view, but once you have found a winning strategy they are not particularly exciting to play – you know right from the beginning who is going to win.
There are many other combinatorial games. Some, like chess, are so complex that we can’t use methods like the ones above. Chess computers don’t try millions of different possibilities – they play very much like a human being would: analysing the current position and following certain strategies.
Another branch of Game Theory is about situations where people have to make decisions. The outcome depends on your own decision but also on everybody else’s decision – which we don’t know in advance. One example where this happens is economics: companies have to make business decisions and “play” against each other in various markets.
Here are a some interesting situations which can arise in game theory:
The Prisoners Dilemma
Nash Equilibrium
Battle of the Sexes
Imagine two prisoners are locked in two separate cells of a prison. They are accused of committing a crime together and are questioned individually. Both prisoners are promised to get away if they betray their accomplice, who will gets the full sentence of 10 years.
If both prisoners stay silent, there is not enough evidence so both get a shorter sentence of 1 year. If both betray each other, each is sentenced to 5 years in prison.
The following table shows the four possible outcomes, depending on the actions of Prisoner A and Prisoner B:
Prisoner A betraysPrisoner A stays silent
Prisoner B betraysA: 5 years
B: 5 years A: 10 years
B goes free
Prisoner B stays silentA goes free
B: 10 years A: 1 year
B: 1 year
Let us suppose we were Prisoner A and thinking about which action to take: betraying of staying silent.
If we knew Prisoner B would betray us (first row), betraying would get us 5 years while staying silent would get us 10 years. Thus we should also betray.
If we knew Prisoner B would stay silent (second row), betraying would get us 0 years while staying silent would get us 1 year. Thus we should betray.
It seems that – no matter what Prisoner B does – betraying will give us a shorter time in jail and thus is the best thing to do.
Prisoner B will be thinking exactly the same and will also betray. Both prisoners will betray each other and will both be sentenced to 5 years in jail.
Notice, however, that if they had cooperated and both stayed silent, they would have managed to force an outcome that would have been better for both of them: just one year each.
John Forbes Nash ( 1928)
In the Prisoners dilemma, the position (A betrays, B betrays) is called a Nash equilibrium: no individual player can improve his/her outcome by changing their strategy.
In 1951, the mathematician John Forbes Nash ( 1928) proved that all “games” of this kind have Nash equilibria (but there can be more than one). These are not necessarily the best outcome for all players (see the Prisoners dilemma), but they are the choices which players will end up making.
John Nash was jointly awarded the 1994 Nobel Prize in Economics for his work, and his biography is depicted in the Academy Award winning movie A Beautiful Mind.
Nash equilibria are of fundamental importance when analysing economic behaviour like that of big companies, as well as wars, arms races or even soccer games. In all these cases, we have to make a decision, taking into account the decisions our opponent(s) could make.
Here is another famous problem in game theory: decision making in marriage.
A couple decides to go out after work. They either go to the Opera (which the wife prefers) or to a football match (which the husband prefers). Their phones are broken and they cannot contact each other where to go.
or?
The following table shows their respective “gains”. If both go to different places, their gain is 0. If they go to the same place, their gain is either 2 or 3, depending on whether they went to their preferred location or not. (The gain is often called the utility function.)
Wife goes to OperaWife goes to Stadium
Husband goes to OperaW: 3
H: 2 W: 0
H: 0
Husband goes to StadiumW: 0
H: 0 W: 2
H: 3
In this case there are two Nash equilibria: both going to the opera or both going to the stadium. This means that it is much harder to make a decision, often involving psychology and behavioural science (not just mathematics).
If we repeat this “experiment” many times we would observe that husband and wife don’t have a fixed strategy of going to a particular place, but that they employ a mixed strategy: both go to each location with a certain probability.
In this example you can calculate that, optimally, the husband goes to the stadium with probability 3/5 and to the opera with probability 2/5. For the wife, these numbers are swapped.
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Definition of [G:C G(x)][G:C G(x)]
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What is the meaning of [G:C G(x)][G:C G(x)] in group theory? Is this equivalent to |G||Z G(x)||G||Z G(x)|, or to |Z G(x)||Z G(x)|?
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edited Dec 1, 2013 at 16:38
Asaf Karagila♦
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asked Dec 1, 2013 at 15:35
solver6solver6
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Do you mean the index of the centralizer of x x?tylerc0816 –tylerc0816 2013-12-01 15:39:22 +00:00 Commented Dec 1, 2013 at 15:39
Maybe, i don't know definition of it.solver6 –solver6 2013-12-01 15:41:16 +00:00 Commented Dec 1, 2013 at 15:41
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C G(x)C G(x) is the centralizer of x x in G G. That is, {g∈G∣g x g−1=x}{g∈G∣g x g−1=x} or equivalently, {g∈G∣g x=x g}{g∈G∣g x=x g}.
The [G:H][G:H] notation means the index of H H in G G, and it is defined as the number of cosets of H H in G G. Lagrange's theorem says that this is equal to G H G H.
By the way, for this particular H=C G(x)H=C G(x), this has a very useful value. It is the size of the conjugacy class of x x, i.e., {g x g−1∣g∈G}{g x g−1∣g∈G}. Are you by chance studying the class equation?
EDIT: C G(x)C G(x) is in fact a subgroup. Obviously e∈C G(x)e∈C G(x). Now, we show that if a,b∈C G(x)a,b∈C G(x), then a b−1∈C G(x)a b−1∈C G(x).
(a b−1)x=a b−1 x(b b−1)=a b−1(x b)b−1=a b−1(b x)b−1=a x b−1=x(a b−1)(a b−1)x=a b−1 x(b b−1)=a b−1(x b)b−1=a b−1(b x)b−1=a x b−1=x(a b−1)
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edited Dec 1, 2013 at 15:54
answered Dec 1, 2013 at 15:40
Henry SwansonHenry Swanson
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But C G C G isn't a subgroup.solver6 –solver6 2013-12-01 15:43:04 +00:00 Commented Dec 1, 2013 at 15:43
So what is [(Z 5:C Z 5(1))][(Z 5:C Z 5(1))]?solver6 –solver6 2013-12-01 15:44:55 +00:00 Commented Dec 1, 2013 at 15:44
First, what is the centralizer of 1 1. (Hint, which elements commute with 1 1?)Henry Swanson –Henry Swanson 2013-12-01 15:54:27 +00:00 Commented Dec 1, 2013 at 15:54
Answer 1 1 is correct?solver6 –solver6 2013-12-01 15:59:26 +00:00 Commented Dec 1, 2013 at 15:59
When does 1+a=a+1 1+a=a+1?Henry Swanson –Henry Swanson 2013-12-01 16:00:34 +00:00 Commented Dec 1, 2013 at 16:00
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You might want to know that Z(G)⊆C G(x)Z(G)⊆C G(x) for all x∈G x∈G. In fact Z(G)=⋂x∈G C G(x)Z(G)=⋂x∈G C G(x).
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answered Dec 2, 2013 at 8:54
Nicky HeksterNicky Hekster
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0Let R R a commutative ring and let a∈R a∈R. What does a R a R mean?
0Prove that |G|=|Z(G)|+∑i′∈I′|G:C G(x i′)||G|=|Z(G)|+∑i′∈I′|G:C G(x i′)|
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15818 | https://www.researchgate.net/publication/41056487_Disulfiram-like_reaction_with_ornidazole | (PDF) Disulfiram-like reaction with ornidazole
Article PDF Available
Disulfiram-like reaction with ornidazole
October 2009
Journal of Postgraduate Medicine 55(4):292-3
DOI:10.4103/0022-3859.58940
Source
PubMed
License
CC BY-NC-SA 4.0
Authors:
Vishal Sharma
Post Graduate Institute of Medical Education and Research
Alka Sharma
Government Medical College & Hospital
Vivek Kumar
Post Graduate Institute of Medical Education and Research
Sourabh Aggarwal
Western Michigan University
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Citations (13)References (5)
Abstract
Many drugs are implicated in causation of disulfiram-like reaction. The disulfiram-like reaction can vary in severity and can occasionally be fatal. The reaction is believed to result from inhibition of metabolism of acetaldehyde to acetate by inhibition of aldehyde dehydrogenase. The increase in serum acetaldehyde results in unpleasant clinical manifestations. Metronidazole is known to cause disulfiram-like reaction. Although no previous report has implicated ornidazole in causation of disulfiram-like reaction, caution has been advised with the use of all imidazoles. We report the case of a 48-year-old male, who was taking ornidazole and developed features of disulfiram-like reaction after taking alcohol. The patient was managed with supportive measures and improved. The report highlights the need for clinicians to advise patients to restrict intake of alcohol if they are being prescribed imidazole derivatives.
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292 J Postgrad Med October 2009 Vol 55 Issue 4
www.jpgmonline.com
Department of Medicine,
University College of
Medical Sciences & GTB
Hospital, Delhi, India
Address for correspondence:
Dr. Vishal Sharma,
E-mail: docvishalsharma@
gmail.com
Received : 06-07-09
Review completed : 12-09-09
Accepted : 29-09-09
PubMed ID :
DOI: 10.4103/0022-3859.58940
J Postgrad Med 2009;55:292-3
A DR Report
Disulfiram-like reaction with ornidazole
Sharma V, Sharma A, Kumar V, Aggarwal S
ABSTRACT
Many drugs are implicate d in caus ation of disulfiram-li ke reaction. The disulf iram-like react ion c an v ary i n se verity
and can occasionally be fatal. The reaction is believed to result from inhibition of metabolism of acetaldehyde
to acetate by inhibition of aldehyde dehydrogenase. The increase in serum acetaldehyde results in unpleasant
clinical manifestations. Metronidazole is known to cause disulfiram-like reaction. Although no previous report
has implicated ornidazole in causation of disulfiram-like reaction, caution has been advised with the use of all
imidazoles. We report the case of a 48-year-old male, who was taking ornidazole and developed features of
disulfiram-like reaction after taking al cohol. The patient was managed with suppo rtive measure s and improved.
The report highlights the need for clinicians to advise patients to restrict intake of alcohol if they are being
prescribed imidazole derivatives.
KEY WORDS: Disulfiram, metronidazole, ornidazole
A lthough metronidazole has been reported to cause
disulfiram-like reaction, no reports concerning such a
reaction with the use of ornidazole exist. W e report the case
of a 48-year-old male who developed disulfiram-like reaction
with ornidazole.
Case Report
A 48-year-old male presented with severe restlessness,
palpitations, facial flushing and sweating. He had a history of
chronic alcohol abuse. The patient had been having diarrhea
and vomiting for the past three days and had been prescribed
a combination of Ofloxacin 200 mg and Ornidazole 500 mg
twice daily. On the day of his pr esentation, he had taken alcohol
almost 2 h after taking two tablets of this combination. The
symptoms started almost 45 min later. The patient presented
with complaint of palpitations. He had no past history of
palpitations, chest pain or any other history. He used to
consume alcohol on almost daily basis for the past 20 years.
He reported no history of any liver disease or cardiac illness
in the past.
On examination, the patient had a blood pressure of 90/64 mm
of Hg, pulse of 127/min, regular and a respiratory rate of
24/ min. He was sweating and his face was flushed. The rest of
systemic examination was normal. His investigations revealed
Hb concentration of 10.1 gm%, total leukocyte count was
5600/mm 3, a few macrocytes wer e seen on peripheral smear
examination. Blood levels of glucose and ur ea were 78 gm%
and 48 mg%, respectively, and serum concentrations of sodium
(139 mEq/l), potassium (4.8 mEq/l), creatinine (0.9 mg%),
bilirubin (0.7 mg%), SGPT(42 IU/l), SGOT (56 IU/l) and
albumin (3.3 gm%) were within the normal range. Arterial
blood gas analysis revealed evidence of respiratory alkalosis
(pH-7.47, P aCO 2-18, P aO 2-108). The th yroid functio n te sts were
normal. His electrocardiograph revealed sinus tachycardia. His
echocardiography and ultrasonography for liver were normal.
Initially, the patient was managed with oxygen inhalation
and intravenous fluids. Oral diazepam was given for relieving
anxiety. The symptoms improved within 2 h and the pati ent was
discharged on third day with advice to quit alcohol and about
the medications he needs to avoid if he drinks. No re-challenge
was undertaken.
Discussion
Disulfiram-like reaction describes the occurrence of
manifestations such as tachycardia, anxiety, throbbing
headaches, facial flushing, weakness, dizziness, anxiety,
naus ea, vomit ing,hy poten sion, d ysrhy thmi a an d pruriti s,
when alcohol is consumed with disulfiram. This reaction
is believed to be a result of incr ease in acetaldehyde levels
because disulfiram inhibits the oxidation of acetaldehyde. The
elevated acetaldehyde causes these manifestations due to both
its direct effects and also histamine release.The r eaction
varies in severity and can even cause mortality, especially
due to dysrthymias. Certain drugs such as metronidazole,
sulfonamides, nitrofurantoin, chloramphenicol have also
been implicated in causation of disulfiram-like reaction.
Metronidazole is believed to cause disulfiram-like reaction
J Postgrad Med October 2009 Vol 55 Issue 4 293
Sharma,
et al
.: Disulfiram-like reaction with ornidazole
which can occasionally be severe enough to cause mortality.
There are no reports implicating ornidazole in causation of
disulfiram reaction, although caution is advised with the use
of all imidazoles.
Even though metronidazole has been believed to cause
disulfiram-li ke reaction, r ecent r eports have questioned the
existence of such an interaction.[2,4] It is believed that these
reactions witnessed with metronidazole may be the result of
some other phenomenon such as ‘serotonin syndrome’ or these
may be peculiar to certain individuals.The present case is
important because a patient had presented with symptoms
fitting with disulfiram reaction after taking Ornidazole. In the
present case the Naranjo adverse drug reaction causality scale
score was 6 which means that the causal r elation between
the adverse reaction and the drug was probable. Inability of
determining acetal dehyde levels and perfor m a re-challenge
could be considered as limitations in determining causality.
Till a consensus emerges about the disulfiram-like effect of
metronidazole, reasonable precautions including clear advice
to abstain from alcohol is warranted when metronidazole or its
congeners are prescribed.
References
Soghoian S, Wiener SW, Díaz-Alcalá JE. Toxicity, Disulfiram.
Emedicine,Medscape.com. 2008 Aug. Available from:
[cited
2009 Jul 6].
Karamanakos PN, Pappas P, Boumba VA, Thomas C, Malamas M,
Vougiouklakis T, et al. Pharmaceutical agents known to produce
disulfiram-like reaction: Effects on hepatic ethanol metabolism and
brain monoamines. Int J T oxicol 2007;26:423-32.
Cina SJ, Russell RA, Conradi SE. Sudden death due to metronidazole/
ethanol interaction. Am J Forensic Med P athol 1996;17:343-6.
Visapää JP, Tillonen JS, Kaihovaara PS, Salaspuro MP. Lack of
disulfiram-like reaction with metronidazole and ethanol. Ann
Pharmacother 2002;36:971-4.
Naranjo CA, Busto U, Sellers EM, Sandor P, Ruiz I, Roberts EA, et al.
A method for estimating the probability of adverse drug reactions.
Clin Pharmacol Ther 1981;30:239-45
Source of Support: Nil, Conict of Interest: None declared.
Citations (13)
References (5)
... Warto także zauważyć, że metronidazol nie jest jedyną pochodną imidazolu, która może wchodzić w interakcje z etanolem . Zgodnie z opisanym przypadkiem przez Guler i wsp., objawy DLR wystąpiły również u pacjenta stosującego ornidazol po wypiciu 3 kieliszków wina . ...
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... Intolerance to alcohol can also be produced by consuming ethanol while receiving nitroimidazoles, such as metronidazole, ornidazole and tinidazole . A death due to an ethanol/ metronidazole interaction and features of disulfiram-like reactions with ornidazole have been reported. And despite the considerable number of case reports describing the association with disulfiram-like reactions secondary to metronidazole and ethanol interaction, reactions do not occur in all patients, suggesting an individual susceptibility . ...
... Besides, the mechanism by which metronidazol induces intolerance to alcohol remains unclear . However, clear advice to abstain from alcohol is warranted when metronidazole or its congeners are prescribed . ...
... A 48-year-old male with severe restlessness, palpitations, facial flushing and sweating, who had taken alcohol almost 2 h after taking ornidazole, was managed with oxygen inhalation, intravenous fluids and oral diazepam for relieving anxiety, with improvement of symptoms improved within 2 h . ...
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Ventricular fibrillation is not known as a complication of metronidazole poisoning. Although some arrhythmias have been reported as a complication of metronidazole intake while taking antiarrhythmic medications, most such arrhythmias are possibly related to co-ingestion of drugs with metronidazole as it affects the metabolism of these drugs. In this case, ventricular fibrillation occurred in a young patient without preexisting medical conditions or any other known drug ingestion, which was never been reported before. We present a case of an 18-year-old male brought in by the ambulance service after attempting to end his life by overdosing on metronidazole. While being transported he developed ventricular fibrillation and received an electric shock, which reverted the episode. Laboratory investigations did not show any clear cause that might have precipitated his arrhythmia.
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The results of recent published studies focusing on the effect of azithromycin as an adjunct to scaling and root planing (SRP) in the treatment of chronic periodontitis are inconsistent. We conducted a meta-analysis of randomized controlled clinical trials to examine the effect of azithromycin combined with SRP on periodontal clinical parameters as compared to SRP alone. An electronic search was carried out on Pubmed, Embase and the Cochrane Central Register of Controlled Trials from their earliest records through December 28, 2014 to identify studies that met pre-stated inclusion criteria. Reference lists of retrieved articles were also reviewed. Data were extracted independently by two authors. Either a fixed- or random-effects model was used to calculate the overall effect sizes of azithromycin on probing depth, attachment level (AL) and bleeding on probing (BOP). Heterogeneity was evaluated using the Q test and I(2) statistic. Publication bias was evaluated by Begg's test and Egger's test. A total of 14 trials were included in the meta-analysis. Compared with SRP alone, locally delivered azithromycin plus SRP statistically significantly reduced probing depth by 0.99 mm (95% CI 0.42-1.57) and increased AL by 1.12 mm (95% CI 0.31-1.92). In addition, systemically administered azithromycin plus SRP statistically significantly reduced probing depth by 0.21 mm (95% CI 0.12-0.29), BOP by 4.50% (95% CI 1.45-7.56) and increased AL by 0.23 mm (95% CI 0.07-0.39). Sensitivity analysis yielded similar results. No evidence of publication bias was observed. The additional benefit of systemic azithromycin was shown at the initially deep probing depth sites, but not at shallow or moderate sites. The overall effect sizes of systemic azithromycin showed a tendency to decrease with time, and meta-regression analysis suggested a negative relation between the length of follow-up and net change in probing depth (r = -0.05, p = 0.02). This meta-analysis provides further evidence that azithromycin used as an adjunct to SRP significantly improves the efficacy of non-surgical periodontal therapy on reducing probing depth, BOP and improving AL, particularly at the initially deep probing depth sites.
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U Busto
Edward M Sellers
D J Greenblatt
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Pharmaceutical Agents Known to Produce Disulfiram-Like Reaction: Effects on Hepatic Ethanol Metabolism and Brain Monoamines
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Sep 2007
Petros Karamanakos
Periklis Pappas
Vassiliki A Boumba
Marios Marselos
Several pharmaceutical agents produce ethanol intolerance, which is often depicted as disulfiram-like reaction. As in the case with disulfiram, the underlying mechanism is believed to be the accumulation of acetaldehyde in the blood, due to inhibition of the hepatic aldehyde dehydrogenases. In the present study, chloramphenicol, furazolidone, metronidazole, and quinacrine, which are reported to produce a disulfiram-like reaction, as well as disulfiram, were administered to Wistar rats and the hepatic activities of alcohol and aldehyde dehydrogenases (1A1 and 2) were determined. The expression of aldehyde dehydrogenase 2 was further assessed by Western blot analysis, while the levels of brain monoamines were also analyzed. Finally, blood acetaldehyde was evaluated after ethanol administration in rats pretreated with disulfiram, chloramphenicol, or quinacrine. The activity of aldehyde dehydrogenase 2 was inhibited by disulfiram, chloramphenicol, and furazolidone, but not by metronidazole or quinacrine. In addition, although well known for metronidazole, quinacrine also did not increase blood acetaldehyde after ethanol administration. The protein expression of aldehyde dehydrogenase 2 was not affected at all. Interestingly, all substances used, except disulfiram, increased the levels of brain serotonin. According to our findings, metronidazole and quinacrine do not produce a typical disulfiram-like reaction, because they do not inhibit hepatic aldehyde dehydrogenase nor increase blood acetaldehyde. Moreover, all tested agents share the common property to enhance brain serotonin, whereas a respective effect of ethanol is well established. Therefore, the ethanol intolerance produced by these agents, either aldehyde dehydrogenase is inhibited or not, could be the result of a "toxic serotonin syndrome," as in the case of the concomitant use of serotonin-active medications.
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Sudden Death Due to Metronidazole/Ethanol Interaction
Article
Jan 1997
Stephen J. Cina
Roger A. Russell
Sandra E. Conradi
Metronidazole (Flagyl), a commonly prescribed antimicrobial agent, can produce a reaction similar to that of disulfiram (Antabuse) when administered to patients drinking ethanol. This drug/chemical interaction results in accumulation of acetaldehyde in the blood. Acetaldehyde is hepatotoxic, cardiotoxic, and arrythmogenic; no lethal serum acetaldehyde level has been established. Sudden death has been reported in patients taking disulfiram while using ethanol; no fatalities have been reported due to ethanol/ metronidazole interactions. Described is a case of a 31-year-old woman who died moments after an assault by a male companion, during which he inflicted minor physical trauma to her upper arm. Toxicologic analysis yielded elevated concentrations of serum ethanol (162 mg/d), acetaldehyde (4.6 mg/d), and metronidazole (0.42 mg/L). The cause of death was reported to be cardiac dysrhythmia due to acetaldehyde toxicity due to an ethanol/ metronidazole interaction. Autonomic stress associated with the assault is likely to have contributed to this woman's death. The mechanism of death is examined.
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Lack of Disulfiram-Like Reaction with Metronidazole and Ethanol
Article
Jul 2002
Jukka-Pekka Visapää
Jyrki Tillonen
Pertti Kaihovaara
Mikko Salaspuro
Metronidazole, an effective antianaerobic agent, has been reported to have aversive properties when ingested with ethanol. This is thought to be due to the blocking of hepatic aldehyde dehydrogenase (ALDH) enzyme followed by the accumulation of acetaldehyde in the blood. However, based on animal studies and on only 10 human case reports, the existence of metronidazole-related disulfiram-like reaction has recently been questioned. To investigate the possible disulfiram-like properties of metronidazole and ethanol in human volunteers. Of 12 healthy male volunteers in this double-blind study, one-half received metronidazole for 5 days and the other half received placebo. All volunteers received ethanol 0.4 g/kg at the beginning of the study. Repeated blood samples were taken every 20 minutes for 4 hours, and blood acetaldehyde and ethanol concentrations were determined. Blood pressure, heart rate, and skin temperature were also measured every 20 minutes for objective signs of a possible disulfiram-like reaction. Volunteers also completed a questionnaire focusing on the subjective signs of disulfiram-like reaction. Metronidazole did not raise blood acetaldehyde or have any objective or subjective adverse effects when used together with ethanol. This study shows that metronidazole does not have an effect on blood acetaldehyde concentrations when ingested with ethanol and does not have any objective or subjective disulfiram-like properties. However, it is possible that disulfiram-like reaction can occur in some subgroups and by other mechanisms than the inhibition of hepatic ALDH.
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Toxicity, Disulfiram. E m e d i c i n e , M e d s c a p e . c o m . 2 0 0 8 A u g
Aug 2009
S Soghoian
Sw Wiener
Je Díaz-Alcalá
Soghoian S, Wiener SW, Díaz-Alcalá JE. Toxicity, Disulfiram. E m e d i c i n e, M e d s c a p e. c o m. 2 0 0 8 A u g. A v a i l a b l e f r o m : [cited 2009 Jul 6].
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Last Updated: 09 Aug 2025
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Confusion about the difference between HN/N and H/N in the second isomorphism theorem
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Im trying to understand the second isomorphism theorem but I am stuck. So in my textbook the author did the following:
Let G be a group and N<G a normal subgroup and H<G a subgroup. Let π:G→G/N π:G→G/N be the canonical homomorphism given by g↦g N g↦g N. If we limit this map to H, then we have homomorphism H↦G/N H↦G/N with kernel H∩N H∩N and image HN/N. I understand why the kernel is equal to H∩N H∩N, but why is the image equal to HN/N?. What I think is the following: The homomorphism H→G/N H→G/N is defined as h↦h N h↦h N which means the image for the whole subgroup H is H/N={h N|h∈H}H/N={h N|h∈H}. Now the problem is that N can not be a subgroup of H so H/N does not make sense. Is that the reason why we have HN/N instead, because they are the same set H N/N={h n N|h∈H}={h N|h∈H}=H/N H N/N={h n N|h∈H}={h N|h∈H}=H/N or are they not? If not what is the difference between them?
abstract-algebra
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group-isomorphism
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asked May 11, 2023 at 20:45
muhammed gunesmuhammed gunes
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They are equal in the sense of SETS, but on the level of groups H/N H/N may not even be a group. But H N/N H N/N is ALWAYS a group and therefore we have to write our underlying set this way to adapt the scenario.Ryan Zhou –Ryan Zhou 2024-05-21 15:07:06 +00:00 Commented May 21, 2024 at 15:07
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Consider an actual example: π:Z→Z/24 Z π:Z→Z/24 Z where G=Z G=Z and N=24 Z N=24 Z. Set H=10 Z H=10 Z. Here we're using additive groups, so the notation becomes additive (H+N H+N instead of H N H N). The image of H H is what you get by repeatedly adding 10 10 to itself modulo 24 24:
{10,20,6,16,2,12,22,8,18,4,14,0}mod 24={0,2,4,6,8,10,12,14,16,18}mod 24=2 Z/24 Z,{10,20,6,16,2,12,22,8,18,4,14,0}mod 24={0,2,4,6,8,10,12,14,16,18}mod 24=2 Z/24 Z,
where 2 Z=10 Z+24 Z 2 Z=10 Z+24 Z (since 2=gcd(10,24)2=gcd(10,24)) =H+N=H+N, so the image of π π is (H+N)/N(H+N)/N. Would you call the image H/N=10 Z/24 Z H/N=10 Z/24 Z?
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answered May 11, 2023 at 22:00
KCdKCd
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You are of course correct that the image of π|H π|H is {h N:h∈H}{h N:h∈H}, which is a subgroup of G/N G/N with group structure given by (h N)(h′N)=(h h′)N(h N)(h′N)=(h h′)N. You could refer to this as H/N H/N if you like, but we tend to strictly reserve the notation of "A/B A/B" when B B is a sub-structure of A A, which, as you point out, is not necessarily the case in this context.
If instead you consider the product group H N H N, then N N really is then a normal subgroup of H N H N, and so the notation (H N)/N(H N)/N follows the standard convention mentioned above. Furthermore, (H N)/N(H N)/N as a group is clearly the same as {h N:h∈H}{h N:h∈H} with the above mentioned product.
To summarise: we use (H N)/N(H N)/N instead of H/N H/N to keep up with standard notation conventions.
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edited May 11, 2023 at 23:06
answered May 11, 2023 at 21:21
Riemann'sPointyNoseRiemann'sPointyNose
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1 I'm not sure if it's really a "matter of taste," since as you stated the notation H/N H/N doesn't really mean anything unless N N is a subgroup of H H.Frank –Frank 2023-05-11 21:51:44 +00:00 Commented May 11, 2023 at 21:51
1 @Frank The notation is meaningful if H is any subset of G G. That is, after all, how we denote individual cosets of N N.Robert Shore –Robert Shore 2023-05-11 22:20:12 +00:00 Commented May 11, 2023 at 22:20
@RobertShore Are you referring to the notation S N S N for a subset S S? I’m talking about the notation S/N S/N, which I’ve only seen used when S S is a group (or other mathematical structure) containing N N as a subgroup (or substructure).Frank –Frank 2023-05-11 22:58:45 +00:00 Commented May 11, 2023 at 22:58
2 @Frank Any notation could have meaning if we decide to give it meaning, and I am saying in this context it certainly would make sense to, given a normal subgroup N N of G G and another subgroup (or even just subset, as Robert Shore above has said) H H, to define H/N H/N as the set of cosets {h N:h∈H}{h N:h∈H}; the point in what I was saying however was this notation is not used and we generally own use A/B A/B if B B is a substructure of A A Riemann'sPointyNose –Riemann'sPointyNose 2023-05-11 23:02:20 +00:00 Commented May 11, 2023 at 23:02
@Frank With that being said, maybe saying it comes down to taste implies that there are a notable group of people who would use this notation, which is not the case. I'll edit my answer to reflect that Riemann'sPointyNose –Riemann'sPointyNose 2023-05-11 23:06:14 +00:00 Commented May 11, 2023 at 23:06
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15820 | https://www.youtube.com/watch?v=R1CM-W8FMRA | What are Composite Numbers? | Math with Mr. J
Math with Mr. J
1730000 subscribers
3304 likes
Description
292434 views
Posted: 1 Feb 2021
Welcome to "What are Composite Numbers?" with Mr. J! Need help with composite numbers? You're in the right place!
Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with composite numbers. Mr. J will go through composite number examples and explain how to determine if a number is composite or not.
✅ What are Prime Numbers?:
✅ Need help with another topic?... Just search what topic you are looking for + "with Mr. J" (for example, "adding fractions with Mr. J".
About Math with Mr. J: This channel offers instructional videos that are directly aligned with math standards. Teachers, parents/guardians, and students from around the world have used this channel to help with math content in many different ways. All material is absolutely free.
Click Here to Subscribe to the Greatest Math Channel On Earth:
Follow Mr. J on Twitter: @MrJMath5
Email: math5.mrj@gmail.com
Music:
Hopefully this video is what you're looking for when it comes to composite numbers.
228 comments
Transcript:
[Music] welcome to math with mr j [Music] in this video i'm going to cover what composite numbers are now composite numbers are whole numbers with more than two factors another way of thinking about it composite numbers are divisible by more than two numbers so let's go through six examples here together in order to better understand what that means and what composite numbers are and we'll jump into number one here where we have 9. so we need to think of the factors of 9 although not the most mathematical or technical way of thinking about it all the numbers that go into 9 or the technical definition of factors the numbers that multiply together to equal 9 so we know 1 times 9 equals 9 so 1 and 9 are factors and we know 3 times 3 equals 9 so 1 3 and 9 are factors of 9. so 9 has more than two factors or it's divisible by more than two numbers nine is divisible by one three and nine or think of it as one three and nine go into nine so nine is composite so let's move on to number two where we have seven and we need to think of the factors of seven so what numbers go into seven or what numbers multiply together to equal seven well we know one times seven equals seven and that's actually the only two factors of 7 7 only has two factors so it's not composite if a number only has two factors it's called a prime number now i have a video going into further depth on prime numbers i'll drop that link down in the description on to number three where we have two now two is a unique number here and we'll find out why in a second so the factors of two well one times 2 equals 2 and there aren't any other factors of 2 so 1 and 2 are the factors of 2. 2 only has 2 factors so 2 is prime just like 7 but 2 is unique because it's the only even number that is prime all other even numbers are automatically composite because we have the factors of 1 and the number itself and then 2 is an automatic factor of all the other even numbers on to number four where we have 23 so 23 we need to think of its factors what numbers can we use to equal 23 multiplied together to equal 23 while 23 only has two factors 1 and 23 so 23 is prime as well it is not composite on to number 5 where we have 24. now we can tell that 24 is composite right away because it's an even number we know that 1 is a factor 24 is a factor and then 2 is a factor so right off the bat we know that 24 has more than two factors therefore composite but we will list them out just to see how many different factors 24 has so we know one and 24 so i'll put 24 over here with a little gap we know 2 times 12 2 and 12. we know 3 times 8 is 24 and then we know 4 and 6. so i'll squeeze the 6 in there so we can see that 24 has quite a bit of factors there um and even if it's three factors it doesn't have to be a lot of factors if it's just more than two it's a composite number so 24 definitely composite and lastly we have number six here 27. now don't think that because 27 is odd it's not composite and is going to be prime well we need to think of the factors 27 is actually composite it's going to be composite because we know 1 and 27 are factors but 3 and 9 are also factors 3 times 9 equals 27 so 27 has more than two factors therefore it's a composite number so there you have it there's um the basics of composite numbers i hope that helped thanks so much for watching until next time peace |
15821 | https://www.quora.com/What-is-an-algorithm-to-calculate-the-average-of-three-numbers-supplied-by-the-user | Something went wrong. Wait a moment and try again.
Computer Science
Input Box
Average (statistics)
General Programming
Data Processing (IT)
Programming Languages
Computer Input
5
What is an algorithm to calculate the average of three numbers supplied by the user?
David Lambert
HS Diploma from Horace Greeley High School (Graduated 1978)
·
Author has 3.6K answers and 2.2M answer views
·
1y
Materials: tape measure, red yarn, green yarn.
Preparation: Using green yarn for positive numbers and red yarn for the negative values, cut a length of appropriate yarn equal in length to the absolute value of the number. For each of the three numbers.
Computation: Individually fold each length of yarn into thirds. Take one of these ropes, lay it alongside the tape measure with one end matching the zero end. (Often has a hook.) If you have a green rope start with it, otherwise you know the sign of the final answer will be negative. Take another rope. Align one end with the end far from the “zero
Materials: tape measure, red yarn, green yarn.
Preparation: Using green yarn for positive numbers and red yarn for the negative values, cut a length of appropriate yarn equal in length to the absolute value of the number. For each of the three numbers.
Computation: Individually fold each length of yarn into thirds. Take one of these ropes, lay it alongside the tape measure with one end matching the zero end. (Often has a hook.) If you have a green rope start with it, otherwise you know the sign of the final answer will be negative. Take another rope. Align one end with the end far from the “zero”. If it’s a green rope lay it out in the same direction as the first. Otherwise if the first were green then this red rope will overlap some or all of the first rope because you’re to lay it out in the opposite direction. Otherwise…etceteras. Got it? Layout the third length ropes end to end, green ropes to the right, red ropes to the left.
Now read the value on the tape measure. That’s the average. If you had to turn the tape measure around, the result is negative.
Related questions
What is the C program to calculate the sum and average of three user numbers?
How do you write an algorithm that will get 3 numbers as input from the users? What is the average and display the three numbers and its resultant average?
What algorithm can I use to calculate the average of five numbers?
What are the addition and average three numbers?
What is an algorithm to find and display the average of 10 numbers?
Sai Sasidhar Paluri
Bachelors of Technology in Computer Science, Atal Bihari Vajpayee Indian Institute of Information Technology and Management, Gwalior (Expected 2027)
·
Author has 92 answers and 32.5K answer views
·
1y
The algorithm to calculate the average of any three numbers supplied by the user is:
Begin procedure main():-
Step 1: PRINT Enter any three number two find average \ n
Step 2: READ a , b , c
Step 3: SET avg := ( a + b + c ) / 3
Step 4: PRINT avg
Step 5: RETURN 0
End procedure
Jeff Erickson
My book (algorithms.wtf) is cheaper than CLRS. And lighter.
·
Author has 3.3K answers and 114.2M answer views
·
2y
Originally Answered: How can I write a basic algorithm and program to compute the average of three numbers?
·
To write the algorithm, write down the definition of the average of three numbers. See that formula? That’s an algorithm. That’s what formulas are.
For example, here is a (hopefully familiar) algorithm to compute the roots of the polynomial ax2+bx+c, given the numbers a, b, and c as input:
b±√b2–4ac2a
If your instructor insists on an algorithm written in pseudocode, then you only need to explain clearly and carefully how to evaluate that formula ,given values for the variables. For example:
The discriminant is bb - 4ac.
Let radical denote the square root of the
To write the algorithm, write down the definition of the average of three numbers. See that formula? That’s an algorithm. That’s what formulas are.
For example, here is a (hopefully familiar) algorithm to compute the roots of the polynomial ax2+bx+c, given the numbers a, b, and c as input:
b±√b2–4ac2a
If your instructor insists on an algorithm written in pseudocode, then you only need to explain clearly and carefully how to evaluate that formula ,given values for the variables. For example:
The discriminant is bb - 4ac.
Let radical denote the square root of the discriminant.
The roots are (b + radical) / (2a) and (b - radical) / (2a).
To write a program, you need to transcribe your pseudocode into your favorite programming language.
import cmath def roots(a,b,c): '''Compute the roots of ax^2 + bx + c = 0.''' discriminant = bb - 4ac radical = cmath.sqrt(discriminant) root1 = (b + radical) / (2a) root2 = (b - radical) / (2a) return root1, root2
Your problem is considerably simpler than this one.
Jay Nabonne
Computer Programmer and Occasional Writer
·
Author has 2.1K answers and 3.4M answer views
·
1y
Originally Answered: How do you write an algorithm that will get 3 numbers as input from the users? What is the average and display the three numbers and its resultant average?
·
You just did:
Get 3 numbers from the user
Compute the average (e.g. sum them up and divide by 3)
Print the the three numbers and the average
That is an algorithm. An algorithm is just a method or process for solving a problem. Long division is an algorithm. A recipe for making a cake is an algorithm (though less precise). And it doesn’t even have to be done on a computer. You can do step 1 by simply asking a person for three numbers. Then calculate the average on a piece of paper. Then tell them “The average of , , and is .”
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Related questions
How do you get the average of three numbers?
What is an algorithm to calculate the average of three numbers?
How do I calculate the average of five numbers?
How can I write a basic algorithm and program to compute the average of three numbers?
What is a program that takes three numbers from a user and find their average?
Thomas Epp
Director of the Math Resource Center
·
Author has 223 answers and 231.8K answer views
·
3y
Originally Answered: What is an algorithm to determine the average of 3 given numbers a, b, and c?
·
That depends on what specifically you mean by the term "average". The most commonly used definition of average is what statisticians call the mean, however the term average can more generally refer to any measure of central tendency, including the median, mode, midrange, etc.
Finding the mean is easy: add up the values, then divide the sum by the number of values. For a, b, and c, you have:
m = (a + b + c) / 3
If you want other measures of central tendency, I'd recommend that you look up their definitions in Wikipedia, Wolfram MathWorld, or some other credible source.
David Schulman
Amateur mathematician
·
Author has 3K answers and 2.8M answer views
·
3y
Originally Answered: What is an algorithm to determine the average of 3 given numbers a, b, and c?
·
Algorithm? Does 13(a+b+c) constitute an algorithm?
More generally,
¯μ≡1nn∑k=1xk
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Martyn Hathaway
BSc in Mathematics, University of Southampton (Graduated 1986)
·
Author has 4.7K answers and 6.7M answer views
·
1y
Originally Answered: What is an algorithm to determine the average of 3 given numbers a, b, and c?
·
You don’t calculate the average of a set of numbers, you calculate an average. There are loads of ways of calculating an average; I bet you have been taught several already.
Mode: The most common value. If two of your numbers are the same, then that is the mode. If all three numbers are different then you have a trimodal data set.
Median: The middle value. If a<b<c then b is the median value.
Arithmetic Mean: a+b+c3
[This is the one that most people think about when they hear/see the word ‘average’, forgetting that they have been told about several other averages in their school life.]
Ge
You don’t calculate the average of a set of numbers, you calculate an average. There are loads of ways of calculating an average; I bet you have been taught several already.
Mode: The most common value. If two of your numbers are the same, then that is the mode. If all three numbers are different then you have a trimodal data set.
Median: The middle value. If a<b<c then b is the median value.
Arithmetic Mean: a+b+c3
[This is the one that most people think about when they hear/see the word ‘average’, forgetting that they have been told about several other averages in their school life.]
Geometric Mean: 3√abc
[Common in financial calculations and demography.]
Harmonic Mean: 31a+1b+1c
Quadratic Mean: √a2+b2+c23
Cubic Mean: 3√a3+b3+c23
There are loads of other means
Manoj Sahani
Computer Engineer, teacher and a freelancer
·
Author has 117 answers and 144K answer views
·
4y
Originally Answered: What is a basic program to find the average of three numbers?
·
The Qbasic program to find the average of three numbers
CLS
INPUT” Enter the first number”; a
INPUT” Enter the second number”; b
INPUT” Enter the third number”; c
LET Average = (a+b+c)/3
PRINT” The average number is ”; Average
END
Qbasic Program To Find The Average Of Three Numbers Using FUNCTION Procedure
DECLARE FUNCTION AVG (a, b, c)
CLS
INPUT” Enter the first number”; a
INPUT” Enter the second number”; b
INPUT” Enter the third number”; c
PRINT” The average number is ”; AVG (a, b, c)
END
FUNCTION AVG (a, b, c)
Average= (a + b + c)/3
AVG= Average
END FUNCTION
Qbasic Program To Find The Average Of Three Numbers
The Qbasic program to find the average of three numbers
CLS
INPUT” Enter the first number”; a
INPUT” Enter the second number”; b
INPUT” Enter the third number”; c
LET Average = (a+b+c)/3
PRINT” The average number is ”; Average
END
Qbasic Program To Find The Average Of Three Numbers Using FUNCTION Procedure
DECLARE FUNCTION AVG (a, b, c)
CLS
INPUT” Enter the first number”; a
INPUT” Enter the second number”; b
INPUT” Enter the third number”; c
PRINT” The average number is ”; AVG (a, b, c)
END
FUNCTION AVG (a, b, c)
Average= (a + b + c)/3
AVG= Average
END FUNCTION
Qbasic Program To Find The Average Of Three Numbers Using SUB Procedure
DECLARE SUB AVG (a, b, c)
CLS
INPUT” Enter the first number”; a
INPUT” Enter the second number”; b
INPUT” Enter the third number”; c
CALL AVG (a, b, c)
END
SUB AVG (a, b, c)
Average= (a + b + c)/3
PRINT " The average is "; Average
END SUB
Note: For more programs, you can visit”csenotez.blogspot.com”
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SharkPause
Knows Indonesian
·
Author has 108 answers and 79.3K answer views
·
3y
Originally Answered: What is a code that would calculate the average of 3 values and display a result?
·
// C++ #include <iostream>using std::cout; int main() { int x = 1, y = 2, z = 3; cout << (x+y+z) / 3; return 0;}
Jeff Hawkins
Technology Instructor (2017–present)
·
Author has 470 answers and 2.1M answer views
·
Updated 6y
Originally Answered: What is a basic program to find the average of three numbers?
·
I hope we were correct in thinking you wanted a basic program, rather than a BASIC program… Here is another basic C++ option, as numbers can be floating point to begin with. This will run as-is at
JDoodle
, etc.
```
include using namespace std; int main() { double a, b, c; cout<<"Enter three numbers, separated by spaces or linefeed: "; cin >> a >> b >> c; cout << "The average is: " << (a + b + c) / 3;}
```
I’m assuming you are very new to this… Also hoping the fact that it is Christmas break means this is not a homework assignment. In any event, the main thing to learn from the small collec
I hope we were correct in thinking you wanted a basic program, rather than a BASIC program… Here is another basic C++ option, as numbers can be floating point to begin with. This will run as-is at JDoodle, etc.
```
include using namespace std; int main() { double a, b, c; cout<<"Enter three numbers, separated by spaces or linefeed: "; cin >> a >> b >> c; cout << "The average is: " << (a + b + c) / 3;}
```
I’m assuming you are very new to this… Also hoping the fact that it is Christmas break means this is not a homework assignment. In any event, the main thing to learn from the small collection of answers here is that data types are important.
The ones and zeros the computer has at it’s disposal can be interpreted many different ways. If you tell the computer that a number is an integer, it will work with it as an integer. If you tell it the number is a float, it will work with fractional numbers. And if you tell it the number is a double-precision float as I’ve done above, it will support very precise and very large numbers.
The different options require different amounts of memory. But as you are only using 3 numbers plus the average, the cost of supporting double precision floats is insignificant, so you might as well do so. If you were working with an array of a billion+ numbers or in a very low memory environment, you would want to use the smallest data type that meets your needs.
Have a great day and a happy New Year!
Mark Gritter
Principal Engineer at Postman
·
Author has 5.7K answers and 11.7M answer views
·
3y
Originally Answered: What is a code that would calculate the average of 3 values and display a result?
·
CREATE TABLE input_values ( value real NOT NULL);INSERT INTO input_values VALUES( 95 );INSERT INTO input_values VALUES( 17 );INSERT INTO input_values VALUES( 23 );SELECT avg(value) from input_values;
print( sum( 95, 17, 23 ) / 3 )
Here’s a good Prolog example from RosettaCode: Averages/Arithmetic mean (there are 187 more examples at the link in other languages!)
mean(List, Mean) :- length(List, Length), sumlist(List, Sum), Mean is Sum / Length. mean( [95, 17, 23], X )?
But maybe you wanted a version specialized to three?
(write (/ (+ 95 17 23) 3))
(mean 95 17 23)
Gilbert Doan
Master (Unattempted) in Mathematics, San Jose State University (Graduated 2018)
·
Author has 29.2K answers and 16.5M answer views
·
Apr 22
Originally Answered: What is an algorithm to determine the average of 3 given numbers a, b, and c?
·
It is linear algorithm based on finite set. You add numbers in such set in any order, as summation commutes. Then divide summation by number of members in set.
It is one measure of middle in set, considered less stable than median or middle number in set, because average is computed number which may or may not be in set, rather than estimated from set.
Its actual use is wanting to know average of very basic stats, which can also be assessed in descriptive stats like median, minimum, maximum, given how many observations or members in finite set.
Sappho Reyes
Author has 941 answers and 169.6K answer views
·
2y
Originally Answered: How can I write a basic algorithm and program to compute the average of three numbers?
·
10 g1 = 0: g2 = 0: g3 = 0: s = 0: a = 0
20 Print “Enter 3 numbers without comma and separated by 1 space:”
30 Read g1, g2, g3
40 s = g1 + g2 + g3
50 a = s/3
60 Print g1. g2, g3, “Sum = “, s, “Average =”, a
70 Print “Continue Y/N?”
80 if r = “Y” then goto 20
90 else end
Related questions
What is the C program to calculate the sum and average of three user numbers?
How do you write an algorithm that will get 3 numbers as input from the users? What is the average and display the three numbers and its resultant average?
What algorithm can I use to calculate the average of five numbers?
What are the addition and average three numbers?
What is an algorithm to find and display the average of 10 numbers?
How do you get the average of three numbers?
What is an algorithm to calculate the average of three numbers?
How do I calculate the average of five numbers?
How can I write a basic algorithm and program to compute the average of three numbers?
What is a program that takes three numbers from a user and find their average?
How do you calculate the average of three numbers in C?
What is an algorithm to find the average of three numbers?
What is the algorithm for finding the greatest of three numbers?
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What is the algorithm for finding the average of multiple numbers?
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15822 | https://twominuteteachersguide.com/2018/11/03/nctm-2018-regional-quadratic-sorting-and-symmetry/ | NCTM 2018 Regional – Quadratic Sorting and Symmetry – Two-Minute Teacher's Guide
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NCTM 2018 Regional – Quadratic Sorting and Symmetry
The TMTG team presented on November 3, 2018, at the Kansas City Regional Conference of the National Council of Teachers of Mathematics. The focus was on quadratic functions — an idea for launching the topic through examples and non-examples of quadratics, some insights about the power of symmetry in thinking about quadratic functions, and an activity that involves graphing the factors of a quadratic function.
Link to Presentation Slides
See also their other presentation on using inciting incidents to spur math lessons.
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15823 | https://www.savemyexams.com/dp/maths/ib/aa/21/hl/revision-notes/geometry-and-trigonometry/vector-planes/angles-between-lines-and-planes/ | IBMathsDPAnalysis & Approaches (AA)HLRevision Notes3. Geometry & TrigonometryVector PlanesAngles Between a Line & a Plane
Angles Between a Line & a Plane (DP IB Analysis & Approaches (AA)): Revision Note
Written by: Amber
Reviewed by: Dan Finlay
Updated on
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Angle between line & plane
What is meant by the angle between a line and a plane?
The angle between a line and a plane is defined to be the angle between:
The line
The projection of the line onto the plane
This is the line of intersection between the plane and a perpendicular plane which contains the line
This is the smallest angle between the line and the plane
It is easiest to think of these two lines making a right-triangle with the normal vector to the plane
The line joining the plane will be the hypotenuse
The line on the plane will be adjacent to the angle
The normal will the opposite the angle
How do I find the angle between a line and a plane?
For example, consider:
The line with equation
The plane with equation
STEP 1
Find the acute angle between the direction vector of the line and the normal vector to the plane
Use the formula
STEP 2
Subtract this angle from 90° to find the acute angle between the line and the plane
Subtract the angle from if working in radians
Examiner Tips and Tricks
Remember that if the scalar product is negative, your answer will result in an obtuse angle. Therefore, taking the absolute value of the scalar product means that you always get the acute angle.
Worked Example
Find the angle in radians between the line L with vector equation and the plane with Cartesian equation .
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Author: Amber
Expertise: Maths Content Creator
Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.
Reviewer: Dan Finlay
Expertise: Maths Subject Lead
Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.
Download notes on Angles Between a Line & a Plane |
15824 | https://math.umd.edu/~immortal/MATH410/lecturenotes/ch3-2.pdf | Math 410 Section 3.2: The Extreme Value Theorem 1. Introduction: The Extreme Value Theorem is one of the two big theorems that emerge from having continuity. The other is the Intermediate Value Theorem.
2. Definition: We say a function f : D →R attains a maximum value provided that f(D) has a maximum, meaning ∃x0 ∈D such that ∀x ∈D, f(x0) ≥f(x).
Such an x0 is a maximizer of f.
Likewise for a minimum value.
Note: This is not the same as f(D) being bounded.
Example: The function f : [−1, 4] →R given by f(x) = 3 −x2 has a maximum value of 3 with maximizer x0 = 0 and has a minimum value of −13 with minimizer x0 = 4.
Example: The function f : (0, 5] →R given by f(x) = x has a maximum value of 5 with maximizer x0 = 5 but has no minimum value. Note however that f(D) is bounded below.
Example: The function f : [0, 5] →R given by f(x) = ( 1 x(x−5) if x ̸= 0, 5 0 if x = 0, 5 has neither a maximum value nor minimum value.
3. Here is the maximum version of the Extreme Value Theorem. The minimum version can be proved either by adjusting this proof or by applying the proof to −f(x).
(a) Lemma: Suppose f : D →R is continuous and D is closed and bounded, then f(D) is bounded above.
Proof: Suppose not. Then for all n ∈N there is some xn ∈D with f(xn) > n. From here we get a sequence {xn} By sequential compactness choose a subsequence {xni} →x0 ∈D (note n1 < n2 < n3 < ... are all integers). By continuity {f(xni)} →f(x0) but this means that {f(xni)} is bounded which contradicts the fact that f(xni) > ni and {ni} is an increasing and unbounded sequence of integers.
QED (b) Theorem (Extreme Value Theorem): Suppose f : D →R is continuous and D is closed and bounded. Then f attains both a maximum and minimum value.
Proof: By the lemma, f(D) is bounded above. Let M = sup(f(D)). We need to find some x0 ∈D with f(x0) = M.
For each n ∈N the value M −1/n is not an upper bound for f(D) and so there exists some xn ∈D with f(xn) > M −1/n. In addition f(xn) ≤M < M + 1/n and so we have M −1/n < f(xn) < M + 1/n or |f(xn) −M| < 1|1/n −0| and so {f(xn)} →M by the Comparison Lemma.
By sequential compactness choose a subsequence {xni} →x0 ∈D. Since {f(xni)} is a subsequence of {f(xn)} we also have {f(xni)} →M but by continuity {f(xni)} →f(x0). Thus f(x0) = M.
QED |
15825 | https://academics.prismahealth.org/academics/education/obgyn-clinical-practice-guidelines/outpatient-obstetrics/placenta-previa | Skip to content
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Placenta Previa
Contributor: Shelley Chapman, MD Last Update: 4/18/17
Definition:
Placental previa- implantation of the placenta over the internal cervical os or just reaching the cervical os by transvaginal scan in the third trimester.
Low lying placenta- placenta edge less than 2 cm from the internal cervical os.
Incidence: 0.3-0.5% of livebirths. The risk of previa following a prior cesarean delivery is 1-5% with a linear increase as the number of prior cesareans increases. With four or more cesareans, the risk for previa approaches 10%. AMA >35 years old increases the risk to 2% and >40 years old to 5%. Multiparity, prior suction curettage and smoking are all associated with an increased risk for previa.
Classical Clinical Presentation: painless bleeding typically in the third trimester as the lower uterine segment develops.
Diagnosis: Any suspicion of placenta previa due to either clinical history or transabdominal ultrasound examination should be confirmed by transvaginal ultrasound.
Management:
Antenatal (Define location of edge of placenta)
Recommend pelvic rest at the time of diagnosis if in the third trimester or vaginal bleeding is present.
Recommend measures to decrease constipation including stool softeners and high-fiber diet.
Counsel the patient to seek immediate medical attention for any vaginal bleeding.
Schedule ultrasound to assess placentation and fetal growth at 32 weeks. The incidence of placenta accreta is increased particularly in the setting of a prior uterine surgery.
If previa at 32 weeks, plan to deliver as below at 36-37 weeks.
If <2 cm from internal cervical os, ie low lying, repeat transvaginal ultrasound at 36 weeks, if patient remains asymptomatic.
If vaginal bleeding occurs, hospitalization may be required with consideration of antenatal corticosteroid therapy and hematocrit assessment with iron therapy or blood transfusion as needed to keep the maternal Hct above 30. These patients will have delivery timing individualized.
If resolution of previa or low-lying placenta, perform transvaginal ultrasound to assess for vasa previa at 34-36 weeks gestation.
Timing of delivery
Placenta previa can result in severe obstetric hemorrhage with subsequent maternal shock, need for transfusion, DIC, hysterectomy, damage to surrounding organs, ICU admission and even death. Placenta previa is likely to result in hemorrhage before delivery of the fetus. Suboptimal timing of delivery can result in decreased resources, fetal/neonatal hypoxemia or acidemia resulting from maternal shock. In a study of 230 cases, the risk of emergent bleed was 4.7% at 35 weeks, 15% at 36 weeks, 30% at 37 weeks, and 59% at 38 weeks. A decision analysis and expert opinion recommend delivery at 36-37 weeks of gestation in patients with uncomplicated placenta previa. Based on this information, we recommend a scheduled cesarean at 36-37 weeks without amniocentesis for FLM if good dating criteria exist. Of note, the incidence of RDS at 36 weeks is 7% and at 37 weeks 3.5%.
Operative planning
Have preoperative HCT >30% if possible.
Type and cross 2-4 units of blood.
Consult anesthesia to plan for possible intraoperative hemodynamic changes, need for blood products and longer operative time.
Have appropriate surgeons available for possible caesarean hysterectomy.
Ninety percent of “low lying placentas” in early pregnancy resolve by the third trimester. Placentas located <2 cm from the internal cervical os as documented by a late third trimester transvaginal scan should be offered cesarean section. Special consideration can be made after informed consent with the non-bleeding patient as to the possibility of vaginal delivery in the patient with a placental edge 1-2 cm from the internal cervical os.
References
Hull AD, Resnik R. Placenta Previa, Placenta Accreta, Abruptio Placentae and Vasa Previa in Creasy and Resnik’s Maternal-Fetal Medicine: Principles and Practice 6th ed. By Saunders, an imprint of Elsevier Inc; 2009.
Mercer B, Verghella V, Foley M et al. Placenta Accreta. Am J Obstet Gynecol 2010 Nov;203(5):430-9.
Robertson PA, Sniderman SH, Laros et al. neonatal morbidity according to gestational age and birth weight from five tertiary care centers in the US 1983-1986. Am J Obstet Gynecol 1992 Jun;166(6 Pt 1): 1629-41; PMID 1615970.
Reddy, U, Abuhamad, A, Levine, D, Saade, G. Fetal Imaging. J Ultrasound Med 2014 May; 33:745-57.
Related Pages
Adnexal Masses in Pregnancy
Antenatal Testing
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Antenatal Guidelines for Narcotic Administration
Assigning EDC
Asthma During Pregnancy
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Depression in Pregnancy
External Cephalic Version (ECV)
Fetal Growth Restriction (FGR)
Gestational Diabetes (Priscilla White Class A DM)
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15826 | https://content.wellspan.org/CPOE/Service%20Line%20Folders/Behavioral%20Health/Clozaril%20Table.pdf | WEEKLY, EVERY OTHER WEEK (BI-WEEKLY) AND EVERY 4 WEEKS (MONTHLY) MONITORING AND ELIGIBILITY On May 12, 2005, the Food and Drug Administration (FDA) approved changes to Clozaril labelling allowing certain qualified patients, under specific conditions to undergo monthly (every 4 weeks) monitoring. The revised labelling states that patients initiated on Clozaril must have a baseline White Blood Cell (WBC) count and baseline Absolute Neutrophil Count (ANC) before initiation of treatment. The WBC must be at least 3500/mm3 and the ANC must be at least 2000/mm3 for initiation of therapy. During the first 6 months of therapy, patient must be monitored on a weekly basis. If acceptable WBC and ANC values [WBC ≥ 3500/mm3 and ANC ≥ 2000/mm3] have been maintained during the first 6 months of continuous therapy, the frequency of monitoring WBC and ANC values may be reduced to every other week (biweekly). After 6 months of every other week monitoring without interruption due to leukopenia, the frequency of monitoring WBC and ANC may be reduced to every 4 weeks (monthly). WBC and ANC values must continue to be monitored weekly for at least 4 weeks after the discontinuation of clozapine regardless of monitoring frequency at the time of discontinuation. Patients with interrupted therapy who are monitored weekly: (no abnormal blood values requiring interruption - WBC < 3000/mm3, ANC<1500/mm3) If a patient has been receiving Clozaril therapy for less than 6 months with acceptable WBC and ANC values (WBC ≥ 3500/mm3 and ANC ≥ 2000/mm3 ), and there is a break in therapy that is less than or equal to one month (30 days), then the patient may continue with weekly monitoring and be eligible to reduce monitoring to every other week after 6 months from the initiation of therapy. If the break in therapy is greater than one month, then the patient must be monitored weekly for another 6 months before being eligible to switch to every other week monitoring. Patients with interrupted therapy who are monitored every other week (biweekly): (no abnormal blood values requiring interruption - WBC < 3000/mm3, ANC<1500/mm3) If a patient has been receiving Clozaril therapy for 6 months or longer with acceptable WBC and ANC values (WBC ≥ 3500/mm3 and ANC ≥ 2000/mm3 ) and there is a break in therapy that is less than or equal to one month (30 days), the patient must undergo weekly WBC and ANC monitoring for 6 weeks, then may return to every other week monitoring. If the break in therapy is greater than one month, then the patient must undergo weekly WBC and ANC monitoring for 6 months before returning to every other week monitoring. Patients with interrupted therapy who are monitored every 4 weeks (monthly): (no abnormal blood values requiring interruption - WBC < 3000/mm3, ANC<1500/mm3) If a patient has been receiving Clozaril therapy for 12 months or longer with acceptable WBC and ANC values (WBC ≥ 3500/mm3 and ANC ≥ to 2000/mm3 ) and there is a break in therapy that is less than or equal to one month (30 days), re-initiation of therapy requires weekly monitoring for 6 weeks, followed by a return to monthly monitoring. If the break in therapy is greater than one month, WBC and ANC monitoring must be performed weekly for 6 months, then every other week for 6 months, then may return to every 4 weeks. Mild Leukopenia/Granulocytopenia during weekly, biweekly, or monthly monitoring: (no abnormal blood values requiring interruption - WBC < 3000/mm3, ANC<1500/mm3) If a patient experiences mild leukopenia and/or granulocytopenia (3500/mm3>WBC ≥ 3000/mm3 and/or 2000/mm3> ANC ≥ 1,500/mm3 ), then the patient must be monitored twice weekly until WBC ≥ 3500/mm3 and ANC ≥ 2000/mm3. These patients are eligible to progress to every other week and every 4 weeks monitoring as outlined above. Moderate and Severe Leukopenia/Granulocytopenia during weekly, biweekly, or monthly monitoring: (abnormal blood values requiring interruption - WBC < 3000/mm3, ANC<1500/mm3) If a patient’s monitoring is weekly, every other week, or every 4 weeks and experiences a decreased WBC or ANC event but remains rechallengeable (3000/mm3 >WBC ≥ 2000/mm3 and/or 1500/mm3 > ANC ≥ 1000/mm3) the patient must be interrupted from Clozaril therapy. The patient may be rechallenged with Clozaril only after the blood values return to a minimum of WBC ≥ 3,500/mm3 and ANC ≥ 2,000/mm3. The patient must be monitored weekly for one year from the date of re-initiation of therapy. If there are no interruptions in therapy due to leukopenia/ granulocytopenia, the patient may proceed to every other week monitoring. After 6 months of every other week monitoring without interruptions in therapy due to leukopenia/granulocytopenia the patient may proceed to every 4 weeks monitoring. Medication supplies: If allowed by the prescriber, patients may receive supplies of medication sufficient for therapy for a period of time equal to that of the monitoring period, i.e. patients monitored weekly may receive a one week (7 day) supply of medication, patients eligible for every other week monitoring may receive a 2 week supply of medication and patients eligible for monthly (4-week) monitoring may receive a 1 month (28 days) supply of medication. Laboratory results: WBC and ANC values may not be more than 7 days old at the time of dispensing, regardless of the monitoring interval allowed. Medication should not be dispensed to patients who have not had WBC and ANC values monitored within 7 days of dispensing. Please verify new patients with the Clozaril National Registry (CNR) at 1.800.448.5938. Patients receiving Clozaril brand clozapine or a generic clozapine product must be registered prior to starting treatment to assure proper monitoring of WBC and ANC values. Click here to download Biweekly to Monthly Conversion Form along with Table 1 (from the PI). |
15827 | https://www.visionlearning.com/en/library/Math-in-Science/62/Exponential-Equations-II/210/ | Jump to main content Jump to website footer
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Equations
Exponential Equations II: The constant e and limits to growth
by Anne E. Egger, Ph.D., Janet Shiver, Ph.D., Teri Willard, Ed.D.
Listen to this reading
00:00
Did you know that by using carbon-14 dating and an exponential equation in math, scientists confirmed that Vikings visited North America 50 years before Columbus arrived? Scientists use a particular form of exponential equation when dealing with natural systems that are continuously changing. This common equation can be used to determine the amount of time that had passed, the degree of growth or decay of something, or the amount of something before or after an amount of time has passed.
A form of exponential equation that is very commonly used in science is N=N0ekt, which describes growth or decay over time.
The constant e is the limit of the expression (1 + 1/n)n with increasing n, and represents the limit of growth for any continuously growing system.
The constant k is a growth constant, whose value depends on the material, process, and environmental conditions of the system.
carbon-14 dating
: also called 14C-dating or radiocarbon dating
constant
: in mathematics, a quantity that has a fixed value; something that does not vary
value
: a number that is assigned based on measurement or a calculation
Perhaps you've heard the expression “multiplying like rabbits,” implying a very rapid increase. There is truth behind this expression: Starting at an age of six months, female rabbits can have a litter of up to 14 baby rabbits every month. If a single female rabbit lived for seven years and maintained that reproduction rate, and all of the female baby rabbits started reproducing at the same rate at the age of six months… well, that would be a lot of rabbits.
The biological limit to how fast the population of rabbits can grow is based on the gestation period, the time to maturity of the rabbits, and the average size of the litter. All of those factors can be combined mathematically to predict the growth rate of a population of rabbits over a given time period. While other factors may reduce the growth rate, that equation would describe the upper limit. The type of equation that describes the type of growth that increases over time is an exponential equation, introduced in our module Exponential Equations in Science I: Growth and Decay. However, when dealing with natural systems that have variability and are continuously changing, the exponential equation takes a particular form that is common in science.
The typical form of exponential equations in science
Many natural phenomena exhibit exponential growth (like population increase) or decay (like the depletion of radioactive isotopes), and thus exponential equations are frequently used in science. In virtually all cases, time is an important variable in these phenomena, so scientists often use a particular exponential equation with the variable of time already built into it. The exponential equation used by many scientists to describe growth or decay events is:
This equation is very similar to y=abx, the equation introduced in our module Exponential Equations in Science I: Growth and Decay. In fact, we can map each component from one equation to the other:
N0 is the amount of something at time 0, which is the same as the initial value a.
e is a constant (of approximate value 2.71828) that replaces the base value b.
k is a constant that determines how quickly the value grows or decays, called the growth or decay rate constant.
t is the variable of time, which replaces the variable x.
N is the amount of something, equivalent to the variable y, which depends on the initial value, the growth rate, and time.
Note that k and t are multiplied by each other in the equation. Because k is a rate, its units are "per unit time," and might be per year (yr-1) or per hour (hr-1). The variable t has the units of time: year, or hour. When k is multiplied by t, therefore, their units cancel and we are left with a unitless exponent. You will see examples of this later in the module.
Where do these constants e and k come from? And why are they present in so many exponential equations that are used in science? First of all, it is important to point out that e and k are very different kinds of constants. Specifically, k is a constant whose value differs for each material or process (for example, the k-value for decay of 14C, a radioactive isotope that decays to 14N, is different than the k-value for 238U, another radioactive isotope that decays to 206Pb). In contrast, e is always e; it always has the exact same value. But what is that value, and why does it show up in exponential equations?
Comprehension Checkpoint
The constant e
You might have learned about the number e before: It is one of the most commonly used irrational numbers, which are numbers that cannot be expressed as fractions. The first 32 digits of e are 2.7182818284590452353602874713527. But that’s only the first 32 digits—in 2010, Shigeru Kondo succeeded in calculating the value of e to 1 trillion digits (Yee, 2011).
But where did e come from, and what does it mean? Throughout the 1600s, many mathematicians in Europe were working with exponents, exploring both the mathematical concepts and the applications of those concepts in everything from astronomy to finance. In 1683, the Swiss mathematician Jacob Bernoulli was studying the expression (1 + 1/n)n. He recognized that this expression was involved in the calculation of compound interest, an important financial topic since Sumerian merchants recorded interest calculations on clay tablets as early as 1700 BCE (Maor, 1994) (see Figure 1 for an example).
Interest is what a bank (or other entity) will pay you on your investment, usually given in the form of an interest rate, like 5% per year. Let’s say you invested $100 in a savings account at your bank at a 5% interest rate. If your bank were to calculate the amount to pay you using simple interest, they would pay you 5% per year on your initial investment, so you would get 5% of $100, or $5, every year. If they use compound interest, however, they calculate your interest on the total amount in your account. So the first year, you would make $5. The second year, you would make 5% interest on $105, or $5.25. The third year, you would make 5% interest on $110.25, or $5.51. In other words, you would make a little bit more every year.
Bernoulli was exploring the idea of paying out that interest more frequently: twice a year, or three times a year. In other words, if you wanted to maximize the interest you were paid on your investment, how frequently should that interest be calculated? In Bernoulli’s expression , the number n refers to the number of times per year that the interest was calculated on your investment. As he worked with this expression, Bernoulli observed that, as n became larger and larger, the value of this expression always fell between 2 and 3, even for very large n values (see Table 1).
Table 1: The value of the expression calculated for increasing n. Notice that the solution to the equation always falls between 2 and 3, even as n gets very large.
| | |
| 1 | 2 |
| 2 | 2.25 |
| 4 | 2.44140625 |
| 6 | 2.52162637 |
| 10 | 2.59374246 |
| 100 | 2.70481382 |
| 1,000 | 2.71692393 |
| 10,000 | 2.71814592 |
| 100,000 | 2.71826823 |
| 1,000,000 | 2.71828046 |
Looking at the values in Table 1, you can see that while the values of the expression are always increasing, the amount of the increase in the solution to the equation gets smaller and smaller as n increases. If you think about this in terms of calculating interest, Bernoulli’s result shows that you don’t gain much more from calculating and paying out the interest ten times a year than you get from calculating it six times a year, and the biggest difference comes between one and three times a year. As larger and larger values of n are substituted, the value of the expression approaches a constant, which is approximately 2.71828—the value reported by Bernoulli as the limit of the expression (Figure 2). This would be the value of the interest paid on your investment if you had a 100% interest rate and if it were paid out constantly, an infinite number of times per year. The value of this limit is the number that we call e today. While Bernoulli formulated the expression and found the value, another mathematician, Leonard Euler, is credited with formalizing the constant with the designation e in the early 1700s.
So e is the limit to Bernoulli’s equation, but why is it used so commonly in exponential equations in science, replacing the base value b? What application does this have beyond calculating interest? To address those questions, it can be helpful to return to an exponential equation such as y = 2x. Table 2 shows a series of y values for that equation.
Table 2: x and corresponding y values for the equation.
| | |
| 0 | 1 |
| 1 | 2 |
| 2 | 4 |
| 3 | 8 |
| 4 | 16 |
If each of the values of x is considered to be a time step of equal length (such as an hour, a day, or a year), then with each time step there is a doubling of y. But the exponential equation y = 2x assumes that all of that doubling occurs right at the time step, rather than gradually. For example, if you had 100 bacterial cells sitting in a petri dish and you knew that they split about once an hour, using this equation would mean that all 100 of those cells would wait 60 minutes and then all split at the same time. That is not realistic, however. Instead, each bacterium splits at a different time, but after an hour has passed, you would expect them all to have split – and some of the bacteria that had split at the beginning of the hour would be starting to split again. In other words, the population of bacteria is continuously growing over that hour, rather than doubling all at once – their growth follows the same mathematical progression as compound interest at an interest rate of 100%, assuming that the interest were being paid out continuously, as we described above.
There is also a limit to how quickly that population can grow. A single bacterium can only split into two, not into four or five. This natural, continuously and gradually growing population is therefore not accurately described by an exponential equation with a base value (b) of 2, but by an exponential equation with a base value (b) of e. The constant e has many uses in math, but in science we consider it the base rate of change (growth or decay) in systems that are continuously changing (growing or decaying) over time. Those systems might be a population of bacteria, a chemical reaction, or the decay of a radioactive material.
In science, the constant e is intimately connected to the constant k, often called the growth rate constant. As an exponent in the equation N = N0ekt, k modifies the base rate e for a specific material, process, or environmental condition. For example, most bacteria split more rapidly at room temperature than at near freezing, so the same bacteria will have a different growth rate constant k at different temperatures since there are different environmental conditions. Both are growth rates, so in both cases k is positive. By comparison, two different radioactive isotopes, such as 14C and 238U , have different values of k even though they both decay radioactively because they are different materials. But because the two isotopes are decaying, or decreasing over time, both growth rate constants are negative – a different process than reproduction.
Comprehension Checkpoint
Solving exponential equations that use e
The equation N = N0ekt can be solved for any of the variables within it, and sometimes scientists are interested in finding t (the amount of time that has passed), k (the growth or decay rate of something in a given set of conditions), N0 (the initial amount of something), or N (the amount of something after a known amount of time has passed). The following sample problems address some of these different possibilities.
Sample Problem 1: Bacterial growth
We are often advised to refrigerate foods after opening them. Leftovers that sit outside the refrigerator overnight might “go bad” quickly even though they will last for several days in the refrigerator. Why is this the case? One reason is the presence of bacteria—bacteria are everywhere, and some are good for us and some are bad for us. Bacteria have been shown to display an exponential growth rate in many cases, but the rate of growth (or the steepness of the curve on the graph) varies significantly depending on temperature.
A study published in 1991 tested growth rates of the bacterium Lactobacillus plantarum (Figure 3), a generally benign bacterium found in fermented foods, at several different temperatures ranging from 6° C to 43° C (Zwietering et al., 1991). After allowing cultures to grow for 24 hours, the authors of the study determined the growth constant (k) for these different temperatures. At 6.0° C (about what you would find in a refrigerator), they found the growth rate constant of k = 0.0164 hr-1, while at 28° C (close to room temperature), they found a growth rate constant of k = 0.8 hr-1. (Remember that the units for k are “per unit time”, and that k can differ by material, process, or, as in this case, environmental variables like temperature.) What does this mean for the number of L. plantarum that would grow in your yogurt if you left it out of the refrigerator overnight for 12 hours?
For reference, here is a key to the exponential equation, N = N0ekt, for this problem:
N = total number of L. plantarum bacteria
N0 = initial amount of L. plantarum bacteria
k = L. plantarum growth rate
t = time passed
e = a constant, approximately 2.71828
Using the equation N = N0ekt, we can determine the number of bacteria at both temperatures after 12 hours. First, let’s figure out how many bacteria there would be if we put the yogurt back into the refrigerator at 6° C. We know that there are some L. plantarum already in the yogurt, so we can assume the following values:
N0 = 10 (we'll assume this initial value, though it's likely much higher)
k = 0.0164 hr-1
t = 12 hr
Substituting these values into the equation, we get:
N = N0ekt
N = 10 e 0.0164(12)
N = 12
As you can see, the number of bacteria does not increase very much in the refrigerator. Let’s try the same equation at the higher temperature, where k = 0.8 h-1:
N = N0ekt
N = 10 e 0.8(12)
N = 147,648 or N = 1.5 x 105
That’s a whole lot more bacteria than you started with – what appears to be a very small change in the value of k results in a very large change in N. This is a characteristic of exponential equations, described in more detail in our module Exponential Equations in Science I: Growth and Decay.
Comprehension Checkpoint
Sample Problem 2: Carbon-14 dating
Carbon-14 dating (or 14C dating) is a technique that can be used to determine the age of anything that was once living that has carbon in it, from pieces of wood and charcoal to bone and skin. 14C is a radioactive isotope present in small amounts in the atmosphere and all living organisms, and it decays exponentially with a decay rate of k = -0.00012 yr-1. While an organism is alive, it is constantly exchanging 14C with the atmosphere, replenishing the decaying isotopes and maintaining an approximately constant amount. Once an organism dies, however, that exchange and replenishment stops, and the 14C that was present at the time of death simply decays. (Note that because this is an exponential decay process, the value of k is negative. In earlier sample problems, we looked at growth processes, and k was positive.) Because we know the decay rate, we can determine how old something is if we know how much 14C is left from the starting amount (for more information about how 14C dating works, see our module on Uncertainty, Error, and Confidence).
In 1995, scientists from the University of Arizona, the US Department of Energy’s Brookhaven National Laboratory, and the Smithsonian Institution used 14C dating to determine the age of a controversial parchment that might be the first-ever map of North America, the Vinland map (shown in Figure 4). Text on the map reads, in part:
By God's will, after a long voyage from the island of Greenland to the south toward the most distant remaining parts of the western ocean sea, sailing southward amidst the ice, the companions Bjarni and Leif Eiriksson discovered a new land, extremely fertile and even having vines, ... which island they named Vinland.
This map, if authentic, suggests that the Vikings knew of North America before Columbus.
The scientists cut a small strip of parchment from the Vinland Map’s lower right corner and ran different parts of the sample through the dating process (Donahue et al., 2002). After running numerous tests the group found that the amount of 14C in the paper was approximately 93.5% of the modern value. Using this information and the equation N = N0ekt, they had what they needed to find the age of the parchment. We know that:
N = N0ekt
N0 = 100% or 1.0
k = -0.00012 yr-1
Substituting these values, we get:
N = N0ekt
0.935 = 1 e -0.00012t
In order to solve for t, we use an operation called the logarithm, and take the natural logarithm (ln) of both sides.
ln0.935 = -0.00012t
-0.0672 = -0.00012t
t = -0.0672 / -0.00012
t = 560 years
According to our calculations, the map was 560 years old in 1995. This suggests that the map was drawn around the year 1435, 50 years before Columbus came to America, suggesting that the Vinland map is authentic.
Comprehension Checkpoint
Applications of exponential equations in science
Equations in the form of N = N0ekt are ubiquitous in science. They are used in Earth science to determine the age of rocks based on the radioactive decay of long-lived isotopes such as 40K, in epidemiology to predict the growth and spread of a virus with a particular incubation period, in chemistry to describe reaction rates, in populations of organisms like rabbits to manage their population, and in countless other ways. All of these systems exhibit not just a simple doubling or tripling (or halving) over time, but are continuously growing or decaying, and are thus more accurately described by a base growth rate of e rather than an integer. |
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FreshLook ColorBlends Contact Lenses 1 box - 6 pack / lenses - 3 month supply
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Amethyst
Blue
Brown
Green
Grey
Honey
Pure Hazel
True Sapphire
Turquoise
Gemstone Green
Brilliant Blue
Sterling Grey
Rating: 9.4/10 335 Reviews
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$83.64Lowest price/box after rebate
Reg: ~~$106.14~~
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FreshLook ColorBlends Specifications & Parameters
| Manufacturer: | Alcon |
| Lens type: | Color Disposable |
| Each box contains: | 6 lenses, a 3-month supply |
| Water Content: | 55% |
| Material type: | Hydrogel |
| Oxygen transmissibility: | 16 Dk/t |
| Can be worn overnight?: | No |
| Recommended wearing time: | Daily |
| Replacement schedule: | 2 Week |
| Availability: | In stock |
| Cost per day: | $1.01 |
| Customer rating: | 9.4/10 |
| Visibility tinted: | Yes |
| Blocks UV: | No |
| Special features: | 3-in-1 color blend technology |
| % Lens Material: | 45% phemfilcon A |
| FDA material group: | Group 4 >50% water ionic polymer |
| Disinfection Method: | Multi-Purpose solution |
| Lens design: | Opaque Color |
| Production method: | M.O.S.T (molded optical surface technology) |
| Optical zone: | 5 mm |
| Diameters: | 14.5 mm |
| Base curves: | 8.6 mm |
| Sphere powers: | -8.00D, -7.50D, -7.00D, -6.50D, -6.00D, -5.75D, -5.50D, -5.25D, -5.00D, -4.75D, -4.50D, -4.25D, -4.00D, -3.75D, -3.50D, -3.25D, -3.00D, -2.75D, -2.50D, -2.25D, -2.00D, -1.75D, -1.50D, -1.25D, -1.00D, -0.75D, -0.50D, -0.25D, +0.00D, +0.25D, +0.50D, +0.75D, +1.00D, +1.25D, +1.50D, +1.75D, +2.00D, +2.50D, +3.00D, +3.50D, +4.00D, +4.50D, +5.00D, +5.50D, +6.00D |
| Lens Colors: | Amethyst, Blue, Brown, Green, Grey, Honey, Pure Hazel, True Sapphire, Turquoise, Gemstone Green, Brilliant Blue, Sterling Grey |
Product Information
FreshLook® Colorblends contact lenses use Alcon’s 3-in-1 Technology to combine three shades in one lens for a natural-looking or bold eye color change. These lenses are designed to enhance your eyes while maintaining a comfortable fit.
Features & Benefits
FreshLook® Colorblends are bi-weekly contact lenses that should be removed nightly and cleaned properly for up to two weeks of wear. Designed for both prescription and cosmetic use, they allow wearers to enjoy enhanced eye color without sacrificing comfort.
Do FreshLook Colorblends Lenses Have UV Protection?
FreshLook® Colorblends lenses do not have built-in UV protection. These lenses are designed primarily for cosmetic enhancement, blending three colors to create a natural-looking eye color change. To help protect your eyes from harmful UV rays while wearing them, it's recommended to wear sunglasses with UV-blocking lenses when outdoors.
| Freshlook Colorblends Contact Lenses Compared |
| Attribute | Freshlook Colorblends | FreshLook Handling Tint | FreshLook One-Day |
| Lens Type | 1-2 Week Disposable | 1-2 Week Disposable | Daily Disposable |
| Water Content | 55% | 55% | 69% |
| Overnight Wear | No | No | No |
| Lens.com Rating | 9.4 | 9.5 | 9.6 |
About FreshLook ColorBlends
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AI Analysis of FreshLook ColorBlends Reviews
FreshLook ColorBlends wins praise for turning natural eyes into believable new shades while staying comfortable enough for everyday wear. Reviewers routinely mention the “wow” factor of the blended pigments, straightforward ordering on Lens.com, and lenses that feel light from morning coffee to late-night outings.
Positive ThemesNatural-Looking Color Enhancement – 37 % of the users love how the tri-tone design blends with the iris for a “your-eyes-but-better” finish that draws compliments.
Convenient Ordering & Fast Delivery – Shoppers describe the Lens.com checkout and shipping experience as “fast,” “easy,” and “hassle-free,” often receiving lenses sooner than expected.
Negative ThemesLens Fragility / Tearing – A small group of users say the lenses can rip if pinched too hard, so gentle handling is advised.
Perceived High Price – 2 % of reviewers call the lenses “pricey,” though many still buy for the standout color.
Quick SummaryMost reviews spotlight believable color change, comfort, and a smooth buying experience, while reports of tearing, dryness, or shade mismatch are infrequent and minor.
Best For: Wearers who want a natural-looking yet noticeable eye-color upgrade without giving up everyday comfort or easy lens care.
Potential Drawbacks: Handle gently to avoid tearing, keep rewetting drops on hand for late-day dryness, and note that colors may vary slightly depending on your natural iris tone.
FreshLook ColorBlends FAQs
Are FreshLook ColorBlends contact lenses suitable for individuals with dark eye colors?
Yes, FreshLook ColorBlends contact lenses are designed to enhance the natural color of your eyes, even if you have dark eye colors.
What FreshLook ColorBlends rebates are available?
FreshLook ColorBlends contacts frequently qualify for rebates at Lens.com. You can qualify with just 4 or 8 boxes, unlike other retailers requiring a minimum of 12. Check our rebate or product pages for the latest offers.
How do I get my FreshLook ColorBlends rebate?
To get a FreshLook ColorBlends rebate, make a qualifying purchase from Lens.com. After your order ships, access the Rebate Center to print your contact lens rebate form(s). The mailing address is on the rebate form.
What if I don't have my prescription? Can I still order FreshLook ColorBlends contacts?
Yes, as long as your prescription is on file with your ECP, Lens.com can verify it for you. If your ECP doesn't respond in time, we'll ship your order according to FTC rules.
How do I check the status of my FreshLook ColorBlends rebates?
You can check the status of your FreshLook ColorBlends rebate by going to the Rebate Center in your account, located in the Customer Service menu. If you need further assistance, please contact our Customer Service Department via phone (1-800 LENS.COM, 24/7) or Live Chat.
How long does it take to adjust to wearing FreshLook ColorBlends contacts?
It can take a few days to weeks to adjust to FreshLook ColorBlends contacts, depending on the lens type, material, and personal factors. Soft lenses typically require less adjustment time.
How long does it take to get my FreshLook ColorBlends rebate?
After submitting your FreshLook ColorBlends rebate forms you can check the status of your rebate by going to Rebate Center. Please allow up to 12 weeks to receive your Lens.com Visa Prepaid Card.
Can I use my eyeglass prescription for FreshLook ColorBlends contact lenses?
No, eyeglass prescriptions don�t include the additional specifications needed for contact lenses, such as base curve and diameter. See your ECP for a contact lens fitting.
Does FreshLook ColorBlends have a rebate?
Rebates can be available from the manufacturer or directly from retailers. Lens.com typically offers rebates for FreshLook ColorBlends as it is a popular content lens. At Lens.com you need to purchase only 4 to 8 boxes of contacts (2/4 per eye) in order to qualify for some of the highest-value rebates in the industry.
Can I buy FreshLook ColorBlends for someone else?
Yes, you can purchase FreshLook ColorBlends contacts for someone else. Simply provide their prescription details and ECP's contact info, and Lens.com will handle the rest.
Are online FreshLook ColorBlends contact lenses the same as from your eye doctor?
Yes, FreshLook ColorBlends contacts from Lens.com are the same lenses sold by your ECP, but at a discounted price due to lower operating costs.
How do I read my FreshLook ColorBlends prescription?
FreshLook ColorBlends prescriptions include power, base curve, diameter, and more. Check out our guide on reading your contact lens prescription for more info.
Do I need an rx/prescription to buy FreshLook ColorBlends contacts online?
Yes, a valid prescription is required to buy FreshLook ColorBlends contacts online, but you don't need a physical copy. Lens.com will verify your prescription with your ECP as required by US federal law, making the process easy for you.
How to tell if my FreshLook ColorBlends contacts are inside out?
FreshLook ColorBlends contacts feel uncomfortable if worn inside out. They may move too much or even pop out. Some lenses have an inside-out mark to help you check.
Can I be allergic to FreshLook ColorBlends contacts?
FreshLook ColorBlends lenses are hypoallergenic, but buildup on the lenses can cause reactions. If you experience symptoms, remove your contacts and consult your ECP.
Can I renew my FreshLook ColorBlends RX/prescription online?
Yes, you can renew your FreshLook ColorBlends prescription online through Lens.com in just 10 minutes. Check if the service is available in your state.
Does FreshLook ColorBlends have rebates?
Yes, FreshLook ColorBlends contacts typically offer rebates through Lens.com. Be sure to check current offers and qualifying purchase requirements.
How should I store my FreshLook ColorBlends contact lenses?
Daily lenses should be discarded after use. Weekly and monthly lenses can be stored in contact solution in a tightly closed case for their recommended duration.
Do I need a prescription, or a copy of my prescription in order to buy FreshLook ColorBlends?
Yes, you need a valid prescription to buy FreshLook ColorBlends contacts, but you don't need a hard copy. Lens.com can verify it with your ECP.
Do FreshLook ColorBlends look natural?
Yes, FreshLook ColorBlends are made to look natural using 3-in-1 color technology-some shades like Blue look subtle, while others like Brilliant Blue are more vibrant.
What colors do FreshLook contacts come in?
FreshLook Colorblends come in Amethyst, Blue, Brown, Green, Grey, Honey, Pure Hazel, True Sapphire, Turquoise, Brilliant Blue, Sterling Grey, and Gemstone Green.
What's the difference between FreshLook Colorblends and FreshLook Colors?
FreshLook Colorblends blend three colors for a natural effect, while FreshLook Colors enhance your natural eye color with a bold iris pattern�both are bi-weekly lenses.
How many boxes of FreshLook ColorBlends do I need to buy to last 6 months, or 12 months?
You need 2 box(es) of FreshLook ColorBlends contacts to last you for 6 months. In order to last you for 12 months, you'll need 4 box(es). These amounts pertain to a single eye, so to cover both eyes, double these amounts.
Can FreshLook ColorBlends contacts expire?
Yes, FreshLook ColorBlends contacts expire and should not be used past the date printed on the box, usually shown as YYYY-MM (year and month).
Can I take a nap in FreshLook ColorBlends contacts?
No, FreshLook ColorBlends contacts are not approved for napping or overnight wear and should be removed before sleep.
FreshLook ColorBlends Reviews
Rating: 9.4/10 335 Reviews
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Anonymous
09/24/2025
Contacts are fine
the review only 3 stars because the lasts times that ive been purshasing contacta here it has take a loong time to receive it. And it always says that takes 3 to 5 days . Service lately has been to slow in the delivery
Anonymous
09/24/2025
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Ana
09/13/2025
Freshlook
Is very comfortable
Ana
09/13/2025
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Lisa Niemann
08/23/2025
Wear time
I want monthly wear these are every 2 weeks
Lisa Niemann
08/23/2025
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Anonymous
08/14/2025
for me to receive it as soon as possible
Excellent service
Anonymous
08/14/2025
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Carmen Gomez
07/21/2025
Quality
Very good customer service
Carmen Gomez
07/21/2025
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Carmen Gomez
07/17/2025
Quality
Very good customer service
Carmen Gomez
07/17/2025
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15829 | https://jamanetwork.com/journals/jamanetworkopen/fullarticle/2821953 | Ferritin Cutoffs and Diagnosis of Iron Deficiency in Primary Care | Pathology and Laboratory Medicine | JAMA Network Open | JAMA Network
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Original Investigation
Pathology and Laboratory Medicine
Ferritin Cutoffs and Diagnosis of Iron Deficiency in Primary Care
Levy Jäger,MD 1; Yael Rachamin,PhD 1,2; Oliver Senn,Prof, MD, MPH 1
et al
Jakob M.Burgstaller,MD, DMD, PhD, EMBA 1; Thomas Rosemann,MD, PhD 1; Stefan Markun,MD 1
Author Affiliations
Article Information
1 Institute of Primary Care, University Hospital Zurich, University of Zurich, Zurich, Switzerland
2 Campus Stiftung Lindenhof Bern (SLB), Bern, Switzerland
Cite This### Citation
Jäger L, Rachamin Y, Senn O, Burgstaller JM, Rosemann T, Markun S. Ferritin Cutoffs and Diagnosis of Iron Deficiency in Primary Care. JAMA Netw Open. 2024;7(8):e2425692. doi:10.1001/jamanetworkopen.2024.25692
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JAMA Netw Open
Published Online: August 5, 2024
2024;7;(8):e2425692. doi:10.1001/jamanetworkopen.2024.25692
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Iron Deficiency in AdultsJAMA Review March 30, 2025
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Supplement 1.
eTable 1. Compiled RECORD (Reporting of Studies Conducted Using Observational Routinely Collected Health Data) checklist
eTable 2. Operationalized criteria for the identification of clinical characteristics
eTable 3. Follow-up times, event counts, and incidence of nonanemic, anemic, and total iron deficiency diagnoses at different ferritin cutoffs
eTable 4. Determinants of ferritin testing (full model)
eTable 5. Determinants of ferritin testing (null model)
eTable 6. Determinants of hemoglobin and C-reactive protein testing accompanying ferritin testing
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Supplement 2.
Data Sharing Statement
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Figure 1. Study Flowchart
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The study period was January 1, 2021, to November 30, 2023. FIRE indicates Family Medicine Research Using Electronic Medical Records; GP, general practitioner.
Figure 2. Determinants of Ferritin Testing
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Results of mixed-effects Cox proportional hazards regression, expressed as adjusted hazard ratios (AHRs) for covariates and a median hazard ratio (MHR, derived from a corresponding null model without general practitioner [GP]–level covariates) to summarize the random-effects distribution. Associations with patient sex-age strata are summarized as follows: for male patients, as AHRs with respect to male patients younger than 25 years; and for female patients, as AHRs with respect to male patients of the respective same age category. A total of 255 119 were studied (to avoid artifacts due to high statistical imbalance, 232 patients were excluded due to unknown or other sex). A total of 56 927 ferritin testing events were studied. CHF indicates congestive heart failure, CKD, chronic kidney disease; IBD, inflammatory bowel disease; PPI proton pump inhibitor; SF, serum ferritin.
Table 1. Characteristics of the Study Cohort a
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| Characteristic | Did not receive ferritin testing during the study period (n = 182 534) | Received ferritin testing during the study period (n = 72 817) |
:---
| Patient sex | | |
| Female | 83 455 (45.7) | 49 543 (68.0) |
| Male | 98 965 (54.2) | 23 270 (32.0) |
| Unknown or other | 114 (0.1) | 4 (<0.01) |
| Patient age at inclusion, y | | |
| <25 | 15 795 (8.7) | 5886 (8.1) |
| 25-34 | 28 116 (15.4) | 9441 (13.0) |
| 35-44 | 29 353 (16.1) | 11 000 (15.1) |
| 45-54 | 29 133 (16.0) | 11 216 (15.4) |
| 55-64 | 25 431 (13.9) | 11 335 (15.6) |
| 65-74 | 25 125 (13.8) | 12 364 (17.0) |
| ≥75 | 29 351 (16.1) | 11 573 (15.9) |
| Unknown | 230 (0.1) | 2 (<0.01) |
| History of clinical factors at inclusion | | |
| Anemia | 11 767 (6.4) | 12 176 (16.7) |
| Iron therapy | 7576 (4.2) | 12 269 (16.8) |
| CKD | 6244 (3.4) | 5946 (8.2) |
| IBD | 861 (0.5) | 826 (1.1) |
| Rheumatic diseases | 1548 (0.8) | 1199 (1.6) |
| CHF | 602 (0.3) | 436 (0.6) |
| Pregnancy | 712 (0.4) | 804 (1.1) |
| Cancer | 2642 (1.4) | 2007 (2.8) |
| PPI therapy | 41 954 (23.0) | 28 709 (39.4) |
| Fatigue | 2701 (1.5) | 3074 (4.2) |
| No. of consultations in the year prior to inclusion | | |
| 0-1 | 114 529 (62.7) | 27 109 (37.2) |
| 2-5 | 33 802 (18.5) | 16 962 (23.3) |
| >5 | 34 203 (18.7) | 28 746 (39.5) |
| Received a ferritin test in the year before inclusion | 7630 (4.2) | 15 884 (21.8) |
| Characteristics of the first ferritin test during the study period | | |
| Was accompanied by a hemoglobin test | NA | 52 499 (72.1) |
| Was accompanied by a CRP test | NA | 36 136 (49.6) |
| Was accompanied by iron studies | | |
| None | NA | 65 781 (90.3) |
| Serum iron | NA | 6095 (8.4) |
| Transferrin | NA | 5820 (8.0) |
| Transferrin saturation | NA | 3856 (5.3) |
| Soluble transferrin receptor | NA | 506 (0.7) |
| Ferritin concentration, median (IQR), ng/mL | NA | 81 (40-155) |
Table 2. Incidences of Iron Deficiency Diagnoses (Nonanemic, Anemic, and Total) at Different Serum Ferritin (Ferritin) Cutoffs
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| Diagnosis | Incidences in cases per 1000 patient-years (95% CI) |
:--- |
| 15 ng/mL | 30 ng/mL | 45 ng/mL |
| Any ferritin test |
| Nonanemic iron deficiency | 4.1 (3.9-4.2) | 14.6 (14.3-15.0) | 25.8 (25.3-26.2) |
| Anemic iron deficiency | 3.5 (3.3-3.7) | 6.0 (5.8-6.2) | 7.5 (7.3-7.7) |
| Total iron deficiency | 10.9 (10.6-11.2) | 29.9 (29.4-30.4) | 48.3 (47.7-48.9) |
| Only ferritin tests not accompanied by elevated CRP concentrations |
| Nonanemic iron deficiency | 4.2 (4.0-4.4) | 14.6 (14.3-15.0) | 25.6 (25.2-26.0) |
| Anemic iron deficiency | 3.1 (3.0-3.3) | 5.3 (5.1-5.5) | 6.6 (6.4-6.8) |
| Total iron deficiency | 9.8 (9.6-10.1) | 26.7 (26.2-27.1) | 43.0 (42.5-43.6) |
Table 3. Determinants of Hemoglobin and C-Reactive Protein Testing Accompanying Ferritin Testing a
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| Determinant | AOR (95% CI) |
:--- |
| Hemoglobin | C-reactive protein |
| Patient sex-age strata | | |
| Male, <25 y | 1.0 [Reference] | 1.0 [Reference] |
| Male, 25-34 y to male, <25 y | 1.20 (0.84-1.72) | 0.91 (0.75-1.09) |
| Male, 35-44 y to male, <25 y | 1.32 (0.93-1.89) | 0.83 (0.69-0.99) |
| Male, 45-54 y to male, <25 y | 1.11 (0.79-1.57) | 0.79 (0.66-0.94) |
| Male, 55-64 y to male, <25 y | 1.24 (0.88-1.73) | 0.77 (0.65-0.91) |
| Male, 65-74 y to male, <25 y | 1.34 (0.96-1.88) | 0.79 (0.67-0.93) |
| Male, ≥75 y to male, <25 y | 1.80 (1.27-2.53) | 0.85 (0.71-1.00) |
| Female to male, <25 y | 0.21 (0.04-1.02) | 0.33 (0.13-0.82) |
| Female to male, 25-34 y | 0.91 (0.70-1.18) | 1.06 (0.93-1.21) |
| Female to male, 35-44 y | 0.83 (0.64-1.06) | 1.12 (0.99-1.27) |
| Female to male, 45-54 y | 1.00 (0.79-1.26) | 1.12 (1.00-1.26) |
| Female to male, 55-64 y | 0.89 (0.71-1.11) | 1.05 (0.94-1.17) |
| Female to male, 65-74 y | 0.81 (0.65-1.01) | 1.06 (0.95-1.18) |
| Female to male, ≥75 y | 1.59 (1.15-2.20) | 0.90 (0.76-1.06) |
| Clinical factors | | |
| Anemia | 1.61 (1.32-1.96) | 0.95 (0.86-1.05) |
| Iron therapy | 0.72 (0.63-0.82) | 0.97 (0.91-1.03) |
| CKD | 1.05 (0.88-1.25) | 0.80 (0.74-0.87) |
| IBD | 1.61 (1.10-2.36) | 1.13 (0.95-1.35) |
| Rheumatic diseases | 0.79 (0.59-1.07) | 1.16 (0.99-1.35) |
| CHF | 0.83 (0.52-1.34) | 0.78 (0.63-0.97) |
| Pregnancy | 1.40 (0.92-2.13) | 0.91 (0.75-1.10) |
| Cancer | 0.94 (0.74-1.21) | 0.97 (0.86-1.09) |
| PPI therapy | 1.09 (0.99-1.20) | 0.99 (0.95-1.04) |
| Fatigue | 1.51 (1.22-1.86) | 1.15 (1.05-1.27) |
| Primary care use, No. of consultations in the past year | | |
| 0-1 | 1.0 [Reference] | 1.0 [Reference] |
| 2-5 | 0.83 (0.74-0.93) | 0.95 (0.90-1.01) |
| >5 | 0.73 (0.65-0.82) | 0.86 (0.82-0.91) |
| GP sex | | |
| Male | 1.0 [Reference] | 1.0 [Reference] |
| Female | 2.94 (0.93-9.25) | 1.58 (0.82-3.04) |
| GP age, y | | |
| <45 | 1.0 [Reference] | 1.0 [Reference] |
| 45-59 | 0.64 (0.21-2.02) | 0.59 (0.31-1.14) |
| ≥60 | 1.23 (0.26-5.80) | 0.81 (0.33-1.95) |
| GP workload, No. of consultations per working week | | |
| <100 | 1.0 [Reference] | 1.0 [Reference] |
| 100-199 | 1.53 (0.43-5.48) | 0.73 (0.35-1.51) |
| ≥200 | 2.23 (0.53-9.47) | 0.39 (0.17-0.90) |
| Practice location | | |
| Nonurban | 1.0 [Reference] | 1.0 [Reference] |
| Urban | 0.30 (0.11-0.87) | 0.51 (0.28-0.93) |
Key Points
QuestionHow is the choice of guideline-recommended cutoffs for ferritin associated with the incidence of iron deficiency diagnoses in primary care?
FindingsIn this cohort study of 255 351 adult primary care patients, ferritin cutoffs of 15, 30, and 45 ng/mL were associated with incidences of iron deficiency diagnoses of 10.9, 29.9, and 48.3 cases per 1000 patient-years, respectively.
MeaningThe results of this study provide useful components for the evaluation of ferritin testing in high-resource primary care settings and call for harmonization of guidelines for iron deficiency.
Abstract
ImportanceFerritin is often measured by general practitioners, but the association of different cutoffs with the rates of iron deficiency diagnoses, particularly nonanemic iron deficiency, is unknown.
ObjectiveTo investigate the association of the ferritin cutoff choice with the incidence of nonanemic and anemic iron deficiency diagnoses in primary care.
Design, Setting, and ParticipantsIn this retrospective cohort study, patients 18 years or older with at least 1 consultation with a general practitioner participating in the Family Medicine Research Using Electronic Medical Records (FIRE) project, an electronic medical records database of Swiss primary care, from January 1, 2021, to November 30, 2023, were evaluated.
ExposuresSex, age, clinical patient characteristics, and professional general practitioner characteristics.
Main Outcomes and MeasuresIncidence of iron deficiency diagnoses (nonanemic and anemic) at ferritin cutoffs of 15, 30, and 45 ng/mL and ferritin testing itself. Time-dependent Cox proportional hazards regression was used to examine associations of patient and general practitioner characteristics with ferritin testing as adjusted hazard ratios (AHRs).
ResultsThe study included 255 351 patients (median [IQR] age, 52 [36-66] years; 52.1% female). Per 1000 patient-years and at ferritin cutoffs of 15, 30, and 45 ng/mL, iron deficiency diagnoses had incidences of 10.9 (95% CI, 10.6-11.2), 29.9 (95% CI, 29.4-30.4), and 48.3 (95% CI, 47.7-48.9) cases, respectively; nonanemic iron deficiency diagnoses had incidences of 4.1 (95% CI, 3.9-4.2), 14.6 (95% CI, 14.3-15.0), and 25.8 (95% CI, 25.3-26.2) cases, respectively; and anemic iron deficiency diagnoses had incidences of 3.5 (95% CI, 3.3-3.7), 6.0 (95% CI, 5.8-6.2), and 7.5 (95% CI, 7.3-7.7) cases, respectively. Ferritin testing showed notable associations with fatigue (AHR, 2.03; 95% CI, 1.95-2.12), anemia (AHR, 1.75; 95% CI, 1.70-1.79), and iron therapy (AHR, 1.50; 95% CI, 1.46-1.54). Ferritin testing was associated with female sex in all age groups, including postmenopausal. Of the patients who received ferritin testing, 72.1% received concomitant hemoglobin testing, and 49.6% received concomitant C-reactive protein testing.
Conclusions and RelevanceIn this retrospective cohort study of primary care patients, ferritin cutoffs of 30 and 45 ng/mL were associated with a substantially higher incidence of iron deficiency compared with 15 ng/mL. These results provide a basis for health system-level evaluation and benchmarking of ferritin testing in high-resource settings and call for a harmonization of diagnostic criteria for iron deficiency in primary care.
Introduction
Iron deficiency is a common condition and a leading cause of years lived with disability worldwide, mainly due to subsequent anemia.1 A depletion of iron stores that does not yet result in anemia, simply referred to as nonanemic iron deficiency, has recently gained attention as a distinct clinical entity.2 Associated with various symptoms, including fatigue, restless legs syndrome, and hair loss,3 nonanemic iron deficiency has been estimated to be more common than anemic iron deficiency.2
Measurement of (serum) ferritin is considered the mainstay of iron deficiency diagnosis,4 and ferritin is a commonly requested laboratory test in various high-resource primary care settings.5-7 In particular, a Swiss study revealed that 27% of the population received serum ferritin testing in 2018,8 and ferritin ranked among the most frequently ordered laboratory tests in Swiss primary care between 2009 and 2018.9 At the same time, iron deficiency guidelines are conflicting as to which populations benefit most from ferritin testing.4 For example, the US Preventive Services Task Force specifically mentions children and pregnant women10 but makes no recommendation for the general population. A prominent Swiss primary care guideline discourages iron deficiency screening in the general population, with exceptions for conditions that require special consideration due to increased risk, such as inflammatory bowel disease (IBD), or when evidence suggests a benefit from replete iron stores, such as chronic kidney disease (CKD).11
The optimal ferritin cutoffs for the diagnosis of iron deficiency, especially nonanemic iron deficiency, are controversial. Different guidelines suggest widely varying cutoffs ranging from 12 to 15 ng/mL through 25 to 30 ng/mL to 45 to 50 ng/mL in the general population (to convert to micrograms per liter, multiply by 1).4 The choice of the ferritin cutoff may have important implications. On the one hand, choosing too low a cutoff could result in withholding therapy from patients affected by the negative health consequences of iron deficiency. On the other hand, choosing too high a cutoff could lead to overtreatment of patients who do not benefit from iron therapy, with potential harm from adverse effects (especially with oral preparations12) and an unjustified waste of health care resources.
Although previous studies have estimated incidences of anemic iron deficiency in primary care,13,14 they have not evaluated the choice of different ferritin cutoffs, and the incidence of nonanemic iron deficiency has not been investigated. However, appropriate data would be essential for a comprehensive evaluation of ferritin testing in primary care. Furthermore, little is known about the factors associated with ferritin testing. Our study aims to fill these gaps by estimating how the incidence of diagnoses of nonanemic and anemic iron deficiency depends on the choice of ferritin cutoff and by examining the determinants of ferritin testing in Swiss primary care.
Methods
Setting and Data Source
We conducted a retrospective cohort study using data from the Family Medicine Research Using Electronic Medical Records (FIRE) project, a database of routine data from Swiss general practitioners hosted by the Institute of Primary Care of the University of Zurich.15 The database contains anonymized data on medication prescriptions, including Anatomical Therapeutic Chemical codes16 and Global Trade Item Numbers, codes of the International Classification of Primary Care (ICPC-2) system,17 body height and weight, sex and year of birth, practice postal code, and laboratory test results.
The local ethics committee of the Canton of Zurich waived ethical approval for this study and the need for informed consent because data from the FIRE project fall outside the scope of the Federal Act on Research involving Human Beings.18 We report the results of this study according to the Reporting of Studies Conducted Using Observational Routinely-Collected Health Data (RECORD) guidelines (eTable 1 in Supplement 1).19
Study Population
We conducted a retrospective cohort study from January 1, 2021, to November 30, 2023, including all patients with at least 1 consultation during the study period. For each patient, we used the date of the first consultation during the study period as the time of inclusion and excluded all patients younger than 18 years at inclusion to obtain the study cohort.
Explanatory Covariates
We defined sex-age strata (combinations of female and male and age <25, 25-34, 35-44, 45-54, 55-64, 65-74, and ≥75 years) and identified the presence of the following clinical factors often mentioned in guidelines as requiring special considerations for ferritin testing4: CKD, IBD, rheumatic diseases, congestive heart failure, pregnancy, cancer, proton pump inhibitor (PPI) therapy, and fatigue (see eTable 2 in Supplement 1 for the operationalized criteria). We further identified anemia as the presence of hemoglobin concentrations less than 12 g/dL for women and less than 13 g/dL for men (to convert to grams per liter, multiply by 10) according to the World Health Organization definition,20 iron therapy as the prescription of Anatomical Therapeutic Chemical codes in subgroup B03A (iron preparations), and defined primary care use as the number of consultations in the year preceding inclusion. For the general practitioners, we retrieved sex, age, workload (in consultations per working week), and urbanity of the practice location (according to the Swiss Federal Statistical Office21).
Statistical Analysis
We report variable summaries as numbers (percentages) or medians (IQRs) as appropriate. Missing data were handled using a complete-case approach in regression analyses. We conducted all analyses from October 2, 2023, to May 29, 2024, using the statistical software R, version 4.3.2 (R Foundation for Statistical Computing).22
We assessed the association of ferritin cutoff choice with the incidence of iron deficiency diagnoses using the following 3 cutoffs recommended by guidelines and expert panels for the adult general population4,11,23-27: 15, 30, and 45 ng/mL. For each cutoff and patient, we defined an iron deficiency event as the first measurement of ferritin concentrations below that cutoff during the study period and calculated patient-time as the number of days from (and including) inclusion to (and including) the earliest of either an iron deficiency event, if any, or the end of the study period. For each ferritin cutoff, we excluded patients with iron deficiency as at least 1 ferritin concentration below that cutoff during the year before inclusion from the respective population at risk. We calculated the total incidence of iron deficiency diagnoses as the number of iron deficiency events per 1000 patient-years. We classified iron deficiency events as indicating nonanemic and anemic iron deficiency when accompanied by hemoglobin concentrations within ±4 weeks excluding and indicating anemia, respectively, and calculated their specific incidences. We report all incidence estimates with exact Poisson mean 95% CIs.28
Comeasurement of C-reactive protein (CRP) with ferritin testing is often recommended to rule out systemic inflammation, which may elevate ferritin independently of iron storages.11,23,25,26 To address this aspect, we conducted a sensitivity analysis in which we repeated the incidence estimations based only on ferritin tests not accompanied by elevated CRP concentrations (within ±1 week, using reported reference ranges).
To assess the incidence and the determinants of ferritin testing, we chose to treat ferritin tests requested after more than 1 year after a previous ferritin test as new instances of ferritin testing, as done in a previous study of the appropriateness of ferritin retesting.29 Therefore, we excluded from this analysis all patients who had received ferritin testing in the year preceding inclusion and defined a ferritin testing event as the first ferritin request during the study period for each of the remaining patients. We calculated patient-time as the number of days from (and including) inclusion to (and including) the earliest of either a ferritin testing event, if any, or the end of the study period and obtained the incidence of ferritin testing as the number of ferritin testing events per 1000 patient-years. We fitted a mixed-effects Cox proportional hazards regression model on the ferritin testing event with patient sex-age strata, primary care use, general practitioner sex and age, general practitioner workload, and practice urbanity as covariates at baseline. Anemia, CKD, IBD, rheumatic disease, congestive heart failure, pregnancy, cancer, PPI therapy, and fatigue were included as time-dependent binary covariates,30 defined as present starting from the date of their first identification in the database. To account for the same general practitioner treating multiple patients, we used normally distributed general practitioner–level random intercepts. We used the R package coxme31 to fit the models and express the associations of the various covariates with ferritin testing as adjusted hazard ratios (AHRs) with corresponding 95% CIs. We quantify general practitioner–level variation as the median hazard ratio (MHR) derived from a null model without general practitioner–level covariates, reported with a Hall percentile 95% CI obtained from 2000 iterations of a cases bootstrap.32 The MHR can be interpreted as the median increase in hazard for ferritin testing in a given patient when randomly selecting 2 general practitioners and comparing the one with a higher propensity to the one with a lower propensity to test for ferritin.33
We also assessed requests for additional iron studies accompanying ferritin testing (within ±1 week), considering iron studies often mentioned in international and local clinical guidelines for iron deficiency4,11: serum iron, transferrin, transferrin saturation, and soluble transferrin receptor. To further examine concomitant testing, we considered all patients of the study cohort who received ferritin testing during the study period and considered the presence of hemoglobin and CRP accompanying the first ferritin test recorded during the study period as 2 distinct binary outcomes. We fitted mixed-effects logistic regression models using the R package lme434 to assess the associations of each outcome with the same determinants and random effects as for ferritin. Associations are expressed as adjusted odds ratios (AORs) with corresponding 95% CIs.
Results
The study cohort consisted of 255 351 patients (median [IQR] age, 52 [36-66] years; 52.1% female and 47.9% male). Figure 1 summarizes the cohort selection process, and Table 1 summarizes the patient characteristics. The cohort patients were seen by 262 general practitioners with a median (IQR) age of 48 (40-57) years and a median (IQR) workload of 139 (86-230) consultations per working week in 80 practices. During the study period, 72 817 patients (28.5%) received ferritin testing. Of the first ferritin tests during the study period, 7036 (9.7%) were accompanied by iron studies, most commonly serum iron in 6095 (8.4%), transferrin in 5820 (8.0%), and transferrin saturation in 3856 (5.3%). Accompanying hemoglobin and CRP tests were present in 52 499 (72.1%) and 36 136 (49.6%) cases, respectively.
Figure 1. Study Flowchart
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The study period was January 1, 2021, to November 30, 2023. FIRE indicates Family Medicine Research Using Electronic Medical Records; GP, general practitioner.
Table 1. Characteristics of the Study Cohort a
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| Characteristic | Did not receive ferritin testing during the study period (n = 182 534) | Received ferritin testing during the study period (n = 72 817) |
:---
| Patient sex | | |
| Female | 83 455 (45.7) | 49 543 (68.0) |
| Male | 98 965 (54.2) | 23 270 (32.0) |
| Unknown or other | 114 (0.1) | 4 (<0.01) |
| Patient age at inclusion, y | | |
| <25 | 15 795 (8.7) | 5886 (8.1) |
| 25-34 | 28 116 (15.4) | 9441 (13.0) |
| 35-44 | 29 353 (16.1) | 11 000 (15.1) |
| 45-54 | 29 133 (16.0) | 11 216 (15.4) |
| 55-64 | 25 431 (13.9) | 11 335 (15.6) |
| 65-74 | 25 125 (13.8) | 12 364 (17.0) |
| ≥75 | 29 351 (16.1) | 11 573 (15.9) |
| Unknown | 230 (0.1) | 2 (<0.01) |
| History of clinical factors at inclusion | | |
| Anemia | 11 767 (6.4) | 12 176 (16.7) |
| Iron therapy | 7576 (4.2) | 12 269 (16.8) |
| CKD | 6244 (3.4) | 5946 (8.2) |
| IBD | 861 (0.5) | 826 (1.1) |
| Rheumatic diseases | 1548 (0.8) | 1199 (1.6) |
| CHF | 602 (0.3) | 436 (0.6) |
| Pregnancy | 712 (0.4) | 804 (1.1) |
| Cancer | 2642 (1.4) | 2007 (2.8) |
| PPI therapy | 41 954 (23.0) | 28 709 (39.4) |
| Fatigue | 2701 (1.5) | 3074 (4.2) |
| No. of consultations in the year prior to inclusion | | |
| 0-1 | 114 529 (62.7) | 27 109 (37.2) |
| 2-5 | 33 802 (18.5) | 16 962 (23.3) |
| >5 | 34 203 (18.7) | 28 746 (39.5) |
| Received a ferritin test in the year before inclusion | 7630 (4.2) | 15 884 (21.8) |
| Characteristics of the first ferritin test during the study period | | |
| Was accompanied by a hemoglobin test | NA | 52 499 (72.1) |
| Was accompanied by a CRP test | NA | 36 136 (49.6) |
| Was accompanied by iron studies | | |
| None | NA | 65 781 (90.3) |
| Serum iron | NA | 6095 (8.4) |
| Transferrin | NA | 5820 (8.0) |
| Transferrin saturation | NA | 3856 (5.3) |
| Soluble transferrin receptor | NA | 506 (0.7) |
| Ferritin concentration, median (IQR), ng/mL | NA | 81 (40-155) |
Table 2 summarizes the incidences of iron deficiency diagnoses at the different ferritin cutoffs (eTable 3 in Supplement 1). At 15, 30, and 45 ng/mL, the total incidences of iron deficiency diagnoses were 10.9 (95% CI, 10.6-11.2), 29.9 (95% CI, 29.4-30.4), and 48.3 (95% CI, 47.7-48.9) cases per 1000 patient-years, respectively. At 15, 30, and 45 ng/mL, the incidences of nonanemic iron deficiency diagnoses were 4.1 (95% CI, 3.9-4.2), 14.6 (95% CI, 14.3-15.0), and 25.8 (95% CI, 25.3-26.2) cases per 1000 patient-years, respectively, and the incidences of anemic iron deficiency diagnoses were 3.5 (95% CI, 3.3-3.7), 6.0 (95% CI, 5.8-6.2), and 7.5 (95% CI, 7.3-7.7) cases per 1000 patient-years, respectively.
Table 2. Incidences of Iron Deficiency Diagnoses (Nonanemic, Anemic, and Total) at Different Serum Ferritin (Ferritin) Cutoffs
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| Diagnosis | Incidences in cases per 1000 patient-years (95% CI) |
:--- |
| 15 ng/mL | 30 ng/mL | 45 ng/mL |
| Any ferritin test |
| Nonanemic iron deficiency | 4.1 (3.9-4.2) | 14.6 (14.3-15.0) | 25.8 (25.3-26.2) |
| Anemic iron deficiency | 3.5 (3.3-3.7) | 6.0 (5.8-6.2) | 7.5 (7.3-7.7) |
| Total iron deficiency | 10.9 (10.6-11.2) | 29.9 (29.4-30.4) | 48.3 (47.7-48.9) |
| Only ferritin tests not accompanied by elevated CRP concentrations |
| Nonanemic iron deficiency | 4.2 (4.0-4.4) | 14.6 (14.3-15.0) | 25.6 (25.2-26.0) |
| Anemic iron deficiency | 3.1 (3.0-3.3) | 5.3 (5.1-5.5) | 6.6 (6.4-6.8) |
| Total iron deficiency | 9.8 (9.6-10.1) | 26.7 (26.2-27.1) | 43.0 (42.5-43.6) |
Among the 231 837 patients (90.8%) in the study cohort who did not receive ferritin testing in the year preceding inclusion, the incidence was 145.6 ferritin testing events per 1000 patient-years (95% CI, 144.4-146.8; median [IQR] follow-up time, 694 [294-956] days). Figure 2 shows the different associations of ferritin testing derived from the Cox proportional hazards regression model (eTables 4-5 in Supplement 1). Ferritin testing showed notable associations with fatigue (AHR, 2.03; 95% CI, 1.95-2.12), anemia (AHR, 1.75; 95% CI, 1.70-1.79), and iron therapy (AHR, 1.50; 95% CI, 1.46-1.54). Ferritin testing was more common in patients with CKD, IBD, PPI therapy, and higher primary care use but was less common with pregnancy. Female patients were more likely to receive ferritin testing than male patients of the same age in all age categories, including postmenopausal. The MHR among general practitioners was 1.76 (95% CI, 1.64-1.82).
Figure 2. Determinants of Ferritin Testing
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Results of mixed-effects Cox proportional hazards regression, expressed as adjusted hazard ratios (AHRs) for covariates and a median hazard ratio (MHR, derived from a corresponding null model without general practitioner [GP]–level covariates) to summarize the random-effects distribution. Associations with patient sex-age strata are summarized as follows: for male patients, as AHRs with respect to male patients younger than 25 years; and for female patients, as AHRs with respect to male patients of the respective same age category. A total of 255 119 were studied (to avoid artifacts due to high statistical imbalance, 232 patients were excluded due to unknown or other sex). A total of 56 927 ferritin testing events were studied. CHF indicates congestive heart failure, CKD, chronic kidney disease; IBD, inflammatory bowel disease; PPI proton pump inhibitor; SF, serum ferritin.
Table 3 summarizes the associations with hemoglobin and CRP tests accompanying ferritin testing (eTable 6 in Supplement 1). Both hemoglobin and CRP tests were positively associated with fatigue and were negatively associated with higher primary care use.
Table 3. Determinants of Hemoglobin and C-Reactive Protein Testing Accompanying Ferritin Testing a
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| Determinant | AOR (95% CI) |
:--- |
| Hemoglobin | C-reactive protein |
| Patient sex-age strata | | |
| Male, <25 y | 1.0 [Reference] | 1.0 [Reference] |
| Male, 25-34 y to male, <25 y | 1.20 (0.84-1.72) | 0.91 (0.75-1.09) |
| Male, 35-44 y to male, <25 y | 1.32 (0.93-1.89) | 0.83 (0.69-0.99) |
| Male, 45-54 y to male, <25 y | 1.11 (0.79-1.57) | 0.79 (0.66-0.94) |
| Male, 55-64 y to male, <25 y | 1.24 (0.88-1.73) | 0.77 (0.65-0.91) |
| Male, 65-74 y to male, <25 y | 1.34 (0.96-1.88) | 0.79 (0.67-0.93) |
| Male, ≥75 y to male, <25 y | 1.80 (1.27-2.53) | 0.85 (0.71-1.00) |
| Female to male, <25 y | 0.21 (0.04-1.02) | 0.33 (0.13-0.82) |
| Female to male, 25-34 y | 0.91 (0.70-1.18) | 1.06 (0.93-1.21) |
| Female to male, 35-44 y | 0.83 (0.64-1.06) | 1.12 (0.99-1.27) |
| Female to male, 45-54 y | 1.00 (0.79-1.26) | 1.12 (1.00-1.26) |
| Female to male, 55-64 y | 0.89 (0.71-1.11) | 1.05 (0.94-1.17) |
| Female to male, 65-74 y | 0.81 (0.65-1.01) | 1.06 (0.95-1.18) |
| Female to male, ≥75 y | 1.59 (1.15-2.20) | 0.90 (0.76-1.06) |
| Clinical factors | | |
| Anemia | 1.61 (1.32-1.96) | 0.95 (0.86-1.05) |
| Iron therapy | 0.72 (0.63-0.82) | 0.97 (0.91-1.03) |
| CKD | 1.05 (0.88-1.25) | 0.80 (0.74-0.87) |
| IBD | 1.61 (1.10-2.36) | 1.13 (0.95-1.35) |
| Rheumatic diseases | 0.79 (0.59-1.07) | 1.16 (0.99-1.35) |
| CHF | 0.83 (0.52-1.34) | 0.78 (0.63-0.97) |
| Pregnancy | 1.40 (0.92-2.13) | 0.91 (0.75-1.10) |
| Cancer | 0.94 (0.74-1.21) | 0.97 (0.86-1.09) |
| PPI therapy | 1.09 (0.99-1.20) | 0.99 (0.95-1.04) |
| Fatigue | 1.51 (1.22-1.86) | 1.15 (1.05-1.27) |
| Primary care use, No. of consultations in the past year | | |
| 0-1 | 1.0 [Reference] | 1.0 [Reference] |
| 2-5 | 0.83 (0.74-0.93) | 0.95 (0.90-1.01) |
| >5 | 0.73 (0.65-0.82) | 0.86 (0.82-0.91) |
| GP sex | | |
| Male | 1.0 [Reference] | 1.0 [Reference] |
| Female | 2.94 (0.93-9.25) | 1.58 (0.82-3.04) |
| GP age, y | | |
| <45 | 1.0 [Reference] | 1.0 [Reference] |
| 45-59 | 0.64 (0.21-2.02) | 0.59 (0.31-1.14) |
| ≥60 | 1.23 (0.26-5.80) | 0.81 (0.33-1.95) |
| GP workload, No. of consultations per working week | | |
| <100 | 1.0 [Reference] | 1.0 [Reference] |
| 100-199 | 1.53 (0.43-5.48) | 0.73 (0.35-1.51) |
| ≥200 | 2.23 (0.53-9.47) | 0.39 (0.17-0.90) |
| Practice location | | |
| Nonurban | 1.0 [Reference] | 1.0 [Reference] |
| Urban | 0.30 (0.11-0.87) | 0.51 (0.28-0.93) |
Discussion
In this study of more than 255 000 patients, we investigated the determinants and variation of ferritin testing and the incidence of iron deficiency diagnoses in Swiss primary care. We observed a substantial association of the choice of ferritin cutoff with the rates of iron deficiency diagnoses, especially nonanemic iron deficiency. In addition, we found a substantial degree of variation in ferritin testing along with interesting associations, such as higher testing rates in postmenopausal women compared with men of the same age and higher testing rates by female and younger general practitioners. We also found gaps in the quality of ferritin testing in terms of a large proportion ordered without accompanying hemoglobin and CRP measurements.
Our most compelling finding regarding the incidence of iron deficiency diagnoses was its strong dependence on the choice of the ferritin cutoff, especially for nonanemic iron deficiency. The use of ferritin cutoffs as treatment thresholds has been widely debated,35 and a recent Cochrane review found insufficient evidence to recommend any specific cutoff in a healthy, asymptomatic population.36 Even within Switzerland, there are conflicting recommendations. A guideline from a prominent Swiss primary care network emphasizes that iron replacement is not warranted at ferritin concentrations above 15 ng/mL,11 whereas a local expert panel has recommended diagnosing iron deficiency at ferritin concentrations below 30 ng/mL.23 Our results show that these controversies affect the management of a considerable number of patients.
The observed incidence of anemic iron deficiency diagnoses of approximately 13 cases per 1000 patient-years is comparable to the results of a similar multinational European study.13 We are not aware of other studies that have estimated the rates of iron deficiency diagnoses based on routine data in high-resource settings. However, the lack of concomitant hemoglobin and CRP measurements has important implications for the interpretation of such incidences. The proportion of ferritin tests without accompanying CRP or hemoglobin tests was surprisingly high because most guidelines explicitly recommend screening for systemic inflammation and anemia in the workup of iron deficiency.4 Hemoglobin and CRP testing was associated with fatigue and with fewer previous primary care visits, suggesting that they were more commonly used in patients in whom a symptom-led search for iron deficiency took place rather than in episodes of care involving routine screening.
Ferritin testing was requested in more than one-fourth of the patients followed up for the 3 years of the study period, which aligns with the findings of a previous Swiss study8 and is a rate only slightly higher than previously observed in Australia.37 Comparisons of ferritin testing rates with estimates available from other countries, such as the UK5 or Canada,7 are complicated by the heterogeneity in the reported measures of testing frequencies and of the populations considered. The paucity of comparable international data suggests a need for updated research.
Among the clinical factors considered, fatigue, anemia, and iron therapy showed the strongest associations with ferritin testing. Fatigue was the clinical factor most strongly associated with ferritin testing, but its prevalence of just more than 2% at inclusion was well below previous annual prevalence estimates of approximately 8% in primary care.38 This observation is consistent with previous conclusions that the documentation of general symptom-related ICPC-2 codes is likely underrepresented in the FIRE database compared with condition-related ICPC-2 codes.15 Nevertheless, our finding suggests that the clinical presentation of the patient plays an important role in the decision to test for ferritin.
Women were invariably more likely to receive ferritin testing than men in the same age group, regardless of anemia or prior iron therapy. Although this sex difference is consistent with a higher risk of iron deficiency due to menstrual blood loss during childbearing age, we have no immediate explanation for this finding in the postmenopausal age group. Symptoms of nonanemic iron deficiency may often have prompted ferritin testing. Many of these symptoms, especially fatigue, are nonspecific but are more prevalent among women as presenting concerns in primary care.2,39 Although we adjusted for fatigue, its low frequency of documentation may have resulted in residual confounding that partially explains the persistent association with female sex.
We also observed notable associations of ferritin testing with general practitioner characteristics, with female and younger general practitioners being more prone to ferritin testing. The sex difference can be interpreted in the context of previous findings that female general practitioners provide more preventive care than their male colleagues,40,41 and the age disparity may be related to the variation in information-seeking behavior of general practitioners of different ages.42 The general practitioner–level variation in ferritin testing expressed by the MHR was striking because it was comparable with several of the AHRs expressing associations with clinical factors. This finding is consistent with previously observed evidence of unwarranted variation in the use of ferritin testing in different primary care settings, which has been interpreted as indicating potential overuse.6,7,43
Most ferritin tests were not accompanied by requests for additional iron studies, suggesting awareness among general practitioners of the recommendation to use ferritin as a first-line test for iron deficiency. This result contrasts with findings from other countries, such as Australia37 and Spain,6 where other iron studies were overrequested compared with ferritin. On the other hand, we observed that serum iron was the most frequently requested additional iron study, contrary to the recommendation of local guidelines to avoid its use.11 This finding suggests a knowledge gap regarding the use of iron studies, in line with results from the international literature.44,45
In summary, our findings can be understood in the context of clinical uncertainty faced by general practitioners regarding the diagnosis of iron deficiency. This uncertainty may be partly due to the lack of consensus among different recommendations for iron deficiency screening and ferritin cutoffs, ultimately calling for more guidance on the management of iron deficiency in primary care.46
Limitations
This study has some limitations. The FIRE database does not allow access to presenting symptoms as documented in the clinical notes, which is an important limitation together with the lack of information on referral. In addition, we had limited access to information on gastrointestinal risk factors, such as celiac disease. Although information on the patients’ diet, which may have prompted screening for iron deficiency, was also unavailable, a previous study found no relevant differences in the prevalence of iron deficiency among Swiss people who were omnivores, vegetarians, or vegans.47
Conclusions
Our study demonstrates a substantial increase in the rate of iron deficiency diagnoses when ferritin cutoffs of 30 and 45 ng/mL, respectively, are chosen over 15 ng/mL. Our results provide an information base for health system–level evaluations of ferritin testing in primary care. In addition, they highlight the need for harmonization of guidelines for the diagnosis of iron deficiency in primary care.
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Article Information
Accepted for Publication: June 6, 2024.
Published: August 5, 2024. doi:10.1001/jamanetworkopen.2024.25692
Open Access: This is an open access article distributed under the terms of the CC-BY License. © 2024 Jäger L et al. JAMA Network Open.
Corresponding Author: Levy Jäger, MD, Institute of Primary Care, University Hospital Zurich, Pestalozzistrasse 24, CH-8091 Zurich, Switzerland (levy.jaeger@usz.ch).
Author Contributions: Dr Jäger had full access to all of the data in the study and takes responsibility for the integrity of the data and the accuracy of the data analysis.
Concept and design: Jäger, Rachamin, Rosemann, Markun.
Acquisition, analysis, or interpretation of data: Jäger, Rachamin, Senn, Burgstaller, Markun.
Drafting of the manuscript: Jäger, Markun.
Critical review of the manuscript for important intellectual content: All authors.
Statistical analysis: Jäger.
Administrative, technical, or material support: Senn, Burgstaller, Rosemann, Markun.
Supervision: Markun.
Conflict of Interest Disclosures: None reported.
Data Sharing Statement: See Supplement 2.
Additional Contributions: Fabio Valeri, MS, database manager of the Institute of Primary Care of the University of Zurich, curated the FIRE database, and the FIRE Study Group of general practitioners contributed data to the FIRE database. Mr Valeri was not specifically compensated for this work.
References
1.
GBD 2021 Anaemia Collaborators. Prevalence, years lived with disability, and trends in anaemia burden by severity and cause, 1990-2021: findings from the Global Burden of Disease Study 2021.Lancet Haematol. 2023;10(9):e713-e734. doi:10.1016/S2352-3026(23)00160-6PubMedGoogle ScholarCrossref
2.
Al-Naseem A , Sallam A , Choudhury S , Thachil J . Iron deficiency without anaemia: a diagnosis that matters.Clin Med (Lond). 2021;21(2):107-113. doi:10.7861/clinmed.2020-0582PubMedGoogle ScholarCrossref
3.
Beatrix J , Piales C , Berland P , Marchiset E , Gerbaud L , Ruivard M . Non-anemic iron deficiency: correlations between symptoms and iron status parameters.Eur J Clin Nutr. 2022;76(6):835-840. doi:10.1038/s41430-021-01047-5PubMedGoogle ScholarCrossref
4.
Peyrin-Biroulet L , Williet N , Cacoub P . Guidelines on the diagnosis and treatment of iron deficiency across indications: a systematic review.Am J Clin Nutr. 2015;102(6):1585-1594. doi:10.3945/ajcn.114.103366PubMedGoogle ScholarCrossref
5.
O’Sullivan JW , Stevens S , Hobbs FDR , et al. Temporal trends in use of tests in UK primary care, 2000-15: retrospective analysis of 250 million tests.BMJ. 2018;363:k4666. doi:10.1136/bmj.k4666PubMedGoogle ScholarCrossref
6.
Salinas M , López-Garrigós M , Flores E , Leiva-Salinas C ; Pilot Group of the Appropriate Utilization of Laboratory Tests (REDCONLAB) working group. Primary care requests for anaemia chemistry tests in Spain: potential iron, transferrin and folate over-requesting.J Clin Pathol. 2017;70(9):760-765. doi:10.1136/jclinpath-2016-204249PubMedGoogle ScholarCrossref
7.
Nguyen LT , Guo M , Hemmelgarn B , et al. Evaluating practice variance among family physicians to identify targets for laboratory utilization management.Clin Chim Acta. 2019;497:1-5. doi:10.1016/j.cca.2019.06.017PubMedGoogle ScholarCrossref
8.
Huber CA , Nagler M , Rosemann T , Blozik E , Näpflin M , Markun S . Trends in micronutrient laboratory testing in Switzerland: a 7-year retrospective analysis of healthcare claims data.Int J Gen Med. 2020;13:1341-1348. doi:10.2147/IJGM.S275406PubMedGoogle ScholarCrossref
9.
Schumacher LD , Jäger L , Meier R , et al. Trends and between-physician variation in laboratory testing: a retrospective longitudinal study in general practice.J Clin Med. 2020;9(6):1787. doi:10.3390/jcm9061787PubMedGoogle ScholarCrossref
10.
Kemper AR , Fan T , Grossman DC , Phipps MG . Gaps in evidence regarding iron deficiency anemia in pregnant women and young children: summary of US Preventive Services Task Force recommendations.Am J Clin Nutr. 2017;106(suppl 6):1555S-1558S. doi:10.3945/ajcn.117.155788PubMedGoogle ScholarCrossref
11.
Chmiel C , Beise U . mediX-Guidelines: Eisenmangel. Verein mediX. 2023. Accessed February 9, 2024.
12.
Tolkien Z , Stecher L , Mander AP , Pereira DI , Powell JJ . Ferrous sulfate supplementation causes significant gastrointestinal side-effects in adults: a systematic review and meta-analysis.PLoS One. 2015;10(2):e0117383. doi:10.1371/journal.pone.0117383PubMedGoogle ScholarCrossref
13.
Levi M , Rosselli M , Simonetti M , et al. Epidemiology of iron deficiency anaemia in four European countries: a population-based study in primary care.Eur J Haematol. 2016;97(6):583-593. doi:10.1111/ejh.12776PubMedGoogle ScholarCrossref
14.
Levi M , Simonetti M , Marconi E , et al. Gender differences in determinants of iron-deficiency anemia: a population-based study conducted in four European countries.Ann Hematol. 2019;98(7):1573-1582. doi:10.1007/s00277-019-03707-wPubMedGoogle ScholarCrossref
15.
Chmiel C , Bhend H , Senn O , Zoller M , Rosemann T ; FIRE study-group. The FIRE project: a milestone for research in primary care in Switzerland.Swiss Med Wkly. 2011;140:w13142. doi:10.4414/smw.2011.13142PubMedGoogle ScholarCrossref
16.
World Health Organization Collaborating Centre for Drug Statistics Methodology. ATC classification index with DDDs. 2023. Accessed January 4, 2024.
17.
World Organization of National Colleges, Academies and Academic Associations of General Practitioners/Family Physicians (WONCA). ICPC-2-R: International Classification of Primary Care. Oxford University Press; 2005. Accessed July 16, 2024.
18.
Federal Act on Research involving Human Beings 30 September 2011 (Status as of 01 January 2014). 2011. Updated September 30, 2011. Accessed February 9, 2024.
19.
Benchimol EI , Smeeth L , Guttmann A , et al; RECORD Working Committee. The Reporting of Studies Conducted Using Observational Routinely-Collected Health Data (RECORD) statement.PLoS Med. 2015;12(10):e1001885. doi:10.1371/journal.pmed.1001885PubMedGoogle ScholarCrossref
20.
World Health Organization. Haemoglobin concentrations for the diagnosis of anaemia and assessment of severity. World Health Organization. 2011. Accessed February 9, 2024.
21.
Gemeindetypologie und Stadt/Land-Typologie 2012. Swiss Federal Statistical Office. 2017. Accessed January 4, 2024.
22.
R Core Team. R: A Language and Environment for Statistical Computing. R Foundation for Statistical Computing, Vienna, Austria. 2023. Accessed July 16, 2024.
23.
Nowak A , Angelillo-Scherrer A , Betticher D , et al. Swiss Delphi study on iron deficiency.Swiss Med Wkly. 2019;149:w20097. doi:10.4414/smw.2019.20097PubMedGoogle ScholarCrossref
24.
Naveed K , Goldberg N , Shore E , et al. Defining ferritin clinical decision limits to improve diagnosis and treatment of iron deficiency: a modified Delphi study.Int J Lab Hematol. 2023;45(3):377-386. doi:10.1111/ijlh.14016PubMedGoogle ScholarCrossref
25.
Fletcher A , Forbes A , Svenson N , Wayne Thomas D ; A British Society for Haematology Good Practice Paper. Guideline for the laboratory diagnosis of iron deficiency in adults (excluding pregnancy) and children.Br J Haematol. 2022;196(3):523-529. doi:10.1111/bjh.17900PubMedGoogle ScholarCrossref
26.
Snook J , Bhala N , Beales ILP , et al. British Society of Gastroenterology guidelines for the management of iron deficiency anaemia in adults.Gut. 2021;70(11):2030-2051. doi:10.1136/gutjnl-2021-325210PubMedGoogle ScholarCrossref
27.
Short MW , Domagalski JE . Iron deficiency anemia: evaluation and management.Am Fam Physician. 2013;87(2):98-104.PubMedGoogle Scholar
28.
Ulm K . A simple method to calculate the confidence interval of a standardized mortality ratio (SMR).Am J Epidemiol. 1990;131(2):373-375. doi:10.1093/oxfordjournals.aje.a115507PubMedGoogle ScholarCrossref
29.
Morgen EK , Naugler C . Inappropriate repeats of six common tests in a Canadian city: a population cohort study within a laboratory informatics framework.Am J Clin Pathol. 2015;144(5):704-712. doi:10.1309/AJCPYXDAUS2F8XJYPubMedGoogle ScholarCrossref
30.
Fisher LD , Lin DY . Time-dependent covariates in the Cox proportional-hazards regression model.Annu Rev Public Health. 1999;20:145-157. doi:10.1146/annurev.publhealth.20.1.145PubMedGoogle ScholarCrossref
31.
coxme: Mixed Effects Cox Models (R package version 2.2-18.1). 2022. Accessed July 16, 2024.
32.
van der Leeden R, Meijer E, Busing FMTA. Resampling multilevel models. In: Leeuw Jd , Meijer E , eds. Handbook of Multilevel Analysis. Springer; 2008:401-433. doi:10.1007/978-0-387-73186-5
33.
Austin PC , Wagner P , Merlo J . The median hazard ratio: a useful measure of variance and general contextual effects in multilevel survival analysis.Stat Med. 2017;36(6):928-938. doi:10.1002/sim.7188PubMedGoogle ScholarCrossref
34.
Bates D , Mächler M , Bolker B , Walker S . Fitting linear mixed-effects models using lme4.J Stat Softw. 2015;67:1-48. doi:10.18637/jss.v067.i01Google ScholarCrossref
35.
Daru J , Colman K , Stanworth SJ , De La Salle B , Wood EM , Pasricha SR . Serum ferritin as an indicator of iron status: what do we need to know?Am J Clin Nutr. 2017;106(suppl 6):1634S-1639S. doi:10.3945/ajcn.117.155960PubMedGoogle ScholarCrossref
36.
Garcia-Casal MN , Pasricha SR , Martinez RX , Lopez-Perez L , Peña-Rosas JP . Serum or plasma ferritin concentration as an index of iron deficiency and overload.Cochrane Database Syst Rev. 2021;5(5):CD011817.PubMedGoogle Scholar
37.
NPS MedicineWise. General practice insights report July 2016-June 2017: a working paper. 2018. Accessed May 24, 2024.
38.
Nicholson K , Stewart M , Thind A . Examining the symptom of fatigue in primary care: a comparative study using electronic medical records.J Innov Health Inform. 2015;22(1):235-243. doi:10.14236/jhi.v22i1.91PubMedGoogle ScholarCrossref
39.
Morgan S , Henderson KM , Tapley A , et al. Investigation of fatigue by Australian general practice registrars: a cross-sectional study.J Prim Health Care. 2015;7(2):109-116. doi:10.1071/HC15109PubMedGoogle ScholarCrossref
40.
Delpech R , Bloy G , Panjo H , Falcoff H , Ringa V , Rigal L . Physicians’ preventive practices: more frequently performed for male patients and by female physicians.BMC Health Serv Res. 2020;20(1):331. doi:10.1186/s12913-020-05136-2PubMedGoogle ScholarCrossref
41.
Krähenmann-Müller S , Virgini VS , Blum MR , et al. Patient and physician gender concordance in preventive care in university primary care settings.Prev Med. 2014;67:242-247. doi:10.1016/j.ypmed.2014.08.004PubMedGoogle ScholarCrossref
42.
Le JV , Pedersen LB , Riisgaard H , et al. Variation in general practitioners’ information-seeking behaviour—a cross-sectional study on the influence of gender, age and practice form.Scand J Prim Health Care. 2016;34(4):327-335. doi:10.1080/02813432.2016.1249057PubMedGoogle ScholarCrossref
43.
Salinas M , López-Garrigós M , Díaz J , et al. Regional variations in test requiring patterns of general practitioners in Spain.Ups J Med Sci. 2011;116(4):247-251. doi:10.3109/03009734.2011.606927PubMedGoogle ScholarCrossref
44.
Read AJ , Waljee AK , Sussman JB , et al. Testing practices, interpretation, and diagnostic evaluation of iron deficiency anemia by US primary care physicians.JAMA Netw Open. 2021;4(10):e2127827. doi:10.1001/jamanetworkopen.2021.27827
ArticlePubMedGoogle ScholarCrossref
45.
Bayen S , Le Grand C , Bayen M , Richard F , Messaadi N . Anemia management in non-menopausal women in a primary care setting: a prospective evaluation of clinical practice.BMC Fam Pract. 2020;21(1):13. doi:10.1186/s12875-020-1086-5PubMedGoogle ScholarCrossref
46.
Jefferds MED , Mei Z , Addo Y , et al. Iron deficiency in the United States: limitations in guidelines, data, and monitoring of disparities.Am J Public Health. 2022;112(S8):S826-S835. doi:10.2105/AJPH.2022.306998PubMedGoogle ScholarCrossref
47.
Schüpbach R , Wegmüller R , Berguerand C , Bui M , Herter-Aeberli I . Micronutrient status and intake in omnivores, vegetarians and vegans in Switzerland.Eur J Nutr. 2017;56(1):283-293. doi:10.1007/s00394-015-1079-7PubMedGoogle ScholarCrossref
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15830 | https://pmc.ncbi.nlm.nih.gov/articles/PMC3737390/ | Pancreatic lipase-related protein 2 digests fats in human milk and formula in concert with gastric lipase and carboxyl ester lipase - PMC
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Pediatr Res
. Author manuscript; available in PMC: 2014 Feb 1.
Published in final edited form as: Pediatr Res. 2013 Jun 3;74(2):127–132. doi: 10.1038/pr.2013.90
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Pancreatic lipase-related protein 2 digests fats in human milk and formula in concert with gastric lipase and carboxyl ester lipase
Karin Johnson
Karin Johnson
1 Department of Pediatrics, Children’s Hospital of Pittsburgh at University of Pittsburgh Medical Center, One Children’s Hospital Drive, 4401 Penn Avenue, Pittsburgh, PA 15224
Find articles by Karin Johnson
1, Leah Ross
Leah Ross
1 Department of Pediatrics, Children’s Hospital of Pittsburgh at University of Pittsburgh Medical Center, One Children’s Hospital Drive, 4401 Penn Avenue, Pittsburgh, PA 15224
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1, Rita Miller
Rita Miller
1 Department of Pediatrics, Children’s Hospital of Pittsburgh at University of Pittsburgh Medical Center, One Children’s Hospital Drive, 4401 Penn Avenue, Pittsburgh, PA 15224
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1, Xunjun Xiao
Xunjun Xiao
1 Department of Pediatrics, Children’s Hospital of Pittsburgh at University of Pittsburgh Medical Center, One Children’s Hospital Drive, 4401 Penn Avenue, Pittsburgh, PA 15224
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1, Mark E Lowe
Mark E Lowe
1 Department of Pediatrics, Children’s Hospital of Pittsburgh at University of Pittsburgh Medical Center, One Children’s Hospital Drive, 4401 Penn Avenue, Pittsburgh, PA 15224
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1,
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1 Department of Pediatrics, Children’s Hospital of Pittsburgh at University of Pittsburgh Medical Center, One Children’s Hospital Drive, 4401 Penn Avenue, Pittsburgh, PA 15224
Corresponding Author: Mark E. Lowe, One Children’s Hospital Drive, 4401 Penn Avenue, Pittsburgh, PA 15224, Tel: 412-692-5412, Fax: 412-692-8906, mark.lowe@chp.edu
Issue date 2013 Aug.
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PMCID: PMC3737390 NIHMSID: NIHMS447524 PMID: 23732775
The publisher's version of this article is available at Pediatr Res
Abstract
INTRODUCTION
Dietary fats must be digested into fatty acids and monoacylglycerols prior to absorption. In adults, colipase-dependent pancreatic triglyceride lipase (PTL) contributes significantly to fat digestion. In newborn rodents and humans, the pancreas expresses low levels of PTL. In rodents, a homologue of PTL, pancreatic lipase related protein 2 (PLRP2) and carboxyl ester lipase (CEL) compensate for the lack of PTL. In human newborns, the role for PLRP2 in dietary fat digestion is unclear. To clarify the potential of human PLRP2 to influence dietary fat digestion in newborns, we determined PLRP2 activity against human milk and infant formula.
METHODS
The activity of purified recombinant PLRP2, gastric lipase and CEL against fats in human milk and formula was measured with each lipase alone and in combination with a standard pH-stat assay.
RESULTS
Colipase added to human milk stimulated fat digestion. PLRP2 and CEL had activity against human milk and formula. Pre-digestion with gastric lipase increased PLRP2 activity against both substrates. Together, CEL and PLRP2 activity was additive with formula and synergistic with human milk.
CONCLUSIONS
PLRP2 can digest fats in human milk and formula. PLRP2 acts in concert with CEL and gastric lipase to digest fats in human milk in vitro.
Background
The digestion of dietary fats into fatty acids and monoacylglycerols is essential for efficient absorption of dietary fat by enterocytes. Digestion of dietary fats, mainly long-chain triglycerides, is accomplished by a variety of lipases (1). In adults, digestion begins in the stomach where gastric lipase releases 15–20% of fatty acids from dietary fat (2). The partially digested fats are released into the duodenum and mixed with bile salts and pancreatic lipases. Colipase-dependent pancreatic triglyceride lipase (PTL) and other pancreatic lipases complete digestion in the duodenum. Of these, PTL is the predominant.
In contrast to its pivotal role in adults, PTL does not contribute significantly to dietary fat digestion in newborns (3). In rodents, mRNA encoding PTL is undetectable at birth (4). Another pancreatic lipase, pancreatic lipase related protein 2 (PLRP2) compensates for the physiological deficiency of PTL in rodent neonates. mRNA encoding PLRP2 is expressed in the pancreas of rodent newborns (4). Importantly, suckling PLRP2-deficient mice have steatorrhea and poor weight gain (5). These studies provide compelling evidence for the essential role of PLRP2 in fat digestion in rodent neonates.
The evidence in humans is not as complete. Like rodents, mRNA encoding PLRP2 is present at adult levels in the pancreas of human newborns whereas the expression of mRNA encoding PTL is low or undetectable (6). Duodenal levels of lipase are lower in premature and term infants than in adults suggesting PTL secretion is diminished or absent (7–8). Taken together these observations suggest that PLRP2 may have an important role in dietary fat digestion in human newborns.
In this study, we provide additional support for PLRP2 mediating dietary fat digestion in human newborns by demonstrating activity of PLRP2 against physiological substrates present in the newborns’ diet. Previous kinetic studies of human PLRP2 have utilized prepared emulsions of a defined triglyceride and bile salt, a much less complex substrate than naturally occurring substrates ingested by newborns (9–10). A single study demonstrated the ability of PLRP2 to hydrolyze the fat globules from bovine milk, a more complex substrate but one that is not generally present in the newborn diet (11). Herein, we studied the ability of human PLRP2 to hydrolyze dietary fats contained in human breast milk fat globules and in infant formula emulsion particles. Because PLRP2 likely acts in concert with gastric lipase and carboxyl ester lipase (CEL), the lipase found in mother’s milk, we measured activity of PLRP2 alone and in combination with gastric lipase and CEL.
Results
Purification of human colipase, PLRP2 and CEL
Recombinant human proteins for these studies were expressed in Pichia pastoris and purified as described in Methods (10, 12–14). The purity of the proteins was assessed by SDS-PAGE. Each protein migrated as a single, predominant band (Figure 1). Several minor, faster-migrating bands are visible in the PLRP2 preparation. These bands are likely degradation products of PLRP2 since they cross-react with PLRP2 antisera (Data not shown).
Figure 1.
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SDS-polyacrylamide gel electrophoresis of the recombinant proteins. The GelCode Blue stained gels are shown. (a) Human colipase on an 18% gel; (b) Human PLRP2 on a 10% gel; (c) Human CEL on a 7.5% gel. Molecular weight markers are included in each panel and the size in kDal is given on the right edge of each panel.
Effect of bile salt activation and freeze-thaw on lipase activity in human milk
To confirm that the mixture of bile salts activated CEL in human milk, we examined the effect of the bile salt mixture on the release of fatty acids using frozen human milk from several different donors. A 4 mM concentration of physiologic bile salts stimulated lipolysis at least 10-fold (P<0.001) (Figure 2a). The level of bile-salt-stimulated activity varied in human milk from individual donors and even with a single donor (Samples 3, 4 and 5). Next, we determined if freeze-thaw affected the lipase activity of human milk. Milk subjected to a single freeze-thaw cycle was compared with fresh milk from the same donor. Fresh human milk had significantly higher activity than did thawed frozen milk from the same donor presumably because freeze-thaw inactivated endogenous milk lipases (Figure 2b). For the rest of our studies we utilized either frozen or fresh human milk from a single donor.
Figure 2.
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Activity of human milk lipases in the presence of bile salts and after freeze-thaw. (a) The release of fatty acids from human milk was measured in the pH-stat with and without a mixture of physiological bile salts as described in Methods. Frozen human milk was obtained from three separate donors. Sample 1 = donor 1; Sample 2 = donor 2; Samples 3, 4 and 5 = donor 3. Black bars, no bile salts; Grey bars, plus bile salts. The difference of the means of no bile salts and plus bile salts for all samples was significant, P<0.001. (b) The release of fatty acids from frozen milk stored less than one week and from fresh human milk. The milk was from donor 3. The assay was done with a mixture of physiological bile salts (4 mM) in the pH-stat as described in Methods. The difference between the means was significant, P = 0.001.
Effect of colipase and PLRP2 on the release of fatty acids from fresh human milk
We next determined if colipase and PLRP2 alone or in combination had any effect on the lipase activity in frozen human milk. When colipase was added to the incubation there was a 22% increase over no colipase with 5 μg of colipase (P = 0.05) and a 48% increase with 25 μg colipase (P = 0.002) (Figure 3a). The 20% difference between 5 and 25 μg was also significant (P = 0.039) indicating a concentration dependence of the colipase effect. When we added 25 μg of PLRP2 without colipase to frozen human milk, the activity (2.4 ± 0.2 μmole fatty acid/min) was not a significant difference in fatty acids released compared to milk alone (2.4 ± 0.2 μmole fatty acid/min). When both colipase and PLRP2 were added in a 5 to 1 molar ratio of colipase to PLRP2, the activity was higher than the activity of human milk alone (P = 0.006 for 15 μg sample versus milk alone) (Figure 3b). The difference between the assays with the combination and the assays with colipase alone was significant for the 10 μg and 15 μg samples.
Figure 3.
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The influence of colipase and PLRP2 on lipolysis of human milk fats in the presence of bile salts. In each experiment activity was measured in the pH-stat in the presence of 4 mM bile salts as described in Methods. The P-value is given for the pair-wise comparisons between the different groups. (a) The release of fatty acids from frozen human milk after addition of various amounts of human colipase. P≤0.05, P≤0.01 (Student’s t-test). (b) Release of fatty acids from frozen human milk by the combination of PLRP2 and colipase at various concentrations. The amount of colipase or PLRP2 or both that was added to the incubation is indicated by the x-axis. White bar, human milk alone; Black bar, colipase alone; Gray bar, colipase and PLRP2; the same mass (μg) of colipase and PLRP2 were added as indicated. This resulted in a 5-fold molar ratio of colipase to PLRP2. P≤0.05 (Student’s t-test). (c) Release of fatty acids from heat-treated human milk by 25 μg of human PLRP2 in the presence of a physiological mixture of bile salts (4 mM) and various concentrations of colipase. P≤0.01; P≤0.001 (Student’s t-test).
To more clearly determine if PLRP2 hydrolyzes fats in human milk, we heat-treated fresh human milk to inactivate endogenous CEL. Heat treatment eliminated the activity from endogenous CEL even in the presence of colipase (Figure 3c). The release of fatty acids from heat-treated human milk was significantly increased by incubation with PLRP2 (P <0.001). The addition of colipase to the incubation increased the activity at both a 1:1 and 5:1 molar ratio to PLRP2 (P < 0.001 for each). The specific activities (U/mg) for 0, 1 and 5-fold molar ratio of colipase to PLRP2 were 8, 12, and 56, respectively.
Release of fatty acids from human milk by various lipases alone and in combination
To determine if there was a synergistic effect of the individual lipases potentially involved in dietary fat digestion, we performed assays on heat-treated human milk with gastric lipase, CEL and PLRP2 alone or in combination in the presence of colipase and bile salts. We first tested the activity of CEL and PLRP2 alone and in combination (Figure 4). In each case, the addition of either lipase increased lipolysis compared to human milk alone. There was no difference in activity between CEL and PLRP2. The combination of CEL and PLRP2 did not have a significant additive or synergistic effect on the release of fatty acids.
Figure 4.
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The dependence of human milk fat digestion on the combined action of gastric lipase, CEL and PLRP2. In each case, 25 μg of each lipase and 5 μg of colipase were added to the incubation except that no colipase was added during the pre-incubation with gastric lipase. The assays were done on heat-treated human milk with a physiological mixture of bile salts (4 mM). The values are the mean ± SD (n = 3). (No gastric lipase) White bar, human milk; Gray bar, CEL; Coarse hatched bar, PLRP2; Fine hatched bar CEL & PLRP2. There is a statistically significant difference among the mean values of the different conditions as determined by one-way ANOVA (F(3,8)= 12.63, P = 0.002). P < 0.05 PLRP2 (0.36 ± 0.2) CEL (0.39 ± 0.03) and CEL & PLRP2 (0.65 ± 0.1) compared to mother’s milk by the Holms-Sidak post-hoc pairwise multiple comparisons method. (Gastric lipase) White bar, human milk; Black bar, gastric lipase; Gray bar, CEL; Coarse hatched bar, PLRP2; Fine hatched bar, CEL & PLRP2. There is a statistically significant difference among the mean values of the different conditions as determined by one-way ANOVA (F(4,10)= 270.15, P = <0.001). P < 0.05 CEL (0.7 ± 0.1) compared to mother’s milk (0.03 ± 0.03), P < 0.05 PLRP2 (2.0 ± 0.2) compared with CEL, gastric lipase and mother’s milk, †P < 0.05 CEL & PLRP2 (4.5 ± 0.3) compared with PLRP2, CEL, gastric lipase and mother’s milk by the Holms-Sidak post-hoc pairwise multiple comparisons method.
To determine if pretreatment with gastric lipase increased the activity of the pancreatic lipases, we repeated the assays after a 20 min pre-incubation with gastric lipase at pH 6.0 as described in the Methods (Figure 4). After the incubation with gastric lipase the pH was raised to 8.0 with NaOH. The mean activity in the sample containing only gastric lipase was not statistically different from the activity in heat-treated human milk alone. Adding CEL to the incubation increased the mean activity slightly. In contrast, adding PLRP2 to the incubation increased the average activity about 6-fold over gastric lipase alone. The average activity of the combination of CEL and PLRP2 was about 13-fold higher than the average activity for gastric lipase alone. The increase was greater than the combined activity of CEL and PLRP2 alone suggesting there is synergy in the activity of these two lipases after gastric lipase has acted on the substrate.
Release of fatty acids from formula by various lipases alone and in combination
As formula is commonly used to substitute mother’s breast milk during infancy, we next investigated whether CEL and PLRP2 had activity against the lipids in an infant formula utilizing the same experimental design described above (Figure 5). Although the formula was brought to pH 8.0 prior to the start of monitoring in the pH-stat, the background was higher than with human milk. The background activity was not eliminated by heat treatment of the formula (Data not shown). Without prior digestion by gastric lipase, the activity of CEL was not statistically different from formula alone. In contrast, PLRP2 had low but detectable activity. The combination of CEL and PLRP2 had the same activity as PLRP2 alone. Preincubation with gastric lipase had a significant effect on the ability of both CEL and PLRP2 to digest the lipids in formula. CEL and PLRP2 had comparable activity when added alone and there was an additive and not a synergistic effect of incubating with both lipases.
Figure 5.
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The dependence of formula fat digestion on the combined action of gastric lipase, CEL and PLRP2. In each case, 25 μg of each lipase and 5 μg of colipase were added to the incubation except that no colipase was added during the pre-incubation with gastric lipase. The assays were done on a human formula, Similac Advance 20, with a physiological mixture of bile salts (4 mM). The values are the mean ± SD (n = 3). (No gastric lipase) White bar, formula; Gray bar, CEL; Coarse hatched bar, PLRP2; Fine hatched bar CEL & PLRP2. There is a statistically significant difference among the mean values of the different conditions as determined by one-way ANOVA (F(3,8)= 32.05, P <0.001). P < 0.05 PLRP2 (0.38 ± 0.08) compared with formula (0.08 ± 0.02) and CEL (0.1 ± 0.04), P < 0.05 CEL & PLRP2 (0.36 ± 0.04) compared with formula and CEL by the Holms-Sidak post-hoc pairwise multiple comparisons method. (Gastric lipase) White bar, formula; Black bar, gastric lipase; Gray bar, CEL; Coarse hatched bar, PLRP2; Fine hatched bar, CEL & PLRP2. There is a statistically significant difference among the mean values of the different conditions as determined by one-way ANOVA (F(4,10)= 220..03, P <0.001). P < 0.05 gastric lipase (0.4 ± 0.2) compared to formula (0.06 ± 0.01), P < 0.05 CEL (1.1 ± 0.1) compared with gastric lipase and formula, †P < 0.05 PLRP2 (1.4 ± 0.1) compared with CEL, gastric lipase and formula, ‡P < 0.05 CEL & PLRP2 (2.3 ± 0.2) compared with PLRP2, CEL, gastric lipase and formula by the Holms-Sidak post-hoc pairwise multiple comparisons method.
Discussion
Our results show that human PLRP2 is able to hydrolyze the fats in human milk and in formula. Furthermore, the activity was increased by colipase. These results agree with the previously reported finding that PLRP2 released fatty acids from heat-treated or homogenized bovine milk and that colipase stimulated the activity (11). Although not surprising, the similar activity against human and bovine milk could not have been predicted with certainty. The composition of milk fat globules differs in protein and fat content. Even though the proteins in the milk of both species are highly conserved there are qualitative and quantitative differences in protein content (15). Importantly, there is variation in milk fat content. Compared to bovine milk, human milk has more 18:1 and 18:2 fatty acids, lower saturated fatty acids, much lower amounts of 4:0 to 8:0 fatty acids and more long-chain polyunsaturated fatty acids (16). The differences in lipid and protein content could influence substrate availability or the physical properties of the fat globule, either of which could affect PLRP2 activity.
The activity of PLRP2 is further enhanced by predigestion of both human milk and infant formula lipids by gastric lipase. In the absence of gastric lipase predigestion, the activity of both CEL and PLRP2 alone or in combination was much lower than after predigestion with gastric lipase. Interestingly, the activity of CEL against infant formula lipids was stimulated 11-fold by gastric lipase predigestion and only 5-fold with human milk fats as the substrate. The activity of PLRP2 was stimulated to the same extent by predigestion with gastric lipase regardless of the substrate. The stimulation of CEL activity by limited digestion with gastric lipase confirms early findings (17). This is the initial report to show that predigestion with gastric lipase increases activity of PLRP2. Presumably, the effect of gastric lipase is to produce fatty acids. Multiple studies have demonstrated that addition of fatty acids to the substrate initiates lipolysis by PTL (17–20). Similarly, adding fatty acids to bovine fat globules or to emulsions of a single triglyceride with bile salts activate the activity of human PLRP2 (10–11). These observations support the hypothesis that gastric lipase initiates PLRP2 activity by generating fatty acids.
When CEL and PLRP2 were both added to human milk after predigestion with gastric lipase, the activity was greater than the sum of the individual activities, suggesting that the two enzymes work synergistically when digesting human milk. A similar result was reported for the digestion of an artificial substrate, triolein, by the combination of CEL and PLRP2 using cellular uptake by Caco-2 cells as a read out (21). A slightly different effect was found with formula as the substrate. The combination of CEL and PLRP2 was not synergistic. Differences in the physicochemical properties of lipid globules in human milk and formula may explain this observation (22–24). Whether the differences in lipolysis of human milk and formula have a significant effect on fat digestion in vivo cannot be addressed by in vitro assays. Premature and Term infants fed formula have slightly lower coefficients of fat absorption than infants fed human milk, but the reason for the difference is not clear (3).
The observation that colipase stimulated lipase activity of human milk above the levels achieved by bile salts alone was unexpected. That colipase did not stimulate fatty acid release in heat-treated milk suggests that it affects the activity of CEL or lipoprotein lipase. Colipase is not predicted to interact with either lipase in the way that it interacts with PTL. CEL does not contain the β-sandwich sheet carboxyl terminal domain that is critical for the interaction of colipase with pancreatic lipases. The observation raises the possibility that colipase has another role in the digestion of dietary fats. In monolayer systems, colipase binds to the surface and alters the lateral distribution of phospholipids and diacylglycerols in the monolayer (25). Perhaps, colipase has surface-active properties that alter the physical aspects of the milk fat globules in a way that improves the activity of lipases. Intriguingly, this observation raises the possibility that colipase stimulates PLRP2 activity through effects on the substrate rather than by anchoring PLRP2 to the substrate interface as proposed for the colipase-PTL complex. This could explain the findings that PLRP2 lipases have activity in the absence of colipase and colipase stimulates that activity to varying degrees depending on the substrate and PLRP2 species (26).
Although the results demonstrate that PLRP2 can hydrolyze fats in human milk, extrapolation to in vivo hydrolysis of human milk is not directly possible from this study. It is difficult to recreate in vivo conditions with an in vitro system and these experiments were not intended to do so. Whereas the pH of 6.0 chosen for gastric lipase incubations is reasonably close to the gastric pH after ingestion of a meal, the pH of 8.0 for the CEL and PLRP2 assays is higher than the pH in the duodenum after a meal which ranges between 6.0 and 6.5 in adults and may be lower in preterm and term infants (2, 27–28). The higher pH was used to increase the sensitivity of the assay with long-chain triglycerides (29). The concentration of bile salts used in the assay is in the range reported for premature and term infants but some infants have concentrations lower than 4 mM (30). Importantly, the amount of each lipase in the gastric or duodenal contents in human newborns is not known with certainty. One study measured a range of 5μg/ml to 80 of gastric lipase in the stomach of premature infants (31). A second study reported of 2 to 4 μg/ml (32). The amount of gastric lipase we added to the assays (1.6 μg/ml) was slightly lower than the reported range. There are no reports of PLRP2 levels in human newborns. A recent study estimated the content of PLRP2 in pancreatic secretions of human adults (around 200 ng/ml) was lower than the concentration of PLRP2 included in our assays (1.6 μg/ml) (33). The concentration we utilized is identical to the concentration of PLRP2 used in the study on bovine milk (11). There are also no reliable estimates of CEL concentrations in the duodenal contents of human newborns. Similarly, the concentration of colipase in the pancreatic secretions of human infants is not known; it is likely that colipase is secreted in excess of PLRP2. In human adults, colipase and PTL are present in a 1:1 molar ratio and the concentration of PLRP2 is about 4-fold lower (27). Thus, the 5-fold molar ratio of colipase to PLRP2 in this study seems reasonable. Lastly, proteases are present in the stomach and duodenum during digestion and proteolysis may influence lipolysis. Even with these caveats, our results are consistent with the hypothesis that the concerted action of gastric lipase, CEL and PLRP2 is required for the efficient digestion of human milk and formula fats.
Heat treatment and freezing human milk may alter the physicochemical properties of milk fat globules by altering the size of the particles and perhaps the organization of the lipids in the droplet (34). The fat content and composition is not significantly affected by heating (34–35). In this study, we used frozen and heat-treated human milk predominantly. Because the physicochemical properties of the substrate influence lipase activity, these treatments may alter the activity of PLRP2 and the other lipases. Even so, heat-treated and frozen human milk is fed to newborns and our observation that PLRP2 can hydrolyze fats from heat-treated and frozen human milk pertains to what happens in practice. We also showed that PLRP2 can release fatty acids from fresh human milk.
Our conclusion that PLRP2 digests dietary fats in human milk and formula raises an interesting speculation about the effect of a nonsense polymorphism in the gene encoding PLRP2. Several studies from different investigators report a high allele frequency (0.33 to 0.50) of this polymorphism (36–37). In an earlier study, we showed that the polymorphism results in a truncated form of PLRP2 that has low activity and is poorly secreted (10). The combination of the results from this study and the earlier study suggests that neonates who are homozygous for the nonsense polymorphism will be PLRP2 deficient and may have decreased dietary fat absorption. Homozygosity for the PLRP2 polymorphism may account for the broad range of fecal fat concentrations found in newborns since the lack of PLRP2 would likely decrease the efficiency of dietary fat digestion in humans as it does in mice (38). If true, fat absorption and nutrition in these newborns may be improved by pancreatic enzyme replacement therapy.
Methods
Materials
Fresh or frozen human breast milk was donated by women with infants in the Neonatal Intensive Care Unit at Magee Women’s Hospital of University of Pittsburgh Medical Center. Mothers on medications were excluded from donation. Fresh milk samples were used within 24 hours of collection. Fresh milk samples from a single donor were used for most experiments. Exceptions are noted in the figure legend. The protocol was approved by the Institutional Review Board of the University of Pittsburgh. Similac Advance 20 calorie infant formula (fat (long-chain safflower, soy and corn oil) 1.11 g, carbohydrate (lactose and galactooligosaccharides) 2.17 g, protein (non-hydrolyzed nonfat milk and whey) 0.42 g in 29.6 ml) was used in some assays. Bile salts were obtained from Sigma (St. Louis, MO). Other chemicals were obtained from standard sources.
Protein methods
The recombinant protein PLRP2, CEL, and colipase (Col) were expressed in Pichia pastoris and purified as previously described (10, 12, 14). The recombinant CEL contained 16 tandem repeats in the hypervariable region. Gastric lipase (GL) was generously provided by Dr. Frederic Carriere at the Laboratoire d’Enzymologie Interfaciale et Physiologie de la Lipolyse in Marseille, France (39). To determine the purity and integrity of the recombinant proteins, 6 μg of each protein was resolved by 7.5, 10 and 18% SDS-PAGE for CEL, PLRP2 and colipase, respectively, and followed by staining with GelCode Blue Stain Reagent (Pierce, Rockford, IL).
Lipase assay method
For some experiments, the endogenous CEL in human milk was inactivated by heat treatment at 73 °C for 20 minutes. Where indicated, a 200 mM mixture containing 84 mM of taurocholate, 44 mM of glycocholate, 52 mM of taurochenodeoxycholate and 20 mM of glycochenodeoxycholate was added to a final concentration of 4 mM (17). The release of fatty acids by the various lipases was determined by continuous titration to pH 8.0 with 50 mM NaOH in a Titra Lab TIM 854 pH-stat at room temperature. The assay contained 3 mL of human milk or formula and 12 mL reaction buffer (1 mM Tris-Cl, pH 8.0, 2.0 mM CaCl2, and 0.15 M NaCl). The amount of bile salts and various lipases in the assay are given in the legends. Assays with single lipases were performed over 15 to 30 minutes. Preincubation with gastric lipase was performed at room temperature for 20 minutes using the pH-stat to maintain the pH at 6.0. The release of fatty acids was measured by back-titrating to pH 8.0 (40). Subsequently, the appropriate amounts of bile salt, lipase and colipase were added and the reaction monitored continuously at pH 8.0 for 15 to 30 minutes. The reactions were linear over the assay period. The results are expressed as μmoles fatty acid released/min. Values are presented as means ± SD of at least 3 determinations.
Statistical methods
Statistical analysis was performed with SigmaStat software (version 3.5, Systat Software, Point Richmond, CA). Pairwise comparisons were analyzed by Student’s t test. Significance was considered P < 0.05. Multiple pairwise comparisons were done by one-way analysis of variance (ANOVA) using the Holm-Sidak method for the pairwise multiple comparisons with alpha = 0.05 and an overall significance level of 0.05.
Acknowledgments
We thank Dr. Ada Youk for advice on statistical methods.
Statement of Financial Support
The studies were supported by National Institutes of Health Grant DK-080820 (to M.E.L.)
References
1.Whitcomb DC, Lowe ME. Human pancreatic digestive enzymes. Dig Dis Sci. 2007;52:1–17. doi: 10.1007/s10620-006-9589-z. [DOI] [PubMed] [Google Scholar]
2.Carriere F, Barrowman JA, Verger R, Laugier R. Secretion and contribution to lipolysis of gastric and pancreatic lipases during a test meal in humans. Gastroenterology. 1993;105:876–88. doi: 10.1016/0016-5085(93)90908-u. [DOI] [PubMed] [Google Scholar]
3.Lindquist S, Hernell O. Lipid digestion and absorption in early life: an update. Curr Opin Clin Nutr Metab Care. 2010;13:314–20. doi: 10.1097/MCO.0b013e328337bbf0. [DOI] [PubMed] [Google Scholar]
4.Li X, Lindquist S, Lowe M, Noppa L, Hernell O. Bile Salt-Stimulated Lipase and Pancreatic Lipase-Related Protein 2 Are the Dominating Lipases in Neonatal Fat Digestion in Mice and Rats. Pediatr Res. 2007;62:537–41. doi: 10.1203/PDR.0b013e3181559e75. [DOI] [PMC free article] [PubMed] [Google Scholar]
5.Lowe ME, Kaplan MH, Jackson-Grusby L, D’Agostino D, Grusby MJ. Decreased neonatal dietary fat absorption and T cell cytotoxicity in pancreatic lipase-related protein 2-deficient mice. J Biol Chem. 1998;273:31215–21. doi: 10.1074/jbc.273.47.31215. [DOI] [PMC free article] [PubMed] [Google Scholar]
6.Yang Y, Sanchez D, Figarella C, Lowe ME. Discoordinate expression of pancreatic lipase and two related proteins in the human fetal pancreas. Pediatr Res. 2000;47:184–8. doi: 10.1203/00006450-200002000-00006. [DOI] [PubMed] [Google Scholar]
7.Boehm G, Bierbach U, Senger H, et al. Activities of lipase and trypsin in duodenal juice of infants small for gestational age. J Ped Gastroenterol Nutri. 1991;12:324–7. doi: 10.1097/00005176-199104000-00006. [DOI] [PubMed] [Google Scholar]
8.Boehm G, Borte M, Muller H, Moro G, Minoli I. Activities of trypsin and lipase in duodenal aspirates of preterm infants: influence of dietary protein and fat composition. Am J Clin Nutr. 1995;61:524–7. doi: 10.1093/ajcn/61.3.524. [DOI] [PubMed] [Google Scholar]
9.Eydoux C, De Caro J, Ferrato F, et al. Further biochemical characterization of human pancreatic lipase-related protein 2 expressed in yeast cells. J Lipid Res. 2007;48:1539–49. doi: 10.1194/jlr.M600486-JLR200. [DOI] [PubMed] [Google Scholar]
10.Xiao X, Mukherjee A, Ross LE, Lowe ME. Pancreatic Lipase-related Protein-2 (PLRP2) Can Contribute to Dietary Fat Digestion in Human Newborns. J Biol Chem. 2011;286:26353–63. doi: 10.1074/jbc.M111.249813. [DOI] [PMC free article] [PubMed] [Google Scholar]
11.Berton A, Sebban-Kreuzer C, Rouvellac S, Lopez C, Crenon I. Individual and combined action of pancreatic lipase and pancreatic lipase-related proteins 1 and 2 on native versus homogenized milk fat globules. Mol Nutr Food Res. 2009;53:1592–602. doi: 10.1002/mnfr.200800563. [DOI] [PubMed] [Google Scholar]
12.Cordle RA, Lowe ME. Purification and characterization of human procolipase expressed in yeast cells. Prot Exp Purif. 1998;13:30–5. doi: 10.1006/prep.1998.0873. [DOI] [PubMed] [Google Scholar]
13.Yang Y, Lowe ME. Human pancreatic triglyceride lipase expressed in yeast cells: purification and characterization. Protein Expr Purif. 1998;13:36–40. doi: 10.1006/prep.1998.0874. [DOI] [PubMed] [Google Scholar]
14.Sahasrabudhe AV, Solapure SM, Khurana R, et al. Production of recombinant human bile salt stimulated lipase and its variant in Pichia pastoris. Protein Expr Purif. 1998;14:425–33. doi: 10.1006/prep.1998.0974. [DOI] [PubMed] [Google Scholar]
15.Lemay DG, Lynn DJ, Martin WF, et al. The bovine lactation genome: insights into the evolution of mammalian milk. Genome Biol. 2009;10:R43. doi: 10.1186/gb-2009-10-4-r43. [DOI] [PMC free article] [PubMed] [Google Scholar]
16.Jensen RG, Ferris AM, Lammi-Keefe CJ, Henderson RA. Lipids of bovine and human milks: a comparison. J Dairy Sci. 1990;73:223–40. doi: 10.3168/jds.S0022-0302(90)78666-3. [DOI] [PubMed] [Google Scholar]
17.Bernback S, Blackberg L, Hernell O. The complete digestion of human milk triacylglycerol in vitro requires gastric lipase, pancreatic colipase-dependent lipase, and bile salt-stimulated lipase. J Clin Invest. 1990;85:1221–6. doi: 10.1172/JCI114556. [DOI] [PMC free article] [PubMed] [Google Scholar]
18.Bernback S, Blackberg L, Hernell O. Fatty acids generated by gastric lipase promote human milk triacylglycerol digestion by pancreatic colipase-dependent lipase. Biochim Biophys Acta. 1989;1001:286–93. doi: 10.1016/0005-2760(89)90113-6. [DOI] [PubMed] [Google Scholar]
19.Gargouri Y, Pieroni G, Riviere C, et al. Importance of human gastric lipase for intestinal lipolysis: an in vitro study. Biochim Biophys Acta. 1986;879:419–23. doi: 10.1016/0005-2760(86)90234-1. [DOI] [PubMed] [Google Scholar]
20.Borgstrom B. Importance of phospholipids, pancreatic phospholipase A2, and fatty acid for the digestion of dietary fat: in vitro experiments with the porcine enzymes. Gastroenterol. 1980;78:954–62. [PubMed] [Google Scholar]
21.Andersson EL, Hernell O, Blackberg L, Falt H, Lindquist S. BSSL and PLRP2: key enzymes for lipid digestion in the newborn examined using the Caco-2 cell line. J Lipid Res. 2011;52:1949–56. doi: 10.1194/jlr.M015685. [DOI] [PMC free article] [PubMed] [Google Scholar]
22.Jensen RG, Ferris AM, Lammi-Keefe CJ. Lipids in human milk and infant formulas. Annu Rev Nutr. 1992;12:417–41. doi: 10.1146/annurev.nu.12.070192.002221. [DOI] [PubMed] [Google Scholar]
23.Michalski MC, Briard V, Michel F, Tasson F, Poulain P. Size distribution of fat globules inhuman colostrum, breast milk, and infant formula. J Dairy Sci. 2005;88:1927–40. doi: 10.3168/jds.S0022-0302(05)72868-X. [DOI] [PubMed] [Google Scholar]
24.Straarup EM, Lauritzen L, Faerk J, Hoy Deceased CE, Michaelsen KF. The stereospecific triacylglycerol structures and Fatty Acid profiles of human milk and infant formulas. J Pediatr Gastroenterol Nutr. 2006;42:293–9. doi: 10.1097/01.mpg.0000214155.51036.4f. [DOI] [PubMed] [Google Scholar]
25.Momsen MM, Dahim M, Brockman HL. Lateral packing of the pancreatic lipase cofactor, colipase, with phosphatidylcholine and substrates. Biochemistry. 1997;36:10073–81. doi: 10.1021/bi9703857. [DOI] [PubMed] [Google Scholar]
26.Lowe ME. The triglyceride lipases of the pancreas. J Lipid Res. 2002;43:2007–16. doi: 10.1194/jlr.r200012-jlr200. [DOI] [PubMed] [Google Scholar]
27.Carriere F, Renou C, Lopez V, et al. The specific activities of human digestive lipases measured from the in vivo and in vitro lipolysis of test meals. Gastroenterol. 2000;119:949–60. doi: 10.1053/gast.2000.18140. [DOI] [PubMed] [Google Scholar]
28.Klumpp TG, Neale AV. The gastric and duodenal contents of normal infants and children. Am J Dis Child. 1930;40:1215–29. [Google Scholar]
29.Lowe ME. Assays for pancreatic triglyceride lipase and colipase. In: Doolittle MH, Reue K, editors. Methods in Molecular Biology: Lipase and Phospholipase Protocols. Totowa NJ: Humana Press Inc; 1998. pp. 59–70. [DOI] [PubMed] [Google Scholar]
30.Boehm G, Braun W, Moro G, Minoli I. Bile acid concentrations in serum and duodenal aspirates of healthy preterm infants: effects of gestational and postnatal age. Biol Neonate. 1997;71:207–14. doi: 10.1159/000244419. [DOI] [PubMed] [Google Scholar]
31.Roman C, Carriere F, Villeneuve P, et al. Quantitative and qualitative study of gastric lipolysis in premature infants: do MCT-enriched infant formulas improve fat digestion? Pediatr Res. 2007;61:83–8. doi: 10.1203/01.pdr.0000250199.24107.fb. [DOI] [PubMed] [Google Scholar]
32.Armand M, Hamosh M, Mehta NR, et al. Effect of human milk or formula on gastric function and fat digestion in the premature infant. Pediatr Res. 1996;40:429–37. doi: 10.1203/00006450-199609000-00011. [DOI] [PubMed] [Google Scholar]
33.Eydoux C, Aloulou A, De Caro J, et al. Human pancreatic lipase-related protein 2: tissue localization along the digestive tract and quantification in pancreatic juice using a specific ELISA. Biochim Biophys Acta. 2006;1760:1497–504. doi: 10.1016/j.bbagen.2006.06.005. [DOI] [PubMed] [Google Scholar]
34.Takahashi K, Mizuno K, Itabashi K. The freeze-thaw process and long intervals after fortification denature human milk fat globules. Am J Perinatol. 2012;29:283–8. doi: 10.1055/s-0031-1295659. [DOI] [PubMed] [Google Scholar]
35.Fidler N, Sauerwald TU, Demmelmair H, Koletzko B. Fat content and fatty acid composition of fresh, pasteurized, or sterilized human milk. Adv Exp Med Biol. 2001;501:485–95. doi: 10.1007/978-1-4615-1371-1_60. [DOI] [PubMed] [Google Scholar]
36.Cao H, Hegele RA. DNA polymorphisms of lipase related genes. J Hum Genet. 2003;48:443–6. doi: 10.1007/s10038-003-0051-1. [DOI] [PubMed] [Google Scholar]
37.Hancock AM, Witonsky DB, Ehler E, et al. Human adaptations to diet, subsistence, and ecoregion are due to subtle shifts in allele frequency. Proc Natl Acad Sci U S A. 2010;107 (Suppl 2):8924–30. doi: 10.1073/pnas.0914625107. [DOI] [PMC free article] [PubMed] [Google Scholar]
38.Fomon SJ, Ziegler EE, Thomas LN, Jensen RL, Filer LJ. Excretion of fat by normal full-term infants fed various milks and formulas. Am J Clin Nutr. 1970;23:1299–313. doi: 10.1093/ajcn/23.10.1299. [DOI] [PubMed] [Google Scholar]
39.Carriere F, Moreau H, Raphel V, et al. Purification and biochemical characterization of dog gastric lipase. Eur J Biochem. 1991;202:75–83. doi: 10.1111/j.1432-1033.1991.tb16346.x. [DOI] [PubMed] [Google Scholar]
40.Gargouri Y, Pieroni G, Riviere C, et al. Kinetic assay of human gastric lipase on short- and long-chain triacylglycerol eulsions. Gastroenterol. 1986;91:919–25. doi: 10.1016/0016-5085(86)90695-5. [DOI] [PubMed] [Google Scholar]
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15831 | https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-geometry/cc-7th-angles/v/complementary-and-supplementary-angles | Use of cookies
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15832 | https://pages.uoregon.edu/adding/courses/431/hw9.pdf | Homework 9 Due Friday, December 6, 2019 1. Which of the following topologies on R are compact? Give a proof either way.
(a) The finite complement topology.
(b) The topology where U is open iffeither U = ∅or 0 ∈U.
(c) The lower semi-continuous topology.
2.
(a) Let X be a compact space, and let F1 ⊃F2 ⊃F3 ⊃· · · be a descending chain of non-empty closed subsets. Show that the intersection F1 ∩F2 ∩F3 ∩· · · is not empty.
Hint: Otherwise X \ F1, X \ F2, . . . is an open cover of X.
(b) Give an example of a non-compact space X and a descending chain of closed subsets F1 ⊃F2 ⊃F3 ⊃· · · whose intersection is empty.
3.
(a) Let X and Y be topological spaces, let A ⊂X, and let U be a neighborhood of A×Y in X ×Y . Show that if Y is compact then there is a neighborhood V of A in X such that V × Y ⊂U.
(Start by drawing a picture!) (b) Give a counterexample when Y is not compact.
4. A continuous map f : X →Y is called proper if the preimage of any compact set K ⊂Y is compact.
(a) Show that the map f : R2 →R given by f(x, y) = x2 + y2 is proper.
(b) Show that the map f : R2 →R given by f(x, y) = x2 −y2 is not proper.
1 (c) If f is proper then the preimage of every point is compact, because points are compact. But give an example of a continuous map f : X →Y for which the preimage of every point is compact, but nonetheless f is not proper.
(d) Show that if X is compact and Y is Hausdorffthen any continuous map f : X →Y is proper.
(e) Let X and Y be topological spaces.
Show that the projection p: X × Y →X is proper if and only if Y is compact.
5. Optional. We define one-point compactification of a topological space X to be ˆ X = X ∪{∞}, with the following topology: if U is open in X, then U is open in ˆ X; and if K ⊂X is compact and closed, then (X \ K) ∪{∞} is open in ˆ X. Equivalently, F ⊂ˆ X is closed if F ∩X is closed, and either (1) ∞∈F or (2) F is compact.
(a) Show that the one-point compactification of [0, 1) is homeomor-phic to [0, 1], and that the one-point compactification of R or (0, 1) is homeomorphic to the circle. Describe the one-point compacti-fication of R2.
(b) Show that ˆ X is compact.
(c) A space X is called locally compact if for every point p ∈X there is a compact set K ⊂X with p ∈int(K).
This is a slightly unusual use of the word “locally,” but it the appropriate one for compactness. For example, R is locally compact, but Q is not.
Show that ˆ X is Hausdorffif and only if X is locally compact and Hausdorff.
(d) A map f : X →Y induces a map ˆ f : ˆ X →ˆ Y . Show that ˆ f is continuous if and only if f is proper.
2 |
15833 | https://www.ebmconsult.com/articles/propranolol-preferred-thyroid-storm-thyrotoxicosis | Why Propranolol Is Preferred to Other Beta-Blockers in Thyrotoxicosis or Thyroid Storm
Summary:
In patients with thyrotoxicosis (thyroid storm), or symptomatic hyperthyroidism, there is an excess of thyroid hormone (T4 & T3) production and secretion that may result in increases in heart rate, tremors and nervousness.
Propranolol is the most widely studied non-selective, beta-1 and beta-2-blocker that can treat the increased heart rate and tremor. Additionally, it may reverse some of the reduced systemic vascular resistance and inhibit the peripheral conversion of T4 to the more biologically active hormone, T3.
The American Association of Clinical Endocrinologists Medical Guidelines for the Evaluation and Treatment of Hyperthyroidism and Hypothyroidism discuss the use of beta blockers in this situation but do not specifically recommend one over another.
Editor-in-Chief: Anthony J. Busti, MD, PharmD, FNLA, FAHA Reviewers: Jon D. Herrington, PharmD, BCPS, BCOP and Donnie Nuzum, PharmD, BCACP, CDE Last Reviewed: October 2015
| Explanation |
| Patients experiencing thyrotoxicosis (thyroid storm), or symptomatic hyperthyroidism, can experience a number of effects that can include tachycardia, palpitations, tremor and/or nervousness. Patient's with this condition are known to have an increased production of the thyroid hormones, thyroxine (T4) and 3,5,3'-triiodothyronine (T3). While the thyroid gland primarily releases T4 into the circulation, T4 is generally metabolized to T3 in the peripheral tissue by two enzymes: monodeiodinase type I (5'D-I) and monodeiodinase type II (5'D-II). The production of T3 is important because it is more biologically potent than T4.1 Increases in T3 result in a number of effects, including an increase in myocardial contractility and speed of diastolic relaxation of the heart.2-,3,4,5 In addition, systemic vascular resistance is reduced, which may put the patient at increased risk for developing high output cardiac failure or even shock.5 The treatment of this potentially emergent situation requires the use of medications that not only inhibit the synthesis of T4 and T3, but also inhibit the peripheral conversion of T4 to T3 by 5'D-I and/or 5'D-II. Propranolol, a non-selective beta-1 and beta-2-blocker, is frequently used to help treat this condition. Propranolol will not only help control the symptomatic tachycardia and tremors associated with thyroid storm, but there is also data that shows propranolol may also known to inhibit the monodeiodinase type I enzyme responsible for conversion of T4 to the more biologically potent T3 hormone.6-10 This reduction in T4's metabolism, via the inhibition of monodeiodinase type I, may cause the T4 to then be shunted through the enzyme monodeiodinase type III (5'D-III) resulting in the production of 3,3',5'-triiodothyronine (reverse T3 or rT3).11,12 Reverse T3 is known to be metabolically inactive. Since blocking beta-2-receptors in blood vessels can result in vasoconstriction, propranolol's beta-2-blocking properties may also treat some of the reduced systemic vascular resistance occurring in this clinical scenario. In addition, propranolol is also a beta-blocker without intrinsic sympathomimetic activity and thus will not mimic the symptoms of thyrotoxicosis. It is for all of these reasons that propranolol has been most studied and is the most commonly used beta-blocker in this setting.6-12 Doses of propranolol of 160 mg or more maybe needed to control symptoms, especially in younger patients with thyrotoxicosis.13 Interestingly, the American Association of Clinical Endocrinologists Medical Guidelines for the Evaluation and Treatment of Hyperthyroidism and Hypothyroidism do not specifically recommend one beta-blocker over another when discussing the use of beta blockers in this situation.14 In patients who have contraindications to propranolol (e.g., asthma or reactive airway disease), the use of diltiazem can be considered as an alternative. If patients have concurrent low-output heart failure during thyrotoxicosis, all negative inotropic medications (including propranolol) should be used with caution.15References: 1. Berry MJ, Larsen PR. The role of selenium in thyroid hormone action. Endocr Rev 1992;13:207-19. 2. Glass CK, Holloway JM. Regulation of gene expression by the thyroid hormone receptor. Biochem Biophys Acta 1990;1032:157-76. 3. Brent GA, Moore DD, Larsen PR. Thyroid hormone regulation of gene expression. Annu Rev Physiol 1991;53:17-35. 4. Dillmann WH. Biochemical basis of thyroid hormone action in the heart. Am J Med 1990;88:626-30. 5. Woeber KA. Thyrotoxicosis and the heart. N Engl J Med 1992;327:94-8. 6. Wiersinga WM, Touber JL. The influence of beta-adrenoreceptor blocking agents on plasma thyroxine and triiodothyronine. J Clin Endocrinol Metab 1977;45:293-8. 7. Verhoeven RP, Visser TJ, Doctor R et al. Plasma thyroxine, 3,3'5-triiodothryonine and 3,3',5'-triiodothyronine during beta-adrenergic blockade in hyperthyroidism. J Clin Endocrinol Metab 1977;44:1002-5. 8. Chambers JB, Pittman CS, Suda AK. The effects of propranolol on thyroxine metabolism and triiodothyronines production in man. J Clin Pharmacol 1982; 22:110-6. 9. Lumholtz IB, Faber J, Kirkegaard C et al: The extrathyroidal effect of D, L-propranolol on 3,3',5'-triiodothyronine, 3',5'-diiodothyronine, 3,3'-diiodothyronine and 3'-monoiodothyronine kinetics. J Clin Endocrinol Metab 1982; 54:1097-100. 10. Wiersinga WM. Propranolol and thyroid hormone production metabolism. Thyroid 1991;1:273-7. 11. Kallner G, Ljunggren JG, Tryselius M. The effect of propranolol on serum levels of T4, T3 and reverse-T3 in hyperthyroidism. Acta Med Scand 1978;204:35-7. 12. Nilsson OR, Karlberg BE, Kagedal B et al. Non-selective and selective beta-1 adrenoreceptor blocking agents in the treatment of hyperthyroidism. Acta Med Scand 1979;206:21-5. 13. Feely J, Forrest A, Gunn A et al. Propranolol dosage in thyrotoxicosis. J Clin Endocrinol Metab 1980;51:658-61. 14. Baskin HJ, Cobin RH, Duick DS et al. American Association of Clinical Endocrinologists medical guidelines for clinical practice for the evaluation and treatment of hyperthyroidism and hypothyroidism. Endocr Pract 2002;8:457-69. 15. Dalan R, Leow MK. Cardiovascular collapse associated with beta-blockade in thyroid storm. Exp Clin Endocrinol Diabetes 2007;115:392-6. |
| Other EBM Consult Related Content |
| + Pharmacology: The Mechanism for Amiodarone Induced Hyperthyroidism + Physical Exam: Thyroid Gland Palpation |
| MESH Terms & Keywords |
| Monodeiodinase Type I, 5'D-I, Monodeiodinase Type II, 5'D-II, Propranolol, Beta Blockers for Thyroid storm, Beta Blockers for Thyrotoxicosis, Propranolol for Thyroid Storm |
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15834 | https://programmingpraxis.com/programming-with-prime-numbers/ | Programming with Prime Numbers
| |
| Contents |
| 1 The Sieve of Eratosthenes |
| 2 Primality Testing by Trial Division |
| 3 Factorization by Trial Division |
| 4 Miller-Rabin Pseudoprimality Checking |
| 5 Factorization by Pollard’s Rho Method |
| 6 Going Further |
| Appendices |
| — C |
| — Haskell |
| — Java |
| — Python |
| — Scheme |
Prime numbers are those integers greater than one that are divisible only by themselves and one; an integer greater than one that is not prime is composite. Prime numbers have fascinated mathematicians since the days of the ancient Greek mathematicians, and remain an object of study today. The sequence of prime numbers begins 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, … and continues to infinity, which Euclid, the famous teacher of geometry, proved about twenty-three centuries ago:
Assume for the moment that the number of primes is finite, and make a list of them: p1, p2, …, pk. Now compute the number n = p1 · p2 · … · pk + 1. Certainly n is not evenly divisible by any of the primes p1, p2, …, pk, because division by any them leaves a remainder of 1. Thus either n is prime, or n is composite but has two or more prime factors not on the list p1, p2, …, pk of prime numbers. In either case the assumption that the number of primes is finite is contradicted, thus proving the infinitude of primes. — Euclid, Elements, Book IX, Proposition 20, circa 300 B.C.
In this essay we will examine three problems related to prime numbers: enumerating the prime numbers, determining if a given number is prime or composite, and factoring a composite number into its prime factors. We describe in detail five relevant functions: one that makes a list of the prime numbers less than a given number using the Sieve of Eratosthenes; two that determine whether a given number is prime or composite, one using trial division and the other using an algorithm developed by Gary Miller and Michael Rabin; and two that find the unique factorization of a given composite number, one using trial division and the other using John Pollard’s rho algorithm. We first describe the algorithms in the body of the essay, then describe actual implementations in five languages — C, Haskell, Java, Python and Scheme — in a series of appendices.
Our goals are modest. Our purpose is pedagogical, so we are primarily interested in the clarity of the code. We describe algorithms that are well known and implement them carefully. And we hope that careful reading will lead you to be a better programmer in addition to learning something about prime numbers. Even so, our functions are genuinely useful for a variety of purposes beyond simple study.
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1 The Sieve of Eratosthenes
The method that is in common use today to make a list of the prime numbers less than a given input n was invented about two hundred years before Christ by Eratosthenes of Cyrene, who was an astronomer, geographer and mathematician, as well as the third chief librarian of Ptolemy’s Great Library at Alexandria; he calculated the distance from Earth to Sun, the tilt of the Earth’s axis, and the circumference of the Earth, and invented the leap day and a system of latitude and longitude. His method begins by making a list of all the numbers from 2 to the desired maximum prime number n. Then the method enters an iterative phase. At each step, the smallest uncrossed number that hasn’t yet been considered is identified, and all multiples of that number are crossed out; this is repeated until no uncrossed numbers remain unconsidered. All the remaining uncrossed numbers are prime.
Thus, the first step crosses out all multiples of 2: 4, 6, 8, 10 and so on. At the second step, the smallest uncrossed number is 3, and multiples of 3 are crossed out: 6, 9, 12, 15 and so on; note that some numbers, such as 6, might be crossed out multiple times. At this point 4 has been crossed out, so the next smallest uncrossed number is 5, and its multiples 10, 15, 20, 25 and so on are also crossed out. The process continues until all uncrossed numbers have been considered. Thus, each prime is used to “sift” its multiples out of the original list, so that only primes are left in the sieve. Or you may prefer the ditty:
Strike the twos and strike the threes,
The Sieve of Eratosthenes!
When the multiples sublime,
The numbers that remain are prime.
Although this is the basic algorithm, there are three optimizations that are routinely applied. First, since 2 is the only even prime, it is best to handle 2 separately and sieve only on odd numbers, reducing the size of the sieve by half. Second, instead of starting the crossing-out at the smallest multiple of the current sieving prime, it is possible to start at the square of the multiple, since all smaller numbers will have already been crossed out; we saw that in the sample when 6 was already crossed out as a multiple of 2 when we were crossing out multiples of 3. Third, as a consequence of the second optimization, sieving can stop as soon as the square of the sieving prime is greater than n, since there is nothing else to do. Here is a formal statement of the sieve of Eratosthenes:
Algorithm 1. Sieve of Eratosthenes: Generate the primes not exceeding n > 1:
1. [Initialization.] Set m ← ⌊(n−1)/2⌋. Create a bitarray B[0..m−1] with each item set to TRUE. Set i ← 0. Set p ← 3. Output the prime 2.
2. [Sieving complete?] If n < p2, go to Step 5.
3. [Found prime?] If B[i] = FALSE, set i ← i + 1, set p ← p + 2, and go to Step 2. Otherwise, output the prime p and set j ← 2i2 + 6i + 3 (or j ← (p2 − 3) / 2).
4. [Sift on p.] If j < m, set B[j] ← FALSE, set j ← j + 2i + 3 (or j ← j + p) and go to Step 4. Otherwise, set i ← i + 1, set p ← p + 2, and go to Step 2.
5. [Terminate?] If i = m, stop.
6. [Report remaining primes.] If B[i] = TRUE, output the prime p. Then, regardless of the value of B[i], set i ← i + 1, set p ← p + 2, and go to Step 5.
The calculation of j in Step 3 is interesting. Since the bitarray contains odd numbers starting from 3, an index i of the bitarray corresponds to the number p = 2i+3; for instance, the fifth item in the bitarray, at index 4, is 11. Sifting starts from the square of the current prime, so to sift the prime 11 at index 4 we start from 112 = 121 at index 59, calculated as (i−3)/2. Thus, to compute the starting index j, we calculate ((2i+3)2−3)/2, which with a little bit of algebra simplifies to the formula in Step 3. You may prefer the alternate calculation (p2−3) / 2 which exploits the identity 2i+3 = p.
As an example, we show the calculation of the primes less than a hundred. The first iteration notes that 3 is the smallest uncrossed number, and crosses out every third number starting from 3 · 3: 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99.
3 5 7 ~~9~~ 11 13 ~~15~~ 17 19 ~~21~~ 23 25 ~~27~~ 29 31 ~~33~~ 35 37 ~~39~~ 41 43 ~~45~~ 47 49 ~~51~~ 53 55 ~~57~~ 59 61 ~~63~~ 65 67 ~~69~~ 71 73 ~~75~~ 77 79 ~~81~~ 83 85 ~~87~~ 89 91 ~~93~~ 95 97 ~~99~~
Now 5 is the next smallest uncrossed number, so we cross out every fifth number starting from 5 · 5: 25, 35, 45, 55, 65, 75, 85, 95. Note that 45 and 75 were previously crossed out, so only six additional numbers are now crossed.
3 5 7 ~~9~~ 11 13 ~~15~~ 17 19 ~~21~~ 23 ~~25~~ ~~27~~ 29 31 ~~33~~ ~~35~~ 37 ~~39~~ 41 43 ~~45~~ 47 49 ~~51~~ 53 ~~55~~ ~~57~~ 59 61 ~~63~~ ~~65~~ 67 ~~69~~ 71 73 ~~75~~ 77 79 ~~81~~ 83 ~~85~~ ~~87~~ 89 91 ~~93~~ ~~95~~ 97 ~~99~~
Now 7 is the next smallest uncrossed number, so we cross out every seventh number starting from 7 · 7: 49, 63, 77, 91. Note that 63 was previously crossed out, so only three additional numbers are crossed.
3 5 7 ~~9~~ 11 13 ~~15~~ 17 19 ~~21~~ 23 ~~25~~ ~~27~~ 29 31 ~~33~~ ~~35~~ 37 ~~39~~ 41 43 ~~45~~ 47 ~~49~~ ~~51~~ 53 ~~55~~ ~~57~~ 59 61 ~~63~~ ~~65~~ 67 ~~69~~ 71 73 ~~75~~ ~~77~~ 79 ~~81~~ 83 ~~85~~ ~~87~~ 89 ~~91~~ ~~93~~ ~~95~~ 97 ~~99~~
Now the next smallest uncrossed number is 11, but 11 · 11 = 121 is greater than 100, so sieving is complete. The complete list of primes, including 2 followed by the remaining uncrossed numbers, is: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.
The algorithm outputs primes in Step 1 (the only even prime 2), Step 3 (the sieving primes), and Step 6 (sweeping up the primes that survived the sieve). The word “output” can mean anything. It is common to collect the primes in a list, but depending on your needs, you could count them, sum them, or use them in many other ways.
The Sieve of Eratosthenes runs in time O(n log log n), and because the only operations in the inner loop (Step 4) are a single comparison, addition and crossing-out, it is very fast in practice.
There are other ways to make lists of prime numbers. If memory is constrained, or if you want only the primes on a limited range from m to n, you may be interested in the segmented Sieve of Eratosthenes, which finds the primes in blocks; the sieving primes are those less than the square root of n, and the minimum multiple of each sieving prime in each segment is reset at the beginning of the next segment. A. O. L. Atkin, an IBM researcher, invented a sieve that is faster than the Sieve of Eratosthenes, as it crosses out multiples of the squares of the sieving primes after some precomputations. Paul Pritchard, an Australian mathematician, has developed several methods of sieving using wheels that have time complexity O(n / log log n), which is asymptotically faster than the Sieve of Eratosthenes, though in practice Pritchard’s sieves are somewhat slower since the bookkeeping in each step of the inner loop is more involved.
Sometimes instead of the primes less than n you need the first x primes. The simplest method is to estimate the value of n using the Prime Number Theorem, first conjectured by Carl Friedrich Gauss in 1792 or 1793 (at the age of fifteen) and proved independently by Jacques Hadamard and Charles Jean de la Vallée-Poussin in 1896, which states that there are approximately x / loge x primes less than x; estimate the value of n that gives the desired value of x, add a buffer of 20% or thereabouts, compute the primes less than n, and throw away the excess.
By the way, many people implement a function to list the prime numbers to a limit that they call the Sieve of Eratosthenes, but really isn’t; their functions use trial division instead. If you use a modulo operator, or division in any form, your algorithm is not the Sieve of Eratosthenes, and will run much slower than the algorithm described above. On a recent-vintage personal computer, a properly-implemented Sieve of Eratosthenes should be able to list the primes less than a million in less than a second.
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2 Primality Testing by Trial Division
We turn next to the problem of classifying a number n as prime or composite. The oldest method, and for nearly two thousand years the only method, was trial division. If n has no remainder when divided by 2, it is composite. Otherwise, if n has no remainder when divided by 3, it is composite. Otherwise, if n has no remainder when divided by 4, it is composite. Otherwise, …. And so on. Iteration stops, and the number is declared prime, when the trial divisor is greater than the square root of n. We can see that this is so because, if n = p · q, then one of p or q must be less than the square root of n while the other is greater than the square root of n (unless p and q are equal, and n is a perfect square). A simple optimization notes that if the number is even it is composite, and if the number is odd any factor must be odd, so it is necessary to divide only by the odd numbers greater than 2, not by the even numbers.
Algorithm 2. Primality Testing by Trial Division: Determine if an integer n > 1 is prime or composite:
1. [Finished if even] If n is even, return COMPOSITE and stop.
2. [Initialize for odd] Set d ← 3.
3. [Terminate if prime] If d 2 > n, return PRIME and stop.
4. [Trial division] If n mod d = 0, return COMPOSITE and stop.
5. [Iterate] Set d ← d + 2. Go to Step 3.
As an example, consider the number 100524249167. It is not even. Dividing by 3 gives a remainder of 2. Dividing by 5 gives a remainder of 2. Dividing by 7 gives a remainder of 7. Dividing by 9 gives a remainder of 5. Dividing by 11 gives a remainder of 1. Dividing by 13 gives a remainder of 4. Dividing by 15 gives a remainder of 2. Dividing by 17 gives a remainder of 8. Dividing by 19 gives a remainder of 3. Dividing by 21 gives a remainder of 20. Dividing by 23 gives a remainder of 23 and, in Step 4, demonstrates that 100524249167 = 23 · 127 · 239 · 311 · 463 is composite.
There are other optimizations possible. If you have a list of prime numbers at hand, you can use only the primes as trial divisors and skip the composites, which makes things go quite a bit faster; this works because, if you’ve already tested its component primes, it is not possible for a composite to be a divisor. If you can’t afford the space to store a list of primes, you can use one of Pritchard’s wheels, just as with the sieve.
Trial division iterates until it reaches either the smallest prime factor of a number or its square root. Most composites are identified fairly quickly; it’s the primes that take longer, as all the trial divisors fail one by one. In general, the time complexity of trial division is O(√n), and if n is large it will take a very long time. Thus, trial division is best limited to cases where n is small, less than a billion or thereabouts. If n is large, it is common to use probabilistic methods to distinguish primes from composites, as the next section will show.
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3 Factorization by Trial Division
The problem of breaking down a composite integer into its prime factors has fascinated mathematicians from the ancient Greeks to modern times. The ancient Greek mathematicians proved the Fundamental Theorem of Arithmetic that the factorization of a number is unique, ignoring the order of the factors. Modern mathematicians use the difficulty of the factorization process to provide cryptographic security for the internet. Factoring integers is a hard and honorable problem.
Euclid gave the proof of the Fundamental Theorem of Arithmetic in his book Elements. We assume without proof that if p divides ab then either p divides a or p divides b; this assertion is known as Euclid’s Lemma and was proved in Elements, Book VII, Proposition 7. The proof is in two parts: first a proof that all positive integers greater than 1 can be written as the product of primes, and second a proof that the factorization is unique.
Suppose that n is the smallest positive integer greater than 1 that cannot be writen as the product of primes. Now n cannot be prime because such a number is the product of a single prime, itself. Thus the composite is n = a · b, where both a and b are positive integers less than n. Since n is the smallest number that cannot be written as the product of primes, both a and b must be able to be written as the product of primes. But then n = a · b can be written as the product of primes simply by combining the factorizations of a and b. But that contradicts our supposition. Therefore, all positive integers greater than 1 can be written as the product of primes.
Furthermore, that factorization is unique, ignoring the order in which the primes are written. Now suppose that s is the smallest positive integer greater than 1 that can be written as two different products of prime numbers, so that s = p1 · p2 · … · pm = q1 · q2 · … · qn. By Euclid’s lemma either p1 divides q1 or p1 divides q2 · … · qn. Therefore p1 = qk for some k. But removing p1 and qk from the initial equivalence leaves a smaller integer that can be factored in two ways, contradicting the initial supposition. Thus there can be no such s, and all integers greater than 1 have a unique factorization. — Euclid, Elements, Book IX, Proposition 14, circa 300 B.C.
There are many algorithms for factoring integers, of which the simplest is trial division. Divide the number being factored by 2, then 3, then 4, and so on. When a factor divides evenly, record it, divide the original number by the factor, then continue the process until the remaining cofactor is prime. A simple optimization notes that once the factors of 2 have been removed from a number, it is odd, and all its factors will be odd, so it is only necessary to perform trial division by the odd numbers starting from 3, thus halving the work required to be done. A second optimization stops the trial division when the factor being tried exceeds the square root of the current cofactor, indicating that the current cofactor is prime and is thus the final factor of the original number. Here is a formal statement of the trial division factorization algorithm:
Algorithm 3. Integer Factorization by Trial Division: Find the prime factors of a composite integer n > 1:
1. [Remove factors of 2] If n is even, output the factor 2, set n ← n ÷ 2, and go to Step 1.
2. [Initialize for odd factors] If n = 1, stop. Otherwise, set f ← 3.
3. [Terminate if prime] If n < f · f, output the factor n and stop.
4. [Trial division] Calculate the quotient q and remainder r when dividing n by f, so that n = q f + r with 0 ≤ r < f.
5. [Loop on odd integers] If r > 0, set f ← f + 2 and go to Step 3. Otherwise, output the factor f, set n ← q, and go to Step 3.
As an example, we find the factors of n = 13195. Since 13195 is odd, Step 1 does nothing and we go to Step 2, where f = 3. Since 3 · 3 < 13195, we calculate q = 13195 ÷ 3 = 4398 and r = 1 in Step 4, so in Step 5 we set f = 3 + 2 = 5 and go to Step 3. Since 5 · 5 < 13195, we calculate q = 13195 ÷ 5 = 2639 and r = 0 in Step 4, so in Step 5 we output the factor 5, set n = 2639, and go to Step 3. Since 5 · 5 < 2639, we calculate q = 2639 ÷ 5 = 527 and r = 4 in Step 4, so in Step 5 we set f = 5 + 2 = 7 and go to Step 3. Since 7 · 7 < 2639, we calculate q = 2639 ÷ 7 = 377 and r = 0 in Step 4, so in Step 5 we output the factor 7, set n = 377, and go to Step 3. Since 7 · 7 < 377, we calculate q = 377 ÷ 7 = 53 and r = 6 in Step 4, so in Step 5 we set f = 7 + 2 = 9 and go to Step 3. Since 9 · 9 < 377, we calculate q = 377 ÷ 9 = 41 and r = 8 in Step 4, so in Step 5 we set f = 9 + 2 = 11 and go to Step 3. Since 11 · 11 < 377, we calculate q = 377 ÷ 11 = 34 and r = 3 in Step 4, so in Step 5 we set f = 11 + 2 = 13 and go to Step 3. Since 13 · 13 < 377, we calculate q = 377 ÷ 13 = 29 and r = 0 in Step 4, so in Step 5 we output the factor 13, set n = 29, and go to Step 3. Since 29 < 13 · 13, we output the factor 29 in Step 3 and stop. The complete factorization is 13195 = 5 · 7 · 13 · 29.
Step 1 removes factors of 2 from n. If you like, you can extend that to other primes: remove factors of 3, then factors of 5, then factors of 7, and so on, never wasting the time to trial-divide by a composite. Of course, that requires you to pre-compute and store the primes up to the square root of n, which may be inconvenient. If you prefer, there is a method known as wheel factorization, akin to Pritchard’s sieving wheels, that achieves most of the benefits of trial division by primes but requires only a small, constant amount of extra space to store the wheel.
The time complexity of trial division is O(√n), as all trial divisors up to the square root of the number being factored must potentially be tried. In practice, you will generally want to choose a bound on the maximum divisor you are willing to test, depending on the speed of your hardware and on your patience; generally speaking, the bound should be fairly small, perhaps somewhere between a thousand and a million. Then, if you have reached the bound without completing the factorization, you can turn to a different, more potent, method of integer factorization.
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4 Miller-Rabin Pseudoprimality Checking
As we saw above, trial division can be used to determine if a number n is prime or composite, but is very slow. Often it is sufficient to show that n is probably prime, which is a very much faster calculation; the French number theorist Henri Cohen calls numbers that pass a probable-prime test industrial-grade primes. The most common method of testing an industrial-grade prime is based on a theorem that dates to 1640.
Pierre de Fermat was a French lawyer at the Parlement in Toulouse, a jurist, and an amateur mathematician who worked in number theory, probability, analytic geometry, differential calculus, and optics. Fermat’s Little Theorem states that if p is a prime number, then for any integer a it is true that ap ≡ a (mod p), which is frequently stated in the equivalent form ap−1 ≡ 1 (mod p) by dividing both sides of the congruence by a, assuming a ≠ 0. As was his habit, Fermat gave no proof for his theorem, which was proved by Gottfried Wilhelm von Leibniz forty years later and first published by Leonhard Euler in 1736. This formula gives us a way to distinguish primes from composites: if we can find an a for which Fermat’s Little Theorem fails, then p must be composite.
But there is a problem. There are some numbers, known as Carmichael numbers, that are composite but pass Fermat’s test for all a; the smallest Carmichael number is 561 = 3 · 11 · 17, and the sequence begins 561, 1105, 1729, 2465, 2821, 6601, 8911, 10585, 15841, 29341, …. In 1976, Gary Lee Miller, a computer science professor at Carnegie Mellon University, developed an alternate test for which there are no strong liars, that is, numbers for which there is no a that distinguishes prime from composite. A strong pseudoprime to base a is an odd composite number n = d · 2s + 1 with d odd for which either ad ≡ 1 (mod n) or ad·2r ≡ −1 (mod n) for some r = 0, 1, …, s−1. This works because n = 2m + 1 is odd, so we can rewrite Fermat’s Little Theorem as a2m − 1 ≡ (am − 1)(am + 1) ≡ 0 (mod n). If n is prime, it must divide one of the factors, but can’t divide both because it would then divide their difference (am + 1) − (am − 1) = 2. Miller’s observation leads to the following algorithm:
Algorithm 4.A. Strong-Pseudoprime Test: Determine if a on the range 1 < a < n is a witness to the compositeness of an odd integer n > 2.
1. [Initialize] Set d ← n − 1. Set s ← 0.
2. [Reduce while even] If d is even, set d ← d / 2, set s ← s + 1, and go to Step 2.
3. [Easy return?] Set t ← ad (mod n). If t = 1 or t = n − 1, output PROBABLY PRIME and stop.
4. [Terminate?] Set s ← s − 1. If s = 0, output COMPOSITE and stop.
5. [Square and test] Set t = t2 (mod n). If t = n − 1, output PROBABLY PRIME and stop. Otherwise, go to Step 4.
The algorithm is stated differently than the math given above, though the result is the same. In the math, we calculated ad·2r, which is initially ad when r = 0, then a2d, then a4d, and so on; in other words, ad is squared at each step. Step 5 thus reduces the modular operation from exponentiation to multiplication; the strength reduction makes the code simpler and faster.
As an example, consider the prime number 73 = 23 · 9 + 1; at the end of Step 2, d = 9 and s = 3. If the witness is 2, then 29 ≡ 1 (mod 73) and 73 is declared PROBABLY PRIME in Step 3. If the witness is 3, then 39 ≡ 46 (mod 73) and the test of Step 3 is indeterminate, but 32·9 ≡ 72 ≡ −1 (mod 73) in the second iteration of Step 5, and 73 is declared PROBABLY PRIME. On the other hand, the composite number 75 = 21 · 37 + 1 is declared COMPOSITE with witness 2 because 237 ≡ 47 (mod 75) in Step 3. One more example is the composite number 2047 = 23 · 89;, which is declared PROBABLY PRIME by the witness 2 but COMPOSITE by the witness 3; 2047 is the smallest number for which 2 is a strong liar to its compositeness.
Miller proved that n must be prime if no a from 2 to 70 (loge n)2 is a witness to the compositeness of n; Eric Bach, a professor at the University of Wisconsin in Madison, later reduced the constant from 70 to 2. Unfortunately, the proof assumes the Riemann Hypothesis and can’t be relied upon because the Riemann Hypothesis remains unproven. However, Michael O. Rabin, an Israeli computer scientist and recipient of the Turing Award, used Miller’s strong pseudo-prime test to build a probabilistic primality test. Rabin proved that for any odd composite n, at least 3/4 of the bases a are witnesses to the compositeness of n; although that’s the proven lower bound, in practice the proportion is much higher than 3/4. Thus, the Miller-Rabin method performs k strong pseudo-prime tests, each with a different a, and if all the tests pass the method concludes that n is prime with probability at least 4−k, and in practice much higher; a common value of k is 25, which gives a maximum probability of 1 error in 1017.
Algorithm 4.B. Miller-Rabin Pseudo-Primality Test: Determine if an odd integer n > 2 is probably prime by performing k strong pseudo-prime tests:
1. [Terminate?] If k = 0, output PROBABLY PRIME and stop.
2. [Strong pseudo-prime test] Choose a random number a such that 1 < a < n. Perform a strong pseudo-prime test using Algorithm 4.A. to determine if a is a witness to the compositeness of n.
3. [Pseudo-prime?] If the strong pseudo-prime test indicates a is a witness to the compositeness of n, output COMPOSITE and stop. Otherwise, set k ← k − 1 and go to Step 1.
Although the algorithm given above specifies random numbers for the bases of the strong pseudo-prime test, it is common to fix the bases in advance, based on the value of n. If n is a 32-bit integer, it is sufficient to test on the three bases 2, 7, and 61; all the odd numbers less than 232 have been tested and no errors in the determination of primality exist. If n is less than a trillion, it is sufficient to test to the bases 2, 13, 23, and 1662803. Gerhard Jaesche used the first seven primes as bases and determined that the first false positive is 341550071728321. And Zhenxiang Zhang plausibly conjectures that there are no errors less than 1036 when using the first twenty primes as bases.
As an example, we determine the primality of 2149−1. Algorithm 4.A returns PROBABLY PRIME for witness 2 but COMPOSITE for witness 3, so 2149−1 = 86656268566282183151 · 8235109336690846723986161 is composite. That determination would be impossible for trial division, at least in any reasonable time frame, as the Prime Number Theorem suggests there are approximately 1.9 · 1018 primes to be tested.
Step 5 of Algorithm 4.A. requires modular exponentiation. Some languages provide modular exponentiation as a built-in function, but others don’t. If your language doesn’t, you will have to write your own function. You should not write your function by first performing the exponentiation and then performing the modulo operation, as the intermediate result of the exponentiation can be very large. Instead, use the square-and-multiply algorithm.
Algorithm 4.C. Modular Exponentiation: Compute be (mod m) with b, e and m all positive integers by the square-and-multiply algorithm:
1. [Initialize] Set r ← 1.
2. [Terminate when finished] If e = 0, return r and stop.
3. [Multiply if odd] If e is odd, set r ← r · b (mod n).
4. [Square and iterate] Set e ← ⌊e / 2⌋. Set b ← b2 (mod n). Go to Step 2.
Consider the calculation 43713 (mod 1741) = 819. Initially b = 437, e = 13, r = 1 and the test in Step 2 fails. Since e is odd, r = 1 · 437 = 437 in Step 3, then e = 13 / 2 = 6 and b = 4372 (mod 1741) = 1200 in Step 4 and the test in Step 2 fails. Since e is even, r = 437 is unchanged in Step 3, then e = 6 / 2 = 3 and b = 12002 (mod 1741) = 193 in Step 4 and the test in Step 2 fails. Since e is odd, r = 437 · 193 (mod 1741) = 773 in Step 3, then e = 3 / 2 = 1 and b = 1932 (mod 1741) = 688 in Step 4 and the test in Step 2 fails. Since e is odd, r = 773 · 688 (mod 1741) = 819 in Step 3, then e = 1 / 2 = 0 and b = 6882 (mod 1741) = 1533 in Step 4. At this point the test in Step 2 succeeds and the result r = 819 is returned. By the way, the intermediate calculation results in the very large number 43713 = 21196232792890476235164446315006597, so you can see why Algorithm 4.C is preferable.
The time complexity of the Miller-Rabin primality checker is O(1), which is vastly better than trial division. The time for the strong pseudo-prime test depends on the number of factors of 2 found in n − 1, which is independent of n. Likewise, the number k of strong pseudo-prime tests is independent of n, so it contributes to the implied constant, not to the overall order. If n is large, the arithmetic takes time O(log log n), but we ignore that in our analysis.
There are other methods for quickly checking the primality of a number, including the Baillie-Wagstaff method that combines a strong pseudoprime test base 2 with a Lucas pseudoprime test and the method of Mathematica that adds a strong pseudoprime test base 3 to the Baillie-Wagstaff method; both methods are faster than the Miller-Rabin method, and also give fewer false positives. If a slight chance of error is too much for you, and you need to prove the primality of a number, you can use the trial division of Algorithm 2, there is a method of Pocklington that uses the factorization of n − 1, a method using Jacobi sums (the APR-CL method), a method using elliptic curves due to Atkin and Morain, and the new AKS method, which operates in proven polynomial time but is not yet practical.
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5 Factorization by Pollard’s Rho Method
In 1975, British mathematician John Pollard invented a method of integer factorization that finds factors in time O(n1/4), which is the square root of the O(√n) time complexity of trial division. The method is simple to program and takes only a small amount of auxiliary space. Before we explain Pollard’s algorithm, we discuss two elements of mathematics on which it relies, the Chinese Remainder Theorem and the birthday paradox.
The original version of the Chinese Remainder Theorem was stated in the third-century by the Chinese mathematician Sun Zi in his book Sun Zi suanjing (literally, “The Mathematical Classic of Sun Zi”), and was proved by the Indian mathematician Aryabhata in the sixth century:
Let r and s be positive integers which are relatively prime and let a and b be any two integers. Then there exists an integer n such that n ≡ a (mod r) and n ≡ b (mod s). Furthermore, n is unique modulo the product r · s.
Sun Zi gave an example: When a number is divided by 3, the remainder is 2. When the same number is divided by 5 the remainder is 3. And when the same number is divided by 7, the remainder is 2. The smallest number that satisfies all three criteria is 23, which you can verify easily. And since the least common multiple of 3, 5, and 7 is 105, any number of the form 23 + 105k, from the arithmetic progression 23, 128, 233, 338, …, is also a solution. Sun Zi used this method to count the men in his emperor’s armies; arrange them in columns of 11, then 12, then 13, take the remainder at each step, and calculate the number of soldiers.
In the theory of probability, the “birthday paradox” calculates the likelihood that in a group of p people two of them will have the same birthday. Obviously, in a group of 367 people the probability is 100%, since there are only 366 possible birthdays. What is surprising is that there is a 99% probability of a matching pair in a group as small as 57 people and a 50% probability of a matching pair in a group as small as 23 people. If, instead of birthdays, we consider integers modulo n, there is a 50% probability that two integers are congruent modulo n in a group of 1.177 √n integers.
Pollard’s rho algorithm uses the quadratic congruential random-number generator x2 + c (mod n) with c ∉ {0, −2} to generate a series of random integers xk. By the Chinese Remainder Theorem, if n = p · q, then x (mod n) corresponds uniquely to the pair of integers x (mod p) and x (mod q). Furthermore, the xk sequence also follows the Chinese Remainder Theorem, so that xk+1 = [xk (mod p)]2 + c (mod p) and xk+1 = [xk (mod q)]2 + c (mod q) so that the sequence of xk falls into a much shorter cycle of length √p by the birthday paradox. Thus p is identified when xk and xk+1 are congruent modulo p, which can be determined when gcd(|xk − xk+1|, n) = p is between 1 and n.
Depending on the values of p, q and c, it is possible that the random-number generator may reach a cycle before a factor is found. Thus, Pollard used Robert Floyd’s tortoise-and-hare cycle-detection method. The sequence of xs starts with two values the same, call them t and h. Then each time t is incremented, h is incremented twice; the hare runs twice as fast as the tortoise. If the hare reaches the tortoise, that is, t = h (mod n), before a factor is found, then a cycle has been reached and further work is pointless. At that point, either the factorization attempt can be abandoned or a new random-number generator can be tried by using a different c.
Pollard called his method “Monte Carlo factorization” because of the use of random numbers. The algorithm is now called the rho algorithm because the sequence of x values has an initial tail followed by a cycle, giving it the shape of the Greek letter rho ρ. Fortunately the algorithm is much simpler than the explanation.
Algorithm 5.A. Pollard’s Rho Method: Find a factor of an odd composite integer n > 1:
1. [Initialization] Set t ← 2, h ← 2, and c ← 1. Define the function f(x) = x2 + c (mod n).
2. [Iteration] Set t ← f(t), h ← f(f(h)), and d ← gcd(t−h, n). If d = 1, go to Step 2.
3. [Termination] If d < n, output d and stop. Otherwise, either stop with failure or continue by setting t ← 2, h ← 2, and c ← c + 1, redefining the function f(x) using the new value of c and going to Step 2.
As an example, we consider the factorization of 8051. Initially, t=2, h=2, and c=1. After one iteration of Step 2, t=22+1=5, h=(22+1)2+1=26, and d=gcd(5−26,8051)=1. After the second iteration of Step 2, t=52+1=26, h=(262+1)2+1=7474, and d=gcd(26−7474,8051)=1. After the third iteration of Step 2, t=262+1=677, h=(74742+1)2+1=871 (mod 8051), and d=gcd(677-871,8051)=97, which is a factor of 8051; the complete factorization is 8051=83·97.
Be sure before you begin that n is composite; if n is prime, then d will always be 1 (because if n is prime it is always coprime to every other number) and the algorithm will loop forever. As with trial division, it is probably wise to set some bound on the maximum number of steps you are willing to take in the iteration of Step 2, because large factors can take a long time to find using this algorithm. You should also be careful not to let c be 0 or -2, because in those cases the random numbers aren’t very random. Note that the factor found in Step 3 may not be prime, in which case you can apply the algorithm again to the reduced factor, using a different c. And of course, once you have one factor, you can continue by factoring the remaining cofactor. Algorithm 5.B gives the complete factorization of a number.
Algorithm 5.B. Integer Factorization by Pollard’s Rho Method: Find all the prime factors of a composite integer n greater than 1:
1. [Remove factors of 2] If n is even, output the factor 2, set n ← n ÷ 2, and go to Step 1.
2. [Terminate if prime] If n is prime by the method of Algorithm 2.B, output the factor n and stop.
3. [Find a factor] Use Algorithm 5.A to find a factor of n and call it f. Output the factor f, set n ← n ÷ f, and go to Step 2.
There are two ways in which Pollard’s algorithm can be improved. First, it should bother you that each number in the random sequence is computed twice; it bothered the Australian mathematician Richard Brent, who devised a cycle-finding algorithm based on powers of 2 that computes each number in the random sequence only once, and it is Brent’s variant that is most often used today. A second improvement notes that for any a, b, and n, gcd(ab,n) > 1 if and only if at least one of gcd(a,n) > 1 or gcd(b,n) > 1, and accumulates the products of the elements of the t and h sequences for several steps (for large n, 100 steps is common) before computing the gcd, thus saving much time; if the gcd is n, then it is possible either that a cycle has been found or that two factors were found since the last gcd, in which case it is necessary to return to values saved from the previous gcd calculation and iterate one step at a time.
The time complexity of Pollard’s rho algorithm depends on the unknown factor d. By the birthday paradox, in the average case it will take 1.177 √d steps to find the factor, or O(√d). Thus, if n is the product of two primes, it will take O(n 1/4) to perform the factorization, assuming the two primes are roughly the same size. In other words, a million iterations of trial division will find factors up to a million, while a million iterations of Pollard’s rho method will find factors up to a trillion; that’s why you want to switch from trial division to Pollard’s rho method at a fairly low bound.
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6 Going Further
Although there is more to programming with prime numbers, we will stop here, since our small library has fulfilled our modest goals. The five appendices give implementations in C, Haskell, Java, Python, and Scheme, and the savvy reader will study all of them, because while they all implement exactly the same algorithms, each does so in a different way, and the differences are enlightening, about both the algorithms and the languages. The C appendix describes the tasteful use of the GMP multi-precision number library. The Haskell and Scheme appendices describe some of the syntax and semantics of those languages, on the assumption that they are unfamiliar to many readers. The Java appendix is the most faithful of all the appendices to the exact structure of the algorithms, including error-checking on the inputs as described in the preambles of each of the algorithms. The Python and Scheme appendices are the most “real-world” implementations, as they include error-checking on the inputs, bounds-checking to stop calculations that take too long, and even a non-mathematical but highly useful extension of the domain of the factoring functions.
Although our goals were modest, we have accomplished much. It’s hard to improve on the Sieve of Eratosthenes, and the Miller-Rabin primality checker will handle inputs of virtually unlimited size. The rho algorithm will find most factors up to a dozen digits or more, regardless of the size of the number being factored.
If Pollard’s rho algorithm won’t crack your composite, there are more powerful algorithms available, though they are beyond our modest aspirations. The elliptic curve method will find factors up to about thirty or forty digits (even fifty or sixty digits if you are patient). The quadratic sieve will split semi-primes up to about 90 digits on a single personal computer or 120 digits on a modest network of personal computers, and the number field sieve will split semi-primes up to about 200 digits on that same network. At the time this of this writing, the current record factorization is 231 decimal digits (768 bits), which took a team of experts about 2000 PC-years, and about eight months of calendar time, on a “network” of computers around the world connected via email.
If your goal isn’t self-study and you really want to factor a large number, and the rho technique fails, you have several options. A good first step is Dario Alpern’s factorization applet at Paul-Zimmermann’s gmp-ecm program at uses a combination of trial division, Pollard’s rho algorithm, another algorithm of Pollard known as p−1, and Hendrik Lenstra’s elliptic curve method to find factors. Jason Papadopoulos’ msieve program at uses both the quadratic sieve of Carl Pomerance and the number field sieve of John Pollard.
There is much more to prime numbers and integer factorization than we have discussed here; for instance, there are methods other than trial division for proving the primality of large numbers (several hundred digits) and methods other than enumeration with a sieve for counting the primes less than a given input number. At the end of each algorithm above was a discussion of alternatives; the interested reader will find that web searches for the topics mentioned will be fruitful and interesting. A superb reference for programmers is the book Prime Numbers: A Computational Perspective by Richard Crandall and Carl B. Pomerance (be sure to look for the second edition, which includes discussion of the new AKS primality prover); beware that although the approach is computational, there is still heavy mathematical content in the book. You may also be interested in the Programming Praxis web site at which has many exercises on the theme of prime numbers.
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Appendix: C
C is a small language with limited data types; integers are limited to what the underlying hardware provides, there are no lists, and there are no bitarrays, so we have to depend on libraries to provide those things for us. Since we are interested in prime numbers and not lists or bitarrays, we will write the smallest libraries that are necessary to get a working program. We begin with bitarrays, which are represented as arrays of characters in which we set and clear individual bits using macros:
`#define ISBITSET(x, i) (( x[i>>3] & (1<<(i&7)) ) != 0)
define SETBIT(x, i) x[i>>3] |= (1<<(i&7));
define CLEARBIT(x, i) x[i>>3] &= (1<<(i&7)) ^ 0xFF;`
To declare a bitarray b of length 8n, say char b[n], and to initialize each bit to 0, say memset(b, 0, sizeof(b)); change the 0 to 255 to set each bit initially to 1. Here is our minimal list library:
typedef struct list {
void data;
struct list next;
} List;
List insert(void data, List next)
{
List new;
new = malloc(sizeof(List));
new->data = data;
new->next = next;
return new;
}
List insert_in_order(void x, List xs)
{
if (xs == NULL || mpz_cmp(x, xs->data) < 0)
{
return insert(x, xs);
}
else
{
List head = xs;
while (xs->next != NULL && mpz_cmp(x, xs->next->data) > 0)
{
xs = xs->next;
}
xs->next = insert(x, xs->next);
return head;
}
}
List reverse(List list) {
List new = NULL;
List next;
`> while (list != NULL)
{
next = list->next;
list->next = new;
new = list;
list = next;
}return new;
}`
int length(List xs)
{
int len = 0;
while (xs != NULL)
{
len += 1;
xs = xs->next;
}
return len;
}
Lists are represented as structs of two members; the empty list is represented as NULL. The trickiest function is reverse, which operates in-place to make each list item point to its predecessor. The list functions leave all notion of memory management to the caller.
With that out of the way, we are ready to begin work on the prime number functions. Our version of the Sieve of Eratosthenes uses b for the bitarray, i indexes into the bitarray, and p = 2i + 3 is the number represented at location i of the bitarray. The first while loop identifies the sieving primes and performs the sieving in an inner while, and the third while loop sweeps up the remaining primes that survive the sieve.
List primes(long n)
{
int m = (n-1) / 2;
char b[m/8+1];
int i = 0;
int p = 3;
List ps = NULL;
int j;
`ps = insert((void ) 2, ps);
memset(b, 255, sizeof(b));
while (pp < n)
{
if (ISBITSET(b,i))
{
ps = insert((void ) p, ps);
j = (pp - 3) / 2;
while (j < m)
{
CLEARBIT(b, j);
j += p;
}
}
i += 1; p += 2;
}
while (i < m)
{
if (ISBITSET(b,i))
{
ps = insert((void ) p, ps);
}
i += 1; p += 2;
}`
return reverse(ps);
}
We look next at the two algorithms that use trial division; we’ll look at them together because they are so similar. We used long integers for the Sieve of Eratosthenes because they are almost certainly big enough, but we will use long long unsigned integers for the two trial division functions because that extends the range of the inputs that we can consider. The function that tests primality using trial division uses an if to identify even numbers, then a while ranges over the odd numbers d from 3 to the square root of n.
int td_prime(long long unsigned n)
{
if (n % 2 == 0)
{
return n == 2;
}
`long long unsigned d = 3;
while (dd <= n)
{
if (n % d == 0)
{
return 0;
}
d += 2;
}`
return 1;
}
The td_factors function is very similar to the td_prime function. The initial if becomes a while, because we no longer want to quit as soon as we find a single factor, and the body of the second while also changes so that it collects all the factors instead of quitting as soon as it finds a single factor; the factors are stacked in increasing order as they are discovered, hence the reversal. The list of factors, which will contain only the original input n if it is prime, is returned as the value of the function.
List td_factors(long long unsigned n)
{
List fs = NULL;
`while (n % 2 == 0)
{
fs = insert((void ) 2, fs);
n /= 2;
}
if (n == 1)
{
return reverse(fs);
}
long long unsigned f = 3;
while (ff <= n)
{
if (n % f == 0)
{
fs = insert((void ) f, fs);
n /= f;
}
else
{
f += 2;
}
}`
fs = insert((void ) n, fs);
return reverse(fs);
}
We used long integers for the Sieve of Eratosthenes and long long unsigned integers for the two trial-division algorithms. Those native integer types are sufficient for those applications; usually only small n are required for the Sieve of Eratosthenes, and trial division is just too slow for large n. But many applications require much larger numbers, so we need a big-integer library, and we choose the GMP library from GNU, which is well-known for its useful interface and fast, bug-free implementation. You can obtain GMP from gmplib.org; to use it in your program, include the line #include <gmp.h> at the top of your program, and link with the option -lgmp.
We look next at Gary Miller’s strong pseudo-prime test. The first while computes d and s, then the if checks for an early return, the second while computes and tests the powers of a, and the default return is composite.
int is_spsp(mpz_t n, mpz_t a)
{
mpz_t d, n1, t;
mpz_inits(d, n1, t, NULL);
mpz_sub_ui(n1, n, 1);
mpz_set(d, n1);
int s = 0;
`while (mpz_even_p(d))
{
mpz_divexact_ui(d, d, 2);
s += 1;
}
mpz_powm(t, a, d, n);
if (mpz_cmp_ui(t, 1) == 0 || mpz_cmp(t, n1) == 0)
{
mpz_clears(d, n1, t, NULL);
return 1;
}
while (--s > 0)
{
mpz_mul(t, t, t);
mpz_mod(t, t, n);
if (mpz_cmp(t, n1) == 0)
{
mpz_clears(d, n1, t, NULL);
return 1;
}
}`
mpz_clears(d, n1, t, NULL);
return 0;
}
Let’s take a moment for a quick lesson in GMP. The datatype of big integers is given by mpz_t, where the mp is for multi-precision, z is for integer (from the German word Zahlen, for number), and _t is to indicate a type variable. All mpz_t variables must be initialized and cleared; in exchange for this effort, GMP takes care of all memory management automatically. The basic operations are given as mpz_add, mpz_mul, mpz_powm and the like, and they all return void, with the result given in the first argument (by analogy to an assignment, which puts the result on the left); the various division operators have their own naming conventions. Most of the operators have _ui variants in which the second operand (third argument) is a long unsigned integer instead of an mpz_t integer. Comparisons take two values and return a negative integer if the first is less than the second, a positive integer is the first is greater than the second, and 0 if the two are equal.
To determine whether a given integer is prime or composite we use 25 random integers, saving the GMP random state in a static variable that persists from one call of the function to the next. The algorithm expects an integer greater than 2, so that is our first test. Then we check that n is odd, and additionally that it is not divisible by 3, 5 or 7; those tests aren’t strictly part of the algorithm, but they eliminate about three-quarters of all positive integers, and if they determine the compositeness of n, they are much cheaper than the full Miller-Rabin test. Finally, if we don’t yet have an answer, we proceed with the full algorithm with k counting down to 0.
int is_prime(mpz_t n)
{
static gmp_randstate_t gmpRandState;
static int is_seeded = 0;
`if (! is_seeded)
{
gmp_randinit_default(gmpRandState);
gmp_randseed_ui(gmpRandState, time(NULL));
is_seeded = 1;
}
mpz_t a, n3, t;
mpz_inits(a, n3, t, NULL);
mpz_sub_ui(n3, n, 3);
int i;
int k = 25;
long unsigned ps[] = { 2, 3, 5, 7 };
if (mpz_cmp_ui(n, 2) < 0)
{
mpz_clears(a, n3, t, NULL);
return 0;
}
for (i = 0; i < sizeof(ps) / sizeof(long unsigned); i++)
{
mpz_mod_ui(t, n, ps[i]);
if (mpz_cmp_ui(t, 0) == 0)
{
mpz_clears(a, n3, t, NULL);
return mpz_cmp_ui(n, ps[i]) == 0;
}
}
while (k > 0)
{
mpz_urandomm(a, gmpRandState, n3);
mpz_add_ui(a, a, 2);
if (! is_spsp(n, a))
{
mpz_clears(a, n3, t, NULL);
return 0;
}
k -= 1;
}`
mpz_clears(a, n3, t, NULL);
return 1;
}
The default GMP random number generator is the Mersenne Twister, which has good randomness properties and a very long period; we initialize the internal state of the random number generator with the current time (number of seconds since the epoch). The function mpz_urandomm returns in its first argument a uniformly-distributed pseudo-random non-negative integer less than its third argument, using and resetting the internal state of the random number generator in its second argument. GMP provides our is_prime function under the name mpz_probab_prime_p, but we give our own implementation anyway, so you can see how it is done.
There are two functions that implement integer factorization by pollard rho: rho_factor finds a single factor, and rho_factors performs the complete factorization. The rho_factor function assumes that n is odd and composite; t is the tortoise, h is the hare, d is the greatest common divisor, and r is a temporary working variable holding the difference between t and h. The function keeps cycling until it finds a prime factor, calling itself recursively with the next greater c if it reaches a cycle or finds a composite factor.
void rho_factor(mpz_t f, mpz_t n, long long unsigned c)
{
mpz_t t, h, d, r;
`mpz_init_set_ui(t, 2);
mpz_init_set_ui(h, 2);
mpz_init_set_ui(d, 1);
mpz_init_set_ui(r, 0);
while (mpz_cmp_si(d, 1) == 0)
{
mpz_mul(t, t, t);
mpz_add_ui(t, t, c);
mpz_mod(t, t, n);
mpz_mul(h, h, h);
mpz_add_ui(h, h, c);
mpz_mod(h, h, n);
mpz_mul(h, h, h);
mpz_add_ui(h, h, c);
mpz_mod(h, h, n);
mpz_sub(r, t, h);
mpz_gcd(d, r, n);
}
if (mpz_cmp(d, n) == 0) / cycle /
{
rho_factor(f, n, c+1);
}
else if (mpz_probab_prime_p(d, 25)) / success /
{
mpz_set(f, d);
}
else / found composite factor /
{
rho_factor(f, d, c+1);
}`
mpz_clears(t, h, d, r, NULL);
}
The rho-factors function extracts factors of 2 in the first while, then calls rho-factor repeatedly in the second while until the remaining cofactor is prime. Rho-factors returns the list of factors in its first argument, like all the GMP functions.
void rho_factors(List fs, mpz_t n)
{
`while (mpz_even_p(n))
{
mpz_t f = malloc(sizeof(f));
mpz_init_set_ui(f, 2);
fs = insert(f, fs);
mpz_divexact_ui(n, n, 2);
}
if (mpz_cmp_ui(n, 1) == 0) return;
while (! (mpz_probab_prime_p(n, 25)))
{
mpz_t f = malloc(sizeof(f));
mpz_init_set_ui(f, 0);
rho_factor(f, n, 1);
fs = insert_in_order(f, fs);
mpz_divexact(n, n, f);
}`
fs = insert_in_order(n, fs);
}
We demonstrate the functions shown above with this main function:
int main(int argc, char argv[])
{
mpz_t n;
mpz_init(n);
List ps = NULL;
List fs = NULL;
`ps = primes(100); / 2 3 5 7 11 13 17 19 23 29 31 37 41 /
while (ps != NULL) / 43 47 53 59 61 67 71 73 79 83 89 97 /
{
printf("%ld%s", (long) ps->data,
(ps->next == NULL) ? "\n" : " ");
ps = ps->next;
}
printf("%d\n", length(primes(1000000))); / 78498 /
printf("%d\n", td_prime(600851475143LL)); / composite /
fs = td_factors(600851475143LL); / 71 839 1471 6857 /
while (fs != NULL)
{
printf("%llu%s", (unsigned long long int) fs->data,
(fs->next == NULL) ? "\n" : " ");
fs = fs->next;
}
mpz_t a;
mpz_init(a);
mpz_set_str(n, "2047", 10);
mpz_set_str(a, "2", 10);
printf("%d\n", is_spsp(n, a)); / pseudo-prime /
mpz_set_str(n, "600851475143", 10); / composite /
printf("%d\n", is_prime(n));
mpz_set_str(n, "2305843009213693951", 10); / prime /
printf("%d\n", is_prime(n));`
mpz_set_str(n, "600851475143", 10);
rho_factors(&fs, n); / 71 839 1471 6857 /
while (fs != NULL) {
printf("%s%s", mpz_get_str(NULL, 10, fs->data),
(fs->next == NULL) ? "\n" : " ");
fs = fs->next;
}
}
To compile the program, say gcc prime.c -lgmp -o prime , and to run the program say ./prime. If you get any warnings about the cast to void you can safely ignore them, as it is always permissible to cast to void. Here is the output from the program:
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
78498
0
71 839 1471 6857
1
0
1
71 839 1471 6857
You can see the program assembled at but you won’t be able to run it because the environment doesn’t provide the GMP library. An abbreviated version of the program that provides only the Sieve of Eratosthenes and the two trial-division algorithms that use native integers is available at
[ Return to top ]
Appendix: Haskell
Haskell is the classic purely functional language, far different from C. We begin our look at Haskell by examining a function that is frequently cited as the Sieve of Eratosthenes. Many texts include a definition something like this:
primes = sieve [2..]
sieve (p:xs) p : sieve [x | x <- xs, xmodp > 0]
But this is not the Sieve of Eratosthenes because it is based on division (the mod operator) rather than addition. It will quickly become slow as the primes grow larger; try, for instance, to extract a list of the primes less than a hundred thousand. Unfortunately, a proper implementation of the Sieve of Eratosthenes is a little bit ugly, since Haskell and arrays don’t easily mix.
Most Haskell programs begin by importing various functions from Haskell’s Standard Libraries, and ours is no exception; all imports must appear before any executable code. ST is the Haskell state-transformer monad, which provides mutable data structures, including Control.Monad.ST and Data.Array.ST. Control.Monad provides imperative-style for and when, and Data.Array.Unboxed provides arrays that store data directly, rather than with a pointer, as long as the data is a suitable type; we will be using arrays of booleans, which are suitable. Finally, Data.List provides a sort function for use by Pollard’s rho algorithm.
import Control.Monad (forM_, when)
import Control.Monad.ST
import Data.Array.ST
import Data.Array.Unboxed
import Data.List (sort)
The Sieve of Eratosthenes is implemented using exactly the same algorithm as all the other languages, though it looks somewhat foreign to imperative-trained eyes. Functions in Haskell optionally begin with a declaration of the type of the function, and we will include one in each of our functions. Thus, sieve :: Int -> UArray Int Bool declares an object named sieve that has type (the double colon) that is a function (the arrow) that takes a value of type Int and returns a value of type UArray Int Bool. Int is some fixed-size integer based on the native machine type. UArray is an unboxed array; Int is the type of its indices and Bool is the type of its values. Note that typenames always begin with a capital letter, as opposed to simple variables or function names that begin with a lower-case letter.
The body of the sieve function is fairly atypical of Haskell code, due to its use of arrays. The first line, runSTUArray $ do (that parses as run, ST for the state-transformer monad, UArray for the unboxed array), sets up the array processing; the array is initialized with indices from 0 to m−1, with all m values True, and assigned to variable bits. An expression like forM_ [0..x] $ \i do would be rendered in C as for (i=0; i<=x; i++); the expression [0 .. x] expands to 0, 1, … x, and is evaluated lazily, as if by a list generator, so the whole list is never reified all at once. Functions readArray and writeArray fetch and store elements of an array. Variable isPrime is assigned either True or False, depending on the value of the element of the bits array with value i. In the inner loop, iteration starts at 2ii+6i+3, and each iteration steps by 2i+3, which is the difference between the first and second elements of the list, continuing until j is greater than m−1.
sieve :: Int -> UArray Int Bool
sieve n = runSTUArray $ do
let m = (n-1)div2
r = floor . sqrt $ fromIntegral n
bits <- newArray (0, m-1) True
forM_ [0 .. rdiv2 - 1] $ \i -> do
isPrime <- readArray bits i
when isPrime $ do
forM_ [2ii+6i+3, 2ii+8i+6 .. (m-1)] $ \j -> do
writeArray bits j False
return bits
The primes function is simple; assocs collects the elements of the bits in order, paired with their index, and those that are True are included in the out-going list of primes. The type signature indicates that the function takes an Int and returns a list of Int values, as indicated by the square brackets. The overall structure of the function is a list which has 2 as its head joined by a colon : to a list comprehension between square brackets [ … ]. The list comprehension has two parts. The expression 2i+3 before the vertical bar | defines the elements of the output list. The generator after the bar assigns to the tuple all of the elements returned by the assocs function and keeps only those where the second element of the tuple is True, binding the first element of the tuple to the variable i that is used in the result expression.
primes :: Int -> [Int]
primes n = 2 : [2i+3 | (i, True) <- assocs $ sieve n]
It is simple to test primality by trial division because Haskell offers a simple way of generating the list of 2 followed by odd numbers. The colon operator, pronounced “cons,” is the list constructor. The expression [3,5..] is a list constructor (anything surrounded by square brackets is a list) with 3 as its first element, 5 as its second element, and so on in an arithmetic progression that increases by 5 − 3 = 2 at each step. The .. operator at the end of the list expression signifies that the list goes on forever; if there is a value after the .. operator, that is the ending value included in list.
tdPrime :: Int -> Bool
tdPrime n = prime (2:[3,5..])
where prime (d:ds)
| n < d d = True
| nmodd == 0 = False
| otherwise = prime ds
Factorization by trial division is expressed more simply in Haskell than in the other languages because Haskell provides an easy way to build the list of trial divisors: the expression 2:[3,5..] is an infinite list generator that returns 2 followed by the odd integers. The guard expressions of the local facts function makes tdFactors very easy to read. Note that we used a where clause, but could equally have used a let … in; in this case the choice is a matter of personal preference, though there are other situations where one or the other is required.
tdFactors :: Int -> [Int]
tdFactors n = facts n (2:[3,5..])
where facts n (f:fs)
| n < f f = [n]
| nmodf == 0 = f : facts (ndivf) (f:fs)
| otherwise = facts n fs
As in C, we have gone as far as we can using native integers, and we’ll switch at this point to big integers; note that Haskell has no long integers, so it’s more restrictive than C. The switch is simpler for Haskell than for C, since big integers are provided directly in the language, in the Integer datatype. For the Miller-Rabin primality test, we first need to write the function to perform modular exponentiation, since Haskell doesn’t provide one in any of its standard libraries:
powmod :: Integer -> Integer -> Integer -> Integer
powmod b e m =
let times p q = (pq)modm
pow b e x
| e == 0 = x
| even e = pow (times b b) (ediv2) x
| otherwise = pow (times b b) (ediv2) (times b x)
in pow b e 1
This function is rather more typical of Haskell than the sieve function that used arrays. The signature indicates that the function takes three Integer values and returns an Integer value. The function is written as if it is three functions because all functions in Haskell are curried, so powmod is actually a function that takes an integer b and returns a function that takes an integer e that returns a function that takes an integer m and returns an integer; thus, it is only a colloquialism, and frankly wrong, to say that powmod is a function that takes three integers and returns an integer. The let … in … defines local values. Local function times performs modular multiplication mod m. Local function pow has three definition, each with a guard (the predicate between the vertical bar | and the equal sign =); the expression corresponding to the first matching guard predicate is calculated and returned as the value of the function. Here, the first guard expression checks for termination and the other two expressions call the pow function recursively.
The strong pseudo-prime test is implemented by three local functions in the isSpsp function: getDandS extracts the powers of 2 from n−1, spsp takes the tuple returned by getDandS as input and performs the easy-return test, and doSpsp computes and tests the powers of a. Note that mod and div are curried prefix functions; the back-quotes turn them into binary infix functions. Note also that if is an expression in Haskell, as opposed to a statement in imperative languages, which means that it returns a value instead of controlling program flow; thus there may be no else-less if, and both consequents of the if must have the same type.
isSpsp :: Integer -> Integer -> Bool
isSpsp n a =
let getDandS d s =
if even d then getDandS (ddiv2) (s+1) else (d, s)
spsp (d, s) =
let t = powmod a d n
in if t == 1 then True else doSpsp t s
doSpsp t s
| s == 0 = False
| t == (n-1) = True
| otherwise = doSpsp ((tt)modn) (s-1)
in spsp $ getDandS (n-1) 0
Haskell makes it difficult to work with random numbers (it’s possible, though inconvenient, in the same way that arrays were inconvenient in the Sieve of Eratosthenes) because they require a state to be maintained from one call to the next, so we use the primes less than a hundred as the bases for the Miller-Rabin primality test.
isPrime :: Integer -> Bool
isPrime n =
let ps = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]
in nelemps || all (isSpsp n) ps
The rhoFactor function finds a single factor by the rho method. Function f is the random-number generator. Function fact implements the tortoise-and-hare loop recursively; the computations in the where clause are done prior to the guard expressions in the body of the function. The rhoFactor function is called recursively if the loop falls into a cycle (the d == n clause) or finds a composite factor (the otherwise clause).
rhoFactor :: Integer -> Integer -> Integer
rhoFactor n c =
let f x = (xx+c)modn
fact t h
| d == 1 = fact t' h'
| d == n = rhoFactor n (c+1)
| isPrime d = d
| otherwise = rhoFactor d (c+1)
where
t' = f t
h' = f (f h)
d = gcd (t' - h') n
in fact 2 2
Function rhoFactors calls rhoFactor repeatedly until it completes the factorization of n. The first two clauses extract factors of 2, the third clause tests primality of a remaining cofactor, and the fourth clause adjusts n and the list of factors after calling rhoFactor. Note that we solved the problem of factoring a perfect power of 2 differently than in the C version of the program, stopping as soon as n is reduced to 2.
rhoFactors :: Integer -> [Integer]
rhoFactors n =
let facts n
| n == 2 =
| even n = 2 : facts (ndiv2)
| isPrime n = [n]
| otherwise = let f = rhoFactor n 1
in f : facts (ndivf)
in sort $ facts n
A main program that exercises the functions defined above is shown below. To compile the program with the GHC compiler, assuming it is stored in file primes.hs, say ghc -o prime prime.hs, and to run the program say ./prime.
main = do
print $ primes 100
print $ length $ primes 1000000
print $ tdPrime 716151937
print $ tdFactors 8051
print $ powmod 437 13 1741
print $ isSpsp 2047 2
print $ isPrime 600851475143
print $ isPrime 2305843009213693951
print $ rhoFactors 600851475143
The output is the same as the C version of the program, except for the change in the input to tdFactors. You can run the program at
[ Return to top ]
Appendix: Java
Java is an object-oriented language, widely used, with a very large collection of libraries. Like Haskell, Java provides big integers, linked lists and bit arrays natively, so we can quickly jump in to the coding. The functions are shown below, but we leave it to you to package them into classes as you wish; in the sample code we put all the functions into class Main. We will be more careful here than in the two prior versions to ensure that we validate all the input arguments. We begin with the Sieve of Eratosthenes, which we limit to ints, but if you prefer a larger data type you are free to change it.
public static LinkedList primes(int n)
{
if (n < 2)
{
throw new IllegalArgumentException("must be greater than one");
}
`int m = (n-1) / 2;
BitSet b = new BitSet(m);
b.set(0, b.size(), true);
int i = 0;
int p = 3;
LinkedList ps = new LinkedList();
ps.add(2);
while (p p < n)
{
if (b.get(i))
{
ps.add(p);
int j = 2ii + 6i + 3;
while (j < m)
{
b.clear(j);
j = j + 2i + 3;
}
}
i += 1; p += 2;
}
while (i < m)
{
if (b.get(i))
{
ps.add(p);
}
i += 1; p += 2;
}`
return ps;
}
We used the built-in exception IllegalArgumentException instead of creating our own exception; it’s easier, and just as clear. We also used the built-in data types BitSet and LinkedList; indeed, it is one of the benefits of programming in Java that the standard libraries provide so much useful code.
Another of the libraries that Java provides is the BigInteger library, and we switch from normal integers to BigInteger for the rest of our functions; int is sufficient for the Sieve of Eratosthenes, because sieving with a large n produces too much output to be useful, but for the other functions BigInteger is definitely useful. The tdPrime function validates its input in the first if, checks for even numbers in the second if statement, and checks for odd divisors in the body of the while.
`public static Boolean tdPrime(BigInteger n)
{
BigInteger two = BigInteger.valueOf(2);
if (n.compareTo(two) < 0)
{
throw new IllegalArgumentException("must be greater than one");
}`
`if (n.mod(two).equals(BigInteger.ZERO))
{
return n.equals(two);
}
BigInteger d = BigInteger.valueOf(3);
while (d.multiply(d).compareTo(n) <= 0)
{
if (n.mod(d).equals(BigInteger.ZERO))
{
return false;
}
d = d.add(two);
}`
return true;
}
The tdFactors function domain-checks the input, removes factors of two, and, if the remaining cofactor is not 1, begins a loop over the odd numbers starting from 3, trying each odd number in turn until it finds factors and the remaining cofactor is greater than the square of the current factor. As with the GMP functions in C, the messiness of doing arithmetic by calling functions hides the simplicity of the algorithm.
public static LinkedList tdFactors(BigInteger n)
{
BigInteger two = BigInteger.valueOf(2);
LinkedList fs = new LinkedList();
`if (n.compareTo(two) < 0)
{
throw new IllegalArgumentException("must be greater than one");
}
while (n.mod(two).equals(BigInteger.ZERO))
{
fs.add(two);
n = n.divide(two);
}
if (n.compareTo(BigInteger.ONE) > 0)
{
BigInteger f = BigInteger.valueOf(3);
while (f.multiply(f).compareTo(n) <= 0)
{
if (n.mod(f).equals(BigInteger.ZERO))
{
fs.add(f);
n = n.divide(f);
}
else
{
f = f.add(two);
}
}
fs.add(n);
}`
return fs;
}
It is annoying that the add method is overloaded, with the same method name referring to the addition of two BigIntegers when used as f.add(two) and to the insertion of an item in a LinkedList when used as fs.add(f). Such usage may not be confusing to the compiler, because it keeps track of the types of all variables, but it can be confusing to the programmer who writes and reads the code and has to make sense of it.
The isSpsp function computes d and s in the first while loop, checks for an early termination in the if, then counts down s in the second while loop. Note that the early termination test is different in the Java version than the Haskell version; the Haskell version separates the early termination test from the n−1 tests, but the Java version combines the early termination test with the first loop of the n−1 tests. Both versions of the function get the right answer; the choice is based on the convenience of the programmer.
private static Boolean isSpsp(BigInteger n, BigInteger a)
{
BigInteger two = BigInteger.valueOf(2);
BigInteger n1 = n.subtract(BigInteger.ONE);
BigInteger d = n1;
int s = 0;
`while (d.mod(two).equals(BigInteger.ZERO))
{
d = d.divide(two);
s += 1;
}
BigInteger t = a.modPow(d, n);
if (t.equals(BigInteger.ONE) || t.equals(n1))
{
return true;
}
while (--s > 0)
{
t = t.multiply(t).mod(n);
if (t.equals(n1))
{
return true;
}
}`
return false;
}
After using a predefined list of bases in the Haskell version of the function, we’re back to using random bases in the isPrime function. The two if tests check the input domain and exit quickly if the input is even, then the while loop performs 25 strong pseudo-prime tests.
public static Boolean isPrime(BigInteger n)
{
Random r = new Random();
BigInteger two = BigInteger.valueOf(2);
BigInteger n3 = n.subtract(BigInteger.valueOf(3));
BigInteger a;
int k = 25;
`if (n.compareTo(two) < 0)
{
return false;
}
if (n.mod(two).equals(BigInteger.ZERO))
{
return n.equals(two);
}
while (k > 0)
{
a = new BigInteger(n.bitLength(), r).add(two);
while (a.compareTo(n) >= 0)
{
a = new BigInteger(n.bitLength(), r).add(two);
}
if (! isSpsp(n, a))
{
return false;
}
k -= 1;
}`
return true;
}
Note that Java’s BigInteger library includes a function isProbablePrime that performs this computation in exactly the same way.
The rhoFactor function races the tortoise and hare until the gcd is greater than 1, then the if–else chain either returns a prime factor or retries the factorization with a different random function.
private static BigInteger rhoFactor(BigInteger n, BigInteger c)
{
BigInteger t = BigInteger.valueOf(2);
BigInteger h = BigInteger.valueOf(2);
BigInteger d = BigInteger.ONE;
while (d.equals(BigInteger.ONE))
{
t = t.multiply(t).add(c).mod(n);
h = h.multiply(h).add(c).mod(n);
h = h.multiply(h).add(c).mod(n);
d = t.subtract(h).gcd(n);
}
if (d.equals(n)) / cycle /
{
return rhoFactor(n, c.add(BigInteger.ONE));
}
else if (isPrime(d)) / success /
{
return d;
}
else / found composite factor /
{
return rhoFactor(d, c.add(BigInteger.ONE));
}
}
The rhoFactors function first validates its input, then extracts factors of 2 in the first while, and, unless the input is a power of 2, calls rhoFactor repeatedly in the second while until the remaining cofactor is prime, sorting the list of factors before returning it. The built-in isProbablePrime function is called rather than the one we defined above.
public static LinkedList rhoFactors(BigInteger n)
{
BigInteger f;
BigInteger two = BigInteger.valueOf(2);
LinkedList fs = new LinkedList();
`if (n.compareTo(two) < 0)
{
return fs;
}
while (n.mod(two).equals(BigInteger.ZERO))
{
n = n.divide(two);
fs.add(two);
}
if (n.equals(BigInteger.ONE))
{
return fs;
}
while (! n.isProbablePrime(25))
{
f = rhoFactor(n, BigInteger.ONE);
n = n.divide(f);
fs.add(f);
}`
fs.add(n);
Collections.sort(fs);
return fs;
}
To show examples of the use of these functions, we have to create a complete program with all of its imports and a class declaration. The program shown below is decidedly simple-minded, sufficient only to show a few examples; you will surely want to do arrange the class differently in your own programs. For sake of brevity, the function bodies are elided below.
`import java.util.LinkedList;
import java.util.BitSet;
import java.math.BigInteger;
import java.lang.Exception;
import java.lang.Boolean;
class Main {
public static LinkedList primes(int n) { ... }
public static Boolean tdPrime(BigInteger n) { ... }
public static LinkedList tdFactors(BigInteger n) { ... }
private static Boolean isSpsp(BigInteger n, BigInteger a) { ... }
public static Boolean isPrime(BigInteger n) { ... }
private static Boolean rhoFactor(BigInteger n, BigInteger c) { ... }
public static LinkedList rhoFactors(BigInteger n) { ... }
public static void main (String[] args)
{
System.out.println(primes(100));
System.out.println(primes(1000000).size());
System.out.println(tdPrime(new BigInteger("600851475143")));
System.out.println(tdFactors(new BigInteger("600851475143")));
System.out.println(isPrime(new BigInteger("600851475143")));
System.out.println(isPrime(new BigInteger("2305843009213693951")));
System.out.println(rhoFactors(new BigInteger("600851475143")));
}
}`
Output from the program is the same as all the other implementations. You can run the program at
[ Return to top ]
Appendix: Python
Python is a commonly-used scripting language with a reputation of being easy to read and write and a mixed imperative/object-oriented flavor. We’ll take the opportunity with Python to extend the domain of integer factorization beyond the integers greater than 1 that mathematicians generally consider. Specifically, we’ll consider −1, 0 and 1 to be prime, so they factor as themselves, and we’ll factor negative numbers by adding −1 to the factors of the corresponding positive number. This isn’t entirely correct, but it isn’t entirely incorrect, either, and is actually useful in some cases. And we’re in good company; Wolfram|Alpha calculates factors the same way we do.
We begin, as we did with Haskell, with a one-liner that purports to be the Sieve of Eratosthenes:
print [x for x in range(2,100) if not [y for y in range(2, int(x0.5)+1) if x%y == 0]]
That prints the primes less than a hundred. But the expression if x%y == 0 at the end gives the game away: it’s really trial division (the % operator for modulo), so it’s not a sieve. Don’t be fooled by cute one-liners!
We begin with the Sieve of Eratosthenes. After validating the input, the first while loop collects the sieving primes and performs the sieving, and the second while loop collects the remaining primes that survived the sieve. Note that append adds an element after the target, unlike the function we wrote in C that inserts an element before the target.
def primes(n):
if type(n) != int and type(n) != long:
raise TypeError('must be integer')
if n < 2:
raise ValueError('must be greater than one')
m = (n-1) // 2
b = [True] m
i, p, ps = 0, 3,
while pp < n:
if b[i]:
ps.append(p)
j = 2ii + 6i + 3
while j < m:
b[j] = False
j = j + 2i + 3
i += 1; p += 2
while i < m:
if b[i]:
ps.append(p)
i += 1; p += 2
return ps
The next function, td_prime, has three possible return values: PRIME, COMPOSITE, and UNKNOWN if the limit is exceeded. We use True and False for PRIME and COMPOSITE, and raise an OverflowError exception if the limit is exceeded, requiring some effort for the user to trap the error and respond accordingly. After validating the input, the if checks even numbers and the while loop checks odd numbers. Python’s optional-argument syntax is simple and convenient; the argument limit is optional, and is given a default value if not specified.
def td_prime(n, limit=1000000):
if type(n) != int and type(n) != long:
raise TypeError('must be integer')
if n % 2 == 0:
return n == 2
d = 3
while d d <= n:
if limit < d:
raise OverflowError('limit exceeded')
if n % d == 0:
return False
d += 2
return True
The td_factors function has the same problem with the limit argument as td_prime, and we solve it the same way, using an OverflowError exception. The if becomes a while on the factors of 2, and the function collects factors instead of returning, but otherwise td_factors is very similar to td_prime.
def td_factors(n, limit=1000000):
if type(n) != int and type(n) != long:
raise TypeError('must be integer')
fs = []
while n % 2 == 0:
fs +=
n /= 2
if n == 1:
return fs
f = 3
while f f <= n:
if limit < f:
raise OverflowError('limit exceeded')
if n % f == 0:
fs += [f]
n /= f
else:
f += 2
return fs + [n]
The Miller-Rabin primality checker makes use of Python’s ability to nest functions to hide the strong pseudo-prime checker inside the is_prime function. It also uses the built-in modular exponentiation function; with two arguments, pow(b,e) computes be, but with three arguments, pow(b,e,m) computes be (mod m). Python offers random numbers, but we prefer to test a fixed set of bases.
def is_prime(n):
if type(n) != int and type(n) != long:
raise TypeError('must be integer')
if n < 2:
return False
ps = [2,3,5,7,11,13,17,19,23,29,31,37,41,
43,47,53,59,61,67,71,73,79,83,89,97]
def is_spsp(n, a):
d, s = n-1, 0
while d%2 == 0:
d /= 2; s += 1
t = pow(a,d,n)
if t == 1:
return True
while s > 0:
if t == n-1:
return True
t = (tt) % n
s -= 1
return False
if n in ps: return True
for p in ps:
if not is_spsp(n,p):
return False
return True
Like is_prime, the rho_factors function reduces namespace pollution by hiding local functions; notice the lambda, which is an alternate way of creating a local function. Again, as with Java, the + function is overloaded for both addition and list construction. The code is similar to Java, even if it looks quite different.
def rho_factors(n, limit=1000000):
if type(n) != int and type(n) != long:
raise TypeError('must be integer')
def gcd(a,b):
while b: a, b = b, a%b
return abs(a)
def rho_factor(n, c, limit):
f = lambda(x): (xx+c) % n
t, h, d = 2, 2, 1
while d == 1:
if limit == 0:
raise OverflowError('limit exceeded')
t = f(t); h = f(f(h)); d = gcd(t-h, n)
if d == n:
return rho_factor(n, c+1, limit)
if is_prime(d):
return d
return rho_factor(d, c+1, limit)
if -1 <= n <= 1: return [n]
if n < -1: return [-1] + rho_factors(-n, limit)
fs = []
while n % 2 == 0:
n = n // 2; fs = fs +
if n == 1: return fs
while not is_prime(n):
f = rho_factor(n, 1, limit)
n = n / f
fs = fs + [f]
return sorted(fs + [n])
We could import the gcd function from the fractions library, but instead we implement it ourselves because it gives us the chance to discuss this famous algorithm. Donald E. Knuth, in Volume 2, Section 4.5.2 of his book The Art of Computer Programming, calls this the “granddaddy” of all algorithms because it is the oldest nontrivial algorithm that has survived to the present day. The algorithm is commonly called the Euclidean algorithm because it was described in Book 7, Propositions 1 and 2 of Euclid’s Elements, but scholars believe the algorithm dates to about two hundred years before Euclid, sometime around 500 B.C. Knuth gives the entire history of the algorithm, and an extensive analysis of its time complexity, which is well worth your time. Euclid’s version of the algorithm worked by repeatedly subtracting the smaller amount from the larger until they are the same; the modern version of the algorithm replaces subtraction with division (the modulo operator).
Here are some sample calls to the functions defined above; the answers are the same as all the other implementations.
print primes(100)
print len(primes(1000000))
print td_prime(600851475143)
print td_factors(600851475143)
print is_prime(600851475143)
print is_prime(2305843009213693951)
print rho_factors(600851475143)
You can run the program at
[ Return to top ]
Appendix: Scheme
Scheme is primarily an academic language, useful for expressing algorithms in imperative, functional, and message-passing styles, with a fully-parenthesized prefix syntax derived from Lisp. Scheme provides big integers natively, and also lists, but has no bit arrays, so our implementation of the Sieve of Eratosthenes uses a vector of booleans, which uses eight bits per element instead of one but works perfectly well.
(define (primes n)
(if (or (not (integer? n)) (< n 2))
(error 'primes "must be integer greater than one")
(let ((len (quotient (- n 1) 2))
(bits (make-vector len #t)))
(let loop ((i 0) (p 3) (ps (list 2)))
(cond ((< n ( p p))
(do ((i i (+ i 1)) (p p (+ p 2))
(ps ps (if (vector-ref bits i) (cons p ps) ps)))
((= i len) (reverse ps))))
((vector-ref bits i)
(do ((j (+ ( 2 i i) ( 6 i) 3) (+ j p)))
((<= len j) (loop (+ i 1) (+ p 2) (cons p ps)))
(vector-set! bits j #f)))
(else (loop (+ i 1) (+ p 2) ps)))))))
Let’s take a moment for a quick lesson in Scheme. An expression like (let ((var1 value1) ...) body) establishes a local binding for each of several var/value pairs that is active in the body of the let; a let expression is the same, except that the bindings are executed left-to-right, and each binding is available to those that follow.
There are two looping constructs. The named-let variant of let, given by (let name ((var1 value1) ...) body), is like let, but additionally binds name to a function with arguments vark whose code is the body of the let, which executes loops when it is called recursively; by convention, the variable name is often called loop, but it is sometimes convenient to use other names.
The other looping construct is do, which is similar to the for of C. The form of the do loop is (do ((var1 value1 next1) ...) (done? ret-value) body ...). Each var/value/next triplet in the first do clause specifies a variable name, a value for the variable when the do is initialized, and an expression evaluated at each step of the do; there may be multiple var/value/next triplets, in which case each is executed simultaneously, rather like a comma operator in a C for statement. The done? predicate terminates the do loop when it becomes true; this is the opposite of a C for loop, which terminates when the condition becomes false. The return value is optional; if it is not given, the return value of the do loop is unspecified. The statements in the body of the do loop are optional, and are evaluated only for their side effects.
Scheme also provides two conditional constructs. The first is (if cond then else), which first evaluates the condition then evaluates one of the two succeeding clauses; like Haskell, an if is an expression, not a control-flow statement. Cond is similar to a nested set of if statements; each clause consists of a condition and body, the conditions are read in order until one is true, when the corresponding body is evaluated.
In the example above, len is the length of the bitarray, called m in the description of the algorithm, and bits is the bitarray itself, a vector of booleans. The cond has three clauses. The first clause is actually the termination clause of Step 5 and Step 6 that is executed last; the body-less do sweeps up the primes after sieving is complete, and the return value is the list of primes, which must be reversed because each newly-found prime is pushed to the front, not the back, of an accumulating list of primes. The second clause sifts each prime, as in Step 4; this do has a body, which clears the jth element of the bitarray, and the return value is an expression that calls the named-let recursively to advance to the next sieving prime. The else clause recurs when p is not prime.
Our td-prime? function follows the Scheme convention that predicates (functions that return a boolean) have names that end in a question mark. An error is signaled if limit is less than the smallest prime factor of n; this isn’t as convenient as raising an exception in Python, because standard Scheme has no way to trap the error, but most implementations provide some kind of error trapping.
(define (td-prime? n . args)
(if (or (not (integer? n)) (< n 2))
(error 'td-prime? "must be integer greater than one")
(let ((limit (if (pair? args) (car args) 1000000)))
(if (even? n) (= n 2)
(let loop ((d 3))
(cond ((< limit d)
(error 'td-prime? "limit exceeded"))
((< n ( d d)) #t)
((zero? (modulo n d)) #f)
(else (loop (+ d 2)))))))))
Td-factors is similar to td-prime?, except that the first if becomes a loop, on twos, and both loops collect the factors that they find instead of stopping on the first factor. Here is a case where the names twos and odds in the named-let provide documentation of the nature of the loop, making the function clearer to the reader.
(define (td-factors n . args)
(if (or (not (integer? n)) (< n 2))
(error 'td-factors "must be integer greater than one")
(let ((limit (if (pair? args) (car args) 1000000)))
(let twos ((n n) (fs '()))
(if (even? n)
(twos (/ n 2) (cons 2 fs))
(let odds ((n n) (d 3) (fs fs))
(cond ((< limit d)
(error 'td-factors "limit exceeded"))
((< n ( d d))
(reverse (cons n fs)))
((zero? (modulo n d))
(odds (/ n d) d (cons d fs)))
(else (odds n (+ d 2) fs)))))))))
We’ve been using lists but haven’t mentioned how they work. A list is either null or is a pair with an item in its car and another list in its cdr; the terms car and cdr are pre-historic. An item x is inserted at the front of a list xs by (cons x xs); the word cons is short for construct. Predicates (null? xs) and (pair? xs) distinguish empty lists from non-empty ones. The null list is represented as '(), and the order of items in a list is reversed by (reverse xs).
The function prime? that implements the Miller-Rabin primality checker illustrates two more features of Scheme. We have been introducing functions with the notation (define (name args …) body), but the alternate notation is (define name (lambda (args …) body)). We use it here because we want variable seed to persist from one invocation of prime? to the next. Since the let is inside the define but outside the lambda, the variable bound by the let retains its value from one call of the function to the next, just like static variables in some programming languages. Thus, prime? is a closure, not just a function, because it encloses the seed variable. And while we’re talking about define, even though it doesn’t apply here, we have been using the dot-notation for some of our argument lists; a construct like (define (f args ... . rest) body) provides a variable-arity argument list, with all arguments after the dot collected into a list rest.
The other Scheme feature is internal-define, which is used to provide local functions that don’t pollute the global namespace. We define three local functions, rand that returns random numbers, expm that performs modular exponentiation (the name is a variant of expt that Scheme provides for the normal powering function) and spsp? that checks if a is a witness to the compositeness of n. And we’re not done; the internal definition expm has its own internal definition times for modular multiplication. An internal-define must appear immediately after another define, a lambda, or a let.
(define prime?
(let ((seed 3141592654))
(lambda (n)
(define (rand)
(set! seed (modulo (+ ( 69069 seed) 1234567) 4294967296))
(+ (quotient ( seed (- n 2)) 4294967296) 2))
(define (expm b e m)
(define (times x y) (modulo ( x y) m))
(let loop ((b b) (e e) (r 1))
(if (zero? e) r
(loop (times b b) (quotient e 2)
(if (odd? e) (times b r) r)))))
(define (spsp? n a)
(do ((d (- n 1) (/ d 2)) (s 0 (+ s 1)))
((odd? d)
(let ((t (expm a d n)))
(if (or (= t 1) (= t (- n 1))) #t
(do ((s (- s 1) (- s 1))
(t (expm t 2 n) (expm t 2 n)))
((or (zero? s) (= t (- n 1)))
(positive? s))))))))
(if (not (integer? n))
(error 'prime? "must be integer")
(if (< n 2) #f
(do ((a (rand) (rand)) (k 25 (- k 1)))
((or (zero? k) (not (spsp? n a)))
(zero? k))))))))
In the prime? function we define our own random number generator, since standard Scheme doesn’t provide one; the static variable seed maintains the current state of the random number generator, which is of a type known as a linear-congruential generator. The multiplier 69069 is due to Knuth. Note that the seed is reset with each call to rand; the witness a is set to the range 1 to n, exclusive.
There are three do loops in the function. The first, the outer do in spsp?, binds two variables, d and s, and iterates until d is odd, performing Step 2 of Algorithm 4.A. The inner do in spsp? binds the two variables s and t and implements the loop of Step 4 and Step 5 of Algorithm 4.A, performing the modular squaring and terminating when s is zero or t is n−1. The result of the do-loop is computed by the predicate (positive? s), which is #t if t ≡ n−1 (mod n) and #f when s reaches 0 without finding t ≡ n−1 (mod n). The do in the main body of the function uses the same idiom of having two terminating conditions and using the finishing predicate to differentiate the two.
Our final function implements integer factorization by Pollard’s rho algorithm. The two internal definitions are cons<, which inserts an item into a list in ascending order instead of at the front, and rho, which implements the rho algorithm. The cons< function turns a vice into a virtue; since standard Scheme lacks a sort function, we insert the factors in order as we find them, instead of writing a sort function to sort them at the end, which gives the virtue of simpler code and is probably faster, given the short length of most lists of factors. The body of the function does error checking, extracts factors of 2, and assembles the complete factorization.
(define (rho-factors n . args)
(define (cons< x xs)
(cond ((null? xs) (list x))
((< x (car xs)) (cons x xs))
(else (cons (car xs) (cons< x (cdr xs))))))
(define (rho n limit)
(let loop ((t 2) (h 2) (d 1) (c 1) (limit limit))
(define (f x) (modulo (+ ( x x) c) n))
(cond ((zero? limit) (error 'rho-factors "limit exceeded"))
((= d 1) (let ((t (f t)) (h (f (f h))))
(loop t h (gcd (- t h) n) c (- limit 1))))
((= d n) (loop 2 2 1 (+ c 1) (- limit 1)))
((prime? d) d)
(else (rho d (- limit 1))))))
(if (not (integer? n))
(error 'rho-factors "must be integer")
(let ((limit (if (pair? args) (car args) 1000)))
(cond ((<= -1 n 1) (list n))
((negative? n) (cons -1 (rho-factors (- n) limit)))
((even? n)
(if (= n 2) (list 2)
(cons 2 (rho-factors (/ n 2) limit))))
(else (let loop ((n n) (fs '()))
(if (prime? n)
(cons< n fs)
(let ((f (rho n limit)))
(loop (/ n f) (cons< f fs))))))))))
Here are some examples:
`> (primes 100)
(2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73
79 83 89 97)
(length (primes 1000000))
78498
(td-prime? 600851475143)
f
(td-factors 600851475143)
(71 839 1471 6857)
(prime? 600851475143)
f
(prime? 2305843009213693951)
t
(rho-factors 600851475143)
(71 839 1471 6857)`
You can run the program at
As your reward for reading this far, we discuss a fast implementation of Pollard’s rho algorithm. We make two changes, one algorithmic and one code-tuning.
The first change is algorithmic: we replace Floyd’s turtle-and-hare cycle-finding algorithm, which Pollard used in his original version of the rho algorithm, with Brent’s powers-of-two cycle-finding algorithm. Each time the step-counter i is a power of two, the value of xi is saved; if a subsequent xj = xi is found before j = 2i, a cycle has been identified. Brent’s cycle-finding algorithm requires only one modular multiplication per step, instead of the three modular multiplications required by Floyd’s cycle-finding algorithm, so even though Brent’s method typically requires more steps that Floyd’s method, in practice the number of modular multiplications is generally about a quarter less than Floyd’s method, giving a welcome speed-up to Pollard’s factoring algorithm.
For instance, consider the sequence 1, 2, 3, 4, 5, 6, 3, 4, 5, 6, …. Initially j = 1, xj = 1, q = 2j = 2 and the saved x = 1. Then j = 2, xj = 2, q is reset to 4 and the saved x is reset to 2. Then j = 3, then j = 4, and q is reset to 8 and the saved x is reset to 4. The iteration continues until j = 8 and xj = 4, which equals the saved x, thus identifying the cycle.
The second change is code-tuning: we replace the gcd at each step with a gcd that is calculated only periodically, performing instead a modular multiplication at each step, which is much faster than a gcd calculation. This is done by taking the product of all the |xi+1 − xi| modulo n for several steps, then taking a gcd of the product at the end of the several steps. If the gcd is 1, then all the intermediate gcd calculations were also 1. If the gcd is prime, it is a factor of n. If the gcd is composite (including the case where the gcd is equal to n) it is necessary to retreat to the saved value of x from the prior gcd calculation and proceed step-by-step through the gcd calculations. The number of steps between successive gcd calculations varies with the size of n (bigger n means less frequent gcd calculations) and the number of trial divisions performed before starting the rho algorithm (more trial divisions means less frequent gcd calculations); values between 10 and 250 may be appropriate depending on the circumstances.
(define (brent n c limit)
(define (f y) (modulo (+ ( y y) c) n))
(define (g p x y) (modulo ( p (abs (- x y))) n))
(let loop1 ((x 2) (y (+ 4 c)) (z (+ 4 c)) (j 1) (q 2) (p 1))
(if (= j limit) (error 'brent "timeout")
(if (= x y) (brent n (+ c 1) (- limit j)) ; cycle
(if (= j q) (let ((t (f y))) (loop1 y (f y) z (+ j 1) ( q 2) (g p y t)))
(if (positive? (modulo j 25)) (loop1 x (f y) z (+ j 1) q (g p x y))
(let ((d (gcd p n)))
(if (= d 1) (loop1 x (f y) y (+ j 1) q (g p x y))
(if (< 1 d n) d ; factor
(let loop2 ((k 1) (z (f z)))
(if (= k 25) (brent n (+ c 1) (- limit j))
(let ((d (gcd (- z x) n)))
(if (= d 1) (loop2 (+ k 1) (f z))
(if (= d n) (brent n (+ c 1) (- limit j))
d))))))))))))))
Our function keeps three values of the random-number sequence: x is the running value, y is the value at the last gcd calculation, and z is the value at the last power of two. The main loop also defines j as the step counter, q as the next power of two, and p as the current product for the short-circuit gcd calculation. Function f delivers the next value in the random-number sequence and function g accumulates the current product for the short-circuit gcd calculation. Loop1 is the main body of the function, and loop2 reruns the short-circuit gcd calculation when necessary; both call the function recursively if they find a cycle. We arbitrarily choose 25 as the number of steps between successive gcd calculations.
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Copyright © 2012 by Programming Praxis of Saint Louis, Missouri, USA. All rights reserved. This work is licensed under the Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License. To view a copy of this license, visit or send a letter to Creative Commons, 171 Second Street, Suite 300, San Francisco, California, 94105, USA.
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The author can be reached at programmingpraxis@gmail.com.
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One Response to “Programming with Prime Numbers”
How To Generate A List Of Prime Numbers In Python | Technology Life said
June 8, 2016 at 12:19 AM
[…] Programming with Prime Numbers | Programming Praxis – Prime numbers are those integers greater than one that are divisible only by themselves and one; an integer greater than one that is not prime is composite. […]
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15835 | https://m.fx361.com/news/2010/1231/29330674.html | 浅析中学物理量的正负号_参考网
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浅析中学物理量的正负号
2010-12-31 刘士勇
中学课程辅导·教师教育(上、下)订阅 2010年11期 收藏
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摘 要：本文简析了中学物理现量的正负号的作用，认识运用上的常见错误及正负号的处理方法。
关键词：物理；正负号；处理方法
中国分类号：G424 文献标识码：A 文章编号：1992-7711（2010）4-041 -01
数学中的正负号，学生很容易理解和掌握，而物理学中的正负号，学生往往对其含义理解不透，应用起来感到困难，也时常常出现差错。为此笔者就中学物理中正负号的含义、常见错误及正确用法进行了简要归纳。
一、 中学物理中正负号常见含义及作用
1.表示矢量的方向
物理量有矢量和标量之分，矢量运算遵循平行四边形定则，因此多个矢量的运算便显得复杂且难于理解。对于同一直线上的矢量，“+”号表示某矢量的方向与规定的正方向（或已默认的正方向）相同，这里的“+”一般省略不写；“－”号表示某矢量的方向与规定的正方向（或已默认的正方向相反，如果某矢量为零矢量，则无方向可言。此类“+”、“－”号只会出现在矢量性物理量中，如位移、速度、力、加速度、动量、电场强度、磁感应强度等矢量，以及这些矢量的变化量。
2.表示大小或高低
“+”号表示比“零”大，“+”号一般省略不写；“－”表示比“零”小，这里的“零”具有相对性。此类“+”“－”只出现在具有相对性的标量性物理量中，如温度、电势、重力势能、电势能、分子势能等标量性物理量。日常生活中我们常用摄氏温标表示物体的温度，温度可以为正，也可以为负，例如今天的气温为+18℃，又如今年冬天最低气温达-40℃。
3.表示物理学中对立的两个方面
物理中的规定“+”号（一般不写）可表示：某力做正功、正电荷、电流的流入，磁感线的穿入，以及某物理量的增加；“－”号可表示：克服某力做功（某力做负功）、负电荷、电流的流出，磁感线的穿出，以及某些物理量的减少等。这里的零表示没有。
二、对物理量的正负在认识上和运用上常见的错误
1.矢量方向性认识不足。矢量是有方向的物理量，许多实际问题中，命题者往往只提供矢量的大小，而不明确方向，学生在解答时经常往往只关注某一方向，而忽视了另一种方向可能性，造成解答不完整或错误。
例：一物体做匀变速直线运动，某时刻速度大小为4m/s，1秒后速度大小变为10m/s，在这1秒内该物体的加速度等于多少？
学生典型的不完整的解答如下：
解：a=ΔV/Δt=(10-4)/1=6m/s
此解就忽视了速度的矢量性，仅考虑了1秒后速度的一种方向可能，而遗漏了速度10m/s与初速度相反的可能性，造成解题不完整。
2.正负号理解的代数化。由于代数学知识的学习，在学生头脑中已形成“正数大于负数”这一思维定势，所以经常会出现如下错误。如对F1=50N，F2=－20N，F3=－90N这三个力的大小进行比较，有相当一部份学生会按照代数比数大小的思路解题，错误地认为F1>F2>F3。
同样这样的思维定势还出现在对物理学中“增量”的理解。物理学中常用“增量”表示物理量的变化，如速度增量，动能增量，内能增量等。但学生常常把“增”与“正”等同，错误地认为增量一定是正值。对于增量为负值的情况，思维上总是难以理解。
3.公式中的加减号与物理量的正负号混淆。在应用加速度计算公式a=(Vt-Vo)/t时，当VO为负值时，有的学生误认为公式中的减号就是该VO的负号，所以不再用负值代入，造成错误。
三、实际解题时正负号的处理方法
1.中学物理中，所应用公式有涉及加减计算的，一般都应带符号计算。如运动学中三个公式：Vt=Vo+at，X=Vot+1/2at2，Vt2－Vo2=2aX，牛顿第二定律、动能定理等等。
2.有些公式只有物理量的乘除运算，而不包含加减运算，运算结果的大小不受各量正负的影响，可以按绝对值带入运算。如E=F/q；f=qvB；F=BIL；W=qU；C=Q/U等等。结果的正负依据相应规则判断即可。
3.如果在具体解决问题时，碰到有些物理量的方向难以确定，也即难以确定其正负值。在这种情况下，可假设物理量为正值带入计算，最终以计算结果再作结论。例：轻杆AB长1m，两端各连接质量1Kg的小球，杆可绕距B端0.2m处的O端在竖直平面内转动，设A球转到最低点时速度为4m/s，则此时相干对O轴的作用力为多大？
分析：解题时必须先分析A、B两球的所力情况，其中B球可能受到杆的拉力，也可能受到杆的支持力，在这种情况下，不妨设B球受到杆的拉力，即与重力同方向，然后按规定符号带入运算，这并不影响解题结果。
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15836 | https://cstheory.stackexchange.com/questions/50478/is-this-a-novel-technique-for-determining-whether-or-not-two-rotated-rectangles | ds.algorithms - Is this a novel technique for determining whether or not two rotated rectangles collide? - Theoretical Computer Science Stack Exchange
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Is this a novel technique for determining whether or not two rotated rectangles collide?
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I was trying to determine whether or not two rectangles rotated around their centers were colliding and randomly thought to try the following algorithm:
Rotate both rectangles by the negative rotation angle value of the first rectangle, such that the first rectangle becomes an AABB (axis-aligned bounding box) rectangle. Now calculate the bounding box of the second rotated rectangle and use the simple "are two axis aligned rectangles colliding" function to see if those two rectangles are colliding:
function rectanglesCollide(r1X, r1Y, r1W, r1H, r2X, r2Y, r2W, r2H) {
return r1X < r2X + r2W && r1X + r1W > r2X && r1Y < r2Y + r2H && r1Y + r1H > r2Y;
}
Now do the same thing but in reverse. That is, rotate both rectangles by the negative rotation angle of the second rectangle, such that now the second rectangle becomes axis-aligned. Then calculate the bounding box for the first rectangle and use the same rectanglesCollide function to determine if those two rectangles are colliding.
From what I can tell, if both of those checks are true, then the rotated rectangles are colliding. I spent a couple days trying to figure out the math and finally got it working! Here's the function for it:
```
function rotatedRectanglesCollide(r1X, r1Y, r1W, r1H, r1A, r2X, r2Y, r2W, r2H, r2A) {
let r1CX = r1X + (r1W / 2);
let r1CY = r1Y + (r1H / 2);
let r2CX = r2X + (r2W / 2);
let r2CY = r2Y + (r2H / 2);
let cosR1A = Math.cos(r1A);
let sinR1A = Math.sin(r1A);
let cosR2A = Math.cos(r2A);
let sinR2A = Math.sin(r2A);
let r1RX = cosR2A (r1CX - r2CX) + sinR2A (r1CY - r2CY) + r2CX;
let r1RY = -sinR2A (r1CX - r2CX) + cosR2A (r1CY - r2CY) + r2CY;
let r2RX = cosR1A (r2CX - r1CX) + sinR1A (r2CY - r1CY) + r1CX;
let r2RY = -sinR1A (r2CX - r1CX) + cosR1A (r2CY - r1CY) + r1CY;
let cosR1AR2A = Math.abs(cosR1A cosR2A + sinR1A sinR2A);
let sinR1AR2A = Math.abs(sinR1A cosR2A - cosR1A sinR2A);
let cosR2AR1A = Math.abs(cosR2A cosR1A + sinR2A sinR1A);
let sinR2AR1A = Math.abs(sinR2A cosR1A - cosR2A sinR1A);
let r1BBH = r1W sinR1AR2A + r1H cosR1AR2A;
let r1BBW = r1W cosR1AR2A + r1H sinR1AR2A;
let r1BBX = r1RX - r1BBW / 2;
let r1BBY = r1RY - r1BBH / 2;
let r2BBH = r2W sinR2AR1A + r2H cosR2AR1A;
let r2BBW = r2W cosR2AR1A + r2H sinR2AR1A;
let r2BBX = r2RX - r2BBW / 2;
let r2BBY = r2RY - r2BBH / 2;
return r1X < r2BBX + r2BBW && r1X + r1W > r2BBX && r1Y < r2BBY + r2BBH && r1Y + r1H > r2BBY &&
r2X < r1BBX + r1BBW && r2X + r2W > r1BBX && r2Y < r1BBY + r1BBH && r2Y + r2H > r1BBY;
}
```
I was wondering if this is novel, known, or even correct? The closest thing I could find was this answer on the original post that I found afterwards that seems to be pretty similar, or maybe it's the exact same thing coded differently? I'm not entirely sure how his code works or whether or not it's a different technique. Also, could this be optimized in any way?
Not sure if my explanation was any good but I ended up making a post about it with more details and a demo if that helps understanding at all.
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edited Sep 16, 2021 at 4:54
Erick WeberErick Weber
asked Sep 13, 2021 at 5:57
Erick WeberErick Weber
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2 Googling for "Separating axis theorem" yields gamedevelopment.tutsplus.com/tutorials/…, which seems similar to what you propose. I think this question probably isn't a research-level question in theoretical CS, so perhaps should be migrated to cs.stackexchange.com.Neal Young –Neal Young 2021-09-14 12:04:26 +00:00 Commented Sep 14, 2021 at 12:04
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If you rotate your rectangles through a common origin, then your method works. Your method works if there always exists a separating line that is parallel to a side of one of your rectangles. Such a line indeed exists. To see this, start from an arbitrary separating line and rotate it until it hits both rectangles, say in vertices p p and q q. For a given angle α α, consider the line l p(α)l p(α) through p p and l q(α)l q(α) through q q. By construction, there exists some α α such that these lines coincide (and pass through both p p and q q) and are both separating lines. Now rotate these separating lines by varying α α until one of them hits a second vertex of a rectangle. If a separating line hits vertices of different rectangles, then the separating lines must coincide (because the region between l p(α)l p(α) and l q(α)l q(α) contains no rectangles), so one of the separating lines hits two vertices of the same rectangle, and hence coincides with an side of a rectangle. Therefore, your method works if you rotate your rectangles through a common origin.
Prior answer: In my original answer, I misinterpreted your method and assumed that you rotated your rectangles around their respective centers. Under that assumption, the method may falsely report that rectangles collide, as illustrated in the figure below. The blue squares do not overlap, but their axis-aligned bounding boxes do. Rotating the squares to make the top right square axis aligned results in the red squares, whose axis-aligned bounding boxes also overlap.
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edited Sep 14, 2021 at 15:47
answered Sep 13, 2021 at 23:48
TimTim
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Sorry, I'm not sure if I'm understanding you but I tried to make a reproduction of the case that you're describing here jsfiddle.net/4rhyp1q5 (move the rectangle with wasd), and I'm not seeing that issue. Maybe you could edit the fiddle with the case that you're describing?Erick Weber –Erick Weber 2021-09-13 23:57:56 +00:00 Commented Sep 13, 2021 at 23:57
This answer looks incorrect to me. When you make the first blue square axis-aligned, there is overlap in the bounding boxes, but when you make the second blue square axis-aligned, there is no overlap in the bounding boxes. Consequently the proposed method does not claim they collide -- the proposed method seems to work fine on the blue squares.D.W. –D.W. 2021-09-14 03:39:39 +00:00 Commented Sep 14, 2021 at 3:39
I find the picture confusing because it shows 4 squares, but the question talks about 2 squares. I suggest you try to draw a picture with two squares, where you show the result of the first check (rotate to make the first square axis-aligned); and a separate picture with two squares, where you show the result of the second check (rotate to make the second square axis-aligned), and I think you'll see why this counterexample doesn't work.D.W. –D.W. 2021-09-14 03:39:41 +00:00 Commented Sep 14, 2021 at 3:39
I updated the figure.Tim –Tim 2021-09-14 08:08:40 +00:00 Commented Sep 14, 2021 at 8:08
@Tim are you still saying that this is indeed a counter-example in some way?Erick Weber –Erick Weber 2021-09-14 08:21:25 +00:00 Commented Sep 14, 2021 at 8:21
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15838 | https://usiena-air.unisi.it/bitstream/11365/1226897/3/entropy.pdf | 29 September 2025 Franzosi, R. (2011). Microcanonical Entropy and Dynamical Measure of Temperature for Systems with Two First Integrals. JOURNAL OF STATISTICAL PHYSICS, 143(4), 824-830 [10.1007/s10955-011-0200-4].
Microcanonical Entropy and Dynamical Measure of Temperature for Systems with Two First Integrals Published: DOI: Terms of use: Open Access (Article begins on next page) The terms and conditions for the reuse of this version of the manuscript are specified in the publishing policy. Works made available under a Creative Commons license can be used according to the terms and conditions of said license.
For all terms of use and more information see the publisher's website.
Availability: This version is available since 2023-02-23T09:20:39Z Original: This is a pre print version of the following article: arXiv:1004.5480v3 [cond-mat.stat-mech] 19 Jul 2011 Microcanonical Entropy and Dynamical Measure of Temperature for Systems with Two First Integrals Roberto Franzosi C.N.I.S.M. UdR di Firenze, Dipartimento di Fisica, Universit` a degli Studi di Firenze, Via Sansone 1, I-50019 Sesto Fiorentino and I.P.S.I.A. C. Cennini, Via dei Mille 12/a, I-53034 Colle di Val d’Elsa (SI), Italy.
(Dated: November 16, 2018) We consider a generic classical many particle system described by an autonomous Hamiltonian H(x1, . . . , xN+2) which, in addition, has a conserved quantity V (x1, . . . , xN+2) = v, so that the Poisson bracket {H, V } vanishes. We derive in detail the microcanonical expressions for entropy and temperature. We show that both of these quantities depend on multidimensional integrals over sub-manifolds given by the intersection of the constant energy hyper-surfaces with those defined by V (x1, . . . , xN+2) = v. We show that temperature and higher order derivatives of entropy are micro-canonical observable that, under the hypothesis of ergodicity, can be calculated as time averages of suitable functions. We derive the explicit expression of the function that gives the temperature.
PACS numbers: 05.20.Gg, 02.40.Vh, 05.20.- y, 05.70.- a Keywords: Statistical Mechanics For an isolated many-body classical system, the ergod-icity makes equivalent thermodynamics and dynamics.
Thus, in these cases, one can measure thermodynamic quantities, like temperature and specific heat, as tem-poral averages of suitable functions along almost each trajectory of a given system, or, equivalently, as an inte-gral over its phase space. The opportunity to pass from the dynamics to the microcanonical-thermodynamics and vice versa, gives the possibility to choose the smarter way to measure a given quantity. Very often the calculation of thermodynamic quantities in the microcanonical en-semble is an impracticable issue, thus one is forced to recur to the canonical ensemble, where these measures are more easily performed by resorting to numeric simula-tions, e.g. by Monte Carlo method. Of course, for a given system this step is performable only if there is equiva-lence between canonical and microcanonical ensembles in the thermodynamic limit. Furthermore, it is worth men-tioning that in many cases, the convergence of thermo-dynamic quantities to their asymptotic values, is much faster if the averages are computed along the dynam-ics, rather than by an important sampling of the canonic phase-space. For these cases therefore, the dynamics is preferred respect to statistics. Furthermore, only for sys-tems described by stable and temperated inter-particle interaction potentials is guaranteed the equivalence of the statistical ensembles in the thermodynamic limit. Nowa-days, several of the most intriguing challenge for modern science, deals with systems of size intermediate between the macroscopic and the microscopic scales. Systems like polymers, DNA-helix, proteins, nanosystems, are large enough to allow a statistical treatment, but are abso-lutely far from the thermodynamic limit. Thus, for these systems, ensemble equivalence is hardly verified and one has no option but performing dynamical averages.
It is in this same spirit that Rugh, in , has presented a dynamical approach for measuring the temperature of a Hamiltonian system in the microcanonical ensemble.
He has shown that for an ergodic classical system, which has only one conserved quantity, i.e. the energy, the in-verse temperature 1/T , can be calculated as a tempo-ral average of a suitable functional along the dynamics.
The Rugh’s seminal work has stimulated several papers [2–4] which have aroused widespread interest, especially among those who simulate the properties of liquids .
The calculation given by Rugh in Ref.
provides a microcanonical definition of temperature that allows its measure also in systems with nonstandard Hamiltonians.
Nevertheless, this calculation works for systems with only one conserved quantity, i.e.
the total energy.
In the present paper we extend the calculation of entropy and of microcanonical temperature to the case of Hamiltonian systems with two first integrals of motion.
In the present paper we consider a classical system of (N + 2) degrees of freedom (with N > 0), which is de-scribed by an autonomous Hamiltonian H, and which has a further independent conserved quantity V , such that {H, V } = 0. We derive in detail the microcanon-ical expressions for entropy, i.e.
the expression of the microcanonical invariant measure, and the temperature, moreover we give a formula to derive, recursively, all order of derivatives of the entropy.
We show that en-tropy and temperature depend on multidimensional in-tegrals over sub-manifolds given by the intersection of the constant energy hyper-surfaces with those defined by V (x1, . . . , xN+2) = v.
In particular, we show that temperature and higher order derivatives of entropy are microcanonical observable that, under the hypothesis of ergodicity, can be calculated as time averages of suitable functions. In Ref. it has been studied the microcanon-ical ensemble of a classical system, whose Hamiltonian is parameter dependent and in presence of other first inte-grals and it has been showed a method, alternative to the present one, that allows one to obtain the first derivative 2 of entropy respect to the conserved quantities. The aim of the present paper is to derive explicitly the functional by means of which the temperature can be calculated as a microcanonical average, in the case of a generic classical system, described by a many-body Hamiltonian with one further conserved quantity. Furthermore, our method al-lowed us to derive an iterative formula that gives the derivatives of S(E) of all orders, for this class of systems.
By this formula, one can measure more general quantities like, e.g. the specific heat.
Let H(x, xN+1, xN+2) be a classical Hamiltonian de-scribing an autonomous many-body system whose co-ordinates and canonical momenta (q1, p1, . . . ) are repre-sented as (N + 2)-component vectors (x, xN+1, xN+2) ∈ RN+2, and let V (x, xN+1, xN+2) be a further conserved quantity which is in involution with H.
We shall assume that V is a smooth function of the coordi-nates.
The system’s motion takes place on the mani-folds M = ΣE ∩Vu, where the ΣE = {(x, xN+1, xN+2) ∈ RN+2|H(x, xN+1, xN+2) = E} are energy level sets and Vu = {(x, xN+1, xN+2) ∈RN+2|V (x, xN+1, xN+2) = u} are subsets of RN+2 where V is constant. Among the equiv-alent expressions allowed for the microcanonical entropy S(E), the surface entropy S(E) = ln Z dNxdxN+1dxN+2δ(H(x, xN+1, xN+2)−E)× δ(V (x, xN+1, xN+2) −u) (1) has an interesting and useful geometric interpretation that we shall derive, following the calculation shown in the chapter on the theory of surfaces of Ref. . Con-sistently with Ref. , we shall assume the level sets of H and V to be non-singular hyper-surfaces. Even if the energy level surfaces (or the level sets of V ) in general constitute a singular foliation, thus for some values of E (or u) the energy surface (or Vu) is not a differential man-ifold, for generic values of E (or u) this is not an issue.
For a generic point x0 ∈M of a non-singular level-set of H and V , ∇H(x0) ̸= 0 and ∇V (x0) ̸= 0. Further-more, since H and V are in involution, almost everywhere ∇H(x0) and ∇V (x0) are independent vectors. Thus, in a neighborhood of x0 we can suppose of reorder the coor-dinate indices in such a way that ∂H/∂xN+1∂V /∂xN+2 − ∂V /∂xN+1∂H/∂xN+2 ̸= 0 for each x of the neighbor-hood. Now, each non-singular manifold M, can be par-titioned by a family F of not overlapping subsets .
With the further condition that, in each subset we can reorder the coordinate indices, as said above, so that ∂H/∂xN+1∂V /∂xN+2 −∂V /∂xN+1∂H/∂xN+2 ̸= 0 every-where in the subset.
Now, by using the first N co-ordinates we give a parametric description of the same subset. Thus, in each of the subsets of F we can choose f α = id for α = 1, . . . , N, and let us set g(y) := f N+1(y) and h(y) := f N+2(y), where y ∈RN . The metric induced by RN+2 on M results ηµν = δµν + ∂µg∂νg + ∂µh∂νh , (2) where ∂α• = ∂• /∂xα, whereas the its determinant can be derived by straightforward calculations and it results η = 1 + N X α=1 h (∂αg)2 + (∂αh)2i + N X µ,ν=1 µ<ν (∂µg∂νh −∂µh∂νg)2 .
(3) The derivatives ∂αg, ∂αh, can be expressed as follows ∂αg = [∂N+2V ∂αH −∂N+2H∂αV ] /D (4) ∂αh = [∂N+1H∂αV −∂N+1V ∂αH] /D , (5) where D =∂N+1H∂N+2V −∂N+1V ∂N+2H . From the expression above we derive the sub-manifold volume form dτ = dNx√η = dNxW D , (6) where W = N+2 X µ,ν=1 µ<ν ∂H ∂xµ ∂V ∂xν −∂H ∂xν ∂V ∂xµ 2 1/2 .
(7) On the other hand, expression (1) can be cast in the following form. For each point x we can introduce the following variables change y1 = h1 x(xN+1, xN+2) , y2 = h2 x(xN+1, xN+2) , (8) with the inverse transformations xN+1 = G1 x(y1, y2) , xN+2 = G2 x(y1, y2) , (9) where h1 x(xN+1, xN+2) = H(x, xN+1, xN+2) −E, h2 x(xN+1, xN+2) = V (x, xN+1, xN+2)−u, and G1 x(y, z) and G2 x(y, z) are the respective inverse transformations. From now on we will suppress the sub-index x. After these def-initions Eq. (1) can be expressed as follows S(E) = ln Z dNxdy1dy2δ(y1)δ(y2)|J(y1, y2)| , where |J(y1, y2)| is the determinant of the Jacobian ma-trix ∂(xN+1, xN+2)/∂(y1, y2). The inverse of the Jacobian matrix can be derived by Eqs. (8)-(9) and it results ∂1G1 = −∂2h2/D = −∂N+2V /D , (10) ∂1G2 = ∂1h2/D = ∂N+1V /D , (11) ∂2G1 = ∂2h1/D = ∂N+2H/D , (12) ∂2G2 = −∂1h1/D = −∂N+2H/D .
(13) Thus, the microcanonical entropy results S(E) = ln Z M dNx 1 D = ln Z M dτ W .
(14) 3 It is worth emphasizing that the invariant measure is in-dependent from the partition of M, since W is invariant under exchange of the indices of the coordinates.
In order to derive the temperature in the micro-canonical ensemble, according to the definition T (E) = (∂S(E)/∂E)−1, we shall use the following generaliza-tion of the Federer-Laurence derivation formula [9– 13].
The flux Φξ with a non-vanishing component in the direction orthogonal to the constant energy hyper-surfaces, but tangent to the level hyper-surfaces of V , can be defined by the vector ξ = nH −(nH · nV )nV , where nH = ▽H/∥▽H∥and nV = ▽V /∥▽V ∥. Let us define nξ = ξ/∥ξ∥, so the generalized derivation formula results ∂k ∂Ek Z M dτψ = Z M dτAk (ψ) , (15) where A(•) = 1 ▽H · nξ [▽(nξ•) −•nV · (nV · ▽)(nξ)] .
(16) The proof of the extention of the Federer-Laurence theo-rem to varieties of codimension two, is rather complicated and lengthy. Furthermore it is outside of the main mo-tivation of the present paper, thus it will be given in a further paper . After Eqs. (17) and (18), we obtain that the inverse temperature is given by 1 T (E, V ) = ⟨Φ(x)⟩µc , (17) where Φ(x)= W ▽H · nξ ▽ nξ W −(nV · ▽) (nξ) W · nV , (18) and ⟨⟩µc stands for the microcanonical average.
When (RN+2, H) is ergodic with respect to the Liou-ville measure dτ/W restricted to a nonsingular manifold M for almost every initial condition x(0) ∈M one has 1 T (E, V ) = lim s→∞ 1 s Z s 0 ds′ [Φ(x(s′))] .
(19) Simple geometric applications.
i) As a simple application we shall derive the invariant metric η in the case of a simple form for H as that one of a four-dimensional hypersphere H = x2 + y2 + z2 + w2 of unit radius, and with a condition given by the hyper-plain V = z + w = 0. By direct calculations we derive ∂xH = 2x, ∂yH = 2y, ∂zH = 2z, ∂wH = 2w, thus it results D = ∂zH∂wV −∂zV ∂wH = 2(z −w). With the notations introduced above, Eqs.
(4) and (5) become ∂xg = x/(z −w), ∂yg = y/(z −w), ∂xh = −x/(z −w), and ∂yh = −y/(z −w), thus the invariant metric results η = 1 + 2 " x z −w 2 + y z −w 2# , (20) and is defined for D ̸= 0 (z ̸= w) that is on the subsets {(x, y, ± r 1 −x2 −y2 2 , ∓ r 1 −x2 −y2 2 )|x2 + y2 < 1}.
The sum of the integrals of √η upon these subsets brings to the result 4π which is the right hyper-surface vol-ume. This solution can be checked noting that in the original problem the condition V = z + w = 0 can be lifted by a suitable change of variables.
Let us intro-duce the variables s = (z −w)/ √ 2 and t = (z + w)/ √ 2.
H and V can be expressed in the new variables as H = x2 + y2 + s2 + t2 = 1 and V = t = 0. Thus the met-ric η is that one of the three-dimensional unitary sphere x2 + y2 + s2 = 1 that is (see ) η = 1 + (x/s)2 + (y/s)2.
This indeed is the expression (20). The sum of the inte-grals of √η upon the two hemispheres, s > 0 and s < 0, gives the right value 4π.
ii) Let us now consider a simple case where all the ge-ometric quantities can be explicitly calculated in order to check the Eqs.
(15) and (16).
We shall consider H = x2 1 + x2 2 + x2 3 = E, a sphere of radius √ E in three dimensions and V = x3/ p x2 1 + x2 2 = u, that is a cone with an angle arctan(u) at the vertex. In this case M is a circle of radius a = p E/(1 + u2).
If we choose ψ = 1, we get easly ∂ ∂E Z M dτ = ∂ ∂E 2πa = πa E .
The terms that appear in Eq. (16) result ▽H·nξ = 2 √ E, ▽nξ = 2/ √ E and nV · (nV · ▽)(nξ) = 1/ √ E.
Thus A(1) = 1/(2E) and consequently R M dτA (1) = πa/E.
By choosing ψ = 1/W we get W = 2E/a2, and ∂ ∂E Z M dτ W = ∂ ∂E a2 2E 2πa = π 2(1 + u2)3/2√ E .
(21) Again, Eq. (16) contains ▽H · nξ = 2 √ E, ▽(nξ/W) = a2/E3/2 and nV · (nV · ▽)(nξ)/W = a2/(2E3/2). Thus A(1/W) = a2/(4E2) and consequently R M dτA (1/W) = πa3/(2E2) which indeed coincides with result (21).
Dynamical system.
Let us now consider a lattice system described by the Hamiltonian H = ν 8 X m (p2 m + q2 m)2 − X m (pmpm+1 + qmqm+1) (22) and the usual Poisson brackets {qm, pn} = δm,n for n, m = 1, . . . , M, with periodic boundary conditions.
The dynamics generated by this Hamiltonian conserves the quantity N = P m(q2 m + p2 m)/2, thus the Eqs. (17) and (18) are in order to calculate the microcanonical tem-perature. The ground-state is achieved by solving the 4 equation δ(H −χN) = 0 in which the Lagrangian multi-plier χ has been introduced to take in account the conser-vation of N. By direct calculations, we got the solution q0m = p 2N/M := q0, p0m = 0 and χ = νN/M −2 .
Small fluctuations around this ground-state cor-respond to T ≳0, in the following we show that this prevision is verified by Eq. (17) and (18), whereas the formula for 1/T given in , that holds in the case of systems with only one first integral (energy), does not work.
The reason of this comparison is to show that the equation derived in , which is valid in the case of system with only one first integral, cannot be used in the case of systems with more than one first inte-grals In Ref.
has been derived an equation sim-ilar to Eqs.
(17) and (18).
The inverse temperature 1/T can be derived by Eqs. (12) and (17) of Ref. after having found two vectors, X0 and X1, such that dH(X0) = 1, dH(X1) = 0, dV (X0) = 0 and dV (X1) = 1.
One can use X0 = cV V ∇H/d −cV H∇V/d and X1 = −cV H∇H/d + cHH∇V/d, where cHH = ∥∇H∥2, cV V = ∥∇V ∥2, cV H = ∇V ∇H, and d = cHHcV V −c2 V H.
Thus it results 1/T = ⟨∇· X0⟩µc. The latter expression seems inequivalent to Eqs. (17) (18). Indeed, it does not contain 1/W which is related to the invariant measure.
It is worth emphasizing that an analogous term 1/∥H∥, which is related to the microcanonical measure of a sys-tem with one first integral, indeed appears in the formula for 1/T derived in . In any case a comparison between these two equations, would require a numerical simula-tion, but this is out of the aim of the present paper. By expanding H in terms of the displacements Qn and Pn from the minimum points qn = q0 and pn = 0, and by cal-culating the terms appearing in (17) and (18) in the limit Qn, Pn →0, after some boring algebra we got the correct low-energy temperature, being 1/T = ⟨Φ⟩µc →∞.
It is worth emphasizing that by using the formula de-rived in Ref. 1/T = ∇ ∇H/∥∇H∥2 µc, which is correct for systems with only one first integral, one ob-tains the erroneous result: T = ν 2q2 0 −2 q0 2 /(2νq2 0).
In conclusion, we have presented a dynamical approach for measuring the temperature of a Hamiltonian system with two first integrals in the microcanonical ensemble.
The formula we have derived allows one to measure the inverse temperature as a time-average, instead of as an average over the phase-space, also in the case of systems with two first integrals.
Furthermore, by Eq. (15) higher orders of derivatives of S(E) can be obtained. Therefore the method here presented allows one to measure g.e. the specific-heat.
I’m indebted with Prof. M. Pettini, Prof. G. Vezzosi and Prof. R. Livi for the usefull discussions.
H.H. Rugh, Phys. Rev. Lett. 78, 772 (1997).
H.H. Rugh, J. Phys. A 31, 7761 (1998).
W. K. den Otter, J. Chem. Phys. 112, 7283 (2000).
G. Rickayzen et al., J. Chem. Phys. 114, 4333 (2001).
B. D. Bluteret al., J. Chem. Phys. 109, 6519 (1998); G.
P. Morriss et al., Phys. Rev. E 59, R5 (1999).
H.H. Rugh, Phys. Rev. E 64, R055101 (2001).
A. I. Khinchin, Mathematical Foundations of Statistical Mechanics, Dover Publications, Inc. New York.
B. A. Dubrovin, S. P. Novikov and A. T. Fomenko, Modern Geometry–Methods and Applications, Part. I The Geometry of Surfaces Transformation Groups, and Fields, GTM Series, Springer-Verlag, New York, 1984.
H. Federer, Geometric Measure Theory, Springer, New York, 1969, p. 249.
P. Laurence, Z. Angew. Math. Phys. 40 (1989) 258.
R. Franzosi et al., Nuclear Physics B 782 (2007) 189-218.
R. Franzosi et al., Nuclear Physics B 782 (2007) 219-240.
M. Pettini, Geometry and Topology in Hamiltonian Dy-namics and Statistical Mechanics, IAM Series n. 33, Springer-Verlag New, 2007.
R. Franzosi, Microcanonical Thermodynamics of Systems with First Integrals, in preparation.
More precisely we can make use of the partition-of-unity of the level set.
Each subset of F can be given in a parametric form of N coordinates if D(x) ̸= 0 for each x in the subset.
The analogous expression valid for systems with only one first integral, derived in , is the following S(E) = ln R ΣE dσ/∥▽H∥.
The Federer-Laurence derivation formula [9, 10] is ∂k( R ΣE ψdΣ)/∂Ek = R ΣE Ak (ψ) dΣ, where A(•) = 1 / ∥▽H ∥▽(▽H/∥▽H∥•).
Here, and in the following nV · (nV · ▽)(nξ) = X j,k nV j nV k ∂k(n ξ j ) .
Another and equivalent solution is given by q0m = 0, p0m = p 2N/M and χ = νN/M −2. |
15839 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4215084/ | The Treatment of Chronic Recurrent Oral Aphthous Ulcers - PMC
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Dtsch Arztebl Int
. 2014 Oct 3;111(40):665–673. doi: 10.3238/arztebl.2014.0665
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The Treatment of Chronic Recurrent Oral Aphthous Ulcers
Andreas Altenburg
Andreas Altenburg, Dr. med.
1 Departments of Dermatology, Venereology, Allergology and Immunology, Dessau Medical Center
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1, Nadine El-Haj
Nadine El-Haj, Dr. med.
1 Departments of Dermatology, Venereology, Allergology and Immunology, Dessau Medical Center
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1, Christiana Micheli
Christiana Micheli, Dipl. med.
1 Departments of Dermatology, Venereology, Allergology and Immunology, Dessau Medical Center
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1, Marion Puttkammer
Marion Puttkammer
2 Pharmacy, Dessau Medical Center: Puttkammer
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2, Mohammed Badawy Abdel-Naser
Mohammed Badawy Abdel-Naser, Prof. Dr. med.
1 Departments of Dermatology, Venereology, Allergology and Immunology, Dessau Medical Center
3 Department of Dermatology and Andrology, Faculty of Medicine, Ain Shams University, Cairo (Egypt)
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1,3, Christos C Zouboulis
Christos C Zouboulis, Prof. Dr. med. Dr. h.c.
1 Departments of Dermatology, Venereology, Allergology and Immunology, Dessau Medical Center
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1,
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1 Departments of Dermatology, Venereology, Allergology and Immunology, Dessau Medical Center
2 Pharmacy, Dessau Medical Center: Puttkammer
3 Department of Dermatology and Andrology, Faculty of Medicine, Ain Shams University, Cairo (Egypt)
Klinik für Dermatologie, Venerologie und Allergologie/Immunologisches Zentrum Städtisches Klinikum Dessau Auenweg 38, 06847 Dessau, Germany christos.zouboulis@klinikum-dessau.de
Received 2014 Jan 14; Accepted 2014 Jun 18; Issue date 2014 Oct.
PMC Copyright notice
PMCID: PMC4215084 PMID: 25346356
See letter "Correspondence (letter to the editor): Acatalasia Was Omitted" in volume 112 on page 222.
See letter "Correspondence (letter to the editor): Fumaric Acid Esters as an Additional Systemic Therapy Option" in volume 112 on page 222.
See letter "Correspondence (reply): In Reply" in volume 112 on page 222.
This article has been corrected. See Dtsch Arztebl Int. 2014 Oct 3;111(40):665.
Abstract
Background
Chronic recurrent oral aphthous ulcers are the most common type of inflammatory efflorescence of the oral mucosa, with a prevalence of 2% to 10% in Caucasian populations. To treat them properly, physicians should know their clinical appearance and course, conditioning factors, underlying causes, and differential diagnosis.
Method
This review is based on pertinent articles that were retrieved by a selective search in PubMed and in the Cochrane Central Register of Controlled Trials.
Results
Hard, acidic, and salty foods and toothpastes containing sodium lauryl sulfate should be avoided, along with alcohol and carbonated drinks. In Germany, the only drugs that have been approved to treat oral aphthous ulcers are corticosteroids, topical antiseptic/anti-inflammatory agents such as triclosan and diclofenac, and local anesthetics such as lidocaine. Antiseptic agents and local anesthetics should be tried first; if these are ineffective, topical corticosteroids should be used. In severe cases, local measures can be combined with systemic drugs, e.g., colchicine, pentoxifylline, or prednisolone. The efficacy of systemic treatment is debated. Other immunosuppressive agents should be given systemically only for refractory or particularly severe oral aphthous ulcers due to Adamantiades-Behçet disease.
Conclusion
The treatment of chronic recurrent oral aphthous ulcers is symptomatic, mainly with topically applied agents. It is tailored to the severity of the problem in the individual case, i.e., the frequency of ulcers, the intensity of pain, and the responsiveness of the lesions to treatment. Effective treatment relieves pain, lessens functional impairment, and lowers the frequency and severity of recurrences.
Oral aphthous ulcers typically present as painful, sharply circumscribed fibrin-covered mucosal defects with a hyperemic border.
Chronic recurrent oral aphthous ulcers occur in three different clinical morphological variants and with two different time courses. Small ulcers of the minor-type (Mikulicz) are less than 1 cm in diameter (usually 2–5 mm) and heal spontaneously in 4–14 days. They account for 80–90% of all recurrent oral aphthous ulcers (1, e1). Scarring occurs in around 8% of cases (1, e2) (Figure 1a). Large ulcers of the major-type (Sutton ulcers) are usually 1–3 cm in diameter, deeply indurated and can last for 10 days to 6 weeks or occasionally even longer (1, e3) (Figure 1b). They account for around 10% of recurrent benign oral ulcers. About 64% of Sutton ulcers heal with scarring. Herpetiform aphthous ulcers are very small (1–2 mm) grouped lesions (1, e4) (Figure 1c). They account for around 5% of recurrent oral aphthous ulcers, are extremely painful and persist for 7–10 days. As many as a 100 ulcers can be present; they may coalesce into larger erosive plagues and about 32% heal with scarring. The three morphologic variants can occasionally appear simultaneously (2).
Figure 1.
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(a) Minor-type oral aphthous ulcers, (b) major-type oral aphthous ulcer, (c) herpetiform oral aphthous ulcers
(Figure 1a modified from Altenburg A, Mahr A, Maldini C, et al.: Epidemiologie und Klinik des Morbus Adamantiades-Behçet in Deutschland: Aktuelle Daten. Ophthalmologe 2012; 109: 531–41.; Figure 1c modified from Altenberg A, et al. Klinik und Therapie chronic rezidivierender Aphthen. Hautarzt 2012; 63: 693–703). Figures 1a and c used with kind permission from Springer-Verlag, Heidelberg
Another classification is based on the time course. The simple chronic recurrent oral aphthous ulcers present with a limited number of small, quickly healing, minimally painful ulcers limited to the oral mucosa and recurring with 3–6 episodes annually. In complex aphthosis, there are a few or many slowly healing intensely painful ulcers on the oral and perhaps genital mucosa (3). The latter may also be perigenital, affecting the scrotum, vulva, anus, perineum and inguinal region. Complex aphthosis features frequently appearing ulcers with either short lesion-free periods or even repeatedly recurrent ulcers, severe pain and even systemic effects such as interference with eating and the resultant problems of inadequate nutrition (3).
Methods
A selective literature search concentrating on randomized controlled therapeutic trials was performed to prepare this review. The literature search employed PubMed and the Cochrane Central Register of Controlled Trials. Letters to the editor and meeting reports were ignored. Because of the small number of published controlled studies, both studies without control groups and studies whose results correlate with our clinical experience were in exceptional cases included.
Epidemiology
Chronic recurrent oral aphthous ulcers are the most common inflammatory disease of the oral mucosa with a prevalence of 2–10% in Caucasian populations; women are more frequently affected (2, 4). A study evaluating 40 693 school children in the USA showed a point prevalence of 1.23% and a lifetime prevalence of 36.5% (5).
Pathogenesis
The etiology of chronic recurrent oral aphthous ulcers is still unclear. A variety of underlying disorders may predispose patients to develop oral aphthous ulcers; they include iron deficiency anemia, neutropenia, and folic acid or vitamin B12 deficiency, as well as a selective vitamin B12 resorption defect (e5– e7). Appropriate replacement therapy has reduced the severity of the disease as documented in case reports (e8) (evidence level [EL] 4). Local mucosal injuries are also possible trigger factors (6, e9). In addition, genetic factors may be important; the family history is positive in up to 40% of patients (7). No consistent association with an HLA haplotype has been shown (e10).
Differential diagnosis
Oral aphthous ulcers are clearly defined in a nosologic sense, but often hard to distinguish clinically from a broad group of similar (aphthoid) erosions and ulcers. The differential diagnosis includes a variety of diseases which can imitate the clinical picture of oral aphthous ulcers (Box).
Box. Important differential diagnostic considerations for oral aphthous ulcers.
Gastrointestinal, mucocutaneous disorders
Ulcerative colitis
Crohn‘s disease
Celiac disease
Infections
Herpes simplex and zoster
Infectious mononucleosis
Hand foot and mouth disease
Herpangina
HIV infection
Syphilis
Acute necrotizing ulcerative gingivitis
Candidiasis
Reactive changes
Morsicatio buccalis
Traumatic eosinophilic ulcer
Malignant diseases
Oral carcinoma
Non-Hodgkin‘s lymphoma
Mucocutaneous rheumatic diseases
Lupus erythematosus
Sweet syndrome
Reactive arthritis
MAGIC syndrome (mouth and genital ulcers with inflamed cartilage)
Sarcoidosis
Bullous and lichenoid dermatoses
Erythema multiforme and its variants, including Stevens-Johnson syndrome and toxic epidermal necrolysis
Bullous autoimmune disorders: pemphigus vulgaris, cicatricial pemphigoid, epidermolysis bullosa acquisita, linear IgA dermatosis
Lichen planus
Other oral disorders
Allergic contact stomatitis
Drug-induced ulcerative stomatitis
Geographic stomatitis
PFAPA syndrome (periodic fever, aphthous ulcers, pharyngitis, cervical adenitis)
(from Altenburg A, et al.: Klinik und Therapie chronisch rezidivierender Aphthen. Hautarzt 2012; 63: 693–703; with kind permission from Springer Verlag, Heidelberg)
Adamantiades-Behçet disease (ABD) is a chronic recurrent systemic vasculitis (e11) in which oral and genital ulcers are major diagnostic criteria. Some include ABD among the autoinflammatory diseases. In ABD, 98.5% of patients have recurrent oral aphthous ulcers; this is the most common manifestation of the disorder (8). Recurrent genital aphthous ulcers are seen in 64.7% (8). In 84.5% of patients, the first manifestation is oral ulcers, while 3.5% start with genital ulcers, which are the second most frequent symptom (8). About 10% of the patients with complex aphthosis in Western Europe and North America develop ABD; the likelihood is higher in the eastern Mediterranean region, Middle East and Asia (9). In order to make the diagnosis, clinical diagnostic criteria are applied, such as those of the International Study Group for Behçet’s Disease (e12), or the new International Criteria for Behçet’s Disease (9) (eBox) which are based on epidemiological data.
eBOX. International criteria for the diagnosis of Adamantiades-Behçet disease (2014) (9).
• Recurrent oral aphthous ulcers 2
• Skin lesions (papulopustules, erythema nodosum, thrombophlebitis)1
• Vascular involvement (arterial or venous thromboses, aneurysms)1
• Recurrent genital aphthous ulcers 2
• Ocular involvement (hypopyon-iritis, uveitis)2
• CNS involvement 1
• Positive pathergy test 1
Adamantiades-Behçet disease: 4 or more points
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Overview of therapy
The studies that we evaluated generally reached an EL 2A because of a variety of limitations including small patient number, reliance on self-reported information, unclear information on randomization, incomplete or lacking blinding, or inadequate information on the nature of the placebo.
Effects on the overall quality of life of the patients was not evaluated in the studies. Undesired effects of topical medications were either mild or not mentioned. In studies on systemic drugs, the undesired effects were not always discussed.
With the exception of the corticosteroids, topical antiseptics and topical anesthetics, all tested substances are used off-label for oral aphthous ulcers in Germany. Rebamipide, clofazimine and camel thorn distillate are not available in Germany.
Dietary and general measures
There are no reliable studies addressing the role of diet in managing aphthous ulcers. Substances that a majority of patients report frequently trigger ulcers should be avoided, especially if the patient in question has noticed an association. In general one should avoid hard, acidic and salty substances such as fruit juices, citrus fruits, tomatoes, and spices like pepper, paprika and curry, as well as alcoholic and carbonated beverages. Avoiding dental care products with sodium lauryl sulfate (SLS) is also desirable. Using a SLS-free toothpaste significantly reduced the healing period and pain of oral aphthous ulcers (10) (EL 1B).
Topical therapy
Topical anesthetics
Topical anesthetics often provide satisfactory pain relief (6). Options include lidocaine as 1% cream (randomized placebo-controlled study; EL2A ), 2% gel or spray; polidocanol as paste; and benzocaine lozenges. There is a pump spray that combines tetracaine 0.5% and polidocanol 0.1%. A mouth wash containing benzocaine and cetylpyridinium chloride is also available.
Antiseptics and anti-inflammatory agents
A mouth wash containing 0.15% triclosan in ethanol and zinc sulfate reduced the number of new aphthous ulcers in 43% of cases, the pain intensity in 45% and extended the ulcer-free interval (12) (Table 1) (EL 1B). Diclofenac 3% in a 2.5% hyaluronic acid gel was superior to a lidocaine 3% gel in reducing pain after 2–6 hours (13) (Table 2) (EL 2A).
Table 1. Therapeutic options to reduce the frequency of recurrence and/or the number of oral aphthous ulcers 1.
| Drug | Dose | Benefits | Controls | Number of probands | Length of study | Evidence level | Reference |
--- --- --- --- |
| Topical medications |
| Triclosan 0.15% in 7.8% w/w ethanol/0.4% w/w zinc sulfate, triclosan 0.15% in 15.6% w/w ethanol/0.4% w/w zinc sulfate | 10 mL for 30 s b.i.d. | Reduction of number of ulcers in 43%, of pain by 45%, increase in ulcer-free days vs. controls (p<0.0001) | Commercially available fluoride mouth wash. Triclosan 0.15% in propylene glycol | 30 | 6 weeks (with each product) | 2A | (12) |
| 2.5% tetracycline solution vs. toothpaste with amyloglucosidase and glucose oxidase | Tetracycline solution 5 mL rinse in mouth for 1 minute q.i.d., toothpaste b.i.d. | 2.5% tetracycline solution: increase (over 40%) in ulcer-free or pain-free days compared to placebo (p<0.05); toothpaste no significant differences from placebo | Placebo mouth wash or placebo toothpaste | 57 | 10 weeks, then switch therapy | 2A | (18) |
| Systemic medications |
| Prednisolone vs. colchicine2 | Prednisolone 5 mg daily p.o., colchicine 0.5 mg daily p.o. | Reduction of pain and number of lesions after 3 months (p<0.001) without significant differences between agents; no placebo | – | 17 (prednisolone), 17 (colchicine) | 3 months | 2A | (32) |
| Rebamipide | 300 mg daily p.o. | ABD: Reduction in number of ulcers and pain in 65% vs. 36% in placebo group (p<0.01) | Placebo | 35 | 12 to 24 weeks | 2A | (e32) |
| Azathioprine | 2.5 mg/kg BW daily p.o. | Reduction in frequency of oral aphthous ulcers from 43% to 11% (p<0.005) | Placebo | 73 (only men) | 2 years | 2A | (35) |
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1 Except for azathioprine, all treatments are off-label. The recommendations are ordered based on disease severity.
2 evaluated as ineffective in (29).
w/w, weight by weight; s, seconds; min, minutes; p. o., per os; ABD, Adamantiades-Behçet disease; BW, body weight
Table 2. Therapeutic options to reduce pain.
| Drug | Dose | Benefits | Controls | Number of probands | Length of study | Evidence level | Reference |
--- --- --- --- |
| Topical medications |
| Diclofenac 3% in 2.5% hyaluronic acid gel | 200 μL once | Less pain 2–6 hours after application of diclofenac-hyaluronic acid gel than after hyaluronic acid gel or lidocaine solution (p= 0.01) | 2.5% hyaluronic acid gel, 2% lidocaine solution | 60 | 8 hours | 2A | (13) |
| Silver nitrate pencil | Once | Less pain at day 1 (p<0.001) | Placebo pencil; prior use of 2% lidocaine solution in both active and control groups | 85 | 7 days | 2A | (15) |
| CO2 laser (2–5 mW) | Once | Reduction of pain immediately after treatment, relief lasting 96 hours (p<0.001) | Inactive laser | 15 | 4 days | 2A | (16) |
| Nd:YAG laser | Once | Less pain immediately and on days 4 and 7 with laser (p<0.05). less exudation with laser (p<0.05) | Triamcinolone 0.1% in oral paste | 14 (laser), 14 (triamcinolone) | 7 days | 2A | (e17) |
| Topical medications with systemic absorption |
| Minocycline 0.2% in aqueous solution | 5 mL q.i.d. | Less pain starting with day 2 (p<0.05) | Tetracycline 0.25% in aqueous solution | 16 (minocycline), 17 (tetracycline) | 10 days, then switch therapy | 2A | (19) |
| Minocycline 0.2% in aqueous solution | 5 mL q.i.d. | Less pain starting with day 2 (p<0.05) | Placebo solution | 18 (minocycline), 15 (placebo) | 10 days | 2A | (e13) |
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Chlorhexidine mouthwash and chamomile extract both reduced the frequency, increased healing speed, and decreased the severity of aphthous ulcers in non-randomized studies (6, 14) (EL 2B). Chlorhexidine gels and sprays are also available. A useful adjuvant therapy is dexpanthenol in a variety of forms (spray, solution and tablets).
Cauterization
Topical application of hydrogen peroxide 0.5% solution or silver nitrate 1–2% solution significantly reduced the pain severity after one day, but did not increase the speed of healing (15) (EL2A).Treatment with a CO 2 (16) or Nd:YAG laser (17, e13) brought immediate pain relief which lasted for 4–7 days (Table 2) (EL 2A).
Topical tetracycline treatment
Using a mouthwash containing chlortetracycline 2.5% increased the number of ulcer-free or pain-free days significantly, by 40% compared to a placebo (18) (Table 1) (EL 2A). In regards to pain reduction, a minocycline 0.2% mouthwash was superior to a tetracycline 0.25% mouthwash (19, e13) (Table 2) (EL 2A).
Tetracycline hydrochloride powder 250 mg can be combined with 10 mL of tap water by the patient immediately before use to avoid stabilization problems. Because of the acid pH value, there may be temporary mucosal burning generally followed by clinical improvement. A stable mixture can also be prepared by neutralizing the tetracycline hydrochloride to create a basic product (6).
Both a standardized formulation—as well as the less-stable freshly prepared solution—can produce rapid healing in some patients, even in those with large ulcers resistant to topical corticosteroids.
Topical corticosteroids
If combined treatment with topical anesthetics and anti-inflammatory agents is not effective, then topical corticosteroids should be employed. In Germany, a registered oral paste containing prednisolone is commonly used, which is applied 1–2 times daily (20). The combination of topical anesthetics (for example, lidocaine gel) during the day with an oral paste containing triamcinolone in the evening is also effective (21). Studies indicate that triamcinolone oral paste is superior to phenytoin syrup (22) (Table 3) (EL 2A). Although both were equally effective in reducing pain, dexamethasone oral paste produced more rapid healing than triamcinolone oral paste (23) (Table 3) (EL 2A).
Table 3. Topical therapeutic options to reduce the duration of illness and size of oral aphthous ulcers.
| Drug | Dose | Benefits | Controls | Number of probands | Length of study | Evidence level | Reference |
--- --- --- --- |
| Amlexanox 5% paste | b.i.d. | Reduction in size and erythema (p<0.05) | Placebo paste | 32 | 4 days | 2A | (26) |
| Amlexanox 2 mg patch | q.i.d. | Smaller thermographically active area on day 4 (p<0.05) | Placebo patch | 26 (amlexanox), 26 (placebo) | 4 days | 2A | (e30) |
| Amlexanox 2 mg adhesive tablets | q.i.d. to one aphthous ulcer for 5 days | Reduction in pain and size on days 4 and 6 (p<0.001) | Placebo adhesive tablet | 104 (amlexanox), 108 (placebo) | 6 days | 2A | (27) |
| 5-aminosalicylic acid 5% cream | t.i.d. | Reduction in duration of aphthous ulcers (7 vs. 11 days; p<0.01) and pain (p<0.05) | Placebo cream | 22 | 14 days | 2A | (25) |
| Sucralfate solution | Apply 5 mL solution with applicator q.i.d. | ABD: Reduction in frequency (p = 0.003) and duration (p = 0.03) | Placebo solution | 40 | 3 months | 2A | (34) |
| Camel thorn distillate (Iranian product) | Rinse mouth with 40 mL for one min q.i.d., then swallow | Size and pain reduced on days 3–7 (p<0.001) and 10 (p<0.02) | Placebo | 49 (camel thorn distillate) and 44 (placebo) | 2 weeks | 2A | (e31) |
| Triamcinolone acetonide 0.1% in oral paste | t.i.d. | ABD: 86.7% response rate compared to 53.3% for phenytoin syrup (p = 0.01) | Phenytoin syrup as mouthwash for 4–5 min t.i.d. | 30 (triamcinolone acetonide), 30 (phenytoin) | 7 days | 2A | (22) |
| Dexamethasone 0.1% in oral paste | q.i.d. | Quicker healing (dexamethasone) (p<0.001) | Triamcinolone acetonide 0.1% in oral paste | 53 (dexamethasone), 37 (triamcinolone) | 14 days | 2A | (23) |
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1 All treatments are off-label. The recommendations are ordered based on disease severity. ABD, Adamantiades-Behçet disease; min, minutes
When topical corticosteroids are used regularly, one should be alert to the possibility of increased numbers of oral yeast infections (24). Especially painful, deep ulcers can be treated with intralesional triamcinolone suspension 0.1–0.5 mL per lesion (21).
Additional topical therapies
A double-blind, placebo-controlled study showed that 5-aminosalicylic acid 5% cream achieved pain reduction and more rapid healing of oral aphthous ulcers (25) (Table 3) (EL 2A). Amlexanox 5% paste or 2 mg tablets, when used in the prodromal stage, led to a reduction in the number and size of oral aphthous ulcers, as well as reduction in pain (26, 27) (Table 3) (EL 2A).
An association between smoking and a reduction in the frequency of recurrences of oral aphthous ulcers has been observed. The number of lesions and the intervals between recurrences appear to be reduced during periods when the patient is smoking versus abstaining from tobacco (28, e14). Experimental evidence indicates that nicotine has an anti-inflammatory effect on keratinocytes (6, 14). Nicotine patches apparently cannot achieve the effects of tobacco smoke (own unpublished data). In a preliminary study with 3 patients, complete remission of recurrent oral aphthous ulcers was achieved with nicotine gum (e15) (EL 4). Neither cyclosporine (70 mg/g oral paste) nor interferon-α-2c gel was effective in treating oral aphthous ulcers (14).
Systemic therapy
A current review of the Cochrane Collaboration analyzed 25 studies (22 of which were placebo-controlled) on systemic therapy of oral aphthous ulcers and found no convincing evidence of efficacy (29) (eTables 1, 2),
Colchicine
Colchicine (0.5–2 mg daily) is helpful for the majority of patients with chronic recurrent oral aphthous ulcers. An off-label trial is recommended for 6 weeks with 1–2 mg daily—followed by long-term therapy depending on how severe the ulcers are and how well-tolerated the medication is (20). In a large open study of Fontes at al. (30), colchicine produced clear improvement in 63% of cases over a period of 3 months and in 37% over many years. 22% of the patients were free of disease, while 41% had at least a 50% reduction in number and duration of aphthous ulcers. In 37% the improvement was maintained for 5 years. In additional controlled studies, colchicine 1–2 mg daily led to significantly fewer oral and genital aphthous ulcers in patients with ABD (e16, e17) (EL 2A). The aphthous ulcers frequently recurred when the treatment was stopped (20). Contraceptive measures after the conclusion of therapy are recommended for 3 months in women and 6 months in men. Up to 45% of patients experienced gastrointestinal symptoms.
If aphthous ulcers fail to respond to colchicine monotherapy, combination approaches are possible. In patients with ABD, treatment with colchicine and benzathine penicillin was superior to colchicine alone in producing a slight improvement in the frequency of ulcers and a clear reduction in their healing time (more than 50%) (e18) (eTable 1) (EL 2A).
etable 1. Systemic therapeutic options to reduce duration of illness and size of aphthous ulcers.
| Drug | Dose | Benefits | Controls | Number of probands | Length of study | Evidence level | Reference |
--- --- --- --- |
| Sucralfate | 1 g q.i.d. | More rapid healing and less pain in 80% of patients compared to 13% with placebo and 38% with antacids (p<0.001) | Placebo solution and antacid solution | 21 | 2 years | 2A | (e19) |
| Colchicine | 1 mg daily p.o. | ABD: fewer oral aphthous ulcers in men and women compared to placebo (p<0.005) | Placebo | 169 | 4 months, then switch therapy | 2A | (e16) |
| Colchicine | 1–2 mg daily p.o. | ABD: no significant difference from placebo | Placebo | 50 (48% women) | 24 weeks | 2A | (e17) |
| Benzathine penicillin plus colchicine | Benzathine penicillin 1.2 IU monthly i.m. plus colchicine 1–1.5 mg daily p.o. | ABD: Reduction in frequency and duration of disease compared to colchicine alone (p<0.005) | Colchicine | 154 | 24 months | 2A | (e18) |
| Pentoxifylline | 400 mg daily p.o. | Reduction in aphthous ulcer size (p = 0.05) | Placebo | 14 (pentoxifylline) and 16 (placebo) | 2 months | 2A | (31) |
| Prednisone vs. montelukast | Prednisone 25 mg daily for 15 days, 12.5 mg daily for 15 days, 6.25 mg daily for 15 days, 6.25 mg q.o.d. for 15 days p.o.; montelukast 10 mg daily for 1 month and then q.o.d. for 1 month p.o. | More rapid healing and reduction in frequency of flares with prednisone vs. montelukast and montelukast vs. placebo (p<0.0001), with both fewer oral aphthous ulcers than with placebo (p<0.01) | Cellulose placebo | 20 (prednisone), 20 (montelukast), and 20 (placebo) | 2 months plus 2 month follow-up | 2A | (33) |
| Dapsone | 100 mg daily p.o. | ABD: Reduction in number, duration (p<0.001) and frequency (p<0.01) | Placebo | 20 | 3 months | 2A | (e20) |
| Zinc sulfate vs. dapsone | Zinc sulfate 300 mg daily p.o.; dapsone 100 mg daily p.o. | Zinc sulfate and dapsone: smaller and fewer lesions compared to placebo | Placebo | 15 (dapsone), 15 (zinc sulfate) and 15 (placebo) | 3 months | 2A | (e29) |
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evaluated as ineffective in (29). ABD, Adamantiades-Behçet disease
In our experience the off-label use of colchicine for chronic recurrent aphthous ulcers is generally approved and reimbursed by insurance companies.
Pentoxifylline
In case reports and older non-controlled studies, both pentoxifylline and oxypentoxifylline; 300 mg 1–3 times daily or 400 mg t.i.d. achieved good response rates (in children 36–50%) (6). In a more recent controlled study, pentoxifylline (400 mg t.i.d.) was only able to reduce the size of oral aphthous ulcers (p = 0.05) (31) (eTable1) (EL 2A).
Systemic corticosteroids
Systemic corticosteroids should be considered if colchicine and pentoxifylline do not produce improvement (20). Prednisolone or prednisone equivalents (10–30 mg daily) can be used on a short-term basis (up to one month) during a flare of the disease to speed healing. In a small controlled study, prednisolone 5 mg daily for 3 months was comparable to colchicine 0.5 mg daily. It produced a clear reduction in pain, as well as in number and size of oral aphthous ulcers (32) (eTables 1and 2) (EL 2A). Prednisone (25 mg daily tapered over 2 months) was more effective than the leukotriene inhibitor montelukast in managing oral aphthous ulcers (33) (eTable1) (EL 2A).
Sucralfate
Sucralfate is used as an antacid in treating gastric and duodenal ulcers. Sucralfate suspension produced more rapid healing and reduced pain of both oral and genital aphthous ulcers (34, e19) (Table 3, eTable 1) (EL 2A).
Dapsone
Dapsone significantly reduced the number and size of oral and genital aphthous ulcers in ABD (e20) (eTable 1) (EL 2A).
Antimetabolites: azathioprine and methotrexate
In a placebo-controlled study, azathioprine reduced the frequency and severity of orogenital aphthous ulcers in ABD; it is approved for this indication in Germany (35) (eTable 1) (EL 1B). In a case series, methotrexate 7.5–20 mg in a single weekly dose was helpful for severe orogenital aphthous ulcers (4) (EL 4).
Cyclosporine
There is information on over 350 patients with ABD treated with variable doses of cyclosporine (1–10 mg/ kg body weight daily) for divergent periods of time (1–77 months) (36). In a controlled study up to 70% of patients experienced improvement in oral aphthous ulcers (37) (eTable 2) (EL 2A).There were more side effects in the cyclosporine group than in the colchicine control group. 92% of the women and 32% of men developed hirsutism, fever, fatigue, and gastrointestinal symptoms, all of which improved when the dose was reduced. In contrast to colchicine, cyclosporine led to increased creatinine and blood urea nitrogen levels. Cyclosporine is approved in Germany for treating uveitis associated with ABD.
etable 2. Systemic therapeutic options to reduce frequency of attacks or number of aphthous ulcers.
| Drug | Dose | Benefits | Controls | Number of probands | Length of study | Evidence level | Reference |
--- --- --- --- |
| Cyclosporine | Cyclosporine 10 mg/kg BW daily p.o., colchicine 1 mg daily p.o. | ABD: Oral aphthous ulcers improved by 70% with cyclosporine vs. 20% with colchicine (p<0.001) | Colchicine | 47 (cyclosporine), 49 (colchicine) | 16 weeks | 2A | (37) |
| Thalidomide | 100 or 300 mg daily p.o. | ABD: Reduction in frequency of oral aphthous ulcers after 4 weeks (p<0.001)—same effects with 100 and 300 mg daily | Placebo | 96 (only men) | 24 weeks | 2A | (e24) |
| Interferon-α-2a | 6 million IU 3 times weekly s.c. | ABD: Reduction in duration (p = 0.02) and pain (p = 0.01) vs. placebo | Placebo | 50 (38% women) | 3 months | 2A | (e26) |
| Interferon-α-2b plus colchicine plus benzathine penicillin | Interferon-α-2b 3 million IU q.o.d., colchicine 1.5 mg daily p.o., benzathine penicillin 1.2 million IU every 3 weeks | ABD: fewer flares of aphthous ulcers (p = 0.007) in the ‧interferon-α-2b group ‧compared to controls | Colchicine plus benzathine penicillin | 65 (interferon-α-2b, colchicine and benzathine penicillin), 65 (colchicine and benzathine penicillin) | 1 year | 2A | (e27) |
| Etanercept | 25 mg twice weekly s.c. | ABD: Reduction in number of lesions and greater likelihood of being free of disease (p = 0.0017) | Placebo | 40 (only men) | 4 weeks | 2A | (e33) |
| Doxycycline | 20 mg b.i.d. p.o. | Reduction in days with new aphthous ulcers (p = 0.04) | Placebo | 25 (doxycycline), 25 (placebo) | 2 months | 2A | (e28) |
| Clofazimine | Clofazimine 100 mg daily p.o. for 30 days, then 100 mg p.o. q.o.d. | More ulcer-free intervals (in 17–44%) than in the ‧other groups | Placebo and colchicine 0.5 mg daily | 23 (clofazimine), 23 (colchicine), and 20 (placebo) | 6 months | 2A | (e34) |
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evaluated as ineffective in (29). BW, body weight; ABD, Adamantiades-Behçet disease
Thalidomide
Thalidomide is considered effective against orogenital aphthous ulcers. In older open or retrospective studies, initial doses of 100–300 mg daily were tapered to 50 mg daily or the medication was discontinued after 3 months, in order to avoid a sensory neuropathy (e21, e22). Thalidomide in a dose of 100 mg daily for an average interval of 5 months was well tolerated by 8 patients with chronic recurrent oral aphthous ulcers (21). Thalidomide should only be used in exceptional cases. Because of its teratogenicity, it is absolutely contraindicated in pregnancy (e23). When it is discontinued, recurrences may develop rapidly (e22, e25). In Germany thalidomide is only approved for treating multiple myeloma.
Interferon-α
Interferon-α can achieve complete or partial remission (reduction in pain, duration and frequency) of recurrent orogenital aphthous ulcers in ABD within 1–4 months (14, 38, e26) (eTable 2) (EL 2A). A low-dose (3 million IU 3 times weekly) maintenance therapy is recommended after 6 months for ABD patients (14). Combination therapy with corticosteroids, colchicine, or benzathine penicillin is possible (e27).
Other systemic agents
In a controlled study, sub-antimicrobial doses of doxycycline (40 mg daily) prolonged the interval between aphthous ulcers (e28) (Table1) (EL 2A). Zinc sulfate 300 mg daily reduced the number and size of aphthous ulcers in comparison to placebo (e29) (eTable1) (EL 2A). In patients with pre-menstrual flares of oral aphthous ulcers, once yearly subcutaneous injections of testosterone helped in some cases (39). Estrogen-dominant oral contraceptives can also be employed (14, 21) (EL 4). An effect is first to be expected after 3 to 6 months.
Summary
Until the etiology of chronic recurrent oral aphthous ulcers is determined, all therapeutic measures are aimed at symptomatic relief. Topical measures should be preferred as first-line therapy because of their low risk for systemic side effects (Figure 2).
Figure 2.
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Algorithm for the treatment of chronic recurrent oral aphthous ulcers to reduce the duration of illness and the size of the ulcers
Systemic measures should only be considered in addition to topical treatment in patients with a severe course and complex aphthosis; options include sucralfate, colchicine, pentoxifylline or prednisolone and combinations thereof. Systemic therapy with other immunosuppressive agents should be reserved for refractory or especially severe aphthous ulcers in patients with ABD.
Key Messages.
The prevalence of chronic recurrent oral aphthous ulcers is 2 to 10%. Women are somewhat more frequently affected.
Iron deficiency anemia, neutropenia, folic acid deficiency, vitamin B12 deficiency, local trauma and other factors all can facilitate the development of oral aphthous ulcers
Hard, acidic and salty food, alcohol and carbonated beverages should be avoided.
In Germany only corticosteroids and topical antiseptics and anti-inflammatory agents are approved for therapy.
In severe cases, topical treatment can be combined with systemic therapy, e.g. with colchicine, pentoxifylline or prednisolone.
Acknowledgments
Translated from the original German by Walter H. C. Burgdorf, MD.
Footnotes
Conflict of interest statement
The authors declare that no conflict of interest exists.
References
1.Altenburg A, Krahl D, Zouboulis CC. Nicht-infektiöse exulzerierende Mundschleimhauterkrankungen. J Dtsch Dermatol Ges. 2008;7:242–257. [Google Scholar]
2.Bork K, Burgdorf W, Hoede N. 3rd. Stuttgart: Schattauer; 2008. Mundschleimhaut und Lippenkrankheiten; pp. 49–58. [Google Scholar]
3.Rogers RS. Complex aphthosis. Adv Exp Med Biol. (3rd) 2003;528:311–316. doi: 10.1007/0-306-48382-3_61. [DOI] [PubMed] [Google Scholar]
4.Hornstein OP. Aphthen und aphthoide Läsionen der Mundschleimhaut. HNO. 1998;46:102–111. doi: 10.1007/s001060050208. [DOI] [PubMed] [Google Scholar]
5.Kleinman DV, Swango PA, Pindborg JJ. Epidemiology of oral mucosal lesions in United States schoolchildren: 1986-87. Community Dent Oral Epidemiol. 1994;22:243–253. doi: 10.1111/j.1600-0528.1994.tb01815.x. [DOI] [PubMed] [Google Scholar]
6.Altenburg A, Abdel-Naser MB, Abdallah M, Seeber H, Zouboulis CC. Practical aspects of management of recurrent aphthous stomatitis. J Eur Acad Dermatol Venereol. 2007;21:1019–1026. doi: 10.1111/j.1468-3083.2007.02393.x. [DOI] [PubMed] [Google Scholar]
7.Chavan M, Jain H, Diwan N, Khedkar S, Shete A, Durkar S. Recurrent aphthous stomatitis: a review. J Oral Pathol Med. 2012;41:577–583. doi: 10.1111/j.1600-0714.2012.01134.x. [DOI] [PubMed] [Google Scholar]
8.Altenburg A, Mahr A, Maldini C, et al. Epidemiologie und Klinik des Morbus Adamantiades-Behçet in Deutschland: Aktuelle Daten. Ophthalmologe. 2012;109:531–541. doi: 10.1007/s00347-012-2601-4. [DOI] [PubMed] [Google Scholar]
9.International Team for the Revision of the International Criteria for Behçet’s Disease (ITR-ICBD): The International Criteria for Behçet’s Disease (ICBD): a collaborative study of 27 countries on the sensitivity and specificity of the new criteria. J Eur Acad Dermatol Venereol. 2014;28:338–347. doi: 10.1111/jdv.12107. [DOI] [PubMed] [Google Scholar]
10.Shim Y, Choi JH, Ahn HJ, Kwon JS. Effect of sodium lauryl sulfate on recurrent aphthous stomatitis: a randomized controlled clinical trial. Oral Dis. 2012;18:655–660. doi: 10.1111/j.1601-0825.2012.01920.x. [DOI] [PubMed] [Google Scholar]
11.Descroix V, Coudert AE, Vigé A, et al. Efficacy of topical 1% lidocaine in the symptomatic treatment of pain associated with oral mucosal trauma or minor oral aphthous ulcer: a randomized, double-blind, placebo-controlled, parallel-group, single-dose study. J Orofac Pain. 2011;25:327–332. [PubMed] [Google Scholar]
12.Skaare AB, Herlofson BB, Barkvoll P. Mouthrinses containing triclosan reduce the incidence of recurrent aphthous ulcers (RAU) J Clin Periodontol. 1996;23:778–781. doi: 10.1111/j.1600-051x.1996.tb00609.x. [DOI] [PubMed] [Google Scholar]
13.Saxen MA, Ambrosius WT, al Rehemtula KF, Russell AL, Eckert GJ. Sustained relief of oral aphthous ulcer pain from topical diclofenac in hyaluronan: a randomized, double-blind clinical trial. Oral Surg Oral Med Oral Pathol Oral Radiol Endod. 1997;84:356–361. doi: 10.1016/s1079-2104(97)90031-7. [DOI] [PubMed] [Google Scholar]
14.Bonitsis NG, Altenburg A, Krause L, Stache T, Zouboulis CC. Current concepts in the treatment of Adamantadies-Behçet’s disease. Drugs Future. 2009;34:749–763. [Google Scholar]
15.Alidaee MR, Taheri A, Mansoori P, Ghodsi SZ. Silver nitrate cautery in aphthous stomatitis: a randomized controlled trial. Br J Dermatol. 2005;153:521–525. doi: 10.1111/j.1365-2133.2005.06490.x. [DOI] [PubMed] [Google Scholar]
16.Zand N, Ataie-Fashtami L, Djavid GE, et al. Relieving pain in minor aphthous stomatitis by a single session of non-thermal carbon dioxide laser irradiation. Lasers Med Sci. 2009;24:515–520. doi: 10.1007/s10103-008-0555-1. [DOI] [PubMed] [Google Scholar]
17.Arabaci T, Kara C, Ciçek Y. Relationship between periodontal parameters and Behçet’s disease and evaluation of different treatments for oral recurrent aphthous stomatitis. J Periodontal Res. 2009;44:718–725. doi: 10.1111/j.1600-0765.2008.01183.x. [DOI] [PubMed] [Google Scholar]
18.Henricsson V, Axéll T. Treatment of recurrent aphthous ulcers with aureomycin mouth rinse or Zendium dentifrice. Acta Odontol Scand. 1985;43:47–52. doi: 10.3109/00016358509064139. [DOI] [PubMed] [Google Scholar]
19.Gorsky M, Epstein J, Rabenstein S, Elishoov H, Yarom N. Topical minocycline and tetracycline rinses in treatment of recurrent aphthous stomatitis: a randomized cross-over study. Dermatol Online J. 2007;13 [PubMed] [Google Scholar]
20.Altenburg A, Zouboulis CC. Current concepts in the treatment of recurrent aphthous stomatitis. Skin Therapy Lett. 2008;13(7):1–4. [PubMed] [Google Scholar]
21.Zouboulis CC. Adamantiades-Behçet’s disease. In: Katsambas AD, Lotti TM, eds.: European Handbook of Dermatological Treatments. 2nd edition. Berlin: Springer. 2003:16–26. [Google Scholar]
22.Fani MM, Ebrahimi H, Pourshahidi S, et al. Comparing the effect of phenytoin syrup and triamcinolone acetonide ointment on aphthous ulcers in patients with Behçet´s syndrome. Iran Red Crescent Med J. 2012;14:75–78. [PMC free article] [PubMed] [Google Scholar]
23.Al-Na’mah ZM, Carson R, Thanoon IA. Dexamucobase: a novel treatment for oral aphthous ulceration. Quintessence Int. 2009;40:399–404. [PubMed] [Google Scholar]
24.Vincent SD, Lilly GE. Clinical, historic, and therpeutic features of aphthous stomatitis. Literature review and open clinical trial employing steroids. Oral Surg Oral Med Oral Pathol. 1992;74:79–86. doi: 10.1016/0030-4220(92)90219-g. [DOI] [PubMed] [Google Scholar]
25.Collier PM, Neill SM, Copeman PW. Topical 5-aminosalicylic acid: a treatment for aphthous ulcers. Br J Dermatol. 1992;126:185–188. doi: 10.1111/j.1365-2133.1992.tb07818.x. [DOI] [PubMed] [Google Scholar]
26.Greer RO, Jr, Lindenmuth JE, Juarez T, Khandwala A. A double-blind study of topically applied 5% amlexanox in the treatment of aphthous ulcers. J Oral Maxillofac Surg. 1993;51:243–248. doi: 10.1016/s0278-2391(10)80164-8. [DOI] [PubMed] [Google Scholar]
27.Liu J, Zeng X, Chen Q, et al. An evaluation on the efficacy and safety of amlexanox oral adhesive tablets in the treatment of recurrent minor aphthous ulceration in a Chinese cohort: a randomized, double-blind, vehicle-controlled, unparallel multicenter clinical trial. Oral Surg Oral Med Oral Pathol Oral Radiol Endod. 2006;102:475–481. doi: 10.1016/j.tripleo.2005.12.014. [DOI] [PubMed] [Google Scholar]
28.Marakoglu K, Sezer RE, Toker HC, Marakoglu I. The recurrent aphthous stomatitis frequency in the smoking cessation people. Clin Oral Investig. 2007;11:149–153. doi: 10.1007/s00784-007-0102-7. [DOI] [PubMed] [Google Scholar]
29.Brocklehurst P, Tickle M, Glenny A-M, et al. Systemic interventions for recurrent aphthous stomatitis (mouth ulcers) Cochrane Database of Systematic Reviews. 2012;9 doi: 10.1002/14651858.CD005411.pub2. CD005411. [DOI] [PMC free article] [PubMed] [Google Scholar]
30.Fontes V, Machet L, Huttenberger B, Lorette G, Vaillant L. Aphtose buccale récidivante: traitement par colchicine (Étude ouverte de 54 cas) Ann Dermatol Venereol. 2002;129:1365–1369. [PubMed] [Google Scholar]
31.Thornhill MH, Baccaglini L, Theaker E, Pemberton MN. A randomized, double-blind, placebo-controlled trial of pentoxifylline for the treatment of recurrent aphthous stomatitis. Arch Dermatol. 2007;143:463–470. doi: 10.1001/archderm.143.4.463. [DOI] [PubMed] [Google Scholar]
32.Pakfetrat A, Mansourian A, Momen-Heravi F, et al. Comparison of colchicine versus prednisolone in recurrent aphthous stomatitis: A double-blind randomized clinical trial. Clin Invest Med. 2010;33:E189–E195. doi: 10.25011/cim.v33i3.13725. [DOI] [PubMed] [Google Scholar]
33.Femiano F, Buonaiuto C, Gombos F, Lanza A, Cirillo N. Pilot study on recurrent aphthous stomatitis (RAS): a randomized placebo-controlled trial for the comparative therapeutic effects of systemic prednisone and systemic montelukast in subjects unresponsive to topical therapy. Oral Surg Oral Med Oral Pathol Oral Radiol Endod. 2010;109:402–407. doi: 10.1016/j.tripleo.2009.10.024. [DOI] [PubMed] [Google Scholar]
34.Alpsoy E, Er H, Durusoy C, Yilmaz E. The use of sucralfate suspension in the treatment of oral and genital ulceration of Behçet disease: a randomized, placebo-controlled, double-blind study. Arch Dermatol. 1999;135:529–532. doi: 10.1001/archderm.135.5.529. [DOI] [PubMed] [Google Scholar]
35.Yazici H, Pazarli H, Barnes CG, et al. A controlled trial of azathioprine in Behçet’s syndrome. N Engl J Med. 1990;322:281–285. doi: 10.1056/NEJM199002013220501. [DOI] [PubMed] [Google Scholar]
36.Zouboulis CC. Morbus Adamantiades-Behçet. In: Mrowietz U, eds.: Ciclosporin in der Dermatologie. Stuttgart: Thieme. 2003:38–51. [Google Scholar]
37.Masuda K, Nakajima A, Urayama A, Nakae K, Kogure M, Inaba G. Double-masked trial of cyclosporine versus colchicine and long-term open study of cyclosporin in Behçet’s disease. Lancet. 1989;1:1093–1096. doi: 10.1016/s0140-6736(89)92381-7. [DOI] [PubMed] [Google Scholar]
38.Zouboulis CC, Orfanos CE. Treatment of Adamantiades-Behçet’s disease with systemic interferon alfa. Arch Dermatol. 1998;134:1010–1016. doi: 10.1001/archderm.134.8.1010. [DOI] [PubMed] [Google Scholar]
39.Misra R, Anderson DC. Treatment of recurrent premenstrual orogenital aphthae with implants of low dose of testosterone. BMJ. 1989;299 doi: 10.1136/bmj.299.6703.834. [DOI] [PMC free article] [PubMed] [Google Scholar]
E1.Schroeder HE, Müller-Glauser W, Sallay K. Stereologic analysis of leukocyte infiltration in oral ulcers of developing Mikulicz aphthae. Oral Surg Oral Med Oral Pathol. 1983;56:629–640. doi: 10.1016/0030-4220(83)90081-6. [DOI] [PubMed] [Google Scholar]
E2.Altenburg A, Papoutsis N, Orawa H, Martus P, Krause L, Zouboulis CC. Epidemiologie und Klinik des Morbus Adamantiades-Behçet in Deutschland - Aktuelle pathogenetische Konzepte und therapeutische Möglichkeiten. J Dtsch Dermatol Ges. 2006;4:49–64. doi: 10.1111/j.1610-0387.2006.05841.x. [DOI] [PubMed] [Google Scholar]
E3.Bruce AJ, Rogers RSW. Acute oral ulcers. Dermatol Clin. (3rd) 2003;21:49–61. doi: 10.1016/s0733-8635(02)00064-5. [DOI] [PubMed] [Google Scholar]
E4.Fox EC. The problem or oral ulcerations in general practice with special reference to herpetic herpetiform lesions. J R Coll Gen Pract. 1970;19:191–200. [PMC free article] [PubMed] [Google Scholar]
E5.Lopez-Jornet P, Camacho-Alonso F, Martos N. Hematological study of patients with aphthous stomatitis. Int J Dermatol. 2014;53:159–163. doi: 10.1111/j.1365-4632.2012.05751.x. [DOI] [PubMed] [Google Scholar]
E6.Chen Y, Fang L, Yang X. Cyclic neutropenia presenting as recurrent oral ulcers and periodontitis. J Clin Pediatr Dent. 2013;37:307–308. doi: 10.17796/jcpd.37.3.n8k0111177074828. [DOI] [PubMed] [Google Scholar]
E7.Koybasi S, Parlak AH, Serin E, Yilmaz F, Serin D. Recurrent aphthous stomatitis: investigation of possible etiologic factors. Am J Otolaryngol. 2006;27:229–232. doi: 10.1016/j.amjoto.2005.09.022. [DOI] [PubMed] [Google Scholar]
E8.Volkov I, Rudoy I, Abu-Rabia U, Masalha T, Masalha R. Case report: Recurrent aphthous stomatitis responds to vitamin B12 treatment. Can Fam Physician. 2005;51:844–845. [PMC free article] [PubMed] [Google Scholar]
E9.Wray D, Graykowski EA, Notkins AL. Role of mucosal injury in initiating recurrent aphthous stomatitis. Br Med J (Clin Res Ed) 1981;283:1569–1570. doi: 10.1136/bmj.283.6306.1569. [DOI] [PMC free article] [PubMed] [Google Scholar]
E10.Jurge S, Kuffer R, Scully C, Porter SR. Mucosal disease series Number VI. Recurrent aphthous stomatitis. Oral Dis. 2006;12:1–21. doi: 10.1111/j.1601-0825.2005.01143.x. [DOI] [PubMed] [Google Scholar]
E11.Jennette JC. 2012 revised International Chapel Hill Consensus Conference nomenclature of vasculitides. Arthritis Rheum. 2013;65:1–11. doi: 10.1002/art.37715. [DOI] [PubMed] [Google Scholar]
E12.International Study Group for Behçet’s Disease. Criteria for diagnosis of Behçet’s disease. Lancet. 1990;335:1078–1980. [PubMed] [Google Scholar]
E13.Gorsky M, Epstein J, Raviv A, Yaniv R, Truelove E. Topical minocycline for managing symptoms of recurrent aphthous stomatitis. Spec Care Dentist. 2008;28:27–31. doi: 10.1111/j.1754-4505.2008.00006.x. [DOI] [PubMed] [Google Scholar]
E14.Rivera-Hidalgo F, Shulman JD, Beach MM. The association of tobacco and other factors with recurrent aphthous stomatitis in an US adult population. Oral Dis. 2004;10:335–345. doi: 10.1111/j.1601-0825.2004.01049.x. [DOI] [PubMed] [Google Scholar]
E15.Bittoun R. Recurrent aphthous ulcers and nicotine. Med J Aust. 1991;154:471–472. doi: 10.5694/j.1326-5377.1991.tb121180.x. [DOI] [PubMed] [Google Scholar]
E16.Davatchi F, Sadeghi Abdollahi B, et al. Colchicine versus placebo in Behçet’s disease: randomized, double-blind, controlled crossover trial. Mod Rheumatol. 2009;19:542–549. doi: 10.1007/s10165-009-0200-2. [DOI] [PubMed] [Google Scholar]
E17.Yurdakul S, Mat C, Tüzün Y, et al. Double-blind trial of colchicine in Behçet’s syndrome. Arthritis Rheum. 2001;44:2686–2692. doi: 10.1002/1529-0131(200111)44:11<2686::aid-art448>3.0.co;2-h. [DOI] [PubMed] [Google Scholar]
E18.Calgüneri M, Kiraz S, Ertenli I, Benekli M, Karaarslan Y, Celik I. The effect of prophylactic penicillin treatment on the course of arthritis episodes in patients with Behçet’s disease. A randomized clinical trial. Arthritis Rheum. 1996;39:2062–2065. doi: 10.1002/art.1780391216. [DOI] [PubMed] [Google Scholar]
E19.Rattan J, Schneider M, Arber N, Gorsky M, Dayan D. Sucralfate suspension as a treatment of recurrent aphthous stomatitis. J Intern Med. 1994;236:341–343. doi: 10.1111/j.1365-2796.1994.tb00805.x. [DOI] [PubMed] [Google Scholar]
E20.Sharquie KE, Najim RA, Abu-Raghif AR. Dapsone in Behçet’s disease: a double-blind, placebo-controlled, cross-over study. J Dermatol. 2002;29:267–279. doi: 10.1111/j.1346-8138.2002.tb00263.x. [DOI] [PubMed] [Google Scholar]
E21.Bonnetblanc JM, Royer C, Bedane C. Thalidomide and recurrent aphthous stomatitis: A follow-up study. Dermatology. 1996;193:321–323. doi: 10.1159/000246279. [DOI] [PubMed] [Google Scholar]
E22.Grinspan D. Significant response of oral aphthosis to thalidomide treatment. J Am Acad Dermatol. 1985;12:85–90. doi: 10.1016/s0190-9622(85)70014-x. [DOI] [PubMed] [Google Scholar]
E23.Grinspan D, Blanco GF, Aguero S. Treatment of aphthae with thalidomide. J Am Acad Dermatol. 1989;20:1060–1063. doi: 10.1016/s0190-9622(89)70132-8. [DOI] [PubMed] [Google Scholar]
E24.Hamuryudan V, Mat C, Saip S, et al. Thalidomide in the treatment of the mucocutaneous lesions of the Behçet syndrome. A randomized, double-blind, placebo-controlled trial. Ann Intern Med. 1998;128:443–450. doi: 10.7326/0003-4819-128-6-199803150-00004. [DOI] [PubMed] [Google Scholar]
E25.Calabrese L, Fleischer AB. Thalidomide: current and potential clinical applications. Am J Med. 2000;108:487–495. doi: 10.1016/s0002-9343(99)00408-8. [DOI] [PubMed] [Google Scholar]
E26.Alpsoy E, Durusoy C, Yilmaz E, et al. Interferon alfa-2a in the treatment of Behçet disease: a randomized placebo-controlled and double-blind study. Arch Dermatol. 2002;138:467–471. doi: 10.1001/archderm.138.4.467. [DOI] [PubMed] [Google Scholar]
E27.Demiroglu H, Ozcebe OI, Barista I, Dündar S, Eldem B. Interferon alfa-2b, colchicine, and benzathine penicillin versus colchicine and benzathine penicillin in Behçet’s disease: a randomised trial. Lancet. 2000;355:605–609. doi: 10.1016/S0140-6736(99)05131-4. [DOI] [PubMed] [Google Scholar]
E28.Preshaw PM, Grainger P, Bradshaw MH, Mohammad AR, Powala CV, Nolan A. Subantimicrobial dose doxycycline in the treatment of recurrent oral aphthous ulceration: a pilot study. J Oral Pathol Med. 2007;36:236–240. doi: 10.1111/j.1600-0714.2007.00507.x. [DOI] [PubMed] [Google Scholar]
E29.Sharquie KE, Najim RA, Al-Hayani RK, Al-Nuaimy AA, Maroof DM. The therapeutic and prophylactic role of oral zinc sulfate in management of recurrent aphthous stomatitis (ras) in comparison with dapsone. Saudi Med J. 2008;29:734–738. [PubMed] [Google Scholar]
E30.Murray B, Biagioni PA, Lamey PJ. The efficacy of amlexanox OraDisc on the prevention of recurrent minor aphthous ulceration. J Oral Pathol Med. 2006;35:117–122. doi: 10.1111/j.1600-0714.2006.00379.x. [DOI] [PubMed] [Google Scholar]
E31.Pourahmad M, Rahiminejad M, Fadaei S, Kashafi H. Effects of camel thorn distillate on recurrent oral aphthous lesions. J Dtsch Dermatol Ges. 2010;8:348–352. doi: 10.1111/j.1610-0387.2010.07316.x. [DOI] [PubMed] [Google Scholar]
E32.Matsuda T, Ohno S, Hirohata S, et al. Efficacy of rebamipide as adjunctive therapy in the treatment of recurrent oral aphthous ulcers in patients with Behçet’s disease: a randomised, double-blind, placebo-controlled study. Drugs. 2003;4:19–28. doi: 10.2165/00126839-200304010-00002. [DOI] [PubMed] [Google Scholar]
E33.Melikoglu M, Fresko I, Mat C, et al. Short-term trial of etanercept in Behçet’s disease: a double blind, placebo controlled study. J Rheumatol. 2005;32:98–105. [PubMed] [Google Scholar]
E34.de Abreu MA, Hirata CH, Pimentel DR, Weckx LL. Treatment of recurrent aphthous stomatitis with clofazimine. Oral Surg Oral Med Oral Pathol Oral Radiol Endod. 2009;108:714–721. doi: 10.1016/j.tripleo.2009.05.009. [DOI] [PubMed] [Google Scholar]
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15840 | https://www.smchealth.org/sites/main/files/file-attachments/guidelines_for_prescribing_and_monitoring_clozapine_8-10-24.pdf?1732576136 | 1
BHRS Guidelines for Prescribing and Monitoring Clozapine
Purpose: To provide a structured approach for the safe and effective use of clozapine in treating treatment-resistant schizophrenia (TRS) and reducing suicidal behavior in patients with schizophrenia or schizoaffective disorder.
Indications
FDA-approved indications:
•
Treatment-resistant schizophrenia
For severely ill patients who do not respond adequately to other antipsychotic treatment.
•
Reducing suicidal behavior in patients with schizophrenia or schizoaffective disorder
Of note: currently the only anti-psychotic FDA-approved for suicide prevention.
Non-FDA uses (with supporting evidence):
•
Psychosis co-occurrent with and due to Parkinson's disease (per Micromedex)
•
Managing aggressive behavior in schizophrenia (per MD Quanbeck)
•
Consider for patients with symptoms partially or fully resistant to other antipsychotics or accompanied by persistent suicidal or self-injurious behavior. Other indications include sensitivity to extrapyramidal symptoms and tardive dyskinesia. Clozapine does not need to be reserved as a treatment of last resort and can be considered after 2-3 failed antipsychotic trials based on clinical judgment.
Initiating Treatment
Conduct a comprehensive baseline assessment, including:
•
General and cardiovascular health status
•
Pro Tip: Utilize
Clozapine Monitoring Guide (Post-REMS): Understanding the Risk of
Neutropenia - A Guide for Healthcare Providers
Clozapine Lab Monitoring Procedures and Workflow with nursing
support
•
CBC with differential, metabolic panel, lipid panel, HbA1c, and vital signs
•
ECG
Particularly in elderly population to document existing abnormalities that might incorrectly be identified as “clozapine-induced” & disrupt treatment.
•
Baseline body habitus: weight, height, & waist circumference
•
Standard of Care for anti-psychotics (i.e., AIMS, serum levels, pregnancy test)
•
Baseline troponin & high sensitivity C-reactive protein (CRP) due to significant risk of myocarditis
Ensure no contraindications are present and obtain informed consent after discussing benefits & risks.
Discontinue contraindicated medications:
•
Benzodiazepines (risk of cardiorespiratory collapse)
•
Carbamazepine (increased risk of neutropenia) 2
•
Ciprofloxacin (dramatically increases clozapine levels, leading to toxicity)
Add “ciprofloxacin allergy while on clozapine due to DDI"
Initiate clozapine at 12.5-25 mg/day (12.5 mg test dose for first-time users), titrating slowly based on tolerability and setting (inpatient vs. outpatient)
•
Continue taper of non-clozapine anti-psychotics while titrating up clozapine
Monitor for therapeutic response and adverse effects over a 6-month trial period
Check clozapine levels 7-14 days after completing titration, aiming for 350-600 ng/mL (minimum therapeutic threshold is 100 ng/mL; point of futility > 1000 ng/mL)
Dosage and Titration
•
Initial target dose: 300-450 mg/day (150 mg/day if co-administered with fluvoxamine)
•
Maximum dose: 900 mg/day
•
Fluvoxamine co-administration with clozapine
•
Clozapine is metabolized into norclozapine by CYP450 1A2 enzymes
•
Norclozapine is thought to be associated with metabolic adverse effects, such as weight gain and dyslipidemia
•
Fluvoxamine is a strong CYP450 1A2 inhibitor
•
To increase the clozapine to norclozapine ratio, initiate fluvoxamine at 25 mg qhs and increase dose by 25 mg as tolerated over 1-3 days to 150 mg
•
Effect: therapeutic levels attained at 150 mg clozapine that are on average equal to 350 mg of clozapine treatment without fluvoxamine.
•
May consider fluvoxamine if clinically appropriate (note different titration schedule for coadministration).
Note : The practice of adding fluvoxamine to clozapine and monitoring the clozapine to norclozapine ratio (CLZ:NDMC) may not be standard across all settings. Potential disruptions can occur if patients are transferred to settings that do not follow this practice.
•
Setting : Non-urgent, outpatient (slow titration)
•
Start at 12.5 – 25 mg daily QHS for 3 days
•
Continue to titrate up 25 mg daily until target dosage is reached
•
Setting : Urgent, inpatient, medically-supervised (fast titration)
•
Without fluvoxamine - Follow below titration schedule:
•
With fluvoxamine - Start at 25 mg and titrate up by 25 mg daily until 150 mg dosage reached
•
Slower titration and lower doses may be necessary for older patients or those with cardiac issues
Day Dosage
7 50 mg am + 100 mg hs
8 50 mg am + 100 mg hs
9 100 mg bid
10 100 mg bid
11 50 mg am + 200 mg hs
12 50 mg am + 200 mg hs
13 100 mg am + 200 mg hs
14 100 mg am + 200 mg hs
Day Dosage
1 12.5 mg bid
2 25 mg am
3 25 mg bid
4 25 mg am + 50 mg hs
5 50 mg bid
6 50 mg am + 75 mg hs 3
•
Adjust dosage for individuals of Asian descent, who typically require lower doses
•
Consider dividing doses if excessive sedation or other side effects (enuresis, AM grogginess) occur
Monitoring
• Hematological monitoring per Clozapine Package Insert
•
Weekly for the first 6 months, biweekly for the second 6 months, then monthly
•
See: Clozapine Lab Monitoring Procedures and Workflow with nursing support
•
Manage neutropenia based on absolute neutrophil count (ANC) thresholds
Note: Refer to Clozapine Monitoring Guide (Post-REMS): Understanding
the Risk of Neutropenia - A Guide for Healthcare Providers for managing
neutropenia. Providers should follow BHRS guidelines and stay informed of any future clinical or regulatory updates to ensure adherence to the most current protocols.
Normal: 1.5 – 8 x 10 3 /μL
Mild: 1 – 1.499 x 10 3 /μL
o Increase monitoring to three times/week until ANC > 1.5 x 10 3 /μL)
Moderate: 0.5 – 0.999 x 10 3 /μL
o Clozapine interruption and daily monitoring until ANC reaches 1 x 10 3 /μL and then clozapine can be resumed
Severe: Less than 0.5 x 10 3 /μL
o Refer to emergency room for further management
o Discontinue clozapine and increase ANC monitoring
o Can resume when ANC > 1.5 x 10 3 /μL if benefits greatly outweigh risks
•
Confirmatory tests within 24 hours if ANC < 1.5 10 3 /μL
•
The formula for the ANC calculation is: ANC = WBC count × (percentage of segs + percentage of bands) or ANC = WBC count × [(number of segs + number of bands) / 100]
WBC count is usually in cells per microliter (cells/μL)
The percentage of segs and bands is given in the differential count as a percentage
•
Treating agranulocytosis: filgrastim can expedite ANC recovery; however, its impact on infection rates and mortality is unclear
Of Note: a significant portion of African-Americans have benign ethnic neutropenia resulting in lower utilization and higher rates of discontinuation despite being less likely to develop severe neutropenia
• Metabolic monitoring:
•
Weight, waist circumference, blood pressure, fasting glucose or HbA1c, and lipid panel at regular intervals
•
Manage abnormalities through lifestyle modifications and pharmacological interventions (metformin and/or topiramate, GLP-1 receptor agonists, or co-administration with fluvoxamine)
Of Note: lower metabolic risks than olanzapine
• Cardiovascular monitoring: 4
•
Monitor for myocarditis, particularly during first 4-8 weeks of treatment
•
Clinical indicators of myocarditis include fever >38°C, chest pain, SBP <100, HR >120, & shortness of breath
•
Refer to ER, if myocarditis or cardiomyopathy is suspected. Diagnostic measures like echocardiogram & troponin levels should be performed in an emergency setting
•
Each visit, assess for symptoms & check vital signs
•
Weekly troponin I & high sensitivity CRP for the first 8 weeks
If Troponin I not available, monitor troponin T (less specific)
CRP > 100 mg/L and troponin > twice the normal limit are critical indicators of clozapine-induced myocarditis in symptomatic patients
Eosinophil count may rise as well, but less consistent and delayed
•
Standard ECG monitoring on clozapine: baseline and annual. Risk for QTc interval prolongation
ECG key for assessing ventricular and valve function with suspected myocarditis/cardiomyopathy
•
Monitor for orthostatic hypotension, bradycardia, and syncope
• Seizure monitoring:
•
Inquire about seizure history and monitor for myoclonic jerking
•
If a seizure occurs, hold clozapine for 24 hours, reduce the dose by 50%, and consider adding an anticonvulsant
Divalproex/Valproic acid (VPA): Load 30 mg/kg over 24 hours. Can exacerbate weight gain and carries an increased risk of developing neutropenia. Effective for both generalized and myoclonic seizures.
Valproic Acid & derivatives have been noted to increase the risk of myocarditis. Use with caution, balancing the risk of seizures with the potential risk of myocarditis. Monitor for increased clozapine toxicities (eg., myocarditis, neutropenia, CNS effects) if combined with valproate. Consider monitoring serum clozapine and/or norclozapine (N-desmethylclozapine) levels, as variable effects on the pharmacokinetics of clozapine have been reported with the combination.
Lamotrigine: Cannot be loaded. Slow titration to reduce risk of developing Steven-Johnson Syndrome. Good tolerability and limited weight gain. Effective for both generalized and myoclonic seizures.
Gabapentin: Effective in preventing generalized seizures but not myoclonic seizures.
Topiramate: Effective in preventing generalized seizures but not myoclonic seizures. Can cause weight loss but cognitive side effects can be problematic.
Levetiracetam: No interactions with clozapine.
Phenytoin and Carbamazepine: Both should be avoided as they will reduce clozapine plasma levels > 50%. Carbamazepine increases the risk of neutropenia.
•
Clozapine titration can be resumed after an anticonvulsant is added
Of Note: 10% risk of generalized tonic-clonic seizures after four years; Risk is dose-dependent, doubling at 300-600 mg/day and tripling at > 600 mg/day 5
• Gastrointestinal monitoring:
•
Prevalence of constipation with clozapine: 32 – 60%
•
Monitor for constipation and treat prophylactically with:
High-fiber foods, adequate fluid intake, & exercise
Stool softeners: DSS 250 mg or Miralax 17 grams at initiation
Laxatives: Dulcolax 10 mg or Senna 17.2 mg daily
If constipation occurs prn lactulose
o For non-responsive cases, start an intestinal secretagogue -lubiprostone or linaclotide)
Of Note: Paralytic ileus can occur if constipation not monitored and treated with fatality rate of 15 - 28%, compared to severe neutropenia fatality rate of only 2 - 4%
•
Prevalence of hypersalivation with clozapine: 90% with greater severity at night
•
Manage hypersalivation with anticholinergic medications or other treatments as needed
Anticholinergics: reduce drooling by 44%; however, long-term carries risk of dementia, falls, and mortality
First-line treatment:
o Atropine 1% ophthalmic drops taken sublingually, initiate at 3 drops qhs.
o Terazosin 2 mg qhs
o Clonidine 0.1 - 0.2 mg qhs
Second-line treatment:
o Glycopyrrolate 2 mg qhs - does not cross blood-brain barrier, reducing anticholinergic effect on cognitive functions observed in other anticholinergics. Still carries risk for constipation
Third-line treatment:
o Botulinum Toxin-B injections into salivary glands
• Therapeutic drug monitoring:
•
Check levels during titration, in cases of poor response, or when non-adherence is suspected
•
Check levels 7-14 days after titration and every 3 months in stable outpatient settings
•
Aim for 350-600 ng/mL, adjusting based on individual response and side effects
Average dose for TRS is ~400 mg/day
Check level in AM, approximately 8-10 hours after evening dose
Monitor clinical response for two weeks after reaching therapeutic level before further dosage increase
Reasonable to consider serum level > 350 ng/mL for refractory symptoms, although evidence of greater efficacy is limited
Variability in clozapine metabolism can lead to unpredictable levels making initial serum monitoring essential.
Continuous monitoring not mandated; however, beneficial in monitoring adherence
• Monitor smoking and caffeine intake due to their impact on clozapine metabolism 6
• Additionally, monitor for Neuroleptic Malignant Syndrome, hepatoxicity, pulmonary embolism, anticholinergic toxicity, and cognitive/motor performance.
• Continuously educate patients on adherence, side effects, and when to seek help
Missed Doses
• Client Education
• Take the missed dose as soon as you remember.
• If it is close to the time for your next dose, skip the missed dose and resume your regular dosing schedule.
• Do not take two doses at the same time or extra doses.
• If you miss one day or more of clozapine, you may need to restart at a lower dose to reduce the risk of side effects. Call your doctor to find out how to restart the medication.
• It’s important to take clozapine every day as prescribed to avoid increased side effects and relapse of symptoms.
• Reinitiation of treatment after interruption
• Reinitiation after an interruption requires dosage reduction to minimize the risk of hypotension, bradycardia, & syncope
1 day's missed dosing: Resume at 40% to 50% of the established dose
2 days' missed dosing: Resume at approximately 25% of the established dose.
Longer interruptions: Re initiate at 12.5 mg once or twice daily.
• If these dosages are well-tolerated, dosage may be increased to the previous dose more quickly than is recommended for initial treatment.
• A clozapine to N-desmethylclozapine (norclozapine) ratio (CLZ:NDMC) greater than 2 is suggestive of a non-trough sample, a recent missed dose, or the saturation/inhibition of the CYP1A2 or other CYP450 enzymes.
Discontinuation
•
Taper clozapine gradually to minimize the risk of cholinergic rebound syndrome or rapid relapse
•
Prescribe anticholinergic medications to prevent withdrawal symptoms, specifically Cholinergic Rebound Syndrome (nightmares, anxiety, insomnia, nausea, diarrhea, sweating, confusion, delirium, and/or catatonia)
•
For nonsmokers, for every 50 mg clozapine add 1 mg benztropine or 25 mg diphenhydramine.
•
For smokers, double the dose
•
Taper anticholinergics slowly after two weeks
•
In cases of severe adverse effects (e.g., agranulocytosis, myocarditis), urgent discontinuation may be necessary
Documentation and Communication
•
Maintain detailed records of all monitoring parameters, interventions, and communication with healthcare providers 7
•
Collaborate with other specialists (e.g., hematology, cardiology) as needed to manage adverse effects
By implementing this policy, healthcare providers can optimize clozapine therapy while minimizing the risk of serious adverse events, ultimately improving outcomes for patients with treatment-resistant schizophrenia and other indicated conditions.
Clinical Pearls, Clozapine Facts and Resources
• Patients with TRS receiving clozapine treatment are more likely to maintain sobriety and reduce substance use compared to patients on other antipsychotics due to its effects on cravings
• MDs often prescribe more than two antipsychotics before using clozapine despite APA guidelines, and many psychiatrists do not use clozapine at all.
• Clozapine benefits vs risks:
Benefits Risks
Treatment resistance (>60% will have a significant response Early severe neutropenia (0.4%)
Decreases suicide risk five-fold Myocarditis (0.18%)
Reduction in all-cause mortality by 45% Seizures (predictor of good response)
Unique anti-aggression properties Chronic sedation
Reduces substance misuse Constipation
Greater functional ability Drooling
Helps stop the “revolving door” of repeated emergency and acute psychiatric care 8
• References for further information
• Clozapine 101 APA 2023
• Clozapine Flowchar t
• Clozapine Underutilization: Addressing the Barriers
• Lexicomp Online. Accessed 27 July 2025.
• Managing Clozapine Side Effects APA 2023
• Micromedex Online. Accessed 27 July 2025.
• Mayo Clinic. "Clozapine (Oral Route)." Mayo Clinic, last updated: 1 July 2025,
Accessed 27 July 2025. 9
• Meyer, Jonathan M., and Jose M. Rubio. " Clozapine Monitoring in the Post-
REMS World: Some Guidance for Clinicians ." The Journal of clinical psychiatry
86.2 (2025): 25ac15898. Accessed 27 July 2025.
• SMI Adviser (an APA & SAMHSA initiative, free)
• The Clozapine Handbook by Jonathan Meyer and Stephen Stahl |
15841 | https://www.youtube.com/watch?v=J8VHAp3AzWk | Central Angles and Inscribed Angles! (theorems AND examples)
You Can Learn Math with Alyssa
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Posted: 28 Apr 2022
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What does central angle mean? What is an example of a central angle? What are inscribed angles? How do you solve for an inscribed angle in a circle? What is an example of an inscribed angle? This video explains the theorems behind central and inscribed angles in circles and how to solve basic and advanced problems using these theorems.
Here are some additional videos on angles related to circles - intersecting chords, angles made from tangents and chords, and angles outside circles:
Angles of Intersecting Chords! (theorem AND examples):
Tangent Chord Angles! (theorem AND examples):
How to Find an Angle Outside of a Circle! (tangents and secants):
Thank you so much for watching! If you have any questions or requests for future videos, please leave a comment.
Keep an eye out for my new Math Merch - it's gonna be fun! :-)
If this was helpful in any way, please like, share, comment, or subscribe - it's the only way I can keep making more of these :-)
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Transcript:
hi everyone it's alyssa and welcome to you can learn math so today we're going to talk about central and inscribed angles in circles now these as a concept central and inscribed angles pretty straightforward pretty simple but hey you should know by now in your math class it's not going to be that easy they're going to throw some wrinkles in there to make you work for it so let's start with what a central or inscribed angle is if i have a circle and i have my my center is right about there and i were to draw an angle like this it's an angle where both corners or both end points of this angle are touching the edge of the circumference of the circle and the vertex of the angle is at the center center central this is a central angle if i were to draw an angle where it doesn't matter where this is this could be here's an example make a more narrow one again both of the end points are on the edge of the circumference of the circle but this time the vertex is also on the circumference on the edge of the circle this is an inscribed angle those are our terms central the vertex is at the center inscribed the vertex is on the edge now these angles can be as wide or as narrow i could have a central angle that's very wide like this i could have an inscribed angle that's very wide like that again the only caveats the only requirements for this would be to have both end points on the circumference the edge of the circle and if it's central the vertex is at the center inscribed the vertex is on the circumference of the edge as well right so what are we going to do with these mathematically well here is here's the rule i say this angle i'm trying to make this as close to there we go as you can tell i have drawn here a 90 degree angle right 90 degrees what if i were to do this a couple more times if i were to make a couple more 90 degree angles so i have four 90 degree angles and they're all at the center and they are intercepting the circumference so it's divided it into four equal parts now each of these parts this would be one-fourth of the circumference this bit would be one-fourth of the circumference this would be one-fourth of the circumference and this would be one-fourth of the circumference now we can represent this as either a actual an actual distance like if i knew that this entire circle all the way around was 20 inches then i would say okay this arc is five inches long this arc is five inches long this is five inches long and this is five inches long or i can make a more general statement like if i didn't know the exact inches around i could say well a circle if i start and then go all the way around i've gone 360 degrees just like with an angle so how much of the circle have i gone here i've gone 90 of that 360. there's another 90 another 90 another 90. so arcs can be measured as degrees just like angles because it's what portion of this 360 degree circle are we talking about so if you have a central angle and here you see this is 90 degrees and this arc up here is 90 degrees because it's one-fourth of the way around this 360 degree circle so what can we assume from that we have our central angle is 90 degrees and our arc is 90 degrees well it's not too much of a leap to say that the central angle and the arc will be the same they will have the same degree measurement and that's always true central angle and arc same degree measurement now this one this next one is a little harder to illustrate and in fact it may seem a little impossible to believe but for inscribed angles i'll try to draw a straight line here there we go an inscribed angle and i tried to get this so that this was the quarter point so that this arc was 90 degrees should be pretty close to 90 degrees the inscribed angle is half of the intercepted arc that's the terminology you're going to hear a lot the intercepted arc the arc it is crossing or intercepting it is half of that so if this arc is 90 this angle is 45 degrees always and i know this is going to seem a little hard to believe but it is true if this angle is over here it's true if this angle is all the way over there it's true if this angle is straight back no matter where this angle is as long as it is intercepting that same arc it will have the same degree measure and you can see that see here all these are you know if you draw them more precisely they will be all precisely half of 90 in this case 45 degrees so that's not super hard to have as a concept to memorize okay if it's in the center it's the same if the vertex is on the edge itself then it's half of the intercepted arc well you know they're not going to leave it that simple for long right right you know this by now so they're going to start throwing different problems at you they're going to use combinations of these different concepts so and they're going to throw x's in x's they come back from algebra they come back to haunt you so here's an example let's say no i make this a straight line okay there we go i love things like this if you're a sci-fi fan you're like hey it looks like a star trek symbol but it's not it's a math problem so let's say they tell you that this angle is 110 degrees and they ask what is this angle what is x you can't look at this as a odd funky looking quadrilateral you have to look at it as two separate angles this angle which is an inscribed angle and this is its intercepted arc and then separately this central angle and this same arc is its intercepted arc so the way you would solve for x is you would use the knowledge that you have that one a central angle and its intercepted arc have the same degree so that means this arc is 110 degrees now we have to kind of push that angle here we have to kind of mentally erase it can't really do that on your paper i know but i'm trying to help here so you have to kind of mentally erase it so all you're looking at is this now this inscribed angle and what's the rule for the inscribed angle it is half of its intercepted arc so half of 110 is 55 degrees so x is equal to 55 degrees they love these sort of overlapping angles they just they really do they really love these here's another example of one and this came from a student i was tutoring let's make those lines a little straighter shall we okay and they very much specify that this is the center and they do not these lines are not going through the center okay so we're calling this a b c and d and they say okay a c this little symbol here means the arc a c is congruent to the arc bd and the ratio of c d to b d is 2 to 1. if the measure that's what this little m means the measure of arc a b equals 80. then they wanted us to find the measure of cd the measure of bd the measure of the angle bcd the measure of ac the measure of the angle abc and the measure of the very large arc a b d you go oh my goodness well we're going to use all of these central and inscribed angles and our knowledge about them to figure this out so first we are told that the arc ac is congruent to the arc bd so this right here is congruent to this all right so these arcs are the same the ratio of c d to b d is two to one i can't really mark that on my drawing right now but i'm going to keep that in mind for the future so i'm gonna say this is two times let's make a little note to myself bd because this is two to one so cd is twice as big as bd so let's just note to myself all right now measure of arc a b is 80 degrees all right this okay arc a b up here i'm actually going to highlight it there we go mark a b is 80 degrees now from this piece of information they love throwing algebra in they really do because you think it's going to be just oh if the central angle is 90 it's 90. i'm done with my homework but they love throwing this stuff in and making it a puzzle so this one was a good example of using some algebra what if i called this x the arc a c is x well then bd is the same it is also x and then we know that cd this arc down here is 2 times bd the ratio is 2 to 1. so this would be 2 x well how do i solve for x well the whole circle is 360 degrees so if i add all these pieces together i can figure it out i have my arc here up top mark arc a b is 80. then i have arc bd is x arc cd is 2x and arc ac or ca is also x and all together those equal 360. i combine my x's to be 4x then i'm going to subtract 80 from both sides and i get 4x equals 280 then i divide both sides by 4 and i get x equals 70. that's very valuable now i know i can fill all these in i know this is 70 this is 70 and this is 140. okay so the first one here what is the measure of arc cd we've solved it's 140 next one what is the measure of arc bd we've solved that's 70. now they want to know the measure of the angle bcd this angle well now we can use our inscribed angle information that we know that see this angle is intercepting this arc that arc is 70. inscribed angles are half of their intercepted arc so half of 70 is 35 degrees so there's our next one the measure of angle bcd is 35 degrees measure of arc ac we know that one too that's also 70 degrees measure of angle a b c that one same thing this is an inscribed angle and its intercepted arc is 70. inscribed angles are half of their intercepted arcs half of 70 is 35 so now i know measure of angle abc is 35 and finally they wanted to know the measure of arc abd from a to b to d that is a hundred and fifty eighty plus seventy so again they love these kind of problems you're gonna see a lot of them where they have overlapping angles and things that look like triangles and quadrilaterals and stuff inside angles but they really are just multiple angles central and intercepted stacked up each other and you kind of have to be a problem-solving detective to solve them i'm going to have videos up shortly that are going to be about other types of interior and exterior angles so look out for those once i post those i will put links to those in the description below i hope you enjoyed that today um if it was helpful useful in any way please like share subscribe you know the drill thank you so much for stopping by hope you have a great day see you later bye you |
15842 | https://math.stackexchange.com/questions/2634557/converting-polar-equations-to-cartesian-equations/2634571 | algebra precalculus - Converting polar equations to cartesian equations. - Mathematics Stack Exchange
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Converting polar equations to cartesian equations.
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Asked 7 years, 7 months ago
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Where
r=sin(3 θ)r=sin(3 θ)
and
y=r sin(θ),x=r cos(θ),r 2=x 2+y 2 y=r sin(θ),x=r cos(θ),r 2=x 2+y 2
I have started by saying that
r=sin(2 θ)cos(θ)+sin(θ)cos(2 θ)r=2 sin(θ)cos 2(θ)+sin(θ)(1−2 sin 2(θ))r=2 sin(θ)cos 2(θ)+sin(θ)−2 sin 3(θ)r=sin(2 θ)cos(θ)+sin(θ)cos(2 θ)r=2 sin(θ)cos 2(θ)+sin(θ)(1−2 sin 2(θ))r=2 sin(θ)cos 2(θ)+sin(θ)−2 sin 3(θ)
simply making the substitutions
sin(θ)=y r,cos(θ)=x r sin(θ)=y r,cos(θ)=x r
noting also that I can square both sides of the above to be substitutions we then can write down
r=2 y x r⋅x 2 r 2+y r−2 y 3 r 3 r=2 y x r⋅x 2 r 2+y r−2 y 3 r 3
then multiplying through by r 3 r 3 we obtain
r 4=2 y x 3+y r 2−2 y 3 r 4=2 y x 3+y r 2−2 y 3
then replacing r 2 r 2 with x 2+y 2 x 2+y 2 we get
(x 2+y 2)2=2 x y 2+y x 2+y 3−2 y 3(x 2+y 2)2=y(3 x 2−y 2)(x 2+y 2)2=2 x y 2+y x 2+y 3−2 y 3(x 2+y 2)2=y(3 x 2−y 2)
However I am unsure of where I have made a mistake as the true answer is
(x 2+y 2)2=4 x 2 y−(x 2+y 2)y(x 2+y 2)2=4 x 2 y−(x 2+y 2)y
working backwards I've so far gotten to the point of asking how I would rearrange
r=sin(3 θ)⇒r=4 cos 2(θ)sin(θ)−sin(θ)r=sin(3 θ)⇒r=4 cos 2(θ)sin(θ)−sin(θ)
so that
sin(3 θ)=4 cos 2(θ)sin(θ)−sin(θ)sin(3 θ)=4 cos 2(θ)sin(θ)−sin(θ)
which I am afraid I'll have to ask help for the next steps. Thanks.
algebra-precalculus
trigonometry
polar-coordinates
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edited Mar 24, 2018 at 1:00
Simply Beautiful Art
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asked Feb 3, 2018 at 19:41
user527827 user527827
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1 Your answer is the same as the "true" answer David Quinn –David Quinn 2018-02-03 19:45:00 +00:00 Commented Feb 3, 2018 at 19:45
Oh, I didn't realise, thank you so much.user527827 –user527827 2018-02-03 19:49:56 +00:00 Commented Feb 3, 2018 at 19:49
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4 Answers 4
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r r r 2 x 2+y 2 r 2(x 2+y 2)(x 2+y 2)2(x 2+y 2)2=sin(3 θ)=sin θ cos 2 θ+cos θ sin 2 θ=r sin θ cos 2 θ+r cos θ sin 2 θ=y cos 2 θ+x sin 2 θ=y r 2(1−2 sin 2 θ)+2 x r 2 sin θ cos θ=y(x 2+y 2)−2 y 3+2 x 2 y=3 x 2 y−y 3 r=sin(3 θ)r=sinθ cos2 θ+cosθ sin2 θ r 2=r sinθ cos2 θ+r cosθ sin2 θ x 2+y 2=y cos2 θ+x sin2 θ r 2(x 2+y 2)=y r 2(1−2 sin 2θ)+2 x r 2 sinθ cosθ(x 2+y 2)2=y(x 2+y 2)−2 y 3+2 x 2 y(x 2+y 2)2=3 x 2 y−y 3
I don't see a difference so far.
r r r 4(x 2+y 2)2(x 2+y 2)2=sin(3 θ)=3 sin θ−4 sin 3 θ=3 r 3 sin θ−4 r 3 sin 3 θ=3 y(x 2+y 2)−4 y 3=3 x 2 y−y 3 r=sin(3 θ)r=3 sinθ−4 sin 3θ r 4=3 r 3 sinθ−4 r 3 sin 3θ(x 2+y 2)2=3 y(x 2+y 2)−4 y 3(x 2+y 2)2=3 x 2 y−y 3
No difference here either. Your answer looks good.
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answered Feb 3, 2018 at 19:52
Awnon BhowmikAwnon Bhowmik
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or you use
sin(3 x)=3 sin(x)−4 sin(x)3 sin(3 x)=3 sin(x)−4 sin(x)3
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answered Feb 3, 2018 at 19:48
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
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Which too also brings me to the same answer as I have at the bottom. @David Quinn has brought to my attention that the model answer and my answer are indeed equivalent. Thank you.user527827 –user527827 2018-02-03 19:50:48 +00:00 Commented Feb 3, 2018 at 19:50
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r=sin(3 t)=I m((cos(t)+i sin(t))3)r=sin(3 t)=I m((cos(t)+i sin(t))3)
=3 cos 2(t)sin(t)−sin 3(t)=3 cos 2(t)sin(t)−sin 3(t)
=3 sin(t)−4 sin 3(t)=3 sin(t)−4 sin 3(t)
=3 y/r−4 y 3/r 3=3 y/r−4 y 3/r 3
thus
r 4=3 y r 2−4 y 3 r 4=3 y r 2−4 y 3
now replace r 2 r 2 by x 2+y 2 x 2+y 2.
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answered Feb 3, 2018 at 19:53
hamam_Abdallahhamam_Abdallah
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r=s i n(3 θ)r=s i n(3 θ)
r=3 s i n(θ)−4(s i n 3(θ))r=3 s i n(θ)−4(s i n 3(θ))
r 2=3 r s i n(θ)−4 r s i n(θ)s i n 2(θ)r 2=3 r s i n(θ)−4 r s i n(θ)s i n 2(θ)
x 2+y 2=3 y−4 y(y 2/(x 2+y 2))x 2+y 2=3 y−4 y(y 2/(x 2+y 2))
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answered Feb 3, 2018 at 20:04
Mohammad Riazi-KermaniMohammad Riazi-Kermani
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15843 | https://www.concepts-of-physics.com/waves/speed-of-sound-in-gases.php | See Our New JEE Book on Amazon
Speed of Sound in Gases
By Jitender Singh
The speed of sound in a fluid is given by v=√B/ρ, where B is bulk modulus of the fluid and ρ is its density. The sound waves travel in a gas by adiabatic compression and rarefaction (expansion). The speed of sound in an ideal gas is given by
where γ is the adiabatic index (ratio of specific heats), R=8.314 J/mol-K is the universal gas constant, T is the absolute temperature and M is the molecular mass. Note that adiabatic bulk modulus of an ideal gas at pressure p is B=γp and its density is ρ=pM/RT.
Newton’s formula for the speed of sound, v=√RT/M, is erroneous because it assumes isothermal compression and rarefaction of the gas (isothermal bulk modulus of an ideal is p). It was corrected by Laplace.
The speed of sound in gases is related to the root mean square speed of particles in the gas vrms=√3RT/M.
For a given gas, γ, R and M are constants. The speed depends only on temperature. It is independent of the pressure. To compute the speed of sound in a gaseous mixture, use mixture's adiabatic index and mean molecular mass.
Air is a mixture of gases with γair=1.4 and mean molecular mass M=28.9×10−3 kg/mol (air is mostly diatomic nitrogen and oxygen). The speed of sound in air at 0℃ is 331 m/s. Its value at temperature T ℃ is approximately equal to
v=(331+0.6T)m/s.
The speed of sound in air increases slightly with an increase in humidity. This is due to a decrease in mean molecular mass of the air due to increase in moisture content (molecular mass of air is 29 g/mol whereas it is 18 g/mol for water). The speed of sound in air is independent of the pressure.
The speed of sound is measured in the laboratory by resonance column method.
Solved Problems from IIT JEE
Problem from IIT JEE 2000
Two monatomic ideal gases 1 and 2 of molecular masses m1 and m2 respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to that in the gas 2 is given by,
√m1/m2
√m2/m1
m1/m2
m2/m1
Solution:
The speed of sound in a gas with molecular mass M and kept at temperature T is given by,
v=√γRT/M.
For the given monatomic gases, γ1=γ2=5/3 and temperature T1=T2. Substitute in above equation to get,
v1/v2=√m2/m1.
Problem from IIT JEE 2004
A source of sound of frequency 600 Hz is placed inside water. The speed of sound in water is 1500 m/s and in air it is 300 m/s. The frequency of sound recorded by an observer who is standing in air is
200 Hz
3000 Hz
120 Hz
600 Hz
Solution:
The frequency is a characteristic of the source and does not change when sound is transmitted from one medium to another. Since both source and observer are stationary, the frequency of sound recorded by the observer is equal to the source frequency, which is 600 Hz.
Problem from IIT JEE 1999
The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300 K is
√2/7
√1/7
√3/5
√6/5
Solution:
The speed of sound in a gas is given by
v=√γRT/M.
The nitrogen is a diatomic gas with γN2=7/5 and MN2=28. The helium is a monatomic gas with γHe=5/3 and MHe=4. Substitute these values in above equation to get
vN2vHe=√γN2γHeMHeMN2=√35.
Problem from IIT JEE 1983
The ratio of the velocity of sound in hydrogen gas (γ=7/5) to that in helium gas (γ=5/3) at the same temperature is √21/5.
Solution: The speed of sound in a gas with molecular mass M, ratio of specific heats γ, and temperature T is given by
v=√γRT/M.
For hydrogen, γH2=7/5 and MH2=2 and for helium γHe=5/3 and MHe=4. The ratio of velocities of sound in hydrogen and helium gases at the same temperature is
vH2vHe=√γH2γHeMHeMH2=√4225.
Problem from IIT JEE 2014
A student is performing an experiment using a resonance column and a tuning fork of frequency 244 Hz. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350±0.005)m, the gas in the tube is (Given: √167RT=640J1/2mol−1/2, √140RT=590J1/2mol−1/2. The molar masses M in grams are given in the options. Take the value of √10/M for each gas as given there.)
Neon (M=20,√1020=710)
Nitrogen (M=28,√1028=35)
Oxygen (M=32,√1032=916)
Argon (M=36,√1036=1732)
Solution: The speed of sound in a gas with molecular mass M, ratio of specific heat γ, and temperature T is given by
v=√γRT/M.
The minimum height of air column for the resonance to occur is
l=λ4=v4ν=14ν√γRTM.
The ratio of specific heat is γm=5/3=1.67 for monatomic gases and γd=7/5=1.4 for diatomic gases. Substitute these values in above equation to get
lNe=14(244)√1.67RT20×10−3=√167RT4(244)√1020=6404(244)710=0.459m,lN2=√140RT4(244)√1028=0.363m,lO2=√140RT4(244)√1032=0.340m,lAr=√167RT4(244)√1036=0.348m.
Thus, only lAr lies in the specified range of (0.350±0.005) m.
Questions on Speed of Sound in Gases
Question 1: If the velocity of sound in air at 0℃ be 330 m/s then the increase in the velocity of sound in air, for 1℃ rise in temperature is?
Question 2: The velocity of sound in air
Question 3: If air molecules are traveling with a root mean square speed of 500 m/s then the speed of sound in air is?
Question 4: What is the speed of sound in a helium gas at a pressure of 150,000 Pa?
Related
Progressive waves
Speed of waves on water surface
Doppler effect
Bulk Modulus of Gases
Kinetic Theory of Gases
References
IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
Concepts of Physics by HC Verma (Link to Amazon)
Physics of Music - Notes (Michigan Technological University)
University Physics (lumenlearning.com)
JEE PYQ+ Physics Books
Solve past year JEE questions with detailed explanations.
📘 Mechanics
🌊 Fluids, Waves & Thermodynamics
⚡ Electricity & Magnetism |
15844 | https://blog.csdn.net/qq_41570269/article/details/106008929 | 工程问题解题策略-CSDN博客
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数量关系--工程问题
最新推荐文章于 2025-08-24 17:00:44 发布
原创 于 2020-05-08 23:12:53 发布·4.1k 阅读
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CC 4.0 BY-SA版权
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行测--数量关系 专栏收录该内容
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数量关系—工程问题
数量关系工程问题分为以下几个类别：
[x] 给出完工时间型
[x] 给出效率比
[x] 经典-牛吃草问题
[x] 给出具体单位型
1.给出完工时间型
解题步骤如下：
赋总量（完工最小时间公倍数)
算效率
根据具体情况列方程或者式子
例题1
(2018 陕西）要完成某项工程，甲施工队单独干需要 30 天才能完成，
乙施工队需要 40 天才能完成。甲、乙合作干了 10 天，因故停工 10 天，再开工
时甲、乙、丙三个施工队一起工作，再干 4 天就可全部完工。那么，丙队单独干
需要大约多少天才能完成这项工程？
A.21 B.22
C.23 D.24
E.25 F.26
G.27 H.28
解：通过阅读题目可知本题目为完工时间型
赋总量：30和40最小公倍数==120
算效率：甲== 4，乙==3
列式子：设丙==X，则有
（3+4) 10+(3+4+X) 4=120
x=5.5
120/5.5约为21.8，所以选B
例题2
（ 2019 北京）录入员小张和小李需要合作完成一项录入任务，这项
任务小李一人需要 8 小时，小张一人需要 10 小时。两人在共同工作了 3 个小时
后，小李因故回了趟家，期间小张一直在工作，小李返回后两个人又用了 1 个小
时就完成了任务。在完成这项任务的过程中，小张比小李多工作了几个小时？
A.1 B.1.5
C.2 D.2.5
解：通过阅读题目可知本题目为完工时间型
赋总量：8和10的最小公倍数为40
算效率：李：5，张：4
列式子：（小张效率+小李效率）共同工作时间+小张效率 单独工作时间==40
（5+4）4+4 所求=40
得到答案为A
2.给出效率比例型
解题步骤如下：
赋效率（满足题目所给比例即可）
算总量，总量==效率比时间
根据题目列式子或者方程
例题1
（ 2019 江西）甲、乙两个工程队共同参与一项建设工程。原计划由
甲队单独施工 30 天完成该项工程三分之一后，乙队加入，两队同时再施工 15
天完成该项工程。由于甲队临时有别的业务，其参加施工的时间不能超过 36 天，
那么为全部完成该项工程，乙队至少要施工多少天？
A.18 B.20
C.24 D.30
解：根据题目 甲队单独施工 30 天完成该项工程三分之一后，乙队加入，两队同时再施工 15天完成该项工程。可以知道这是效率比型，如果是完工时间型，那么应该说甲用30天完成全部工作，或者说，乙用20天完成全部工作，或者说，甲乙用20天合作完成工作
假设甲的效率为“1”
算总量：那么，301=总量/3，总量为90，在这一步根据“乙队加入，两队同时再施工 15
天完成该项工程”得到
设乙的效率为“X”，那么，(1+X)15=90 (2/3),得到X=3
列式子：
361+( ? ) 3==90,解得18，所以选A
例题2
2016 河南）甲、乙、丙三个植树队同时各种 400 棵树，当甲队把
400 棵树全部种完时，乙队还有 150 棵树没种，丙队才种了 220 棵树。当乙队全
部种完时，丙队还有多少棵树没种？
A.48 B.52
C.66 D.74
解：这道题问“当乙队全部种完时，丙队还有多少棵树没种”但是在题目中没有给出乙和丙对的完工时间，所以是效率比问题。
根据题目可以知道当乙对种植250棵树时，丙对种植了220棵树并且他们花费的时间相同，效率时间=工程量
所以乙效率：丙效率==25:22
当乙种植完400棵树，花费时间16，丙种植了1622=352棵，所以答案为400-352=48
3.牛吃草问题
牛吃草问题只要记住公式就成
Y==(N 牛吃草的效率-X)_T_
Y==草的总数
N==牛的数量
X==草生长的效率
T==时间
一般默认牛吃草的效率为“1”，所以公式为Y==(N -X)_T_
问题识别：总量是随时间变化的，就好像牛吃草的过程中，草也会生长；
形式上是排比句，如:一片草地，20 头牛 10 天吃完，10 头牛 30 天吃完，问 30 头牛几天吃完？
例题1
2019 安徽）某河道由于淤泥堆积影响到船只航行安全，现由工程
队使用挖沙机进行清淤工作，清淤时上游河水又会带来新的泥沙。若使用 1 台挖
沙机 300 天可完成清淤工作，使用 2 台挖沙机 100 天可完成清淤工作。为了尽快
让河道恢复使用，上级部门要求工程队 25 天内完成河道的全部清淤工作，那么
工程队至少要有多少台挖沙机同时工作？
A.4 B.5
C.6 D.7
解: "清淤时上游河水又会带来新的泥沙"以及“1 台挖沙机 300 天可完成清淤工作，使用 2 台挖沙机 100 天可完成清淤工作”可以知道是牛吃草问题，在本题中没有给出清理淤泥的 效率，所以默认为一
列方程：
Y==(1-X)300
Y==(2-x)200
解得X == 0.5，Y==150
(挖机数量-0.5)25>=150,解得6.5，所以答案为7
扩展
如果问放几台挖机，泥沙永远挖不完，允许是小数，那么答案是0.5泥沙永远清理不完
这种问题的答案==X，记住即可
例题2
2018 深圳）某轮船发生漏水事故，漏洞处不断地匀速进水，船员
发现险情后立即开启抽水机向外抽水。已知每台抽水机每分钟抽水 20 立方米，
若同时使用 2 台抽水机 15 分钟能把水抽完，若同时使用 3 台抽水机 9 分钟能把
水抽完。当抽水机开始向外抽水时，该轮船已进水多少立方米？
A.360 B.450
C.540 D.600
解：本题由于给出了“每台抽水机每分钟抽水 20 立方米”，相当于给出了牛吃草的效率，所以不可以再默认为“1”
Y==(220-X) 15
Y==(320-X) 9
解得：X== 10，Y==450，所以选B
4.给出具体单位型
特征：：如做 1000 个蛋糕、挖 1200 米的公路、做 500套零件。“个、米、套”属于具体单位。
解题步骤：
设未知数，设小不设大，设中间量
找到等量关系列方程
例题1
2019 陕西）制作一批风筝，甲需要 12 天完成，乙需要 18 天完成，
两人共同制作，完成时甲比乙多制作了 72 个，如果按“甲制作一天、乙制作两
天”的方式重复下去，当制作完成时，甲制作的风筝有多少个
A.140 B.145
C.150 D.155
E.160 F.165
G.170 H.175
刚读到“甲需要 12 天完成，乙需要 18 天完成”以为是普通的工程问题，赋值总量为 36。但是读到“完成时甲比乙多制作了 72 个”，出现具体单位，问题是“甲制作的风筝有多少个”，也出现了具体单位，求的是具体个数。
解：
设风筝总量为甲乙的最小公倍数，即36X，那么甲的效率== 3X，乙的效率==2X
设时间为T那么：(3X+2X)T=36X 那么T== 7.2，又“完成时甲比乙多制作了 72 个”
有（3X-2X）7.2== 72，所以X== 10，并且可算得风筝总数==360
一个周期所做风筝总数== 甲的效率1天+乙的效率 2天==70，所以做5个周期后还有10个，刚好由甲来完成
所以，甲制作风筝数==305+20 ==160所以选E
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谢三岁的博客
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行程 问题 形成 问题 可以分为： 基本行程 问题：S=VT 相对行程 问题：直线或者环形的追击相遇 比例行程：S=VT 当S一定时，V与T成反比，V一定时S与T成正比 基本行程 问题 牢牢记住S=VT，有时条件太多拎不清可以列表 例题： 2018 国考）一辆汽车第一 天 行驶了 5 个小时，第二 天 行驶了 6 00 公里，第三 天 比第一 天 少行驶 200 公里，三 天 共行驶了 18 个小时。已知第一 天 的平均速度与三 天 全程的平均速度相同，问三 天 共行驶了多少公里？ A . 800 B . 900 C . 1000 D . 11
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5
How do you find the area under a cosine curve?
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Bernd Leps
Former scientific official; retired · Author has 5.8K answers and 1.3M answer views
·2y
I wonder, what a prompt generator will do with a cosine curve.
But be it so.
The are under a cosine curve can be found by integration as other areas-under-curve also.
The integral function of a cosine function can easily obtained, because the derivates of the sine function and the cosine function will reiterate: (cos x)’ = -sin x; (cos x)’’ = (-sin x)’ = - cos x; (cos x)’’‘ = (-sin x)’’= (-cos x)’ = sin x; (cos x)’’’’ = (-sin x)’’’ = (-cos x)’’ = (sin x)’ = cos x. (If we need a why, see below.)
We can read from this, that the integral function of cos x is sin x.
The integral under the cosine curve
Continue Reading
I wonder, what a prompt generator will do with a cosine curve.
But be it so.
The are under a cosine curve can be found by integration as other areas-under-curve also.
The integral function of a cosine function can easily obtained, because the derivates of the sine function and the cosine function will reiterate: (cos x)’ = -sin x; (cos x)’’ = (-sin x)’ = - cos x; (cos x)’’‘ = (-sin x)’’= (-cos x)’ = sin x; (cos x)’’’’ = (-sin x)’’’ = (-cos x)’’ = (sin x)’ = cos x. (If we need a why, see below.)
We can read from this, that the integral function of cos x is sin x.
The integral under the cosine curve from x = a to x= b is therefore sin(b)-sin(a).
BUT
“Area under a curve” may not be the same as “result of the integral”, if we are interested in areas as absolute (always positive) values. An integral will result in a negative value, if the curve is below the x axis and we have an “area above the curve” rather than “below…”. If that is of importance, we have to check, whether the cosine curve changes its sign somewhere between a and b. (That will happen at x = (2n+1) times pi, n an integer.)
If so, we have to split the integrating way from a to b at such points and sum up the area parts, each taken positive.
And, here apart from the main answer, the derivative of a sine or a cosine can be obtained in the classical way: f’(x) = limit of [ (f(x + Δx) + f(x))/Δx ] for Δx approaching zero.
Here, (cos x)’ = lim [ (cos(x+Δx) - cos(x))/Δx] = lim [ (cos x cos Δx - sin x sin Δx - cos x)/ Δx]
But cos Δx is approaching 1, if Δx is approaching zero, so we have
(cos x)’ = lim [ (cos x - sin x sin Δx - cos x)/Δx] = lim [ (-sin x sin Δx) /Δx]
But sin Δx is approaching Δx for small Δx: Look at a very small angle in the unit circle and see, that the opposite cathetus representing sin(angle) is more and more similar to the bow length of the circle (which is “the” angle, if the radius is 1). So lim sin (Δx)/Δx = 1 and (cos x)’ = -sin x.
Same way for (sin x)’ = cos x.
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More answers below
What is the area of a cosine curve?
Does the area under the cosine curve equal one?
How do you calculate the area under the cosine curve from zero to two pi?
What are some ways that finding the area under a curve or between two or more curves is useful?
What are the practical uses of calculus, other than calculating the area under curves?
Mark Regan
Former Math, Chemistry, and Calculus Tutor at Bakersfield College (1990–1992)
·Aug 3
Related
How do you calculate the area under the cosine curve from zero to two pi?
The area under the cosine curve from zero to two pi is going to be zero because there is the same area above the x-axis as below it.
Continue Reading
The area under the cosine curve from zero to two pi is going to be zero because there is the same area above the x-axis as below it.
Upvote ·
Ajay Sreenivas
Former Aerospace Engineer/ Staff Consultant at Ball Aerospace (1980–2010) · Author has 5.3K answers and 1.7M answer views
·2y
Related
How do you calculate the area under the cosine curve from zero to two pi?
Area = A1 + A2 + A3 + A4 where
A1 = abs [integral from 0 to (pi/2) of cos(theta) dtheta = abs[- si1n(pi/2) - sin(0)] = 1
A2 = abs [integral from pi/2 to pi of cos(theta) dtheta = abs-[ -si1n(pi) - sin(pi/2)] = 1
A3 = abs [integral from pi to (3pi/2) of cos(theta) dtheta = abs[ - si1n(3pi/2) + si...
Upvote ·
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·4y
Related
How do you find the area under a curve using Riemann sums?
It involves finding an expression for the sum of the areas of n strips then finding what this limit approaches as n approaches infinity.
It is not easy!
Thank heavens for the discovery of the Fundamental Theorem of Calculus!
Now we just do this…
Continue Reading
It involves finding an expression for the sum of the areas of n strips then finding what this limit approaches as n approaches infinity.
It is not easy!
Thank heavens for the discovery of the Fundamental Theorem of Calculus!
Now we just do this…
Upvote ·
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More answers below
Why is the area under a normal curve 1?
What is the derivative of the area under a curve?
How do I find the area between two curves?
What is the relationship between sine and cosine?
How do you calculate the area under a curve on Desmos?
Haresh Sagar
Studied Science&Mathematics (Graduated 1988) · Author has 6.2K answers and 7M answer views
·2y
Related
What is the area enclosed by a sine curve?
This is the graph of sine curve. The area of all the curves below and above [math]x[/math] axis is equal and the distance between intersection points is [math]\pi[/math] units.
Let us take one of the curve which is in first quadrant and whose intersection points are math[/math] and math[/math].
Since this curve is not regular like circle, it doesn't have a specific area formula so it's quite difficult to determine exact area. To avoid approximation, the exact area has to be determined through integration method.
The integration method involves summation of areas of infinitely thin imaginary rectangles fitted into curve whose height
Continue Reading
This is the graph of sine curve. The area of all the curves below and above [math]x[/math] axis is equal and the distance between intersection points is [math]\pi[/math] units.
Let us take one of the curve which is in first quadrant and whose intersection points are math[/math] and math[/math].
Since this curve is not regular like circle, it doesn't have a specific area formula so it's quite difficult to determine exact area. To avoid approximation, the exact area has to be determined through integration method.
The integration method involves summation of areas of infinitely thin imaginary rectangles fitted into curve whose height is denoted by [math]f(x)[/math] and width by [math]dx[/math]. We can imagine that the imaginary rectangles fitted inside sine curve have infinitely small heights beginning at math[/math] and ending at math[/math]. These heights get bigger and bigger and reach maximum at [math]\pi/2[/math] where [math]sin(x)=1[/math].
The summation of these rectangles is denoted by,
[math]A=\displaystyle\int\limits_a^b\;f(x)dx[/math]
Here elongated s([math]\int[/math]) indicates summation, [math]a[/math] and [math]b[/math] are boundaries between which imaginary rectangles are fitted.
Taking anti derivative of [math]f(x)dx[/math] and evaluating, we get exact area. Thus area enclosed by sine curve is,
[math]A=\displaystyle\int\limits_0^{\pi}\;Sin(x)dx[/math]
[math]=|-Cos(x)|_0^{\pi}=1--1=2\;\;square\;\;units[/math]
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Daniel Shapiro
I think, therefore I am... better than most people · Author has 233 answers and 574.1K answer views
·8y
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Why is the antiderivative the area under the curve?
My teacher explained it in terms of the velocity function v(t) and its antiderivative, the position function s(t). Let’s assume that v(t) is continuous on whatever interval we care about. Therefore, it’s also integrable on that interval.
What does an integral do? It sums infinitely many infinitesimally small things.
When we find the area under a curve, we are actually taking the sum of infinitely many rectangles with infinitesimal width.
So let’s think about a rectangle under the velocity curve. It has some width close to zero, but not actually zero. This width represents some change in time.
The
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My teacher explained it in terms of the velocity function v(t) and its antiderivative, the position function s(t). Let’s assume that v(t) is continuous on whatever interval we care about. Therefore, it’s also integrable on that interval.
What does an integral do? It sums infinitely many infinitesimally small things.
When we find the area under a curve, we are actually taking the sum of infinitely many rectangles with infinitesimal width.
So let’s think about a rectangle under the velocity curve. It has some width close to zero, but not actually zero. This width represents some change in time.
The rectangle also has a height equal v(t) for that t-value. The height represents some velocity. That’s change in position/time.
So now we have a rectangle, and we know the units of it’s sides. Let’s find the area of this rectangle. We multiply some time by some position/time. Our units cancel and we are left with just some number representing our position. This number will change depending on where our rectangle is on the x-axis. That means our position is a function of time. We’ll call that function s(t).
So now we know that the area under the velocity curve represents the position. But how do we know that s(t) is the antiderivative of v(t)? Well, if we prove that v(t) is the derivative of s(t), then s(t) is the antiderivative of v(t). We know that by the definition of the words.
So let’s prove that v(t) is the derivative of s(t). The word “derivative” actually means “instantaneous rate of change”. And what does velocity mean? Velocity is the instantaneous change in position. So that means v(t) is the derivative of s(t), so s(t) is the antiderivative of v(t).
So now we have shown that the area under the velocity function v(t) is represented by s(t), which happens to be the antiderivative of v(t).
But I did not mention anything specific about v(t) or s(t). My only assumption was that v(t) was continuous and therefore integrable. So for any velocity function, the area underneath is given by the antiderivative.
But I only called v(t) a velocity function and s(t) a position function because it gives the math a real-world application that we already understand. Really, the area under any function is given by its antiderivative, as I showed above.
Don’t believe that this applies to a non-velocity function? Okay. say you have some continuous function f(x) that does NOT describe a velocity. Suppose you want to find the area under f(x). I will create a particle whose velocity “coincidentally” is described by f(t). Therefore, the area under f(t) is described by its antiderivative, using the logic shown above. So the area under f(x), regardless of whether f(x) is a velocity function, is described by the antiderivative.
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Gary Ward
MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views
·2y
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How do you find the area under a curve when given two functions?
How do you find the area under a curve when given two functions.
You must have limits, a and b, and I believe you mean the area between the two functions. It is not much different than finding the area between a function and another function, y = 0, the x-axis.
Set the two functions equal to each other and solve to find any intersections between a and b. If there are you will have to work it piecemeal.
Subtract the lower function from the higher function and integrate for each piece, then sum up the parts.
Example:
Set x^3 - 5x + 5 = x² equal to 0 and solve.
x^3 - x² - 5x + 5 = 0 x = -2.236, 1, and
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How do you find the area under a curve when given two functions.
You must have limits, a and b, and I believe you mean the area between the two functions. It is not much different than finding the area between a function and another function, y = 0, the x-axis.
Set the two functions equal to each other and solve to find any intersections between a and b. If there are you will have to work it piecemeal.
Subtract the lower function from the higher function and integrate for each piece, then sum up the parts.
Example:
Set x^3 - 5x + 5 = x² equal to 0 and solve.
x^3 - x² - 5x + 5 = 0 x = -2.236, 1, and 2.236
Since ±2.236 is outside our limits of ±2, they can be ignored, but 1 cannot be ignored.
From -2 to 1 (x^3 - 5x + 5) - x² will give us the positive area between the curves.
From 1 to 2 x² - (x^3 - 5x + 5) will give us the positive area between the curves.
[math]\displaystyle \int_{-2}^1 \, x^3 - x^2- 5x + 5 \, dx = \frac{63}{4}[/math]
[math]\displaystyle \int_1^2 \, -x^3 + x^2 +5x - 5 \, dx = \frac{13}{12}[/math]
[math]\displaystyle \frac{63}{4} + \frac{13}{12} = \frac{189 +13}{12} = \frac{101}{6} \approx 16.83 \, units^2[/math]
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Subhasish Debroy
Former SDE at Bharat Sanchar Nigam Limited (BSNL) · Author has 6.6K answers and 5.8M answer views
·2y
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What is the area enclosed by a sine curve?
Ans.
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Ans.
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Robert Wilson
Studied Philosophy&Psychology at University of California, Irvine · Author has 323 answers and 4.1M answer views
·Updated 5y
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Why is the definite integral the area under the curve?
I am legitimately shocked no one mentioned the mean value theorem. It’s only the reason why the fundamental theorem of calculus even works in the first place.
I assume you’re asking this question not out of confusion about sums of rectangles, surely you understand that. Rather, I’m assuming you want to know why Area is computed by f(b) - f(a). And that is an outstanding question.
The reason is pretty simple. It’s the mean value theorem, which says that for any continuous interval [a, b], say, there exists a ‘c’ in [a, b] s.t. if f(x) is differentiable on the open interval (a, b),
then, f’(c) = (f
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I am legitimately shocked no one mentioned the mean value theorem. It’s only the reason why the fundamental theorem of calculus even works in the first place.
I assume you’re asking this question not out of confusion about sums of rectangles, surely you understand that. Rather, I’m assuming you want to know why Area is computed by f(b) - f(a). And that is an outstanding question.
The reason is pretty simple. It’s the mean value theorem, which says that for any continuous interval [a, b], say, there exists a ‘c’ in [a, b] s.t. if f(x) is differentiable on the open interval (a, b),
then, f’(c) = (f(b) - f(a))/(b-a).
Okay, now at first glance you might want to call bullshit. That’s fair. It does look suspect upon first glance.
But consider this, it is certainly true that for a smooth curve there exists f(c) st. for c in [a, b] f(a) <\ f(c) <\ f(b). Easy enough.
Now, what the mean value theorem says is something different. What it says is that if you drove 60 miles in one hour, you must have driven exactly 60mph at least once during your trip.
Now if you don’t believe me that’s fine. Just think about it, if you drove less than 60mph the entire way you would not have driven 60 miles in one hour; not possible.
You could drive a variety of speeds and have it come out to 60 miles over 1 hour; that’s fine too. But here’s the thing, if you always drove less than 60mph, you would not have gone 60 miles in 1 hour, since this is not possible as we said. But if you drove sometimes slower than 60mph and sometimes faster than 60mph, but never stopped your spedometer at exactly 60mph, you still drove 60mph at least once, since you have to have gone 60mph when accelerating to a speed faster than 60mph. After all, a speedometer can only get to 70mph if the needle first touches and passes 60mph, and to slow down from 70mph to, say 50mph, the needle must go back down and touch and cross 60mph again, therefore you must have gone 60mph at least once.
The only other option is to go 60mph the entire way, which hammers home the point nicely.
Now, doings a bit of algebra, we see that f’(c)(b-a) = f(b) - f(a), and well what do you know? That looks riemann-sum-ish, except it isn’t since we’re using f’(c) instead of f(c). Riemann used the actual function (which makes more sense, mind you, to me anyway), but the smart-ass who figured out that the mean value theorem applies here is actually more accurate in getting the area right.
Turns out that f(b) - f(a) = the infinite sum of (f(Xn) - f(Xn-1)) = f’(c)(b - a) = the infinite sum f’(Cn)(Xn - Xn-1).
Altogether then, the integral of g(x) from a to b = f’(c)(b-a) = f(b) - f(a), where f’(c) = g(c).
And that’s why you find the anti-derivative before calculating the area of the function from a to b. It all seems like a lucky coincidence doesn’t it! I bet that’s not too far from the truth.
I’d still rather take Riemann sums manually myself. They make more sense, ¯_(ツ)_/¯.
But wait! We want the integral for f(x) not g(x)!
Yeah, so find an anti-derivative for f(x). Call that r(x), (isn’t ‘r’ such a cool letter for a function? I like it.) then the integral of f(x) from a to b = r’(c)(b-a) = r(b) - r(a).
So, I mean, yeah it’s slightly weird. What the hell does r(x) evaluated at a and b have to do with f(x)?! You might protest.
Well, it just happens that f’(x) isn’t what we want. We want r(x) so that r’(x) = f(x) not f’(x).
Perhaps, less perversely, then, the integral of f(x) from a to b = f(c)(b -a) = r(b) - r(a). Since r’(c) = f(c). And that is the relationship going on here. And why the definite integral gives area.
Note, that car example I have was not my original example but I lost the link it came from. I’ll add it here if I an find it again. Here’s a pretty good alternative link.
Note that this is not the definition of the definite integral. Integrable functions do not need to have continuous domains. The fundamental theorem of calculus only works if the function you want to integrate is both continuous on the interval you want to integrate over and differentiable over the interior of that interval.
A function is Riemann integrable, in general, when the sup{L(f, P) : P is a partition of [a, b]} = inf{U(f, P) : P is a partition of [a, b]}. This, is the real criterion for differentiability with respect to Darboux integrals, at least. There are other kinds of integrals, but these are the ones studied in a calculus class.
Don’t forget this as it’s very important to understand conceptually. People often think the integral is defined as F(b) - F(a), no it is not.
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Terry Moore
PhD in statistics · Author has 16.6K answers and 29.4M answer views
·Updated 3y
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Why is the antiderivative the area under the curve?
It’s best not to think primarily of derivatives and integrals geometrically. They say a picture is worth a thousand words. That’s true, and the graphical representations of functions and their derivatives and integrals are very helpful. Visualisation is good.
But my point is: an integral is not defined to be the area under a curve and the derivative is not defined to be the gradient of a curve. If you plot a graph of a function, these properties of the curve represent the integrals and derivatives, but that does not mean that they are derivatives and integrals.
If you have a problem that natural
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It’s best not to think primarily of derivatives and integrals geometrically. They say a picture is worth a thousand words. That’s true, and the graphical representations of functions and their derivatives and integrals are very helpful. Visualisation is good.
But my point is: an integral is not defined to be the area under a curve and the derivative is not defined to be the gradient of a curve. If you plot a graph of a function, these properties of the curve represent the integrals and derivatives, but that does not mean that they are derivatives and integrals.
If you have a problem that naturally involves the integral of a function then you already understand without forcing some visualisation upon it. For example, if you know the acceleration of a body as a function of time, the antederivative is the velocity, and the antederivative of that is the distance. One way to picture this is to draw two separate graphs, a graph of acceleration against time, and then a graph of velocity against time. Alternatively, the velocity-time graph represents the distance (via the area under the curve) and the acceleration (via the gradient).
Exactly the same could be said of the derivative of a derivative. It is not so easy to visualise on the graph of the original function. It is related to curvature, but it is not exactly that (except where the derivative is zero). But then how would you visualise the third derivative?
You have already hit upon the problem with visualisation. It limits you to one particular point of view. It works for some problems, but not for others. For example the area under what curve corresponds to the volume of a cone? It’s the graph of the cross sectional area versus the height, but it is not helpful. You already have a picture of the cone and you have to understand how to find the volume before you could plot the graph of the cross section; then you would have to see why the area gives the volume. What’s the point?
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George Ivey
Former Math Professor at Gallaudet University · Author has 23.7K answers and 2.6M answer views
·2y
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What is the area under a cosine curve from zero to pi/two?
The area is [math]\int_0^{\pi/2} cos(x)dx=\left[sin(x)\right]_0^{\pi/2}= sin(\pi/2)= 1[/math].
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Simon Bridge
teacher, all levels (NZ) - PGdipEd · Author has 86.2K answers and 38.8M answer views
·3y
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How do you find the area under a sine curve?
Integrate it between the limits.
Over 2-pi, of course, the net area is zero. Similarly for equal distances across x=0.
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Marco Biagini
MSc in Mathematics, Eidgenössische Technische Hochschule (Graduated 1982) · Author has 5.4K answers and 5.8M answer views
·2y
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How do you calculate the area under a curve?
Have you heard of Internet searches? Integration!
Here is a simple example:
Find the area between the following two curves:
y = 0 and y = -x^2 + 2 x + 3
Set up the integral to compute the area between 0 and -x^2 + 2 x + 3:
A = integral_lower^upper abs(-(-x^2 + 2 x + 3) + 0) dx
Plot y = 0 and y = -x^2 + 2 x + 3 to observe that the curves intersect to create one bounded region, whose area is represented by A.
Solve the equation 0 = -x^2 + 2 x + 3 to get the points of intersection.
x = -1 or x = 3
These points of intersection indicate the lower and upper bounds for integration for the area between 0 and -
Continue Reading
Have you heard of Internet searches? Integration!
Here is a simple example:
Find the area between the following two curves:
y = 0 and y = -x^2 + 2 x + 3
Set up the integral to compute the area between 0 and -x^2 + 2 x + 3:
A = integral_lower^upper abs(-(-x^2 + 2 x + 3) + 0) dx
Plot y = 0 and y = -x^2 + 2 x + 3 to observe that the curves intersect to create one bounded region, whose area is represented by A.
Solve the equation 0 = -x^2 + 2 x + 3 to get the points of intersection.
x = -1 or x = 3
These points of intersection indicate the lower and upper bounds for integration for the area between 0 and -x^2 + 2 x + 3. Thus we have:
A = integral_(-1)^3 abs(-(-x^2 + 2 x + 3) + 0) dx
Compute integral_(-1)^3 abs(-(-x^2 + 2 x + 3) + 0) dx noting that for -1<x<3, -x^2 + 2 x + 3 is strictly greater than 0 so we can simplify abs(-(-x^2 + 2 x + 3) + 0) to -x^2 + 2 x + 3.
Answer: A = integral_(-1)^3 (-x^2 + 2 x + 3) dx = 32/3
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15846 | https://counter2015.com/2020/09/30/fibnacci/ | 5分钟学不会斐波那契数列
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本文的写作灵感来自与凤凰木的面试题 《我们为什么要考斐波那契数列──一道上机笔试题的解析，兼谈技术面试与编程基本功》,先行致谢。
UPDATE:2024-04 上面这篇文章已经有了更新的版本 凤凰木的笔记:面试中的斐波那契数列问题
同时，也是填之前的坑
要求计算斐波那契数列的第50 10 ^6 项，前四位和后四位连起来就是本次的key
了解基本的大学数学知识(数列，等比公式，待定系数法，矩阵…)
基本的编程知识
了解简单的Scala语法(非必须)
问题的引出
斐波那契数列（The Fibonacci sequence， OEIS A000045 ）的定义如下：
奥数题：爬楼梯，铺砖块
欧几里得算法求最大公约数时，最坏情况为两个连续的 Fibonacci 数列
斐波那契回调线，在金融交易市场中被用做确认支撑位和阻力位的一种工具
一道简单的小学数学题
现在有长度为10， 宽度为2的长方形地板需要铺设地砖，地砖有两种
一种是长度为1，宽度为2的
一种是长度为2，宽度为1的
这就是一个典型的求斐波那契数列的问题，和爬楼梯类似。
f(1) = 1, f(2) = 2, f(3) = 3
最右边的地砖是竖着放的，剩下需要摆放的长度减少至n-1，那么这个问题可以被化归成f(n-1)
最右边的地砖是横着放的，那么这个地砖下面的一个位置必然也是横着放的，这两个地砖一共占用了2的长度，剩下需要拜访的长度减少至n-2, 那么这个问题可以被化归成f(n-2)
所以有状态转移方程 f(n) = f(n-1) + f(n-2)
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| 1 2 3 4 5 6 7 8 9 10 | f(1)=1 1 1f(2)=2 2 2f(3)=3 3 3f(4)=5 4 5f(5)=8 5 8f(6)=13 6 13f(7)=21 7 21f(8)=34 8 34f(9)=55 9 55f(10)=89 10 89 |
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| 1 2 3 4 5 6 7 8 9 | def fib(n: Int): Int = {def fib def fib Int Int var (a, b) = (0, 1) var 0 1 for (i <- 1 to n) {for 1 val temp = b val b = a + b a = temp } b} |
一点简单的扩展
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| 1 | def fib (n: Int): Int = if (n <= 1) n else fib(n-1) + fib(n-2) def fib def fib Int Int if 1 else -1 -2 |
这个问题在大学C语言教材里4.3.2 斐波那契序列（计算与时间）也有提到，下面是书中的图
我们可以看到，很多步骤被多次重复计算，比如fib(2)就被计算了3次
更严重的问题是，参数值增加 1，函数 fib 的计算时间将为原来的 1.6 倍左右（这是理
论估计，实际时间可能更多）。
我们来对比一下这两段代码，第一段代码确实比第二段代码好吗？
不知道你知不知道大O表示法，这是由某不知名高老头提出的记法，用来估算算法的时间复杂度的一种符号。
当问题规模为n时，用这种表示方法，前者的时间复杂度为O(n)，后者的时间复杂度近似为O(2^n)
之所以说是近似，后者的时间复杂度其实和fib(n)相关，感兴趣的话可以查阅这篇论文和讨论以及这篇知乎文章
这样看来，从时间上来看，第一种方案显然更好，但是从代码的美感上来说，前者需要维护的状态过多，不如递归解法直观
什么是代码的美呢？
这其实是一个很主观的感受，以树的后序遍历为例，下面是递归和非递归的写法
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| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 | class TreeNode(var _value: Int = 0) {class TreeNode(var _value: Int = 0) class TreeNodevar _value: Int = 0 Int var value: Int = _value var Int var left: TreeNode = _ var TreeNode var right: TreeNode = _ var TreeNode} def postorderTraversal(root: TreeNode): List[Int] = {def postorderTraversal def postorderTraversal TreeNode List Int if (root == null) Nil if null Nil else postorderTraversal(root.left) ++ postorderTraversal(root.right) :+ root.value else} def postorderTraversal2(root: TreeNode): List[Int] = {def postorderTraversal2 def postorderTraversal2 TreeNode List Int import scala.collection.mutable import val s = mutable.StackTreeNode val Stack TreeNode val res = mutable.ListBufferInt val ListBuffer Int if (root == null) return res.toList if null return s.push(root, root) while (s.nonEmpty) {while val node = s.pop() val if (s.nonEmpty && node == s.top) {if if (node.right != null) if null s.push(node.right, node.right) if (node.left != null) if null s.push(node.left, node.left) } else {else res += node.value } } res.toList} |
我个人觉得前者可读性更强。
优美和性能，我全都要
有没有办法可以兼顾递归的简洁和循环迭代的效率呢？
其实用尾递归可以让编译器帮我们自动把它转换成循环的形式，同时还不影响性能(附带地还避免了栈溢出的错误)
~~scala编译这么慢怎么想都是你们的错，自己偷懒让编译器干活~~
只要简单的做个『包装』就行了
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| 1 2 3 4 5 6 7 8 | def fib(n: Int): Int = {def fib def fib Int Int @scala.annotation.tailrec @scala def f(n: Int, a: Int = 0, b: Int = 1): Int = {def f def f Int Int 0 Int 1 Int if (n == 0) a if 0 else f(n-1, b, a+b) else -1 } f(n)} |
关于尾递归的的实现，很难找到相关资料，下面是我用Intellij IDEA 2020.1.4 的反编译上面的代码到Java，只摘取关键部分
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| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 | public int fib(final int n) {public int fib(final int n) final int return this.f$1(n, f$default$2$1(), f$default$3$1()); return this 1 default 2 1 default 3 1} private final int f$1(final int n, final int a, final int b) {private final int 1 final int final int final int while(n != 0) {while 0 int var10000 = n - 1; int var10000 = 1 int var10001 = b; int var10001 = b += a; a = var10001; n = var10000; } return a; return} private static final int f$default$2$1() {private static final int default 2 1 return 0; return 0} private static final int f$default$3$1() {private static final int default 3 1 return 1; return 1} |
通过阅读反编译后的代码，我们发现，编译器帮我们把递归形式的代码，等价展开成迭代形式的代码
所谓外递归内迭代，这样就两全其美了
什么是奇技淫巧啊
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| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 | def fib(n: Int): Int = {def fib def fib Int Int case class MemoA, B extends (A => B) {caseclass MemoA, B extends (A => B) class Memo A Bf: A => B A B extendsA => B A B import scala.collection.mutable import private val cache = mutable.MapA, B private val Map A B def apply(a: A): B = cache.getOrElseUpdate(a, f(a)) def apply def apply A B } lazy val f: Memo[Int, Int] = Memo {lazy val Memo Int Int Memo case n: Int if n < 2 => n case Int if 2 case n: Int => f(n-1) + f(n-2) case Int -1 -2 } f(n)} |
这段代码实现了不使用尾递归的方式，也能把递归的代码优化到O(n)时间复杂度
思路是计算过程中，额外开辟一个空间存放中间结果
后续计算时，如果需要的结果已经计算完成，就能直接取来用
一点范围的扩展
目前为止，我们的代码对于较小的测试样例表现良好，但是对于更大的输入呢？
fib(100)的准确值是多少？
如果你把上面的代码照搬，直接输入的话
| | |
--- |
| 1 2 | scala> fib(100) 100val res4: Int = -980107325 val Int -980107325 |
~~显然，计算的结果溢出了Int能表示的范围~~
这里还是稍微花点时间，讲解下显然背后的原理
所谓补码
计算机的底层存储，是按位(bit)来存的，每一个位有两个状态，0和1，在电路上分别对应低电平和高电平
假如我们用2bit来表示一个整数
| 首位 | 末位 | 代表的数 |
---
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 2 |
| 1 | 1 | 3 |
这就是无符号整数，只能表示非负数
如果想要表示负数，我们可以简单的偏移一个量(-2)，这样可以让正负数的量尽可能均匀
| 首位 | 末位 | 代表的数 |
---
| 0 | 0 | -2 |
| 0 | 1 | -1 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
这样表示上是没有问题了，但是计算上不方便
举个例子，这套系统应该能表示-2 ~ 1范围内的运算，那么
-1(01) + -1(01) = -2(00)
括号内为我们使用的编码，括号外为需要计算的数字
01 + 01 = 10 才对，编码对不上
不过，这难不倒聪明的前人，只要稍加修改下编码方案，就能让符号位也能正常参与运算
| 首位 | 末位 | 代表的数 |
---
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | -2 |
| 1 | 1 | -1 |
-1(11) + -1(11) = -2(10)
这就是所谓的补码
我们使用的Int数据类型，也是用补码来实现的，使用了32bit来存储，所以能表示的范围是 -2^31 ~ 2^31 -1
所谓溢出，就是指运算超过能表示的范围，观察下下面的例子你就明白了
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 | scala> val a = 1<<16 val 1 16val a: Int = 65536 val Int 65536 scala> a aval res70: Int = 0 val Int 0 scala> val b = 1<<31 val 1 31val b: Int = -2147483648 val Int -2147483648 scala> b - 1 1val res71: Int = 2147483647 val Int 2147483647 |
~~好了，显然内的内容讲完了~~
事实上，fib(47)就溢出了
| | |
--- |
| 1 2 3 4 5 | scala> fib(46) 46val res9: Int = 1836311903 val Int 1836311903 scala> fib(47) 47val res10: Int = -1323752223 val Int -1323752223 |
知道问题就好办了，我们只需要增加数据能表示的范围就行了
Scala 内置的BigInt很方便就能做到这一点
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 12 | def fib(n: Int): BigInt = {def fib def fib Int BigInt var (a, b) = (BigInt(0), BigInt(1)) var BigInt 0 BigInt 1 for (i <- 1 to n) {for 1 val temp = b val b = a + b a = temp } b} //scala> fib(100) //scala> fib(100)//val res11: BigInt = 573147844013817084101 //val res11: BigInt = 573147844013817084101 |
一个思考题：Long能存下fib(100)的结果而不溢出吗？
花枪其二
或者我们可以稍微耍个花枪，用LazyList来构造一个生成器
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 | val fibs: LazyList[BigInt] = val LazyList BigInt BigInt(0) #:: BigInt(1) #:: (fibs zip fibs.tail).map{ t => t._1 + t._2 } BigInt 0 BigInt 1 scala> fibsval res0: LazyList[BigInt] = LazyList(<not computed>) val LazyList BigInt LazyList scala> fibs(4) 4val res1: BigInt = 3 val BigInt 3 scala> fibsval res2: LazyList[BigInt] = LazyList(0, 1, 1, 2, 3, <not computed>) val LazyList BigInt LazyList 0 1 1 2 3 |
这样定义的fibs有个好处，就是懒，不问不算，但是一旦算好，结果就保存起来了
寻找失去的精度
为什么不用Double呢？根据IEEE754,双精度浮点数的精度有限
在IEEE 754标准中，用64位来存储双精度浮点数
其中分配了52位来存储浮点数的有效数字，11位存储指数，1位存储正负号
一旦需要确保精度，就会出现问题
| | |
--- |
| 1 2 | scala> val a = 1.99999999999999999999999 val1.99999999999999999999999val a: Double = 2.0 val Double2.0 |
我们可以试下Double版本的代码
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | def fib(n: Int): Double = {def fib def fib Int Double var (a, b) = (0.0, 1.0) var0.01.0 for (i <- 1 to n) {for 1 val temp = b val b = a + b a = temp } b} //scala> fib(100) //scala> fib(100)//val res73: Double = 5.731478440138172E20 //val res73: Double = 5.731478440138172E20 // 对比BigInt版本的结果 // 对比BigInt版本的结果//scala> fib(100) //scala> fib(100)//val res11: BigInt = 573147844013817084101 //val res11: BigInt = 573147844013817084101 |
为了加深理解，我们来手推一遍Double的编码过程
| | |
--- |
| 1 2 3 4 5 6 7 | def convertRadix(str: String, radix: Int, buf: String): String = {def convertRadix def convertRadix String Int String String val i = BigInt(str) val BigInt if (i == 0) buf.reverse if 0 else if (i % 2 != 0) {else if 2 0 convertRadix((i/2).toString, radix, buf + "1") 2 "1" } else convertRadix((i/2).toString, radix, buf + "0") else 2 "0"} |
| | |
--- |
| 1 2 3 | 令 a = 573147844013817084101a的二进制表示 a2 = 1 1111 0001 0010 0000 0110 0010 1111 0111 0110 1001 0000 1001 0000 0011 1000 1100 0101 |
图源：wikipedia
对应的公式为
(−1)sign(1+52∑i=1b52−i2−i)∗2e−1023
符号位 S： 0， 表示非负数
二进制表示：
1 1111 0001 0010 0000 0110 0010 1111 0111 0110 1001 0000 1001 0000 0011 1000 1100 0101
小数点移动后
1. 1111 0001 0010 0000 0110 0010 1111 0111 0110 1001 0000 1001 0000 0011 1000 1100 0101
移动位数E1（左移为正，右移为负）: 68
存储的阶码E，11位： 68+1023 = 1091 = (10001000011)b
尾数M，52位 （超过的截断，不足的补零）：
1111 0001 0010 0000 0110 0010 1111 0111 0110 1001 0000 1001 0000
实际存储的数据
| | |
--- |
| 1 2 | S E M 0 10001000011 1111 0001 0010 0000 0110 0010 1111 0111 0110 1001 0000 1001 0000 |
我们逆向还原下
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 | def recover(s: Int, e: String, m: String): BigDecimal = {def recover def recover Int String String BigDecimal val e1 = Integer.parseInt(e, 2) - 1023 val Integer 2 1023 val sign = math.pow(-1, s) val -1 require(m.length == 52) 52 val num = 1 + (1 to 52).map(i=>(m(i-1)-'0')BigDecimal(2).pow(-i)).sum val 1 1 52 -1 0 BigDecimal 2 println(e1, sign, num) BigDecimal(2).pow(e1) sign num BigDecimal 2} / /scala> recover(0, "10001000011", "1111000100100000011000101111011101101001000010010000") scala> recover(0, "10001000011", "1111000100100000011000101111011101101001000010010000")(68,1.0,1.941900430109885888896315009333193) (68,1.0,1.941900430109885888896315009333193)val res53: BigDecimal = 573147844013817069567.9999999999999 val res53: BigDecimal = 573147844013817069567.9999999999999 scala> recover(0, "10001000011", "1111000100100000011000101111011101101001000010010000").toDouble scala> recover(0, "10001000011", "1111000100100000011000101111011101101001000010010000").toDouble(68,1.0,1.941900430109885888896315009333193) (68,1.0,1.941900430109885888896315009333193)val res54: Double = 5.731478440138171E20 val res54: Double = 5.731478440138171E20 // 对比之前的结果 完全吻合 // 对比之前的结果 完全吻合 scala> fib(100) scala> fib(100)val res73: Double = 5.731478440138172E20 val res73: Double = 5.731478440138172E20 / / |
所以，用Double前，首先需要问自己一个问题，这个数据需不需要确保完整的精度。
所以，要二分吗
之前的问题都是求值，现在换一个问题
斐波那契数列中，第一个有 1000 位数字的是第几项（从零开始算的话）？
考虑略微修改后，复用之前的代码
| | |
--- |
| 1 2 3 4 5 6 7 8 9 | def fib(n: Int): BigInt = {def fib def fib Int BigInt var (a, b) = (BigInt(0), BigInt(1)) var BigInt 0 BigInt 1 for (i <- 1 to n) {for 1 val temp = b val b = a + b a = temp } a} |
一个朴素的做法是逐个试探
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | scala> fib(200).toString.length 200val res80: Int = 42 val Int 42 scala> fib(700).toString.length 700val res81: Int = 146 val Int 146 scala> fib(3000).toString.length 3000val res82: Int = 627 val Int 627 scala> fib(5000).toString.length 5000val res83: Int = 1045 val Int 1045 scala> fib(4000).toString.length 4000val res84: Int = 836 val Int 836 ... |
之所以能这样做，是因为fib(n)的长度是一个随着n单调非递减的函数，我们只要对比结果
比1000大，调小n再试一次
比1000小，调大n再试一次
正好是1000，此时n就是答案
我们使用二分法，需要确定上界和下界
显然fib(n)的长度小于n，这确定了下界
上界可以用放缩法来确定
我们知道
∵fib(n)=fib(n−1)+fib(n−2)>2×fib(n−2) ∴fib(n)>16×fib(n−8)>10×fib(n−8)
所以n增加8时，对应的十进制长度至少增加1
len(fib(8n))>n
上界可以选择为8n
我们可以用二分法来替代手动试探结果的过程
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 | def fibLen(x: Int): Int = {def fibLen def fibLen Int Int var left = x var var right = 8 x var 8 def len(guess: Int) = fib(guess).toString.length def len def len Int while (left < right) { while var guess = left + (right-left)/2 var 2 var l = len(guess) var if (l < x) {if left = guess + 1 1 } else if (l >= x) {else if right = guess } println(guess, l, left, right) } left} / /scala> fibLen(1000) scala> fibLen(1000)(6500,1359,1000,6500) (6500,1359,1000,6500)(3750,784,3751,6500) (3750,784,3751,6500)(5125,1071,3751,5125) (5125,1071,3751,5125)(4438,928,4439,5125) (4438,928,4439,5125)(4782,1000,4439,4782) (4782,1000,4439,4782)(4610,964,4611,4782) (4610,964,4611,4782)(4696,982,4697,4782) (4696,982,4697,4782)(4739,991,4740,4782) (4739,991,4740,4782)(4761,995,4762,4782) (4761,995,4762,4782)(4772,997,4773,4782) (4772,997,4773,4782)(4777,998,4778,4782) (4777,998,4778,4782)(4780,999,4781,4782) (4780,999,4781,4782)(4781,999,4782,4782) (4781,999,4782,4782)val res108: Int = 4782 val res108: Int = 4782/ / |
花枪其三
我们也可以暴力遍历，这个时间复杂度是可以接受的
| | |
--- |
| 1 2 3 4 | val fibs: LazyList[BigInt] = val LazyList BigInt BigInt(0) #:: BigInt(1) #:: (fibs zip fibs.tail).map{ t => t._1 + t._2 } BigInt 0 BigInt 1 fibs.zipWithIndex.dropWhile(_._1.toString.length<1000).head._2 // 4782 1000 // 4782 |
输入范围扩充
如果我们要求一个大一些的数据，比如fib(10^9)
这个数太大了，所以我们求的是 fib(10^9) % (10^9 + 7)来检验结果是否正确
即求
fib(109)
首先分析下问题的规模，10^9的情况下，O(n)的做法过慢，最好能有更好的算法
~~如果你是布鲁特佛斯的爱好者，那当我没说~~
普通青年：矩阵快速幂
首先需要介绍两个前置知识
矩阵
快速幂
矩阵
由 m × n 个数a[i][j]排成的m行n列的数表称为m行n列的矩阵，简称m × n矩阵。记作
这里会用到一点简单的矩阵乘法
快速幂
考虑这么一个问题，求解
An mod m
模运算的乘法规则和基本四则运算类似
比较朴素的做法需要O(n)的时间
| | |
--- |
| 1 2 3 4 5 6 7 | def pow(a: Long, n: Long, m: Long): Long = {def pow def pow Long Long Long Long var res = 1L var 1 for (i <- 1 to n) {for 1 res = (res a) % m } res} |
其实我们可以利用上一步的结果来加速运算
比如，对于求A^10
也可以写作
| | |
--- |
| 1 2 3 4 | pow(a, 10) = pow(a^2, 5) 10 2 5 = a^2 pow(a^4, 2) 2 4 2 = a^2 pow(a^8, 1) 2 8 1 = a^2 a^8 2 8 |
每次对需要计算的幂除以二向下取整，通过这样的方法，每次需要计算的幂次都是原来的一半
这样可以把时间复杂度从O(n)降低到O(logn)
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 12 13 | def pow(a: Long, n: Long, m: Long): Long = {def pow def pow Long Long Long Long var res = 1L var 1 var i = n var var base = a var while (i > 0) {while 0 if (i % 2 != 0) {if 2 0 res = (res base) % m } base = (base base) % m i = i / 2 2 } res} |
矩阵快速幂
有了矩阵和快速幂的前置科技，矩阵快速幂的科技被点亮就是水到渠成的了
用F(n)表示斐波那契数列的第n项
考虑构造一个矩阵
如果能找(gou)到(zao)这样的矩阵A，把左右两边看成递推数列，有
依据矩阵的乘法，可知
由斐波那契数列的递归式，可知
Fn+1=Fn+Fn−1， Fn=Fn
比较系数后，可以求出矩阵A
于是，可以得出
因此，只要求出A^n 就能知道第N项的斐波那契数列
这个求幂运算可以利用之前快速幂的算法，依葫芦画瓢
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 | final case class Matrix(elements: Array[Array[Long]], M: Long = 1000000007) {final caseclass Matrix(elements: Array[Array[Long]], M: Long = 1000000007) class Matrixelements: Array[Array[Long]], M: Long = 1000000007 Array Array Long M Long val rows: Int = elements.length val Int val cols: Int = elements.headOption.getOrElse(Array.empty[Long]).length val Int Array Long def (that: Matrix): Matrix = {def def Matrix Matrix require(this.cols == that.rows, "invalid matrix size") this "invalid matrix size" val res = Array.fill(this.rows, that.cols)(0L) val Array this 0 for (i <- 0 until this.rows; j <- 0 until that.cols) for 0 this 0 for (k <- 0 until this.cols) {for 0 this res(i)(j) = (res(i)(j) + this.elements(i)(k) that.elements(k)(j)) % M this M } Matrix(res) Matrix } def ^(power: Int): Matrix = {def ^ def ^ Int Matrix require(power >= 0, "power should not be negative") 0 "power should not be negative" val unitElements = Array.fill(this.rows, this.cols)(0L) val Array this this 0 for ((i, j) <- (0 until this.rows) zip (0 until this.cols)) unitElements(i)(j) = 1L for 0 this 0 this 1 val unitMatrix = Matrix(unitElements) val Matrix @scala.annotation.tailrec @scala def fastPow(res: Matrix, p: Int, base: Matrix): Matrix = {def fastPow def fastPow Matrix Int Matrix Matrix if (p == 0) res if 0 else if (p % 2 == 0) fastPow(res, p/2, base base) else if 2 0 2 else fastPow(res base, p/2, base base) else 2 } fastPow(unitMatrix, power, this) this }} |
简单测试下,速度还是很快的
| | |
--- |
| 1 2 3 4 5 6 7 8 | scala> val A = Matrix(Array(Array(1L,1L), Array(1L,0L))) val A Matrix Array Array 1 1 Array 1 0val A: Matrix = Matrix([[J@5e85c21b,1000000007) val A Matrix Matrix J 5e85 1000000007 scala> def fib(n: Int) = (A^n).elements(1)(0) def fib def fib Int A 1 0def fib(n: Int): Long def fib def fib Int Long scala> fib(1000000000) 1000000000val res7: Long = 21 val Long 21 |
文艺青年：快速倍增法
矩阵快速幂的算法的时间复杂度是O(m^3logn)级别的，这里的m计算两个m长度的数乘积所需要的时间
其实，有一个速度更快的方案，我们可以略过矩阵计算这一步骤
一个朴素的思路是，观察并猜出斐波那契数列的22矩阵表达形式，然后用数学归纳法证明
对上述式子做一个变换
由左右两个矩阵间的对应关系，可得
我们需要用到也就是
有了这两个式子，我们能从F(n), F(n-1) 计算出 F(2n)，F(2n-1)
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 | def fastFib(n: Int, M: Long = 1000000007): (Long, Long) = {def fastFib def fastFib Int M Long 1000000007 Long Long if (n == 0) (0L, 1L) if 0 0 1 else {else val p = fastFib(n / 2) val 2 val a = p._1 ((2 p._2 - p._1) % M + M) % M val 2 M M M val b = (p._1 p._1 + p._2 p._2) % M val M if (n % 2 == 0) (a, b) if 2 0 else (b, (a + b) % M) else M }} |
这里之所以要(... % M + M)，是为了防止中间过程出现负数
简单试验下
| | |
--- |
| 1 2 | scala> fastFib(1000000000)._1 1000000000val res64: Long = 21 val Long 21 |
二逼青年： 扩充数域
一个化归的笑话
如果不懂什么矩阵，也不懂什么数学归纳法，更不懂什么生成函数，能不能用初等方法求出答案呢？
首先我们需要前置知识：斐波那契数列的通项公式推导
在此之前，我们先来解决递推式的一般推导问题
xn=axn−1,x0=C,a≠0
这个式子很好求通项公式，观察到下一项是上一项的倍数，所以有
一天，数学家觉得自己已受够了数学，于是他跑到消防队去宣布他想当消防员。
消防队长说：「您看上去不错，可是我得先给您一个测试。」
消防队长带数学家到消防队后院小巷，巷子里有一个货栈，一只消防栓和一卷软管。
消防队长问：「假设货栈起火，您怎么办？」
数学家回答：「我把消防栓接到软管上， 打开水龙头，把火浇灭。」
消防队长说：「完全正确。最后一个问题：假设您走进小巷，而货栈没有起火，您怎么办？」
数学家疑惑地思索了半天，终于答道：「我就把货栈点着。」
消防队长大叫起来：「什么？太可怕了，您为什么要把货栈点着？」
数学家回答：「这样我就把问题化简为一个我已经解决过的问题了。」
好了，有了这一问的基础，我们加大点难度，尝试更加一般的结论
yn=ayn−1+b,y0=C
如果这个式子的左右两边能化归到第一问等比的形式，就好求了
(yn+t)=a(yn−1+t)
展开后可以求出
∵yn=ayn−1+(a−1)t ∴b=(a−1)t,t=ba−1
这里要注意一个问题，分母不能为0，所以对a=1的情况要额外讨论，当a=1时，原式便退化成了等差数列
∵yn=yn−1+b∴yn=C+nb
当a不为1时
yn+ba−1yn−1+ba−1=a 由第一问的结论可知 yn+ba−1=(C+ba−1)∗an
综上所述
yn={(C+ba−1)∗an−ba−1,a≠1C+nb,a=1
说了这么多，让我们回到原来的问题
Fn=Fn−1+Fn−2,F0=0,F1=1
我们希望求出上面这个问题的一般解
有了两问的铺垫，我们知道可以用构(cou)造的方法化归到已知的问题
对于
yn=ayn−1+b,y0=C
我们知道其通项公式为
我们的目标希望整理出如下的式子
使用待定系数法
由一元二次方程，可知
⎧⎨⎩p=−1−√52q=1−√52或者⎧⎨⎩p=√5−12q=√5+12
接着
右边不是常数，无法直接使用第二问的结论，没关系，我们继续凑
所以
为了便于书写，我们做如下约定
a=1+√52,b=1−√52
那么
Fn=1√5(an−bn)
一点简单的数论
对于求模运算，由于a, b的分母都为2，首先需要找到2在1000000007的模逆元
方法有很多，这里直接给出结果500000004
我们知道，斐波那契的每一项都是整数，因而尽管上面的式子中有无理数，但是这一部分都是会被互相抵消的
为了不丢失精度，将无理数看作类似复数的东西，设
m+n√5记作数对，其中m,n都为整数 M=1000000007 t=500000004
那么
承前所言，无理数部分会抵消，所以c一定为0,我们关心的就是d的值
接下来我们需要定义数对在模意义下的运算
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 | final case class ExPair(a: Long, b: Long, M: Long = 1000000007) {final caseclass ExPair(a: Long, b: Long, M: Long = 1000000007) class ExPaira: Long, b: Long, M: Long = 1000000007 Long Long M Long def +(that: ExPair): ExPair = {def + def + ExPair ExPair ExPair((this.a + that.a) % M, (this.b + that.b) % M) ExPair this M this M } def (that: ExPair): ExPair = {def def ExPair ExPair val na = (this.a that.a + 5 this.b that.b) % M val this 5 this M val nb = (this.a that.b + this.b that.a) % M val this this M ExPair(na, nb) ExPair } def -(that: ExPair): ExPair = {def - def - ExPair ExPair ExPair((this.a - that.a + M) % M, (this.b - that.b + M) % M) ExPair this M M this M M } def ^(power: Int): ExPair = {def ^ def ^ Int ExPair require(power >= 1, "power should be positive") 1 "power should be positive" @scala.annotation.tailrec @scala def fastPow(res: Option[ExPair], p: Int, base: ExPair): ExPair = {def fastPow def fastPow Option ExPair Int ExPair ExPair if (p == 0) res.get if 0 else if (p % 2 == 0) {else if 2 0 fastPow(res, p/2, base base) 2 } else {else if (res.isDefined) fastPow(Some(res.get base), p/2, base base) if Some 2 else fastPow(Some(base), p/2, base base) else Some 2 } } fastPow(None, power, this) None this }} |
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | scala> val T = 500000004 val T 500000004val T: Int = 500000004 val T Int 500000004 scala> val A = ExPair(T, T) val A ExPair T Tval A: ExPair = ExPair(500000004,500000004,1000000007) val A ExPair ExPair 500000004 500000004 1000000007 scala> val B = ExPair(T, -T) val B ExPair T Tval B: ExPair = ExPair(500000004,-500000004,1000000007) val B ExPair ExPair 500000004 -500000004 1000000007 scala> def fib(n: Int) = ((A^n)-(B^n)).b def fib def fib Int A Bdef fib(n: Int): Long def fib def fib Int Long scala> (1 to 10).map(fib) 1 10val res47: IndexedSeq[Long] = Vector(1, 1, 2, 3, 5, 8, 13, 21, 34, 55) val IndexedSeq Long Vector 1 1 2 3 5 8 13 21 34 55 scala> fib(1000000000) 1000000000val res48: Long = 21 val Long 21 |
Googol的斐波那契
某著名搜索引擎的来源，就是单词Googol，起名字还真是要有品味的，不然就像这样
图片来源
F(10100)mod(109+7)
Ap+q=ApAq Ap⋅q=(Ap)q Apq=Apq−1⋅p=(Apq−1)p=… =(((Ap)p)…)p)pq个
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 | // 复用之前方法 // 复用之前方法final case class Matrix(elements: Array[Array[Long]], M: Long = 1000000007) {final caseclass Matrix(elements: Array[Array[Long]], M: Long = 1000000007) class Matrixelements: Array[Array[Long]], M: Long = 1000000007 Array Array Long M Long val rows: Int = elements.length val Int val cols: Int = elements.headOption.getOrElse(Array.empty[Long]).length val Int Array Long def (that: Matrix): Matrix = {def def Matrix Matrix require(this.cols == that.rows, "invalid matrix size") this "invalid matrix size" val res = Array.fill(this.rows, that.cols)(0L) val Array this 0 for (i <- 0 until this.rows; j <- 0 until that.cols) for 0 this 0 for (k <- 0 until this.cols) {for 0 this res(i)(j) = (res(i)(j) + this.elements(i)(k) that.elements(k)(j)) % M this M } Matrix(res) Matrix } def ^(power: Int): Matrix = {def ^ def ^ Int Matrix require(power >= 0, "power should not be negative") 0 "power should not be negative" val unitElements = Array.fill(this.rows, this.cols)(0L) val Array this this 0 for ((i, j) <- (0 until this.rows) zip (0 until this.cols)) unitElements(i)(j) = 1L for 0 this 0 this 1 val unitMatrix = Matrix(unitElements) val Matrix @scala.annotation.tailrec @scala def fastPow(res: Matrix, p: Int, base: Matrix): Matrix = {def fastPow def fastPow Matrix Int Matrix Matrix if (p == 0) res if 0 else if (p % 2 == 0) fastPow(res, p/2, base base) else if 2 0 2 else fastPow(res base, p/2, base base) else 2 } fastPow(unitMatrix, power, this) this } // 新加的 // 新加的 def ^^(p: Int, q: Int): Matrix = {def ^^ def ^^ Int Int Matrix require(p>0 && q>0, "power parameters, p and q, should be positive") 0 0"power parameters, p and q, should be positive" (1 until q).foldLeft(this ^ p)((cur: Matrix, _) => cur ^ p) 1 this Matrix }} // fib(p^q) // fib(p^q)def fibPQ(p: Int, q: Int) = fastFib(p)._1 def fibPQ def fibPQ Int Int |
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 | scala> val A = Matrix(Array(Array(1L,1L), Array(1L,0L))) val A Matrix Array Array 1 1 Array 1 0val A: Matrix = Matrix([[J@73ba068e,1000000007) val A Matrix Matrix J 73 1000000007 scala> def fibPQ(p: Int, q: Int) = (A^^(p,q)).elements(1)(0) def fibPQ def fibPQ Int Int A 1 0def fibPQ(p: Int, q: Int): Long def fibPQ def fibPQ Int Int Long scala> fibPQ(10, 9) 10 9val res49: Long = 21 val Long 21 scala> fibPQ(10, 100) 10 100val res50: Long = 175077019 val Long 175077019 scala> fibPQ(10, 1000) 10 1000val res51: Long = 552179166 val Long 552179166 |
扩展阅读
现有的浮点运算是基于二进制的，有没有十进制的呢
基于斐波那契的数据结构，Fib堆
基于斐波那契的算法，Fib搜索
求解逆元时常用的算法，扩展欧几里得(ex-gcd)以及费马小定理
求解常系数线性齐次递推关系，参考《离散数学》8.2.2小节
Karatsuba乘法
P.S 图有点糊，因为部分公式是直接贴的图而不是渲染的
我现在暂时还没想到好办法来处理 mathjax和hexo冲突导致的渲染问题
本文对应的pdf版可以在这里获取
参考链接
WikiPedia: Fibonacci number
OI-wiki: fibonacci
Github: 我们为什么要考斐波那契数列？一道上机笔试题的解析，兼谈技术面试和编程基本功
博客园: 【学习笔记】斐波那契数列的简单性质
Fast Fibonacci algorithms
CSDN: float和double的编码示例
知乎：斐波那契数列通项公式是怎样推导出来的？ - Lancewu的回答
人民教育出版社-课程教材研究所：斐波那契数列的通项公式推导
除另有声明外，本博客文章均采用 知识共享(Creative Commons) 署名-非商业性使用-相同方式共享 3.0 中国大陆许可协议 进行许可。
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15847 | https://www.geeksforgeeks.org/dsa/sum-arithmetic-geometric-sequence/ | Sum of Arithmetic Geometric Sequence
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07 Aug, 2022
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In mathematics, an arithmetico–geometric sequence is the result of the term-by-term multiplication of a geometric progression with the corresponding terms of an arithmetic progression.
is an arithmetico–geometric sequence.
Given the value of a(First term of AP), n(Number of terms), d(Common Difference), b(First term of GP), r(Common ratio of GP). The task is find the sum of first n term of the AGP.
Examples:
Input : First term of AP, a = 1,
Common difference of AP, d = 1,
First term of GP, b = 2,
Common ratio of GP r = 2,
Number of terms, n = 3
Output : 34
Explanation
Sum = 12 + 222 + 323
= 2 + 8 + 24
= 34
The nth term of an arithmetico–geometric sequence is the product of the n-th term of an arithmetic sequence and the nth term of a geometric one. Arithmetico–geometric sequences arise in various applications, such as the computation of expected values in probability theory. For example Counting Expected Number of Trials until Success.
n-th term of an AGP is denoted by: tn = [a + (n - 1) d] (b rn-1)
Method 1: (Brute Force)
The idea is to find each term of the AGP and find the sum.
Below is the implementation of this approach:
C++
````
// CPP Program to find the sum of first n terms.
include
using namespace std;
// Return the sum of first n term of AGP
int sumofNterm(int a, int d, int b, int r, int n)
{
// finding the each term of AGP and adding
// it to sum.
int sum = 0;
for (int i = 1; i <= n ; i++)
sum += ((a + (i -1) d) (b pow(r, i - 1)));
return sum;
}
// Driven Program
int main()
{
int a = 1, d = 1, b = 2, r = 2, n = 3;
cout << sumofNterm(a, d, b, r, n) << endl;
return 0;
}
````
// CPP Program to find the sum of first n terms.
// CPP Program to find the sum of first n terms.
```
include#include
```
using namespace std; using namespace std
// Return the sum of first n term of AGP
// Return the sum of first n term of AGP
int sumofNterm(int a, int d, int b, int r, int n) int sumofNterm int a int d int b int r int n
{
// finding the each term of AGP and adding
// finding the each term of AGP and adding
// it to sum.
// it to sum.
int sum = 0; int sum = 0
for (int i = 1; i <= n ; i++) for int i = 1 i<= n i ++
sum += ((a + (i -1) d) (b pow(r, i - 1))); sum += a + i - 1 d b pow r i - 1
return sum; return sum
}
// Driven Program
// Driven Program
int main() int main
{
int a = 1, d = 1, b = 2, r = 2, n = 3; int a = 1 d = 1 b = 2 r = 2 n = 3
cout << sumofNterm(a, d, b, r, n) << endl; cout<< sumofNterm a d b r n<< endl
return 0; return 0
}
Java
````
// Java Program to find the sum of first n terms.
import java.io.;
class GFG {
// Return the sum of first n term of AGP
static int sumofNterm(int a, int d, int b, int r, int n)
{
// finding the each term of AGP and adding
// it to sum.
int sum = 0;
for (int i = 1; i <= n ; i++)
sum += ((a + (i -1) d) (b Math.pow(r, i - 1)));
return sum;
}
// Driven Program
public static void main(String args[])
{
int a = 1, d = 1, b = 2, r = 2, n = 3;
System.out.println(sumofNterm(a, d, b, r, n));
}
}
// This code is contributed by Nikita Tiwari.
````
Python3
````
Python3 code to find the
sum of first n terms.
import math
Return the sum of first
n term of AGP
def sumofNterm( a , d , b ,
r , n ):
# finding the each term
# of AGP and adding it to sum.
sum = 0
for i in range(1,n+1):
sum += ((a + (i -1) d)
(b math.pow(r, i - 1)))
return int(sum)
Driven Code
a = 1
d = 1
b = 2
r = 2
n = 3
print(sumofNterm(a, d, b, r, n))
This code is contributed by "Sharad_Bhardwaj".
````
C#
````
// C# Program to find the sum of first n terms.
using System;
class GFG {
// Return the sum of first n term of AGP
static int sumofNterm(int a, int d, int b, int r, int n)
{
// Finding the each term of AGP
// and adding it to sum.
int sum = 0;
for (int i = 1; i <= n ; i++)
sum += (int)((a + (i -1) d)
(b Math.Pow(r, i - 1)));
return sum;
}
// Driver Code
public static void Main()
{
int a = 1, d = 1, b = 2, r = 2, n = 3;
Console.Write(sumofNterm(a, d, b, r, n));
}
}
// This code is contributed by vt_m.
````
PHP
````
php
// PHP program to find the
// sum of first n terms.
// Return the sum of first
// n term of AGP
function sumofNterm($a, $d, $b, $r, $n)
{
// finding the each term
// of AGP and adding
// it to sum.
$sum = 0;
for ($i = 1; $i <= $n ; $i++)
$sum += (($a + ($i -1) $d)
($b pow($r, $i - 1)));
return $sum;
}
// Driver Code
$a = 1; $d = 1; $b = 2; $r = 2; $n = 3;
echo(sumofNterm($a, $d, $b, $r, $n));
// This code is contributed by Ajit.
?
````
JavaScript
````
// javascript program to find the
// sum of first n terms.
// Return the sum of first
// n term of AGP
function sumofNterm(a, d, b, r, n)
{
// finding the each term
// of AGP and adding
// it to sum.
let sum = 0;
for (let i = 1; i <= n ; i++)
sum += ((a + (i -1) d)
(b Math.pow(r, i - 1)));
return sum;
}
// Driver Code
let a = 1;
let d = 1;
let b = 2;
let r = 2;
let n = 3;
document.write(sumofNterm(a, d, b, r, n));
// This code is contributed by sravan kumar (vignan)
````
Output:
34
Time Complexity: O(nlogn) since using a inbuilt pow function inside a loop
Auxiliary Space: O(1) as using constant variables
Method 2: (Using Formula)
Proof,
```
Series,
Sn = ab + (a+d)br + (a+2d)br2 + ..... + (a + (n-1)d)brn-1
Multiplying Sn by r,
rSn = abr + (a+d)br2 + (a+2d)br3 + ..... + (a + (n-1)d)brn
Subtract rSn from Sn,
(1 - r)Sn = [a + (a + d)r + (a + 2d)r2 + ...... + [a + (n-1)d]rn-1]
- [ar + (a + d)r2 + (a + 2d)r3 + ...... + [a + (n-1)d]rn]
= b[a + d(r + r2 + r3 + ...... + rn-1)
- [a + (n-1)d]rn]
(Using sum of geometric series Sn = a(1 - rn-1)/(1-r))
= b[a + dr(1 - rn-1)/(1-r) - [a + (n-1)d]rn]
```
Below is the implementation of this approach:
CPP
````
// CPP Program to find the sum of first n terms.
include
using namespace std;
// Return the sum of first n term of AGP
int sumofNterm(int a, int d, int b, int r, int n)
{
int ans = 0;
ans += a;
ans += ((d r (1 - pow(r, n-1)))/(1-r));
ans -= (a + (n-1)d)pow(r, n);
return (ansb)/(1-r);
}
// Driven Program
int main()
{
int a = 1, d = 1, b = 2, r = 2, n = 3;
cout << sumofNterm(a, d, b, r, n) << endl;
return 0;
}
````
// CPP Program to find the sum of first n terms.
// CPP Program to find the sum of first n terms.
```
include#include
```
using namespace std; using namespace std
// Return the sum of first n term of AGP
// Return the sum of first n term of AGP
int sumofNterm(int a, int d, int b, int r, int n) int sumofNterm int a int d int b int r int n
{
int ans = 0; int ans = 0
ans += a; ans += a
ans += ((d r (1 - pow(r, n-1)))/(1-r)); ans += d r 1 - pow r n - 1/ 1 - r
ans -= (a + (n-1)d)pow(r, n); ans -= a + n - 1 d pow r n
return (ansb)/(1-r); return ans b/ 1 - r
}
// Driven Program
// Driven Program
int main() int main
{
int a = 1, d = 1, b = 2, r = 2, n = 3; int a = 1 d = 1 b = 2 r = 2 n = 3
cout << sumofNterm(a, d, b, r, n) << endl; cout<< sumofNterm a d b r n<< endl
return 0; return 0
}
Java
````
// Java Program to find the sum of first n terms.
import java.io.;
import java.math.;
class GFG {
// Return the sum of first n term of AGP
static int sumofNterm(int a, int d, int b, int r, int n)
{
int ans = 0;
ans += a;
ans += ((d r (1 - (int)(Math.pow(r, n-1))))/(1-r));
ans -= (a + (n-1)d)(int)(Math.pow(r, n));
return (ansb)/(1-r);
}
// Driven Program
public static void main(String args[])
{
int a = 1, d = 1, b = 2, r = 2, n = 3;
System.out.println(sumofNterm(a, d, b, r, n));
}
}
// This code is contributed by Nikita Tiwari.
````
Python3
````
Python3 code to find
the sum of first n terms.
import math
Return the sum of
first n term of AGP
def sumofNterm( a , d , b ,
r , n ):
ans = 0
ans += a
ans += ((d r (1 - math.pow(r, n-1))
)/(1-r))
ans -= (a + (n-1)d)math.pow(r, n)
return int((ansb)/(1-r))
Driven Code
a = 1
d = 1
b = 2
r = 2
n = 3
print(sumofNterm(a, d, b, r, n) )
This code is contributed by "Sharad_Bhardwaj".
````
C#
````
// C# Program to find the sum of first n terms.
using System;
class GFG {
// Return the sum of first n term of AGP
static int sumofNterm(int a, int d, int b, int r, int n)
{
int ans = 0;
ans += a;
ans += ((d r (1 - (int)(Math.Pow(r, n-1))))
/ (1-r));
ans -= (a + (n-1) d)
(int)(Math.Pow(r, n));
return (ans b) / (1 - r);
}
// Driver Code
public static void Main()
{
int a = 1, d = 1, b = 2, r = 2, n = 3;
Console.Write(sumofNterm(a, d, b, r, n));
}
}
// This code is contributed by vt_m.
````
PHP
````
php
// PHP program to find the
// sum of first n terms.
// Return the sum of first
// n term of AGP
function sumofNterm($a, $d, $b, $r, $n)
{
// finding the each term
// of AGP and adding
// it to sum.
$sum = 0;
for ($i = 1; $i <= $n ; $i++)
$sum += (($a + ($i -1) $d)
($b pow($r, $i - 1)));
return $sum;
}
// Driver Code
$a = 1; $d = 1; $b = 2; $r = 2; $n = 3;
echo(sumofNterm($a, $d, $b, $r, $n));
// This code is contributed by Ajit.
?
````
JavaScript
````
// JavaScript Program to find the sum of first n terms.
// Return the sum of first n term of AGP
function sumofNterm(a, d, b, r, n)
{
let ans = 0;
ans += a;
ans += ((d r (1 - (Math.pow(r, n-1))))/(1-r));
ans -= (a + (n-1)d)(Math.pow(r, n));
return (ansb)/(1-r);
}
// Driver code
let a = 1, d = 1, b = 2, r = 2, n = 3;
document.write(sumofNterm(a, d, b, r, n));
````
Output:
34
Time Complexity: O(logn)
Auxiliary Space: O(1)
A
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Cut The Knot - Probability Riddles
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Probability Riddles
Alexander Bogomolny
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Foreword by Nassim Nicholas Taleb viiPreface ix1 | Intuitive Probability 1
Bridge Hands
Secretary’s Problem
Pencil Logo
Balls of Two Colors I
by 396
Fair Dice
Selecting a Medicine
Probability of Random Lines Crossing
Numbers, One Inequality
Metamorphosis of a Quadratic Function
Playing with
Balls of Two Colors
Linus Pauling’s Argument
Galton’s Paradox
Solutions (p. 5)
2 | What Is Probability?133 | Likely Surprises 15
Balls of Two Colors II
Coin Tossing Contest
Probability of Visiting
Probability of Increasing Sequence
Probability of Two Integers
Being Coprime
Overlapping Random Intervals
Random Clock Hands
River after Heavy Rain
The Most Likely Position
Coin Tossing Surprises I
Solutions (p. 18)
4 | Basic Probability 33
Admittance to a Tennis Club
Black Boxes in a Chain
A Question of
Concerning Even Number of Heads
Getting Ahead by Two
Recollecting Forgotten Digit
Two Balls of the Same Color
To Bet or
Not to Bet
Lights on a Christmas Tree
Drawing Numbers and Summing Them
Up
Numbers in a Square
Converting Temperature from °C to °F
Probability of
No Distinct Positive Roots
Playing with Integers and Limits
Given Probability,
Find the Sample Space
Gladiator Games
In Praise of Odds
Probability in
Scoring
Probability of 2
n
Beginning with Digit 1
Probability of First Digits in
a Sequence of Powers
Probability of Four Random Integers Having a Common
Factor
Probability of a Cube Ending with 11
Odds and Chances in Horse Race
Betting
Acting As a Team
Sum of Two Outcomes of Tossing Three Dice
Chess
Players Truel
Two Loaded Dice
Crossing Bridge in Crowds
Solutions (p. 39)
5 | Geometric Probability 73
Three Numbers
Three Points on a Circle I
Three Points on a Circle II
Two Friends
Meeting
Hitting a Dart Board
Circle Coverage
Points in a Semicircle
Flat
Probabilities on a Sphere
Four Random Points on a Sphere
Random Numbers and
Obtuse Triangle
Random Intervals with One Dominant
Distributing Balls of Two
Colors in Two Bags
Hemisphere Coverage
Random Points on a Segment
Probability
of First Digit in Product
Birds on a Wire
Lucky Times at a Moscow Math
Olympiad
Probability of a Random Inequality
Points on a Square Grid
A
Triangle out of Three Broken Sticks
Probability in Dart Throwing
Probability in
Triangle
Solutions (p. 78)
6 | Combinatorics 121
Shuing Probability
Probability with Factorials
Two Varsity Divisions
Probability
of Having 5 in the Numerator
Random Arithmetic Progressions
Red Faces of a
Cube
Red and Green Balls in Red and Green Boxes
Probability of Equal Areas
on a Chessboard
Random Sum
Probability of Equilateral Triangle
Shelving an
Encyclopedia
Loaded Dice I
Loaded Dice II
Dropping Numbers into a 3 x 3
Square
Probability of Matching Socks
Numbered Balls Out of a Box
Planting Trees
in a Row
Six Numbers, Two Inequalities
Six Numbers, Three Inequalities
Tying
Knots in Brazil
Tying Knots in Russia
Guessing Hat Numbers
Probability of an
Odd Number of Sixes
Probability of Average
Marking and Breaking Sticks
Bubbling
of Sorts
Probability of Successive Integers
Solutions (p. 128)
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Acute Tubular Necrosis Mnemonic for Effective Learning
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Acute Tubular Necrosis
Acute-sign Tuba Necrosis-crow
Picmonic
Acute tubular necrosis (ATN) refers to the death of renal tubule cells in response to a variety of different insults, leading to acute kidney injury.
12 KEY FACTS
ETIOLOGY
Ischemic Injury
Ice-ischemia Injury
One group of etiologies of ATN is ischemic injury, where blood flow to the kidney is reduced, leading to cellular hypoxia and necrosis. This can occur during episodes of severe hypotension (e.g., sepsis, hemorrhage, etc) or surgeries where the renal vasculature is clamped.
Nephrotoxic Injury
Kidney with Toxic-green-glow Injury
ATN can also occur after the kidneys are exposed to nephrotoxins, resulting in tubule cell damage and death. Nephrotoxins include medications like aminoglycoside antibiotics, NSAIDs, or contrast dyes used for radiologic procedures.
HISTOLOGY
Granular Muddy Brown Casts
Grains and Muddy Brown Casts
ATN can be diagnosed by looking at the patient’s urine under microscopy. Granular casts will be revealed, representing necrotic tubule cells. They classically have a muddy brown appearance.
SIGNS & SYMPTOMS
Intrinsic Renal Failure
N-triscuit Dead Kidney
ATN leads to intrinsic renal failure, meaning the renal failure is caused by injury to the kidney tissue itself. After the initial insult, patients first go through an oliguric phase of severe kidney dysfunction, followed later by a polyuric phase while the kidneys begin to recover.
Oliguric Phase
Old-gopher
The first phase of ATN is the oliguric phase, where the kidney damage is so severe that the kidneys can’t properly form normal amounts of urine, leading to decreased urine output or oliguria. Patients will have other signs of acute kidney injury or renal failure, including hypertension, fluid retention, and lab abnormalities.
Metabolic Acidosis
Metal-ball Acidic-lemon
During the oliguric phase, patients will have a metabolic acidosis due to an inability of the kidney to secrete hydrogen from the body, and impaired bicarbonate buffering. In severe cases, uremia can result.
Hyperkalemia
Hiker-banana
Hyperkalemia results in part from the kidney’s being unable to secrete excess potassium from the blood into the urine.
Increase in BUN and Creatinine
Up-arrow BUN and Cr-eam
As with any acute kidney injury, the kidney cannot function to filter the blood properly, and so waste products build up in the bloodstream. This includes BUN and creatinine. The fractional excretion of sodium (FeNa) will also increase since the kidney cannot function normally to reclaim excreted sodium.
Polyuria Phase
Polly-urinates
After the oliguric phase of ATN, patients enter the polyuric phase, where urine output increases. This occurs because, even though the kidney is beginning to recover, it still cannot concentrate urine effectively.
Decrease in BUN and Creatinine
Down-arrow BUN and Cr-eam
In this phase, the previously elevated BUN and creatinine levels will begin to decrease.
Hypokalemia
Hippo-banana
Patients in the polyuric phase are at risk for developing hypokalemia due to potassium loss in the urine.
TREATMENT
Supportive Care
Supportive IV bags
Treatment for ATN revolves around supportive care for the clinical manifestations of renal failure. This includes control of hypertension, fluid overload, and electrolyte abnormalities. The underlying insult should be identified and corrected. In the case of nephrotoxic medications, they should be stopped. In severe cases of ATN, dialysis could be required.
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15850 | https://www.collegesidekick.com/study-docs/15774510 | BOOK Inorganic Chemistry - Catherine E. Housecroft And Alan G. Sharpe 2nd Edition_549.pdf - 512 Chapter 18 ¢ Organometallic compounds of s- and p-block elements Polymerization - College Sidekick
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BOOK Inorganic Chemistry - Catherine E. Housecroft And Alan G. Sharpe 2nd Edition_549.pdf
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Zumdahl Ninth Edition Test Bank (1).pdf Test bank Dear colleagues, This book was bought as a gift to our intimate Family at STEM that we hope would be of some assistance. We wish you all a fruitful learning and a resounding success. Yours sincerely, STEM Freelancer's Association Psst.The book October 6 University BIO 10
MTLE - CC (REVISED).pdf MEDICAL TECHNOLOGY LICENSURE EXAM REVIEW CLINICAL CHEMISTRY Lecturer: Roderick Balce Notes by: Xiao - The Conqueror of Demons, The Vigilant Yaksha, & Alatus, the Golden-Winged King 1. EQUIPMENTS AND SUPPLIES GLASS PIPETTES PIPET 1. Volumetric (most accura San Pedro College - Davao City CHEMISTRY ANALYTICAL
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Science-S8-End-of-Unit-2-Test.pdf lOMoARcPSD|38846755 S8 End-of-Unit 2 Test - Summary food and beverage management food and beverage management (Food Institute of Malaysia) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Vasanthakuma Singapore Campus Science 2023
512 Chapter 18 ¢ Organometallic compounds of s- and p-block elements Polymerization of alkenes 1s of great industrial importance, and one key issue 1s the production of stereoregular poly- mers; 1sotactic polypropene has a higher melting point, density and tensile strength than the atactic form. Use of the Ziegler—Natta heterogeneous catalyst controls not only the stereospecificity of the polymer but also allows the poly- merization of RCH=CH, (R = H or CH;) to be carried out at 298 K and under atmospheric pressure. This 1s 1n contrast to early processes which used both higher temperatures and pressures. The Ziegler—Natta catalyst consists of TiCly; and Et;:Al with Et,AlICl as co-catalyst, and its development carned K. Ziegler and G. Natta the 1963 Nobel Prize in Chemistry. Ziegler was also responsible for initiating the use of aluminium trialkyls as catalysts for the growth of unbranched alkane chains: EI3A1 S ¢ IICH2=CH2 oo Et:Al(CH:CH:)nEt 18.26). On an industrial scale. the direct reaction of Al with a terminal alkene and H, (equation 18.27) 1s employed. 2Al1 + 3R,Hg — 2R;Al + 3Hg (18.25) AICl; + 3RMgCl — R;Al + 3MgCl, (18.26) Al +3H, + 3R,C=CH, — (R,CHCH,);Al (18.27) Reactions between Al and alkyl halides yield alkyl aluminium halides (equation 18.28); note that 18.7 1s in equilibrium with 'R,Al(p-X),AIR,| and [RXAI(p-X),AIRX] via a redistribu- tion reaction, but 18.7 predominates in the mixture. X R it \Al\n\R 2Al+3RX — L~ . s Wy (18.28) X (18.7) Al +3iH, + 2R;Al — 3R,AlH (18.29) Alkyl alummium hydrides are obtained by reaction 18.29. These compounds, although unstable to both air and water, are important catalysts for the polymerization of alkenes and other unsaturated organic compounds. Ziegler—Natta catalysts containing trialkyl alummium compounds are introduced 1n Box 18.5. Hi 212 pm Crd H}C ll,," \Al\\\CH3 HCY N/ TNCH. H3 195 pm (18.8) Earlier we noted that R3;B compounds are monomeric. In contrast, aluminium trialkyls form dimers. Although this resembles the behaviour of the halides discussed in Section Subsequent conversion to long-chain alcohols 1s of commercial importance in the detergent industry. Ziegler—Natta catalysts are discussed further in Box 23.7 and Section 26.7. For further discussion of relevant industries, see: S. Dobson (1995) Chemistry &Industry, p. 870 — 'Man-made fibre markets: Recent agitation and change'. R.G. Harvan (1997) Chemistry &Industry, p. 212 - 'Polyethylene: New directions for a commodity thermo- plastic'. D.F. Oxley (1996) Chemistry &Industry, p. 535 —"The world market for polypropylene'. 12.6, there are differences in bonding. Trimethylaluminium (mp 313K) possesses structure 18.8 and so bonding schemes can be developed in like manner as for B,H. The fact that Al—Cyigge >Al—Ciermina 18 consistent with 3c-2e bonding in the AI-C—Al bridges, but with 2¢-2e terminal bonds. Equilibria between dimer and monomer exist in solution, with the monomer becoming more favoured as the steric demands of the alkyl group increase. Mixed alkyl halides also dimerize as exemplified 1n structure 18.7, but with particularly bulky R groups, the monomer (with trigonal planar Al) is favoured, e.g. (2.4.6-' Bu;C4H,)AICI, (Figure 18.8a). Triphenylaluminium also exists as a dimer, but 1n the mesityl derivative (mesityl = 2.4,6-Me3yCgH»), the steric demands of the substituents stabilize the monomer. Figure 18.8b shows the structure of Me,Al(u-Ph),AlMe,. and the orientations of the bridging phenyl groups are the same as i Ph,Al(u-Ph),AlPh,. This orientation is sterically favoured and places each ipso-carbon atom in an approximately tetrahedral environment. The ipso-carbon atom of a phenyl ring 1s the one to which the substituent 1s attached: e.g. in PPh,, the ipso-C of each Ph ring 1s bonded to P. In dimers containing RC=C-bridges, a different type of bonding operates. The structure of Ph,Al(PhC=C),AlPh, (18.9) shows that the alkynyl bridges lean over towards one of the Al centres. This 1s interpreted in terms of their behaving as o.7-ligands: each forms one Al-C o-bond and interacts with the second Al centre by using the C=C 7- bond. Thus, each alkynyl group 1s able to provide three electrons for bridge bonding in contrast to one electron being supplied by an alkyl or aryl group; the bonding is shown schematically in 18.10.
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15851 | https://math.mit.edu/classes/18.785/2016fa/LectureNotes18.pdf | 18.785 Number theory I Lecture #18 Fall 2016 11/10/2016 18 Dirichlet L-functions, primes in arithmetic progressions Having proved the prime number theorem, we would like to prove an analogous result for primes in arithmetic progressions.
We begin with Dirichlet’s theorem on primes in arithmetic progressions, a result that predates the prime number theorem by sixty years.
Theorem 18.1 (Dirichlet 1837). For all coprime integers a and m there are infinitely many primes p ≡a mod m.
In fact Dirichlet proved more than this. In a sense that we will make precise below, he proved that for every fixed modulus m the primes are equidistributed among the residue classes in (Z/mZ)×. The equidistribution statement that Dirichlet was able to prove is a bit weaker than one might like, but it is more than enough to establish Theorem 18.1.
Remark 18.2. Many of the standard tools of complex analysis we take for granted were not available to Dirichlet in 1837. Riemann was the first to seriously study ζ(s) as a function of a complex variable, some twenty years after Dirichlet proved Theorem 18.1. We will work in a more modern setting, but our approach still follows the spirit of Dirichlet’s proof.
18.1 Infinitely many primes To motivate Dirichlet’s method of proof, let us consider the following (admittedly clumsy) proof that there are infinitely many primes. It is sufficient to show that the Euler product ζ(s) = Y p (1 −p−s)−1 diverges as s →1+. Of course we know ζ(s) has a pole at s = 1 (by Theorem 16.3), but let us suppose for the moment that we did not already know this. Taking logarithms yields log ζ(s) = − X p log(1 −p−s) = X p p−s + O(1), (1) as s →1+, where we have used the asymptotic bounds −log(1 −x) = x + O(x2) (as x →0) and X p O(p−2s) = O(1) (Re(s) > 1/2).
We can estimate P p≤x 1 p via Mertens’ second theorem, one of three he proved in .
Theorem 18.3 (Mertens 1874). As x →∞we have (1) X p≤x log p p = log x + R(x), where |R(x)| < 2.1 (2) X p≤x 1 p = log log x + B + O 1 log x , where B = 0.261497 . . . is Mertens’ constant; (3) X p≤x log 1−1 p = −log log x −γ + O 1 log x , where γ = 0.577216 . . . is Euler’s constant.
1In fact, R(x) = −B3 + o(1) where B3 = 1.332582 . . . is an explicit constant.
Lecture by Andrew Sutherland Proof. See Problem Set 9.
Thus not only does P p−s diverge as s →1+, we can say with a fair degree of precision how quickly this happens.
We should note, however, that Mertens’ estimate is not as strong as the prime number theorem. Indeed, as you will prove on Problem Set 9, the Prime Number Theorem is equivalent to the statement X p≤x 1 p = log log x + B + o 1 log x , which is (ever so slightly) sharper than Mertens’ estimate.2 18.1.1 Infinitely many primes congruent to 1 modulo 4 To demonstrate how the argument above generalizes to primes in arithmetic progressions, let us prove there are infinitely many primes congruent to 1 mod 4.
We might initially consider Y p ≡1 mod 4 (1 −p−s)−1 = X n≥1 p|n ⇒p ≡1 mod 4 n−s, but the sum on the RHS is a bit awkward. Let us instead define a Dirichlet character χ(n) := 1 if n ≡1 mod 4, −1 if n ≡−1 mod 4, 0 otherwise, and consider the Dirichlet L-function L(s, χ) := Y p (1 −χ(p)p−s)−1 = X n≥1 χ(n)n−s = 1 −3−s + 5−s −7−s + 9−s + · · · , which converges absolutely on Re(s) > 1. As s →1+ we have log L(s, χ) = − X p log(1 −χ(p)p−s) = X p χ(p)p−s + O(1) = X p ≡1 mod 4 p−s − X p ≡3 mod 4 p−s + O(1), and log ζ(s) = X p ≡1 mod 4 p−s + X p ≡3 mod 4 p−s + O(1), thus log ζ(s) + log L(s, χ) 2 = X p ≡1 mod 4 p−s + O(1).
Provided log L(s, χ) = O(1) as s →1+, the LHS (and hence the RHS) must tend to infinity as s →1+, since ζ(s) →∞as s →1+. It thus suffices to show that L(s, χ) has an analytic 2The error term in the PNT actually implies P p≤x 1 p = log log x + B + O 1 x , but an o( 1 log x) bound is already enough to show π(x) ∼x/ log x. That the difference between a little-o and a big-O is the difference between proving the PNT and not proving it demonstrates how critical it is to understand error terms.
18.785 Fall 2016, Lecture #18, Page 2 continuation to a neighborhood of s = 1 with L(1, χ) ̸= 0 (in which case there is a branch of the complex logarithm holomorphic on a neighborhood of L(1, χ)). We will prove this in the next lecture. Assuming this for the moment, we then have X p ≤x p ≡1 mod 4 1 p = 1 2 log log x + O(1).
Mertens’ second theorem implies that the same holds if we instead sum over p ≡3 mod 4.
The primes are thus equidistributed modulo m = 4 in the sense that for all integers a coprime to m we have X p≤x p ≡a mod m 1 p ∼ 1 φ(m) X p≤x 1 p ∼ 1 φ(m) log log x.
We should note that this statement is weaker than the prime number theorem for arithmetic progressions, which states that π(x; m, a) ∼ 1 φ(m)π(x), where π(x; m, a) counts the primes p ≤x for which p ≡a mod m (see Problem Set 9).
Dirichlet did not have Mertens’ asymptotic bounds so he stated his results in a different way, using what is now called the Dirichlet density of a set of primes S, d(S) := lim s→1+ P p∈S p−s P p p−s , defined whenever this limit exists (one can also define notions of lower and upper Dirichlet density using lim inf and lim sup that are always defined and agree whenever d(S) is defined).
This definition differs from the more common notion of natural density δ(S) := lim x→∞ #{p ≤x : p ∈S} #{p ≤x} .
Dirichlet proved that for all coprime integers a and m the set of primes p ≡a mod m has Dirichlet density 1/φ(m), whereas the prime number theorem for arithmetic progressions states that this set has natural density 1/φ(m). If a set of primes S has a natural density then it has a Dirichlet density and the two are equal, but the converse need not hold: there are sets of primes that have a Dirichlet density but no natural density (see Problem Set 9).
In order to complete our proof that there are infinitely many primes p ≡1 mod 4, we still need to show L(1, χ) ̸= 0. We will achieve this in the next lecture, but for now let us show that this reduces to understanding the behavior of the Dedekind zeta function3 ζQ(i)(s) at s = 1. In general the Dedekind zeta function of a number field K is defined by ζK(s) := X I N(I)−s = Y p (1 −N(p)−s)−1, 3The Dedekind zeta function is named after Richard Dedekind, the last doctoral student of Gauss. He received his Ph.D. in 1854, the same year as Riemann, another student of Gauss. Dedekind and Riemann both studied under Dirichlet as well.
18.785 Fall 2016, Lecture #18, Page 3 where the sum ranges over nonzero ideals of the ring of integers OK, the product ranges over nonzero prime ideals of OK (primes of K), and N(I) := [OK : I] is the absolute norm.
Note that ζQ(s) = ζ(s), so this is a natural generalization of the Riemann zeta function.
That the Euler product for ζK(s) converges for Re(s) > 1 follows easily from the case ζQ(s) = ζ(s) proved in Theorem 16.2. We use unique factorization of ideals in the Dedekind domain OK to convert the sum over ideals I into a product over prime ideals p. The sum of the residue field degrees of primes p|p is bounded by P p|p epfp = [K :Q] = n, (Theorem 5.31) and N(p) = pfp (Theorem 6.9), so for Re(s) > 1 we have Q p|p |N(p)−s| ≤|p−ns| for each rational prime p. We thus have X p log(1 −N(p)−s) ≤n X p log(1 −p−s) .
The sum on the LHS converges on Re(s) > 1, so the sum on the RHS must as well.
For K = Q(i) we can rewrite the Euler product for ζK(s) as ζK(s) = Y p (1 −N(p)−s)−1 = Y p Y p|p (1 −N(p)−s)−1 = (1 −2−s)−1 Y p≡1 mod 4 (1 −p−s)−1(1 −p−s)−1 Y p≡3 mod 4 (1 −p−2s)−1 = (1 −2−s)−1 Y p≡1 mod 4 (1 −p−s)−1(1 −p−s)−1 Y p≡3 mod 4 (1 −p−s)−1(1 + p−s)−1 = Y p (1 −p−s)−1 Y p (1 −χ(p)p−s)−1 = ζ(s)L(s, χ), where we have used the fact that we have • one prime p of norm N(p) = 2 above the single prime p = 2 that ramifies in Q(i); • two primes p, ¯ p of norm N(p) = N(¯ p) = p above each prime p that spits in Q(i), equivalently, the primes p ≡1 mod 4; • one prime p of norm N(p) = p2 above each prime p that remains inert in Q(i), equivalently, the primes p ≡3 mod 4.
We know that ζ(s) has a simple pole at s = 1, so if we can show that ζK(s) extends to a meromorphic function with simple pole at s = 1 then we will know that L(s, χ) extends to a meromorphic function that is holomorphic and nonvanishing at s = 1; indeed, we must then have L(1, χ) ̸= 0, since ords=1L(s, χ) = ords=1ζK(s) −ords=1ζ(s) = −1 −(−1) = 0.
In fact, ζK(s) extends to a meromorphic function on Re(s) > 1 2 with a simple pole at s = 1; this can be proved directly, but it follows from a much more general and striking result, the analytic class number formula, which was also proved by Dirichlet (at least for quadratic fields). We will prove the analytic class number formula in the next lecture. For the remainder of this lecture we focus on generalizing our approach to handle arbitrary moduli m.
18.785 Fall 2016, Lecture #18, Page 4 18.2 Characters of finite abelian groups We want to generalize the Dirichlet character χ that we defined above. To do this we first recall some facts about characters of finite abelian groups; the domain Z of the Dirichlet character χ we used in the case m = 4 is not a finite abelian group, but χ restricts to a character of the multiplicative group (Z/4Z)×.
Definition 18.4. A character of a finite abelian group G is a homomorphism χ: G →U(1), where U(1) := {z ∈C : |z| = 1} is the unitary group. The character group (or dual group) of G is the abelian group b G := hom(G, U(1)) with pointwise multiplication: (χ1χ2)(g) := χ1(g)χ2(g). The inverse of χ is given by com-plex conjugation: χ−1(g) = χ(g) := χ(g) and the identify of b G is the trivial character.
Remark 18.5. This definition generalizes to locally compact abelian groups G, in which case each character χ: G →U(1) is a homomorphism of topological groups and the dual group b G is locally compact under the compact-open topology which has a basis of neigh-borhoods of the identity the sets U(C, V ) := {χ ∈b G : χ(C) ⊆V }, where C ranges over compact subsets of G and V ranges over open neighborhoods of the identity in U(1). The locally compact group b G is called the Pontryagin dual of G.4 When G is finite it necessarily has the discrete topology (since it must be Hausdorff), every homomorphism G →U(1) is automatically continuous, and the compact-open topology on b G is also discrete.
Proposition 18.6. Let G be a finite abelian group with character group ˆ G. Then G ≃ˆ G.
Proof. As a finite abelian group we can write G as a direct product of cyclic groups G = ⟨g1⟩× · · · × ⟨gn⟩≃Z/n1Z × · · · × Z/nrZ, with ni = |gi|, and each g ∈G can be uniquely written as g = Q i gei i with ei ∈[0, ni −1].
Now fix (not necessarily distinct) primitive ni-th roots of unity αi ∈U(1) and define χi ∈ˆ G by χi(gi) = αi and χi(gj) = 1 for j ̸= i. Then |χi| = |αi| = ni, and each χ ∈b G can be written uniquely as Q i χei i with ei ∈[0, ni −1], where χ(gi) = αei i . Therefore b G = ⟨χ1⟩× · · · × ⟨χn⟩≃Z/n1Z × · · · × Z/nrZ.
Corollary 18.7. Let G be a finite abelian group. Then g ∈G is the identity if and only if χ(g) = 1 for all χ ∈b G and χ ∈b G is the identity if and only if χ(g) = 1 for all g ∈G.
The isomorphism in Proposition 18.6 is not canonical.
Indeed, there are #Aut(G) distinct ways to choose the αi used to construct the isomorphism G ≃b G. But there is a canonical isomorphism from G to the character group of b G, the double dual of G.
Corollary 18.8. Let G be a finite abelian group. The evaluation map g 7→(χ 7→χ(g)) is a canonical isomorphism from G to its double dual.
4Some authors define the topology on the Pontryagin duality using uniform convergence on compact sets; for topological groups this is equivalent to the compact-open topology. The unitary group U(1) ≃R/Z is also referred to as the 1-torus or circle group and may be denoted T or S1 and viewed as an additive group.
18.785 Fall 2016, Lecture #18, Page 5 Proof. It is clear that the map above is a homomorphism, and Proposition 18.6 implies that G is isomorphic to its dual group b G, which is in turn isomorphic to its dual group, the double dual of G). So it suffices to show the map is injective, which follows from Corollary 18.7: if g is in the kernel then χ(g) = 1 for all χ ∈b G and g = 1G, by Corollary 18.7, Corollary 18.8 allows us to view G as the character group of G∧by defining g(χ) := χ(g).
Remark 18.9. Corollary 18.8 is a special case of Pontryagin duality, which applies to any locally compact abelian group G. For infinite groups, G and b G need not be isomorphic; for example, the character group of Z is isomorphic to U(1) (but in some cases they are, as when G is R or Qp, or any local field, see [3, XV, Lemma 2.2.1]). But the canonical isomorphism between G and its double dual always holds.
This is analogous to the situation with vector spaces: a finite dimensional vector space is isomorphic to its dual space while an infinite dimensional vector space need not be, but every vector space V is canonically isomorphic to its double-dual, and the isomorphism is given by evaluation. We should note that for a locally compact topological vector space V over a field k, the Pontryagin dual is not the same thing as the vector space dual: the Pontryagin dual corresponds to Hom(V, U(1)) (morphisms of locally compact groups) while the vector space dual corresponds to Homk(V, k) (morphisms of topological k-vector spaces).
For example, the vector space dual of Q is isomorphic to Q but the Pontryagin dual of Q is uncountable (as we shall see in later lectures).
Proposition 18.10. Let G be a finite abelian group. For all g1, g2 ∈G we have ⟨g1, g2⟩:= 1 #G X χ∈b G χ(g1)χ(g2) = ( 1 if g1 = g2, 0 if g1 ̸= g2, and for all χ1, χ2 ∈b G we have ⟨χ1, χ2⟩:= 1 #G X g∈G χ1(g)χ2(g) = ( 1 if χ1 = χ2, 0 if χ1 ̸= χ2.
Proof. By duality it suffices to consider ⟨g1, g1⟩. If g1 = g2 then χ(g1)χ(g2) = 1 for all χ ∈b G and ⟨g1, g2⟩= # b G/#G = 1. If g1 ̸= g2 then by Corollary 18.7 there exists λ ∈b G for which α := λ(g1)λ1(g2) = λ(g1g−1 2 ) ̸= 1. We then have α⟨g1, g2⟩= 1 #G X χ∈b G (λχ)(g1)(λχ)(g2) = 1 #G X χ∈λ b G χ(g1)χ(g2) = ⟨g1, g2⟩, which implies ⟨g1, g2⟩= 0, since α ̸= 1.
Corollary 18.11. For χ ∈b G we have P g∈G χ(g) ̸= 0 if and only χ is the trivial character.
Remark 18.12. The orthogonality of characters given by Proposition 18.10 is a special case of the orthogonality of characters one encounters in Fourier analysis on compact groups; since G is finite the weighted sum over G corresponds to integrating against its Haar measure (the counting measure µ normalized so that µ(G) = 1).
We conclude our discussion of character groups with a theorem analogous to the funda-mental theorem of Galois theory.
18.785 Fall 2016, Lecture #18, Page 6 Proposition 18.13. Let G be a finite abelian group. There is an inclusion reversing bijec-tion ϕ between subgroups H of G and subgroups K of b G defined by ϕ(H) := {χ ∈b G : χ(h) = 1 for all h ∈H}.
The inverse bijection φ is given by φ(K) := {g ∈G : χ(g) = 1 for all χ ∈K}, and b H ≃b G/ϕ(H) and K ≃G/φ(K); in particular, #H = [ b G:ϕ(H)] and #K = [G:φ(K)].
Proof. Its clear from the definitions that ϕ and φ are inclusion reversing. Let H be a sub-group of G. The group K = ϕ(H) consists of the characters of G whose kernel contains H.
It is clear that H′ := φ(K) contains H, since it is equal to the intersection of these kernels, and by duality it is similarly clear that K′ := ϕ(H′) contains K. We then have H ⊆H′ and ϕ(H) ⊆ϕ(H′), but ϕ is inclusion reversing so H = H′; thus φ ◦ϕ is the identity map, and by duality, so is ϕ ◦φ.
The restriction map b G →b H defined by χ 7→χ|H is a group homomorphism with kernel K = ϕ(H). It is surjective because if we let χ1 := 1 b G then we have #H#K = X h∈H X χ∈K χ(h) = X h∈H X χ∈K χ(h)χ1(h) = X g∈G X χ∈K χ(g)χ1(g) = #G, by Proposition 18.10, and therefore # b H#K = # b G (by Proposition 18.6). It follows that b H ≃b G/ϕ(H), and by duality, K ≃G/φ(K).
18.3 Dirichlet characters We now define the notion of a Dirichlet character. Historically, these preceded the notion of a group character; they were introduced by Dirichlet in 1831, decades before the notion of an abstract group had been formalized.5 Definition 18.14. A function f : Z →C is called an arithmetic function. The function f is multiplicative if f(1) = 1 and f(mn) = f(m)f(n) for all coprime m, n ∈Z, and it is totally multiplicative (or completely multiplicative) if f(1) = 1 and f(mn) = f(m)f(n) for all m, n ∈Z. For m ∈Z>0 we say that f is m-periodic if f(n + m) = f(n) for all n ∈Z, and we call m the period of f it is the least m > 0 for which this holds.
Definition 18.15. A Dirichlet character is an arithmetic function χ: Z →C that is peri-odic and totally multiplicative.
The image of a Dirichlet character is a multiplicatively closed subset of C, hence the union of a finite subgroup of U(1) and a subset of {0}. The constant function 1(n) := 1 is the trivial Dirichlet character; it is the unique Dirichlet character of period 1. Each m-periodic Dirichlet character χ restricts to a group character χ on (Z/mZ)×. Conversely, every group character χ of (Z/mZ)× can be extended to a Dirichlet character χ by defining χ(n) = 0 for n ̸∈(Z/mZ)×;6 this is called extension by zero.
5Indeed, Galois’ original paper was rejected that same year; his seminal work wasn’t published until 1846.
6When we write n ̸∈(Z/mZ)× we of course refer to the image of n under the quotient map Z →Z/mZ.
18.785 Fall 2016, Lecture #18, Page 7 Definition 18.16. A Dirichlet character of modulus m is an m-periodic Dirichlet char-acter χ that is the extension by zero of a group character on (Z/mZ)×; equivalently, an m-periodic Dirichlet character for which n ∈(Z/mZ)× ⇐ ⇒χ(n) ̸= 0.
Remark 18.17. Some authors only define Dirichlet characters of modulus m, thereby baking m into the definition of a Dirichlet character; we simply view Dirichlet characters as functions Z →C that satisfy certain properties. Note that a single Dirichlet character may be a Dirichlet character of modulus m for infinitely many m (for example, the unique Dirichlet character of modulus 2 is also a Dirichlet character of modulus 2k for all k ≥1).
The Dirichlet characters of modulus m form a group under pointwise multiplication that is canonically isomorphic to the character group of (Z/mZ)×. Not every m-periodic Dirichlet character χ is a Dirichlet character of modulus m, since an m-periodic Dirichlet character need not vanish on n ∈(Z/mZ)×. More generally, we have the following lemma.
Lemma 18.18. Let χ be a Dirichlet character of period m. Then χ is a Dirichlet character of modulus m′ if and only if m|m′|mk for some k (which holds in particular for m′ = m).
Proof. Suppose for the sake of contradiction that χ(n) ̸= 0 for some n ∈Z that has a prime factor p in common with m. Then χ(p) ̸= 0, since χ(p)χ(n/p) = χ(n) ̸= 0, and for r ∈Z, χ(r)χ(p) = χ(rp) = χ(rp + m) = χ(r + m/p)χ(p), which implies χ(r) = χ(r+m/p), since χ(p) ̸= 0. Thus χ is m/p-periodic, which contradicts the minimality of the period m. Therefore χ(n) = 0 for all n ̸∈(Z/mZ)×. Conversely, if n ∈(Z/mZ)× then we can pick a = ne ≡1 mod m and χ(1) = χ(a) = χ(ne) = χ(n)e cannot be 0, so χ(n) ̸= 0. Thus χ is a Dirichlet character of modulus m.
If m|m′|mk then the prime factors of m′ coincide with those of m. It follows that n ∈(Z/m′Z)× ⇐ ⇒n ∈(Z/mZ)× ⇐ ⇒χ(n) ̸= 0, and χ is clearly m′-periodic (since m|m′), so χ is a Dirichlet character of modulus m′.
Conversely, if χ is a Dirichlet character of modulus m′, then χ is m′-periodic, and therefore m|m′, since m is the period of χ. And since χ is a Dirichlet character of modulus m and of modulus m′, for each prime p we have p ̸∈(Z/mZ)× ⇐ ⇒χ(p) = 0 ⇐ ⇒p ̸∈(Z/m′Z)×, thus the prime divisors of m and m′ coincide and m′ must divide some power mk of m.
18.3.1 Primitive Dirichlet characters Given a Dirichlet character χ1 of modulus m1 dividing m2, we can always create a Dirichlet character χ2 of modulus m2 by defining χ2(n) := χ1(n) for n ∈(Z/m2Z)× and χ2(n) := 0 otherwise (so χ2 is the extension by zero of the restriction of χ1 to (Z/m2Z)×). If m2 is divisible by a prime p that does not divide m1, the Dirichlet characters χ1 and χ2 will not be the same (χ2(p) = 0 ̸= χ1(p), for example), they will agree on n ∈(Z/m2Z)× but not n ∈(Z/m1Z)×.7 We can create infinitely many new Dirichlet characters from χ1 in this way, but they will differ from χ1 only in a rather trivial sense. We would like to to distinguish the Dirichlet characters that arise in this way from those that do not.
7Note that while the group (Z/m1Z)× is smaller than the group (Z/m2Z)× the set of integers n ∈ (Z/m1Z)× (the n coprime to m1) is larger than the set of integers n ∈(Z/m2Z)× (the n coprime to m2).
18.785 Fall 2016, Lecture #18, Page 8 Definition 18.19. Let χ1 and χ2 be Dirichlet characters of modulus m1 and m2, respec-tively, with m1|m2. If χ2(n) = χ1(n) for n ∈(Z/m2Z)× then χ2 is induced by χ1.
Lemma 18.20. A Dirichlet character χ2 of modulus m2 is induced by a Dirichlet character of modulus m1|m2 if and only if χ2 is constant on residue classes in (Z/m2Z)× that are congruent modulo m1.
When this holds, the Dirichlet character χ1 of modulus m1 that induces χ2 is uniquely determined.
Proof. If χ2 is induced by χ1 then it must be constant on residue classes in (Z/m2Z)× that are congruent modulo m1, since χ1 is. To prove the converse we first show that the surjective ring homomorphism Z/m2Z →Z/m1Z given by reduction modulo m1 induces a surjective homomorphism π: (Z/m2Z)× →(Z/m1Z)× of unit groups,8 Suppose u1 ∈Z is a unit modulo m1. Let a be the product of all primes dividing m2/m1 but not u1. Then u2 = u1 + m1a is not divisible by any prime p|m1 (since u1 isn’t), nor is it divisible by any prime p|(m2/m1): by construction, such a p divides exactly one of u1 and m1a. Thus u2 is a unit modulo m2 that reduces to u1 modulo m1 and π is surjective.
If χ2 is a Dirichlet character of modulus m2 constant on fibers of π we can define a Dirichlet character χ1 of modulus m1 via χ1(n1) := χ2(n2) for n1 ∈(Z/m1Z)× with n2 ∈π−1(n1) (any such n2 will do). This χ1 induces χ2, and if χ′ 1 also induces χ2 it must satisfy the same condition χ1(n1) = χ2(n2) that uniquely determines χ1.
Definition 18.21. A Dirichlet character is primitive if it is not induced by any Dirichlet character other than itself. A Dirichlet character χ induced by 1 is called principal (and is then primitive if only if χ = 1).
For m ∈Z>0 we use 1m to denote the principal Dirichlet character of modulus m; it corresponds to the identity element under the canonical isomorphism between Dirichlet characters of modulus m and the character group of (Z/mZ)×.
Lemma 18.22. Let χ be a Dirichlet character of modulus m. Then X n ∈Z/mZ χ(n) ̸= 0 ⇐ ⇒ χ = 1m.
Proof. We have χ(n) = 0 for n ̸∈(Z/mZ)×, and the sum over (Z/mZ)× is nonzero if and only if χ restricts to the trivial character on (Z/mZ)×, by Corollary 18.11.
Note that the principal Dirichlet characters 1m and 1m′ necessarily coincide when m|m′|mk; for example the principal Dirichlet character of modulus 2 (the parity function) is the same as the principal Dirichlet character of modulus 4 (and every power of 2).
Theorem 18.23. Every Dirichlet character χ is induced by a primitive Dirichlet charac-ter e χ that is uniquely determined by χ.
Proof. Let us define a partial ordering ⪯on the set of all Dirichlet characters by defining χ1 ⪯χ2 if χ1 induces χ2. The relation ⪯is clearly reflexive, and it follows from Lemma 18.20 that it is transitive.
Let χ be a Dirichlet character of period m and consider the set X = {χ′ : χ′ ⪯χ}. Each χ′ ∈X necessarily has period m′ dividing m and there is at most one χ′ of period m′ for each divisor m′ of m, by Lemma 18.20. Thus X is finite, and nonempty (since χ ∈X).
8In fact, one can show that every surjective homomorphism of finite rings induces a surjective homomor-phism of unit groups, but this does not hold in general (consider Z →Z/5Z, for example).
18.785 Fall 2016, Lecture #18, Page 9 Suppose χ1, χ2 ∈X have periods m1 and m2, respectively.
Then m1 and m2 both divide m, as does m3 = gcd(m1, m2). We have a commutative square of surjective unit group homomorphisms induced by reduction maps: (Z/mZ)× (Z/m1Z)× (Z/m2Z)× (Z/m3Z)× ← ↠ ← ↠ ← ↠ ← ↠ From Lemma 18.20 we know that χ is constant on residue classes in (Z/mZ)× that are con-gruent modulo either m1 or m2, and therefore χ is constant on residue classes in (Z/mZ)× that are congruent modulo m3, as are χ1 and χ2 (which are determined by χ). It follows that there is a unique Dirichlet character χ3 of modulus m3 that induces χ, χ1, and χ2.
Thus every pair χ1, χ2 ∈X has a lower bound χ3 under the partial ordering ⪯that is compatible with the total ordering of X by period. This implies that X contains a unique element e χ that is minimal, both with respect to the partial ordering ⪯and with respect to the total ordering by period; it must be primitive, by the transitivity of ⪯.
Definition 18.24. The conductor of a Dirichlet character χ is the period of the unique primitive Dirichlet character e χ that induces χ.
Corollary 18.25. For a Dirichlet character χ of modulus m we have P n∈Z/mZ χ(n) ̸= 0 if and only if χ has conductor 1.
Proof. This follows immediately from Lemma 18.22.
Corollary 18.26. Let Z(m) denote the set of Dirichlet characters of modulus m, let X(m) denote the set of primitive Dirichlet characters of conductor dividing m, and let b G(m) denote the character group of (Z/mZ)×. We have canonical bijections Z(m) ∼ − →X(m) ∼ − →b G(m) χ 7− →e χ 7− →(m 7→e χ(m)).
Proof. By Theorem 18.23, the map χ →e χ is injective, and it is also surjective: each e χ ∈X(m) induce the character χ ∈Z(m) by setting χ(n) := e χ(n) for n ∈(Z/mZ)× and extending by zero. As previously noted, the map χ →(m 7→χ(m)) defines a bijection Z(m) →b G(m) (a group isomorphism, in fact), and this bijection factors through the map χ 7→e χ, since e χ(n) = χ(n) for n ∈(Z/mZ)×.
Remark 18.27. Corollary 18.26 implies that we can make X(m) a group by defining e χ1e χ2 := ] χ1χ2. Note that ] χ1χ2 is not the pointwise product of e χ1 and e χ2 (which is typically not primitive), it is the unique primitive character that induces the pointwise product.
Example 18.28. 12-periodic Dirichlet characters, ordered by period m and conductor c.
m c 0 1 2 3 4 5 6 7 8 9 10 11 mod-12 principal primitive 1 1 1 1 1 1 1 1 1 1 1 1 1 1 no yes yes 2 1 0 1 0 1 0 1 0 1 0 1 0 1 no yes no 3 1 0 1 1 0 1 1 0 1 1 0 1 1 no yes no 3 3 0 1 -1 0 1 -1 0 1 -1 0 1 -1 no no yes 4 4 0 1 0 -1 0 1 0 -1 0 1 0 -1 no no yes 6 1 0 1 0 0 0 1 0 1 0 0 0 1 yes yes no 6 3 0 1 0 0 0 -1 0 1 0 0 0 -1 yes no no 12 4 0 1 0 0 0 1 0 -1 0 0 0 -1 yes no no 12 12 0 1 0 0 0 -1 0 -1 0 0 0 1 yes no yes 18.785 Fall 2016, Lecture #18, Page 10 18.4 Dirichlet L-functions Definition 18.29. The Dirichlet L-function associated to a Dirichlet character χ is L(s, χ) := Y p (1 −χ(p)p−s)−1 = X n≥1 χ(n)n−s.
The sum and product converge absolutely for Re s > 1, since |χ(n)| ≤1, thus L(s, χ) is holomorphic on Re(s) > 1.
For the trivial Dirichlet character 1 have L(s, 1) = ζ(s). For the principal character 1m of modulus m induced by 1 we have ζ(s) = L(s, 1m) Y p|m (1 −p−s)−1.
The product on the RHS is finite, hence bounded and nonzero as s →1+, so the L-function L(s, 1m) has a simple pole at s = 1 with residue ress=1 L(s, 1m) = lim s→1 (s −1)ζ(s) Y p|m (1 −p−s) = Y p|m (1 −p−1) = φ(m) m .
But the L-functions of non-principal Dirichlet characters do not have a pole at s = 1.
Proposition 18.30. Let χ be a non-principal Dirichlet character of modulus m. Then L(s, χ) extends to a holomorphic function on Re s > 0.
Proof. Define the function T : R≥0 →C by T(x) := X 0 0, since it is the limit of the uniformly converging se-quence of functions φn(s) := s R n 0 T(x)x−s−1dx (here we use the fact that T(x) is bounded), and is thus the analytic continuation of L(x, χ) to Re(s) > 0.
Remark 18.31. In fact, L(s, χ) extends to a holomorphic function on C whenever χ is non-principal.
18.785 Fall 2016, Lecture #18, Page 11 18.5 Primes in arithmetic progressions We now return to our goal of proving Dirichlet’s theorem on primes in arithmetic progres-sions. Let a and m be coprime integers. We want to show that the sum X p ≡a mod m p−s is unbounded as s →1+. To convert this to a sum over all primes we use Proposition 18.10 to construct the indicator function 1 φ(m) X χ χ(p/a) = ( 1 if p ≡a mod m, 0 otherwise where p/a is computed modulo m and χ ranges over primitive Dirichlet characters of conduc-tor dividing m (which we identify with the character group of (Z/mZ)× via Corollary 18.26).
As s →1+ we have X p ≡a mod m p−s = X p p−s 1 φ(m) X χ χ(p/a) = X χ χ(1/a) φ(m) X p χ(p)p−s = X χ χ(1/a) φ(m) log L(s, χ) + O(1) = log ζ(s) φ(m) + X χ̸=1 χ(1/a) φ(m) log L(s, χ) + O(1).
We now make the key claim that so long as χ is not principal, we have L(1, χ) ̸= 0.
This implies that log L(1, χ) = O(1) as s →1+ and therefore X p ≡a mod m p−s = log ζ(s) φ(m) + O(1) is unbounded as s →1+, since ζ(s) is. Moreover, Mertens’ second theorem implies X p ≤x p ≡a mod m 1 p ∼log log x φ(m) , and we can compute the Dirichlet density of S := {p ≡a mod m}: d(S) = lim s→1+ P p∈S p−s P p p−s = 1 φ(m).
We will prove that L(1, χ) ̸= 0 when χ is non-principal in the next lecture by showing that the Dedekind zeta function ζK(s) of the mth cyclotomic field K = Q(ζm) can be written as ζK(s) = Y χ L(s, χ) = ζ(s) Y χ̸=1 L(s, χ), where χ ranges over the primitive Dirichlet characters of conductor dividing m. We will then use the analytic class number formula for ζK(s) to deduce that the second product on the RHS is holomorphic and nonzero at s = 1, hence all its factors are.
18.785 Fall 2016, Lecture #18, Page 12 18.6 Stieltjes integrals For the benefit of those who have not seen them before, we recall a few facts about Stieltjes integrals (also called Riemann-Stieltjes integrals), taken from [1, Ch. 7]. These generalize the Riemann integral but are less general than the Lebesgue integral; they provide a handy way for converting sums to integrals that is often used in analytic number theory.
Definition 18.32. Let f and g be (real or complex valued) functions defined on a nonempty real interval [a, b]. For any partition P = (x0, . . . , xn) of [a, b] and sequence T = (t1, . . . , tk) with tk ∈[xk−1, xk], we define the Riemann-Stieltjes sum S(P, T, f, g) := n X k=1 f(tk) g(xk) −g(xk−1) We say that f is Riemann-Stieltjes integrable with respect to g and write f ∈S(g) if there is a (real or complex) number S such that for every ϵ > 0 there is a partition Pϵ of [a, b] such that for every refinement P = (x0, . . . , xn) of Pϵ and every sequence T = (t1, . . . , tn) with tk ∈[xk−1, xk] we have |S(P, T, f, g) −S| < ϵ.9 When such an S exists it is necessarily unique and we denote it by R b a f dg, the Riemann-Stieltjes integral of f with respect to g. Improper Riemann-Stieltjes integrals are then defined as limits Z ∞ a f dg := lim b→∞ Z b a f dg (and similarly for the lower limit), and we define R a b f dg = − R b a f dg and R a a f dg = 0.
Taking g(x) = x yields the Riemann integral. The Riemann-Stieltjes integral satisfies the usual properties of linearity, summability, and integration by parts.
Proposition 18.33. Let f, g, and h be functions on [a, b] and let c1 and c2 be constants.
The following hold: • If f, g ∈S(h) then R b a (c1f + c2g) dh = c1 R b a f dh + c2 R b a g dh.
• If f ∈S(g), S(h) then R b a f d(c1g + c2h) = ci R b a f dg + c2 R b a f dh.
• If f ∈S(g) then for any c ∈[a, b] we have R b a f dg = R c a f dg + R b c f dg.
• If f ∈S(g) then g ∈S(f) and R b a f dg + R b a g d f = f(b)g(b) −f(a)g(a).
• If f = f1 + if2 and g = g1 + ig2 with f1, f2 ∈S(g1), S(g2) then Z b a f dg = Z b a f1 dg1 − Z b a f2 dg2 + i Z b a f2 dg1 + Z b a f1 dg2 .
Proof. See [1, Thm. 7.2-7,7.50].
The last identity allows us to reduce complex-valued integrals to real-valued integrals.
The following proposition allows us to reduce Stieltjes integrals to Riemann integrals.
9This definition (due to Pollard) is more general than that originally given by Stieltjes but is now standard.
18.785 Fall 2016, Lecture #18, Page 13 Proposition 18.34. Let f and g be real-valued functions on [a, b] and suppose g has a continuous derivative g′ on [a, b]. Then Z b a f dg = Z b a f(x)g′(x)dx.
Proof. See [1, Thm. 7.8].
A key advantage of the Stieltjes integral R b a f dg is that neither the integrand f nor the integrator g is required to be continuous.
It suffices for f and g to be of bounded variation and not share any discontinuities (and they can even share certain discontinuities, see Theorem 18.36).
Definition 18.35. Let f be a (real or complex valued) function defined on a nonempty real interval [a, b]. Then f is of bounded variation if there exists a (real or complex) number M such that n−1 X i=0 |f(xi+1) −f(xi)| < M for every partition P = (x0, . . . , xn) of [a, b]. If f has a continuous derivative f′ on [a, b] this is equivalent to requiring R b a |f′(x)|dx < ∞. Every piecewise monotone function is of bounded variation. In particular, any step function with finitely many discontinuities on [a, b] is of bounded variation.
Theorem 18.36. Let f and g be functions on [a, b] of bounded variation such that for every c ∈[a, b] the function f is continuous from the left at c and the function g is continuous from the right at c. Then R b a f dg and R b a g d f both exist.
Proof. See [2, Thm. 3.7].
Corollary 18.37. Let f and g be functions on [a, b] such that f and g are not both discon-tinuous from the left or from the right at integers n ∈[a, b], and let G(x) = P a<n≤x g(n).
Then X a<n≤b f(n)g(n) = Z b a f(x) dG(x).
In particular, the integral on the RHS always exists.
Proof. See [1, Thm. 7.11].
As an example of using Stieltjes integrals, let us derive an asymptotic estimate for the the harmonic sum H(x) := X 1≤n≤x 1 n.
Theorem 18.38. For x ∈R≥1, as x →∞we have H(x) = log x + γ + O 1 x where γ = limx→∞(H(x) −log x) = 0.577216 . . . is Euler’s constant.
18.785 Fall 2016, Lecture #18, Page 14 Proof. Let [t] denote the greatest integer function. Applying Corollary 18.37 with g(t) = 1 and G(t) = P 1≤n≤x 1 = [t], we have H(x) = X 1≤n≤x 1 n = Z x 1− 1 t d[t] = [t] t x 1−− Z x 1−[t] d1 t = [x] x + Z x 1− [t] t2 dt = [x] x + Z x 1− 1 t dt − Z x 1− t −[t] t2 dt = [x] x + log x − Z x 1− t −[t] t2 dt, where we used integration by parts in the second line and applied Proposition 18.34 to get the third line. Now let γ = 1 − R ∞ 1−(t −[t])/t2 dt. Then H(x) = [x] x + log x −1 + γ + Z ∞ x t −[t] t2 dt = log x + γ + [x] −x x + Z ∞ x t −[t] t2 dt .
(2) Both summands in the parenthesized quantity in (2) are clearly O( 1 x); thus γ = lim x→∞(H(x) −log x) , and the theorem follows.
Remark 18.39. We can refine this estimate by applying a similar analysis to the paren-thesized quantity in (2); the key point is that the error term is an exact expression, not an asymptotic estimate, and we can continue this process until we obtain an asymptotic expansion to whatever precision we require. For example, one finds that H(x) = log x + γ + 1 2x − 1 2x2 + 1 120x4 + O 1 x6 .
References Tom Apostol, Mathematical analysis, 2nd edition, Addison-Wesley, 1974.
Paul Bateman and Harold Diamond, Analytic number theory: An introductory course, World Scientific, 2004.
J.W.S. Cassels and A. Fr¨ ohlich, Algebraic number theory, 2nd edition, London Mathe-matical Society, 2010.
Franz Mertenz, Ein Beitrag zur analytischen Zahlentheorie, J. reine agnew. Math., 78 (1874), 46–62.
18.785 Fall 2016, Lecture #18, Page 15 |
15852 | https://phys.libretexts.org/Courses/Coalinga_College/Physical_Science_for_Educators_(CID%3A_PHYS_14)/08%3A_Energy_Physics_and_Chemistry/8.04%3A_Work_and_Energy/8.4.01%3A_Potential_Energy-_Gravity_and_Springs/8.4.1.01%3A_Spring_Potential_Energy | 8.4.1.1: Spring Potential Energy - Physics LibreTexts
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8.4.1: Potential Energy- Gravity and Springs
8.4: Work and Energy
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8.4.1.1: Spring Potential Energy
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8.4.1: Potential Energy- Gravity and Springs
8.4.2: Forms of Energy
Page ID 98516
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Learning Objectives
2. Connecting Ideas
3. Example 8.4.1.1.1: Calculating Stored Energy: A Tranquilizer Gun Spring
1. Strategy for a
2. Solution for a
3. Strategy for b
4. Solution for b
5. Discussion
Exercise 8.4.1.1.1
Exercise 8.4.1.1.2
Section Summary
Glossary
Learning Objectives
Explain the work done in deforming a spring.
Describe the potential energy stored in a deformed spring.
Hooke's Law, F=−kx, describes force exerted by a spring being deformed. Here, F is the restoring force, x is the displacement from equilibrium or deformation, and k is a constant related to the difficulty in deforming the system. The minus sign indicates the restoring force is in the direction opposite to the displacement.
In order to produce a deformation, work must be done. That is, a force must be exerted through a distance, whether you pluck a guitar string or compress a car spring. If the only result is deformation, and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy. The potential energy stored in a spring is PE el=1 2kx 2. Here, we generalize the idea to elastic potential energy for a deformation of any system that can be described by Hooke’s law. Hence,
PE el=1 2kx 2,
where PE el is the elastic potential energy stored in any deformed system that obeys Hooke’s law and has a displacement x from equilibrium and a force constant k.
Connecting Ideas
Hooke's Law is the same equation that was used to develop themotion of waves. If we think about the oscillating motion of a compressed spring after the depressing force is released, this illustrates the reason for this connection.
It is possible to find the work done in deforming a system in order to find the energy stored. This work is performed by an applied force F app. The applied force is exactly opposite to the restoring force (action-reaction), and so F app=kx. Figure8.4.1.1.1 shows a graph of the applied force versus deformation x for a system that can be described by Hooke’s law. Work done on the system is force multiplied by distance, which equals the area under the curve or (1/2)kx 2 (Method A in the figure). Another way to determine the work is to note that the force increases linearly from 0 to kx , so that the average force is (1/2)kx, the distance moved is x, and thus W=F appd=[(1/2)kx](x)=(1/2)kx 2 (Method B in the figure).
Figure 8.4.1.1.1: A graph of applied force versus distance for the deformation of a system that can be described by Hooke’s law is displayed. The work done on the system equals the area under the graph or the area of the triangle, which is half its base multiplied by its height, or W=(1/2)kx 2.
Example 8.4.1.1.1: Calculating Stored Energy: A Tranquilizer Gun Spring
We can use a toy gun’s spring mechanism to ask and answer two simple questions: (a) How much energy is stored in the spring of a tranquilizer gun that has a force constant of 50.0 N/m and is compressed 0.150 m? (b) If you neglect friction and the mass of the spring, at what speed will a 2.00-g projectile be ejected from the gun?
Figure 8.4.1.1.2: (a) In this image of the gun, the spring is uncompressed before being cocked. (b) The spring has been compressed a distance x, and the projectile is in place. (c) When released, the spring converts elastic potential energy PE el into kinetic energy.
Strategy for a
(a): The energy stored in the spring can be found directly from elastic potential energy equation, because k and x are given.
Solution for a
Entering the given values for k and x yields
PE el=1 2kx 2=1 2(50.0N/m)(0.150m)2=0.563N⋅m=0.563J
Strategy for b
Because there is no friction, the potential energy is converted entirely into kinetic energy. The expression for kinetic energy can be solved for the projectile’s speed.
Solution for b
Identify known quantities:
KE f=PE el or 1/2mv 2=(1/2)kx 2=PE el=0.563J
Solve for v:
v=[2PE el m]1/2=[2(0.563J)0.002kg]1/2=23.7(J/kg)1/2
Convert units: 23.7m/s
Discussion
(a) and (b): This projectile speed is impressive for a tranquilizer gun (more than 80 km/h). The numbers in this problem seem reasonable. The force needed to compress the spring is small enough for an adult to manage, and the energy imparted to the dart is small enough to limit the damage it might do. Yet, the speed of the dart is great enough for it to travel an acceptable distance.
Exercise 8.4.1.1.1
Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations of the ruler. In what way could you modify this simple experiment to increase the rigidity of the system?
Answer
You could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment.
Exercise 8.4.1.1.2
If you apply a deforming force on an object and let it come to equilibrium, what happened to the work you did on the system?
Answer
It was stored in the object as potential energy.
Section Summary
Hooke’s law describes force exerted by a spring being deformed,
F=−kx,
where F is the restoring force, x is the displacement from equilibrium or deformation, and k is the force constant of the system.
Elastic potential energy PE el stored in the deformation of a system that can be described by Hooke’s law is given by
PE el=(1/2)kx 2.
Glossary
deformationdisplacement from equilibriumelastic potential energypotential energy stored as a result of deformation of an elastic object, such as the stretching of a spring
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15853 | https://www.youtube.com/watch?v=SpEnXMzDZAI | Perpendicular Vectors: Dot Product
Wuoti Math
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Posted: 28 Feb 2013
Determine if two vectors are perpendicular by checking if the inner product (dot product) is equal to zero.
11 comments
Transcript:
greetings and welcome uh in today's lesson we're going to talk about perpendicular vectors uh both in a plane and in space and uh we're going to have uh pretty much two formulas there's technically three um that we're going to look at and the first one is called the inner product uh it's sometimes also referred to as the dot product so these are going to behave somewhat like multiplication I guess in a sense but both of them are these the two formulas that we're doing are unique from one another so there's kind of like two types of uh multiplication so to speak and the dot product or inner product is defined as such so if so if I have vectors A and B are two vectors um specifically A1 A2 and Vector B1 B2 um then the inner product is the following and this is our formula so it will be uh denoted as the a DOT product right so that's where the dot name comes from where it's the multiplication dot symbol and it is A1 B1 plus A2 B2 so I want to point out that this formula when I multiply all of these uh coordinates of these vectors it's going to give me an actual number um so it's not going to give me a vector back it's just giving me a number and this actually it reminds me of um the determinant back when we did matrices uh how the determinant was just a particular number uh and there's a fact about the dotproduct that uh I guess I'll say this way is if um a do b equals z then the vectors are perpendicular I guess I'll abbreviate with a perpendicular symbol there upside down capital t is what it looks like so uh this ends up serving as a test to determine the perpendicularity of vectors and in this case uh specific specifically in a plane um now so this is kind of like the reason why we use this um and that's that's for in a plane and then there's also a formula that if I have uh if Vector a was say um A1 A2 A3 so this is where the variation comes in I really consider this the same formula though uh and Vector B is B1 B2 B3 then the dotproduct for vectors in space three-dimensional uh would be A1 B1 uh plus A2 B2 plus A3 B3 so it's pretty much the same formula it just adds that one more component for the third dimension at the very tail end there so here's our first uh formula so to speak so let's do an example in which we'll determine uh if the vectors are perpendicular and let's say I've got these vectors so let's say I've got Vector uh p is 7 and 14 and let's say I've got Vector Q is 2 and -1 uh if I want to determine if these are perpendicular uh what I would do is use my little dot product formula [Music] so p do Q is going to equal uh 7 2 + 14 -1 and I believe um back when we did determinant of matrices it was subtract so make sure you don't mix up um these two formulas uh so let's multiply this out so that's 14 + uh -14 which equals 0 and since that is equal to zero I could say therefore uh p is perpendicular to Q right uh this little triple dot is just a symbol for therefore uh so that's one of the ways we can use this is just as a test to determine per perpendicularity if uh it was any number other than zero uh then they would not have been perpendicular and let's just do a quick one um uh for a three-dimensional one so let's say uh Vector a is is -31 1 and let's say Vector B is uh let's say 2 8 and - 1 so now I'm in three dimensions so let's uh figure out what the dotproduct of these two are and uh in order to do that I just multiply their corresponding coordinates essentially uh so -3 2 + 1 8 + 1 -1 so I'll get -6 + 8 -1 uh so that's equal to one so since this is not zero uh therefore Vector a is is not perpendicular to be right so it's a quick little test as easy as that um the next Formula I'll show you in the next video thanks for watching |
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1983 AIME Problems/Problem 5
Contents
1 Problem
2 Solutions
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4
3 See Also
Problem
Suppose that the sum of the squares of two complex numbers and is and the sum of the cubes is . What is the largest real value that can have?
Solutions
Solution 1
One way to solve this problem is by substitution. We have
and
Hence observe that we can write and .
This reduces the equations to and .
Because we want the largest possible , let's find an expression for in terms of .
.
Substituting, , which factorizes as (the Rational Root Theorem may be used here, along with synthetic division).
The largest possible solution is therefore .
Solution 2
An alternate way to solve this is to let and .
Because we are looking for a value of that is real, we know that , and thus .
Expanding will give two equations, since the real and imaginary parts must match up.
Looking at the imaginary part of that equation, , so , and and are actually complex conjugates.
Looking at the real part of the equation and plugging in , , or .
Now, evaluating the real part of , which equals (ignoring the odd powers of , since they would not result in something in the form of ):
Since we know that , it can be plugged in for in the above equation to yield:
Since the problem is looking for to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, all fail, but does work. Thus, the real part of both numbers is , and their sum is .
Solution 3
Begin by assuming that and are roots of some polynomial of the form , such that by Vieta's Formulas and some algebra (left as an exercise to the reader), and . Substituting , we deduce that , whose roots are , , and . Since is the sum of the roots and is maximized when , the answer is .
Solution 4
Also, Substituting our above into this, we get . Letting , we have that . Testing , we find that this is a root, to get
See Also
1983 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 4Followed by
Problem 6
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
All AIME Problems and Solutions
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15855 | https://nrich.maths.org/articles/introduction-modular-arithmetic | An introduction to modular arithmetic | NRICH
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|Article by
Vicky Neale
|PublishedTue, 01/02/2011 - 00:00
An introduction to modular arithmetic
The best way to introduce modular arithmetic is to think of the face of a clock.
Image
The numbers go from 1 to 12, but when you get to "13 o'clock", it actually becomes 1 o'clock again (think of how the 24 hour clock numbering works). So 13 becomes 1, 14 becomes 2, and so on.
This can keep going, so when you get to "25 o'clock'', you are actually back round to where 1 o'clock is on the clock face (and also where 13 o'clock was too).
Image
So in this clock world, you only care where you are in relation to the numbers 1 to 12. In this world, 1,13,25,37,… are all thought of as the same thing, as are 2,14,26,38,… and so on.
What we are saying is "13=1+ some multiple of 12", and "38=2+ some multiple of 12", or, alternatively, "the remainder when you divide 13 by 12 is 1" and "the remainder when you divide 38 by 12 is 2''. The way we write this mathematically is 13≡1 mod 12, 38≡2 mod 12, and so on. This is read as "13 is congruent to 1 mod (or modulo) 12" and "38 is congruent to 2 mod 12".
But you don't have to work only in mod 12 (that's the technical term for it). For example, you could work mod 7, or mod 46 instead if you wanted to (just think of clocks numbered from 1 to 7 and 1 to 46 respectively; every time you get past the biggest number, you reset to 1 again).
Image
Let's go back to the normal clock face with the numbers 1 to 12 on it for a moment. Mathematicians usually prefer to put a 0 where the 12 would normally be, so that you would usually write (for example) 24≡0 mod 12 rather than 24≡12 mod 12, although both of these are correct. That is, we think of a normal clock face as being numbered from 0 to 11 instead. This makes sense: we'd normally say that 24 leaves a remainder of 0 when we divide by 12, rather than saying it leaves a remainder of 12 when we divide by 12!
Let's be a bit more formal. In general, if you are working in mod n (where n is any whole number), we write a≡b mod n if a and b leave the same remainder when you divide them by n. This is the same as saying that we write a≡b mod n if n divides a−b. (Look at what we did earlier to see that this definition fits with our examples above.)
So far, we've only talked about notation. Now let's do some maths, and see how congruences (what we've described above) can make things a bit clearer.
Here are some useful properties. We can add congruences. That is, if a≡b mod n and c≡d mod n, then a+c≡(b+d)mod n. Why is this? Well, a≡b mod n means that a=b+k n, where k is an integer. Similarly, c≡d mod n means that c=d+l n, where l is an integer. So a+c=(b+k n)+(d+l n)=(b+d)+(k+l)n, so a+c≡(b+d)mod n. For example, 17≡4 mod 13, and 42≡3 mod 13, so 17+42≡4+3≡7 mod 13. Note that both of the congruences that we're adding are mod n, and so is the answer - we don't add the moduli.
Now you prove that if a≡b mod n and c≡d mod n then a−c≡(b−d)mod n. Also, prove that we can do something similar for multiplication: if a≡b mod n and c≡d mod n, then a c≡b d mod n. You can prove this in the same way that we used above for addition. Again, both of the congruences that we're multiplying are mod n, and so is the answer - we don't multiply the moduli. Can you come up with an example to disprove the claim that a≡b mod n and c≡d mod m means that a c≡b d mod m n?
Division is a bit more tricky: you have to be really careful. Here's an example of why. 10≡2 mod 8. But if we "divide both sides by 2", we'd have 5≡1 mod 8, which is clearly nonsense! To get a true congruence, we'd have to divide the 8 by 2 as well: 5≡1 mod 4 is fine. Why? Well, a≡b mod n means that a=b+k n for some integer n. But now this is a normal equation, and if we're going to divide a by something, then we have to divide all of the right-hand side by 2 as well, including k n. In general, it's best not to divide congruences; instead, think about what they really mean (rather than using the shorthand) and work from there.
Things are quite special if we work mod p, where p is prime, because then each number that isn't 0 mod p has what we call an inverse (or a multiplicative inverse , if we're being fancy). What that means is that for each a≢0 mod p, there is a b such that a b≡1 mod p.
Let's think about an example. We'll work mod 7. Then really the only non-zero things are 1,2,3,4,5 and 6 (because every other whole number is equivalent to one of them or 0). So let's find inverses for them. Well, 1 is pretty easy: 1×1≡1 mod 7. What about 2? 2×4≡1 mod 7. So 4 is the inverse of 2. In fact, we can also see from this that 2 is the inverse of 4 - so that's saved us some work! 3×5≡1 mod 7, so 3 and 5 are inverses. And finally, 6×6≡1 mod 7, so 6 is the inverse of itself. So yes, each of the non-zero elements mod 7 has an inverse. Try some primes out yourself: 11 and 13 are fairly small! If you're feeling confident, see whether you can discover which numbers have inverses mod 4, or mod 6, or mod 8. What about mod 15? Do you notice any patterns?
To prove this, things are going to get a tiny bit more tricky, so I'm going to save the proof for the end and first give an example of using congruences to do useful mathematics.
Suppose we're given the number 11111111 and someone asks us whether it's divisible by 3. We could try to actually divide it. But you probably know a much easier method: we add up the digits and see whether that's divisible by 3. There's a whole article about this sort of divisibility test here . Let's prove this using congruence notation.
Suppose that our number is a n 10 n+a n−1 10 n−1+…+10 a 1+a 0, so it looks like a n a n−1…a 1 a 0. Then the sum of its digits is a n+a n−1+…+a 1+a 0. We'd like to prove that a n 10 n+a n−1 10 n−1+…+10 a 1+a 0 is divisible by 3 if and only if a n+a n−1+…+a 1+a 0 is divisible by 3. Now we notice that 10≡1 mod 3, so 10×10≡1 mod 3, and more generally 10 k≡1 mod 3 for all k. Using our results from earlier about adding and multiplying congruences, we discover
a n 10 n+a n−1 10 n−1+…+10 a 1+a 0≡a n+a n−1+…+a 1+a 0 mod 3 So if our number is divisible by 3 (that is, if a n 10 n+a n−1 10 n−1+…+10 a 1+a 0≡0 mod 3), then certainly so is the sum of its digits, and vice versa, as we wanted! The congruence notation hasn't really done any of the maths for us, but it's hopefully made it a bit easier to write out the proof clearly. See whether you can use the notation to prove any of the other divisibility tests in that article.
Now for the proof I promised you earlier. We're going to show that if a and n have no common factors, then a has a multiplicative inverse mod n (reminder: that means a number b such that a×b≡1 mod n). In particular, if n is prime, then its only factor apart from 1 is itself, so saying "a and n share no common factors'' is just the same as saying "a isn't divisible by n'', that is, a≢0 mod p: this is what we had above. I'm going to assume that you know about using Euclid's algorithm to solve equations of the form a x+b y=1 where a and b have no common factors. There's a really good article explaining this here , so have a read, and then come back.
If you're reading this far, then hopefully you'll agree that if a and n share no common factors, then we can find x and y such that a x+n y=1. (The fancy name for this is Bezout's Theorem .) So we've got our x and y such that a x+n y=1. We can rewrite this as a x=1−y n, and now let's use the congruence notation from earlier: we have a x≡1 mod n. So now x is the multiplicative inverse of a mod n, and we're done!
Here's a challenge: use this technique based on Euclid's algorithm to find the inverse of 14 mod 37.
Understanding this is the beginning of a branch of mathematics called Number Theory, which contains some beautiful, fascinating and amazing theorems. Enjoy!
This article is based on a contribution to Ask NRICH by Matthew Buckley.
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15856 | https://www.ciaaw.org/abridged-atomic-weights.htm | Home
Atomic Weights
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Atomic Weights
Abridged &
Historical Atomic Weights
Natural Variations
Monoisotopic &
Radioactive Elements
Atomic Masses
Abridged Standard Atomic Weights
The Commission does acknowledge that the full Table of Standard Atomic Weights might provide too many details. Abridged standard atomic weights are quoted to five significant figures unless such precision cannot be attained due to the variability of isotopic composition in normal materials or due to the limitations of the measurement capability.
Abridged Standard Atomic Weights 2024
| Z | Symbol | Element | Abridged Standard Atomic Weight | Notes |
--- ---
| 1 | H | hydrogen | 1.0080 ± 0.0002 | m |
| 2 | He | helium | 4.0026 ± 0.0001 | g r |
| 3 | Li | lithium | 6.94 ± 0.06 | m |
| 4 | Be | beryllium | 9.0122 ± 0.0001 |
| 5 | B | boron | 10.81 ± 0.02 | m |
| 6 | C | carbon | 12.011 ± 0.002 |
| 7 | N | nitrogen | 14.007 ± 0.001 | m |
| 8 | O | oxygen | 15.999 ± 0.001 | m |
| 9 | F | fluorine | 18.998 ± 0.001 |
| 10 | Ne | neon | 20.180 ± 0.001 | g m |
| 11 | Na | sodium | 22.990 ± 0.001 |
| 12 | Mg | magnesium | 24.305 ± 0.002 |
| 13 | Al | aluminium | 26.982 ± 0.001 |
| 14 | Si | silicon | 28.085 ± 0.001 |
| 15 | P | phosphorus | 30.974 ± 0.001 |
| 16 | S | sulfur | 32.06 ± 0.02 |
| 17 | Cl | chlorine | 35.45 ± 0.01 | m |
| 18 | Ar | argon | 39.95 ± 0.16 |
| 19 | K | potassium | 39.098 ± 0.001 |
| 20 | Ca | calcium | 40.078 ± 0.004 | g |
| 21 | Sc | scandium | 44.956 ± 0.001 |
| 22 | Ti | titanium | 47.867 ± 0.001 |
| 23 | V | vanadium | 50.942 ± 0.001 |
| 24 | Cr | chromium | 51.996 ± 0.001 |
| 25 | Mn | manganese | 54.938 ± 0.001 |
| 26 | Fe | iron | 55.845 ± 0.002 |
| 27 | Co | cobalt | 58.933 ± 0.001 |
| 28 | Ni | nickel | 58.693 ± 0.001 | r |
| 29 | Cu | copper | 63.546 ± 0.003 | r |
| 30 | Zn | zinc | 65.38 ± 0.02 | r |
| 31 | Ga | gallium | 69.723 ± 0.001 |
| 32 | Ge | germanium | 72.630 ± 0.008 |
| 33 | As | arsenic | 74.922 ± 0.001 |
| 34 | Se | selenium | 78.971 ± 0.008 | r |
| 35 | Br | bromine | 79.904 ± 0.003 |
| 36 | Kr | krypton | 83.798 ± 0.002 | g m |
| 37 | Rb | rubidium | 85.468 ± 0.001 | g |
| 38 | Sr | strontium | 87.62 ± 0.01 | g r |
| 39 | Y | yttrium | 88.906 ± 0.001 |
| 40 | Zr | zirconium | 91.222 ± 0.003 | g |
| 41 | Nb | niobium | 92.906 ± 0.001 |
| 42 | Mo | molybdenum | 95.95 ± 0.01 | g |
| 43 | Tc | technetium | — |
| 44 | Ru | ruthenium | 101.07 ± 0.02 | g |
| 45 | Rh | rhodium | 102.91 ± 0.01 |
| 46 | Pd | palladium | 106.42 ± 0.01 | g |
| 47 | Ag | silver | 107.87 ± 0.01 | g |
| 48 | Cd | cadmium | 112.41 ± 0.01 | g |
| 49 | In | indium | 114.82 ± 0.01 |
| 50 | Sn | tin | 118.71 ± 0.01 | g |
| 51 | Sb | antimony | 121.76 ± 0.01 | g |
| 52 | Te | tellurium | 127.60 ± 0.03 | g |
| 53 | I | iodine | 126.90 ± 0.01 |
| 54 | Xe | xenon | 131.29 ± 0.01 | g m |
| 55 | Cs | caesium | 132.91 ± 0.01 |
| 56 | Ba | barium | 137.33 ± 0.01 |
| 57 | La | lanthanum | 138.91 ± 0.01 | g |
| 58 | Ce | cerium | 140.12 ± 0.01 | g |
| 59 | Pr | praseodymium | 140.91 ± 0.01 |
| 60 | Nd | neodymium | 144.24 ± 0.01 | g |
| 61 | Pm | promethium | — |
| 62 | Sm | samarium | 150.36 ± 0.02 | g |
| 63 | Eu | europium | 151.96 ± 0.01 | g |
| 64 | Gd | gadolinium | 157.25 ± 0.01 | g |
| 65 | Tb | terbium | 158.93 ± 0.01 |
| 66 | Dy | dysprosium | 162.50 ± 0.01 | g |
| 67 | Ho | holmium | 164.93 ± 0.01 |
| 68 | Er | erbium | 167.26 ± 0.01 | g |
| 69 | Tm | thulium | 168.93 ± 0.01 |
| 70 | Yb | ytterbium | 173.05 ± 0.02 | g |
| 71 | Lu | lutetium | 174.97 ± 0.01 | g |
| 72 | Hf | hafnium | 178.49 ± 0.01 |
| 73 | Ta | tantalum | 180.95 ± 0.01 |
| 74 | W | tungsten | 183.84 ± 0.01 |
| 75 | Re | rhenium | 186.21 ± 0.01 |
| 76 | Os | osmium | 190.23 ± 0.03 | g |
| 77 | Ir | iridium | 192.22 ± 0.01 |
| 78 | Pt | platinum | 195.08 ± 0.02 |
| 79 | Au | gold | 196.97 ± 0.01 |
| 80 | Hg | mercury | 200.59 ± 0.01 |
| 81 | Tl | thallium | 204.38 ± 0.01 |
| 82 | Pb | lead | 207.2 ± 1.1 |
| 83 | Bi | bismuth | 208.98 ± 0.01 |
| 84 | Po | polonium | — |
| 85 | At | astatine | — |
| 86 | Rn | radon | — |
| 87 | Fr | francium | — |
| 88 | Ra | radium | — |
| 89 | Ac | actinium | — |
| 90 | Th | thorium | 232.04 ± 0.01 | g |
| 91 | Pa | protactinium | 231.04 ± 0.01 |
| 92 | U | uranium | 238.03 ± 0.01 | g m |
| 93 | Np | neptunium | — |
| 94 | Pu | plutonium | — |
| 95 | Am | americium | — |
| 96 | Cm | curium | — |
| 97 | Bk | berkelium | — |
| 98 | Cf | californium | — |
| 99 | Es | einsteinium | — |
| 100 | Fm | fermium | — |
| 101 | Md | mendelevium | — |
| 102 | No | nobelium | — |
| 103 | Lr | lawrencium | — |
| 104 | Rf | rutherfordium | — |
| 105 | Db | dubnium | — |
| 106 | Sg | seaborgium | — |
| 107 | Bh | bohrium | — |
| 108 | Hs | hassium | — |
| 109 | Mt | meitnerium | — |
| 110 | Ds | darmstadtium | — |
| 111 | Rg | roentgenium | — |
| 112 | Cn | copernicium | — |
| 113 | Nh | nihonium | — |
| 114 | Fl | flerovium | — |
| 115 | Mc | moscovium | — |
| 116 | Lv | livermorium | — |
| 117 | Ts | tennessine | — |
| 118 | Og | oganesson | — |
| Z | Symbol | Element | Abridged Standard Atomic Weight | Notes |
Footnotes Back to Top g Geological materials are known in which the element has an isotopic composition outside the limits for normal material. The difference between the atomic weight of the element in such materials and that given in the table may exceed the stated uncertainty. m Modified isotopic compositions may be found in commercially available material because the material has been subjected to an undisclosed or inadvertent isotopic fractionation. Substantial deviations in atomic weight of the element from that given in the table can occur. r Range in isotopic composition of normal terrestrial material prevents a more precise standard atomic weight being given; the tabulated atomic-weight value and uncertainty should be applicable to normal materials.
Citation
The most recent Standard Atomic Weights are presented in this Table and they are based on the "Atomic Weights 2021" report. Standard Atomic Weights of gadolinium, lutetium, and zirconium have been revised by the CIAAW in 2024 and these revisions are included in this table.
In 2013, the Commission recommended that standard atomic weights are best abridged to five digits.
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15857 | https://www.sciencedirect.com/topics/food-science/penicillium | Skip to Main content
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Penicillium
In subject area:Food Science
Penicillium refers to a genus of fungi that typically produces a blue-green, powdery mold, commonly found on corn kernels, with several species capable of causing ear rot and some associated with mycotoxin production, such as ochratoxins.
AI generated definition based on: Corn (Third Edition), 2019
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Chapter
PENICILLIUM | and
2014, Encyclopedia of Food Microbiology (Second Edition)J.I. Pitt
Abstract
The genus Penicillium is a very important fungal genus because of its ubiquity and the role of many species in food spoilage and mycotoxin production. Recent changes to nomenclature mean that henceforth species classified in the sexual genus Eupenicillium will now be incorporated in Penicillium, while species in Penicillium subgenus Biverticillium will be classified in Talaromyces, the other sexual genus associated with Penicillium. This article provides an outline of the taxonomy of Penicillium and Talaromyces, an overview of the most important species in food spoilage, and of the most important mycotoxins produced by Penicillium species.
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Reference work2014, Encyclopedia of Food Microbiology (Second Edition)J.I. Pitt
Chapter
Fungi
2016, Encyclopedia of Food and HealthA. Moretti, S. Sarrocco
The Penicillium Genus
The genus Penicillium (Eurotiales, Trichocomaceae) is maybe the most spread fungal genus in nature. Many of its species are ubiquitous and colonize a wide range of habitats. Generally, Penicillium species are considered saprotrophic fungi living on plant parts, soil, decaying organic substances, and plant residues. Moreover, many of the most common Penicillium species are associated not only with stored food and feed but also with animal dung and building materials, indoor air, and several other habitats. With respect to other mycotoxigenic genera, Penicillium species are less adapted to high temperatures. Many Penicillium show psychrophilic behavior and can therefore contaminate food stored at low temperatures, being able to grow at temperature around 0 °C. Some Penicillium species are plant pathogens acting on several fruit trees and producing harmful mycotoxins that can accumulate on several food products.
The taxonomy of this genus is difficult, mainly because of the enormous variability within the genus. Therefore, a polyphasic approach based on phenotypic characteristics (morphology and physiology), phylogeny, and secondary metabolite profiles (extrolites) is currently applied to create sections, which are useful taxonomic tools for the ecological and phylogenetic grouping of consistently related species. The genus Penicillium includes around 100 toxigenic species, and the range of known mycotoxins produced is wider than those of other toxigenic fungal genera such as Aspergillus and Fusarium. The Penicillium subgenus Penicillium includes the highest number of toxigenic species able to produce a large number of bioactive extrolites, including several mycotoxins. OTA is produced only by P. verrucosum and P. nordicum, with the first species being very common on cereals, mostly in cold environments, and the second species being a known contaminant of protein-rich foods, primarily found on dry-cured meat products.
Patulin, another important mycotoxin produced by Penicillium species, is a water-soluble polyketide lactone that occurs most often in apples spoiled by P. expansum or in products made from spoiled apples, such as apple juice, pies, and conserves. It has also been found in other fruits, including pears and grapes, and in vegetables, cereals, and cheese. Most of the information on the toxicity of patulin is derived from animal studies. At relatively high doses, patulin is acutely toxic in mice and rats, causing gastrointestinal lesions, distension, and hemorrhagia in the stomach and small intestine. It has been suggested that patulin is carcinogenic at low levels in the diet, but IARC has stated that there is insufficient evidence of carcinogenicity in animals or humans, other than at extremely high doses. Many other mycotoxins are produced by species of Penicillium. The most common are citrinin, considered nephrotoxic, penicillic acid, cyclopiazonic acid, citreoviridin, and PR toxin.
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Reference work2016, Encyclopedia of Food and HealthA. Moretti, S. Sarrocco
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and related genera
2006, Food Spoilage MicroorganismsJ.I. Pitt
16.2Taxonomy
Penicillium is a large genus. One hundred and fifty species were recognized in the last complete taxonomy (Pitt, 1979), but subsequent studies indicate that this number is conservative. The most recent general compilation of species names (Pitt et al., 2000) lists about 220 species, and since that time 30 or more further species have been published (Samson and Frisvad, 2004). At least 50 species are of common occurrence (Pitt, 2000). All common species grow and sporulate well on synthetic or semisynthetic media, and are usually readily recognizable to genus level.
Classification within Penicillium is based primarily on microscopic morphology (Fig. 16.1). The genus is divided into subgenera based on the number and arrangement of phialides (elements producing conidia) and metulae and rami (elements supporting phialides) on the main stalk cells (stipes). The classification of Pitt (1979) includes four subgenera: Aspergilloides, where phialides are borne directly on the stipes without intervening supporting elements; Furcatum and Biverticillium, where phialides are supported by metulae; and Penicillium, where both metulae and rami are usually present. The majority of important food spoilage (and toxigenic) species are found in subgenus Penicillium.
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Book2006, Food Spoilage MicroorganismsJ.I. Pitt
Chapter
SPOILAGE | Fungi in Food – An Overview
2003, Encyclopedia of Food Sciences and Nutrition (Second Edition)L.B. Bullerman
Penicillium
Penicillium is also a widespread genus that is important in foods. Penicillium species contaminate a wide variety of foods and are capable of growing at refrigeration temperatures. Thus they often spoil refrigerated foods, especially cheese. They are also common on grains, breads, cakes, fruits, preserves, cured and aged hams and sausages, and in the spoilage of certain fruits. The penicillia produce conidia from conidiophores that branch near the apex, forming a brush-like structure or penicillus (Figure 2). At the apex of the conidiophore are somewhat enlarged cells known as metulae. From the metulae arise the sterigma or phialides. It is in these structures that the conidia are produced and pushed out in chains. The conidia of the penicillia are colored, but mostly in shades of gray to blue to blue-green. The colors are not as distinctive for the various species as for the aspergilli, and are therefore not as helpful in the identification of species. Some species form ascospores in cleistothecia and are also placed in the teleomorphic genera of Talaromyces or Eupenicillium. There are a number of important Penicillium species. P. verrucosum, P. viridicatum, and P. aurantiogriseum are common in grains and some can also occur on cheese. They can produce a number of mycotoxins, including ochratoxin, penicillic acid, and others. P. martensii, a synonym for P. aurantiogriseum, has been found growing in high-moisture corn and can produce penicillic acid. P. expansum causes rots of fruits, especially apples, and produces patulin. P. digitatum, with green-colored conidia, causes soft rot of citrus fruit, usually at ambient temperatures, whereas P. italicum, which has blue spores, causes soft rot of citrus at refrigerated temperatures. P. roqueforti has blueish-colored conidia and is used in the ripening of blue-veined cheeses. However, wild types of P. roqueforti often occur in dairy environments and also contaminate other types of cheeses, such as Cheddar and Swiss, and grow and cause spoilage under refrigerated storage. P. camemberti produces grayish spores, and is used for surface ripening of Camembert and Brie cheeses. Many other Penicillium species are known to produce various toxic substances, many of which also have antibiotic properties, but appear to be too toxic for therapeutic use. The most famous antibiotic, penicillin, which has been used to cure countless bacterial infections, is produced by P. notatum. (See CHEESES | Chemistry and Microbiology of Maturation.)
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Reference work2003, Encyclopedia of Food Sciences and Nutrition (Second Edition)L.B. Bullerman
Chapter
Mycotoxins in Corn: Occurrence, Impacts, and Management
2019, Corn (Third Edition)Gary P. Munkvold, ... Christiane Gruber-Dorninger
Penicillium
Penicillium species typically produce a blue-green, powdery mold on corn kernels, although the appearance differs among species. Like Fusarium, taxonomy in the genus Penicillium has undergone numerous changes and overlapping species designations exist. Several species of Penicillium are toxigenic, but the most common species found in corn are not associated with major mycotoxin problems. Several species of Penicillium infect corn in storage, but Penicillium oxalicum is the primary species known to cause an ear rot in the field (Payne, 1999). Like Fusarium and Aspergillus, infection in the field by Penicillium often occurs in kernels damaged by insects or other agents. Penicillium species such as P. aurantiogriseum and P. viridicatum can grow at low water activity and fare well in stored corn. Penicillium is more common in cooler climates than is Aspergillus. Some Penicillium species can grow at grain moisture contents of 16% to 17%. Kernels with "blue eye" have embryos infected with Penicillium. Most species of Penicillium are not associated with mycotoxin problems in corn, but Penicillium verrucosum and P. viridicatum can produce ochratoxins, cyclopiazonic acid, penicillic acid, or citrinin (Mislivec and Tuite, 1970; Wilson and Abramson, 1992; Abramson, 1997; Payne, 1999) (Table 9.2). At higher latitudes, such as in Canada, P. verrucosum is the primary producer of ochratoxin A, which has received considerable attention in recent years as a potential health hazard in corn grain. Several other mycotoxins can be produced by Penicillium species, and these are covered in more detail in other publications (Wilson and Abramson, 1992; Scott, 1994; Abramson, 1997).
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Book2019, Corn (Third Edition)Gary P. Munkvold, ... Christiane Gruber-Dorninger
Chapter
STARTER CULTURES | Importance of Selected Genera
2014, Encyclopedia of Food Microbiology (Second Edition)W.M.A. Mullan
Penicillium
Penicillium species are used widely in food fermentations, and through the secretion of pectinases, amylases, proteinases, lipases, and other enzymes, they can break down complex structural elements of foods releasing substrates for growth. It is this ability that makes molds so useful in food fermentations.
Important dairy species include Penicillium roqueforti, which produce the blue veins in cheeses like Roquefort, and P. camemberti, the white mold covering Brie and Camembert. Both of these molds secrete lipases and proteases that markedly influence the texture, aroma, and taste of these cheeses. Several Penicillium species are important in meat fermentations. These include Penicillium chrysogenum and Penicillium nalgiovense. These molds promote flavor development during ripening. Mold metabolites also prevent the growth of surface contaminants in fermented sausages.
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Reference work2014, Encyclopedia of Food Microbiology (Second Edition)W.M.A. Mullan
Chapter
and related genera
2006, Food Spoilage MicroorganismsJ.I. Pitt
16.1Introduction
Fungi belonging to the genus Penicillium are among the most ubiquitous organisms on earth. They have been found wherever they have been sought, from the Antarctic across the tropics to Greenland. It is hard to find a sample of soil or decaying vegetation free of Penicillium spores. Because of this, the common viewpoint is that Penicillia are just contaminants, present in low numbers, difficult to speciate and probably not worth speciating.
Work carried out during the past 20 years, however, has shown that Penicillium species are often remarkably substrate-specific, though usually for reasons we do not understand. Improvements in taxonomy have shown that certain commonly isolated species are really soil fungi, seen in foods only as contaminants, while others are consistently foodborne, with a high potential to cause spoilage if conditions prove favourable.
Most Penicillium species do not cause such overt spoilage as Aspergillus species often do, as growth of Penicillium species is less rampant and usually self-limiting. Nevertheless, Penicillium species are an important cause of food spoilage. The ability to recognize common spoilage species can be very helpful, both in efforts to elucidate the problem, and in answering the perennial questions about possible mycotoxin contamination.
In general terms, speciation of Penicillium isolates is not easy. However, if account is taken of the substrate on which a species is found, a great deal of information can often be obtained relatively easily. So in the sections that follow, species have been grouped according to the substrate or substrates on which they are likely to cause spoilage.
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Book2006, Food Spoilage MicroorganismsJ.I. Pitt
Chapter
(Blue Mold)
2014, Postharvest DecayDeena Errampalli
Taxonomy and Morphology
The genus Penicillium was first described in the literature by John H.F. Link (1767–1851), a mycologist and a professor of Botany in Rostock, Breslau and Berlin in 1809. In the paper, Link described three species of genus Penicillium, P. candidum (Link), P. expansum (Link), and P. glaucum (Link), all of which produced brush-like conidiophores (Link, 1809). The species P. expansum, which causes blue mold in apple was selected as the type species, to which the name of the species is permanently linked.
Since its first description by Link in 1809, researchers have been refining the taxonomy of the genus Penicillium. Historically, the identification of species in the genus Penicillium relied on the morphorological characteristics, the branching of the conidiophores. Conidiophores are the dense brush-like conidia-bearing structures. The conidiophores are simple or branched and are terminated by clusters of flask-shaped phialides. The conidia are produced in dry chains from the tips of the phialides, with the youngest spore at the base of the chain, and are nearly always green. Based on the type of spore-bearing systems (penicilli) P. expansum was classified as terverticillate, branches that bear a second order of branches, bearing in turn a cluster of phialides. A polyphasic approach to the taxonomy based on morphology, growth pattern, ecology, extrolites, and partial β-tubulin sequences, placed P. expansum, P. marinum and P. sclerotigenum in the section penicillium and series expansa (Samson and Pitt, 2000; Frisvad and Samson, 2004). Partial RPB1, RPB2 (RNA polymerase II genes), Tsr1 (putative ribosome biogenesis protein) and Cct8 (putative chaperonin complex component TCP-1) gene sequences were applied to distinguish three families in Trichocomacea (Houbraken and Samson, 2011). Identification of species of Penicillium, including P. expansum that produce mycotoxins is critical and, in spite of the availability of modern methods, its identification remains difficult (Paterson et al., 2004). Proficiency testing is rare and conventional identification methods do not inform reliably as to whether mycotoxins were detected/produced. Paterson et al. (2004) described a solution which consists of identifying a Penicillium strain as terverticillate and then undertaking mycotoxin analysis. As the new technologies in DNA and secondary metabolite research characterize new identifiers, the refining of the classification of Penicillium continues (Table 6.1).
Table 6.1. Taxonomy of Penicillium expansum (Link)
| | |
--- |
| Kingdom | Fungi |
| Phylum | Ascomycota |
| Order | Eurotiales |
| Class | Eurotiomycetes |
| Family | Trichocomaeae |
| GenusSubgenusSeriesSpeices | PenicilliumPenicilliumexpansaexpansum |
Morphological characterization of P. expansum was based on the macroscopic study, growth rate, color texture and topography of the colony using two standard media namely potato dextrose agar (PDA) and Czapek Dox and microscopic study, observation of hyphae and conidia under a light microscope. Molecular characterization was based on molecular markers, (RAPD and ITS) and DNA sequence of β-tubulin gene (Peterson, 2000; Czerwinski et al., 2009). On potato dextrose agar at 25°C, the colonies of P. expansum grow rapidly forming a velvety carpet of surface filamentous mycelium up to 2 mm deep. The colonies are initially white becoming dull yellow green to blue green in 3–4 days (Fig. 6.1A). The plate reverse is usually pale to yellowish. As to micromorphology, the conidial heads are asymmetric and once or twice branched (terverticillate) and conidiophores are smooth, 400–700 μm long and with smooth elliptical conidia at 4–5 × 2.5–3.5 μm (Fig. 6.1B).
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Book2014, Postharvest DecayDeena Errampalli
Chapter
Studies in Natural Products Chemistry
2016, Studies in Natural Products ChemistryMahesh S. Majik, ... Satu G. Gawas
Metabolites From Penicillium sp
The genus Penicillium is probably the most well known of the filamentous fungi. Although spoiled food is the most common instance in which they are observed, penicillia have a wide variety of habitats, including plants, animals, and in some cases other fungi. Penicillium sp. are primarily known as saprophytes growing on organic substrates including dead plant material; however, a few species have evolved for growth on living tissue. Penicillium sp. is widely studied from associated marine organisms. An isolate of P. citrinum obtained from an unidentified Japanese demosponge yielded JBIR-59 (41) (Fig. 9.8), along with a series of known sorbicillinoids . Investigation of Penicillium sp., from the red alga Laurencia sp. (Bahamas Islands) resulted in isolation of penilumamide (42), an unusual alkaloid combining a lumazine system with l-methionine sulfoxide and an anthranilic acid ester . Compound (42) was inactive when tested for cytotoxic and antimicrobial properties, and also did not affect cellular Ca2+ signaling in neuroendocrine cells. The fungus Penicillium sp., isolated from the tree Clerodendrum inerme growing in the intertidal zone of the South China Sea, produced three new esters of desoxypatulinic acid (43–45) .
The endophytic fungus P. commune, isolated from the Chinese pseudomangrove plant Hibiscus tiliaceus was found to contain a new ester of orsellinic acid with glycerol (46) and other structurally simple known metabolites . Whereas, Chinese mangrove (Excoecaria agallocha) derived fungus P. expansum produced expansols A (47) and B (48), and two new phenolic bisabolane sesquiterpenoids, (S)-(+)-11-dehydrosydonic acid (49) and (7S,11S)-(+)-12-acetoxysydonic acid (50) (Fig 9.9). Compound (47) exhibited moderate cytotoxicity against the HL-60 cell line, while (48) inhibited the proliferation of A549 and HL-60 cells, while (49) and (50) were inactive. Yet another Penicillium sp., isolated from the Chinese mangrove plant Acanthus ilicifolius, was found to contain isomeric pyrrolyl 4-quinolinone alkaloid penicinoline (51) which exhibited potent cytotoxicity toward the 95-D and HepG2 cell lines, and also displayed strong insecticidal activity against Aphis gossypii .
Penicillium aurantiogriseum isolated from mud of Bohai sea (China) led to the isolation of three new aromatic skeleton auranomides A–C (52–53) . While the tumonoic acids K (54a), L (54b) and methyl 2-(2-acetyl-3,5-dihydroxy-4,6-dimethylphenyl)acetate (55) (Fig. 9.10) were isolated from P. citrinum (sediment, Fujian, China) .
In 2012, Wang et al. isolated seven polyketides, communols (56–59) (Fig. 9.11) from Penicillium commune (gorgonian Muricella abnormalis, China) . Among this series of compounds communols (56), (59c), and (59d) showed moderate antibacterial activity against E. coli and E. aerogenes.
The sesquiterpene-γ-pyrone, a novel skeletal class, has been reported for the first time from marine-derived fungus Penicillium sp. [marine sediments at the depth of 25 m collected from Geomundo Island, Korea] . Penicillipyrones A and B (60–61) are meroterpenoids which were found to be inactive against K562 and A549 cell line (Fig. 9.12). However, penicillipyrone (61) exhibited significant induction of quinone reductase (QR) in Hepa 1c1c7 murine hepatoma cells over concentration of 5–40 μM, indicating the importance of the presence of C-2 keto functionality in its QR-inducing activity. Thus, penicillipyrone B (61) plays potential role in cancer prevention via induction of QR. Sorbicillinoids are members of larger family of polyketides which has been previously reported from marine as well as terrestrial sources with their broad spectrum of biological activities . In 2014, Li and coworkers investigated the culture of marine sediment-derived fungus Penicillium chrysogenum PJX-17 which led to the isolation of two novel sorbicillinoids, that is, sorbicatechols A and B (62–63) (Fig. 9.12) along with caffeic acid methyl ester (64) . Furthermore, sorbicatechols A and B with bicyclo[2.2.2]octane and 2-methoxy phenol moiety exhibited antiviral activity against H1N1 with IC50 85 and 113 μM, respectively.
More recently, Afiyatullov and coworkers investigated the strains Penicillium thomii KMM 4645 and Penicillium lividum KMM 4663 (associated with the marine brown alga Sargassum miyabei, the Sea of Japan) . Overall, 10 new austalide meroterpenoids (65–74) (Fig. 9.13) were resulted from this species wherein (65–66) and (72–73) were able to inhibit AP-1-dependent transcriptional activity in JB6 C141 cell line.
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Book series2016, Studies in Natural Products ChemistryMahesh S. Majik, ... Satu G. Gawas
Chapter
PENICILLIUM | /Penicillia in Food Production
2014, Encyclopedia of Food Microbiology (Second Edition)J.C. Frisvad
Introduction
Some Penicillium species have been used for centuries for production of certain popular food products, such as white mold cheese, blue mold cheese, and mold-fermented salami. Apart from these well-known applications, Penicillium species occur as contaminants on many cheese and meat products, and sometimes they are accepted on raw milk cheeses as an essential part of the product. In other cases, Penicillium growth on foods is entirely undesirable, especially as many Penicillium species produce mycotoxins and volatile secondary metabolites that could be regarded as off-flavors. Most mold-fermented products are produced in two steps. A lactic acid bacterial fermentation will be essential for cheese and salami quality and aroma, but fungi that can grow at lower water activities can tolerate the lactic acid–fermented products and grow after salting and drying. Penicillium roqueforti is especially tolerant to the metabolic products of lactic acid bacteria, such as lactic acid, acetic acid, and carbon dioxide, and is thus inside the cheese products, while Penicillium camemberti and Penicillium nalgiovense are more salt tolerant and will grow on the surface of the products. Sometimes other fungi including yeasts and smear bacteria, including Brevibacterium linens, will also contribute to the aroma of cheeses. The different microorganisms also produce extracellular enzymes, changing the texture of the food products.
Penicillium species are also known for their production of bioactive secondary metabolites used as drugs, such as penicillin, griseofulvin, compactin, fumagillin, fumitremorgin C, and mycophenolic acid. Other biotechnological applications include colorants, volatile aromatic compounds, organic acids, vitamins, and many other products. However, most species of Penicillium are regarded as spoilage and mycotoxin-producing organisms. These fungi grow well on most foods and produce a series of mycotoxins, including ochratoxin A, citrinin, patulin, citreoviridin, and secalonic acids.
Mold-fermented cheeses and sausages are, however, very popular in many countries, and are regarded as valuable and tasty delicacy foods. Fungi of the genera Aspergillus, Eurotium, Neurospora, Rhizopus, and several others are also used for fermenting foods, especially in Asia, but in Europe the most popular fermented foods involve P. camemberti, P. nalgiovense, and P. roqueforti.
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Reference work2014, Encyclopedia of Food Microbiology (Second Edition)J.C. Frisvad
Related terms:
Cereal
Patulin
Aflatoxin
Ochratoxin
Aspergillus
Mycotoxin
Fusarium
Agar
Blue Cheese
Rhizopus
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Exercise 5.1 Solutions 14 Questions
Exercise 5.2 Solutions 8 Questions
Exercise 5.3 Solutions 10 Questions
Miscellaneous Exercise on Chapter 5 Solutions 20 Questions; the summarisation of the topics discussed in Chapter 5 of the Class 11 NCERT curriculum is listed below.
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Exercise 5.1 Page No: 103
Express each of the complex numbers given in Exercises 1 to 10 in the form a + ib.
1. (5i) (-3/5i)
(5i) (-3/5i) = 5 x (-3/5) x i2
= -3 x -1 [i2 = -1]
= 3
(5i) (-3/5i) = 3 + i0
2. i9 + i19
Solution:
i9 + i19 = (i2)4. i + (i2)9. i
= (-1)4 . i + (-1)9 .i
= 1 x i + -1 x i
= i – i
= 0
Hence,
i9 + i19 = 0 + i0
3. i-39
Solution:
i-39 = 1/ i39 = 1/ i4 x 9 + 3 = 1/ (19 x i3) = 1/ i3 = 1/ (-i) [i4 = 1, i3 = -I and i2 = -1]
Now, multiplying the numerator and denominator by i we get
i-39 = 1 x i / (-i x i)
= i/ 1 = i
Hence,
i-39 = 0 + i
4. 3(7 + i7) + i(7 + i7)
Solution:
3(7 + i7) + i(7 + i7) = 21 + i21 + i7 + i2 7
= 21 + i28 – 7 [i2 = -1]
= 14 + i28
Hence,
3(7 + i7) + i(7 + i7) = 14 + i28
5. (1 – i) – (–1 + i6)
Solution:
(1 – i) – (–1 + i6) = 1 – i + 1 – i6
= 2 – i7
Hence,
(1 – i) – (–1 + i6) = 2 – i7
6.
Solution:
7.
Solution:
8. (1 – i)4
Solution:
(1 – i)4 = [(1 – i)2]2
= [1 + i2 – 2i]2
= [1 – 1 – 2i]2 [i2 = -1]
= (-2i)2
= 4(-1)
= -4
Hence, (1 – i)4 = -4 + 0i
9. (1/3 + 3i)3
Solution:
Hence, (1/3 + 3i)3 = -242/27 – 26i
10. (-2 – 1/3i)3
Solution:
Hence,
(-2 – 1/3i)3 = -22/3 – 107/27i
Find the multiplicative inverse of each of the complex numbers given in Exercises 11 to 13.
11. 4 – 3i
Solution:
Let’s consider z = 4 – 3i
Then,
= 4 + 3iand
|z|2 = 42 + (-3)2 = 16 + 9 = 25
Thus, the multiplicative inverse of 4 – 3i is given by z-1
12. √5 + 3i
Solution:
Let’s consider z = √5 + 3i
|z|2 = (√5)2 + 32 = 5 + 9 = 14
Thus, the multiplicative inverse of √5 + 3i is given by z-1
13. – i
Solution:
Let’s consider z = –i
Thus, the multiplicative inverse of –i is given by z-1
14. Express the following expression in the form of a + ib:
Solution:
Exercise 5.2 Page No: 108
Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.
1. z = – 1 – i √3
Solution:
2. z = -√3 + i
Solution:
Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:
3. 1 – i
Solution:
4. – 1 + i
Solution:
5. – 1 – i
Solution:
6. – 3
Solution:
7. 3 + i
Solution:
8. i
Solution:
Exercise 5.3 Page No: 109
Solve each of the following equations:
1. x2 + 3 = 0
Solution:
Given the quadratic equation,
x2 + 3 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 1, b = 0, and c = 3
So, the discriminant of the given equation will be
D = b2 – 4ac = 02 – 4 × 1 × 3 = –12
Hence, the required solutions are
2. 2x2 + x + 1 = 0
Solution:
Given the quadratic equation,
2x2 +x+ 1 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 2, b = 1, andc= 1
So, the discriminant of the given equation will be
D = b2 – 4ac = 12 – 4 × 2 × 1 = 1 – 8 = –7
Hence, the required solutions are
3. x2 + 3x + 9 = 0
Solution:
Given the quadratic equation,
x2 + 3x + 9 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 1, b = 3, and c = 9
So, the discriminant of the given equation will be
D = b2 – 4ac = 32 – 4 × 1 × 9 = 9 – 36 = –27
Hence, the required solutions are
4. –x2 + x – 2 = 0
Solution:
Given the quadratic equation,
–x2 + x– 2 = 0
On comparing it with ax2 + bx + c = 0, we have
a = –1, b = 1, and c = –2
So, the discriminant of the given equation will be
D = b2 – 4ac = 12 – 4 × (–1) × (–2) = 1 – 8 = –7
Hence, the required solutions are
5. x2 + 3x + 5 = 0
Solution:
Given the quadratic equation,
x2 + 3x + 5 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 1, b = 3, and c = 5
So, the discriminant of the given equation will be
D = b2 – 4ac = 32 – 4 × 1 × 5 =9 – 20 = –11
Hence, the required solutions are
6. x2 – x + 2 = 0
Solution:
Given the quadratic equation,
x2 – x + 2 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 1, b = –1, and c = 2
So, the discriminant of the given equation is
D = b2 – 4ac = (–1)2 – 4 × 1 × 2 = 1 – 8 = –7
Hence, the required solutions are
7. √2x2 + x + √2 = 0
Solution:
Given the quadratic equation,
√2x2 + x + √2 = 0
On comparing it with ax2 + bx + c = 0, we have
a = √2, b = 1, and c = √2
So, the discriminant of the given equation is
D = b2 – 4ac = (1)2 – 4 × √2 × √2 = 1 – 8 = –7
Hence, the required solutions are
8. √3x2 – √2x + 3√3 = 0
Solution:
Given the quadratic equation,
√3x2 – √2x + 3√3 = 0
On comparing it with ax2 + bx + c = 0, we have
a = √3, b = -√2, and c = 3√3
So, the discriminant of the given equation is
D = b2 – 4ac = (-√2)2 – 4 × √3 × 3√3 = 2 – 36 = –34
Hence, the required solutions are
9. x2 + x + 1/√2 = 0
Solution:
Given the quadratic equation,
x2 + x + 1/√2 = 0
It can be rewritten as,
√2x2 + √2x + 1 = 0
On comparing it with ax2 + bx + c = 0, we have
a = √2, b = √2, and c = 1
So, the discriminant of the given equation is
D = b2 – 4ac = (√2)2 – 4 × √2 × 1 = 2 – 4√2 = 2(1 – 2√2)
Hence, the required solutions are
10. x2 + x/√2 + 1 = 0
Solution:
Given the quadratic equation,
x2 + x/√2 + 1 = 0
It can be rewritten as,
√2x2 + x + √2 = 0
On comparing it with ax2 + bx + c = 0, we have
a = √2, b = 1, and c = √2
So, the discriminant of the given equation is
D = b2 – 4ac = (1)2 – 4 × √2 × √2 = 1 – 8 = -7
Hence, the required solutions are
Miscellaneous Exercise Page No: 112
1.
Solution:
2. For any two complex numbersz1 and z2, prove that
Re (z1z2)= Re z1Re z2 – Im z1 Im z2
Solution:
3. Reduce to the standard form.
Solution:
4.
Solution:
5. Convert the following into the polar form:
(i) , (ii)
Solution:
Solve each of the equations in Exercises 6 to 9.
6. 3x2 – 4x + 20/3 = 0
Solution:
Given the quadratic equation, 3x2 – 4x + 20/3 = 0
It can be re-written as: 9x2 – 12x + 20 = 0
On comparing it with ax2 + bx + c = 0, we get
a = 9, b = –12, and c = 20
So, the discriminant of the given equation will be
D = b2 – 4ac = (–12)2 – 4 × 9 × 20 = 144 – 720 = –576
Hence, the required solutions are
7. x2 – 2x + 3/2 = 0
Solution:
Given the quadratic equation, x2 – 2x + 3/2 = 0
It can be re-written as 2x2 – 4x + 3 = 0
On comparing it with ax2 + bx + c = 0, we get
a = 2, b = –4, and c = 3
So, the discriminant of the given equation will be
D = b2 – 4ac = (–4)2 – 4 × 2 × 3 = 16 – 24 = –8
Hence, the required solutions are
8. 27x2 – 10x + 1 = 0
Solution:
Given the quadratic equation, 27x2 – 10x + 1 = 0
On comparing it with ax2 + bx + c = 0, we get
a = 27, b = –10, and c = 1
So, the discriminant of the given equation will be
D = b2 – 4ac = (–10)2 – 4 × 27 × 1 = 100 – 108 = –8
Hence, the required solutions are
9. 21x2 – 28x + 10 = 0
Solution:
Given the quadratic equation, 21x2 – 28x + 10 = 0
On comparing it with ax2 + bx+ c= 0, we have
a = 21, b = –28, and c = 10
So, the discriminant of the given equation will be
D = b2 – 4ac = (–28)2 – 4 × 21 × 10 = 784 – 840 = –56
Hence, the required solutions are
10. If z1 = 2 – i, z2 = 1 + i, find
Solution:
Given, z1 = 2 – i, z2 = 1 + i
11.
Solution:
12. Let z1 = 2 – i, z2 = -2 + i. Find
(i) , (ii)
Solution:
13. Find the modulus and argument of the complex number.
Solution:
14. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i.
Solution:
Let’s assume z = (x – iy) (3 + 5i)
And,
(3x + 5y) – i(5x – 3y) = -6 -24i
On equating real and imaginary parts, we have
3x + 5y = -6 …… (i)
5x – 3y = 24 …… (ii)
Performing (i) x 3 + (ii) x 5, we get
(9x + 15y) + (25x – 15y) = -18 + 120
34x = 102
x = 102/34 = 3
Putting the value of x in equation (i), we get
3(3) + 5y = -6
5y = -6 – 9 = -15
y = -3
Therefore, the values of xand y are 3 and –3, respectively.
15. Find the modulus of
Solution:
16. If (x + iy)3 = u + iv, then show that
Solution:
17. If α and β are different complex numbers with |β| = 1, then find
Solution:
18. Find the number of non-zero integral solutions of the equation |1 – i|x = 2x.
Solution:
Therefore, 0 is the only integral solution of the given equation.
Hence, the number of non-zero integral solutions of the given equation is 0.
19. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2
Solution:
20. If, then find the least positive integral value of m.
Solution:
Thus, the least positive integer is 1.
Therefore, the least positive integral value of m is 4 (= 4 × 1).
| |
| Also Access |
| NCERT Exemplar for Class 11 Maths Chapter 5 |
| CBSE Notes for Class 11 Maths Chapter 5 |
NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations
Chapter 5 of Class 11 Complex Numbers and Quadratic Equations has 3 exercises and a miscellaneous exercise to help the students practise the required number of problems to understand all the concepts. The topics and sub-topics discussed in the PDF of NCERT Solutions for Class 11 of this chapter include
5.1 Introduction
We know that some of the quadratic equations have no real solutions. That means the solution of such equations includes complex numbers. Here, we have found the solution of a quadratic equation ax2 + bx + c = 0 where D = b2 – 4ac < 0.
5.2 Complex Numbers
Definition of complex numbers, examples and explanations about the real and imaginary parts of complex numbers have been discussed in this section. Class 11 Maths NCERT Supplementary Exercise Solutions PDF helps the students to understand the questions in detail.
5.3 Algebra of Complex Numbers
5.3.1 Addition of two complex numbers
5.3.2 Difference of two complex numbers
5.3.3 Multiplication of two complex numbers
5.3.4 Division of two complex number
5.3.5 Power of i
5.3.6 The square roots of a negative real number
5.3.7 Identities
After studying these exercises, students are able to understand the basic BODMAS operations on complex numbers, along with their properties, power of i, square root of a negative real number and identities of complex numbers.
5.4 The Modulus and the Conjugate of a Complex Number
The detailed explanation provides the modulus and conjugate of a complex number with solved examples.
5.5 Argand Plane and Polar Representation
5.5.1 Polar representation of a complex number
In this section, it has been explained how to write the ordered pairs for the given complex numbers, the definition of a Complex plane or Argand plane and the polar representation of the ordered pairs in terms of complex numbers.
A number of the form a + ib, where a and b are real numbers, is called a complex number, “a” is called the real part, and “b” is called the imaginary part of the complex number
Let z1 = a + ib and z2 = c + id. Then
z1 + z2 = (a + c) + i (b + d)
z1z2 = (ac – bd) + i (ad + bc)
For any non-zero complex number z = a + ib (a ≠ 0, b ≠ 0), there exists a complex number, denoted by 1/z or z–1, called the multiplicative inverse of z
For any integer k, i4k = 1, i4k + 1 = i, i4k + 2 = – 1, i4k + 3 = – i
The polar form of the complex number z = x + iy is r (cosθ + i sinθ)
A polynomial equation of n degree has n roots.
Disclaimer –
Dropped Topics –
5.5.1 Polar Representation of a Complex Number
5.6 Quadratic Equation
Example 11 and Exercise 5.3
Examples 13, 15, 16
Ques. 5–8, 9 and 13 (Miscellaneous Exercise)
Last three points in the Summary
5.7 Square-root of a Complex Number
Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 5
Q1
What are the topics covered under each exercise of NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations?
The NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations has exercise-wise problems, which cover equations related to quadratic equations and complex numbers. These solutions provide the students with clarity on concepts and theorems of complex numbers. This chapter has three exercises for which the solutions are provided here in PDF format. The first exercise deals with determining the multiplicative inverse, the second exercise deals with the modulus and argument of a given set of numbers, and the third exercise deals with solving quadratic equations.
Explain the marks distribution in NCERT Solutions for Class 11 Maths.
There are about 6 units in the NCERT Solutions for Class 11 Maths in which the marks are distributed.
First unit – Sets and Functions (60 marks)
Second unit – Algebra (30 marks)
Third unit – Coordinate Geometry (10 marks)
Fourth unit – Calculus (30 marks)
Fifth unit – Mathematical Reasoning, Statistics (10 marks)
Sixth unit – Probability (30 marks)
Does BYJU’S give the most reliable answers in Chapter 5 of NCERT Solutions for Class 11 Maths?
The most accurate and reliable NCERT Solutions for Class 11 Maths Chapter 5 are available on BYJU’S. Students can easily download the solutions, which are present in PDF format and access them during their exam preparation. The solutions are framed and compiled by a set of expert faculty who possess numerous years of experience in the respective subjects. The most detailed solutions to the exercise-wise problems are curated with the aim of helping students ace the exam without fear.
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15859 | http://globalcognition.net/ThirdGradeBigIdeas/thirdgradebigideas_files/UG_TilingToUnderstandArea.html | Tiling to Understand Area
Year Three - Tiling to Understand Area
Home - Tiling to Understand Area Map
About this Big Idea
In this Big Idea students explore area through tiling visualisations so as to develop a deep understanding of this fundamental mathematical concept.
Understanding Goals:
Students will understand:
the concept of area and its relationship to two-dimensional space.
why different shapes might have the same area.
the distinction between perimeter and area.
how creativity and innovation are essential parts of mathematics.
Background:
It is important that students have a clear understanding of the distinction between perimeter and area. 'Perimeter' is derived from the Greek words that mean to measure around the outside: peri, meaning 'around', and metron, meaning 'measure'. Area relates to the measurement of two-dimensional space in the same way that volume and capacity relate to the measurement of three-dimensional space.
Core Content from the Syllabus:
Working Mathematically
Area
estimate the areas of rectangles (including squares) in square centimetres
discuss strategies used to estimate area in square centimetres, eg visualising repeated units (Communicating, Problem Solving)
compare the areas of regular and irregular shapes by informal means
measure the areas of irregular shapes using a square-centimetre grid overlay
compare how different placements of a grid overlay make measuring area easier or harder (Problem Solving)
explain why two students may obtain different measurements of the area of the same irregular shape (Communicating, Reasoning)
Language:
length, distance, metre, centimetre, millimetre, ruler, measure, estimate, perimeter
area, irregular area, measure, grid, row, column, parts of (units), square centimetre, square metre
Connected to:
Being Flexible with Number
Thinking Around Shapes
Seeing Multiplication as Area
Understanding 1/2
Thinking in Equal Groups
Mindset Mathematics Learning Activities
Visualise
Students use geoboards and square tiles to develop an understanding of area as covering with square units. Students make shapes on the geoboard to figure out, "How big is this shape?"- See page 101
Questions for reflection:
What shapes did you make?
How did you figure out how much space the shape takes up?
How did you record your thinking?
What is area? Provide examples to show what area is?
Can you distinguish between perimeter and area?
Play
Students play with area by building pixelated letters out of square units to see what letters could have an area of 4, 5, 6 and up to 12 square units. - See page 111
Questions for reflection:
What area can make the most letters? Why?
What letters can be made with many different areas? Why?
Were there any letters that you couldn't make? Why?
Do you think you have found all the possible letters that can be made with these areas? Why or Why not?
If you were to make the digits 0-9 on a grid, which digit do you think would have the greatest area? The least? Why?
Digital Technology Extension
Students can apply their knowledge of letter area and designs to the challenge of displaying letters on small LED displays such as that on the BBC MicroBit. Students will need to use their knowledge of the area required for each letter and translate this to the 5 x 5 grid of LED's on the MicroBit or similar display.
Image - BBC MicroBit - an easily programmed system on a chip computer with LED display.
Design & Production Extension
Students could investigate how a display board could be made using coloured paper, cardboard and craft supplies. Students would need to consider the size of the matrix they would need to display all letters and numbers and devise a method for changing the display by perhaps flipping pixels over or covering unused pixels. Displays could be back lit by placing on a light box or by attaching to a window.
Investigate
Students create rectangles with the same area and look for patterns opening the door to connecting area and multiplication. - See page 119
Questions for reflection:
What evidence did we find regarding our conjectures? What support did we find?
What ideas need revision?
What other evidence might we need to gather?
What patterns do you notice?
What new questions do you have?
What did you notice in the rectangles that surprised you? What are you wondering now?
Credit:
Boaler, Munson & Williams (2018) - Mindset Mathematics: Visualizing and investigating big ideas Grade 5
NESA - Mathematics K-10 - 2012
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. |
15860 | https://www.edx.org/learn/statistics/edx-intro-course-spreadsheets-and-statistics | edX: Intro Course: Spreadsheets and Statistics | edX
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15861 | https://www.asccp.org/Assets/b2263c88-ec67-4ab0-9f07-6f112e76f8d7/637269576182030000/2019-asccp-risk-based-management-consensus-3-5-pdf | Downloaded from by BhDMf5ePHKav1zEoum1tQfN4a+kJLhEZgbsIHo4XMi0hCywCX1AWnYQp/IlQrHD3wX04VDhDA6562ASvZQF6y+i/9vPBQOaKGvZHVBTglws= on 06/05/2020
2019 ASCCP Risk-Based Management Consensus Guidelines for Abnormal Cervical Cancer Screening Tests and Cancer Precursors
Rebecca B. Perkins, MD, MSc, 1 Richard S. Guido, MD, 2 Philip E. Castle, PhD, 3 David Chelmow, MD, 4Mark H. Einstein, MD, MS, 5 Francisco Garcia, MD, MPH, 6 Warner K. Huh, MD, 7 Jane J. Kim, PhD, MSc, 8Anna-Barbara Moscicki, MD, 9 Ritu Nayar, MD, 10 Mona Saraiya, MD, MPH, 11 George F . Sawaya, MD, 12 Nicolas Wentzensen, MD, PhD, MS, 13 and Mark Schiffman, MD, MPH 14 for the 2019 ASCCP Risk-Based Management Consensus Guidelines Committee
Key Words: cervical cytology, HPV testing, management of abnormal cervical cancer screening tests, guidelines (J Low Genit Tract Dis 2020;24: 102 –131) SECTION A. EXECUTIVE SUMMARY B. INTRODUCTION C. GUIDING PRINCIPLES D. METHODS SUB-SECTION D.1 Process and Timeline D.2 Choice of CIN3+ as Main Clinical Endpoint for Risk Estimates D.3 Multiple Data Sets Used to Validate Risks D.4 Estimation of Risks D.5 Assigning Combinations of Test Results to Clinical Actions D.6 Rating the Recommendations Table of Contents E. PARADIGM SHIFT: CLINICAL ACTION THRESHOLDS E.1 Clinical Action Thresholds Leading to Recommendation of Surveillance E.2 Clinical Action Threshold Leading to Recommendation of Colposcopy E.3 Clinical Action Thresholds Leading to Recommendations of Treatment E.4 Clinical Situations Leading to Management Recommendation F. UPDATES RELATED TO PATHOLOGY REPORTING AND LABORATORY TESTS F.1 Statement on the Use of a 2-Tier Terminology (Histologic LSIL/HSIL) for Reporting Histopathology of Squamous Lesions of the Lower Anogenital Tract F.2 Updated Management of Primary HPV Screening (Replaces Interim Guidance) F.3 Statement on HPV Tests Used in Management From the 1Boston University School of Medicine/Boston Medical Center, Boston, MA; 2University of Pittsburgh/Magee-Women's Hospital, Pittsburgh, PA; 3Albert Einstein College of Medicine, New York, NY; 4Virginia Common-wealth University School of Medicine, Richmond, VA; 5Rutgers, New Jersey Medical School, Newark, NJ; 6Pima County Health & Community Services, Tucson, AZ; 7UAB School of Medicine, Birmingham, AL; 8Harvard T.H. Chan School of Public Health Boston, MA; 9University of California, Los Angeles, CA; 10 Northwestern University, Feinberg School of Medicine-Northwestern Memo-rial Hospital, Chicago, IL; 11 Division of Cancer Prevention and Control, Cen-ters for Disease Control and Prevention, Atlanta, GA; 12 University of California, San Francisco, San Francisco, CA; 13 Division of Cancer Epidemiol-ogy and Genetics and Division of Cancer Prevention, National Cancer Institute, Bethesda, MD; and 14 Division of Cancer Epidemiology and Genetics and Divi-sion of Cancer Prevention, National Cancer Institute, Bethesda, MD This article is open access, and reprints are available for download at asccp.org, jlgtd.com, or via pubmed. R.B.P. and R. S. G. contributed equally to the development of this manuscript and are co-first authors. The guidelines effort received support from the National Cancer Institute and ASCCP. Participating organizations supported travel for their participating representatives. All participating consensus organizations, including the primary funders, had equal and balanced roles in the consensus process including data analysis and interpretation, writing of manuscript, and decision to submit for publication. No industry funds were used in the development of these guidelines. The corresponding authors had final responsibility for the submission decision. The National Cancer Institute (including M.S. and N.W.) receives cervical screening results at reduced or no cost from commercial research partners (Qiagen, Roche, BD, MobileODT, Arbor Vita) for independent evaluations of screening methods and strategies. A.-B.M. is an advisory board member of Merck and GSK. R.S.G. is an ASCCP consultant of Inovio Pharmaceuticals DSMB. W.K.H. is connected with Inovio Pharmaceuticals DSMB. P.E.C. has received HPV tests and assays at a reduced or no cost from Roche, Becton Dickinson, Arbor Vita Corporation, and Cepheid for research. M.H.E. has advised companies and participated in educational activities but does not receive any honoraria or payments for these activities, In some cases, his employer, Rutgers, receives payment for his time for these activities from Papivax, Cynvec, Merck, Hologic, and PDS Biotechnologies. He has been the overall PI or local PI for clinical trials from Johnson&Johnson, Pfizer, Iovance, and Inovio. Funding for these activities is for the research related costs of the trials. The other authors have declared they have no conflicts of interest. Disclaimer: The conclusions, findings, and opinions expressed by authors contributing to this journal do not necessarily reflect the official position of the US Department of Health and Human Services, the Public Health Service, the Centers for Disease Control and Prevention, or the National Cancer Institute. Copyright © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP. This is an open-access article distributed under the terms of the Creative Commons Attribution-Non Commercial-No Derivatives License 4.0 (CCBY -NC-ND), where it is permissible to download and share the work provided it is properly cited. The work cannot be changed in any way or used commercially without permission from the journal. DOI: 10.1097/LGT.0000000000000525 ORIGINAL RESEARCH ARTICLE : C ERVIX AND HPV
102 Journal of Lower Genital Tract Disease • Volume 24, Number 2, April 2020 A. EXECUTIVE SUMMARY
Updated US consensus guidelines for management of cervical screening abnormalities are needed to accommodate the 3 available cervical screening strategies: primary human papillomavirus (HPV) screening, cotesting with HPV testing and cervical cytology, and cer-vical cytology alone. New data indicate that a patient's risk of devel-oping cervical precancer or cancer can be estimated using current screening test results and previous screening test and biopsy results, while considering personal factors such as age and immunosuppres-sion. Routine screening applies only to asymptomatic individuals who do not require surveillance for prior abnormal screening results. The 2012 consensus guidelines were the first to be based on the principle of equal management for equal risk, specifically, the risk of a patient developing cervical cancer, estimated by the surro-gate end point of the 5-year risk of cervical intraepithelial neoplasia (CIN) grade 3 (CIN 3) or more severe diagnoses (CIN 3+), regard-less of which test combinations yielded this risk level. Introduction of risk-based guidelines in 2012 was a conceptual breakthrough, but the recommendations retained a continued reliance on compli-cated algorithms and insufficiently incorporated screening history. With a more nuanced understanding of how previous results affect risk, and more variables to consider, the 2019 guidelines further align management recommendations with current understanding of HPV natural history and cervical carcinogenesis. More frequent surveillance, colposcopy, and treatment are recommended for pa-tients at progressively higher risk, whereas those at lower risk can defer colposcopy, undergo follow-up at longer surveillance inter-vals, and, when at sufficiently low risk, return to routine screening. Clearly defined risk thresholds to guide management are designed to continue functioning appropriately when population-level preva-lence of CIN 3+ decreases because of HPV vaccination and also as new screening and triage tests are introduced. The revised guidelines provide a framework for incorporating new data and technologies as ongoing incremental recommendation revisions, minimizing the time needed to implement changes that are beneficial to patient care.
B. INTRODUCTION
This is the fourth American Society of Colposcopy and Cer-vical Pathology (ASCCP)-sponsored consensus guidelines for management of cervical cancer screening abnormalities, after the original consensus conferences in 2001 1 and subsequent up-dates in 2006 2 and 2012. 3 An interim guidance publication provid-ing management recommendations for primary HPV screening was released in 2015. 4 This document updates and replaces all previous guidance. The key difference between 2019 guidelines and previous versions is the change from primarily test results –based algorithms (e.g., “Colposcopy is recommended for patients with HPV-positive atypical squamous cells of undetermined significance [ASC-US], low-grade squamous intraepithelial lesion [LSIL], ” etc.) to primarily
“risk-based ” guidelines ( e.g., “Colposcopy is recommended for any combination of history and current test results yielding a 4.0% or greater probability of finding CIN 3+, ” etc.). See Box 1 for essential changes. Tables of risk estimates for possible combinations of current screening test results and screening history (including unknown his-tory) have been generated from a prospective longitudinal cohort of more than 1.5 million patients followed for more than a decade at Kaiser Permanente Northern California (KPNC). All KPNC esti-mates of risk underlying guideline decisions are detailed in the ac-companying article by Egemen et al .5 The applicability of these risk estimates to other United States regions and populations has been confirmed in other data sets from screening programs and clinical tri-als. 6 Many patients, especially those with minor abnormalities, can be managed by identifying their risk level using Tables 1A to 5B in Egemen et al 5 and linking it to a recommended clinical action (return to routine screening, surveillance with repeat testing at 1- or 3-year G. RARE CYTOLOGY RESULTS G.1 Evaluation of cytology interpreted as atypical glandular cells (AGC) or adenocarcinoma in situ (AIS) G.2 Unsatisfactory Cytology G.3 Absent Transformation Zone on Cytology G.4 Benign Endometrial Cells in Premenopausal Patients or Benign Glandular Cells in Post-Hysterectomy Patient H. COLPOSCOPY PRACTICE STANDARDS AND EXCEPTIONS TO COLPOSCOPY CLINICAL ACTION THRESHOLD H.1 ASCCP Colposcopy Standards H.2 Exceptions to Colposcopy Threshold I. MANAGING HISTOLOGY RESULTS I.1 Histologic HSIL, Not Further Specified or Qualified I.2 Management of Histologic HSIL (CIN 2 or CIN 3) I.3 Management of CIN 2 in Those Who Are Concerned About the Potential Effect of Treatment on Future Pregnancy Outcomes I.4 Management of Histologic LSIL (CIN 1) or less Preceded by ASC-H or HSIL Cytology I.5 Histologic LSIL (CIN 1) Diagnosed Repeatedly for at Least 2 Years. I.6 Management of AIS: Adoption of Society of Gynecologic Oncology Recommendations J. SURVEILLANCE AFTER ABNORMALITIES J.1 Specific Tests and Testing Intervals When Managing Abnormal Results J.2 Short-Term Follow-up After Treatment for Histologic HSIL J.3 Guidance for Long-Term Follow-up After Treatment for High-Grade Histology or Cytology J.4 Guidance for Long-Term Follow-up After Low-Grade Cytology (HPV-Positive NILM, ASC-US, or LSIL) or Histologic LSIL (CIN 1) Abnormalities Without Evidence of Histologic or Cytologic High-Grade Abnormalities K. SPECIAL POPULATIONS K.1 Management of Patients Younger Than 25 Years K.2 Managing Patients During Pregnancy K.3 Managing Patients With Immunosuppression K.4 Managing Patients After Hysterectomy K.5 Managing Patients Older Than 65 Years With a History of Prior Abnormalities L. CURRENT CONSIDERATIONS AND FUTURE DIRECTIONS L.1 Current Considerations L.2 Future Directions
Journal of Lower Genital Tract Disease •Volume 24, Number 2, April 2020 2019 Consensus Guidelines
© 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .
103 Box 1. Essential Changes From Prior Management Guidelines
1) Recommendations are based on risk, not results.
• Recommendations of colposcopy, treatment, or surveillance will be based on a patient's risk of CIN 3+ determined by a combination of current results and past history (including unknown history). The same current test results may yield dif-ferent management recommendations depending on the history of recent past test results. 2) Colposcopy can be deferred for certain patients.
• Repeat HPV testing or cotesting at 1 year is recommended for patients with minor screening abnormalities indicating HPV infection with low risk of underlying CIN 3+ ( e.g., HPV-positive, low-grade cytologic abnormalities after a docu-mented negative screening HPV test or cotest). 3) Guidance for expedited treatment is expanded ( i.e., treatment without colposcopic biopsy).
• Expedited treatment was an option for patients with HSIL cytology in the 2012 guidelines; this guidance is now better defined.
• For non-pregnant patients 25 years or older, expedited treatment, defined as treatment without preceding colposcopic biopsy demonstrating CIN 2+, is preferred when the immediate risk of CIN 3+ is ≥60%, and is acceptable for those with risks between 25% and 60%. Expedited treatment is preferred for nonpregnant patients 25 years or older with high-grade squamous intraepithelial lesion (HSIL) cytology and concurrent positive testing for HPV genotype 16 (HPV 16) ( i.e.,
HPV 16 –positive HSIL cytology) and never or rarely screened patients with HPV-positive HSIL cytology regardless of HPV genotype.
• Shared decision-making should be used when considering expedited treatment, especially for patients with concerns about the potential impact of treatment on pregnancy outcomes. 4) Excisional treatment is preferred to ablative treatment for histologic HSIL (CIN 2 or CIN 3) in the United States. Excision is recommended for adenocarcinoma in situ (AIS). 5) Observation is preferred to treatment for CIN 1. 6) Histopathology reports based on Lower Anogenital Squamous Terminology (LAST)/World Health Organization (WHO) rec-ommendations for reporting histologic HSIL should include CIN 2 or CIN 3 qualifiers, i.e., HSIL(CIN 2) and HSIL (CIN 3). 7) All positive primary HPV screening tests, regardless of genotype, should have additional reflex triage testing performed from the same laboratory specimen ( e.g., reflex cytology).
• Additional testing from the same laboratory specimen is recommended because the findings may inform colposcopy practice. For example, those HPV-16 positive HSIL cytology qualify for expedited treatment.
• HPV 16 or 18 infections have the highest risk for CIN 3 and occult cancer, so additional evaluation ( e.g., colposcopy with biopsy) is necessary even when cytology results are negative.
• If HPV 16 or 18 testing is positive, and additional laboratory testing of the same sample is not feasible, the patient should proceed directly to colposcopy. 8) Continued surveillance with HPV testing or cotesting at 3-year intervals for at least 25 years is recommended after treatment and initial post-treatment management of histologic HSIL, CIN 2, CIN 3, or AIS. Continued surveillance at 3-year intervals be-yond 25 years is acceptable for as long as the patient's life expectancy and ability to be screened are not significantly compro-mised by serious health issues.
• The 2012 guidelines recommended return to 5-year screening intervals and did not specify when screening should cease. New evidence indicates that risk remains elevated for at least 25 years, with no evidence that treated patients ever return to risk levels compatible with 5-year intervals. 9) Surveillance with cytology alone is acceptable only if testing with HPV or cotesting is not feasible. Cytology is less sensitive than HPV testing for detection of precancer and is therefore recommended more often. Cytology is recommended at 6-month intervals when HPV testing or cotesting is recommended annually. Cytology is recommended annually when 3-year intervals are recommended for HPV or cotesting. 10) Human papillomavirus assays that are Food and Drug Administration (FDA)-approved for screening should be used for management according to their regulatory approval in the United States. ( Note: all HPV testing in this document refers to testing for high-risk HPV types only).
• For all management indications, HPV mRNA and HPV DNA tests without FDA approval for primary screening alone should only be used as a cotest with cytology, unless sufficient, rigorous data are available to support use of these par-ticular tests in management.
Perkins et al. Journal of Lower Genital Tract Disease •Volume 24, Number 2, April 2020
104 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .intervals, colposcopy, or treatment). To facilitate use of these tables, the same information will be accessible via smartphone app (for purchase) and web (no cost) through Deci-sion aids may facilitate use of the tables. 7 Common abnormalities are managed using risk estimates outlined in Section E, and rare abnormalities are managed via the result-specific consensus rec-ommendations outlined in Sections G-K. The minimum amount of data required to generate a recom-mendation will include the patient's age and current test results, as we recognize that previous screening history is often not known. Increased precision of management guidance will be possible if information is available on test results within the past 5 years and previous precancer treatment within the past 25 years. 3 Cur-rent results and past history are designed to generate the patient's risk estimate from data tables. 5 Risk estimates are available for the following clinical situations: abnormal screening test results with unknown history, abnormal screening test results with medical record documentation of a preceding negative HPV test or cotest, surveillance of previous abnormal screening test results that did not require immediate colposcopic referral ( e.g., follow-up after an HPV-positive cytology negative result), colposcopy/biopsy re-sults, and follow-up surveillance tests after colposcopy or after treat-ment for, or resolution of, high-grade abnormalities ( e.g., CIN 2+). The recognition that persistent HPV infection is necessary for developing precancer and cancer ( defined as CIN 3+, which includes diagnoses of CIN 3, AIS, and cancer ) underlies the 2019 guideline update. Prospective longitudinal data indicate that when a new abnormal screening test result follows a negative HPV test or cotest within the past 5 years, the estimated risk of CIN 3+ is reduced by approximately 50%. 8 A negative cytology result within 3 years of a new abnormal screening test, however, does not confer a similar reduction in risk. 9 The 2019 guidelines also recognize that a colposcopic examination performed according to accepted standards ( e.g., using the KPNC colposcopy protocol or the ASCCP Colposcopy Standards 10 ) confirming low-grade or normal histology reduces a patient's estimated risk of having precancer/cancer in the next 2 years. 11 This allows patients with an HPV-positive ASC-US or LSIL result at their 1-year follow-up visit after a colposcopy confirming normal- or low-grade his-tology to return for repeat HPVor cotesting in 1 more year, rather than immediately return to colposcopy. Thus, incorporating a patient's history of previous HPV tests and colposcopy/biopsy re-sults will permit detection and treatment of CIN 3+ while avoiding unnecessary interventions for patients with new HPV infections who are at lower risk. 12
C. GUIDING PRINCIPLES
Guidelines are based on several guiding principles. The first 4 guiding principles are new for 2019, whereas the others are from the 2012 guidelines. As the 2012 guidelines are familiar to pro-viders, we changed management recommendations only when new evidence favored an altered management strategy. Note that management guidelines apply only to patients with current or pre-vious abnormal screening test results; screening guidelines for in-dividuals in the general population, that are not being followed for a screening abnormality, are addressed elsewhere. 13,14
New 2019 Principles
HPV –based testing is the basis for risk estimation. The term HPV-based testing is used throughout this document and refers to use of either primary HPV testing alone or HPV testing in conjunction with cervical cytology (cotesting).
Characteristics of HPV infections, including HPV type and the duration of infection, determine a patient's risk of CIN 3+. 15 –18 Although cytology has high specificity (apart from ASC-US) and can be helpful when estimating immediate risk, its lower sen-sitivity and lower negative predictive value compared with HPV testing reduces its utility for long-term risk prediction. 9 The re-sults of HPV tests alone or in conjunction with cytology are used to guide recommendations that allow lengthening of follow-up in-tervals and deferral of colposcopy for low-risk results. Of note, risk estimates underlying the 2019 management guidelines are based on HPV DNA testing.
Personalized risk-based management is possible with knowledge of current results and past history. A patient's risk of having or developing CIN 3+ is estimated based on current and previous results, as well as history of previous precancer treatment. Management recommendations use thresholds of risk. 19 Recommen-dations of routine screening, 1-year or 3-year surveillance, colpos-copy, or treatment correspond to a risk stratum, a range of risk for CIN 3+. The lower threshold of each risk stratum, called Clinical Action Threshold, defines the level at which the management rec-ommendation changes. The Clinical Action Thresholds for each risk stratum were determined through the consensus process. Risks were estimated for all combination of current results and past his-tory (including unknown history) for which adequate data were available at KPNC. Management can be determined via look-up ta-bles, 5 and use of the tables can be facilitated using decision aids.
Guidelines must allow updates to incorporate new test methods as they are validated, and to adjust for decreasing CIN3+ risks as more patients who received HPV vaccination reach screening age . The field of cervical cancer prevention is rapidly evolving, with new technologies being continually validated. Data on the validation of new technologies are being published fre-quently, and risk reduction from HPV vaccination is increasing as vaccine coverage increases and vaccinated individuals age into screening cohorts. Up to now, guideline revisions have required full consensus conferences, which are time-consuming, expensive, and not compatible with the rapid evolution of the field. The 2019 guide-lines build a framework that allows incorporation of new technologies and modified strategies without requiring full consensus conferences, so that revisions may rapidly incorporate new findings and be quickly disseminated to optimize patient care. Clinical Action Thresholds for management created through the 2019 consensus process will remain in place, but as new tests become available and more long-term data accrue, the test combi-nations used to reach these thresholds will change. For example, at the 2019 consensus conference, HPV vaccination levels in the United States population currently 25 years or older were deemed too low to warrant incorporating HPV vaccination into the 2019 management recommendations. However, this is expected to change in the near future as more vaccinated patients, who have lower CIN 3+ risk, reach the age of 25 years and additional data accrue demonstrating the impact of vaccination on the CIN 3+ risk associated with abnormal test result combinations. The framework outlined here will allow guideline modification as robust data become available and are publicized. Because Clinical Action Thresholds re-main constant, new data can be added while the Clinical Action Thresholds remain unchanged. This design is intentional to reduce clinician confusion associated with frequently changing guidelines.
Colposcopy practice must follow guidance detailed in the ASCCP Colposcopy Standards. 10 Colposcopy with targeted biopsy remains the primary method of detecting precancers requiring treat-ment. Because patients are managed less aggressively after a colposcopic examination where CIN grade 2 or higher (CIN 2+) is not found, maximizing detection of CIN 2+ at each colposcopy visit is paramount. Evidence-based practice recommends that bi-opsies be taken of all discrete acetowhite areas, usually 2 to 4 bi-opsies at each colposcopic examination. For those at lowest risk, defined as less than HSIL cytology, no evidence of HPV 16/18
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105 infection, and a completely normal colposcopic impression ( i.e. ,no acetowhitening, metaplasia, or other visible abnormality, and a fully visualized squamocolumnar junction), untargeted (ran-dom) biopsies are not recommended and patients with acompletely normal colposcopic impression can be observed with-out biopsy. To ensure that CIN 2+ is not missed, the ASCCP Colposcopy Standards emphasize the need for biopsies even when the colposcopic impression is normal but any degree of acetowhitening, metaplasia, or other abnormality is present.
2012 Principles Carried Forward
The primary goal of screening and management is cancer prevention through detection and treatment of cervical precancer.
Numerous population-level studies indicate that incidence and mortality from cervical cancer decrease as detection and treatment of high-grade histologic cervical abnormalities (generally defined as CIN 2+) increases. 20,21 Timely detection and treatment of the highest grade of precancers (CIN 3/AIS) have been the bench-mark used for previous guidelines 3 and remain the primary goal of the 2019 management guideline; a secondary goal (because of the relative rarity of this finding in the United States) is early diagnosis of cervical cancer to reduce related morbidity and mor-tality. A patient's risk of having or developing CIN 3+ is estimated based on current and previous results, as well as history of previ-ous precancer treatment. Management recommendations are guided by risk thresholds. 19 Recommendations of routine screen-ing, 1- or 3-year surveillance, colposcopy, or treatment each cor-respond to a risk stratum. These risk strata (ranges of risk for CIN 3+) are defined by Clinical Action Thresholds that were deter-mined through the consensus process (Section E).
Guidelines apply to all individuals with a cervix. Guidelines apply to women and transgender men with a cervix, including indi-viduals who have undergone supracervical hysterectomy. Risk esti-mates were validated in individuals of diverse racial, ethnic, and socioeconomic backgrounds and shown to be comparable. 6Though not the primary focus of the 2019 guidelines, management recommendations are also provided for patients who have under-gone hysterectomy with removal of the cervix and who have a pre-vious diagnosis of histologic HSIL, CIN 2, CIN 2/3, CIN 3, and/or AIS, irrespective of whether the hysterectomy was performed for precancer treatment or another indication.
Equal management for equal risk. History and current test results are used to calculate a patient's current and future risk of CIN 3+. Similar risks are managed similarly, regardless of the combination of results/history used to estimate the risk.
Balancing benefits and harms. Providing the best care means balancing cancer prevention with overtesting and overtreat-ment. Preventing all cervical cancers is unfortunately not an achiev-able goal. Interventions to prevent cervical cancer can cause harm. The 2019 guidelines are designed to maximize cervical cancer pre-vention and minimize harms from overtesting and overtreating by managing patients according to their current and future risks of CIN 3+. High-risk patients require closer follow-up to maximize detection of CIN 3+, whereas low-risk patients require fewer tests and procedures.
Guidelines apply to asymptomatic patients that require management of abnormal cervical screening test results. Patients with symptoms such as abnormal uterine or vaginal bleeding or a vis-ibly abnormal-appearing cervix require appropriate diagnostic testing as this may be a sign of cancer. 22 This evaluation may in-clude cervical cytology, colposcopy, diagnostic imaging, and cervical, endocervical, or endometrial biopsy. Guidelines cannot cover all clinical situations and clinical judgment is advised, es-pecially in those circumstances which are not covered by the 2019 guidelines.
Guidelines are intended for use in the United States . Ap-propriate management may differ in countries with limited follow-up capabilities, less availability of colposcopy, limited pathology infrastructure, or different views of the trade-offs between cancer risk, cost, and overtesting/overtreatment.
D. METHODS D.1 Process and Timeline
The ASCCP and National Cancer Institute (NCI) established a Memorandum of Understanding in January 2017 to undertake the work of this guideline update. As with the previous 2001, 2006, and 2012 guidelines, 1–3 NCI produced risk data and other scientific support for the consensus guideline process. The ASCCP sponsored the consensus effort to develop and ratify the guidelines. Stakeholder organizations representing best practice in the United States were identified and invited to participate. These included medical professional societies, patient advocacy groups, and federal agencies integral to cervical cancer screening and management of abnormal results (see Table 1). Participation of the stakeholder organizations included identifying organization representatives and, for nongovernment participants, sponsoring their travel to consensus conferences. Representatives from 19 or-ganizations attended the initial meeting in February 2018. At that time, 7 working groups were convened. In previous consensus conferences, working groups considered specific test outcomes (e.g., ASC-US/HPV-positive) and special populations. In contrast, the 7 working groups for the 2019 guidelines were organized with the goal of establishing consensus Clinical Action Thresholds. 1. The treatment group evaluated which risk levels of CIN 3+ war-rant expedited treatment without confirmatory biopsy, as well as addressing treatment-related issues. 2. The colposcopy group considered the threshold for colposcopy referral. 3. The surveillance group created a hierarchy of retesting at shorter intervals than currently recommended for routine
TABLE 1. Participating Organizations
Medical Professional Societies
•ASCCP
•American Academy of Family Physicians
•American Cancer Society
•American College of Nurse-Midwives
•American College of Obstetricians and Gynecologists
•American Society for Clinical Pathology
•American Society of Cytopathology
•College of American Pathologists
•Nurses for Sexual and Reproductive Health
•Nurse Practitioners in Women's Health
•Papanicolaou Society of Cytopathology
•Society of Gynecologic Oncology
•Women Veterans Health Strategic Healthcare Group Patient Advocacy Organizations
•American Sexual Health Association
•Cervivor
•Latino Cancer Institute
•Team Maureen Federal Agencies
•Centers for Disease Control and Prevention
•National Cancer Institute
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106 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .screening with either HPV primary testing or cotesting (5 years) and also examined when patients could return to routine screening. Patients undergoing surveillance include those with minimally abnormal screening results not requiring colposcopy (e.g., HPV-positive Negative for Intraepithelial Lesion or Ma-lignancy [NILM]), after colposcopy with low-grade results, or after treatment for high-grade abnormality. 4. The risk modification group evaluated factors that might change a patient's estimated risk or management, focusing on pregnancy and immunosuppression. 5. The high value care group performed decision analyses related to proposed management strategies and will continue to assess value as the 2019 guidelines are implemented. 6. The new technologies group evaluated laboratory terminology and emerging technologies specifically related to management. 7. The communications group created and reviewed relevant con-tent for public communication to both clinicians and the lay public about the guidelines and the development process. Working groups were composed of 2 to 8 members, includ-ing representatives of participating stakeholder organizations, content experts, and nonclinician representatives of patient ad-vocacy organizations. Working groups met regularly from sum-mer 2018 through fall 2019 to review data and develop guidelines for management. The consensus process was overseen by a 23-person steering committee convened by the ASCCP and was directed by a leadership team consisting of 1 NCI representative (M.S.) and 2 ASCCP representatives (R.G., R.P.). Because the guidelines represent a paradigm shift, the guidelines process in-cluded a deliberate and extensive process of stakeholder engage-ment. These included patient and provider surveys, a consensus meeting to review preliminary guidelines, and a 6-week open public comment period before the final consensus voting meeting in October 2019. 23
D.2 Choice of CIN 3+ as Main Clinical End Point for Risk Estimates
For the management guidelines, we chose CIN 3+ as the best surrogate for cancer risk. The definition of CIN 3+ as used in these guidelines includes CIN 3, AIS, and the rare cases of invasive cervical cancer that are found in screening programs. These management guidelines consider CIN 3+ risk at the time point relevant for the clinical action being considered —Clinical Action Thresholds for colposcopy and treatment consider im-mediate risks of CIN 3+, whereas longer-term surveillance rec-ommendations use 5-year risks. CIN3+ was chosen as an endpoint instead of cancer because cancer is uncommon in the United States, and risk is profoundly decreased by precursor treatment. Cancers that are found in robust screening programs may represent cancers already prevalent at first screening, rare instances of aggressive or HPV-negative tu-mors not detectable by screening, or false negative results. 24 CIN 3+ was chosen instead of CIN 2+ because it is a more path-ologically reproducible diagnosis, 25 the HPV type distribution in CIN 3+ lesions more closely approximates that of invasive cervi-cal cancers than the larger range of types found in CIN 2, 15 –18,26 and CIN 2 has appreciable regression rates in the ab-sence of treatment. 27 –29 The choice of CIN 3+ does have some limitations, as even among CIN 3/AIS lesions, risks of progres-sion to cancer differ. Glandular lesions including AIS, lesions with HPV 16 and 18 infections, and those occurring in older pa-tients have higher cancer risks than HPV-negative lesions and those occurring in younger patients. 30 Different nomenclatures for cervical histopathology are in use in the United States. The LAST Project and the WHO recommend a 2-tiered terminology (histologic LSIL/HSIL) for reporting histo-pathology of HPV-associated squamous lesions, similar to the Bethesda system used for reporting cervical cytology. 31,32 How-ever, the CIN nomenclature is still commonly used, and data used to generate this set of guidelines relied on CIN nomenclature. Al-though no direct correlation is possible without use of the p16 bio-marker, histologic HSIL is similar but not identical to CIN 2/3. 33
D.3 Multiple Data Sets Used to Validate Risks
Prior guidelines relied heavily on a large prospective data set including results of cytology, HPV testing, colposcopy, histology, and follow-up outcomes from KPNC, which adopted triennial cotesting as standard practice in 2003. The KPNC data continue to be the largest, most comprehensive data source in the United States for risk estimation of combinations of HPV DNA testing and cytology. For the 2019 guidelines, several additional databases were analyzed to ensure that results are applicable to patients of di-verse racial, ethnic, and socioeconomic strata. Risk estimates were compared using screening and follow-up data from clinical trials (BD Onclarity registrational trials), 34,35 a state registry (New Mexico HPV Pap Registry 36,37 ), and the Centers for Disease Control and Prevention's (CDC's) National Breast and Cervical Cancer Early Detection Program, a national program that includes many low-income and minority patients. 38 The populations vary in rates of abnormal screening results and the prevalence of CIN 3+. Nonetheless, the comparison showed that the risks of CIN 3+ for the specific combination of current results and screening history were similar in that they fell within the same risk bands for management. Cheung et al 6 demonstrates the similarity of CIN 3+ risks associated with screening test result combinations among the different populations of screened patients from these data sets.
In summary, different populations within the United States have higher or lower rates of CIN 3+ due to factors including access to screening and HPV infection prevalence. Nonetheless, patients with similar test results and screening history combinations have largely similar CIN 3+ risk, regardless of their geographic location, race, ethnicity, or socioeconomic status.
D.4 Estimation of Risks
Details of how risks of CIN 3+ were calculated for the many combinations of test results, including longitudinal series of tests over time, are described in the accompanying Methods article. 6In brief, for each combination of past and current test results, the risk of CIN 3+ was estimated using prevalence-incidence mixture models, 39 which consist of joint estimation of prevalent CIN 3+ at the time of the current testing using a logistic regression model, and incident CIN 3+ at subsequent testing using a proportional hazards model. These joint models are designed to handle verifi-cation bias and interval censoring. Verification bias in this context means that histopathologic outcomes are only available for pa-tients referred to colposcopy; thus, CIN3+ cases that occur in the setting of false negative screening or abnormal screening tests that were not referred for colposcopy will not be detected. Interval censoring in this context means that the CIN 3+ is diagnosed at colposcopy visits, but the actual time of onset of incident CIN 3+ cannot be determined as it is typically asymptomatic and oc-curs between testing visits. These flexible models are designed to provide risk estimates without forcing the data into a rigid dis-tribution assumption ( e.g. , Weibull).
D.5 Assigning Combinations of Test Results to Clinical Actions
For each combination of current test results and screening history (including unknown history), recommended management
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107 was determined by first estimating immediate and 5-year risk of CIN 3+. The estimated risk was compared with the proposed Clinical Action Thresholds to determine management recommendation, un-der the principle of “equal management for equal risk. ” For example, HPV-positive ASC-US and LSIL cytology have very similar risks of CIN 3+ and are therefore managed similarly. For some rare combina-tions of test results, too few patients developed CIN 3+ to estimate risk with statistical certainty. In these situations, a combination of published literature, previous guidelines, and expert consensus opin-ion were used to develop recommendations.
D.6 Rating the Recommendations
Recommendation strength (A –E) and quality of evidence (I –
III) were graded using the system that has been used for previous consensus guidelines (Table 2). Two types of evidence were con-sidered to be strong enough to permit a level A recommendation: (a) systematic literature reviews of trials and observational studies, evaluated by the new technologies group using the QUADAS-2 adapted criteria to inform risk estimates for the guidelines 40 and (b) reliable risk estimates from the KPNC prospective longitudinal cohort study. Reliable point estimates are defined as having an 80% certainty of falling within the risk bounds for the recom-mended management (based on the standard errors of the immedi-ate and 5-year risk estimates) ( e.g., colposcopy and surveillance respectively) 6 High-quality evidence from systematic reviews and reliable risk estimates from KPNC are considered level 2 evidence. Strong recommendations against a management option (level E) rarely had substantial evidence because the obvious risk of harm precluded a clinical trial ( e.g., endometrial biopsy in pregnancy). When neither primary data nor literature provided high-level ev-idence, previous guidelines or newly developed expert consensus opinions were used (level 3 evidence), usually leading to a C recom-mendation. Some recommendations are endorsements of guidelines from other organizations, which were not rated. When considering specific guideline recommendations, each group reviewed evidence derived from systematic reviews of published evidence and primary data from the KPNC cohort, assessed the strength and consistency of this evidence, and made recommendations based on quality of data and a balance of benefits and harms.
E. PARADIGM SHIFT: CLINICAL ACTION THRESHOLDS
This section explains the paradigm shift from results-based to risk-based guidelines. We describe the primary Clinical Action Thresholds on which management recommendations are based and the clinical situations in which these Clinical Action Thresh-olds are applied. For most abnormal screening results and subse-quent management visits, the recommendations are based on risks estimated and validated by prospective data from large co-horts. Clinicians can use the 2019 guidelines to manage their pa-tients via the tables in Egemen et al 5 or by using an app or website designed to facilitate navigation of the tables available at http:// www.asccp.org, including a no cost version. Sections G to K de-scribe recommendations for rare clinical situations where man-agement is based on factors other than risk estimates. Management recommendations are based on Clinical Action Thresholds and correspond to risk strata (see Figure 1):
• The 5-year return Clinical Action Threshold approximates the risk for a patient after a negative screening test using HPV test-ing or cotesting in the general population, for whom retesting in 5 years is recommended by national screening guidelines. 13,14 Patients with risks at or below this threshold are recommended to receive routine screening at 5-year intervals with HPV-based testing (Section E.1).
• The 3-year return Clinical Action Threshold approximates the risk for a patient after a negative cervical cytology screen in the general population, for whom retesting in 3 years is recom-mended by national screening guidelines. 13,14 Patients with risks at or below this threshold but above the 5-year threshold are rec-ommended to receive HPV-based testing in 3 years (Section E.1).
• One-year return is recommended for patients with risks above the 3-year threshold but below the Clinical Action Threshold for colposcopy (Section E.1).
• The colposcopy Clinical Action Threshold approximates the risk for a patient after an HPV-positive ASC-US or LSIL screening result in the general population, for whom colposcopy is recom-mended in the 2012 guidelines. 3 Patients with risks at or above this threshold but below the expedited treatment threshold are rec-ommended to receive colposcopy (Section E.2).
• The expedited treatment or colposcopy acceptable Clinical Action Threshold approximates the risk for a patient after an HPV-positive atypical squamous cells cannot exclude HSIL (ASC-H) cytology screening result in the general pop-ulation. Patients with risks at or above this threshold but below the expedited treatment preferred threshold are recommended to receive counseling from their providers to choose between evaluation with colposcopy and biopsy or expedited treatment (Section E.3). Expedited treatment is defined as treatment with-out confirmatory colposcopic biopsy.
• The expedited treatment preferred Clinical Action Threshold approximates the risk for a patient after an HPV 16 –positive HSIL cytology screening result in the general population. It is preferred that patients with risks at or above this threshold re-ceive expedited treatment unless they are pregnant, younger than 25 years, or have concerns about the potential effects of
TABLE 2. Rating the Recommendations
Strength of recommendation A. Good evidence for efficacy and substantial clinical benefit sup-port recommendation for use. B. Moderate evidence for efficacy or only limited clinical benefit supports recommendation for use. C. Evidence for efficacy is insufficient to support a recommendation for or against use, but recommendations may be made on other grounds. D. Moderate evidence for lack of efficacy or for adverse outcome supports a recommendation against use. E. Good evidence for lack of efficacy or for adverse outcome supports a recommendation against use. Quality of evidence I. Evidence from at least one randomized, controlled trial. II. Evidence from at least one clinical trial without randomization, from cohort or case-controlled analytic studies (preferably from more than one center), or from multiple time-series studies, or dramatic results from uncontrolled experiments. III. Evidence from opinions of respected authorities based on clinical experience, descriptive studies, or reports of expert committees. Terminology used for recommendations Recommended. Good data to support use when only one option is available Preferred. Option is the best (or one of the best) when there are multiple options Acceptable. One of multiple options when there is either data indicating that another approach is superior or when there are no data to favor any single option Not recommended. Weak evidence against use and marginal risk for adverse consequences Unacceptable. Good evidence against use
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108 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .treatment on future pregnancy outcomes that outweigh concerns about cancer (Section E.3).
• Of note, patients with histologic HSIL (CIN 2) who have chosen observation are recommended to receive colposcopy and HPV-based testing at 6-month intervals (Section I.3).
E.1 Clinical Action Thresholds Leading to Recommendation of Surveillance
Introduction. Surveillance is defined as follow-up testing at a shorter interval than that currently recommended for routine screening with either HPV primary testing or cotesting (5 years). Surveillance is recommended for patients whose risk of CIN 3+ based on current test results and screening history is higher than the risk for the general screening population, but lower than the risk at which colposcopy is recommended. Unlike colposcopy and treatment, which are performed as soon as possible after a qualifying abnormal result, surveillance entails retesting at intervals of 1 to less than 5 years. Therefore, we used the 5-year risk of CIN 3+ as the estimated risk level when assigning surveillance Clinical Action Thresholds. Surveillance intervals are defined in Figure 1 and explained in detail hereinafter. Surveillance thresholds are based on the principle of equal management for equal risks and were designed to support current screening and surveillance recommendations, which are generally accepted as a reasonable balance of benefits and harms. 3 In the 2012 guidelines, intervals of 1 and 3 years were used for surveillance, with return to routine HPV-based screening at 5 years. 3 Because clinicians and patients are familiar with these intervals, and review of evidence did not reveal a compelling reason to change these intervals, these intervals are retained. Note that for observation in very high-risk patients ( e.g., untreated CIN2, AIS treated with conization) colposcopy and repeat testing at 6-month intervals is recommended.
Guideline: When patients have an estimated 5-year CIN 3+ risk of less than 0.15% based on past history and current test re-sults, return to routine screening at 5-year intervals using HPV-based testing is recommended (AII).
Rationale : Using the principle of equal management for equal risk, this Clinical Action Threshold corresponds to the 5-year CIN 3+ risk after negative HPV-based screening (HPV testing or cotesting) in the general population (see Table 1A in Egemen
et al 5) for whom national guidelines recommend a 5-year re-turn. 13,14 Estimated 5-year CIN 3+ risks in the KPNC database after a negative HPV test and cotest are 0.14% (95% CI = 0.13% –
0.15%) and 0.12% (95% CI = 0.12% –0.13%), respectively. Note that cytology alone is never recommended at 5-year intervals.
Guideline: When patients have an estimated 5-year CIN 3+ risk of 0.15% or greater but less than 0.55% based on history and current test results, repeat testing in 3 years with HPV-based testing is recommended (AII).
Rationale: Using the principle of equal management for equal risk, the 3-year return Clinical Action Threshold corre-sponds to the 5-year CIN 3+ risk after negative cervical cytology in the general population, for whom national guidelines recom-mend a 3-year return. 13,14 Estimated 5-year CIN 3+ risks after a negative cytology result without HPV testing ranged from 0.33% in the KPNC population to 0.52% in the New Mexico HPV Pap Registry, to an estimated 0.45% in the screened popula-tion of the CDC's National Breast and Cervical Cancer Early De-tection Program. Thus, 0.55% was considered an appropriate
FIGUR'E 1. This figure demonstrates how patient risk is evaluated. For a given current results and history combination, the immediate CIN 3+ risk is examined. If this risk is 4% or greater, immediate management via colposcopy or treatment is indicated. If the immediate risk is less than 4%, the 5-year CIN 3+ risk is examined to determine whether patients should return in 1, 3, or 5 years.
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109 value for the Clinical Action Threshold. Three-year surveillance is recommended for patients whose risk falls between the 3- and 5-year follow-up thresholds. Consistent with the 2012 guidelines, patients with a low-grade cotest result ( e.g., HPV-positive ASC-US or LSIL) followed by a colposcopy with results of less than CIN 2, followed in turn by a negative follow-up HPV test or cotest reach the 3-year return threshold (see Figure 2). Also consistent with pre-vious guidelines, patients with an HPV-negative ASC-US screening result in the setting of an unknown history can return at 3 years (estimated 5 year CIN 3+ risk 0.40%). 5
Guideline: When patients have an estimated risk of CIN 3+ based on history and current results that is below the threshold for immediate colposcopy (4.0% immediate risk) and above the 3-year follow-up threshold ( ≥0.55% at 5 years), repeat test-ing in 1 year with HPV-based testing is recommended (AII).
Rationale: One-year surveillance implies close follow-up for those whose risks fall between the Clinical Action Thresholds for colposcopy and 3-year follow-up. Consistent with the 2012 consen-sus recommendations, 3 follow-up at 1 year is recommended after screening tests showing minimal abnormalities: HPV-positive/ NILM or HPV-negative/LSIL with unknown previous screening history (immediate risks 2.1% and 1.1% respectively 5); 1-year sur-veillance is also recommended after colposcopy with biopsies of histologic LSIL (CIN 1) or less preceded by a low-grade cotest result (defined as HPV-positive LSIL, HPV-positive ASC-US, or repeated HPV-positive NILM). New data for these guidelines find that the risk of CIN 3+ is substantially reduced after a doc-umented negative HPV primary screening test or cotest or nor-mal colposcopic examination with biopsy confirmation of less than CIN 2. 5 Based on lower CIN 3+ risks, 1-year surveillance, not colposcopy, is recommended for most patients with new HPV-positive ASC-US or LSIL results after a documented neg-ative HPV test or cotest within an appropriate screening interval (approximately 5 years) or colposcopic examination less than CIN 2 within the past year (see Figure 2). Of note, a previous negative cytology result alone does not reduce subsequent risk like a negative HPV-based screen; therefore, cytology alone is not used to modify subsequent management recommendations.
E.2 Clinical Action Threshold Leading to Recommendation of Colposcopy
Guideline: When patients have an estimated immediate risk of diagnosis of CIN 3+ of 4.0% or greater based on history and current results, referral to colposcopy is recommended (AII).
Rationale: The following principles were used to develop the Clinical Action Threshold for referral to colposcopy: ( a) colposcopy visits recommended by the threshold should yield information useful for clinical decision-making. Thus, the threshold was based on the
FIGURE 2. This figure demonstrates how a patient with a common low-grade screening abnormality (HPV-positive ASC-US) would be managed based on risk estimates. The initial screening result would lead to colposcopy (immediate risk 4.2%). Colposcopy of less than CIN 2 has a 5-year risk of 3.2% (1-year return). At the 1-year return visit, a second HPV-positive ASC-US result has an immediate risk of 3.1% (1-year return). If the patient has a repeat abnormal screen at the next follow-up, colposcopy is recommended. If the HPV-based test is negative, return in 3 years is recommended. NA, not applicable because stable risk estimates are not available.
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110 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .risk of diagnosing CIN 3+ upon immediate referral to colposcopy. ( b)In the absence of a compelling rationale, the colposcopy threshold should be similar to 2012 referral recommendations that are generally accepted as an appropriate balance of benefits and harms. The 2001 consensus guidelines 1 were the first to standardize the colposcopy referral threshold, referring patients with LSIL and HPV-positive ASC-US to colposcopy. This recommendation has been car-ried forward through revisions in 2006 and 2012. 2,3 The workgroup reviewed frequently cited studies and noted that immediate risk (CIN 3+ found among patients referred directly to colposcopy) ranged from 3% to 7%. 41 –44 Current KPNC data were reviewed, 5and it was noted that immediate CIN 3+ risk clustered in 3 groups: (a) high-grade test results (defined as cytology ASC-H, atypical glandular cell [AGC], HSIL, or higher) having high (>25%) risk; (b) low-grade results (HPV-positive ASC-US or HPV-positive LSIL cytology with unknown previous screening history and HPV-positive NILM cytology occurring at 2 consecutive annual visits) having just over 4.0% risk; and ( c) result combinations for which colposcopy has historically not been performed having risks below 4% (HPV-positive NILM cytology, HPV-negative LSIL cytology, and HPV-negative ASC-US cytology with unknown previous screening histories). The Clinical Action Threshold of a 4% immediate CIN 3+ risk was considered a reasonable balance of benefits and harms as, in a population with unknown screening history, it led to referral of HPV-positive patients with ASC-US or LSIL cytology, but not the large group of patients with HPV-positive NILM cytology. To validate the 4.0% Clinical Action Threshold for colpos-copy, the KPNC CIN 3+ prevalent risk estimates were compared with those from other study populations with more diversity in sociodemographic characteristics including the New Mexico HPV Pap Registry, 45 CDC's National Breast and Cervical Cancer Early Detection Program, and the BD Onclarity registrational trials. The 4% threshold functioned similarly. 3,6 The 4.0% immediate risk Clinical Action Threshold has im-portant implications for patients with at least 1 previous negative HPV-based test because surveillance is recommended rather than immediate colposcopy for low-grade abnormalities (HPV-positive ASC-US or LSIL) in patients whose preceding screening result was a negative HPV test or cotest within a routine screening inter-val (approximately 5 years). 5 This additional information reduces the immediate CIN 3+ risk to approximately 2%, leading to a rec-ommendation of 1-year surveillance instead of immediate colpos-copy. Adoption of the 4.0% Clinical Action Threshold reduces the number of patients referred for colposcopy over 2 rounds of screen-ing from an estimated 9.8%, using the 2012 ASCCP recommenda-tions, to 8.3% using the 2019 recommendations. Exceptions to the 4.0% threshold, encompassing results with cancer risk dispropor-tionately higher than CIN 3+ risk, are discussed in Section H.2.
E.3 Clinical Action Thresholds Leading to Recommendations of Treatment
The primary goal of treatment is cancer prevention through destruction or excision of precancerous lesions (CIN 3, AIS) to prevent the development of invasive cancer. In the only known ob-servational study of untreated CIN 3, the long-term risk of develop-ing invasive cancer was as high as 30% for 30 years 46 ; progression rates could not be estimated at KPNC because of high rates of timely treatment. Because treatment is generally recommended as soon as possible after the identification of a precancerous lesion, the immediate CIN 3+ risk was used when evaluating potential thresholds. Historically, the treatment threshold has been histologic CIN 2. The LAST guidelines reports both p16-positive CIN 2 and CIN 3 as histologic HSIL. Consistent with previous guidelines, the threshold for treatment remains histologic HSIL/AIS (by LAST terminology) or CIN 2+ (by 3-tiered terminology) except in special circumstances (Sections I.3, K.1, and K.2). When considering ex-pedited treatment versus colposcopy with biopsy, clinicians should have a thorough discussion with patients regarding the risks and benefits. Treatment without preceding histologic confirmation can be conducted in one visit among those at high immediate risk of CIN 3+. Reasons for choosing expedited treatment vary and may include personal preference, limited healthcare access, finan-cial concerns, and cancer-related anxiety. The age cutoff of 25 years or older for recommending expedited treatment was cho-sen as an appropriate balance of benefits and harms due to very low cancer rates and high rates of regression of precancers among women in this age group. 27,47
Guideline: For nonpregnant patients 25 years or older with an estimated immediate risk of CIN 3+ of 60% or greater based on history and current results, treatment using an excisional proce-dure without previous biopsy confirmation is preferred but col-poscopy with biopsy is acceptable (BII).
Rationale: In the 2012 guidelines, expedited treatment ( i.e. ,without biopsy confirmation) was an acceptable management op-tion for HSIL cytology. 3 Patients with HSIL cytology undergoing expedited treatment are diagnosed with CIN 3+ in 49% to 75% of cases. 48 –52 The KPNC data show similar risks: HPV-positive HSIL cytology has immediate risks of CIN 3+ and CIN 2+ of 49% and 77%, respectively. 5 Two clinical situations currently ex-ceed the 60% threshold where expedited treatment is preferred. HSIL cytology that is HPV 16 –positive has an immediate CIN 3+ of 60%, CIN 2+ risks of 77%, and immediate cancer risks of 8.1%. 53 In the CDC's National Breast and Cervical Cancer Early Detection Program, women with HPV-positive HSIL cytology (re-gardless of genotype) who were underscreened (generally defined as no screening in >5 years) had an immediate CIN 3+ risk of 64% and CIN 2+ risks of 82% (cancer risk not available). Based on the KPNC data, for clinical situations that exceed the 60% threshold, 1.7 patients will receive diagnostic excisional procedures for every CIN 3+ treated, a low rate of overtreatment.
Guideline: For nonpregnant patients 25 years or older with an estimated immediate risk of CIN 3+ 25% or greater and less than 60% based on history and current results, treatment using an excisional procedure without previous biopsy confirmation or histologic evaluation with colposcopy and biopsy are both acceptable (AII).
Rationale: The 2012 guidelines allow treatment without biopsy-proven histologic confirmation include patients who have HSIL cytology independent of HPV status. In the KPNC data set, the 25% to 59% risks strata includes patients with the fol-lowing results and immediate CIN 2+/CIN 3+ risks, respectively: (a) HPV-negative HSIL cytology: 47%/25%; ( b) HPV-positive ASC-H cytology: 50%/26%; ( c) HPV-positive AGC (all catego-ries): 40%/26%; and ( d) HPV-positive HSIL cytology: 77%/ 49%. Using this threshold, 2.8 patients will undergo excisional procedures for every CIN 3+ treated.
E.4 Clinical Situations Leading to Management Recommendations
Patients with abnormal cervical cancer screening results enter management via 5 common clinical situations: ( a) initial management of an abnormal screening test result (see Tables 1A, B; Egemen et al 5); ( b) return visit for surveillance of a previous abnormal result that did not lead to colposcopy referral ( e.g.,
HPV-negative ASC-US), with consideration of whether to continue surveillance or refer to colposcopy (see Tables 2A –C; Egemen
et al 5); ( c) evaluation of the colposcopic biopsy results with con-sideration of whether to treat or begin postcolposcopy surveil-lance (see Table 3; Egemen et al 5); ( d) managing test results at
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111 the return visit for surveillance after a colposcopic biopsy showing less than CIN 2 (Tables 4a, b; Egemen et al 5); and ( e) follow-up af-ter treatment of CIN2 or CIN3 (see Tables 5a, 5b; Egemen et al 5). Recommendations are based on risks of immediate and future CIN 3+ diagnoses in light of current and past results. Regardless of the pathway by which patients enter management, equivalent risks are managed similarly. For each of the 5 clinical situations, risk ta-bles and recommendations based on the Clinical Action Thresholds are detailed in the accompanying article by Egemen et al .5 The reader is directed to the definitive updated source of risk tables, which are freely available online ( RiskTables). A small percentage of patients will present with a combination of results and personal characteristics requiring con-sideration outside of the available risk data. Management of these special situations is described in Sections G to K.
F. UPDATES RELATED TO PATHOLOGY REPORTING AND LABORATORY TESTS
Although most of the 2019 guidelines describe clinical man-agement of patients by providers, the consensus process also ad-dressed laboratory considerations that directly relate to results reporting and use of ancillary tests.
F.1 Statement on the Use of a 2-Tier Terminology (Histologic LSIL/HSIL) for Reporting Histopathology of Squamous Lesions of the Lower Anogenital Tract
Guideline: It is important to use p16 immunohistochemical staining according to the guidance provided by the CAP-ASCCP LAST Project. 31 p16 immunohistochemistry should be used for specific indications as recommended by the LAST guidelines when interpreting the hematoxylin and eosin (H&E) slide. A pos-itive p16 immunostain supports the diagnosis of histologic HSIL if the morphological assessment of H&E slides is consistent with CIN 2 or CIN 3. There is a risk of overcalling cervical histology results when p16 is used incorrectly. Most importantly, a mor-phologic CIN 1 on H&E should not be upgraded to histologic HSIL (CIN 2) even if p16 positive. For epidemiologic and clinical management purposes, it is strongly recommended to qualify a histologic HSIL result by CIN 2 or CIN 3, according to the options given by the LAST guidelines (example histologic HSIL [CIN 2]).
Rationale: This CIN qualification can have clinical importance (e.g., to identify cases of CIN 2 in patients for whom conservative management is an acceptable option). It is also important for postvaccine surveillance studies and quality control assessments of cervical precancer that have historically relied on CIN 2 and CIN 3 end points. Furthermore, it is important for future research efforts to distinguish diagnoses of histologic HSIL (CIN 2) from HSIL (CIN 3) so that diagnostic categories are compatible with the histologic end points used for current guidelines. In 2012, consensus recommendations were published on the use of a 2-tiered terminology for reporting histopathology of squa-mous lesions of the anogenital tract by the College of American Pa-thologists and the ASCCP. 31 The central components of the LAST guidelines include a 2-tiered nomenclature that distinguishes histo-logic LSIL and histologic HSIL and recommendations for the use of adjunctive p16 immunohistochemistry to assist interpretation of anogenital histology. p16 is a tissue marker of HPVoncogene overex-pression and transformation and can support histologic assessment. Current guidelines are based on CIN 3 end points, the most reliable correlate of a cervical precancer. Currently, there are insuffi-cient data to evaluate risk estimates with histologic HSIL end points. Recent studies have shown that distinguishing CIN 2 and CIN 3 within the LAST histologic HSIL group is biologically and clinically meaningful. 33 Although some studies have shown that p16 immuno-histochemistry improves interpretation of cervical biopsies, others have raised concerns about overuse and overdiagnosis. 54 –59
F.2 Updated Management of Primary HPV Screening (Replaces Interim Guidance)
Guideline: When primary HPV screening is used, performance of an additional reflex triage test ( e.g., reflex cytology) for all positive HPV tests regardless of genotype is preferred (this includes tests pos-itive for genotypes HPV 16/18) (CIII). However, if primary HPV screening test genotyping results are HPV 16 or HPV 18 positive and reflex triage testing from the same laboratory specimen is not fea-sible, referral for colposcopy before obtaining additional testing is ac-ceptable (CIII). If genotyping for HPV 16 or HPV 18 is positive, and triage testing is not performed before the colposcopy, collection of an additional triage test ( e.g., cytology) at the colposcopy visit is recommended (CIII).
Rationale: The US FDA approved the cobas HPV test (Roche, Indianapolis, IN), in March 2014, and the Onclarity HPV Test (Becton Dickinson, Franklin Lakes, NJ), in April 2018, for primary HPV testing for screening for patients 25 years or older. 60 Both these tests offer and are approved for partial HPV genotyping. Use of primary HPV screening will likely increase in the future, as it is more effective than screening with cytology alone and performs similarly to and with lower costs than screening with cotesting. 4,42 Because HPV –16 positive and HPV 18 –positive test results have the highest risk of CIN 3 and occult cancers, ad-ditional diagnostic procedures are recommended for all positive test results ( e.g., colposcopy with biopsy for NILM and low-grade cytology and expedited treatment for HSIL cytology that is positive for HPV type 16). This guideline replaces interim guidance (2015) for the management of a positive result for HPV primary screening, which recommended direct referral to colposcopy for HPV test results positive for HPV 16 and/or HPV 18, and performance of cytology for positive results due to other (non-16/18) high-risk HPV types. 4 The immediate risk of CIN3+ in patients with HPV 16 –positive and HSIL cytology ex-ceeds the treatment threshold of 60%; therefore, these patients should be given the option for expedited treatment without preced-ing confirmatory biopsy (see Section E.3). Expedited treatment is only possible if cytology is performed. Therefore, reflex cytology is recommended for all HPV-positive primary screening results, re-gardless of HPV genotype. If reflex testing from the same labora-tory specimen as the HPV test is not feasible, patients should proceed directly to colposcopy. 4 In this situation, collection of an additional triage test ( e.g., cytology) is recommended at the time of colposcopy to provide further information for risk-based management ( e.g., if HPV 16 –positive HSIL cytology is identified, treatment may be considered even if CIN 2+ is not identified on biopsy). Combining a test with high specificity (e.g., cytology when it is interpreted as HSIL) with a test with high sensitivity ( i.e. , HPV test) allows more precise, risk-based management of these patients.
F.3 Statement on HPV Tests Used in Management
Guideline: HPV assays should be used for management ac-cording to their regulatory approval for screening, unless there are sufficient data to support use of the assay differently (AI).
Rationale: Several HPV assays have been approved in the United States for clinical use in screening and triage. 61 None of these assays have specific indications for management, but they are widely used for postcolposcopy and posttreatment surveil-lance. For these indications, HPV assays approved for screening should be used according to their regulatory approval. For example, when an HPV test has been approved for cotesting, it should be
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112 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .used in management in the context of cotesting, unless there are suf-ficient, exceptionally rigorous data to support use of the assay dif-ferently ( e.g., as outlined in Clarke et al .40 ). Approved assays include target- and signal-amplification assays of HPV DNA, as well as HPV mRNA. Most FDA-approved HPV DNA assays have similar performance characteristics. 62 Most assays are approved for adjunct testing with cytology (also referred to as cotesting), whereas a subset of HPV DNA assays have also been approved for primary HPV testing alone, without concomitant cytology.
G. RARE CYTOLOGY RESULTS G.1 Evaluation of Cytology Interpreted as AGC or AIS
Guideline: For nonpregnant patients of all ages with all sub-categories of AGC and AIS, except when atypical endometrial cells are specified, colposcopy is recommended regardless of HPV test result; endocervical sampling is recommended at initial colposcopy except in pregnancy (for management in pregnancy, see Section K.2) (AII). Accordingly, triage by reflex HPV testing is not recommended, and triage by repeat cytology is unacceptable (DII). Endometrial sampling is recommended in conjunction with colposcopy and endocervical sampling in nonpregnant patients 35 years or older with all categories of AGC and AIS (AII). Endome-trial sampling is also recommended for nonpregnant patients younger than 35 years at increased risk of endometrial neoplasia based on clinical indications ( e.g., abnormal uterine bleeding, con-ditions suggesting chronic anovulation, or obesity) (AII). For patients with atypical endometrial cells specified, initial evaluation limited to endometrial and endocervical sampling is preferred, with colposcopy acceptable at the time of initial evaluation. If colposcopy was deferred and no endometrial pathol-ogy is identified, additional evaluation with colposcopy is then recommended (see Figure 3).
Subsequent Management. Guideline: For patients with cytology showing AGC not otherwise specified or atypical endocervical cells not otherwise specified in whom histologic HSIL (CIN 2+) or AIS/cancer is not identified, cotesting at 1 and 2 years is recommended. If both cotests are negative, repeat cotesting at 3 years is recommended. If any test is abnormal, then colposcopy is recommended (BII). If CIN 2 or CIN 3 but no glandular lesion is identified histologically for patients with cytology atypical glandular, endocervical, or endometrial cells not otherwise specified, management should be according to the 2019 guidelines for the lesion diagnosed (Section I) (CII). For patients with atypical glandular or endocervical cells “favor neoplasia ” or endocervical AIS cytology, if invasive disease is not identified during initial colposcopic workup, a diagnostic excisional procedure is recommended. The diagnostic excisional procedure used in this setting should provide an intact specimen with interpretable margins (BII). Endocervical sampling above the excisional bed is preferred (BII) (see Figure 4).
Rationale: Atypical glandular cells on cytology is a poorly reproducible diagnostic category. 63 Positive HPV test results, es-pecially when positive for HPV type 18, can be indicative of higher risk of CIN 2+ lesions. However, colposcopy is recom-mended for all patients regardless of HPV result. Literature is lim-ited, and comparisons between studies are difficult because of inconsistent use of the Bethesda system for classification of AGC. 64 Atypical glandular cells can be associated with polyps and metaplasia as well as adenocarcinomas of the cervix; cancers of the endometrium, fallopian tube, ovary, and other sites are also found, especially in older women who test HPV negative. 65,66 Using the Bethesda terminology, AGC, favor neoplasia, or adeno-carcinoma cytology is frequently indicative of invasive or preinvasive disease. 64 For this reason, diagnostic excisional proce-dures are recommended even when histologic HSIL or AIS has not been identified. Cytologic AGC results are associated with a histologic diagnosis of AIS in 3% to 4%, CIN 2+ in 9%, and inva-sive cancer in 2% to 3%. 67 –69 In the KPNC data, HPV-positive AGC (all categories) had an immediate CIN 3+ risk of 26% and HPV-negative AGC had an immediate CIN 3+ risk of 1.1%. Con-sistent with other literature, cotest results of HPV-positive AGC fa-vor neoplasia or adenocarcinoma had an immediate CIN 3+ risk of 55%, whereas other HPV-positive AGC categories had immediate CIN 3+ risks of approximately 20%. Although endometrial cancer is rare in premenopausal patients without risk factors, the prevalence of premenopausal endometrial cancer is increasing, underscoring the importance of endometrial sampling when indicated. 70,71
FIGURE 3. This figure describes the initial workup of AGC found on cervical cytology.
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113 G.2 Unsatisfactory Cytology
Guideline: For patients with an unsatisfactory cytology result and no, unknown, or a negative HPV test result, repeat age-based screening (cytology, cotest, or primary HPV test) in 2 to 4 months is recommended (BIII). Triage using HPV testing is not recom-mended (DIII). Before repeat cytology, treatment to resolve atrophy or obscuring inflammation when a specific infection is present is acceptable (CIII). For patients 25 years and older who are cotested and have unsatisfactory cytology and a positive HPV test without genotyping, repeat cytology in 2 to 4 months or colposcopy is ac-ceptable (BII). If a positive HPV test with partial genotyping is pos-itive for HPV 16 or HPV 18, direct referral for colposcopy is recommended (BII) (see Figure 5).
Rationale: Literature was reviewed from 2012 to 2019, and no evidence was found to change recommendations. 72 –82 When cotesting is performed, a negative HPV test in the setting of an un-satisfactory cytology may reflect an inadequate sample. Although
FIGURE 4. This figure describes follow-up management that should occur after the diagnostic examinations described in Figure 3.
FIGURE 5. This figure describes the steps involved in clinical management of unsatisfactory cytology. Note that “unknown genotype ”refers to both HPV testing without genotyping, and HPV testing where genotyping is negative for HPV 16 and 18 but positive for other high-risk HPV types.
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114 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .a negative HPV test (performed from the same vial as the cytol-ogy) may be adequate for testing even when the cytology cellular-ity is inadequate for diagnosis, interpreting the HPV result in the setting of insufficient cellularity has not been validated, which is of concern given that repeat testing is not recom-mended for up to 5 years after a negative HPV screen. Negative results on HPV tests that are not FDA approved for primary cervical cancer screening should not be considered valid in the absence of adequate cytology (Section F.3). In summary, a negative HPV result from a cotest with inadequate cellularity on cytology should not be interpreted as negative primary HPV test and should be repeated.
G.3 Absent Transformation Zone on Screening Cytology
Guideline: For patients aged 21 to 29 years with negative screening cytology and absent endocervical cells/transformation zone component ( i.e. , endocervical cells or squamous metaplastic cells), routine screening is recommended (BIII). When cervical cytology alone is performed for screening, HPV testing as a triage test after negative cytology and absent endocervical cells/ transformation zone component in this age group is unacceptable (DIII). For patients 30 years or older with NILM cytology and ab-sent endocervical cells/transformation zone component and no or unknown HPV test result, HPV testing is preferred (BIII). Repeat cytology in 3 years is acceptable if HPV testing is not per-formed (BIII). If HPV testing is performed, manage using Clin-ical Action Thresholds according to 2019 consensus guidelines (see Figure 6).
Rationale: Literature reviewed for the 2012 guidelines indi-cated a lower risk of CIN 3+ for patients with absent transformation zone/endocervical cells than those with cells present, leading to a recommendation to manage these results similarly. 3 The HPV test-ing is preferred in women 30 years or older to facilitate subsequent risk-based management. A review of the literature from 2012 to 2019 on whether the absence of a transformation zone component (TZ/EC, i.e. , endocervical cells or squamous metaplastic cells) on NILM cytology slides affected patients' subsequent risks of histo-logic HSIL (CIN 2, CIN 3) diagnoses showed no evidence to change the 2012 recommendations. 83,84
G.4 Benign Endometrial Cells in Premenopausal Patients or Benign Glandular Cells in Posthysterectomy Patients
Guideline: For asymptomatic premenopausal patients with benign endometrial cells, endometrial stromal cells, or histiocytes, no further evaluation is recommended (BII). For postmenopausal patients with benign endometrial cells, endometrial assessment is recommended (BII). For posthysterectomy patients with a cytol-ogy report of benign glandular cells, no further evaluation is recommended (BII).
Rationale: In the Bethesda system for reporting cervical cy-tology, cytologically benign-appearing endometrial cells are re-ported in women 45 years or older under the “other ” general category, and follow-up left to the clinical provider. Benign glandu-lar cells in women after hysterectomy are reported in the negative (NILM) Bethesda category. Literature review for the 2012 guide-lines indicated increased risk of endometrial pathology in postmen-opausal patients with endometrial cells on cytology but did not indicate increased endometrial cancer risk for premenopausal pa-tients with benign endometrial cells in the absence abnormal uterine bleeding. 3 The literature review was updated using a PubMed search for recent publications since 2012 that address benign-appearing endometrial cells in postmenopausal and glandular cells in posthysterectomy individuals. References were reviewed and no evidence was found to change the 2012 recommendations. 85 –93
H. COLPOSCOPY PRACTICE STANDARDS AND EXCEPTIONS TO COLPOSCOPY CLINICAL ACTION THRESHOLD H.1 ASCCP Colposcopy Standards
The ASCCP Risk-Based Management Consensus Guide-lines reaffirm that colposcopy should be practiced according to
FIGURE 6. This figure describes the steps involved in clinical management of cytology that is negative for intraepithelial lesion or malignancy, but with absent transformation zone or endocervical cells.
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115 the ASCCP Colposcopy Standards. 10,94 For those at lowest risk, defined as less than HSIL cytology, no evidence of HPV 16/18 in-fection, and a completely normal colposcopic impression ( i.e. , no acetowhitening, metaplasia, or other visible abnormality, and a fully visualized squamocolumnar junction), untargeted (random), biopsies are not recommended and patients with a completely nor-mal colposcopic impression can be observed without biopsy. For those not meeting the lowest risk criteria, multiple targeted biopsies, at least 2 and up to 4, are recommended targeting all acetowhite areas to improve detection of prevalent precancers. The ASCCP Colposcopy Standards emphasize the need for biopsies even when the colposcopic impression is normal but any degree of acetowhitening, metaplasia, or other abnormality is present to ensure that CIN 2+ is not missed. 94 As more patients are allowed to defer colposcopy under the ASCCP Risk-Based Management Consensus guidelines, obtaining adequate biop-sies to effectively rule out CIN 2+ at each colposcopy examina-tion is paramount. Note that the KPNC colposcopy protocols precede the Col-poscopy Standards and are based on 4-quadrant biopsies and an ECC that were widely conducted in KPNC. The recommendations against untargeted biopsies are based on the risk of occult CIN 2+ of 1% to 7% and CIN 3+ of less than 1% among patients with less than HSIL cytology, HPV 16/18 negative, and normal colposcopic impression. This indicates that management recommendations using the ASCCP Colposcopy Standards would be equivalent to those using KPNC protocols in nearly all cases. The most recent recommendations pertaining to the use of ECC are from the 2012 guidelines, restated here for clarity: ECC is preferred for non-pregnant patients when colposcopy is inadequate, in those not at lowest risk in whom no lesion is identified, and is acceptable when a lesion is seen.
H.2 Exceptions to Colposcopy Threshold
Guideline: For patients with ASC-H cytology, colposcopy is recommended regardless of HPV result (AII).
Rationale: In the KPNC data, HPV-negative ASC-H and HPV-positive ASC-H had very different CIN 3+ rates, but similar cancer rates. The HPV –positive ASC-H had an immediate CIN 3+ risk of 26% and a cancer risk of 0.92%, whereas HPV-negative ASC-H had an immediate CIN 3+ risk of 3.4%, but an immediate cancer risk of 0.69%. Because the immediate cancer risk for ASC-H is disproportionately high compared with the CIN 3+ risk, the working group carried forward the 2012 recom-mendations and recommended colposcopy for all patients with ASC-H, regardless of HPV test results. 3
Guideline: For patients with HPV 18 –positive NILM, colpos-copy is recommended (AII). (Note colposcopy is also recommended for HPV 16 –positive NILM, repeated here for clarity.)
Rationale: HPV 18 –positive NILM had a 3.0% prevalent CIN 3+ risk, less than the Clinical Action Threshold for colpos-copy. However, HPV 18 –positive NILM had a disproportionately high cancer risk compared with other results: 0.2% immediately and 0.56% at 5 years. This suggests that HPV 18-related CIN 3 or AIS may be difficult to diagnose and/or more apt to rapidly progress from precancer to cancer. The elevated cancer prevalence with HPV 18 positivity has been previously noted, 95 and HPV 18 is one of the most common HPV types found in invasive cervical cancers. 96 Given the elevated cancer risk, referral to colposcopy is recommended.
Guideline: Colposcopy should be performed after 2 consec-utive unsatisfactory screening tests (CIII).
Rationale: No new evidence was found, so the 2012 guide-line was carried forward. 3
FIGURE 7. This figure describes the steps involved in clinical management of histologic HSIL.
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116 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .I. MANAGING HISTOLOGY RESULTS Treatment Considerations for Patients 25 Years or Older
Individuals who exceed treatment thresholds may undergo expedited treatment, defined as excisional treatment without pre-ceding histologic confirmation. However, most patients will require both screening test and colposcopic biopsy results to determine the next step in management. The following section outlines guiding principles to consider when managing these results. Treatment guidelines are dichotomized by younger than 25 years or 25 years or older because of high spontaneous regression rates of HPV infec-tion and CIN 2 and low incidence of cancer in those younger than 25 years. Individuals younger than 25 years are discussed under Special Populations (Section K). The term “young women ” is no longer used. The consensus guidelines recognize that patients of various ages are concerned with the potential impact of treatment on future pregnancy outcomes. Shared decision-making is espe-cially critical when individuals consider treatment of histologic HSIL (CIN 2) and abnormalities with a relatively low likelihood of underlying CIN 3+, such as histologic LSIL (CIN 1) preceded by HSIL or ASC-H cytology, or persistent histologic LSIL (CIN 1).
I.1 Management of Histologic HSIL, not Further Specified or Qualified
Histologic reporting of cervical biopsies has moved to the LAST/WHO criteria, but its uptake by pathologists has not been uni-versal. The consensus recommendation of the LAST guidelines (Sec-tion F.1) is to qualify histologic HSIL using the CIN nomenclature (CIN 2 or CIN 3). Because of measurable regression rates for CIN 2, 26 the present guidelines subdivide treatment options based on the CIN qualifiers of CIN 2 and CIN 3. However, pathology reports incorporating the LAST criteria may not specify a CIN diagnosis.
Guideline: Treatment is preferred if histologic HSIL cannot be specified ( e.g., reported as histologic HSIL or histologic HSIL [CIN 2,3]) (CIII) (see Figure 7).
Rationale: CIN 3 is considered a direct cervical cancer pre-cursor. If CIN 3 cannot be excluded, managing the patient as if CIN 3 is present is preferred. This conservative approach was con-sidered safest for patients. Alternatively, the clinician could call the pathologist to further qualify the CIN equivalent and issue an additional report, then manage using the revised diagnosis.
I.2 Management of Histologic HSIL (CIN 2 or CIN 3)
Guideline: In all nonpregnant patients with a diagnosis of histologic HSIL (CIN 3), treatment is recommended and obser-vation is unacceptable (AII). In nonpregnant patients with histo-logic HSIL (CIN 2), treatment is recommended, unless the patient's concerns about the effect of treatment on future preg-nancy outweigh concerns about cancer (BII). Observation is un-acceptable when the squamocolumnar junction or the upper limit of the lesion is not fully visualized or when the results of an en-docervical sampling, if performed, is CIN 2+ or ungraded (EIII) (see Figure 7). 3
Rationale: As CIN 3 is considered an immediate cancer pre-cursor, treatment is always recommended and observation is never acceptable, except during pregnancy (Section K.2). Observation is acceptable for CIN 2 in patients concerned about the potential ef-fects of treatment on future pregnancy outcomes.
Guideline: When treatment of histologic HSIL is planned, excisional treatment is preferred, and treatment with ablation is ac-ceptable (BI). Outside of the setting of a clinical research trial, nonsur-gical therapies, including topical agents, therapeutic vaccines, and other biologics, are unacceptable for the treatment of histologic HSIL
FIGURE 8. This figure describes management of CIN 2 in patients whose concerns about the effects of treatment on a future pregnancy outweigh their concerns about cancer. Also addressed is the management of histologic HSIL not further specified in women younger than 25 years, for whom observation is acceptable, and for women 25 years or older for whom treatment is preferred.
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117 (CIN 2 or CIN 3) (DIII). Hysterectomy is unacceptable as primary therapy solely for the treatment of histologic HSIL (CIN 2, CIN 3, or unqualified) (EII). When considering ablative therapy, in particular cryotherapy, ablation is unacceptable in the following circumstances. as defined by the WHO: ( a) the lesion extends into the canal and ( b)when the lesion covers more than 75% of the surface area of the ectocervix or extends beyond the cryotip being used. 97 Additional sit-uations for which cryotherapy is not recommended include the fol-lowing: ( a) the squamocolumnar junction or the upper limit of any lesion is not fully visualized; ( b) endocervical canal sample is diag-nosed as CIN 2+ or CIN that cannot be graded; ( c) after previous treatment for CIN 2+; ( d) in the setting of inadequate biopsies of the cervix to confirm histologic diagnosis; and ( e) if cancer is suspected (EIII).
Rationale: The WHO recommends LEEP over cryotherapy in settings where LEEP is “available and accessible. ”97 In the United States, excisional treatment is used more commonly than ablation treatment for the treatment of histologic HSIL. Excisional therapy consists of loop electrosurgical excision procedure (LEEP or LLETZ), cold knife conization, and laser cone biopsy. Ablation treatment includes cryotherapy, laser ablation, and thermoablation. 98 Few recent data have compared the effectiveness of excisional and ablative therapy. Most recent studies evaluating ablative ther-apies have been performed outside of the United States, primarily in low-resource settings. A meta-analysis of randomized trials demonstrated a CIN recurrence rate of 26.6% at 12 months after LEEP compared 31.0% for cryotherapy. 99 However, another meta-analysis calculated that the recurrence rate of CIN 2 –3 was 5.3% after both cryotherapy and LEEP and 1.4% after cold knife conization. More adverse events were noted with cold knife conization than with LEEP, and more with LEEP than with cryo-therapy. 100 A Cochrane review comparing surgical techniques for treatment of CIN concluded that no technique was clearly superior in terms of treatment failure or associated morbidity. 101 However, for high-grade abnormalities, LEEP has the benefit of providing a histologic specimen, which may reveal a higher grade of squa-mous abnormality or a glandular abnormality, and also provides information on margin status, a predictor of CIN 2+ persistence or recurrence. 102,103 Laser ablation differs from other ablative techniques and, when performed by highly experienced providers, may be appropriate in special circumstances including treatment of large cervical lesions or when lesion extends to the vagina, pro-vided all other criteria for ablation are met.
I.3 Management of CIN 2 in Those Who Are Concerned About the Potential Effect of Treatment on Future Pregnancy Outcomes
Guideline: For patients with a diagnosis of histologic HSIL (CIN 2) whose concerns about the effects of treatment on a future pregnancy outweigh their concerns about cancer, either observation or treatment is acceptable provided the squamocolumnar junction is visible and CIN 2+ or ungraded CIN is not identified on endo-cervical sampling (CII) (see Figure 8). If the histologic HSIL cannot be specified as CIN 2, treatment is preferred, but ob-servation is acceptable (CIII). For patients 25 years or older, observation includes colposcopy and HPV-based testing at 6-month intervals for up to 2 years (See Section K.1 for man-agement age of younger than 25 years). If during surveillance, all evaluations demonstrate less than CIN 2 and less than ASC-H on 2 successive occasions, 6 months apart, subsequent surveil-lance should occur at 1 year after the second evaluation and use HPV-based testing. If negative on 3 consecutive annual surveil-lance tests, proceed to long-term surveillance (Section J.3). If CIN 2 remains present for a 2-year period, treatment is recom-mended (CII) (see Figure 8).
Rationale: Unlike CIN 3, which is considered a direct cancer precursor, CIN 2 has an appreciable regression rate. A systematic review and meta-analysis of studies from 1973 to 2016 indicated that among CIN 2 managed conservatively, 50% regressed, 32% persisted, and 18% progressed to CIN 3+. Notably, most regression occurred within the first 12 months, whereas rates of progression continued to increase over time. Regression rates were higher (60%) in women younger than 30 years. 29 A recent study at the KPNC of 2,417 patients followed for a median of 48 months with colposcopy and cotesting at 6-month intervals found similar results: 50% regressed to CIN 1 or less, though remained in intensive sur-veillance for persistent HPV positivity, 30% were treated for persis-tence or progression, and 20% returned to routine screening. Six patients in the KPNC cohort developed cervical cancer, half of whom had significant follow-up delays. 27 The primary rationale for deferring treatment of CIN 2 is the potential risk of adverse obstetric outcomes after excisional or ablative therapy; however, the magnitude of this risk is debated. 104 Studies are complicated by the finding that patients with untreated CIN have a higher risk of premature delivery than the general pop-ulation. 105,106 Although several studies have concluded that exci-sion is associated with increased risk of preterm birth, especially as excision depth increases, 104,105,107 –109 others have found no such association after adjustment for potential confounding factors. 110 –113 Ablation treatment seems to have little or no effect on adverse pregnancy outcomes. 105,107,108,114 A Cochrane Re-view concluded that results should be interpreted with caution be-cause of data being of low or very low quality. 105
I.4 Management of LSIL (CIN 1) or Less Preceded by ASC-H or HSIL Cytology
Guideline: When CIN 2+ is not identified histologically after an ASC-H or HSIL cytology result, it is acceptable to review the cytologic, histologic, and colposcopic findings. If the review yields a revised interpretation, management should follow guide-lines for the revised diagnosis (CIII). When CIN 2+ is not identi-fied, HSIL cytology is managed more aggressively than ASC-H cytology. For cytology showing HSIL, but biopsy showing histo-logic LSIL (CIN 1) or less, either an immediate diagnostic exci-sional procedure or observation with HPV-based testing and colposcopy at 1 year is acceptable, provided in the latter case that the initial colposcopic examination fully visualized the squamocolumnar junction and the upper limit of any lesion, and that the endocervical sampling, if collected, was less than CIN 2 (BII). For ASC-H, if the colposcopic examination can fully visual-ize the squamocolumnar junction and the upper limit of any lesion and that the endocervical sampling, if collected, is negative, obser-vation at 1 year with HPV-based testing is recommended; a diag-nostic excisional procedure is not recommended (BII). For both HSIL and ASC-H cytology, if observation is elected, and all tests are negative at the 1-year visit, repeat HPV-based testing is recom-mended in 1 year (at 2 years from the original cytology). If all tests are negative at both the 1- and 2-year follow-up visits, return for retesting with HPV-based testing in 3 years is recommended, then proceed with long-term surveillance (Section J.3). If any test is abnormal during the observation period, repeat colposcopy is recommended, and management based on resulting biopsies is recommended. A diagnostic excisional procedure is recom-mended for patients with HSIL cytology results at either the 1- or 2-year visit, or ASC-H results that persist at the 2-year visit (CIII) (see Figures 9, 10).
Rationale: Patients with a diagnosis of histologic LSIL (CIN 1) after HSIL and ASC-H cytology have 1-year CIN 3+ risks of 3.9% and 1.4%, respectively. 5 Because HSIL cytology is associated with a higher risk than ASC-H cytology, colposcopy is
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118 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .recommended in addition to HPV-based testing at the 1-year follow-up if excision is not elected. Failure to detect CIN 2+ at col-poscopy in patients with HSIL cytology does not mean that a CIN 2+ lesion has been excluded, although occult carcinoma is unlikely. As a result, patients with HSIL cytology who do not have immedi-ate diagnostic excision require close follow-up. Few studies of HSIL cytology managed without treatment have been reported, and follow-up in those is limited; management relies on expert opinion. 3 Of note, at all colposcopic examination when no lesion is identified on the cervix, the vagina and vulva must be examined for vaginal or vulvar intraepithelial neoplasia.
I.5 Histologic LSIL (CIN 1) Diagnosed Repeatedly for at Least 2 Years
Guideline: For patients 25 years or older with histologic LSIL (CIN 1) who is diagnosed at consecutive visits for at least 2 years, observation is preferred (BII) but treatment is
FIGURE 9. This figure describes management of histologic LSIL (CIN 1) preceded by HSIL cytology.
FIGURE 10. This figure describes management of histologic LSIL (CIN 1) preceded by ASC-H cytology.
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119 acceptable (CIII). If treatment is selected and the entire squamocolumnar junction and all lesions were fully visual-ized during colposcopic examination, either excision or abla-tion treatments are acceptable (CII).
Rationale: Histologic LSIL (CIN 1) is the histologic mani-festation of HPV infection. CIN 1 may be associated with onco-genic (high-risk) or low-risk HPV infections and may be due to persistent infection with 1 type or sequential infections with differ-ent types. HPV 16 is less common in CIN 1 than in CIN 3. 3 His-tologic LSIL (CIN 1) and cytologic ASC-US/HPV+ and LSIL are the same biologically and thus should be managed similarly. Re-gression rates are high, especially in younger patients, and subse-quent diagnosis of CIN 2+ is uncommon regardless of whether CIN 1 is found on endocervical sampling or a biopsy of the trans-formation zone. 3,115 The KPNC data showed a similar, relatively low 5-year risk of CIN 3+ of approximately 2% when CIN 1 or no lesion was found on colposcopy/biopsy after HPV-positive cytologic ASC-US or LSIL. In the KPNC data set of individ-uals with CIN 1 on biopsy on 2 consecutive visits, the subsequent follow-up demonstrated that 52% were HPV negative, 48% were HPV positive, and of the HPV-positive group, 92% had NILM, ASC-US, or LSIL cytology. A study of 126 women undergoing LEEP for CIN 1 diagnosed at consecutive visits for 2 years found that 87% had CIN 1 or negative pathology, whereas 13% had his-tologic HSIL (CIN 2+). 116 Based on these data, and considering the potential harms of treatment, the present recommendations prefer continued observation of those with histologic LSIL (CIN1) diagnosed on consecutive visits for at least 2 years. Treat-ment is an acceptable option based on patient preference, after shared decision-making. Because the immediate estimated CIN3 + risk is less than the 25% treatment threshold, this is considered a special situation.
I.6 Management of AIS: Adoption of Society of Gynecologic Oncology Recommendations
The Society of Gynecologic Oncology recently completed guidelines on the management of AIS; recommendations were adopted by the 2019 ASCCP Risk-Based Management Guide-lines consensus committee and are summarized below. Evidence is not graded as the consensus committee did not perform primary data review.
Guideline: A diagnostic excisional procedure is recommended for all patients with a diagnosis of AIS on cervical biopsy to rule out invasive adenocarcinoma, even when definitive hysterectomy is planned. Excisional procedures should optimally remove an in-tact specimen to facilitate accurate interpretation of margin status. Although there is no preference for cold knife conization versus LEEP, intentional disruption of the specimen by performance of a LEEP followed by a “top hat ” endocervical excision to achieve the desired specimen length is unacceptable. An excisional spec-imen length of at least 10 mm is preferred, and this can be in-creased to 18 to 20 mm for patients who are not concerned about the effect of treatment on future pregnancy. These dimen-sions are preferred regardless of whether hysterectomy is planned. After the initial diagnostic procedure, hysterectomy is the preferred management for all patients who have a histologic diag-nosis of AIS, although fertility-sparing management for appropri-ately selected patients is acceptable. For patients with confirmed AIS with negative margins on the excisional specimen, simple hysterectomy is preferred. For patients with confirmed AIS with positive margins on the excisional specimen, re-excision to achieve negative margins is preferred, even if hysterectomy is planned. For patients with AIS and persistent positive margins for whom additional excisional procedures are not feasible, either
FIGURE 11. This figure describes management of AIS. This management algorithm was developed by the Society of Gynecologic Oncology and endorsed by the ASCCP Risk-Based Management Consensus process.
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120 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .a simple or modified radical hysterectomy is acceptable. After hysterectomy, surveillance per the ASCCP surveillance guidelines for treated CIN 2+ is recommended (Section J.3). For patients of reproductive age who desire future pregnancy, fertility-sparing management with an excisional procedure is ac-ceptable provided that negative margins have been achieved on the excisional specimen, and the patient is willing and able to ad-here to surveillance recommendations. If negative margins cannot be achieved after maximal excisional attempts, fertility-sparing management is not recommended. For patients who undergo fertility-sparing management, surveillance with cotesting and en-docervical sampling is recommended every 6 months for at least 3 years, then annually for at least 2 years, or until hysterectomy is performed. For patients who have consistently negative cotesting and endocervical sampling results for 5 years, extending the surveillance interval to every 3 years starting in the sixth year of surveillance is acceptable. Small retrospective studies have shown HPV test results to be the best predictor for recurrent dis-ease. Therefore, for patients who have consistently negative cotesting and endocervical sampling results, continued surveil-lance is acceptable after completion of childbearing. For patients who have had positive HPV test results or abnormal cytology/ histologic results during surveillance, hysterectomy at the comple-tion of childbearing is preferred (see Figure 11).
Rationale: The Society of Gynecologic Oncology recently conducted a literature review and is publishing recommendations for management of AIS. The ASCCP recommendations adopted the Society of Gynecologic Oncology recommendations, and ad-ditional details are provided in the Society of Gynecologic Oncol-ogy reference. 117 A brief summary of the rationale is provided below. Hysterectomy is recommended for AIS for several reasons. Adenocarcinoma in situ is frequently located within the endocer-vical canal and colposcopic changes may be minimal; therefore, determination of the necessary length of a cervical excisional specimen may be difficult. Adenocarcinoma in situ also has a higher risk of being multifocal, so negative margins on an exci-sional procedure specimen do not ensure complete excision of disease. Importantly, in the setting of histologic AIS on biopsy, in-vasive cancer cannot be excluded without a diagnostic excisional procedure. Finally, although increased detection and treatment of squamous cell cancer precursors ( e.g., CIN 3) is associated with a decrease in the incidence of invasive squamous cell car-cinoma, the same has not been demonstrated for AIS. 118 Be-cause of the challenges in diagnosing and monitoring AIS, hysterectomy remains the standard treatment for AIS for pa-tients who do not desire future pregnancy. For patients desiring future pregnancy, observation after an excisional procedure re-mains an option, but this carries a less than 10% risk of recur-rent AIS and a small risk of invasive cancer even with negative margins. Both margin status and endocervical sam-pling performed at the time of excisional procedure predict re-sidual disease and risk of invasive cancer on hysterectomy specimen. After treatment, HPV tests results are the strongest predictor for recurrent AIS. 119 –122
J. SURVEILLANCE AFTER ABNORMALITIES J.1 Guidance for Specific Tests and Testing Intervals When Managing Abnormal Results
Guideline: After abnormal cervical cancer screening test re-sults for patients 25 years or older, colposcopic biopsy results, or treatment of histologic HSIL, surveillance with either HPV testing alone or cotesting is preferred (AI). Surveillance with cervical cytology alone is acceptable only if testing with HPV or cotesting is not feasible (CIII). Cytology is recommended at 6-month intervals when 1-year intervals are recommended for HPV or cotesting, and annually when 3-year intervals are recom-mended for HPV or cotesting (AII). Cytology should be used for patients younger than 25 years, with transition to HPV-based test-ing at 25 years or older (AII).
Rationale: Individuals treated for histologic HSIL or with a recent abnormal screening test result have an elevated risk of cer-vical precancer warranting close follow-up. 5,123 HPV testing and cotesting are more sensitive than cytology alone in detecting CIN 2+ in both the postcolposcopy and posttreatment settings. 124 –126 As there is marginal difference between cotesting and HPV testing alone in detection of recurrent or persistent CIN 2+, either test may be used for surveillance. 126,127 Because cytology is less sen-sitive than HPV or cotesting, cytology must be performed more frequently to achieve similar sensitivity for the detection of CIN 3+. For example, in cases of low-grade cytology followed by colposcopy/biopsy less than CIN 2, follow-up testing at 1 year is recommended. If the follow-up test is an HPV test with negative results, the 5-year CIN 3+ risk is 0.51%, consistent with a 3-year return. However, if the follow-up test is cytology only with nega-tive results, the 5-year CIN 3+ risk is 1.5%, consistent with a 1-year return.
J.2 Short-Term Follow-up After Treatment for Histologic HSIL
Guideline: After treatment, HPV-based testing at 6 months is preferred regardless of the margin status of the excisional specimen (BII) (see Figure 7). If HPV-based tests are positive, colposcopy and appropriate biopsies should be performed (AII). Follow-up at 6 months with colposcopy and ECC is acceptable (BIII). When margins are positive for CIN 2+ or ECC performed at the time of the excisional procedure shows CIN 2+ in patients 25 years or older who are not concerned about the potential effect of treatment on future pregnancy outcomes, repeat excision or ob-servation is acceptable. For observation, HPV-based testing in 6 months is preferred; it is also acceptable to perform a colpos-copy and ECC at 6 months. For patients younger than 25 years or those who are concerned about the potential effect of treatment on future pregnancy outcomes, observation is recommended. (See Section J.3 for subsequent management). If recurrent histologic HSIL (CIN 2+) develops after excisional treatment, and repeat ex-cision is not feasible or not desired, hysterectomy is recommended (see Figure 7).
Rationale: The preferential use of HPV-based testing (cotesting or HPV primary testing) is supported by evidence that posttreatment HPV testing is the most accurate predictor of treat-ment outcome. 125 Although the relative risk of persistent or recur-rent histologic HSIL (CIN 2+) is almost 5 times higher after excisional treatment with positive margins compared with negative margins (RR = 4.8; 95% CI = 3.2 –7.2), 103 only 56% (95% CI, 49 –66%) of persistent/recurrent precancer was predicted by positive margin status. The poor ability for margin status to predict persistent/recurrent precancer argues against dif-ferentiating follow-up testing by margin status alone. In con-trast, the ability of HPV-based testing to predict persistent/ recurrent histologic HSIL (CIN 2+) is 91% (95% CI = 82% –96%) and does not differ significantly between pa-tients with positive versus negative margins. The absolute risk of persistent/recurrent histologic HSIL (CIN 2+) after exci-sion with positive margins is 17% (95% CI = 13 –22%). How-ever, repeat excisional treatment without repeat testing is considered acceptable for certain patients after appropriate counseling and consideration of age, likelihood of subsequent resolution of histologic HSIL/HPV infection, concern for the
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121 effect of treatment on future pregnancy, and ability to adhere to surveillance recommendations.
J.3 Guidance for Long-Term Follow-up After Treatment for High-Grade Histology or Cytology
Guideline: In patients treated for histologic or cytologic HSIL, after the initial HPV-based test at 6 months, annual HPV or cotesting is preferred until 3 consecutive negative tests have been obtained (AII). After the initial intensive surveillance period, continued surveillance at 3-year intervals is recommended for at least 25 years after treatment of high-grade histology (histologic HSIL, CIN 2, CIN 3, or AIS) or high-grade cytology (HSIL or persistent ASC-H) even if this is beyond the age of 65 years (BII). When patients with a history of treated high-grade histology or cytology reach the age of 65 years, if they have completed the initial 25-year surveillance period, continued surveillance at 3-year intervals is acceptable and may continue as long as the pa-tient is in reasonably good health (BIII). Discontinuation of screening is recommended if a patient has a limited life expec-tancy. Management according to the highest-grade abnormality found on histology or cytology is recommended.
Rationale: According to KPNC data, the 5-year CIN 3+ risks after treatment of CIN 3 for 1, 2, and 3 negative cotests/primary HPV tests were 1.7%/2.0%, 0.68%/0.91%, and 0.35%/0.44%, re-spectively. 5 Therefore, annual surveillance by cotesting or HPV testing is recommended until 3 negative annual HPV-based tests have been obtained. After a third negative HPV-based test, KPNC data suggest that the 5-year CIN 3+ risk remains above the 0.15% threshold for return to routine, 5-year HPV-based cervical screen-ing. Long-term population studies support this finding, as they demonstrate a persistent twofold increase in cervical cancer risk after treatment of histologic HSIL. Risk persists for at least 25 years and seems to be increased for patients older than 50 years. 123,128,129 Therefore, continued 3-year surveillance is recom-mended for a minimum of 25 years. As cervical cancer risk seems to remain above general population levels, 123 continued screening for as long the patient remains in good health is acceptable.
J.4 Guidance for Long-Term Follow-up After Low-Grade Cytology (HPV-Positive NILM, ASC-US, or LSIL) or Histologic LSIL (CIN 1) Abnormalities Without Evidence of Histologic or Cytologic High-Grade Abnormalities
Guideline: Among patients initially diagnosed with low-grade cytology or histologic abnormalities or HPV infections, continued surveillance according to risk estimation using available data is recommended (CIII).
Rationale: The 5-year CIN 3+ risks for abnormal screening test results without evidence of cytologic or histologic HSIL followed by negative HPV-based testing were 0.51% after the first negative test and 0.23% after the second negative test. Thus, pa-tients reach criteria for a 3-year return after the second negative HPV-based test. 5 The ability to perform accurate risk estimation for 3 or more rounds of negative testing after abnormalities is lim-ited by very small numbers of CIN 3+ diagnoses in patients with persistently negative follow-up testing after low-grade cytologic or histologic abnormalities. We estimated risk for two common scenarios related to long-term negative follow-up. The first was HPV+/NILM followed by 3 rounds of negative cotesting. At KPNC, the estimated 5-year CIN 3+ risk was 0.17% (95% CI = 0.14% –0.44%), therefore continued testing at 3-year intervals is recommended at this time. The second group included patients with low-grade abnormalities, who underwent colposcopy at which CIN2+ was not found, and then had 3 rounds of negative cotesting. This group had an estimated 5-year CIN3+ risk of 0.03% (95% CI= 0.0 –0.19%), and thus does qualify for return to a 5-year interval. The 5-year CIN3+ risks for various clinical scenarios will be re-estimated as either longer-term follow-up ac-crue or risk modification based on genotyping are available, and publicly available tables will be modified accordingly (https:// CervixCa.nlm.nih.gov/RiskTables).
K. SPECIAL POPULATIONS
Introduction: Guidelines described previously apply to the average risk individual with an intact cervix and are based primarily
FIGURE 12. This figure describes management of cytologic abnormalities in patients younger than 25 years.
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122 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .on screening and management data from patients aged 25 to 65 years in the KPNC population. However, several populations re-quire special management considerations. Management of patients who are younger than 25 years, pregnant, immunosuppressed, posthysterectomy, and older than 65 years are detailed hereinafter.
K.1 Management of Patients Younger Than 25 Years
In the 2012 guidelines, patients aged 21 to 24 years were consid-ered to be a special population. In the current guidelines, the consen-sus was to reference this group as “patients younger than 25 years. ”
Initial Management After an Abnormal Screening Test Result. Guideline: In patients younger than 25 years with low-grade cytology screening results of LSIL, ASC-US HPV-positive, or ASC-US without HPV testing, repeat cytology alone at 1 and 2 years after the initial abnormal result is recommended (BII). Colposcopy is recommended if high-grade cytology is found at any point (HSIL, ASC-H, AGC, AIS) or if low-grade cytology persists at the 2-year follow-up visit (BII). If reflex HPV testing for ASC-US is performed and the results are negative, repeat cytology in 3 years is recommended (BII). After 2 consecutive negative cytology results, return to routine age-based screening is recommended (BII). If colposcopy is performed and the results are less than CIN 2 ( i.e. , histologic LSIL [CIN 1] or less), repeat cytology in 1 year (BII), and manage as above ( e.g., repeat cytology for ASC-US/LSIL, colposcopy for ASC-H or higher). Clinicians should switch to using risk estimates when patients reach the age of 25 years (see Figures 12, 13).
Rationale: HPV vaccination became available in the United States in 2006, and patients at the target age for vaccination have now entered the younger than 25-year age group. 130 Conse-quently, population-level risks of CIN 3+ for a given screening re-sults are expected to decrease through a combination of individual and herd immunity. Observation is indicated for low-grade cytol-ogy results (ASC-US, LSIL), which are likely to represent non-16/ 18 HPV infections with a high probability for regression and low risk for rapid progression to cancer. Accurate risk estimation for this age group is very difficult because vaccination is rapidly changing population-level CIN 3+ risk and the conservative 2012 manage-ment guidelines recommend against colposcopy/biopsy for lesser cytology abnormalities, which limits the ability to accurately mea-sure CIN 3+ rates in this age group. Therefore, in the absence of new compelling data to change management in this age group, the 2012 algorithms are carried forward. The guidelines outlined in this document are designed to adapt to changes in population vaccination coverage as well as new technologies, and we antici-pate that incorporating HPV vaccination effects on the population-level prevalence of HPV infections will affect management recommendations in the near future.
Management of Cytology ASC-H and HSIL in Patients Younger Than 25 Years. Guideline: Colposcopy is recommended for patients younger than 25 years with ASC-H or HSIL cytology (AII). Immediate treatment without histologic confirmation is not recommended (see Figure 13).
Rationale: Although overall CIN 3+ prevalence is lower, cy-tology results of ASC-H are associated with higher risks of CIN 3+ than ASC-US, even in patients younger than 25. 3 Therefore, colposcopy is warranted to evaluate the cervix for CIN 3+. Imme-diate treatment without histologic confirmation is not warranted in this population because of the high rate of resolution of CIN 2+ and the potential harms of treatment.
Management of Histology of Less Than CIN 2 Preceded by Cytology ASC-H and HSIL in Patients Younger Than 25 Years. Guideline: Observation is recommended and diagnostic excisional procedures are not recommended for patients younger than 25 years with a preceding cytology of ASC-H or HSIL and a colposcopy with biopsy of CIN 1 or less as long as the squamocolumnar junction and the upper limit of
FIGURE 13. This figure describes management of histologic LSIL (CIN 1) in patients younger than 25 years.
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123 all lesions are fully visualized, the endocervical sampling is less than CIN 2, and review of histology/cytology does not change the diagnosis. Observation with colposcopy and cytology in 1 and 2 years is recommended for those with HSIL cytology. Cytology at 1 and 2 years is recommended for those with ASC-H cytology, with colposcopy recommended for ASC-US or above on repeat testing. If CIN 2+ is diagnosed, this is managed per guidelines described in the following section. If a high-grade cytologic abnormality (HSIL, ASC-H) without histologic HSIL persists for 2 years, a diagnostic excisional procedure is recommended (unless the patient is pregnant). A diagnostic excisional procedure is recommended in patients when the squamocolumnar junction or the upper limit of all lesions are not fully visualized (see Figures 9, 10).
Rationale : CIN 1 or less preceded by cytologic ASC-H or HSIL is a rare diagnosis and not well represented in the KPNC population. The rationale for conservative management of this clinical situation is discussed in Section I.4.
Management of Histologic HSIL (CIN 2 or CIN 3) for Patients Younger Than 25 Years. Guideline: In patients younger than 25 years with histologic HSIL (CIN 3), treatment is recommended, and observation is unacceptable (EII). In patients younger than 25 years with histologic HSIL (CIN 2), observation is preferred, and treatment is acceptable (BII). In patients younger than 25 years with histologic HSIL unspecified as CIN 2 or CIN 3, observation or treatment is acceptable. Observation includes colposcopy and cytology at 6-month intervals. If during surveillance of histologic HSIL, all cytology results are less than ASC-H and histology results are less than CIN 2 at 6 and 12 months, subsequent surveillance should be at 1 year after the second evaluation. If CIN 2 or unspecified histologic HSIL persists for a 2-year period, treatment is recommended. Excisional treatment is recommended when the squamocolumnar junction or the lesion(s) are not fully visualized (see Figure 8).
Rationale: Cervical cancer is uncommon in patients younger than 25 years despite the high prevalence of HPV infections and high-grade histologic lesions (especially CIN 2). 16,131 Younger patients have higher rates of regression for histologic HSIL (par-ticularly CIN 2) and lower risks of progression to invasive can-cer. 26,27,132,133 Therefore, less intensive management strategies that do not include HPV testing are appropriate for this popula-tion. The exception is CIN 3, which is considered a direct cervical cancer precursor and should be treated at any age.
K.2 Managing Patients During Pregnancy
Guideline: In pregnancy, management of abnormal screen-ing results using the same Clinical Action Thresholds for surveil-lance and colposcopy established for nonpregnant patients is recommended (CIII). Endocervical curettage, endometrial biopsy, and treatment without biopsy are unacceptable during pregnancy (EIII). A diagnostic excisional procedure or repeat biopsy is recom-mended only if cancer is suspected based on cytology, colposcopy, or histology (BII). If histologic HSIL (CIN 2 or CIN 3) is diagnosed at the first colposcopy examination during pregnancy, surveillance colposcopy and testing (diagnostic cytology/HPV depending on age) is preferred every 12 to 24 weeks, but deferring colposcopy to the postpartum period is acceptable (BII). Repeat biopsy is rec-ommended if invasion is suspected or the appearance of the lesion worsens (BII). Treatment of histologic HSIL (CIN 2 or CIN 3) dur-ing pregnancy is not recommended (DII). If AIS is diagnosed dur-ing pregnancy, referral to a gynecologic oncologist is preferred, but management by a gynecologist skilled in the colposcopic diagnosis and treatment of AIS is acceptable (CIII). In the postpartum period, colposcopy is recommended no earlier than 4 weeks after delivery (BII). In patients diagnosed with histologic HSIL (CIN2 or CIN3) during pregnancy, if a le-sion is detected at postpartum colposcopy, an excisional treatment procedure or full diagnostic evaluation (cervical cytology, HPV ,and biopsy) is acceptable (BII). In the absence of a lesion on col-poscopy, a full diagnostic evaluation is recommended; expedited treatment is not recommended (BII).
Rationale: Pregnancy was considered as a special population in which to consider management and treatment options that weigh the risk to fetus and mother versus the risk of missing cancer. Rate of progression to cancer is not thought to be different in preg-nancy. The 2012 management guidelines for pregnant patients were considered, 3 and literature published since 2012 was reviewed. 134 –139 The adoption of Clinical Action Thresholds in 2019 necessitated modification of the 2012 guidelines, which were based on test results. Although the risk of precancer is not known to be elevated among pregnant patients, cervical hyperemia and other physiologic changes of pregnancy may impact the likelihood of precancer and cancer detection. Colposcopist experience, specifically in the evaluation of the pregnant patient, is known to affect the ability to visually distinguish cancers from pregnancy-related changes, increasing the risk of a missed cancer diagnosis. Colposcopy by an experienced provider during preg-nancy is preferred. The intervals recommended for follow-up are relatively wide taking into consideration the experience and comfort level of the colposcopist, gestational age of the fetus, and the potential for loss to follow-up. Pregnancy does not seem to alter the risk for or rate of progression from cervical precancer to cancer, and colposcopy-directed biopsies in pregnant patients seem to be safe. The 2019 guidelines allow deferral of colposcopy for minor abnormalities in women with prior negative HPV testing or colposcopic ex-aminations at which CIN2+ was not found. Therefore, women referred for colposcopy under the 2019 guidelines will have higher risk of prevalent CIN3+ due to either lack of prior screening or persistent HPV infections. In general, data in preg-nancy are limited, however, and shared decision-making taking into account both the pregnant patient and the fetus is critical for management. Although the risk for progression to cancer during a preg-nancy is low, an estimated 11% of new mothers lose their health insurance in the postpartum period. This loss of healthcare access disproportionately affects those most at risk for cervical cancer; rates of noninsurance may be 2 to 3 times as high among low-income and minority patients, as well as those living in states that did not expand Medicaid. 140 For individuals who do not qualify for health insurance before pregnancy, pregnancy care is a unique event that facilitates entry into health care coverage. However, Medicaid coverage often terminates at the end of the calendar month in which the delivery occurred or at 6 to 8 weeks postpar-tum. Because most deliveries in the United States are to individ-uals with Medicaid, pregnancy may provide an opportunity to identify cancer precursors and even cancer in this population. Healthcare access was considered when developing guidelines. Individuals who are screened infrequently or are unable to com-plete appropriate follow-up are at increased risk for developing cervical cancer. 141
K.3 Managing Patients With Immunosuppression
Immunocompromised patients include those with HIV , solid organ transplant, or allogeneic hematopoietic stem cell transplant, as well as those with systemic lupus erythematous, and those with inflammatory bowel disease or rheumatologic disease requiring current immunosuppressive treatments. The cervical cancer
Perkins et al. Journal of Lower Genital Tract Disease •Volume 24, Number 2, April 2020
124 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .screening guidelines for persons living with HIV have been sup-ported by an increasing number of publications, including prospec-tive studies. Although the literature for other immunosuppressed populations remains limited, these other conditions that suppress cell-mediated immunity have also been associated with virally in-duced cancers, including cervical cancer. 142,143 Therefore, cervical cancer screening and abnormal result management recommenda-tions for immunocompromised individuals without HIV use the guidelines developed for people living with HIV 144 : screening should begin within 1 year of first insertional sexual activity and continue throughout a patient's lifetime: annually for 3 years, then every 3 years (cytology only) until the age of 30 years, and then ei-ther continuing with cytology alone or cotesting every 3 years after the age of 30 years.
Guideline: In immunocompromised patients of any age, col-poscopy referral is recommended for all results cytology results of HPV-positive ASC-US or higher. If HPV testing is not performed on ASC-US results, then repeat cytology in 6 to 12 months is recom-mended, with colposcopy referral for ASC-US or higher. For any re-sult of ASC-US or higher on repeat cytology or if HPV positive, referral to colposcopy is recommended. For all cytology results of LSIL or worse (including ASC-H, AGC, AIS, and HSIL), referral to colposcopy is recommended regardless of HPV test result if done.
Rationale: Because of higher risk of CIN 3+ with low-grade cytologic abnormalities among HIV+ individuals, colposcopic re-ferral is recommended for HPV-positive ASC-US. 145 Lack of data at KPNC precludes risk estimation for immunosuppressed pa-tients. Sexually active patients with HIV infection who are younger than 21 years may have a high rate of progression to precancer. No similar prospective data are available for adoles-cents who acquired HIV during the perinatal period, but as many as 30% of adolescents perinatally infected had ASC-US or greater on their first cervical cytology. Because of the rela-tively high HPV prevalence before age 30 years, HPV cotesting is not recommended for patients younger than 30 years of age with HIV. 144
K.4 Managing Patients After Hysterectomy
Guideline: After a diagnosis of high-grade histology or cy-tology, patients may undergo hysterectomy for reasons related or unrelated to their cervical abnormalities. If hysterectomy is performed for treatment, patients should have 3 consecutive annual HPV-based tests before entering long-term surveil-lance. Long-term surveillance after treatment for histologic HSIL (CIN 2 or CIN 3) or AIS involves HPV-based testing at 3-year intervals for 25 years, regardless of whether the pa-tient has had a hysterectomy either for treatment or at any point during the surveillance period (CIII). Among patients who have undergone hysterectomy but either have no previ-ous diagnosis of CIN 2+ within the previous 25 years or have completed the 25 year surveillance period, screening is generally not recommended. However, if performed, abnormal vaginal screening test results should be managed according to published recommendations (BII). 146
Rationale: The risk of high-grade vaginal intraepithelial neo-plasia is elevated among patients who have had a hysterectomy for treatment of histologic HSIL. 146 Although HPV testing is not FDA approved for vaginal samples, sensitivity of HPV-based test-ing in the setting of posthysterectomy for histologic HSIL seems superior to cytology alone. 147 For patients who have undergone a hysterectomy for benign disease and are screened with cytology and/or HPV testing, ASC-US HPV-positive and LSIL cytology should be managed with follow-up in 12 months and only those with high-grade cytology (HSIL, ASC-H, AGC) should be re-ferred immediately for vaginal colposcopy. 148
K.5 Managing Patients Older Than 65 Years With a History of Prior Abnormalities
Guideline: If patients over age 65 years undergo HPV testing, cotesting, or cytology, management according to guidelines for patients aged 25 to 65 years is recommended (CII). If surveillance testing is recommended for either a his-tory of abnormal screening results or treatment for precancer, discontinuing surveillance is unacceptable if the patient is in reasonably good health and testing is feasible (DII). Discon-tinuation of surveillance is recommended for patients with a lim-ited life expectancy (EIII).
Rationale: Screening for patients older than 65 years should follow national guidelines. 14,149 However, approximately 20% of cervical cancers occur in patients older than 65 years. 150,151 To mitigate cancer risk in patients older than age 65 years, previous consensus management guidelines included continued testing in patients with abnormal results as well as those who do not meet exit criteria. 13,14,152 Although the sensitivity of cytology, HPV testing, and colposcopy seem to be higher in premenopausal than postmenopausal patients, evidence indicates that screening in pa-tients older than 65 years is associated with a lower risk of the sub-sequent development of cervical cancer. 153 Because cessation of routine screening is recommended in adequately screened patients at the age of 65 years, data on the prognostic value of specific screening test results in older patients are limited. However, as cancer rates remain appreciable beyond the age of 65 years, 150,151 and cancer diagnostic procedures such as mammography, breast biopsy, and colonoscopy are recommended beyond the age of 65 years, 154 –156 the consensus decision was to use the guidelines for patients aged 25 to 65 years in evaluating older individuals with abnormal results but without limited life expectancy. Patients with previous CIN 3+ seem to have an elevated lifetime risk of developing cervical or vaginal cancer and thus may require sur-veillance testing beyond the age of 65 years. 123 However, patient comfort and the limitations of positioning and examining older patients should enter into the shared decision-making conver-sation about when to discontinue screening. Vaginal estrogen use for a limited time (3 weeks) can be considered to obtain adequate sampling. 157
L. CURRENT CONSIDERATIONS AND FUTURE DIRECTIONS L.1 Current Considerations
The 2019 guidelines are designed to take into account factors that influence Clinical Action Thresholds. Working groups con-sidered risk factors to determine their importance for inclusion in clinical applications of the guidelines, taking into account both the magnitude of effect on the estimated risk, as well as the feasi-bility of collecting accurate data in clinical practice to inform man-agement. Screening history profoundly influenced risk estimates, specifically current HPV and cytology test results, previous HPV test results, and history of histologic HSIL. Patient screening his-tory is often not known; therefore, unknown history is considered separately as a risk factor. Additional factors were considered be-cause of their association with cervical cancer in the literature: HPV vaccination, age, hormonal contraception use, history of sexually transmitted infection, parity, cigarette smoking, obesity, and sexual behaviors including age of first intercourse and multi-ple partners. HPV vaccination in adolescence (generally before the age of 18 years) does seem to reduce the risk of HPV 16/18 infections and associated histologic HSIL. 158,159 However, HPV vaccination status was omitted from this revision of the guidelines because ( a) management guidelines are already very conservative
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© 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .
125 in the population younger than 25 years, ( b) the population prev-alence of on-time HPV vaccination in the 25- to 29-year-old pop-ulation is currently lower than that needed for herd immunity, 160 thus changing recommendations for this population as a whole is not yet warranted, and ( c) making person-specific recommen-dations based on age at vaccine series initiation and number of doses received is impractical in the United States in the absence of linkable, comprehensive, state-based immunization registries. Overall, none of the other factors contributed clinically meaning-ful risk beyond that afforded by the screening factors noted previ-ously. Therefore, additional factors were not included in risk estimates. Analyses were limited for heavy smoking history and younger than 30 years.
L.2 Future Directions
A successor to the new technologies group will be proposed to continue the consensus process, and to provide continuous future updates to guidelines as new tests become available for manage-ment. Decreases in the overall population prevalence of HPV infec-tion, especially HPV 16/18 genotypes, are expected as individuals vaccinated as adolescents reach screening age. The guidelines outlined in this document are designed to adapt to decreases in on-cogenic HPV prevalence because of HPV vaccination as well as new screening and management technologies. As data on the CIN 3+ risks associated with screening test results become available for individuals aged 25 to 29 years who received timely vaccination, we anticipate that decreases in population-level prevalence of HPV infections will affect management rec-ommendations for this age group in the near future. In addition, new technologies that enter the market will be evaluated for their utility in improving the diagnosis and management of CIN 3+. Examples of clinically useful products would be those with in-creased specificity for detecting high-grade abnormalities or the ability during longitudinal follow-up to distinguish incident (new) from prevalent (persistent) HPV infections. No specific new technologies are listed as creating a comprehensive list of products in development is beyond the scope of this article. In the near future, we will also complete analyses related to costs, benefits, and effectiveness. The high value care group laid out a future research agenda that includes simulation modeling to estimate the quality-of life and economic effects of proposed changes to managing those with abnormal cervical cancer screen-ing test results over multiple rounds of screening. Finally, we are tasked with disseminating these guidelines within the United States to create a new national standard of care for management of abnormal cervical cancer screening test re-sults. Changing from recommendations that could be easily mem-orized by clinicians to guidelines that incorporate both current results and history is a major undertaking. However, the result of successful adoption should be reduction of unnecessary testing and invasive procedures in low-risk patients and identification of high-risk patients who will benefit from more intensive surveillance. Maximizing cancer prevention benefits while minimizing the harms of overtesting and overtreatment is a worthwhile but lofty goal, and these guidelines require more robust implementation plans than previous iterations. The process of guidelines dissem-ination will involve a comprehensive communications and dis-semination plan using best practices for risk communication and health promotion. Components include the following: pre-sentations at national, regional and local meetings, social media outreach to engage clinicians and medical societies, and devel-opment of promotional materials to answer frequently asked questions. Additional areas for future research include develop-ment of an evaluation and impact process for these new recom-mendations on clinical practices. Because low-income and minority women bear the greatest burden of cervical cancer, par-ticular emphasis will be placed on working with these communi-ties and the providers who serve them.
GLOSSARY
CIN 2+: this term includes CIN 2, CIN 3, AIS, and cancer
CIN 3+: this term includes CIN 3, AIS, and cancer
Clinical Action Threshold: this term refers to risk levels that prompt different clinical management strategies. For example, an immediate CIN 3+ risk of 4% is the Clinical Action Threshold for colposcopy; risks below this threshold undergo surveillance, whereas risks above this threshold, but below the expedited treat-ment threshold, undergo colposcopy.
Colposcopy Standards: this term refers to the ASCCP Col-poscopy Standards that provide evidence-based recommendations for the practice of colposcopy
Cotesting: this term refers to screening or surveillance per-formed with both cytology and HPV testing.
Expedited treatment: this term means treatment without confirmatory colposcopic biopsy ( e.g., see and treat).
Excisional treatment: this term includes procedures that remove the transformation zone and produce a specimen for histologic analysis, such as loop electrosurgical excision proce-dure (LEEP), laser cone biopsy, large loop excision of the transformation zone (LLETZ), and cold knife conization.
HPV: this term refers to human papillomavirus. Within this text, HPV refers specifically to high-risk HPV as defined by IARC, including the 12 types that are considered class 1 carcino-gens, plus type 68 which is considered a class 2A carcinogen ( i.e. ,HPV types 16, 18, 31, 33, 35, 39, 45, 51, 52, 56, 58, 59, and 68).
HPV-based testing: this term is used in this document to de-scribe the use of either cotesting or primary HPV screening for surveillance after abnormalities. It does not apply to reflex HPV testing for triage of ASC-US cytology in this document. The HPV testing and positive HPV results discussed throughout this document refer to high-risk HPV types only.
Lower Anogenital Squamous Terminology (LAST): this term refers to 2-tiered pathology criteria for evaluating histologic specimens obtained via colposcopic biopsy
Primary HPV testing: testing with HPV testing alone as a screening or surveillance test.
Reflex testing : this means that laboratories should perform a specific additional triage test in the setting of a positive screening test to inform the next steps in management. For example, an ASC-US cytology should trigger a reflex HPV test. New for these guidelines, a positive a positive primary HPV screening test should trigger both a reflex genotyping test (to determine the presence/ absence of HPV 16/18 if that information is not included in the initial primary test result) and also a reflex cytology test to determine whether the patient would be a candidate for expedited management.
Surveillance: this term refers to repeat testing (HPV primary screening, cotesting, or cytology alone) that occurs at shorter in-tervals than those recommended for routine screening. For exam-ple, HPV primary testing or cotesting at intervals of less than 5 years, or cytology alone at intervals of less than 3 years. Additional contributing authors for the ASCCP Risk Based Management Consensus Guidelines Committee Deborah Arrindell, Washington DC Pelin Batur, MD, Cleveland OH Alicia Carter, MD, Burlington NC Patty Cason, MS, FNP, Los Angeles, CA Xiaojian Chen MS, Bethesda, MD Li Cheung PhD, Bethesda, MD
Perkins et al. Journal of Lower Genital Tract Disease •Volume 24, Number 2, April 2020
126 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .Kim Choma, DNP, Teaneck NJ Megan Clarke, PhD, MHS, Rockville MD Christine Conageski, MD, Aurora CO Miriam Cremer, MD, MPH, Cleveland, OH Barbara Crothers, DO, Silver Spring MD Teresa Darragh, MD, San Francisco CA Maria Demarco, PhD, Rockville MD Eileen Duffey-Lind, MSN, Boston, MA Ysabel Duron, BA, San Jose CA Didem Egemen PhD, Bethesda, MD Carol Eisenhut, MD, MBA, Indianapolis IN Tamika Felder, Upper Marlboro MD Sarah Feldman, MD, MPH, Boston MA Michael Gold, MD, Tulsa OK Robert Goulart, MD, Springfield MA Paul Han, MD, Portland ME Sally Hersh, DNP, Portland OR Aimee Holland, DNP, Birmingham AL Eric Huang, MD, Seattle, WA Michelle Khan, MD, MPH, San Leandro CA Rachel Kupets, MD, Toronto ON, Canada Margaret Long, MD, Rochester MN Thomas Lorey MD, Berkeley, CA Jennifer Loukissas, MPP, Bethesda MD Jeanne Murphy, PhD, Washington DC Amber Naresh, MD, MPH, New Orleans LA Erin Nelson, MD, San Antonio TX Akiva Novetsky, MD, MS, Newark NJ Jeffrey Quinlan, MD, Bethesda, MD Debbie Saslow, PhD, Atlanta GA Kathryn Sharpless, MD, PhD, Portland ME Katie Smith, MD, MS, Oklahoma City OK Elizabeth Stier, MD, Boston MA Colleen Stockdale, MD, MS, Iowa City IA Sana Tabbara, MD, Washington DC Deanna Teoh, MD, MS, Minneapolis MN Elizabeth Unger, PhD, MD, Atlanta GA Alan Waxman, MD, MPH, Albuquerque NM Kelly Welch, North Falmouth, MA Claudia Werner, MD, Dallas TX Amy Wiser, MD, Portland OR Rosemary Zuna, MD, Oklahoma City OK
REFERENCES
Wright TC, Cox JT, Massad LS, et al. 2001 Consensus Guidelines for the Management of Women with Cervical Cytological Abnormalities. J Low Genit Tract Dis 2002;6:127 –43. 2. Wright TC, Massad LS, Dunton CJ, et al. 2006 consensus guidelines for the management of women with abnormal cervical cancer screening tests.
Am J Obstet Gynecol 2007;197:346 –55. 3. Massad LS, Einstein MH, Huh WK, et al. 2012 updated consensus guidelines for the management of abnormal cervical cancer screening tests and cancer precursors. Obstet Gynecol 2013;121:829 –46. 4. Huh WK, Ault KA, Chelmow D, et al. Use of primary high-risk human papillomavirus testing for cervical cancer screening: interim clinical guidance. Gynecol Oncol 2015;136:178 –82. 5. Egemen D, Cheung LC, Chen X, et al. Risk estimates supporting the 2019 ASCCP Risk-Based Management Consensus Guidelines. J Low Genit Tract Dis 2020;24:132 –43. 6. Cheung LC, Egemen, Chen et al. A 2019 ASCCP Risk-Based Management Consensus Guidelines: methods for risk estimation, recommended management, and validation. J Low Genit Tract Dis 2020; 24:90 –101. 7. Drug and Therapeutics Bulletin. An introduction to patient decision aids.
BMJ 2013;347:f4147 –7. 8. Castle PE, Kinney WK, Xue X, et al. Effect of Several negative rounds of human papillomavirus and cytology co-testing on safety against cervical cancer: an observational cohort study. Ann Intern Med
2018;168:20 –9. 9. Schiffman M, Kinney WK, Cheung LC, et al. Relative performance of HPV and cytology components of cotesting in cervical screening. J Natl Cancer Inst 2018;110:501 –8. 10. Wentzensen N, Massad LS, Mayeaux EJ, et al. Evidence-based consensus recommendations for colposcopy practice for cervical cancer prevention in the United States. J Low Genit Tract Dis 2017;21:216 –22. 11. Demarco M, Cheung LC, Kinney WK, et al. Low risk of cervical cancer/precancer among most women under surveillance postcolposcopy.
J Low Genit Tract Dis 2018;22:97 –103. 12. Perkins RB, Schiffman M, Guido RS. The next generation of cervical cancer screening programs: making the case for risk-based guidelines.
Curr Probl Cancer 2018;42:521 –6. 13. US Preventive Services Task Force, Curry SJ, Krist AH, Owens DK, et al. Screening for cervical cancer: US preventive services task force recommendation statement. JAMA 2018;320:674 –86. 14. Saslow D, Solomon D, Lawson HW, et al. American cancer society, american society for colposcopy and cervical pathology, and american society for clinical pathology screening guidelines for the prevention and early detection of cervical cancer. Am J Clin Pathol 2012; 137:516 –42. 15. Elfgren K, Elfström KM, Naucler P, et al. Management of women with human papillomavirus persistence: long-term follow-up of a randomized clinical trial. Am J Obstet Gynecol 2017;216:264.e1 –7. 16. Aro K, Nieminen P, Louvanto K, et al. Age-specific HPV type distribution in high-grade cervical disease in screened and unvaccinated women.
Gynecol Oncol 2019;154:354 –9. 17. Wentzensen N, Schiffman M, Dunn T, et al. Multiple human papillomavirus genotype infections in cervical cancer progression in the study to understand cervical cancer early endpoints and determinants. Int J Cancer 2009;125:2151 –8. 18. Wentzensen N, Schiffman M, Dunn ST, et al. Grading the severity of cervical neoplasia based on combined histopathology, cytopathology, and HPV genotype distribution among 1,700 women referred to colposcopy in Oklahoma. Int J Cancer 2009;124:964 –9. 19. Pauker SG, Kassirer JP. The Threshold Approach to Clinical Decision Making. N Engl J Med 1980;302:1109 –17. 20. Aareleid T, Pukkala E, Thomson H, et al. Cervical cancer incidence and mortality trends in Finland and Estonia: a screened vs. an unscreened population. Eur J Cancer 1993;29A:745 –9. 21. Andrae B, Andersson TM, Lambert PC, et al. Screening and cervical cancer cure: population based cohort study. BMJ 2012;344:e900. 22. ASCCP. Committee opinion: evaluation of the cervix in patients with abnormal vaginal bleeding. 2017. Available at: Accessed November 20, 2019. 23. Perkins RB, Fuzzell LN, Lake P, et al. Incorporating Stakeholder Feedback in Guidelines Development for the Management of Abnormal Cervical Cancer Screening Tests. J Low Genit Tract Dis 2020;24:167 –77. 24. Castle PE, Kinney WK, Cheung LC, et al. Why does cervical cancer occur in a state-of-the-art screening program? Gynecol Oncol 2017;146:546 –53. 25. Carreon JD, Sherman ME, Guillén D, et al. CIN2 is a much less reproducible and less valid diagnosis than CIN3: results from a histological review of population-based cervical samples. Int J Gynecol Pathol 2007;26:441 –6. 26. Wentzensen N, Wilson LE, Wheeler CM, et al. Hierarchical clustering of human papillomavirus genotype patterns in the ASCUS-LSIL triage study. Cancer Res 2010;70:8578 –86.
Journal of Lower Genital Tract Disease • Volume 24, Number 2, April 2020 2019 Consensus Guidelines
© 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP . 127 27. Silver MI, Gage JC, Schiffman M, et al. Clinical outcomes after conservative management of cervical intraepithelial neoplasia grade 2 (CIN2) in women ages 21-39 years. Cancer Prev Res (Phila) 2018;11: 165 –70. 28. Moscicki AB, Ma Y, Wibbelsman C, et al. Rate of and risks for regression of cervical intraepithelial neoplasia 2 in adolescents and young women.
Obstet Gynecol 2010;116:1373 –80. 29. Tainio K, Athanasiou A, Tikkinen KAO, et al. Clinical course of untreated cervical intraepithelial neoplasia grade 2 under active surveillance: systematic review and meta-analysis. BMJ 2018;360:k499. 30. Dalla Palma P, Giorgi Rossi P, Collina G, et al. The risk of false-positive histology according to the reason for colposcopy referral in cervical cancer screening: a blind revision of all histologic lesions found in the NTCC trial. Am J Clin Pathol 2008;129:75 –80. 31. Darragh TM, Colgan TJ, Cox JT, et al. The Lower Anogenital Squamous Terminology standardization project for HPV-associated lesions: background and consensus recommendations from the College of American Pathologists and the American Society for Colposcopy and Cervical Pathology. J Low Genit Tract Dis 2012;16:205 –42. 32. WHO Classification of Tumours of Female Reproductive Organs. Fourth Edition - WHO - OMS . Available at: anglais/detart1.jsp?codlan=1&codcol=70&codcch=4006. Accessed January 10, 2020. 33. Castle PE, Adcock R, Cuzick J, et al. Relationships of p16 immunohistochemistry and other biomarkers with diagnoses of cervical abnormalities: implications for LAST terminology. Arch Pathol Lab Med
doi:10.5858/arpa.2019-0241-OA. [Epub ahead of print]. 34. Wright TC, Stoler MH, Behrens CM, et al. The ATHENA human papillomavirus study: design, methods, and baseline results.
Am J Obstet Gynecol 2012;206:46.e1 –11. 35. Stoler MH, Wright TC, Parvu V , et al. The onclarity human papillomavirus trial: design, methods, and baseline results.
Gynecol Oncol 2018;149:498 –505. 36. Gage JC, Schiffman M, Hunt WC, et al. Cervical histopathology variability among laboratories: a population-based statewide investigation. Am J Clin Pathol 2013;139:330 –5. 37. Wheeler CM, Hunt WC, Cuzick J, et al. A population-based study of human papillomavirus genotype prevalence in the United States: baseline measures prior to mass human papillomavirus vaccination. Int J Cancer
2013;132:198 –207. 38. Ekwueme DU, Uzunangelov VJ, Hoerger TJ, et al. Impact of the National Breast and Cervical Cancer Early Detection Program on cervical cancer mortality among uninsured low-income women in the U.S., 1991-2007.
Am J Prev Med 2014;47:300 –8. 39. Cheung LC, Pan Q, Hyun N, et al. Mixture models for undiagnosed prevalent disease and interval-censored incident disease: applications to a cohort assembled from electronic health records. Stat Med 2017;36:3583 –95. 40. Clarke MA, Darragh TM, Nelson E, et al. Reporting and assessing the quality of diagnostic accuracy studies for cervical cancer screening and management. J Low Genit Tract Dis 2020;24:157 –66. 41. Schiffman M, Solomon D. Findings to date from the ASCUS-LSIL Triage Study (ALTS). Arch Pathol Lab Med 2003;127:946 –9. 42. Wright TC, Stoler MH, Behrens CM, et al. Primary cervical cancer screening with human papillomavirus: end of study results from the ATHENA study using HPV as the first-line screening test. Gynecol Oncol
2015;136:189 –97. 43. Stoler MH, Wright TC Jr., Sharma A, et al. High-risk human papillomavirus testing in women with ASC-US cytology: results from the ATHENA HPV study. Am J Clin Pathol 2011;135:468 –75. 44. Arbyn M, Roelens J, Simoens C, et al. Human papillomavirus testing versus repeat cytology for triage of minor cytological cervical lesions.
Cochrane Database Syst Rev 2013;Mar 28;(3):CD008054. doi: 10.1002/14651858.CD008054.pub2. 45. Gage JC, Hunt WC, Schiffman M, et al. Similar risk patterns after cervical screening in two large U.S. populations: implications for clinical guidelines. Obstet Gynecol 2016;128:1248 –57. 46. McCredie MRE, Sharples KJ, Paul C, et al. Natural history of cervical neoplasia and risk of invasive cancer in women with cervical intraepithelial neoplasia 3: a retrospective cohort study. Lancet Oncol 2008;9:425 –34. 47. SEER. Available at: html#incidence-mortality. Accessed October 7, 2019. 48. Silver MI, Andrews J, Cooper CK, et al. Risk of cervical intraepithelial neoplasia 2 or worse by cytology, human papillomavirus 16/18, and colposcopy impression: a systematic review and meta-analysis. Obstet Gynecol 2018;132:725 –35. 49. Smith HJ, Leath CA, Huh WK, et al. See-and-treat for high-grade cytology: do young women have different rates of high-grade histology?
J Low Genit Tract Dis 2016;20:243 –6. 50. Kuroki LM, Bergeron LM, Gao F, et al. See-and-treat loop electrosurgical excision procedure for high-grade cervical cytology: are we overtreating?
J Low Genit Tract Dis 2016;20:247 –51. 51. Numnum TM, Kirby TO, Leath CA, et al. A prospective evaluation of ‘see and treat ’ in women with HSIL Pap smear results: is this an appropriate strategy? J Low Genit Tract Dis 2005;9:2 –6. 52. Cho H, Kim JH. Treatment of the patients with abnormal cervical cytology: a
“see-and-treat ” versus three-step strategy. J Gynecol Oncol 2009;20:164 –8. 53. Demarco M, Egemen D, Raine-Bennett TR, et al. A Study of Partial Human Papillomavirus Genotyping in Support of the 2019 ASCCP Risk-Based Management Consensus Guidelines. J Low Genit Tract Dis
2020;24:144 –7. 54. Bergeron C, Ordi J, Schmidt D, et al. Conjunctive p16INK4a testing significantly increases accuracy in diagnosing high-grade cervical intraepithelial neoplasia. Am J Clin Pathol 2010;133:395 –406. 55. Stoler MH, Wright TC, Ferenczy A, et al. Routine use of adjunctive p16 immunohistochemistry improves diagnostic agreement of cervical biopsy interpretation: results from the CERTAIN study. Am J Surg Pathol 2018; 42:1001 –9. 56. Thrall MJ. Effect of Lower Anogenital Squamous Terminology recommendations on the use of p16 immunohistochemistry and the proportion of high-grade diagnoses in cervical biopsy specimens.
Am J Clin Pathol 2016;145:524 –30. 57. Reuschenbach M, Wentzensen N, Dijkstra MG, et al. p16INK4a immunohistochemistry in cervical biopsy specimens: a systematic review and meta-analysis of the interobserver agreement. Am J Clin Pathol 2014;142:767 –72. 58. Maniar KP, Sanchez B, Paintal A, et al. Role of the biomarker p16 in downgrading -IN 2 diagnoses and predicting higher-grade lesions.
Am J Surg Pathol 2015;39:1708 –18. 59. Torres S, Wentzensen N, Stoler M, et al. Estimating the benefits and harms of p16 utilization on cervical biopsy interpretation in routine clinical practice. Under Review .60. FDA Executive Summary New Approaches in the Evaluation for High-Risk Human Papillomavirus Nucleic Acid Detection Devices Prepared for the March 8, 2019 meeting of the Microbiology Devices Panel of the Medical Devices Advisory Committee. Available at: https:// www.fda.gov/media/122799/download. Accessed November 2, 2019. 61. FDA. Meeting materials of the microbiology devices panel. FDA .Bethesda, MD; 2014. Available at: AdvisoryCommittees/CommitteesMeetingMaterials/MedicalDevices/ MedicalDevicesAdvisoryCommittee/MicrobiologyDevicesPanel/ UCM388564.pdf. Accessed November 10, 2019. 62. Arbyn M, Snijders PJ, Meijer CJ, et al. Which high-risk HPV assays fulfil criteria for use in primary cervical cancer screening? Clin Microbiol Infect
2015;21:817 –26. 63. Lee KR, Darragh TM, Joste NE, et al. Atypical glandular cells of undetermined significance (AGUS): interobserver reproducibility in
Perkins et al. Journal of Lower Genital Tract Disease • Volume 24, Number 2, April 2020
128 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP .cervical smears and corresponding thin-layer preparations. Am J Clin Pathol 2002;117:96 –102. 64. Levine L, Lucci JA, Dinh TV . Atypical glandular cells: new Bethesda Terminology and Management Guidelines. Obstet Gynecol Surv 2003;58: 399 –406. 65. Zhao C, Florea A, Onisko A, et al. Histologic follow-up results in 662 patients with Pap test findings of atypical glandular cells: Results from a large academic womens hospital laboratory employing sensitive screening methods. Gynecol Oncol 2009;114:383 –9. 66. Verdoodt F, Jiang X, Williams M, et al. High-risk HPV testing in the management of atypical glandular cells: a systematic review and meta-analysis. Int J Cancer 2016;138:303 –10. 67. Katki HA, Schiffman M, Castle PE, et al. Five-year risks of CIN 3+ and cervical cancer among women with HPV-positive and HPV-negative high-grade pap results. J Low Genit Tract Dis 2013;17:S50 –5. 68. Davey DD, Neal MH, Wilbur DC, et al. Bethesda 2001 implementation and reporting rates: 2003 practices of participants in the College of American Pathologists Interlaboratory Comparison Program in Cervicovaginal Cytology. Arch Pathol Lab Med 2004;128:1224 –9. 69. Schnatz PF, Guile M, O'Sullivan DM, et al. Clinical significance of atypical glandular cells on cervical cytology. Obstet Gynecol 2006;107:701 –8. 70. Castle PE, Fetterman B, Poitras N, et al. Relationship of atypical glandular cell cytology, age, and human papillomavirus detection to cervical and endometrial cancer risks. Obstet Gynecol 2010;115:243 –8. 71. Wise MR, Jordan V , Lagas A, et al. Obesity and endometrial hyperplasia and cancer in premenopausal women: a systematic review. Am J Obstet Gynecol 2016;214:689.e1 –17. 72. Fontaine D, Narine N, Naugler C. Unsatisfactory rates vary between cervical cytology samples prepared using ThinPrep and SurePath platforms: a review and meta-analysis. BMJ Open 2012;2:e000847. 73. Selvaggi SM. Factors contributing to high ThinPrep® Pap test unsatisfactory rates in an academic medical center laboratory. Diagn Cytopathol 2014;42:380 –3. 74. Randolph ML, Wu HH, Crabtree WN. Reprocessing unsatisfactory ThinPrep papanicolaou tests using a modified SurePath preparation technique. Cancer Cytopathol 2014;122:343 –8. 75. Mirzamani N, Chau K, Rafael O, et al. Quality assessment and improvement of ‘Unsatisfactory ’ liquid-based cervicovaginal papanicolaou smears. Diagn Cytopathol 2017;45:873 –7. 76. Zhao L, Wentzensen N, Zhang RR, et al. Factors associated with reduced accuracy in Papanicolaou tests for patients with invasive cervical cancer.
Cancer Cytopathol 2014;122:694 –701. 77. Quiroga-Garza G, Satrum LS, Trujillo CJ, et al. Common causes for unsatisfactory Pap tests in a high-risk population: insights into a yet unresolved problem in gynecologic cytology. J Am Soc Cytopathol 2014; 3:256 –60. 78. Nygård JF, Sauer T, Nygård M, et al. CIN 2/3 and cervical cancer in an organised screening programme after an unsatisfactory or a normal Pap smear: a seven-year prospective study of the Norwegian population-based screening programme. J Med Screen 2004;11:70 –6. 79. Carozzi FM, Del Mistro A, Cuschieri K, et al. HPV testing for primary cervical screening: laboratory issues and evolving requirements for robust quality assurance. J Clin Virol 2016;76(suppl 1):S22 –8. 80. Preisler S, Rebolj M, Ejegod DM, et al. Cross-reactivity profiles of hybrid capture II, cobas, and APTIMA human papillomavirus assays: split-sample study. BMC Cancer 2016;16:510. 81. Ghofrani M, Zhao C, Davey DD, et al. Update on the College of American Pathologists experience with high-risk human papillomavirus proficiency testing for cytology. Arch Pathol Lab Med 2016;140:1371 –4. 82. College of American Pathologists Commission on Laboratory Accreditation. Laboratory Accreditation Program: Cytopathology Checklist 2017. checklists/cytopathology_08.21.2017.pdf (behind firewall). 83. Sultana F, English DR, Simpson JA, et al. High-grade cervical abnormalities and cervical cancer in women following a negative Pap smear with and without an endocervical component: a cohort study with 10 years of follow-up. Int J Cancer 2014;135:1213 –9. 84. Polanco Jacome EC, Maerki J, Chau K, et al. Lack of transformation zone in cervical Pap tests, should it be a concern? A quality assurance initiative.
Diagn Cytopathol 2018;46:584 –8. 85. Hastings JW, Alston MJ, Mazzoni SE, et al. Frequency of adequate endometrial biopsy in evaluation of postmenopausal women with benign endometrial cells on Pap test. J Low Genit Tract Dis 2017;21: 258 –60. 86. Moyer AB, El-Zaatari ZM, Thrall MJ. The effects of the Bethesda System 2014 on endometrial cell reporting and follow-up endometrial biopsies in women 45 years of age and over. J Am Soc Cytopathol 2018;7:201 –4. 87. Fischer G, Haddad M, Cormier K. Endometrial cells on Pap tests: ideal reporting is more complex than just finding the right age. Diagn Cytopathol 2017;45:587 –91. 88. Grada Z, Paquette C, Eklund CM, et al. Evaluating the age cutoff criterion for reporting benign-appearing endometrial cells in routine pap tests: an 8-year retrospective review. Acta Cytol 2017;61:194 –8. 89. Colletti SM, Tranesh GA, Nassar A. Significance of finding benign endometrial cells in women 40-45 versus 46 years or older on Papanicolaou tests and histologic follow-up. Cytojournal
2017;14:22. 90. Weiss VL, Cate F, Bloom L, et al. Age cut-off for reporting endometrial cells on a Papanicolaou test: 50 years may be more appropriate than 45 years. Cytopathology 2016;27:242 –8. 91. Izadi-Mood N, Sarmadi S, Sanii S, et al. Normal-appearing endometrial cells in Pap tests of women aged forty years or older and cytohistological correlates. Acta Cytol 2015;59:175 –9. 92. Kir G, Gocmen A, Cetiner H, et al. Clinical significance of benign endometrial cells found in papanicolaou tests of Turkish women aged 40 years and older. J Cytol 2013;30:156 –8. 93. Ramdall RB, Wallach RC, Cangiarella J, et al. Origin, frequency and clinical significance of glandular cells in liquid-based pap tests from patients posthysterectomy. Acta Cytol 2009;53:1 –9. 94. Wentzensen N, Schiffman M, Silver MI, et al. ASCCP Colposcopy Standards: risk-based colposcopy practice. J Low Genit Tract Dis 2017; 21:230 –4. 95. Kurman RJ, Schiffman MH, Lancaster WD, et al. Analysis of individual cervical human papillomavirus types in neolasia: a possible role for type 18 in rapid progression. Am J Obstet Gynecol 1988;159:293 –6. 96. Clifford G, Franceschi S, Diaz M, et al. Chapter 3: HPV type-distribution in women with and without cervical neoplastic diseases. Vaccine 2006; 24(suppl 3):S26 –34. 97. WHO Guidelines: Use of Cryotherapy for Cervical Intraepithelial Neoplasia . Geneva: World Health Organization; 2011. Available at: http:// www.ncbi.nlm.nih.gov/books/NBK138476/. Accessed January 13, 2020. 98. WHO guidelines for the use of thermal ablation for cervical pre-cancer lesions. Available at: Accessed January 13, 2020. 99. D'Alessandro P, Arduino B, Borgo M, et al. Loop electrosurgical excision procedure versus cryotherapy in the treatment of cervical intraepithelial neoplasia: a systematic review and meta-analysis of randomized controlled trials. Gynecol Minim Invasive Ther 2018;7:145 –51. 100. Santesso N, Mustafa RA, Wiercioch W, et al. Systematic reviews and meta-analyses of benefits and harms of cryotherapy, LEEP, and cold knife conization to treat cervical intraepithelial neoplasia. Int J Gynecol Obstet
2016;132:266 –71. 101. Martin-Hirsch PPL, Paraskevaidis E, Bryant A, et al. Surgery for cervical intraepithelial neoplasia. Cochrane Database Syst Rev
2013;CD001318.
Journal of Lower Genital Tract Disease • Volume 24, Number 2, April 2020 2019 Consensus Guidelines
© 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP . 129 102. Kalogirou D, Antoniou G, Karakitsos P, et al. Predictive factors used to justify hysterectomy after loop conization: increasing age and severity of disease. Eur J Gynaecol Oncol 1997;18:113 –6. 103. Arbyn M, Redman CWE, Verdoodt F, et al. Incomplete excision of cervical precancer as a predictor of treatment failure: a systematic review and meta-analysis. Lancet Oncol 2017;18:1665 –79. 104. Kyrgiou M, Athanasiou A, Paraskevaidi M, et al. Adverse obstetric outcomes after local treatment for cervical preinvasive and early invasive disease according to cone depth: systematic review and meta-analysis.
BMJ 2016;354:i3633. 105. Kyrgiou M, Koliopoulos G, Martin-Hirsch P, et al. Obstetric outcomes after conservative treatment for intraepithelial or early invasive cervical lesions: systematic review and meta-analysis. Lancet 2006;367:489 –98. 106. Reilly R, Paranjothy S, Beer H, et al. Birth outcomes following treatment for precancerous changes to the cervix: a population-based record linkage study: birth outcomes following colposcopy treatments. BJOG 2012;119: 236 –44. 107. Bruinsma F, Quinn M. The risk of preterm birth following treatment for precancerous changes in the cervix: a systematic review and meta-analysis: cervical dysplasia and preterm birth: a meta-analysis.
BJOG 2011;118:1031 –41. 108. Bjørge T, Skare GB, Bjørge L, et al. Adverse pregnancy outcomes after treatment for cervical intraepithelial neoplasia. Obstet Gynecol 2016;128: 1265 –73. 109. Miller ES, Sakowicz A, Grobman WA. The association between cervical dysplasia, a short cervix, and preterm birth. Am J Obstet Gynecol 2015; 213:543.e1 –4. 110. Bruinsma F, Lumley J, Tan J, et al. Precancerous changes in the cervix and risk of subsequent preterm birth. BJOG 2006;114:70 –80. 111. Shanbhag S, Clark H, Timmaraju V , et al. Pregnancy outcome after treatment for cervical intraepithelial neoplasia. Obstet Gynecol 2009;114: 727 –35. 112. Castanon A, Brocklehurst P, Evans H, et al. Risk of preterm birth after treatment for cervical intraepithelial neoplasia among women attending colposcopy in England: retrospective-prospective cohort study. BMJ
2012;345:e5174. 113. Werner CL, Lo JY, Heffernan T, et al. Loop electrosurgical excision procedure and risk of preterm birth. Obstet Gynecol 2010;115: 605 –8. 114. Weinmann S, Naleway A, Swamy G, et al. Pregnancy outcomes after treatment for cervical cancer precursor lesions: an observational study.
PLoS One 2017;12:e0165276. 115. Fukuchi E, Fetterman B, Poitras N, et al. Risk of cervical precancer and cancer in women with cervical intraepithelial neoplasia grade 1 on endocervical curettage. J Low Genit Tract Dis 2013;17:255 –60. 116. Lueng SOA, Vitonis A, Feldman S. Yield of loop electrosurgical excision procedure (LEEP) among patients with and without known high-grade cervical dysplasia. Soc Gynecol Oncologists 2020. 117. Teoh D, Musa F, Salani R, et al. Diagnosis and management of adenocarcinoma in situ: a society of gynecologic oncology evidence-based review and recommendations. Obstet Gynecol
Wang SS, Sherman ME, Hildesheim A, et al. Cervical adenocarcinoma and squamous cell carcinoma incidence trends among white women and black women in the United States for 1976-2000. Cancer 2004;100: 1035 –44. 119. Salani R, Puri I, Bristow RE. Adenocarcinoma in situ of the uterine cervix: a metaanalysis of 1278 patients evaluating the predictive value of conization margin status. Am J Obstet Gynecol 2009;200: 182.e1 –5. 120. Lea JS, Shin CH, Sheets EE, et al. Endocervical curettage at conization to predict residual cervical adenocarcinoma in situ. Gynecol Oncol 2002;87: 129 –32. 121. Costa S, Venturoli S, Negri G, et al. Factors predicting the outcome of conservatively treated adenocarcinoma in situ of the uterine cervix: an analysis of 166 cases. Gynecol Oncol 2012;124:490 –5. 122. Costa S, Venturoli S, Origoni M, et al. Performance of HPV DNA testing in the follow-up after treatment of high-grade cervical lesions, adenocarcinoma in situ (AIS) and microinvasive carcinoma.
Ecancermedicalscience 2015;9:528. 123. Strander B, Andersson-Ellström A, Milsom I, et al. Long term risk of invasive cancer after treatment for cervical intraepithelial neoplasia grade 3: population based cohort study. BMJ 2007;335:1077. 124. Kocken M, Uijterwaal MH, de Vries ALM, et al. High-risk human papillomavirus testing versus cytology in predicting post-treatment disease in women treated for high-grade cervical disease: a systematic review and meta-analysis. Gynecol Oncol 2012;125:500 –7. 125. Arbyn M, Ronco G, Anttila A, et al. Evidence regarding human papillomavirus testing in secondary prevention of cervical cancer. Vaccine
2012;30(suppl 5):F88 –99. 126. Clarke M, Unger ER, Zuna R, et al. A systematic review of tests for post-colposcopy and post-treatment surveillance. J Low Genit Tract Dis
2020;24:148 –56. 127. Cuschieri K, Bhatia R, Cruickshank M, et al. HPV testing in the context of post-treatment follow up (test of cure). J Clin Virol 2016;76(suppl 1): S56 –61. 128. Melnikow J, McGahan C, Sawaya GF, et al. Cervical intraepithelial neoplasia outcomes after treatment: long-term follow-up from the British Columbia Cohort Study. J Natl Cancer Inst 2009;101:721 –8. 129. Kalliala I, Anttila A, Pukkala E, et al. Risk of cervical and other cancers after treatment of cervical intraepithelial neoplasia: retrospective cohort study. BMJ 2005;331:1183 –5. 130. Williams WW, Lu P-J, O'Halloran A, et al. Surveillance of vaccination coverage among adult populations - United States, 2015. MMWR Surveill Summ 2017;66:1 –28. 131. Benard VB, Watson M, Castle PE, et al. Cervical carcinoma rates among young females in the United States. Obstet Gynecol 2012;120:1117 –23. 132. Loopik DL, Doucette S, Bekkers RLM, et al. Regression and progression predictors of CIN2 in women younger than 25 years. J Low Genit Tract Dis 2016;20:213 –7. 133. Lee MH, Finlayson SJ, Gukova K, et al. Outcomes of conservative management of high grade squamous intraepithelial lesions in young women. J Low Genit Tract Dis 2018;22:212 –8. 134. He Y, Wu YM, Wang T, et al. Perinatal outcomes of pregnant women with cervical intraepithelial neoplasia. Arch Gynecol Obstet 2013;288: 1237 –42. 135. Kärrberg C, Brännström M, Strander B, et al. Colposcopically directed cervical biopsy during pregnancy; minor surgical and obstetrical complications and high rates of persistence and regression. Acta Obstet Gynecol Scand 2013;92:692 –9. 136. Mailath-Pokorny M, Schwameis R, Grimm C, et al. Natural history of cervical intraepithelial neoplasia in pregnancy: postpartum histo-pathologic outcome and review of the literature. BMC Pregnancy Childbirth 2016;16:74. 137. Schaefer K, Peters D, Aulmann S, et al. Value and feasibility of LLETZ procedures for pregnant women with suspected high-grade squamous intraepithelial lesions and microinvasive cervical cancer. Int J Gynecol Obstet 2012;118:141 –4. 138. Schuster S, Joura E, Kohlberger P. Natural history of squamous intraepithelial lesions in pregnancy and mode of delivery.
Anticancer Res 2018;38:2439 –42. 139. Wu YM, Wang T, He Y, et al. Clinical management of cervical intraepithelial neoplasia in pregnant and postpartum women.
Arch Gynecol Obstet 2014;289:1071 –7. 140. McMorrow, Stacey and Kenney, Genevieve. Despite progress under the ACA, many new mothers lack insurance coverage. 2018. Available at:
Perkins et al. Journal of Lower Genital Tract Disease • Volume 24, Number 2, April 2020
130 © 2020 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the ASCCP . Accessed November 22, 2019. 141. Downs LS, Smith JS, Scarinci I, et al. The disparity of cervical cancer in diverse populations. Gynecol Oncol 2008;109:S22 –30. 142. Moscicki AB, Flowers L, Huchko MJ, et al. Guidelines for Cervical Cancer Screening in Immunosuppressed Women Without HIV Infection.
J Low Genit Tract Dis 2019;23:87 –101. 143. Chin-Hong PV , Reid GE, AST Infectious Diseases Community of Practice. Human papillomavirus infection in solid organ transplant recipients: Guidelines from the American Society of Transplantation Infectious Diseases Community of Practice. Clin Transplant 2019; 33:e13590. 144. US Department of Health and Human Services. Guidelines for the prevention and treatment of opportunistic infections in adults and adolescents with HIV . 2018. Available at: guidelines/html/4/adult-and-adolescent-opportunistic-infection/343/ human-papillomavirus. Accessed November 25, 2019. 145. Robbins HA, Strickler HD, Massad LS, et al. Cervical cancer screening intervals and management for women living with HIV: a risk benchmarking approach. AIDS 2017;31:1035 –44. 146. Khan MJ, Massad LS, Kinney W, et al. A common clinical dilemma: management of abnormal vaginal cytology and human papillomavirus test results. Gynecol Oncol 2016;141:364 –70. 147. Schockaert S, Poppe W, Arbyn M, et al. Incidence of vaginal intraepithelial neoplasia after hysterectomy for cervical intraepithelial neoplasia: a retrospective study. Am J Obstet Gynecol 2008;199:113.e1 –5. 148. Cong Q, Song Y, Wang Q, et al. A retrospective study of cytology, high-risk HPV , and colposcopy results of vaginal intraepithelial neoplasia patients. Biomed Res Int 2018;2018:5894801. 149. USPSTF. Screening for Cervical Cancer. Available at: gov/clinic/uspstf/uspscerv.htm 2003. Accessed April 10, 2012. 150. Gravitt PE, Landy R, Schiffman M. How confident can we be in the current guidelines for exiting cervical screening? Prev Med 2018;114: 188 –92. 151. Feldman S, Cook E, Davis M, et al. Cervical cancer incidence among elderly women in massachusetts compared with younger women.
J Low Genit Tract Dis 2018;22:314 –7. 152. Gage JC, Katki HA, Schiffman M, et al. The low risk of precancer after a screening result of human papillomavirus-negative/atypical squamous cells of undetermined significance papanicolaou and implications for clinical management: low precancer risk after HPV-negative/ASC-US. Cancer Cytopathol 2014;122:842 –50. 153. Rosenblatt KA, Osterbur EF, Douglas JA. Case-control study of cervical cancer and gynecologic screening: a SEER-Medicare analysis. Gynecol Oncol 2016;142:395 –400. 154. Siu AL, U.S. Preventive Services Task Force. Screening for breast cancer: U.S. Preventive Services Task Force Recommendation Statement.
Ann Intern Med 2016;164:279 –96. 155. American Cancer Society. American Cancer Society Guideline for Colorectal Cancer Screening. 2018. Available at: cancer/colon-rectal-cancer/detection-diagnosis-staging/ acs-recommendations.html. Accessed November 17, 2019. 156. US Preventive Services Task Force, Bibbins-Domingo K, Grossman DC, et al. Screening for colorectal cancer: US Preventive Services Task Force Recommendation Statement. JAMA 2016;315: 2564 –75. 157. Mayeux EJ, Cox JT. Modern Colposcopy Textbook and Atlas - New Edition - ASCCP . 3rd ed. Rockville, MD: ASCCP; 2011. 158. Castle PE, Xie X, Xue X, et al. Impact of human papillomavirus vaccination on the clinical meaning of cervical screening results.
Prev Med 2019;118:44 –50. 159. Gertig DM, Brotherton JM, Budd AC, et al. Impact of a population-based HPV vaccination program on cervical abnormalities: a data linkage study.
BMC Med 2013;11:227. 160. Drolet M, Benard E, Boily MC, et al. Population-level impact and herd effects following human papillomavirus vaccination programmes: a systematic review and meta-analysis. Lancet Infect Dis 2015;15: 565 –80.
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15862 | https://www.quora.com/Explain-why-is-phenoxide-ion-more-stable-than-phenol-towards-electrophilic-substitution-reaction | Explain why is phenoxide ion more stable than phenol towards electrophilic substitution reaction? - Quora
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Organic Chemistry
Resonating Structure
Phenoxide Ion
Biochemistry
Electrophilic Substitutio...
Stability of Compounds
Resonances
Phenol
Organic Reactions
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Explain why is phenoxide ion more stable than phenol towards electrophilic substitution reaction?
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The stability of the phenoxide ion compared to phenol in electrophilic substitution reactions can be understood through the concepts of resonance and charge distribution.
Resonance Stabilization
Phenol Structure: Phenol (C₆H₅OH) has a hydroxyl (-OH) group attached to a benzene ring. In phenol, the lone pair of electrons on the oxygen can participate in resonance with the π system of the benzene ring, stabilizing the molecule. However, the hydroxyl group is an electron-withdrawing group through its inductive effect.
Phenoxide Ion Formation: When phenol loses a proton (H⁺), it forms the phenoxide i
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The stability of the phenoxide ion compared to phenol in electrophilic substitution reactions can be understood through the concepts of resonance and charge distribution.
Resonance Stabilization
Phenol Structure: Phenol (C₆H₅OH) has a hydroxyl (-OH) group attached to a benzene ring. In phenol, the lone pair of electrons on the oxygen can participate in resonance with the π system of the benzene ring, stabilizing the molecule. However, the hydroxyl group is an electron-withdrawing group through its inductive effect.
Phenoxide Ion Formation: When phenol loses a proton (H⁺), it forms the phenoxide ion (C₆H₅O⁻). In this ion, the negative charge on the oxygen can also delocalize into the benzene ring through resonance. The resonance structures of the phenoxide ion include several forms where the negative charge is shared between the oxygen and the carbon atoms of the benzene ring.
Charge Distribution
In the phenoxide ion, the negative charge is stabilized by the resonance with the aromatic ring, leading to a more distributed charge across the molecule. This delocalization reduces the overall energy and increases the stability of the ion.
In contrast, in phenol, the -OH group is not as effective in stabilizing the positive charge that can develop during electrophilic substitution due to its electron-withdrawing nature.
Electrophilic Substitution Reaction
During electrophilic substitution, the aromatic ring undergoes temporary formation of a carbocation (a positively charged intermediate). The stability of this intermediate is crucial for the reaction to proceed.
The phenoxide ion, due to its resonance stabilization, allows for better distribution of charge, making the intermediate formed during the reaction more stable compared to that of phenol.
Conclusion
In summary, the phenoxide ion is more stable than phenol in the context of electrophilic substitution reactions due to:
Enhanced resonance stabilization that delocalizes the negative charge.
Better charge distribution during the formation of the electrophilic substitution intermediate.
This increased stability makes the phenoxide ion a more favorable substrate for electrophilic attacks compared to phenol.
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Purna Chandra Sahu
M. Sc. in Pure Chemistry&Inorganic Chemistry, Calcutta Uniiversity (Graduated 1987) · Author has 2.9K answers and 6M answer views
·4y
In phenoxide ion, there are three lone electron pairs on O atom and it carries a negative charge, whereas in phenol, O atom carries two lone electron pairs and it carries no charge. In both cases, negative charge develops in ortho and para positions. But electron density of negative charge in phenoxide ion is greater than that in phenol.
As electrophiles are electron deficient species, they are more attracted towards ortho and para positions in phenoxide than phenol.
Hope this helps.
Image: Google.
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In phenoxide ion, there are three lone electron pairs on O atom and it carries a negative charge, whereas in phenol, O atom carries two lone electron pairs and it carries no charge. In both cases, negative charge develops in ortho and para positions. But electron density of negative charge in phenoxide ion is greater than that in phenol.
As electrophiles are electron deficient species, they are more attracted towards ortho and para positions in phenoxide than phenol.
Hope this helps.
Image: Google.
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Barbara Murray
PhD in Organic Chemistry, University of Illinois at Urbana-Champaign (Graduated 1984) · Upvoted by
Michael David Wiley
, Ph.D. Organic Chemistry, University of Washington (1969) · Author has 636 answers and 683.1K answer views
·6y
Originally Answered: Why is phenoxide more stable than phenol? ·
I think there is some confusion here. Phenol is a ordinary, stable acid. If it gives up a proton, it becomes a base, phenoxide. Phenol is a weak acid which means that in solution, the equilibrium between the phenol and the phenoxide favors phenol. If phenoxide was more stable than phenol, the equilibrium would favor the phenoxide. When someone talks about phenoxide being resonance stabilized, that explains why phenol is more acidic that a non-aromatic alcohol. A non-aromatic alcohol is not acidic because the conjugate base is not resonance stabilized. Read the following section carefully, and
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I think there is some confusion here. Phenol is a ordinary, stable acid. If it gives up a proton, it becomes a base, phenoxide. Phenol is a weak acid which means that in solution, the equilibrium between the phenol and the phenoxide favors phenol. If phenoxide was more stable than phenol, the equilibrium would favor the phenoxide. When someone talks about phenoxide being resonance stabilized, that explains why phenol is more acidic that a non-aromatic alcohol. A non-aromatic alcohol is not acidic because the conjugate base is not resonance stabilized. Read the following section carefully, and I think you should see where you are comparing the wrong two things. You are comparing phenoxide with phenol when you should be comparing phenoxide with other conjugate bases of other alcohols. I hope that makes sense.
19.9 Acidity of Carboxylic Acids and Phenols
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More answers below
Why is phenoxide more stable than phenol?
Why is an acetate ion more stable than a phenoxide ion even though the latter is aromatic?
Why is a phenoxide ion more reactive than phenol towards an electrophilic substitution reaction?
What is an ortho-para directing effects of the groups in electrophilic aromatic substitution reactions? Explain with examples.
What are the characteristics of electrophilic substitution reactions?
P R A G Y A
B.sc from C. S. J. M. University, Kanpur (Graduated 2019)
·5y
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Why is an acetate ion more stable than a phenoxide ion even though the latter is aromatic?
Rasonating structure of acetate ion-
Rasonating structure of phenoxide ion-
Hence we can see that in acetate ion electron density 100% on oxygen atom.and in phenoxide ion only 25% electron density present on oxygen atom although 75% electron density present on carbon atom that means electron spend more time on carbon atom rather than oxygen atom.
As we know - ve charge is stable on more electronegative atom but in phenoxide ion -ve charge spends more time on less electronegative atom and in acetate ion -ve charge 💯% present on more electronegative atom. That's why acetate ion is more stable than
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Rasonating structure of acetate ion-
Rasonating structure of phenoxide ion-
Hence we can see that in acetate ion electron density 100% on oxygen atom.and in phenoxide ion only 25% electron density present on oxygen atom although 75% electron density present on carbon atom that means electron spend more time on carbon atom rather than oxygen atom.
As we know - ve charge is stable on more electronegative atom but in phenoxide ion -ve charge spends more time on less electronegative atom and in acetate ion -ve charge 💯% present on more electronegative atom. That's why acetate ion is more stable than phenoxide ion.
There is no role of aromaticity because -ve charge present outside the ring, and ring is aromatic because of its 3 double bond.-ve charge doesn't participate in aromaticity yaa it's increase reactivity toward electophile or activate the ring for electrophilic substitution reaction.
Thanku..
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Related questions
Why is phenoxide more stable than phenol?
Why is an acetate ion more stable than a phenoxide ion even though the latter is aromatic?
Why is a phenoxide ion more reactive than phenol towards an electrophilic substitution reaction?
What is an ortho-para directing effects of the groups in electrophilic aromatic substitution reactions? Explain with examples.
What are the characteristics of electrophilic substitution reactions?
What are some examples of electrophilic substitution reactions?
Why are phenols more reactive towards an electrophilic substitution reaction than anisole?
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15863 | https://thirdspacelearning.com/gcse-maths/number/rationalise-the-denominator/ | What is rationalising the denominator?
How to rationalise the denominator
Rationalising the denominator worksheet
Rationalising the denominator examples
Example 1: integer on top, surd on the bottom Example 2: integer on top, surd on the bottom Example 3: surd expression on the top and bottom Example 4: surd expression on the top and bottom, some simplification required Example 5: combining with adding or subtracting surds
Common misconceptions
Related lessons
Practice rationalising the denominator questions
Rationalising the denominator GCSE questions
Learning checklist
Next lessons
Still stuck?
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In order to access this I need to be confident with:
Equivalent fractions Rational numbers Irrational numbers Simplifying surds Adding and subtracting surds Multiplying and dividing surds
This topic is relevant for:
Introduction
What is rationalising the denominator?
How to rationalise the denominator
Rationalising the denominator worksheet
Rationalising the denominator examples
↓
Example 1: integer on top, surd on the bottom Example 2: integer on top, surd on the bottom Example 3: surd expression on the top and bottom Example 4: surd expression on the top and bottom, some simplification required Example 5: combining with adding or subtracting surds
Common misconceptions
Related lessons
Practice rationalising the denominator questions
Rationalising the denominator GCSE questions
Learning checklist
Next lessons
Still stuck?
GCSE Maths Number Surds
Rationalise The Denominator
Rationalise the Denominator
Here we will learn about rationalising the denominator of surd expressions including how to rationalise the denominator for simple surd expressions, and then extend these skills to more complicated surd fractions.
There are also rationalising the denominator worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
What is rationalising the denominator?
Rationalising the denominator is where we convert the denominator of a fraction from an irrational number to a rational number.
E.g.
[\frac{8}{\sqrt{2}}=\frac{8\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}=\frac{8\sqrt{2}}{2}=4\sqrt{2}]
A number that can be written as an integer (whole number) or a simple fraction is called a rational number.
E.g.
2, 100, −30 are all rational numbers.
So are numbers like \frac{1}{2} , \frac{3}{4} and \frac{1}{9} .
Rational numbers can also be terminating decimals like 0.5 or recurring decimals like 0.111…
Any number that can’t be written in this form is called irrational. In decimal form, these are infinite, with no recurring or repeating pattern. Surds are roots which give irrational numbers; remember that at GCSE, we only deal with square root.
All divisions can be written as fractions.
4\div 2 can be written as \frac{4}{2}
In a similar way:
[4 \div \sqrt{2}=\frac{4}{\sqrt{2}}]
However, it’s much easier in mathematics to divide by an integer where possible, so it is useful to be able to convert surd fractions with irrational denominators (surds on the bottom) to fractions with rational denominators.
We do this by using the ideas associated with equivalent fractions:
If the numerator and denominator are both multiplied by the same number or expression, the fraction remains equivalent to the original.
The process of changing the denominator of a fraction to a rational number in this way is called rationalising (or rationalizing) the denominator.
On this page, we will look at cases where the denominator is a single surd.
Step by step guide: Rationalising surds (coming soon)
What is rationalising the denominator?
How to rationalise the denominator
In order to rationalise the denominator:
Simplify any surds, if necessary.
Multiply both the numerator and the denominator by the surd in the denominator.
Simplify the answer fully.
How to rationalise the denominator
Rationalising the denominator worksheet
Get your free rationalising the denominator worksheet of 20+ questions and answers. Includes reasoning and applied questions.
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Rationalising the denominator worksheet
Get your free rationalising the denominator worksheet of 20+ questions and answers. Includes reasoning and applied questions.
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Rationalising the denominator examples
Example 1: integer on top, surd on the bottom
Rationalise the denominator:
[\frac{4}{\sqrt{2}}]
Simplify any surds, if necessary.
In this example this is already done.
2Multiply both the numerator and the denominator by the surd in the denominator.
So here we multiply the top and the bottom of the fraction by root 2:
[\frac{4 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}]
Numerator:
[4 \times \sqrt{2}=4 \sqrt{2}]
Denominator:
[\sqrt{2} \times \sqrt{2}=2]
So the full expression becomes:
[\frac{4 \sqrt{2}}{2}]
The denominator is now rationalised, because 2 is a rational number.
3Simplify the answer fully.
[4\div2=2]
So the final answer actually simplifies to
[2 \sqrt{2}]
Example 2: integer on top, surd on the bottom
Rationalise the denominator:
[\frac{6}{\sqrt{45}}]
Simplify any surds, if necessary.
Root 45 will simplify:
[\begin{aligned} \sqrt{45} &=\sqrt{9 \times 5} \ &=\sqrt{9} \times \sqrt{5} \ &=3 \sqrt{5} \end{aligned}]
Multiply both the numerator and the denominator by the surd in the denominator.
So here we multiply the top and the bottom of the fraction by root 5:
[\frac{6 \times \sqrt{5}}{3 \sqrt{5} \times \sqrt{5}}]
Numerator:
[6\times\sqrt{5}=6\sqrt{5}]
Denominator:
[3 \sqrt{5} \times \sqrt{5}=3 \times 5=15]
So the full expression becomes:
[\frac{6 \sqrt{5}}{15}]
The denominator is now rationalised, because 15 is a rational number.
Simplify the answer fully.
[6 \div 15=\frac{6}{15}=\frac{2}{5}]
So the final answer simplifies to
[\frac{2 \sqrt{5}}{5}]
Example 3: surd expression on the top and bottom
Rationalise the denominator:
[\frac{4\sqrt{3}}{\sqrt{7}}]
Simplify any surds, if necessary.
Both surds are already in their simplest forms.
Multiply both the numerator and the denominator by the surd in the denominator.
So here we multiply the top and the bottom of the fraction by root 7:
[\frac{4 \sqrt{3} \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}}]
Numerator:
[4 \sqrt{3} \times \sqrt{7}=4 \sqrt{21}]
Denominator:
[\sqrt{7} \times \sqrt{7}=7]
So the full expression becomes:
[\frac{4 \sqrt{21}}{7}]
The denominator is now rationalised, because 7 is a rational number.
Simplify the answer fully.
[\frac{4 \sqrt{21}}{7}]
is already in its simplest form; there are no common factors of the integers and there is no square factor of 21 to simplify the root.
Example 4: surd expression on the top and bottom, some simplification required
Rationalise the denominator:
[\frac{6\sqrt{8}}{\sqrt{3}}]
Simplify any surds, if necessary.
Root 8 will simplify:
[\begin{aligned} 6 \sqrt{8} &=6 \times(\sqrt{4 \times 2}) \ &=6 \times(\sqrt{4} \times \sqrt{2}) \ &=6 \times(2 \sqrt{2}) \ &=12 \sqrt{2} \end{aligned}]
Multiply both the numerator and the denominator by the surd in the denominator.
So here we multiply the top and the bottom of the fraction by root 3:
[\frac{12 \sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}]
Numerator:
[12 \sqrt{2} \times \sqrt{3}=12 \sqrt{6}]
Denominator:
[\sqrt{3} \times \sqrt{3}=3]
So the full expression becomes:
[\frac{12 \sqrt{6}}{3}]
The denominator is now rationalised, because 3 is a rational number.
Simplify the answer fully
12÷3=4
So the final answer simplifies to:
4 \sqrt{6}
Example 5: combining with adding or subtracting surds
Rationalise the denominator:
[\frac{3 \sqrt{12}+\sqrt{108}}{\sqrt{6}}]
Simplify any surds, if necessary.
Root 12 will simplify:
[\begin{aligned} 3 \sqrt{12} &=3 \times(\sqrt{4 \times 3}) \ &=3 \times(\sqrt{4} \times \sqrt{3}) \ =& 3 \times(2 \sqrt{3}) \ =& 6 \sqrt{3} \end{aligned}]
Root 108 will also simplify:
[\begin{aligned} \sqrt{108} &=\sqrt{36 \times 3} \ =& \sqrt{36} \times \sqrt{3} \ =& 6 \sqrt{3} \end{aligned}]
So the numerator is:
[6 \sqrt{3}+6 \sqrt{3}]
and because these are like surds, it will simplify to
[12 \sqrt{3}]
Multiply both the numerator and the denominator by the surd in the denominator.
So here we multiply the top and the bottom of the fraction by root 6:
[\frac{12 \sqrt{3} \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}}]
Numerator:
[12 \sqrt{3} \times \sqrt{6}=12 \sqrt{18}]
Denominator:
[\sqrt{6} \times \sqrt{6}=6]
So the full expression becomes:
[\frac{12 \sqrt{18}}{3}]
The denominator is now rationalised, because 6 is a rational number.
Simplify the answer fully.
Root 18 simplifies further:
[\begin{aligned} \sqrt{18} &=\sqrt{9 \times 2} \ &=\sqrt{9} \times \sqrt{2} \ &=3 \sqrt{2} \end{aligned}]
So the expression becomes:
[\frac{12 \times 3 \sqrt{2}}{6}=\frac{36 \sqrt{2}}{6}]
Finally, 36\div6=6 , so the fully simplified answer is:
[6\sqrt{2}]
Common misconceptions
Multiplying by the incorrect surd, if there are surds in both numerator and denominator
Always make sure you multiply both top and bottom of the fraction by the surd in the denominator of the fraction.
On step 3, if dividing integers to simplify, check that the factor you want to divide by is common to all of the integers in the numerator, as well as the denominator.
E.g.
[\frac{9+6 \sqrt{2}}{3}=3-2 \sqrt{2}]
This can be simplified, because 3 is a factor of 3 , 9 and 6 . However, \frac{4+6 \sqrt{2}}{3} cannot be simplified further, because 3 isn’t a factor of 4 .
Related lessons
Rationalise the denominator is part of our series of lessons to support revision on surds. You may find it helpful to start with the main surds lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
Surds
Rationalising surds
Multiplying and dividing surds
Simplifying surds
Adding and subtracting surds
Practice rationalising the denominator questions
Rationalise the denominator \frac{1}{\sqrt{5}}
\frac{5}{\sqrt{5}}
\frac{1}{5}
\frac{\sqrt{5}}{5}
\sqrt{5}
Multiply the numerator and denominator by root 5 .
Rationalise the denominator \frac{7}{\sqrt{3}}
\frac{7 \sqrt{3}}{3}
\frac{\sqrt{3}}{7}
\frac{1}{5}
\sqrt{21}
Multiply the numerator and denominator by root 3 .
Rationalise the denominator \frac{20}{\sqrt{40}}
\frac{20 \sqrt{40}}{40}
\frac{\sqrt{40}}{2}
\sqrt{20}
\sqrt{10}
Simplify root 40 first, then multiply the top and bottom by root 10 . Alternatively, you could multiply the top and bottom by root 40 , but remember to simplify your answer fully at the end.
Rationalise the denominator \frac{5 \sqrt{3}}{\sqrt{10}}
\sqrt{15}
\frac{5 \sqrt{30}}{10}
\frac{\sqrt{30}}{2}
\frac{\sqrt{30}}{5}
Multiply the top and bottom by root 10 , using the multiplication rule of surds: \sqrt{3} \times \sqrt{10}=\sqrt{30} Finally, simplify the integers: \frac{5}{10}=\frac{1}{2}
Rationalise the denominator \frac{5}{4-\sqrt{2}}
\frac{20+5 \sqrt{2}}{14}
7+5 \sqrt{2}
20-5 \sqrt{2}
\frac{20+5 \sqrt{2}}{4}
Multiply the top and bottom by 4-\sqrt{2} to give: \frac{5(4-\sqrt{2})}{(4+\sqrt{2})(4-\sqrt{2})} The denominator simplifies to 16-2=14 , and the numerator expands to become 20-5 \sqrt{2}
Rationalise the denominator \frac{2 \sqrt{5}}{3-\sqrt{5}}
\frac{6 \sqrt{5}-10}{4}
\frac{2}{3}
\frac{5+3 \sqrt{5}}{2}
\frac{5 \sqrt{5}}{4}
Multiply the top and bottom by 3-\sqrt{5} to give: \frac{2\sqrt{5}(3-\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})} The denominator simplifies to 9-5=4 , and the numerator expands to become 6 \sqrt{5}-10 . Divide through by 2 to give the fully simplified final answer.
Rationalising the denominator GCSE questions
Rationalise the denominator: \frac{12}{2\sqrt{2}}
(2 marks)
Show answer
\frac{12\sqrt{2}}{4} or \frac{6\sqrt{2}}{2}
(1)
3\sqrt{2}
(1)
Simplify fully \frac{(2\sqrt{6}-2)(2\sqrt{6}+2)}{\sqrt{10}}
(3 marks)
Show answer
(2\sqrt{6}-2)(2\sqrt{6}+2)=24+4\sqrt{6}-4\sqrt{6}-4
Any two terms correct
(1)
All four correct
(1)
\frac{20}{\sqrt10}=\frac{20\sqrt{10}}{10}=2\sqrt{10}
(1)
Show that \frac{\sqrt{28}-8}{\sqrt{7}+2} can be written as 10-4\sqrt{7}
(5 marks)
Show answer
\sqrt{28}=2\sqrt{7}
(1)
(1)
\frac{30-12\sqrt{7}}{3} ( 1 mark for numerator, 1 mark for denominator)
(1)
Learning checklist
You have now learned how to:
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Rationalise the denominator when it is a single surd
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Rationalise the denominator when it contains a binomial
The next lessons are
Standard form
Simple interest and compound interest
Types of numbers
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Introduction
What is rationalising the denominator?
How to rationalise the denominator
Rationalising the denominator worksheet
Rationalising the denominator examples
↓
Example 1: integer on top, surd on the bottom Example 2: integer on top, surd on the bottom Example 3: surd expression on the top and bottom Example 4: surd expression on the top and bottom, some simplification required Example 5: combining with adding or subtracting surds
Common misconceptions
Related lessons
Practice rationalising the denominator questions
Rationalising the denominator GCSE questions
Learning checklist
Next lessons
Still stuck?
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15864 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10334856/ | Parathyroid Crisis: A Case of Elective Parathyroidectomy - PMC
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. 2023 Jun 11;15(6):e40251. doi: 10.7759/cureus.40251
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Parathyroid Crisis: A Case of Elective Parathyroidectomy
Uchechi Adeniran
Uchechi Adeniran
1 Internal Medicine, Woodhull Medical Center, Brooklyn, USA
Find articles by Uchechi Adeniran
1, Beisi Ji
Beisi Ji
2 Internal Medicine, Downstate Health Science University of New York - Downstate Medical Center, Brooklyn, USA
Find articles by Beisi Ji
2,✉, Israa Hussein
Israa Hussein
3 Endocrinology, Diabetes and Metabolism, Downstate Health Science University of New York - Downstate Medical Center, Brooklyn, USA
Find articles by Israa Hussein
3, Lina Soni
Lina Soni
4 Endocrinology, Downstate Health Science University of New York - Downstate Medical Center, Brooklyn, USA
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4
Editors: Alexander Muacevic, John R Adler
Author information
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Copyright and License information
1 Internal Medicine, Woodhull Medical Center, Brooklyn, USA
2 Internal Medicine, Downstate Health Science University of New York - Downstate Medical Center, Brooklyn, USA
3 Endocrinology, Diabetes and Metabolism, Downstate Health Science University of New York - Downstate Medical Center, Brooklyn, USA
4 Endocrinology, Downstate Health Science University of New York - Downstate Medical Center, Brooklyn, USA
✉
Beisi Ji beisi.ji@downstate.edu
✉
Corresponding author.
Accepted 2023 Jun 11; Collection date 2023 Jun.
Copyright © 2023, Adeniran et al.
This is an open access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
PMC Copyright notice
PMCID: PMC10334856 PMID: 37440804
Abstract
A parathyroid crisis is characterized by a severe elevation in calcium, usually above 14-15 mg/dl alongside acute signs and symptoms of hypercalcemia. It is a rare but potentially life-threatening complication of primary hyperparathyroidism (PHPT). Among all primary hyperparathyroidism cases, parathyroid carcinoma accounts for only less than 1%. Although the definitive management is surgical parathyroidectomy, the exact timing of surgery is not well-established. We describe a case of a patient with abrupt onset of severe hypercalcemia who was managed medically and discharged for elective parathyroidectomy. This was because her workup was suspicious for parathyroid carcinoma, and there was a need to pursue a positron emission tomography (PET)-computed tomography (CT) scan to evaluate for other malignancies before proceeding with parathyroidectomy. The patient experienced the resolution of her symptoms of acute encephalopathy and improvement in her calcium levels from 22.3 mg/dl (8.8-10.2 mg/dl) on admission to 9.1 mg/dl on day 13 of hospitalization and discharge.In this report, we review the literature on the timing of parathyroidectomy in patients with a parathyroid crisis and how this has evolved over time with the use of new hypocalcemic agents. We discuss that parathyroidectomy performed emergently within 72 hours vs after 72 hours has not shown any significant impact on long-term survival in patients with parathyroid crisis. Moreover, medical management is crucial while waiting for surgery.
Keywords: primary hyperparathyroidism, hypercalcemia, elective parathyroidectomy, parathyroid carcinoma, parathyroid crisis
Introduction
A parathyroid crisis is a rare endocrine emergency for which modern management has substantially improved outcomes in mortality and morbidity. A parathyroid crisis is defined as hyperparathyroidism with a marked elevation of parathyroid hormone (PTH), severe hypercalcemia (usually >14 - 15 mg/dl), and acute symptoms such as acute encephalopathy, anorexia, vomiting, cardiac arrhythmia, and acute renal injury [1-4]. More than 80% of PHPT cases are caused by a single parathyroid adenoma while parathyroid hyperplasia accounts for 10%, double adenoma for 4%, and parathyroid carcinoma for less than 1% .
The treatment of a parathyroid crisis is to normalize calcium levels. The initial and most important step in the management of hypercalcemia is to administer intravenous isotonic saline for extracellular volume expansion and promote hypercalciuria, with or without the usage of loop diuretics to prevent volume overload. Biphosphonate and calcitonin can also be used to treat parathyroid crises . For parathyroid carcinoma, the definitive treatment remains surgical parathyroidectomy, however, the exact timing is not well-established.
Here, we present a case of a patient who presented with confusion and weakness, found to have a parathyroid crisis, who was medically optimized and discharged with follow-up for adequate surgical planning prior to elective parathyroidectomy.
Case presentation
A 72-year-old female was brought in by Emergency Medical Services after she was found with confusion and extreme weakness in her apartment. The patient was last seen well by her family two days prior. She had a history of hypertension and hyperlipidemia and was taking antihypertensives medications (lisinopril and amlodipine) and atorvastatin. She also had a remote history of kidney stones. A review of systems was significant for polyuria, polydipsia, urinary incontinence, and muscle weakness.
Laboratory studies were significant for serum calcium level of 22.3 mg/dl (8.8-10.2 mg/dl, the baseline was 13.6 mg/dl one week prior), ionized calcium of 2.94 mmol/L (1.16-1.32 mmol/L), albumin of 4.6 g/dl (2.8-5.7 g/dl), chloride/phosphate ratio of 35. PTH of 1,077 pg/ml (15.0-65.0 pg/ml), 25-OH vitamin D of 56.68 ng/ml (>=30.00 ng/ml), 1, 25-OH vitamin D of 75.2 pg/ml (19.9-79.3 pg/ml), phosphorus of 2.9 mg/dl (2.5-4.5 mg/dl), estimated glomerular filtration rate (eGFR) of 37.0 ml/min/1.73m 2 (>60 ml/min/1.73 m 2, was normal one week prior), blood urea nitrogen (BUN) of 19.0 mg/dl(8.0-23.0 mg/dl), creatinine of 1.50 mg/dl (0.50-0.90 mg/dl, the baseline was 0.57 mg/dl one week prior). The patient was treated with intravenous normal saline, zoledronic acid, and calcitonin. We closely monitored her calcium levels and mental status. On day 4 of hospitalization, her calcium level went down to 10.1 mg/dl, and she was alert and oriented to person, place, and time and was back to her baseline function.
Ultrasound of the thyroid gland showed a heterogeneous multinodular thyroid gland with the largest nodule measuring 1.3 cm in the left thyroid lobe and 1.2 cm in the right thyroid lobe. The nuclear medicine parathyroid scan revealed foci of abnormal tracer uptake in the region of the right and left poles, suggestive of parathyroid hyperplasia and multiple foci adenomas and demonstrated increased uptake in the right chest and right axilla suspicious of malignancy (Figure 1). The computed tomography (CT) scan of the soft tissue neck with contrast showed bilateral retro thyroid nodules with enhancement - a 12 x 9 x 17 mm nodule in the left and a 15 x 11 x 20 mm nodule in the right, consistent with parathyroid adenomas.
Figure 1. Nuclear medicine parathyroid scan .
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Nuclear medicine parathyroid scan showing foci of abnormal tracer uptake in the region of the right and left poles suggestive of parathyroid hyperplasia and multiple foci adenomas and increased uptake in the right chest and right axilla suspicious of malignancy (arrow)
The patient was discharged on day 13 with a calcium level of 9.1 mg/dl (Table 1), and she was discharged on cinacalcet 30 mg twice a day. She was instructed to follow up for a positron emission tomography (PET)-computed tomography (CT) scan of the whole body to assess for carcinoma metastasis (due to the parathyroid scan showing increased uptake in the right chest and right axilla suspicious of malignancy)and evaluate for other malignancy prior to surgical parathyroidectomy. The patient was also scheduled to follow up outpatient with Otolaryngology and Endocrinology.
Table 1. Routine biochemical profile, including the metabolic panel and blood count one week before admission, admission day 1, admission day 3, discharge, and latest follow-up.
NA: not available; WBC: white blood cell; BUN: blood urine nitrogen; GFR: estimated glomerular filtration rate; PTH: parathyroid hormone; PTHrP: parathyroid hormone-related protein
Variable One week before admission Admission day 1 Admission day 3 Discharge Latest follow-up Reference range
WBC (K/uL)7.37 12.98 9.17 8.34 NA 4.5 – 10.9
Hemoglobin (g/dL)12.6 14.5 12.9 12.0 NA 12.0 – 16.0
Calcium (mg/dL)13.5 22.3 10.1 9.1 10.0 8.8 – 10.2
Albumin (g/dL)4.0 4.6 4.1 4.0 4.3 2.8 – 5.7
BUN (mg/dL)6.0 19.0 15.0 14.0 13.0 8.0 – 23.0
Creatinine (mg/dL)0.64 1.5 1.19 0.95 0.67 0.5 – 0.9
GFR (ml/min/1.73m2)>60.0 37.0 48.9>60.0>60.0>60.0
Magnesium (mg/dL)NA 1.48 1.72 1.65 NA 1.6 – 2.6
Phosphorus (mg/dL)NA 2.9 2.1 3.1 NA 2.5 – 4.5
PTH (pg/mL)NA 1077.0 NA NA 61.0 15.0 - 65.0
25-Hydroxy Vitamin D (ng/mL)NA 56.68 NA NA 59.67>= 30
1,25-Hydroxy Vitamin D (pg/mL)NA 75.2 NA NA NA 19.9 – 79.3
PTHrP (pmol/L)NA<2.0 NA NA NA<2.0
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Discussion
A parathyroid crisis is a potentially fatal complication of primary hyperparathyroidism. Severe hypercalcemia occurs in 1-2% of patients with primary hyperparathyroidism. Other complications, such as renal stones, renal insufficiency, proximal muscle weakness, brown tumor, and neuropsychiatric syndrome, are common in patients with parathyroid crises .
Although the parathyroid hormone was demonstrated in 1926 when Collip injected his dog with large doses of parathyroid extract, leading to a rise in serum calcium to 20 mg/100 ml and death,preceded by anorexia and vomiting, the parathyroid crisis was not recognized in man until 1932. A boy was treated with repeated injections of parathyroid extract for purpura, which resulted in persistent vomiting and lethargy, and the boy was close to dying. However, he recovered when the parathyroid extracts were discontinued . The first case of the parathyroid crisis was reported by Hanes in 1938 and was caused by a parathyroid adenoma.
The management strategies for parathyroid crises have evolved over time with the advent of new hypocalcemic drugs. Although the definitive treatment remains surgical parathyroidectomy, the exact timing is not well-established. Wang and Guyton conducted a study that included 14 patients with a parathyroid crisis at Massachusetts General Hospital between 1964 and 1978 and concluded that prompt surgical intervention was the ideal treatment for parathyroid crises, preferably within 72 hours of acute onset of symptoms. However, medical treatment at the time was limited to the use of loop diuretics (furosemide), prednisone, potassium phosphate/sodium acid phosphate, and calcitonin in most patients . Phytayakorn and McHenry, later on, described the use of bisphosphonate as an effective bridge to parathyroidectomy in patients with parathyroid crises . Lew et al. carried out a study on the long-term results of parathyroidectomy for hypercalcemic crises. They reviewed 1055 consecutive patients who underwent parathyroidectomy at their institution from January 1, 1969, to October 31, 2004. They identified 43 patients with a hypercalcemic crisis. All the patients were noted to have received medical optimization with Intravenous fluids, loop diuretics, and hypocalcemic drugs (including bisphosphonate), followed by surgical excision. The average time between admission and parathyroidectomy was five days (range of 1-29 days). They concluded that there was no significant difference in long-term survival in patients treated with emergency parathyroidectomy in less than 72 hrs vs. parathyroidectomy done 72 hrs or more. Early surgical intervention was proposed [8,9]. Recent case series are in favor of early surgery after medical optimization rather than emergency surgery. Priority should be on medical management while workup and diagnostic tests are being pursued, and surgery should be expedited based on the patient's suitability and comorbidity .
In some patients, like our case, in which there was a need for more imaging studies to evaluate for malignancy owing to suspicious findings of increased uptake in the right chest and right axilla on the parathyroid scan; this raises the challenge of balancing between having all study reports available for optimal surgical planning vs. pursuing an emergent surgical intervention. Our patient was medically optimized, with improvement in serum calcium from 22.3 mg/dl on admission to 9.1 mg/dl and improvement in mental status and kidney function. The patient was subsequently scheduled to follow up for elective surgical parathyroidectomy after adequate imaging studies had been performed.
Conclusions
A parathyroid crisis carries a high potential for mortality and morbidity if not treated promptly. With the advent of more medical therapeutic options, studies have shown no significant difference in long-term survival in patients who received an emergency parathyroidectomy performed within 72 hours vs. parathyroidectomy done after 72 hours. The focus is on the medical optimization of the patient while they await surgery.
The authors have declared that no competing interests exist.
Human Ethics
Consent was obtained or waived by all participants in this study
References
1.Hyperparathyroid crisis: clinical and pathologic studies of 14 patients. Wang CA, Guyton SW. Ann Surg. 1979;190:782–790. doi: 10.1097/00000658-197912000-00019. [DOI] [PMC free article] [PubMed] [Google Scholar]
2.Hyperparathyroid crisis. A collective review. MacLeod WA, Holloway CK. Ann Surg. 1967;166:1012–1015. doi: 10.1097/00000658-196712000-00021. [DOI] [PMC free article] [PubMed] [Google Scholar]
3.A review of primary hyperparathyroidism. Owens BB. J Infus Nurs. 2009;32:87–92. doi: 10.1097/NAN.0b013e318198d483. [DOI] [PubMed] [Google Scholar]
4.Hyperparathyroid crisis: the timing of surgery. Yu HH, Lou SY, Chou YH, Chan HM, Chen HT, Huang SM. Asian J Surg. 2011;34:147–152. doi: 10.1016/j.asjsur.2011.11.004. [DOI] [PubMed] [Google Scholar]
5.Hypercalcaemic crisis due to primary hyperparathyroidism: report of two cases. Sala TD, Mureşan S, Roman R, Lazăr A, Ion R, Paşcanu I. J Crit Care Med (Targu Mures) 2019;5:34–39. doi: 10.2478/jccm-2019-0004. [DOI] [PMC free article] [PubMed] [Google Scholar]
6.Parathyroid crisis in a 20 year old--an unusual cause of hypercalcaemic crisis. Wong P, Carmeci C, Jeffrey RB, Weigel RJ. Postgrad Med J. 2001;77:468–470. doi: 10.1136/pmj.77.909.468. [DOI] [PMC free article] [PubMed] [Google Scholar]
7.Hyperparathyroid crisis: use of bisphosphonates as a bridge to parathyroidectomy. Phitayakorn R, McHenry CR. J Am Coll Surg. 2008;206:1106–1115. doi: 10.1016/j.jamcollsurg.2007.11.010. [DOI] [PubMed] [Google Scholar]
8.Hypercalcemic crisis: a clinical review. Ahmad S, Kuraganti G, Steenkamp D. Am J Med. 2015;128:239–245. doi: 10.1016/j.amjmed.2014.09.030. [DOI] [PubMed] [Google Scholar]
9.Long-term results of parathyroidectomy for hypercalcemic crisis. Lew JI, Solorzano CC, Irvin GL 3rd. Arch Surg. 2006;141:696–699. doi: 10.1001/archsurg.141.7.696. [DOI] [PubMed] [Google Scholar]
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15865 | https://fiveable.me/key-terms/hs-honors-geometry/obtuse-triangle-circumcenter-location | Obtuse Triangle Circumcenter Location - (Honors Geometry) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Obtuse Triangle Circumcenter Location
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key term - Obtuse Triangle Circumcenter Location
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Definition
The circumcenter of an obtuse triangle is the point where the perpendicular bisectors of the sides intersect, and it lies outside the triangle. This is a significant characteristic since it distinguishes obtuse triangles from acute and right triangles, whose circumcenters are located inside or on the triangle, respectively. The location of the circumcenter helps in understanding various properties related to triangles, such as the relationship between angles and the lengths of sides.
5 Must Know Facts For Your Next Test
In an obtuse triangle, the circumcenter lies outside the triangle because one angle is greater than 90 degrees.
The circumcenter can be found by constructing the perpendicular bisectors of at least two sides of the triangle, which will intersect at this point.
This property of the circumcenter helps in solving various geometric problems involving triangle constructions and calculations.
Unlike acute triangles, where the circumcenter is always inside, obtuse triangles have a unique location that can affect certain geometric relationships.
The circumradius (the radius of the circumcircle) of an obtuse triangle can be larger than that of an acute triangle due to its circumcenter being farther from the vertices.
Review Questions
How does the location of the circumcenter in an obtuse triangle compare to that in acute and right triangles?
In an obtuse triangle, the circumcenter is located outside the triangle due to one of its angles being greater than 90 degrees. In contrast, for acute triangles, the circumcenter lies inside because all angles are less than 90 degrees. For right triangles, the circumcenter is found at the midpoint of the hypotenuse. Understanding these differences helps in analyzing various properties and relationships within different types of triangles.
What role do perpendicular bisectors play in determining the circumcenter of an obtuse triangle?
Perpendicular bisectors are essential for finding the circumcenter as they intersect at this point. For an obtuse triangle, you can construct perpendicular bisectors for any two sides, and their intersection will yield the circumcenter outside the triangle. This technique highlights how side lengths and angles relate in determining key triangle centers.
Evaluate how understanding the location of a circumcenter in obtuse triangles can aid in solving real-world problems involving geometry.
Knowing that the circumcenter of an obtuse triangle is outside helps in real-world applications like construction and design where triangles are involved. This knowledge allows for better planning in layouts that rely on precise distances from vertices to a central point, such as in creating parks or buildings with triangular plots. Moreover, it assists in calculating distances in navigation and even in computer graphics where triangle meshes are common.
Related terms
Circumcircle:
A circumcircle is the circle that passes through all three vertices of a triangle, with its center being the circumcenter.
Perpendicular Bisector:
A perpendicular bisector is a line that divides a segment into two equal parts at a right angle, crucial for locating the circumcenter.
Acute Triangle:An acute triangle is a triangle where all three angles are less than 90 degrees, with its circumcenter located inside the triangle.
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15866 | https://blogs.30dayscoding.com/blogs/dsa/advanced-algorithms/number-theory-algorithms/modular-exponentiation/ | Data structure and Algorithm
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Modular Exponentiation
In the field of number theory, modular exponentiation is a fundamental
algorithm used to efficiently calculate large exponents modulo a given number.
This algorithm plays a crucial role in various cryptographic systems and
mathematical calculations. In this article, we will dive deep into the concept
of modular exponentiation and explore its step-by-step implementation.
Understanding Modular Exponentiation
To understand modular exponentiation, we need to first grasp the concept of
modular arithmetic. In modular arithmetic, numbers “wrap around” after
reaching a certain value called the modulus. The result is always within the
range of 0 to (modulus - 1). In the case of modular exponentiation, we are
interested in calculating the result of a number raised to a large exponent
modulo a given number.
Algorithm Steps:
Step-by-Step Implementation
Let’s go through a detailed step-by-step implementation of the modular
exponentiation algorithm.
`while (exponent > 0) { if (exponent % 2 === 1) { result = (result base) %
modulus; } base = (base base) % modulus; exponent = Math.floor(exponent /
2); } return result; } ```
The code above demonstrates a simple implementation of the modular
exponentiation algorithm using JavaScript. Let's break it down:
We define a function named c modularExponentiation which takes three parameters: the base, the exponent, and the modulus.
We initialize the result to 1.
We enter a while loop that continues until the exponent is greater than 0.
Inside the loop, we check if the current bit of the exponent is 1 (using the modulus operator).
If the current bit is 1, we multiply the result by the base and take the modulus to ensure the result stays within the range of the modulus.
We square the base and take the modulus to calculate the next iteration's value.
We divide the exponent by 2 using integer division to move to the next bit of the binary representation.
Once the while loop completes, we return the final calculated result.
By following the above steps, we can efficiently calculate the modular
exponentiation of a number. This algorithm greatly reduces the computational
complexity involved in performing exponentiation operations on large integers.
Benefits and Applications
The modular exponentiation algorithm offers several benefits and finds
applications in various fields:
Cryptographic Systems: Modular exponentiation is a key component in many cryptographic systems, including public key encryption algorithms like RSA. It enables secure communication and data encryption by providing a fast and efficient way to perform modular exponentiation operations.
Number Theory: Modular exponentiation plays a significant role in number theory research, especially when dealing with prime numbers, discrete logarithms, and primality testing algorithms. It allows mathematicians to explore and analyze properties of numbers and develop new mathematical theories.
Computational Mathematics: Many computational mathematics problems involve large exponents and modular operations. The modular exponentiation algorithm provides an efficient way to solve such problems without encountering performance issues due to the size of the numbers involved.
FAQ
Q1: What is the purpose of modular exponentiation?
A1: The purpose of modular exponentiation is to efficiently calculate the
result of a number raised to a large exponent modulo a given number. It is
extensively used in cryptography, number theory, and computational
mathematics.
Q2: Can modular exponentiation be performed with negative numbers?
A2: Yes, modular exponentiation can be performed with negative numbers.
However, the negative numbers should adhere to the rules of modular
arithmetic, where the result always falls within the range of 0 to (modulus -
1).
Q3: What happens if the base or exponent is negative?
A3: If the base or exponent is negative, the modular exponentiation algorithm
can still be applied. However, the negative sign is typically ignored during
the calculation, as the modular arithmetic principle ensures the result
remains within the defined range.
Q4: Are there any optimized techniques for modular exponentiation?
A4: Yes, several optimized techniques exist for modular exponentiation, such
as the Montgomery reduction algorithm and the sliding window technique. These
techniques further enhance the efficiency of the algorithm for specific
scenarios and use cases.
Q5: Can modular exponentiation be used for non-integer exponents?
A5: No, the modular exponentiation algorithm is designed specifically for
integer exponents. It operates on the binary representation of the exponent,
utilizing the bit values to efficiently perform the exponentiation
calculation.
🎉 Congrats! You've now gained a solid understanding of the modular
exponentiation algorithm and its applications. Feel free to explore further
and experiment with the code provided to deepen your knowledge in this
fascinating field of number theory. Happy coding! ✨
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15867 | https://www.webelements.com/compounds/barium/barium_dihydride.html | | Rb | Sr | Y |
---
| Cs | Ba | La |
| Fr | Ra | Ac |
Barium dihydride
The following are some synonyms of barium dihydride:
The oxidation number of barium in barium dihydride is 2.
Synthesis
Barium hydride is made by the reaction of barium metal and hydrogen gas at high temperature and pressure.
Ba(s) + H2(g) → BaH2(s)
Solid state structure
Element analysis
The table shows element percentages for BaH2 (barium dihydride).
| Element | % |
--- |
| Ba | 98.55 |
| H | 1.45 |
Isotope pattern for BaH2
The chart below shows the calculated isotope pattern for the formula BaH2 with the most intense ion set to 100%.
References
The data on these compounds pages are assembled and adapted from the primary literature and several other sources including the following.
Explore periodic propertes from these links
WebElements: THE periodic table on the WWW [www.webelements.com]
Copyright 1993-2025 Mark Winter [ The University of Sheffield and WebElements Ltd, UK]. All rights reserved.
You can reference the WebElements periodic table as follows:
"WebElements, accessed September 2025." |
15868 | https://advancedmathyoungstudents.com/blog/?p=71 | Skip to primary content
Advanced Math for Young Students
A high school physics teacher's mathematical digressions
Rectangles and the Golden Ratio
Posted on by Administrator
These may all be digressions, but (as I reassure my students all the time), stay with me. I’m heading somewhere with all of this. For now, let’s just keep going.
Here is a rectangle:
And, yes, in Your First Book of Shapes or its equivalent, that was the shape on the “Square” page. But then in geometry class, you learned that the squares are a subset of the rectangles. They just happen to have equal length and width. Still, when I say “picture a rectangle”, you probably don’t picture a square. So I’ll try again.
Here is a rectangle:
OK, that is surely a rectangle and not a square. But is it what you picture when I say “rectangle”? I’m guessing that my rectangle is skinnier than the one in your mind. Some rectangles are more “rectangular-y” than others. We could say they have higher “rectangularity”. But is there a way to put a number on that? Actually, there is. We can use ratios.
Ratios are perfect for this job. A ratio is a comparison using division. To find the ratio of A to B, we calculate A÷B. The answer tells us how many times bigger A is than B.
So let’s define the “Rectangularity” of a rectangle as the ratio of its length to its width. In that case, my first rectangle had rectangularity of 1 which made it a square. And my second rectangle had rectangularity of 12. Its length was 12 times as great as its width.
We can keep calling this ratio “rectangularity”, a name that I made up, but in fact it already has a name: the length-to-width ratio of a rectangle is called its “aspect ratio”. And in classical design, there is an aspect ratio that is considered ideal. It just looks right: not too square-y, not too rectangular-y …just right. It is called the Golden Ratio. You can find tons of information about it on-line so I won’t try to replicate it all here. I just want to look at one interesting feature of a golden rectangle and show how we can use that feature to calculate the Golden Ratio.
Start with a Golden Rectangle…
It looks nice. But here’s the interesting property.
Slice off a square on one end…that little rectangle that remains is also a golden rectangle. And yes, if you slice off the end square of this little rectangle, you get another even littler golden rectangle.
Repeat as many times as you like to produce an endless stream of ever-smaller golden rectangles.
Let’s start again. Suppose the width of the rectangle = 1 and the length = x.
Since the ratio x/1=x, x must also be the value of the golden ratio. But what is that value?
Well, for the little rectangle, the length is 1 and the width is x-1. So that rectangle has an aspect ratio of 1/(x-1). But that aspect ratio is the same as the aspect ratio of the bigger rectangle we started with. We can write:
To find the value of the golden ratio, we just have to solve that equation. When you cross-multiply, you get:
x(x-1) = 1
You can distribute to get:
x2 – x = 1
But now what? If you use trial and error with a calculator, you will soon find that the golden ratio is a little more than 1.6. But if you want an exact solution, you need to know how to handle quadratic equations.
And that’s what I really want to talk about. Quadratic equations pop up in physics. Not every day, but occasionally. And they usually can’t be factored. (Neither can the one we have here.) My students deal with them a variety of ways. Some grab their TI89s and let the calculator do the work. Some of them use the quadratic formula. Almost nobody “completes the square”.
I admit that I used to make fun of completing the square as a classic piece of cook-book math – all procedure, no understanding. But I think I was wrong about that. I wasn’t picturing the procedure geometrically. And, as I have mentioned in a few other posts, it’s better with pictures. I’ll explain in the next post.
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A distinctive feature of this shape is that when a square section is removed, the remainder is another golden rectangle ; that is, with the same aspect ratio as the first. Square removal can be repeated infinitely, in which case corresponding corners of the squares form an infinite sequence of points on the golden spiral , the unique logarithmic spiral with this property.
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15869 | https://www.cs.ubc.ca/~schmidtm/Documents/2016_540_Argmax.pdf | Argmax and Max Calculus Mark Schmidt January 6, 2016 1 Argmax, Max, and Supremum We define the argmax of a function f defined on a set D as argmax x∈D f(x) = {x|f(x) ≥f(y), ∀y ∈D}.
In other words, it is the set of inputs x from the domain D that achieve the highest function value. For example, argmaxx∈R −x2 = {0}, since −x2 is maximized when x = 0. Note that the output of the argmax function is a set, since more than one value might achieve the maximum. The result of the argmax function might even be an infinite set. For example, the argmax of cos(x) is the set containing all integer multiples of 2π.
We can define the max of a function f defined on a set D as max x∈D f(x) = f(x∗), for any x∗∈argmax x∈D f(x).
The max function gives the largest possible value of f(x) for any x in the domain, which is the function value achieved by any element of the argmax. Unlike the argmax, the max function is unique since all elements of the argmax achieve the same value. However, the max may not exist because the argmax may be empty.
For example, if f(x) = x then f just keeps growing as we increase x (there is no largest value). A more subtle issue happens if we consider f(x) = −ex. In this case, the function is never larger than 0 but the maximum doesn’t exist since we can always move it closer to 0 by decreasing x.
To handle functions like f(x) = −ex, we define the sup function (‘supremum’) as the smallest value of the set {y|y ≥f(x), ∀x ∈D}. That is, it’s the smallest value that is greater than or equal to f(x) for any x in D. Often the sup is equal to the max, but the sup is sometimes defined even when the max is not defined.
For example, supx∈R −x2 = maxx∈R −x2 = 0 but supx∈R −ex = 0 even though the max is not defined.
If we are talking about smallest elements instead of largest elements, we replace argmax/max/sup by argmin/min/inf. We can relate minimization to maximization (or even define it) by maximizing the negation of f, argmin x∈D f(x) = argmax x∈D −f(x), min x∈D f(x) = −max x∈D −f(x), inf x∈D f(x) = −sup x∈D −f(x).
When the variable/domain are obvious from context, we sometimes use a simpler notation for the max and argmax, argmax f(x) = {x|f(x) ≥f(y), ∀y ∈D} max f(x) = f(x∗), for any x∗∈argmax x∈D f(x) sup f(x) = min y|y≥f(x),∀x∈D y.
1 2 Operations that Preserve Argmax There are a variety of operations that do not change the argmax set. Since the argmax is defined by an inequality, these are basically operations tha preserve inequalities. Below are some examples.
1. If θ is a constant then we have argmax f(x) = argmax f(x) + θ.
2. If θ > 0 we have argmax f(x) = argmax θf(x).
3. If θ < 0 we have argmax f(x) = argmin θf(x).
4. If argmax f(x) > 0, then argmax f(x) = argmin 1 f(x).
5. If g is strictly monotonic, meaning that α > β implies g(α) > g(β), then argmax g(f(x)) = argmax f(x).
The last property is used frequently when doing calculations involving probabilities, since the logarithm (a strictly monotonic function) transforms multiplication of probabilities into addition of log-probabilities, argmax n Y i=1 pi(x) = argmax n X i=1 log pi(x).
3 Max Identities We also have a set of related identities for the max function.
1. If θ is a constant we have max{f(x) + θ} = θ + max f(x).
2. If θ > 0 we have max θf(x) = θ max f(x).
3. If θ < 0 we have max θf(x) = θ min f(x).
4. If argmax f(x) > 0, then max 1 f(x) = 1 min f(x).
5. If g is strictly monotonic then max g(f(x)) = g(max f(x)).
As before, we assume that the max functions exist. Similar identities hold for the sup funciton.
2 4 Inequalities A variety of inequalities follow from the definition of the max function.
1. If y ∈D, then f(y) ≤max f(x), and similarly for any x∗∈argmaxf(x) we have f(y) ≤f(x∗).
A generalization of this is that for a subset D′ ⊆D we have max x∈D′ f(x) ≤max x∈D f(x).
2. We also have a triangle-like inequality max{f(x) + g(x)} ≤max{f(x)} + max{g(x)}.
3. A looser version of the triangle inequality is max{f(x) + g(x)} ≤2 max{max f(x), max g(x)}.
If you have n functions instead of 2, you replace the 2 by n and take the outer max over the maxima of all functions. A notable variant of this is that if we have n numbers xi (for i = 1, 2, . . . , n), then we have n X i=1 xi ≤n max i xi, or written another way, that the maximum is larger than the average, max i xi ≥1 n n X i=1 xi.
4. We have a special case of the Cauchy-Schwartz inequality, max{f(x)g(x)} ≤max{|f(x)|} max{|g(x)|}, 5. If f(x) > 0 for all x and g(x) > 0 for all x then we have a reciprocal version, max f(x) g(x) ≤max f(x) min g(x) .
6. A more subtle inequality is that if we have n positive numbers xi and n numbers yi then max xi yi ≥ Pn i=1 xi Pn i=1 yi .
3 |
15870 | https://byjus.com/maths/square-root-of-8/ | The square root of 8 in radical form is represented as √8 which is also equal to 2√2 and as a fraction, it is equal to 2.828 approximately. Squares root of a number is the number, which, on multiplying by itself gives the original number. Since 8 is not a perfect square, hence the value is represented in root form.
The value 2√2 is said to be surd, as it cannot be further simplified. As we all know, 8 = 2 × 2 × 2, thus we can see the number is a perfect cube of 2. So, we can determine the cube root of 8 which is equal to 2, such as 3√8 = 2. But for numbers which are perfect squares such as 4, 9, 16, 25, etc. the root for them can be determined easily, such as;
√4 = √(2 x 2) = 2
√9 = √(3 x 3) = 3
√16 = √(4 x 4) = 4
√25 = √(5 x 5) = 5
| |
| Square Root of 8 √8 = 2.82842712475 √8 = 2.828 up to three places of decimal |
Now that you know the value of root 8, let us calculate its value here. Although finding the square root of a number by using calculators is very easy, but students should know how to calculate it manually.
The square root is represented by the symbol, ‘√’. This symbol ‘√’ is called a radical symbol or radix. The number underneath the radical symbol or radix is called as radicand.
Method to Find Value of Root 8
To give you an idea, the square of 2 is 4 and the square of 3 is 9. So, the square root of 8 essentially lies between these two digits. But, since 3 square is 9, which is greater than 8, it most probably lies between 2.8 or 2.9. The actual answer for the square root of 8 is 2.82842712475. And this is really close to our estimate. Similarly, you could use this technique to find out square roots of other numbers.
Do you know what the square root of 8 to the power of 3 is? To solve the problem, we need to take the square root of 8. Then, we take that value and multiply by itself 3 times.
So essentially, square root of 8 to the power of 3 = (2.828 x 2.828 x 2.828) = 22.627.
Now, it’s easy to find the square root of 8, and we also learned what square root of 8 to the power of 3 is.
Also learn to find the square root of 6, 7, 9, 10, 11 and 12, by reaching to the below-mentioned links.
| |
| Square Root Of 6 Square Root Of 7 Square Root Of 9 Square Root Of 10 Square Root Of 11 Square Root Of 12 |
Square root Table From 1 to 15
In the table given here, find the values of root of numbers from 1 to 15 to use it for mathematical calculations.
| | | |
---
| Number | Squares | Square Root (Upto 3 places of decimal) |
| 1 | 12 = 1 | √1 = 1 |
| 2 | 22 = 4 | √2 = 1.414 |
| 3 | 32 = 9 | √3 = 1.732 |
| 4 | 42 = 16 | √4 = 2.000 |
| 5 | 52 = 25 | √5 = 2.236 |
| 6 | 62 = 36 | √6 = 2.449 |
| 7 | 72 = 49 | √7 = 2.646 |
| 8 | 82 = 64 | √8 = 2.828 |
| 9 | 92 = 81 | √9 = 3.000 |
| 10 | 102 = 100 | √10 = 3.162` |
| 11 | 112 = 121 | √11 = 3.317 |
| 12 | 122 = 144 | √12 = 3.464 |
| 13 | 132 = 169 | √13 = 3.606 |
| 14 | 142 = 196 | √14 = 3.742 |
| 15 | 152 = 225 | √15 = 3.873 |
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15871 | https://empossible.net/wp-content/uploads/2018/03/Topic-3-Linear-Wire-Antennas-Part-0.pdf | Linear Wire Antennas EE-4382/5306 - Antenna Engineering Outline • Introduction • Infinitesimal Dipole • Small Dipole • Finite Length Dipole • Half-Wave Dipole • Ground Effect 2 Linear Wire Antennas Constantine A. Balanis, Antenna Theory: Analysis and Design 4th Ed., Wiley, 2016.
Stutzman, Thiele, Antenna Theory and Design 3rd Ed., Wiley, 2012.
Introduction 3 Wire Antennas - Introduction Slide 4 Wire antennas are the simplest, cheapest, and many times most effective antennas for many applications.
The easiness of configuration and analysis is why we begin with these types of antennas.
Linear Wire Antennas Infinitesimal Dipole 5 Infinitesimal Dipole Slide 6 An infinitesimal dipole is in the order of 𝑙≪λ, 𝑎≪𝜆, where 𝑙is the length of the antenna, and 𝑎is the thickness.
The simplest antenna, it is just an open wire fed at the center with an alternating source.
To simplify the mathematical analysis, we will assume an infinitesimal vertical dipole placed along the z-axis, and the current is constant throughout the wire.
𝐈z′ = ෝ 𝒂𝒛𝐼0 𝐿≪𝜆 z 𝐼0 𝐼 Linear Wire Antennas Infinitesimal Dipole Slide 7 Linear Wire Antennas Infinitesimal Dipole - Fields Slide 8 For an electric current source, the magnetic field is equal to 𝐻𝜃= 𝐻𝑟= 0 𝐻𝜙= 𝑗𝑘𝐼0𝑙sin 𝜃 4𝜋𝑟 1 + 1 𝑗𝑘𝑟𝑒−𝑗𝑘𝑟 The electric field is equal to 𝐸𝜙= 0 𝐸𝑟= 𝜂𝐼0𝑙cos 𝜃 2𝜋𝑟2 1 + 1 𝑗𝑘𝑟𝑒−𝑗𝑘𝑟 𝐸𝜃= 𝑗𝜂𝑘𝐼0𝑙sin 𝜃 4𝜋𝑟 1 + 1 𝑗𝑘𝑟− 1 𝑘𝑟2 𝑒−𝑗𝑘𝑟 Linear Wire Antennas Infinitesimal Dipole - Fields Slide 9 Linear Wire Antennas Infinitesimal Dipole – Near-Field Slide 10 At the near-field region 𝑘𝑟≪1, the fields with higher order terms dominate: 𝐸𝜙= 𝐻𝜃= 𝐻𝑟= 0 𝐻𝜙= 𝑗𝑘𝐼0𝑙sin 𝜃 4𝜋𝑟 1 + 1 𝑗𝑘𝑟𝑒−𝑗𝑘𝑟= 𝐼0𝑙𝑒−𝑗𝑘𝑟 4𝜋𝑟2 sin(𝜃) 𝐸𝑟= 𝜂𝐼0𝑙cos 𝜃 2𝜋𝑟2 1 + 1 𝑗𝑘𝑟𝑒−𝑗𝑘𝑟= −𝑗𝜂𝐼0𝑙𝑒−𝑗𝑘𝑟 2𝜋𝑘𝑟3 cos(𝜃) 𝐸𝜃= 𝑗𝜂𝑘𝐼0𝑙sin 𝜃 4𝜋𝑟 1 + 1 𝑗𝑘𝑟− 1 𝑘𝑟2 𝑒−𝑗𝑘𝑟= −𝑗𝜂𝐼0𝑙𝑒−𝑗𝑘𝑟 2𝜋𝑘𝑟3 sin(𝜃) Linear Wire Antennas Infinitesimal Dipole - Far-field Slide 11 At the far-field region 𝑘𝑟≫1, the fields with lower order terms dominate: 𝐸𝜙= 𝐻𝜃= 𝐻𝑟= 0 𝐻𝜙= 𝑗𝑘𝐼0𝑙sin 𝜃 4𝜋𝑟 1 + 1 𝑗𝑘𝑟𝑒−𝑗𝑘𝑟= 𝑗𝑘𝐼0𝑙𝑒−𝑗𝑘𝑟 4𝜋𝑟 sin(𝜃) 𝐸𝑟= 𝜂𝐼0𝑙cos 𝜃 2𝜋𝑟2 1 + 1 𝑗𝑘𝑟𝑒−𝑗𝑘𝑟≅0 𝐸𝜃= 𝑗𝜂𝑘𝐼0𝑙sin 𝜃 4𝜋𝑟 1 + 1 𝑗𝑘𝑟− 1 𝑘𝑟2 𝑒−𝑗𝑘𝑟= 𝑗𝜂𝑘𝐼0𝑙𝑒−𝑗𝑘𝑟 4𝜋𝑟 sin(𝜃) Linear Wire Antennas Power Density: For a lossless antenna, the real part of the input resistance is designated as radiation resistance.
Through the mechanism of radiation resistance, the power from guided waves transfers to free-space waves.
Recall the complex Poynting Vector: 𝐖= 1 2 𝐄× 𝐇∗= 1 2 ෝ 𝒂𝒓𝐸𝑟+ ෝ 𝒂𝜽𝐸𝜃× ෝ 𝒂𝝓𝐻𝜙 ∗ = 1 2 (ෝ 𝒂𝒓𝐸𝜃𝐻𝜙 ∗−ෝ 𝒂𝜽𝐸𝑟𝐻𝜙 ∗) Thus we get two components for power: 𝑊 𝑟= 𝜂 8 𝐼0𝑙 𝜆 2 sin2(𝜃) 𝑟2 1 −𝑗 1 𝑘𝑟3 𝑊 𝜃= 𝑗𝜂𝑘𝐼0𝑙2 cos 𝜃sin 𝜃 16𝜋2𝑟3 1 + 1 𝑘𝑟2 Infinitesimal Dipole – Power Density Slide 12 Linear Wire Antennas Infinitesimal Dipole – Power Density Slide 13 Power Density: The complex power in the radial direction is obtained by integrating the components over the closed sphere: 𝑃= 𝑆 𝐖∙𝑑𝒔= න 0 2𝜋 න 0 𝜋 ෝ 𝒂𝒓𝑊 𝑟+ ෝ 𝒂𝜽𝑊 𝜃∙ෝ 𝒂𝒓𝑟2 sin 𝜃𝑑𝜃𝑑𝜙 And we obtain 𝑃= 𝜂𝜋 3 𝐼0𝑙 𝜆 2 1 −𝑗 1 𝑘𝑟3 For real power in the far field, we only care about the first term. Reactive (imaginary power) is more intense in the near-field region, but does not contribute to the far-field power.
Linear Wire Antennas Infinitesimal Dipole – Power Density and Radiation Resistance Slide 14 Power Density: From the power calculated we obtain 𝑃 𝑟𝑎𝑑= 𝜂𝜋 3 𝐼0𝑙 𝜆 2 = 1 2 𝐼0 2𝑅𝑟 Radiation Resistance: Where 𝑅𝑟is the radiation resistance given by 𝑅𝑟= 𝜂2𝜋 3 𝑙 𝜆 2 = 80𝜋2 𝑙 𝜆 2 Assumes antenna is in free-space Linear Wire Antennas Infinitesimal Dipole – Directivity Slide 15 We can calculate the directivity from the average power.
𝑊 𝑟𝑎𝑑= 𝑊 𝑎𝑣𝑒= 𝜂 8 𝐼0𝑙 𝜆 2 sin2(𝜃) 𝑟2 = 𝜂 2 𝑘𝐼0𝑙 4𝜋 2 sin2(𝜃) 𝑟2 𝑈= 𝑟2𝑊 𝑟𝑎𝑑= 𝜂 8 𝐼0𝑙 𝜆 2 sin2(𝜃) 𝐷0 = 𝑈𝑚𝑎𝑥 𝑈0 = 4𝜋𝑈𝑚𝑎𝑥 𝑃 𝑟𝑎𝑑 = 4𝜋𝜂 8 𝑘𝐼0 𝜆 2 𝜂𝜋 3 𝐼0𝑙 𝜆 2 = 3 2 𝐷0 = 3 2 = 1.76 dBi 𝑈𝑛= sin2(𝜃) Linear Wire Antennas Infinitesimal Dipole – Radiation Pattern Slide 16 Linear Wire Antennas Infinitesimal Dipole – Radiation Resistance Example Slide 17 Example 4.1 (page 150 Balanis) Find the radiation resistance of an infinitesimal dipole whose length is 𝑙= 𝜆 50 𝑅𝑟= 80𝜋2 𝑙 𝜆 2 = 80𝜋2 1 50 2 = 0.316 Ω Linear Wire Antennas Short Dipole 18 Short (Small) Dipole Slide 19 The short dipole is a more practical, ‘real’ antenna, usually has lengths from 𝜆 50 to 𝜆 10. A better representation of current distribution of wire antennas is the triangular distribution.
Linear Wire Antennas Short Dipole Current Distribution Slide 20 𝐈e x, y, z = ෝ 𝒂𝑧𝐼0 1 −2 𝑙𝑧, 0 ≤𝑧≤𝑙 2 ෝ 𝒂𝑧𝐼0 1 + 2 𝑙𝑧, −𝑙 2 ≤𝑧≤0 𝑙 z 𝐼0 𝐼 Linear Wire Antennas Short Dipole Geometry Slide 21 Linear Wire Antennas Far-Field E- and H- Fields Slide 22 For an electric current source, the magnetic field is equal to 𝐻𝑟= 𝐻𝜃= 0 𝐻𝜙= 𝑗𝑘𝐼0𝑙 8𝜋𝑟𝑒−𝑗𝑘𝑟sin 𝜃 The electric field is equal to 𝐸𝑟= 𝐸𝜙= 0 𝐸𝜃= 𝑗𝜂𝑘𝐼0𝑙 8𝜋𝑟𝑒−𝑗𝑘𝑟sin 𝜃 The fields of the short dipole are one-half of the infinitesimal dipole Linear Wire Antennas Short Dipole - Directivity and Radiation Resistance Slide 23 The power radiated by the short dipole is one-fourth 1 4 of the infinitesimal dipole.
𝑅𝑟= 20𝜋2 𝑙 𝜆 2 𝐷0 = 3 2 = 1.76 dBi 𝑈𝑛= sin2(𝜃) The radiation pattern is the same for the short and infinitesimal dipole. Since the directivity is controlled by the power pattern of the antenna, it is also the same as the infinitesimal dipole. Linear Wire Antennas Short Dipole – Radiation Pattern Slide 24 Linear Wire Antennas |
15872 | https://www.chegg.com/homework-help/questions-and-answers/standard-atmospheric-pressure-14696-lbf-gas-constant-dry-air-53352-ft-lbf-lbm-r-specific-h-q61720753 | Your solution’s ready to go!
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Question: Standard Atmospheric Pressure = 14.696 lbf/in? Gas Constant of Dry Air = 53.352 (ft-lbf)/(lbm-ºR) Specific Heat Capacity of Dry Air (Cpa) = 0.240 BTU/(Ibm-°F) Specific Heat Capacity of Water Vapor (Cpv) = 0.444 BTU/(lbm-°F) Gas Constant for Water Vapor = 85.78 (ft-lbf)/(Ibm-ºR) 2. Please use the following table to answer the following questions: Degree of
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Order of Operations Worksheets
In mathematics, the order of operations is a set of rules in mathematics. This order sets the sequence in which multiple operations should be performed. In order to solve questions correctly, it is important to practice questions based on the order of operations with the help of the order of operation worksheets. These worksheets consist of questions based on the different types of order of operations.
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Worksheets based on the order of operations can make it easier to recognize different orders and solve questions correctly. With the help of the worksheets based on the order of operations, students can easily get used to the rules of the order of operationsoid mistakes while solving questions.
These math worksheets can also help students get familiar with different mnemonics like PEMDAS. These mnemonics can be beneficial when it comes to recognizing the order of operations.
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15874 | https://demonstrations.wolfram.com/OneHalfTheApothemTimesThePerimeter/ | One-Half the Apothem Times the Perimeter | Wolfram Demonstrations Project
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One-Half the Apothem Times the Perimeter
n u m b e r o f s i d e s3 s h o w a l l a p o t h e m s- [x] h i g h l i g h t a s e c t o r- [x] s h o w v a l u e s- [x] c i r c u m r a d i u s=1 a p o t h e m=0.5 0 0 0 p e r i m e t e r=5.1 9 6 2 a r e a=1.2 9 9 0
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An apothem of a regular polygon is the length of the segment from the center to the midpoint of a side. The area of a regular polygon can be calculated as one-half the apothem multiplied by the perimeter. This Demonstration explains why that formula works.
Contributed by: Daniel Tokarz(2016)
Open content licensed under CC BY-NC-SA
Snapshots
Details
Snapshot 1: each triangle is identical, bounded on the sides by the green circumradii
Snapshot 2: the area is equal to 1/2 the area multiplied by the perimeter
Snapshot 3: for as few or as many sides as an
n
-gon has, this formula works
Imagine a regular
n
-sided polygon divided into a set of
n
identical isosceles triangles.One such triangle is the sector shown in light blue in the Demonstration.Each triangle has a base of side length
s
and a height
h
equal to the apothem.Thus the area of each triangle is
1/2 s h
.If we add up all the triangle areas,the area of the entire polygon can be written as
1/2 p h
,where
p
is the perimeter.As
n→∞
,the polygon approaches a circle.An apothem equals the radius
r
,while the perimeter equals
2 π r
,giving the well-known result for the area of a circle:
1
2
r×2 π r=π
2
r
.
Permanent Citation
Cite this as
Daniel Tokarz (2016), "One-Half the Apothem Times the Perimeter" Wolfram Demonstrations Project. demonstrations.wolfram.com/OneHalfTheApothemTimesThePerimeter/
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15875 | https://radiopaedia.org/cases/snowman-sign-1?lang=us | Snowman sign | Radiology Case | Radiopaedia.org
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Snowman sign
Case contributed by Ian Bickle
Diagnosis almost certain
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Citation, DOI, disclosures and case data
Citation:
Bickle I, Snowman sign. Case study, Radiopaedia.org (Accessed on 29 Sep 2025)
DOI:
Permalink:
rID:
36858
Case published:
13 May 2015, Ian Bickle
Revisions:
1 time, by 1 contributor - see full revision history and disclosures
Systems:
Cardiac
Quiz mode:
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Presentation
Patient with total anomalous pulmonary venous return (TAPVR). .
Patient Data
Age: 23
Gender: Female
From the case:Snowman sign
Annotated image
Download
Info
Snowman image from www.clipartlord.com. (Free for public use and domain).
The shape of the snowman superimposed over the CXR demonstrates the origin of this sign.
CXR image contributed by Dr M Sanal Kumar:
Case Discussion
Snowman signrefers to the configuration of the heart and its superior borders resembling a snowman. This is seen in total anomalous pulmonary venous return (TAPVR) type I i.e. supra cardiac type.
2 articles feature images from this case
Snowman sign (total anomalous pulmonary venous return)
Snowman sign (disambiguation)
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Snowman sign (total anomalous pulmonary venous return)
Total anomalous pulmonary venous return
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15876 | https://www.geeksforgeeks.org/problems/convert-from-any-base-to-decimal3736/1 | Convert from any base to decimal | Practice | GeeksforGeeks
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ProblemEditorialSubmissionsComments
Convert from any base to decimal
Difficulty: BasicAccuracy: 36.12%Submissions: 4K+Points: 1Average Time: 7m
Given a number N and its base b, convert it to decimal. The base of number can be anything such that all digits can be represented using 0 to 9 and A to Z. Value of A is 10, value of B is 11 and so on.
Example 1:
Input: b = 2, N = 1100 Output: 12 Explaination: It is a binary number whose decimal equivalent is 12.
Example 2:
Input: b = 16, N = A Output: 10 Explaination: It's a hexadecimal number whose decimal equivalent is 10.
Your Task:
You do not need to read input or print anything. Your task is to complete the function decimalEquivalent() which takes N and b as input parameters and returns their decimal equivalent. Otherwise return -1 if N is not possible.
Expected Time Complexity:O(|N|)[|N| means the length of the number N]
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ b ≤ 16
1 ≤ N < decimal equivalent 10 9
Company Tags
Adobe
Topic Tags
MathematicalAlgorithms
Related Articles
Convert Base Decimal Vice Versa
Discussions ( 25 Threads )
Most Recent
Commenting as Anonymous_GeekSubmit
Bhupesh Dewangan5 months ago Apr 28, 2025 23:19 (GMT +5:30)
C++ Solution
int basePower = 1; // Equivalent to b^0
int decimalValue = 0;
int n = N.size();
for (int i = n - 1; i >= 0; i--) {
int digitValue;
if (isalpha(N[i])) {
digitValue = toupper(N[i]) - 'A' + 10;
} else if (isdigit(N[i])) {
digitValue = N[i] - '0';
} else {
return -1; // Invalid character
}
if (digitValue >= b) {
return -1; // Digit not valid in base 'b'
}
decimalValue += digitValue basePower;
basePower = b;
}
return decimalValue;
1
Reply
Gayatri Patil8 months ago Jan 16, 2025 22:44 (GMT +5:30)
class Solution{
static int decimalEquivalent(String N, int b)
{
int i=0,num,sum=0;
while(i<N.length()){
if(N.charAt(i)>= '0' && N.charAt(i) <= '9'){
num= N.charAt(i)-'0';
}
else {
num= N.charAt(i)-'A'+10;
}
if(num>=b){
return -1;
}
sum= sum+ (num(int)(Math.pow(b,(N.length()-1-i))));
i++;
}
return sum;
}
}
0
Reply
Abhishek Kumar9 months ago Jan 01, 2025 00:40 (GMT +5:30)
```
def decimalEquivalent(self, N, b):
hashmap={
'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9,
'A': 10, 'B': 11, 'C': 12, 'D': 13, 'E': 14, 'F': 15}
ans=0
temp=str(N)
k=0
for i in temp[::-1]:
if i not in hashmap or hashmap[i]>=b:
return -1
ans+=(bk)(hashmap[i])
k+=1
return ans
```
0
Reply
Abhishek Kumar1 year ago Sep 17, 2024 15:38 (GMT +5:30)
def decimalEquivalent(self, N, b):
res=0
k=0
for i in N[::-1]:
if i.isdigit() and int(i)<b:
res+=(bk)(int(i))
elif not i.isdigit() and int(ord(i)-55)<b:
res+=(bk)(int(ord(i)-55))
else :
return -1
k+=1
return res
0
Reply
Manshi Rani1 year ago Jul 25, 2024 18:57 (GMT +5:30)
please upvote!!!!!!!!!
int decimalEquivalent(string N, int b){
int a=1;
int dec=0;int n=N.size();
for (int i =n-1;i>=0;i--){
if(isalpha(N[i])){
int c =N[i]-55;
if (c>=b) return -1;
else dec=dec+a(c);
}
else{
int c =N[i]-48;
if (c>=b) return -1;
else dec=dec+a(c);
}
a=ab;
}
return dec;
}
0
Reply
(Show 1 Replies)
Timofey1 year ago Jul 15, 2024 11:05 (GMT +5:30)
Python3 solution:
```
User function Template for python3
class Solution:
def decimalEquivalent(self, N, b):
# code here
res = 0
p = 0
dc = {"A":10, "B":11, "C":12, "D": 13, "E":14, "F": 15 }
for sym in N[::-1]:
if sym in dc:
num = dc[sym]
else:
num = int(sym)
if num > b-1:
return -1
res += num (bp)
p+=1
return res
```
1
Reply
Cellul4r1 year ago Jun 09, 2024 17:34 (GMT +5:30)
int decimalEquivalent(string N, int b){
// code here
int len = N.length(), ans = 0;
for(int i= len-1;i>=0;--i) {
char c = N[i];
int val;
if(c >= 'A' && c <= 'F') {
val = c - 'A' + 10;
} else {
val = c - '0';
}
if(val >= b) return -1;
// cout << val << " " << len - i - 1 << endl;
ans += (val pow(b,len - i - 1));
}
return ans;
}
0
Reply
Akash Roushan1 year ago May 23, 2024 22:51 (GMT +5:30)
python 1 line
try:
return int(str(n),s)
except:
return -1
0
Reply
W L1 year ago Nov 21, 2023 15:43 (GMT +5:30)
int decimalEquivalent(string N, int b){
// code here
int rst=0;
int base = 1;
for(int i=N.size()-1;i>=0;i--){
char c = N[i];
int v = 1;
if(c>='0'&&c<='9'){
v = c - '0';
}
else if(c>='A'&&c<='F'){
v = c - 'A' + 10;
}
else if(c=='-'){
rst = -rst;
continue;
}
else{
return -1;
}
if(v>=b){
return -1;
}
rst += v base;
base = b;
}
return rst;
}
0
Reply
iR080T1 year ago Oct 21, 2023 10:37 (GMT +5:30)
C++
int decimalEquivalent( string n, int b ) {
for ( const char c : n )
if ( c - ( isdigit( c ) ? '0' : 'A' ) >= b )
return -1;
return stoi( n, nullptr, b );
} // 0.01+
2
Reply
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15877 | https://www.quora.com/How-does-one-explain-the-truth-table-for-p-implies-not-p-Shouldn-t-one-expect-a-logical-contradiction | How does one explain the truth table for: p implies not p? Shouldn’t one expect a logical contradiction? - Quora
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How does one explain the truth table for: p implies not p? Shouldn’t one expect a logical contradiction?
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Dan Christensen
Creator of proof-checking freeware DC Proof 2.0 · Upvoted by
David Joyce
, Professor Emeritus of Mathematics and Computer Science · Author has 2.7K answers and 816.4K answer views
·4y
If you are familiar with proofs by contradiction, this proof based on a form of natural deduction should help explain this result (screenshot from my proof checker):
Line 5 is derived using a proof by contradiction, the the latest undischarged premise there being P on line 2, with the contradiction being on line 4.
Contrary to the claims of some here, when dealing with logical propositions that are unambiguously either true or false in the present, there really is no difference between logical and material implication. See my recent answer of this topic here.
Continue Reading
If you are familiar with proofs by contradiction, this proof based on a form of natural deduction should help explain this result (screenshot from my proof checker):
Line 5 is derived using a proof by contradiction, the the latest undischarged premise there being P on line 2, with the contradiction being on line 4.
Contrary to the claims of some here, when dealing with logical propositions that are unambiguously either true or false in the present, there really is no difference between logical and material implication. See my recent answer of this topic here.
Upvote ·
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David Joyce
Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979) · Author has 9.9K answers and 67.8M answer views
·Updated 3y
The truth table for p→¬p p→¬p is not the same as the truth table for the contradiction p∧¬p.p∧¬p.
p¬p p→¬p p∧¬p⊤⊥⊥⊥⊥⊤⊤⊥p¬p p→¬p p∧¬p⊤⊥⊥⊥⊥⊤⊤⊥
The column for p→¬p p→¬p is the same as the column for ¬p.¬p. That means that p→¬p p→¬p is logically equivalent to ¬p.¬p. The column for p∧¬p p∧¬p contains only ⊥,⊥, and that means it’s a contradiction.
(The symbol ⊤⊤ is often used for “true” while the symbol ⊥⊥ is used for “false”.)
Should you expect a logical contradiction? No, I since p→¬p p→¬p is not
Continue Reading
The truth table for p→¬p p→¬p is not the same as the truth table for the contradiction p∧¬p.p∧¬p.
p¬p p→¬p p∧¬p⊤⊥⊥⊥⊥⊤⊤⊥p¬p p→¬p p∧¬p⊤⊥⊥⊥⊥⊤⊤⊥
The column for p→¬p p→¬p is the same as the column for ¬p.¬p. That means that p→¬p p→¬p is logically equivalent to ¬p.¬p. The column for p∧¬p p∧¬p contains only ⊥,⊥, and that means it’s a contradiction.
(The symbol ⊤⊤ is often used for “true” while the symbol ⊥⊥ is used for “false”.)
Should you expect a logical contradiction? No, I since p→¬p p→¬p is not a contradiction. It’s just a convoluted way of saying p p is false.
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To analyze the expression p⟹¬p p⟹¬p (read as "p implies not p"), we can create a truth table to examine its truth values under different conditions for p p.
Truth Table for p⟹¬p p⟹¬p
| p p | ¬p¬p | p⟹¬p p⟹¬p |
---------------
| T | F | F |
| F | T | T |
Explanation:
When p p is True (T):
¬p¬p (not p) is False (F).
The implication p⟹¬p p⟹¬p translates to "True implies False," which is False.
When p p is False (F):
¬p¬p is True (T).
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To analyze the expression p⟹¬p p⟹¬p (read as "p implies not p"), we can create a truth table to examine its truth values under different conditions for p p.
Truth Table for p⟹¬p p⟹¬p
| p p | ¬p¬p | p⟹¬p p⟹¬p |
---------------
| T | F | F |
| F | T | T |
Explanation:
When p p is True (T):
¬p¬p (not p) is False (F).
The implication p⟹¬p p⟹¬p translates to "True implies False," which is False.
When p p is False (F):
¬p¬p is True (T).
The implication p⟹¬p p⟹¬p translates to "False implies True," which is True.
Understanding the Logical Contradiction:
The expression p⟹¬p p⟹¬p indeed creates a logical contradiction when p p is true, as it asserts that if p p is true, then p p must also be false, which is contradictory.
However, when p p is false, the implication holds true by the definition of implication in logic. An implication A⟹B A⟹B is considered true if either A A is false or both A A and B B are true.
Thus, while it might seem that p⟹¬p p⟹¬p should always be false due to a contradiction when p p is true, the overall expression is not a contradiction in a broader sense since it is true when p p is false. This leads to the conclusion that p⟹¬p p⟹¬p is not a tautology (always true) nor a contradiction (always false), but rather a statement that has mixed truth values depending on the truth of p p.
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More answers below
How can p implies not p be true?
Consider the logical proposition: p implies not p. One would expect a logical contradiction, but this is not so. What is the truth table for this?
In logic, why is it that (p & not-p) is a contradiction while (if p then not-p) is not a contradiction?
Is the statement "If p then not-p" always false?
What is the reasoning behind the truth table for 'implies' (p→q p→q)? In particular, why does "p→q p→q" = true when p is false?
Frank Davies
Associate Applied Science in Computer Science, South Puget Sound Community College · Author has 5.1K answers and 1.1M answer views
·1y
Related
How can p implies not p be true?
It can be true because of p being false.
In fact, this is how one would prove p to be false.
Initially we would declare p to be true, …not really knowing one way or the other.
But perhaps we also know for certain that: if p is true, then q must be true.
And we know for certain that: if q is true, then so is r also true.
And we know for certain that: if r is true, then p must be false.
So we arrive at: if p is true then p is not true.
(p implies not p) = true
…and that is a true statement: p does indeed imply not p.
…so we should not have a problem with this as a whole statement.
What you have noti
Continue Reading
It can be true because of p being false.
In fact, this is how one would prove p to be false.
Initially we would declare p to be true, …not really knowing one way or the other.
But perhaps we also know for certain that: if p is true, then q must be true.
And we know for certain that: if q is true, then so is r also true.
And we know for certain that: if r is true, then p must be false.
So we arrive at: if p is true then p is not true.
(p implies not p) = true
…and that is a true statement: p does indeed imply not p.
…so we should not have a problem with this as a whole statement.
What you have noticed… is…
This is a paradox or contradiction, ONLY if p is true.
Because a false statement may imply a true statement, without effecting the true statement.
“If I ate ice cream for breakfast, then, the sun is bright.”
…The sun is bright regardless of “I ate Ice cream…” being true or not, right?
There are no paradoxes in reality.
So, we reach the conclusion the p can only be false.
However, with the computer, we can take the output and use it as data the next cycle.
…Looping forever.
‘Start’: go to next instruction.
if (p = true) make p false. Go to next instruction.
Print the value of p. Then go to next instruction.
if (p = false) make p true. Go to next instruction.
Print the value of p. Then go to next instruction.
Go back to ‘start’.
Or something like that.
Or you could label each ‘p’. p1, p2, p3 etc.
I mentioned this because then you could check if p1 = p2 etc.
…And find a contradiction if they were not equal.
…So possibly be able to say p is false.
I can’t use logic to prove p is true.
But I can use logic to prove p is false.
“This is the way.”
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Alex Heitzman
Studied at University of Nebraska-Lincoln · Author has 931 answers and 845.2K answer views
·6y
Related
What is the reasoning behind the truth table for 'implies' (p→q p→q)? In particular, why does "p→q p→q" = true when p is false?
Here’s how I finally understood it. I can guarantee you that it will make sense after this.
Suppose you want to examine the statement “all red cars in town have a sunroof”. This is an implication statement p⟹q p⟹q, where
p=p= “a car in this town is red”
q=q= “this car has a sunroof”
How do you prove or disprove this statement? You would have to start checking cars in your town. Look at each car, and consider:
Check p p: Is this car red?
Check q q: Does this car have a sunroof?
If you come across just one counter-example (a case where p⟹q p⟹q is false), then the whole statement is false. If you ch
Continue Reading
Here’s how I finally understood it. I can guarantee you that it will make sense after this.
Suppose you want to examine the statement “all red cars in town have a sunroof”. This is an implication statement p⟹q p⟹q, where
p=p= “a car in this town is red”
q=q= “this car has a sunroof”
How do you prove or disprove this statement? You would have to start checking cars in your town. Look at each car, and consider:
Check p p: Is this car red?
Check q q: Does this car have a sunroof?
If you come across just one counter-example (a case where p⟹q p⟹q is false), then the whole statement is false. If you check every car in town and it’s always true, then the overall statement is true.
Now what do you do if you come across a car that happens to be black? Could it be a counter-example to the statement we are checking? Of course not! It doesn’t matter whether the black car has a sunroof or not. A real counter-example would be a red car that does not have a sunroof. Nothing else.
Therefore, if the car is not red, you must consider the statement you are checking to be true in this case.
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Mark Bradley
Economist
·9mo
What are the dumbest financial mistakes most Americans make?
Where do I start?
I’m a huge financial nerd, and have spent an embarrassing amount of time talking to people about their money habits.
Here are the biggest mistakes people are making and how to fix them:
Not having a separate high interest savings account: Having a separate account allows you to see the results of all your hard work and keep your money separate so you're less tempted to spend it. Plus with rates above 5.00%, the interest you can earn compared to most banks really adds up.
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Where do I start?
I’m a huge financial nerd, and have spent an embarrassing amount of time talking to people about their money habits.
Here are the biggest mistakes people are making and how to fix them:
Not having a separate high interest savings account: Having a separate account allows you to see the results of all your hard work and keep your money separate so you're less tempted to spend it. Plus with rates above 5.00%, the interest you can earn compared to most banks really adds up.
Here is a list of the top savings accounts available today. Deposit $5 before moving on because this is one of the biggest mistakes and easiest ones to fix.
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Philip Calabrese
PhD in Mathematics&Minor in Physics, Illinois Institute of Technology (Graduated 1968) · Author has 261 answers and 132.3K answer views
·4y
Originally Answered: Consider the logical proposition: p implies not p. One would expect a logical contradiction, but this is not so. What is the truth table for this? ·
If p -> not p is interpreted as “material implication” it becomes not p or not p = not p, which is not an inconsistency. That is one of the fallacies of rendering “p implies q” as “q or not p”, so called “material implication“. Rather, p -> q is better rendered as “all models (examples) of p are models of q. Then p -> not p means all models of p are models of not p, which is a contradiction.
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Related questions
More answers below
How do you prepare the truth table p → (pvq)?
What is the truth table of P implies Q?
What is the truth value of p ∨ q when p → q is false?
If statements p and q are true and r and s are false, what are the truth values of the following: (p ∧ ~r) ∧ (~q ∨ s)?
How do you show that (p q) and (p q) are logically equivalent without using a truth table (discrete mathematics, math)?
Tom Wetzel
PhD in Philosophy, University of California, Los Angeles (Graduated 1978) · Author has 11K answers and 10.8M answer views
·4y
Originally Answered: Consider the logical proposition: p implies not p. One would expect a logical contradiction, but this is not so. What is the truth table for this? ·
This is a way of saying that P is false. If a proposition implies its own negation, then indeed it is false.
Thus if you suppose P were true, then if (If P then not-P) were true, it would follow that not-P would be false. Hence, If P then Not-P will be false, according to the truth table. A conditional is false if it has a true antecedent and a false consequent.
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What makes a cover letter stand out?
When you’re writing a cover letter, the little things can make a big difference. The better your letter is, the better your chances are of getting a job interview. Follow these tips for composing a winning cover letter, and you will increase your chances of landing the job you want.
Research and brainstorm
Before you start writing your cover letter, familiarize yourself with the role requirements. Read the job listing carefully and pull out the most important information. Then, spend time on the company’s website to get a strong sense of the company’s culture, values, and mission. Brainstorm the
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When you’re writing a cover letter, the little things can make a big difference. The better your letter is, the better your chances are of getting a job interview. Follow these tips for composing a winning cover letter, and you will increase your chances of landing the job you want.
Research and brainstorm
Before you start writing your cover letter, familiarize yourself with the role requirements. Read the job listing carefully and pull out the most important information. Then, spend time on the company’s website to get a strong sense of the company’s culture, values, and mission. Brainstorm the most effective way to communicate your suitability for the role in your cover letter. Brainstorming is a key part of the writing process.
Personalize your greeting
The first thing the recruiter or hiring manager will notice in your cover letter is whether you addressed it to them personally. It’s not always easy to find the recruiter’s or hiring manager’s name, but it’s always worth your time to directly address the person who will be making the hiring decision.
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Sign up for a free Grammarly account and ensure your cover letter is readable, clear, and concise with real-time suggestions for stronger, more precise language.
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Eugene Bucamp
Knows French · Author has 4K answers and 1.1M answer views
·Updated 3y
The question relates in fact to the material implication, p ⊃ q, and not to the logical implication p → q.
According to the truth table of the material implication, p ⊃ q is false when p is true and q false, and p ⊃ q is true in the three remaining cases.
Applied to p ⊃ ¬p, we get that p ⊃ ¬p is false when p is true, and it is true when p is false.
The crucial point is that material implication is not at all an implication, despite the misleading label. The material implication p ⊃ q means ¬p ∨ q. It does not mean p implies q.
From that it follows that p ⊃ ¬p is equivalent to ¬p ∨ ¬p, which is equ
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The question relates in fact to the material implication, p ⊃ q, and not to the logical implication p → q.
According to the truth table of the material implication, p ⊃ q is false when p is true and q false, and p ⊃ q is true in the three remaining cases.
Applied to p ⊃ ¬p, we get that p ⊃ ¬p is false when p is true, and it is true when p is false.
The crucial point is that material implication is not at all an implication, despite the misleading label. The material implication p ⊃ q means ¬p ∨ q. It does not mean p implies q.
From that it follows that p ⊃ ¬p is equivalent to ¬p ∨ ¬p, which is equivalent to ¬p, which is of course false when p is true, and true when p is false.
So p ⊃ ¬p does not mean that p implies ¬p. It just means ¬p!
The problem, then, is not with the truth table of the material implication, it is with the label “material implication” itself. This label of course suggests that the material implication is a mathematical model of the logical implication, and so suggests that p ⊃ q means that p implies q, as the question itself suggests. Well, no, p ⊃ q just means ¬p ∨ q, and p ⊃ ¬p just means ¬p.
This is of course extremely misleading. This is why many students of mathematical logic are terminally confused. Logic textbooks often allude to this confusion without however ever acknowledging that the problem is entirely caused by the misleading label “material implication”.
This isn’t a mistake, more like systematic deception. Mathematicians also use for example the expression “classical logic” to refer to the logic system based on the material implication, and this even though this system is contradictory with Aristotelian logic, which surely is classical logic if anything is. This label then misleads students into believing that the system of mathematical logic based on the material implication is in line with Aristotelian logic, and this is false.
You have to realise how misleading that is. The reality is that mathematical logic cannot provide even one correct model of the logical implication. There is a large collection of logic systems in mathematical logic and nowhere will you find in it any mathematical model of the logical implication. Thus, calling the material implication “implication” can only be interpreted by students as clear suggestion that it is a model of the logical implication. But it is not. The material implication p ⊃ q just means ¬p ∨ q. It does not mean p implies q.
Mathematicians haven’t a clue how the logical implication really works and so they haven’t a clue how logic really works.
The deception extends further. Since they haven’t a clue how logic really works, many will say that mathematical logic systems are not meant to be models of the logic of human deductive reasoning. In fact, many mathematicians will go further and cast doubt on the very existence of the logic of human deductive reasoning. And once they have done that, nobody can reproach them of not understanding something that they say does not even exist. And who is going to prove otherwise? Well, clearly, no mathematician will.
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9 2
Nicholas Cooper
Author has 1.6K answers and 2.9M answer views
·5y
Related
How can p implies not p be true?
First things first, if a logic state is true for all possible inputs, it is a tautology. If it is false for all possible inputs, it is a contradiction. Finally, if there are some true results, then it is contingent.
The statement is contingent.
For analogy, let’s consider implication p→q p→q to mean that p is at most as true as q (p≤q p≤q) since p can cause q to be true, but so can something else, and so q can be true when p is not, but q must be true when p is true.
Now, let’s assume our negation switches between 0 and 1, but we really could demonstrate that doing logic with the values {−1,+1{−1,+1
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First things first, if a logic state is true for all possible inputs, it is a tautology. If it is false for all possible inputs, it is a contradiction. Finally, if there are some true results, then it is contingent.
The statement is contingent.
For analogy, let’s consider implication p→q p→q to mean that p is at most as true as q (p≤q p≤q) since p can cause q to be true, but so can something else, and so q can be true when p is not, but q must be true when p is true.
Now, let’s assume our negation switches between 0 and 1, but we really could demonstrate that doing logic with the values {−1,+1}{−1,+1} is equivalent (isomorphic) leaving logical negation to look like numeric negation.
With all of those (structural) analogies in mind, the comparison of truth values p≤−p p≤−p is true in the case that p is false (-1) and false in the case p is true (+1).
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D.K. Johnston
PhD in Philosophy · Author has 200 answers and 208.1K answer views
·4y
Originally Answered: Consider the logical proposition: p implies not p. One would expect a logical contradiction, but this is not so. What is the truth table for this? ·
The truth table is only two lines long, so I’m sure you can work it out for yourself. I’ll give you one hint: in traditional bivalent truth-functional logic, a conditional is always true when its antecedent is false, regardless of what the the consequent might be.
Upvote ·
Michiel Bogaert
Unable to carry all the maths books he studied · Author has 3.3K answers and 2.8M answer views
·5y
Related
How can p implies not p be true?
It’s true, only if p is false. The statement looks wonkey, but that’s because it’s a special case
Consider “if it’s a square, it’s blue” . That statement holds up for blue square, and even holds up for blue circles and even green circles. Because green or blue circles aren’t squares, it doesn’t matter: whatever’s writte after “if it’s a square” is irrelevant.
So, when we got a green circle, “if it’s a square, the pope will win the worldcup basketball” is still true …
And likewise, when we got a circle, “if it’s a square, the thing we just said is a square isn’t a square” is also true. it’s a circ
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It’s true, only if p is false. The statement looks wonkey, but that’s because it’s a special case
Consider “if it’s a square, it’s blue” . That statement holds up for blue square, and even holds up for blue circles and even green circles. Because green or blue circles aren’t squares, it doesn’t matter: whatever’s writte after “if it’s a square” is irrelevant.
So, when we got a green circle, “if it’s a square, the pope will win the worldcup basketball” is still true …
And likewise, when we got a circle, “if it’s a square, the thing we just said is a square isn’t a square” is also true. it’s a circle, we don’t care we create a paradox or contradication in a senario we never get to.
… and as such, p implies not p is true, if p is false. Because if p is false, we don’t care that if p would have been true, we created a contradiction.
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Frank Davies
Associate Applied Science in Computer Science, South Puget Sound Community College · Author has 5.1K answers and 1.1M answer views
·1y
Yes, that’s right!
And thus we prove the p is false.
…Because we know it cannot be both true and false.
…And we know it cannot be true or it implies that it is false.
…And we also know that any false statement may imply a true one.
“Bo ate a million bananas, so the moon is in orbit about the Earth.”
Well, Bo didn’t, but the moon still orbits us anyway.
We cannot prove something to be true.
…We can assume it, or infer it from other statements.
But we can prove a statement to be false.
If p implies not p, then p is false.
Upvote ·
Steven
Computer scientist and programmer · Author has 8.5K answers and 38.4M answer views
·5y
Related
How can p implies not p be true?
Lets go over the possibilities!
The definition of implication states that “true implies false” is false, and all other implications are true.
If p is true, then the statement is “true implies false”, which is false by the definition above.
If p is false, then the statement is “false implies true”, which by the quirks of the definition, is true. From a falsehood, you can logically derive everything!
To conclude, the statement is true provided that p is false.
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Related questions
How can p implies not p be true?
In logic, why is it that (p & not-p) is a contradiction while (if p then not-p) is not a contradiction?
Consider the logical proposition: p implies not p. One would expect a logical contradiction, but this is not so. What is the truth table for this?
What is the reasoning behind the truth table for 'implies' (p→q p→q)? In particular, why does "p→q p→q" = true when p is false?
In a truth table of conditional statement in Mathematical logic, how p implies q is true when p is false and q is true?
How do you prepare the truth table p → (pvq)?
What is the truth table of P implies Q?
What is the truth value of p ∨ q when p → q is false?
If statements p and q are true and r and s are false, what are the truth values of the following: (p ∧ ~r) ∧ (~q ∨ s)?
How do you show that (p q) and (p q) are logically equivalent without using a truth table (discrete mathematics, math)?
Is the statement "If p then not-p" always false?
What is the truth values of p or negation of a if the truth values of p bi implies q is false?
If P implies Q, and not-Q implies not-P, what does this imply about the truth value of P?
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How can p implies not p be true?
Consider the logical proposition: p implies not p. One would expect a logical contradiction, but this is not so. What is the truth table for this?
In logic, why is it that (p & not-p) is a contradiction while (if p then not-p) is not a contradiction?
Is the statement "If p then not-p" always false?
What is the reasoning behind the truth table for 'implies' (p→q p→q)? In particular, why does "p→q p→q" = true when p is false?
In a truth table of conditional statement in Mathematical logic, how p implies q is true when p is false and q is true?
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5: Functions
MTH 220 Discrete Math
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Sun, 17 Nov 2019 04:59:12 GMT
5.5: Inverse Functions and Composition
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5.4: Onto Functions and Images/Preimages of Sets
5.6: Infinite Sets and Cardinality
Page ID 26045
Harris Kwong
State University of New York at Fredonia via OpenSUNY
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Table of contents
1. Definition: Inverse Function
1. How to find f−1
Composite Function
Identity Function relates to Inverse Functions
Summary and Review
Exercises
A bijection(or one-to-one correspondence) is a function that is both one-to-one and onto. Naturally, if a function is a bijection, we say that it is bijective. If a function f:A→B is a bijection, we can define another function g that essentially reverses the assignment rule associated with f. Then, applying the function g to any element y from the codomain B, we are able to obtain an element x from the domain A such that f(x)=y. Let us refine this idea into a more concrete definition.
Definition: Inverse Function
Let f:A→B be a bijective function. Its inverse function is the function f−1:B→A with the property that (5.5.1)f−1(b)=a⇔b=f(a). The notation f−1 is pronounced as “f inverse.” See figure below for a pictorial view of an inverse function.
Why is f−1:B→A a well-defined function? For it to be well-defined, every element b∈B must have a unique image. This means given any element b∈B, we must be able to find one and only one element a∈A such that f(a)=b. Such an a exists, because f is onto, and there is only one such element a because f is one-to-one. Therefore, f−1 is a well-defined function.
How to find f−1
If a function f is defined by a computational rule, then the input value x and the output value y are related by the equation y=f(x). In an inverse function, the role of the input and output are switched. Therefore, we can find the inverse function f−1 by following these steps:
f−1(y)=x⟺y=f(x), so write y=f(x), using the function definition of f(x).
Solve for x. That is, express x in terms of y. The resulting expression is f−1(y).
Be sure to write the final answer in the form f−1(y)=…. Do not forget to include the domain and the codomain, and describe them properly.
Example 5.5.1
To find the inverse function of f:R→R defined by f(x)=2x+1, we start with the equation y=2x+1. Solving for x, we find x=1 2(y−1). Therefore, the inverse function is (5.5.2)f−1:R→R,f−1(y)=1 2(y−1). It is important to describe the domain and the codomain, because they may not be the same as the original function.
Example 5.5.2
The function s:[−π 2,π 2]→[−1,1] defined by s(x)=sinx is a bijection. Its inverse function is
(5.5.3)s−1:[−1,1]→[−π 2,π 2],s−1(y)=arcsiny.
The function arcsiny is also written as sin−1y, which follows the same notation we use for inverse functions.
hands-on Exercise 5.5.1
The function f:[−3,∞)→[0,∞) is defined as f(x)=x+3. Show that it is a bijection, and find its inverse function
hands-on Exercise 5.5.2
Find the inverse function of g:R→(0,∞) defined by g(x)=e x.
Remark
Exercise caution with the notation. Assume the function f:Z→Z is a bijection. The notation f−1(3) means the image of 3 under the inverse function f−1. If f−1(3)=5, we know that f(5)=3. The notation f−1({3}) means the preimage of the set {3}. In this case, we find f−1({3})={5}. The results are essentially the same if the function is bijective.
If a function g:Z→Z is many-to-one, then it does not have an inverse function. This makes the notation g−1(3) meaningless. Nonetheless, g−1({3}) is well-defined, because it means the preimage of {3}. If g−1({3})={1,2,5}, we know g(1)=g(2)=g(5)=3.
In general, f−1(D) means the preimage of the subset D under the function f. Here, the function f can be any function. If f is a bijection, then f−1(D) can also mean the image of the subset D under the inverse function f−1. There is no confusion here, because the results are the same.
Example 5.5.3
The function f:R→R is defined as (5.5.4)f(x)={3x if x≤1,2x+1 if x>1. Find its inverse function.
Solution
Since f is a piecewise-defined function, we expect its inverse function to be piecewise-defined as well. First, we need to find the two ranges of input values in f−1. The images for x≤1 are y≤3, and the images for x>1 are y>3. Hence, the codomain of f, which becomes the domain of f−1, is split into two halves at 3. The inverse function should look like (5.5.5)f−1(x)={???if x≤3,???if x>3. Next, we determine the formulas in the two ranges. We find
(5.5.6)f−1(x)={1 3 x if x≤3,1 2(x−1)if x>3. The details are left to you as an exercise.
hands-on Exercise 5.5.3
Find the inverse function of g:R→R defined by (5.5.7)g(x)={3x+5 if x≤6,5x−7 if x>6. Be sure you describe g−1 properly.
Example 5.5.4
Find the inverse function of f:Z→N∪{0} defined by (5.5.8)f(n)={2n if n≥0,−2n−1 if n<0.
Solution
In an inverse function, the domain and the codomain are switched, so we have to start with f−1:N∪{0}→Z before we describe the formula that defines f−1. Writing n=f(m), we find (5.5.9)n={2m if m≥0,−2m−1 if m<0. We need to consider two cases.
If n=2m, then n is even, and m=n 2.
If n=−2m−1, then n is odd, and m=−n+1 2.
Therefore, the inverse function is defined by f−1:N∪{0}→Z by:
(5.5.10)f−1(n)={2 n if n is even,−n+1 2 if n is odd.
Verify this with some numeric examples.
hands-on Exercise 5.5.5
The function f:Z→N is defined as (5.5.11)f(n)={−2n if n<0,2n+1 if n≥0. Find its inverse.
Let A and B be finite sets. If there exists a bijection f:A→B, then the elements of A and B are in one-to-one correspondence via f. Hence, |A|=|B|. This idea will be very important for our section on Infinite Sets and Cardinality.
Composite Function
Given functions f:A→B′ and g:B→C where B′⊆B , the composite function, g∘f, which is pronounced as “g after f”, is defined as (5.5.12)g∘f:A→C,(g∘f)(x)=g(f(x)). The image is obtained in two steps. First, f(x) is obtained. Next, it is passed to g to obtain the final result. It works like connecting two machines to form a bigger one, see first figure below. We can also use an arrow diagram to provide another pictorial view, see second figure below.
Numeric value of (g∘f)(x) can be computed in two steps. For example, to compute (g∘f)(5), we first compute the value of f(5), and then the value of g(f(5)). To find the algebraic description of (g∘f)(x), we need to compute and simplify the formula for g(f(x)). In this case, it is often easier to start from the “outside” function. More precisely, start with g, and write the intermediate answer in terms of f(x), then substitute in the definition of f(x) and simplify the result.
Example 5.5.5
Assume f,g:R→R are defined as f(x)=x 2, and g(x)=3x+1. We find
(5.5.13)(g∘f)(x)=g(f(x))=3[f(x)]+1=3x 2+1,(f∘g)(x)=f(g(x))=[g(x)]2=(3x+1)2.
Therefore,
(5.5.14)g∘f:R→R,(g∘f)(x)=3x 2+1
(5.5.15)f∘g:R→R,(f∘g)(x)=(3x+1)2
We note that, in general, f∘g≠g∘f.
hands-on Exercise 5.5.6
If p,q:R→R are defined as p(x)=2x+5, and q(x)=x 2+1, determine p∘q and q∘p. Do not forget to describe the domain and the codomain
Example 5.5.6
Define f,g:R→R as
(5.5.16)f(x)={3x+1 if x<0,2x+5 if x≥0,
and g(x)=5x−7. Find g∘f.
Solution
Since f is a piecewise-defined function, we expect the composite function g∘f is also a piecewise-defined function. It is defined by (5.5.17)(g∘f)(x)=g(f(x))=5f(x)−7={5(3x+1)−7 if x<0,5(2x+5)−7 if x≥0.
After simplification, we find g∘f:R→R, by:(5.5.18)(g∘f)(x)={15x−2 if x<0,10x+18 if x≥0. In this example, it is rather obvious what the domain and codomain are. Nevertheless, it is always a good practice to include them when we describe a function.
hands-on Exercise 5.5.7
The functions f:R→R and g:R→R are defined by (5.5.19)f(x)=3x+2,and g(x)={x 2 if x≤5,2x−1 if x>5. Determine f∘g
Example 5.5.7
Let R∗ denote the set of nonzero real numbers. Suppose
(5.5.20)f:R∗→R,f(x)=1 x
(5.5.21)g:R→(0,∞),g(x)=3x 2+11.
Determine f∘g and g∘f. Be sure to specify their domains and codomains.
Solution
To compute f∘g, we start with g, whose domain is R. Hence, R is the domain of f∘g. The result from g is a number in (0,∞). The interval (0,∞) contains positive numbers only, so it is a subset of R∗. Therefore, we can continue our computation with f, and the final result is a number in R. Hence, the codomain of f∘g is R. The image is computed according to f(g(x))=1/g(x)=1/(3x 2+11). We are now ready to present our answer:
f∘g:R→R, by:
(5.5.22)(f∘g)(x)=1 3x 2+11.
In a similar manner, the composite function g∘f:R∗(0,∞) is defined as (5.5.23)(g∘f)(x)=3 x 2+11. Be sure you understand how we determine the domain and codomain of g∘f.
Identity Function relates to Inverse Functions
Recall the definition of the Identity Function:
The identity function on any nonempty set A maps any element back to itself:(5.5.24)I A:A→A,I A(x)=x. .
Theorem 5.5.1
For a bijective function f:A→B,
(5.5.25)f−1∘f=I A,and f∘f−1=I B,
where i A and i B denote the identity function on A and B, respectively.
Proof
To prove that f−1∘f=I A, we need to show that (f−1∘f)(a)=a for all a∈A. Assume f(a)=b. Then, because f−1 is the inverse function of f, we know that f−1(b)=a. Therefore,
(5.5.26)(f−1∘f)(a)=f−1(f(a))=f−1(b)=a,
which is what we want to show. The proof of f∘f−1=I B procceds in the exact same manner, and is omitted here.
Example 5.5.8
Show that the functions f,g:R→R defined by f(x)=2x+1 and g(x)=1 2(x−1) are inverse functions of each other.
Solution
The problem does not ask you to find the inverse function of f or the inverse function of g. Instead, the answers are given to you already. You job is to verify that the answers are indeed correct, that the functions are inverse functions of each other.
Form the two composite functions f∘g and g∘f, and check whether they both equal to the identity function:
(5.5.27)(f∘g)(x)=f(g(x))=2g(x)+1=2[1 2(x−1)]+1=x,(g∘f)(x)=g(f(x))=1 2[f(x)−1]=1 2[(2x+1)−1]=x.
We conclude that f and g are inverse functions of each other.
hands-on Exercise 5.5.8
Verify that f:R→R+ defined by f(x)=e x, and g:R+→R defined by g(x)=lnx, are inverse functions of each other
Theorem 5.5.2
Suppose f:A→B and g:B→C. Let I A and I B denote the identity function on A and B, respectively. We have the following results.
f∘I A=f and I B∘f=f.
If both f and g are one-to-one, then g∘f is also one-to-one.
If both f and g are onto, then g∘f is also onto.
If g∘f is bijective, then (g∘f)−1=f−1∘g−1.
Proof of (a)
To show that f∘I A=f, we need to show that (f∘I A)(a)=f(a) for all a∈A. This follows from direct computation: (5.5.28)(f∘I A)(a)=f(I A(a))=f(a). The proofs of I B∘f=f and (b)–(d) are left as exercises.
Summary and Review
A bijection is a function that is both one-to-one and onto.
The inverse of a bijection f:A→B is the function f−1:B→A with the property that (5.5.29)f(x)=y⇔x=f−1(y).
In brief, an inverse function reverses the assignment rule of f. It starts with an element y in the codomain of f, and recovers the element x in the domain of f such that f(x)=y.
Given B′⊆B, the composition of two functions f:A→B′ and g:B→C is the function g∘f:A→C defined by (g∘f)(x)=g(f(x)).
If f:A→B is bijective, then f−1∘f=I A and f∘f−1=I B.
To check whether f:A→B and g:B→A are inverse of each other, we need to show that
(g∘f)(x)=g(f(x))=x for all x∈A, and
(f∘g)(y)=f(g(y))=y for all y∈B.
Exercises
Exercise 5.5.1
Find the inverse of each of the following bijections.
u:Q→Q, u(x)=3x−2.
v:Q−{1}→Q−{2}, v(x)=2x x−1.
w:Z→Z, w(n)=n+3.
Solution
(a) u−1:Q→Q, u−1(x)=(x+2)/3
Exercise 5.5.2
Find the inverse of the function r:(0,∞)→R defined by r(x)=4+3lnx.
Exercise 5.5.3
The images of the bijection α:{1,2,3,4,5,6,7,8}→{a,b,c,d,e,f,g,h} are given below. (5.5.30)x 1 2 3 4 5 6 7 8 α(x)g a d h b e f c Find its inverse function.
Solution
The images under α−1:{a,b,c,d,e,f,g,h}→{1,2,3,4,5,6,7,8} are given below. (5.5.31)x a b c d e f g h α−1(x)2 5 8 3 6 7 1 4
Exercise 5.5.4
The function h:(0,∞)→(0,∞) is defined by h(x)=x+1 x. Determine h∘h. Simplify your answer as much as possible.
Exercise 5.5.5
The functions g,f:R→R are defined by f(x)=1−3x and g(x)=x 2+1. Evaluate f(g(f(0))).
Solution
We do not need to find the formula of the composite function, as we can evaluate the result directly: f(g(f(0)))=f(g(1))=f(2)=−5.
Exercise 5.5.6
The functions f,g:Z→Z are defined by (5.5.32)f(n)={2n−1 if n≥0 2n if n<0 and g(n)={n+1 if n is even 3n if n is odd Determine g∘f
Exercise 5.5.7
Describe g∘f.
f:Z→N, f(n)=n 2+1; g:N→Q, g(n)=1 n.
f:R→(0,1), f(x)=1/(x 2+1); g:(0,1)→(0,1), g(x)=1−x.
f:Q−{2}→Q∗, f(x)=1/(x−2); g:Q∗→Q∗, g(x)=1/x.
f:R→[1,∞),f(x)=x 2+1; g:[1,∞)→[0,∞)g(x)=x−1.
f:Q−{10/3}→Q−{3},f(x)=3x−7; g:Q−{3}→Q−{2}, g(x)=2x/(x−3).
Solution
(a) g∘f:Z→Q, (g∘f)(n)=1/(n 2+1)
(b) g∘f:R→(0,1), (g∘f)(x)=x 2/(x 2+1)
Exercise 5.5.8
If f:A→B and g:B→C are functions and g∘f is one-to-one, must g be one-to-one? Prove or give a counter-example.
Exercise 5.5.9
If f:A→B and g:B→C are functions and g∘f is onto, must f be onto? Prove or give a counter-example.
Solution
No. Consider f:{2,3}→{a,b,c} by {(2,a),(3,b)} and g:{a,b,c}→{5} by {(a,5),(b,5),(c,5)}.
Then f∘g:{2,3}→{5} is defined by{(2,5),(3,5)}. Clearly f∘g is onto, while f is not onto.
Exercise 5.5.10
If f:A→B and g:B→C are functions and g∘f is one-to-one, must f be one-to-one? Prove or give a counter-example.
Exercise 5.5.11
If f:A→B and g:B→C are functions and g∘f is onto, must g be onto? Prove or give a counter-example.
Answer
Yes, if f:A→B and g:B→C are functions and g∘f is onto, then g must be onto.
Proof
If g is not onto, then ∃c∈C such that there is no b∈B such that g(b)=c.
However, since g∘f is onto, we know∃a∈A such that(g∘f)(a)=c. This means g(f(a))=c.
f(a)∈B and g(f(a))=c; let b=f(a)and now there is a b∈B such that g(b)=c.
Since every element in set C does have a pre-image in set B, by the definition of onto,g must be onto.
Exercise 5.5.12
Given the bijections f and g, find f∘g, (f∘g)−1 and g−1∘f−1.
f:Z→Z, f(n)=n+1; g:Z→Z, g(n)=2−n.
f:Q→Q, f(x)=5x; g:Q→Q, g(x)=x−2 5.
f:Q−{2}→Q−{2}, f(x)=3x−4; g:Q−{2}→Q−{2}, g(x)=x x−2.
Exercise 5.5.13
Prove part (b) of Theorem 5.5.2.
Statement of Theorem 5.5.2b
Given f:A→B and g:B→C, if both f and g are one-to-one, then g∘f is also one-to-one.
Proof
Suppose(g∘f)(a 1)=(g∘f)(a 2) for some a 1,a 2∈A. WMST a 1=a 2.
By definition of composition of functions,we have (5.5.33)g(f(a 1))=g(f(a 2)).
f(a 1)∈B and f(a 2)∈B. Let b 1=f(a 1) and b 2=f(a 2). Substituting into equation 5.5.3,(5.5.34)g(b 1)=g(b 2).
Since g is one-to-one, we know b 1=b 2 by definition of one-to-one. Since b 1=b 2 we have f(a 1)=f(a 2).
Now, since f is one-to-one, we know a 1=a 2 by definition of one-to-one.
Thus we have demonstrated if(g∘f)(a 1)=(g∘f)(a 2) then a 1=a 2 and therefore by the definition of one-to-one,g∘f is one-to-one.
Exercise 5.5.14
Prove part (c) of Theorem 5.5.2
Exercise 5.5.15
Prove part (d) of Theorem 5.5.
This page titled 5.5: Inverse Functions and Composition is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Harris Kwong (OpenSUNY) .
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5.4: Onto Functions and Images/Preimages of Sets
5.6: Infinite Sets and Cardinality
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Hand
Hand Introduction
Anatomy
High-Yield Topics
Extensor Tendon Compartments
Ligaments of the Fingers
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Hand Tumors & Lesions
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Raynaud's Syndrome
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Digital Artery Aneurysm
Nail Bed
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Updated: Dec 18 2023
PIN Compression Syndrome
Evan Watts MD
Topic Podcast
PIN Compression Syndrome
Experts
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Sumary
PIN compression syndrome is a compressive neuropathy of the PIN which affects the nerve supply of the forearm extensor compartment.
Diagnosis is made clinically with weakness of thumb and wrist extensors without sensory deficits.
Treatment is a course of conservative management with splinting and surgical decompression reserved for persistent cases lasting > 3 months.
Epidemiology
Incidence
~ 3 per 100,000 annually
Demographics
more common in manual laborers, males and bodybuilders
Etiology
Pathophysiology
mechanism of injury
microtrauma
from repetitive pronosupination movements
trauma
fracture/dislocation (e.g., monteggia fx, radial head fx, etc)
space filling lesions
e.g. ganglion, lipomas, etc
inflammation
e.g. rheumatoid synovitis of radiocapitellar joint
iatrogenic (surgery)
pathoanatomy:
five potential sites of compression include
fibrous tissue anterior to the radiocapitellar joint
between the brachialis and brachioradialis
“leash of Henry”
are recurrent radial vessels that fan out across the PIN at the level of the radial neck
extensor carpi radialis brevis edge
medio-proximal edge of the extensor carpi radialis brevis
"arcade of Fröhse"
which is the proximal edge of the superficial portion of the supinator
supinator muscle edge
distal edge of the supinator muscle
Anatomy
PIN
origin
PIN is a branch of the radial nerve that provides motor innervation to the extensor compartment
course
passes between the two heads of origin of the supinator muscle
direct contact with the radial neck osteology
passes over abductor pollicis longus muscle origin to reach interosseous membrane
transverses along the posterior interosseous membrane
innervation
motor
common extensors
ECRB (often from radial nerve proper, but can be from PIN)
Extensor digitorum communis (EDC)
Extensor digiti minimi (EDM)
Extensor carpi ulnaris (ECU)
deep extensors
Supinator
Abductor pollicis longus (APL)
Extensor pollicus brevis (EPB)
Extensor pollicus longus (EPL)
Extensor indicis proprius (EIP)
sensory
sensory fibers to dorsal wrist capsule
provided by terminal branch which is located on the floor of the 4th extensor compartment
no cutaneous innervation
Presentation
Symptoms
insidious onset, often goes undiagnosed
defining symptoms
pain in the forearm and wrist
location depends on site of PIN compression
e.g., pain just distal to the lateral epicondyle of the elbow may be caused by compression at the arcade of Frohse
weakness with finger, wrist and thumb movements
Physical exam
inspection
chronic compression may cause forearm extensor compartment muscle atrophy
motion
weakness
finger metacarpal extension weakness
wrist extension weakness
inability to extend wrist in neutral or ulnar deviation
the wrist will extend with radial deviation due to intact ECRL (radial n.) and absent ECU (PIN).
provocative tests
resisted supination
will increase pain symptoms
normal tenodesis test
tenodesis test is used to differentiate from extensor tendon rupture from RA
Evaluation
Radiographs
indications
not commonly needed for the diagnosis of PIN compression syndrome
MRI
indications
not commonly needed for the diagnosis of PIN compression syndrome
may be help to site and delineate the soft tissue mass responsible for compression
helpful for surgical planning of mass resection
Studies
EMG
indications
may help identify the level of nerve compression
may be used to rule out differential diagnoses of neuropathy
Differential
Cervical spine nerve compression
Brachial plexus compression
Peripheral neuropathy
Diagnosis
Clinical
diagnosis is made with careful history and physical examination
Treatment
Nonoperative
rest, activity modification, stretching, splinting, NSAIDS
indications
recommended as first-line treatment for all cases
lidocaine/corticosteroid injection
indications
a compressive mass, such as lipoma or ganglion, has been ruled out
isolated tenderness distal to lateral epicondyle
trial of rest, activity modification, anti-inflammatories were not effective
technique
single injection 3-4 cm distal to lateral epicondyle at site of compression
surgical decompression
indications
symptoms persist for greater than three months of nonoperative treatment
compressive mass detected on imaging
outcomes
results are variable
spontaneous recovery of motor function was seen in 75 - 97% of non-traumatic case series
may continue to improve for up to 18 months
Technique
Surgical decompression
approach
anterolateral approach to elbow is most common approach
may also consider posterior approach
decompression
decompression should begin with release of
fibrous bands connecting brachialis and brachioradialis
leash of Henry
fibrous edge of ECRB
radial tunnel, including arcade of Frosche and distal supinator
Complications
Neglected PIN compression syndrome
muscle fibrosis of PIN innervated muscles
resulting in tendon transfer procedures to re-establish function
Chronic pain
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15880 | https://tamucc.pressbooks.pub/appliedstatswithjamovi/chapter/12-dispersion/ | Skip to content
Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.
12: Dispersion
12.1 Measuring Variability
Dispersion refers to how data points in a dataset vary in relation to a measure of central tendency, such as the mean, median, or mode. While central tendency summarizes the center of a distribution, measures of dispersion describe the spread or variability of the data. Dispersion is essential for understanding how consistent or scattered data points are around the central value. High dispersion indicates that the data points are widely spread, while low dispersion suggests they are closely clustered.
Understanding dispersion is crucial in quantitative research because it informs how precise estimates are and how much observed values deviate from typical values. It also supports accurate interpretation of patterns and differences within datasets. The most common measures of dispersion include standard deviation, interquartile range (IQR), variance, range, and frequencies (for nominal variables).
12.2 Standard Deviation
The standard deviation is the square root of the variance and is one of the most commonly used measures of dispersion. It provides an intuitive sense of variability because it is expressed in the same units as the original data, making it easier to interpret than variance. The standard deviation indicates how far, on average, each data point lies from the mean of the dataset. It is most appropriate when the data is approximately normally distributed.
Standard deviation is especially useful when you need a measure of variability that accounts for every value in the dataset. However, it is sensitive to outliers, which can inflate the standard deviation and make the data appear more variable than it actually is for most observations.
12.3 Interquartile Range (IQR)
The interquartile range (IQR) measures the spread of the middle 50% of a dataset. It is calculated by subtracting the first quartile from the third quartile, capturing the range within which the central half of the data lies. The IQR is a robust measure of dispersion because it is less affected by outliers than the full range, making it especially useful for skewed datasets or those with extreme values. It is commonly used when the goal is to focus on typical variability while excluding the influence of outliers.
The IQR is particularly valuable when the standard deviation is misleading due to skewness, as it provides a clearer picture of the central spread without being distorted by extreme values.
12.4 Variance
Variance measures the average squared deviation of each data point from the mean. It reflects how much the data values differ from the mean on average, but because the deviations are squared, the result is expressed in squared units, which makes interpretation less intuitive. For this reason, variance is often not reported as part of basic descriptive statistics, even though it is a foundational concept in statistics.
12.5 The Range
The range is the simplest measure of dispersion, calculated by subtracting the minimum value in a dataset from the maximum value. It provides a quick sense of the overall spread between the lowest and highest values. However, the range is highly sensitive to outliers, meaning that a single extreme value can greatly distort the result.
The range is most appropriate for small or clean datasets where outliers are not present. In larger or more complex datasets, the range may be misleading and is typically supplemented with more robust measures like the interquartile range or standard deviation.
12.6 Nominal Variable Dispersion
For nominal variables, which are categorical and have no inherent order, frequencies serve as a practical way to assess dispersion. Frequencies represent the count of occurrences for each category in the dataset, helping researchers understand how the data is distributed across categories, whether it is concentrated in one or a few categories or spread more evenly.
A high frequency in a single category indicates low dispersion, as most of the data fall into that category. In contrast, a more even spread of frequencies suggests higher dispersion, with the data more equally divided among categories. Frequencies provide insight into the variability of categorical data and help identify how concentrated or diverse the responses are.
12.7 Dispersion in Jamovi
Jamovi can generate values for key measures of dispersion such as standard deviation, interquartile range (IQR), variance, and range, allowing you to assess how much your data varies around the central tendency.
How To: Dispersion
To calculate standard deviation, IQR, variance, and range in Jamovi, go to the Analyses tab, select Exploration, then Descriptives.
Move variables into the Variables box.
Under the Statistics drop-down, check Standard deviation, IQR, Variance, and Range under Dispersion.
For nominal variables, Jamovi provides a frequency distribution that helps you assess the variability of categorical data.
How To: Frequencies
To calculate frequencies in Jamovi, go to the Analyses tab, select Exploration, then Descriptives.
Move nominal variables into the Variables box.
Check Frequency tables under the Split by box.
Uncheck the pre-selected options under the Statistics drop-down.
12.8 Choosing the Right Measure of Dispersion
The choice of dispersion measure depends on the nature of the data and the type of analysis being conducted. For symmetric distributions without outliers, the standard deviation is typically preferred, as it provides a detailed view of how data points vary around the mean. For skewed distributions or datasets with outliers, the interquartile range (IQR) is often more informative because it focuses on the central 50% of the data and is less affected by extreme values. In research, it is often helpful to report both standard deviation and IQR when analyzing continuous variables. This dual approach gives a more complete picture of variability, especially when the data are not normally distributed. For categorical (nominal) data, frequencies are the appropriate measure, indicating how evenly or unevenly the data is distributed across categories.
Chapter 12 Summary Key Takeaways
Dispersion describes the variability or spread of data around a central value. Measures of dispersion help researchers understand how consistent or scattered values are within a dataset. Key measures include standard deviation, interquartile range (IQR), variance, range, and frequencies for nominal variables. The appropriate measure depends on the type and distribution of the data. Standard deviation is ideal for normally distributed data, while IQR is more robust for skewed data or datasets with outliers. For categorical data, frequencies help evaluate how evenly responses are distributed across categories. Jamovi support the calculation and interpretation of these measures.
The standard deviation shows how spread out continuous data is around the mean and is best used with normally distributed data.
The interquartile range (IQR) measures the spread of the middle 50% of the data and is helpful when the data is skewed or contains outliers.
The variance quantifies how much data points deviate from the mean on average, using squared units that are common in statistical modeling.
The range indicates the distance between the smallest and largest values but can be distorted by extreme values.
Frequencies reveal how often each category appears in nominal data, helping to assess whether responses are concentrated or evenly distributed.
The choice of which measure of dispersion to use depends on the type of data, its distribution, and whether outliers are present.
License
Applied Statistics for Quantitative Research: A Practical Guide with Jamovi Copyright © by Christopher Benedetti is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.
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15881 | https://periodicos.uem.br/actascitechnol/index.php/ActaSciTechnol/article/download/56088/751375153444/751375239677 | Acta Scientiarum
ISSN on -line: 1807 -8664
Doi: 10.4025/ actascitechnol .v44i1. 56088
BIOTECHNOLOGY
Acta Scientiarum. Technology , v. 4 4, e 56088 , 20 22
Production and market comparison of urokinase and
streptokinase as effective and cheap fibrinolytic agents for
treatment of cardiovascular diseases
Ali Nawaz , Ikram ul Haq, Haseeb Amin, Zinnia Shah, Ammar Javed, Hamid Mukhtar and Fatima Akram
1Institute of Industrial Biotechnology, G overnment College University Lahore , Katchery Road, Anarkali Bazaar Lahore, Punjab 54000, Pakistan. Author for
correspondence. E -mail: ali.nawaz@gcu.edu.pk
ABSTRACT . Failure of hemostasis and the forma tion of blood clots in th e arteries are the main reasons that
provoke the onset of cardiov ascular diseases (CVDs) such as myocardial infarction and ischemic stroke.
Cardiovascular diseases have become the primary cause of deaths and disabilities across the globe.
Therefore, this problem nee ds to be addressed with urgency . The disintegration of blood clots requires
fibrinolytic agents, whic h are involved in thrombolysis . Streptokinase and urokina se are fibrinolytic
enzymes; the former is primarily produced from microbial sources and the latter is isolated from urine,
respectively. Streptokinase and urokinase h ave been in use for a long time to treat cardiovascular diseases.
This review explains in detail the comparison of employing stre ptokinase and urokinase for the said purpose
in a cost -effective manner. The recombinant production of both the agents has been discussed in detail.
Furthermore, the effic acy of both the agents has been compared based upon their side effects and retention
time in the body. A thorough study has been made to compare the influence of using both the agents on the
health of cardiovascular patients in the last decade .
Keywords: Streptokinase; uroquinase; thrombolytic agentes; fibrinolysis; plasminogen activator .
Received on Octobe r5, 2020 .
Accepted on Ma rch 30 , 202 1.
Introduction
Cardiovascular diseases (CVDs) account for almost 17.9 million deaths every year and have been reported
to be the leading cause of death by the World Health Organization . Medical complications such as myocardial
infarction, ischemic stroke, peripheral arterial thrombosis, pulmonary embolism, and deep vein thrombosis
(Ruscica, Corsini, Ferri, Banach, & Sirtori, 2020) , which are majorly caused by atherosclerosis, are all in cluded
in the set of CVDs. Clot (blood) dissolving agents or fibrinol ytic agents have been proved as efficient means
to treat CVDs (Ghosh, Pulicherla, Rekha, Venkat Rao, & Sambasiva Rao, 2012) . Various fibrinolytic agents
are commercially available amongs t which Streptokinase and urokinase are mostly preferred. Generally,
fibrinolytic agents activate the fibrinolytic system by converting the proenzyme plasminogen (inactive form)
into its active form plasmin. Specific activators (tissue type activators and the vascular type activators)
present in various tissues and body fluids are required for the activation of the fibrinolytic system (Arnetz et
al., 2020) . Thrombolytic treatment is mainly directed towards the activation of plasminogen to plasmin at the
location of thrombus, on the surface of fibrin (Stephani et al., 2017). Plasmin is a serine protease that acts on
both the fibrin and the fibrinogen, degrading them and thus dissolving the blood clots (Bell, 2002).
Streptokinase is an extracellular enzyme that is produced by several strains of β- hemolytic Streptococci ,
flourishing mainly in the upper respiratory tract (Bhardwaj & Angayarkanni, 2015a) , whereas urokinases are
produced by the human kidney cells and are later isolatedfrom urine (Mahmood, Mihal cioiu, & Rabbani, 2018) .
Where streptokinases are exogenous enzymes and may provoke allergic reactions, urokinases are indigenous to
human origin and hence can be repeatedly administered without causing any unwanted antigenic response.
However, removing th e antigenic sites in Streprococcus through modern rec ombinant engineering techniques
deprives the streptokinase of its antigenicity (Akbar, Zia, Ahmad, Arooj, & Nusrat, 2020) , which can then be
produced as a reliable choice for the treatment of CVDs . Strep tokinases are neither proteases nor true plasminogen
activators like urokinases (Avgerinos, 2017) . They bind to one of the circulating plasminogen molecules to form a
streptokinase -plasminogen complex (Figure 1), the resulting conformational changes (Figur e 2) favor the
conversion of surrounding plasminogen to active plasmin (Aghaeepoor et al., 2019; Sahoo & Sahoo, 2020) .Page 2 of 12 Nawaz et al.
Acta Scientiarum. Technology , v. 4 4, e 56088 , 20 22
Urokinase, unlike streptokinase, is a true plasmi nogen activator and having been isolated from urine, it is often
called UPA: urinary -typ e plasminogen activator. Mc Farlane and Pilling (Mahmood et al., 2018) first isolated
urokinase in 1947; after which in 1972, UPA was first licensed for use in France (Kunamneni, Ogaugwu, & Goli,
2018) . It is a serine protease which also functions as a thr ombolytic agent by hydrolyzing the peptide link in
plasminogen, transforming it into plasmin that can, in turn, hydrolyze the fibrin mesh in the blood clots (Figure 3)
(Tan et al., 2017) . Urokinase activates both, the fibrin specific plasminogen and the fi brin non -specific
plasminogen (Kadir & Bayraktutan, 20 20 ). Activation of fibrin non -specific plasminogen results in excessive
bleeding making the process undesirable. However, the activation of fibrin -specific plasminogen is favorable,
resulting in the bre akdown of blood clots into soluble peptides (Urano, Castellino, & Suzuki, 2018) .
Figure 1. Role of streptokinase in localized fibrinolysis and systemic lytic state.
Figure 2. Streptokinase mode of action in blood clot lysis. Fibrinolytic agent’s role in cardiovascular diseases Page 3of 12
Acta Scientiarum. Technology , v. 4 4, e 56088 , 20 22
Figure 3 .Urokinase mode of action in blood clot lysis.
As a fibrinolytic agent, urokinase is favored over streptokinase essentially because it is a human
indigenous protein and thus can be administered repeatedly without the fear of causing any allergic response
in the body. It has high fibrin specificity and fewer side effects (Nawaz et al., 2020) and works more or less in
the same way as streptokinase. Urokinase can also be isolated from plasma, seminal fluid and tissue cultures
of kidney cells (Fedan, 2019) ; Howe ver, the production expenses limit the productio n feasibility to a great
extent; f or instance, 1500L of urine is only sufficient to harvest a single dose of urokinase . This has led to the
uncovering of bioengineered microbes as potential sources of commer cial urokinase (Agrawal & Patil, 2020) .
Comparison between the characteristics of streptokinase and urokinase has been depicted in Table 1. This
article has compared the feasibility of industrial production and the market values of these two agents.
Tabl e 1. Comparison between the characteristics of streptokinase and uroquinase.
Characteristics Streptokinase Urokinase
Molecular Weight (kD) 48 32/54
Plasminogen Activation ID
Fibrin Specificity Yes Yes
Fibrinogeno -lysis Yes Yes
Half -life (min) 10/90 2
Elimination Kidney Kidney
Immunogenicity Yes No
Cost Less More
Source: (Baruah, Dash, Chaudhari, & Kadam, 2006) .
Microbial production of streptokinase and urokinase
Streptokinases and urokinases are the products of bioprocesses which are highly specific to some microbial
species, as is also evident from Table 2. The search for cost -effective and medically sound procedures for their
commercial production demands an in -depth insight into the potential of novel microbial sources, and their
pathways involved in the synthesis of these enzymes.
Table 2. Microbial sources for the production of streptokinase and urokinase along with their enzyme activity.
Sr.
No. Microbial Source Streptokinase
Units
Urokinase
Units Reference
1. β-hemolytic streptococci 467.73 -(Dubey, Kumar, Agrawala, Char, & Pusp, 2011)
2. Pseudomonas sp. -785.73 (Dubey et al., 2011)
3. Streptococcus equisimilis 474.56 -(Banerjee, Chisti, & Banerjee, 2004)
4. Streptococcus agalactiae 147.08 -(Naeem, Sadia, Awan, & Zia, 2018)
5. Enterococcus gallinarum -81.3 (Nawaz et al., 2020) Page 4of 12 Nawaz et al.
Acta Scientiarum. Technology , v. 4 4, e 56088 , 20 22
Streptokinase
There are several microbial strains which can be used for the production of streptokinase. It is obtained
mainly from Streptococcus sp. isolated from the upper respiratory tract, throat, or sputum samples of infected
persons (Hatami & Eghdami, 2018) . The other Streptococcus sp. reported for streptokinase production include
S. dysgalactiae, S. equinus, S. mutans, S. feacalis, S. uberis, S. equisimilis, S. sanguis, S. lactis, and S. pyogenes (C,
V, Babu, Ethiraj, & Naine, 2013; El -Mongy & Taha, 2012; Gupta, Saxena, & Meshram, 2020) . The production
of streptokinase from microbial sources is an economical process that can be carried out on a mass -scale by
maintaining the necessary growth parameters and optimizing media in the industrial fermentations to meet
the global demand (Ghosal et al., 2017) . This establishes the fact that the optimization of fermentation
parameters plays an essential role in enhancing the production rate for commercial consumption (Cheng &
Kim, 2015) . Despite the ease with which many scientists had been successfully producing commercial
streptokinase, various side effects associated to its use declined its industrial value (Chakravarti et al., 2013) .
The typical undesirable outcome has been the onset of allergic reactions including fever, chills, nausea,
itching, skin rashes, and respiratory trouble. This has been linked to the presence of several antigenic epitopes
on the streptokinase mol ecule (Chandran, Singh, Thomas, Basu, & Brahmachari, 2016) .
One obvious remedy to overcome the problem caused by the presence of these immunogenic sites was to
remove them resulting in a non -immunogenic streptokinase molecule (A kbar , 2020) . Another way of
producing streptokinase with little or no immunogenicity is through recombinant production or development
of a mutant strain. Deitcher and Jaff, for instance, amplified a streptokinase gene from S. pyogenes C1, cloned
it into pET28a(+) vector and transformed it to a non -pathogenic microorganism like E. coli BL21 (DE3)
(Deitcher & Jaff, 2002) . Next, by inducing the expression vector, high expression of the recombinant protein recSK
was produced which held little or no capacity to react as an immunogen (Deepak & Geetha, 2019) . Another study
carried out by (Arshad, Zia, Asghar, & Joyia, 2019) in an attempt to achieve the said objective, enhanced production
of streptokinase was obtained with little immunogenicity by chemical mutagenesis of S. agalactiae.
Urokinase
The search for an economical production procedure to s ynthesize a high fibrin -specific -urokinase still
continues. Being immunologically unreactive, unlike streptokinase, urokinases have gained a comparative
preference (Fajardo -Espinoza, Romero -Rojas, & Hernández -Sánchez, 2019) . There are many sources to
obtain urokinases but micr oorganisms are the best sources because they are easy to grow and the extraction
and purification of urokinase is relatively easier. Primitively, urokinase is produced from urine, but the
procedure is quite lengthy, complicated and costly (Gupta et al., 2020) . These limitations have been countered
by utilizing human cell cultures for the production of urokinase. Human embryonic kidney cells, human
fibroblasts, human myeloma cells, and human umbilical vein endothelial cells are currently being used for the
production of urokinase. Studies show that the quantity of urokinase produced by cultured cells is
approximately 50 -100 ng mL -1, which is higher as compared to that pr oduced from urine i.e. 10 -15 ng mL -1
(Roychoudhury, Khaparde, Mattiasson, & Kumar, 2006) .
Fermentation technology happens to be the leading application for employing different microorganisms
including bacteria (E. Coli , Pseudomonas sp., B. subtilis, Halobacillus sp., Enterococcos sp.), fungi
(Asperigillus sp., Penicillium sp., Trichothecium sp., S. cerevisiae), algae, and actinomycetes in the process of
urokinase -production (Agrawal & Patil, 2020; Nawaz et al., 2020) . T he evident attainment of high yields of
urokinase from microbial sources validate their acceptance over pre -existing sources of UPA (Sahoo & Sahoo,
2020) . In addition, the availability of factors including microbial source, physical parameters, and the cultural
conditions make the process sensitive to multi -faceted optimization strategies. One of the significant
concerns in the fermentation process is to look for a cost -effective media (Ghaffar, Ahmed, Munir, Faisal, &
Mahmood, 2015) .
Appropriate media formulation can also help produce high yields of urokinase. Minor changes in the media
formulation, cultural conditions, and physical parameters can affect enzyme production to a great extent
(Zhu, Mollet, Hubert, Kyung, & Zhang, 2017) . Submerged and solid -state fermentation procedures are famous
with respect to urokinase production. Submerged type is preferred as it provides high yield due to máximum
nutrient availability and accessible approach for downstream processing (Sharma, Sharma, & Shivlata, 2015) . The
most significant concern about fermentation technology is the cost involved. In this regard, recombinant
technology proves to be a cost -effective and reliable alternative. For instance, thrombolytic agents are being Fibrinolytic agent’s role in cardiovascular diseases Page 5 of 12
Acta Scientiarum. Technology , v. 4 4, e 56088 , 20 22
produced using various expression platforms like bacteria (Adivitiya & Khasa, 2017) . Dubey has r eported the
production of 12.5U 50 μ L-1 urokinase by the recombinant method. Low molecular weight fragment of DNA,
isolated from Pseudomonas sp., was ligated into pET28a(+) vector (Du bey et al., 2011) . It was transformed to
E.coli BL21 - RIL and induced for expression regulated by the T7 promoter. Not only did this method prove d
to be feasible, but was also found much efficient in contrast to fermentation (Graor, Young, Risius, &
Rusc hhaupt, 1987) .
Industrial production and componential costs
Isolation
The isolation process of a potential microorganism for the production of thrombolytic agents takes up a
significant fraction of the total production cost. As streptokinase can quickly be produced from microbes that
are isolated from sputum samples or dental plaque (Qiao et al., 2018) , it makes the process cost -effective.
However, the production process can be made more cost -effective by optimizing the media required for cell
growth and en zyme production. Production of urokinase is also possible by the isolation of microorganisms
(bacteria) from various sources like soil, seawater, and fermented food samples such as Sardine, pickle,
chickpea, soybean, tuna, soya sauce, olive, hot sauce, cor n, red kidney beans, sweet corn, and tomato ketchup
(Nawaz et al., 2020) . Isolation of microorganisms from these sources produces urokinase that is much cost -
effective and efficient as compared to its production by recombinant techniques and tissue culture methods
(Huish, Thelwell, & Longstaff, 2017) .
Production
Streptokinase is preferably produced through continuous culture methods that contain glucose as an
energy source (Dubey et al., 2011) . Producing streptokinase on a large scale using industrial ferme ntation can
significantly decrease the production cost by optimizing different factors like pH, inoculum size, temperature
of the fermentation medium, fermentation time, agitation speed, and aeration rate (Singh , Saxena, & Yadav,
2017) . The microbial synthesis of streptokinase has been further reported to increase using mutant microbial
species and opti mization of fermentation medium . Arshad et al., (2019) reported a 4.14 -fold increase in
streptokinase production by optimization of ferm entation media. Similarly, Wu et al. , (2019) developed a
cheap fermentation media for the cost -effec tive synthesis of streptokinase.
Contrarily, separation of urokinase from urine is not cost -effective as it produces very low yields with
equally low effici ency as well. Therefore, recombinant methods stand more valid for its production (Adivitiya &
Khasa, 2017) . High secretion of urokinase by cells is an essential factor in using cell cultures for the production of
urokinase. Keeping in view the therapeutic importance and cost efficiency of producing a plasminogen activator
from human source, cell lines like human kidney cell lines are used for urokinase production (Roychoudhury et al.,
2006) . These cell lines are supplied with a complex growth medium includi ng sugars, amino acids, and growth
factors which contribute majorly towards high yield in lesser input (Li, Yuan, & Yuan, 2020) .
Shelf -life
Shelf life and reusability after the opening of the thrombolytic agent’s vials are the noteworthy factors
considered when calculating the cost -effectiveness of both, streptokinase and urokinase. Comparing the
shelf -life of streptokinase and urokinase, (Hasanpour, Esmaeili, Hosseini, & Amani, 2021) stated that
unopened vials could remain stable for 3 years at l ess than 25 oC, and 4 years at below freezing temperatures,
respectively. The reconstituted solution should not be stored in a refrigerator for more than 24 hours at +2 to +8 °C.
However, if urokinase is diluted to 5000 IU mL -1 in sterile water or 0.9% sodiu m chloride, it retains its in vitro
fibrinolytic activity for up to 6 m onths when stored at -20 to -70 °C (Hickman, Pawlowski, Sekhon, Marks, & Gupta,
2018) . These studies provide yet another striking advantage of urokinase over streptokinase.
Recovery
Various mechanisms have been explained for the recovery of streptokinase, either from fermentation broths or
crude samples that are commercially available. According to a study (Dubey et al., 2011) , a five to six -fold increase
in purity was observ ed by first performing column chromatography and then column electrophoresis in sucrose
density gradients. In a likewise study, Xu et al., (2019) obtained a final specific activity of 100,000 units of
streptokinase per mg of protein by coupling ion -exchang e and gel -permeation chromatography. Page 6 of 12 Nawaz et al.
Acta Scientiarum. Technology , v. 4 4, e 56088 , 20 22
Unlike streptokinase (obtained from a microbial source), urokinase (obtained from a human source) is
recovered in shallow concentrations which makes the process costly (Mousa & Broce, 2017) . Various methods
are being ap plied for the recovery of urokinase, including ion -exchange chromatography, ammonium sulfate
precipitation, affinity chromatography, and gel permeation chromatography (Badhe & Nanda, 2018) . Among
all of these, affinity chromatography has emerged as the most powerful method for purification of urokinase
involving synthetic and biological ligands that bind specifically to urokinase. The recovery methods for
urokinase are mostly multi -step processes that produce lower yield but require high operating costs (Lijnen
& Collen, 1995) . The development of new methods however, does produce enhanced recovery at reasonable
cost. Xu et al., (2019) coupled processes involving foam fraction ation and silica gel adsorption which increased
urokinase activity and purificati on yield by 25.3% and 79.2%, respectively, and rendered it cost -effective.
Efficacy
Severe allergic reactions which may progress to fatality are very likely with streptokinase dosages. As
streptokinases are of bacterial origin, these molecules are coated with several antigens which can initiate an
allergic response within the patient’s body (Krishnamurthy, Belur, & Subramanya, 2018) . If such a response
does occur, the administration of streptokinase is immediately ceased and is not recommended to be
consumed again by the patient for a minimum duration of the following four years (Bell, 2002) . In such cases,
the patient is then treated with alternative agents like alteplase or urokinase (Gurewich, 2019) . Dr axler ,
Sashindranath, & Medcalf , (2017) showe d that patients previously infected with streptococci required high
doses for thrombolysis due to the extent of its antibodies within the patient’s body.
In addition to the violent allergic response, streptokinase is also associated with a higher risk of d eveloping
hemorrhage (Banerjee et al., 2004; Sawhney, Katare, & Sahni, 2016) . A recent study subjected a cohort of 80
patients to fibrinolytic therapy; 15 among which showed hemorrhage effect includ ed five patients who were
given urokinase and ten who were given streptokinase (Picard et al., 2019) . Results by Yazdi et al., (2017) have
shown that bleeding was the most significant complication seen in patients injected with streptokinase as it
binds with both circulating and non -circulating plasminogen. Bleeding may occur spontaneously or at the
puncture site. Intracranial hemorrhage and hemorrhage stroke are of significant concern in this case (Prasad,
Singh, Kanabar, & Vijayvergiya, 2020) . Different factors including tumor, aneurysm, infarc tion, bleeding
diathesis, advanced age, uncontrolled hypertension, severe heart disease, low body weight, trauma, or
surgical intervention in the cerebral system may increase the risk of bleeding (Aslanabadi, Safaie, Talebi,
Dousti, & Entezari -Maleki, 2018 ). Minor bleeding from the venous entry port can be stopped by applying
pressure but in case of severe bleeding, discontinuation of the drug as soon as possible remains the best
approach (Skottrup, Dahlén, Baelum, & Lopez, 2018) .
Thrombolysis with urokinase is a more potent alternative to streptokinase. The advantageous effects of
urokinase promoting revascularization and nerve regeneration make it a much efficient fibrinolytic agent
(Kadir & Bayraktutan, 2020) . In a study, as an example, 27% ± 8% o f thrombus was dissolved by streptokinase,
while 64% ± 9% of an equivalent thrombus sample was dissolved by urokinase within the same time frame of
60 minutes (Ouriel, 2002) . A recent study falls in agreement, stating that streptokinase partially consumes
plasminogen (and plasmin) while forming the activator complex, whereas urokinase converts all of the
plasminogen to plasmin, making it more efficient (Xu et al., 2019) . Furthermore, for the local thrombolytic
therapy, improved efficacy of lysis and a decre ased percentage of systemic fibrinolytic effect and bleeding
problems make urokinase preferable over streptokinase again (Tan et al., 2017) . Treatment of cardiovascular
diseases CVDs require a balance to be maintained between fibrinolytic and thrombolytic activity that is also
efficiently sustained by urokinase, while streptokinase only poorly manages it (Ouriel, 2002) . Moreover,
considering the antigenic effects, being indigenous to human origin, urokinase makes itself perfectly
applicable for repeated adm inistration with little or no complications (Stump, Lijnen, & Collen, 1986) .
Despite all the merits of urokinase discussed above, studies also suggest that it seemingly takes longer time
in comparison for streptokinase to dissolve the same amount of clot (Tanaka, Key, & Levy, 2009) .
Market Potential
The first industry for streptokinase was established in 2001 (Zhong et al., 2020) . Being cost -effective,
streptokinase production through recombinant techniques has a more significant market potential (Oh,
Modi ano, Bachanova, & Vallera, 2020) . The cost of streptokinase for 250,000 units is approximately $21 as
compared to urokinase that costs for $133 for the same amount, which is much expensive. The Chinese market Fibrinolytic agent’s role in cardiovascular diseases Page 7 of 12
Acta Scientiarum. Technology , v. 4 4, e 56088 , 20 22
for urokinase progressed to 50.56 CNY million from 2013 to 2017, representing that market for urokinase is
further developing in the coming years (Kapoor, 2016) . The strategies of domestic production, import and
export, and consumption have helped businessmen to analyze and capitalize on potential opportunities (Tripon
et al., 2020) . The major industries for urokinase are NDPHARM, Wanhua Biochem, and LIVZON which account ed
for 22.35 percent, 10.11 percent, and 21.86 percent of revenue in 2019, respectively. The urokinase market across
the globe is estimated to reach USD 56 million by 2026 (Gurewich, 2019) .
However, to meet the globally increasing demand of urokinase, various researches are being conducted to
increase the bioreactor volumetric productivity by using various methods like medi a manipulation, novel
bioreactor design, and feeding strategies for perfusion culture (Kabir, Moreino, & Siam, 2019) . The recent
significant developments done by the vendors in the United States Urokinase Market include ImaRX
Therapeutics, Jiangsu Aidea Ph armaceutica, Jiangxi Haoran Bio -Pharma, NDPharm, and Wanhua Biochem
(Opacic, Paefgen, Lammers, & Kiessling, 2017) . Growing cases of thrombolytic disorders a cross the world is
the driving force of the urokinase market – however, the lack of proper personnel monitoring the condition
of patients while diagnosing hampers the growth. The challenge for cost -effective production hence remains
valid (Vimal & Kumar, 2019) . The market potential of thrombolytic therapy using drugs like streptokinase and
urokinase has been divided into segments, classified based on drug type, application, distribution channel,
and region (Arshad et al., 2019) (Table 3).
Table 3. Worldwide market potential of streptokinase and urokinase in terms of market price.
Sr no .Trade name Thrombolytic agent Quantity Price (USD) Company
1Bharat Serum and Vaccines Ltd Streptokinase 100 uL18.13 Bharat Serum
2TTK Healthcare Ltd Streptokinase -26.60 Wdrugs/streptokinase
3TTK Healthcare Ltd Urokinase -33.02 Ndrugs
4MERCK Urokinase -40.25 Sigma Aldrich
5United States Biological Streptokinase 100 uL247 Usbio
6Medical Isotopes Streptokinase 5mg 950 Medical Isotopes
7Congruent Pharma Streptokinase 5mL25 Congruent Pharma
8Salvavidas Streptokinase 4mL27 Salvavidas Pharmaceutical
9Yogeshwari Streptokinase 5mL32 Yogeshwari Medicals
10 Alfa Aesar Urokinase 1mg 281 Alfa Aesar
11 Apollo Urokinase 1mg 174 Apollo Scientific Ltd.
12 American Custom Urokinase 1mg 728 American Custom Corporation
Source: Bhardwaj and Angayarkanni (2015b ); Ghosh, Saha, and Sahoo (2021 ); Mendieta et al. (2019 ); Ponnada, Pulicherla, and Rao (2012) .
Conclusion
Streptokinase and urokinase are widely used across the globe as fibrinolytic agents. Urokinase appears to
be costly but more useful than streptokinase. Streptokinase possesses specific side effects due its bacterial
origin, but they can be overcome through recombinant techniques. Recombinant production of streptokinase
and urokinase has a significant market potential that will further develop in upcoming years. Generally,
wha tever method implied, reducting or optimizating the process steps involved can lead to an overall decrease
in the operational costs and a relative increase in the produ ct yield. The previous trends and recent
experiences suggest that fibrinolytic therapy has been progressively incorporated into the routine
management of patients with thrombolytic disorders. The analysis of the comparative efficacy of different
thrombolyt ic agents in combination and the inspection of different methods of administration of the
thrombolytic agents are the two facets that need be explored further in future researches to identify a
considerably efficient production process for these agents.
Re ferences
Adivitiya, & Khasa, Y. P. (2017). The evolution of recombinant thrombolytics: current status and future
directions. Bioengineered , 8(4), 331 –358. DOI : 10.1080/21655979.2016.1229718
Aghaeepoor, M., Akbarzadeh, A., Kobarfard, F., Shabani, A. A., Dehnavi, E., Aval, S. J., & Eidgahi, M. R. A.
(2019). Optimization and high level production of recombinant synthetic streptokinase in E. Coli using
response surface methodology. Iranian Journa l of Pharmaceutical Research , 18 (2), 961 –973.
DOI : 10.22037/ijpr.2019.1100636 Page 8 of 12 Nawaz et al.
Acta Scientiarum. Technology , v. 4 4, e 56088 , 20 22
Agrawal, N. V. , & Patil, N. A. (2020). Enzyme technology prospects and their biomedical applications . In R.
Vyas (Ed.), Advances in Bioengineering (p. 147 -159). Si ngapore , SI : Springer.
DOI : 10.1007/978 -981 -15 -2063 -1_9
Akbar, G., Zia, M. A., Ahmad, A., Arooj, N., & Nusrat, S. (2020). Review on streptokinase with its antigenic
determinants and perspectives to develop its recombinant enzyme with minimum immunogenicity .
Journal of Innovative Sciences , 6(1), 17 -23 . DOI: 10.17582/journal.jis/2020/6.1.17.23
Akbar, G., Zia, M. A., Ahmad, A., Arooj, N., & Nusrat, S. (2020). Review on streptokinase with its antigenic
determinants and perspectives to develop its recombinant enzyme with minimum immunogenicity .
Journal of Innovative Sciences , 6(1) , 17 -23. DOI : 10.17582/journal.jis/2020/6.1.17.23
Arnetz, B. B., Goetz, C. M., Arnetz, J. E., Sudan, S., Van Schagen, J., Piersma, K., & Reyelts, F. (2020).
Enhancing healthcare efficiency to achieve the quadruple aim: an exploratory study. BMC Research
Notes , 13 (362 ), 1 –6. DOI : 10.1186/s13104 -020 -05199 -8
Arshad, A., Zia, M. A. , Asghar, M., & Joyia, F. A. (2019). Enhanced production of streptokinase by chemical
mutagenesis of streptococcus agalactia e EBL -20. Brazilian Archives of Biology and Technology , 62 .
DOI : 10.1590/1678 -4324 -2019170813
Aslanabadi, N., Safaie, N., Talebi, F., Doust i, S., & Entezari -Maleki, T. (2018). The streptokinase therapy
complications and its associated risk factors in patients with acute ST elevation myocardial infarction .
IJPR - Iranian Journal of Pharmaceutical Research , 17 , 53 –63.
Avgerinos, E. (2017). Thr ombolytic therapy. Society for Vascular Surgery , 48 (10), 1363 –1369.
Badhe, R. V, & Nanda, R. K. (2018). Novel method for recovery of purified protein from non -denatured slab gel
using electrophoretic column . Pharmaceutical Resonance, 1(1), 38 –41.
Banerjee, A., Chisti, Y., & Banerjee, U. C. (2004). Streptokinase --a clinically useful thrombolytic agent.
Biotechnology Advances , 22 (4), 287 –307. DOI : 10.1016/j.biotechadv.2003.09.004
Baruah, D. B., Dash, R. N., Chaudhari, M. R., & Kadam, S . S. (2006). Plasminogen activators: a comparison.
Vascular Pharmacology , 44 (1), 1 –9. DOI : 10.1016/j.vph.2005.09.003
Bell, W. R. (2002). Th erapeutic agents —pharmacokinetics and pharmacodynamics . Reviews in Cardiovascular
Medicine , 3(S2), 34 –44.
Bhardwaj, S., & Angayarkanni, J. (2015a). Streptokinase production from Streptococcus dysgalactiae subsp.
equisimilis SK -6 in the presence of surfactants, growth factors and trace elements. 3 Biotech , 5(2), 187 –
DOI : 10.1007/s13205 -014 -0209 -x
Bhardwaj, S., & Angayarkanni, J. (2015b). Streptokinase production from Streptococcus dysgalactiae subsp.
equisimilis SK -6 in the presence of surfactants, growth factors and trace elements. 3 Biotech , 5(2), 187 –
DOI: 3205 -014 -0209 -x
C, S. D., V, M., Babu, V., Ethiraj, S., & Naine, J. (2013). Optimization studies for enhanced production of
streptokinase by Streptococcus equisimilis UVM6. Journal of Pure and Applied Microbiology , 7(3) , 2337 –2341.
Chakravarti, N., Peddareddigari, V. G. R., Warneke, C. L., Johnson, M. M., Overwijk, W. W., Hwu, P., &
Prieto, V. G. (2013). Differential expression of the G -protein -coupled formyl peptide receptor in
melanoma associates with aggressive phenotype. The Ame rican Journal of Dermatopathology , 35 (2), 184 –
DOI: 10.1097/DAD.0b013e31825b2506
Chandran, N. , Singh, M. G., Thomas, Z., Basu, D., & Brahmachari, S. K. (2016). Thrombolytic treatment for
myocardial infarction: Affordability and drug mark et segmentation. Economic and Political Weekly , LI(38),
18 –21.
Deepak, K., & Geetha, A. (2019). Cost variation analysis of various brands of anticoagulants, fibrinolytics,
and antiplatelet drugs currently available in Indian pharmaceutical market. Nationa l Journal of
Physiology, Pharmacy and Pharmacology , 9(5), 368 -372 .
DOI: 10.5455/njppp.2019.9.0205602032019
Deitcher, S. R., & Jaff, M. R. (2002). Pharmacologic and clinical characteristics of thrombolytic agents.
Reviews in Cardiovascular Me dicine , 3(Suppl 2 ), 25 -33.
Cheng, N. T., & Kim, A. S. (2015). Intravenous thrombolysis for acute ischemic stroke within 3 hours versus
between 3 and 4.5 hours of symptom onset . The Neurohospitalist , 5(3), 101 –109. DOI :
10.1177/19418744155831 16 Fibrinolytic agent’s role in cardiovascular diseases Page 9 of 12
Acta Scientiarum. Technology , v. 4 4, e 56088 , 20 22
Draxler, D. F., Sashindranath, M., & Medcalf, R. L. (2017). Plasmin: a modulator of immune fun ction.
Seminars in Thrombosis and Hemostasis , 43 (2), 143 –153. DOI : 10.1055/s -0036 -1586227
Dubey, R., Kumar, J., Agrawala, D., Char, T., & Pusp, P. (2011). Isolation, production, purification, assay and
characterization of fibrinolytic enzymes (nattokinase, streptokinase and urokinase) from bacterial
sources. African Journal of Biotechnology, 10 (8), 1408 –1420. DOI : 10.5897/AJB10.1268
El -Mongy, M. A., & Taha, T. M. (2012). In vitro detection and optimization of streptokinase production by
two streptococcal strains in a relatively low cost growth medium. International Research Journa l of
Microbiology , 3(4) , 153 –163.
Fajardo -Espinoza, F. S., Romero -Rojas, A., & Hernández -Sánchez, H. (2019). Production of bioactive
peptides from bovine colostrum whey using enzymatic hydrolysis . Revista Mexicana de Ingeniería
Química , 19 (1 ), 1 -9. DOI : ht tp://doi.org/ 10.24275/rmiq/Alim525
Fedan, J. S. (2019). Fibrinolytic drug. Encyclop edia Britannica, 14 Aug. 2019 Inc. , N.P. Retrieved from
-drug
Ghaffar, A., Ahmed, B., Munir, B., Faisal, R., & Mahmood, Z. (20 15). Production and characterization of
streptokinase enzyme by using Streptococcus mutans strain in liquid state fermentation through corn
steep liquor (CSL) substrate. Biochem istry & Physiol ogy: Open Acess , 4(178), 2.
DOI: -9652.1000178
Ghosal, T., Augustine, N., Siddapur, A., Babu, V., Samuel, M. K., & Chandrasekaran, S. D. (2017). Strain
improvement, optimization and purification studies for enhanced production of streptokinase from
Strepto coccus uberis TNA -M1. Frontiers in Biology , 12 (5), 376 –384. DOI : 10.1007/s11515 -
017 -1467 -x
Ghosh, M., Pulicherla, K. K., Rekha, V. P. B., Venkat Rao, G., & Sambasiva Rao, K. R. S. (2012). A review on
successive generations of streptokinase based thrombolytic agents. International Journal of Pharmacy and
Pharmaceutical Sciences , 4, 38 –42.
Ghosh, S., Saha, S., & Sahoo, S. (2021). Production of thrombolytic and fibrinolytic proteases: current
advances and future prospective . In H. Thatoi, S. Mohapatra & S. K. Das (Eds.), Bioprospecting of Enzymes
in Industry, Healthcare and Sustainable Environment (p. 363 –381). Singapore, SI: Springer.
Graor, R. A., Young, J. R., Risius, B., & Ruschhaupt, W. F. (1987). Comparison of cost effectiveness of
streptokinase and urokinase in the treatment of deep vein thrombosis. Annals of Vascular Surgery , 1(5),
524 –528. DOI : 10.1016/S0890 -5096(06)61434 -9
Gupta, M., Saxena, S., & Meshram, V. (2020). Microbial clot busters: an overview of source, production,
properties and fibrinolytic activity. In J. Singh, V. Meshram & M. Gupta (Eds.), Bioactive Natural prod ucts
in Drug Discovery (p. 485 –516). Singapore , SI: Springer.
Gurewich, V. (2019). Fibrinolysis: a misunderstood natural defense whose therapeutic potential is unknown .
Cardiovascular Drugs and Therapy , 33 (6), 749 –753. DOI : 10.1007/s10557 -01 9-06923 -8
Hasanpour, A., Esmaeili, F., Hosseini, H., & Amani, A. (2021). Use of mPEG -PLGA nanoparticles to improve
bioactivity and hemocompatibility of streptokinase: In -vitro and in -vivo studies. Materials Science and
Engineering: C , 118 , 111427. DOI:
Hatami, A. A. , & Eghdami, A. (2018). Streptokinase enzyme extraction from Streptococcus agalactiae isolated
from mucous membrane and investigation of fibrinolytic effects and its toxicity on healthy cells Hu02.
Medical Sciences Journal , 28 (3) , 220 –229. DOI : 10.29252/iau.28.3.220
Hickman, D. A., Pawlowski, C. L., Sekhon, U. D . S., Marks, J., & Gupta, A. S . (2018). Biomaterials and
advanced technologies for hemostatic management of bleed ing . Advanced Materials , 30 (4) , 1700859 .
DOI: 10.1002/adma.201700859
Huish, S., Thelwell, C., & Longstaff, C. (2017). Activity regulation by fibrinogen and fibrin of streptokinase from
Streptococcus Pyogenes . PloS One , 12 (1), e0170936 . DOI: 10.1371/journal.pone.0170936
Kabir, E. R., Moreino, S. S., & Siam, M. K. S. (2019). The breakthrough of biosimilars: a twist in the narrative
of biological therapy . Biomolecules , 9(9) , 410 . DOI : 10.3390/biom9090410
Kadir, R. R. A., & Bayraktutan, U. (20 20 ). Urokinase plasminogen activator: a potential thrombolytic agent
for ischaemic stroke . Cellular and Molecular Neurobiology , 40 (3), 347 -355 .
DOI: 10.1007/s10571 -019 -00737 -wPage 10 of 12 Nawaz et al.
Acta Scientiarum. Technology , v. 4 4, e 56088 , 20 22
Kadir, R. R. A., & Bayraktutan, U. (2020). Urokinase plasminogen activator: a potential thrombolytic agent
for ischaemic stroke . Cellular and Molecular Neurobiology , 40 (3), 347 –355.
DOI: 10.1007/s10571 -019 -00737 -w
Kapoor, P. M. (2016). From lactat e to soluble urokinase p lasminogen activator receptor: t he journe y for ideal
cardiac biomarker: a re we there in 2016? Annals of Cardiac Anaesthesia , 19 (2), 211 –213.
DOI: 10.4103/0971 -9784.179587
Krishnamurthy, A., Belur, P. D., & Subramanya, S. B. (2018). Methods available to assess therapeutic
potential of fibrinolytic enzymes of microbial origin: a review. Journal of Analytical Science and
Technology , 9(1 0) . DOI : 10.1186/s40543 -018 -0143 -3
Kunamneni, A., Ogaugwu, C., & Goli, D . (2018). Enzymes as therapeutic agents. In C. S. Nunes & V. Kumar
(Eds.), Enzymes in Human and Animal Nutrition : Principles in Human and Animal Nutrition (p. 301 –312).
Cambridge, MA: Elsevier.
Li, T., Yuan, D., & Yuan, J. (2020). Antithrombotic drugs —pharmacology and perspectives . In M. Wang (Ed.),
Coronary Artery Disease: Therapeutic and Drug Discovery. Advances in Experimental Medicine and B iology
(Vol. 1177, p. 101 –131). Singapore, SI: Spring er.
Lijnen, H. R., & Collen, D. (1995). Fibrinolytic agents: mechanisms of activity and pharmacology.
Thrombosis and Haemostasis , 74 (1), 387 –390.
Mahmood, N., Mihalcioiu, C., & Rabbani, S. A. (2018). Multifaceted role of the urokinase -type plasminogen
acti vator (uPA) and its receptor (uPAR): diagnostic, prognostic, and therapeutic applications. Frontiers in
Oncology , 8. DOI : 10.3389/fonc.2018.00024
Mendieta, I., Nuñez -Anita, R. E., Nava -Villalba, M., Zambrano -Estrada, X., Delgado -González, E. , Anguiano,
B., & Aceves, C. (2019). Molecular iodine exerts antineoplastic effects by diminishing proliferation and
invasive potential and activating the immune response in mammary cancer xenografts. BMC Cancer ,
19 (261 ), 1 –12. DOI: s12885 -019 -5437 -3
Mousa, A. Y., & Broce, M. (2017 ). Catheter -directed thrombolysis versus full anticoagulation alone in
treating proximal iliofemoral deep venous thrombosis. Vascular , 25 (1), 111 –112.
DOI: 10.1177/1708538116655957
Naeem, M., Sadia, B., Awan, F. S. , & Zia, M. A. (2018). Enhanced production of streptokinase by UV - and
ethidium bromide -treated streptococus equisimilis muta nt. Pakistan Journal of Zoology , 50 (2), 655 -661 .
DOI : 10.1 7582/journal.pjz/2018.50.2.655.661
Nawaz, A., Gillani, S. Q., Tahir, S. F., Shah, K. A., Ashraf, S., Mukhtar, H., & Haq, I. U. (2020). Biosynthesis of
fibrinolytic agent urokinase by Enterococcus gallinarum isolated from sardine. Revista Mexicana de
Ingeni ería Química , 19 (1 ). DOI : 10.24275/rmiq/Bio1654
Oh, F., Modiano, J. F., Bachanova, V., & Vallera, D. A. (2020). Bispecific targeting of EGFR and urokinase
receptor (uPAR) using ligand -targeted toxins in solid tumors. Biomolecules , 10 , 956.
DOI: 10.3390/biom10060956
Opacic, T., Paefgen, V., Lammers, T., & Kiessling, F. (2017). Status and trends in the development of clinical
diagnostic agents. WIRES Nanomedicine and Nanobiotechnology , 9(4) , e1441 .
DOI: 10.1002/wnan.1441
Ouriel, K. (2002). Comparison of safety and efficacy of the various thrombolytic agents. Reviews in
Cardiovascular Medicine , 3( 2), S17 -24.
Picard, F., Llitjos, J. -F., Diefenbronn, M., Laghlam, D., Seret, G., Sokoloff, A., … Varenne, O. (2019). The
balance of thrombosis and hemorrhage in STEMI patients with or without associated cardiac arrest: a n
observational study. Resuscitation , 145 , 83 –90. DOI : 10.1016/j.resuscitation.2019.10.022
Ponnada, P. S. K. , Pu licherla, K. K., & Rao, K. R. S. S. (2012). Current status of production, clinical usage and
market scenario of s treptokinase. Journal of Pharmacy Research , 5 (8), 4223 -4229.
Prasad, K., Singh, P., Kanabar, K., & Vijayvergiya, R. (2020). Pulmonary haemorrha ge following
thrombolysis wi th streptokinase in myocardial infarction. BMJ Case Reports , 13 (1) , e232308 .
DOI: 10.1136/bcr -2019 -232308
Singh , P., Saxena, A., & Yadav, M. (2017). Review of streptokinase isolation studies from various technique s.
International Journal of Pharmacy and Life Sciences , 8(6), 5526 –5530. Fibrinolytic agent’s role in cardiovascular diseases Page 11 of 12
Acta Scientiarum. Technology , v. 4 4, e 56088 , 20 22
Qiao, Y., Wu, M., Feng , Y., Zhou, Z., Chen, L., & Chen, F. (2018). Alterations of oral microbiota distinguish
children with autism spectrum disorders from healthy controls. Scientific Reports , 8(1) .
DOI: 10.1038/s41598 -018 -19982 -y
Roychoudhury, P. K., Khaparde, S. S., Mattiasson, B., & Kumar, A. (2006). Synthesis, regulation and
production of urokinase using mammalian cell culture: a comprehensive review. Biotechnology Advances ,
24 (5), 514 –528. DOI : adv.2006.05.002
Ruscica, M., Corsini, A., Ferri, N., Banach, M., & Sirtori, C. R. (2020). Clinical approach to the inflammatory
etiology of cardiovascular diseases. Pharmacological Research , 159 , 104916.
DOI: 10.1016/j.phrs.2020.104916
Sahoo , S. K., & Sahoo, S. (2020). Exploitation of fibrinolytic enzymes in combating blood clotting disorders -
recent advances and strategies: a comprehensive review . In J. K. Patra, A. C. Shukla & G. Das (Eds.),
Advances in Pharmaceutical Biotechnology: Recent P rogress and Future Applications (p. 383 –395).
Singapore , SI: Springer .
Sawhney, P., Katare, K., & Sahni, G. (2016). PEGylation of truncated streptokinase leads to for mulation of a
useful drug with ameliorated attributes . PloS One , 11 (5), e0155831.
DOI: ht tp://doi.org/ 10.1371/journal.pone.0155831
Sharma, A., Sharma, A., & Shivlata, L. (2015). Optimization of medium components for enhanced
production of extracellular fibrinolytic protease from Citrobacter braakii . International Journal of Current
Microbiolog y and Applied Sciences, 4(7), 248 –259.
Skottrup, P. D., Dahlén, G., Baelum, V., & Lopez, R. (2018). Soluble urokinase -type plasminogen activator
receptor is associated with signs of periodontitis in adolescents. European Journal of Oral Sciences ,
126 (4), 292 –299. DOI : 10.1111/eos.12424
Stephani, L., Tjandrawinata, R. R., Afifah, D. N., Lim, Y., Ismaya, W. T., & Suhartono, M. T. (2017). Food
origin fibrinolytic enzyme with multiple actions . HAYATI Journal of Biosciences , 24 (3), 124 –130.
DOI : 10.1016/j.hjb.2017.09.003
Stump, D. C., Lijnen, H. R., & Collen, D. (1986). Purification and characterization of single -chain urokinase -
type plasminogen activator from human cell cultures. The Journal of Biological Chemistry , 261 (3), 1274 –
DOI: -9258(17) 36086 -6
Tan, Q., Chen, Q., Niu, Y., Feng, Z., Li, L., Tao, Y., … Chen, Z. (2017). Urokinase, a promising candidate for
fibrinol ytic therapy for intracerebral hemorrhage. Journal of Neurosurgery , 126 (2), 548 –557.
DOI: 10.3171/2016.1.JNS152287
Tanaka, K. A., Key, N. S ., & Levy, J. H. (2009). Blood coagulation: hemostasis and thrombin r egulation.
Anesthesia & Analgesia , 108 (5) , 1433 -1446 . DOI:
Tripon, S., Mayer, P., Svab, A., Habersetzer, F., Pessaux, P., Baumert, T. F., & Ene scu, I. (2020 ).
Intraabdominal urokinase in the treatment of loculated infected ascites in cirrhosis. Clinics and Research
in Hepatology and Gastroenterology , 45(2), 101486. DOI : 10.1016/j.clinre.2020.06.012
Urano, T ., Castellino, F. J., & Suzuki, Y. (2018). Regulation of plasminogen activation on cell surfaces and
fibrin. JTH - Journal of Thrombosis and Haemostasis , 16 (8), 1487 –1497.
DOI: 10.1111/jth.14157
Vimal, A., & Kumar, A. (2019). Transforming th e healthcare system through therapeutic enzymes . In M.
Kuddus (Ed.), Enzymes in Food Biotechnology. Production, Applications, and Future Prospects (p. 603 -625).
Cam bridge, MA: Academic Press.
Wu, R., Chen, G., Pan, S., Zeng, J., & Liang, Z. (2019). Cost -effective fibrinolytic enzyme production by
Bacillus subtilis WR350 using medium supplemented with corn steep powder and sucrose. Scientific
Reports , 9(1), 6824. DOI : 10.1038/s415 98 -019 -43371 -8
Xu, N., Wang, L., Dou, N., Zhang, L., Guan, J., Chang, Y., & Li, R. (2019). Foam fractionation for enhancing
silica gel adsorption of urokinase from human urine. Asia -Pacific Journal of Chemical Engineering , 14 (4),
e-2334 . DOI : g/ 10.1002/apj.2334
Yazdi, A. H., Khalilipur, E., Zahedmehr, A., Pouya, S. A., Pakrou, M., Ghaznavi, M. A., … Rouzitalab, M.
(2017). Fibrinolytic therapy with streptokinase vs tenecteplase for patients with St -elevation MI not
amenable to primary PCI . Irani an Heart Journal , 18 (2) , 43 –49. Page 12 of 12 Nawaz et al.
Acta Scientiarum. Technology , v. 4 4, e 56088 , 20 22
Zhong, Y., Gong, W. J., Gao, X. H., Li, Y. N., Liu, K., Hu, Y. G., & Qi, J. S. (2020). Synthesis and evaluation of
a novel nanoparticle carrying urokinase used in targeted thrombolysis. Journal of Biomedical Materials
Resea rch. Part A , 108 (2), 193 –200. DOI : 10.1002/jbm.a.36803
Zhu, M. M., Mollet, M., Hubert, R. S., Kyung, Y. S., & Zhang, G. G. (2017). Industrial production of
therapeutic proteins: cell lines, cell culture, and purification. In J. A. Kent, T. V Bommaraju & S. D.
Barnicki (Eds.), Handbook of Industrial Chemistry and Biotechnology (p . 1639 –1669). Cham , SW: Springer. |
15882 | https://people.math.harvard.edu/archive/21b_spring_09/cookbook/cookbook.pdf | ODE COOKBOOK Math 21b, 2009 O. Knill x′ −λx = 0 x(t) = Ceλt This first order ODE is by far the most important differential equation. It is the mother of all differential equations. A linear system of differential equation x′(t) = Ax(t) reduces to this after diagonalization. We can rewrite the differential equation as (D −λ)x = 0. That is x is in the kernel of D −λ. An other interpretation is that x is an eigenfunction of D belonging to the eigenvalue λ. This differential equation describes exponential growth or exponential decay.
x′′ + k2x = 0 x(t) = C1 cos(kt) + C2 sin(kt)/k This second order ODE is by far the second most important differential equation. It is the father of all differential equations. Any linear system of differential equations x′′(t) = Ax(t) reduces to this after diagonalization. We can rewrite the differential equation as (D2+k2)x = 0.
That is x is in the kernel of D2 + k2. An other interpretation is that x is an eigenfunction of D2 belonging to the eigenvalue −k2. This differential equation describes oscillations or waves.
OPERATOR METHOD. A general method to find solutions to p(D)x = g is to factor the polynomial p(D) = (D −λ1) · · · (D −λn)x = g, then invert each factor to get x = (D −λn)−1.....(D −λ1)−1g where (D −λ)−1g = Ceλt + eλt R t 0 e−λsg(s) ds COOKBOOK METHOD. The operator method always works. But it can produce a consider-able amount of work. Engineers therefore rely also on cookbook recipes. The solution of an inhomogeneous differential equation p(D)x = g is found by first finding the homogeneous solution xh which is the solution to p(D)x = 0. Then a particular solution xp of the system p(D)x = g found by an educated guess. This method is often much faster but it requires to know the ”recipes”. Fortunately, it is quite easy: as a rule of thumb: feed in the same class of functions which you see on the right hand side and if the right hand side should contain a function in the kernel of p(D), try with a function multiplied by t. The general solution of the system p(D)x = g is x = xh + xp.
FINDING THE HOMOGENEOUS SOLUTION. p(D) = (D −λ1)(D −λ2) = D2 +bD +c. The next table covers all cases for homogeneous second order differential equations x′′ +px′ +q = 0.
λ1 ̸= λ2 real C1eλ1t + C2eλ2t λ1 = λ2 real C1eλ1t + C2teλ1t ik = λ1 = −λ2 imaginary C1 cos(kt) + C2 sin(kt) λ1 = a + ik, λ2 = a −ik C1eat cos(kt) + C2eat sin(kt) FINDING AN INHOMOGENEOUS SOLUTION. This can be found by applying the operator inversions with C = 0 or by an educated guess. For x′′ = g(t) we just integrate twice, otherwise, check with the following table: table: g(t) = a constant x(t) = A constant g(t) = at + b x(t) = At + B g(t) = at2 + bt + c x(t) = At2 + Bt + C g(t) = a cos(bt) x(t) = A cos(bt) + B sin(bt) g(t) = a sin(bt) x(t) = A cos(bt) + B sin(bt) g(t) = a cos(bt) with p(D)g = 0 x(t) = At cos(bt) + Bt sin(bt) g(t) = a sin(bt) with p(D)g = 0 x(t) = At cos(bt) + Bt sin(bt) g(t) = aebt x(t) = Aebt g(t) = aebt with p(D)g = 0 x(t) = Atebt g(t) = q(t) polynomial x(t) = polynomial of same degree EXAMPLE 1: f ′′ = cos(5x) This is of the form D2f = g and can be solved by inverting D which is integration: integrate a first time to get Df = C1 + sin(5x)/5.
Integrate a second time to get f = C2 + C1t −cos(5t)/25 This is the operator mathod in the case λ = 0.
EXAMPLE 2: f ′ −2f = 2t2 −1 This homogeneous differential equation f ′ −5f = 0 is hardwired to our brain.
We know its solution is Ce2t.
To get a homogeneous solution, try f(t) = At2 + Bt + C.
We have to compare coefficients of f ′ −2f = −2At2 + (2A −2B)t + B −2C = 2t2 −1.
We see that A = −1, B = −1, C = 0.
The special solution is −t2 −t.
The complete solution is f = −t2 −t + Ce2t EXAMPLE 3: f ′ −2f = e2t In this case, the right hand side is in the kernel of the operator T = D −2 in equation T(f) = g. The homogeneous solution is the same as in example 2, to find the inhomogeneous solution, try f(t) = Ate2t. We get f ′ −2f = Ae2t so that A = 1. The complete solution is f = te2t + Ce2t EXAMPLE 4: f ′′ −4f = et To find the solution of the homogeneous equation (D2 −4)f = 0, we factor (D−2)(D+2)f = 0 and add solutions of (D −2)f = 0 and (D + 2)f = 0 which gives C1e2t + C2e−2t. To get a special solution, we try Aet and get from f ′′ −4f = et that A = −1/3. The complete solution is f = −et/3 + C1e2t + C2e−2t EXAMPLE 5: f ′′ −4f = e2t The homogeneous solution C1e2t + C2e−2t is the same as before. To get a special solution, we can not use Ae2t because it is in the kernel of D2 −4. We try Ate2t, compare coefficients and get f = te2t/4 + C1e2t + C2e−2t EXAMPLE 6: f ′′ + 4f = et The homogeneous equation is a harmonic oscillator with solution C1 cos(2t) + C2 sin(2t).
To get a special solution, we try Aet compare coefficients and get f = et/5 + C1 cos(2t) + C2 sin(2t) EXAMPLE 7: f ′′ + 4f = sin(t) The homogeneous solution C1 cos(2t) + C2 sin(2t) is the same as in the last example. To get a special solution, we try A sin(t) + B cos(t) compare coefficients (because we have only even derivatives, we can even try A sin(t)) and get f = sin(t)/3 + C1 cos(2t) + C2 sin(2t) EXAMPLE 8: f ′′ + 4f = sin(2t) The solution C1 cos(2t)+C2 sin(2t) is the same as in the last example. To get a special solution, we can not try A sin(t) because it is in the kernel of the operator. We try At sin(t) + Bt cos(2t) instead and compare coefficients f = sin(2t)/16 −t cos(2t)/4 + C1 cos(2t) + C2 sin(2t) EXAMPLE 9: f ′′ + 8f ′ + 16f = sin(5t) The homogeneous solution is C1e−4t + C2te−4t.
To get a spe-cial solution, we try A sin(5t) + B cos(5t) compare coefficients and get f = −40 cos(5t)/412 + −9 sin(t)/412 + C1e−4t + C2te−4t EXAMPLE 10: f ′′ + 8f ′ + 16f = e−4t The homogeneous solution is still C1e−4t + C2te−4t.
To get a special solution, we can not try e−4t nor te−4t because both are in the kernel.
Add an other t and try with At2e−4t.
f = t2e−4t/2 + C1e−4t + C2te−4t EXAMPLE 11: f ′′ + f ′ + f = e−4t By factoring D2+D+1 = (D−(1+ √ 3i)/2)(D−(1− √ 3i)/2) we get the homogeneous solution C1e−t/2 cos( √ 3t) + C2e−t/2 sin( √ 3t). For a special solution, try Ae−4t. Comparing coefficients gives A = 1/13. f = e−4t/13 + C1e−t/2 cos( √ 3t) + C2e−t/2 sin( √ 3t) |
15883 | https://en.wikipedia.org/wiki/Hyperboloid_structure | Jump to content
Hyperboloid structure
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From Wikipedia, the free encyclopedia
Type of unbounded quadratic surface-shaped building or work
Hyperboloid structures are architectural structures designed using a hyperboloid in one sheet. Often these are tall structures, such as towers, where the hyperboloid geometry's structural strength is used to support an object high above the ground. Hyperboloid geometry is often used for decorative effect as well as structural economy. The first hyperboloid structures were built by Russian engineer Vladimir Shukhov (1853–1939), including the Shukhov Tower in Polibino, Dankovsky District, Lipetsk Oblast, Russia.
Properties
[edit]
Hyperbolic structures have a negative Gaussian curvature, meaning they curve inward rather than curving outward or being straight. As doubly ruled surfaces, they can be made with a lattice of straight beams, hence are easier to build than curved surfaces that do not have a ruling and must instead be built with curved beams.
Hyperboloid structures are superior in stability against outside forces compared with "straight" buildings, but have shapes often creating large amounts of unusable volume (low space efficiency). Hence they are more commonly used in purpose-driven structures, such as water towers (to support a large mass), cooling towers, and aesthetic features.
A hyperbolic structure is beneficial for cooling towers. At the bottom, the widening of the tower provides a large area for installation of fill to promote thin film evaporative cooling of the circulated water. As the water first evaporates and rises, the narrowing effect helps accelerate the laminar flow, and then as it widens out, contact between the heated air and atmospheric air supports turbulent mixing.[citation needed]
Work of Shukhov
[edit]
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In the 1880s, Shukhov began to work on the problem of the design of roof systems to use a minimum of materials, time and labor. His calculations were most likely derived from mathematician Pafnuty Chebyshev's work on the theory of best approximations of functions. Shukhov's mathematical explorations of efficient roof structures led to his invention of a new system that was innovative both structurally and spatially. By applying his analytical skills to the doubly curved surfaces Nikolai Lobachevsky named "hyperbolic", Shukhov derived a family of equations that led to new structural and constructional systems, known as hyperboloids of revolution and hyperbolic paraboloids.
The steel gridshells of the exhibition pavilions of the 1896 All-Russian Industrial and Handicrafts Exposition in Nizhny Novgorod were the first publicly prominent examples of Shukhov's new system. Two pavilions of this type were built for the Nizhni Novgorod exposition, one oval in plan and one circular. The roofs of these pavilions were doubly curved gridshells formed entirely of a lattice of straight angle-iron and flat iron bars. Shukhov himself called them azhurnaia bashnia ("lace tower", i.e., lattice tower). The patent of this system, for which Shukhov applied in 1895, was awarded in 1899.
Shukhov also turned his attention to the development of an efficient and easily constructed structural system (gridshell) for a tower carrying a large load at the top – the problem of the water tower. His solution was inspired by observing the action of a woven basket supporting a heavy weight. Again, it took the form of a doubly curved surface constructed of a light network of straight iron bars and angle iron. Over the next 20 years, he designed and built nearly 200 of these towers, no two exactly alike, most with heights in the range of 12m to 68m.
At least as early as 1911, Shukhov began experimenting with the concept of forming a tower out of stacked sections of hyperboloids. Stacking the sections permitted the form of the tower to taper more at the top, with a less pronounced "waist" between the shape-defining rings at bottom and top. Increasing the number of sections would increase the tapering of the overall form, to the point that it began to resemble a cone.
By 1918 Shukhov had developed this concept into the design of a nine-section stacked hyperboloid radio broadcasting tower in Moscow. Shukhov designed a 350m tower, which would have surpassed the Eiffel Tower in height by 50m, while using less than a quarter of the amount of material. His design, as well as the full set of supporting calculations analyzing the hyperbolic geometry and sizing the network of members, was completed by February 1919. However, the 2200 tons of steel required to build the tower to 350m were not available. In July 1919, Lenin decreed that the tower should be built to a height of 150m, and the necessary steel was to be made available from the army's supplies. Construction of the smaller tower with six stacked hyperboloids began within a few months, and Shukhov Tower was completed by March 1922.
Other architects
[edit]
Antoni Gaudi and Shukhov carried out experiments with hyperboloid structures nearly simultaneously, but independently, in 1880–1895. Antoni Gaudi used structures in the form of hyperbolic paraboloid (hypar) and hyperboloid of revolution in the Sagrada Família in 1910. In the Sagrada Família, there are a few places on the nativity facade – a design not equated with Gaudi's ruled-surface design, where the hyperboloid crops up. All around the scene with the pelican, there are numerous examples (including the basket held by one of the figures). There is a hyperboloid adding structural stability to the cypress tree (by connecting it to the bridge). The "bishop's mitre" spires are capped with hyperboloids.[citation needed]
In the Palau Güell, there is one set of interior columns along the main facade with hyperbolic capitals. The crown of the famous parabolic vault is a hyperboloid. The vault of one of the stables at the Church of Colònia Güell is a hyperboloid. There is a unique column in the Park Güell that is a hyperboloid. The famous Spanish engineer and architect Eduardo Torroja designed a thin-shell water tower in Fedala and the roof of Hipódromo de la Zarzuela in the form of hyperboloid of revolution. Le Corbusier and Félix Candela used hyperboloid structures (hypar).[citation needed]
A hyperboloid cooling tower by Frederik van Iterson and Gerard Kuypers was patented in the Netherlands on August 16, 1916. The first Van Iterson cooling tower was built and put to use at the Dutch State Mine (DSM) Emma in 1918. A whole series of the same and later designs would follow.
The Georgia Dome (1992) was the first Hypar-Tensegrity dome to be built.
Gallery
[edit]
See also: List of hyperboloid structures
The hyperbolic paraboloid is a doubly ruled surface so it may be used to construct a saddle roof from straight beams.
The Warszawa Ochota railway station has a hyperbolic paraboloid saddle roof. Warsaw, Poland, 1962.
The Scotiabank Saddledome arena has a hyperbolic paraboloid saddle roof, Calgary, Canada, 1983.
Stackable Pringles chips are hyperbolic paraboloids.
3D-printed dual-use pen/toothbrush holder-cup. Printed on Ultimaker 2, 2015
The first 1918 Van Iterson cooling tower, 1984
Five cooling towers in a row at DSM Emma, around 1930
The Corporation Street Bridge is a horizontal doubly ruled hyperboloid structure, Manchester, England, 1999.
See also
[edit]
Architecture portal
Geodesic dome
Lattice mast
List of thin-shell structures
Sam Scorer
Tensile structure
World's first hyperboloid structure
Notes
[edit]
^ "Hyperboloid water tower". International Database and Gallery of Structures. Nicolas Janberg, ICS. 2007. Retrieved 2007-11-28.
^ Cowan, Henry J. (1991), Handbook of architectural technology, Van Nostrand Reinhold, p. 175, ISBN 9780442205256, It is easier to build timber formwork for a concrete structure or to fabricate a steel structure if a surface is singly ruled, and even more so if it is doubly ruled.
^ Reid, Esmond (1988). Understanding Buildings: A Multidisciplinary Approach. The MIT Press. p. 35. ISBN 978-0-262-68054-7. Retrieved 2009-08-09.
^ Burry, M.C., J.R. Burry, G.M. Dunlop and A. Maher (2001). "Drawing Together Euclidean and Topological Threads (pdf)" (PDF). Presented at SIRC 2001 – the Thirteenth Annual Colloquium of the Spatial Information Research Center. Dunedin, New Zealand: University of Otago. Archived from the original (PDF) on 2007-10-31. Retrieved 2007-11-28.{{cite web}}: CS1 maint: multiple names: authors list (link)
^ "Fedala Reservoir". International Database and Gallery of Structures. Nicolas Janberg, ICS. 2007. Retrieved 2007-11-28.
^ "Zarzuela Hippodrome". International Database and Gallery of Structures. Nicolas Janberg, ICS. 2007. Retrieved 2007-11-28.
^ NL/GB Patent No. 108,863: "GB108863A Improved Construction of Cooling Towers of Reinforced Concrete". Espacenet, Patent search. Retrieved 2023-12-03.
^ "Koeltorens van de Staatsmijn Emma". Glück Auf (in Dutch). Retrieved 2023-12-03.
^ Castro, Gerardo and Matthys P. Levy (1992). "Analysis of the Georgia Dome Cable Roof". Proceedings of the Eighth Conference of Computing in Civil Engineering and Geographic Information Systems Symposium. Housing The Spectacle. Retrieved 2007-11-28.
References
[edit]
"The Nijni-Novgorod exhibition: Water tower, room under construction, springing of 91 feet span", "The Engineer", № 19.3.1897, pp. 292–294, London, 1897.
William Craft Brumfield, "The Origins of Modernism in Russian Architecture", University of California Press, 1991, ISBN 0-520-06929-3.
Elizabeth Cooper English: “Arkhitektura i mnimosti”: The origins of Soviet avant-garde rationalist architecture in the Russian mystical-philosophical and mathematical intellectual tradition”, a dissertation in architecture, 264p., University of Pennsylvania, 2000.
"Vladimir G. Suchov 1853–1939. Die Kunst der sparsamen Konstruktion.", Rainer Graefe, Jos Tomlow und andere, 192 pp., Deutsche Verlags-Anstalt, Stuttgart, 1990, ISBN 3-421-02984-9.
External links
[edit]
Wikiquote has quotations related to Hyperboloid structure.
Wikimedia Commons has media related to Hyperboloid structure.
The research of the Shukhov's World's First Hyperboloid structure, Prof. Dr. Armin Grün
International campaign to save the Shukhov Tower
Anticlastic hyperboloid shells
Shells: Hyperbolic paraboloids (hypar)
Hyperbolic Paraboloids & Concrete Shells
Special Structures
Rainer Graefe: “Vladimir G. Šuchov 1853–1939 – Die Kunst der sparsamen Konstruktion.”,
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15884 | https://www.khanacademy.org/math/ka-math-class-8/x8d1919797820695b:comparing-quantities-ncert-new/x8d1919797820695b:recalling-ratios-and-percentage/v/intro-to-ratios-indian-accent | Intro to ratios (video) | Khan Academy
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KA Math Class 8
Course: KA Math Class 8>Unit 7
Lesson 1: Recalling ratios and percentage
Percent word problem: penguins
Percent word problems
Intro to ratios
Unitary method
Finding ratios
Unit rates
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KA Math Class 8>
Comparing quantities>
Recalling ratios and percentage
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Intro to ratios
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The video explains ratios, which show the relationship between two quantities. Using apples and oranges as an example, it demonstrates how to calculate and reduce ratios (6:9 to 2:3) and how to reverse the ratio (9:6 to 3:2).Created by Kalakrit (Dubbing).
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15885 | https://www.ck12.org/flexi/cbse-math/multiply-and-divide-rational-numbers/what-is-additive-and-multiplicative-inverse-in-maths/ | Flexi answers - What is additive and multiplicative inverse in maths? | CK-12 Foundation
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Question
What is additive and multiplicative inverse in maths?
Flexi Says:
Additive Inverse:
The additive inverse of a number a is the number that, when added to a,gives the additive identity, which is 0. For any number a,its additive inverse is denoted as −a.
Mathematically, the additive inverse is defined as:
a+(−a)=0.
For example, the additive inverse of 5 is -5, because 5+(−5)=0.
Multiplicative Inverse:
The multiplicative inverse (or reciprocal) of a non-zero number a is the number that, when multiplied by a,gives the multiplicative identity, which is 1. The multiplicative inverse of a is denoted as 1 a.
Mathematically, the multiplicative inverse is defined as:
a×1 a=1.
For example, the multiplicative inverse of 3 is 1 3,because 3×1 3=1.
Both the additive and multiplicative inverses are important concepts in mathematics and are used in various operations and calculations.
Analogy / Example
Try Asking:
What are additive and multiplicative inverses?What is the multiplicative inverse of -15?Identify the rational number which is different from the other three Unexpected text node: ': \frac{2}{3}, \frac{-4}{5}, \frac{1}{2}, \frac{1}{3}' . Explain your reasoning.
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15887 | https://ecampusontario.pressbooks.pub/introstats/chapter/5-5-calculating-probabilities-for-a-normal-distribution/ | Skip to content
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5.5 Calculating Probabilities for a Normal Distribution
LEARNING OBJECTIVES
Recognize the normal probability distribution and apply it appropriately.
Calculate probabilities associated with a normal distribution.
Probabilities for a normal random variable equal the area under the corresponding normal distribution curve. The probability that the value for falls in between the values and is the area under the normal distribution curve to the right of and to the left of .
CALCULATING NORMAL PROBABILITIES IN EXCEL
To calculate probabilities associated with normal random variables in Excel, use the norm.dist(x,,,logic operator) function.
For x, enter the value for x.
For , enter the mean of the normal distribution.
For , enter the standard deviation of the normal distribution.
For the logic operator, enter true. Note: Because we are calculating the area under the curve, we always enter true for the logic operator.
The output from the norm.dist function is the probability that . That is, the output from the norm.dist function is the area to the left of value of x.
Visit the Microsoft page for more information about the norm.dist function.
NOTE
The norm.dist function always tells us the area to the left of the value entered for x.
To find the area to the right of the value of x, we use 1-norm.dist(x,,,true). This corresponds to the probability that .
To find the area in between x1 and x2 with , we use norm.dist(x2,,,true)-norm.dist(x1,,,true).This corresponds to the probability that .
An alternative approach in Excel is to use the norm.s.dist(z,true) function. In the norm.s.dist function, we enter the z-score for the corresponding value of x and the output will be the area to the left x.
CALCULATING -VALUES FOR A NORMAL DISTRIBUTION IN EXCEL
Given the area to the left of an (unknown) x-value, use the norm.inv(probability,,) function.
For probability, enter the area to the left of x.
For , enter the mean of the normal distribution.
For , enter the standard deviation of the normal distribution.
The output from the norm.inv function is the value of x so that the area to left of x equals the given probability. That is, the output from the norm.inv function is the value of x so that the .
Visit the Microsoft page for more information about the norm.inv function.
NOTE
The norm.inv function requires that we enter the area to the left of the unknown x-value. If we are given the area to the right of the unknown x-value, we enter 1-area to the right for the probability in the norm.inv function. That is, given the area to the right of the x-value, we use norm.inv(1-area,,).
EXAMPLE
The final exam scores in a statistics class are normally distributed with a mean of 63 and a standard deviation of 5.
Find the probability that a randomly selected student scored more than 65 on the exam.
Find the probability that a randomly selected student scored less than 75.
90% of the students scored less than what value?
30% of the students scored more than what value?
Solution:
Let be the scores on the final exam.
We want to find :
| | | |
---
| Function | 1-norm.dist | Answer |
| Field 1 | 65 | 0.3446 |
| Field 2 | 63 | |
| Field 3 | 5 | |
| Field 4 | true | |
The probability that a student scores more than 65 is 0.3446 (or 34.46%)
2. We want to find :
| | | |
---
| Function | norm.dist | Answer |
| Field 1 | 75 | 0.9918 |
| Field 2 | 63 | |
| Field 3 | 5 | |
| Field 4 | true | |
The probability that a student scores less than 75 is 0.9918 (or 99.18%).
3. We want to find the value of so that the area to the left of is 0.9.
| | | |
---
| Function | norm.inv | Answer |
| Field 1 | 0.9 | 69.41 |
| Field 2 | 63 | |
| Field 3 | 5 | |
90% of the students scored below 69.41 points on the exam.
4. We want to find the value of so that the area to the right of is 0.3. This is the same as finding the value of so that the area to left of is 0.7 (1-0.3).
| | | |
---
| Function | norm.inv | Answer |
| Field 1 | 0.7 | 65.62 |
| Field 2 | 63 | |
| Field 3 | 5 | |
30% of the students scored more 65.62 points on the exam.
TRY IT
The golf scores for a school team are normally distributed with a mean of 68 and a standard deviation of 3.
Find the probability that a randomly selected golfer scored less than 65.
Find the probability that a randomly selected golfer scored more than 72.
Click to see Solution
| | | |
---
| Function | norm.dist | Answer |
| Field 1 | 65 | 0.1587 |
| Field 2 | 68 | |
| Field 3 | 3 | |
| Field 4 | true | |
| | | |
---
| Function | 1-norm.dist | Answer |
| Field 1 | 72 | 0.0912 |
| Field 2 | 68 | |
| Field 3 | 3 | |
| Field 4 | true | |
EXAMPLE
A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is 2 hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is 0.5 hour.
Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.
Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.
Solution:
Let be the amount of time (in hours) a household personal computer is used for entertainment.
We want to find .
| | | | |
--- --- |
| Function | norm.dist | -norm.dist | Answer |
| Field 1 | 2.75 | 1.8 | 0.5886 |
| Field 2 | 2 | 2 | |
| Field 3 | 0.5 | 0.5 | |
| Field 4 | true | true | |
The probability a household computer is used for entertainment between 1.8 and 2.75 hours a day is 0.5886 (or 58.86%).
2. We need to find the value x so that 25% of the number of hours as less than this value.
| | | |
---
| Function | norm.inv | Answer |
| Field 1 | 0.25 | 1.66 |
| Field 2 | 2 | |
| Field 3 | 0.5 | |
25% of the value are less than 1.66 hours.
TRY IT
The golf scores for a school team are normally distributed with a mean of 68 and a standard deviation of 3. Find the probability that a golfer scored between 66 and 70.
Click to see Solution
| | | | |
--- --- |
| Function | norm.dist | -norm.dist | Answer |
| Field 1 | 70 | 66 | 0.4950 |
| Field 2 | 68 | 68 | |
| Field 3 | 3 | 3 | |
| Field 4 | true | true | |
EXAMPLE
There are approximately one billion smartphone users in the world today. In the United States the ages of smartphone users from 13 to 55+ follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.
Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.
Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.
80% of the users in the age range 13 to 55+ are less than what age?
40% of the ages that range from 13 to 55+ are at least what age?
Solution:
| | | | |
--- --- |
| Function | norm.dist | -norm.dist | Answer |
| Field 1 | 64.7 | 23 | 0.8186 |
| Field 2 | 36.9 | 36.9 | |
| Field 3 | 13.9 | 13.9 | |
| Field 4 | true | true | |
The probability a smartphone user is between 23 and 64.7 years of age is 0.8186 (or 81.86%).
2. | | | |
---
| Function | norm.dist | Answer |
| Field 1 | 50.8 | 0.8413 |
| Field 2 | 36.9 | |
| Field 3 | 13.9 | |
| Field 4 | true | |
The probability that a smartphone user is less than 50.8 years of age is 0.8413 (or 84.13%).
3. | | | |
---
| Function | norm.inv | Answer |
| Field 1 | 0.8 | 48.6 |
| Field 2 | 36.9 | |
| Field 3 | 13.9 | |
80% of the smartphone users in the age range 13 – 55+ are 48.6 years old or less.
4. | | | |
---
| Function | norm.inv | Answer |
| Field 1 | 0.6 | 40.42 |
| Field 2 | 36.9 | |
| Field 3 | 13.9 | |
40% of the smartphone users in the age range 13 – 55+ are older than 40.42 years of age.
TRY IT
There are approximately one billion smartphone users in the world today. In the United States the ages of smartphone users from 13 to 55+ follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.
30% of smartphone users are older than what age?
What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old.
Click to see Solution
| | | |
---
| Function | norm.inv | Answer |
| Field 1 | 0.7 | 44.19 |
| Field 2 | 36.9 | |
| Field 3 | 13.9 | |
| | | |
---
| Function | norm.dist | Answer |
| Field 1 | 27 | 0.2382 |
| Field 2 | 36.9 | |
| Field 3 | 13.9 | |
| Field 4 | true | |
EXAMPLE
A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.
Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm.
90% of the diameters of the mandarin oranges are less than what value?
35% of the diameters of the mandarin oranges are greater than what value?
Solution:
| | | |
---
| Function | 1-norm.dist | Answer |
| Field 1 | 6 | 0.2660 |
| Field 2 | 5.85 | |
| Field 3 | 0.24 | |
| Field 4 | true | |
The probability an orange has a diameter greater than 6 cm is 0.2660 (or 26.60%).
2. | | | |
---
| Function | norm.inv | Answer |
| Field 1 | 0.9 | 6.16 |
| Field 2 | 5.85 | |
| Field 3 | 0.24 | |
90% of the diameters of the oranges are less than 6.16 cm.
3. | | | |
---
| Function | norm.inv | Answer |
| Field 1 | 0.65 | 5.94 |
| Field 2 | 5.85 | |
| Field 3 | 0.24 | |
35% of the diameters of the oranges are greater than 5.94 cm.
Watch this video: Excel 2013 Statistical Analysis #39: Probabilities for Normal (Bell) Probability Distribution by ExcelIsFun [24:07]
Concept Review
The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. The probability that the value for a normal random variable falls in between the values and is the area under the normal distribution curve to the right of and to the left of .
Attribution
“6.2 Using the Normal Distribution“ in Introductory Statistics by OpenStax is licensed under a Creative Commons Attribution 4.0 International License. |
15888 | https://math.stackexchange.com/questions/829621/intersection-of-ellipse-and-hyperbola-at-a-right-angle | calculus - Intersection of ellipse and hyperbola at a right angle - Mathematics Stack Exchange
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Intersection of ellipse and hyperbola at a right angle [closed]
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Need to show that two functions intersect at a right angle.
Show that the ellipse
x 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1
and the hyperbola
x 2 α 2−y 2 β 2=1 x 2 α 2−y 2 β 2=1
will intersect at a right angle if
α 2≤a 2 and a 2−b 2=α 2+β 2 α 2≤a 2 and a 2−b 2=α 2+β 2
Not sure how to tackle this question, graphing didn't help.
calculus
multivariable-calculus
conic-sections
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edited Jan 17, 2017 at 12:09
Martin Sleziak
56.3k 20 20 gold badges 211 211 silver badges 391 391 bronze badges
asked Jun 10, 2014 at 18:54
user152739user152739
365 1 1 gold badge 3 3 silver badges 6 6 bronze badges
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1 I've tidied up your math expressions. Please double check that I didn't do any mistakes. Also, click the edit link at the bottom of your question if you want to know how it's done.Arthur –Arthur 2014-06-10 19:27:19 +00:00 Commented Jun 10, 2014 at 19:27
Thank you so much! I'll try to use it next time :)user152739 –user152739 2014-06-10 19:32:28 +00:00 Commented Jun 10, 2014 at 19:32
max(a,b)2−min(a,b)2=α 2+β 2 max(a,b)2−min(a,b)2=α 2+β 2 Takahiro Waki –Takahiro Waki 2017-01-21 14:09:19 +00:00 Commented Jan 21, 2017 at 14:09
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Note that if a 2=b 2+α 2+β 2 a 2=b 2+α 2+β 2 holds, then a 2≥α 2 a 2≥α 2 holds since
a 2=(b 2+β 2)+α 2≥0+α 2=α 2.a 2=(b 2+β 2)+α 2≥0+α 2=α 2.
There are four intersection points (x,y)(x,y) where
x 2=a 2 α 2(b 2+β 2)α 2 b 2+a 2 β 2,y 2=β 2 b 2(a 2−α 2)α 2 b 2+a 2 β 2=β 2 b 2(b 2+β 2)α 2 b 2+a 2 β 2(1)(1)x 2=a 2 α 2(b 2+β 2)α 2 b 2+a 2 β 2,y 2=β 2 b 2(a 2−α 2)α 2 b 2+a 2 β 2=β 2 b 2(b 2+β 2)α 2 b 2+a 2 β 2
Now
x 2 a 2+y 2 b 2=1⟹y y′e b 2=−x a 2 x 2 a 2+y 2 b 2=1⟹y y e′b 2=−x a 2
x 2 α 2−y 2 β 2=1⟹y y′h β 2=x α 2 x 2 α 2−y 2 β 2=1⟹y y h′β 2=x α 2
giving
y y′e b 2⋅y y′h β 2=−x 2 a 2 α 2(2)(2)y y e′b 2⋅y y h′β 2=−x 2 a 2 α 2
From (1)(2)(1)(2), we have
y′e y′h=−x 2 a 2 α 2⋅b 2 β 2 y 2=−b 2 β 2 a 2 α 2⋅a 2 α 2 b 2 β 2=−1 y e′y h′=−x 2 a 2 α 2⋅b 2 β 2 y 2=−b 2 β 2 a 2 α 2⋅a 2 α 2 b 2 β 2=−1
Therefore, the two curves intersect at right angle.
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answered Jan 17, 2017 at 10:08
mathlovemathlove
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So, this should imply the stronger proposition: the two curves intersect at a right angle⟺if and only if a 2−b 2=α 2+β 2 the two curves intersect at a right angle⟺if and only if a 2−b 2=α 2+β 2 Is It?Hazem Orabi –Hazem Orabi 2017-01-18 09:42:34 +00:00 Commented Jan 18, 2017 at 9:42
@HazemOrabi: Yes, you are right. (I proved only ⇐⇐ since that is to be proved and some answers here prove ⇒⇒.)mathlove –mathlove 2017-01-18 09:50:56 +00:00 Commented Jan 18, 2017 at 9:50
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The two curves are an ellipse and a hyperbola, having coordinate axes as symmetry axes. It is well known (and easy to prove) that if A A is a point on an ellipse and F F, G G are its foci, then the bisector of angle F A G F A G is the line normal to the ellipse at A A; and if A A is a point on a hyperbola and F′F′, G′G′ are its foci, then the bisector of angle F′A G′F′A G′ is the line tangent to the hyperbola at A A.
For the two curves to intersect at right angles, it is necessary these two lines to be the same, and that happens only if they have the same foci. The squared distance of a focus from the center is given by a 2−b 2 a 2−b 2 for the ellipse and by α 2+β 2 α 2+β 2 for the hyperbola, meaning that the curves have the same foci if a 2−b 2=α 2+β 2 a 2−b 2=α 2+β 2.
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edited Jan 17, 2017 at 10:38
answered Jan 17, 2017 at 10:25
Intelligenti paucaIntelligenti pauca
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Apparently, if α 2>a 2 α 2>a 2, the two curves never intersect in the first place. Now let's just find the derivatives at the point of intersection and show that their product is −1−1.
Ellipse:
y=b 1−(x a)2−−−−−−−−√y′=b⋅−2 x/a 2 2 1−(x a)2−−−−−−−√=−b 2 x a 2 y(1)(1)y=b 1−(x a)2 y′=b⋅−2 x/a 2 2 1−(x a)2=−b 2 x a 2 y
Hyperbola:
y=β(x α)2−1−−−−−−−−√y′=β⋅2 x/α 2 2(x α)2−1−−−−−−−√=β 2 x α 2 y(2)(2)y=β(x α)2−1 y′=β⋅2 x/α 2 2(x α)2−1=β 2 x α 2 y
Point of intersection:
x 2 α 2−y 2 β 2=1 x 2 a 2+y 2 b 2=1 x 2 α 2−y 2 β 2=1 x 2 a 2+y 2 b 2=1
Let's multiply the first equation by β 2 β 2, the second by b 2 b 2, and add them together.
β 2 x 2 α 2+b 2 x 2 a 2=β 2+b 2 x 2=β 2+b 2 β 2 α 2+b 2 a 2=a 2 α 2 β 2+b 2 a 2 β 2+α 2 b 2 y 2=b 2(1−x 2 a 2)=b 2⋅a 2 β 2+α 2 b 2−α 2 β 2−α 2 b 2 a 2 β 2+α 2 b 2=b 2 β 2⋅a 2−α 2 a 2 β 2+α 2 b 2(3)(3)β 2 x 2 α 2+b 2 x 2 a 2=β 2+b 2 x 2=β 2+b 2 β 2 α 2+b 2 a 2=a 2 α 2 β 2+b 2 a 2 β 2+α 2 b 2 y 2=b 2(1−x 2 a 2)=b 2⋅a 2 β 2+α 2 b 2−α 2 β 2−α 2 b 2 a 2 β 2+α 2 b 2=b 2 β 2⋅a 2−α 2 a 2 β 2+α 2 b 2
Now back to the derivatives:
−b 2 x a 2 y⋅β 2 x α 2 y=−1 b 2 β 2 x 2 a 2 α 2 y 2=1−b 2 x a 2 y⋅β 2 x α 2 y=−1 b 2 β 2 x 2 a 2 α 2 y 2=1
Plug in x 2 x 2 and y 2 y 2 from (3). Then nearly everything magically cancels out, and we're left with...
β 2+b 2 a 2−α 2=1 β 2+b 2 a 2−α 2=1
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answered Jan 17, 2017 at 9:45
Ivan NeretinIvan Neretin
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Well, yes; what's wrong with that?Ivan Neretin –Ivan Neretin 2017-01-17 10:32:07 +00:00 Commented Jan 17, 2017 at 10:32
1 The converse comes for free. Suppose we are given the condition on the coefficients and work all the way back from there. The intersection points are still the same, and the derivatives are what they are, and when we multiply the derivatives, we get -1, which means...Ivan Neretin –Ivan Neretin 2017-01-17 10:38:59 +00:00 Commented Jan 17, 2017 at 10:38
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The gradient vector (∂f∂x,∂f∂y)(∂f∂x,∂f∂y) is normal to the curve f(x,y)=0 f(x,y)=0.
Then
(2 x a 2,2 y b 2)⋅(2 x α 2,−2 y β 2)=4 x 2 a 2 α 2−4 y 2 b 2 β 2=0.(2 x a 2,2 y b 2)⋅(2 x α 2,−2 y β 2)=4 x 2 a 2 α 2−4 y 2 b 2 β 2=0.
We can eliminate x 2 x 2 and y 2 y 2 from the three equations by
∣∣∣∣∣∣∣∣1 a 2 1 α 2 1 a 2 α 2 1 b 2−1 β 2−1 b 2 β 1 1 0∣∣∣∣∣∣∣∣=β 2+α 2+b 2−a 2 a 2 b 2 α 2 β 2=0.|1 a 2 1 b 2 1 1 α 2−1 β 2 1 1 a 2 α 2−1 b 2 β 0|=β 2+α 2+b 2−a 2 a 2 b 2 α 2 β 2=0.
Hence the claim.
I have not investigated the inequality α 20.x 2,y 2>0. [Confirmed by the expressions provided by @mathlove.]
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answered Jan 17, 2017 at 10:06
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There is absolutely no need to indulge y' and tedious calculations. We can simply utilise the reflection properties of ellipse and hyperbola to note that confocal ellipse and hyperbola shall always intersect orthogonally. Here, ellipse focus is sqrt(a^2-b^2) and that of hyperbola is sqrt(alpha^2+beta^2) which are equal (given). Hence the result. An important takeaway from this answer is that the best methods to solve conics questions is to involve BOTH Euclidean as well as analytical geometry. A video on YouTube by Mathologer on parabolas explains this in the best possible manner.
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answered Aug 2, 2019 at 4:48
aditya guptaaditya gupta
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Wiki link explains the situation clearly in the paragraph Confocal ellipses and hyperbolas.
Note that when ray is correctly traced,
Normal to ellipse is angle bisector between foci and enables internal reflection from one focus to the other as shown.(α=α)(α=α)
Normal to hyperbola is angle bisector between foci and enables external reflection from one focus to the other as shown.(β=β)(β=β)
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answered Aug 2, 2019 at 5:07
NarasimhamNarasimham
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15889 | https://www.merriam-webster.com/thesaurus/cultivate! | Est. 1828
Synonyms of cultivate
as in to develop
as in to promote
as in to grow
as in to harvest
as in to develop
as in to promote
as in to grow
as in to harvest
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Definition of cultivate
as in to develop
to come to have gradually cultivated a taste for opera in order to fit in with his new circle of friends
Synonyms & Similar Words
take on
take in
Antonyms & Near Antonyms
abandon
forsake
desert
shed
reject
discard
unload
ditch
dump
cast
jettison
throw away
throw out
scrap
slough
fling (off or away)
shuck (off)
junk
sluff
2
as in to promote
to help the growth or development of cultivated a passion for learning among his students over the years
Synonyms & Similar Words
promote
encourage
nurture
foster
advance
nourish
forward
further
assist
incubate
support
advocate
uphold
nurse
fund
endorse
underwrite
finance
aid
subsidize
patronize
advertise
endow
back
indorse
publicize
work (for)
boost
champion
campaign (for)
stake
abet
plug
tout
agitate (for)
Antonyms & Near Antonyms
prevent
hinder
discourage
inhibit
prohibit
frustrate
enjoin
forbid
oppose
proscribe
fight
combat
bar
outlaw
counter
ban
interdict
contend (with)
suppress
subdue
stifle
repress
check
battle
impede
halt
retard
arrest
squelch
interfere (with)
obstruct
squash
encumber
snuff (out)
hobble
shackle
fetter
manacle
3
as in to grow
to look after or assist the growth of by labor and care in an attempt to produce New World counterparts of the wines that he had enjoyed in Europe, Jefferson cultivated several varieties of grapes at Monticello
Synonyms & Similar Words
grow
produce
promote
culture
raise
plant
tend
harvest
crop
propagate
dress
breed
rear
reap
sow
gather
glean
ripen
sprout
germinate
root
quicken
Antonyms & Near Antonyms
kill
pick
dig
pull (up)
cut
pluck
uproot
hay
extirpate
mow
4
as in to harvest
to work by plowing, sowing, and raising crops on we ought to cultivate the field out back
Synonyms & Similar Words
harvest
tend
farm
plant
till
crop
reap
hoe
sharecrop
harrow
Example Sentences
Examples are automatically compiled from online sources to show current usage. Read More Opinions expressed in the examples do not represent those of Merriam-Webster or its editors. Send us feedback.
Recent Examples of cultivate Teaching children to appreciate beauty in the small details can cultivate their sense of wonder and awe. —Katie Grant, Parents, 24 Sep. 2025 Since launching in 2023, the group has cultivated a loyal following and now averages over 100 rides per month, according to Alba. —Ladan Anoushfar, CNN Money, 23 Sep. 2025 Because Celium is cultivated and finished individually, natural variations in texture and translucency make every piece unique. —Alexandra Harrell, Sourcing Journal, 23 Sep. 2025 This proactive approach positions USD among national leaders in the critical effort to cultivate positive group environments that prevent hazing and enhance student belonging and leadership development. —Elizabeth Allan, San Diego Union-Tribune, 23 Sep. 2025 See All Example Sentences for cultivate
Recent Examples of Synonyms for cultivate
develop
promote
grow
harvest
acquire
encourage
produce
tend
Verb
It was developed in 2016 by Vicente Esteban, who was a masters student at the Royal College of Art in London.
— Lisa Deaderick, San Diego Union-Tribune, 20 Sep. 2025
What was your approach to developing and bringing him to life?
— Giana Levy, Variety, 20 Sep. 2025
Definition of develop
Verb
In past years, the event has served as a platform for Beijing to promote its national security agenda.
— Evelyn Cheng, CNBC, 24 Sep. 2025
Across the Oscars, the BAFTAs, and the SAG awards, glitzy galas, and glam expeditions to Paris to promote The Substance, Moore and her stylist Brad Goreski delighted in gold and sparkles.
— Anna Cafolla, Vogue, 24 Sep. 2025
Definition of promote
Verb
In the midst of the life-or-death chaos, Arisu grows closer to Usagi (Tao Tsuchiya), an athletic loner mourning the death of her mountain climber father.
— Kayti Burt, Time, 26 Sep. 2025
For comparison, current industry leader Cheniere aims to grow to about 75 MTPA near 2030big growth, but not keeping pace with Venture Globals plans.
— Jordan Blum, Fortune, 25 Sep. 2025
Definition of grow
Verb
It's long been harvested for its carrageenan, which is used in the food industry as a thickening agent and emulsifier in products like ice cream and puddings, even nut milks.
— Maria Godoy, NPR, 22 Sep. 2025
Hunters in Kentucky are limited to harvesting one antlered deer per person statewide for the season.
— Marina Johnson, Louisville Courier Journal, 22 Sep. 2025
Definition of harvest
Verb
In an expansive canvass during the investigation, police acquired footage from more than 17 locations around the community in Moscow and Pullman to piece together the whereabouts of the vehicle the day of the murders, and tie it back to Kohberger.
— Darin Oswald, Idaho Statesman, 26 Sep. 2025
Before the suspension, Nexstar announced plans to acquire Tegna, a rival broadcast company, for more than $6 billion.
— Brenton Blanchet, PEOPLE, 26 Sep. 2025
Definition of acquire
Verb
The resort's Star Camp kids club encourages youngsters to be stewards of the land and sea, emphasizing sustainable initiatives such as recycling and ocean ecology.
— Allison Tibaldi, USA Today, 27 Sep. 2025
In honor of their service and selfness, Hobbs has ordered all state buildings to be at half-staff the days of their burials and encourages individuals, organizations and businesses to participate in the tribute.
— Paige Moore, AZCentral.com, 26 Sep. 2025
Definition of encourage
Verb
Now, Montblanc wants to enter the tech sphere with a new digital writing pad, inspired by its century of experience producing pens.
— James Peckham, PC Magazine, 22 Sep. 2025
The Chinese firm, which assembles devices and counts Apple as a client, is reportedly producing a consumer AI device for OpenAI.
— Yeo Boon Ping, CNBC, 22 Sep. 2025
Definition of produce
Verb
Both Jagloms personality and his films tended to divide people.
— Carmel Dagan, Variety, 25 Sep. 2025
While the participation rate for alcoholic consumption has picked up among Gen Zers of legal drinking age, jumping from 46% in 2023 to 70% in 2025, Lodewijks said younger Americans tend to drink less often than previous generations.
— Bailey Schulz, USA Today, 25 Sep. 2025
Definition of tend
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15890 | https://math.stackexchange.com/questions/127300/maximum-range-of-a-projectile-launched-from-an-elevation | Stack Exchange Network
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Maximum range of a projectile (launched from an elevation)
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$\begingroup$
If a projectile is launched at a speed $u$ from a height $H$ above the horizontal axis, $g$ is the acceleration due to gravity, and air resistance is ignored, its trajectory is $$y=H+x \tan θ-x^2\frac g{2u^2}\left(1+\tan ^2\theta\right),$$ and its maximum range is $$R_{\max }=\frac ug\sqrt{u^2+2gH}.$$
I would like to derive the above $R_{\max},$ and here's what I've done:
substitute $(x,y)=(R,0)$ into the trajectory equation;
differentiate the result with respect to $\theta;$
substitute $\left(R,\frac {\mathrm dR}{\mathrm d\theta}\right)=\left(R_{\max},0\right).$
However, this eliminates $H$ and fails to lead to the desired expression for $R_{\max }.$ How to actually derive the above $R_{\max }?$
P.S. This is the context; in the above, I've replaced all occurrences of $L$ below with $\frac{u^2}g$.
physics
classical-mechanics
kinematics
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edited Oct 19, 2021 at 9:45
ryangryang
asked Apr 2, 2012 at 17:15
ryangryang
1
$\endgroup$
3
$\begingroup$ Cross-posted to physics.stackexchange.com/q/23186/2451 $\endgroup$
Qmechanic
– Qmechanic
2012-04-02 20:54:59 +00:00
Commented Apr 2, 2012 at 20:54
1
$\begingroup$ possible duplicate of What is the optimum angle of projection when throwing a stone off a cliff? $\endgroup$
Ross Millikan
– Ross Millikan
2012-04-02 21:52:44 +00:00
Commented Apr 2, 2012 at 21:52
$\begingroup$ Note: this current Question is in fact premised on the previous link's Question (so, they are not duplicates), whose angle sought is in fact already obtained in Step 3 above. $\endgroup$
ryang
– ryang
2024-10-02 01:41:51 +00:00
Commented Oct 2, 2024 at 1:41
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4 Answers 4
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5
$\begingroup$
We are given the trajectory of a projectile: $$ y = H + x\tan(\theta) - \frac{g}{2u^2}x^2(1+\tan^2(\theta)), $$ where $H$ is the initial height, $g$ is the (positive) gravitational constant and $u$ is the initial speed. Since we are looking for the maximum range we set $y=0$ (i.e. the projectile is on the ground). If we let $L=u^2/g$, then $$ H + x\tan(\theta) - \frac1{2L}x^2(1+\tan^2(\theta)) = 0 $$ Differentiate both sides with respect to $\theta$. $$ \frac{dx}{d\theta}\tan(\theta) + x\;\sec^2(\theta) - \left[\frac1L x \frac{dx}{d\theta}(1+\tan^2(\theta)) + \frac{1}{2L}x^2(2\tan(\theta)\sec^2(\theta))\right] = 0 $$ Solving for $\frac{dx}{d\theta}$ yields $$ \frac{dx}{d\theta} = \frac{x \sec^2(\theta)[\frac{x}{L}\tan(\theta)-1]}{\tan(\theta)-\frac{x}{L}(1+\tan^2(\theta))} $$ This derivative is $0$ when $\tan(\theta) = \frac{L}{x}$ and hence this corresponds to a critical number $\theta$ for the range of the projectile. We should now show that the $x$ value it corresponds to is a maximum, but I'll just assume that's the case. It pretty obvious in the setting of the problem. Finally, we replace $\tan(\theta)$ with $\frac{L}{x}$ in the second equation from the top and solve for $x$. $$ H + L - \frac{1}{2L}x^2 - \frac{L}2 = 0. $$ This leads immediately to $x = \sqrt{L^2 + 2LH}$. The angle $\theta$ can now be found easily.
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edited Apr 3, 2012 at 2:32
answered Apr 2, 2012 at 17:35
PatrickPatrick
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$\endgroup$
2
$\begingroup$ Why on earth did you lose $g$? $\endgroup$
TonyK
– TonyK
2012-04-02 17:39:47 +00:00
Commented Apr 2, 2012 at 17:39
$\begingroup$ I replaced my original answer with this derivation of the maximum range formula. I think this is what you wanted to see. $\endgroup$
Patrick
– Patrick
2012-04-03 02:36:13 +00:00
Commented Apr 3, 2012 at 2:36
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2
$\begingroup$
its trajectory is $$y=H+x \tan θ-x^2\frac g{2u^2}\left(1+\tan ^2\theta\right)$$
I would like to derive the above $R_{\max},$ and here's what I've done:
substitute $(x,y)=(R,0)$ into the trajectory equation;
differentiate the result with respect to $\theta;$
substitute $\left(R,\frac {\mathrm dR}{\mathrm d\theta}\right)=\left(R_{\max},0\right).$
This is correct, and gives $$0=R_{\max}\sec^2\theta-R_{\max}^2\frac{g}{2u^2}2\tan\theta\sec^2\theta$$$$\tan\theta=\frac{u^2}{gR_{\max}}.\tag A$$
Step 1 also directly gives $$0=H+R_{\max}\tan θ-R_{\max}^2\frac g{2u^2}\left(1+\tan ^2\theta\right)$$$$\frac {gR_{\max}^2}{2u^2}\left(1+\tan ^2\theta\right)=H+R_{\max}\tan θ.\tag B$$
Substituting $(A)$ into $(B)$ then gives \begin{align}\frac{gR_{\max}^2}{2u^2}\left(1+\frac{u^4}{g^2R_{\max}^2}\right)&=H+\frac{u^2}g\ \frac{gR_{\max}^2}{2u^2}+\frac{u^2}{2g}&=H+\frac{u^2}g\ R_{\max}&=\sqrt{\frac{2u^2}g\left(H+\frac{u^2}{2g}\right)}\ &=\sqrt{\frac{u^2}{g^2}\left(2gH+u^2\right)}\ &=\frac ug\sqrt{u^2+2gH},\end{align} as desired.
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edited Feb 23, 2022 at 3:02
answered Oct 19, 2021 at 9:45
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$\begingroup$ P.S. This self-Answer is adapted from physics.stackexchange.com/a/23187/8446. $\endgroup$
ryang
– ryang
2021-10-19 10:04:00 +00:00
Commented Oct 19, 2021 at 10:04
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There. is actually a much "classier, old school solution" to this problem. I came across it as a question in an older A level M2 textbook by a remarkably inventive author D. Quadling . I confess that I used the brute force differentiation method to get through it but realised that Quadling had laid out the problem with a strong hint that a rather lovely simple solution was possible without calculus. Here are some hints, do shout if you need more( one needs to unravel this oneself to really appreciate its elegance.
Take the standard cartesian trajectory equation but with the emuzzle speed, u,written as u= sqrt(ag).Now complete the square eg form ( xtan(theta)-a)^2 etc Iam sure you will agree the result is stunning .
Chris
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answered Mar 23, 2017 at 20:15
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Adapting concepts from the question and solutions here, we have
$$R_\text{max}=\frac {uw}g=\frac {u\sqrt{u^2+2gH}}g=\color{red}{\frac ug\sqrt{u^2+2gH}}$$ and $$\tan\phi=\frac {\ell-H}{\sqrt{\ell^2-H^2}} =\frac {\frac {u^2}g}{\frac ug \sqrt{u^2+2gH}}=\color{red}{\frac u{\sqrt{u^2+2gH}}}$$ where $\ell$ is the linear distance between the launch and end points of the projectile.
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answered Mar 28, 2018 at 17:36
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15891 | https://www2.math.upenn.edu/~jhaglund/preprints/jh3.pdf | Catalan Paths and q, t-Enumeration ∗ J. Haglund † Department of Mathematics University of Pennsylvania, Philadelphia, PA 19104-6395 jhaglund@math.upenn.edu August 12,2014 This chapter contains an account of a two-parameter version of the Catalan numbers, and corresponding two-parameter versions of related objects such as parking functions and Schr¨ oder paths, which have become important in algebraic combinatorics and other areas of mathematics as well. Although the original motivation for the definition of these objects was the study of Macdonald polynomials and the representation theory of diagonal harmonics, in this account we focus only on the combinatorics associated to their description in terms of lattice paths. Hence this chapter can be read by anyone with a modest background in combinatorics. In Section 1 we include basic facts involving q-analogues, permutation statistics, and symmetric functions which we need in later sections. Sections 2, 3, and 4 contain the results on the q, t-versions of the Catalan numbers, parking functions, and Schr¨ oder paths, respectively. Section 5 contains a brief account of the recent exciting extensions of these objects which have arisen in the study of string theory, knot invariants, and the Hilbert scheme from algebraic geometry.
1 Introduction to q-Analogues and Catalan Numbers Permutation Statistics and Gaussian Polynomials In combinatorics a q-analogue of a counting function is typically a polynomial in q which reduces to the function in question when q = 1, and furthermore satisfies versions of some or all of the algebraic properties, such as recursions, of the function.
We sometimes regard q as a real parameter satisfying 0 < q < 1. We define the q-analogue of the real number x, denoted [x] as [x] = 1 −qx 1 −q .
By l’Hˆ opital’s rule, [x] →x as q →1−. Let N denote the nonnegative integers. For n ∈N, we define the q-analogue of n!, denoted [n]! as [n]! = n Y i=1 [i] = (1 + q)(1 + q + q2) · · · (1 + q + . . . + qn−1).
∗Much of this chapter is a condensed version of Chapters 1, 3, 4, and 5 of the author’s book The q, t-Catalan Numbers and the Space of Diagonal Harmonics: With an Appendix on the Combinatorics of Macdonald Polyno-mials, c ⃝2008 American Mathematical Society (AMS) and is reused here with the kind permission of the AMS.
†Work supported by NSF grant DMS-1200296 1 We let |S| denote the cardinality of a finite set S.
By a statistic on a set S we mean a combinatorial rule which associates an element of N to each element of S. A permutation statistic is a statistic on the symmetric group Sn. We use the one-line notation σ1σ2 · · · σn for the element σ = 1 2 . . .
n σ1 σ2 . . .
σn of Sn. More generally, a word (or multiset permutation) σ1σ2 · · · σn is a linear list of the elements of some multiset of nonnegative integers. (The reader may wish to consult [Sta12, Chapter 1] for more background on multiset permutations.) An inversion of a word σ is a pair (i, j), 1 ≤i < j ≤n such that σi > σj. A descent of σ is an integer i, 1 ≤i ≤n −1, for which σi > σi+1. The set of such i is known as the descent set, denoted Des(σ). We define the inversion statistic inv(σ) as the number of inversions of σ and the major index statistic maj(σ) as the sum of the descents of σ, i.e.
inv(σ) = X iσj 1, maj(σ) = X i σi>σi+1 i.
For example, inv(613524) = 8, while Des(613524) = {1, 4} and maj(613524) = 5.
A permutation statistic is said to be Mahonian if its distribution over Sn is [n]!.
Theorem 1 Both inv and maj are Mahonian, i.e.
X σ∈Sn qinv(σ) = [n]! = X σ∈Sn qmaj(σ).
(1) Proof. Given β ∈Sn−1, let β(k) denote the permutation in Sn obtained by inserting n between the (k −1)st and kth elements of β.
For example, 2143(3) = 21543.
Clearly inv(β(k)) = inv(β) + n −k, so X σ∈Sn qinv(σ) = X β∈Sn−1 (1 + q + q2 + . . . + qn−1)qinv(β) (2) and thus by induction inv is Mahonian.
A modified version of this idea works for maj. Say the descents of β ∈Sn−1 are at places i1 < i2 < · · · < is. Then maj(β(n)) = maj(β), maj(β(is + 1)) = maj(β) + 1, . . . , maj(β(i1 + 1)) = maj(β) + s, maj(β(1)) = s + 1.
If the non-descents less than n −1 of β are at places α1 < α2 < · · · < αn−2−s, then maj(β(α1 + 1)) = maj(β) + s −(α1 −1) + α1 + 1 = maj(β) + s + 2.
To see why, note that s −(α1 −1) is the number of descents of β to the right of α1, each of which will be shifted one place to the right by the insertion of n just after βα1. Also, we have a new descent at α1 + 1. By similar reasoning, maj(β(α2)) = maj(β) + s −(α2 −2) + α2 + 1 = maj(β) + s + 3, .
.
.
maj(β(αn−2−s)) = maj(β) + s −(αn−2−s −(n −2 −s)) + αn−2−s + 1 = maj(β) + n −1.
2 Thus X σ∈Sn qmaj(σ) = X β∈Sn−1 (1 + q + . . . + qs + qs+1 + . . . + qn−1)qmaj(β) (3) and again by induction maj is Mahonian.
2 Major P. MacMahon introduced the major-index statistic and proved it is Mahonian [Mac60].
Foata [Foa68] found a map Φ which sends a permutation with a given major index to another with the same value for inv. Furthermore, if we denote the descent set of σ−1 by Ides(σ), then Ides(φ(σ)) = Ides(σ). The image φ of the map Φ can be described as follows. If n ≤2, φ(σ) = σ.
If n > 2, we add a number to φ one at a time; begin by setting φ(1) = σ1, φ(2) = σ1σ2 and φ(3) = σ1σ2σ3. Then if σ2 > σ3, draw a bar after each element of φ(3) which is greater than σ3, while if σ2 < σ3, draw a bar after each element of φ(3) which is less than σ3. Also add a bar before φ(3) 1 . For example, if σ = 4137562 we now have φ(3) = |41|3. Now regard the numbers between two consecutive bars as “blocks”, and in each block, move the last element to the beginning, and finally remove all bars. We end up with φ(3) = 143.
Proceeding inductively, we begin by letting φ(i) be the result of adding σi to the end of φ(i−1). Then if σi−1 > σi, draw a bar after each element of φ(i) which is greater than σi, while if σi−1 < σi, draw a bar after each element of φ(i) which is less than σi. Also draw a bar before φ(i) 1 . Then in each block, move the last element to the beginning, and finally remove all bars. If σ = 4137562 the successive stages of the algorithm yield φ(3) = 143 φ(4) = |1|4|3|7 →1437 φ(5) = |1437|5 →71435 φ(6) = |71|4|3|5|6 →174356 φ(7) = |17|4|3|5|6|2 →7143562 and so φ(4137562) = 7143562.
Proposition 1 We have maj(σ) = inv(φ(σ)). Furthermore, Ides(σ) = Ides(φ(σ)), and also φ(σ) and σ have the same last letter.
Proof. We claim inv(φ(k)) = maj(σ1 · · · σk) for 1 ≤k ≤n. Clearly this is true for k ≤2. Assume it is true for k < j, where 2 < j ≤n. If σj−1 > σj, maj(σ1 · · · σj) = maj(σ1 · · · σj−1) + j −1. On the other hand, for each block arising in the procedure creating φ(j), the last element is greater than σj, which creates a new inversion, and when it is moved to the beginning of the block, it also creates a new inversion with each element in its block. It follows that inv(φ(j)) = inv(φ(j−1)) + j −1. Similar remarks hold if σj−1 < σj. In this case maj(σ1 · · · σj−1) = maj(σ1 · · · σj). Also, each element of φ which is not the last element in its block is larger than σj, which creates a new inversion, but a corresponding inversion between this element and the last element in its block is lost when we cycle the last element to the beginning. Hence inv(φ(j−1)) = inv(φ(j)) and the claim follows.
Note that Ides(σ) equals the set of all i, 1 ≤i < n such that i + 1 occurs before i in σ. In order for the φ map to change this set, at some stage, say when creating φ(j), we must move i 3 from the end of a block to the beginning, passing i −1 or i + 1 along the way. But this could only happen if σj is strictly between i and either i −1 or i + 1, an impossibility.
2 We now show that the map Φ is invertible by constructing the permutation β = Φ−1(σ).
Begin by setting β(1) = σ. Then if σn > σ1, draw a bar before each number in β(1) which is less than σn, and also before σn. If σn < σ1, draw a bar before each number in β(1) which is greater than σn, and also before σn. Next move each number at the beginning of a block to the end of the block.
The last letter of β is now fixed. Next set β(2) = β(1), and compare the n −1st letter with the first, creating blocks as above, and draw an extra bar before the n−1st letter. For example, if σ = 7143562 the successive stages of the algorithm to construct β yield β(1) = |71|4|3|5|6|2 →1743562 β(2) = |17|4|3|5|62 →7143562 β(3) = |7143|562 →1437562 β(4) = |1|4|3|7562 →1437562 β(5) = |14|37562 →4137562 β(6) = β(7) = 4137562 and so Φ−1(7143562) = 4137562. Notice that at each stage we are reversing the steps of the algorithm to compute φ, and it is easy to see this holds in general.
An involution on a set S is a bijective map from S to S whose square is the identity.
Foata and Sch¨ utzenberger [FS78] showed that the map iΦiΦ−1i, where i is the inverse map on permutations, is an involution on Sn which interchanges inv and maj.
For n, k ∈N with 0 ≤k ≤n, let n k = n k q = [n]!
[k]![n −k]! = (1 −qn)(1 −qn−1) · · · (1 −qn−k+1) (1 −qk)(1 −qk−1) · · · (1 −q) (4) denote the Gaussian polynomial.
These are special cases of more general objects known as q-binomial coefficients, which are defined for x ∈R as x k = (qx−k+1; q)k (q; q)k , (5) where (a; q)k = (a)k = (1 −a)(1 −qa) · · · (1 −qk−1a) is the “q-rising factorial”.
A partition λ is a nonincreasing finite sequence λ1 ≥λ2 ≥. . . of positive integers. λi is called the ith part of λ. We let ℓ(λ) denote the number of parts, and |λ| = P i λi the sum of the parts.
For various formulas it will be convenient to assume λj = 0 for j > ℓ(λ). The Ferrers graph of λ is an array of unit squares, called cells, with λi cells in the ith row, where the first cell in each row is left-justified. We often use λ to refer to its Ferrers graph. We define the conjugate partition, λ′ as the partition whose Ferrers graph is obtained from λ by reflecting across the diagonal x = y, as in Figure 1. Here (i, j) ∈λ refers to a cell with (column, row) coordinates (i, j), with the lower left-hand-cell of λ having coordinates (1, 1). The notation x ∈λ means x is a cell in λ. For technical reasons we say that 0 has one partition, the empty set ∅, with ℓ(∅) = 0 = |∅|.
The following result shows the Gaussian polynomials are in fact polynomials in q, which is not obvious from their definition.
4 Figure 1: On the left, the Ferrers graph of the partition (4, 3, 2, 2), and on the right, that of its conjugate (4, 3, 2, 2)′ = (4, 4, 2, 1).
Theorem 2 For n, k ∈N, n + k k = X λ⊆nk q|λ|, (6) where the sum is over all partitions λ whose Ferrers graph fits inside a k × n rectangle, i.e., for which λ1 ≤n and ℓ(λ) ≤k.
Proof.
Let P(n, k) = X λ⊆nk q|λ|.
Clearly P(n, k) = X λ⊆nk λ1=n q|λ| + X λ⊆nk λ1≤n−1 q|λ| = qnP(n, k −1) + P(n −1, k).
(7) On the other hand qn n + k −1 k −1 + n −1 + k k = qn [n + k −1]!
[k −1]![n]! + [n −1 + k]!
[k]![n −1]!
= qn[k][n + k −1]! + [n −1 + k]![n] [k]![n]!
= [n + k −1]!
[k]![n]!
(qn(1 + q + . . . + qk−1) + 1 + q + . . . + qn−1) = [n + k]!
[k]![n]! .
Since P(n, 0) = P(0, k) = 1, both sides of (6) thus satisfy the same recurrence and initial conditions.
2 Given α = (α0, . . . , αs) ∈Ns+1, let {0α01α1 · · · sαs} denote the multiset with αi copies of i, where α0 + . . . + αs = n. We let Mα denote the set of all permutations of this multiset and refer to α as the weight of any given one of these words.
Also let n α0, . . . , αs = [n]!
[α0]! · · · [αs]!
(8) 5 denote the q-multinomial coefficient.
The following result is due to MacMahon [Mac60].
Theorem 3 Both inv and maj are multiset Mahonian, i.e. given α ∈Ns+1, X σ∈Mα qinv(σ) = n α0, . . . , αs = X σ∈Mα qmaj(σ).
(9) Remark 1 Foata’s map also proves Theorem 3 bijectively. To see why, given a multiset permu-tation σ of M(β) let σ′ denote the standardization of σ, defined to be the permutation obtained by replacing the β0 0’s by the numbers 1 through β0, in increasing order as we move left to right in σ, then replacing the β1 1’s by the numbers β0 + 1 through β0 + β1, again in increasing order as we move left to right in σ, etc. For example, the standardization of 31344221 is 51678342.
Note that Ides(σ′) ⊆{β1, β1 + β2, . . .} (10) and in fact standardization gives a bijection between elements of M(β) and permutations sat-isfying (10). Since the map Φ fixes the inverse descent set, Φ maps M(β) to itself bijectively, sending maj to inv.
Exercise 1 If σ is a word of length n define the co-major index of σ as follows.
comaj(σ) = X σi>σi+1 1≤i<n n −i.
(11) Show that Foata’s map φ implies there is a bijective map coφ on words of fixed weight such that comaj(σ) = inv(coφ(σ)).
(12) The Catalan Numbers and Dyck Paths A lattice path is a sequence of North N(0, 1) and East E(1, 0) steps in the first quadrant of the xy-plane, starting at the origin (0, 0) and ending at say (m, n). We let Lm,n denote the set of all such paths, and L+ m,n the subset of Lm,n consisting of paths which never go below the line y = n mx. A Dyck, sometimes called a Catalan path, is an element of L+ n,n for some n.
Let Cn = 1 n+1 2n n denote the nth Catalan number, so C0, C1, . . . = 1, 1, 2, 5, 14, 42, . . . .
There are now over 200 known combinatorial interpretations for the Catalan numbers. (See [Sta99, Ex. 6.19, p. 219] for a list of 66 of these interpretations.) One of these is the number of elements of L+ n,n. For 1 ≤k ≤n, the number of Dyck paths from (0, 0) to (k, k) which touch the line y = x only at (0, 0) and (k, k) is Ck−1, since such a path must begin with a N step, end with an E step, and never go below the line y = x + 1 as it goes from (0, 1) to (k −1, k). The number of ways to extend such a path to (n, n) and still remain a Dyck path is clearly Cn−k. It follows that Cn = n X k=1 Ck−1Cn−k, n ≥1.
(13) 6 There are two natural q-analogues of Cn. Given π ∈Ln,m, let σ(π) be the element of M(m,n) resulting from the following algorithm. First initialize σ to the empty string. Next start at (0, 0), move along π and add a 0 to the end of σ(π) every time a N step is encountered, and add a 1 to the end of σ(π) every time an E step is encountered. Similarly, given σ ∈M(m,n), let π(σ) be the element of Ln,m obtained by inverting the above algorithm. We call the transformation of π to σ or its inverse the coding of π or σ. For π ∈L+ n,n, let ai(π) denote the number of complete squares, in the ith row from the bottom of π, which are to the right of π and to the left of the line y = x. We refer to ai(π) as the length of the ith row of π. Furthermore call (a1(π), a2(π), . . . , an(π)) the area vector of π, and set area(π) = P i ai(π). For example, the path in Figure 2 has area vector (0, 1, 1, 2, 1, 2, 0), and σ(π) = 00100110011101. By convention we say L+ 0,0 contains one path, the empty path ∅, with area(∅) = 0.
1 2 1 1 0 0 2 Figure 2: A Dyck path, with row lengths on the right. The area statistic is 1+1+2+1+2 = 7.
Let M+ (m,n) denote the elements σ of M(m,n) corresponding to paths in L+ n,m. Words in M+ n,n are thus characterized by the property that in any initial segment there are at least as many 0’s as 1’s. The first q-analogue of Cn is given by the following.
Theorem 4 (MacMahon [Mac60, p. 214]) X π∈L+ n,n qmaj(σ(π)) = 1 [n + 1] 2n n .
(14) Proof.
We give a bijective proof, taken from [FH85]. Let M− (m,n) = M(m,n) \ M+ (m,n), and let L− n,m = Ln,m \ L+ n,m be the corresponding set of lattice paths. Given a path π ∈L− n,n, let P be the point with smallest x-coordinate among those lattice points (i, j) in π for which j −i is maximal, i.e. whose distance from the line y = x in a SE direction is maximal. (Since π ∈L− n,n, this maximal value of i −j is positive.) Let P ′ be the lattice point on π before P. There must be an east step connecting P ′ to P (preceded by another east step unless P ′ is the origin).
Change this east step into a north step and shift the remainder of the path after P up one unit and left one unit. We now have a path φ(π) from (0, 0) to (n −1, n + 1), and moreover 7 maj(σ(φ(π))) = maj(σ(π)) −1. For example, if π is the path on the left in Figure 3, then φ(π) is the path on the right.
Figure 3: On the left, a path π in L− 7,7, with its image φ(π) on the right.
It is easy to see that this map is invertible. Given a lattice path π′ from (0, 0) to (n−1, n+1), let P ′ be the point with maximal x-coordinate among those lattice points (i, j) in π′ for which j −i is maximal. Thus X σ∈M− (n,n) qmaj(σ) = X σ′∈M(n+1,n−1) qmaj(σ′)+1 = q 2n n + 1 , (15) using (9). Hence X π∈L+ n,n qmaj(σ(π)) = X σ∈M(n,n) qmaj(σ) − X σ∈M− (n,n) qmaj(σ) (16) = 2n n −q 2n n + 1 = 1 [n + 1] 2n n .
(17) 2 The second natural q-analogue of Cn was studied by Carlitz and Riordan [CR64]. They define Cn(q) = X π∈L+ n,n qarea(π).
(18) For example, the paths in L+ 3,3 given in Figure 4 have area, from left-to-right, 3, 2, 1, 1, 0, so C3(q) = 1 + 2q + q2 + q3.
Proposition 2 Cn(q) = n X k=1 qk−1Ck−1(q)Cn−k(q), n ≥1.
(19) 8 Figure 4: The paths in L+ 3,3 have Carlitz-Riordan area weights q3, q2, q, q, 1.
Proof.
As in the proof of (13), we break up our path π according to the “point of first return” to the line y = x. If this occurs at (k, k), then the area of the part of π from (0, 1) to (k −1, k), when viewed as an element of L+ k−1,k−1, is k −1 less than the area of this portion of π when viewed as a path in L+ n,n.
2 Exercise 2 Define a co-inversion of σ to be a pair (i, j) with i < j and σi < σj. Show Cn(q) = X π∈L+ n,n qcoinv(σ(π))−(n+1 2 ), (20) where coinv(σ) is the number of co-inversions of σ.
The q-Vandermonde Convolution Let p+1φp a1, a2, . . . , ap+1 b1, . . . , bp ; q; z = ∞ X k=0 (a1)k · · · (ap+1)k (q)k(b1)k · · · (bp)k zk (21) denote the basic hypergeometric series. A good general reference for this subject is [GR04]. The following result is known as Cauchy’s q-binomial series.
Theorem 5 1φ0 a − ; q; z = ∞ X n=0 (a)n (q)n zn = (az)∞ (z)∞ , |z| < 1, |q| < 1, (22) where (a; q)∞= (a)∞= Q∞ i=0(1 −aqi).
Proof.
The following proof is based on the proof in [GR04, Chap. 1]. Let F(a, z) = ∞ X n=0 (a)n (q)n zn.
Then F(a, z) −F(a, qz) = (1 −a)zF(aq, z) (23) 9 and F(a, z) −F(aq, z) = −azF(aq, z).
(24) Eliminating F(aq, z) from (23) and (24) we get F(a, z) = (1 −az) (1 −z) F(a, qz).
Iterating this n times, then taking the limit as n →∞we get F(a, z) = lim n→∞ (az; q)n (z; q)n F(a, qnz) = (az; q)∞ (z; q)∞ F(a, 0) = (az; q)∞ (z; q)∞ .
(25) 2 Corollary 1 The q-binomial theorem: n X k=0 q(k 2) n k zk = (−z; q)n (26) and ∞ X k=0 n + k k zk = 1 (z; q)n+1 .
(27) Proof.
To prove (26), set a = q−n and z = −zqn in (22) and simplify. To prove (27), let a = qn+1 in (22) and simplify.
2 For any function f(z), let f(z)|zk denote the coefficient of zk in the Maclaurin series for f(z).
Corollary 2 h X k=0 q(n−k)(h−k) n k m h −k = m + n h .
(28) Proof.
By (26), q(h 2) m + n h = m+n−1 Y k=0 (1 + zqk)|zh = n−1 Y k=0 (1 + zqk) m−1 Y j=0 (1 + zqnqj)|zh = ( n−1 X k=0 q(k 2) n k zk)( m−1 X j=0 q(j 2) m j (zqn)j)|zh = h X k=0 q(k 2) n k q(h−k 2 ) m h −k (qn)h−k.
10 The result now reduces to the identity k 2 + h −k 2 + n(h −k) − h 2 = (n −k)(h −k).
2 Corollary 3 h X k=0 q(m+1)k n −1 + k k m + h −k h −k = m + n + h h .
(29) Proof.
By (27), m + n + h h = 1 (z)m+n+1 |zh = 1 (z)m+1 1 (zqm+1)n |zh = h X j=0 zj m + j j q h X k=0 (zqm+1)k n −1 + k k q !
|zh = h X k=0 q(m+1)k n −1 + k k m + h −k h −k q .
2 We note that (28) and (29) have alternative, elementary proofs based on q-counting lattice paths. Both identities are special cases of the following result, known as the q-Vandermonde convolution. For a proof see [GR04, Chap. 1].
Theorem 6 Let n ∈N. Then 2φ1 a, q−n c ; q; q = (c/a)n (c)n an.
(30) Exercise 3 By reversing summation in (30), show that 2φ1 a, q−n c ; q; cqn/a = (c/a)n (c)n .
(31) Exercise 4 Show Newton’s binomial series ∞ X n=0 a(a + 1) · · · (a + n −1) n!
zn = 1 (1 −z)a , |z| < 1, a ∈R (32) can be derived from (22) by replacing a by qa and letting q →1−. For simplicity you can assume a, z ∈R.
11 Symmetric Functions The Basics Given f(x1, . . . , xn) ∈K[x1, x2, . . . , xn] for some field K, and σ ∈Sn, let σf = f(xσ1, . . . , xσn).
(33) We say f is a symmetric function if σf = f for all σ ∈Sn. It will be convenient to work with more general functions f depending on countably many indeterminates x1, x2, . . ., indicated by f(x1, x2, . . .), in which case we view f as a formal power series in the xi, and say it is a symmetric function if it is invariant under any permutation of the variables. The standard references on this topic are [Sta99, Chap. 7] and [Mac95]. We will often let Xn and X stand for the set of variables {x1, . . . , xn} and {x1, x2, . . .}, respectively.
We let Λ denote the ring of symmetric functions in x1, x2, . . . and Λn the vector subspace consisting of symmetric functions which are homogeneous of degree n. The most basic symmetric functions are the monomial symmetric functions, which depend on a partition λ in addition to a set of variables. They are denoted by mλ(X) = mλ(x1, x2, . . .). In a symmetric function it is typical to leave offexplicit mention of the variables, with a set of variables being understood from context, so mλ = mλ(X). We illustrate these first by means of examples. We let Par(n) denote the set of partitions of n, and use the notation λ ⊢n as an abbreviation for λ ∈Par(n).
Example 1 m1,1 = X i<j xixj m2,1,1(X3) = x2 1x2x3 + x1x2 2x3 + x1x2x2 3 m2(X) = X i x2 i .
In general, mλ(X) is the sum of all distinct monomials in the xi whose multiset of exponents equals the multiset of parts of λ.
Any element of Λ can be expressed uniquely as a linear combination of the mλ.
We let 1n stand for the partition consisting of n parts of size 1. The function m1n is called the nth elementary symmetric function, which we denote by en. Then ∞ Y i=1 (1 + zxi) = ∞ X n=0 znen, e0 = 1.
(34) Another important special case is mn = P i xn i , known as the power-sum symmetric function, de-noted pn. We also define the complete homogeneous symmetric functions hn, by hn = P λ⊢n mλ, or equivalently 1 Q∞ i=1(1 −zxi) = ∞ X n=0 znhn, h0 = 1.
(35) 12 For λ ⊢n, we define eλ = Q i eλi, pλ = Q i pλi, and hλ = Q i hλi. For example, e2,1 = X i<j xixj X k xk = m2,1 + 3m1,1,1 p2,1 = X i x2 i X j xj = m3 + m2,1 h2,1 = ( X i x2 i + X i<j xixj) X k xk = m3 + 2m2,1 + 3m1,1,1.
Assuming we have at least n variables, it is known that {eλ, λ ⊢n} forms a basis for Λn, and so do {pλ, λ ⊢n} and {hλ, λ ⊢n}.
Definition 1 Two simple functions on partitions we will often use are n(λ) = X i (i −1)λi = X i λ′ i 2 zλ = Y i imimi!, where mi = mi(λ) is the number of parts of λ equal to i. For example, n(54331) = 4+6+9+4 = 23, and z(5433111) = 5 ∗4 ∗32 ∗2! ∗3! = 2160.
Exercise 5 Use (34) and (35) to show that en = X λ⊢n (−1)n−ℓ(λ)pλ zλ , hn = X λ⊢n pλ zλ .
We let ω denote the ring endomorphism ω : Λ →Λ defined by ω(pk) = (−1)k−1pk.
(36) Thus ω is an involution with ω(pλ) = (−1)|λ|−ℓ(λ)pλ, and by Exercise 5, ω(en) = hn, and more generally ω(eλ) = hλ.
For f ∈Λ, the special value f(1, q, q2, . . . , qn−1) is known as the principal specialization (of order n) of f.
Theorem 7 em(1, q, . . . , qn−1) = q(m 2) n m q (37) hm(1, q, . . . , qn−1) = n −1 + m m q (38) pm(1, q, . . . , qn−1) = 1 −qnm 1 −qm .
(39) 13 Proof.
The principal specializations for em and hm follow directly from (26), (27), (34) and (35).
2 Remark 2 The principal specialization of mλ doesn’t have a particularly simple description, although if ps1 n denotes the set of n variables, each equal to 1, then [Sta99, p. 303] mλ(ps1 n) = n m1, m2, m3, . . .
, (40) where again mi is the multiplicity of the number i in the multiset of parts of λ.
Remark 3 Λ is known to be isomorphic to K[p1, p2, . . .]. Hence, although identities like h2,1 = m3 + 2m2,1 + 3m1,1,1 appear at first to depend on a set of variables, it is customary to view them as polynomial identities in the pλ. Since the pk (in infinitely many variables) are algebraically independent, we can specialize them to whatever we please, forgetting about the original set of variables X.
We define the Hall scalar product, a bilinear form from Λ × Λ to Q, by ⟨pλ, pβ⟩= zλχ(λ = β), (41) where for any logical statement L χ(L) = ( 1 if L is true 0 if L is false.
(42) Clearly ⟨f, g⟩= ⟨g, f⟩. Also, ⟨ωf, ωg⟩= ⟨f, g⟩, which follows from the definition if f = pλ, g = pβ, and by bilinearity for general f, g since the pλ form a basis for Λ.
Theorem 8 The hλ and the mβ are dual with respect to the Hall scalar product, i.e.
⟨hλ, mβ⟩= χ(λ = β).
(43) Proof.
See [Mac95] or [Sta99].
2 For any f ∈Λ, and any basis {bλ, λ ∈Par} of Λ, let f|bλ denote the coefficient of bλ when f is expressed in terms of the bλ. Then (43) implies Corollary 4 ⟨f, hλ⟩= f|mλ.
(44) 14 1 3 2 1 2 1 1 2 2 3 1 1 3 1 1 3 3 2 1 2 2 1 2 3 3 2 2 3 3 3 Figure 5: Some SSYT of shape (3, 2).
Tableaux and Schur Functions Given λ, µ ∈Par(n), a semi-standard Young tableaux (or SSYT) of shape λ and weight µ is a filling of the cells of the Ferrers graph of λ with the elements of the multiset {1µ12µ2 · · · }, so that the numbers weakly increase across rows and strictly increase up columns. Let SSY T(λ, µ) denote the set of these fillings, and Kλ,µ the cardinality of this set. The Kλ,µ are known as the Kostka numbers. Our definition also makes sense if our weight is a weak composition of n , i.e. any finite sequence of nonnegative integers whose sum is n. For example, K(3,2),(2,2,1) = K(3,2),(2,1,2) = K(3,2),(1,2,2) = 2 as in Figure 5.
If the Ferrers graph of a partition β is contained in the Ferrers graph of λ, denoted β ⊆λ, let λ/β refer to the subset of cells of λ which are not in β. This is referred to as a skew shape.
Define a SSYT of shape λ/β and weight ν, where |ν| = |λ| −|β|, to be a filling of the cells of λ/β with elements of {1ν12ν2 · · · }, again with weak increase across rows and strict increase up columns. The number of such tableaux is denoted Kλ/β,ν.
Let wcomp(µ) denote the set of all weak compositions whose multiset of nonzero parts equals the multiset of parts of µ.
It follows easily from Figure 5 that K(3,2),α = 2 for all α ∈wcomp(2, 2, 1). Hence X α,T Y i xαi i = 2m(2,2,1), (45) where the sum is over all tableaux T of shape (3, 2) and weight some element of wcomp(2, 2, 1).
This is a special case of a more general phenomenon. For λ ∈Par(n), define sλ = X α,T Y i xαi i , (46) where the sum is over all weak compositions α of n, and all possible tableaux T of shape λ and weight α. Then sλ = X µ⊢n Kλ,µmµ, (47) 15 i.e. Kλ,β = Kλ,α for all compositions β, α whose multiset of parts is the same (we leave it to the interested reader to prove this fact bijectively). The sλ are called Schur functions, and are fundamental to the theory of symmetric functions. Two special cases of (47) are sn = hn (since Kn,µ = 1 for all µ ∈Par(n)) and s1n = en (since K1n,µ = χ(µ = 1n)).
A SSYT of weight 1n is called standard, or a SYT. The set of SYT of shape λ is denoted SY T(λ). For (i, j) ∈λ, let the content of (i, j), denoted c(i, j), be i −j. Also, let h(i, j) denote the “hook length” of (i, j), defined as the number of cells to the right of (i, j) in row j plus the number of cells above (i, j) in column i plus 1. For example, if λ = (5, 5, 3, 3, 1), h(2, 2) = 6.
It is customary to let f λ denote the number of SYT of shape λ, i.e. f λ = Kλ,1n. There is a beautiful formula for f λ, namely f λ = n!
Q (i,j)∈λ h(i, j).
(48) Below we list some of the important properties of Schur functions. See [Sta99, Chapter 7] for proofs of these well-known identities, and how (48) can be derived from (49).
Theorem 9 Let λ, µ ∈Par. Then 1. The Schur functions are orthonormal with respect to the Hall scalar product, i.e.
< sλ, sµ >= χ(λ = µ).
Thus for any f ∈Λ, < f, sλ >= f|sλ.
2. Action by ω: ω(sλ) = sλ′.
3. Principal Specialization: For λ ∈Par, sλ(1, q, q2, . . . , qn−1) = qn(λ) Y (i,j)∈λ [n + a −l] [a + l + 1] , (49) where for a given square (i, j) ∈λ, we define the coarm a′ and coleg l′ as in Figure 6 from Section 2.
4. Cauchy Identities: For any two alphabets of variables X, Y , let XY denote the set of variables {xiyj}. Then en(XY ) = X λ∈Par(n) sλ(X)sλ′(Y ) (50) hn(XY ) = X λ∈Par(n) sλ(X)sλ(Y ).
(51) 16 Statistics on Tableaux There is a q-analogue of the Kostka numbers, denoted by Kλ,µ(q), which has many applications in representation theory and the combinatorics of tableaux. Originally defined algebraically in an indirect fashion, the Kλ,µ(q) are polynomials in q which satisfy Kλ,µ(1) = Kλ,µ. Foulkes [Fou74] conjectured that there should be a statistic stat(T) on SSYT T of shape λ and weight µ such that Kλ,µ(q) = X T∈SSY T(λ) qstat(T).
(52) This conjecture was resolved by Lascoux and Sch¨ utzenberger [LS78], who found a statistic charge to generate these polynomials. Butler [But94] provided a detailed account of their proof, filling in a lot of missing details. A short proof, based on the combinatorial formula for Macdonald polynomials, is contained in [Hag08][Appendix A].
Assume we have a tableau T ∈SSY T(λ, µ) where µ ∈Par. It will be more convenient for us to describe a slight modification of charge(T), called cocharge(T), which is defined as n(µ) −charge. The reading word read(T) of T is obtained by reading the entries in T from left to right in the top row of T, then continuing left to right in the second row from the top of T, etc. For example, the tableau in the upper-left of Figure 5 has reading word 22113. To calculate cocharge(T), perform the following algorithm on read(T).
Algorithm 1 1. Start at the end of read(T) and scan left until you encounter a 1 - say this occurs at spot i1, so read(T)i1 = 1. Then start there and scan left until you encounter a 2. If you hit the end of read(T) before finding a 2, loop around and continue searching left, starting at the end of read(T). Say the first 2 you find equals read(T)i2. Now iterate, start at i2 and search left until you find a 3, etc. Continue in this way until you have found 4, 5, . . . , µ1, with µ1 occurring at spot iµ1. Then the first subword of textread(T) is defined to be the elements of the set {read(T)i1, . . . , read(T)iµ1}, listed in the order in which they occur in read(T) if we start at the beginning of read(T) and move left to right. For example, if read(T) = 21613244153 then the first subword equals 632415, corresponding to places 3, 5, 6, 8, 9, 10 of read(T).
Next remove the elements of the first subword from read(T) and find the first subword of what’s left. Call this the second subword. Remove this and find the first subword in what’s left and call this the third subword of read(T), etc. For the word 21613244153, the subwords are 632415, 2143, 1.
2. The value of charge(T) will be the sum of the values of charge on each of the subwords of rw(T). Thus it suffices to assume rw(T) ∈Sm for some m, in which case we set cocharge(rw(T)) = comaj(rw(T)−1), where read(T)−1 is the usual inverse in Sm, with comaj as in (11).
(Another way of describing cocharge(read(T)) is the sum of m−i over all i for which i+1 occurs before i in read(T).) For example, if σ = 632415, then σ−1 = 532461 and cocharge(σ) = 5+4+1 = 10, and finally cocharge(21613244153) = 10 + 4 + 0 = 14.
17 Note that to compute charge, we could create subwords in the same manner, and count m−i for each i with i + 1 occurring to the right of i instead of to the left. For λ, µ ∈Par(n) we set ˜ Kλ,µ(q) = qn(µ)Kλ,µ(1/q) (53) = X T∈SSY T(λ,µ) qcocharge(T).
In addition to the cocharge statistic, there is a major index statistic on SYT which is often useful. Given a SYT T of shape λ, define a descent of T to be a value of i, 1 ≤i < |λ|, for which i + 1 occurs in a row above i in T. Let maj(T) = X i (54) comaj(T) = X |λ| −i, (55) where the sums are over the descents of T. Then [Sta99, p.363] sλ(1, q, q2, . . .) = 1 (q)n X T∈SY T(λ) qmaj(T) (56) = 1 (q)n X T∈SY T(λ) qcomaj(T).
Representation Theory Let G be a finite group. A (matrix) representation of G is a group homomorphism from G to GLn(C), the set of invertible n × n matrices with entries in C. See [Sag01] and [JL01] for a detailed discussion of the representation theory of the symmetric group and other finite groups.
We include an informal discussion of some of the main ideas here, in order to motivate the combinatorial problems we will be discussing.
We will identify a representation with the image of a homomorphism from G to GLn(C), namely the set of square invertible matrices {M(g), g ∈G} with the property that M(g)M(h) = M(gh) for all g, h ∈G.
(57) On the left-hand-side of (57) we are using ordinary matrix multiplication, and on the right-hand-side, to define gh, multiplication in G. The number of rows of a given M(g) is called the dimension of the representation.
An action of G on a set S is a map from G×S to S, denoted by g(s) for g ∈G, s ∈S, which satisfies g(h(s)) = (gh)(s) ∀g, h ∈G, s ∈S, (58) with e(s) = s for all s ∈S, where e is the identity in G. Let V be a finite dimensional C vector space, with basis w1, w2, . . . wn. Any linear action of G on V makes V into a CG module. A module is called irreducible if it has no submodules other than {0} and itself. Maschke’s theorem [JL01] says that every nonzero CG-module V can be expressed as a direct sum of irreducible submodules.
18 If we form a matrix M(g) whose ith row consists of the coefficients of the wj when expanding g(wi) in the w basis, then {M(g), g ∈G} is a representation of G. In general {M(g), g ∈G} will depend on the choice of basis, but the trace of the matrices will not. The trace of the matrix M(g) is called the character of the module (under the given action), which we denote char(V ).
If V = Ld j=1 Vj, where each Vj is irreducible, then an ordered basis of V can be chosen so that the matrix M will be in block-diagonal form, where the sizes of the blocks are the dimensions of the Vj. Clearly char(V ) = Pd j=1 char(Vj). It turns out that there are only a certain number of possible functions which occur as characters of irreducible modules, namely one for each conjugacy class of G. These are called the irreducible characters of G.
In the case G = Sn, the conjugacy classes are in one-to-one correspondence with partitions λ ∈Par(n), and the irreducible characters are denoted χλ. The dimension of a given Vλ with character χλ is known to be f λ. The value of a given χλ(σ) depends only on the conjugacy class of σ. For the symmetric group the conjugacy class of an element is determined by rearranging the lengths of the disjoint cycles of σ into nonincreasing order to form a partition called the cycle-type of σ. Thus we can talk about χλ(β), which means the value of χλ at any permutation of cycle type β. For example, χ(n)(β) = 1 for all β ⊢n, so χ(n) is called the trivial character.
Also, χ1n(β) = (−1)n−ℓ(β) for all β ⊢n, so χ1n is called the sign character, since (−1)n−ℓ(β) is the sign of any permutation of cycle type β.
One reason Schur functions are important in representation theory is the following (see [Sta99, p.347], [Mac95, Chapter 1]).
Theorem 10 When expanding the pµ into the sλ basis, the coefficients are the χλ. To be exact pµ = X λ⊢n χλ(µ)sλ sλ = X µ⊢n z−1 µ χλ(µ)pµ.
Let C[Xn] = C[x1, . . . , xn]. Given f(x1, . . . , xn) ∈C[Xn] and σ ∈Sn, then σf = f(xσ1, . . . , xσn) (59) defines an action of Sn on C[Xn].
Assume V is a homogeneous subspace of C[Xn] which can be decomposed as V = ∞ M i=0 V (i), (60) where V (i) is the subspace consisting of all elements of V which are homogeneous of degree i in the xj, and is finite dimensional. This gives a grading of the space V , and we define the Hilbert series H(V ; q) of V to be the sum H(V ; q) = ∞ X i=0 qidim(V (i)), (61) where dim indicates the dimension as a C vector space. If in addition V is fixed by the Sn action, we define the Frobenius series F(V ; q) of V to be the symmetric function ∞ X i=0 qi X λ∈Par(i) Mult(χλ, V (i))sλ, (62) 19 where Mult(χλ, V (i)) is the multiplicity of the irreducible character χλ in the character of V (i) under the action. In other words, if we decompose V (i) into irreducible Sn-submodules, Mult(χλ, V (i)) is the number of these submodules whose trace equals χλ.
A polynomial in C[Xn] is alternating, or an alternant, if σf = (−1)inv(σ)f ∀σ ∈Sn.
(63) The set of alternants in V forms a subspace called the subspace of alternants, or anti-symmetric elements, denoted V ǫ. This is also an Sn-submodule of V .
Proposition 3 The Hilbert series of V ǫ equals the coefficient of s1n in the Frobenius series of V , i.e.
H(V ǫ; q) = ⟨F(V ; q), s1n⟩.
(64) Proof.
Let B be a basis for V (i) with the property that the matrices M(σ) are in block form.
Then b ∈B is also in (V ǫ)(i) if and only if the column of M(σ) corresponding to b has entries (−1)inv(σ) on the diagonal and 0’s elsewhere, i.e. is a block corresponding to χ1n. Thus ⟨F(V ; q), s1n⟩= ∞ X i=0 qi dim((V ǫ)(i)) = H(V ǫ; q).
(65) 2 Remark 4 Since the dimension of the representation corresponding to χλ equals f λ, which by (44) equals < sλ, h1n >, we have ⟨F(V ; q), h1n⟩= H(V ; q).
(66) Example 2 Since a basis for C[Xn] can be obtained by taking all possible monomials in the xi, H(C[Xn]; q) = (1 −q)−n.
(67) Taking into account the Sn-action, it is known [Hai94, Section 1.4] that F(C[Xn]; q) = X λ∈Par(n) sλ P T∈SY T(λ) qmaj(T) (q)n (68) = X λ∈Par(n) sλsλ(1, q, q2, . . .) = Y i,j 1 (1 −qixjz)|zn.
The Ring of Coinvariants and the Space of Diagonal Harmonics The set of symmetric polynomials in the xi, denoted C[Xn]Sn, which is generated by 1, e1, . . . en, is called the ring of invariants. The quotient ring Rn = C[x1, . . . , xn]/ < e1, e2, . . . , en >, or equivalently C[x1, . . . , xn]/ < p1, p2, . . . , pn >, obtained by forming the quotient by the ideal generated by all symmetric polynomials of positive degree, is known as the ring of coinvariants.
20 It is known that Rn is finite dimensional as a C-vector space, with dim(Rn) = n!, and more generally that H(Rn; q) = [n]!.
(69) E. Artin [Art76] derived a specific basis for Rn, namely the cosets of { Y 1≤i≤n xαi i , 0 ≤αi ≤i −1}.
(70) Also, F(Rn; q) = X λ∈Par(n) sλ X T∈SY T(λ) qmaj(T), (71) a result that Stanley [Sta79], [Sta03] attributes to unpublished work of Lusztig. This shows the Frobenius series of Rn is (q)n times the Frobenius series of C[Xn].
Let Vn = det 1 x1 . . .
xn−1 1 1 x2 . . .
xn−1 2 .
.
.
1 xn . . .
xn−1 n = Y 1≤i<j≤n (xj −xi) be the Vandermonde determinant. The space of harmonics Hn can be defined as the C vector space spanned by Vn and its partial derivatives of all orders. Haiman [Hai94] provides a detailed proof that Hn is isomorphic to Rn as an Sn module, and notes that an explicit isomorphism α is obtained by letting α(h), h ∈Hn, be the element of C[Xn] represented modulo < e1, . . . , en > by h. Thus dim(Hn) = n! and moreover the character of Hn under the Sn-action is given by (71). He also argues that (71) follows immediately from (68) and the fact that Hn generates C[Xn] as a free module over C[Xn]Sn.
There is a natural extension of this construction to two sets of variables, which has a very rich algebraic and combinatorial structure. Let the ring of diagonal coinvariants DRn be defined as DRn = C[Xn, Yn]/ n X i=1 xh i yk i , ∀h + k > 0 + .
(72) By analogy we also define the space of diagonal harmonics DHn by DHn = ( f ∈C[Xn, Yn] : n X i=1 (∂xi)h(∂yi)kf = 0, ∀h + k > 0 ) .
(73) Many of the properties of Hn and Rn carry over to two sets of variables.
For example the symmetric group acts “diagonally” on DRn and DHn by permuting the X and Y variables in the same way, which turns DHn and DRn into finite-dimensional isomorphic Sn-modules.
We can decompose DHn by homogeneous X and Y degree, and the Sn action respects this bi-grading. Hence we can talk about the Hilbert series H(DHn; q, t) and and the Frobenius series F(DHn; q, t).
21 2 The q, t-Catalan Numbers Given a cell x ∈λ, let the arm a = a(x), leg l = l(x), coarm a′ = a′(x), and coleg l′ = l′(x) be the number of cells strictly between x and the border of λ in the east, north, west, and south directions, respectively, as in Figure 6. Also, define Bµ = Bµ(q, t) = X x∈µ qa′tl′, Πµ = Πµ(q, t) = ′ Y x∈µ (1 −qa′tl′), (74) where a prime symbol ′ above a product or a sum over cells of a partition µ indicates we ignore the corner (1, 1) cell, and B∅= 0, Π∅= 1. For example, B(2,2,1) = 1 + q + t + qt + t2 and Π(2,2,1) = (1 −q)(1 −t)(1 −qt)(1 −t2). Note that n(µ) = X x∈µ l′ = X x∈µ l.
(75) (i,j) a / (1,1) l l / a Figure 6: The arm a, coarm a′, leg l, and coleg l′ of a cell.
In 1996 Garsia and Haiman [GH96] introduced an amazing two-parameter Catalan sequence, Cn(q, t), which they defined as the following sum of rational functions: Cn(q, t) = X µ⊢n T 2 µMΠµBµ wµ , (76) where M = (1 −q)(1 −t), Tµ = tn(µ)qn(µ′), wµ = Y x∈µ (qa −tl+1)(tl −qa+1).
(77) The definition of Cn(q, t) was motivated by ideas involving algebraic geometry, and Garsia and Haiman conjectured that Cn(q, t) ∈N[q, t]. More specifically, they conjectured that Cn(q, t) is the sign character in the Frobenius series for DHn, i.e.
Cn(q, t) = ⟨F(DHn; q, t), s1n⟩.
(78) Mark Haiman proved this conjecture in 2001 by obtaining the following expression for F(DHn; q, t) in terms of Macdonald polynomials.
22 Theorem 11 [Hai02] F(DHn; q, t) = X µ⊢n TµMΠµBµ ˜ Hµ(X; q, t) wµ , (79) where the ˜ Hµ(X; q, t) form the modified Macdonald polynomial symmetric function basis. Al-though we will not describe these polynomials explicitly here, the interested reader can find a combinatorial description of them in [HHL05a] or [Hag08][Appendix A].
We mention that ⟨˜ Hµ(X; q, t), s1n(X)⟩= Tµ, (80) so (79) implies (78). The right-hand-side of (79) can be expressed more compactly as ∇en(X), where ∇is the linear operator on symmetric functions satisfying ∇˜ Hµ(X; q, t) = Tµ ˜ Hµ(X; q, t).
(81) Hence we can refer to F(DHn; q, t) and ∇en interchangeably.
Around the same time Haiman proved Theorem 11, Garsia and Haglund proved indepen-dently that Cn(q, t) can be expressed combinatorially in terms of statistics on Dyck paths which we now describe.
The Bounce Statistic Our combinatorial formula for Cn(q, t), the q, t-Catalan number, involves a new statistic on Dyck paths we call bounce.
Definition 2 Given π ∈L+ n,n, define the bounce path of π to be the path described by the following algorithm.
Start at (0, 0) and travel North along π until you encounter the beginning of an E step. Then turn East and travel straight until you hit the diagonal y = x. Then turn North and travel straight until you again encounter the beginning of an E step of π, then turn East and travel to the diagonal, etc. Continue in this way until you arrive at (n, n).
We can think of our bounce path as describing the trail of a billiard ball shot North from (0, 0), which “bounces” right whenever it encounters a horizontal step and “bounces” up when it encounters the line y = x. The bouncing ball will strike the diagonal at places (0, 0), (j1, j1), (j2, j2), . . . , (jb−1, jb−1), (jb, jb) = (n, n) . We define the bounce statistic bounce(π) to be the sum bounce(π) = b−1 X i=1 n −ji, (82) and we call b the number of bounces, with j1 the length of the first bounce, j2 −j1 the length of the second bounce, etc. The lattice points where the bouncing billiard ball switches from traveling North to East are called the peaks of π. The first peak is the peak with smallest y coordinate, the second peak the one with next smallest y coordinate, etc. For example, for the path π in Figure 7, there are 5 bounces of lengths 3, 2, 2, 3, 1 and bounce(π) = 19. The first two peaks have coordinates (0, 3) and (3, 5).
23 (3,3) (5,5) (7,7) (10,10) Figure 7: The bounce path (dotted line) of a Dyck path (solid line). The bounce statistic equals 11 −3 + 11 −5 + 11 −7 + 11 −10 = 8 + 6 + 4 + 1 = 19.
Let Fn(q, t) = X π∈L+ n,n qarea(π)tbounce(π).
(83) Theorem 12 Cn(q, t) = Fn(q, t).
(84) Combining Theorem 12 with Theorem 11 we have the following.
Corollary 5 H(DHǫ n; q, t) = X π∈L+ n,n qarea(π)tbounce(π).
(85) Theorem 12 was first Conjectured by Haglund in 2000 [Hag03] after a prolonged study of tables of Cn(q, t). It was then proved by Garsia and Haglund [GH01], [GH02]. At the present time there is no known way of proving Corollary 5 without using both Theorems 12 and 11.
The proof of Theorem 12 is based on a recursive structure underlying Fn(q, t).
Definition 3 Let L+ n,n(k) denote the set of all π ∈L+ n,n which begin with exactly k N steps followed by an E step. By convention L+ 0,0(k) consists of the empty path if k = 0 and is empty otherwise. Set Fn,k(q, t) = X π∈L+ n,n(k) qarea(π)tbounce(π), Fn,0 = χ(n = 0).
(86) 24 Theorem 13 [Hag03]. For 1 ≤k ≤n, Fn,k(q, t) = n−k X r=0 r + k −1 r q tn−kq(k 2)Fn−k,r(q, t).
(87) Proof.
Given β ∈L+ n,n(k), with first bounce k and second bounce say r, then β must pass through the lattice points with coordinates (1, k) and (k, k + r) (the two large dots in Figure 8).
Decompose β into two parts, the first part being the portion of β starting at (0, 0) and ending at (k, k + r), and the second the portion starting at (k, k + r) and ending at (n, n). If we adjoin a sequence of r N steps to the beginning of the second part, we obtain a path β′ in L+ n−k,n−k(r). It is easy to check that bounce(β) = bounce(β′) + n −k. It remains to relate area(β) and area(β′).
Clearly the area inside the triangle below the first bounce step is k 2 . If we fix β′, and let β vary over all paths in L+ n,n(k) which travel through (1, k) and (k, k + r), then the sum of qarea(β) will equal qarea(β′)q(k 2) k + r −1 r q (88) by (6). Thus Fn,k(q, t) = n−k X r=0 X β′∈L+ n−k,n−k(r) qarea(β′)tbounce(β′)tn−kq(k 2) k + r −1 r q (89) = n−k X r=0 r + k −1 r q tn−kq(k 2)Fn−k,r(q, t).
(90) 2 Corollary 6 Fn(q, t) = n X b=1 X α1+α2+...+αb=n αi>0 tα2+2α3+...+(b−1)αbq Pb i=1 (αi 2 ) b−1 Y i=1 αi + αi+1 −1 αi+1 q , (91) where the inner sum is over all compositions α of n into b positive integers.
Proof. This follows by iterating the recurrence in Theorem 13. The inner term in the sum over b equals the sum of qareatbounce over all paths π whose bounce path has b steps of lengths α1, . . . , αb.
For such a π, the contribution of the first bounce to bounce(π) is n −α1 = α2 + . . . + αb, the contribution of the second bounce is n −α1 −α2 = α3 + . . . + αb, et cetera, so bounce(π) = α2 + 2α3 + . . . + (b −1)αb.
2 25 β r k Figure 8: A path whose first two bounce steps are k and r.
The Special Values t = 1 and t = 1/q Garsia and Haiman proved that Cn(q, 1) = Cn(q) (92) q(n 2)Cn(q, 1/q) = 1 [n + 1] 2n n q , (93) which shows that both the Carlitz-Riordan and MacMahon q-Catalan numbers are special cases of Cn(q, t). In this section we derive analogous results for Fn,k(q, 1) and Fn,k(q, 1/q).
By definition we have Fn(q, 1) = Cn(q).
(94) It is perhaps worth mentioning that the Fn,k(q, 1) satisfy the simple recurrence Fn,k(q, 1) = n X m=k qm−1Fm−1,k−1(q, 1)Fn−m(q, 1).
(95) This follows by grouping paths in L+ n,n(k) according to the first time they return to the diagonal, at say (m, m), then arguing as in the proof of Proposition 19.
The Fn,k(q, 1/q) satisfy the following refinement of (93).
Theorem 14 For 1 ≤k ≤n, q(n 2)Fn,k(q, 1/q) = [k] [n] 2n −k −1 n −k q q(k−1)n.
(96) 26 Proof.
Since Fn,n(q, t) = q(n 2), Theorem 14 holds for k = n. If 1 ≤k < n, we start with Theorem 13 and then use induction on n: q(n 2)Fn,k(q, q−1) = q(n 2)q−(n−k 2 ) n−k X r=1 q(n−k 2 )Fn−k,r(q, q−1)q(k 2)−(n−k) r + k −1 r q (97) = q(n 2)+(k 2)−(n−k)q−(n−k 2 ) n−k X r=1 r + k −1 r q [r] [n −k] 2(n −k) −r −1 n −k −r q q(r−1)(n−k) (98) = q(k−1)n n−k X r=1 r + k −1 r q [r] [n −k] 2(n −k) −2 −(r −1) n −k −1 −(r −1) q q(r−1)(n−k) (99) = q(k−1)n [k] [n −k] n−k−1 X u=0 k + u u q qu(n−k) 2(n −k) −2 −u n −k −1 −u q .
(100) Using (27) we can write the right-hand side of (100) as q(k−1)n [k] [n −k] 1 (zqn−k)k+1 1 (z)n−k zn−k−1 = q(k−1)n [k] [n −k] 1 (z)n+1 zn−k−1 (101) = q(k−1)n [k] [n −k] n + n −k −1 n q .
(102) 2 Corollary 7 q(n 2)Fn(q, 1/q) = 1 [n + 1] 2n n q .
(103) Proof.
N. Loehr has pointed out that we can use Fn+1,1(q, t) = tnFn(q, t), (104) which by Theorem 14 implies q(n+1 2 )Fn+1,1(q, 1/q) = [n + 1] 2(n + 1) −2 n + 1 −1 q = [n + 1] 2n n q (105) = q(n+1 2 )−nFn(q, 1/q) = q(n 2)Fn(q, 1/q).
2 The Symmetry Problem and the dinv Statistic From its definition, it is easy to show Cn(q, t) = Cn(t, q), since the arm and leg values for µ equal the leg and arm values for µ′, respectively, which implies q and t are interchanged when comparing terms in (76) corresponding to µ and µ′. This also follows from the theorem that Cn(q, t) = H(DHǫ n; q, t). Thus we have X π∈L+ n,n qarea(π)tbounce(π) = X π∈L+ n,n qbounce(π)tarea(π), (106) 27 a surprising statement in view of the apparent dissimilarity of the area and bounce statistics.
At present there is no other known way to prove (106) other than as a corollary of Theorem 12.
Open Problem 1 Prove (106) by exhibiting a bijection on Dyck paths which interchanges area and bounce.
A solution to Problem 1 should lead to a deeper understanding of the combinatorics of DHn.
We now give a combinatorial proof from [Hag03] of a very special case of (106), by showing that the marginal distributions of area and bounce are the same, i.e. Fn(q, 1) = Fn(1, q).
Theorem 15 X π∈L+ n,n qarea(π) = X π∈L+ n,n qbounce(π).
(107) Proof.
Given π ∈L+ n,n, let a1a2 · · · an denote the sequence whose ith element is the ith co-ordinate of the area vector of π, i.e.
the length of the ith row (from the bottom) of π.
A moment’s thought shows that such a sequence is characterized by the property that it begins with zero, consists of n nonnegative integers, and has no 2-ascents, i.e. values of i for which ai+1 > ai + 1. To construct such a sequence we begin with an arbitrary multiset of row lengths, say {0α11α2 · · · (b −1)αb} and then choose a multiset permutation τ of {0α11α2} which begins with 0 in α1−1+α2 α2 ways. Next we will insert the α3 twos into τ, the requirement of having no 2-ascents translating into having no consecutive 02 pairs. This means the number of ways to do this is α2−1+α3 α3 , independent of the choice of τ. The formula Fn(q, 1) = n X b=1 X α1+...+αb=n αi>0 q Pb i=2 αi(i−1) b−1 Y i=1 αi −1 + αi+1 αi+1 (108) follows, since the product above counts the number of Dyck paths with a specified multiset of row lengths, and the power of q is the common value of area for all these paths. Comparing (108) with the q = 1, t = q case of (91) completes the proof.
2 There is another pair of statistics for the q, t-Catalan discovered by M. Haiman [Hai00].
It involves pairing area with a different statistic we call dinv, for “diagonal inversion” or “d-inversion”. It is defined, with ai the length of the ith row from the bottom, as follows.
Definition 4 Let π ∈L+ n,n. Let dinv(π) = |{(i, j) : 1 ≤i < j ≤n ai = aj}| + |{(i, j) : 1 ≤i < j ≤n ai = aj + 1}|.
In words, dinv(π) is the number of pairs of rows of π of the same length, or which differ by one in length, with the longer row below the shorter. For example, for the path on the left in Figure 9, with row lengths on the right, the inversion pairs (i, j) are (3, 7), (4, 7), (5, 7), (6, 8) (corre-sponding to rows which differ by one in length) and (2, 7), (3, 4), (3, 5), (3, 8), (4, 5), (4, 8), (5, 8) (corresponding to pairs of rows of the same length), thus dinv = 11. We call inversion pairs between rows of the same length “equal-length” inversions, and the other kind “offset-length” inversions.
28 2 2 2 2 1 0 1 3 Figure 9: A path π with row lengths to the right, and the image ζ(π).
Theorem 16 X π∈L+ n,n qdinv(π)tarea(π) = X π∈L+ n,n qarea(π)tbounce(π).
(109) Proof.
We will describe a bijective map ζ on Dyck paths with the property that dinv(π) = area(ζ(π)) (110) area(π) = bounce(ζ(π)).
Say b −1 is the length of the longest row of π. The lengths of the bounce steps of ζ will be α1, . . . , αb, where αi is the number of rows of length i −1 in π. To construct the actual path ζ, place a pen at the lattice point (α1, α1 + α2) (the second peak of the bounce path of ζ).
Start at the end of the area sequence and travel left. Whenever you encounter a 1, trace a South step with your pen. Whenever you encounter a 0, trace a West step. Skip over all other numbers. Your pen will end up at the top of the first peak of ζ. Now go back to the end of the area sequence, and place your pen at the top of the third peak. Traverse the area sequence again from right to left, but this time whenever you encounter a 2 trace out a South step, and whenever you encounter a 1, trace out a West step. Skip over any other numbers. Your pen will end up at the top of the second peak of ζ. Continue at the top of the fourth peak looking at how the rows of length 3 and 2 are interleaved, etc.. See Figure 9.
It is easy to see this map is a bijection, since given ζ, from the bounce path we can determine the multiset of row lengths of π. We can then build up the area sequence of π just as in the proof of Theorem 15. From the portion of the path between the first and second peaks we can see how to interleave the rows of lengths 0 and 1, and then we can insert the rows of length 2 into the area sequence, etc.
Note that when tracing out the part of ζ between the first and second peaks, whenever we encounter a 0 and trace out a West step, the number of area squares directly below this West step and above the bounce path of ζ equals the number of 1’s to the left of this 0 in the area sequence, which is the number of offset-length inversion pairs involving the corresponding row of length 0. Since the area below the bounce path clearly counts the total number of equal-length inversions, it follows that dinv(π) = area(ζ(π)).
29 Now by direct calculation, bounce(ζ) = n −α1 + n −α1 −α2 + . . . n −α1 −. . . −αb−1 (111) = (α2 + . . . + αb) + . . . + (αb) = b−1 X i=1 iαi+1 = area(π).
2 Remark 5 The construction of the bounce path for a Dyck path occurs in an independent con-text, in work of Andrews, Krattenthaler, Orsina and Papi [AKOP02] on the enumeration of ad-nilpotent ideals of a Borel subalgebra of sl(n + 1, C). They prove the number of times a given nilpotent ideal needs to be bracketed with itself to become zero equals the number of bounces of the bounce path of a certain Dyck path associated to the ideal. Another of their results is a bijective map on Dyck paths which sends a path with b bounces to a path whose longest row is of length b −1. The ζ map above is just the inverse of their map. Because they only considered the number of bounces, and not the bounce statistic per se, they did not notice any connection between Cn(q, t) and their construction.
Theorem 12 now implies Corollary 8 Cn(q, t) = X π∈L+ n,n qdinv(π)tarea(π).
(112) We also have Fn,k(q, t) = X π∈L+ n,n π has exactly k rows of length 0 qdinv(π)tarea(π), (113) since under the ζ map, paths with k rows of length 0 correspond to paths whose first bounce step is of length k.
Remark 6 N. Loehr has noted that if one can find a map which fixes area and sends bounce to dinv, by combining this with the ζ map one would have a map which interchanges area and dinv, solving Problem 1.
Exercise 6 Let Gn,k(q, t) = X π∈L+ n,n π has exactly k rows of length 0 qdinv(π)tarea(π).
(114) Without referencing any results on the bounce statistic, prove combinatorially that Gn,k(q, t) = tn−kq(k 2) n−k X r=0 r + k −1 r q Gn−k,r(q, t).
(115) 30 Exercise 7 Haiman’s conjectured statistics for Cn(q, t) actually involved a different description of dinv. Let λ(π) denote the partition consisting of the n 2 −area(π) squares above π but inside the n × n square. (This is the Ferrers graph of a partition in the so-called English convention, which is obtained from the French convention of Figure 1 by reflecting the graph about the x-axis.
In this convention, the leg l(s) of a square s is defined as the number of squares of λ below s in the column and above the lower border π, and the arm a(s) as the number of squares of λ to the right and in the row.) Then Haiman’s original version of dinv was the number of cells s of λ for which l(s) ≤a(s) ≤l(s) + 1.
(116) Prove this definition of dinv is equivalent to Definition 4.
q-Lagrange Inversion q-Lagrange inversion is useful when analyzing the special case t = 1 of F(DHn; q, t). In this section we derive a general q-Lagrange inversion theorem based on work of Garsia and Haiman.
We will be working in the ring of formal power series, and we begin with a result of Garsia [Gar81].
Theorem 17 If (F ◦q G)(z) = X n fnG(z)G(qz) · · · G(qn−1z), (117) where F = P n fnzn, then for F and G without constant term, F ◦q G = z and G ◦q−1 F = z (118) are equivalent to each other and also to (Φ ◦q−1 F) ◦q G = Φ = (Φ ◦q G) ◦q−1 F for all Φ.
(119) Given π ∈L+ n,n, let β(π) = β1(π)β2(π) · · · denote the partition consisting of the vertical step lengths of π (i.e. the lengths of the maximal blocks of consecutive 0’s in σ(π)), arranged in nonincreasing order. For example, for the path on the left in Figure 9 we have β = (3, 2, 2, 1).
By convention we set β(∅) = ∅. Define H(z) via the equation 1/H(−z) := P∞ k=0 ekzk. Using Theorem 17, Haiman [Hai94, pp. 47-48] derived the following.
Theorem 18 There is a unique solution h∗ n(q), n ≥0 to the equation ∞ X k=0 ekzk = ∞ X n=0 q−(n 2)h∗ n(q)znH(−q−1z)H(−q−2z) · · · H(−q−nz), h∗ 0(q) = 1.
(120) For n > 0, h∗ n(q) has the explicit expression h∗ n(q) = X π∈L+ n,n qarea(π)eβ(π).
(121) For example, we have h∗ 3(q) = q3e3 + q2e2,1 + 2qe2,1 + e13.
(122) 31 We now derive a slight generalization of Theorem 18 which stratifies Dyck paths according to the length of their first bounce step.
Theorem 19 Let ck, k ≥0 be a set of variables. Define h∗ n(c, q), n ≥0 via the equation ∞ X k=0 ekckzk = ∞ X n=0 q−(n 2)h∗ n(c, q)znH(−q−1z)H(−q−2z) · · · H(−q−nz), h∗ 0(cq) = c0.
(123) Then for n ≥0, h∗ n(c, q) has the explicit expression h∗ n(c, q) = n X k=0 ck X π∈L+ n,n(k) qarea(π)eβ(π).
(124) For example, we have h∗ 3(c, q) = q3e3c3 + (q2e2,1 + qe2,1)c2 + (qe2,1 + e13)c1.
(125) Proof.
Our proof follows Haiman’s proof of Theorem 18 closely. Set H∗(z, c; q) := P∞ n=0 h∗ n(c, q)zn, H∗(z; q) := P∞ n=0 h∗ n(q)zn, Φ = H∗(zq, c; q), F = zH(−z), and G = zH∗(qz; q). Replacing z by zq in (123) we see that Theorem 19 is equivalent to the statement ∞ X k=0 ekckqkzk = ∞ X n=0 qnh∗ n(c, q)zH(−z)zq−1H(−q−1z) · · · zq1−nH(−q1−nz) (126) = Φ ◦q−1 F.
On the other hand, Theorem 18 can be expressed as 1 H(−z) = ∞ X n=0 q−(n 2)h∗ n(q)znH(−z/q) · · · H(−z/qn), (127) or z = ∞ X n=0 q−(n 2)h∗ n(q)zn+1H(−z)H(−z/q) · · · H(−z/qn) (128) = ∞ X n=0 qnh∗ n(q) {zH(−z)} z q H(−z/q) · · · z qn H(−z/qn) = G ◦q−1 F.
(129) Thus, using Theorem 17, we have Φ = (Φ ◦q−1 F) ◦q G (130) = ( ∞ X k=0 ekµkqkzk) ◦q G.
32 Comparing coefficients of zn in (130) and simplifying we see that Theorem 19 is equivalent to the statement qnh∗ n(c, q) = n X k=0 q(k 2)+kekck X n1+...+nk=n−k ni≥0 qn−k k Y i=1 q(i−1)nih∗ ni(q).
(131) To prove (131) we use the “factorization of Dyck paths” as discussed in [Hai94]. This can be be represented pictorially as in Figure 10. The terms multiplied by ck correspond to π ∈L+ n,n(k).
The area of the parallelogram region whose left border is on the line y = x + i −1 is (i −1)ni, where ni is the length of the left border of the parallelogram. Using the fact that the path to the left of this parallelogram is in L+ ni,ni, (131) now becomes transparent.
2 k Figure 10: A Dyck path factored into smaller paths.
Letting ek = 1, cj = χ(j = k) and replacing q by q−1 and z by z/q in Theorem 19 we get the following.
Corollary 9 For 1 ≤k ≤n, zk = X n≥k q(n 2)−n+kFn,k(q−1, 1)zn(z)n.
(132) Theorem 19 is a q-analogue of the general Lagrange inversion formula [AAR99, p.629] f(x) = f(0) + ∞ X n=1 xn n!φ(x)n dn−1 dxn−1 (f ′(x)φn(x)) x=0 , (133) 33 where φ and f are analytic in a neighborhood of 0, with φ(0) ̸= 0. To see why, assume WLOG f(0) = 1, and set f(x) = P∞ k=0 ckxk and φ = 1 H(−x) = P∞ k=0 ekxk in (133) to get ∞ X k=1 ckxk = ∞ X n=1 xn n! H(−x)n dn−1 dxn−1 ∞ X k=1 kckxk−1( ∞ X m=0 emxm)n !
|x=0 (134) = ∞ X n=1 xn n! H(−x)n n X k=1 kck(n −1)!( ∞ X m=0 emxm)n|xn−k (135) = ∞ X n=1 xnH(−x)n n X k=1 ck k n X j1+j2+...+jn=n−k ji≥0 ej1ej2 · · · ejn (136) = ∞ X n=1 xnH(−x)n n X k=1 ck k n X α⊢n−k eα n k −ℓ(α), n1(α), n2(α), . . .
.
(137) The equivalence of (137) to the q = 1 case of Theorem 19 (with ck replaced by ck/ek) will follow if we can show that for any fixed α ⊢n −k, X π∈L+ n,n(k) β(π)−k=α 1 = k n n k −ℓ(α), n1(α), n2(α), . . .
, (138) where β −k is the partition obtained by removing one part of size k from β. See [Hag03] for an inductive proof of (138).
In [Hai94] Haiman includes a discussion of the connection of Theorem 18 to q-Lagrange in-version formulas of Andrews, Garsia, and Gessel [And75], [Gar81], [Ges80]. Further background on these formulas is contained in [Sta88]. Garsia and Haiman used q-Lagrange inversion to obtain the interesting identity ∇en|t=1 = X π∈L+ n,n qarea(π)eβ(π) (139) for the t = 1 case of the Frobenius series.
Garsia and Haiman were also able to obtain the t = 1/q case of F(DHn; q, t), which can be expressed as follows.
Theorem 20 q(n 2)F(DHn; q, 1/q) = 1 [n + 1]en(XY ), (140) where Y = {1, q, . . . , qn}. Equivalently, by the Cauchy identity (50), D q(n 2)F(DHn; q, 1/q), sλ E = 1 [n + 1]sλ′(1, q, . . . , qn).
(141) Note that by Theorem 7, the special case λ = 1n of (141) reduces to (93), the formula for MacMahon’s maj-statistic q-Catalan.
Open Problem 2 Find a q, t-version of the Lagrange inversion formula which will yield an identity for F(DHn; q, t), and which reduces to Theorem 19 when t = 1 and incorporates (140).
34 3 Parking Functions and the Hilbert Series Extension of the dinv Statistic Another beautiful corollary of Haiman’s formula for F(DHn; q, t) is that the dimension of DHn equals (n + 1)n−1, which is the number of parking functions on n cars. See the chapter in this volume on parking functions for more information on these important combinatorial objects.
In this chapter we will view parking functions geometrically, as a Dyck path π together with a placement of the numbers, or “cars”, 1 through n in the squares just to the right of N steps of π, with strict decrease down columns. See Figure 11.
2 3 8 7 5 1 4 6 Figure 11: A parking function P.
We now describe an extension of the dinv statistic to parking functions. Let Pn denote the set of all parking functions on n cars. Given P ∈Pn with associated Dyck path π = π(P), if car i is in row j we say occupant(j) = i. Let dinv(P) be the number of pairs (i, j), 1 ≤i < j ≤n such that dinv(P) = |{(i, j) : 1 ≤i < j ≤n, ai = aj, and occupant(i) < occupant(j)}| + |{(i, j) : 1 ≤i < j ≤n, ai = aj + 1, and occupant(i) > occupant(j)}|.
Thus dinv(P) is the number of pairs of rows of P of the same length, with the row above containing the larger car, or which differ by one in length, with the longer row below the shorter, and the longer row containing the larger car. For example, for the parking function in Figure 11, the inversion pairs (i, j) are (1, 7), (2, 7), (2, 8), (3, 4), (4, 8) and (5, 6), so dinv(P) = 6.
We define area(P) = area(π), and also define the reading word of P, denoted read(P), to be the permutation obtained by reading the cars along diagonals in a southwest direction, starting with the diagonal farthest from the line y = x, then working inwards. For example, the parking function in Figure 11 has area 9 and reading word 64781532.
Remark 7 Note that read(P) = n · · · 21 if and only if dinv(P) = dinv(π). We call this parking function the Maxdinv parking function for π, which we denote by Maxdinv(π).
35 Recall that Remark 4 implies H(DHn; q, t) = ⟨F(DHn; q, t), h1n⟩.
(142) In [HL05] N. Loehr and the author advance the following Conjectured combinatorial formula for the Hilbert series, which is still open.
Conjecture 1 H(DHn; q, t) = X P ∈Pn qdinv(P )tarea(P ).
(143) Conjecture 1 has been verified in Maple for n ≤11. The truth of the Conjecture when q = 1 follows easily from (139) and (142). Later in this section we will show (Corollary 11) that dinv has the same distribution as area over Pn, which implies the Conjecture is also true when t = 1.
An Explicit Formula Given τ ∈Sn, with descents at places i1 < i2 < . . . < ik, we call the first i1 letters of τ the first run of τ, the next i2 −i1 letters of τ the second run of τ, . . ., and the last n −ik letters of τ the (k + 1)st run of τ. For example, the runs of 58246137 are 58, 246 and 137. It will prove convenient to call element 0 the k+2nd run of τ. Let cars(τ) denote the set of parking functions whose cars in rows of length 0 consist of the elements of the (k + 1)st run of τ (in any order), whose cars in rows of length 1 consist of the elements of the kth run of τ (in any order), . . ., and whose cars in rows of length k consist of the elements of the first run of τ (in any order).
For example, the elements of cars(31254) are listed in Figure 12.
Let τ be as above, and let i be a given integer satisfying 1 ≤i ≤n. If τi is in the jth run of τ, we define wi(τ) to be the number of elements in the jth run which are larger than τi, plus the number of elements in the j +1st run which are smaller than τi. For example, if τ = 385924617, then the values of w1, w2, . . . , w9 are 1, 1, 3, 3, 3, 2, 1, 2, 1.
Theorem 21 Given τ ∈Sn, X P ∈cars(τ) qdinv(P )tarea(P ) = tmaj(τ) n Y i=1 [wi(τ)]q.
(144) Proof.
We will build up elements of cars(τ) by starting at the end of τ, where elements go in rows of length 0, and adding elements right to left. We define a partial parking function to be a Dyck path π ∈L+ m,m for some m, together with a placement of m distinct positive integers (not necessarily the integers 1 through m) to the right of the N steps of π, with strict decrease down columns. Say τ = 385924617 and we have just added car 9 to obtain a partial parking function A with cars 1 and 7 in rows of length 0, cars 2, 4 and 6 in rows of length 1, and car 9 in a row of length 2, as in the upper left grid of Figure 13. The rows with ∗’s to the right are rows above which we can insert a row of length 2 with car 5 in it and still have a partial parking function.
Note the number of starred rows equals w3(τ), and that in general wi(τ) can be defined as the number of ways to insert a row containing car τi into a partial parking function containing cars τi+1, . . . , τn, in rows of the appropriate length, and still obtain a partial parking function.
36 4 5 5 4 3 2 5 4 1 2 3 5 4 2 1 3 3 2 1 1 Figure 12: The elements of cars(31254).
Consider what happens to dinv as we insert the row with car 5 above a starred row to form A′. Pairs of rows which form inversions in A will also form inversions in A′. Furthermore the rows of length 0 in A, or of length 1 with a car larger than 5, cannot form inversions with car 5 no matter where it is inserted. However, a starred row will form an inversion with car 5 if and only if car 5 is in a row below it. It follows that if we weight insertions by qdinv, inserting car 5 at the various places will generate a factor of [wi(τ)] times the weight of A, as in Figure 13.
Finally note that for any path π corresponding to an element of cars(τ), maj(τ) = area(π).
2 By summing Theorem 21 over all τ ∈Sn we get the following.
Corollary 10 X P ∈Pn qdinv(P )tarea(P ) = X τ∈Sn tmaj(τ) n Y i=1 [wi(τ)]q.
(145) Open Problem 3 Prove X P ∈Pn qdinv(P )tarea(P ) (146) is symmetric in q, t.
Remark 8 Beyond merely proving the symmetry of (146), one could hope to find a bijective proof. It is interesting to note that by (144) the symmetry in q, t when one variable equals 0 reduces to the fact that both inv and maj are Mahonian. Hence any bijective proof of symmetry may have to involve generalizing Foata’s bijective transformation of maj into inv.
37 1 2 6 9 5 4 7 2 1 6 9 4 7 dinv = dinv (A) + 1 A 1 2 6 9 4 5 7 dinv = dinv (A) + 0 1 2 5 6 9 4 7 dinv = dinv (A) + 2 Figure 13: Partial parking functions occurring in the proof of Theorem 21.
The Statistic area′ By a diagonal labeling of a Dyck path π ∈L+ n,n we mean a placement of the numbers 1 through n in the squares on the main diagonal y = x in such a way that for every consecutive EN pair of steps of π, the number in the same column as the E step is smaller than the number in the same row as the N step. Let An denote the set of pairs (A, π) where A is a diagonal labeling of π ∈L+ n,n. Given such a pair (A, π), we let area′(A, π) denote the number of area squares x of π for which the number on the diagonal in the same column as x is smaller than the number in the same row as x. Also define bounce(A, π) = bounce(π).
The following result appears in [HL05].
Theorem 22 There is a bijection between Pn and An which sends (dinv, area) to (area′, bounce).
Proof.
Given P ∈Pn with associated path π, we begin to construct a pair (A, ζ(π)) by first letting ζ(π) be the same path formed by the ζ map from the proof of Theorem 16. The length α1 of the first bounce of ζ is the number of rows of π of length 0, etc.. Next place the cars which occur in P in the rows of length 0 in the lowest α1 diagonal squares of ζ, in such a way that the 38 order in which they occur, reading top to bottom, in P is the same as the order in which they occur, reading top to bottom, in ζ. Then place the cars which occur in P in the rows of length 1 in the next α2 diagonal squares of ζ, in such a way that the order in which they occur, reading top to bottom, in P is the same as the order in which they occur, reading top to bottom, in ζ.
Continue in this way until all the diagonal squares are filled, resulting in the pair (A, ζ). See Figure 14 for an example.
The properties of the ζ map immediately imply area(π) = bounce(A, ζ) and that (A, ζ) ∈An.
The reader will have no trouble showing that the equation dinv(P) = area′(A, ζ) is also implicit in the proof of Theorem 16.
2 2 2 2 2 1 0 1 3 5 2 6 1 7 8 2 4 3 3 4 5 8 1 7 6 x x x x x ( dinv, area ) = ( 5, 13) ( area , bounce ) = ( 5, 13 ) Figure 14: The map in the proof of Theorem 22. Squares contributing to area′ are marked with x’s.
Remark 9 Drew Armstrong [Arm13] has found an interpretation for the area′ statistic, as well as the bounce and dinv statistics, in terms of hyperplane arrangements. See also [AR12].
The pmaj Statistic We now define a statistic on parking functions called pmaj, due to Loehr and Remmel [LR04], [Loe05a] which generalizes the bounce statistic. Given P ∈Pn, we define the pmaj-parking order, denoted β(P), by the following procedure. Let Ci = Ci(P) denote the set of cars in column i of P, and let β1 be the largest car in C1. We begin by parking car β1 in spot 1. Next we perform the “dragnet” operation, which takes all the cars in C1 \ {β1} and combines them with C2 to form C′ 2. Let β2 be the largest car in C′ 2 which is smaller then β1. If there is no such car, let β2 be the largest car in C′ 2. Park car β2 in spot 2 and then iterate this procedure.
Assuming we have just parked car βi−1 in spot i −1, 3 ≤i < n, we let C′ i = C′ i−1 \ {βi−1} and let βi be the largest car in C′ i which is smaller than βi−1, if any, while otherwise βi is the largest car in C′ i. For the example in Figure 15, we have C1 = {5}, C2 = {1, 7}, C3 = {}, etc. and C′ 2 = {1, 7}, C′ 3 = {7}, C′ 4 = {2, 4, 6}, C′ 5 = {2, 3, 4}, etc., with β = 51764328.
Now let rev(β(P)) = (βn, βn−1, . . . , β1) and define pmaj(P) = maj(rev(β(P))).
For the parking function of Figure 15 we have rev(β) = 82346715 and pmaj = 1 + 6 = 7.
Given π ∈L+ n,n, it is easy to see that if P is the parking function for π obtained by placing car i in row 39 5 7 1 2 4 6 3 8 5 6 4 3 2 8 β = 7 1 Figure 15: A parking function P with pmaj parking order β = 51764328.
i for 1 ≤i ≤n, then pmaj(P) = bounce(π). We call this parking function the primary pmaj parking function for π.
We now describe a bijection Γ from Pn to Pn from [LR04] which sends (area, pmaj) → (dinv, area).
The crucial observation behind it is this.
Fix γ ∈Sn and consider the set of parking functions which satisfy rev(β(P)) = γ. We can build up this set recursively by first forming a partial parking function consisting of car γn in column 1. If γn−1 < γn, then we can form a partial parking function consisting of two cars whose pmaj parking order is γnγn−1 in two ways. We can either have both cars γn and γn−1 in column 1, or car γn in column 1 and car γn−1 in column 2. If γn−1 > γn, then we must have car γn in column 1 and car γn−1 in column 2. In the case where γn−1 < γn, there were two choices for columns to insert car γn−1 into, corresponding to the fact that wn−1(γ) = 2. When γn−1 > γn, there was only one choice for the column to insert γn−1 into, and correspondingly wn−1(γ) = 1.
More generally, say we have a partial parking function consisting of cars in the set {γn, . . . , γi+1} whose pmaj parking order is γn · · · γi+2γi+1. It is easy to see that the number of ways to insert car γi into this so the new partial parking function has pmaj parking order γn · · · γi+1γi is exactly wi(γ). Furthermore, as you insert car γi into columns n −i + 1, n −i, . . . , n −i −wi(γ) + 2 the area of the partial parking function increases by 1 each time. It follows that X P ∈Pn qarea(P )tpmaj(P ) = X γ∈Sn tmaj(γ) n Y i=1 [wi(γ)]q.
(147) Moreover, we can identify the values of (area, pmaj) for individual parking functions by con-sidering permutations γ ∈Sn and corresponding n-tuples (u1, . . . , un) with 0 ≤ui < wi(γ) for 1 ≤i ≤n. (Note un always equals 0). Then maj(γ) = pmaj(P), and u1 + . . . + un = area(P).
(For those familiar with the description of parking functions in terms of preference functions, we have f(βn+1−i) = n + 1 −i −ui for 1 ≤i ≤n.) Now given such a pair γ ∈Sn and corresponding n-tuple (u1, . . . , un), from the proof of Theorem 21 we can build up a parking function Q recursively by inserting cars γn, γn−1, . . .
40 one at a time, where for each j the insertion of car γj adds uj to dinv(Q). Thus we end up with a bijection Γ : P 7→Q with (area(P), pmaj(P)) = (dinv(Q), area(Q)). The top of Figure 16 gives the various partial parking functions in the construction of P, and after those are the various partial parking functions in the construction of Q, for γ = 563412 and (u1, . . . , u6) = (2, 0, 1, 0, 1, 0).
2 2 1 2 1 4 2 1 4 3 6 4 3 2 1 6 5 4 3 2 1 4 2 1 2 1 2 4 3 2 1 6 4 3 2 1 6 5 4 3 2 1 P Q Figure 16: The recursive construction of the P and Q parking functions in the Γ correspondence for γ = 563412 and u = (2, 0, 1, 0, 1, 0).
Corollary 11 The marginal distributions of pmaj, area, and dinv over Pn are all the same, i.e.
X P ∈Pn qpmaj(P ) = X P ∈Pn qarea(P ) = X P ∈Pn qdinv(P ).
(148) Exercise 8 Notice that in Figure 16, the final correspondence is between parking functions which equal the primary pmaj and Maxdinv parking functions for their respective paths. Show that this is true in general, i.e. that when P equals the primary pmaj parking function for π then the bijection Γ reduces to the inverse of the bijection ζ from the proof of Theorem 16.
The Cyclic-Shift Operation Given S ⊆{1, . . . , n}, let Pn,S denote the set of parking functions for which C1(P) = S. If x ∈{1, 2, . . . , n}, define CY Cn(x) = ( x + 1 if x < n 1 if x = n.
(149) 41 For any set S ⊆{1, 2, . . . , n}, let CY Cn(S) = {CY Cn(x) : x ∈S}. Assume S = {s1 < s2 < · · · < sk} with sk < n. Given P ∈Pn,S, define the cyclic-shift of P, denoted CY Cn(P), to be the parking function obtained by replacing Ci, the cars in column i of P, with CY Cn(Ci), for each 1 ≤i ≤n. Note that the column of P containing car n will have to be sorted, with car 1 moved to the bottom of the column. The map CY Cn(P) is undefined if car n is in column 1.
See the top portion of Figure 17 for an example.
2 5 1 3 4 5 4 3 2 1 1 3 2 5 4 4 3 2 1 P CYC(P) CYC2(P) R(P) Figure 17: The map R(P).
Proposition 4 [Loe05a] Suppose P ∈Pn,S with S = {s1 < s2 < · · · < sk}, sk < n. Then pmaj(P) = pmaj(CY C(P)) + 1.
(150) Proof.
Imagine adding a second coordinate to each car, with car i initially represented by (i, i). If we list the second coordinates of each car as they occur in the pmaj parking order for P, by definition we get the sequence β1(P)β2(P) · · · βn(P). We now perform the cyclic-shift operation to obtain CY C(P), but when doing so we operate only on the first coordinates of each car, leaving the second coordinates unchanged. The reader will have no trouble verifying that if we now list the second coordinates of each car as they occur in the pmaj parking order for CY C(P), we again get the sequence β1(P)β2(P) · · · βn(P). It follows that the pmaj parking order of CY C(P) can be obtained by starting with the pmaj parking order β of P and performing the cyclic-shift operation on each element of β individually. See Figure 18.
Say n occurs in the permutation rev(β(P)) in spot j. Note that we must have j < n, or otherwise car n would be in column 1 of P. Clearly when we perform the cyclic-shift operation on the individual elements of the permutation rev(β(P)) the descent set will remain the same, except that the descent at j is now replaced by a descent at j −1 if j > 1, or is removed if j = 1.
In any case the value of the major index of rev(β(P)) is decreased by 1.
2 42 (5,5) (2,2) (3,3) (1,1) (4,4) (1,5) (3,2) (4,3) (2,1) (5,4) (1,1) (2,2) (5,5) (4,3) (3,2) (2,1) (1,5) (5,4) (4,4) (3,3) Figure 18: The cyclic-shift operation and the pmaj parking order.
Using Proposition 4, Loehr derives the following recurrence.
Theorem 23 Let n ≥1 and S = {s1 < · · · < sk} ⊆{1, . . . , n}. Set Pn,S(q, t) = X P ∈Pn,S qarea(P )tpmaj(P ).
(151) Then Pn,S(q, t) = qk−1tn−sk X T⊆{1,...,n}\S Pn−1,CY C n−sk n (S∪T{sk})(q, t), (152) with the initial conditions Pn,∅(q, t) = 0 for all n and P1,{1}(q, t) = 1.
Proof.
For P ∈Pn,S, let Q = CY Cn−sk n (P). Then pmaj(Q) + n −sk = pmaj(P) (153) area(Q) = area(P).
Since car n is in the first column of Q, in the pmaj parking order for Q car n is in spot 1. By definition, the dragnet operation will then combine the remaining cars in column 1 of Q with the cars in column 2 of Q. Now car n being in spot 1 translates into car n being at the end of rev(β(Q)), which means n −1 is not in the descent set of rev(β(Q)). Thus if we define R(P) to be the element of Pn−1 obtained by parking car n in spot 1, performing the dragnet, then truncating column 1 and spot 1 as in Figure 17, we have pmaj(R(P)) = pmaj(P) −(n −sk).
(154) Furthermore, performing the dragnet leaves the number of area cells in columns 2, 3, . . . , n of Q unchanged but eliminates the k −1 area cells in column 1 of Q. Thus area(R(P)) = area(P) −k + 1 (155) and the recursion now follows easily.
2 Loehr also derives the following compact formula for Pn,S when t = 1/q.
43 Theorem 24 For n ≥0 and S = {s1 < · · · < sk} ⊆{1, . . . , n}, q(n 2)Pn,S(1/q, q) = qn−k[n]n−k−1 X x∈S qn−x.
(156) Proof.
Our proof is, for the most part, taken from [Loe05a]. If k = n, Pn,S(1/q, q) = Pn,{1,...,n}(1/q, q) = q−(n 2), (157) while the right-hand-side of (156) equals [n]−1[n] = 1. Thus (156) holds for k = n. It also holds trivially for n = 0 and n = 1. So assume n > 1 and 0 < k < n. From (152), Pn,S(1/q, q) = qn+1−sk−k X T⊆{1,...,n}\S Pn−1,CY C n−sk n (S∪T{sk})(1/q, q) (158) = qn+1−sk−k n−k X j=0 X T⊆{1,...,n}\S |T|=j Pn−1,CY C n−sk n (S∪T{sk})(1/q, q).
The summand when j = n −k equals Pn−1,{1,...,n−1}(1/q, q) = q−(n−1 2 ).
(159) For 0 ≤j < n −k, by induction the summand equals qn−1−(j+k−1)−(n−1 2 )[n −1]n−1−(j+k−1)−1 X x∈CY C n−sk n (S∪T{sk}) qn−1−x, (160) since j + k −1 = |CY Cn−sk n (S ∪T \ {sk})|. Plugging (160) into (158) and reversing summation we now have q−2n+sk+kq(n 2)Pn,S(1/q, q) = 1 + n−k−1 X j=0 qn−k−j[n −1]n−k−j−1 n−1 X x=1 qn−1−x (161) × X T χ(T ⊆{1, . . . , n} \ S, |T| = j, CY Csk−n n (x) ∈S ∪T \ {sk}).
To compute the inner sum over T above, we consider two cases.
1. x = n−(sk −si) for some i ≤k. Since x < n, this implies i < k, and since CY Csk−n n (x) = si, we have CY Csk−n n (x) ∈S ∪T \ {sk}. Thus the inner sum above equals the number of j-element subsets of {1, . . . , n} \ S, or n−k j .
2. x ̸= n −(sk −si) for all i ≤k. By Exercise 9 below, the inner sum over T in (161) equals n−k−1 j−1 .
44 Applying the above analysis to (161) we now have q−2n+sk+kq(n 2)Pn,S(1/q, q) = 1 + n−k−1 X j=0 qn−k−j[n −1]n−k−j−1 (162) × X x satisfies (1) n −k −1 j + n −k −1 j −1 qn−1−x + n−k−1 X j=0 qn−k−j[n −1]n−k−j−1 X x satisfies (2) n −k −1 j −1 qn−1−x.
Now x satisfies (1) if and only if n −1 −x = sk −si −1 for some i < k, and so q−2n+sk+kq(n 2)Pn,S(1/q, q) = 1 + n−k−1 X j=0 qn−k−j[n −1]n−k−j n −k −1 j −1 (163) + n−k−1 X j=0 qn−k−j−1[n −1]n−k−j−1 n −k −1 j k−1 X i=1 qsk−si = n−k−1 X m=0 (m=j−1) n −k −1 m (q[n −1])n−k−m−1 + k−1 X i=1 qsk−si n−k−1 X j=0 (q[n −1])n−k−j−1 n −k −1 j = (1 + q[n −1])n−k−1 + k−1 X i=1 qsk−si(1 + q[n −1])n−k−1 = [n]n−k−1(1 + k−1 X i=1 qsk−si).
Thus q(n 2)Pn,S(1/q, q) = qn−k[n]n−k−1 X x∈S qsk−x+n−sk.
(164) 2 Exercise 9 Show that if x ̸= n −(sk −si) for all i ≤k, the inner sum over T in (161) equals n−k−1 j−1 .
As a corollary of his formula for F(DHn; q, t), Haiman proves a Conjecture he attributes in [Hai94] to Stanley, namely that q(n 2)H(DHn; 1/q, q) = [n + 1]n−1.
(165) Theorem 24 and (165) together imply Conjecture 1 is true when t = q and q = 1/q. To see why, first observe that Pn+1,{n+1}(q, t) = X P ∈Pn qarea(P )tpmaj(P ).
(166) 45 Hence by Theorem 24, q(n 2) X P ∈Pn q−area(P )qpmaj(P ) = q(n 2)Pn+1,{n+1}(1/q, q) (167) = q−nq(n+1 2 )Pn+1,{n+1}(1/q, q) = q−nqn[n + 1]n−1q0 = [n + 1]n−1.
The main impediment to proving Conjecture 1 seems to be the lack of a recursive decompo-sition of the Hilbert series along the lines of (87).
Tesler Matrices For an n × n upper triangular matrix, we define the jth hook sum, where 1 ≤j ≤n, to be the the sum of all the entries in the jth row of the matrix, minus the sum of all the entries in the jth column strictly above the diagonal. A Tesler matrix of order n is an n × n upper-triangular matrix of nonnegative integers, such that all the hook sums equal 1. Let Tes(n) denote the set of Tesler matrices of order n. For example, the elements of Tes(3) are 1 0 0 0 1 0 0 0 1 , 1 0 0 0 0 1 0 0 2 , 0 1 0 0 2 0 0 0 1 , 0 1 0 0 1 1 0 0 2 , (168) 0 1 0 0 0 2 0 0 3 , 0 0 1 0 1 0 0 0 2 , 0 0 1 0 0 1 0 0 3 .
Let [k]q,t = (tk −qk)/(t −q) denote the q, t-analog of the integer k, and recall that M = (1 −q)(1 −t). To each Tesler matrix C we associate the weight wt(C) = (−M)pos(C)−n Y cij>0 [cij]q,t, where pos(C) is the number of positive entries in C. For example, the weight of 0 1 0 0 1 1 0 0 2 is (t + q)(−M) = −(t + q)(1 −q)(1 −t).
Using the theory of Macdonald polynomials and (79), Haglund [Hag11] proved the following result, which gives a formula for H(DHn; q, t) in terms of Tesler matrices.
Theorem 25 [Hag11] H(DHn; q, t) = X C∈Tes(n) wt(C).
(169) 46 Example 3 When n = 3, the terms in (168), with weights, give H(DH3; q, t) = 1 + (t + q) + (t + q) −(1 −q)(1 −t)(t + q)+ (170) (t + q)(t2 + tq + q2) + (t + q) + (t2 + tq + q2).
Note that Formula (169) is clearly a polynomial, and clearly symmetric in q, t. It gives a possible way of attacking Conjecture 1 without the use of symmetric function theory, since in principle one could figure out how to cancel the negative terms in the right-hand-side of (169), leaving a positive expression as in Conjecture 1. P. Levande [Lev11], [Lev12] has shown how to do this cancellation when t = 1 and when t = 0.
4 The q, t-Schr¨ oder Polynomial The Schr¨ oder Bounce and Area Statistics In this section we develop the theory of the q, t-Schr¨ oder polynomial, which gives a combinatorial interpretation, in terms of statistics on Schr¨ oder lattice paths, for the coefficient of a Schur hook shape in F(DHn; q, t). A Schr¨ oder path is a lattice path from (0, 0) to (n, n) consisting of N(0, 1), E(1, 0) and diagonal D(1, 1) steps which never goes below the line y = x. We let L+ n,n,d denote the set of Schr¨ oder lattice paths consisting of n −d N steps, n −d E steps, and d D steps. We refer to a triangle whose vertex set is of the form {(i, j), (i+1, j), (i+1, j +1)} for some (i, j) as a “lower triangle”, and define the area of a Schr¨ oder path π to be the number of lower triangles below π and above the line y = x. Note that if π has no D steps, then the Schr¨ oder definition of area agrees with the definition of the area of a Catalan path. We let ai(π) denote the length of the ith row, i.e., the number of lower triangles between the path and the diagonal in the ith row from the bottom of π, so area(π) = Pn i=1 ai(π).
Given π ∈L+ n,n,d, let σ(π) be the word of 0’s, 1’s and 2’s obtained in the following way.
Initialize σ to be the empty string, then start at (0, 0) and travel along π to (n, n), adding a 0, 1, or 2 to the end of σ(π) when we encounter a N, D, or E step, respectively, of π. (If π is a Dyck path, then this definition of σ(π) is the same as the previous definition from Section 1, except that we end up with a word of 0’s and 2’s instead of 0’s and 1’s. Since all our applications involving σ(π) depend only on the relative order of the elements of σ(π), this change is only a superficial one.) We define the statistic bounce(π) by means of the following algorithm.
Algorithm 2 1. First remove all D steps from π, and collapse to obtain a Catalan path Γ(π). More precisely, let Γ(π) be the Catalan path for which σ(Γ(π)) equals σ(π) with all 1’s removed. Recall the ith peak of Γ(π) is the lattice point where the bounce path for Γ(π) switches direction from N to E for the ith time. The lattice point at the beginning of the corresponding E step of π is called the ith peak of π.
2. For each D step x of π, let nump(x) be the number of peaks of π below x. Then define bounce(π) = bounce(Γ(π)) + X x nump(x), (171) where the sum is over all D steps of π. For example, if π is the Schr¨ oder path on the left in Figure 19, with Γ(π) on the right, then bounce(π) = (3 + 1) + (0 + 1 + 1 + 2) = 8. Note that if π has no D steps, this definition of bounce(π) agrees with the previous definition from Section 2.
47 Figure 19: On the left, a Schr¨ oder path π with the peaks marked by large dots. On the right is Γ(π) and its bounce path and peaks.
We call the vector whose ith coordinate is the length of the ith bounce step of Γ(π) the bounce vector of π. Say Γ(π) has b bounce steps, and call the set of rows of π between peaks i and i + 1 section i of π for 1 ≤i < b. In addition we call section 0 the set of rows below peak 1, and section b the set of rows above peak b. If π has βi D steps in section i, 0 ≤i ≤b, we refer to (β0, β1, . . . , βb) as the shift vector of π.
For example, the path on the left in Figure 19 has bounce vector (2, 2, 1) and shift vector (1, 2, 1, 0). We refer to the portion of σ(π) corresponding to the ith section of π as the ith section of σ(π).
Given n, d ∈N, we define the q, t-Schr¨ oder polynomial Sn,d(q, t) as follows.
Sn,d(q, t) = X π∈L+ n,n,d qarea(π)tbounce(π).
(172) These polynomials were introduced by Egge, Haglund, Killpatrick and Kremer [EHKK03]. They Conjectured the following result, which was subsequently proved by Haglund using plethystic results involving Macdonald polynomials [Hag04].
Theorem 26 For all 0 ≤d ≤n, Sn,d(q, t) = ⟨F(DHn; q, t), en−dhd⟩.
(173) Since Sn,0(q, t) = Fn(q, t) = Cn(q, t), (174) the d = 0 case of Theorem 26 reduces to Theorem 12.
48 Let ˜ L+ n,n,d denote the set of paths π which are in L+ n,n,d and also have no D step above the highest N step, i.e. no 1’s in σ(π) after the rightmost 0. Define ˜ Sn,d(q, t) = X π∈˜ L+ n,n,d qarea(π)tbounce(π).
(175) Then we have Theorem 27 Theorem 26 is equivalent to the statement that for all 0 ≤d ≤n −1, ˜ Sn,d(q, t) = F(DHn; q, t), sd+1,1n−d−1 .
(176) Proof.
Given π ∈˜ L+ n,n,d, we can map π to a path α(π) ∈L+ n,n,d+1 by replacing the highest N step and the following E step of π by a D step. By Exercise 10 below, this map leaves area and bounce unchanged. Conversely, if α ∈L+ n,n,d has a D step above the highest N step, we can map it to a path π ∈˜ L+ n,n,d−1 in an area and bounce preserving fashion by changing the highest D step to a NE pair. It follows that for 1 ≤d ≤n, Sn,d(q, t) = X π∈˜ L+ n,n,d qarea(π)tbounce(π) + X π∈˜ L+ n,n,d−1 qarea(π)tbounce(π) (177) = ˜ Sn,d(q, t) + ˜ Sn,d−1(q, t).
Since Sn,0(q, t) = ˜ Sn,0(q, t), en−dhd = sd+1,1n−d−1 + sd,1n−d for 0 < d ≤n −1, and Sn,n(q, t) = 1 = ˜ Sn,n−1(q, t), the result follows by a simple inductive argument.
2 Exercise 10 Given π and α(π) in the proof of Theorem 27, show that area(π) = area(α(π)) (178) bounce(π) = bounce(α(π)).
(179) Define q, t-analogues of the big Schr¨ oder numbers rn and little Schr¨ oder numbers ˜ rn as follows.
rn(q, t) = n X d=0 Sn,d(q, t) (180) ˜ rn(q, t) = n−1 X d=0 ˜ Sn,d(q, t).
(181) The numbers rn(1, 1) count the total number of Schr¨ oder paths from (0, 0) to (n, n). The ˜ rn(1, 1) are known to count many different objects [Sta99, p.178], including the number of Schr¨ oder paths from (0, 0) to (n, n) which have no D steps on the line y = x. From our comments above we have the simple identity rn(q, t) = 2˜ rn(q, t), and using Haiman’s formula for F(DHn; q, t) we get the polynomial identities n X d=0 wdSn,d(q, t) = X µ⊢n Tµ Q x∈µ(w + qa′tl′)MΠµBµ wµ (182) n−1 X d=0 wd ˜ Sn,d(q, t) = X µ⊢n Tµ Q x∈µ, x̸=(0,0)(w + qa′tl′)MΠµBµ wµ .
(183) 49 An interesting special case of (183) is ˜ rn,d(q, t) = X µ⊢n Tµ Q′ x∈µ(1 −q2a′t2l′)MBµ wµ .
(184) Recurrences and Explicit Formulae We begin with a useful lemma about area and Schr¨ oder paths.
Lemma 1 (The “boundary lemma”). Given a, b, c ∈N, let boundary(a, b, c) be the path whose σ word is 2c1b0a. Then X π qarea′(π) = a + b + c a, b, c q , (185) where the sum is over all paths π from (0, 0) to (c + b, a + b) consisting of a N steps, b D steps and c E steps, and area′(π) is the number of lower triangles between π and boundary(a, b, c).
Proof.
Given π as above, we claim the number of coinversions of σ(π) equals area′(π). To see why, start with π as in Figure 20, and note that when consecutive ND, DE, or NE steps are interchanged, area′ decreases by 1. Thus area′(π) equals the number of such interchanges needed to transform π into boundary(a, b, c), or equivalently to transform σ(π) into 2c1b0a. But this is just coinv(σ(π)). Thus X π qarea′(π) = X σ∈M(a,b,c) qcoinv(σ) (186) = X σ∈M(a,b,c) qinv(σ) = a + b + c a, b, c q by (9).
2 Given n, d, k ∈N with 1 ≤k ≤n, let L+ n,n,d(k) denote the set of paths in L+ n,n,d which have k total D plus N steps below the lowest E step. We view L+ n,n,n(k) as containing the path with n D steps if k = n and L+ n,n,n(k) as being the empty set if k < n. Define Sn,d,k(q, t) = X π∈L+ n,n,d(k) qarea(π)tbounce(π), (187) with Sn,n,k(q, t) = χ(k = n). There is a recursive structure underlying the Sn,d,k(q, t) which extends that underlying the Fn,k(q, t). The following result is derived in [Hag04], and is similar to recurrence relations occurring in [EHKK03]. For any two integers n, k we use the notation δn,k = χ(n = k).
(188) 50 π boundary (4,3,5) Figure 20: The region between a path π and the corresponding boundary path. For this region area′ = 27.
Theorem 28 Let n, k, d ∈N with 1 ≤k ≤n. Then Sn,n,k(q, t) = δn,k, (189) and for 0 ≤d < n, Sn,d,k(q, t) = tn−k min(k,n−d) X p=max(1,k−d) k p q q(p 2) n−k X j=0 p + j −1 j q Sn−k,d+p−k,j(q, t), (190) with the initial conditions S0,0,k = δk,0, Sn,d,0 = δn,0δd,0.
(191) Proof.
If d = n then (189) follows directly from the definition. If d < n then π has at least one peak. Say π has p N steps and k −p D steps in section 0. First assume p < n −d, i.e. Π has at least two peaks. We now describe an operation we call truncation, which takes π ∈L+ n,n,d(k) and maps it to a π′ ∈L+ n−k,n−k,d−k+p with one less peak. Given such a π, to create π′ start with σ(π) and remove the first k letters (section 0). Also remove all the 2’s in section 1. The result is σ(π′).
For the path on the left in Figure 19, σ(π) = 10020201120212, k = 3 and σ(π′) = 001120212.
We will use Figure 21 as a visual aid in the remainder of our argument. Let j be the total number of diagonal and north steps of π in section 1 of π. By construction the bounce path for Γ(π′) will be identical to the bounce path for Γ(Π) except the first bounce of Γ(π) is truncated.
This bounce step hits the diagonal at (p, p), and so the contribution to bounce(π′) from the bounce path will be n −d −p less than to bounce(π). Furthermore, for each D step of π above peak 1 of Π, the number of peaks of π′ below it will be one less than the number of peaks of π below it. It follows that bounce(π) = bounce(π′) + n −d −p + d −(k −p) = bounce(π′) + n −k.
51 p p peak 1 peak 2 j Region 1 Region 0 Figure 21: A path π decomposed into various regions under truncation.
Since the area below the triangle of side p from Figure 21 is p 2 , area(π) = area(π′) + p 2 + area0 + area1, (192) where area0 is the area of section 0 of π, and area1 is the portion of the area of section 1 of π not included in area(π′). When we sum qarea0(π) over all π ∈L+ n,n,d(k) which get mapped to π′ under truncation, we generate a factor of k p q (193) by the c = 0 case of the boundary lemma.
From the proof of the boundary lemma, area1 equals the number of coinversions of the 1st section of σ(π) involving pairs of 0’s and 2’s or pairs of 1’s and 2’s. We need to consider the sum of q to the number of such coinversions, summed over all π which map to π′, or equivalently, summed over all ways to interleave the p 2’s into the fixed sequence of j 0’s and 1’s in section 0 of π′. Taking into account the fact that such an interleaving must begin with a 2 but is otherwise unrestricted, we get a factor of p −1 + j j q , (194) 52 since each 2 will form a coinversion with each 0 and each 1 occurring before it. It is now clear how the various terms in (190) arise.
Finally, we consider the case when there is only one peak, so p = n −d. Since there are d −(k −p) = d −k + n −d = n −k D steps above peak 1 of π, we have bounce(π) = n −k.
Taking area into account, by the above analysis we get Sn,d,k(q, t) = tn−kq(n−d 2 ) k n −d q n −d −1 + n −k n −k q (195) which agrees with the p = n −d term on the right-hand-side of (190) since Sn−k,n−k,j(q, t) = δj,n−k from the initial conditions.
2 The following explicit formula for Sn,d(q, t) was obtained in [EHKK03].
Theorem 29 For all 0 ≤d < n, Sn,d(q, t) = n−d X b=1 X α1+...+αb=n−d, αi>0 β0+β1+...+βb=d, βi≥0 β0 + α1 β0 q βb + αb −1 βb q q(α1 2 )+...+(αb 2 ) (196) tβ1+2β2+...+bβb+α2+2α3+...+(b−1)αb b−1 Y i=1 βi + αi+1 + αi −1 βi, αi+1, αi −1 q .
Proof.
Consider the sum of qareatbounce over all π which have bounce vector (α1, . . . , αb), and shift vector (β0, β1, . . . , βb). For all such π the value of bounce is given by the exponent of t in (196). The area below the bounce path generates the q(α1 2 )+...+(αb 2 ) term. When computing the portion of area above the bounce path, section 0 of π contributes the β0+α1 β0 q term. Similarly, section b contributes the βb+αb−1 βb q term (the first step above peak b must be an E step by the definition of a peak, which explains why we subtract 1 from αb). For section i, 1 ≤i < b, we sum over all ways to interleave the βi D steps with the αi+1 N steps and the αi E steps, subject to the constraint we start with an E step. By the boundary lemma, we get the βi+αi+1+αi−1 βi,αi+1,αi−1 q term.
2 The Special Value t = 1/q Theorem 30 For 1 ≤k ≤n and 0 ≤d ≤n, q(n 2)−(d 2)Sn,d,k(q, 1/q) = q(k−1)(n−d) [k] [n] 2n −k −d −1 n −k q n d q .
(197) Proof. (Sketch). The result can be obtained by induction, as in the case of the proof of Theorem 14. The details of this argument can be found in [Hag08][pp. 54-56].
2 Corollary 12 For 0 ≤d ≤n, q(n 2)−(d 2)Sn,d(q, 1/q) = 1 [n −d + 1] 2n −d n −d, n −d, d q .
(198) 53 Proof.
It is easy to see combinatorially that Sn+1,d,1(q, t) = tnSn,d(q, t). Thus q(n 2)−(d 2)Sn,d(q, 1/q) = q(n 2)+n−(d 2)Sn+1,d,1(q, 1/q) (199) = q(n+1 2 )−(d 2)Sn+1,d,1(q, 1/q) (200) = q(1−1)(n+1−d) [n + 1] 2(n + 1) −1 −d −1 n q n + 1 d q (201) = 1 [n −d + 1] 2n −d n −d, n −d, d q .
(202) 2 Remark 10 Corollary 12 proves that Sn,d(q, t) is symmetric in q, t when t = 1/q. For we have Sn,d(q, 1/q) = q−(n 2)+(d 2) 1 [n −d + 1] 2n −d n −d, n −d, d q , (203) and replacing q by 1/q we get Sn,d(1/q, q) = q(n 2)−(d 2) qn−d [n −d + 1] 2n −d n −d, n −d, d q q2(n−d 2 )+(d 2)−(2n−d 2 ), (204) since [n]!1/q = [n]!q−(n 2). Now n 2 − d 2 + n −d + 2 n −d 2 + d 2 − 2n −d 2 = d 2 − n 2 , (205) so Sn,d(q, 1/q) = Sn,d(1/q, q). It is of course an open problem to show Sn,d(q, t) = Sn,d(t, q) bijectively, since the d = 0 case is Open Problem 1.
A Schr¨ oder dinv-Statistic Let Cn(q, t, w) = n X d=0 wdSn,d(q, t), (206) and for π ∈L+ n,n, let a′ i(π) equal the number of area squares in the ith column from the right.
Also set a′ 0(π) = −1 for all π. For example, for the path on the right in Figure 9, we have (a′ 0, a′ 1, a′ 2, . . . , a′ 8) = (−1, 0, 1, 1, 2, 3, 3, 1, 0).
The (q, t)-Schr¨ oder can be expressed strictly in terms of Dyck paths as follows.
Theorem 31 Cn(q, t, w) = X π∈L+ n,n qarea(π)tbounce(π) Y 1≤i≤n a′ i>a′ i−1 (1 + w/qa′ i).
(207) 54 Proof.
(sketch) Show the coefficient of wd in (207) satisfies the same recurrence as (190).
2 Next define the reading order of the rows of π to be the order in which the rows are listed by decreasing value of ai, where if two rows have the same ai-value, the row above is listed first.
For the path on the left in Figure 9, the reading order is row 6, row 8, row 5, row 4, row 3, row 7, row 2, row 1.
(208) Finally let bk = bk(π) be the number of inversion pairs as in Definition 4 which involve the kth row in the reading order and rows before it in the reading order, and set b0 = −1. For the path on the left in Figure 9, we have (b0, b1, . . . , b8) = (−1, 0, 1, 1, 2, 3, 3, 1, 0).
(209) Note that in this example a′ i(ζ(π)) = bi(π) for all i, and it is not hard to see that this is true in general. Hence we have Cn(q, t, w) = X π∈L+ n,n qdinv(π)tarea(π) Y 1≤i≤n bi>bi−1 (1 + w/qbi).
(210) It is also easy to see that bi > bi−1 if and only if rowi contains a column top, i.e. the N step of the path in that row is followed immediately by an E step.
Now to get a term in (210) contributing to Sn,d(q, t) we make a selection of d column tops, and for the rows containing those column tops we subtract the corresponding bi contribution to dinv. If we start with the path π and replace those column tops by D steps, we get a Schr¨ oder path, and we can define dinv of this path as dinv(π) minus the sum of the chosen bi.
The Shuffle Conjecture In [HHL+05b] a nice conjecture for the expansion of F(DHn; q, t) into monomials is introduced.
It is often referred to as the shuffle conjecture, since it can be phrased in the following simple way: decompose {1, 2, . . . , n} into increasing sequences of consecutive integers α1, α2, . . . , αk of lengths |α1|, |α2|, . . . and decreasing sequences of consecutive integers β1, . . . , βs of lengths |β1|, |β2|, . . .. For example, if n = 8 we might have k = 1, s = 2, α1 = {6, 7, 8}, β1 = {5, 4, 3}, β2 = {2, 1}.
(211) Given such a decomposition, we say a permutation σ ∈Sn is an α, β-shuffle if for each i, j all the terms in αi occur in increasing order in σ, and all the terms of βj occur in decreasing order in σ. For example, for α, β as in (211), the permutation 26571483 is an α, β-shuffle.
Conjecture 2 (The Shuffle Conjecture: [HHL+05b]) ⟨F(DHn; q, t), h|α1|h|α2| · · · h|αk|e|β1|e|β2| · · · e|βs|⟩= X P ∈Pn read(P ) is an α, β-shuffle qdinv(P )tarea(P ).
(212) If our decomposition of {1, 2, . . . , n} is into n sequences consisting of only one element each, then the set of α, β-shuffles is the set of all parking functions. By Remark 4 the left-hand-side of (212) gives H(DHn; q, t), and so in this case the shuffle conjecture reduces to Conjecture 1.
55 On the other hand, consider the case k = s = 1, α1 = n −d + 1, . . . , n, β1 = n −d, . . . , 1.
The left-hand-side of (212) reduces to ⟨F(DHn; q, t), hden−d⟩, the q, t-Schr¨ oder. If read(P) is an α, β-shuffle, we must have all the elements of α1 occurring at the top of columns. It is easy to see that none of these cars can form dinv-pairs with any other car which occurs before them in the reading order. We can identify columns containing elements of α1 at the top with selections in the right-hand-side of (210) of the d-column tops to be regarded as D steps, and one finds in fact that the Schr¨ oder dinv-statistic in the right-hand-side of (210) is the same as dinv(P).
For a subset D of {1, 2, . . . , n −1}, let Qn,D(Z) = X a1≤a2≤···≤an ai=ai+1 = ⇒i/ ∈D za1za2 · · · zan (213) denote Gessel’s fundamental quasisymmetric function. Another equivalent form of the Shuffle Conjecture appearing in [HHL+05b] is the statement that F(DHn; q, t) = X P ∈Pn qdinv(P )tarea(P )Qn,Ides(read(P ))(x1, . . . , xn), (214) where recall that for any permutation σ ∈Sn, Ides(σ) is the set of all i for which i + 1 occurs before i in σ. The authors show that the sum Aπ(x1, . . . , xn; q) = X P ∈Pn π fixed qdinv(P )Qn,Ides(read(P ))(x1, . . . , xn), (215) obtained by restricting the right-hand-side of (214) to a fixed path π, is a special case of a family of symmetric functions introduced by Lascoux, Leclerc, and Thibon [LLT97] commonly called LLT polynomials. In [LLT97] it is conjectured that these polynomials are always Schur positive, and two recent preprints claim to give independent proofs of this conjecture. One of these is by Grojnowski and Haiman [GH06], and uses the representation theory of Hecke algebras. The other, by S. Assaf [Ass13], is a purely combinatorial construction involving objects called dual equivalence graphs.
In [Hag08][Theorem 6.8, p. 98] it is shown that if π is a path with the property that each non-empty column has its base on the diagonal x = y (a so called “balanced” path), then the Schur coefficients of Aπ(x1, . . . , xn; q) can be expressed in terms of the charge statistic of Algorithm 1. Hence one could hope that for general LLT polynomials, or at least for the type of LLT’s corresponding to Dyck paths, there is some similar formula. Note that the product eβ(π) in (139) can be easily expanded in terms of Schur functions, and in fact the coefficient of sλ in eβ is Kλ′,β. Thus there may be a way of associating powers of q with SSYT, depending on the shape of π in some way, to generate the LLT polynomial Aπ(x1, . . . , xn; q).
Open Problem 4 Find a nice formula, perhaps in terms of a generalized charge statistic, for the coefficients in the Schur expansion of the LLT polynomial corresponding to a given Dyck path, i.e., the right-hand-side of (215).
Assaf’s construction in terms of dual equivalence graphs does yield a combinatorial formula of sorts for an arbitrary LLT polynomial, but her construction is rather involved, and it is not yet known how to reproduce explicit formulas, like the one for balanced paths in terms of charge, from her formula. Note that the shuffle conjecture gives a prediction for the coefficient of a monomial symmetric function in F(DHn; q, t), but does not (except in the hook case) give a nice conjecture for the more desirable Schur coefficients.
56 5 Rational Catalan Combinatorics The Superpolynomial Invariant of a Torus Knot Throughout this section (m, n) is a fixed pair of relatively prime, positive integers.
In the last few years an exciting generalization of the q, t-Schr¨ oder and the shuffle conjecture has been introduced. This generalization depends on an arbitrary pair (m, n), and has an interpretation in terms of knot theory. By a knot we mean an embedding of a circle in R3, and by a knot invariant a polynomial which is the same on equivalent knots. Two classical knot invariants on a knot K are the Jones polynomial VK(t) and the HOMFLY polynomial PK(a, q). Dunfield, Gukov, and Rasmussen [DGR06] hypothesized the existence of a superpolynomial knot invariant PK(a, q, t) which would contain the HOMFLY and Jones polynomials as limiting cases, as well as having other desirable properties. Possible definitions of the superpolynomial for torus knots T(m,n) have recently been suggested by Angnanovic and Shakirov [AS11] (see also [AS12]), Cherednik [Che13], and Oblomkov, Rasmussen, and Shende [ORS12]. All three methods seem to give the same polynomial, and in fact Gorsky and Negut [GN13] have proved the descriptions in [AS11] and [Che13] do in fact give the same polynomial.
The description in [AS11] is in terms of Macdonald polynomials, and Gorsky first realized that when m = n+1, if we use the Cherednik parametrization, then the superpolynomial can be expressed as the function Cn(q, t, −a) from (210), giving a completely new interpretation for the q, t-Schr¨ oder.
In [ORS12] a conjectured combinatorial expression for the superpolynomial of T(m, n) in terms of weighted lattice paths is given, which we now describe. Let Grid(m, n) be the n × m grid of labelled squares whose upper-left-hand corner square is labelled with (n −1)(m −1) −1, and whose labels decrease by m as you go down columns and by n as you go across rows. For example, Grid(3, 7) = 11 4 −3 8 1 −6 5 −2 −9 2 −5 −12 −1 −8 −15 −4 −11 −18 −7 −14 −21 (216) To the corners of the squares of Grid(m, n) we associate Cartesian coordinates, where the lower-left-hand corner of the grid has coordinates (0, 0), and the upper-right-hand-corner of the grid (m, n). Let L+ (m,n) denote the set of lattice paths π for which none of the squares with negative labels are above π. (This agrees with our definition of L+ (m,n) from Section 1 as paths which stay above the line my = nx). For a given π, we let area(π) denote the number of squares in Grid(m, n) with positive labels which are below π. Furthermore, let dinv(π) denote the number of squares in Grid(m, n) which are above π and whose arm and leg lengths satisfy a l + 1 < m n < a + 1 l .
(217) For example, if (m, n) = (3, 7) and π = NNNNNEENNE, then area(π) = 2 (corresponding to the squares with labels 2 and 5). Also, dinv(π) = 2; the squares with labels 11, 8, 4, 1 have 57 a = l = 1, a = 1, l = 0, a = 0, l = 1, a = l = 0, respectively, and so the squares with labels 8 and 11 do not satisfy (217), while the squares with labels 1 and 4 do.
Next we define a generalization of the formula (210) for general (m, n). Given π ∈L+ (m,n), let R(π) denote the set of labels of squares which are at the top of some column of π. Say these labels occur in columns c1, c2, . . . , ck as we move left to right. Then for 1 ≤i ≤k, let ti denote the label of the square which is in the same row as the square at the top of column ci, and also in column ci+1, and set T(π) = {t1, t2, . . . , tk−1}. For example, if π is the path on the left of Figure 23, then R(π) = {−3, 1, 5} (218) T(π) = {−6, −2}.
(219) Now form a vector α(π) = (α1, . . . , αk) consisting of the elements of R(π) in decreasing order, and let ci(π) denote the number of elements of R(π) which are larger than αi, minus the number of elements of T(π) which are larger than αi. For the example of (218), we have α = (5, 1, −3), and so c1 = 0 −0 = 0, c2 = 1 −0 = 1, c3 = 2 −1 = 1. Furthermore set c0 = −1.
Conjecture 3 [ORS12] For any pair (m, n) of positive, relatively prime integers, PT(m,n)(−w, q, t) = X π∈L+ m,n qdinv(π)tarea(π) Y ci>ci−1 1≤i≤k (1 + w/qci).
(220) Here T(m, n) is the (m, n) torus knot, which winds around the torus m times in one direction and n times in the other before returning to the starting point, and we use the parametrization of the superpolynomial occurring in [Che13][p. 18, eq. (2.12)].
Exercise 11 Show that if m = n + 1, (220) reduces to (210).
There is also a version of the shuffle conjecture for any (m, n). Let an (m, n)-parking function be a path π ∈L+ (m,n) together with a placement of the integers 1 through n (called cars) just to the right of the N steps of π, with strict decrease down columns. For such a pair P, we let rank(j) be the label of the square that contains j, and we set tdinv(P) = |{(i, j) : 1 ≤i < j ≤n and rank(i) < rank(j) < rank(i) + m}|.
(221) Furthermore we let the reading word read(P) be the permutation obtained by listing the cars by decreasing order of their ranks. For example, for the (3, 7)-parking function of Figure 22, tdinv = 3, with inversion pairs formed by pairs of cars (6, 7), (4, 6), and (2, 4), and the reading word is 7642531. Let maxtdinv(π) be tdinv of the parking function for π whose reading word is the reverse of the identity, and for any parking function P for π set dinv(P) = dinv(π) + tdinv(P) −maxtdinv(π).
(222) Then the combinatorial side of the (m, n) shuffle conjecture is the function B(m,n)(x1, . . . , xn; q, t) = X (m, n) parking functions P qdinv(P )tarea(π)Qn,Ides(read(P ))(x1, . . . , xn). (223) 58 5 2 1 7 6 4 3 Figure 22: A (3, 7)-parking function.
Gorsky and Negut [GN13] show how the results of Aganagic and Shakirov on torus knot invariants can be expressed in terms of Macdonald polynomials using advanced objects such as the Hilbert scheme. Bergeron, Garsia, Leven, and Xin [BGLX14a], [BGLX14b] have shown how this Macdonald polynomial construction can be done combinatorially with plethystic symmetric function operators, and in fact they define operators Q(m,n) for any relatively prime (m, n) by a recursive procedure. The rational shuffle conjecture can then be phrased as Q(m,n)(−1)n = B(m,n)(x1, . . . , xn; q, t).
The symmetric function Q(n+1,n)(−1)n reduces to ∇en, and so the rational shuffle conjecture reduces to the original shuffle conjecture when m = n + 1.
Gorsky and Negut also conjecture that Q(m,n)(−1)n is the Frobenius series of the unique finite-dimensional irreducible representation of the rational Cherednik algebra with parameter m/n, with respect to a certain bigrading. Hikita [Hik14] has shown that B(m,n)(x1, . . . , xn; q, t) is the bigraded Frobenius series of certain Sn-modules arising in the study of the homology of type A affine Springer fibers, which gives another possible way of attacking the rational shuffle conjecture. We mention that the portion of the right-hand-side of (223) corresponding to a fixed path π is an LLT polynomial, and hence is Schur positive.
Many of the results in this chapter involving the special cases t = 1 and t = 1/q have elegant extensions to general, relatively prime (m, n). For example, Gorsky and Negut [GN13] prove that when t = 1/q the q, t-Catalan for (m, n), obtained by taking the coefficient of s1n in Q(m,n)(−1)n, reduces to 1 [m] m + n −1 n .
(224) Another example is that the total number of (m, n)-parking functions is mn−1.
Tesler Matrices and the Superpolynomial There is a general formula for Q(m,n)(−1)n in terms of Tesler matrices.
Theorem 32 (Gorsky, Negut 2013) For any pair of positive, relatively prime integers (m, n), Qm,n(−1)n = X C∈Tes(m,n) m Y i=1 ecii Y 1≤i0 ([ci,i+1 + 1]q,t −[ci,i+1]q,t) Y 2≤i+10 (−M)[ci,j]q,t. (225) 59 Here Tes(m, n) is the set of m × m upper-triangular matrices C satisfying ci,i + X j>i ci,j − X j0 (1 −a) Y 1≤i<m ([ci,i+1 + 1]q,t −[ci,i+1]q,t) Y 2≤i+10 (−M)[ci,j]q,t.
(231) There are a number of intriguing open problems involving the combinatorics of (m, n)-Catalan paths. For example, there is a candidate extension of the zeta map of Figure 9 which can be described as follows. Given a path π ∈L+ (m,n), call the set of corners of grid squares which are touched by π the “vertices” of π. Next define S(π) to be the set consisting of the labels of those squares whose upper-left-hand-corners are vertices of π. A given label in S(π) is called an N label if the vertex associated to it is the start of an N step, otherwise it is called an E label. For example, if π is the path on the left in Figure 23, then π = NNNNNENENE S(π) = {−10, −7, −4, −1, 2, 5, −2, 1, −6, −3}.
(232) We now define the “sweep map” of [ALW14], denoted ζ, from L+ (m,n) to L+ (m,n) as follows: or-der the elements of S(π) in increasing order to create a vector of labels D(π) = (d1, d2, . . . , dm+n).
60 −10 2 5 1 −3 0 0 0 0 1 1 1 ζ 0 0 0 0 0 1 1 −6 −2 −1 −4 −7 Figure 23: The sweep map.
Then create a path φ(π) by defining the ith step of φ(π) to be an N step if di is an N label, and an E step if di is an E label. For the example in (232), we have D(π) = (−10, −7, −6, −4, −3, −2, −1, 1, 2, 5) φ(π) = NNNNENNENE.
(233) Exercise 12 Show that when m = n + 1, paths in L+ (m,n) are in bijection with paths in L+ (n,n), and that the sweep map reduces to the ζ map of Figure 9.
Open Problem 5 Prove that for general coprime (m, n) the sweep map is a bijection from L+ (m,n) →L+ (m,n).
This problem has been studied by Gorsky, Mazin, and Vazirani [GMV14] and Armstrong, Loehr, and Warrington [ALW14]. See also [AHJ14]. In [GMV14] it is shown that the sweep map is a bijection whenever m = kn + 1 or m = kn −1 for some positive integer k. We note that in the case m = kn + 1 Loehr [Loe03], [Loe05b] (see also [Hag08][pp. 108-109]) has defined an extension of the bounce statistic, which when combined with area generates the q, t-Catalan for m = kn + 1. For this “paths in a n × kn rectangle” case there is also an interpretation for the rational shuffle conjecture in terms of a generalization of diagonal harmonics (see [HHL+05b]).
References [AAR99] G. E. Andrews, R. Askey, and R. Roy. Special Functions, volume 71 of Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 1999.
[AHJ14] Drew Armstrong, Christopher R. H. Hanusa, and Brant C. Jones.
Results and conjectures on simultaneous core partitions.
European J. Combin., 41:205–220, 2014.
[AKOP02] G. E. Andrews, C. Krattenthaler, L. Orsina, and P. Papi. ad-Nilpotent b-ideals in sl(n) having a fixed class of nilpotence: combinatorics and enumeration. Trans.
Amer. Math. Soc., 354(10):3835–3853 (electronic), 2002.
61 [ALW14] D. Armstrong, N. Loehr, and G. Warrington. Rational parking functions and Cata-lan numbers. Math arXiv:1403.1845, 2014.
[And75] G. E. Andrews. Identities in combinatorics, II: a q-analog of the Lagrange inversion theorem. Proc. Amer. Math. Soc., 53:240–245, 1975.
[AR12] Drew Armstrong and Brendon Rhoades. The Shi arrangement and the Ish arrange-ment. Trans. Amer. Math. Soc., 364(3):1509–1528, 2012.
[Arm13] Drew Armstrong. Hyperplane arrangements and diagonal harmonics. J. Comb., 4(2):157–190, 2013.
[Art76] E. Artin. Galois Theory. Notre Dame Mathematical Lectures. Notre Dame Ind., 1976. Reprint of the 1942 original.
[AS11] M. Aganagic and S. Shakirov. Knot homology from refined Chern-Simons theory.
arXiv:1105.5117, 2011.
[AS12] Mina Aganagic and Shamil Shakirov. Refined Chern-Simons theory and knot ho-mology. In String-Math 2011, volume 85 of Proc. Sympos. Pure Math., pages 3–31.
Amer. Math. Soc., Providence, RI, 2012.
[Ass13] S. Assaf. Dual equivalence graphs and a combinatorial proof of LLT and Macdonald positivity. arXiv:1005.3759 dated Oct. 23, 2013, 2013.
[BGLX14a] F. Bergeron, A. M. Garsia, E. Leven, and G. Xin. Compositional (km, kn)-shuffle conjectures. math arXiv:1404.4616, 2014.
[BGLX14b] F. Bergeron, A. M. Garsia, E. Leven, and G. Xin. Some remakable new plethystic operators in the thoery of Macdonald polynomials. math arXiv:1405.0316, 2014.
[But94] L. M. Butler. Subgroup lattices and symmetric functions. Mem. Amer. Math. Soc., 112(539):vi+160, 1994.
[Che13] Ivan Cherednik. Jones polynomials of torus knots via DAHA. Int. Math. Res. Not.
IMRN, (23):5366–5425, 2013.
[CR64] L. Carlitz and J. Riordan. Two element lattice permutation numbers and their q-generalization. Duke J. Math., 31:371–388, 1964.
[DGR06] Nathan M. Dunfield, Sergei Gukov, and Jacob Rasmussen. The superpolynomial for knot homologies. Experiment. Math., 15(2):129–159, 2006.
[EHKK03] E. Egge, J. Haglund, K. Killpatrick, and D. Kremer. A Schr¨ oder generalization of Haglund’s statistic on Catalan paths. Electron. J. Combin., 10:Research Paper 16, 21 pp. (electronic), 2003.
[FH85] J. F¨ urlinger and J. Hofbauer.
q-Catalan numbers.
J. Combin. Theory Ser. A, 40:248–264, 1985.
[Foa68] D. Foata. On the Netto inversion number of a sequence. Proc. Amer. Math. Soc., 19:236–240, 1968.
62 [Fou74] H. O. Foulkes.
A survey of some combinatorial aspects of symmetric functions.
In Permutations (Actes Colloq., Univ. Ren´ e-Descartes, Paris, 1972), pages 79–92.
Gauthier-Villars, Paris, 1974.
[FS78] D. Foata and M.-P. Sch¨ utzenberger. Major index and inversion number of permu-tations. Math. Nachr., 83:143–159, 1978.
[Gar81] A. M. Garsia. A q-analogue of the Lagrange inversion formula. Houston J. Math., 7:205–237, 1981.
[Ges80] I. Gessel. A noncommutative generalization and q-analog of the Lagrange inversion formula. Trans. Amer. Math. Soc., 257:455–482, 1980.
[GH] A. M. Garsia and J. Haglund. A polynomial expression for the character of diagonal harmonics. Annals of Combinatorics. to appear.
[GH96] A. M. Garsia and M. Haiman. A remarkable q, t-Catalan sequence and q-Lagrange inversion. J. Algebraic Combin., 5(3):191–244, 1996.
[GH01] A. M. Garsia and J. Haglund.
A positivity result in the theory of Macdonald polynomials. Proc. Nat. Acad. Sci. U.S.A., 98:4313–4316, 2001.
[GH02] A. M. Garsia and J. Haglund. A proof of the q, t-Catalan positivity conjecture.
Discrete Math., 256:677–717, 2002.
[GH06] I. Grojnowski and M. Haiman. Affine Hecke algebras and positivity of LLT and Mac-donald polynomials. 2006. Preprint at mhaiman/.
[GMV14] E. Gorsky, M. Mazin, and M. Vazirani.
Affine permutations and rational slope parking functions. math arXiv:1403.0303, 2014.
[GN13] E. Gorsky and A. Negut. Refined knot invariants and Hilbert schemes. Preprint on the math arXiv: 2013.
[GR04] G. Gasper and M. Rahman. Basic hypergeometric series, second ed., volume 96 of Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 2004. With a foreword by Richard Askey.
[Hag03] J. Haglund.
Conjectured statistics for the q, t-Catalan numbers.
Adv. Math., 175(2):319–334, 2003.
[Hag04] J. Haglund. A proof of the q, t-Schr¨ oder conjecture. Int. Math. Res. Not., 11:525– 560, 2004.
[Hag08] James Haglund. The q,t-Catalan numbers and the space of diagonal harmonics, vol-ume 41 of University Lecture Series. American Mathematical Society, Providence, RI, 2008. With an appendix on the combinatorics of Macdonald polynomials.
[Hag11] J. Haglund. A polynomial expression for the Hilbert series of the quotient ring of diagonal coinvariants. Adv. Math., 227:2092–2106, 2011.
63 [Hai94] M. Haiman. Conjectures on the quotient ring by diagonal invariants. J. Algebraic Combin., 3:17–76, 1994.
[Hai00] M. Haiman. Private communication, 2000.
[Hai02] M. Haiman. Vanishing theorems and character formulas for the Hilbert scheme of points in the plane. Invent. Math., 149:371–407, 2002.
[HHL05a] J. Haglund, M. Haiman, and N. Loehr. A combinatorial formula for Macdonald polynomials. Jour. Amer. Math. Soc., 18:735–761, 2005.
[HHL+05b] J. Haglund, M. Haiman, N. Loehr, J. B. Remmel, and A. Ulyanov. A combinatorial formula for the character of the diagonal coinvariants. Duke Math. J., 126:195–232, 2005.
[Hik14] Tatsuyuki Hikita. Affine Springer fibers of type A and combinatorics of diagonal coinvariants. Adv. Math., 263:88–122, 2014.
[HL05] J. Haglund and N. Loehr.
A conjectured combinatorial formula for the Hilbert series for Diagonal Harmonics. Discrete Math. (Proceedings of the FPSAC 2002 Conference held in Melbourne, Australia), 298:189–204, 2005.
[JL01] G. James and M. Liebeck. Representations and characters of groups. Cambridge University Press, New York, second edition, 2001.
[Lev11] P. Levande. Special cases of the parking functions conjecture and upper-triangular matrices.
DMTCS, Proceedings of the 23rd Internat. Conf. on Formal Power Series and Algebraic Combinatorics (FPSAC 2011), pages 635–644, 2011.
See wwww.dmtcs.org/dmtcs-ojs/index.php/proceedings/issue/view/119.
[Lev12] Paul Levande. Combinatorial structures and generating functions of Fishburn num-bers, parking functions, and Tesler matrices. ProQuest LLC, Ann Arbor, MI, 2012.
Thesis (Ph.D.)–University of Pennsylvania.
[LLT97] Alain Lascoux, Bernard Leclerc, and Jean-Yves Thibon. Ribbon tableaux, Hall-Littlewood functions, quantum affine algebras, and unipotent varieties. J. Math.
Phys., 38(2):1041–1068, 1997.
[Loe03] N. Loehr.
Multivariate Analogues of Catalan Numbers, Parking Functions, and their Extensions. Ph.D. Thesis, University of California at San Diego, 2003.
[Loe05a] Nicholas A. Loehr. Combinatorics of q, t-parking functions. Adv. in Appl. Math., 34(2):408–425, 2005.
[Loe05b] Nicholas A. Loehr.
Conjectured statistics for the higher q, t-Catalan sequences.
Electron. J. Combin., 12:Research Paper 9, 54 pp. (electronic), 2005.
[LR04] Nicholas A. Loehr and Jeffrey B. Remmel. Conjectured combinatorial models for the Hilbert series of generalized diagonal harmonics modules. Electron. J. Combin., 11(1):Research Paper 68, 64 pp. (electronic), 2004.
64 [LS78] A. Lascoux and M.-P. Sch¨ utzenberger. Sur une conjecture de H. O. Foulkes. C. R.
Acad. Sci. Paris S´ er. A-B, 286:A323–A324, 1978.
[Mac60] P. A. MacMahon. Combinatory Analysis. Two volumes (bound as one). Chelsea Publishing Co., New York, 1960.
[Mac95] I. G. Macdonald. Symmetric Functions and Hall Polynomials. Oxford Mathematical Monographs, second ed., Oxford Science Publications. The Clarendon Press Oxford University Press, New York, 1995.
[ORS12] Alexei Oblomkov, Jacob Rasmussen, and Vivek Shende.
The Hilbert scheme of a plane curve singularity and the HOMFLY homology of its link.
math arXiv:1201.2115, 2012.
[Sag01] Bruce E. Sagan. The symmetric group, volume 203 of Graduate Texts in Mathe-matics. Springer-Verlag, New York, second edition, 2001. Representations, combi-natorial algorithms, and symmetric functions.
[Sta79] R. P. Stanley. Invariants of finite groups and their applications to combinatorics.
Bull. Amer. Math. Soc. (new series), 1:475–511, 1979.
[Sta88] D. Stanton. Recent results for the q-Lagrange inversion formula. In Ramanujan revisited (Urbana-Champaign, Ill., 1987), pages 525–536. Academic Press, Boston, MA, 1988.
[Sta99] R. P. Stanley. Enumerative Combinatorics, volume 2. Cambridge University Press, Cambridge, United Kingdom, 1999.
[Sta03] R. P. Stanley. Recent progress in algebraic combinatorics. Bull. Amer. Math. Soc., 40:55–68, 2003.
[Sta12] Richard P. Stanley. Enumerative combinatorics. Volume 1, volume 49 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, second edition, 2012.
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Proof of $ |a-b| = |b-a| $
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While working out some intriguing qualities of absolute values for my studies of calculus, I frequently used the formula below.
I know that the formula below is clearly correct but how would I prove it? $$ |a-b| = |b-a| $$ $$ a,b \in\mathbb R $$
Believing that I require an actual proof for the formula I used so often I attempted to prove that formula "by cases". It appeared, however, that there is a more elegant proof somewhere out there.
Thanks in advance.
absolute-value
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edited May 24, 2015 at 22:44
EugeneEugene
asked May 22, 2015 at 15:35
EugeneEugene
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$\begingroup$ Write it as $(a-b)=(-1)(b-a)$ $\endgroup$
Someone
– Someone
2015-05-22 15:37:18 +00:00
Commented May 22, 2015 at 15:37
1
$\begingroup$ sqare both sides of thze equation $\endgroup$
Dr. Sonnhard Graubner
– Dr. Sonnhard Graubner
2015-05-22 15:37:47 +00:00
Commented May 22, 2015 at 15:37
1
$\begingroup$ The answer depends on how you define the absolute value. If it is defined by cases, the proof is likely to work that way, too. $\endgroup$
Joonas Ilmavirta
– Joonas Ilmavirta
2015-07-04 16:01:07 +00:00
Commented Jul 4, 2015 at 16:01
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I like to use that $|x| = \sqrt{x^{2}}$. Then $$|a-b|=\sqrt{(a-b)^{2}}=\sqrt{(a^2-2ab+b^2)}=\sqrt{(b-a)^{2}}=|b-a|.$$
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answered May 22, 2015 at 16:01
LythiaLythia
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$\begingroup$ Elegent . +1 ^^ $\endgroup$
Someone
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Use the definition of absolute value. If $a-b \geq 0$, then: $$|a-b| = a-b = -(b-a) = |b-a|,$$ where the last step is given because $a-b \geq 0 \implies b-a \leq 0$ and so $-(b-a) = |b-a|$. You treat the other case similarly.
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answered May 22, 2015 at 15:37
Ivo TerekIvo Terek
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Hint:
$$|a-b| = \begin{cases} a-b, & a - b > 0 \ -(a-b), & a - b \leq 0 \end{cases}$$
$$|b-a| = \begin{cases} b-a, & b-a > 0 \ -(b-a), & b-a \leq 0 \end{cases}$$
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answered May 22, 2015 at 15:37
ClarinetistClarinetist
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If $a=b$, this is trivial. WLOG, suppose $a
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answered May 22, 2015 at 15:37
SloanSloan
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You just need to have a look, how the absolute value is defined: For $x \in \Bbb R$, $$ \vert x \vert := \begin{cases} x \; , & \text{if $x \geq 0$} \ -x \; , & \text{if } x < 0\end{cases} \; .$$
So let $a, b \in \Bbb R$. Let's first assume that $a > b$, then $a - b > 0$, and by the defintion of the absolute value we get $$ \vert a - b \vert = a - b \; . $$ Since $b - a < 0$, we get by the definition of the absolute value $$\vert b - a \vert = -(b-a) = a - b \; ,$$ so we conclude that $$\vert a-b \vert = \vert b - a\vert \; ,$$ if $a > b$. Do the same for the case $b > a$ and note, that the case $a=b$ is trivial.
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answered May 22, 2015 at 15:46
aexlaexl
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$|a - b|$
= $|-b + a|$ (commutative)
= $|-(b - a)|$ (distributive)
= $||-(b - a)||$ (identity)
= $|(b - a)|$ (absolute value)
= $|b - a|$ (associative)
Q.E.D.
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edited Dec 11, 2021 at 4:57
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answered Dec 11, 2021 at 4:27
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I recently proved this statement using only a few basic properties of real numbers and the definition of absolute value. So it may be a bit easier to grasp than some of the other proofs presented here.
Proof:
Suppose a and b are real numbers. By trichotomy law only one of the following relation holds: a = b, a > b, or a < b.
Case 1: a = b
|a - b| = |0 - 0| = |0| = 0.
|b - a| = |0 - 0| = |0| = 0.
Because |0| = 0 by definition of absolute value.
Hence |a - b| = |b - a|
Case 2: a > b
Starting with a > b,
a - b > b - b
a - b > 0
Starting with a > b,
a - a > b - a
0 > b - a
By definition of absolute value, |a - b| = a - b because a - b > 0. And |b - a| = -(b - a) = a - b because b - a < 0.
Hence |a - b| = |b - a|
Case 3: a < b
Starting with a < b,
a - b < b - b
a - b < 0
Starting with a < b,
a - a < b - a
0 < b - a
By definition of absolute value, |a - b| = -(a - b) = b - a because a - b < 0. And |b - a| = b - a because b - a > 0.
Hence |a - b| = |b - a|
In all three cases |a - b| = |b - a|, therefore for any real numbers a and b, |a - b| = |b - a|.
Q.E.D.
References:
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edited Apr 6, 2022 at 20:12
answered Apr 6, 2022 at 17:40
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English (UK)
Meaning of mean in English
mean verb (EXPRESS)
Add to word list
Add to word list
A2 [ T ]
to express or represent something such as an idea, thought, or fact:
What does this word mean?
[ + that ] These figures mean that almost six percent of the working population is unemployed.
mean by What do you mean by that remark?
She's kind of strange though. Do you know what I mean?
"They all showed up." "You mean the entire family?"
[ T ]
used to add emphasis to what you are saying:
and I mean I want you home by midnight. And I mean midnight.
mean it Give it back now! I mean it.
More examplesFewer examples
When he said three o'clock, I thought he meant in the afternoon.
Charges are made on a sliding scale, which means that the amount you must pay increases with the level of your income.
If you don't know what the word means, look it up in a dictionary.
You should take it as a compliment when I fall asleep in your company - it means I'm relaxed.
A continuous white line in the middle of the road means no overtaking.
SMART Vocabulary: related words and phrases
Meaning & significance
acceptation
add
add up to something phrasal verb
backspin
be a badge of something idiom
i.e.
inauspiciously
interpretable
locution
locutionary
meaning
meaningfully
purport
sense
significant
speak
subtext
subtextually
use
what's with something? idiom
See more results »
mean verb (HAVE RESULT)
B1 [ T ]
to have a particular result:
Lower costs mean lower prices.
[ + that ] Advances in electronics mean that the technology is already available.
[ + -ing verb ] If we want to catch the 7.30 train, that will mean leaving the house at 6.00.
More examplesFewer examples
High acidity levels in the water mean that the fish are not so large.
Shortages mean that even staples like bread are difficult to find.
In a way I'd prefer it if they didn't come, because it would mean extra work.
Let's say that the journey takes three hours, that means you'll arrive at two o'clock.
The irregularity of English spelling means that it is easy to make mistakes.
SMART Vocabulary: related words and phrases
Meaning & significance
acceptation
add
add up to something phrasal verb
backspin
be a badge of something idiom
i.e.
inauspiciously
interpretable
locution
locutionary
meaning
meaningfully
purport
sense
significant
speak
subtext
subtextually
use
what's with something? idiom
See more results »
mean verb (INTEND)
B1 [ I or T ]
to intend:
mean any harm I'm sorry if I offended you - I didn't mean any harm.
be meant for The books with large print are meant for partially sighted readers.
[ + to infinitive ] I've been meaning to call you all week.
mean to Do you think she meant to say 9 a.m. instead of 9 p.m.?
mean for someone to do something They didn't mean for her to read the letter.
be meant to do sth
to be intended to:
These batteries are meant to last for a year.
See more
More examplesFewer examples
I didn't mean to be rude - it just came out like that.
He doesn't really mean it - he's just being contrary.
He's always making flattering remarks, but he doesn't really mean them.
I didn't mean to upset her - it was just a bit of fun.
He didn't mean it - he said it in the heat of the moment.
SMART Vocabulary: related words and phrases
Planning, expecting and arranging
accidentally
accidentally on purpose idiom
advertent
advisably
advisedly
conscious
cue
game fixing
game something out phrasal verb
half expect someone/something to do something
hatch
have something in mind idiom
have something up your sleeve idiom
muster
pencil
provide
provide against something phrasal verb
provide for someone phrasal verb
provident
providently
See more results »
mean verb (HAVE IMPORTANCE)
B1 [ T ]
to have an important emotional effect on someone:
mean a lot, nothing, something, etc. to It wasn't a valuable picture but it meant a lot to me.
mean nothing to Possessions mean nothing to him.
More examplesFewer examples
Her children mean all the world to her.
Gerald means nothing to me now.
Her career means everything to her.
Nothing means more to me than my children's happiness.
Her approval meant a lot to me.
SMART Vocabulary: related words and phrases
Meaning & significance
acceptation
add
add up to something phrasal verb
backspin
be a badge of something idiom
i.e.
inauspiciously
interpretable
locution
locutionary
meaning
meaningfully
purport
sense
significant
speak
subtext
subtextually
use
what's with something? idiom
See more results »
Grammar
Mean
We use mean to explain or ask what a word or phrase refers to. We form questions with mean with the auxiliary verb do: …
I mean
We use I mean very commonly in speaking as a discourse marker. We use it when we want to add to what we have just said, to make a point clearer or to correct what we have just said: …
You know what I mean
We often use the phrase you know what I mean (or if you know what I mean or do you know what I mean?) in speaking, to check that our listener understands what we are saying or to show that we assume the listener has the same opinion about something: …
Nouns
Nouns are one of the four major word classes, along with verbs, adjectives and adverbs. Nouns are the largest word class. …
Be meant to
Be meant to is used to talk about what is desirable, expected or intended: …
Idioms
be meant for each other
I mean
mean business
mean well
(do you) see what I mean?
what do you mean?
mean
adjective
uk
/miːn/ us
/miːn/
mean adjective (NOT GENEROUS)
B2 mainly UK
not willing to give or share things, especially money:
He's too mean to buy her a ring.
mean with My landlord's very mean with the heating - it's only on for two hours each day.
Synonyms
stingy informal disapproving
tight (NOT GENEROUS) informal disapproving
tight-fisted informal disapproving
More examplesFewer examples
She's really quite unpleasant about other people and she's as mean as hell.
He's a mean old scrooge!
"That was amazingly generous of you!" "Well, that was a two-edged comment - are you saying I'm usually mean?"
He's too mean to buy any new clothes.
She only gave you 50p? That was a bit mean.
SMART Vocabulary: related words and phrases
Mean with money
begrudge
cheapskate
cheeseparing
chintzy
frugal
frugally
miser
miserliness
miserly
money grabber
money-grubber
money-grubbing
penuriously
pinchpenny
scrooge
skinflint
stingily
tightwad
ungenerous
ungenerously
See more results »
mean adjective (NOT KIND)
B2
unkind or unpleasant:
mean to Stop being so mean to me!
She just said it to be mean.
Thesaurus: synonyms, antonyms, and examples
not kind to someone or something and causing pain
cruelTeasing them for being overweight is cruel.
callousHe had a callous disregard for the feelings of others.
cold-bloodedThe budget is based on a cold-blooded analysis of the markets.
ruthlessHe was a ruthless dictator.
heartlessHe has been described as a heartless boss by several employees.
See more results »
More examplesFewer examples
He's as mean as they come.
You shouldn't have been so mean to your mother - she deserves better.
And she didn't invite him? That was a bit mean!
Stop being so mean to your brother!
It was mean of him to make her stay late.
SMART Vocabulary: related words and phrases
Unkind, cruel & unfeeling
acerbic
acerbically
acerbity
acidly
anti-cruelty
cruel
cruelly
cruelty
cruelty to someone/something
cutthroat
meanness
mental cruelty
merciless
mercilessly
mordant
trenchantly
uncharitable
uncharitably
unchristian
uncompassionate
See more results »
mean adjective (VIOLENT)
mainly US
frightening and likely to become violent:
a mean and angry mob
a mean-looking youth
SMART Vocabulary: related words and phrases
Violent or aggressive
abusively
aggressive
aggressively
aggressiveness
aggro
ferocious
ferociously
ferocity
fierce
fierceness
outrage
proactive aggression
pugnacious
pugnaciously
pugnacity
untameable
vicious
viciously
viciousness
violence
See more results »
mean adjective (GOOD)
[ before noun ] informal
very good:
She's a mean piano player.
She plays a mean piano (= she plays very well).
SMART Vocabulary: related words and phrases
Informal words for good
A-OK
amazeballs
apple pie
awesomesauce
badass
dank
insane
juicy
killer
knock
knock spots off something idiom
like a boss idiom
lit
out of sight idiom
rule OK idiom
safe
shabby
shit hot
sight
smashing
See more results »
mean adjective (BAD QUALITY)
[ before noun ] informal
poor, dirty, and of bad quality:
He was living in a mean little hut.
SMART Vocabulary: related words and phrases
Not of good quality
am-dram
appallingly
atrocious
atrociously
awfulness
cheapo
cheesy
hacky
inadequacy
inadequate
inadequately
janky
paltry
schlock
schlocky
shitty
shocking
shockingly
skanky
trinket
See more results »
mean adjective (MATHEMATICS)
C2 [ before noun ]
mathematics
specialized
a mean number is an average number:
a mean value
Their mean weight was 76.4 kilos.
SMART Vocabulary: related words and phrases
Statistics
actuarial
actuary
aggregately
aggregatively
analysis of covariance
background variable
convenience sampling
correction factor
correlation coefficient
covariance matrix
covariate
infant mortality
infant mortality rate
interpoint
interquartile range
intervening variable
unadjusted
unsmoothed
unweighted
variate
See more results »
You can also find related words, phrases, and synonyms in the topics:
Averages
Idioms
no mean something
no mean achievement/feat
mean
noun [ S ]
uk
/miːn/ us
/miːn/
mean noun [S] (MATHEMATICS)
mathematics
specialized (also the arithmetic mean)
the result you get by adding two or more amounts together and dividing the total by the number of amounts:
the mean of The mean of 5, 4, 10, and 15 is 8.5.
Compare
average noun (AMOUNT)
SMART Vocabulary: related words and phrases
Statistics
actuarial
actuary
aggregately
aggregatively
analysis of covariance
background variable
convenience sampling
correction factor
correlation coefficient
covariance matrix
covariate
infant mortality
infant mortality rate
interpoint
interquartile range
intervening variable
unadjusted
unsmoothed
unweighted
variate
See more results »
You can also find related words, phrases, and synonyms in the topics:
Averages
mean noun [S] (METHOD)
formal
a quality or way of doing something that is in the middle of two completely different qualities or ways of doing something:
a mean between We need to find a mean between test questions that are too difficult and those that are too easy.
SMART Vocabulary: related words and phrases
Ways of achieving things
actively
aid
another
avenue
bases
basis
formula
media
method
methodological
methodologically
methodologist
procedure
standard operating procedure
strategy
styleless
stylistic
stylistically
thus
vehicle
See more results »
You can also find related words, phrases, and synonyms in the topics:
Averages
(Definition of mean from the Cambridge Advanced Learner's Dictionary & Thesaurus © Cambridge University Press)
mean | American Dictionary
mean
verb
us
/min/ past tense and past participle meant us/ment/
mean verb (EXPRESS)
Add to word list
Add to word list
[ T ]
to represent or express something intended, or to refer to someone or something:
"What does ’rough’ mean?" "It means ’not smooth.’"
[ + that clause ] These figures mean that almost 7% of the population is unemployed.
"Do you see that girl over there?" "Do you mean the one with short blond hair?"
mean verb (HAVE RESULT)
[ T ]
to have as a result:
Lower costs mean higher profits.
[ + (that) clause ] If she doesn’t answer the phone, it means (that) she’s out in the garden.
mean verb (HAVE IMPORTANCE)
[ T ]
to have the importance or value of:
My grandmother’s ring wasn’t valuable, but it meant a lot to me.
mean verb (INTEND)
[ I/T ]
to say or do something intentionally; intend:
[ T ] I think she meant 8 o’clock, although she said 7 o’clock.
[ I ] I’ve been meaning to call you but I’ve been so busy I never got around to it.
[ I/T ]
Mean can also be used to add emphasis to what you are saying:
[ T ] She means what she says.
Idioms
mean business
mean well
mean
noun [ C ]
us/min/
mean noun [C] (AVERAGE)
mathematics
a number that is the result of adding a group of numbers together and then dividing the result by how many numbers were in the group
mean
adjective [ -er/-est only ]
us
/min/
mean adjective [-er/-est only] (NOT KIND)
unkind or not caring:
I felt a little mean when I said I couldn’t visit her in the hospital until Saturday.
mean adjective [-er/-est only] (GOOD)
slang
very good:
She plays a mean bass fiddle.
(Definition of mean from the Cambridge Academic Content Dictionary © Cambridge University Press)
mean | Business English
mean
noun [ S ]
uk
/miːn/ us
Add to word list
Add to word list
MEASURES
(also arithmetic mean)
the result you get by adding two or more amounts together and dividing the total by the number of amounts:
The mean of 5, 4, 10, and 15 is 8.5.
a quality or way of doing something that is in the middle of two completely different qualities or ways of doing something:
a mean between sth and sth This description doesn't give enough information, and this one is too long – we need to find a mean between the two.
mean
verb [ T ]
uk
/miːn/ us
present participle meant | past tense and past participle meant
to express or represent something such as an idea, thought, or fact:
What does this word mean?
mean sth by sth What do you mean by 'rightsizing the department'?
to have a particular result:
Lower costs mean lower prices.
mean (that) Advances in electronics mean that the technology is already available.
mean doing sth If we increased our workforce, that would mean finding larger premises.
mean
adjective
uk
/miːn/ us
MEASURES
a mean number is the result you get by adding two or more amounts together and dividing the total by the number of amounts:
The mean weight of the crates is 76.4 kilos.
The table above shows the mean price per dozen of large grade A eggs.
unkind:
be mean to sb If she's ever mean to staff, she always apologizes afterwards.
not generous:
be mean with sth My boss is well known for being mean with money.
(Definition of mean from the Cambridge Business English Dictionary © Cambridge University Press)
Examples of mean
mean
Mean values should, where possible, be accompanied by standard errors or similar statistical indications of variance.
From the Cambridge English Corpus
However, the publisher reserves the right to typeset material by conventional means if an author's disk proves unsatisfactory.
From the Cambridge English Corpus
This symmetry means that any of the cells in the pair can star t firing before the other.
From the Cambridge English Corpus
What the heroine most likely means here, is that she doesn't wish to become involved outside her chosen province of literature.
From the Cambridge English Corpus
Particularly troubling to us is that no means is offered for determining at what time(s) a given parameter reflects either planning or control.
From the Cambridge English Corpus
Order is maintained by two means: direct management of the relations of the players and indirect structuring of the terms of play.
From the Cambridge English Corpus
This means that at the outset, we will deal with the context of ordinary differential equations.
From the Cambridge English Corpus
Their restriction to shallow marine environments suggests that photosynthesis was the primary means of organic matter production for these benthic communities.
From the Cambridge English Corpus
Physically, this is because, on average, the mean shear causes larger separation between particle pairs so that larger eddies disperse them.
From the Cambridge English Corpus
On the other hand, each form has its own unique onset and rime, which means that no form benefits from phonetic consistency.
From the Cambridge English Corpus
This task could be carried out entirely by means of (purely) natural-scientific (naturwissenschaftlichen) methods.
From the Cambridge English Corpus
In the 1920s, the school was eager to distinguish itself from the government schools as a means of attracting more pupils.
From the Cambridge English Corpus
Observation and interviews with the children and the classroom teachers were used as a means to gather data on the potentially psychological impact of singing.
From the Cambridge English Corpus
Each point represents the mean values of these 2 variables measured on 2 large hooks from each of 10 protoscoleces from a single isolate.
From the Cambridge English Corpus
Having only one parent might have meant a household necessity of replacing the loss/absence of one parent.
From the Cambridge English Corpus
These examples are from corpora and from sources on the web. Any opinions in the examples do not represent the opinion of the Cambridge Dictionary editors or of Cambridge University Press or its licensors.
Collocations with mean
mean
These are words often used in combination with mean.
Click on a collocation to see more examples of it.
adequate means
Consequently, the patient's current ability to cope depends, in part, on whether he achieved adequate means for adaptation when he experienced stress as a child.
From the Cambridge English Corpus
alternate means
The ethics consultant stressed to the attending physician that fluids need not be administered as long as the baby is kept comfortable by alternate means.
From the Cambridge English Corpus
alternative means
Disarmament without providing soldiers with training, as well as help in finding alternative means of livelihood, leads to instability and internal crises.
From the Cambridge English Corpus
These examples are from corpora and from sources on the web. Any opinions in the examples do not represent the opinion of the Cambridge Dictionary editors or of Cambridge University Press or its licensors.
See all collocations with mean
What is the pronunciation of mean?
What is the pronunciation of meant?
Translations of mean
in Chinese (Traditional)
表達, 意思是, 意味著…
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in Chinese (Simplified)
表达, 意思是, 意味着…
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in Spanish
significa, querer decir, significar…
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in Portuguese
querer dizer, significar, querer fazer algo…
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in more languages
in Marathi
in Japanese
in Turkish
in French
in Catalan
in Dutch
in Tamil
in Hindi
in Gujarati
in Danish
in Swedish
in Malay
in German
in Norwegian
in Urdu
in Ukrainian
in Russian
in Telugu
in Arabic
in Bengali
in Czech
in Indonesian
in Thai
in Vietnamese
in Polish
in Korean
in Italian
म्हणजे, हवेच असणे, इच्छा असणे…
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～を意味する, ～をもたらす, ～と（意味して）言う…
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demek, anlamına gelmek, demek istemek…
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signifier, vouloir dire, avoir l’intention de…
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significar, comportar, voler dir…
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krenterig, gemeen, kwaadaardig…
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ஒரு யோசனை, சிந்தனை அல்லது உண்மை போன்ற ஒன்றை வெளிப்படுத்த அல்லது பிரதிநிதித்துவப்படுத்த, நீங்கள் சொல்வதற்கு முக்கியத்துவம் சேர்க்க பயன்படுகிறது…
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(किसी विचार या तथ्य का) मतलब, अभिप्राय, अपने कथन पर बल देने के लिए प्रयुक्त…
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અણસાર આપવો, સૂચવવું, તમે જે કહો છો તે ભારપૂર્વક કહેવા માટે વપરાય છે.…
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nærig, fedtet, ondskabsfuld…
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snål, simpel, gemen…
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kedekut, jahat, kejam…
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knauserig, gemein, bösartig…
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bety, mene, ekkel…
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معنی ہونا, مراد ہونا, مطلب ہونا…
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скупий, скнарий, низький…
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значить, иметь в виду, означать…
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ఆర్డము, మీరు చెప్పా దానికి ఒత్తు వేయటం, ఒక నిర్దిష్ట ఫలితాన్ని పొందు…
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يَعْني, يُؤَدّي إلى, يَقصُد…
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অর্থ প্রকাশ করা, কোনো ভাবনা, চিন্তা বা তথ্য ব্যক্ত করা বা উপস্থাপিত করা…
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lakomý, hanebný, nečestný…
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kikir, jahat, kejam…
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ใจแคบ, ทำให้เสียหาย, อารมณ์เสีย…
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keo kiệt, có vẻ là, tầm thường…
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znaczyć, oznaczać, mieć na myśli…
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-를 의미하다, -(결과를) 의미하다, -를 의도하다…
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significare, voler dire, intendere…
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mealtime
mealy
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BETA
mean
mean business idiom
mean curvature
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mean free path
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hold forth
UK
/həʊld/
US
/hoʊld/
to talk about a particular subject for a long time, often in a way that other people find boring
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Contents
English
mean (EXPRESS)
mean (HAVE RESULT)
mean (INTEND)
be meant to do sth
mean (HAVE IMPORTANCE)
mean (NOT GENEROUS)
mean (NOT KIND)
mean (VIOLENT)
mean (GOOD)
mean (BAD QUALITY)
mean (MATHEMATICS)
mean (MATHEMATICS)
mean (METHOD)
Verb
mean (EXPRESS)
mean (HAVE RESULT)
mean (HAVE IMPORTANCE)
mean (INTEND)
Noun
mean (AVERAGE)
Adjective
mean (NOT KIND)
mean (GOOD)
Noun
Verb
Adjective
Examples
Collocations
Translations
Grammar
All translations
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15894 | https://www.ebay.com/itm/167740607074?itmmeta=01K3J5Y4KR8XCNEN2EB2DWZ84K&hash=item270e1e8a62:g:VRoAAeSwkM5oq6pZ&itmprp=enc%3AAQAKAAAA4MHg7L1Zz0LA5DYYmRTS30m8PzNXPXbUkhcNxsT94MdOPE1TUVBaNrHrq1yfWAY%2BJ9ApYHoVFj44nUO4MpN7YFnHCDag%2Fzy1yhtPoxNFR%2FonZ016xDTtKhOnVKJ24yTvfiXMZCJO5Ip0Yb039BVnmTJrZ3ccUUJ69OP76rYpCE%2FiCCfisGd5i%2Bp%2FaNZ4IxhoMGfqykCA7aRUKOH0RJbM1vAUGvTkKcCK4yNuCbaEnjONMv1z0AcOFzaiu7Y8zHZwCQzROXtDinkTdxlk9F45wRgwvlzt%2FG2BGuWk%2FX5g1Jjz%7Ctkp%3ABk9SR4DK-MWcZg | Pratesi classic three lines Bath sheet Luxury Cotton Made In Italy. Beautiful | eBay
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9m (1020)- Feedback left by buyer. Past 6 months Verified purchase Excellent seller item came exactly as described super fast shipping 5⭐️ Pratesi Original Classic Ivory Federico 2standard Shams 20x28”Paradise Fabric (#167327214195)
c6 (179)- Feedback left by buyer. Past 6 months Verified purchase These sheets are the best! Everything with the transaction went fine. Pratesi bed sheets queen set (4pieces) Top Sheet , Bottom Fitted , 2 Std. Shams (#165842043081)
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sr (42)- Feedback left by buyer. Past year Verified purchase So very beautiful. Thank you so much. Highly recommend this seller. Pratesi Original Queen sheet set (4pieces)Classic Chain w/orange , Iconic (#166283914157)
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nt (63)- Feedback left by buyer. More than a year ago Verified purchase beautiful quality and great packaging. responsive seller! great A++++ Pratesi Original queen set Raso white Top Sheet , Bottom Fitted , 2 Std. Shams (#165992190027)
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9m (1020)- Feedback left by buyer. Past 6 months Verified purchase Excellent seller item came exactly as described super fast shipping 5⭐️ Pratesi Original Classic Ivory Federico 2standard Shams 20x28”Paradise Fabric (#167327214195)
c6 (179)- Feedback left by buyer. Past 6 months Verified purchase These sheets are the best! Everything with the transaction went fine. Pratesi bed sheets queen set (4pieces) Top Sheet , Bottom Fitted , 2 Std. Shams (#165842043081)
ee (30)- Feedback left by buyer. Past year Verified purchase Quick shipping, pleasant seller, beautiful product. Pratesi King fitted , Original Angel Skin Cotton 78”x80”x15” White (#167166957727)
jn (23)- Feedback left by buyer. More than a year ago Verified purchase Great seller. Received item as listed, perfect condition, prompt shipping. Pratesi King fitted , Original Angel Skin Cotton 78”x80”x15” White (#166760417565)
sr (42)- Feedback left by buyer. Past year Verified purchase So very beautiful. Thank you so much. Highly recommend this seller. Pratesi Original Queen sheet set (4pieces)Classic Chain w/orange , Iconic (#166283914157)
ax (150)- Feedback left by buyer. More than a year ago Verified purchase Great seller. Item was accurately described. Fast shipper. Would purchase from this seller again Pratesi King fitted , Original Angel Skin Cotton 78”x80”x15” White (#166774408362)
ne (601)- Feedback left by buyer. Past year Verified purchase Seamless transaction. #1 e-Bayer will buy from again Pratesi King fitted , Original Angel Skin Cotton 78”x80”x15” White (#167088425641)
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15895 | https://walshbr.com/textanalysiscoursebook/book/issues/google-ngram/ | Google NGram Viewer
Introduction to Text Analysis
Table of Contents Preface Acknowledgements Introduction Issues in Digital Text Analysis Close Reading Crowdsourcing Digital Archives Data Cleaning Cyborg Readers Reading at Scale Topic Modeling Classifiers Sentiment Analysis Conclusion
License and Copyright
Introduction to Text Analysis
A Coursebook
Google NGram Viewer
The Google NGram Viewer is often the first thing brought out when people discuss large-scale textual analysis, and it serves nicely as a basic introduction into the possibilities of computer-assisted reading.
The Google NGram Viewer provides a quick and easy way to explore changes in language over the course of many years in many texts. Provide a word or comma-separated phrase, and the NGram viewer will graph how often these search terms occur over a given corpus for a given number of years. You can specify a number of years as well as a particular Google Books corpus.
The tool allows you to search hundreds of thousands of texts quickly and, by tracking a few words or phrases, draw inferences about cultural and historical shifts. If we search on ‘science’ and ‘religion,’ for example, we could draw conclusions about their relative importance at various points in last few centuries.
Looking at the graph, one could see evidence for an argument about the increasing secularization of society in the last two centuries. The steady increase of usage of the word science over the last 200 years accompanied by the precipitous decline of the word religion beginning in the mid-nineteenth century could provide concrete evidence for what might otherwise be anecdotal. But not so fast: what is actually being measured here? We need to ask questions about a number of pieces of this argument, including ones regarding:
Corpus
Methodology
Interpretation
Corpus
With any large-scale text analysis like this, the underlying data is everything. Imagine running the same word search for ‘science’ and ‘religion’ over 1000 texts used in religious schools or services. It would probably look quite different! The same would hold true if we targeted only biology, botany, and physics textbooks over the same time period. While these are fairly stark examples, the same principle holds true: the input affects the output. The data we choose for a study can skew our conclusions, and it is important for us to think carefully about their selection as a part of the process.
What is the corpus, or set of texts, being used to generate this data?
Where is this data coming from?
The Google NGram Viewer offers a dropdown menu where you can select a corpus to study. Our results would look a lot different depending on which corpus we selected. The corpora for these options are pulled from the Google Books scanning project (to see similar visualizations of your own corpus, you could try working with Bookworm, a related tool). This raises a number of difficulties. As Eitan Adam Pechenick, Christopher M. Danforth, and Peter Sheridan Dodds have noted, the corpus only has one copy of each book in its dataset. So things do not get scaled for circulation or popularity. A book that only sells one copy is weighted the same as a book that sells a thousand copies: they are both a single copy according to Google’s methods.
The Google Books corpus has also, at times, been criticized for its heavy reliance on poor quality scans of texts to generate their data (more on this in later chapters). The computer can’t infer, for example, that the mispelling ‘scyience’ should be lumped in with the results for ‘science.’ Any underlying problems in scanning or uploading texts will skew the results. In addition, the results are better after 1820. There were far fewer books published before then, and even fewer are on Google Books.
As Ted Underwood suggests, when approached with a healthy sense of skepticism, many of these issues do not discount the use of the tool for “relative comparisons between words and periods” after 1820 or so. We can’t know direct truths through the viewer, but we can still use the data for analysis. For now, just remember that graphs can appear to express fact when, in fact, the data is murky, subject for debate, or skewed.
Methodology
Even with a perfect corpus, our choices can make a big difference in the results we produce. The above search only accounts for single words, but there are more nuanced ways of using the NGram Viewer. An n-gram is another name for a sequence of words of length n. Take this short phrase:
‘a test sentence.’
We have three n-grams of length 1 (“a”, “test” and “sentence”), two n-grams of length 2 (“a test” and “test sentence”), and 1 n-gram of length 3 (“a test sentence”). Or, we could use shorthand: we have 3 unigrams or tokens, 2 bigrams, and 1 trigram. These are just fancy ways to describe different ways of chunking up a piece of text so that we can work with it. And we can do the same thing in the NGram Viewer. Take this NGram for the token ‘scandal’ in an English corpus:
It appears like something fairly dramatic happened around 1660 that caused a massive spike in the usage of ‘scandal.’ This in itself could be significant, but we might be interested in more nuanced readings of this data. We might want to see, say, bigrams containing scandal like ‘political scandal’ and ‘religious scandal’ to observe when certain types of scandals come into prominence. The NGram Viewer allows for a number of nuanced searches that you can read about here. For now, let’s try out a wildcard search - ‘ scandal’:
The asterisk in searches like this matches anything, so it will return all two-word phrases containing ‘scandal’ as a second word. And, handy for us, it will show us the top ten uses. In this case, they’re almost all articles or prepositions: ‘the scandal,’ ‘a scandal,’ ‘of scandal,’ etc. And they all seem to spike around 1660 as well. We would need more information about this time period to tell exactly what is going on here, and to do so we might want to specifically exclude these common usages. Given the relative unreliability of N-Grams before 1820, this dramatic uptick might be due to just a few works that used the term “scandal” around this time – and might not be representative of larger patterns.
We might also want to look at different forms of the same word. After all, the above search only captures the singular form of ‘scandal’, but any word can occur in multiple forms over the course of a corpus. The NGram Viewer can account for this as well:
That massive spike we see in the use of ‘scandal’ is not quite matched by other forms of the word. In particular, the adjective form ‘scandalous’ enjoys more usage until the mid-nineteenth century. Maybe scandal as a noun, as an idea, as a thing unto itself explodes onto the scene in the mid-nineteenth century, wereas before it was something more a thing attached to other people, places, and events.
To drill down more deeply into another term relevant to this course, check out this ngram of the word ‘crime’ in the English corpus:
According to this chart, after a drop during the early-eighteenth century, English writers discussed crime more consistently and ubiquitously than ever before. But what about authors writing in other languages? Here is the same search in French.
The general trend of more mentions of crime in the 19th century than the 20th holds true in both the French and English corpora. However, if you pay careful attention to the y-axis you will note that French authors actually are mentioning crime far more frequently relative to the rest of the writing at the time. The trends are similar, but the percentage of times ‘crime’ shows up is much higher in France. In England during this time, uses of the word hover around 0.0045. French writing mentioning ‘crime’ is over double that percentage during the same period, and it does not dip down to that number until 1880.
You will also note a different trajectory to these two N-Grams. In the English-language corpus, mentions of “crime” go down gradually over the course of the nineteenth century. In contrast, there is a big spike in the French corpus which starts going down quite dramatically in the 1830s. When you read portions of Louis Chevalier’s Laboring Classes and Dangerous Classes in Paris during the First Half of the Nineteenth Century later in the term, you’ll get a sense of why this interest in crime surges in the early nineteenth century and then dies down.
If we were using N-Grams for more than just a demonstration, we would want to do a lot more research and thinking about both language and history. For instance, in the above example, is the fact that French authors seem to be using “crime” more often than English-language ones due to a difference in language and usage? Perhaps English authors often use a synonym for crime, whereas French ones do not? Or does it reflect the fact that French authors were more concerned with crime than English ones?
Or, let’s say we were interested in history of scientific racism in European and American thought. In that case, we might want to know about the trajectory of the word “race” over time.
This N-Gram shows an increasing use of this term over the course of the eighteenth and nineteenth century, peaking around 1890 and then gradually declining in the twentieth century, albeit with some upswings. On the one hand, scientific racism had one of its heydays in the late nineteenth century, so maybe this N-Gram shows this historical trend. But as soon as you think more about this topic, you would realize that it’s a lot more complicated that this. For one, it might strike you as a little odd that the line dips in the late 1950s and early 1960s, an era when the Civil Rights movement was emerging. Were people really writing less about race then than before? Alternatively, the term “race” can mean a lot of different things; in this case, the results we are interested in (the categorization of people according to their phenotypes) are undoubtedly getting jumbled in with references to sporting events and elections. And as soon as we started doing research on the history of scientific racism, we would learn that writers used the term “race” to refer to groups of people in different ways in the eighteenth century than they did at the end of the nineteenth century. We would also want to think about terms that are associated with or used as synonyms for race. Maybe authors in the twentieth century were using other words to talk about race? If so, what might those be?
So, language changes over time. A single word might radically change in usage over the centuries in ways that skew our results. We also use different terms over time to describe the same phenomena. These are all things we would want to say a lot more in any interpretations using N-Grams. We would also need to consider what they can (and cannot) tell us and think about potential problems in my reading. But hopefully the implications of the technology will be exciting to you nonetheless.
Digital methods can allow us to make observations about vast numbers of texts. Far more than you would be able to read yourself.
That last phrase should cause some alarm: we haven’t actually read any of these texts, but we are making observations about them nonetheless. We hope you will think deeply about the implications about such an act.
What does this form of reading lose?
What does it gain?
How it can it be approached in ways that minimize the former and maximize the latter?
Interpretation
Of course, these graphs mean nothing on their own. It is our job to look at the results and describe them in meaningful ways. But be critical of what you see. You might find something interesting, but you might be looking at nonsense. It is your job to tell the difference. Beware of apophenia, the all to human urge to look at random data and find meaningful patterns in it. You can find wild patterns in anything if you look hard enough. After all, visualizations can confuse as much as clarify. Numbers and graphs do not carry objective meaning. Miriam Posner summarized it pithily on Twitter once:
Always think. Never let a graph think for you.
Further Resources
Ted Underwood on “How not to do things with words” for helpful criticisms of studies employing the Ngram Viewer.
Danny Sulivan on “When OCR Goes Bad: Google’s Ngram Viewer & The F-Word” for more on Google NGram and OCR.
Geoff Nunberg on “Google Books: A Metadata Train Wreck” for more problems with the viewer. |
15896 | https://www.vin.com/apputil/content/defaultadv1.aspx?id=3850235&pid=8768&print=1 | Principles and Application of Cerclage Wire Full and Hemicerclage Wiring - WSAVA 2003 Congress - VIN
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Principles and Application of Cerclage Wire Full and Hemicerclage Wiring
World Small Animal Veterinary Association World Congress Proceedings, 2003
Roger Clarke, BVSc, MRCVS, FACVSc, Registered Veterinary Specialist, Small Animal Surgery
Associate Professor, Bundoora Veterinary Clinic and Hospital
Bundoora, Victoria, Australia
Introduction
Orthopaedic wire is commonly used in small animal fracture repair to hold fragments of bone together by static inter-fragmentary compression. This may be achieved by compressing individual fragments together as in inter-fragmentary wiring, or by cerclage wiring,where wire is wrapped around a bundle of fragments arranged in a stable anatomic conformation and compressing the bundle together.
Orthopaedic wire is a non-reactive stainless steel alloy, which is far more malleable than the stainless steel alloy used to make bone plates or pins. There are three basic forms of cerclage wiring, full cerclage, hemicerclage wiring and tension band wiring, which is a specialised form of hemicerclage. This paper will deal with the first two.
Cerclage wire has an important part to play in small animal orthopaedic surgery and the majority of failures can be attributed to a neglect of the basic principles, which govern the use of cerclage wires.
Full circlage wiring
Full circlage wiring utilizes a 360° circumferential wire placed around the bone at a fracture site. This use is generally restricted to the diaphyseal segments of long bones. The fracture is carefully reconstructed and the fragments are wired in place prior to applying the definitive form of fixation. Full 360° anatomic reconstruction of the fracture at the level of the cerclage wire is mandatory, otherwise the fragments will move and collapse and the wire will loosen. The analogy of wooden barrel staves as the fracture fragments and the metal rings as the cerclage wires is often used to illustrate the need for stable anatomic reconstruction. Full cerclage wiring is only suitable for long oblique diaphyseal fractures where the length of the fracture is greater than twice the diameter of the bone at the fracture site (>45 0). If the fracture line is greater than twice the diameter of the bone at the fracture site, the wire will achieve inter-fragmentary compression. If the length of the fracture line is less than twice the bone diameter (<45 0) then shearing forces will be produced which will disrupt the fracture. Full cerclage wires can be used in conjunction with a K-wire or Steinmann pin to stabilise oblique diaphyseal fractures of less then 45 0, where full circlage would fail. An inter-fragmentary Kirschner wire is placed perpendicular to the fracture line (and thus oblique to the long axis of the bone) and this wires is then straddled with a full cerclage wire such that the K-wire prevents the cerclage wire from slipping. The wire is placed proximal to the proximal end of the K-wire and distal to the distal part. Another alternative in such fractures is to use a hemicerclage wire placed perpendicular to the fracture line in place of a full cerclage wire.
When the shaft of the bone tapers, there is a natural tendency for full cerclage wire to slip to a narrower part of the bone thereby becoming loose. This can be countered by placing a K-wire through the bone, across the fracture site and wiring around the k-wire in such a way that the wire cannot slip down the shaft of the bone. Alternatively the bone can be notched where the wire encircles the shaft, this can result in fractures unless it is done carefully. Another alternative to prevent slippage, is hemicerclage wiring.
Hemicerclage wiring
Hemicerclage is where the wire is placed through, rather than around one of the main fragments of bone. The other components of the fracture are then enclosed in the wire to hold them firmly to the main fragment. Hemicerclage wires may be used in the repair of long bone fractures, particularly at or near the metaphysis where the circumference of the bone changes in diameter. Holes are drilled through the cortex and the wire is then passed such that it crosses across the fracture line on the outer cortex of the bone. It may or may not cross the fracture line in the medulla. Because the wire is physically fixed to the bone it is prevented from sliding up or down the shaft of the bone.
In some instances, the hemicerclage wire may be passed around an intramedullary pin to compress the pin against the inner cortex. This has the advantage of improving the stability of the repair when using intra-medullary bone pins, but the distinct disadvantage of making the wire loose when the pin is removed, thereby necessitating removal of the loose wires.
Many different patterns have been described for hemicerclage wires. The figure-of-eight anti-rational wire designed for use in transverse fractures is one example. This technique is useful in smaller animals and may prove very cost-effective when used in conjunction with an intra-medullary pin.
Many of the more exotic configurations may exceed the limit of the wire's strength and they should be used with caution.
Key points to remember
Orthopaedic wire is usually not strong enough to be used as the sole means of repair of diaphyseal long bone fractures and should never be used in this fashion.
Orthopaedic Wiring is used as an adjunct to maintain fracture reduction (either temporarily or permanently), while the bone is primarily stabilised by a stronger mechanism such as a plate, intramedullary pin or external fixation frame. Multiple cerclage wires provide adjunctive fixation against compressive shear, bending & rotational forces
PRINCIPLES OF PROPER CERCLAGE WIRE APPLICATION
Use monofilamentwire of the correct size for the patient. The wire must be monofilament and of adequate diameter for the size (weight) of the animal--0.8mm or 20 gauge (cats and small dogs), 1mm or 18 gauge (medium to large dogs), 1.2mm and 1.25mm (16 gauge) can be used in large to giant breeds. When in doubt use a larger size wire. Wire size is determined by the size of the patient and the anatomical location of the fracture. One of the most common errors in wiring is that too small a wire is used and it cannot withstand the loads of weight bearing. In long bone fractures all wires should be of a similar diameter. A paper 1 in 1974 suggested that cerclage implants may lead to circulatory compromise and lysis of the underlying bone.
However, Rhinelander,2 in 1968 showed that the use of tightly placed cerclage wires of small diameter did not disturb the centripetal flow of blood from the medulla to the periosteum of the bone and other clinical reports have since confirmed this finding.3, 4, 5. 1.25 mm is probably the maximum diameter that can be safely used. Wide cerclage implants, such as the now discontinued Parham bands, will disrupt periosteal circulation and cause necrosis of the underlying bone. These have obviously fallen out of favour.
The bone must have stable anatomical reduction over a full 3600at the level of application or the compression produced by the wire will cause the bone to collapse or fragment further. The practical limit for cerclage to work well is usually three large well-reduced fragments of bone.
Cerclage wiring is most suited to long oblique fractures where the length of the fracture is greater than twice the diameter of the bone at the fracture site. If the fracture line is greater than twice the diameter of the bone at the fracture site, the wire will achieve inter-fragmentary compression. If the length of the fracture line is less than twice the bone diameter then shearing forces will be produced which will disrupt the fracture.
If the area of the bone is tapering then the wire must be prevented from migrating form its implantation site, either by notching the bone, using hemicerclage or an inter-fragmentary K-wire.
Wires should be at least 1 cm apart, as there is no mechanical advantage to closer spacing. Wires are placed at least 5 mm from the ends of the fracture to avoid crushing the fracture ends and to provide mechanical advantage. AVOID placing wires into or near the fracture site.
More than one wire should be used.One wire is not sufficient and is biomechanically unsound as it concentrates bending and rotational forces on the one wire and leads to early breakage..
Wires should be placed perpendicular to the long axis of the bone unless specific procedures are taken to ensure that the wire maintains its oblique position.
Do not entrap soft tissues between wire and the bone when passing the wire around the bone.
AVOID placing the wires around both bones, in two bone systems such as the radius and ulna and tibia and fibula.
The wires used must be tight and must not loosen as other wires are placed, or as the bone heals or as the animal moves. Loose wires disrupt the capillary circulation to the periosteum, cause bone necrosis and lysis and interfere with callus formation.
Wiring Techniques
Passing cerclage wires around the bone
Wires are passed around the bone using specially constructed hollow wire guides or solid wire guides with an eye in one end. The wire passer or guide is passed around the bone as close to the outer cortex as possible taking care not to trap soft tissue between the wire and the bone. When using hollow wire passers, the wire is inserted into the tubing until it appears on the other side of the bone and the passer is then withdrawn. If looped wires are used the wire should be passed through the passer so that the loop does not become trapped in the tube. Wires passed with a wire passer which utilises an eye, should be pulled through the tissues in such a way as to minimise trauma and to avoid entrapment of soft tissues. Where the soft tissues are attached to the bone, the wire passer should be passed THROUGH the soft tissues as close to the bone as possible and not over the muscles.
Once passed around the fracture, the wire is carefully aligned perpendicular to the long axis of the bone and tightened. Where possible the fracture should be perfectly reduced and the fracture site compressed using bone forceps before the wire is tightened. Wire should be placed and tightened and if the tightening of one wire causes the wires previously placed to loosen, then the loose wires should be tightened or replaced with tight wires. Twisted wires can be tightened further should this occur, but looped wires have to be replaced.
Tightening cerclage wires
There are two main methods used to tighten cerclage wires, twisting wires or utilising a preformed loop on one end. The oldest means is by simply twisting the two ends of the wire together. As the wire ends are twisted together symmetrically, traction is applied to keep the spiral of wire even on both sides. If a spiral does not form and one wire coils around the other, the repair is weakened considerably. Wires can be held in wire twisting forceps or special wire twisting devices designed to keep the wires in a uniform position. While tightening, the wire must be constantly adjusted to keep the cerclage section perpendicular to the long axis of the bone. The twist must be started close to the bone and excessive twists avoided. Three twists are sufficient. Any twists over three is excessive and excessive twisting will weaken the wire without tightening the cerclage. Once the wire is tight, the twisted ends should NOT be bent over to lie flat against the bone, as this has been shown to considerably decrease the amount of pressure generated in the twisted wire. The excess wire is cut off leaving a minimum of three full twists. Wires should be twisted in such a way that the twisted ends do not irritate essential soft tissues such as nerves and blood vessels. Muscle tissue will form a fibrotic capsule over the wire ends to protect the tissues from further damage.
Advantages of twisted cerclage wires
Can re-tighten wires
Simple to apply and have great versatility
More economical to use
Disadvantages of twisted cerclage wires
Wires do not align perpendicular to the long axis and tend to twist as they are tightened
Lower final cerclage wire tension compared to looped wires
Twisted ends protrude into soft tissues and if twisted over are weakened considerably
Pre-formed loop cerclage wires
The other method of wire tightening uses cerclage wires with pre-formed loops. These can be home-made or purchased commercially. They utilise a specially designed wire tightener. The non-looped end of the cerclage wire is placed around the bone, using a hollow wire guide, and is then threaded through the pre-formed loop at the end of the wire. The free end is then placed in the wire tightener and attached to a crank handle. Traction is applied to the wire by wrapping the free end around the crank handle. When the wire is tight, tension is maintained while the tightener device is bent over at 180°. The crank handle is then reversed, exposing a short length of the wire so that it may be cut approximately 1 cm from the loop. The end of the wire lies flat against the bone and does not project into the soft tissues. The amount of tension required is just enough to distort the loop and the expansile forces required to disrupt these wires, although slightly less than twisted wires, exceed the requirements for bone healing.
Advantages of loop wires
The final cerclage tension generated by pre-formed loop wires is superior to those of the twist type
The end of the wire can be laid flat against the bone
When tightening these types of wire they remain perpendicular to the long axis of the bone and compress the fracture site equally. There is no tendency to twist at an angle to the long axis of the bone
Disadvantages of loop wires
Loop tightened wires cannot be re-tightened once the tightening device has been removed and require replacement
Loop tightened wires are more expensive than rolls of ordinary wire
Key points to remember
The final amount of tension on a wire is dependent on the reduction and compression of the fracture before wire is applied; the instrument used to tighten the wire, the type of knot used to secure the wire, and the experience of the surgeon.
Removing cerclage wires
Loose orthopaedic wires usually need to be removed. Tight wires can usually be left in place. If wires are overgrown with bone, it can be more traumatic to remove the bone than to leave the wire in place. When placing wires it is helpful to record the location of the knot on the radiograph or medical record. If all wires are placed in a similar fashion it greatly simplifies later removal. The wire is cut on one side of the knot using side-cutters and the wire is withdrawn by pulling on the knotted or looped end. The end of the wire that is pulled through the tissue should be as smooth and atraumatic as possible. Excessive pulling can fracture bone, so leverage over another instrument is a useful way of spreading the tension and gently removing wires.
References
1.Newton C.D., Hohn, R.B. (1974) Fracture non-union resulting from cerclage appliances. J Am Vet Med Assoc. 164:503.
2.Rhinelander F.W. (1968), The normal microcirculation of diaphyseal cortex and its response to fracture. J Bone Joint surg 50A:784
3.Hinko P.J., Rhinelander F.W. (1975) Effective use of cerclage in the treatment of long-bone fractures in dogs. J Am Vet Med Assoc 166:520.
4.Withrow S.J., Holmberg D.L. (1977) Use of full cerclage wires in the fixation of 18 consecutive longbone fractures in small animals. J Am Anim Hosp Assoc. 13:735.
5.Withrow S.J. (1978) Use and misuse of full cerclage wires in fracture repair. Vet Clin North Am. 8:201.
Further reading
1.Brinker WO, Piermattei DL, Flo GL. Handbook of Small Animal Orthopedics and Fracture Repair. 3rd ed. 1997, Philadelphia, WB Saunders, pp. 104-112.
2.Wilson JW, Belloli DM, Robbins T. Resistance of cerclage to knot failure. JAVMA 1985: 187:389-391
3.Blass CE, et al. Static and dynamic cerclage wire analysis. Vet Surg 1986:15; 181-184
Speaker Information
(click the speaker's name to view other papers and abstracts submitted by this speaker)
Roger Clarke, BVSc, MRCVS, FACVSc
Registered Veterinary Specialist, Small Animal Surgery
Associate Professor, Bundoora Veterinary Clinic and Hospital
Bundoora, Victoria, Australia
URL:
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15898 | https://en.wikipedia.org/wiki/Counting | Jump to content
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Contents
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1 Forms of counting
2 Inclusive counting
3 Education and development
4 Counting in mathematics
5 See also
6 References
Counting
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From Wikipedia, the free encyclopedia
Finding the number of elements of a finite set
For its application to music, see Counting (music).
Counting is the process of determining the number of elements of a finite set of objects; that is, determining the size of a set. The traditional way of counting consists of continually increasing a (mental or spoken) counter by a unit for every element of the set, in some order, while marking (or displacing) those elements to avoid visiting the same element more than once, until no unmarked elements are left; if the counter was set to one after the first object, the value after visiting the final object gives the desired number of elements. The related term enumeration refers to uniquely identifying the elements of a finite (combinatorial) set or infinite set by assigning a number to each element.
Counting sometimes involves numbers other than one; for example, when counting money, counting out change, "counting by twos" (2, 4, 6, 8, 10, 12, ...), or "counting by fives" (5, 10, 15, 20, 25, ...).
There is archaeological evidence suggesting that humans have been counting for at least 50,000 years. Counting was primarily used by ancient cultures to keep track of social and economic data such as the number of group members, prey animals, property, or debts (that is, accountancy). Notched bones were also found in the Border Caves in South Africa, which may suggest that the concept of counting was known to humans as far back as 44,000 BCE. The development of counting led to the development of mathematical notation, numeral systems, and writing.
Forms of counting
[edit]
Further information: Prehistoric numerals and Numerical digit
Verbal counting involves speaking sequential numbers aloud or mentally to track progress. Generally such counting is done with base 10 numbers: "1, 2, 3, 4", etc. Verbal counting is often used for objects that are currently present rather than for counting things over time, since following an interruption counting must resume from where it was left off, a number that has to be recorded or remembered.
Counting a small set of objects, especially over time, can be accomplished efficiently with tally marks: making a mark for each number and then counting all of the marks when done tallying. Tallying is base 1 counting.
Finger counting is convenient and common for small numbers. Children count on fingers to facilitate tallying and for performing simple mathematical operations. Older finger counting methods used the four fingers and the three bones in each finger (phalanges) to count to twelve. Other hand-gesture systems are also in use, for example the Chinese system by which one can count to 10 using only gestures of one hand. With finger binary it is possible to keep a finger count up to 1023 = 210 − 1.
Various devices can also be used to facilitate counting, such as tally counters and abacuses.
Inclusive counting
[edit]
Inclusive/exclusive counting are two different methods of counting. For exclusive counting, unit intervals are counted at the end of each interval. For inclusive counting, unit intervals are counted beginning with the start of the first interval and ending with end of the last interval. This results in a count which is always greater by one when using inclusive counting, as compared to using exclusive counting, for the same set. Apparently, the introduction of the number zero to the number line resolved this difficulty; however, inclusive counting is still useful for some things.
Refer also to the fencepost error, which is a type of off-by-one error.
Modern mathematical English language usage has introduced another difficulty, however. Because an exclusive counting is generally tacitly assumed, the term "inclusive" is generally used in reference to a set which is actually counted exclusively. For example; How many numbers are included in the set that ranges from 3 to 8, inclusive? The set is counted exclusively, once the range of the set has been made certain by the use of the word "inclusive". The answer is 6; that is 8-3+1, where the +1 range adjustment makes the adjusted exclusive count numerically equivalent to an inclusive count, even though the range of the inclusive count does not include the number eight unit interval. So, it's necessary to discern the difference in usage between the terms "inclusive counting" and "inclusive" or "inclusively", and one must recognize that it's not uncommon for the former term to be loosely used for the latter process.
Inclusive counting is usually encountered when dealing with time in Roman calendars and the Romance languages. In the ancient Roman calendar, the nones (meaning "nine") is 8 days before the ides; more generally, dates are specified as inclusively counted days up to the next named day.
In the Christian liturgical calendar, Quinquagesima (meaning 50) is 49 days before Easter Sunday. When counting "inclusively", the Sunday (the start day) will be day 1 and therefore the following Sunday will be the eighth day. For example, the French phrase for "fortnight" is quinzaine (15 [days]), and similar words are present in Greek (δεκαπενθήμερο, dekapenthímero), Spanish (quincena) and Portuguese (quinzena).
In contrast, the English word "fortnight" itself derives from "a fourteen-night", as the archaic "sennight" does from "a seven-night"; the English words are not examples of inclusive counting. In exclusive counting languages such as English, when counting eight days "from Sunday", Monday will be day 1, Tuesday day 2, and the following Monday will be the eighth day.[citation needed] For many years it was a standard practice in English law for the phrase "from a date" to mean "beginning on the day after that date": this practice is now deprecated because of the high risk of misunderstanding.
Similar counting is involved in East Asian age reckoning, in which newborns are considered to be 1 at birth.
Musical terminology also uses inclusive counting of intervals between notes of the standard scale: going up one note is a second interval, going up two notes is a third interval, etc., and going up seven notes is an octave.
Education and development
[edit]
Main article: Pre-math skills
Learning to count is an important educational/developmental milestone in most cultures of the world. Learning to count is a child's very first step into mathematics, and constitutes the most fundamental idea of that discipline. However, some cultures in Amazonia and the Australian Outback do not count, and their languages do not have number words.
Many children at just 2 years of age have some skill in reciting the count list (that is, saying "one, two, three, ..."). They can also answer questions of ordinality for small numbers, for example, "What comes after three?". They can even be skilled at pointing to each object in a set and reciting the words one after another. This leads many parents and educators to the conclusion that the child knows how to use counting to determine the size of a set. Research suggests that it takes about a year after learning these skills for a child to understand what they mean and why the procedures are performed. In the meantime, children learn how to name cardinalities that they can subitize.
Counting in mathematics
[edit]
Main article: Combinatorics
See also: Countable set
In mathematics, the essence of counting a set and finding a result n, is that it establishes a one-to-one correspondence (or bijection) of the subject set with the subset of positive integers {1, 2, ..., n}. A fundamental fact, which can be proved by mathematical induction, is that no bijection can exist between {1, 2, ..., n} and {1, 2, ..., m} unless n = m; this fact (together with the fact that two bijections can be composed to give another bijection) ensures that counting the same set in different ways can never result in different numbers (unless an error is made). This is the fundamental mathematical theorem that gives counting its purpose; however you count a (finite) set, the answer is the same. In a broader context, the theorem is an example of a theorem in the mathematical field of (finite) combinatorics—hence (finite) combinatorics is sometimes referred to as "the mathematics of counting."
Many sets that arise in mathematics do not allow a bijection to be established with {1, 2, ..., n} for any natural number n; these are called infinite sets, while those sets for which such a bijection does exist (for some n) are called finite sets. Infinite sets cannot be counted in the usual sense; for one thing, the mathematical theorems which underlie this usual sense for finite sets are false for infinite sets. Furthermore, different definitions of the concepts in terms of which these theorems are stated, while equivalent for finite sets, are inequivalent in the context of infinite sets.
The notion of counting may be extended to them in the sense of establishing (the existence of) a bijection with some well-understood set. For instance, if a set can be brought into bijection with the set of all natural numbers, then it is called "countably infinite." This kind of counting differs in a fundamental way from counting of finite sets, in that adding new elements to a set does not necessarily increase its size, because the possibility of a bijection with the original set is not excluded. For instance, the set of all integers (including negative numbers) can be brought into bijection with the set of natural numbers, and even seemingly much larger sets like that of all finite sequences of rational numbers are still (only) countably infinite. Nevertheless, there are sets, such as the set of real numbers, that can be shown to be "too large" to admit a bijection with the natural numbers, and these sets are called "uncountable". Sets for which there exists a bijection between them are said to have the same cardinality, and in the most general sense counting a set can be taken to mean determining its cardinality. Beyond the cardinalities given by each of the natural numbers, there is an infinite hierarchy of infinite cardinalities, although only very few such cardinalities occur in ordinary mathematics (that is, outside set theory that explicitly studies possible cardinalities).
Counting, mostly of finite sets, has various applications in mathematics. One important principle is that if two sets X and Y have the same finite number of elements, and a function f: X → Y is known to be injective, then it is also surjective, and vice versa. A related fact is known as the pigeonhole principle, which states that if two sets X and Y have finite numbers of elements n and m with n > m, then any map f: X → Y is not injective (so there exist two distinct elements of X that f sends to the same element of Y); this follows from the former principle, since if f were injective, then so would its restriction to a strict subset S of X with m elements, which restriction would then be surjective, contradicting the fact that for x in X outside S, f(x) cannot be in the image of the restriction. Similar counting arguments can prove the existence of certain objects without explicitly providing an example. In the case of infinite sets this can even apply in situations where it is impossible to give an example.[citation needed]
The domain of enumerative combinatorics deals with computing the number of elements of finite sets, without actually counting them; the latter usually being impossible because infinite families of finite sets are considered at once, such as the set of permutations of {1, 2, ..., n} for any natural number n.
See also
[edit]
Calculation
Card reading (bridge)
Cardinal number
Combinatorics
Count data
Counting (music)
Counting problem (complexity)
Counting sheep
Counting-out game
Developmental psychology
Elementary arithmetic
Finger counting
History of mathematics
Jeton
Level of measurement
List of numbers
Mathematical quantity
Ordinal number
Particle number
Subitizing and counting
Tally mark
Unary numeral system
Yan tan tethera (Counting sheep in Britain)
References
[edit]
^ An Introduction to the History of Mathematics (6th Edition) by Howard Eves (1990) p.9
^ "Early Human Counting Tools". Math Timeline. Retrieved 2018-04-26.
^ Macey, Samuel L. (1989). The Dynamics of Progress: Time, Method, and Measure. Atlanta, Georgia: University of Georgia Press. p. 92. ISBN 978-0-8203-3796-8.
^ a b Evans, James (1998). "4". The History and Practice of Ancient Astronomy. Oxford University Press. p. 164. ISBN 019987445X.
^ "Drafting bills for Parliament". gov.uk. Office of the Parliamentary Counsel. 18 June 2020. See heading 8.
^ Butterworth, B., Reeve, R., Reynolds, F., & Lloyd, D. (2008). Numerical thought with and without words: Evidence from indigenous Australian children. Proceedings of the National Academy of Sciences, 105(35), 13179–13184.
^ Gordon, P. (2004). Numerical cognition without words: Evidence from Amazonia. Science, 306, 496–499.
^ Fuson, K.C. (1988). Children's counting and concepts of number. New York: Springer–Verlag.
^ Le Corre, M., & Carey, S. (2007). One, two, three, four, nothing more: An investigation of the conceptual sources of the verbal counting principles. Cognition, 105, 395–438.
^ Le Corre, M., Van de Walle, G., Brannon, E. M., Carey, S. (2006). Re-visiting the competence/performance debate in the acquisition of the counting principles. Cognitive Psychology, 52(2), 130–169.
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15899 | https://rcer.econ.rochester.edu/RCERPAPERS/rcer_541.pdf | UNIVERSITY OF ROCHESTER Bertrand's price competition in markets with fixed costs Saporiti, Alejandro and Colomaz, Germán Working Paper No. 541 September 2008 Bertrand’s price competition in markets with fixed costs∗ Alejandro Saporiti† Germ´ an Coloma‡ September 2008 Abstract We analyze Bertrand’s price competition in a homogenous good market with a fixed cost and an increasing marginal cost (i.e., with variable returns to scale). If the fixed cost is avoid-able, we show that the non-subadditivity of the cost function at the output corresponding to the oligopoly break-even price, denoted by D(pL(n)), is sufficient to guarantee that the market supports an equilibrium in pure strategies with two or more active firms supplying at least D(pL(n)). Conversely, the existence of a pure strategy equilibrium ensures that the cost function is not subadditive at every output greater than or equal to D(pL(n)). As a by-product, the latter implies that the average cost cannot be decreasing over the range of outputs mentioned before. In addition, we also prove that the existence of a price-taking equilibrium is sufficient, but not necessary, for Bertrand’s price competition to possess an equilibrium in pure strategies. This provides a simple existence result for the case where the fixed cost is fully unavoidable. JEL Classification: D43, L13.
1 Introduction In Industrial Organization, the simplest model of price competition, called ‘Bertrand’s price competition’ in honor of its initiator the French mathematician Joseph Bertrand, studies the market of a homogenous good in which a small number of firms simultaneously post a price and commit to sale the quantity of the firm’s product that consumers demand given those posted prices. The classical result in the literature on Bertrand’s competition is the well known Bertrand’s paradox, which says that, if firms are identical, the average cost is constant, and total revenues are bounded, all Nash equilibria in the mixed extension of the pricing game are characterized by two or more firms charging the marginal cost (Harrington, 1989).1 With unbounded revenues, there are also mixed strategy equilibria where prices always excess the marginal cost (Baye and Morgan, 1999; Kaplan and Wettstein, 2000). However, such equilibria are ruled out by the usual assumptions on the demand function, namely, continuity and a finite choke-offprice. Thus, under reasonable market conditions, the message coming ∗We thank Leandro Arozamena, Krishmendu Dastidar, and Alex Dickson for useful comments and suggestions.
†University of Manchester, Manchester M13 9PL, UK. alejandro.saporiti@manchester.ac.uk.
‡Universidad del CEMA, Av. C´ ordoba 374, C1054AAP Buenos Aires, Argentina. gcoloma@ucema.edu.ar.
1If one firm has an absolute cost advantage over its rivals, it prices at the marginal cost of the next to lowest cost firm and captures the entire market. All other firms earn zero profit.
1 out from Bertrand’s competition is that the perfectly competitive outcome, with price equal to marginal cost and zero equilibrium profits, is achieved independently of the number of firms in the market.2 In recent years, there has been a renewed interest for examining the Bertrand’s paradox under different cost conditions. A remarkable work within this literature is Dastidar (1995), who has shown that with decreasing returns to scale, i.e., with an increasing average cost, the result does not hold. Specifically, Dastidar has proved that, for firms with identical, continuous, and convex cost functions, price competition a la Bertrand typically leads to multiple pure strategy Nash equilibria.3 Furthermore, if the cost function is sufficiently convex, even the joint profit-maximizing price can be in the range of equilibrium prices; and, it may actually be easier to arrive to that outcome when there are more firms in the market (Dastidar, 2001).
The reason behind the existence of multiple pure strategy equilibria is simple. With de-creasing returns to scale, being the only firm charging the lowest price and supplying the whole market leads to lower profits because the average cost increases too fast. Therefore, there is an incentive to join the group. This explains why undercutting the rest is not profitable in equilibrium even if the other firms price above the marginal cost. Hoernig (2002) has also found that the same logic applies in the mixed extension. Consequently, there is a continuum of mixed strategy Nash equilibria with continuous support. Moreover, any finite set of pure equilibrium prices that lead to positive equilibrium profits can be supported in a mixed strategy equilibrium.
Unbounded returns are not necessary for this result.
Interestingly, the set of equilibria gets drastically smaller when there is the possibility of lim-ited cooperation among the firms. Indeed, if firms possess an identical and increasing average cost, Bertrand’s price competition admits a unique and symmetric coalition-proof Nash equi-librium (Chowdhury and Sengupta, 2004). The equilibrium price is decreasing in the number of firms; and, in the limit, it converges to the competitive price. If firms have asymmetric costs and they share the market according to capacity, i.e., according to the competitive supply of each firm, a coalition-proof Nash equilibrium always exists. Moreover, if firms do not use weakly dominated strategies, the minimum price charged in any of such equilibria is always above the competitive price, but it converges to the marginal cost as the number of firms increases.4 Regarding Bertrand’s competition with increasing returns to scale, the literature indicates that the existence of a Nash equilibrium, either in pure or mixed strategies, is problematic. On one hand, if the marginal cost is decreasing, Dastidar (2006) has recently shown that, under the usual ‘equal sharing’ tie-breaking rule, which roughly means that consumers split equally among the firms that charge the lowest price, Bertrand competition does not possess a Nash 2If firms cannot observe their rivals’ costs, the precision to price slightly below the rivals disappears. Thus, Spulber (1995) showed that all firms pricing above the marginal cost and getting positive expected profits is an equilibrium. However, as the number of firms increases equilibrium prices converge to the average cost.
3For firms with asymmetric costs, pure strategy equilibria always exists; it could be unique or non-unique; and in any equilibrium all firms with positive sales charge the same price.
4The limiting properties of the set of coalition-proof Nash equilibria are interesting because Novshek and Chowdhury (2003) have shown that the multiplicity of pure strategy Nash equilibria holds even when the market is large. In particular, if the average cost is increasing, they have proved that the limit equilibrium set includes the perfectly competitive price, but it is not a singleton.
2 equilibrium in pure strategies. The existence of mixed equilibria remains an open question.5 On the other hand, when the marginal cost is constant, but there exists an avoidable fixed cost, existing works, including Vives (1999, pg. 118) and Baye and Kovenock (2008), point out that pure strategy Nash equilibria may or may not exist. The reason is firms as usual have the incentive to undercut each other in order to increase their sales; but they may prefer to exit rather than to pay the fixed cost and produce a positive amount of output. Since oligopoly theory is most relevant in markets with significant scale economies, Shapiro (1989, pgs. 344-345) reckoned that the nonexistence of an equilibrium is a serious drawback of the model.
Nonetheless, Bertrand’s price competition with a constant marginal cost and an avoidable fixed cost does indeed possess a pure strategy Nash equilibrium when prices vary over a grid.
Moreover, for a symmetric duopoly with linear demand, Chaudhuri (1996) has shown that, in the limit, as the size of the grid becomes very small, there is a unique equilibrium that converges to the contestable outcome; that is, in the limit, there is average cost pricing with a single firm supplying the whole market and earning zero profits. This result has been extended later on by Chowdhury (2002) to the case with asymmetric firms, finding among other things that as the size of the grid approaches zero, the equilibrium prices converge to the limit-pricing outcome where the price charged by the most efficient firm is just low enough to prevent entry.
With a sunk entry cost, instead of an avoidable fixed cost, a two stage, simultaneous move game of entry and pricing also shows nonexistence of pure strategy Nash equilibria (Sharkey and Sibley, 1993). By contrast, mixed strategy equilibria always exist. If firms are symmetric, as more firms become potential competitors, the equilibrium price distribution places greater weight on high prices, contradicting the usual intuition from perfect competition. Instead, if entrants face different sunk costs, the equilibrium of the game lead to blockaded entry for higher cost firms (Marquez, 1997).
Surprisingly, the analysis of Bertrand’s competition under the more familiar case of variable returns to scale has not received enough attention in the literature. To the best of our knowledge, there are only two papers that deal with this matter. The first article, due to Novshek and Chowdhury (2003), finds that, with a continuous and U-shaped average cost, as the market becomes large, (i.e., as the number of firms increases or, alternatively, as the market size is taken to infinity), the equilibrium set is empty for some parameter values, and it comprises a whole interval of prices for others. The lower bound of this interval is bounded away from the minimum average cost. No conditions are provided to ensure equilibrium existence.
The second article, due to Yano (2006a), studies a pricing game with a more complex set of strategies. Specifically, the strategy of each firm is a paring of a unit price and the set of quantities that the firm is indifferent to sell at that unit price.6 The unsatisfied demand is then proportional rationed; that means, if the total amount that buyers wish to acquire at a given price is different from that which firms offer to sell at that price, each agent on the long side 5Under the less common ‘winner-take-all’ tie-breaking rule, a zero profit Nash equilibrium exists if and only if the monopoly profit function has an initial break-even price. In addition, if the function is left lower semi-continuous and bounded from above, the zero profit’s outcome is unique (Baye and Morgan, 1999).
6The firm is indifferent between any two quantities at a given price if they give rise to the same profit.
3 gets to trade proportionately to the amount that he desires in such a way that the equilibrium between demand and supply is reestablished. Yano argues that, by incorporating this rationing process, the resulting pricing game may be thought of as belonging to the family of Bertrand-Edgeworth price games.7 Several equilibria arise in this framework, including the equilibrium of the standard Bertrand’s price competition and the contestable outcome (see also Yano, 2006b).
Taking Dastidar’s (1995) model as the benchmark, in this paper we reexamine Bertrand’s price competition under variable returns to scale. Like in Dastidar, we suppose that the total cost function C(·) exhibits an increasing marginal cost. However, following Grossman (1981), we assume that the total cost is the sum of a continuous and convex variable cost, V C(·), and a fixed cost, F ≥0.
Telser (1991) calls this type of markets, with U-shaped average cost, ‘Viner industries’. Since we do not restrict a priori the nature of the fixed cost, the paper accommodates cases where the fixed cost is (i) completely avoidable, i.e., C(0) = 0; (ii) partly avoidable, i.e., C(0) ∈(0, F); and (iii) unavoidable, i.e., C(0) = F, in which case we are back to Dastidar’s (1995) scenery. In contrast with the latter case, the first two situations give rise to discontinuities and non-convexities in the firms’ payofffunctions, making the analysis of equilibrium existence a nontrivial exercise.
Within the framework briefly depicted above, this paper investigates necessary and suf-ficient conditions to guarantee the existence of pure strategy Nash equilibria.
Remarkably, when the fixed cost is fully avoidable, we find an interesting and unexplored relationship be-tween Bertrand’s competition and cost subadditivity.8 That relationship indicates that the non-subadditivity of the cost function at the output corresponding to the oligopoly break-even price, denoted by D(pL(n)), is sufficient to guarantee that the market supports a (not necessarily symmetric) equilibrium in pure strategies with two or more firms supplying at least D(pL(n)).
Conversely, the existence of a pure strategy equilibrium ensures that the cost function is not subadditive at every output greater than or equal to D(pL(n)). As a by-product, the latter implies that the average cost cannot be decreasing over the mentioned range of outputs.
In addition to the previous analysis, under the cost conditions specified before, this work also reexamines the relationship between the existence of a pure strategy equilibrium in Bertrand’s competition and of a price-taking or competitive equilibrium in the market. We find that the latter is sufficient but not necessary for the Bertrand price game to possess an equilibrium in pure strategies. In particular, since in our framework the former always exists when the fixed cost is unavoidable, this provides an existence result much simpler than Dastidar (1995).
The rest of the paper is organized as follows. Section 2 describes the model and the equi-librium concept, referred to as Bertrand equilibrium.
Section 3 deals with the relationship between price-taking equilibria and Bertrand equilibria. Section 4 contains the main results of the article, linking symmetric and nonsymmetric Bertrand equilibria with cost subadditivity.
For expositional convenience, some of the proofs of this section are relegated to the Appendix, which is displayed as usual at the end of the paper. Final remarks are done in Section 5.
7For the difference between Bertrand and Bertrand-Edgeworth competition, see Vives (1999, Chap. 5).
8A cost function C(·) is subadditive at q ∈R if the cost of producing q with a single firm is smaller than the sum of the costs of producing it separately with a group of two or more identical firms.
4 2 The model Consider the market of a homogenous good, with a unit price P and an aggregate demand D(P). Let N = {1, 2, . . . , n}, n ≥2, be the set of firms operating in the market. Suppose each firm i ∈N competes for the market demand D(·) by simultaneously and independently proposing to the costumers a price pi from the interval [0, ∞). Let qi = qi(pi, p−i) denote firm i’s output supply as a function of (pi, p−i), where p−i = (p1, . . . , pi−1, pi+1, . . . , pn) is the list of prices chosen by the other firms.
The following assumptions complete the description of the model.
Assumption 1 The aggregate demand D(·) is bounded on R+; that is, there exist K > 0 and P > 0 such that D(0) = K and D(P) = 0 for all P ≥P. In addition, D(·) is twice continuously differentiable and decreasing on (0, P); i.e., ∀P ∈(0, P), D′(P) < 0.
Assumption 2 For each firm i ∈N, the production cost associated with any qi ∈R+ is C(qi) = ( V C(qi) + F if qi > 0, C(0) if qi = 0, where F ≥0 represents a fixed cost, C(0) ∈[0, F], and V C(·) is a variable cost function, which is twice continuously differentiable, increasing and convex on R+, with V C(0) = 0 and 0 ≤V C′(0) < P.
Even though Assumption 2 does not specify the nature of the fixed cost and, consequently, the exact value of C(0), in the rest of the paper we consider two possibilities. The first case takes place when F is unavoidable, meaning that C(0) = F. In this case, the cost function C(·) is continuous and convex on R+. The second possibility occurs when F is positive and can be completely or partially avoided by producing no output, so that C(0) < F. In contrast with the first case, in the second the cost function C(·) is not only discontinuous at 0, but also non-convex around the origin. As we show in Section 3 these two scenarios result in quite different predictions regarding the existence of equilibria.
Our next assumption determines the individual demand faced by each firm for every possible profile of prices. To do that, we adopt the standard market sharing rule used in the literature on price competition, according to which the market demand is equally split between the firms that charge the lowest price, and the remaining firms sell nothing.9 Assumption 3 For each firm i ∈N and every (pi, p−i) ∈[0, ∞)n, the individual demand of i at (pi, p−i), denoted by di(pi, p−i), is defined as follows: di(pi, p−i) = D(pi) if pi < pj ∀j ∈N\{i}, 1 m D(pi) if pi ≤pj ∀j ∈N\{i} & pi = pkt ∀t = 1, . . . , m −1, 0 if pi > pj for some j ∈N\{i}.
(1) 9For price competition under alternative sharing rules, see among others Baye and Morgan (2002), Dastidar (2006), and a recent article by Hoernig (2007).
5 As usual in Bertrand’s competition, we assume that each firm always meets all the demand at the price it announces. More formally, Assumption 4 For all i ∈N, and all (pi, p−i) ∈[0, ∞)n, qi(pi, p−i) = di(pi, p−i).
Let H : [0, P] × N →R be such that, for all p ∈[0, P] and all m ∈N, H(p, m) = p D(p) m −C µD(p) m ¶ .
Assumption 5 For each m ∈N, H(· , m) is strictly quasi-concave on (0, P), with ph(m) = arg max p∈(0,P) H(p, m); and, for all m ̸= 1, 0 < H(ph(m), m) < H(pM, 1), where pM = ph(1).
Assumption 5 guarantees that, for every m ∈N, H( · , m) has an interior maximum. This is because H(0, m) = −V C(K/m) −F < 0 and H(P, m) = −C(0) ≤0. In addition, it also ensures that the monopoly receives the greatest maximal benefits.
The model of price competition described above follows Dastidar (1995). The only difference is that in our framework F is a fixed cost which may or may not be avoided by producing zero output. On the contrary, in Dastidar (1995) only unavoidable fixed costs are considered, although it is not explicitly stated in that way. Apart from this, the two models are similar.
Let πi(pi, p−i) = pi di(pi, p−i) −C(di(pi, p−i)) be firm i’s profit function. We denote by Gn = ⟨[0, ∞), πi⟩i∈N the price competition game defined by Assumptions 1 – 5.
A pure strategy Bertrand equilibrium (PSBE) for Gn is a profile of prices (pi, p−i) ∈[0, ∞)n such that, for each i ∈N and all ˆ pi ∈[0, ∞), πi(pi, p−i) ≥πi(ˆ pi, p−i). We denote by B(Gn) the set of all such equilibria, and by S(Gn) ⊆B(Gn) the subset of symmetric pure strategy equilibria, where for all (p1, . . . , pn) ∈S(Gn) and all i, j ∈N, i ̸= j, pi = pj.
3 Price-taking equilibrium and Bertrand equilibrium We begin this section by showing that, independently of the nature of the fixed cost, the existence of a price-taking equilibrium (yet to be defined) in the homogenous good market described in Section 2 is a sufficient condition for a pure strategy Bertrand equilibrium to exist; i.e., it is sufficient for B(Gn) ̸= ∅.
Let En = ⟨N, D(·), C(·)⟩represent the homogenous good market where every firm i ∈N maximizes the function Πi(P, Qi) = P Qi −C(Qi) with respect to Qi ∈R+ taking the price P > 0 as given. Suppose as before D(·) and C(·) satisfy Assumptions 1 and 2, respectively.
Then, a price-taking equilibrium (PTE) for En is a price P C ∈(0, P) and a profile of outputs (QC 1 , . . . , QC n ) ∈Rn + with the property that, for each firm i ∈N, QC i ∈arg max Qi∈R+ Πi(P C, Qi), (2) and n X i=1 QC i = D(P C).
(3) 6 Notice that, by Assumption 2, for all i ∈N and any P C ∈(0, P), Πi(P C, Qi) = P C Qi − C(Qi) is concave on R++, and Πi(P C, 0) = P C 0 −C(0) = −C(0). Hence, a unique output QC i ∈R+ satisfying (2) always exists. Moreover, since firms are identical, QC 1 = . . . = QC n .
Denote this common value by QC. By equation (3), QC = D(P C)/n. Hence, abusing the notation, in what follows we denote a PTE by the pair (P C, QC).
Proposition 1 If (P C, QC) is a price-taking equilibrium for En = ⟨N, D(·), C(·)⟩, then (p1, . . . , pn) = (P C, . . . , P C) is a pure strategy Bertrand equilibrium for Gn = ⟨[0, ∞), πi⟩i∈N.
Proof.
Let (P C, QC) be a PTE for En = ⟨N, D(·), C(·)⟩, where P C ∈(0, P) and QC = D(P C)/n.
By Assumption 1, QC > 0.
Therefore, (2) implies that P C QC −C(QC) ≥ P C 0 −C(0) = −C(0).
Consider the game Gn = ⟨[0, ∞), πi⟩i∈N and the strategy profile pC = (P C, . . . , P C). Notice that, for all i ∈N, πi(P C, . . . , P C) = P C D(P C) n −C ³ D(P C) n ´ .
Hence, for all i ∈N, πi(P C, . . . , P C) ≥−C(0). Suppose, by contradiction, pC ̸∈B(Gn). Then, there must exist a firm i ∈N and a price ˆ pi ∈[0, ∞) such that πi(ˆ pi, pC −i) > πi(P C, pC −i). If ˆ pi > P C, then di(ˆ pi, pC −i) = 0, meaning that πi(ˆ pi, pC −i) = ˆ pi 0 −C(0) = −C(0), which stands in contradiction with the fact that πi(ˆ pi, pC −i) > πi(P C, pC −i).
Therefore, ˆ pi < P C and, by (1), di(ˆ pi, pC −i) = D(ˆ pi) > 0. If ˆ pi = 0, then ˆ qi = K and πi(ˆ pi, pC −i) = −C(K) < −C(0), a contradiction. Thus, ˆ pi > 0. Let ˆ Qi = arg maxQ∈R+ Πi(ˆ pi, Q). Note that, since ˆ pi < P C, Πi(ˆ pi, ˆ Qi) = maxQ∈R+{ˆ pi · Q −C(Q)} ≤maxQ∈R+{P C · Q −C(Q)} = Πi(P C, QC).
How-ever, Πi(P C, QC) = πi(P C, pC −i) < πi(ˆ pi, pC −i).
Therefore, maxQ∈R+{ˆ pi · Q −C(Q)} < ˆ pi · D(ˆ pi) −C(D(ˆ pi)), a contradiction. Thus, (P C, . . . , P C) ∈B(Gn). ■ The previous proposition shows that the existence of a price-taking equilibrium in a homoge-nous good market, with a finite number of identical firms and demand and cost functions that satisfy our assumptions, is a sufficient condition to guarantee the existence of a pure strategy equilibrium in the market when firms compete in prices a la Bertrand, instead of taking the market price as given. A similar result has been previously stated by Vives (1999, pg. 120) for the case where all firms have identical, increasing, smooth and convex cost functions. The contribution of Proposition 1 is to show that that assertion also holds in markets with fixed costs, regardless of whether the fixed cost is unavoidable or (totally or partially) avoidable.10 Thus, a natural implication is that, in our framework, the continuity and convexity of C(·) at the origin are not required to ensure the validity of the assertion.
Another consequence of Proposition 1 is that, when F is an unavoidable fixed cost, the set of (symmetric) pure strategy Bertrand equilibria is always nonempty.11 Corollary 1 If C(0) = F, the set of symmetric pure strategy equilibria S(Gn) is nonempty.
Proof. Let C(0) = F. Consider the homogenous good market En = ⟨N, D(·), C(·)⟩introduced above, where each firm i ∈N maximizes Πi(P, Qi) with respect to Qi ∈R+ taking the price 10This paper allows F to be equal to 0. Thus, Proposition 1 and Corollary 1 below also hold in the more familiar case where there are decreasing returns to scale and no fixed cost.
11Actually, as Dastidar (1995) have shown, a whole interval of prices can be typically supported as pure strategy equilibrium. See also Klaus and Brandts (2008) for experimental results.
7 P > 0 as given. Suppose D(·) and C(·) satisfy Assumptions 1 and 2, respectively. We wish to prove that En has a PTE.
Fix any price P ∈(0, P) and any firm i ∈N, and let Q∗ i (P) = arg maxQi>0 Πi(P, Qi). By Assumption 2, Q∗ i (P) exists and is unique (recall that Πi(P, ·) is concave on R++). Moreover, by the first order condition, Q∗ i (P) = MC−1(P), where MC−1(·) denotes the inverse of V C′(·), which exists because V C′(·) is increasing on R+. Notice that, Πi(P, Q∗ i (P)) = Q∗ i (P) · P −V C(Q∗ i (P)) Q∗ i (P) ¸ −F > −F = Πi(P, 0), because, by the first order condition, P = V C′(Q∗ i (P)) and, by Assumption 2, V C′(Q∗ i (P)) > V C(Q∗ i (P)) Q∗ i (P) . Therefore, for every price P ∈(0, P) and every firm i ∈N, the optimal output supply of i at P is given by Q∗ i (P) = MC−1(P). Since firms are identical, the market supply is S(P) = P i∈N Q∗ i (P) = n [MC−1(P)]. Thus, the equilibrium price is obtained by solving the equation D(P) = n [MC−1(P)], which has a solution on (0, P) due to our assumptions on the demand and cost functions. Denote this value by P ∗. It is immediate to see that (P ∗, Q∗ i (P ∗)) constitutes a PTE for En. Hence, by Proposition 1, the profile (p1, . . . , pn) = (P ∗, . . . , P ∗) ∈ S(Gn). ■ Now, the reader may wonder what happens when F is an avoidable fixed cost. Indeed, it is relatively simple to construct examples where neither a price-taking equilibrium nor a pure strategy Bertrand equilibrium exist. Here is one. Let D(P) = 10 −P and N = {1, 2}. Suppose C(qi) = 1/2q2 i + F if qi > 0, and let C(0) = 0 otherwise.
If a PTE exists, then (2) and (3) imply that P C = 10/3 and QC = 10/3. However, Πi(P C, QC) ≥0 (= Πi(P C, 0)) only if F ≤50/9 (≈5.55 ′). Thus, if F > 50/9, the market does not possess a PTE.
Regarding Bertrand equilibria, a pair of prices (p∗ 1, p∗ 2) ∈[0, ∞)2 constitutes a symmetric pure strategy equilibrium if and only if p∗ 1 = p∗ 2 and, for all i ∈N, (a) H(p∗ i , 2) ≥0, and (b) for all ˆ pi < p∗ i , H(p∗ i , 2) ≥H(ˆ pi, 1). Notice that the latter condition requires that 1/2p∗ i (10 −p∗ i ) − 3/8(10 −p∗ i )2 ≤0, which is satisfied whenever p∗ i ≤30/7 (≈4.2858). (See Figure 1 for the case where F = 6.) On the other hand, from (a), it follows that p∗ i ≥6 − p 16 −8/5 F. Hence, a price p∗ i simultaneously satisfying both conditions exists if and only if F ≤400/49 (≈8.1633).
For instance, when F = 6, the price p∗ i = 4 is a solution of (a) and (b). Therefore, the profile (p∗ 1, p∗ 2) = (4, 4) ∈S(G2). Actually, if the fixed cost F ≤50/9, then the set of symmetric Bertrand equilibria includes the price-taking equilibrium; that is, (10/3, 10/3) ∈S(G2). How-ever, when F ∈(50/9, 400/49), a PTE does not exist, but the game possesses multiple PSBE.12 Indeed, as Figure 1 illustrates, when F = 6 any price between the lower bound pL = 6 − p 32/5 (≈3.4702) and the upper bound pH = 30/7 satisfies conditions (a) and (b) and, therefore, constitutes a symmetric PSBE.
12A similar result appears in Grossman (1981), but associated with a different equilibrium concept. Using a model similar to ours, Grossman has shown that a PTE, if it exists, is a supply function equilibrium. However, the latter may exist even if there is no PTE in the market.
8 0 2 4 6 8 10 −5 0 5 10 15 p(10 −p) −1 2(10 −p)2 −6 p (10−p) 2 −1 2 ¡ 10−p 2 ¢2 −6 pH = 30/7 pL = 6 − p 32/5 set of psbe Figure 1: Existence of Bertrand equilibria (F = 6) In addition, it also comes to light from the previous example that, if F ∈ ¡400 49 , 10 ¤ , then a symmetric equilibrium in pure strategies does not exist. For F = 9, this is illustrated in Figure 2, where it can be easily seen that, for any price p for which the solid curve H(p, 2) is over the horizontal axis, the dashed curve H(p, 1) lies above. This implies that, whenever both firms choose any price p ∈[0, P] satisfying the condition H(p, 2) ≥0, there is a deviation ˆ pi < p for one firm, say for firm i, such that H(p, 2) < H(ˆ pi, 1). In effect, for the case in which p−i = pL, the diagram shows that firm i’s strategy ˆ pi < pL dominates pL, in the sense that πi(ˆ pi, pL) = ˆ pi (10 −ˆ pi)−1 2 (10 −ˆ pi)2 −9 > > pL (10 −pL) 2 −1 2 µ10 −pL 2 ¶2 −9 = πi(pL, pL).
Therefore, (pL, pL) ̸∈S(G2). And, since the same reasoning applies for every price p ∈ (pL, p′], it follows that S(G2) = ∅. (Notice that (P, P) = (10, 10) is not an equilibrium either, because any of the two firms can profitably deviate to the monopoly price pM = 20/3, which renders a payoffof H(pM, 1) ≈7.66 ′ > 0 = πi(10, 10).) Finally, observe that when F = 9 our example not only fails to possess a symmetric pure strategy equilibrium, but also a PSBE with p1 ̸= p2. To see this, assume, by contradiction, such equilibrium exists. Without loss of generality, suppose that p1 < p2. Note that p1 ≤pM = 20/3.
Otherwise, firm 1 can profitably deviate to pM. Then, by Assumption 3, π1(p1, p2) = H(p1, 1) and π2(p1, p2) = 0. Suppose first H(p1, 1) > 0. Then, q1(p1, p2) > 0; and, by continuity of H(p, 1) = p (10 −p) −1 2 (10 −p)2 −9 in p at p1, there is a price p′ 2 < p1 such that H(p′ 2, 1) > 0 = π2(p1, p2), contradicting that p2 is firm 2’s best response to p1.
Next, observe that, if (p1, p2) ∈B(G2), then H(p1, 1) cannot be negative. This is because 9 0 2 4 6 8 10 −5 0 5 10 p(10 −p) −1 2(10 −p)2 −9 p (10−p) 2 −1 2 ¡ 10−p 2 ¢2 −9 p′ pL = 6 − p 8/5 ˆ pi Figure 2: Nonexistence of Bertrand equilibria (F = 9) π1(10, ˆ p2) = 0 for all ˆ p2 ∈[0, ∞). Therefore, H(p1, 1) = 0; and, given the shape of H(·, 1) displayed in Figure 2, it has to be that p1 = (20 − √ 46)/3 (≈4.4059).
If H(p2, 1) > 0, then using the continuity of H(p, 1) in p at p2, there must be a price p′ 1 < p2 such that H(p′ 1, 1) > 0 = π1(p1, p2), which would contradict that firm 1 is playing his best response against p2. Thus, H(p2, 1) ≤0. But, since p2 > p1 and H(pM, 1) > 0, this implies p2 > pM.
Hence, firm 1 can profitably deviate to pM, meaning that (p1, p2) is not a PSBE for G2.13 4 Cost subadditivity and Bertrand equilibrium The example discussed above shows that, if the fixed cost can be completely avoided by produc-ing no output, then depending upon its value the set of pure strategy Bertrand equilibria may be empty. Under a constant marginal cost, this problem has been previously noted by Shapiro (1989), Vives (1999) and Baye and Kovenock (2008), among others.14 A sensible question to ask is therefore what conditions (if any) prevent this from happening. Finding these conditions will occupy the remainder of the paper.
To begin to analyze this matter requires us to define a key property of the cost function, namely, subadditivity. Following Panzar (1989, pg. 23), we say that a cost function C(·) is subadditive at q ∈R+ if for every list of outputs q1, . . . , qn, with qi ∈R+ and qi ̸= q for all i = 1, . . . , n, it is the case that C(q) < Pn i=1 C(qi) whenever Pn i=1 qi = q.
In words, C(·) is subadditive at q if the cost of producing q with a single firm is smaller than the sum of the costs of producing it separately with a group of two or more identical 13Incidentally, note that the example shows that, in our framework, Baye and Morgan’s (1999) condition, i.e., the existence of an initial break-even price in the profit function H(·, 1), is not sufficient to guarantee a PSBE.
14In fact, Baye and Kovenock (2008) have also shown that, with a constant marginal cost, an avoidable fixed cost may preclude the existence of mixed strategy equilibria as well.
10 firms. As Baumol (1977) pointed out, subadditivity of the cost function is a necessary and sufficient condition for a natural monopoly to exist. Notice, however, that subadditivity is a local property in that it refers to a particular point on the cost curve. Thus, it is possible for a market to be a natural monopoly for a certain output, but not for others.
When the cost function C(·) is twice continuously differentiable and the marginal cost is increasing, there is a simple necessary condition for subadditivity. In effect, under the cost conditions stated before, any output q is divided in positive portions most cheaply among n identical firms if each firm produces the same amount qi = q/n. Hence, since the minimized cost corresponding to output q for a n-firm market is Pn i=1 C(qi) = n C(q/n), it follows that C(·) is subadditive at q ∈R+ only if C(q) < n C(q/n). If there are only two firms, then this condition is also sufficient. This is because the requirement embedded in the definition of subadditivity that qi ̸= q for all i ∈N implies, when n = 2, that qi ̸= 0 for all i = 1, 2.
We claim now that, if the fixed cost is fully avoidable, then a necessary condition for the existence of a symmetric pure strategy Bertrand equilibrium in the price competition game defined in Section 2 is for the cost function not to be subadditive at the output corresponding to the oligopoly break-even price. To formally prove this assertion, the following preliminary results will be useful.
Lemma 1 For every m ∈N, there is a price ˆ p(m) ∈(0, P) such that H(ˆ p(m), m) = −C(0).
Proof.
Fix any m ∈N.
By Assumptions 1 and 2, H(0, m) = −V C(K/m) −F.
Thus, H(0, m) < −F; and, since C(0) ≤F, we have that H(0, m) < −C(0). On the other hand, by Assumption 5, H(ph(m), m) > 0 ≥−C(0). Hence, by the intermediate value theorem, there is a price ˆ p(m) ∈(0, ph(m)) such that H(ˆ p(m), m) = −C(0). ■ Fix m ∈N and let pL(m) = min{ˆ p(m) ∈(0, P) : H(ˆ p(m), m) = −C(0)}. By Lemma 1, pL(m) is well defined. By Assumption 1, D(pL(m)) > 0. Suppose now pL(m) > pM. Then, m ̸= 1; and, by Assumption 5, H(· , 1) is non-increasing at pL(m); i.e., ∂H(pL(m),1) ∂p ≤0. Hence, pL(m) ≥V C′(D(pL(m))) −D(pL(m)) D′(pL(m)).
(4) Similarly, by Assumption 5 and the fact that, by definition, pL(m) < ph(m), H(· , m) is non-decreasing at pL(m); i.e., ∂H(pL(m),m) ∂p ≥0. Therefore, pL(m) ≤V C′ µD(pL(m)) m ¶ −D(pL(m)) D′(pL(m)).
(5) Finally, since V C′(·) is increasing and −D(pL(m)) D′(pL(m)) is positive, V C′ µD(pL(m)) m ¶ −D(pL(m)) D′(pL(m)) < V C′(D(pL(m))) −D(pL(m)) D′(pL(m)); and, by (4) and (5), we get that pL(m) < pL(m), a contradiction.
Thus, for all m ∈N, pL(m) ≤pM.
11 Lemma 2 For all p < pM, H(p, 1) −H(p, n) = 0 implies that ∂[H(p,1)−H(p,n)] ∂p > 0.
Proof. For every price p < pM, we have that H(p, 1) −H(p, n) = n −1 n p D(p) −V C(D(p)) + V C µD(p) n ¶ .
(6) Taking the derivative of (6) with respect to p, ∂[H(p, 1) −H(p, n)] ∂p = n −1 n D(p)+ + [p −V C′(D(p))] D′(p) − · p −V C′ µD(p) n ¶¸ D′(p) n .
(7) Consider any price p ∈[0, pM) with the property that H(p, 1) −H(p, n) = 0. Then, n −1 n p D(p) = V C(D(p)) −V C µD(p) n ¶ .
(8) By convexity of V C(·), V C(D(p)) −V C µD(p) n ¶ < n −1 n D(p) V C′(D(p)), (9) and V C(D(p)) −V C µD(p) n ¶ > n −1 n D(p) V C′ µD(p) n ¶ .
(10) Thus, combining (8) and (9), we have that p < V C′(D(p)); and, from the expressions in (8) and (10), it also follows that p > V C′ ³ D(p) n ´ . Therefore, since (n −1)/n D(p) > 0 and D′(p) < 0, the right hand side of (7) is greater than zero; i.e., ∂[H(p,1)−H(p,n)] ∂p > 0. ■ Lemma 3 If S(Gn) ̸= ∅, then the strategy profile (p1, . . . , pn) = (pL(n), . . . , pL(n)) ∈S(Gn).
Proof.
Suppose, by contradiction, the strategy profile (p1, . . . , pn) = (pL(n), . . . , pL(n)) ̸∈ S(Gn). Then, there must be a firm i ∈N and a price ˜ pi < pL(n) such that πi(˜ pi, (pL(n))−i) > πi(pL(n), (pL(n))−i), where (pL(n))−i denotes the sub-profile of prices in which everybody ex-cept firm i chooses pL(n). Notice that πi(˜ pi, (pL(n))−i) = H(˜ pi, 1) and πi(pL(n), (pL(n))−i) = H(pL(n), n). Thus, H(˜ pi, 1) −H(pL(n), n) > 0. Moreover, since H(pL(n), n) = −C(0) and H(˜ pi, n) < −C(0),15 it also follows that H(˜ pi, 1) −H(˜ pi, n) > 0.
Therefore, given that H(0, 1) −H(0, n) = −V C(K) + V C(K/n) < 0 and H(· , 1) −H(· , n) is continuous on [0, ˜ pi], there must be a price p′ ∈(0, ˜ pi) such that H(p′, 1) −H(p′, n) = 0.
Next, recall that, by hypothesis, S(Gn) ̸= ∅. That is, there is a price p∗∈(pL(n), pM) such that H(p∗, n) ≥−C(0) and, for all p < p∗, H(p, 1) −H(p∗, n) ≤0. Since p can be chosen 15Note that H(˜ pi, n) ̸= −C(0), because ˜ pi < pL(n) and, by definition, pL(n) is the smallest price for which H(· , n) equals −C(0).
On the other hand, since H(0, n) < −C(0), H(˜ pi, n) cannot be greater than −C(0).
Otherwise, there would be a price p ∈(0, ˜ pi) with the property that H(p, n) = −C(0), which again contradicts the definition of pL(n). Thus, H(˜ pi, n) < −C(0).
12 arbitrarily close to p∗, by continuity, it must be that H(p∗, 1) −H(p∗, n) ≤0. On the other hand, by Assumption 5, H(pM, 1) −H(pM, n) > 0. So, there must be a price p′′ ∈(pL(n), pM) such that H(p′′, 1) −H(p′′, n) = 0.
In summary, if S(Gn) ̸= ∅and (pL(n), . . . , pL(n)) ̸∈S(Gn), the previous two paragraphs indicate that the curves H(· , 1) and H(· , n) must intersect each other at least twice on (0, pM).
Therefore, in order to show that (pL(n), . . . , pL(n)) is indeed a symmetric pure strategy Bertrand equilibrium for Gn, it is enough to prove that there is only one such intersection; i.e., it is sufficient to show that there is a unique price p ∈(0, pM) for which H(p, 1) −H(p, n) = 0.
Without of generality, assume that there is a pair of prices pα, pβ ∈(0, pM), pα < pβ, such that H(pα, 1) −H(pα, n) = 0 and H(pβ, 1) −H(pβ, n) = 0. Notice that, by Lemma 2, for ϵ1 > 0 small enough, H(pα, 1) −H(pα, n) = 0 implies that H(pα + ϵ1, 1) −H(pα + ϵ1, n) > 0.
In the same way, by Lemma 2, for δ > 0 small enough, H(pβ, 1) −H(pβ, n) = 0 implies that H(pβ −δ, 1) −H(pβ −δ, n) < 0.
Hence, since H(· , 1) −H(· , n) is continuous on (0, pM), there must be a price pα+1 ∈(pα, pβ) such that H(pα+1, 1) −H(pα+1, n) = 0.
Repeating the previous argument, for ϵ2 > 0 small enough, H(pα+1, 1) −H(pα+1, n) = 0 implies that H(pα+1 + ϵ2, 1) −H(pα+1 + ϵ2, n) > 0. Hence, there must be a price pα+2 ∈(pα+1, pβ) such that H(pα+2, 1) −H(pα+2, n) = 0.
Repeating these steps over and over again, we get a sequence of prices {pα+s}∞ s=1 ⊂(pα, pβ) with the property that H(pα+s, 1) −H(pα+s, n) = 0 for all s = 1, . . . , ∞. Observe that, by construction, each term pα+s of the sequence is closer to pβ than what it was pα+s−1. Therefore, by Lemma 2, for some s ≥1 sufficiently high, there must exist ϵ ∈(0, δ) and a price ¯ ¯ p ∈ (pα+s + ϵ, pβ −ϵ) such that H(¯ ¯ p, 1) −H(¯ ¯ p, n) > 0 and H(¯ ¯ p, 1) −H(¯ ¯ p, n) < 0, which provides the desired contradiction. ■ Now, we are ready to state and prove Proposition 2.
Proposition 2 Suppose C(0) = 0. If the set of symmetric pure strategy equilibrium S(Gn) is nonempty, then the cost function C(·) is not subadditive at D(pL(n)).
Proof.
Suppose, by contradiction, that C(·) is subadditive at D(pL(n)).
(Recall that, by definition of pL(n), D(pL(n)) > 0.) Then, it must be that producing D(pL(n)) with a single firm is cheaper than producing it with n identical firms; that is, C(D(pL(n))) < n C µD(pL(n)) n ¶ .
(11) Adding the term −pL(n) D(pL(n)) to both sides of (11), it follows that −pL(n) D(pL(n)) + C(D(pL(n))) < −pL(n) D(pL(n)) + n C µD(pL(n)) n ¶ , which can be rewritten as pL(n) D(pL(n)) −C(D(pL(n))) > n · pL(n) D(pL(n)) n −C µD(pL(n)) n ¶¸ .
(12) 13 By definition of pL(n), the right hand side of (12) is equal to −n C(0).
Hence, if C(0) = 0, then (12) implies that pL(n) D(pL(n)) −V C(D(pL(n))) −F > 0.
By conti-nuity of p D(p) −V C(D(p)) −F in p at pL(n), there is a price p′ < pL(n) such that p′ D(p′) −V C(D(p′)) −F > 0. Fix any firm i ∈N, and consider firm i’s strategy p′ i = p′.
By Assumption 3, πi(p′ i, (pL(n))−i) = p′ i D(p′ i) −V C(D(p′ i)) −F. Hence, πi(p′ i, (pL(n))−i) > 0.
On the other hand, πi(pL(n), . . . , pL(n)) = H(pL(n), n) = 0.
Thus, firm i can profitably deviate at (pL(n), (pL(n))−i), from pL(n) to p′ i, contradicting that, by Lemma 3, the profile (pL(n), (pL(n))−i) ∈S(Gn). Therefore, C(·) is not subadditive at D(pL(n)). ■ Proposition 2 formalizes the intuitive idea that, if the fixed cost is avoidable (or, there is no fixed cost at all), then a necessary condition for the existence of a symmetric pure strategy Bertrand equilibrium in a homogenous good market is for the market not to be a natural monopoly at the output corresponding to the oligopoly break-even price. Unfortunately, this result does not hold if C(0) ̸= 0. To see this, suppose that n = 3 and D(P) = 10 −P, and let C(qi) = 3 22q2 i + 15 2 , if qi > 0, and C(0) = 15 2 otherwise. Routine calculations show that pL(3) = 10 23 and H(10/23, 1) ≈−15.82. Hence, (p1, p2, p3) = ¡10 23, 10 23, 10 23 ¢ is a symmetric PSBE. However, C(·) is subadditive at D(10/23), because C(D(10/23)) ≈19.98, 2 C(D(10/23)/2) ≈21.24, and 3 C(D(10/23)/3) ≈26.66.
Inspired by the example examined at the end of Section 3, where the lack of a symmetric pure strategy equilibrium and the subadditivity of the cost function at D(pL) occur for the same range of values of F, (namely, for all F ∈ ¡400 49 , 10 ¤ ),16 a natural and interesting question to ask is whether or not the converse of Proposition 2 holds. As we state in Proposition 3, if there is a price-taking equilibrium in the market, then the answer to this question is obviously affirmative, simply because Proposition 1 ensures that the set of symmetric pure strategy equilibria is always nonempty. More interestingly, it also holds in a duopoly, independently of the nature of the fixed cost (i.e., regardless of the value of C(0)). The reason behind this last result can be found in the following lemma.
Lemma 4 If C(D(pL(n))) ≥ n C ³ D(pL(n)) n ´ , then the strategy profile (p1, . . . , pn) = (pL(n), . . . , pL(n)) constitutes a pure strategy Bertrand equilibrium for Gn.
Proof. Suppose, by contradiction, (pL(n), . . . , pL(n)) ̸∈S(Gn). Then, there must be a price ˜ p ∈(0, pL(n)) such that H(˜ p, 1) > −C(0) = H(pL(n), n). By definition of pL(n), D(pL(n)) > 0.
Thus, the hypothesis in Lemma 4, i.e., C(D(pL(n))) ≥n C ³ D(pL(n)) n ´ , can be rewritten as V C(D(pL(n))) ≥n V C µD(pL(n)) n ¶ + (n −1)F.
(13) Using the definition of pL(n), it is easy to see that the right hand side of (13) is equal to pL(n) D(pL(n)) −F + n C(0).
(14) 16In the example in question, the cost function is subadditive at D(pL) = 10 −(6 − p 16 −8/5F) if and only if V C(D(pL)) < 2V C(D(pL)/2) −F. Routine calculations show that this inequality is satisfied if and only if F ∈ ¡ 400 49 , 10 ¤ .
14 Hence, substituting (14) into (13), it follows that H(pL(n), 1) ≤−C(0).
However, this contradicts that, by Assumption 5, H(·, 1) is quasi-concave on (0, P), because pL(n) ∈(˜ p, pM) and H(pL(n), 1) ≤−C(0) < min{H(˜ p, 1), H(pM, 1)}. ■ Proposition 3 Suppose that either (P C, QC) is a price-taking equilibrium for En = ⟨N, D(·), C(·)⟩, or that there are only two firms in the market. Then, if C(·) is not subad-ditive at D(pL(n)), the set of symmetric pure strategy equilibria S(Gn) is nonempty.
Proof. If (P C, QC) is a PTE for En = ⟨N, D(·), C(·)⟩, then the desired result follows from Proposition 1. On the other hand, if n = 2, then C(·) is not subadditive at D(pL(2)) if and only if C(D(pL(2))) ≥2 C ³ D(pL(2)) 2 ´ . Hence, by Lemma 4, (pL(2), pL(2)) ∈S(G2). ■ In short, Proposition 3 tells us that, if a homogenous good market is not a natural monopoly and either, there is a price-taking equilibrium, or there are only two firms, then the market supports a symmetric pure strategy equilibrium where firms compete in prices ` a la Bertrand and supply a total output which leaves each of them indifferent between staying in operation and exit the market. In particular, this holds when the market has an unavoidable fixed cost, because in that case a price-taking equilibrium always exists.
By contrast, if there are more than two firms and the fixed cost is completely or partially avoidable, then the converse of Proposition 2 is not true. To illustrate this, consider again the demand and cost function corresponding to the example analyzed in Section 3. Assume that n = 5, and let F = 4.3. Then, pL(5) ≈4.3983 and H(pL(5), 1) ≈4.65 > 0 = H(pL(5), 5).
Therefore, (p1, . . . , p5) = (pL(5), . . . , pL(5)) is not a (symmetric) PSBE for G5; and, by Lemma 3, we can conclude that S(G5) = ∅. However, it is easy to verify in this numerical example that C(·) is not subadditive at D(pL(5)) ≈5.6017. Indeed, producing D(pL(5)) with a single firm generates a cost equal to C(D(pL(5))) ≈19, 9895, whereas producing it with two identical firms costs 2 · C(D(pL(5))/2) ≈16.4447.
So, is there something to say about Bertrand’s price competition when the n-firm market is not a natural monopoly and there is no price-taking equilibrium? Indeed, we show next that, if the fixed cost is avoidable, then the non-subadditivity of the cost function C(·) at D(pL(n)) is sufficient to guarantee that the market supports a (not necessarily symmetric) pure strategy Bertrand equilibrium where two or more identical firms jointly supply at least D(pL(n)). And, conversely, the existence of a pure strategy Bertrand equilibrium ensures that the market is not a natural monopoly at every output greater than or equal to D(pL(n)). In particular, the latter implies that the average cost cannot be decreasing on [D(pL(n)), K), (see Corollary 2 below and the discussion following this result).17 Theorem 1 Suppose C(0) = 0.
If the cost function C(·) is not subadditive at D(pL(n)), then there exist a pure strategy Bertrand equilibrium (p1, . . . , pn) ∈B(Gn) where P i∈N qi(p1, . . . , pn) ≥D(pL(n)). Conversely, if a pure strategy Bertrand equilibrium exists, then the cost function C(·) is not subadditive on the interval [D(pL(n)), K).
17An example where under increasing returns to scale and equal sharing rule neither pure nor mixed strategy equilibria exist is exhibited in Hoernig (2007, pg. 582).
15 Proof. See the Appendix. ■ Given any output q > 0, the average cost at q is defined as AC(q) = C(q)/q. The average cost function AC(·) is decreasing at q if there exists a δ > 0 such that for all q′, q′′ ∈(q −δ, q + δ), with q′ < q′′, AC(q′′) < AC(q′). Additionally, AC(·) is said to decrease through q if for all q′, q′′ ∈(0, q], with q′ < q′′, AC(q′′) < AC(q′), (Panzar, 1989, pg. 24). If, like in our case, C(·) is twice continuously differentiable on R++, then AC(·) is decreasing at q if ∂AC(q) ∂q < 0; and AC(·) is decreasing through q if for all q′ ∈(0, q], ∂AC(q′) ∂q < 0, (i.e., if AC(·) is decreasing on (0, q]).
Lemma 5 If the average cost AC(·) is decreasing through q, then the cost function C(·) is subadditive at q, but not conversely.
Proof. The proof is based on Panzar (1989, pg. 25). Fix any q > 0 and assume AC(·) is decreasing through q. Consider any division q1, . . . , qn of q, with the property that (i) ∀i ∈N, 0 ≤qi < q, and (ii) P i∈N qi = q.
Let N+ = {i ∈N : qi > 0}.
Then, for all i ∈N+, AC(q) < AC(qi), which is equivalent to C(qi) > (qi/q) · C(q). Summing over N+, we have P i∈N+ C(qi) > C(q). Therefore, since C(0) ≥0, it follows that P i∈N C(qi) > C(q). Finally, since q1, . . . , qn was arbitrarily chosen, this implies that C(·) is subadditive at q.
To show that subadditivity does not imply decreasing average costs, consider the cost func-tion C(q) = 1/2 · q2 + 100 for all q ≥0. It is easy to see that AC(·) is not decreasing at q = 15.
However, if n = 2, then C(·) is subadditive at 15. ■ Corollary 2 Suppose C(0) = 0.
The set of pure strategy Bertrand equilibrium B(Gn) is nonempty only if the average cost AC(·) is not decreasing on [D(pL(n)), K).
Proof. Immediate from Theorem 1 and Lemma 5. ■ The second part of Theorem 1 and its implication in Corollary 2 are closely related with Dastidar’s (2006) Proposition 3, which says that the set of Bertrand equilibria B(Gn) is nonempty only if C(·) is not concave. Hence, before closing this section, it may be worthy to underline some differences between these results.
First of all, let’s emphasize that the necessary condition for equilibrium existence stated in the second part of Theorem 1 considerably sharpens Dastidar’s (2006) condition, because concavity implies subadditivity, but not conversely. Thus, we could have a cost function which is non-concave and subadditive at the same time. A function like that would violate our necessary condition for existence, whereas it wouldn’t do so with Dastidar’s. Secondly, in Theorem 1 we allow for avoidable fixed costs and, therefore, for discontinuities in the cost function around the origin. On the contrary, in Dastidar (2006) the cost function is continuous and F = 0.
Finally, Theorem 1 provides not only a necessary condition for B(Gn) ̸= ∅, but also a sufficient condition. Instead, Dastidar (2006) only gives a necessary condition.
16 5 Concluding remarks The main conclusions of this paper can be summarized by restating Propositions 1, 2 and 3, and Theorem 1. By looking at Proposition 1 we see that, in a market with convex variable costs and fixed costs, the existence of a price-taking equilibrium is a sufficient but not a necessary condition for a pure strategy Bertrand equilibrium to exist. That means it may be perfectly the case that a PTE does not exist, while the set of symmetric PSBE is nonempty, (see, for instance, the example at the end of Section 3).
The nonexistence of a price-taking equilibrium in markets with convex variable costs and fixed costs has been examined by the literature on ‘empty-core markets’, (Telser, 1991). It has also appeared in Grossman (1981), who studied supply function equilibria for oligopolies with avoidable fixed costs and convex variable costs. Our results in Section 3 are to some extent similar to Grossman’s, because he has shown that in a setting similar to ours supply function equilibria may exist even when a PTE does not. However, neither Telser nor Grossman have addressed the existence of Bertrand equilibria in those markets.
As we show in Section 4, the existence of a symmetric PSBE when variable costs are convex and the fixed cost is avoidable is related with the subadditivity of the cost function at the oligopoly break-even price pL(n). In Proposition 2, we demonstrate that, if C(0) = 0 and a symmetric PSBE exists, then firms’ cost functions cannot be subadditive at D(pL(n)). This is equivalent to say that, under the previous conditions, the market cannot be a natural monopoly when firms break-even and demand is in equilibrium. As Proposition 3 points out, the reverse of that statement is also true if there are only two firms in the market, or if there is a PTE.
Unfortunately, numerical examples show that it does not hold in other situations.
The last result of this work, and perhaps the most important, is Theorem 1, which generalize Propositions 2 and 3 to cases where equilibria are not symmetric, there are no PTE and n > 2.
In short, Theorem 1 says that, when the fixed cost is fully avoidable and the cost function C(·) is not subadditive at D(pL(n)), there always exists a Bertrand equilibrium in pure strategies, though it need not be a symmetric one. Conversely, if a PSBE exists, then C(·) cannot be subadditive for all quantities greater than or equal to D(pL(n)).
The results of this article relate the existence of Bertrand equilibrium with the nonexistence of natural monopoly. By doing so, the paper explores a relationship which has not been previ-ously considered in the literature. Our findings can also be taken as a contribution to the theory of endogenous industry structure. This is because pL(n) is typically increasing in the number of firms that operate in the market. Hence, under variable returns to scale, it is possible that a cost function be subadditive at D(pL(n + 1)) and not at D(pL(n)). In that case, a market with n firms or less might have a symmetric PSBE, but a market with n+1 firms or more might not.
That could be useful to figure out the maximum number of active firms that a homogenous good market can support under price competition. The analysis of this conjecture and the study of more general forms of non-convexities and discontinuities of the cost function and their impact on Bertrand’s competition are left for a future research.
17 6 Appendix: Proof of Theorem 1 In order to prove Theorem 1, first we show the following auxiliary result: Lemma 6 For all p < P, ∂[H(p,1)−m H(p,m)] ∂p > 0.
Proof.
For every price p < P, we have that H(p, 1) −m H(p, m) = −V C(D(p)) −F + m V C ³ D(p) m ´ + m F. Taking the derivative with respect to p, ∂[H(p, 1) −m H(p, m)] ∂p = D′(p) · V C′ µD(p) m ¶ −V C′(D(p)) ¸ , (15) which is positive because D′(p) < 0 and V C′(·) is increasing. ■ Proof of Theorem 1. Assume the cost function C(·) is not subadditive at D(pL(n)). Then, there must be a m ∈{2, . . . , n} such that C(D(pL(n))) ≥m C µD(pL(n)) m ¶ .
(16) If m = n, we are done. By Lemma 4, (p1, . . . , pn) = (pL(n), . . . , pL(n)) ∈S(Gn) ⊆B(Gn).
Moreover, P i∈N qi(p1, . . . , pn) = n D(pL(n)) n = D(pL(n)). So, suppose that C(D(pL(n))) < n C µD(pL(n)) n ¶ .
(17) Adding the term −pL(n) D(pL(n)) to both sides of (16), it follows that −pL(n) D(pL(n)) + C(D(pL(n))) ≥−pL(n) D(pL(n)) + m C ³ D(pL(n)) m ´ , which implies that H(pL(n), 1) ≤m H(pL(n), m).
(18) Following the same steps, it is easy to see from (17) that H(pL(n), 1) > n H(pL(n), n).
(19) Therefore, since C(0) = 0, combining (18) and (19), we get that both H(pL(n), 1) > 0 and m H(pL(n), m) > 0.
Recall that, by Lemma 1, there is a price p(m) ∈(0, ph(m)) such that H(p(m), m) = 0.
Since pL(m) is the smallest of such prices, and H(pL(n), m) > 0 and H(0, m) < 0, it follows that pL(m) < pL(n). Indeed, pL(m) is also smaller than pM, because Assumption 5 implies that pL(n) ≤pM. Suppose, by contradiction, there is a price ˆ p < pL(m) such that H(ˆ p, 1) > H(pL(m), m).
(20) Since H(· , 1) is quasi-concave on (0, P), pL(m) ∈(ˆ p, pM) implies that H(pL(m), 1) ≥ min{H(ˆ p, 1), H(pM, 1)} = H(ˆ p, 1). Hence, using (20), H(pL(m), 1) > H(pL(m), m). More-18 over, since H(pL(m), m) = 0, m H(pL(m), m) = H(pL(m), m). Therefore, H(pL(m), 1) > m H(pL(m), m).
(21) Given that H(· , 1) −m H(· , m) is continuous on [0, P), the expressions in (18) and (21) imply that there is a price pα ∈(pL(m), pL(n)] such that H(pα, 1) −m H(pα, m) = 0.
(22) Thus, by Lemma 6 and (22), there must exist ϵ > 0 small enough with the property that H(pα −ϵ, 1) −m H(pα −ϵ, m) < 0. But then, using (21) once again, it follows that there is a price pα+1 ∈(pL(m), pα −ϵ) such that H(pα+1, 1) −m H(pα+1, m) = 0. And repeating the argument over and over again, we get a sequence of prices {pα+s}∞ s=1 ⊂(pL(m), pα) with the property that for all s = 1, . . . , ∞, H(pα+s, 1) −m H(pα+s, m) = 0.
(23) Notice that each term pα+s of the sequence is closer to pL(m) than what it was pα+s−1.
Therefore, invoking Lemma 6 together with the expressions in (21) and (23), we conclude that for some s ≥1 sufficiently high, there must exist ϵ > 0 and a price ¯ ¯ p ∈(pL(m), pα+s −ϵ) for which H(¯ ¯ p, 1)−m H(¯ ¯ p, m) must simultaneously be positive and negative, a contradiction. This contradiction was obtained by assuming the existence of a price ˆ p < pL(m) that verifies (20).
Hence, for all ˆ p < pL(m), H(ˆ p, 1) ≤0 = H(pL(m), m).
Let pL(m∗) ≡min{pL(s), s ∈{2, . . . , n}}. Clearly, pL(m∗) ≤pL(m). Thus, since by defi-nition H(pL(m∗), m∗) = 0, for all ˆ p ≤pL(m∗), H(ˆ p, 1) ≤0 = H(pL(m∗), m∗). Next, suppose, by contradiction, there is a s ∈{m∗+ 1, . . . , n} such that H(pL(m∗), s) > 0 = H(pL(m∗), m∗).
Since H(0, s) < 0 and H(· , s) is continuous on [0, P), there must exist p′ ∈(0, pL(m∗)) such that H(p′, s) = 0, contradicting the definition of pL(m∗). Therefore, for all s ∈{m∗+ 1, . . . , n}, H(pL(m∗), s) ≤0.
Finally, we claim that the strategy profile p = (p1, . . . , pn) ∈[0, P)n, with the property that (i) for all i = 1, . . . , m∗, pi = pL(m∗), and (ii) for all j = m∗+ 1, . . . , n, pj > pL(m∗), constitutes a PSBE for Gn. Indeed, if i ∈{1, . . . , m∗}, then πi(pi, p−i) = H(pL(m∗), m∗) = 0.
Consider a deviation ˆ pi ̸= pi for firm i. If ˆ pi > pi, then πi(ˆ pi, p−i) = 0. Instead, if ˆ pi < pi, then πi(ˆ pi, p−i) = H(ˆ pi, 1) ≤0, where the last inequality follows from the fact that, according with the analysis in the previous paragraph, for all ˆ p ≤pL(m∗), H(ˆ p, 1) ≤0.
On the other hand, if i ∈{m∗+ 1, . . . , n}, then πi(pi, p−i) = 0.
Once again, consider a deviation ˆ pi ̸= pi for firm i. If ˆ pi > pL(m∗), then πi(ˆ pi, p−i) = 0. If ˆ pi < pL(m∗), then πi(ˆ pi, p−i) = H(ˆ pi, 1) ≤0. Lastly, if ˆ pi = pL(m∗), then πi(ˆ pi, p−i) = H(pL(m∗), m∗+ 1), which we have already shown is smaller than or equal to 0. Therefore, p = (p1, . . . , pn) ∈B(Gn). And, since pL(m∗) ≤pL(n), P i∈N qi(p1, . . . , pn) = m∗D(pL(m∗)) m∗ ≥D(pL(n)).
Now, let’s prove the second part of Theorem 1. That is, let’s show that if B(Gn) ̸= ∅, then the assertion “the cost function C(·) is subadditive at every output q ∈[D(pL(n)), K)” is false.
19 Clearly, if S(Gn) ̸= ∅, the result follows from Proposition 2. Hence, assume S(Gn) = ∅.
Fix any equilibrium profile p = (p1, . . . , pn) ∈B(Gn) and suppose, by contradiction, there is a firm k ∈N whose reported price pk < pj for all j ∈N{k}. Without loss of generality, denote by ph, h ̸= k, the second smallest price among the announced prices (p1, . . . , pn); i.e., let ph = min{pj; j ̸= k}. Then, (a) If pk > pM, firm k can profitably deviate to the monopoly profit-maximizing price pM; (b) If pk = pM, Assumption 5 and continuity of H(· , 1) at pM imply that there exists ϵ > 0 such that H(pk −ϵ, 1) > 0. But, since πh(p1, . . . , pn) = 0, this means that firm h can do better by proposing pk −ϵ instead of ph, a contradiction; (c) Finally, if pk < pM, depending upon the location of ph the following happens. If ph > pM, firm k can profitably deviate to pM as before. Otherwise, if ph ≤pM, then by Assumption 5 there is ϵ > 0 such that H(ph −ϵ, 1) > H(pk, 1) = πk(pk, p−k), which contradicts that (p1, . . . , pn) ∈B(Gn).
Hence, using (a)-(c), we conclude that, if (p1, . . . , pn) ∈B(Gn) and S(Gn) = ∅, there must be at least two firms which tie at the lowest price, say p∗, and another firm proposing a price above p∗.18 That is, there must exist m ∈{2, . . . , n −1}, n > 2, and p∗∈[0, ∞) such that (i) For all i ∈{i1, . . . , im} ⊂N, pi = p∗; and (ii) For all j ̸∈{i1, . . . , im}, pj > p∗. Following a similar reasoning than in (a) and (b), it is easy to see that p∗< pM.
Notice that, since S(Gn) = ∅, H(pL(n), n) < H(pL(n), 1), which implies that H(pL(n), 1) > 0. Thus, given that H(0, 1) = −C(K) < 0, it follows that pL(1) < pL(n). Next, using the argument behind the proof of Lemma 3, we show that pL(m) ≤pL(1).
In effect, assume, by contradiction, pL(m) > pL(1). (Recall that before Lemma 2 we proved pL(m) ≤pM.) By Assumption 5, H(pL(m), 1) > H(pL(1), 1) = 0. Thus, H(pL(m), 1) −H(pL(m), m) > 0.
(24) On the other hand, H(0, 1) −H(0, m) = −C(K) + C(K/m) < 0.
(25) Therefore, from (24) and (25) and continuity of H(· , 1) −H(· , m) on [0, pM], there exists a price pα ∈(0, pL(m)) such that H(pα, 1) −H(pα, m) = 0.
(26) Recall that, at the equilibrium (p1, . . . , pn) ∈B(Gn), m firms tie at the lowest price p∗; hence, H(p∗, m) ≥0. Otherwise, any of these firms can profitably deviate to P. Moreover, p∗≥pL(m), because H(· , m) is negative below pL(m).
In addition, (p1, . . . , pn) ∈B(Gn) 18Recall that, since by supposition S(Gn) = ∅, at most n −1 firms can tie at p∗.
20 implies H(p∗, m) ≥H(p∗, 1), which is equivalent to H(p∗, 1) −H(p∗, m) ≤0. Thus, using (24), we conclude that p∗̸= pL(m). Finally, by Assumption 5, H(pM, 1) −H(pM, m) > 0. Therefore, there exists a price pβ ∈[p∗, pM) such that H(pβ, 1) −H(pβ, m) = 0.
(27) Summarizing, by assuming that pL(m) > pL(1), (26) and (27) indicate that the curves H(· , 1) and H(· , m) must intersect each other at least twice on (0, pM). An argument analogous to the one used in the proof of Lemma 3 shows that this assertion is false.19 Thus, pL(m) ≤ pL(1). Furthermore, since we have already shown that pL(1) < pL(n), we get pL(m) < pL(n); and, more importantly, D(pL(m)) > D(pL(n)).
So, it remains to show that C(·) is not subadditive at D(pL(m)). To do that, note that H(· , 1) is negative below pL(m), because pL(1) ≥pL(m). That means 0 = m H(pL(m), m) > H(pL(m), 1), which renders the desired result; i.e., m C ³ D(pL(m)) m ´ < C(D(pL(m))). Therefore, C(·) is not subadditive on [D(pL(n)), K). ■ References Baumol, W. (1977) On the proper cost tests for natural monopoly in a multiproduct industry, American Economic Review 67, 809-822.
Baye, M., and D. Kovenock (2008) Bertrand competition, The New Palgrave Dictionary of Economics (2nd Ed.), by Steven Durlauf and Lawrence Blume, Eds., Palgrave Macmillan.
Baye, M., and J. Morgan (2002) Winner-take-all price competition, Economic Theory 19, 271-282.
Baye, M., and J. Morgan (1999) A folk theorem for one-shot Bertrand games, Economics Letters 65, 59-65.
Chaudhuri, P. (1996) The contestable outcome as a Bertrand equilibrium, Economics Letters 50, 237-242.
Chowdhury, P., and K. Sengupta (2004) Coalition-proof Bertrand equilibria, Economic Theory 24, 307-324.
Chowdhury, P. (2002) Limit-pricing as Bertrand equilibrium, Economic Theory 19, 811-822.
Dastidar, K. (2006) Existence of pure strategy Bertrand equilibrium revisited, manuscript.
Dastidar, K. (2001) Collusive outcomes in price competition, Journal of Economics 73, 81-93.
Dastidar, K. (1995) On the existence of pure strategy Bertrand equilibrium, Economic Theory 5, 19-32.
19The argument exploits a claim similar to Lemma 2, obtained by replacing n with m. The complete proof is available upon request.
21 Grossman, S. (1981) Nash equilibrium and the industrial organization of markets with large fixed costs, Econometrica 49, 1149-1172.
Harrington, J. (1989) A re-evaluation of perfect competition as the solution to the Bertrand price game, Mathematical Social Science 17, 315-328.
Hoernig, S. (2007) Bertrand games and sharing rules, Economic Theory 31, 573-585.
Hoernig, S. (2002) Mixed Bertrand equilibria under decreasing returns to scale: An embarrass-ment of riches, Economics Letters 74, 359-362.
Kaplan, T., and D. Wettstein (2000) The possibility of mixed strategy equilibria with constant returns to scale technology under Bertrand competition, Spanish Economic Review 2, 65-71.
Klaus, A., and J. Brandts (2008) Pricing in Bertrand competition with increasing marginal costs, Games and Economic Behavior 63, 1-31.
Marquez, R. (1997) A note on Bertrand competition with asymmetric fixed costs, Economics Letters 57, 87-96.
Novshek, W., and P. Chowdhury (2003) Bertrand equilibria with entry: Limit results, Interna-tional Journal of Industrial Organization 21, 795-808.
Panzar, J. (1989) Technological determinants of firm and industry structure, Handbook of In-dustrial Organization, by R. Schmalensee and R. Willig, Eds., Elsevier: Vol. 1, 4-59.
Sharkey, W., and D. Sibley (1993) A Bertrand model of pricing and entry, Economics Letters 41, 199-206.
Shapiro, C. (1989) Theories of oligopoly behavior, Handbook of Industrial Organization, by R.
Schmalensee and R. Willig, Eds., Elsevier: Vol. 1, 329-414.
Spulber, D. (1995) Bertrand competition when rivals’ costs are unknown, Journal of Industrial Economics 43 (1), 1-11.
Telser, L. (1991) Industry total cost functions and the status of the core, Journal of Industrial Economics 39, 225-240.
Vives, X. (1999) Oligopoly pricing: old ideas and new tools, Cambridge, MA: MIT Press.
Yano, M. (2006a) A price competition game under free entry, Economic Theory 29, 395-414.
Yano, M. (2006b) The Bertrand equilibrium in a price competition game, Advances in Mathe-matical Economics 8, 449-465.
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