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16500	https://apollo.nvu.vsc.edu/classes/met130/notes/chapter3/spring.html	Seasons in the Northern Hemisphere - Spring Seasons in the Northern Hemisphere - Spring begins on the Vernal Equinox (20 March) 12 hours of sunlight everywhere sun angle is 90� everywhere ends on the Summer Solstice (21 June) images adopted from The Atmosphere by F.K. Lutgens and E.J. Tarbuck, �1998 by Prentice-Hall, Inc Test your knowledge with this interactive tool that determines the amount of solar radiation at the top of the atmosphere anytime during the year and at any latitude QUESTIONS FOR THOUGHT: Consider two scenarios: (a) The tilt of the Earth decreased to 10�. (b) The tilt of the Earth increased to 40�. How would this change the summer and winter temperatures here at LSC? Where would you expect to experience the smallest variation in temperature from year to year and from month to month? Why? At the top of the earth's atmosphere during the early summer (Northern Hemisphere), above what latitude would you expect to receive the most solar radiation in one day? During the same time of year, where would you expect to receive the most solar radiation at the surface?
16501	https://pubmed.ncbi.nlm.nih.gov/6318576/	Relative peripheral and central chemosensory responses to metabolic alkalosis - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Relative peripheral and central chemosensory responses to metabolic alkalosis M Pokorski,S Lahiri PMID: 6318576 DOI: 10.1152/ajpregu.1983.245.6.R873 Item in Clipboard Relative peripheral and central chemosensory responses to metabolic alkalosis M Pokorski et al. Am J Physiol.1983 Dec. Show details Display options Display options Format Am J Physiol Actions Search in PubMed Search in NLM Catalog Add to Search . 1983 Dec;245(6):R873-80. doi: 10.1152/ajpregu.1983.245.6.R873. Authors M Pokorski,S Lahiri PMID: 6318576 DOI: 10.1152/ajpregu.1983.245.6.R873 Item in Clipboard Cite Display options Display options Format Abstract We investigated the relative contribution of peripheral and central chemosensory mechanisms to ventilatory responses to metabolic alkalosis in anesthetized cats by simultaneously measuring steady-state carotid body chemosensory activity and ventilation. The effects of graded steady-state levels of metabolic alkalosis at constant levels of arterial O2 and CO2 partial pressure (PaO2 and PaCO2, respectively) were studied first. Then the responses to isocapnic hypoxia and hyperoxic hypercapnia before and after the induction of a given level of metabolic alkalosis were studied. From the relationship between the carotid chemosensory activity and ventilation, the contribution of the two chemosensory mechanisms was estimated. The depression of ventilation that could not be accounted for by a decrease in the carotid chemosensory activity is attributed to the central effect. We found that metabolic alkalosis decreased both carotid chemosensory activity and ventilation at all levels of PaO2 or PaCO2. The ventilatory effect of alkalosis increased during hypoxia due to suppression of both peripheral chemosensory input and its interaction with the central CO2-H+ drive. During hyperoxia the central effect of alkalosis was predominant, although the peripheral effect increased with hypercapnia. We conclude that acute metabolic alkalosis suppresses both peripheral and central chemosensory drives, and its ventilatory effect grows larger with decreasing PaO2. PubMed Disclaimer Similar articles Inhibition of aortic chemoreceptor responses by metabolic alkalosis in the cat.Pokorski M, Lahiri S.Pokorski M, et al.J Appl Physiol Respir Environ Exerc Physiol. 1982 Jul;53(1):75-80. doi: 10.1152/jappl.1982.53.1.75.J Appl Physiol Respir Environ Exerc Physiol. 1982.PMID: 6811527 Carotid body chemoreceptor and ventilatory responses to sustained hypoxia and hypercapnia in the cat.Andronikou S, Shirahata M, Mokashi A, Lahiri S.Andronikou S, et al.Respir Physiol. 1988 Jun;72(3):361-74. doi: 10.1016/0034-5687(88)90094-1.Respir Physiol. 1988.PMID: 2970107 Efferent inhibition of carotid body chemoreception in chronically hypoxic cats.Lahiri S, Smatresk N, Pokorski M, Barnard P, Mokashi A.Lahiri S, et al.Am J Physiol. 1983 Nov;245(5 Pt 1):R678-83. doi: 10.1152/ajpregu.1983.245.5.R678.Am J Physiol. 1983.PMID: 6416087 Correlative contribution of carotid and aortic afferences to the ventilatory chemosensory drive in steady-state normoxia and to the ventilatory chemoreflexes induced by transient hypoxia.Eugenin J, Larraín C, Zapata P.Eugenin J, et al.Arch Biol Med Exp. 1989 Dec;22(4):395-408.Arch Biol Med Exp. 1989.PMID: 2488538 Review. Metabolic alkalosis.Cogan MG, Liu FY, Berger BE, Sebastian A, Rector FC Jr.Cogan MG, et al.Med Clin North Am. 1983 Jul;67(4):903-14. doi: 10.1016/s0025-7125(16)31184-1.Med Clin North Am. 1983.PMID: 6410135 Review.No abstract available. See all similar articles Cited by A role for xanthine oxidase in the control of fetal cardiovascular function in late gestation sheep.Herrera EA, Kane AD, Hansell JA, Thakor AS, Allison BJ, Niu Y, Giussani DA.Herrera EA, et al.J Physiol. 2012 Apr 15;590(8):1825-37. doi: 10.1113/jphysiol.2011.224576. Epub 2012 Feb 13.J Physiol. 2012.PMID: 22331413 Free PMC article. Peripheral chemoreceptors: function and plasticity of the carotid body.Kumar P, Prabhakar NR.Kumar P, et al.Compr Physiol. 2012 Jan;2(1):141-219. doi: 10.1002/cphy.c100069.Compr Physiol. 2012.PMID: 23728973 Free PMC article.Review. Publication types Research Support, U.S. Gov't, P.H.S. Actions Search in PubMed Search in MeSH Add to Search MeSH terms Action Potentials Actions Search in PubMed Search in MeSH Add to Search Alkalosis / physiopathology Actions Search in PubMed Search in MeSH Add to Search Animals Actions Search in PubMed Search in MeSH Add to Search Bicarbonates Actions Search in PubMed Search in MeSH Add to Search Carbon Dioxide / blood Actions Search in PubMed Search in MeSH Add to Search Carotid Sinus / physiopathology Actions Search in PubMed Search in MeSH Add to Search Cats Actions Search in PubMed Search in MeSH Add to Search Chemoreceptor Cells / physiopathology Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Hypercapnia / physiopathology Actions Search in PubMed Search in MeSH Add to Search Hypoxia / physiopathology Actions Search in PubMed Search in MeSH Add to Search Oxygen / blood Actions Search in PubMed Search in MeSH Add to Search Partial Pressure Actions Search in PubMed Search in MeSH Add to Search Sodium Bicarbonate Actions Search in PubMed Search in MeSH Add to Search Substances Bicarbonates Actions Search in PubMed Search in MeSH Add to Search Carbon Dioxide Actions Search in PubMed Search in MeSH Add to Search Sodium Bicarbonate Actions Search in PubMed Search in MeSH Add to Search Oxygen Actions Search in PubMed Search in MeSH Add to Search Related information MedGen PubChem Compound (MeSH Keyword) Grants and funding HL-08899-19/HL/NHLBI NIH HHS/United States HL-19737-07/HL/NHLBI NIH HHS/United States [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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16502	https://nls-mapper.iab.keio.ac.jp/cgi-bin/NLS_Mapper_help.cgi	NLS Mapper cNLS Mapper Help Page About cNLS Mapper cNLS Mapper accurately predicts nuclear localization signals (NLSs) specific to the importin αβ pathway by calculating NLS scores (levels of NLS activities), but not by the conventional sequence similarity search or by the machine learning strategy. The NLS scores are calculated with four NLS profiles (for class 1/2, class 3, class 4, and bipartite NLSs), each of which represents a contribution of every amino acid residue at every position within an NLS class to the entire NLS activity (1) . These profiles are generated by an extensive amino acid replacement analysis for each NLS class in budding yeast. We have found that each residue within an NLS, most cases, additively and independently contributes to the entire activity (1, 2) . Thus, NLS activities (scores) can be estimated by serially adding positive or negative contribution scores given in an NLS profile. The additive property of NLS motifs also allowed us to design potent peptide inhibitors specific to the importin α/β nuclear import pathway by serially selecting amino acids with a high contribution score in an NLS profile (2) . It should be noted that the NLS profiles were generated by nuclear import assays in budding yeast and, thus, NLS prediction for other species may be less effective than that for yeast, although the importin α/β pathway is highly conserved in eukaryotes. In addition, cNLS Mapper cannot predict importin α-independent NLSs (NLSs directly recognized by several importin β members). How to use cNLS Mapper #### (a) Input sequense Any sequence forms including plain text are acceptable. Any characters except alphabet are ignored, but non-amino acid codes, such as b, x, and z, are prohibited. A minimal and maximal acceptable length of imput sequence is 11 and 4999 amino acids, respectively. A single sequence can be pasted in the text box or a text file containing a sequence in fasta or row format. (b) Cut-off score cNLS Mapper extracts putative NLS sequences with a score equal to or more than the selected cut-off score. Higher scores indicate stronger NLS activities. Briefly, a GUS-GFP reporter protein fused to an NLS with a score of 8, 9, or 10 is exclusively localized to the nucleus, that with a score of 7 or 8 partially localized to the nucleus, that with a score of 3, 4, or 5 localized to both the nucleus and the cytoplasm , and that with a score of 1 or 2 localized to the cytoplasm (see Supplemental Figure S1 in refs. 1 or 3 for detail). (c) Search for bipartite NLSs with a long linker The optimal length of the linker of bipartite NLSs is 10-12 amino acids, as demonstrated earlier. However, we have observed that bipartite NLSs with linkers at least up to 20 amino acids are functional in an unstructured flexible region, although a longer linker decreases the NLS activity. Because unstructured regions of proteins are generally found in the N- and C-terminal regions, the protein regions searched for bipartite NLSs with a long linker are restricted, as default, to the terminal 60-amino acid regions. But you can select an option by which the entire region of a protein is searched for the long bipartite NLSs. This option may also help to identify a structure-dependent bipartite NLS, whose terminal basic cores are located close to each other in the tertiary structure. #### (d) Search result A result page will appear immediately after you click the "Predict NLS" (submit)button. In the upper part of result page, your input sequence is presented , where, if there is a predicted NLS(s), it is highlighted in red. In the lower part, the predicted NLSs are devided into two classes, monopartite and bipartite NLSs, and their sequence, positions in the protein, and calculated scores are listed . The program eliminates the flanking residues that do not affect the NLS activity from the initially predicted sequences to display only essential sequences. cNLS Mapper often displays several similar overlapping NLSs with different scores. Because the program scans the protein sequence with a window size of 16 amino acid residues for monopartite NLSs and a window size of 26-28 amino acid residues for bipartite NLSs (26-36 for long bipartite NLSs) and with a shift size of one amino acid and calculates the score for every sequence, basic-rich regions often exhibit high scores at multiple different scan windows. Thus, when multiple overlapping NLS sequences are displayed, the sequence with the highest score is, in most cases, considered to be the true NLS. APPENDIX: a brief overview of nuclear import and export of proteins #### Importin α-dependent and -independent NLSs Nuclear import and export of macromolecules are mediated by receptors called karyopherins, also known as importins and exportins. Nuclear import of proteins is generally initiated by the formation of a ternary complex with importin α, importin β1 and a cargo, where impotin α serves as an adapter for importin β1 and recognizes nuclear localization signals (NLSs) within the cargos. On the other hand, several importin β members including importin β1 and importin β2 (also known as Transportin) directly binds some specific classes of NLSs. #### Classical NLS The classical NLSs rich in basic amino acids are known as NLSs recognized by importin α, and are classified into two major classes, monopartite and bipartite NLSs. Monopartite NLSs contain a single cluster of basic residues and are divided into two subclasses; one with at least 4 consecutive basic amino acids (class 1) and the other with three basic amino acids, represented by K(K/R)X(K/R) as a putative consensus sequence where X indicates any amino acid (class 2). These two classes are exemplified by the SV40 large T antigen NLS (PKKKRKV) and the c-Myc NLS (PAAKRVKLD). Bipartite NLSs contain two clusters of basic residues separated by a 10-12 amino acid linker and are exemplified by the nucleoplasmin NLS (KRPAATKKAGQAKKKK). A putative consensus sequence of the bipartite NLS has been defined as (K/R)(K/R)X10-12(K/R)3/5 where X indicates any amino acid and (K/R)3/5 represents at least three of either lysine or arginine out of five consecutive amino acids. Although these consensus sequences have been approved, there are a considerable number of non-functional NLSs that match the consensus sequences. Translated ORFs (5869 proteins) from the budding yeast genome were searched using the consensus NLS patterns, over 3000 and 1500 independent proteins that perfectly matched the monopartite and bipartite patterns, respectively, have been found. This substantially exceeds the expected number of nuclear protein which is estimated to be ~30% of all the yeast proteins. Thus, the current prevailing consensus for the classical NLSs is incomplete and there must be more specific consensus sequences. #### Noncanonical NLSs recognized by importin α Many noncanonical NLSs that do not match the classical NLSs have been defined. Kosugi et al. have been reported three classes of noncanonical monpartite NLSs (classes 3-5) that bind directly to importin α(3) . Class 3 and class 4 NLSs have KRX(W/F/Y)XXAF and (P/R)XXKR(K/R) core sequences, respectively, and specifically bind to the minor binding pocket of importin α. This is contrasting with the classical class 1 and class 2 NLSs that specifically bind to the major binding pocket of importin α. It is suggested that a bipartite NLS is a hybrid sequence consisted of imperfect class 1/2 and class 3/4 NLSs because the N-terminal and C-terminal basic regions of bipartite NLSs bind to the minor and major binding pockets of importin α, respectively (4) . Class 5 NLSs has a LGKR(K/R)(W/F/Y) core sequence and is specific only to plants, in contrast with the other classes, all of which are functional in yeast, plants and mammals (3) . Noncanonical NLSs recognized by importin β members Although this NLS prediction program deals only with importin α-dependent NLSs, there are a number of noncanonical NLSs that directly bind to importin βs (5) . These NLSs are generally longer and more variable than the classical NLSs. The consensus sequence of the NLSs recognized by importin β1 has not been established but many of the NLSs are rich in arginine, as exemplified by the HTLV-1 Rex NLS (MPKTRRRPRRSQRKRPPT). On the other hand, it has been reported that the consensuses of importin β2 (Transportin)-dependent NLSs are represented by patterns with overall basic character that contain a central hydrophobic or basic motif followed by a C-terminal R/H/KX(2-5)PY consensus sequence (6) . Nuclear export signals Many nuclear proteins conatin both NLSs and nuclear export signals (NESs) and shuttle between the nucleus and the cytoplasm through coordination of these transport signals. Nuclear export of proteins occurs through the classical nuclear export pathway mediated by an evolutionarily conserved CRM1/exportin protein and through the nonclassical export pathways mediated by other importin β members such as Msn5 (7) . For the classical nuclear export pathway, the CRM1-Ran-GTP complex binds directly to the NES contained in the cargo and directs the export of the ternary complex from the nucleus. NESs that are recognized by CRM1 (i.e., leucine-rich NESs) typically contain large hydrophobic conserved residues separated by a variable number of amino acids, given by the traditional consensus sequence L-X(2,3)-[LIVFM]-X(2,3)-L-X-[LI], where X(2,3) represents any two or three amino acids. A recent study has demonstrated that the classical NESs are classified into six patterns according to the hydrophobic spacing and has established more reliable NES consensus sequences, which fit most experimentally confirmed NESs, than the traditional consensus sequence (8) . Kosugi S., Hasebe M., Tomita M., and Yanagawa H. (2009) Systematic identification of yeast cell cycle-dependent nucleocytoplasmic shuttling proteins by prediction of composite motifs. Proc. Natl. Acad. Sci. USA 106, 10171-10176. 2. Kosugi S., Hasebe M., Entani T., Takayama S., Tomita M., and Yanagawa H. (2008) Chem. Biol. 15, 940-949. 3. Kosugi S., Hasebe M., Matsumura N., Takashima H., Miyamoto-Sato E., Tomita M., and Yanagawa H. (2009) J. Biol. Chem. 284, 478-485. 4. Conti E., Uy M., Leighton L., Blobel G., and Kuriyan J. (1998) Cell 94, 193-204. 5. Harel, A., and Forbes, D.J. (2004) Mol. Cell 16, 319-330. 6. Lee B.J., Cansizoglu A.E., Suel K.E., Louis T.H., Zhang Z., and Chook Y.M. (2006) Cell 126, 543-558. 7. Ossareh-Nazari, B., Gwizdek, C., and Dargemont, C. (2001) Traffic 2, 684-689. 8. Kosugi, S., Hasebe M., Tomita M., and Yanagawa H. (2008) Traffic 9, 2053-2062.
16503	https://openstax.org/books/university-physics-volume-3/pages/10-4-nuclear-reactions	Skip to Content Go to accessibility page Keyboard shortcuts menu Log in University Physics Volume 3 10.4 Nuclear Reactions University Physics Volume 3 10.4 Nuclear Reactions Search for key terms or text. Learning Objectives By the end of this section, you will be able to: Describe and compare three types of nuclear radiation Use nuclear symbols to describe changes that occur during nuclear reactions Describe processes involved in the decay series of heavy elements Early experiments revealed three types of nuclear “rays” or radiation: alpha (α) rays, beta (β) rays, and gamma (γ) rays. These three types of radiation are differentiated by their ability to penetrate matter. Alpha radiation is barely able to pass through a thin sheet of paper. Beta radiation can penetrate aluminum to a depth of about 3 mm, and gamma radiation can penetrate lead to a depth of 2 or more centimeters (Figure 10.11). Figure 10.11 A comparison of the penetration depths of alpha (α), beta (β), and gamma (γ) radiation through various materials. The electrical properties of these three types of radiation are investigated by passing them through a uniform magnetic field, as shown in Figure 10.12. According to the magnetic force equation F⃗ =qv⃗ ×B⃗ , positively charged particles are deflected upward, negatively charged particles are deflected downward, and particles with no charge pass through the magnetic field undeflected. Eventually, α rays were identified with helium nuclei (4He),β rays with electrons and positrons (positively charged electrons or antielectrons), and γ rays with high-energy photons. We discuss alpha, beta, and gamma radiation in detail in the remainder of this section. Figure 10.12 The effect of a magnetic field on alpha (α), beta (β), and gamma (γ) radiation. This figure is a schematic only. The relative paths of the particles depend on their masses and initial kinetic energies. Alpha Decay Heavy unstable nuclei emit α radiation. In α-particle decay (or alpha decay), the nucleus loses two protons and two neutrons, so the atomic number decreases by two, whereas its mass number decreases by four. Before the decay, the nucleus is called the parent nucleus. The nucleus or nuclei produced in the decay are referred to as the daughter nucleus or daughter nuclei. We represent an α decay symbolically by AZX→A−4Z−2X+42He 10.21 where AZX is the parent nucleus, A−4Z−2X is the daughter nucleus, and 42He is the α particle. In α decay, a nucleus of atomic number Z decays into a nucleus of atomic number Z−2 and atomic mass A−4. Interestingly, the dream of the ancient alchemists to turn other metals into gold is scientifically feasible through the alpha-decay process. The efforts of the alchemists failed because they relied on chemical interactions rather than nuclear interactions. Interactive Watch alpha particles escape from a polonium nucleus, causing radioactive alpha decay. See how random decay times relate to the half-life. To try a simulation of alpha decay, visit alpha particles An example of alpha decay is uranium-238: 23892U→23490X+42He. The atomic number has dropped from 92 to 90. The chemical element with Z=90 is thorium. Hence, Uranium-238 has decayed to Thorium-234 by the emission of an α particle, written 23892U→23490Th+42He. Subsequently, 23490Th decays by β emission with a half-life of 24 days. The energy released in this alpha decay takes the form of kinetic energies of the thorium and helium nuclei, although the kinetic energy of thorium is smaller than helium due to its heavier mass and smaller velocity. Example 10.7 Plutonium Alpha Decay Find the energy emitted in the α decay of 239Pu. Strategy The energy emitted in the α decay of 239Pu can be found using the equation E=(Δm)c2. We must first find Δm, the difference in mass between the parent nucleus and the products of the decay. Solution The decay equation is 239Pu→235U+4He. Thus, the pertinent masses are those of 239Pu, 235U, and the α particle or 4He, all of which are known. The initial mass was m(239Pu)=239.052157u. The final mass is the sum m(235U)+m(4He)=235.043924u+4.002602u=239.046526u. Thus, Δm=m(239Pu)−[m(235U)+m(4He)]=239.052157u−239.046526u=0.0005631u. Now we can find E by entering Δm into the equation: E=(Δm)c2=(0.005631u)c2. We know 1u=931.5MeV/c2, so we have E=(0.005631)(931.5MeV/c2)(c2)=5.25MeV. Significance The energy released in this α decay is in the MeV range, many times greater than chemical reaction energies. Most of this energy becomes kinetic energy of the α particle (or 4He nucleus), which moves away at high speed. The energy carried away by the recoil of the 235U nucleus is much smaller due to its relatively large mass. The 235U nucleus can be left in an excited state to later emit photons (γ rays). Beta Decay In most β particle decays (or beta decay), either an electron (β−) or positron (β+) is emitted by a nucleus. A positron has the same mass as the electron, but its charge is +e. For this reason, a positron is sometimes called an antielectron. How does β decay occur? A possible explanation is the electron (positron) is confined to the nucleus prior to the decay and somehow escapes. To obtain a rough estimate of the escape energy, consider a simplified model of an electron trapped in a box (or in the terminology of quantum mechanics, a one-dimensional square well) that has the width of a typical nucleus (10−14m). According to the Heisenberg uncertainty principle in Quantum Mechanics, the uncertainty of the momentum of the electron is: Δp>hΔx=6.6×10−34m2⋅kg/s10−14m=6.6×10−20kg⋅m/s. Taking this momentum value (an underestimate) to be the “true value,” the kinetic energy of the electron on escape is approximately (Δp)22me=(6.6×10−20kg⋅m/s)22(9.1×10−31kg)=2.0×10−9J=12,400MeV. Experimentally, the electrons emitted in β− decay are found to have kinetic energies of the order of only a few MeV. We therefore conclude that the electron is somehow produced in the decay rather than escaping the nucleus. Particle production (annihilation) is described by theories that combine quantum mechanics and relativity, a subject of a more advanced course in physics. Nuclear beta decay involves the conversion of one nucleon into another. For example, a neutron can decay to a proton by the emission of an electron (β−) and a nearly massless particle called an antineutrino (ν–): 10n→11p+0−1e+v–. The notation 0−1e is used to designate the electron. Its mass number is 0 because it is not a nucleon, and its atomic number is −1 to signify that it has a charge of −e. The proton is represented by 11p because its mass number and atomic number are 1. When this occurs within an atomic nucleus, we have the following equation for beta decay: AZX→AZ+1X+0−1e+v–. 10.22 Enrico Fermi proposed a theory of beta decay in 1934, but his work was initially rejected. Other physicists' experiments to prove it were unsuccessful, casting further doubt on the theory. Chinese-born physicist Chien-Shiung Wu, who had developed a number of processes critical to the Manhattan Project and related research, identified a number of flaws in the earlier experimental methods and materials. She designed a new method, verified Fermi's theory, and later went on to establish the core principles of beta decay. As discussed in another chapter, this process occurs due to the weak nuclear force. Interactive Watch beta decay occur for a collection of nuclei or for an individual nucleus. As an example, the isotope 23490Th is unstable and decays by β− emission with a half-life of 24 days. Its decay can be represented as 23490Th→23491X+0−1e+v–. Since the chemical element with atomic number 91 is protactinium (Pa), we can write the β− decay of thorium as 23490Th→23491Pa+0−1e+v–. The reverse process is also possible: A proton can decay to a neutron by the emission of a positron (β+) and a nearly massless particle called a neutrino (v). This reaction is written as 11p→10n+0+1e+v. The positron 0+1e is emitted with the neutrino v, and the neutron remains in the nucleus. (Like β− decay, the positron does not precede the decay but is produced in the decay.) For an isolated proton, this process is impossible because the neutron is heavier than the proton. However, this process is possible within the nucleus because the proton can receive energy from other nucleons for the transition. As an example, the isotope of aluminum 2613Al decays by β+ emission with a half-life of 7.40×105y. The decay is written as 2613Al→2612X+0+1e+v. The atomic number 12 corresponds to magnesium. Hence, 2613Al→2612Mg+0+1e+v. As a nuclear reaction, positron emission can be written as AZX→AZ−1X+0+1e+v. The neutrino was not detected in the early experiments on β decay. However, the laws of energy and momentum seemed to require such a particle. Later, neutrinos were detected through their interactions with nuclei. Example 10.8 Bismuth Alpha and Beta Decay The 21183Bi nucleus undergoes both α and β− decay. For each case, what is the daughter nucleus? Strategy We can use the processes described by Equation 10.21 and Equation 10.22, as well as the Periodic Table, to identify the resulting elements. Solution The atomic number and the mass number for the α particle are 2 and 4, respectively. Thus, when a bismuth-211 nucleus emits an α particle, the daughter nucleus has an atomic number of 81 and a mass number of 207. The element with an atomic number of 81 is thallium, so the decay is given by 21183Bi→20781Ti+42He. In β− decay, the atomic number increases by 1, while the mass number stays the same. The element with an atomic number of 84 is polonium, so the decay is given by 21183Bi→21184Po+0−1e+v–. Check Your Understanding 10.4 In radioactive beta decay, does the atomic mass number, A, increase or decrease? Gamma Decay A nucleus in an excited state can decay to a lower-level state by the emission of a “gamma-ray” photon, and this is known as gamma decay. This is analogous to de-excitation of an atomic electron. Gamma decay is represented symbolically by AZX→AZX+γ where the asterisk () on the nucleus indicates an excited state. In γ decay, neither the atomic number nor the mass number changes, so the type of nucleus does not change. Radioactive Decay Series Nuclei with Z>82 are unstable and decay naturally. Many of these nuclei have very short lifetimes, so they are not found in nature. Notable exceptions include 23290Th (or Th-232) with a half-life of 1.39×1010 years, and 23892U (or U-238) with a half-life of 7.04×108 years. When a heavy nucleus decays to a lighter one, the lighter daughter nucleus can become the parent nucleus for the next decay, and so on. This process can produce a long series of nuclear decays called a decay series. The series ends with a stable nucleus. To illustrate the concept of a decay series, consider the decay of Th-232 series (Figure 10.13). The neutron number, N, is plotted on the vertical y-axis, and the atomic number, Z, is plotted on the horizontal x-axis, so Th-232 is found at the coordinates (N,Z)=(142,90). Th-232 decays by α emission with a half-life of 1.39×1010 years. Alpha decay decreases the atomic number by 2 and the mass number by 4, so we have 23290Th→22888Ra+42He. The neutron number for Radium-228 is 140, so it is found in the diagram at the coordinates (N,Z)=(140,88). Radium-228 is also unstable and decays by β− emission with a half-life of 5.76 years to Actinum-228. The atomic number increases by 1, the mass number remains the same, and the neutron number decreases by 1. Notice that in the graph, α emission appears as a line sloping downward to the left, with both N and Z decreasing by 2. Beta emission, on the other hand, appears as a line sloping downward to the right with N decreasing by 1, and Z increasing by 1. After several additional alpha and beta decays, the series ends with the stable nucleus Pb-208. The relative frequency of different types of radioactive decays (alpha, beta, and gamma) depends on many factors, including the strength of the forces involved and the number of ways a given reaction can occur without violating the conservation of energy and momentum. How often a radioactive decay occurs often depends on a sensitive balance of the strong and electromagnetic forces. These forces are discussed in Particle Physics and Cosmology. Figure 10.13 In the thorium 23290Th decay series, alpha (α) decays reduce the atomic number, as indicated by the red arrows. Beta (β−) decays increase the atomic number, as indicated by the blue arrows. The series ends at the stable nucleus Pb-208. As another example, consider the U-238 decay series shown in Figure 10.14. After numerous alpha and beta decays, the series ends with the stable nucleus Pb-206. An example of a decay whose parent nucleus no longer exists naturally is shown in Figure 10.15. It starts with Neptunium-237 and ends in the stable nucleus Bismuth-209. Neptunium is called a transuranic element because it lies beyond uranium in the periodic table. Uranium has the highest atomic number (Z=92) of any element found in nature. Elements with Z>92 can be produced only in the laboratory. They most probably also existed in nature at the time of the formation of Earth, but because of their relatively short lifetimes, they have completely decayed. There is nothing fundamentally different between naturally occurring and artificial elements. Figure 10.14 In the Uranium-238 decay series, alpha (α) decays reduce the atomic number, as indicated by the red arrows. Beta (β−) decays increase the atomic number, as indicated by the blue arrows. The series ends at the stable nucleus Pb-206. Notice that for Bi (21), the decay may proceed through either alpha or beta decay. Figure 10.15 In the Neptunium-237 decay series, alpha (α) decays reduce the atomic number, as indicated by the red arrows. Beta (β−) decays increase the atomic number, as indicated by the blue arrows. The series ends at the stable nucleus Bi-209. Radioactivity in the Earth According to geologists, if there were no heat source, Earth should have cooled to its present temperature in no more than 1 billion years. Yet, Earth is more than 4 billion years old. Why is Earth cooling so slowly? The answer is nuclear radioactivity, that is, high-energy particles produced in radioactive decays heat Earth from the inside (Figure 10.16). Figure 10.16 Earth is heated by nuclear reactions (alpha, beta, and gamma decays). Without these reactions, Earth’s core and mantle would be much cooler than it is now. Candidate nuclei for this heating model are 238U and40K, which possess half-lives similar to or longer than the age of Earth. The energy produced by these decays (per second per cubic meter) is small, but the energy cannot escape easily, so Earth’s core is very hot. Thermal energy in Earth’s core is transferred to Earth’s surface and away from it through the processes of convection, conduction, and radiation. Previous Next Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Samuel J. Ling, Jeff Sanny, William Moebs Publisher/website: OpenStax Book title: University Physics Volume 3 Publication date: Sep 29, 2016 Location: Houston, Texas Book URL: Section URL: © Jul 8, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . 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16504	https://www.geogebra.org/m/Tjmq295g	Reflections in the Plane – GeoGebra Google Classroom GeoGebra Classroom Sign in Search Google Classroom GeoGebra Classroom Home Resources Profile Classroom App Downloads Reflections in the Plane Author:Karen Cunningham Topic:Reflection New Resources Transformaciones usando coordenadas [z`]]]( Forming Similar Triangles גיליון אלקטרוני להעלאת נתוני בעיה ויצירת גרף בהתאם רישום חופשי Discover Resources Sketch 3 Binomial and Geometric Distributions Rope around Earth Zlati rez, krog, kvadrat, potenca Hình 2.61 trang 72 Discover Topics Diagrams Arithmetic Operations Scatter Plot Interest Calculation Expected Value AboutPartnersHelp Center Terms of ServicePrivacyLicense Graphing CalculatorCalculator SuiteMath Resources Download our apps here: English / English (United States) © 2025 GeoGebra®
16505	https://www.nationwidechildrens.org/conditions/health-library/sinusitis-in-children	Skip to Content Nationwide Children’s Hospital Sinusitis in Children What is sinusitis in children? Sinusitis is an inflammation of the sinuses. Inflammation often happens after a cold or with allergies. There are three types of sinusitis: Short-term (acute). Symptoms of this type of infection last less than 12 weeks and get better with the correct treatment. Long-term (chronic). These symptoms last longer than 12 weeks. Recurrent. This means the infection comes back again and again. It means 3 or more episodes of acute sinusitis in a year. The sinuses are air-filled spaces (cavities) near the nose. They are lined with mucous membranes. There are four different sinuses: Ethmoid sinus. Located around the bridge of the nose. This sinus is present at birth and continues to grow. Maxillary sinus. Located around the cheeks. This sinus is also present at birth and continues to grow. Frontal sinus. Located in the area of the forehead. This sinus does not develop until around age 7. Sphenoid sinus. Located deep behind the nose. This sinus does not develop until the teen years. What causes sinusitis in a child? The cause of sinusitis varies depending on which type your child has. Acute sinusitis is mainly caused by viruses, but it can also be caused by bacteria. The cause of chronic sinusitis may not be known. Or it may be related to allergies, infections, or nasal polyps. What are the symptoms of sinusitis in a child? The most common symptoms of sinusitis include: Stuffy nose. Thick, colored drainage in the nose. Drainage down the back of the throat (postnasal drip). Headache. Cough. Swelling around the eyes. Fever. Irritability or fatigue. If your child has any of these symptoms for more than 10 to 14 days, make sure they see their doctor. Which children are at risk for sinusitis? A sinus infection sometimes happens after an upper respiratory infection or common cold. The cold causes swelling that can block the opening of the sinuses. This can cause a sinus infection. Allergies can also lead to sinusitis because of swelling and increased mucus. Other possible conditions that can lead to sinusitis include: Exposure to secondhand smoke. Nose injury. Birth defect with abnormality of the roof of the mouth (cleft palate). Problem with stomach acids (gastroesophageal reflux disease, or GERD). Cystic fibrosis and immunodeficiency syndromes. Large adenoids (tissue behind the nose). Immune problems or antibody deficiencies are risks for chronic sinus conditions. How is sinusitis diagnosed in a child? The doctor will ask about your child’s symptoms and health history. They will give your child a physical exam. Your child may also have tests, such as: Sinus X-rays. An X-ray exam of the sinuses may help with the diagnosis. CT scan of the sinuses. A CT scan shows detailed images of any part of the body. They are more detailed than X-rays. Cultures from the sinuses. A swab of discharge from the nose may be taken. The sample is checked for bacteria or other germs. How is sinusitis treated in a child? Treatment will depend on your child’s symptoms, age, and general health. It will also depend on how severe the condition is. Acute sinusitis Acute sinusitis may get better on its own. If it does not, your child’s doctor may prescribe: Antibiotics. If your child’s sinuses are infected with bacteria, antibiotics are given to kill the bacteria. If your child’s symptoms haven’t improved after 3 to 5 days, their doctor may try a different antibiotic. Allergy medicines. For sinusitis caused by allergies, antihistamines and other allergy medicines can reduce swelling. Don’t use over-the-counter decongestant nasal sprays without checking with your child’s doctor. These sprays may make symptoms worse. Recurrent sinusitis is also treated with antibiotic and allergy medicines. Your child’s doctor may refer you to an ear, nose, and throat doctor (ENT or otolaryngologist) for testing and treatment. Chronic sinusitis Treatment for chronic sinusitis may include: ENT specialist visit. Your child may be referred to an ENT doctor. Antibiotics. Your child may need to take antibiotics for a longer time. If bacteria aren’t the cause, antibiotics won’t help. Inhaled corticosteroid medicine. Nasal sprays or drops with steroids are often prescribed. Saline wash. This helps keep the sinuses and nose moist. Ask your child’s doctor or nurse for instructions. Allergy shots or immunotherapy. If your child has nasal allergies, shots may help reduce their reaction to allergens such as pollen, dust mites, or mold. Surgery. Surgery for chronic sinusitis is an option. But it is not done very often in children. Care may also include: Fluids. A glass of water or juice every hour or two is a good rule. Fluids help thin mucus, allowing it to drain more easily. Fluids also help prevent dehydration. Other medicines. Nasal sprays with antihistamines and decongestants, saline sprays or drops, or medicines to loosen and clear mucus may be prescribed. Warm compresses. Apply a warm, moist towel to your child’s nose, cheeks, and eyes to help ease pain in the face. What are possible complications of sinusitis in a child? In very rare cases, acute sinusitis can result in a brain infection. What can you do to prevent sinusitis in your child? There are things that can help prevent sinusitis in your child. They include: Support a healthy lifestyle by providing a healthy diet, encouraging good fluid intake, and making time for physical exercise. Have your child use saline sprays, washes, or both. Use these often to keep the nose as moist as possible. Use a humidifier in dry indoor environments. Follow the directions for cleaning the machine. Keep your child away from smoke and the aerosol mist from vaping. Do not smoke or vape around your child or in your house. Avoid being around others who are smoking or vaping. Keep your child away from things that cause allergy symptoms. Don't force water into the sinuses. For example, your child should not jump into water. Limit time in chlorinated pools. The chlorine can irritate the nose and sinuses. Teach your child when and how to wash their hands. Keep you and your child up-to-date on immunizations. Don't have close contact with people who have colds or other upper respiratory infections. Key points about sinusitis in children Sinusitis is an inflammation of the sinuses. When discharge from the nose is blocked, germs may start to grow. This leads to a sinus infection or sinusitis. Acute sinusitis may get better on its own. But if it doesn’t, medicine can be prescribed. For chronic sinusitis, the doctor may refer your child to an ear, nose, and throat doctor (ENT) for testing and treatment. To help prevent sinusitis, encourage a healthy lifestyle and have your child use saline sprays or washes to keep the nose moist. Use a humidifier in dry inside areas. Next steps Tips to help you get the most from a visit to your child’s doctor: Know the reason for the visit and what you want to happen. Before your visit, write down questions you want answered. At the visit, write down the name of a new diagnosis and any new medicines, treatments, or tests. Also write down any new instructions the doctor gives you for your child. Know why a new medicine or treatment is prescribed and how it will help your child. Also know what the side effects are and when they should be reported. Ask if your child’s condition can be treated in other ways. Know why a test or procedure is recommended and what the results could mean. Know what to expect if your child does not take the medicine or have the test or procedure. If your child has a follow-up appointment, write down the date, time, and purpose for that visit. Know how you can contact your child’s doctor after office hours. This is important if your child becomes ill and you have questions or need advice. Related Specialties Ear Nose & Throat Services (Otolaryngology) Online Medical Reviewer: Melinda Murray Ratini DORiley Holtz RN BS Date Last Reviewed: 8/1/2025 © 2000-2025 The StayWell Company, LLC. All rights reserved. This information is not intended as a substitute for professional medical care. Always follow your healthcare professional's instructions. © 2000-2025 The StayWell Company, LLC. All rights reserved. This information is not intended as a substitute for professional medical care. Always follow your healthcare professional's instructions. Related Conditions Sinusitis in Children Upper Respiratory Disorders What's Up with Sinusitis? Show More
16506	https://eudml.org/doc/123936	Login \| Register \| (Why Register?) A new proof of the monotonicity of power means. Witkowski, Alfred [JIPAM. Journal of Inequalities in Pure & Applied Mathematics [electronic only]]( /journal/10079) (2004) Access Full Article Access to full text Full (PDF) How to cite Witkowski, Alfred. "A new proof of the monotonicity of power means.." JIPAM. Journal of Inequalities in Pure & Applied Mathematics [electronic only] 5.1 (2004): Paper No. 6, 2 p.-Paper No. 6, 2 p.. @article{Witkowski2004, author = {Witkowski, Alfred}, journal = {JIPAM. Journal of Inequalities in Pure & Applied Mathematics [electronic only]}, keywords = {weighted power mean; Bernoulli inequality; convex functions; monotonicity}, language = {eng}, number = {1}, pages = {Paper No. 6, 2 p.-Paper No. 6, 2 p.}, publisher = {Victoria University, School of Communications and Informatics}, title = {A new proof of the monotonicity of power means.}, url = { volume = {5}, year = {2004}, } TY - JOUR AU - Witkowski, Alfred TI - A new proof of the monotonicity of power means. JO - JIPAM. Journal of Inequalities in Pure & Applied Mathematics [electronic only] PY - 2004 PB - Victoria University, School of Communications and Informatics VL - 5 IS - 1 SP - Paper No. 6, 2 p. EP - Paper No. 6, 2 p. LA - eng KW - weighted power mean; Bernoulli inequality; convex functions; monotonicity UR - ER - NotesEmbed ? You must be logged in to post comments. To embed these notes on your page include the following JavaScript code on your page where you want the notes to appear. Only the controls for the widget will be shown in your chosen language. Notes will be shown in their authored language. Tells the widget how many notes to show per page. You can cycle through additional notes using the next and previous controls. Note: Best practice suggests putting the JavaScript code just before the closing </body> tag. </body> Article Keywords weighted power mean, Bernoulli inequality, convex functions, monotonicity Subjects You must be logged in to add subjects. Miscellaneous topics Inequalities Functions of one variable From the Journal [JIPAM. Journal of Inequalities in Pure & Applied Mathematics [electronic only]]( /journal/10079) (2004) In Other Databases On Google Scholar You must be logged in to use personal lists. Infobox To add items to a personal list choose the desired list from the selection box or create a new list. To close, click the Close button or press the ESC key. Close Infobox This article was successfully added to the collection. To close, click the Close button or press the ESC key. Close Share this Article
16507	https://www.scribd.com/document/311194253/Order-Statistic	Order Statistic: From Wikipedia, The Free Encyclopedia \| PDF \| Median \| Probability Distribution Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for days 0 ratings 0% found this document useful (0 votes) 151 views 6 pages Order Statistic: From Wikipedia, The Free Encyclopedia The document discusses order statistics, which are the values of a data set arranged in order of magnitude. It defines order statistics and provides examples. It then examines the probabilit… Full description Uploaded by hienluong293 AI-enhanced title and description Download Save Save Order Statistic For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Report 0 ratings 0% found this document useful (0 votes) 151 views 6 pages Order Statistic: From Wikipedia, The Free Encyclopedia The document discusses order statistics, which are the values of a data set arranged in order of magnitude. It defines order statistics and provides examples. It then examines the probability distributions of order statistics, including their joint distributions. It discusses how order statistics can be used to create confidence intervals for quantiles like the median. Large sample sizes cause order statistics like the sample median to approximate a normal distribution. Read more Download 0 ratings 0% found this document useful (0 votes) 151 views 6 pages Order Statistic: From Wikipedia, The Free Encyclopedia The document discusses order statistics, which are the values of a data set arranged in order of magnitude. It defines order statistics and provides examples. It then examines the probabilit… Full description Uploaded by hienluong293 AI-enhanced title and description Go to previous items Go to next items Download Save Save Order Statistic For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Report Download Save Order Statistic For Later You are on page 1/ 6 Search Fullscreen Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial Footer menu Back to top About About Scribd, Inc. Everand: Ebooks & Audiobooks Slideshare Join our team! Contact us Support Help / FAQ Accessibility Purchase help AdChoices Legal Terms Privacy Copyright Do not sell or share my personal information Social Instagram Instagram Facebook Facebook Pinterest Pinterest Get our free apps About About Scribd, Inc. Everand: Ebooks & Audiobooks Slideshare Join our team! Contact us Legal Terms Privacy Copyright Do not sell or share my personal information Support Help / FAQ Accessibility Purchase help AdChoices Social Instagram Instagram Facebook Facebook Pinterest Pinterest Get our free apps Documents Language: English Copyright © 2025 Scribd Inc. We take content rights seriously. Learn more in our FAQs or report infringement here. We take content rights seriously. Learn more in our FAQs or report infringement here. Language: English Copyright © 2025 Scribd Inc. 576648e32a3d8b82ca71961b7a986505
16508	https://doc.quark.cn/preview/xingyeziliao-qikanlunwen-qikanlunwen/4FF60DF1829D46F46D9517659F6EEC2E	正确认识中学物理中的正、负号搜索海量文档登录首页写作记录浏览历史云文档收藏会员客户端正确认识中学物理中的正、负号会员专享 PDF 105.4KB 1页 1/1 展开阅读剩余文档内容复制下一篇对中学物理中正负号的理解 2006年第2期河北理科教学研究短文集锦1m口一1mm口吉mn225EE一E。1m”一1m6由56式可以看出，在不同的惯性坐标系中动能增量是不同的，它们来自于位移和速度的相对性2外力功的相对性根据功的定义，在固定系的观测者观察到恒力F作的功为：WFreV1口maV丢mn7在运动系的观测者观察到恒力的功为：FF1at1mat8由78两式可以看出，在不同惯性系中，虽然牛顿定律的表达形式相同，但功的数值不同，它们是由于位移的不同引起的由WAE知，动能定理在s系成立但因为WSE，所以有动能定理在s系也是成立的即有动能定理在不同的惯性系中保持形式不变性由上面讨论可以得出：位移速度动能动能增量和外力作的功都具有相对性在不同的惯性系中有不同的量值，但是并不影响相对性原理在动能定理上的成立，动能定理在惯性系的变换中保持形式不变 2页 · 102KB · 1000次浏览教育人文相关资料 7.3万次浏览 1.5万人收藏下载全部 13篇选择正确认识中学物理中的正、负号 1页 105.4KB 对中学物理中正负号的理解 2页 102KB 浅析中学物理中的正负号 2页 510KB 正确辩识物理量的正负号 5页 840.1KB 浅析中学物理中正负号的含义 1页 126.6KB 浅谈中学物理中的正负号 1页 1.6M 浅谈中学物理中正负号的意义 1页 87.8KB 中学物理学中的正负号 2页 86KB 刍议中学物理中正负号的意义 2页 67.2KB 中学物理中的正负号 3页 107.4KB 物理量正、负号的正确使用 2页 87.3KB 试论物理学中正负号的含义及其处理 2页 82.5KB 辨析物理量中正负号的含义 2页 102.7KB 转为Word 收藏分享下载文档免费获取文档安装夸克
16509	https://chemistry.stackexchange.com/questions/115966/what-is-the-ph-at-both-equivalence-points-of-titration-between-diprotic-tartaric	What is the pH at both equivalence points of titration between diprotic tartaric acid and NaOH? - Chemistry Stack Exchange Join Chemistry By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Chemistry helpchat Chemistry Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more What is the pH at both equivalence points of titration between diprotic tartaric acid and NaOH? Ask Question Asked 6 years, 4 months ago Modified6 years, 3 months ago Viewed 993 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Here's all the info I have and my attempt in solving the question: The chemical equation for the titration: C X 4 H X 4 O X 6+2 N a O H⟶N a X 2 C X 4 H X 4 O X 6+2 H X 2 O C X 4 H X 4 O X 6+2 N a O H⟶N a X 2 C X 4 H X 4 O X 6+2 H X 2 O Here are the 1st and 2nd K a K a value for tartaric acid: K a 1=9.2×10−4,K a 2=4.3×10−5 K a 1=9.2×10−4,K a 2=4.3×10−5 Molarity of tartaric acid = 5×10−4 5×10−4 Standard solution of N a O H N a O H was made using 0.98 g 0.98 g of solid N a O H N a O H dissolved in 250 m l 250 m l of water. The molar mass of N a O H N a O H is 39.997 g/m o l 39.997 g/m o l. This is what I have done so far. n (C4H4O6) = cv = (5 × 10-4 M) (25ml ÷ 1000L) = 0.02 mol n (NaOH) = m ÷ M = (0.98g) ÷ (39.997 g/mol) = 0.0245 mol (4dp). C (NaOH) = n ÷ v = (0.0245 mol) ÷ (250ml ÷ 1000L) = 0.098M N (NaOH) at equivalence point = cv = (0.098M) (0.204L) = 0.02 mol C4H4O6(aq)+ 2NaOH(aq) → Na2C4H4O6(aq) + 2H2O(l) I 0.02 0.02 0 - C - 0.02 - 0.02 + 0.02 - E 0 0 0.02 - C (Na2C4H4O6) = n ÷ v = 0.02 ÷ (0.025L + 0.25L) = 0.0727M (4dp) Na2C4H4O6(aq) + H2O(l) → C4H4O4Na2^+(aq) + OH^-(aq) below is ice table, stackexchange won't allow me to seperate the values. I 0.0727 - 0 0 C - x - + x + x E 0.0727 – x - x x Kb = Kw ÷ Ka = (1 × 10-14) ÷ (9.2 × 10-4) = 1.087 × 10-11 Kb = [C4H4O4Na2+] [OH-] ÷ [Na2C4H4O6] 1.087 × 10-11 = x2 ÷ 0.0727 x = √ ((1.087 × 10-11) (0.0727)) x = 8.89 × 10-7 pOH = - log [OH-] = - log (8.89 × 10-7) = 1.9 pH = 14 – pOH = 14 – 1.9 = 12.1 acid-base Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jun 10, 2019 at 22:50 Karsten♦ 44.1k 8 8 gold badges 81 81 silver badges 204 204 bronze badges asked May 26, 2019 at 16:44 LisaLisa 11 1 1 silver badge 4 4 bronze badges 2 The question is based on the wrong terminology. Rather than the endpoint, the question would seemingly be asking about the equivalence point.MaxW –MaxW 2019-05-26 16:56:03 +00:00 Commented May 26, 2019 at 16:56 Thank you, you are right. I've just changed it. Would you by any chance know how to approach this question, as I still don't quite understand it Lisa –Lisa 2019-05-26 16:59:10 +00:00 Commented May 26, 2019 at 16:59 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. For the 1st equivalence point, the tartaric acid H X 2 A H X 2 A is titrated to N a H A N a H A For the solutions of acid salts, we can with advantage use the formula p H=p K a 1+p K a 2 2 p H=p K a 1+p K a 2 2 that can be derived from equilibrium 2 H A X−↽−−⇀H X 2 A+A X 2−2 H A X−↽−−⇀H X 2 A+A X 2− For the 2nd equivalence point there is N a X 2 A N a X 2 A solution, which is hydrolyzed with equilibrium A X 2−+H X 2 O↽−−⇀H A X−+O H X−A X 2−+H X 2 O↽−−⇀H A X−+O H X− With neglecting OH- from water hydrolysis, [H A X−]=[O H X−][H A X−]=[O H X−] K a 2 p K a 2 p H=[H X+][A X 2−][H A X−]=p H−log[A X 2−]+log[H A X−]=p H−log[A X 2−]+log[O H X−]=p H−log[A X 2−]−p O H=p H−log[A X 2−]−14+p H=7+1 2(p K a 2+log c N a X 2 A)K a 2=[H X+][A X 2−][H A X−]p K a 2=p H−log⁡[A X 2−]+log⁡[H A X−]=p H−log⁡[A X 2−]+log⁡[O H X−]=p H−log⁡[A X 2−]−p O H=p H−log⁡[A X 2−]−14+p H p H=7+1 2(p K a 2+log⁡c N a X 2 A) The molarity of N a O H N a O H solution c N a O H=m M×V=0.098 01 m o l/L c N a O H=m M×V=0.098 01 m o l/L The volume of N a O H N a O H solution for given volume of tartaric acid solution V N a O H=V H X 2 A×2×c H X 2 A c N a O H=0.0102×V H X 2 A V N a O H=V H X 2 A×2×c H X 2 A c N a O H=0.0102×V H X 2 A The final concentration [N a X 2 A]=[H X 2 A X i n i t]⋅V H X 2 A V H X 2 A+V N a O H=[H X 2 A X i n i t]⋅V H X 2 A V H X 2 A+0.0102⋅V H X 2 A=[H X 2 A X i n i t]1+0.0102=5×10−4/1.0102=4.9495×10−4 m o l/L=r o u n d i n g 4.95×10−4 m o l/L[N a X 2 A]=[H X 2 A X i n i t]⋅V H X 2 A V H X 2 A+V N a O H=[H X 2 A X i n i t]⋅V H X 2 A V H X 2 A+0.0102⋅V H X 2 A=[H X 2 A X i n i t]1+0.0102=5×10−4/1.0102=4.9495×10−4 m o l/L=r o u n d i n g 4.95×10−4 m o l/L Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited May 27, 2019 at 14:59 answered May 26, 2019 at 18:31 PoutnikPoutnik 48k 4 4 gold badges 59 59 silver badges 118 118 bronze badges 7 Thanks so much. I understand now. Just two questions though. For the volume, wouldn't it be 0.0102 VH2A. Also, why is it 1.0103 in the formula with final concentration?Lisa –Lisa 2019-05-27 11:38:48 +00:00 Commented May 27, 2019 at 11:38 It was, perhaps unluckily autowrapped. The acid solution is diluted by NaOH addition. Plus, perhaps rounding issue. Anyway 5 valid digits is excessive precision.Poutnik –Poutnik 2019-05-27 12:36:14 +00:00 Commented May 27, 2019 at 12:36 Hello, sorry just a little confused. So volume of NaOH is meant to be 0.0102 instead of 0.0103?Lisa –Lisa 2019-05-27 12:46:55 +00:00 Commented May 27, 2019 at 12:46 and why is 1 added to 0.0103? Lastly, where is the last equation for final concentration derived from?Lisa –Lisa 2019-05-27 12:47:52 +00:00 Commented May 27, 2019 at 12:47 @Lisa As I have said, rounding error. See the fixed and updated answer.Poutnik –Poutnik 2019-05-27 14:15:50 +00:00 Commented May 27, 2019 at 14:15 \|Show 2 more comments Your Answer Reminder: Answers generated by AI tools are not allowed due to Chemistry Stack Exchange's artificial intelligence policy Thanks for contributing an answer to Chemistry Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. 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16510	https://forum.wordreference.com/threads/salvage-vs-save.4081705/	Salvage vs save \| WordReference Forums WordReference.comLanguage Forums ForumsRules/Help/FAQHelp/FAQ MembersCurrent visitors Interface Language Dictionary search: Log inRegister What's newSearch Search [x] Search titles and first posts only [x] Search titles only By: Search Advanced search… Rules/Help/FAQHelp/FAQ MembersCurrent visitors Interface Language Menu Log in Register Install the app Install How to install the app on iOS Follow along with the video below to see how to install our site as a web app on your home screen. Note: This feature may not be available in some browsers. English Only English Only Salvage vs save Thread starterzoraastr3133 Start dateMar 4, 2024 zoraastr3133 New Member USA Tajik, Russian, German Mar 4, 2024 #1 Hi there! I was wondering if the two words can be used interchangeably without a clash in a sentence? I am more inclined to say save, knowing that salvage is used for unfortunate cases. But I came across this sentence in Longman dictionary "She tried to salvage the relationship." So, why not to say "She tried to save the relationship"? So, what is the difference? The Newt Senior Member New England English - US Mar 4, 2024 #2 "Salvage" is saving part of something when other parts are given up for lost. So without explanatory context it would seem like the wrong word. Reactions:zoraastr3133 and london calling kentix Senior Member English - U.S. Mar 4, 2024 #3 To me, the difference is that salvage means very significant damage has already been done that needs to be reversed or all hope is gone. Ships are a good analogy. You can (try to) save a ship or you can salvage a ship, but those are two different things. For instance, to save a ship might mean to keep it from sinking. You do the things necessary to prevent that. In the case of salvage, the ship has already sunk and the point of salvage is to pull it out of the water and repair so it can be used again. If you can't salvage it, it will stay where it is and rust away. In a relationship, to me, salvage indicates something so serious has happened that the relationship will die or is effectively dead unless drastic action is taken. It might be possible to save a relationship with less dramatic changes. It might still survive anyway even if it's not great. Last edited: Mar 4, 2024 Reactions:zoraastr3133, sdgraham, Aguas Claras and 1 other person dojibear Senior Member Fresno CA English (US - northeast) Mar 5, 2024 #4 The Newt said: "Salvage" is saving part of something when other parts are given up for lost. Click to expand... Different people might use it differently. To me, "salvage" is about saving part of something, even though you can't save all of it. For example: _It's too late to save the marriage, but maybe they can salvage a friendship. The car was destroyed in the accident. All they could salvage was two good tires._ Reactions:The Newt lingobingo Senior Member London English - England Mar 5, 2024 #5 It would be natural to talk of salvaging what was left of a relationship, or of anything else. Reactions:sdgraham and The Newt U USMeg Senior Member Virginia, USA English/USA Mar 5, 2024 #6 lingobingo said: It would be natural to talk of salvaging what was left of a relationship, or of anything else. Click to expand... Maybe in all these instances we could say that salvage is keeping the bits of what is left that are still worth something. kentix Senior Member English - U.S. Mar 5, 2024 #7 I think that works well enough with cars and ships but it doesn't seem realistic with a serious relationship. That doesn't come in bits and pieces that are separable and individually saveable. And with emotions involved, things don't really work that way. U USMeg Senior Member Virginia, USA English/USA Mar 5, 2024 #8 She tried to find something worth saving... Reactions:zoraastr3133 lingobingo Senior Member London English - England Mar 5, 2024 #9 USMeg said: Maybe in all these instances we could say that salvage is keeping the bits of what is left that are still worth something. Click to expand... My point was simply that to “salvage what’s left of” something (in either a practical or a figurative sense) is a common collocation. Reactions:zoraastr3133 and velisarius U USMeg Senior Member Virginia, USA English/USA Mar 5, 2024 #10 lingobingo said: My point was simply that to “salvage what’s left of” something (in either a practical or a figurative sense) is a common collocation. Click to expand... Common perhaps, but I think it's redundant. zoraastr3133 New Member USA Tajik, Russian, German Mar 6, 2024 #11 kentix said: To me, the difference is that salvage means very significant damage has already been done that needs to be reversed or all hope is gone. Ships are a good analogy. You can (try to) save a ship or you can salvage a ship, but those are two different things. For instance, to save a ship might mean to keep it from sinking. You do the things necessary to prevent that. In the case of salvage, the ship has already sunk and the point of salvage is to pull it out of the water and repair so it can be used again. If you can't salvage it, it will stay where it is and rust away. In a relationship, to me, salvage indicates something so serious has happened that the relationship will die or is effectively dead unless drastic action is taken. It might be possible to save a relationship with less dramatic changes. It might still survive anyway even if it's not great. Click to expand... Thank you so much for your detailed explanation! So, it means, if the write precursor context was given along the way to describe the falling apart marriage and then add the word, it wouldn't sound odd? D dedicatedenglish Senior Member Vietnamese Mar 6, 2024 #12 zoraastr3133 said: Thank you so much for your detailed explanation! So, it means, if the write precursor context was given along the way to describe the falling apart marriage and then add the word, it wouldn't sound odd? Click to expand... I think it would sound dramatic but not odd. Reactions:zoraastr3133 You must log in or register to reply here. Share: BlueskyLinkedInWhatsAppEmailShareLink English Only English Only WR styleSystemLightDark English (EN-us) Contact us Terms and rules Privacy policy Help RSS Community platform by XenForo®© 2010-2025 XenForo Ltd. Back TopBottom
16511	https://www.teacherspayteachers.com/browse/free?search=decimal%20and%20fraction%20conversion%20games	Decimal and Fraction Conversion Games \| TPT Log InSign Up Cart is empty Total: $0.00 View Wish ListView Cart Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Adult education Resource type Student practice Independent work packet Worksheets Assessment Graphic organizers Task cards Flash cards Teacher tools Classroom management Teacher manuals Outlines Rubrics Syllabi Unit plans Lessons Activities Games Centers Projects Laboratory Songs Clip art Classroom decor Bulletin board ideas Posters Word walls Printables Seasonal Holiday Black History Month Christmas-Chanukah-Kwanzaa Earth Day Easter Halloween Hispanic Heritage Month Martin Luther King Day Presidents' Day St. Patrick's Day Thanksgiving New Year Valentine's Day Women's History Month Seasonal Autumn Winter Spring Summer Back to school End of year ELA ELA by grade PreK ELA Kindergarten ELA 1st grade ELA 2nd grade ELA 3rd grade ELA 4th grade ELA 5th grade ELA 6th grade ELA 7th grade ELA 8th grade ELA High school ELA Elementary ELA Reading Writing Phonics Vocabulary Grammar Spelling Poetry ELA test prep Middle school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep High school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep Math Math by grade PreK math Kindergarten math 1st grade math 2nd grade math 3rd grade math 4th grade math 5th grade math 6th grade math 7th grade math 8th grade math High school math Elementary math Basic operations Numbers Geometry Measurement Mental math Place value Arithmetic Fractions Decimals Math test prep Middle school math Algebra Basic operations Decimals Fractions Geometry Math test prep High school math Algebra Algebra 2 Geometry Math test prep Statistics Precalculus Calculus Science Science by grade PreK science Kindergarten science 1st grade science 2nd grade science 3rd grade science 4th grade science 5th grade science 6th grade science 7th grade science 8th grade science High school science By topic Astronomy Biology Chemistry Earth sciences Physics Physical science Social studies Social studies by grade PreK social studies Kindergarten social studies 1st grade social studies 2nd grade social studies 3rd grade social studies 4th grade social studies 5th grade social studies 6th grade social studies 7th grade social studies 8th grade social studies High school social studies Social studies by topic Ancient history Economics European history Government Geography Native Americans Middle ages Psychology U.S. History World history Languages Languages American sign language Arabic Chinese French German Italian Japanese Latin Portuguese Spanish Arts Arts Art history Graphic arts Visual arts Other (arts) Performing arts Dance Drama Instrumental music Music Music composition Vocal music Special education Speech therapy Social emotional Social emotional Character education Classroom community School counseling School psychology Social emotional learning Specialty Specialty Career and technical education Child care Coaching Cooking Health Life skills Occupational therapy Physical education Physical therapy Professional development Service learning Vocational education Other (specialty) Decimal and Fraction Conversion Games 24+results Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Search Grade Subject Free Format All filters (1) Filters Free Clear all Grade Elementary 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade Higher education Not grade specific More Subject Math Algebra Arithmetic Basic operations Decimals Fractions Math test prep Mental math Numbers More Price Free Under $5 $5-10 $10 and up Format Microsoft Microsoft Word PDF Video More Resource type Printables Hands-on activities Activities Centers Projects Games Student assessment Test preparation Student practice Worksheets Graphic organizers Homework Task cards More Standard Theme Holiday Halloween Thanksgiving More Audience Homeschool Parents Staff & administrators More Language English (UK) More Converting between Decimals, Fractions and Percentages - Task Card REVIEW GAME Created by Teach With Fergy This FREE 16 conversion Task Card Resource is unique. It can be utilized in a clue gathering, puzzle solving, highly active and engaging activity or as an alternative to worksheet review. Thank you very much for visiting my store. If you'd like to receive 50% off your next TeachWithFergy purchase, (includes your entire shopping cart) please CLICK HERE. A new page will open but don't worry, you won't lose this one. Do you like these Task Card Activities and want to save some money? I've bund 3 rd - 7 th Decimals, Fractions, Math FREE Rated 4.83 out of 5, based on 48 reviews 4.8(48) Log in to Download Wish List Fall Themed Math Games (9 games) Created by Teacher Playground Fall Themed Math Games (over 180 pages) BUNDLES of FUN!! You and your students will love these Math games that can be used from 2nd-6th grade with different levels. There are 9 different topics with over 170 pages of pure educational fun. There are even template pages if you want to add more or create more games. Get the BUNDLE...Save Money! 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How I use them:My students LOVE playing Bingo, especially when they're able to work at their own pace (rather than waiting for the math problems to be given to 7 th - 11 th Algebra, Basic Operations, Math CCSS 8.G.B.6 , 8.G.B.7 , 8.G.B.8 +4 FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List Fraction, Decimal, & Percent Conversion Bingo Created by MsMaryMath This activity will help students have fun while practicing converting fraction, decimals, and percents. This includes fractions (not mixed numbers) and repeating decimals. 5 th - 7 th Decimals, Fractions, Math FREE Rated 4.5 out of 5, based on 4 reviews 4.5(4) Log in to Download Wish List GO FISH! Fraction, Decimal, and Percent Conversions Created by Jill Hawkins 42 cards total. All have fish graphic on them. Looks amazing printed on cardstock and laminated for repeated use. 5 th - 8 th Math FREE Rated 4.67 out of 5, based on 15 reviews 4.7(15) Log in to Download Wish List Convert between Fractions, Decimals and Percent Created by Retro Solutions Bring the fun of the classic card games Rummy, Blackjack, and War into your math classroom. This set of cards has numerous cards with fractions, decimals and percents. In “Rummy” the students try to get a set of three cards that are equivalent (a fraction, a decimal and a percent). In Blackjack the students try to get a sum as close to 2 as possible without exceeding 2. The students to compare values in ‘War”. The cards can be sorted so students can practice one type of number at a time or can 5 th - 8 th Decimals, Fractions, Math CCSS 5.NBT.A.3b , 4.NF.A.2 , 6.RP.A.3c +2 FREE Log in to Download Wish List Aliens on Mars Impostor Game - review fractions, decimals, mixed numbers Created by Mind Bender This material reviews students on conversion of fractions, decimals, and mixed numbers. Materials include PPT, worksheets, and teacher materials. The game can be used for any topic. You will just need 7 sets of 5 questions. No time to prep for a different set of questions? Send me your questions, and I will prep the rest of the materials to customize the game just for $5. Student Mission: Solve the problems to collect clues about your neighbors on Mars. When you’ve foun 3 rd - 6 th Decimals, Fractions, Math CCSS 4.NF.C.5 , 4.NF.C.6 , 4.NF.C.7 FREE Log in to Download Wish List Fractions Percentages Decimals Conversion Game Created by Nigel Cooper I have .... Who has....? This game is designed to reinforce the basic conversions between fractions, decimals and percentages. All students receive one card. There are 28 in the bundle. The students are asked to stand up and the person with START on their card will start. They read out what is below. Then they sit down. The next person to speak will be the student with the answer. You can play it a few times and time the students to give them a sense of competition. This game builds teamwork an 6 th - 7 th Decimals, Fractions, Math FREE Rated 4 out of 5, based on 1 reviews 4.0(1) Log in to Download Wish List Thanksgiving Themed Fraction/Decimal Conversion Game Created by Teacher Playground You and your students will love this Fraction/Decimal Conversion Math game with varying levels. Start at the beginning and see how far you can go! There are even template pages if you want to add more or create more games. 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Adding Fractions - pairs of students use cards to create fractions, then add them together and simplify answersDecimal Doozy - pairs of students create decimals of different p 5 th - 8 th Decimals, Fractions, Math CCSS 5.NBT.A.3 , 5.NBT.A.3b , 4.NF.A.2 +3 FREE Rated 5 out of 5, based on 9 reviews 5.0(9) Log in to Download Wish List Fraction, Decimal, Percent Conversion Activity Created by Smiles and Sproutlings Language Learning Looking for a fun way to engage students as they learn to convert fractions, decimals and percents? Engage and Educate with a Classic Twist!Transform math practice into an interactive activity with Math Tic Tac Toe: Fractions, Decimals, and Percents Edition. Perfect for Grade 7 and Grade 8 students, this no-prep game blends classic gameplay with essential math skills, making learning conversions fun and less daunting. 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Fraction Percentage Decimal Dominoes by Dana Weld is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 United States License. 4 th - 8 th Fractions FREE Rated 4.71 out of 5, based on 32 reviews 4.7(32) Log in to Download Wish List 1:55 Converting Decimals To Percents Song: Common Core Math Video ★ Math Conversions Created by Math Lessons by NUMBEROCK You are going to love this creative Decimals to Percents Song that teaches kids how to convert decimal numbers to percentages in the most unique of ways in this FREEBIE! CLICK HERE to play NUMBEROCK’s printable game, create a colorful anchor chart, and get illustrated worksheets, word problems, and more for inquiring students who will be unusually motivated after watching our engaging video freebie. Resources not included in the free version of this video: ✦ Animated Music Video in HD 5 th - 7 th Decimals, Fractions, Math CCSS 6.RP.A.3c FREE Rated 5 out of 5, based on 6 reviews 5.0(6) Watch Full Video Wish List Percent, Decimal, Fraction Scavenger Hunt Created by Mrs VDL Students will practice their basic conversion skills with this scavenger hunt you can set up around your classroom. There are 28 questions for students to work through and where they start is where they end so they can all start at different locations. There are questions that are similar to each other so students can work on understanding common misconceptions. There are no repeating decimals or negative numbers. 5 th - 7 th Decimals, Fractions, Math FREE Log in to Download Wish List Fraction, Decimal, Percent Conversion Puzzle Created by Victoria Winland Fraction, Decimal, and Percent conversion puzzle practice. Each puzzle has a matching fraction, decimal, and percent. There are 39 pieces to match. Perfect for group work and large classes! Use cardstock and laminate for long-lasting puzzle pieces. Includes teacher instructions 5 th - 9 th, Higher Education Decimals, Fractions, Math CCSS 6.RP.A.3a , 6.RP.A.3c , 7.RP.A.2 +1 FREE Log in to Download Wish List I Have, Who Has Game Created by Common Core Math Fun Conversions. I Have, Who Has is a game to support our ESL learners. You divide the cards into the whatever size group you would like. Someone has a start card. They read aloud their card. The next period will go that has their match, and then reads their question card. Example: I have the start card. Who has -6 + 2? I have -4. Who has -10 - 7? I have -17. Who has ...etc 7 th Basic Operations, Decimals, Fractions FREE Log in to Download Wish List Paper Football Statistics Created by I Got You Math Play a game and practice your fraction, decimal, and percent conversions! In this activity students complete trials using the game of paper football. There are four scenarios: the slide, the touchdown, and the distance. Students complete trials in groups of 2 to 4. Students first write fractions to represent their data, then convert these fractions to decimals and percents. In the end, students use their data to articulate who will the best in a game of paper football. 5 th - 7 th Decimals, Fractions, Math CCSS 6.RP.A.3c FREE Rated 5 out of 5, based on 3 reviews 5.0(3) Log in to Download Wish List Using Baseball Cards to Convert Fractions, Decimals and Percents Created by mrsLsquared This project practices conversion between fractions, decimals and percents. It also uses a 10 X 10 grid to represent percentages. This grid is later used as a coordinate plane in a classroom game of baseball! 5 th - 7 th Decimals, Fractions, Math FREE Rated 4.7 out of 5, based on 10 reviews 4.7(10) Log in to Download Wish List Halloween Math and ELA Bundle Created by Free Resources Mostly Math Your students will love the fun games and activities included in this fun resource! Halloween Math and ELA Bundle includes: fraction, decimal, percent conversion game PBL activities on planning a Halloween party division by two digit divisors fraction activities multiplication and conversion in metric system parts of speech tree maps and fun elimination game a booklet on commonly misused homophones These activities will keep your students entertained throughout the wee 4 th - 6 th Decimals, Fractions CCSS 5.MD.A.1 , 5.NBT.B.5 , 5.NBT.B.6 +7 FREE Rated 5 out of 5, based on 4 reviews 5.0(4) Log in to Download Wish List Fraction Circles Created by Middle School Frenzy This file is a variety of circles split into different fractional pieces. I use this in my classroom as a group set of manipulatives. I use the fraction circles along with the decimal and percent circles to help teach equivalency and conversions. A good idea is to print cut out and laminate for a group set, I keep one set on my dry erase board I simply attached magnet buttons to the back of each laminated fractional piece. Enjoy! 4 th - 8 th Fractions, Math FREE Rated 4.71 out of 5, based on 15 reviews 4.7(15) Log in to Download Wish List Equivalent Fraction Dominoes Created by MrsPotatoHead Ready to use; just print off (preferably on card stock), laminate, and cut out individual dominoes. 144 dominoes included These can be used for many different activities. Simplified fractions include: 1/20, 1/8, 1/5, 1/4, 1/3, 3/8, 2/5, 1/2, 3/5, 5/8, 2/3, 3/4, 4/5. A simple version of this game would be to have students draw 10 dominoes and start with one of the "doubles". They will then take turns trying to create equivalent fractions from the starting domino. Another alternat 6 th - 9 th Fractions, Math, Numbers FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List FREEBIE "Find Someone Who..." 5th Grade End of Year Math Review Activity Created by Engaged in Mathematics ➪ This product aligns to the 5th grade CCSS!Looking for a fun and interactive way for your students to prepare for state or end of the year math tests? Download this FREE and highly engaging "Find Someone Who..." math review activity! Note: This product was updated as of July 2020 to include a DIGITAL \| DISTANCE LEARNING version! What's Included: ➪ 1 paper version of "Find Someone Who..." ➪ A link to the digital \| distance learning version of "Find Someone Who..." ➪ Answer key Topics Incl 5 th Math, Math Test Prep FREE Rated 4.78 out of 5, based on 91 reviews 4.8(91) Log in to Download Wish List 1 Showing 1-24 of 24+results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. 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16512	https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations	Jump to content Search Contents 1 Lorentz transformation 2 Historical background 3 Physical principles 3.1 The second postulate 3.2 Invariance of interval 3.2.1 Rigorous Statement and Proof of Proportionality of ds2 and ds′2 4 Standard configuration 5 The solutions 6 Using the geometry of spacetime 6.1 Landau & Lifshitz solution 6.2 Hyperbolic rotation 7 From physical principles 7.1 Time dilation and length contraction 7.2 Spherical wavefronts of light 7.3 Galilean and Einstein's relativity 7.3.1 Galilean reference frames 7.3.2 Using the principle of relativity 7.3.3 Determining the constants of the first equation 7.3.4 Determining the constants of the second equation 7.4 Einstein's popular derivation 8 Using group theory 8.1 From group postulates 8.2 Boost from generators 9 From experiments 10 See also 11 Notes 12 References Derivations of the Lorentz transformations Oʻzbekcha / ўзбекча Русский Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia \| \| \| Part of a series on \| \| Spacetime \| \| Special relativity General relativity \| \| Spacetime concepts Spacetime manifold Equivalence principle Lorentz transformations Minkowski space \| \| General relativity Introduction to general relativity Mathematics of general relativity Einstein field equations \| \| Classical gravity Introduction to gravitation Newton's law of universal gravitation \| \| Relevant mathematics Four-vector Derivations of relativity Spacetime diagrams Differential geometry Curved space Curved spacetime Mathematics of general relativity Spacetime topology \| \| Physics portal Category \| \| v t e \| There are many ways to derive the Lorentz transformations using a variety of physical principles, ranging from Maxwell's equations to Einstein's postulates of special relativity, and mathematical tools, spanning from elementary algebra and hyperbolic functions, to linear algebra and group theory. This article provides a few of the easier ones to follow in the context of special relativity, for the simplest case of a Lorentz boost in standard configuration, i.e. two inertial frames moving relative to each other at constant (uniform) relative velocity less than the speed of light, and using Cartesian coordinates so that the x and x′ axes are collinear. Lorentz transformation [edit] Main article: Lorentz transformation In the fundamental branches of modern physics, namely general relativity and its widely applicable subset special relativity, as well as relativistic quantum mechanics and relativistic quantum field theory, the Lorentz transformation is the transformation rule under which all four-vectors and tensors containing physical quantities transform from one frame of reference to another. The prime examples of such four-vectors are the four-position and four-momentum of a particle, and for fields the electromagnetic tensor and stress–energy tensor. The fact that these objects transform according to the Lorentz transformation is what mathematically defines them as vectors and tensors; see tensor for a definition. Given the components of the four-vectors or tensors in some frame, the "transformation rule" allows one to determine the altered components of the same four-vectors or tensors in another frame, which could be boosted or accelerated, relative to the original frame. A "boost" should not be conflated with spatial translation, rather it's characterized by the relative velocity between frames. The transformation rule itself depends on the relative motion of the frames. In the simplest case of two inertial frames the relative velocity between enters the transformation rule. For rotating reference frames or general non-inertial reference frames, more parameters are needed, including the relative velocity (magnitude and direction), the rotation axis and angle turned through. Historical background [edit] Further information: History of Lorentz transformations The usual treatment (e.g., Albert Einstein's original work) is based on the invariance of the speed of light. However, this is not necessarily the starting point: indeed (as is described, for example, in the second volume of the Course of Theoretical Physics by Landau and Lifshitz), what is really at stake is the locality of interactions: one supposes that the influence that one particle, say, exerts on another can not be transmitted instantaneously. Hence, there exists a theoretical maximal speed of information transmission which must be invariant, and it turns out that this speed coincides with the speed of light in vacuum. Newton had himself called the idea of action at a distance philosophically "absurd", and held that gravity had to be transmitted by some agent according to certain laws. Michelson and Morley in 1887 designed an experiment, employing an interferometer and a half-silvered mirror, that was accurate enough to detect aether flow. The mirror system reflected the light back into the interferometer. If there were an aether drift, it would produce a phase shift and a change in the interference that would be detected. However, no phase shift was ever found. The negative outcome of the Michelson–Morley experiment left the concept of aether (or its drift) undermined. There was consequent perplexity as to why light evidently behaves like a wave, without any detectable medium through which wave activity might propagate. In a 1964 paper, Erik Christopher Zeeman showed that the causality-preserving property, a condition that is weaker in a mathematical sense than the invariance of the speed of light, is enough to assure that the coordinate transformations are the Lorentz transformations. Norman Goldstein's paper shows a similar result using inertiality (the preservation of time-like lines) rather than causality. Physical principles [edit] Einstein based his theory of special relativity on two fundamental postulates. First, all physical laws are the same for all inertial frames of reference, regardless of their relative state of motion; and second, the speed of light in free space is the same in all inertial frames of reference, again, regardless of the relative velocity of each reference frame. The Lorentz transformation is fundamentally a direct consequence of this second postulate. The second postulate [edit] Assume the second postulate of special relativity stating the constancy of the speed of light, independent of reference frame, and consider a collection of reference systems moving with respect to each other with constant velocity, i.e. inertial systems, each endowed with its own set of Cartesian coordinates labeling the points, i.e. events of spacetime. To express the invariance of the speed of light in mathematical form, fix two events in spacetime, to be recorded in each reference frame. Let the first event be the emission of a light signal, and the second event be it being absorbed. Pick any reference frame in the collection. In its coordinates, the first event will be assigned coordinates , and the second . The spatial distance between emission and absorption is , but this is also the distance traveled by the signal. One may therefore set up the equation Every other coordinate system will record, in its own coordinates, the same equation. This is the immediate mathematical consequence of the invariance of the speed of light. The quantity on the left is called the spacetime interval. The interval is, for events separated by light signals, the same (zero) in all reference frames, and is therefore called invariant. Invariance of interval [edit] For the Lorentz transformation to have the physical significance realized by nature, it is crucial that the interval is an invariant quantity for any two events, not just for those separated by light signals. To establish this, one considers an infinitesimal interval, as recorded in a system . Let be another system assigning the interval to the same two infinitesimally separated events. Since if , then the interval will also be zero in any other system (second postulate), and since and are infinitesimals of the same order, they must be proportional to each other, On what may depend? It may not depend on the positions of the two events in spacetime, because that would violate the postulated homogeneity of spacetime. It might depend on the relative velocity between and , but only on the speed, not on the direction, because the latter would violate the isotropy of space. Now bring in systems and , From these it follows, Now, one observes that on the right-hand side that depend on both and ; as well as on the angle between the vectors and . However, one also observes that the left-hand side does not depend on this angle. Thus, the only way for the equation to hold true is if the function is a constant. Further, by the same equation this constant is unity. Thus, for all systems . Since this holds for all infinitesimal intervals, it holds for all intervals. Most, if not all, derivations of the Lorentz transformations take this for granted.[clarification needed] In those derivations, they use the constancy of the speed of light (invariance of light-like separated events) only. This result ensures that the Lorentz transformation is the correct transformation.[clarification needed] Rigorous Statement and Proof of Proportionality of ds2 and ds′2 [edit] Theorem: Let be integers, and a vector space over of dimension . Let be an indefinite-inner product on with signature type . Suppose is a symmetric bilinear form on such that the null set of the associated quadratic form of is contained in that of (i.e. suppose that for every , if then ). Then, there exists a constant such that . Furthermore, if we assume and that also has signature type , then we have . Remarks. In the section above, the term "infinitesimal" in relation to is actually referring (pointwise) to a quadratic form over a four-dimensional real vector space (namely the tangent space at a point of the spacetime manifold). The argument above is copied almost verbatim from Landau and Lifshitz, where the proportionality of and is merely stated as an 'obvious' fact even though the statement is not formulated in a mathematically precise fashion nor proven. This is a non-obvious mathematical fact which needs to be justified; fortunately the proof is relatively simple and it amounts to basic algebraic observations and manipulations. The above assumptions on means the following: is a bilinear form which is symmetric and non-degenerate, such that there exists an ordered basis of for which An equivalent way of saying this is that has the matrix representation relative to the ordered basis . If we consider the special case where then we're dealing with the situation of Lorentzian signature in 4-dimensions, which is what relativity is based on (or one could adopt the opposite convention with an overall minus sign; but this clearly doesn't affect the truth of the theorem). Also, in this case, if we assume and both have quadratics forms with the same null-set (in physics terminology, we say that and give rise to the same light cone) then the theorem tells us that there is a constant such that . Modulo some differences in notation, this is precisely what was used in the section above. Proof of Theorem (index notation) For convenience, let us agree in this proof that Greek indices like range over while Latin indices like range over . Also, we shall use the Einstein summation convention throughout. Fix a basis of relative to which has the matrix representation . Also, for each and having unit Euclidean norm consider the vector . Then, by bilinearity we have , hence by our assumption, we have as well. Using bilinearity and symmetry of , this is equivalent to Since this is true for all of unit norm, we can replace with to get Now, we subtract these two equations and divide by 4 to obtain that for all of unit norm, So, by choosing and (i.e with 1 in the specified index and 0 elsewhere), we see that As a result of this, our first equation is simplified to This is once again true for all and of unit norm. As a result all the off-diagonal terms vanish; in more detail, suppose are distinct indices. Consider . Then, since the right side of the equation doesn't depend on , we see that and hence . By an almost identical argument we deduce that if are distinct indices then . Finally, by successively letting range over and then letting range over , we see that , or in other words, has the matrix representation , which is equivalent to saying . So, the constant of proportionality claimed in the theorem is . Finally, if we assume that both have signature types and then (we can't have because that would mean , which is impossible since having signature type means it is a non-zero bilinear form. Also, if , then it means has positive diagonal entries and negative diagonal entries; i.e it is of signature , since we assumed , so this is also not possible. This leaves us with as the only option). This completes the proof of the theorem. Proof of Theorem (conceptual) Fix a basis of relative to which has the matrix representation . The point is that the vector space can be decomposed into subspaces (the span of the first basis vectors) and (then span of the other basis vectors) such that each vector in can be written uniquely as for and ; moreover , and . So (by bilinearity) Since the first summand on the right in non-positive and the second in non-negative, for any and , we can find a scalar such that . From now on, always consider and . By bilinearity If , then also and the same is true for (since the null-set of is contained in that of ). In that case, subtracting the two expression above (and dividing by 4) yields As above, for each and , there is a scalar such that , so , which by bilinearity means . Now consider nonzero such that . We can find such that . By the expressions above, Analogically, for , one can show that if , then also . So it holds for all vectors in .[clarification needed] For , if , for some , we can (scaling one of the if necessary) assume , which by the above means that . So . Finally, if we assume that both have signature types and then (we can't have because that would mean , which is impossible since having signature type means it is a non-zero bilinear form. Also, if , then it means has positive diagonal entries and negative diagonal entries; i.e. it is of signature , since we assumed , so this is also not possible. This leaves us with as the only option). This completes the proof of the theorem. Proof of Theorem (conceptual and broken down) By Sylvester's law of inertia, we can fix a basis of relative to which has the matrix representation . The point is that the vector space can be decomposed into subspaces (the span of the first basis vectors) and (then span of the other basis vectors) such that each vector in can be written uniquely as for and ; moreover , and . We will write for from now on. Lemma: There exists a constant such that for any and , (a) (b) , where Proof of Lemma Let a = .By bilinearity: Since the null set of is contained in that of : So By 6, , proving (a),By 7 and 2, and so . Keeping fixed and varying , we see that this ratio does not depend and . Similarly, it does not depend on . Call this ratio . Now for , For all , we have So Standard configuration [edit] The invariant interval can be seen as a non-positive definite distance function on spacetime. The set of transformations sought must leave this distance invariant. Due to the reference frame's coordinate system's cartesian nature, one concludes that, as in the Euclidean case, the possible transformations are made up of translations and rotations, where a slightly broader meaning should be allowed for the term rotation. The interval is quite trivially invariant under translation. For rotations, there are four coordinates. Hence there are six planes of rotation. Three of those are rotations in spatial planes. The interval is invariant under ordinary rotations too. It remains to find a "rotation" in the three remaining coordinate planes that leaves the interval invariant. Equivalently, to find a way to assign coordinates so that they coincide with the coordinates corresponding to a moving frame. The general problem is to find a transformation such that To solve the general problem, one may use the knowledge about invariance of the interval of translations and ordinary rotations to assume, without loss of generality, that the frames F and F′ are aligned in such a way that their coordinate axes all meet at t = t′ = 0 and that the x and x′ axes are permanently aligned and system F′ has speed V along the positive x-axis. Call this the standard configuration. It reduces the general problem to finding a transformation such that The standard configuration is used in most examples below. A linear solution of the simpler problem solves the more general problem since coordinate differences then transform the same way. Linearity is often assumed or argued somehow in the literature when this simpler problem is considered. If the solution to the simpler problem is not linear, then it doesn't solve the original problem because of the cross terms appearing when expanding the squares. The solutions [edit] As mentioned, the general problem is solved by translations in spacetime. These do not appear as a solution to the simpler problem posed, while the boosts do (and sometimes rotations depending on angle of attack). Even more solutions exist if one only insist on invariance of the interval for lightlike separated events. These are nonlinear conformal ("angle preserving") transformations. One has Lorentz transformations ⊂ Poincaré transformations ⊂ conformal group transformations. Some equations of physics are conformal invariant, e.g. the Maxwell's equations in source-free space, but not all. The relevance of the conformal transformations in spacetime is not known at present, but the conformal group in two dimensions is highly relevant in conformal field theory and statistical mechanics. It is thus the Poincaré group that is singled out by the postulates of special relativity. It is the presence of Lorentz boosts (for which velocity addition is different from mere vector addition that would allow for speeds greater than the speed of light) as opposed to ordinary boosts that separates it from the Galilean group of Galilean relativity. Spatial rotations, spatial and temporal inversions and translations are present in both groups and have the same consequences in both theories (conservation laws of momentum, energy, and angular momentum). Not all accepted theories respect symmetry under the inversions. Using the geometry of spacetime [edit] Landau & Lifshitz solution [edit] These three hyperbolic function formulae (H1–H3) are referenced below: The problem posed in standard configuration for a boost in the x-direction, where the primed coordinates refer to the moving system is solved by finding a linear solution to the simpler problem The most general solution is, as can be verified by direct substitution using (H1), \| \| \| --- \| \| \| 1 \| To find the role of Ψ in the physical setting, record the progression of the origin of F′, i.e. x′ = 0, x = vt. The equations become (using first x′ = 0), Now divide: where x = vt was used in the first step, (H2) and (H3) in the second, which, when plugged back in (1), gives or, with the usual abbreviations, This calculation is repeated with more detail in section hyperbolic rotation. Hyperbolic rotation [edit] Main article: Hyperbolic rotation The Lorentz transformations can also be derived by simple application of the special relativity postulates and using hyperbolic identities. Relativity postulates Start from the equations of the spherical wave front of a light pulse, centred at the origin: which take the same form in both frames because of the special relativity postulates. Next, consider relative motion along the x-axes of each frame, in standard configuration above, so that y = y′, z = z′, which simplifies to Linearity Now assume that the transformations take the linear form: where A, B, C, D are to be found. If they were non-linear, they would not take the same form for all observers, since fictitious forces (hence accelerations) would occur in one frame even if the velocity was constant in another, which is inconsistent with inertial frame transformations. Substituting into the previous result: and comparing coefficients of x2, t2, xt: Hyperbolic rotation The equations suggest the hyperbolic identity Introducing the rapidity parameter ϕ as a hyperbolic angle allows the consistent identifications where the signs after the square roots are chosen so that x' and t' increase if x and t increase, respectively. The hyperbolic transformations have been solved for: If the signs were chosen differently the position and time coordinates would need to be replaced by −x and/or −t so that x and t increase not decrease. To find how ϕ relates to the relative velocity, from the standard configuration the origin of the primed frame x′ = 0 is measured in the unprimed frame to be x = vt (or the equivalent and opposite way round; the origin of the unprimed frame is x = 0 and in the primed frame it is at x′ = −vt): and hyperbolic identities leads to the relations between β, γ, and ϕ, From physical principles [edit] The problem is usually restricted to two dimensions by using a velocity along the x axis such that the y and z coordinates do not intervene, as described in standard configuration above. Time dilation and length contraction [edit] The transformation equations can be derived from time dilation and length contraction, which in turn can be derived from first principles. With O and O′ representing the spatial origins of the frames F and F′, and some event M, the relation between the position vectors (which here reduce to oriented segments OM, OO′ and O′M) in both frames is given by: OM = OO′ + O′M. Using coordinates (x,t) in F and (x′,t′) in F′ for event M, in frame F the segments are OM = x, OO′ = vt and O′M = x′/γ (since x′ is O′M as measured in F′): Likewise, in frame F′, the segments are OM = x/γ (since x is OM as measured in F), OO′ = vt′ and O′M = x′: By rearranging the first equation, we get which is the space part of the Lorentz transformation. The second relation gives which is the inverse of the space part. Eliminating x′ between the two space part equations gives that, if , simplifies to: which is the time part of the transformation, the inverse of which is found by a similar elimination of x: Spherical wavefronts of light [edit] The following is similar to that of Einstein. As in the Galilean transformation, the Lorentz transformation is linear since the relative velocity of the reference frames is constant as a vector; otherwise, inertial forces would appear. They are called inertial or Galilean reference frames. According to relativity no Galilean reference frame is privileged. Another condition is that the speed of light must be independent of the reference frame, in practice of the velocity of the light source. Consider two inertial frames of reference O and O′, assuming O to be at rest while O′ is moving with a velocity v with respect to O in the positive x-direction. The origins of O and O′ initially coincide with each other. A light signal is emitted from the common origin and travels as a spherical wave front. Consider a point P on a spherical wavefront at a distance r and r′ from the origins of O and O′ respectively. According to the second postulate of the special theory of relativity the speed of light is the same in both frames, so for the point P: The equation of a sphere in frame O is given by For the spherical wavefront that becomes Similarly, the equation of a sphere in frame O′ is given by so the spherical wavefront satisfies The origin O′ is moving along x-axis. Therefore, x′ must vary linearly with x and t. Therefore, the transformation has the form For the origin of O′ x′ and x are given by so, for all t, and thus This simplifies the transformation to where γ is to be determined. At this point γ is not necessarily a constant, but is required to reduce to 1 for v ≪ c. The inverse transformation is the same except that the sign of v is reversed: The above two equations give the relation between t and t′ as: or Replacing x′, y′, z′ and t′ in the spherical wavefront equation in the O′ frame, with their expressions in terms of x, y, z and t produces: and therefore, which implies, or Comparing the coefficient of t2 in the above equation with the coefficient of t2 in the spherical wavefront equation for frame O produces: Equivalent expressions for γ can be obtained by matching the x2 coefficients or setting the tx coefficient to zero. Rearranging: or, choosing the positive root to ensure that the x and x' axes and the time axes point in the same direction, which is called the Lorentz factor. This produces the Lorentz transformation from the above expression. It is given by The Lorentz transformation is not the only transformation leaving invariant the shape of spherical waves, as there is a wider set of spherical wave transformations in the context of conformal geometry, leaving invariant the expression . However, scale changing conformal transformations cannot be used to symmetrically describe all laws of nature including mechanics, whereas the Lorentz transformations (the only one implying ) represent a symmetry of all laws of nature and reduce to Galilean transformations at . Galilean and Einstein's relativity [edit] Galilean reference frames [edit] In classical kinematics, the total displacement x in the R frame is the sum of the relative displacement x′ in frame R′ and of the distance between the two origins x − x′. If v is the relative velocity of R′ relative to R, the transformation is: x = x′ + vt, or x′ = x − vt. This relationship is linear for a constant v, that is when R and R′ are Galilean frames of reference. In Einstein's relativity, the main difference from Galilean relativity is that space and time coordinates are intertwined, and in different inertial frames t ≠ t′. Since space is assumed to be homogeneous, the transformation must be linear. The most general linear relationship is obtained with four constant coefficients, A, B, γ, and b: The linear transformation becomes the Galilean transformation when γ = B = 1, b = −v and A = 0. An object at rest in the R′ frame at position x′ = 0 moves with constant velocity v in the R frame. Hence the transformation must yield x′ = 0 if x = vt. Therefore, b = −γv and the first equation is written as Using the principle of relativity [edit] According to the principle of relativity, there is no privileged Galilean frame of reference: therefore the inverse transformation for the position from frame R′ to frame R should have the same form as the original but with the velocity in the opposite direction, i.o.w. replacing v with -v: and thus Determining the constants of the first equation [edit] Since the speed of light is the same in all frames of reference, for the case of a light signal, the transformation must guarantee that t = x/c when t′ = x′/c. Substituting for t and t′ in the preceding equations gives: Multiplying these two equations together gives, At any time after t = t′ = 0, xx′ is not zero, so dividing both sides of the equation by xx′ results in which is called the "Lorentz factor". When the transformation equations are required to satisfy the light signal equations in the form x = ct and x′ = ct′, by substituting the x and x'-values, the same technique produces the same expression for the Lorentz factor. Determining the constants of the second equation [edit] The transformation equation for time can be easily obtained by considering the special case of a light signal, again satisfying x = ct and x′ = ct′, by substituting term by term into the earlier obtained equation for the spatial coordinate giving so that which, when identified with determines the transformation coefficients A and B as So A and B are the unique constant coefficients necessary to preserve the constancy of the speed of light in the primed system of coordinates. Einstein's popular derivation [edit] In his popular book Einstein derived the Lorentz transformation by arguing that there must be two non-zero coupling constants λ and μ such that that correspond to light traveling along the positive and negative x-axis, respectively. For light x = ct if and only if x′ = ct′. Adding and subtracting the two equations and defining gives Substituting x′ = 0 corresponding to x = vt and noting that the relative velocity is v = bc/γ, this gives The constant γ can be evaluated by demanding c2t2 − x2 = c2t′2 − x′2 as per standard configuration. Using group theory [edit] From group postulates [edit] Following is a classical derivation (see, e.g., and references therein) based on group postulates and isotropy of the space. Coordinate transformations as a group The coordinate transformations between inertial frames form a group (called the proper Lorentz group) with the group operation being the composition of transformations (performing one transformation after another). Indeed, the four group axioms are satisfied: Closure: the composition of two transformations is a transformation: consider a composition of transformations from the inertial frame K to inertial frame K′, (denoted as K → K′), and then from K′ to inertial frame K′′, [K′ → K′′], there exists a transformation, [K → K′] [K′ → K′′], directly from an inertial frame K to inertial frame K′′. Associativity: the transformations ( [K → K′] [K′ → K′′] ) [K′′ → K′′′] and [K → K′] ( [K′ → K′′] [K′′ → K′′′] ) are identical. Identity element: there is an identity element, a transformation K → K. Inverse element: for any transformation K → K′ there exists an inverse transformation K′ → K. Transformation matrices consistent with group axioms Consider two inertial frames, K and K′, the latter moving with velocity v with respect to the former. By rotations and shifts we can choose the x and x′ axes along the relative velocity vector and also that the events (t, x) = (0,0) and (t′, x′) = (0,0) coincide. Since the velocity boost is along the x (and x′) axes nothing happens to the perpendicular coordinates and we can just omit them for brevity. Now since the transformation we are looking after connects two inertial frames, it has to transform a linear motion in (t, x) into a linear motion in (t′, x′) coordinates. Therefore, it must be a linear transformation. The general form of a linear transformation is where α, β, γ and δ are some yet unknown functions of the relative velocity v. Let us now consider the motion of the origin of the frame K′. In the K′ frame it has coordinates (t′, x′ = 0), while in the K frame it has coordinates (t, x = vt). These two points are connected by the transformation from which we get Analogously, considering the motion of the origin of the frame K, we get from which we get Combining these two gives α = γ and the transformation matrix has simplified, Now consider the group postulate inverse element. There are two ways we can go from the K′ coordinate system to the K coordinate system. The first is to apply the inverse of the transform matrix to the K′ coordinates: The second is, considering that the K′ coordinate system is moving at a velocity v relative to the K coordinate system, the K coordinate system must be moving at a velocity −v relative to the K′ coordinate system. Replacing v with −v in the transformation matrix gives: Now the function γ can not depend upon the direction of v because it is apparently the factor which defines the relativistic contraction and time dilation. These two (in an isotropic world of ours) cannot depend upon the direction of v. Thus, γ(−v) = γ(v) and comparing the two matrices, we get According to the closure group postulate a composition of two coordinate transformations is also a coordinate transformation, thus the product of two of our matrices should also be a matrix of the same form. Transforming K to K′ and from K′ to K′′ gives the following transformation matrix to go from K to K′′: In the original transform matrix, the main diagonal elements are both equal to γ, hence, for the combined transform matrix above to be of the same form as the original transform matrix, the main diagonal elements must also be equal. Equating these elements and rearranging gives: The denominator will be nonzero for nonzero v, because γ(v) is always nonzero; If v = 0 we have the identity matrix which coincides with putting v = 0 in the matrix we get at the end of this derivation for the other values of v, making the final matrix valid for all nonnegative v. For the nonzero v, this combination of function must be a universal constant, one and the same for all inertial frames. Define this constant as δ(v)/v γ(v) = κ, where κ has the dimension of 1/v2. Solving we finally get and thus the transformation matrix, consistent with the group axioms, is given by If κ > 0, then there would be transformations (with κv2 ≫ 1) which transform time into a spatial coordinate and vice versa. We exclude this on physical grounds, because time can only run in the positive direction. Thus two types of transformation matrices are consistent with group postulates: with the universal constant κ = 0, and with κ < 0. Galilean transformations If κ = 0 then we get the Galilean-Newtonian kinematics with the Galilean transformation, where time is absolute, t′ = t, and the relative velocity v of two inertial frames is not limited. Lorentz transformations If κ < 0, then we set which becomes the invariant speed, the speed of light in vacuum. This yields κ = −1/c2 and thus we get special relativity with Lorentz transformation where the speed of light is a finite universal constant determining the highest possible relative velocity between inertial frames. If v ≪ c the Galilean transformation is a good approximation to the Lorentz transformation. Only experiment can answer the question which of the two possibilities, κ = 0 or κ < 0, is realized in our world. The experiments measuring the speed of light, first performed by a Danish physicist Ole Rømer, show that it is finite, and the Michelson–Morley experiment showed that it is an absolute speed, and thus that κ < 0. Boost from generators [edit] Using rapidity ϕ to parametrize the Lorentz transformation, the boost in the x direction is likewise for a boost in the y-direction and the z-direction where ex, ey, ez are the Cartesian basis vectors, a set of mutually perpendicular unit vectors along their indicated directions. If one frame is boosted with velocity v relative to another, it is convenient to introduce a unit vector n = v/v = β/β in the direction of relative motion. The general boost is Notice the matrix depends on the direction of the relative motion as well as the rapidity, in all three numbers (two for direction, one for rapidity). We can cast each of the boost matrices in another form as follows. First consider the boost in the x direction. The Taylor expansion of the boost matrix about ϕ = 0 is where the derivatives of the matrix with respect to ϕ are given by differentiating each entry of the matrix separately, and the notation \|ϕ = 0 indicates ϕ is set to zero after the derivatives are evaluated. Expanding to first order gives the infinitesimal transformation which is valid if ϕ is small (hence ϕ2 and higher powers are negligible), and can be interpreted as no boost (the first term I is the 4×4 identity matrix), followed by a small boost. The matrix is the generator of the boost in the x direction, so the infinitesimal boost is Now, ϕ is small, so dividing by a positive integer N gives an even smaller increment of rapidity ϕ/N, and N of these infinitesimal boosts will give the original infinitesimal boost with rapidity ϕ, In the limit of an infinite number of infinitely small steps, we obtain the finite boost transformation which is the limit definition of the exponential due to Leonhard Euler, and is now true for any ϕ. Repeating the process for the boosts in the y and z directions obtains the other generators and the boosts are For any direction, the infinitesimal transformation is (small ϕ and expansion to first order) is the generator of the boost in direction n. It is the full boost generator, a vector of matrices K = (Kx, Ky, Kz), projected into the direction of the boost n. The infinitesimal boost is Then in the limit of an infinite number of infinitely small steps, we obtain the finite boost transformation which is now true for any ϕ. Expanding the matrix exponential of −ϕ(n ⋅ K) in its power series we now need the powers of the generator. The square is but the cube (n ⋅ K)3 returns to (n ⋅ K), and as always the zeroth power is the 4×4 identity, (n ⋅ K)0 = I. In general the odd powers n = 1, 3, 5, ... are while the even powers n = 2, 4, 6, ... are therefore the explicit form of the boost matrix depends only the generator and its square. Splitting the power series into an odd power series and an even power series, using the odd and even powers of the generator, and the Taylor series of sinh ϕ and cosh ϕ about ϕ = 0 obtains a more compact but detailed form of the boost matrix where 0 = −1 + 1 is introduced for the even power series to complete the Taylor series for cosh ϕ. The boost is similar to Rodrigues' rotation formula, Negating the rapidity in the exponential gives the inverse transformation matrix, In quantum mechanics, relativistic quantum mechanics, and quantum field theory, a different convention is used for the boost generators; all of the boost generators are multiplied by a factor of the imaginary unit i = √−1. From experiments [edit] Main article: Test theories of special relativity Howard Percy Robertson and others showed that the Lorentz transformation can also be derived empirically. In order to achieve this, it's necessary to write down coordinate transformations that include experimentally testable parameters. For instance, let there be given a single "preferred" inertial frame in which the speed of light is constant, isotropic, and independent of the velocity of the source. It is also assumed that Einstein synchronization and synchronization by slow clock transport are equivalent in this frame. Then assume another frame in relative motion, in which clocks and rods have the same internal constitution as in the preferred frame. The following relations, however, are left undefined: differences in time measurements, differences in measured longitudinal lengths, differences in measured transverse lengths, depends on the clock synchronization procedure in the moving frame, then the transformation formulas (assumed to be linear) between those frames are given by: depends on the synchronization convention and is not determined experimentally, it obtains the value by using Einstein synchronization in both frames. The ratio between and is determined by the Michelson–Morley experiment, the ratio between and is determined by the Kennedy–Thorndike experiment, and alone is determined by the Ives–Stilwell experiment. In this way, they have been determined with great precision to and , which converts the above transformation into the Lorentz transformation. See also [edit] Lorentz group Noether's theorem Poincaré group Proper time Relativistic metric Spinor Notes [edit] ^ "Newton's Philosophy". stanford.edu. 2021. ^ Zeeman, Erik Christopher (1964), "Causality implies the Lorentz group", Journal of Mathematical Physics, 5 (4): 490–493, Bibcode:1964JMP.....5..490Z, doi:10.1063/1.1704140 ^ Goldstein, Norman (2007). "Inertiality Implies the Lorentz Group" (PDF). Mathematical Physics Electronic Journal. 13. ISSN 1086-6655. Retrieved 14 February 2016. ^ a b c d (Landau & Lifshitz 2002) ^ University Physics – With Modern Physics (12th Edition), H.D. Young, R.A. Freedman (Original edition), Addison-Wesley (Pearson International), 1st Edition: 1949, 12th Edition: 2008, ISBN 978-0-321-50130-1 ^ Greiner & Bromley 2000, Chapter 16 ^ Weinberg 2002, Footnote p. 56 ^ Relativity DeMystified, D. McMahon, Mc Graw Hill (USA), 2006, ISBN 0-07-145545-0 ^ An Introduction to Mechanics, D. Kleppner, R.J. Kolenkow, Cambridge University Press, 2010, ISBN 978-0-521-19821-9 ^ Levy, Jean-Michel (2007). "A simple derivation of the Lorentz transformation and of the related velocity and acceleration formulae" (PDF). p. 2. Retrieved 2017-01-11. ^ Einstein, Albert (1916). "Relativity: The Special and General Theory" (PDF). Retrieved 2012-01-23. ^ Stauffer, Dietrich; Stanley, Harry Eugene (1995). From Newton to Mandelbrot: A Primer in Theoretical Physics (2nd enlarged ed.). Springer-Verlag. p. 80,81. ISBN 978-3-540-59191-7. ^ "Albert Einstein: First of all, it is clear that these equations must be linear because of the properties of homogeneity that we attribute to space and time. ^ Born, Max (2012). Einstein's Theory of Relativity (revised ed.). Courier Dover Publications. pp. 236–237. ISBN 978-0-486-14212-8. Extract of page 237 ^ Gupta, S. K. (2010). Engineering Physics: Vol. 1 (18th ed.). Krishna Prakashan Media. pp. 12–13. ISBN 978-81-8283-098-1. Extract of page 12 ^ Einstein, Albert (1916). "Relativity: The Special and General Theory" (PDF). Retrieved 2012-01-23. ^ Robertson, H. P. (1949). "Postulate versus Observation in the Special Theory of Relativity" (PDF). Reviews of Modern Physics. 21 (3): 378–382. Bibcode:1949RvMP...21..378R. doi:10.1103/RevModPhys.21.378. ^ Mansouri R., Sexl R.U. (1977). "A test theory of special relativity. I: Simultaneity and clock synchronization". Gen. Rel. Gravit. 8 (7): 497–513. Bibcode:1977GReGr...8..497M. doi:10.1007/BF00762634. S2CID 67852594. References [edit] Greiner, W.; Bromley, D. A. (2000). Relativistic Quantum Mechanics (3rd ed.). springer. ISBN 9783540674573. Landau, L.D.; Lifshitz, E.M. (2002) . The Classical Theory of Fields. Course of Theoretical Physics. Vol. 2 (4th ed.). Butterworth–Heinemann. ISBN 0-7506-2768-9. Weinberg, S. (2002), The Quantum Theory of Fields, vol. 1, Cambridge University Press, ISBN 0-521-55001-7 Retrieved from " Categories: General relativity Special relativity Hidden categories: Articles with short description Short description with empty Wikidata description Wikipedia articles needing clarification from October 2021 Wikipedia articles needing clarification from October 2023 Derivations of the Lorentz transformations Add topic
16513	https://www.doubtnut.com/qna/13656576	Two fixed charges 4Q (positive) and Q (negative) are located at A and Bthe distance AB being 3m Two fixed charges 4Q (positive) and Q (negative) are located at A and Bthe distance AB being 3m The point P where the resultant field due to both is zero is on AB outside AB The point P where the resutlant field due to both is zero is on AB inside AB If a positive charge is placed at P and displaced slightly along AB it will execute oscillations. If a negative charge is placed at P and displaced slightly along AB it will execute oscillations. The correct Answer is:A,D Use coulomb's law under equilibrium concept Topper's Solved these Questions Explore 3 Videos Explore 3 Videos Explore 14 Videos Explore 14 Videos Explore 6 Videos Similar Questions Two point charges Q1 and Q2 are positioned at points I and 2. The field intensity to the right of the charge Q2 on the line that passes through the two charges varies according to a law that is represented in the figure. The field intensity is assumed to be positive if its direction coincides with the positive direction on the x-axis. The distance between the charges is l. (a) Find the sign of each charge (b) Find the ratio of the absolute values of the charges ∣∣∣Q1Q2∣∣∣ (c) Find the value of b where the field intensity is maximum [(a) Q2 is negative and Q1 is positive (b) (l+aa)2, (c) l(l+aa)2/3−1⎤⎥ ⎥ ⎥⎦ Two positive charges q1 and q2 are located at the points with redius vectors r1 and r2. Find a negative charge q3 and a radius vector r3 of the point at which it has to be placed for the force acting on each of the three charges to be equal to zero. Knowledge Check Two fixed charges 4Q (positive ) and q (negative ) are located at A and B , ther distance AB being 3m. ocr_image. Two fixed charges −2Q and Q are locatted at the points with coordinates ( -3 a, 0) and (+3 a, 0) prespectively in the X−Y plane . The expression for the potential V(x) at a general point on the x-axis is given by . Two charges +Q and −2Q are located at points A and B on a horizontal line as shown below The electric field is zero at a point which is located at a finite distance ocr_image Two fixed, equal, positive charges, each of magnitude 5×10−5 coul are located at points A and B separated by a distance of 6m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB. The moving charge, when it reaches the point C at a distance of 4m from O, has a kinetic energy of 4 joules. Calculate the distance of the farthest point D which the negative charge will reach before returning towards C. Two negative charges of a unit magnitude and a positive charge q are placed along a straight line. At what position and value of q will the system be in equilibrium? (Negative charges are fixed). A particle of mass m and charge +q is located midway between two fixed charged particles each having a charge +q and at a distance 2L apart. Assuming that the middle charge moves along the line joining the fixed charges, calculate the frequency of oscillation when it is displaced slightly. Two fixed charges −2Q and +Q are located at points (−3a,0) and (+3a,0) respectively. Then which of the following statement is correct? A particle of mass m and charge q is located midway between two fixed charged particles each having a charge q and at a distance 2L apart so that distance of this charge from each charge located at end is L. Assuming that the middle charge moves along the line joining the fixed charges. The frequency of oscillation when it is displaced slightly is NARAYNA-ELECTROSTATICS AND GAUSS LAW-Level V A negative point charge placed at the point A is Four charges of 1mC,2mC,3mC and -6mC are placed one at each corner of ... Two fixed charges 4Q (positive) and Q (negative) are located at A and ... Two identical charges +Q are kept fixed some distance apart.A small pa... Select the correct statement :(Only force on a particle is due to elec... Two point charges Q and - Q //4 are separated by a distance X. Then ... Three point charge Q,4Q and 16Q are placed on a straight line 9cm long... Two infinite sheets of uniform charge density + sigma and - sigma a... If we use permitivity epsilon, resistance R, gravitational constant G ... A thin walled spherical conducting shells S of radius R is given charg... A hollow closed conductor of irregular shape is given some charge . Wh... three points charges are placed at the cornors of an equilibrium trian... An electric dipole is placed at the centre of a sphere. Mark the corre... An electric field converges at the origin whose magnitude is given by ... A conducting sphere A of radius a ,with charge Q , is placed concentri... Two thin conducting shells of radii R and 3R are as shown in the fig. ... X and Y are large, parallel conducting plates close to each other. Eac... A particle of mass m and charge q is thrown in a region where uniform ... Two large parallel conducting plates are placed close to each other ,t... the electric potential in the region of space is given by :V(x)=A+Bx+C... 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16514	https://dokan.co/blog/488099/discount-pricing-strategy/	Skip to content 7+ Most Effective Discount Pricing Strategy (+Examples) Shams Sumon Marketing, Dokan Cloud \| 20 min read Human Written Outline What is Discount Pricing Strategy? When You Should Offer Discount Pricing 7+ Most Effective Discount Pricing Strategies for 2025 Pros and Cons of Discount Pricing Strategy 3 Types of Discount Pricing Discount Pricing Strategy Examples for Your Inspiration Final Words on Discount Pricing Strategy Subscribe to Dokan blog Share on Discount pricing isn’t a new idea at all. Coca-Cola was the first brand to try it out way back in 1887, and it was a huge success. Since then, discounts have been a go-to move for businesses and customers alike. Using the discount policy, companies boost their sales, and customers get great deals. According to a research by CapitalOneShopping shows that; 91% of Americans check for discounts before making a purchase online. If you want to accelerate your sales by offering discounts, you need to know how you should do that to get maximum sales. Keeping that in mind, in this guide, we are going to discuss everything about discount pricing strategy including what it is, when you should do it, and examples of successful discount strategies. Let’s get started! Quick Summary: Discount pricing is when businesses lower prices to sell more, attract new customers, or reward loyal ones. Popular strategies include BOGO, flash sales, and special offers. It helps grow sales and keep customers happy. What is Discount Pricing Strategy? A discount pricing strategy is when a business lowers the price of its products or services to encourage more people to buy. It’s like when you see a “50% off” sign at your favorite store—it’s designed to grab your attention and make you feel like you’re getting a great deal. Businesses use discount pricing strategies for many reasons. They might want to clear out old stock, attract more customers, or boost sales during a slow period. The key to a good discount pricing strategy is to find the right balance. You want to offer a deal that makes customers excited to buy but still ensures your business makes a profit. When done right, discount pricing can be a powerful tool to increase sales and grow your customer base. When You Should Offer Discount Pricing Offering discount pricing can be a smart move, but timing is everything. Here are some situations when it makes sense to offer discounts: Clearing Out Old Inventory: If you have products that aren’t selling well or are out of season, offering a discount can help clear out that inventory and make room for new stock. Attracting New Customers: Discounts can be a great way to bring in new customers. A special offer or a first-time buyer discount can encourage people to try your products or services for the first time. Boosting Sales During Slow Periods: Every business has slow periods. Offering a discount during these times can help maintain steady sales and keep your cash flow healthy. Launching a New Product: When introducing a new product, offering a discount can create buzz and encourage people to give it a try. It’s a way to generate interest and get feedback from customers. Encouraging Bulk Purchases: If you want to sell more of a product, offering a discount for buying in bulk can be effective. This not only increases sales but also helps with customer retention. Rewarding Loyal Customers: Offering discounts to your regular customers can make them feel appreciated and encourage them to keep coming back. Loyalty programs often use this strategy. If you are having any of these occasions, you can launch your discount pricing strategy. But before showing you the details of the discount pricing strategies, let’s discuss types of discount pricing. 7+ Most Effective Discount Pricing Strategies for 2025 Now you know what discount pricing strategy is and the types of discount pricing strategies. It’s time to discuss the 7 most effective discount pricing strategies for online businesses. First, let’s take a quick look at the list of the discount pricing strategies: Buy One, Get One Free (BOGO) New Customer Discount Flash Sales Early Bird Discount Loyalty Discount Free Gifts with Purchase Personalized Discounts Retargeting Discounts Now we will discuss each of the above-mentioned points in detail! 01. Buy One, Get One Free (BOGO) The Buy One, Get One Free (BOGO) strategy is a classic and effective way to boost sales. With BOGO, customers get an additional item for free when they purchase one at full price. It’s a simple concept but incredibly appealing because it feels like an amazing deal—who doesn’t love getting something for free? Here is an example of a Buy One, Get One Free offer. BOGO works well because it encourages customers to buy more than they initially planned. For example, if someone was thinking about buying just one item, the promise of a free second item might nudge them to make the purchase. It’s especially effective for clearing out inventory and increasing sales volume. 02. New Customer Discount A New Customer Discount is a powerful way to attract fresh faces to your business. This strategy involves offering a special discount to people who are buying from you for the first time. It’s like rolling out the red carpet and giving them a warm welcome to your brand. Here is an example of how the new customer discount strategy works. New Customer Discounts work because they reduce the risk for first-time buyers. If someone is on the fence about trying your product or service, a discount can be the nudge they need to make that first purchase. It’s also a great way to stand out from competitors—potential customers might choose your brand over another simply because you’re offering a better deal. 03. Flash Sales Flash Sales are short-term, high-impact promotions designed to create a sense of urgency among customers. These sales typically last for a few hours or, at most, a day, offering significant discounts on selected items. The limited-time nature of Flash Sales plays on the fear of missing out (FOMO), driving customers to make quick purchasing decisions. Flash Sales are incredibly effective for several reasons. First, they generate excitement and buzz around your brand. Customers know they have to act fast, so they’re more likely to buy immediately rather than wait. This can lead to a surge in sales within a very short period. 04. Early Bird Discount An Early Bird Discount is a pricing strategy that rewards customers who make a purchase or sign up for an event early. The idea is simple: the sooner someone buys, the better the deal they get. This strategy is commonly used for events, product launches, or limited-time offers where businesses want to encourage quick commitments. For example, an event organizer might offer a lower price to the first 100 people who buy tickets, then gradually increase the price as the event date gets closer. This not only boosts early ticket sales but also creates buzz and excitement around the event. 05. Loyalty Discount A Loyalty Discount is a special offer given to repeat customers as a way to thank them for their continued support. This strategy is all about building and maintaining strong relationships with your customers. By offering discounts to those who regularly buy from you, you’re showing that you appreciate their loyalty and encouraging them to keep coming back. Here is an example of what a loyalty discount pricing strategy could look like. Loyalty Discounts can take various forms, such as a percentage off on future purchases, exclusive deals, or even points that can be redeemed for discounts. For example, a coffee shop might offer a free drink after a customer buys ten, or an online store might give loyal customers early access to sales with additional discounts. 06. Free Gifts with Purchase Offering Free Gifts with Purchase is a popular discount strategy where customers receive an additional item for free when they buy a specific product or spend a certain amount. This strategy adds extra value to a purchase, making the overall deal more attractive to customers. This is an example of a free gifts with purchase discount pricing strategy. Free Gifts with Purchases work well because they make customers feel like they’re getting something extra for their money. It’s a simple way to enhance the shopping experience and encourage customers to buy more. For example, a beauty brand might offer a free makeup bag with the purchase of certain products, or a tech store might include free accessories with the purchase of a new gadget. 07. Personalized Discounts Personalized Discounts are special offers made just for you based on what you like or what you’ve bought before. Instead of a one-size-fits-all deal, personalized discounts give you something that matches your interests. For example, if today is your birthday, you might get a discount for your birthday. Or, if you haven’t shopped in a while, a store might send you a “We miss you” discount to get you back. These discounts work well because they give you something you want. When a deal feels made for you, you’re more likely to use it. 08. Retargeting Discount A Retargeting Discount is a smart strategy to bring back customers who have shown interest in your products but did not complete a purchase. With this strategy, you offer a special discount to customers who abandoned their carts, visited a product page, or engaged with your website. For example, a customer adds a pair of shoes to their cart but leaves without buying. Later, they receive an email or ad offering 10% off that exact pair if they complete the purchase within 24 hours. Retargeting discounts work well because they remind customers of what they were interested in. They create a sense of urgency and make the offer more appealing. This strategy is especially effective for recovering lost sales and increasing conversions on items that might otherwise have been forgotten. Pros and Cons of Discount Pricing Strategy Applying a discount pricing strategy has both advantages and disadvantages. It will get you more sales. At the same time, you will make less money on each thing you sell. Interesting, right? Let’s find more pros and cons of discount pricing: Pros of Discount Pricing First, we will discuss the positive sides of the discount pricing: Boosts Sales: Discounts quickly attract more customers, especially during slow times or new product launches. Attracts New Customers: A good discount can encourage first-time buyers to try your brand. Clears Out Inventory: Discounts help move unsold or outdated products, making room for new stock. Builds Loyalty: Offering discounts can make customers feel valued, encouraging repeat business. Increases Brand Awareness: A well-promoted discount can get people talking about your brand. Cons of Discount Pricing Discount pricing strategy won’t always be beneficial. Let’s check the flip side of the discount pricing: Reduces Profits: Offering discounts means earning less on each sale, which can hurt overall profits. People Think Your Stuff is Cheap: People might think your stuff isn’t worth much. Limits Long-Term Growth: Relying too much on discounts can shift focus to price, not brand value. While discount pricing can boost sales and attract customers, you should always use it wisely to avoid potential downsides. 3 Types of Discount Pricing There are many types of discounts. We have categorized these discounts into three main types. These types are: Seasonal Discount Pricing Clearance Discount Pricing Volume Discount Pricing Now let’s focus on each type and their characteristics! 1. Seasonal Discount Pricing Seasonal discount pricing is when businesses lower their prices during certain times of the year. Such as: holiday sales, back-to-school deals, or summer clearance events. These discounts are tied to specific seasons or holidays and are a great way to boost sales during those periods. 2. Clearance Discount Pricing Clearance discount pricing is all about getting rid of products that are no longer needed. This might be: end-of-season items, discontinued products, or overstocked goods. The goal is to clear out space for new inventory, even if it means selling at a lower price. 3. Volume Discount Pricing Volume discount pricing is when customers get a discount for buying in bulk. The more they buy, the bigger the discount. This strategy is common in wholesale businesses but can be used in retail as well. For example, a store might offer “buy 3, get 1 free” deals or lower the price per unit when customers buy in larger quantities. Discount Pricing Strategy Examples for Your Inspiration Companies around the world including big and small leverage discount pricing strategy. Here are some inspiring examples of how companies use discounts to drive sales and build customer loyalty: 1. Amazon Prime Day Amazon’s Prime Day is a massive annual event where Prime members get access to exclusive discounts on thousands of products. This not only boosts sales but also drives new sign-ups for Amazon Prime. The limited-time nature and wide range of discounts create excitement and urgency, encouraging customers to make purchases they might not have considered otherwise. 2. Black Friday and Cyber Monday Retailers worldwide participate in Black Friday and Cyber Monday, offering huge discounts on products both online and in stores. These events are famous for attracting massive crowds and generating significant sales in a short period. The anticipation and deep discounts make these days some of the busiest shopping times of the year. 3. Starbucks Rewards Program Starbucks uses a loyalty program to offer personalized discounts and free drinks to repeat customers. For example, members can earn stars with each purchase, which can be redeemed for free items. Starbucks also offers special discounts on customers’ birthdays or after reaching certain milestones, which keeps customers engaged and coming back for more. 4. Old Navy’s Seasonal Sales Old Navy frequently offers seasonal sales, such as back-to-school or end-of-season clearances, with deep discounts on clothing and accessories. These sales help the brand clear out old inventory while driving foot traffic to stores and increasing online sales. 5. Apple’s Student Discounts Apple offers discounts on its products for students and educators, making expensive technology more affordable for this group. This strategy not only boosts sales but also builds brand loyalty among younger customers who may continue using Apple products long-term. These examples show how diverse and effective discount pricing strategies can be, whether you’re a small business or a global brand. Use these ideas as inspiration to create your discount campaigns that resonate with your customers and drive your business forward. Final Words on Discount Pricing Strategy Now you know what is a discount pricing strategy when you should apply discount pricing, what types of discounts, and the 7+ most effective discount strategies. So, what are you waiting for? Follow this guide and offer discounts to your customers to boost sales. If you want to increase the conversion rate of your online business, you can check our blog on increasing the conversion rate like a pro. That being said, if you have any further queries related to discount pricing strategies, don’t hesitate to share them with us using the comment box below. We are all ears! Subscribe to Dokan blog We send weekly newsletters, no spam for sure! 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16515	https://jejjohnson.github.io/uncertain_gps/Taylor/error_propagation/	Error Propagation - [x] - [x] Skip to content Uncertain Gaussian Processes Error Propagation Initializing search jejjohnson/uncertain_gps Uncertain Gaussian Processes jejjohnson/uncertain_gps Home Literature [x] Taylor Taylor Linearized GP [x] Error Propagation Error Propagation Table of contents Taylor Series Expansion Law of Error Propagation Proof: Mean Function Proof: Variance Function Resources [x] Notebooks Notebooks [x] JAX JAX Basics Refactor Objax [x] Numpyro Numpyro Numpyro (SVI) Numpyro (MCMC) [x] GPyTorch GPyTorch Uncertain Inputs (Taylor) Table of contents Taylor Series Expansion Law of Error Propagation Proof: Mean Function Proof: Variance Function Resources Source Error Propagation¶ Taylor Series Expansion¶ A Taylor series is representation of a function as an infinite sum of terms that are calculated from the values of the functions derivatives at a single point - Wiki Often times we come across functions that are very difficult to compute analytically. Below we have the simple first-order Taylor series approximation. Let's take some function f(x) where x∼N(μ x,Σ x) described by a mean μ x and covariance Σ x. The Taylor series expansion around the function f(x) is: z=f(x)≈f(μ x)+∂f∂x\|x=μ x(x−μ x) Law of Error Propagation¶ This results in a mean and error covariance of the new distribution z defined by: μ z=f(μ x) Σ z=∇x f(μ x)Σ x∇x f(μ x)⊤ Proof: Mean Function¶ Given the mean function: E[x]=1 N∑i=1 x i We can simply apply this to the first-order Taylor series function. μ z=E x[f(μ x)+∂f∂x\|x=μ x(x−μ x)]=E x[f(μ x)]+E x[∂f∂x\|x=μ x(x−μ x)]=f(μ x)+E x[∂f∂x\|x=μ x x]−E x[∂f∂x\|x=μ x μ x]=f(μ x)+∂f∂x\|x=μ x μ x−∂f∂x\|x=μ x μ x=f(μ x) Proof: Variance Function¶ Given the variance function V[x]=E[x−μ x]2 σ z 2=E[f(μ x)−∂f∂x\|x=μ x(x−μ x)−μ x]=E[∂f∂x\|x=μ x(x−μ x)]2=(∂f∂x\|x=μ x)2 E[x−μ x]2=(∂f∂x\|x=μ x)2 Σ x I've linked a nice tutorial for propagating variances below if you would like to go through the derivations yourself. Resources¶ Essence of Calculus, Chapter 11 \| Taylor Series - 3Blue1Brown - youtube Introduction to Error Propagation: Derivation, Meaning and Examples - PDF Statistical uncertainty and error propagation - Vermeer - PDF Previous Linearized GP Copyright © 2020 J. Emmanuel Johnson Made with Material for MkDocs
16516	https://pmc.ncbi.nlm.nih.gov/articles/PMC6463807/	Needle aspiration versus incision and drainage for the treatment of peritonsillar abscess - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Cochrane Database Syst Rev . 2016 Dec 23;2016(12):CD006287. doi: 10.1002/14651858.CD006287.pub4 Search in PMC Search in PubMed View in NLM Catalog Add to search Needle aspiration versus incision and drainage for the treatment of peritonsillar abscess Brent A Chang Brent A Chang 1 University of British Columbia, Division of Otolaryngology Head & Neck Surgery, 2775 Laurel St. 4th Floor Otolaryngology (ENT), Vancouver, BC, Canada, V5Z 1M9 Find articles by Brent A Chang 1,✉, Andrew Thamboo Andrew Thamboo 1 University of British Columbia, Division of Otolaryngology Head & Neck Surgery, 2775 Laurel St. 4th Floor Otolaryngology (ENT), Vancouver, BC, Canada, V5Z 1M9 Find articles by Andrew Thamboo 1, Martin J Burton Martin J Burton 2 UK Cochrane Centre, Summertown Pavilion, 18 ‐ 24 Middle Way, Oxford, UK, OX2 7LG Find articles by Martin J Burton 2, Chris Diamond Chris Diamond 1 University of British Columbia, Division of Otolaryngology Head & Neck Surgery, 2775 Laurel St. 4th Floor Otolaryngology (ENT), Vancouver, BC, Canada, V5Z 1M9 Find articles by Chris Diamond 1, Desmond A Nunez Desmond A Nunez 1 University of British Columbia, Division of Otolaryngology Head & Neck Surgery, 2775 Laurel St. 4th Floor Otolaryngology (ENT), Vancouver, BC, Canada, V5Z 1M9 Find articles by Desmond A Nunez 1 Editor: Cochrane ENT Group Author information Article notes Copyright and License information 1 University of British Columbia, Division of Otolaryngology Head & Neck Surgery, 2775 Laurel St. 4th Floor Otolaryngology (ENT), Vancouver, BC, Canada, V5Z 1M9 2 UK Cochrane Centre, Summertown Pavilion, 18 ‐ 24 Middle Way, Oxford, UK, OX2 7LG ✉ Corresponding author. Collection date 2016 Dec. Copyright © 2016 The Cochrane Collaboration. Published by John Wiley & Sons, Ltd. PMC Copyright notice PMCID: PMC6463807 PMID: 28009937 This article is an update of the article with doi: 10.1002/14651858.CD006287.pub3. Abstract Background Peritonsillar abscess is a common infection presenting as a collection of pus in the peritonsillar area. The condition is characterised by a severe sore throat, difficulty in swallowing and pain on swallowing, fever and malaise, and trismus. Needle aspiration and incision and drainage are the two main treatment modalities currently used in the treatment of this condition. The effectiveness of one versus the other has not been clearly demonstrated and remains an area of debate. Objectives To assess the effectiveness and risks of needle aspiration versus incision and drainage for the treatment of peritonsillar abscess in older children (eight years of age or older), adolescents and adults. Search methods The Cochrane ENT Information Specialist searched the ENT Trials Register; Central Register of Controlled Trials (CENTRAL 2016, Issue 7); Ovid MEDLINE; Ovid Embase; CINAHL; Web of Science; ClinicalTrials.gov; ICTRP and additional sources for published and unpublished trials. The date of the search was 25 August 2016. Selection criteria Randomised controlled trials comparing needle aspiration with incision and drainage. Data collection and analysis We used the standard methodological procedures expected by Cochrane. Our primary outcomes were recurrence rate (proportion of patients needing repeat intervention) and adverse effects associated with the intervention. Secondary outcomes were time to resumption of normal diet, complications of the disease process and symptom scores. We used GRADE to assess the quality of evidence for each outcome; this is indicated in italics. Main results We included 11 studies (674 participants). The risk of bias was high or unclear in all of the included studies. All studies compared needle aspiration to incision and drainage. All but one of the 11 studies reported on the primary outcome of recurrence. When we pooled data from the 10 studies the recurrence rate was higher in the needle aspiration group compared with incision and drainage: risk ratio (RR) 3.74 (95% confidence interval (CI) 1.63 to 8.59; 612 participants). We detected moderate heterogeneity in this analysis (I 2 = 48%). In interpreting the pooled result it is important to note that the evidence for this outcome was of very low quality. None of the other outcomes (adverse effects of the intervention, time to resumption of normal diet, complications of the disease process and symptom scores) were consistently measured across all studies. Only three studies reported on adverse effects/events associated with the intervention and only one such event in a single patient was reported (post‐procedure bleeding following incision and drainage: 1/28, 3.6%) (very low‐quality evidence). Time to resumption of normal diet was compared in two studies; neither found an obvious difference between needle aspiration and incision and drainage (very low‐quality evidence). Only three studies stated that they would report complications of the disease process. In these three studies, the only complication reported was admission to hospital for dehydration in two patients who underwent incision and drainage (2/13, 6.7%). Symptom scores were measured in four studies; three evaluated pain using different scales and one other symptoms. The data could not be pooled in a meta‐analysis. Two studies evaluating procedural pain reported this to be lower in the needle aspiration groups. One study found comparable rates of pain resolution at five days post‐intervention between groups. The quality of the evidence for symptom scores was very low. Authors' conclusions Although a number of studies have sought to evaluate whether or not needle aspiration or incision and drainage is more effective in patients with peritonsillar abscess, there is no high‐quality evidence to allow a firm conclusion to be drawn and the answer remains uncertain. Very low‐quality evidence suggests that incision and drainage may be associated with a lower chance of recurrence than needle aspiration. There is some very low‐quality evidence to suggest that needle aspiration is less painful. Plain language summary Needle aspiration compared to incision and drainage for the treatment of peritonsillar abscess (quinsy) Review question This review compared the effectiveness of the two main treatment options for peritonsillar abscess: needle aspiration and incision and drainage. Background Peritonsillar abscesses are infections at the back of the throat in which a collection of pus (abscess) has formed next to the tonsil. The condition is characterised by a severe sore throat, difficulty in swallowing and pain on swallowing, fever and malaise, and trismus (inability to open the mouth completely). Treatment is usually by one of two methods. The first is needle aspiration (sucking the pus out using a syringe and needle) and the second 'incision and drainage' (putting a small knife into the abscess to let the pus drain out). It remains unclear whether one type of treatment is better than the other. Study characteristics We included 11 studies with a total of 674 participants. The participants in the studies were aged from 8 to 79. The studies were conducted in a number of countries (six from Pakistan, two from the USA, one from Taiwan and two from South Africa). All but one of the 11 studies reported the difference in recurrence rate between needle aspiration and incision and drainage. Four studies compared symptom scores associated with the procedure and two studies compared time to resumption of normal diet. Three studies reported adverse effects/events associated with the intervention. Two studies reported complications of the disease process itself. The evidence is current to August 2016. Key results Ten studies reported on the recurrence of peritonsillar abscess (our main outcome). Most of them did not clearly define 'recurrence' and they varied in the timing of its assessment, however we were able to combine (pool) the data from these studies. When we pooled the data the recurrence rate was higher in the needle aspiration group compared with incision and drainage. It is important to note that the evidence for this outcome was of very low quality. Some studies found that patients had more pain when they had incision and drainage. Quality of the evidence We identified problems or potential problems in all of the included studies. The most important of these was that the studies did not all assess recurrence in the same way, at the same time, using the same criteria. The quality of the evidence for all of the outcomes that we looked at was very low. Summary of findings Summary of findings for the main comparison. Needle aspiration versus incision and drainage for the treatment of peritonsillar abscess. Needle aspiration versus incision and drainage for the treatment of peritonsillar abscess Patient or population: patients older than 8 years with peritonsillar abscess Setting: inpatients and outpatients Intervention: incision and drainage Comparison: needle aspiration OutcomesAnticipated absolute effects (95% CI)Relative effect№ of participants (studies)Quality of the evidence (GRADE)Comments Risk with incision and drainageRisk with needle aspiration Primary outcome: recurrence rate Study population RR 3.74, 95% CI 1.63 to 8.59 612 (10 RCTs)⊕⊝⊝⊝ very low1,2— 47 per 1000 245 per 1000 Primary outcome: adverse effects/events associated with the interventions One study reported post‐procedural bleeding in 1 patient (3.6%) in the incision and drainage group, with no adverse effects/events reported in the needle aspiration group. Two studies stated that no complications were seen in either group.—226 (3 RCTs)⊕⊝⊝⊝ very low2,3 Adverse effects/events were not mentioned as a pre‐specified outcome measure in any of the studies. Secondary outcome: time to resumption of normal diet One study found no difference in the time to resumption of normal diet (mean 3.7 days in both groups, no confidence intervals provided). Another study found that a similar percentage of patients returned to solid food within 4 days (87%: needle aspiration, 88%: incision and drainage).—124 (2 RCTs)⊕⊝⊝⊝ very low2,4— Secondary outcome: complications of the disease process One study described a complication of 2 patients requiring admission to hospital for dehydration in the incision and drainage group and no complications in the needle aspiration group. One study stated that no complications were seen in either group.—170 (2 RCTs)⊕⊝⊝⊝ very low2,5 Complications of the disease process were not mentioned as a pre‐specified outcome measure in any of the studies. Secondary outcome: symptom scores (Multiple different outcome scales used)Procedural pain Study 1 Pain was less in the needle aspiration group: MD ‐0.8, 95% CI ‐1.16 to ‐0.44 (10‐point scale) Study 2 Reported less pain in the needle aspiration group Pain resolution Study 3 Pain resolution was similar between groups at 5 days post‐intervention Other symptoms Study 4 Reported comparable symptom scores between groups at presentation and 48 hours—Study 1 110 participants Study 2 56 participants Study 3 62 participants Study 4 52 participants⊕⊝⊝⊝ very low2,6— The risk in the intervention group (and its 95% confidence interval) is based on the assumed risk in the comparison group and the relative effect of the intervention (and its 95% CI). CI: confidence interval; MD: mean difference; RCT: randomised controlled trial; RR: risk ratio GRADE Working Group grades of evidenceHigh quality: We are very confident that the true effect lies close to that of the estimate of the effect Moderate quality: We are moderately confident in the effect estimate: The true effect is likely to be close to the estimate of the effect, but there is a possibility that it is substantially different Low quality: Our confidence in the effect estimate is limited: The true effect may be substantially different from the estimate of the effect Very low quality: We have very little confidence in the effect estimate: The true effect is likely to be substantially different from the estimate of effect Open in a new tab 1 Downgraded once due to serious risk of inconsistency (unexplained heterogeneity). 2 Downgraded twice due to very serious risk of bias (limitations in study design). 3 Adverse event (post‐procedural bleeding) was not well described. 4 Incomplete data (no standard deviations or confidence intervals provided). 5 Admission to hospital for rehydration is inherently subjective and depends on multiple clinical variables. 6 Downgraded once due to imprecision and differences in data reporting. Background Description of the condition Quinsy, better known as peritonsillar abscess, is one of the most common abscesses treated by otolaryngologists. It is one of the most common ear, nose and throat (ENT) emergencies seen in acute ENT clinics and emergency departments in hospitals. Peritonsillar abscesses are a collection of pus between the fibrous capsule of the tonsil and the superior constrictor muscles of the pharynx. A peritonsillar abscess tends to be unilateral (on one side) and is believed to arise from obstructed Weber glands within the superior pole of the tonsil or from acute tonsillitis (Johnson 2005). In the United States, the incidence is estimated to be 30 people per 100,000 per year, which accounts for about 45,000 cases per year (Herzon 1995). Peritonsillar abscess affects people of all ages (Herzon 1995), and has a significant resource cost (Johnson 2005; Powell 2012). Patients with a peritonsillar abscess present with fever, dysphagia, trismus, otalgia, a change in the voice and ipsilateral throat pain. On examination there is a swelling in the oropharynx with medialisation of the tonsil. Physical examination usually reveals uvular deviation to the contralateral side, tonsillar exudate, trismus and jugulodigastric lymphadenopathy. Patients with peritonsillar abscesses are at risk of extension of the abscess into deeper neck spaces and airway obstruction. A combination of symptoms and signs is the accepted basis for diagnosis. Further diagnostic investigations are not generally used unless the peritonsillar abscess presents with neck involvement (Powell 2012). In these circumstances the most common diagnostic test is the computed tomography (CT) scan. CT scans can accurately diagnose peritonsillar abscesses with 100% sensitivity and help to determine the extent of the disease. Ultrasound is less useful, with a diagnostic accuracy of 89% to 95% sensitivity and 79% to 100% specificity, based on level 3 evidence (Powell 2012). Description of the intervention Treatment commonly involves drainage of the abscess together with antibiotics, but there is no agreement on the optimal technique for initial drainage of a peritonsillar abscess (Hall 1990; Herzon 1995; Johnson 2003). Physicians generally have a choice between needle aspiration and incision and drainage, although quinsy tonsillectomy and antibiotic therapy alone are also less commonly used treatment options. Needle aspiration uses a large‐bore needle inserted through the palatoglossus muscles into the abscess. Several insertions of the needle in different locations may be performed during a single treatment episode. The incision and drainage method uses a guarded scalpel to incise the palatoglossus muscle and enter the peritonsillar space/abscess. The peritonsillar space is then opened widely by dissection with blunt forceps to promote drainage of the abscess. A patient with a peritonsillar abscess is sometimes treated with a combination of these methods. Why it is important to do this review Patients with a quinsy often present 'out of hours' and in many healthcare settings are first assessed and managed by doctors in training. A survey in the UK showed that 60% of otolaryngologists would use needle aspiration as their primary method for draining a peritonsillar abscess.If the needle aspiration failed, 52% would then perform scalpel incision and drainage.The survey also showed geographic differences in the management of peritonsillar abscess (Mehanna 2002).In Singapore, a retrospective review showed that most patients were treated with scalpel incision and drainage (66%) (Ong 2004). There is no consensus regarding the best drainage procedure and each method has risks and benefits. Needle aspiration may potentially be less painful, cheaper and technically easier to perform; it can also double as a diagnostic method. Incision and drainage that includes blunt dissection theoretically promotes more effective drainage of the abscess by dissecting through the tissue barriers (septations) that divide the abscess cavity into micro‐cavities or loculations. The resulting wide pathway to the oral cavity allows air to enter the depths of the abscess cavity, increasing the oxygen tension that in turn reduces the survival of anaerobic bacteria exposed to air. However, it is a more invasive method and it has been suggested that it may carry a risk of aspiration of purulent material or incisional injury to underlying structures (Khayr 2005; Spires 1987). The method of drainage used may be associated with varying degrees of abscess recurrence, pain and haemorrhage.A review performed by Johnson et al concluded that needle aspiration was the best initial treatment, followed by incision and drainage if needle aspiration failed (Johnson 2003).Since then, additional studies on this subject have been undertaken, making it important to re‐evaluate the available evidence.If there is an optimal drainage procedure for peritonsillar abscesses, then it should be adopted widely. Objectives To assess the effectiveness and risks of needle aspiration versus incision and drainage for the treatment of peritonsillar abscess in older children (eight years of age or older), adolescents and adults. Methods Criteria for considering studies for this review Types of studies Randomised controlled trials (RCTs). Types of participants Adult or child patients (of eight years of age or older) with a clinical diagnosis of peritonsillar abscess. Types of interventions Needle aspiration versus incision and drainage. We defined needle aspiration as either a single insertion or multiple insertions of a needle with aspiration during the same clinical procedure. We defined needle aspirations undertaken on subsequent days or during a separate attendance at a healthcare facility on the same day as repeat interventions. We defined incision and drainage as incision of the pharyngeal mucosa with or without any additional wound exploration to promote drainage. We considered 'confirmatory' needle aspiration immediately prior to this procedure as part of the incision and drainage intervention. We considered antibiotic therapy as part of the intervention (needle aspiration or incision and drainage), as long as it was available to both treatment groups. Types of outcome measures Primary outcomes We analysed the following outcomes in the review, but we did not use them as a basis for including or excluding studies. Recurrence rate (proportion of patients needing repeat intervention) Adverse effects/events associated with the interventions Secondary outcomes Time to resumption of normal diet Complications of the disease process Symptom scores Search methods for identification of studies The Cochrane ENT Information Specialist conducted systematic searches for randomised controlled trials and controlled clinical trials. There were no language, publication year or publication status restrictions. The date of the search was 25 August 2016. Electronic searches The Information Specialist searched: Cochrane ENT Trials Register (searched 25 August 2016); Cochrane Central Register of Controlled Trials (CENTRAL 2016, Issue 7); Ovid MEDLINE (1946 to 25 August 2016): Ovid MEDLINE (In‐Process & Other Non‐Indexed Citations); PubMed (as a top up to searches in Ovid MEDLINE); Ovid Embase (1974 to 2016 week 34); EBSCO CINAHL (1982 to 25 August 2016); LILACS, lilacs.bvsalud.org (searched 25 August 2016); KoreaMed (searched via Google Scholar 25 August 2016); IndMed, www.indmed.nic.in (searched 25 August 2016); PakMediNet, www.pakmedinet.com (searched 25 August 2016); Web of Knowledge, Web of Science (1945 to 25 August 2016); CNKI, www.cnki.com.cn (searched via Google Scholar 25 August 2016); ClinicalTrials.gov (searched via the Cochrane Register of Studies 25 August 2016); World Health Organization (WHO) International Clinical Trials Registry Platform (ICTRP), www.who.int/ictrp (searched 25 August 2016); ISRCTN, www.isrctn.com (searched 25 August 2016); Google Scholar, scholar.google.co.uk (searched 25 August 2016); Google, www.google.com (searched 25 August 2016). The Information Specialist modelled subject strategies for databases on the search strategy designed for CENTRAL. Where appropriate, they were combined with subject strategy adaptations of the highly sensitive search strategy designed by Cochrane for identifying randomised controlled trials and controlled clinical trials (as described in the Cochrane Handbook for Systematic Reviews of Interventions Version 5.1.0, Box 6.4.b. (Handbook 2011). Search strategies for major databases including CENTRAL are provided in Appendix 1. Searching other resources We scanned the reference lists of identified publications for additional trials and contacted trial authors where necessary. In addition, the Information Specialist searched Ovid MEDLINE, the Cochrane Library and Google to retrieve existing systematic reviews relevant to this systematic review, so that we could scan their reference lists for additional trials. Data collection and analysis Selection of studies Two authors (BAC and AT) independently reviewed all retrieved articles and determine eligibility based on the inclusion criteria. We resolved disagreements by consensus discussion. Data extraction and management Two authors (BAC and AT) independently extracted data from studies using standardised data forms. Briefly, extracted data items included information on study design, study participants, study characteristics, interventions and outcomes. We extracted data so as to allow an intention‐to‐treat analysis. Where data were missing, we wrote to the authors of the study to request further information. Assessment of risk of bias in included studies BAC, AT and DAN undertook assessment of the risk of bias of the trials being considered for inclusion independently, with the following taken into consideration, as guided by the Cochrane Handbook for Systematic Reviews of Interventions (Handbook 2011): sequence generation; allocation concealment; blinding; incomplete outcome data; selective outcome reporting; and other sources of bias. We used the Cochrane 'Risk of bias' tool in RevMan 5.3 (RevMan 2014), which involves describing each of these domains as reported in the trial and then assigning a judgement about the adequacy of each entry: 'low', 'high' or 'unclear' risk of bias. Each review author independently determined whether a study was sufficiently free of bias in each domain and a majority decision determined the ratings given. Measures of treatment effect Recurrence (primary outcome) was measured as dichotomous data. We used risk ratios (RR) with 95% confidence intervals to compare data. We planned to use the mean difference to assess continuous data. We had planned to use hazard ratios to assess time‐to‐event data. Dealing with missing data We made attempts to contact original investigators to obtain any missing data. We had planned sensitivity analysis to assess the impact of data that could not be obtained. Assessment of heterogeneity We assessed heterogeneity using the I² statistic. Roughly, we used the following thresholds as a guide to heterogeneity (RevMan 2014): 0% to 40%: might not be important; 30% to 60%: may represent moderate heterogeneity; 50% to 90%: may represent substantial heterogeneity; 75% to 100%: considerable heterogeneity. We planned meta‐analysis for outcomes in which heterogeneity was less than 50%. Assessment of reporting biases We had intended the use of funnel plots to assess the potential for reporting (publication) bias, if necessary. However, this was not possible. Data synthesis We had intended to use a fixed‐effect model if there was minimal heterogeneity in the included studies (less than 30%); otherwise we used a random‐effects model. Subgroup analysis and investigation of heterogeneity We had planned no subgroup analysis. Sensitivity analysis If there was any ambiguity regarding whether studies would meet the inclusion criteria or if study quality was insufficient, then we had intended the use of sensitivity analysis (repetition of the analysis with the inclusion of different data or trials due to decisions that may have been arbitrary or unclear) (Handbook 2011). GRADE and 'Summary of findings' table Two authors independently used the GRADE approach to rate the overall quality of evidence. The quality of evidence reflects the extent to which we are confident that an estimate of effect is correct and we applied this in the interpretation of results. There are four possible ratings: high, moderate, low and very low. A rating of high quality of evidence implies that we are confident in our estimate of effect and that further research is very unlikely to change our confidence in the estimate of effect. A rating of very low quality implies that any estimate of effect obtained is very uncertain. The GRADE approach rates evidence from RCTs that do not have serious limitations as high quality. However, several factors can lead to the downgrading of the evidence to moderate, low or very low. The degree of downgrading is determined by the seriousness of these factors: study limitations (risk of bias); inconsistency; indirectness of evidence; imprecision; and publication bias. We included a 'Summary of findings' table, constructed according to the recommendations described in Chapter 10 of the Cochrane Handbook for Systematic Reviews of Interventions (Handbook 2011). We included the following outcomes in the 'Summary of findings' table: recurrence rate, adverse effects/events associated with the intervention, time to resumption of normal diet, complications of the disease process and symptom scores. Results Description of studies Results of the search The literature search identified 370 potential records after duplicates were removed. Of these, we deemed 17 eligible for full‐text review. We identified one additional article from searching the reference lists of these publications, yielding 18 total studies that we reviewed in full. Eleven publications fulfilled the review's inclusion criteria (Chi 2014; Khan 2011; Khan 2012; Khokhar 2015; Maharaj 1991; Nwe 2000; Rafi 2007; Sheikh 2012; Spires 1987; Stringer 1988; Younas 2015). We included these 11 studies in the review. We excluded six studies (Fry 1987; Herzon 1995; Khayr 2005; Kulkarni 2013; Stringer 1986; Wolf 1994). See Excluded studies and Characteristics of excluded studies. One study, Tyagi 2011, is awaiting classification (see Characteristics of studies awaiting classification). Currently, there are no ongoing studies identified. See Figure 1 for a PRISMA flow diagram depicting the search and study selection process. 1. Open in a new tab Process for sifting search results and selecting studies for inclusion. Included studies See Characteristics of included studies. Design All of the included studies were randomised controlled trials; however, three of the studies were quasi‐randomised (Khan 2012; Maharaj 1991; Spires 1987). Maharaj 1991 and Khan 2012 allocated patients on an alternate basis. Spires 1987 only randomised a portion of the patients. None of the trials were obviously blinded in terms of the study personnel collecting follow‐up data. Intention‐to‐treat analysis was not specifically carried out. Sample sizes The numbers of participants in the included studies were as follows: 62 (Spires 1987), 52 (Stringer 1988), 60 (Maharaj 1991), 75 (only 50 patients were in groups compared in this review) (Nwe 2000), 50 (Rafi 2007), 62 (Khan 2011), 56 (Khan 2012), 50 (Sheikh 2012), 110 (Chi 2014), 62 (Younas 2015) and 60 (Khokhar 2015). Other than Khokhar 2015, no mention was made of sample size calculations in any of the studies to ensure that they were adequately powered. Setting Nwe 2000 did not specify the setting of the study, although it was published in South Africa. Otherwise, all included studies took place in a hospital setting. Khan 2011, Khan 2012, Khokhar 2015, Rafi 2007, Sheikh 2012 and Younas 2015 all took place in Pakistan and were undertaken with participants in an inpatient hospital setting. Chi 2014 took place in Taiwan and also had participants admitted to hospital in an inpatient setting. The remaining studies treated patients in an outpatient hospital setting: Maharaj 1991 (South Africa), Spires 1987 (USA) and Stringer 1988 (USA). Participants Spires 1987 included patients aged 12 to 53, Stringer 1988 included patients aged 13 to 60, Nwe 2000 included patients aged 15 to 43, Khan 2011 included patients aged 15 to 35, Khan 2012 included patients aged 16 to 50 and Chi 2014 included patients aged 12 to 79. Rafi 2007 included patients aged 22 to 43 and Sheikh 2012 included patients aged 18 to 51. Khokhar 2015 included patients aged 17 to 53. Younas 2015 included patients aged 8 to 57. Maharaj 1991 included patients ranging from "under 14" to "over 40", and did not state how many were male or female. Nwe 2000 had a population that was 67% female; all the other studies had more male than female participants. Interventions All 11 studies compared needle aspiration with incision and drainage (Chi 2014; Khan 2011; Khan 2012; Khokhar 2015Maharaj 1991; Nwe 2000; Rafi 2007; Sheikh 2012; Spires 1987; Stringer 1988, Younas 2015). One study also compared a third group treated only with intravenous antibiotics (Nwe 2000). Outcomes Primary outcomes All studies except one (Younas 2015) reported on recurrence rate (Chi 2014; Khan 2011; Khan 2012; Khokhar 2015; Maharaj 1991; Nwe 2000; Rafi 2007; Sheikh 2012; Spires 1987; Stringer 1988). However, the definition of recurrence was not well described. This is summarised in Table 2. Less than half of the studies provided a specific definition for recurrence. There was also wide variability in the method and timing of assessment for recurrence. 1. Definition of recurrence and timing. Study IDDefinition of recurrence or criteria for re‐intervention describedTiming of assessment of recurrence Spires 1987No 2, 7 days (2x returned day 1) Stringer 1988"Failure to improve symptom scale score; visual evidence of a persistent abscess"1, 2 days (24, 48 hours) Maharaj 1991"reaccumulation of pus"1, 7 days Nwe 2000"patients in whom the trismus and pyrexia persisted 48 hours after the initial treatment"2 days (48 hours) Rafi 2007No Not stated Khan 2011No Not stated Khan 2012No Not stated Sheikh 2012Yes0, 1, 2 days Chi 2014No Not stated Khokhar 2015No"during the course of the study", 7, 14 days Younas 2015N/A N/A Open in a new tab "Improvement in patients was determined by examining the patient the next day after the procedure, a reduction in supra tonsillar swelling along with decrease in pain and also improvement in odynophagia were taken as criteria of improvement and termination of surgical attempts." N/A: not available Only three studies made reference to plans to identify and report any adverse events/effects associated with the intervention (Chi 2014; Khan 2012; Khokhar 2015). Secondary outcomes Time to resumption of normal diet was reported by two studies (Spires 1987; Younas 2015). Complications of the disease process were mentioned in two studies, with Maharaj 1991 reporting two complications (dehydration requiring hospitalisation) and Chi 2014 reporting no complications. Symptom scores were reported by four studies (Chi 2014; Khan 2012; Stringer 1988; Younas 2015), each using different scales. Chi 2014 measured pain intensity one hour after the procedure using a visual analogue score. Khan 2012 had patients grade their score as mild, moderate or severe and measured proportions. Younas 2015 compared the percentage of patients who had resolution of pain by five days after the intervention. Stringer 1988 scored patients according to a scale out of 5 (1 = eating normally, minimal to moderate pain; 2 = eating impaired, moderate pain; 3 = unable to eat solids, moderate to severe pain; 4 = unable to eat solids or liquids, severe pain; 5 = unable to eat solids or liquids, severe pain, volume depletion). Excluded studies After initial screening six studies did not meet the inclusion criteria (see Characteristics of excluded studies). One was classified as a retrospective study (Wolf 1994). Two were classified as review papers (Herzon 1995; Khayr 2005). One study was classified as a commentary (Fry 1987). We considered one study (Stringer 1986) an earlier publication of data in one of the included studies (Stringer 1988). One study was found to be non‐randomised (Kulkarni 2013). Risk of bias in included studies The risk of bias of the included studies is shown in the Characteristics of included studies table and summarised in Figure 2. 2. Open in a new tab 'Risk of bias' summary: review authors' judgements about each risk of bias item for each included study. Allocation All 11 included studies were reportedly randomised. However, it was not clear that proper random sequence generation had been used in any of the studies (Chi 2014; Khan 2011; Khan 2012; Maharaj 1991; Nwe 2000; Rafi 2007; Sheikh 2012; Spires 1987; Younas 2015), other than Stringer 1988 and Khokhar 2015 who used a random numbers table. Additionally, two studies allocated participants on an alternate basis (Khan 2012; Maharaj 1991), and one based on hospital number (Spires 1987). Allocation concealment was not specified to have been used in any of the studies (Chi 2014; Khan 2011; Khan 2012; Khokhar 2015; Maharaj 1991; Nwe 2000; Rafi 2007; Sheikh 2012; Spires 1987; Younas 2015). Stringer 1988 reported not using any allocation concealment. Blinding Blinding of participants or study personnel was either not used (Stringer 1988), or not specifically mentioned in any of the included studies (Chi 2014; Khan 2011; Khan 2012; Khokhar 2015; Maharaj 1991; Nwe 2000; Rafi 2007; Sheikh 2012; Spires 1987; Younas 2015). Incomplete outcome data Most studies did not specifically report losses to follow‐up. Several studies had patients admitted to hospital and we assumed in these cases that there were no losses unless specifically stated (Chi 2014; Khan 2011; Khan 2012; Rafi 2007; Sheikh 2012). Maharaj 1991 lost 18% of patients to follow‐up on day one and 37% at day seven. Patient losses were unclear in Spires 1987 and Nwe 2000. Khokhar 2015 excluded patients who failed to return for follow‐up. No mention of intention‐to‐treat analysis was made in any of the studies (Chi 2014; Khan 2011; Khan 2012; Khokhar 2015; Maharaj 1991; Nwe 2000; Rafi 2007; Sheikh 2012; Spires 1987; Stringer 1988; Younas 2015). In Younas 2015, there was a discrepancy between the number of patients reportedly randomised (62) and the number reported in the results (64); therefore, there is a possibility of incomplete or missing data. Selective reporting None of the included studies had published study protocols with pre‐specified outcomes. Stringer 1988 communicated that all measured outcomes were reported in the published manuscript and, therefore, we deemed it at low risk of bias in this category. We judged all the other studies to have an unclear risk of bias in this category (Chi 2014; Khan 2011; Khan 2012; Khokhar 2015; Maharaj 1991; Nwe 2000; Rafi 2007; Sheikh 2012; Spires 1987; Stringer 1988; Younas 2015). Other potential sources of bias We assessed all studies except one (Maharaj 1991) as having an unclear risk of bias in this category due to concerns about methodology. Specifically, five studies did not provide a definition for recurrence or criteria for re‐intervention (Chi 2014; Khan 2011; Khan 2012; Khokhar 2015; Rafi 2007; Spires 1987). Two studies had subjective definitions of recurrence, including persistent trismus (Nwe 2000), "failure to improve visual symptoms score", "visual evidence of a persistent abscess" (Stringer 1988) and "a reduction in supra tonsillar swelling along with decrease in pain and also improvement in odynophagia" (Sheikh 2012). Four studies did not specify the timing of assessment for recurrence (Chi 2014; Khan 2011; Khan 2012; Rafi 2007). Effects of interventions See: Table 1 See Table 1. Primary outcomes Recurrence rate (proportion of patients needing repeat intervention) Ten out of 11 of the included studies reported recurrence rate following needle aspiration versus incision and drainage (Chi 2014; Khan 2011; Khan 2012; Khokhar 2015; Maharaj 1991; Nwe 2000; Rafi 2007; Sheikh 2012; Spires 1987; Stringer 1988). There was wide variability in the recurrence rate in both groups. Recurrence in the needle aspiration group ranged from 4.9% to 80.0%. In the incision and drainage group recurrence ranged from 0% to 20%. The studies that reported recurrence all reported a comparable or higher recurrence rate with needle aspiration. Timing of assessment for recurrence was inconsistent. Specifically, timing of assessment for recurrence ranged from one day to seven days for initial post‐intervention assessment. The timing of follow‐up was not actually specified in four of the studies (Chi 2014; Khan 2011; Khan 2012; Rafi 2007). One study did not specifically report recurrence rate (Younas 2015). An "initial success rate" was reported; however, this was not defined. When we pooled data from the 10 studies the recurrence rate was higher in the needle aspiration group compared with incision and drainage: risk ratio (RR) 3.74 (95% confidence interval (CI) 1.63 to 8.59; 612 participants). We detected moderate heterogeneity in this analysis (I 2 = 48%). In interpreting the pooled result it is important to note that the quality of evidence for this outcome was very low. Adverse effects/events associated with the interventions Only one study reported an adverse outcome associated with the intervention (Khan 2012). In this study "reactionary haemorrhage" was described in one of 28 patients (3.6%) in the incision and drainage group. Chi 2014 and Khokhar 2015 stated that no complications were seen in either group. No other studies reported characterising adverse effects/events associated with the intervention. Secondary outcomes Time to resumption of normal diet Two studies specifically compared time to resumption of normal diet (Spires 1987; Younas 2015). Spires 1987 found no difference in the average time to resumption of a normal diet between the needle aspiration versus the incision and drainage group (3.7 days, range 1 to 14 days in the needle aspiration group versus 3.7 days, range 1 to 10 days in the incision and drainage group; no standard deviation or statistics provided). Younas 2015 compared the percentage of patients that had returned to a semisolid or solid diet within a certain time frame. They found that 87% of patients treated with needle aspiration returned to semisolid food within two days and solid food within four days. Comparatively, they found that 88% of patients treated with needle aspiration returned to semisolid food by two days and solid food by four days. Two studies described "duration of symptoms" (Stringer 1988) or "period of recovery" (Rafi 2007), but did not specifically define these terms. Complications of the disease process Complications relating to the disease process were generally not reported. Only one study described a complication in two patients (6.7%) who required admission to hospital for dehydration (following incision and drainage) (Maharaj 1991). Chi 2014 stated that no complications were seen in either group. No other study specifically reported measuring complications of the disease process. Symptom scores Pain Two studies reported on pain associated with the intervention. Chi 2014 reported pain intensity one hour after the procedure with visual analogue scores (out of 10) and found a statistically lower amount of pain in the needle aspiration group (4.5 ± 0.8 in the needle aspiration group versus 5.3 ± 1.1 in the incision and drainage group; mean difference (MD) ‐0.8, 95% CI ‐1.16 to ‐0.44, 10‐point scale). Khan 2012 compared postoperative pain as measured by a "mild/moderate/severe" subjective grade. They reported a statistically significant higher proportion of patients with higher pain scores in the incision and drainage group (needle aspiration: 50.0% mild, 28.6% moderate, 21.4% severe; incision and drainage: 17.9% mild, 21.4% moderate, 60.7% severe; P < 0.01). Outcome measures for these two studies were not comparable and, therefore, meta‐analysis was not possible. Younas 2015 reported on pain five days after the intervention: 75% of patients in the needle aspiration group had no pain, whereas 78% of patients treated with incision and drainage had no pain. Other symptoms One study reported symptom scores that were not directly related to procedural pain (Stringer 1988). This study reported symptom scores at presentation and 48 hours (1 = eating normally, minimal to moderate pain; 2 = eating impaired, moderate pain; 3 = unable to eat solids, moderate to severe pain; 4 = unable to eat solids or liquids, severe pain; 5 = unable to eat solids or liquids, severe pain, volume depletion). They found averages of 3.4 (needle aspiration) versus 3.3 (incision and drainage) at presentation and 1.3 (needle aspiration) versus 1.4 (incision and drainage), and complete resolution of symptoms at 10 days post‐treatment in all groups (no statistics or confidence intervals provided). Discussion Summary of main results Peritonsillar abscess is one of the most common ear, nose and throat (ENT) emergencies presenting to acute ENT services and more often than not requires a procedural intervention as the mode of treatment. Despite this, there remains uncertainty about which technique ‐ needle aspiration or incision and drainage ‐ is more effective. There is an absence of high‐quality evidence to show whether or not one technique is superior to the other. There is very low‐quality evidence to suggest that there is a lower recurrence rate with incision and drainage compared with needle aspiration. At the same time, very low‐quality evidence also suggests that needle aspiration may be less painful than incision and drainage. There is an absence of evidence to answer questions about adverse effects/events associated with the intervention, time to resumption of normal diet, complications of the disease process and symptom scores for factors other than pain. Overall completeness and applicability of evidence Data from the studies identified are insufficient to allow us to address all of the objectives of this review and produce a clear and decisive answer. The included studies have enrolled the relevant types of participants. Whilst the studies do evaluate the interventions we sought to include, in some cases those interventions were part of a treatment plan involving periods of hospitalisation and courses of intravenous or oral antibiotics (or both) of varying length. This is relevant because in some healthcare settings patients are almost invariably managed as outpatients without being admitted to hospital. Not all outcomes that we felt were important had been investigated in the included studies. More significantly, the failure to define 'recurrence' and to include comprehensive information about the timing of any recurrence or its identification makes the evidence base incomplete. In all cases, for those outcomes for which we found data, the evidence was of very low quality. There are likely to be geographic or institutional differences in the presentation of peritonsillar abscess. Six out of 11 studies originated from the same country (Pakistan) (see Characteristics of included studies). Different microbial patterns and choice of antimicrobial agents in different geographic areas could theoretically influence recurrence rates. The different settings in which patients were treated in the included studies highlights the potential for variation through different management protocols; over half of the included studies managed patients in an inpatient setting. Other factors of relevance include the likelihood of seeking medical attention, access to medical care and costs or thresholds for performing a repeat intervention. In this review, we have not compared other treatment options for the management of patients with a peritonsillar abscess. So called 'quinsy tonsillectomy' or 'hot tonsillectomy' is another option available for treatment. We have not compared quinsy tonsillectomy, given that it is less commonly performed as a first‐line treatment (Qureshi 2015), and that different practical considerations guide the decision to perform it (such as patient age and operating room resources). We have also not compared non‐interventional options, such as antibiotic therapy alone. This was done in order to answer a specific clinical question of interest, as opposed to exploring all the available options for treating peritonsillar abscesses. Quality of the evidence Overall, the quality of evidence is very low for all the included outcomes (assessed using the GRADE criteria, see Table 1). The high risk of bias and flawed design of the studies reviewed is apparent when they are analysed using the Cochrane 'Risk of bias' tool (Figure 2). The descriptions of the methods used in most studies lacked the detail required to assess risk of bias completely. None of this review's predetermined outcome measures were well assessed by any of the trials. The specific criteria for determining recurrence were poorly defined in all of the studies. A number of studies included in this review had recurrence rates higher than we might normally expect from day‐to‐day practice. In one study, the needle aspiration group had a recurrence rate as high as 80% (Rafi 2007). Khan 2012 had a recurrence rate as high as 64% in the needle aspiration group. Several studies had recurrence rates over 20% in the needle aspiration group (Khan 2011; Nwe 2000; Sheikh 2012). These recurrence rates differ significantly from the recurrence rates reported in other published studies and from our experience. In the review by Powell 2012, an overall recurrence rate for peritonsillar abscess of ˜10% was estimated with either needle aspiration or incision and drainage. This raises the question of the competence of the clinician or practitioner managing the condition and the criteria for defining recurrence, as well as the question of regional differences in peritonsillar abscess presentation and management. Ideally, recurrence should be confirmed by evidence of pus at a repeat incision or needle aspiration procedure. The absence of clearly stated diagnostic criteria for peritonsillar abscess recurrence in the studies makes it impossible to determine whether the same patients presenting to another clinician would be diagnosed as suffering from a recurrence or the more common post‐inflammatory oedema associated with a recent abscess. The heterogeneity encountered in our assessment of recurrence rate, whilst only moderate (I 2 = 48%), merits further discussion. As mentioned earlier, most of the studies did not clearly define 'recurrence' and varied widely in the timing of post‐intervention recurrence assessment. Only four of the 10 studies that measured recurrence provided a definition for what constituted 'recurrence' or specified criteria for re‐intervention. Definitions of recurrence included visual evidence of swelling (Stringer 1988), persistence of pyrexia/trismus (Nwe 2000), or persistent odynophagia (Sheikh 2012). Only one study defined recurrence with reference to the review's definition of an abscess, that is as "reaccumulation of pus" (Maharaj 1991). This study did not describe how the presence of pus was determined. Therefore, 'recurrence' in the studies reviewed could be anything along a spectrum from simply some residual symptoms to true reaccumulation of pus. Additionally, the timing of post‐treatment assessment for recurrence was not stated in four of the studies and was variable amongst the remaining studies (ranging from post‐procedure day 1 to 7). The inter‐study variation in the timing of post‐treatment review and the criteria used to diagnose peritonsillar abscess recurrence raises the question of whether 'recurrence' – unless much more clearly defined – is a valid outcome metric. Inter‐study methodological variation could account for the majority of the heterogeneity in recurrence rates. However, other factors may have contributed; it would be reasonable to postulate that clinical variability also contributed to the heterogeneity, given that there was wide inter‐study variation in treatment setting and adjunctive variables. There is inherent operator variability in the delivery of the primary intervention itself. In many settings care for patients with quinsy is delivered by relatively junior doctors. We have defined the interventions as above (Types of interventions). However, differences in operator skill may partly explain differences in the outcome of the interventional studies. Variation in the timing of patient presentation for health care, co‐treatment factors such as antibiotic choice, and access and compliance with the prescribed antibiotic regimen in these studies, which are drawn from countries with divergent healthcare resources, may also account for heterogeneous outcomes. Limited methodological descriptions and our inability to gain further information from the authors make it difficult to explore these possible explanations. One of the theoretical arguments against incision and drainage is that it is a more painful procedure for patients. Given that incision and drainage often involves a diagnostic needle aspiration in addition to further procedural intervention, it is understandable that this theory exists. In our review, two studies (n = 166) did suggest that subjective pain scores were higher in patients who underwent incision and drainage compared to needle aspiration (Chi 2014; Khan 2012). One study showed comparable resolution of pain at five days post‐treatment (Younas 2015). Given that this was assessed at five days after the procedure, it is difficult to separate the pain associated with the procedure from the resolution of pain associated with improvement of the disease process. Different measurement scales made pooled analysis impractical. However, the data are of limited quality for the reasons mentioned above and, therefore, this remains unclear. Potential biases in the review process We think it is unlikely that there have been significant biases in the review process itself. Whilst it is likely that all or most relevant studies have been identified, our failure to identify useable data from each included study has been universal. A few minor post hoc changes to the methods we had planned at protocol stage (Chang 2014) are described in Differences between protocol and review. Authors' conclusions Implications for practice. Currently, only very low‐quality evidence is available to determine whether needle aspiration or incision and drainage is most effective for treating peritonsillar abscess. This suggests that incision and drainage is associated with a lower recurrence rate. The absence of high‐quality evidence results in a dilemma for practitioners and patients. Through shared and informed decision‐making, practitioners and patients must balance the potential ‐ but uncertain ‐ benefit of an incision and drainage procedure, which may be more painful, against needle aspiration, which may have of a greater chance of recurrence but be potentially less painful. Implications for research. A sufficiently powered, well‐designed, high‐quality study is required to compare the effectiveness of needle aspiration with incision and drainage for the treatment of peritonsillar abscesses. If a randomised controlled clinical trial were conducted, it would be useful to compare recurrence (in a strictly defined manner), adverse events/effects associated with the intervention, time until resumption of normal diet, complications of the disease process and symptom scores. History Protocol first published: Issue 4, 2006 Review first published: Issue 12, 2016 \| Date \| Event \| Description \| --- \| 30 June 2014 \| New citation required and major changes \| New authors. Withdrawn protocol redrafted, updated and republished. \| Open in a new tab Notes The original protocol was withdrawn from Issue 11, 2011 of the Cochrane Library onwards as the authors were unable to continue with the review. A new protocol by new authors was published in 2014 (Chang 2014). Acknowledgements We would like to acknowledge Cochrane ENT for their helpful guidance and input. This project was supported by the National Institute for Health Research, via Cochrane Infrastructure, Cochrane Programme Grant or Cochrane Incentive funding to Cochrane ENT. The views and opinions expressed therein are those of the authors and do not necessarily reflect those of the Systematic Reviews Programme, NIHR, NHS or the Department of Health. Keir J, Almeyda R, Bowyer D and Wilbourn M were authors of the original protocol (withdrawn in 2011). New authors took over the review in 2014, publishing a new protocol (Chang 2014). Appendices Appendix 1. Search strategies CENTRALOvid MEDLINEEMBASE 1 MeSH descriptor: [Peritonsillar Abscess] explode all trees 2 abscess near tonsil or abscess near peritonsil or abscess near retrotonsil or abscess near peri‐tonsil 3 suppurat near tonsil or suppurat near peritonsil or suppurat near retrotonsil or suppurat near peri‐tonsil 4 sepsis near tonsil or sepsis near peritonsil or sepsis near retrotonsil or sepsis near peri‐tonsil 5 septic near tonsil or septic near peritonsil or septic near retrotonsil or septic near peri‐tonsil 6 pus near tonsil or pus near peritonsil or pus near retrotonsil or pus near peri‐tonsil 7 infect near peritonsil or infect near retrotonsil or infect near peri‐tonsil 8 acute near peritonsil or acute near retrotonsil or acute near peri‐tonsil 9 quinsy or "interval tonsil" 10 #1 or #2 or #3 or #4 or #5 or #6 or #7 or #8 or #9 1 exp Peritonsillar Abscess/ 2 ((abscess adj5 tonsil) or (abscess adj5 peritonsil) or (abscess adj5 retrotonsil)).ab,ti. 3 ((suppurat adj5 tonsil) or (suppurat adj5 peritonsil) or (suppurat adj5 retrotonsil) or (suppurat adj5 peri‐tonsil)).ab,ti. 4 ((sepsis adj5 tonsil) or (sepsis adj5 peritonsil) or (sepsis adj5 retrotonsil) or (sepsis adj5 peri‐tonsil)).ab,ti. 5 ((septic adj5 tonsil) or (septic adj5 peritonsil) or (septic adj5 retrotonsil) or (septic adj5 peri‐tonsil)).ab,ti. 6 ((pus adj5 tonsil) or (pus adj5 peritonsil) or (pus adj5 retrotonsil) or (pus adj5 peri‐tonsil)).ab,ti. 7 ((infect adj5 peritonsil) or (infect adj5 retrotonsil) or (infect adj5 peri‐tonsil)).ab,ti. 8 ((acute adj5 peritonsil) or (acute adj5 retrotonsil) or (acute adj5 peri‐tonsil)).ab,ti. 9 (quinsy or "interval tonsil").ab,ti. 10 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 1 exp Peritonsillar Abscess/ 2 ((abscess adj5 tonsil) or (abscess adj5 peritonsil) or (abscess adj5 retrotonsil)).ab,ti. 3 ((suppurat adj5 tonsil) or (suppurat adj5 peritonsil) or (suppurat adj5 retrotonsil) or (suppurat adj5 peri‐tonsil)).ab,ti. 4 ((sepsis adj5 tonsil) or (sepsis adj5 peritonsil) or (sepsis adj5 retrotonsil) or (sepsis adj5 peri‐tonsil)).ab,ti. 5 ((septic adj5 tonsil) or (septic adj5 peritonsil) or (septic adj5 retrotonsil) or (septic adj5 peri‐tonsil)).ab,ti. 6 ((pus adj5 tonsil) or (pus adj5 peritonsil) or (pus adj5 retrotonsil) or (pus adj5 peri‐tonsil)).ab,ti. 7 ((infect adj5 peritonsil) or (infect adj5 retrotonsil) or (infect adj5 peri‐tonsil)).ab,ti. 8 ((acute adj5 peritonsil) or (acute adj5 retrotonsil) or (acute adj5 peri‐tonsil)).ab,ti. 9 (quinsy or "interval tonsil").ab,ti. 10 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 CINAHLWeb of ScienceClinicalTrials.gov S10 S1 OR S2 OR S3 OR S4 OR S5 OR S6 OR S7 OR S8 OR S9 S9 quinsy or "interval tonsil" S8 acute N5 peritonsil or acute N5 retrotonsil or acute N5 peri‐tonsil S7 infect N5 peritonsil or infect N5 retrotonsil or infect N5 peri‐tonsil S6 pus N5 tonsil or pus N5 peritonsil or pus N5 retrotonsil or pus N5 peri‐tonsil S5 TX septic N5 tonsil or septic N5 peritonsil or septic N5 retrotonsil or septic N5 peri‐tonsil S4 TX sepsis N5 tonsil or sepsis N5 peritonsil or sepsis N5 retrotonsil or sepsis N5 peri‐tonsil S3 TX suppurat N5 tonsil or suppurat N5 peritonsil or suppurati N5 retrotonsil or suppurat N5 peri‐tonsil S2 TX abscess N5 tonsil or abscess N5 peritonsil or abscess N5 retrotonsil or abscess N5 peri‐tonsil S1 (MH "Peritonsillar Abscess")#1 TOPIC: (abscess near/5 tonsil or abscess near/5 peritonsil or abscess near/5 retrotonsil or abscess near/5 peri‐tonsil) 2 TOPIC: (suppurat near/5 tonsil or suppurat near/5 peritonsil or suppurat near/5 retrotonsil or suppurat near/5 peri‐tonsil) 3 TOPIC: (sepsis near/5 tonsil or sepsis near/5 peritonsil or sepsis near/5 retrotonsil or sepsis near/5 peri‐tonsil) 4 TOPIC: (septic near/5 tonsil or septic near/5 peritonsil or septic near/5 retrotonsil or septic near/5 peri‐tonsil) 5 TOPIC: (pus near/5 tonsil or pus near/5 peritonsil or pus near/5 retrotonsil or pus near/5 peri‐tonsil) 6 TOPIC: (infect near/5 peritonsil or infect near/5 retrotonsil or infect near/5 peri‐tonsil) 7 TOPIC: (acute near/5 peritonsil or acute near/5 retrotonsil or acute near/5 peri‐tonsil) 8 TOPIC: (quinsy or "interval tonsil") 9 #8 OR #7 OR #6 OR #5 OR #4 OR #3 OR #2 OR #1((abscess OR sepsis OR septic OR pus OR infect OR acute) AND (tonsil OR peritonsil OR retrotonsil OR per‐tonsil)) OR quinsy OR "interval tonsil" Open in a new tab Data and analyses Comparison 1. Needle aspiration versus incision and drainage. \| Outcome or subgroup title \| No. of studies \| No. of participants \| Statistical method \| Effect size \| --- --- \| 1 Rate of recurrence \| 10 \| 612 \| Risk Ratio (M‐H, Random, 95% CI) \| 3.74 [1.63, 8.59] \| Open in a new tab 1.1. Analysis. Open in a new tab Comparison 1 Needle aspiration versus incision and drainage, Outcome 1 Rate of recurrence. Characteristics of studies Characteristics of included studies [ordered by study ID] Chi 2014. MethodsAllocation: randomised trial Design: parallel‐group ParticipantsNumber randomised: 110 Age range: 12 to 79 Gender: 89% males Setting: hospital, Taiwan Inclusion criteria: diagnosis of peritonsillar abscess Exclusion criteria: none stated Participant characteristics: mean age = 31.0 ± 15.0 years, days of symptoms prior to presentation = 4.7 ± 2.8 Interventions Needle aspiration versus incision and drainage Intervention group: needle aspiration n = 55 Comparator group: incision and drainage n = 55 Use of additional interventions: antibiotics, intravenous hydration Outcomes 1. Recurrence rate Length of hospital stay Pain score (Outcomes not specified as primary/secondary) Funding sources Not specified Declarations of interest Not specified NotesParticipants lost to follow‐up: none Risk of bias BiasAuthors' judgementSupport for judgement Random sequence generation (selection bias)Unclear risk Quote: "Patients were randomly divided into two groups..." Comment: specific method of randomisation not stated Allocation concealment (selection bias)Unclear risk Allocation concealment not specified Blinding of participants and personnel (performance bias) All outcomes High risk No mention of blinding made; probably not done Blinding of outcome assessment (detection bias) All outcomes High risk No mention of blinding made; probably not done Incomplete outcome data (attrition bias) All outcomes Low risk All patients presumably accounted for (admitted as inpatients); no dropouts stated Selective reporting (reporting bias)Unclear risk Insufficient information to permit a judgement of high or low risk Other bias Unclear risk Not enough data to assess other risk of bias. Incomplete methodological description (definition of recurrence/criteria for re‐intervention not described, timing of assessment for recurrence not described). Open in a new tab Khan 2011. MethodsAllocation: randomised trial, alternation Design: parallel‐group ParticipantsNumber randomised: 62 Age range: 15 to 35 Gender: 74% males Setting: hospital, Pakistan Inclusion criteria: not specifically stated Exclusion criteria: other associated illness Participant characteristics: mean age = 24.6 years Interventions Needle aspiration versus incision and drainage Intervention group: needle aspiration n = 31 Comparator group: incision and drainage n = 31 Use of additional interventions: antibiotics (injection benzyl penicillin and metronidazole), analgesics Outcomes 1. Recurrence rate Length of hospital stay (Outcomes not specified as primary/secondary) Funding sources Not specified Declarations of interest Not specified Notes Needle aspiration failures treated with incision and drainage Participants lost to follow‐up: none Risk of bias BiasAuthors' judgementSupport for judgement Random sequence generation (selection bias)Unclear risk Randomisation method not specified Allocation concealment (selection bias)Unclear risk No allocation concealment specified Blinding of participants and personnel (performance bias) All outcomes High risk No mention of blinding made; probably not done Blinding of outcome assessment (detection bias) All outcomes High risk No mention of blinding made; probably not done Incomplete outcome data (attrition bias) All outcomes Low risk All patients presumably accounted for (admitted as inpatients); no dropouts stated Selective reporting (reporting bias)Unclear risk Insufficient information to permit a judgement of high or low risk Other bias Unclear risk Not enough data to assess other risk of bias. Incomplete methodological description (definition of recurrence/criteria for re‐intervention not described, timing of assessment for recurrence not described). Open in a new tab Khan 2012. MethodsAllocation: randomised trial, alternation Design: parallel‐group ParticipantsNumber randomised: 56 Age range: 16 to 50 Gender: 71% males Setting: hospital, Pakistan Eligibility criteria: age > 15 years with peritonsillar abscess Exclusion criteria: patients with bleeding disorders, acute follicular tonsillitis Participant characteristics: mean age = 31.2 years Interventions Needle aspiration versus incision and drainage Intervention group: needle aspiration n = 28 Comparator group: incision and drainage n = 28 Use of additional interventions: antibiotics (intravenous amoxicillin/clavulanate and metronidazole), povidone‐iodine (Pyodine) mouth wash, intravenous crystalloid (if necessary) Outcomes 1. Recurrence rate Symptom score Length of hospital stay (Outcomes not specified as primary/secondary) Funding sources None stated Declarations of interest None declared Notes Alternate basis randomisation Participants lost to follow‐up: none Risk of bias BiasAuthors' judgementSupport for judgement Random sequence generation (selection bias)High risk Alternate basis randomisation Allocation concealment (selection bias)High risk Alternate basis randomisation Blinding of participants and personnel (performance bias) All outcomes High risk No mention of blinding made; probably not done Blinding of outcome assessment (detection bias) All outcomes High risk No mention of blinding made; probably not done Incomplete outcome data (attrition bias) All outcomes Low risk All patients presumably accounted for (admitted as inpatients); no dropouts stated Selective reporting (reporting bias)Unclear risk Insufficient information to permit a judgement of high or low risk Other bias Unclear risk Not enough data to assess other risk of bias. Incomplete methodological description (definition of recurrence/criteria for re‐intervention not described, timing of assessment for recurrence not described) Open in a new tab Khokhar 2015. MethodsAllocation: randomised trial Design: parallel‐group ParticipantsNumber randomised: 70 (data reported on 60) Age range: 17 to 53 Gender: 76.7% males Setting: military hospital, Pakistan Inclusion criteria: "clinical diagnosis of PTA was made on clinical features of unitonsillar erythema, swelling, odynophagia (pain on swallowing), uvular deviation towards the opposite direction and trismus", age ≥ 15 Exclusion criteria: "patients with a history of bleeding disorders or diabetes mellitus, on anticoagulant drugs or diagnosed with immunodeficiency disorders or who refused to undergo the procedure under local anesthesia"; "Patients failing to follow up were excluded from this study" Participant characteristics: mean age 32.7 Interventions Needle aspiration versus incision and drainage Intervention group: needle aspiration n = 35 Comparator group: incision and drainage n = 35 Use of additional interventions: Use of additional interventions: antibiotics and analgesia ((a) co‐amoxiclav 1.2 g intravenously 8‐hourly and metronidazole 500 mg intravenously 8‐hourly for 3 days, followed by oral co‐amoxiclav 1 g twice daily and metronidazole 400 mg 3 times a day for the next 4 days, (b) paracetamol 1 g 8‐hourly orally "for fever and analgesia") Outcomes 1. Time of resolution of odynophagia (days) Fever – time to resolution (days) Above information used to define "Recovery period" as "time taken to settle both odynophagia and fever" Recurrence Complications Funding sources Not specified Declarations of interest None declared NotesParticipants lost to follow‐up: "Patients failing to follow up were excluded from this study" Risk of bias BiasAuthors' judgementSupport for judgement Random sequence generation (selection bias)Low risk Used a "random numbers table" Allocation concealment (selection bias)Unclear risk Not certain if random numbers table was open Blinding of participants and personnel (performance bias) All outcomes High risk No mention of blinding; probably not done Blinding of outcome assessment (detection bias) All outcomes High risk No mention of blinding; probably not done Incomplete outcome data (attrition bias) All outcomes Unclear risk 1. "Patients, who had cultured organisms resistant to [the] antibiotics [named above] were excluded from the study" "Patients failing to follow up were excluded from this study" Selective reporting (reporting bias)Unclear risk Insufficient information to permit a judgement of high or low risk Other bias Unclear risk Not enough data to assess other risk of bias Open in a new tab Maharaj 1991. MethodsAllocation: randomised, alternation Design: parallel‐group ParticipantsNumber randomised: 60 Age range: "under 14" to "over 40" Gender: not specified Setting: hospital, South Africa Inclusion criteria: positive needle aspiration Exclusion criteria: negative needle aspiration Participant characteristics: mean age not provided, "all patients presented with some degree of odynophagia and drooling of saliva", 47% had trismus and 27% had pyrexia Interventions Needle aspiration versus incision and drainage Intervention group: needle aspiration n = 30 Comparator group: incision and drainage n = 30 Use of additional interventions: antibiotics (penicillin), analgesics, mouthwash (unspecified) Outcomes 1. Recurrence rate (Outcomes not specified as primary/secondary) Funding sources Not specified Declarations of interest Not specified NotesParticipants lost to follow‐up: 11 (18%) at day 1, 22 (37%) at day 7 Risk of bias BiasAuthors' judgementSupport for judgement Random sequence generation (selection bias)High risk Randomised on alternate basis Allocation concealment (selection bias)High risk Randomised on alternate basis Blinding of participants and personnel (performance bias) All outcomes High risk No mention of blinding made; probably not done Blinding of outcome assessment (detection bias) All outcomes High risk No mention of blinding made; probably not done Incomplete outcome data (attrition bias) All outcomes High risk 82% of patients followed up on day 1 post‐treatment and 63% followed up on day 7 Selective reporting (reporting bias)Unclear risk Insufficient information to permit a judgement of high or low risk Other bias Low risk No other sources of bias identified Open in a new tab Nwe 2000. MethodsAllocation: randomised trial Design: parallel‐group ParticipantsNumber randomised: 75 (only 50 patients were in groups compared in this review) Age range: 15 to 43 Gender: 33% males Setting: South Africa Inclusion criteria: not described Exclusion criteria: none stated Participant characteristics: "unilateral swelling of the tonsil and soft palate, and medial displacement of the uvula", "all patients were pyrexial" Interventions Needle aspiration versus incision and drainage versus intravenous antibiotics alone Intervention group: intravenous antibiotics n = 25 Comparator group 1: needle aspiration n = 25 Comparator group 2: incision and drainage n = 25 Use of additional interventions: single dose of intravenous antibiotics Outcomes 1. Distance between upper and lower incisor teeth (degree of trismus) Body temperature Ability to drink water Microbiology cultures Treatment failures (recurrence) Funding sources None stated Declarations of interest None declared NotesParticipants lost to follow‐up: unclear Risk of bias BiasAuthors' judgementSupport for judgement Random sequence generation (selection bias)Unclear risk Randomisation method not specified Allocation concealment (selection bias)Unclear risk No allocation concealment specified Blinding of participants and personnel (performance bias) All outcomes High risk No mention of blinding made; probably not done Blinding of outcome assessment (detection bias) All outcomes High risk No mention of blinding made; probably not done Incomplete outcome data (attrition bias) All outcomes Unclear risk Attrition not clear, dropouts not mentioned Selective reporting (reporting bias)Unclear risk Insufficient information to permit a judgement of high or low risk Other bias Unclear risk Not enough data to assess other risk of bias Open in a new tab Rafi 2007. MethodsAllocation: randomised trial Design: parallel‐group ParticipantsNumber randomised: 50 Age range: 22 to 43 Gender: 94% males Setting: hospital, Pakistan Inclusion criteria: "peritonsillar abscess" Exclusion criteria: none stated Participant characteristics: "all the patients were otherwise healthy and young with no immune compromising disease", no patients with previous peritonsillar abscess Interventions Needle aspiration versus incision and drainage Intervention group: needle aspiration n = 25 Comparator group: incision and drainage n = 25 Use of additional interventions: antibiotics (lincomycin) Outcomes 1. Recurrence rate Length of hospital stay "Period of recovery" (Outcomes not specified as primary/secondary) Funding sources Not specified Declarations of interest Not specified NotesParticipants lost to follow‐up: none Risk of bias BiasAuthors' judgementSupport for judgement Random sequence generation (selection bias)Unclear risk Randomisation method not specified Allocation concealment (selection bias)Unclear risk No allocation concealment specified Blinding of participants and personnel (performance bias) All outcomes High risk No mention of blinding made; probably not done Blinding of outcome assessment (detection bias) All outcomes High risk No mention of blinding made; probably not done Incomplete outcome data (attrition bias) All outcomes Low risk All patients presumably accounted for (admitted as inpatients); no dropouts stated Selective reporting (reporting bias)Unclear risk Insufficient information to permit a judgement of high or low risk Other bias Unclear risk Not enough data to assess other risk of bias. Incomplete methodological description (definition of recurrence/criteria for re‐intervention not described, timing of assessment for recurrence not described) Open in a new tab Sheikh 2012. MethodsAllocation: randomised trial Design: parallel‐group ParticipantsNumber randomised: 50 Age range: 18 to 51 Gender: 62% males Setting: hospital, Pakistan Inclusion criteria: "presented with peritonsillar abscess" Exclusion criteria: diabetics, less than 18 years old Participant characteristics: mean age = 32.7 years Interventions Needle aspiration versus incision and drainage Intervention group: needle aspiration n = 25 Comparator group: incision and drainage n = 25 Use of additional interventions: antibiotics, chlorhexidine mouth wash OutcomesPrimary outcome: recurrence rate Secondary outcomes: none Funding sources Not specified Declarations of interest Not specified Notes Needle aspiration failures treated with incision and drainage Participants lost to follow‐up: none Risk of bias BiasAuthors' judgementSupport for judgement Random sequence generation (selection bias)Unclear risk Randomisation method not specified Allocation concealment (selection bias)Unclear risk No allocation concealment specified Blinding of participants and personnel (performance bias) All outcomes High risk No mention of blinding made; probably not done Blinding of outcome assessment (detection bias) All outcomes High risk No mention of blinding made; probably not done Incomplete outcome data (attrition bias) All outcomes Low risk All patients presumably accounted for (admitted as inpatients); no dropouts stated Selective reporting (reporting bias)Unclear risk Insufficient information to permit a judgement of high or low risk Other bias Unclear risk Not enough data to assess other risk of bias. Subjective definition of recurrence/criteria for re‐intervention Open in a new tab Spires 1987. MethodsAllocation: randomised trial Design: parallel‐group ParticipantsNumber randomised: 62 Age range: 12 to 53 Gender: "2:1 male predilection" Setting: hospital, USA Inclusion criteria: not specified Exclusion criteria: not specified Participant characteristics: median age = 24; 1 patient with bilateral abscesses Interventions Needle aspiration versus incision and drainage Intervention group: needle aspiration n = 41 Comparator group: incision and drainage n = 21 Use of additional interventions: analgesics, antibiotics (penicillin V or cephalexin or erythromycin) Outcomes 1. Recurrence rate Time to resumption of normal diet (Outcomes not specified as primary/secondary) Funding sources Not specified Declarations of interest Not specified Notes Only partial randomisation Participants lost to follow‐up: unclear Risk of bias BiasAuthors' judgementSupport for judgement Random sequence generation (selection bias)High risk First 15 participants treated with needle aspiration, then patients were subsequently randomised Allocation concealment (selection bias)High risk Randomised by hospital number Blinding of participants and personnel (performance bias) All outcomes High risk No mention of blinding made; probably not done Blinding of outcome assessment (detection bias) All outcomes High risk No mention of blinding made; probably not done Incomplete outcome data (attrition bias) All outcomes Unclear risk Attrition not clear, dropouts not mentioned Selective reporting (reporting bias)Unclear risk Insufficient information to permit a judgement of high or low risk Other bias Unclear risk Not enough data to assess other risk of bias. Incomplete methodological description (definition of recurrence/criteria for re‐intervention not described) Open in a new tab Stringer 1988. MethodsAllocation: randomised trial Design: parallel‐group ParticipantsNumber randomised: 52 Age range: 13 to 60 Gender: 60% males Setting: hospital, USA Inclusion criteria: positive needle aspiration Exclusion criteria: negative needle aspiration Participant characteristics: mean age = 27 years average duration of symptoms prior to presentation = 5.3 days Interventions Needle aspiration versus incision and drainage Intervention group: needle aspiration n = 24 Comparator group: incision and drainage n = 28 Use of additional interventions: antibiotics (initial dose of intramuscular penicillin G followed by intramuscular penicillin G or oral penicillin V for 10 days), erythromycin/cephalosporin/clindamycin if allergic Outcomes 1. Recurrence rate Symptom score (Outcomes not specified as primary/secondary) Funding sources Not specified Declarations of interest Not specified Notes Authors contacted and further information/clarification obtained Participants lost to follow‐up: unclear Risk of bias BiasAuthors' judgementSupport for judgement Random sequence generation (selection bias)Low risk Random numbers table Allocation concealment (selection bias)High risk No allocation concealment performed Blinding of participants and personnel (performance bias) All outcomes High risk Blinding not performed Blinding of outcome assessment (detection bias) All outcomes High risk No mention of blinding made; probably not done Incomplete outcome data (attrition bias) All outcomes Low risk No dropouts mentioned Selective reporting (reporting bias)Low risk All measured outcomes stated to have been reported Other bias Unclear risk Not enough data to assess other risk of bias. Subjective definition of recurrence/criteria for re‐intervention. Open in a new tab Younas 2015. MethodsAllocation: randomised trial Design: parallel‐group ParticipantsNumber randomised: 62 (discrepant total = 64 in results) Age range: 8 to 57 Gender: 61.3% male Setting: hospital, Pakistan Eligibility criteria: "clinically diagnosed for peritonsillar abscess" Exclusion criteria: none stated Participant characteristics: average duration of symptoms prior to presentation = 6 days; 84% treated with antibiotics prior to presentation InterventionsIntervention group: needle aspiration n = 32 Intervention group: incision and drainage n = 32 Use of additional interventions: not stated Outcomes 1. "Success" – not defined Length of time to return to semisolid/solid food (days) Proportion with "no pain by 05 days" (%) Length of hospital stay (days) Funding sources Not specified Declarations of interest Not specified NotesParticipants lost to follow‐up: unclear Risk of bias BiasAuthors' judgementSupport for judgement Random sequence generation (selection bias)Unclear risk Randomisation method not specified Allocation concealment (selection bias)Unclear risk No allocation concealment specified Blinding of participants and personnel (performance bias) All outcomes High risk No mention of blinding; probably not done Blinding of outcome assessment (detection bias) All outcomes High risk No mention of blinding; probably not done Incomplete outcome data (attrition bias) All outcomes Unclear risk Attrition not clear; dropouts not mentioned; numbers in methods and results do not match Selective reporting (reporting bias)Unclear risk Insufficient information to permit a judgement of high or low risk Other bias Unclear risk Not enough data to assess other risks of bias Open in a new tab Characteristics of excluded studies [ordered by study ID] \| Study \| Reason for exclusion \| --- \| \| Fry 1987 \| Commentary \| \| Herzon 1995 \| Review paper \| \| Khayr 2005 \| Review paper \| \| Kulkarni 2013 \| Participants were not randomised: "All patients were divided in two groups according to surgical procedures carried out". While the study states that this was a prospective study, it seems that outcomes were assessed retrospectively. \| \| Stringer 1986 \| Considered a duplicate publication of Stringer 1988 \| \| Wolf 1994 \| Retrospective study \| Open in a new tab Characteristics of studies awaiting assessment [ordered by study ID] Tyagi 2011. Methods— Participants— Interventions— Outcomes— Notes Study not available at this time. This study was inaccessible by all attempted means, including contact with the journal editor Open in a new tab Differences between protocol and review We widened the age criteria for participants in included studies based on post hoc review. A number of studies included older child (generally > 8 years old), adolescent and adult participants. Stratified data by age group (adults versus adolescents) were not available in any of the studies. Given that adolescents and older children with peritonsillar abscess are often managed similarly to adults, we thought that the review would also be applicable to this age group. Younger patients typically require a general anaesthetic for management and, therefore, have different practical considerations for intervention. We therefore made a decision to include trials with older children (age > 8), adolescents and adults. We added the methods for the creation of a 'Summary of findings' table and GRADE assessment to the review after the protocol was published. We also added a standard statement to clarify the role of outcomes in the review (Types of outcome measures). Contributions of authors Brent A Chang: literature searching, study selection, data collection, data management, 'Risk of bias' assessment, data analysis, data interpretation, drafting the protocol and review. Andrew Thamboo: literature searching, study selection, data collection, data management, 'Risk of bias' assessment, data analysis, data interpretation, drafting the protocol and review. Chris Diamond: protocol design, content expertise, revising the protocol and review, critical appraisal and study selection. Martin J Burton: content expertise, data interpretation, revising the review. Desmond A Nunez: protocol design, 'Risk of bias' assessment, content expertise, revising the protocol and review. Sources of support Internal sources No sources of support supplied External sources National Institute for Health Research, UK. Infrastructure funding for Cochrane ENT Declarations of interest Brent A Chang: none to declare. Andrew Thamboo: none to declare. Chris Diamond: none to declare. Martin J Burton: Professor Martin Burton is joint Co‐ordinating Editor of Cochrane ENT, but had no role in the editorial sign‐off this review, which was carried out by Cochrane ENT's second Co‐ordinating Editor. Desmond A Nunez: none to declare. New References References to studies included in this review Chi 2014 {published data only} Chi TH, Yuan CH, Tsao YH. Comparison of needle aspiration with incision and drainage for the treatment of peritonsillar abscess. WIMJ Open 2014;1(1):11‐3. [Google Scholar] Khan 2011 {published data only} Khan Q, Hamza A, Haleem G, Wahid FI. Comparative study of management of peritonsillar abscess by needle aspiration versus incision and drainage. Pakistan Journal of Medical & Health Sciences 2011;5(4):699‐701. [Google Scholar] Khan 2012 {published data only} Khan MI, Iqbal K, Muhammad. Peritonsillar abscess: comparison of outcome of incision and drainage versus needle aspiration. Gomal Journal of Medical Sciences 2012;10(2):205‐8. [Google Scholar] Khokhar 2015 {published data only} Khokhar MR, Raza SN, Ayaz SB. Comparison of efficacy of needle aspiration versus incision and drainage in management of peritonsillar abscess. Pakistan Armed Forces Medical Journal 2015;65:172‐7. [Google Scholar] Maharaj 1991 {published data only} Maharaj D, Rajah V, Hemsley S. Management of peritonsillar abscess. Journal of Laryngology and Otology 1991;105:743‐5. [DOI] [PubMed] [Google Scholar] Nwe 2000 {published data only} Nwe TT, Singh B. Management of pain in peritonsillar abscess. Journal of Laryngology and Otology 2000;114:765‐7. [DOI] [PubMed] [Google Scholar] Rafi 2007 {published data only} Rafi T, Farooq MU, Sheikh HU. Comparative study of management of peritonsillar abscess by needle aspiration versus incision and drainage. Journal of Surgery Pakistan 2007;12(3):126‐8. [Google Scholar] Sheikh 2012 {published data only} Sheikh ZA, Najam A, Ahmed A. Needle aspiration an equal to incision and drainage in management of peritonsillar abscess. Pakistan Armed Forces Medical Journal 2012;62(2):online. [Google Scholar] Spires 1987 {published data only} Spires JR, Owens JJ, Woodson GE, Miller RH. Treatment of peritonsillar abscess. Archives of Otolaryngology ‐ Head & Neck Surgery 1987;113:984‐6. [DOI] [PubMed] [Google Scholar] Stringer 1988 {published data only} Stringer SP, Schaefer SD, Close LG. A randomized trial for outpatient management of peritonsillar abscess. Archives of Otolaryngology ‐ Head & Neck Surgery 1988;114(3):296‐8. [DOI] [PubMed] [Google Scholar] Younas 2015 {published data only} Younas M. Satisfaction of patients with peritonsillar abscess with permucosal needle aspiration versus incision & drainage. Medical Forum Monthly 2015;26(3):37‐9. [Google Scholar] References to studies excluded from this review Fry 1987 {published data only} Fry TL. A randomized trial for outpatient management of peritonsillar abscess. Archives of Otolaryngology ‐‐ Head and Neck Surgery 1987;113(1):19. [DOI] [PubMed] [Google Scholar] Herzon 1995 {published data only} Herzon FS. Peritonsillar abscess: incidence, current management practices, and a proposal for treatment guidelines. Laryngoscope 1995;105:1‐17. [DOI] [PubMed] [Google Scholar] Khayr 2005 {published data only} Khayr W, Taepke J. Management of peritonsillar abscess: needle aspiration versus incision and drainage versus tonsillectomy. American Journal of Therapeutics 2005;12:344‐50. [PubMed] [Google Scholar] Kulkarni 2013 {published data only} Kulkarni V, Patel T. Management of peritonsillar abscess: comparative prospective study of needle aspiration and incision & drainage in central Indian population. Asian Journal of Biomedical and Pharmaceutical Sciences 2013;3(18):61‐3. [Google Scholar] Stringer 1986 {published data only} Stringer SP, Schaefer SD, Close LG. A randomized trial for outpatient management of peritonsillar abscess. Otolaryngology ‐ Head and Neck Surgery 1986;114(3):296‐8. [DOI] [PubMed] [Google Scholar] Wolf 1994 {published data only} Wolf M, Even‐Chen I, Kronenberg J. Peritonsillar abscess: repeated needle aspiration versus incision and drainage. Annals of Otology, Rhinology & Laryngology 1994;103(7):554‐7. [DOI] [PubMed] [Google Scholar] References to studies awaiting assessment Tyagi 2011 {published data only} Tyagi V, Kaushal A, Garg D, De S, Nagpure P. Treatment of peritonsillar abscess: a prospective study of aspiration versus incision and drainage. Calcutta Medical Journal 2011;9(3):e3. [Google Scholar] Additional references Hall 1990 Hall SF. Peritonsillar abscess: the treatment options. Journal of Otolaryngology 1990;19:226‐9. [PubMed] [Google Scholar] Handbook 2011 Higgins JPT, Green S (editors). Cochrane Handbook for Systematic Reviews of Interventions Version 5.1.0 [updated March 2011]. The Cochrane Collaboration, 2011. Available from www.cochrane‐handbook.org. Johnson 2003 Johnson RF, Stewart MG, Wright CC. An evidence based review of the treatment of peritonsillar abscess. Otolaryngology ‐ Head and Neck Surgery 2003;128(3):332‐43. [DOI] [PubMed] [Google Scholar] Johnson 2005 Johnson RF, Stewart MG. The contemporary approach to diagnosis and management of peritonsillar abscess. Current Opinion in Otolaryngology & Head and Neck Surgery 2005;13:157‐60. [DOI] [PubMed] [Google Scholar] Mehanna 2002 Mehanna HM, Al‐Bahnasawi L, White A. National audit of the management of peritonsillar abscess. Postgraduate Medical Journal 2002;78(923):545‐8. [DOI] [PMC free article] [PubMed] [Google Scholar] Ong 2004 Ong YK, Goh YH, Lee YL. Peritonsillar infections: local experience. Singapore Medical Journal 2004;45(3):105‐9. [PubMed] [Google Scholar] Powell 2012 Powell J, Wilson JA. An evidence‐based review of peritonsillar abscess. Clinical Otolaryngology 2012;37(2):136‐45. [PUBMED: 22321140] [DOI] [PubMed] [Google Scholar] Qureshi 2015 Qureshi H, Ference E, Novis S, Pritchett CV, Smith SS, Schroeder JW. Trends in the management of pediatric peritonsillar abscess infections in the U.S., 2000‐2009. International Journal of Pediatric Otorhinolaryngology 2015;79(4):527‐31. [DOI] [PubMed] [Google Scholar] RevMan 2014 [Computer program] The Nordic Cochrane Centre, The Cochrane Collaboration. Review Manager (RevMan). Version 5.3. Copenhagen: The Nordic Cochrane Centre, The Cochrane Collaboration, 2014. References to other published versions of this review Chang 2014 Chang BA, Thamboo A, Diamond C, Nunez DA. Needle aspiration versus incision and drainage for the treatment of peritonsillar abscess. Cochrane Database of Systematic Reviews 2014, Issue 7. [DOI: 10.1002/14651858.CD006287.pub3] [DOI] [PMC free article] [PubMed] [Google Scholar] Articles from The Cochrane Database of Systematic Reviews are provided here courtesy of Wiley ACTIONS View on publisher site PDF (419.9 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Plain language summary Summary of findings Background Objectives Methods Results Discussion Authors' conclusions History Notes Acknowledgements Appendices Data and analyses Characteristics of studies Differences between protocol and review Contributions of authors Sources of support Declarations of interest References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
16517	https://ictgames.com/mobilePage/decimalDemonstrator/	Decimal Demonstrator Discover more Multiplication games Maths games online buy educational kids games Best video game consoles statistics learning tools for kids Maths Decimal Demonstrator An interactive place value and decimals display. Aimed at children 8yrs and older. Learning Objectives: Y3: To count up and down in tenths; recognise that tenths arise from dividing an object into 10 equal parts and in dividing one-digit numbers or quantities by 10 Pupils connect tenths to place value, decimal measures and to division by 10. Y4: To count up and down in hundredths; recognise that hundredths arise when dividing an object by one hundred and dividing tenths by ten. Pupils should connect hundredths to tenths and place value and decimal measure Ways to use it: Partition: Press on each card to partition. What is the value of this digit? Hide & Count: Hide the numbers, the teacher adds tenths, hundreths etc. When the teacher stops ask the children to write the number on their drywipe boards. Count Backwards: Start from 1 and count backwards in tenths or hundredths. Count Forwards: Start from 99.90 and count forwards in hundredths.Build on the learning with these games: Discover more Education counting and ordering game purchase Counting games Teaching tools Kids learning games hire UK teacher for tutoring buy shape recognition games buy educational kids games This game is © copyright 2023, J.Barrett, ictgames.com All Rights Reserved. Made using the and libraries.
16518	https://artofproblemsolving.com/wiki/index.php/Set_of_Complex_Numbers?srsltid=AfmBOoqyn1Y3Mi2AoEA9xExOby_mPgpvpOhjFwoOwvqDyytI9sJKxNxI	Art of Problem Solving Set of Complex Numbers - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Set of Complex Numbers Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Set of Complex Numbers The set of complex numbers is a set of all numbers ever existed, from pi to 1. Contents 1 Complex Numbers 1.1 Real Numbers 1.2 Imaginary Numbers 2 Set of all Real Numbers 2.1 Irrational Numbers 2.2 Rational Numbers 3 Set of Rational Numbers 3.1 Integers 3.2 Natural Numbers Complex Numbers Complex numbers are numbers which have a real part and an imaginary part, usually written in the from a + b. Real Numbers Real numbers is the set of all numbers which aren't imaginary, eg. don't have the b counterpart and can be squared into some nonnegative number. Imaginary Numbers Imaginary numbers is the set of numbers which do have the b counterpart. They will always be squared into a negative number. Set of all Real Numbers The set of all real numbers can be classified into two major sub sets, the set of irrational and rational numbers. Irrational Numbers Irrational numbers are all numbers which cannot be simplified into the from a/b where b isn't 0. Rational Numbers Rational numbers are all numbers which can be simplified into a/b. The rational numbers can be further split into to more subsets, subsubsets, and so on. Set of Rational Numbers Integers Integers are all numbers which aren't fractions but can be simplified into a/b. For example, 2 is an integer and 2/3 isn't. Usually integers are numbers we use for our daily counting. They are both negative and nonnegative. Natural Numbers Natural numbers are all positive integers. Examples are 2 and 3. 0 is nonnegative, but isn't positive so is therefore excluded. Natural numbers are a subset of integers, which are a subset of rational numbers, which combine with irrational numbers to make real numbers, which combine with imaginary numbers to make the set of all complex numbers. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16519	https://math.stackexchange.com/questions/2247070/binomial-expansion-where-is-the-mistake	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Binomial expansion, where is the mistake? Ask Question Asked Modified 8 years, 5 months ago Viewed 165 times 0 $\begingroup$ I attacked the Question and the marking scheme Question 3. Question Find the term independent of $x$ in the series expansion of $$\left(4x^3+\frac1{2x}\right)^8$$ First I expanded inside the bracked, then took the 2x out and got: $$\dfrac{1}{256x^8}(4x^4+1)^8$$ Now from here we can see that we get $x^8$ when $(4x^4)$ will be on the power of 2, which is at $8C6$ Now that will be: $$8C6(\dfrac{1}{256x^8})(4x^4)^2$$ which is $$2816x^8(\dfrac{1}{256x^8})$$ which will be $$\dfrac{7}{4}$$ but the marking scheme shows just 7. Where did I make a mistake? binomial-coefficients Share edited Apr 22, 2017 at 21:14 John Doe 14.8k11 gold badge3030 silver badges5454 bronze badges asked Apr 22, 2017 at 20:34 JohnFireJohnFire 42944 silver badges1212 bronze badges $\endgroup$ 5 1 $\begingroup$ Those are both marking schemes aren't they? $\endgroup$ John Doe – John Doe 2017-04-22 20:37:25 +00:00 Commented Apr 22, 2017 at 20:37 $\begingroup$ @JohnDoe My bad, edited $\endgroup$ JohnFire – JohnFire 2017-04-22 20:38:48 +00:00 Commented Apr 22, 2017 at 20:38 2 $\begingroup$ Also, in future, could you please type up the question here, rather than making people go find it in a document? :) $\endgroup$ John Doe – John Doe 2017-04-22 20:39:33 +00:00 Commented Apr 22, 2017 at 20:39 1 $\begingroup$ Your mistake is, when you factored out the $2x$, the $4x^3$ becomes $8x^4$ not $4x^4$. $\endgroup$ John Doe – John Doe 2017-04-22 20:43:17 +00:00 Commented Apr 22, 2017 at 20:43 $\begingroup$ @JohnDoe Yes, sorry. Ahh, I even wrote it as $$8x^4$$ for the first time on paper, then changed it to 4... thank you $\endgroup$ JohnFire – JohnFire 2017-04-22 20:44:39 +00:00 Commented Apr 22, 2017 at 20:44 Add a comment \| 2 Answers 2 Reset to default 2 $\begingroup$ When you took $2x$ out, it would be $$\frac{1}{256x^8}(8x^4+1)^8$$ instead of $$\dfrac{1}{256x^8}(4x^4+1)^8$$ Now, your answer will be multiplied by a factor of$2^2$ which will finally give the answer to be $7$. Share edited Apr 23, 2017 at 3:04 answered Apr 22, 2017 at 20:43 Jaideep KhareJaideep Khare 19.6k44 gold badges4444 silver badges8080 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ The binomial expansion will have a constant term when the powers of $x$ $$ \left( 4x^3\right)^{p} \left( \frac{1}{2x} \right)^{q} $$ total to $0$. That is $$ x^{3p-q} = x^{0}. $$ Therefore $$ q = 3p. \tag{1} $$ Each term in the binomial expansion has order 8. Using Pascal's triangle $$ \left( 4x^{3} \right)^{8} + 8 \left( 4x^{3} \right)^{7}\left( \frac{1}{2x} \right)^{1} + \color{blue}{28 \left( 4x^{3} \right)^{6} \left( \frac{1}{2x} \right)^{2}} + \dots $$ In other words, $$ p + q = 8 \tag{2} $$ Which values of $p$ and $q$ satisfy (1) and (2)? $$ p =2, \quad q = 6. $$ This is $\color{blue}{blue}$ term in the expansion: $$ \boxed{ \color{blue}{28 \left( 4x^{3} \right)^{6} \left( \frac{1}{2x} \right)^{2}} = 28 \frac{16}{64} = 7 } $$ Pascal's triangle Share edited Apr 22, 2017 at 22:27 answered Apr 22, 2017 at 21:00 dantopadantopa 10.8k1010 gold badges4646 silver badges8585 bronze badges $\endgroup$ 1 1 $\begingroup$ $$28 \frac{16}{64} =7 \neq \frac{1}{7}$$ $\endgroup$ Jaideep Khare – Jaideep Khare 2017-04-22 22:23:44 +00:00 Commented Apr 22, 2017 at 22:23 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions binomial-coefficients See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related Squared binomial coefficient 0 Is this just a version of the binomial theorem? 2 Binomial Theorem Question (Expansion of Three Terms) 1 Proving an inequality that looks related to the Binomial series, 0 A sum related to binomial theorem 1 Find the sum of the series . Problem related to series of binomial coefficients 1 Using binomial theorem to express $ \frac{1}{\sqrt{5}} \ [((\sqrt{5}/2)+(1/2))^{10} - (-(\sqrt{5}/2)+(1/2))^{10}]$ as a single finite series Connection Between Derivations of Finite and Infinite Binomial Expansion Hot Network Questions How long would it take for me to get all the items in Bongo Cat? Is direct sum of finite spectra cancellative? 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16520	https://math.stackexchange.com/questions/3477137/using-gap-method-while-selecting	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Using GAP method while selecting Ask Question Asked Modified 5 years, 9 months ago Viewed 438 times 0 $\begingroup$ I am having some trouble in questions like selecting $r$ people from $n$ people such that no two people are together. I know that we have to use the GAP method but I don't understand why. I know how to use it while arranging but I don't understand how to use it when we are selecting people. Eg. We have to select 2 letters from $CHILDREN$ such that they are not together. When we use gap method we have to place two objects in 7 places. But why? $\$ x $\$ x $\$ x $\$ x $\$ x $\$ x $\_$ Another doubt that I have is that the two letters that we could place could also be $CH$ like shown: $C$ x $H$ x $\$ x $\$ x $\$ x $\$ x $\_$ But in the original word, they are not together. This is another place where using this method seems confusing and wrong to me... combinatorics permutations Share edited Dec 15, 2019 at 11:40 HSBHSB asked Dec 15, 2019 at 11:29 HSBHSB 1333 bronze badges $\endgroup$ 3 $\begingroup$ Here is how to use GAP - well, I wish that were what you need. $\endgroup$ Dietrich Burde – Dietrich Burde 2019-12-15 12:07:50 +00:00 Commented Dec 15, 2019 at 12:07 $\begingroup$ Here GAP is the gap method... Not what you are talking about... $\endgroup$ HSB – HSB 2019-12-15 12:08:41 +00:00 Commented Dec 15, 2019 at 12:08 $\begingroup$ Oh sorry, i misread your comment $\endgroup$ HSB – HSB 2019-12-15 12:12:14 +00:00 Commented Dec 15, 2019 at 12:12 Add a comment \| 1 Answer 1 Reset to default 0 $\begingroup$ Let's use the gap method for selecting two letters from the word CHILDREN so that the selected letters are not adjacent. For now, ignore the actual letters. First, we will arrange six blue and two green balls so that the green balls are not adjacent. Place six blue balls in a row. This creates seven spaces, five between successive blue balls and two at the ends of the row. To ensure that no two green balls are adjacent, we choose two of these spaces in which to insert a green ball. For instance, if we choose the second and fifth spaces above, we obtain the arrangement shown below. Finally, we label the balls from left to right with the letters of the word CHILDREN. The letters on the green balls are the desired selections of nonadjacent letters from the word CHILDREN. For the example above, we obtain so the nonadjacent letters we selected were H and R. Since there are $$\binom{7}{2} = 21$$ ways to choose where the green balls will be placed, there are $21$ ways to select two letters from the word CHILDREN that are not adjacent. Notice that when we applied the gap method, we ignored the actual letters until after we made the selections. That said, you can verify that the answer is correct by listing all the selections of two nonadjacent letters of the word CHILDREN. Share answered Dec 15, 2019 at 12:40 N. F. TaussigN. F. Taussig 79.3k1414 gold badges6262 silver badges7777 bronze badges $\endgroup$ 2 $\begingroup$ I got the solution... I was just wondering where you got those images from... or if you made them, then using what software? $\endgroup$ HSB – HSB 2019-12-15 14:12:42 +00:00 Commented Dec 15, 2019 at 14:12 $\begingroup$ I made them using the LaTeX graphics program tikz. $\endgroup$ N. F. Taussig – N. F. Taussig 2019-12-15 14:56:04 +00:00 Commented Dec 15, 2019 at 14:56 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics permutations See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked -1 Choosing non-adjacent chairs Related 1 Gap and string method in permutations and combinations Total no. of words formed when similar letters are involved Combinatorics/Probability question about selecting people 0 A Problem on circular permutation. 1 Reverse gap method in permutations and combinations 3 select sequence without "ABCD" 1 Selecting 3 objects from n objects placed circularly Number of ways in which 5 girls and 5 boys can be arranged in a line such that only 4 girls stand adjacent to one another Hot Network Questions Is encrypting the login keyring necessary if you have full disk encryption? How can I get Remote Desktop (RD) to scale properly AND set maximum windowed size? Two calendar months on the same page Matthew 24:5 Many will come in my name! 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16521	https://www.youtube.com/watch?v=l-Rr7dZB7Ow	GeoGebra Tutorial: How to Graph Piecewise Functions GeoGebraist 1100 subscribers 165 likes Description 21045 views Posted: 4 Apr 2020 In this GeoGebra tutorial, we are going to learn how to graph piecewise functions. In doing this activity, we learn the following: 1.) graph functions with specific domains 2.) plot points using their coordinates 3.) change the color of points and graphs Visit our Facebook page: 14 comments Transcript: What's up guys. The GeoGebraist here. Welcome to another GeoGebra tutorial. In this video, we will be learning how to graph piecewise functions using GeoGebra. We will graph p(x) = -2x when x is less than or equal to -1, x^3 - 2x + 2 when x is between -1 and 2 excluding both ends, and 6 when x is between 2 and 5 excluding 2 and including 5. So there are several ways to do this, but I think what I'm going to teach you is the simplest: we will graph each function separately specifying their domains. For the first function, we have -2x, so we type f(x) = -2x and then a comma, and then we specify the domain inside the parentheses. So x is less than and then equal to -1. When you type less than and then equal GeoGebra automatically converts the symbols to this symbol. And then we press the Enter key. For the second function, we have g(x) = x^3 - 2x +2 g(x) = x^3 - 2x +2 then comma. For the domain we have between negative 1 and 2 and then we press the Enter key. And for the last function, we have h(x) = 6, and then the domain is between 2 and 5. So, x is less than or equal to 5 but x is greater than 2. So we just press Enter key. Next we don't want the labels of the function so we hide the labels and at the same time we change all the color [of the graphs] to black. To do that we click one of the graphs and then change the color to black and then we want the labels hidden so we click one of the functions and then - if you didn't see it - go to the Style bar and then [click on] the Gear button and then click on Show Label so you have to uncheck this checkbox. We do that to all functions. Now what about the points? We know that for -2x, for this graph, x is less than or equal to - 1, meaning -1 is included. We can just emphasize a a point here by plotting a point at (-1,2). We just type (-1,2) and then we'll have another point here that's the white point, the point with a white interior, that's (-1,3), right? Since for f(x) = -2x, x = -1 is included, we can just change the color of this point to black and then remove the label. However, for this point, it's not included, therefore we change the color to white. We can go to the Gear button and then choose white. And for this point here at (2,6) is that included or not? Well it's not included at x = 2 the point is undefined so we click on C and then then change color to white. And lastly, we have at (5,6) here the the point is defined so we can just change the color to black and hide the label. Okay so that's it. Our function p(x) = -2x when x is less than or equal to -1 x^3 - 2x + 2 between -1 and 2, and 6 between 2 and 5. So if you have another trick, if you have an easier way to do this, maybe you can share your method in the comments below.
16522	https://studywell.com/trigonometry/reciprocal-trig-functions/	Skip to content Reciprocal Trigonometric functions When we take 1 divided by each of the original trigonometric functions we obtain the reciprocal trigonometric functions: An easy way to remember which one is which is to look at the 3rd letter in each reciprocal function. For example, the third letter of cosec is s and so this is one over sin. We can solve equations involving reciprocal trigonometric functions in the same way we solve trigonometric equations. See Example 1. You might also notice that . These reciprocal functions can be evaluated for angles given in degrees and radians. For example, or . However, like the graph of tan(x), all of the reciprocal trigonometric functions have periodic asymptotes. We cannot evaluate them when the denominator is 0. See graphs below. Graphs of the Reciprocal Trigonometric Functions Here are the graphs of the , and respectively. As you can see the angles are given in radians: Consider the graph of , for example. It is defined by and so has asymptotes whenever . As you can see from the graph, this occurs for multiples of i.e, when . As tends towards these values, the curve grows infinitely positive if is positive and infinitely negative if is negative. This asymptotic behaviour can be seen in the other two curves as well. Now consider the graph of . This graph has periodic maxima and minima at multiples of . For example, there is a maximum at since . Notice that takes values between and and so will take values less than or greater than . This follows from the fact that dividing by a number less than 1 gives a result greater than 1, etc. Hence, the domain and range for are: DOMAIN of RANGE of where we note that we cannot evaluate nor at odd multiples of . See more on domain and range. The domain of both and is . The range of is the same as that of . Since the range of is with no exceptions, the range of is also . See Example 2 for a transformation example. Reciprocal Trigonometric Function Identities There are two more trigonometric identities you should learn. See other trigonometric identities. These two new identities are useful when solving certain trigonometric equations: See Example 3. There is a quirky way to remember these identities. ‘The one with the tan is sexy’ and ‘the one with the coat is cosy’. They both come from the trigonometric identity . Dividing both sides by gives the first identity and by gives the second. Examples Solve for . Solution: By dividing both sides by 5 and writing as we get the equation: which can be rearranged to . Let so that we are looking for solutions to for . See more on solving trigonometric equations. The solutions are: and so the solutions are . Sketch and state the domain and range of the graph of . Solution: Recall from the transformations page that replacing with shrinks the original graph by half in the -direction. Adding 1 translates the graph upwards by 1. Hence the graph of is as follows: The domain of is . The range is . Given that and is obtuse, find the value of . Show that Solution: Using the identity we have . Hence, . is negative when is obtuse and so . Using the definitions: Extra Resources Questions by Topic A2ReciprocalTrigExamQuestions Open Reciprocal Trigonometric Functions in a New Window AS Maths Trigonometry Trigonometry Non Right-Angled Triangles Trigonometric Equations Trigonometric Graphs Trigonometric Identities A2 Maths Trigonometry Trigonometry Radians, Arc Length & Area of a Sector Small Angle Approximations Reciprocal Trigonometric Functions Inverse Trigonometric Functions Double & Compound Angle Formulae Return To The Maths Index A-Level Mathematics About StudyWell is a website for students studying A-Level Maths (or equivalent. course). 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16523	https://blog.csdn.net/2301_76884115/article/details/138000612	博客下载学习社区 GitCode InsCodeAI 会议 AI 搜索最新推荐文章于 2025-09-26 22:58:42 发布原创于 2024-04-20 15:17:23 发布 · 1.8k 阅读 · 11 · 6 · CC 4.0 BY-SA版权版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。文章标签： #数学分析 #笔记文章目录三角函数的周期性本篇文章适合个人复习翻阅，不建议新手入门使用三角函数的周期性本节的主题是研究三角函数的周期性，我们之前已经解析地定义三角函数为 cos ⁡ x = ∑ k = 0 ∞ ( − 1 ) k x 2 k ( 2 k ) ! , sin ⁡ x = ∑ k = 0 ∞ ( − 1 ) k x 2 k + 1 ( 2 k + 1 ) ! \cos{x}=\sum\limits_{k=0}^{\infty}\frac{(-1)^kx^{2k}}{(2k)!},\sin{x}=\sum\limits_{k=0}^{\infty}\frac{(-1)^kx^{2k+1}}{(2k+1)!} cosx=k=0∑∞(2k)!(−1)kx2k,sinx=k=0∑∞(2k+1)!(−1)kx2k+1 设函数 F : R → R 2 , x ↦ F ( x ) = ( sin ⁡ x cos ⁡ x ) F:\mathbb{R}\to\mathbb{R}^2,x\mapsto F(x)=(sinxcosx) F:R→R2,x↦F(x)=(sinxcosx) 记 S ( x ) = sin ⁡ x , C ( x ) = cos ⁡ x S(x)=\sin{x},C(x)=\cos{x} S(x)=sinx,C(x)=cosx 记矩阵 J = ( 0 1 − 1 0 ) J=(01−10) J=(0 最低0.47元/天解锁文章福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用关注关注 11 踩 6 收藏觉得还不错? 一键收藏 0 评论分享复制链接分享到 QQ 分享到新浪微博扫一扫举报举报专栏目录数字信号处理 --- 周期信号的三角函数表示一（三角函数的性质和三角波的合成）热门推荐松下J27录放机 05-17 3万+ 三角函数的性质一系列三角函数谐波（harmonic sinusoids）是傅里叶分析的基石，我们可以用这些不同频率的谐波构建各种各样的信号/波形。谐波（harmonics）：现在我们选择一个频率为f0的任意频率（arbitrary frequency）的正弦/余弦函数为基波（fundamental frequency）。则有一系列的基于该波的谐波（har...... 参与评论您还未登录，请先登录后发表或查看评论商业数学:三角函数与周期性现象分析 8-21 三角函数在商业数学中的应用是多方面的,它们不仅可以帮助我们描述周期性现象,还可以通过积分等数学工具进行实际的商业预测和分析。通过学习如何运用三角函数,商业分析师和决策者能够更好地理解市场动态,预测销售趋势,并制定更加精确的商业策略。本文通过解析《商务、经济、生命科学和社会科学的微积分》第九章的内容,旨在... 三角函数周期性讲解,求三角函数周期_xcosx周期函数的步骤资源... 9-13 在本节中,我们将深入探讨三角函数的周期性,特别是如何求解三角函数的周期。让我们定义周期函数。一个函数 ( f(x) ) 被称为周期函数,如果存在一个非零常数 ( T ),对定义域内的任意 ( x ),都有 ( f(x + T) = f(x) ) 成立。这个非零常数 ( T ) 就是函数 ( f(x) ) 的... 三角函数积分相关总结 m0_56112405的博客 01-06 3770 浅显总结一些三角函数积分相关的性质，三角函数周期的求法静雅斋数学 08-02 8113 总结高中阶段三角函数周期性的几乎所有求解方法和思路初中数学:三角函数的周期性质资源 9-8 ### 二、三角函数的周期性三角函数是最常见的周期函数之一,主要包括正弦函数 (\sin x)、余弦函数 (\cos x) 和正切函数 (\tan x)。这些函数具有特定的周期性: 1. 正弦函数和余弦函数:(\sin x) 和 (\cos x) 的最小正周期均为 (2\pi)。 2. 正切函数:(\tan x... 077 三角函数定积分性质_三角函数定积分的性质 9-26 077 三角函数定积分性质 077 三角函数定积分性质高等数学基础知识总结（1） qq_70019265的博客 10-09 2159 定积分表示函数 f(x)在区间 [a,b]上的累积效应或面积。分割区间：将区间 [a,b]分割成 n 个小区间，每个小区间的长度为 Δxi，其中，且。取样本点：在每个小区间内取一个样本点。构造黎曼和：构造黎曼和，表示函数 f(x) 在区间 [a,b]上的近似累积效应或面积。取极限：当分割的区间数 n 趋向于无穷大，且每个小区间的长度 Δxi趋向于零时，黎曼和的极限即为定积分：设函数 f(x,y) 在点 (a,b) 的某个去心邻域内有定义。【Python】《书虫大冒险》长单词查询 Barry_Wu_的博客 04-22 2470 书虫大冒险三角函数(trigonometric function)_三角函数英文 9-25 三角函数(英语:Trigonometric functions)是数学中常见的一类关于角度的函数。三角函数将直角三角形的内角和它的两个边的比值相关联,也可以等价地用与单位圆有关的各种线段的长度来定义。三角函数在研究三角形和圆等几何形状的性质时有重要作用,也是研究周期性现象的基础数学工具。在数学分析中,三角函数也被定义为无穷级... 面向初学者的 MQL4 语言系列之3——技术指标和内置函数 9-15 三角函数是基于角度的数学函数。它们在分析周期性过程时非常重要。与之密切相关的是反三角函数。更多信息可见于: 中,所有这些函数接受以弧度而非度数表示的参数。即,如果要求 20 度的正弦,必须先将 20 度转换为... 三角函数解题中周期的选择技巧 weixin_34258838的博客 02-19 1012 前言三角函数是高中阶段比较常见的周期函数，研究其性质或者解有关三角函数不等式时，肯定少不了周期性的考量。一般情况下基本周期我们都选[0，2π]来研究，但不是所有问题都这样选取周期就简单，以下举例说明。一、研究单调性例1 研究函数y=2sin(3x+π4)+1的单调区间。(整体思想：X=3x+π4) 分析：由于函数... 高一数学期末复习练习：三角函数精选.doc 09-26 5. 三角函数的性质：包括周期性（如sinx的周期是2π，cosx的周期是2π），单调性（在特定区间内的增减性），最值（如sinx和cosx在[-π, π]上的最大值为1，最小值为-1）。 6. 三角函数的图象与参数变化：... 高中数学三角函数公式大全记忆_正切函数资源 9-23 3. 正割和余割函数:正割函数secθ=1/cosθ,余割函数cscθ=1/sinθ,它们分别是余弦和正弦函数的倒数,提供了解决某些问题的另一种视角。 4. 基本性质:三角函数的周期性,sinθ和cosθ的周期是2π,tanθ的周期是π。此外,它们还具有奇偶性,如sin(-θ)=-sinθ,cos(-θ)=cosθ,体现出正弦为奇函数,余弦为偶... Excel函数应用之数学与三角函数资源 9-16 在Excel中,数学与三角函数是数据分析和处理中不可或缺的一部分,它们可以帮助我们执行各种复杂的计算,从简单的加减乘除到高级的三角运算和对数转换。以下将详细阐述这些函数的应用及其重要性。 ABS函数用于计算任何数值的绝对值。无论输入的数字是正还是负,ABS都会返回其非负值。例如,ABS(-5)将返回5,而ABS(3)也将... 高一数学期末复习练习：三角函数总复习精选.doc 09-26 以下是复习三角函数时需要掌握的关键知识点： 1. 函数图像的平移： - 函数图像的平移规律是：向右平移h个单位，函数表达式变为f(x+h)；向左平移h个单位，函数表达式变为f(x-h)。 - 例如题目中提到的`y=... 高中数学试题：三角函数单元复习试题(二).doc 10-12 3. 三角函数的图像与周期性：虽然题目中未直接画图，但题目中的角度变化和函数值的计算涉及到三角函数的周期性和图像的对称性，比如sin(α±β)和cos(α±β)的计算。 4. 三角函数的符号判断：例如sinα+... ...微积分读本》第二章三角学回顾_普林斯三角函数法则 9-7 2.3 三角函数的图像正弦和余弦函数的图像大家都很熟悉了,周期为2π,就不放出来了,正切函数的周期为π,为x的奇函数。渐近线为-π/2和π/2之间。 y=sec(x),y=csc(x)及y=cot(x)的函数图像如y = sec(x), y = csc(x)及 y = cot(x)的函数图像如y=sec(x),y=csc(x)及y=cot(x)的函数图像如... 江苏省2015届高三数学一轮复习备考试题：三角函数.doc 11-27 2. 三角函数的周期性：如题目中询问函数的最小正周期，这是由基本三角函数的周期性决定的，一般而言，正弦和余弦函数的周期是2π。 3. 三角恒等变换：例如通过已知的三角比关系，求解未知的三角比，如题中... cos（nx）的转化公式 baodream的博客 09-16 1万+ cos（nx）有可转化公式：转化如图：例题： 2017 ACM-ICPC 亚洲区（西安赛区）网络赛 F 链接如下：点击打开链接注意：上下双阶乘可约掉一部分除法用逆元还有判断一下m和n的一些特殊值代码如下： include include include include include in 三角函数的周期的求法 weixin_33871366的博客 12-05 2676 三角函数的周期的求法【定义法，变换法，公式法，图像法】例1求f(x)=(asinx+cosx)2的周期；【法1】：赋值法，令a=1，则函数f(x)=1+sin2x，故T=2π2=π；【法2】：将函数变形为(f(x)=[\sqrt{a^2+1}sin(x+\phi)]^2=(a^2+1)\cdot \cfra... 从周期函数的三角形式傅里叶级数引申到指数形式的傅里叶级数集成电路设计那些事儿 01-29 1万+ 微信公众号： xiaoshi-IC，小石谈IC 名字比较拗口，实际上是：对于一个周期函数，从三角形式傅里叶级数引申到指数形式的傅里叶级数，为什么引申到指数形式，因为加入指数形式后，运算更加简洁，同时引入了方向的概念，因此也有了负频率。下面是详细的推导过程：基本思路：主要就是利用欧拉公式，把三角形式中从cos sin函数分别用欧拉公式代入，然后令 F(nw1)=（an - jbn）/2... 【MuJoCo学习笔记】#2 接触动力学腱系统执行器传感器 mojin2005的博客 09-23 1746 edgetrackcomdmin, dmaxdmaxdminwidthwidthexponentestiffnesskdFdk⋅dmax−dmind−dminetrack在遍历总数为nframe的视频帧的过程中，原速下每一帧截取的间隔为1/fps，在n倍速下则应为n/fps，故第i帧的时刻为ni/fps。上一篇末尾我们提到如何记录分析仿真运动数据，由于这一步代码冗长、重复度高，且matplotlib。【OpenGL】LearnOpenGL学习笔记28 - 延迟渲染 Deferred Rendering 最新发布 Duo1J的随笔 09-26 947 LearnOpenGL学习笔记 - 延迟渲染 Deferred Rendering MySQL学习笔记02：安装配置与程序结构深度解析 2301_76657443的博客 09-24 1366 在上一篇文章中，我分享了重新开始系统学习MySQL的计划。今天开始第一课时的学习：MySQL安装与程序结构。虽然之前已经安装过MySQL，但这次我要更深入地理解MySQL的程序架构和配置细节。问题1：服务名称不匹配错误现象：sc query MySQL80 失败，提示"指定的服务未安装"解决方法：使用 sc query type= service \| find /i "mysql" 查找正确的服务名我的实际服务名：MySQL80_42问题2：端口占用冲突# 检查端口占用情况。 Mysql DBA学习笔记（客户端常用工具） leo_yty的博客 09-24 961 本文介绍了MySQL常用客户端工具的使用方法，包括mysql客户端连接本地/远程数据库、mysqladmin管理工具、mysqlbinlog日志查看工具、mysqlshow对象查找工具等。重点讲解了数据库备份工具mysqldump的语法和参数选项，以及数据恢复工具mysqlimport和source命令的使用方法。文章还提供了数据库备份与恢复的具体操作示例，如将数据库备份为SQL文件或文本文件，以及从备份文件恢复数据的过程。这些工具涵盖了MySQL数据库管理的主要操作场景，适合数据库管理员参考使用。 K8s学习笔记(八) K8s资源对象 m0_66372733的博客 09-26 785 在 Kubernetes（K8s）中，是集群的核心抽象，用于定义和描述集群的期望状态（Desired State），例如 “运行哪些应用”“应用需要多少资源”“如何访问应用” 等。K8s 的控制平面（如 kube-controller-manager、kube-scheduler）会持续监控这些资源对象，将集群的调整为期望状态，这一过程称为 “调和（Reconciliation）”。 Universal bovine identification via depth data and deep metric learning 阅读笔记 qq_52108058的博客 09-23 1457 随着牛群规模扩大,农场中牛与人比例失衡使得个体人工监测更具挑战性。因此,实时牛只识别对农场至关重要,也是迈向精准畜牧业的关键一步深度度量学习方法,利用现成3D摄像头获取的深度数据作为新型生物特征进行牛只识别深度作为生物特征可能将我们方法的实际应用范围扩展到英国其余68%缺乏独特被毛图案的牛品种。依托卷积神经网络CNN和多层感知机MLP为主干网络，通过提取泛化能力强的嵌入空间来区分个体。网络嵌入通过k-邻近 KNN 等算法通过简单的聚类实现高精度的识别。关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号京网文〔2020〕1039-165号经营性网站备案信息北京互联网违法和不良信息举报中心家长监护网络110报警服务中国互联网举报中心 Chrome商店下载账号管理规范版权与免责声明版权申诉出版物许可证营业执照 ©1999-2025北京创新乐知网络技术有限公司评论被折叠的条评论为什么被折叠? 到【灌水乐园】发言查看更多评论添加红包发出的红包实付元钱包余额 0 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。余额充值确定取消举报包含不实信息涉及个人隐私请选择具体原因（必选）侮辱谩骂诽谤请选择具体原因（必选）搬家样式博文样式 AI 搜索智能体 AI 编程 AI 作业助手专业的中文 IT 技术社区，与千万技术人共成长客服返回顶部
16524	https://www.merriam-webster.com/dictionary/junta	JUNTA Definition & Meaning - Merriam-Webster Chatbot Chatbot Games Word of the Day Grammar Word Finder Slang NewNewsletters Wordplay Rhymes Thesaurus Join MWU More Games Word of the Day Grammar Wordplay Slang Rhymes Word Finder Newsletters New Thesaurus Join MWU Shop Books Merch Log In Username My Words Recents Account Log Out Est. 1828 Dictionary Definition Definition Example Sentences Word History Entries Near Related Articles Cite this Entry Citation Share Kids Definition Kids More from M-W Show more Show more Citation Share Kids More from M-W Save Word To save this word, you'll need to log in.Log In junta noun jun·taˈhu̇n-təˈjən- ˈhən- Synonyms of junta 1 :a council or committee for political or governmental purposes especially:a group of persons controlling a government especially after a revolutionary seizure of power a military junta 2 :junto Examples of junta in a Sentence Recent Examples on the Web 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. 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Kuti was an outspoken critic of Nigeria’s military juntas in the 1970s.—Peter White, Deadline, 17 Sep. 2025 Reuters could not independently verify the reports, as internet and mobile services in parts of Rakhine have been cut off by the Myanmar junta and attempts to reach Kyauktaw residents were unsuccessful.—Reuters, NBC news, 16 Sep. 2025 Instead of being handed to the military in Tanintharyi, the 40 Rohingya were later passed on to a local armed group – one of the dozens that have sprung up across the country to battle the junta.—Esha Mitra, CNN Money, 14 Sep. 2025 Anti-Western sentiment has taken hold under the nearly three-year old junta of Burkinabe leader Ibrahim Traoré.—Yinka Adegoke, semafor.com, 27 Aug. 2025 See All Example Sentences for junta Word History Etymology Spanish, from feminine of junto joined, from Latin junctus, past participle of jungere to join — more at yoke First Known Use 1622, in the meaning defined at sense 1 Time Traveler The first known use of junta was in 1622 See more words from the same year Browse Nearby Words junt junta junto See all Nearby Words Articles Related to junta ### Forms of Government Quiz How do people take and hold power? Cite this Entry Style “Junta.” Merriam-Webster.com Dictionary, Merriam-Webster, Accessed 28 Sep. 2025. Copy Citation Share Kids Definition junta noun jun·taˈhu̇n-təˈjənt-ə ˈhən-tə 1 :a group of persons controlling a government especially after a revolution 2 :junto More from Merriam-Webster on junta Thesaurus: All synonyms and antonyms for junta Nglish: Translation of junta for Spanish Speakers Britannica.com: Encyclopedia article about junta Last Updated:20 Sep 2025 - Updated example sentences Love words? Need even more definitions? Subscribe to America's largest dictionary and get thousands more definitions and advanced search—ad free! Merriam-Webster unabridged More from Merriam-Webster ### Can you solve 4 words at once? Play Play ### Can you solve 4 words at once? Play Play Top Lookups Word of the Day kerfuffle See Definitions and Examples » Get Word of the Day daily email! 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16525	https://chem.libretexts.org/Workbench/Chemistry_LHS_Bridge/10%3A_Acid-Base_Equilibria/10.02%3A_pH_and_pOH	Search x Text Color Text Size Margin Size Font Type selected template will load here Error This action is not available. 10.2: pH and pOH ( \newcommand{\vecs}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } ) ( \newcommand{\vecd}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash {#1}}} ) ( \newcommand{\id}{\mathrm{id}}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\kernel}{\mathrm{null}\,}) ( \newcommand{\range}{\mathrm{range}\,}) ( \newcommand{\RealPart}{\mathrm{Re}}) ( \newcommand{\ImaginaryPart}{\mathrm{Im}}) ( \newcommand{\Argument}{\mathrm{Arg}}) ( \newcommand{\norm}{\\| #1 \\|}) ( \newcommand{\inner}{\langle #1, #2 \rangle}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\id}{\mathrm{id}}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\kernel}{\mathrm{null}\,}) ( \newcommand{\range}{\mathrm{range}\,}) ( \newcommand{\RealPart}{\mathrm{Re}}) ( \newcommand{\ImaginaryPart}{\mathrm{Im}}) ( \newcommand{\Argument}{\mathrm{Arg}}) ( \newcommand{\norm}{\\| #1 \\|}) ( \newcommand{\inner}{\langle #1, #2 \rangle}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\AA}{\unicode[.8,0]{x212B}}) ( \newcommand{\vectorA}{\vec{#1}} % arrow) ( \newcommand{\vectorAt}{\vec{\text{#1}}} % arrow) ( \newcommand{\vectorB}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } ) ( \newcommand{\vectorC}{\textbf{#1}} ) ( \newcommand{\vectorD}{\overrightarrow{#1}} ) (\newcommand{\ket}{\left\| #1 \right>} ) ( \newcommand{\bra}{\left< #1 \right\|} ) ( \newcommand{\braket}{\left< #1 \vphantom{#2} \right\| \left. #2 \vphantom{#1} \right>} ) ( \newcommand{\qmvec}{\mathbf{\vec{#1}}} ) ( \newcommand{\op}{\hat{\mathbf{#1}}}) ( \newcommand{\expect}{\langle #1 \rangle}) ( \newcommand{\vectorDt}{\overrightarrow{\text{#1}}} ) ( \newcommand{\vectE}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} ) ( \newcommand{\vecs}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } ) ( \newcommand{\vecd}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash {#1}}} ) Learning Objectives As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water ((K_w)). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions. A common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “X” is the quantity of interest and “log” is the base-10 logarithm: [\mathrm{pX=−\log X} \label{1} ] The pH of a solution is therefore defined as shown here, where [H3O+] is the molar concentration of hydronium ion in the solution: [\mathrm{pH=-\log[H_3O^+]}\label{(\PageIndex{2})} ] Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression: [\mathrm{[H_3O^+]=10^{−pH}}\label{(\PageIndex{3})} ] Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH: [\mathrm{pOH=-\log [OH^−]}\label{(\PageIndex{4})} ] or [\mathrm{[OH^-]=10^{−pOH}} \label{(\PageIndex{5})} ] Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the (K_w) expression: [K_\ce{w}=\ce{[H_3O^+][OH^- ]} \label{(\PageIndex{6})} ] [-\log K_\ce{w}=\mathrm{-\log([H_3O^+][OH^−])=-\log[H_3O^+] + -\log[OH^-]}\label{(\PageIndex{7})} ] [\mathrm{p\mathit{K}_w=pH + pOH} \label{(\PageIndex{8})} ] At 25 °C, the value of (K_w) is (1.0 \times 10^{−14}), and so: [\mathrm{14.00=pH + pOH} \label{(\PageIndex{9})} ] The hydronium ion molarity in pure water (or any neutral solution) is ( 1.0 \times 10^{-7}\; M) at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore: [\mathrm{pH=-\log[H_3O^+]=-\log(1.0\times 10^{−7}) = 7.00} \label{10} ] [\mathrm{pOH=-\log[OH^−]=-\log(1.0\times 10^{−7}) = 7.00} \label{11} ] And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than ( 1.0 \times 10^{-7}\; M) and hydroxide ion molarities less than ( 1.0 \times 10^{-7}\; M) (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than ( 1.0 \times 10^{-7}\; M) and hydroxide ion molarities greater than ( 1.0 \times 10^{-7}\; M) (corresponding to pH values greater than 7.00 and pOH values less than 7.00). When (pH=7) Solutions are not Neutral Since the autoionization constant (K_w) is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, the hydronium molarity of pure water at 80 °C is 4.9 × 10−7 M, which corresponds to pH and pOH values of: [\begin{align} pH &=-\log[\ce{H_3O^+}] \[4pt] &= -\log(4.9\times 10^{−7}) \[4pt] &=6.31 \label{12} \end{align} ] [\begin{align} pOH &=-\log[\ce{OH^-}]\[4pt] & =-\log(4.9\times 10^{−7}) \[4pt] &=6.31 \label{13}\end{align} ] At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at nonstandard temperatures, such as enzyme reactions in warm-blooded organisms. Unless otherwise noted, references to pH values are presumed to be those at standard temperature (25 °C) (Table (\PageIndex{1})). Table (\PageIndex{1}): Summary of Relations for Acidic, Basic and Neutral Solutions \| Classification \| Relative Ion Concentrations \| pH at 25 °C \| --- \| acidic \| [H3O+] > [OH−] \| pH < 7 \| \| neutral \| [H3O+] = [OH−] \| pH = 7 \| \| basic \| [H3O+] < [OH−] \| pH > 7 \| Figure (\PageIndex{1}) shows the relationships between [H3O+], [OH−], pH, and pOH, and gives values for these properties at standard temperatures for some common substances. Example (\PageIndex{1}): Calculation of pH from (\ce{[H_3O^+]}) What is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of (1.2 \times 10^{−3}\; M)? Solution [\begin{align} pH &=-\log [H_3O^+] \[4pt] &= -\log(1.2 \times 10^{−3}) \[4pt] &=−(−2.92) \[4pt]&=2.92 \end{align} \nonumber ] Exercise (\PageIndex{1}) Water exposed to air contains carbonic acid, H2CO3, due to the reaction between carbon dioxide and water: [\ce{CO2(aq) + H2O (l) \rightleftharpoons H2CO3(aq)} \nonumber ] Air-saturated water has a hydronium ion concentration caused by the dissolved (\ce{CO_2}) of (2.0 \times 10^{−6}\; M), about 20-times larger than that of pure water. Calculate the pH of the solution at 25 °C. 5.70 Example (\PageIndex{2}): Calculation of Hydronium Ion Concentration from pH Calculate the hydronium ion concentration of blood, the pH of which is 7.3 (slightly alkaline). Solution [\mathrm{pH=-\log[H_3O^+]=7.3} \nonumber ] [\mathrm{\log[H_3O^+]=−7.3} \nonumber ] [\mathrm{[H_3O^+]=10^{−7.3}} \nonumber ] or [[\ce{H_3O^+}]=\textrm{antilog of} −7.3 \nonumber ] [[\ce{H_3O^+}]=5\times 10^{−8}\;M \nonumber ] (On a calculator take the antilog, or the “inverse” log, of −7.3, or calculate 10−7.3.) Exercise (\PageIndex{2}) Calculate the hydronium ion concentration of a solution with a pH of −1.07. 12 M Environmental Science Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO2 which forms carbonic acid: [\ce{H2O (l) + CO2(g) ⟶ H2CO3(aq)} \label{14} ] [\ce{H2CO3(aq) \rightleftharpoons H^+(aq) + HCO3^- (aq)} \label{15} ] Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO2, SO2, SO3, NO, and NO2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here: [\ce{H2O (l) + SO3(g) ⟶ H2SO4(aq)} \label{16} ] [\ce{H2SO4(aq) ⟶ H^+(aq) + HSO4^- (aq)} \label{17} ] Carbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine. Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure (\PageIndex{2})). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India. Example (\PageIndex{3}): Calculation of pOH What are the pOH and the pH of a 0.0125-M solution of potassium hydroxide, KOH? Solution Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH−] = 0.0125 M: [\mathrm{pOH=-\log[OH^− ]=-\log 0.0125} \nonumber ] [=−(−1.903)=1.903 \nonumber ] The pH can be found from the (\ce{pOH}): [\mathrm{pH+pOH=14.00} \nonumber ] [\mathrm{pH=14.00−pOH=14.00−1.903=12.10} \nonumber ] Exercise (\PageIndex{3}) The hydronium ion concentration of vinegar is approximately (4 \times 10^{−3}\; M). What are the corresponding values of pOH and pH? pOH = 11.6, pH = 14.00 - pOH = 2.4 The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure (\PageIndex{3})). The pH of a solution may also be visually estimated using colored indicators (Figure (\PageIndex{3})). Summary The concentration of hydronium ion in a solution of an acid in water is greater than ( 1.0 \times 10^{-7}\; M) at 25 °C. The concentration of hydroxide ion in a solution of a base in water is greater than ( 1.0 \times 10^{-7}\; M) at 25 °C. The concentration of H3O+ in a solution can be expressed as the pH of the solution; (\ce{pH} = -\log \ce{H3O+}). The concentration of OH− can be expressed as the pOH of the solution: (\ce{pOH} = -\log[\ce{OH-}]). In pure water, pH = 7.00 and pOH = 7.00 Key Equations Glossary This page titled 10.2: pH and pOH is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. 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16526	https://inria.hal.science/hal-02969790/document	NON EXISTENCE RESULT OF NONTRIVIAL SOLUTIONS TO THE EQUATION −∆u = f (u) SALVADOR L ´OPEZ-MART ´INEZ AND ALEXIS MOLINO Abstract. In this paper we prove the nonexistence of nontrivial solu-tion to { −∆u=f(u)in Ω ,u= 0 on ∂Ω, being Ω ⊂RN(N≥1) a bounded domain and flocally Lispchitz with non-positive primitive. As a consequence, we discuss the long-time behavior of solutions to the so-called sine-Gordon equation. Introduction In this paper we are interested in nonexistence results of nontrivial solu-tions for semilinear elliptic differential equations. Specifically, given f : R → R any locally Lipschitz function and Ω ⊂ RN (N ≥ 1) a bounded domain with smooth boundary, we obtain neccesary conditions to Ω and f for the nonexistence of nontrivial solutions for the Dirichlet problem (P ) { −∆u = f (u) in Ω ,u = 0 on ∂Ω. Along this note, a classical solution to ( P ) ( solution from now on) will be a function u ∈ C 2(Ω) ∩ C 1,α (Ω), for some α ∈ (0 , 1), satisfying ( P ) pointwise. Observe that, by regularity results, every bounded weak solution is a solution to this problem (see e.g. Struwe (2008)). When studying any kind of problem involving differential equations, it is always useful to know necessary conditions for the existence of solution. In this way, it follows immediately that a necessary condition for the existence of a solution u to ( P ) is that u must satisfy the identity (1) ∫ Ω \|∇ u\|2 = ∫ Ω f (u)u. As a consequence, a straightforward nonexistence result for problem ( P )states that if (2) f (s)s ≤ 0, for all s ∈ R, MSC :35J05, 35J15, 35J25. Keywords: Elliptic Equation, Dirichlet Problem, Nonexistence. 12SALVADOR L ´OPEZ-MART ´INEZ AND ALEXIS MOLINO there exists no nontrivial solution to ( P ). In addition, the well-known Pohoˇ zaev identity (Pohoˇ zaev (1965)) yields a sort of generalization of this simple result. To be more precise, every solution u to ( P ) must satisfy the following identity: (3) 12 ∫ ∂Ω \|∇ u(x)\|2 x · ν(x)dx + N − 22 ∫ Ω \|∇ u(x)\|2dx = N ∫ Ω F (u(x)) dx, where F (s) = ∫ s 0 f (t)dt and ν denotes the unit vector normal to ∂Ω pointing outwards. Observe that if Ω is starshaped with respect to 0 (i.e., x · ν(x) > 0on ∂Ω) and N ≥ 3, the left hand side of (3) is non-negative. Therefore, if Ω is starshaped, N ≥ 3 and (4) F (s) ≤ 0, for all s ∈ R, there exists no nontrivial solution to ( P ). Note that (4) implies that f (0) = 0. Consequently, the trivial solution is always a solution. Condition sf (s) ≤ 0 clearly guarantees F (s) ≤ 0, but not conversely. In this way, a natural question is whether the condition Ω is starshaped is essential for the nonexistence of nontrivial solution to ( P ), for any bounded domain Ω and f satisfying (4). A similar situation arises when one analyzes the well-known supercritical case result, also derived from (3). In fact, if f (s) = λ\|s\|p−2s, for λ > 0 and p ≥ 2∗, there exists no nontrivial solution to ( P ) provided N ≥ 3 and Ω is starshaped. However, there are examples of non-starshaped domains for which, surprisingly, there exist nontrivial solutions for p ≥ 2∗. For instance, positive solutions have been found when the domain is an annulus (see the seminal paper Kazdan and Warner (1975) and references therein) or for domains with small holes (del Pino et al. (2002)). Nevertheless, much less is known about the influence of the geometry of Ω in the existence of solution to problem ( P ) in the case F (s) ≤ 0 and the literature contains only partial nonexistence results. Observe that for functions f globally Lipschitz, with L−Lipschitz constant, it follows that \|f (s)\| ≤ L\|s\|. Thus, applying Poincar´ e inequality in (1), we obtain λ1 ∫ Ω u2 ≤ ∫ Ω \|∇ u\|2 = ∫ Ω f (u)u ≤ L ∫ Ω u2. Therefore, this simple computation gives the nonexistence of nontrivial so-lutions as long as L < λ 1, being λ1 the first eingenvalue for the Laplacian operator with zero Dirichlet boundary conditions. In this line, in Ricceri (2008) and Fan (2009) the authors prove the nonexistence of nontrivial so-lution provided that N ≥ 2 and L < 3λ1 or L ≤ 3λ1, respectively, without assuming any geometric condition on ∂Ω. Recently, in Goubet and Ricceri (2019), the nonexistence of nontrivial solutions is shown if either ∂Ω has non-negative mean curvature or Ω is an annulus, also for functions f glob-ally Lipschitz and N ≥ 2. On the other hand, in Cl´ ement and Sweers (1987) (see also Dancer and Schmitt (1987)), a condition similar to F (s) ≤ 0 for positive solutions is imposed. Specifically, the result states as follows. NON EXISTENCE RESULT OF NONTRIVIAL SOLUTIONS TO THE EQUATION −∆u = f (u)3 Theorem 1.1 (Cl´ ement and Sweers (1987)) . Let f be a C1 function. Sup-pose that there are two numbers s1 < s 2, with s2 > 0, such that f (s1) = f (s2) = 0 and f > 0 in (s1, s 2). In additon, assume that there is s ∈ [0 , s 2) such that F (s) ≥ F (s2). Then, there is no positive solution u of (P ) satis-fying max u ∈ (s1, s 2). In the present note, inspired by the above result, we prove that there is no nontrivial solution to problem ( P ) (not necessarily positive) provided ∫ s 0 f (t)dt ≤ 0, being f a locally Lispchitz function (Theorem 2.1). Here, no additional hypotheses on Ω and N are required. This exposes the unexpected fact that there is no geometric assumption on Ω that gives a nontrivial solution. As a consequence, solution to the sine-Gordon equation (11) tends to 0 as t → ∞ (Proposition 3.1). 2. main result Theorem 2.1. Let Ω ⊂ RN (N ≥ 1) a bounded domain with smooth bound-ary and f : R → R a locally Lipschitz function satisfying the condition F (s) = ∫ s 0 f (t)dt ≤ 0 for all s ∈ R. Then, u ≡ 0 is the unique solution to (P ).Proof. Clearly, zero is a solution. We argue by contradiction and assume that there exists a nontrivial solution u to ( P ). First of all, notice that −u is a solution to { −∆u = −f (−u) in Ω ,u = 0 on ∂Ω. Since the function −f (−s) satisfies the hypotheses of the theorem, there is no loss of generality in assuming that u∞ := max x∈Ω u(x) > 0. On the other hand, since f is locally Lipschitz and the value of f (s) for s > u ∞ is irrelevant, we can also assume that f is globally Lipschitz, with Lipschitz constant L > 0, and that lim s→+∞ f (s) = −∞ .It is easy to check that f (u∞) > 0. Indeed, arguing by contradiction, assume that f (u∞) ≤ 0. Then, (5) − ∆u∞ + Lu ∞ ≥ f (u∞) + Lu ∞ in Ω . Moreover, (6) − ∆u + Lu = f (u) + Lu in Ω . Subtracting (6) from (5), and using that f (s) + Ls is non-decreasing, we obtain −∆( u∞ − u) + L(u∞ − u) ≥ f (u∞) + Lu ∞ − f (u) − Lu ≥ 0 in Ω . Since u∞ > u on ∂Ω, the strong maximum principle implies that u∞ > u in Ω, which is a contradiction. Thus, the fact that f (u∞) > 0 implies that there are s1, s 2 > 0 such that s1 < u ∞ < s 2 and (7) f (s) > 0 ∀s ∈ (s1, s 2).4 SALVADOR L ´OPEZ-MART ´INEZ AND ALEXIS MOLINO Moreover, since F (s) ≤ 0 and lim s→+∞ f (s) = −∞ , we can choose respec-tively s1 and s2 such that f (s1) = f (s2) = 0. Further, we can assume that F (s2) < 0 since, otherwise (i.e., if F (s2) = 0), we can modify f to another L-Lipschitz function f ∗ such that f (s) > f ∗(s) > 0 for s ∈ (u∞, s 2)and f = f ∗ elsewhere. In this way, u is still a solution to ( P ), but now F (s2) < 0. Now we will find a family of supersolutions to ( P ) which will lead to a contradiction by comparison with u. For this purpose, we follow the original reasoning in Cl´ ement and Sweers (1987), which in principle is performed for f ∈ C 1(R). Here we adapt the proof to our setting and check that it also works for Lipschitz functions. Indeed, consider the following initial value problem  −w′′ (r) = f (w(r)) , ∀r > 0,w(0) = s2,w′(0) = −√−F (s2). Since f is Lipschitz there is a unique solution w ∈ C 2([0 , +∞)). Multiplying the equation by w′(r) and integrating, we obtain (w′(r)) 2 = −F (s2) + 2 ∫ s2 w(r) f (s)ds = F (s2) − 2F (w(r)) .(8) Thus, using (7) we get that (9) (w′(r)) 2 > 0 for w(r) ∈ [s1, s 2]. Now, since w(0) = s2 and w′(0) < 0, we deduce easily that w(r) ∈ (s1, s 2)for all r > 0 small enough. We claim now that there exists r0 > 0 such that w(r0) = s1. Indeed, assume by contradiction that w(r) > s 1 for all r > 0. Then, by (9) we have that w is decreasing in (0 , +∞). Hence, there exists s3 ∈ [s1, s 2) such that lim r→+∞ w(s) = s3. But this is impossible since w′′ (r) = −f (w(r)) < 0 for all r > 0, i.e. w is concave. In consequence, since w(r0) = s1 and w′(r0) < 0, we deduce that inf r≥0 w(r) 0. Indeed, assuming oth-erwise, there exists a sequence {rn} ⊂ [0 , +∞) such that lim n→∞ w(rn) = 0. Then, for n large enough, we deduce from (8) that ( w′(rn)) 2 < F (s2)2 < 0, a contradiction. Thus, we have proved that (10) 0 < inf w < s 1. Next, we define W (r) = { s2, r ∈ (−∞ , 0 ] , min {w(r), s 2}, r ∈ (0 , ∞). Since we can assume that f (s) < 0 for s > s 2, it follows that w is convex if w(r) > s 2. This implies that, if w(r2) = s2 for some r2 > 0, then W (r) = s2NON EXISTENCE RESULT OF NONTRIVIAL SOLUTIONS TO THE EQUATION −∆u = f (u)5 for all r ≥ r2. Otherwise, w(r) < s 2 for all r > 0, so W (r) = w(r) for all r > 0. For every t ∈ R, consider the family of parametric functions vt(x) = W (x1 −t) for all x = ( x1, ..., x N ) ∈ RN . We will prove now that u(x) ≤ vt(x)for all x ∈ Ω and for all t ∈ R using the sweeping principle by Serrin. Indeed, let U = {t ∈ R : u(x) ≤ vt(x) for all x ∈ Ω}. Note that vt = s2 for t large enough, and u < s 2 in Ω, so U is nonempty. Notice also that W is a globally Lipschitz function, so the function t 7 → vt(x)is continuous uniformly in x. In particular, U is closed. Let us now take t ∈ U . Observe that vt ∈ W 1,∞(Ω) and −∆vt ≥ f (vt)in Ω (in the weak sense). Then, since s 7 → f (s) + Ls is non-decreasing and u ≤ vt in Ω, we have that −∆( vt − u) + L(vt − u) ≥ 0 in Ω. Notice that u(x) = 0 < inf w ≤ vt(x) ∀x ∈ ∂Ω, so vt 6 ≡ u. Then, the strong maximum principle implies that u(x) < v t(x)for all x ∈ Ω. Therefore, the uniform continuity of s 7 → vs implies that there exists T > 0, independent of x, such that u(x) < v s(x) for all x ∈ Ω and for all s ∈ (t − T, t + T ). That is to say, ( t − T, t + T ) ⊂ U , so U is open. In conclusion, U = R, and thus, u ≤ vt for all t ∈ R. In consequence, u(x) ≤ inf t∈R vt(x) = inf r> 0 w(r) < s 1, ∀x ∈ Ω, which is a contradiction with the fact that u∞ ∈ (s1, s 2). Sine-Gordon equation In this section we study the asymptotic behavior, as t → ∞ , of global and bounded solutions of the second order evolution problem (11)  utt + αu t − ∆u + β sin u = 0 in R+ × Ω,u = 0 on R+ × ∂Ω,u(0 , x ) = u0(x) x in Ω ,ut(0 , x ) = u1(x) x in Ω , also called the sine-Gordon equation. Here, α, β > 0, the unknown is a scalar function u(t, x ) which maps [0 , ∞) × Ω into R, ut and utt denote the first and second derivatives of u with respect to the variable t and the initial data (u0, u 1) belongs to H10 (Ω) × L2(Ω). In physics the sine-Gordon equation is used to model, for instance, the dynamics of a Josephson junction driven by a current source. It is well known the existence and uniqueness of solution u ∈ C ([0 , T ], H 10 (Ω)) and ut ∈ C ([0 , T ], L 2(Ω)) for any T > 0 (Temam, 1997, IV. Theorem 2.1.). Proposition 3.1. Let N ≥ 1, Ω be a bounded smooth domain of RN and u be the solution to (11) with initial data (u0, u 1) belonging to the energy space H10 (Ω) × L2(Ω) . Then lim t→∞ ‖ut(t, ·)‖L2(Ω) = lim t→∞ ‖u(t, ·)‖H10 (Ω) = 0 .6 SALVADOR L ´OPEZ-MART ´INEZ AND ALEXIS MOLINO Proof. Directly applying (Haraux and Jendoubi, 1999, Th. 1.2) we obtain that lim t→∞ ‖ut(t, ·)‖L2(Ω) = lim t→∞ ‖u(t, ·) − ϕ(·)‖H10 (Ω) = 0 , where ϕ ∈ H2(Ω) ∩ H10 (Ω) is a solution of −∆ϕ = −β sin ϕ, see also (Haraux and Jendoubi, 1999, Ex. 4.1.1). On the other hand, since F (s) = β(cos s − 1) is non-positive for all s ∈ R,it follows by Theorem 2.1 that ϕ ≡ 0 and the proof is finished. Remark 3.2 . Proposition 3.1 improves the recent result by Goubet (2019), which arrives at the same asymptotic behavior of the solution but under the restrictions of the dimension ( N ≥ 2) and, either the domain Ω has non-negative mean curvature (Goubet, 2019, Th. 2.1), or Ω is an annulus of RN (Goubet, 2019, Prop. 2.1). Acknowledgements First and second author are supported by PGC2018-096422-B-I00 (MCIU / AEI / FEDER, UE) and by Junta de Andaluc´ ıa FQM-116 (Spain). First author is also supported by Programa de Contratos Predoctorales del Plan Propio de la Universidad de Granada. References Cl´ ement, P. and Sweers, G. Existence and multiplicity results for a semi-linear elliptic eigenvalue problem. Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) , 14(1):97–121, 1987. ISSN 0391-173X. Dancer, E. N. and Schmitt, K. On positive solutions of semilinear elliptic equations. Proc. Amer. Math. Soc. , 101(3):445–452, 1987. ISSN 0002-9939. del Pino, M., Felmer, F., and Musso, M. Multi-peak solutions for super-critical elliptic problems in domains with small holes. Journal of Differ-ential Equations , 182(2):511 – 540, 2002. ISSN 0022-0396. Fan, X. On Ricceri’s conjecture for a class of nonlinear eigenvalue problems. Appl. Math. Lett. , 22(9):1386–1389, 2009. ISSN 0893-9659. Goubet, O. Remarks on some dissipative sine-gordon equa-tions. Complex Variables and Elliptic Equations , 0(0):1–7, 2019, .Goubet, O. and Ricceri, B. Non-existence results for an eigenvalue problem involving Lipschitzian non-linearities with non-positive primitive. Bull. Lond. Math. Soc. , 51(3):531–538, 2019. ISSN 0024-6093. Haraux, A. and Jendoubi, M. A. Convergence of bounded weak solutions of the wave equation with dissipation and analytic nonlinearity. Calc. Var. Partial Differential Equations , 9(2):95–124, 1999. ISSN 0944-2669. NON EXISTENCE RESULT OF NONTRIVIAL SOLUTIONS TO THE EQUATION −∆u = f (u)7 Kazdan, J. L. and Warner, F. W. Remarks on some quasilinear elliptic equations. Comm. Pure Appl. Math. , 28(5):567–597, 1975. ISSN 0010-3640. Pohoˇ zaev, S. I. On the eigenfunctions of the equation ∆ u+λf (u) = 0. Dokl. Akad. Nauk SSSR , 165:36–39, 1965. ISSN 0002-3264. Ricceri, B. A remark on a class of nonlinear eigenvalue problems. Nonlinear Analysis: Theory, Methods & Applications , 69(9):2964 – 2967, 2008. ISSN 0362-546X. Struwe, M. Variational methods , volume 34 of Series of Modern Surveys in Mathematics . Springer-Verlag, Berlin, fourth edition, 2008. ISBN 978-3-540-74012-4. Applications to nonlinear partial differential equations and Hamiltonian systems. Temam, R. Infinite-dimensional dynamical systems in mechanics and physics , volume 68 of Applied Mathematical Sciences . Springer-Verlag, New York, second edition, 1997. ISBN 0-387-94866-X.
16527	https://www.summitcountyutah.gov/DocumentCenter/View/20863/History-of-Urban-Development-US	18 Many factors have affected the development of cities in the United States, such as social, economic, and environmental concerns. Some of these are outlined below. Teacher Reading City and Suburb: An Overview of Urbanization in the United States 1700–1800 The early economy of the American colo - nies was based on farming, where indi - viduals owned and controlled their own land. As trade and industry grew, so did cities, like Boston, Philadelphia, and New York. These early cities seemed crowded and dirty compared to the open spaces of the farm. So, even as the size and number of cities increased in America over the next two centuries, the notion of living on one’s own land in the country remained a powerful ideal. Travel in cities was limited by how far people could walk or use animal power to travel, so living close to work and stores saved time and effort. In early cities, homes, stores, and places of employment were clustered together for convenience. Early cities also relied on rivers and water- ways for transportation and shipping of supplies between communities. 1800–1900 As a result of the Industrial Revolution, cit - ies grew and changed in structure. People moved to cities from farms and from over - seas to find jobs in the city factories. City populations increased. Factories filled the air with smoke. Residents and planners became concerned about issues such as air pollution, minimal sewage treatment, and the growing city population. Poorer city residents lived in crowded apartment buildings called tenements. Apartments were usually about 300 square feet with three rooms, and typically housed about seven people. The earliest tenements lacked indoor plumbing and did not have much fresh air circulation. These crowded and unsanitary living con - ditions often created an unhealthy envi - ronment, resulting in high levels of illness among residents. As a reaction to these concerns, planners looked for solutions by proposing new plans for cities and plans for the newly appearing suburbs, located outside of the city. Linked to the city by train, these planned communities, allowed some residents to live in a country-like place, away from the city’s pollution and disease, but still commute to the city to shop or work. The single-family house, cul-de-sac, and tree- lined curving roads were characteristics of these early suburbs. Planners also sought to address the growing slums and lack of green space in cities like New York and Boston. Reform - ers called for park systems that would serve not just neighborhoods but entire cities, as places for all classes to use. 19 The late 1800’s saw the birth of the skyscraper, a building type that has pro - foundly influenced urban development. With the continued growth of cities, the available land for new buildings was less and less available; the only alternative was to build up. Before the 1880s, buildings of more than a few stories had to be sup - ported by thick brick or stone walls— the taller the building, the thicker the masonry base. In the 1880’s two technological advances allowed for the development of the skyscraper: steel framework and safer elevators. Later improvements in steel frame construction meant exterior walls did not have to support the building, allowing architects to build taller buildings and to replace masonry walls with glass. Improvements in elevator safety and per - formance also allowed buildings to reach greater heights. Taller buildings meant less land being used and greater concen - trations of people, thereby creating denser cities and conserving land. 1900–1920 In the early 1900s efforts were made to improve conditions for the poor and re - form local government. At the same time, the City Beautiful Movement became a powerful force in shaping cities. It used city beautification—civic art, architecture, parks and open spaces—to instill civic pride and improve the physical condi - tion of cities. The World’s Columbian Exposition in Chicago in 1893, the first economically successful US World’s Fair, was seen as an opportunity to highlight the rebirth of Chicago just 22 years after a great fire destroyed the city. The design for the Columbian Exposition, by Frederick Law Olmsted and Daniel Burnham, be - came the inspiration for the City Beautiful Movement. It made Chicago seem like an ideal place with its classically inspired architecture, formal plan, and monumental buildings. The plan emphasized civic- centers, tree-lined boulevards, and public spaces with plants and trees. Concepts from the City Beautiful Move - ment were incorporated into the 1901 McMillan Plan for Washington, D.C., which redefined the National Mall as the nation’s civic center of monumental ar - chitecture and public open space. Other cities followed suit with their own plans inspired by the City Beautiful Movement. 20 As the City Beautiful Movement reshaped urban centers, the “garden city” model of suburban development proposed a new way to combine features of both the city and the suburb. Developed by writer Ebenezer Howard in response to the pollution in London, England and shaped by ideas in his book Garden Cities of Tomorrow (1902), Howard’s ideal plan included a series of satellite cities, which were connected to each other and a central city by a train system. A green belt around the city would limit city expansion outward and provide areas for recreation. In Howard’s plan the cities would have a limit to the number of residents able to live and work there. He wanted to combine the best features of city and country life. Americans experimented with the Garden City model in the 1920’s and 1930’s. Another suburban model was the street - car suburb. With the development of the streetcar in the late 1800s, by the early 1900s streetcar suburbs were growing along streetcar lines leading out of the city. These new communities, developed on small lots and offering one, two, and three-family houses, attracted the middle class with their relatively low-cost housing and easy access to transportation. Between 1900 and 1920, city planning was firmly established as an orderly way to shape the physical appearance of the city and to direct urban growth. In 1903, President Theodore Roosevelt appointed a Public Lands Commission to study and recommend regulations for orderly land development and management. 1920–1940 The Standard State Zoning Enabling Act (1922) and the Standard City Plan - ning Enabling Act (1928), both prepared by the U.S. Department of Commerce, formed the basis for zoning laws across the country. These Acts enabled local communities to plan and zone uses within their jurisdiction. Zoning meant predict - ability—property owners would know what could be built on and, importantly, next to their land. Zoning also created value— land assigned for commercial uses, for example, was seen as more valuable than land assigned for agriculture. Separating uses was also seen as a way to improve people’s health and safety by keeping residential uses away from polluting in - dustries like factories and landfills. Zoning by use would have a profound affect on planning in the 20th century. By the 1920s, cars were fast becoming an accepted and desired part of American life. As car ownership grew, transit use declined. Driving in the city meant deal - ing with traffic jams and difficult parking. In the 1920s, planners began to address how cars had begun to affect everyday life. Radburn, New Jersey, for example, de - veloped in 1920s, had separated pedes - trian and automobile routes. The 1920’s also saw the first car-oriented shopping centers, which provided off-street park - ing. The Park & Shop, still in existence in Washington, D.C., was an early example, with parking provided in a lot in front of the strip of stores. 21 In the 1920s, the members of the Re - gional Planning Association of America (RPAA) began to see unchecked expan - sion of the suburbs or sprawl as a poten - tial threat to the undeveloped natural envi - ronment. Members of the RPAA combined the idea of the garden city with the idea of conservation of land to create regional ecological planning. These ideas stemmed from developments in the field of ecolo - gy—the study of relationships within natu - ral systems. Planners started to consider how humans could fit, in a balanced way, into these natural systems. Natural sys - tems, like watersheds, often exist across a wide geography that includes multiple municipalities. Lewis Mumford, an histori - an and writer on science and architecture, promoted the idea of regional ecological planning as a way to balance nature and the built environment across broad areas such as New York State. Beginning in the 1930s the Federal Government began to play a larger role in planning, mostly through the tool of environmental policies. For example, dur - ing the 1930s the Civilian Conservation Corps (CCC) was established with the dual purpose of providing work for the unemployed and converting unused or underused land into forested areas for recreation. The CCC constructed the roads, campgrounds, and trails for many of the local, state, and national parks in existence today. In 1934, during the Great Depression, the federal government passed the National Housing Act and created the Federal Housing Administration (FHA) to support housing development. The FHA insured long-term mortgages by private lenders, making it easier for citizens to borrow money for homes. FHA regulations also provided very low-risk financing to build - ers of large suburban subdivisions. 1940–1960 The 1940s saw more federal acts that continued to greatly affect planning, in - cluding the creation of new highways that would move traffic from the city center to the suburbs and the GI Bill, which un - derwrote mortgages for new homes to returning soldiers for, making the dream of owning your own home accessible to even more people. A suburban housing boom followed World War II. A strong economy, the growing number of young families, and the de - sire—and economic ability— to own a home pulled the middle class away from the city. Developers met the demand for new housing with mass-produced large, single-family subdivisions on inexpensive farmland outside the city. New highways put these developments within an easy automobile commute of the city. Levit - town, NY and Park Forest, IL, both built in 1947, are early examples of these new developments. By the end of the 1950s, America’s suburban pattern of develop - ment—shopping malls and office parks serving single-family subdivisions— was firmly established. The 42,000 miles of new highways cre - ated under the Interstate Highway Act opened up vast amounts of land for development. Construction of highways through cities, together with urban re - newal, also accelerated the demolition of urban neighborhoods. 22 With more people moving to the subur - ban neighborhoods, many city neighbor - hoods became run down. The Housing Act authorized the use of federal money to tear down blighted city neighborhoods in order to encourage redevelopment, a process known as urban renewal. Much of the public housing built to replace neigh - borhood housing was badly designed and poorly maintained, encouraging neglected buildings and crime. 1960- 1980 As the suburbs prospered, cities declined. The loss of businesses and the middle class tax base made it difficult for cities to provide services. Urban riots in the 1960s fed the perception that cities were dangerous places. Public transportation could not compete with cars to serve the far away suburbs. Most city streetcars were removed from service by the end of the 1960s, and ridership on the remaining forms of public transit declined. The 1960s and 1970s saw the growth of sprawl—the unchecked development of rural areas on the edges of urban areas. Sprawl was characterized by the low den - sity of development, long distances be - tween work and home, and a reliance on cars for transportation. This type of plan often led to increased traffic, air pollution, and the loss of open space. Concerns about sprawl were voiced as early as the 1960s and 1970s. In the 1960s a series of laws and acts were written to set aside and protect land as wilderness areas. The 1970s saw the birth of the modern environmental plan - ning movement. During the 1970s the government became more involvement in environmental issues. The first Earth Day brought attention to the poor state of the environment. Activ - ists wanted the government to acknowl - edge that some problems were too big to be dealt with at the city, regional, or state level and needed to be addressed with national solutions. The federal government began to create national agencies to address and control environmental issues and create national standards to limit pollution. Economics became a bigger factor in controlling or limiting environmental ef - fects especially from private companies. The federal government began to offer financial incentives to encourage, rather than command, private companies to implement more environmentally friendly practices. During the early 1970s a sharp increase in the price of oil prompted some people to question our dependence on cars and oil from other countries. Interest grew in improving public transportation, and the federal government began to fund these improvements. In 1964, federal money first became available to improve existing subway systems. A few cities, like Wash - ington, D.C., had the density and potential for growth to support new subway sys - tems. In the 1980s and 1990s, new light rail systems, descendants of the streetcar, began in cities like San Diego, Portland, and Dallas. 23 1980 – 2000 There was a resulting backlash from orga - nizations against government regulations that they thought were too costly, inflex - ible, and burdensome to private industry. Development of Smart Growth planning, which considers economics in combi - nation with environmental and planning concerns, allowed for economic and population growth while protecting the environment. During the 1980s and 1990s “smart growth” evolved as an approach to devel - opment that counters some of the ef - fects of sprawl. Its goals include focusing growth in areas of existing infrastructure, creating transportation choices and walk - able communities, mixing housing types and uses, and protecting open space. Some local and state governments began using land use policy, tax incentives, and updated planning and zoning tools to pro - mote smart growth. At the same time the number of Land Trusts was on the rise. A greater number of Land Trusts, non-profit organizations, were being formed to acquire land and conservation easements to help preserve land for natural areas, farmland or forest - land. 2000 – Present In the 21st century for the first time in his - tory there are more people living in cities than in villages or rural areas. Urban areas are increasing in size and population. Urban ecological planning looks at the city as a part of an ecosystem and how hu - mans can be a part of the natural ecosys - tem without negatively affecting it. New Urbanism, also called traditional neighborhood design, is a design ap - proach similar to the smart growth frame - work. It is derived from the form and function of traditional American towns. It emphasizes a network of interconnected, narrow streets, buildings set close to the street, walkable neighborhoods, mixed uses, green spaces and transit oriented development as a way to promote better air quality and healthier people. Today, there is a growing movement for sustainable planning which is a holistic view of city and region. In this view lead - ers consider environmental, economic and social factors when planning settlements for people.
16528	https://www.youtube.com/watch?v=NQIbsPgoN_A	Optimization: Cone Minimum Surface Area JediMathster 67 subscribers 12 likes Description 1853 views Posted: 29 Oct 2020 A partial solution to minimize the Lateral surface area of a cone. Chapters 0:00 Intro 0:51 Problem 1:30 Formula for Surface Area 2:27 Partial Solution Transcript: Intro hello hope you're doing well this is another video about optimization in this problem i'm going to talk about a paper cup you know the paper cups that you get from a water dispenser that's shaped like a cone how are they constructed why do they have that certain shape so this is what i want to talk about if you want to dispense a certain amount of water or you want that cup to hold a certain amount of water how should you construct a cup so that you have the least amount of paper possible in other words you want to be a little bit sustainable here you want to be economical so here's a precise wording of the problem Problem suppose i have a cone-shaped paper drinking cup to hold 27 cubic centimeters of water i want to find the height and radius of the cup in other words kind of the shape of the cup how tall is it how wide is it that uses the smallest amount of paper up until now we've talked about the volume of a cone but we've never talked about the surface area because so to make our life a little bit easier here i did a little bit of searching i just typed in cone and i typed in an s Formula for Surface Area and lo and behold i have the surface area of a cone and in fact there's lots of different formulas that you can choose here now this is the surface area of a cone and this is actually the one for a closed bottom on the cone what we really want is something called the lateral surface of this in other words just the sloped part of it not the bottom and i know i'm going to turn this upside down because i've got to put water inside of it but there's a closed surface area and an open surface area a closed top surface area and a closed top surface area so the question if i go back i have 27 cubic centimeters to use here so i know a volume equation is going to come into play so let's see how i do this okay Partial Solution so i draw my picture i'm really not going to use this picture that much because well i have an h and an r on here and what's given in the problem well i know the volume is 27 cubic centimeters so i know that my volume formula one third part pi r squared h if i combine this with the 27 27 times 3 is 81 if i solve for h i get 81 divided by pi r cubed that's what h is i could also solve for r as well but i'm choosing to solve for h in this problem it is totally up to you now the surface area of a cone there's a closed top and an open top we want the open top formula we want this one because we want to put water inside the cup we don't want a lid on this thing pi r times the square root of h squared plus r squared there's two variables there i want to convert this all into one variable what does that mean it means solve oh solving for one of the variables and making a substitution so if i make that substitution i get pi r times h squared but h is 81 over pi r squared plus h squared now i have not plus h squared plus r squared plus r squared now i have a function for the surface area in one variable this is an open interval problem because i can make r as large as i want and if i make r say a thousand h is going to be really really tiny but it's still going to have 27 cubic centimeters along it will be unusable but it still exists so this is an open interval problem so we cannot use a closed interval method here so at this point i hope you know how to finish this problem you're going to have to take a derivative i suggest using some software to do this setting that derivative equal to zero finding the critical values and then going through a first or second derivative test for this i suggest using the first derivative test to actually classify your critical values in terms of min's and max and that's what i'm going to do for this problem i'm going to stop it here this is the complete setup i'm going to leave it up to you to finish the calculus on it i hope you found this helpful and if you have any questions just send me an email have a good day be safe bye
16529	https://webspace.maths.qmul.ac.uk/p.j.cameron/C50/no7.pdf	C50 Enumerative & Asymptotic Combinatorics Notes 7 Spring 2003 This part of the notes is about species. We begin with two preliminary topics: labelled and unlabelled objects; and Cayley’s formula for the number of trees on n vertices. Labelled and unlabelled Group actions can be used to clarify the difference between two types of counting of combinatorial objects, namely counting labelled and unlabelled objects. Typically, we are counting structures “based on” a set of n points: these may be partitions or permutations, or more elaborate relational structures such as graphs, trees, partially ordered sets, etc. An isomorphism between two such objects is a bijection between their base sets which preserves the structure. A labelled object is simply an object whose base set is {1,2,...,n}. Two objects count as different unless they are identical. On the other hand, for unlabelled objects, we wish to count them as the same obtain one from the other by re-labelling the points of the base set. In other words, an unlabelled object is an isomorphism class of objects. For example, for graphs on three vertices, there are eight labelled objects, but four unlabelled ones. Now the symmetric group Sn acts on the set of all labelled objects on the set {1,...,n}; its orbits are the unlabelled objects. So counting unlabelled objects is equivalent to counting orbits of Sn in an appropriate action. A given object A has an automorphism group Aut(A), consisting of all permu-tations of the set of points which map the object to itself. The number of different labellings of A is n!/\|Aut(A)\|, since of the n! labellings, two are the same if and only if they are related by an automorphism of A. (More formally, labellings correspond bijectively to cosets of Aut(A) in the symmetric group Sn.) So the number of labelled objects is ∑ A n! \|Aut(A)\|, 1 where the sum is over the unlabelled objects on n points. The cycle index method can be applied to give more sophisticated counts. For example, let us count graphs on 4 vertices. The number of pairs of vertices is 6, and each pair is either an edge or a non-edge. So the number of labelled graphs is 26 = 64, and the number of labelled graphs with k edges is 6 k for k = 0,...,6. In order to count orbits, we must let S4 act on the set of 64 graphs. But we can think of a graph as the set of 4 2 = 6 pairs of vertices with a ﬁgure (either an edge or a non-edge) attached to each. So we must compute the cycle index of S4 acting on pairs of vertices. Table 1 gives details. The notation 1221, for example, means “two ﬁxed points and one 2-cycle”. Such an element, say the transposition (1,2), ﬁxes the two pairs {1,2} and {3,4}, and permutes the other four pairs in two 2-cycles; so its cycle structure on pairs is 1222. Cycles on Cycles on Number vertices pairs 14 16 1 1221 1222 6 22 1222 3 13 32 8 4 24 6 Table 1: Cycle index of S4 So the cycle index of the permutation group G induced on pairs by S4 is Z(G) = 1 24(s6 1 +9s2 1s2 2 +8s2 3 +6s2s4). Now if we take edges to have weight 1 and non-edges to have weight 0 (that is, ﬁgure-counting series A(x) = 1+x), the function-counting series is B(x) = 1+x+2x2 +3x3 +2x4 +x5 +x6, the generating function for unlabelled graphs on four vertices by number of edges. We conclude by summarising some of our earlier results on counting labelled and unlabelled structures. Table 2 gives the numbers of labelled and unlabelled structures on n points; B(n) and p(n) are the Bell and partition numbers. We see from the table that it is possible, even in very natural cases, to have the same number of labelled objects but different numbers of unlabelled ones, or vice versa. 2 Structure Labelled Unlabelled Subsets 2n n+1 Partitions B(n) p(n) Permutations n! p(n) Total orders n! 1 Table 2: Labelled and unlabelled Cayley’s Theorem We begin with a particular species where there is a simple but unexpected formula for the labelled counting problem. A tree is a connected graph with no cycles. It is straightforward to show that a tree on n vertices contains n −1 edges, and that any connected graph has a spanning tree (that is, some set of n −1 of its edges forms a tree). Moreover, any tree has a vertex lying on only one edge (since the average number of edges per vertex is 2(n −1)/n < 2). Such a vertex is called a leaf. If we remove from a tree a leaf and its incident edge, the result is still a tree. Cayley’s Theorem states: Theorem 1 The number of labelled trees on n vertices is nn−2. There are many different proofs of this theorem. Below, we will see two proofs which are made clearer by means of the concept of species. But ﬁrst, one of the classics: Pr¨ ufer’s proof of Cayley’s Theorem We construct a bijection between the set of all trees on the vertex set {1,...,n} and the set of all (n−2)-tuples of elements from this set. The tuple associated with a tree is called its Pr¨ ufer code. First we describe the map from trees to Pr¨ ufer codes. Start with the empty code. Repeat the following procedure until only two vertices remain: select the leaf with smallest label; append the label of its unique neighbour to the code; and then remove the leaf and its incident edge. Next, the construction of a tree from a Pr¨ ufer code P. We use an auxiliary list L of vertices added as leaves, which is initially empty. Now, while P is not empty, we join the ﬁrst element of P to the smallest-numbered vertex v which is not in either P or L, and then add v to L and remove the ﬁrst element of P. When P is empty, two vertices have not been put into L; the ﬁnal edge of the tree joins these two vertices. I leave to the reader the task of showing that these two constructions deﬁne inverse bijections. The method actually gives much more information: 3 Proposition 2 In the tree with Pr¨ ufer code P, the valency of the vertex i is one more than the number of occurrences of i in P. For, at the conclusion of the second algorithm, if we add in the last two vertices to L, then L contains each vertex precisely once; and edges join each of the ﬁrst n−2 vertices of L to the corresponding vertex in P, together with an edge joining the last two vertices of L. Using this, one can count labelled trees with any prescribed degree sequence. Species and counting Species, invented by Andr´ e Joyal in 1980, provide an attempt to unify some of the many structures and techniques which appear in combinatorial enumeration. I don’t attempt to be too precise about what a species is. Think of it as a set of “points” carrying some structure (a graph, a poset, a permutation, etc.) We can ask for the number of labelled or unlabelled structures on n points in a given species. Almost the only thing we assume about a species G is that, for each n, there are only a ﬁnite number of G-objects on n points (so that we can count them). The only property we use of the objects in a species is that we “know” whether a bijective map between the point sets of two objects is an isomorphism between them (and hence we know the automorphism group of each object). We make one further (inessential but convenient) assumption, namely that there is a unique object on the empty set of points. We say that two species are equivalent (written G ∼H ) if there is a bijection between the objects of the two species on a given point set such that the automorphism groups of corresponding objects are equal. The most important formal power series associated with a species is its cycle index, which is deﬁned by the rule ˜ Z(G) = ∑ A∈G Z(Aut(A)), where Aut(A) is the automorphism group of A. Clearly, equivalent objects have the same cycle index. The cycle index is well-deﬁned since a monomial sa1 1 ···sar r arises only from cycle indices involving n = ∑r i=1 iai points, and by assumption there are only ﬁnitely many of these. There are two important specialisations of the cycle index of a species G; these are the exponential generating function G(x) = ∑ n≥0 Gnxn n! 4 for the number Gn of labelled n-element G-objects (that is, objects on the point set {1,...,n}); and the ordinary generating function g(x) = ∑ n≥0 gnxn for the number gn of unlabelled n-element G-objects (that is, isomorphism classes). Theorem 3 Let G be a species. Then (a) G(x) = ˜ Z(G;s1 ←x,si ←0 for i > 1); (b) g(x) = ˜ Z(G;si ←xi). Proof The number of different labellings of an object A on n points is clearly n!/\|Aut(A)\|. So it is enough to show that, for any permutation group G, we have Z(G;s1 ←x,si ←0 for i > 1) = xn/\|G\|, Z(G;si ←xi) = xn. The ﬁrst equation holds because putting si = 0 for all i > 1 kills all permutations except the identity. The second holds because, with this substitution, each group ele-ment contributes xn, and the result is 1/\|G\|∑g∈G xn = xn. Examples of species There are a few simple species for which we can do all the sums explicitly. Example: Sets The species S has as its objects the ﬁnite sets, with one set of each cardinality up to isomorphism. Its cycle index was calculated in Chapter 5: ˜ Z(S) = ∑ n≥0 (Sn) = exp ∑ i≥1 si i ! . Hence we ﬁnd that S(x) = exp(x), s(x) = exp ∑ i≥1 xi i ! = exp(−log(1−x)) = 1 1−x, in agreement with the fact that Sn = sn = 1 for all n ≥0. 5 Example: Total orders Let L be the species of total (or linear) orders. Each n-set can be totally ordered in n! ways, all of which are isomorhic, and each of which is rigid (that is, has the trivial automorphism group). We have ˜ Z(L) = ∑ n≥0 sn 1 = 1 1−s1 , so that L(x) = l(x) = 1 1−x. Example: Circular orders The species C consists of circular orders. An element of this species corresponds to placing the points of the object around a circle, where only the relative positions are considered, and there is no distinguished starting point. Thus, there is just one unlabelled n-element object in C for all n, and the number of labelled objects is equal to the number (n−1)! of cyclic permutations for n ≥1. The unique n-element structure has φ(m) automorphisms each with n/m cycles of length m for all m dividing n, where φ is Euler’s function. Hence ˜ Z(C) = 1−∑ m≥1 φ(m) m log(1−sm), C(x) = 1+ ∑ n≥1 xn n = 1−log(1−x), c(x) = ∑xn = 1 1−x. Example: Permutations An object of the species P consists of a set carrying a permutation. We will see later how P can be expressed as a composition, from which its cycle index can be deduced (Exercise 2 on Sheet 7). We have ˜ Z(P) = ∏ n≥1 (1−sn)−1, P(x) = 1 1−x, p(x) = ∏ n≥1 (1−xn)−1. The function p(x) is the generating function for number partitions. For, as we saw earlier, an unlabelled permutation is the same as a conjugacy class of permutations; and conjugacy classes are determined by their cycle structure. 6 Operations on species There are several ways of building new species from old; only a few important ones are discussed here. Products Let G and H be species. We deﬁne the product K = G ×H as follows: an object of K on a set X consists of a distinguished subset Y of X, a G-object on Y, and a H -object on X \Y. Since these objects are chosen independently, it is easy to check that ˜ Z(G ×H ) = ˜ Z(G) ˜ Z(H ). Since the generating functions for labelled and unlabelled structures are specialisa-tions of the cycle index, we have similar multiplicative formulae for them. For example, if S, G and G◦are the species of sets, graphs, and graphs with no isolated vertices respectively, then G ∼S ×G◦. Substitution Let G and H be species. We deﬁne the substitution K = G[H ] as follows: an object of K on a set X consists of a partition of X, an H -object on each part of the partition, and a G-object on the set of parts of the partition. Alternatively, we may regard it as a G-object in which every point is replaced by a non-empty H -object. The cycle index is obtained from that of G by the substitution si ←˜ Z(H ;sj ←sij)−1 for all i. (The −1 in the formula corresponds to removing the empty H -structure before substituting.) From this, we see that the exponential generating functions for labelled structures obey the simple substitution law: K(x) = G(H(x)−1). The situaation for unlabelled structures is more complicated, and k(x) cannot be ob-tained from g(x) and h(x) alone. Instead, we have k(x) = ˜ Z(G;si ←h(xi)−1). This equation also follows from the Cycle Index Theorem, since we are counting functions on G-structures where the ﬁgures are non-empty H -structures with weight equal to cardinality. 7 For example, if S, P and C are the species of sets permutations, and circular orders, then the standard decomposition of a permutation into disjoint cycles can be written P ∼S[C]. The counting series for labelled structures are given by S(x) = ∑ n≥0 xn n! = exp(x), P(x) = ∑ n≥0 n!xn n! = 1 1−x, C(x) = 1+ ∑ n≥0 (n−1)!xn n! = 1−log(1−x); so the equation above becomes 1 1−x = exp(−log(1−x)), So the decomposition of a permutation into cycles is the combinatorial equivalent of the fact that exp and log are inverse functions! Rooted (or pointed) structures Given a species G, let G∗be the species of rooted G-structures: such a structure consists of a G-structure with a distinguished point. We have ˜ Z(G∗) = s1 ∂ ∂s1 ˜ Z(G), and so G∗(x) = x d dxG(x). Sometimes it is convenient to remove the distinguished point. This just removes the factors s1 and t in the above formulae, so that this operation corresponds to differenti-ation. As a result, we denote the result by G′. For example, if C is the class of cycles, then C ′ corresponds to the class L of total (linear) orders. We have L(x) = d dxC(x) = d dx(1−log(1−x)) = 1 1−x, in agreement with the preceding example. 8 Cayley’s Theorem revisited The notion of species can be used to give two further proofs of Cayley’s Theorem. First proof Let L and P be the species of total (or linear) orders and permutations, respectively. These species are quite different, but have the property that the numbers of labelled objects on n points are the same (namely n!). Hence the numbers of labelled objects in the two species L[T ∗] and P[T ∗] are equal. (Here T ∗is the species of rooted trees.) Consider an object in L[T ∗]. This consists of a linear order (x1,...,xr), with a rooted tree Ti at xi for all i. I claim that this is equivalent to a tree with two distin-guished vertices. Take edges {xi,xi+1} for i = 1,...,r −1, and identify xi with the root of Ti for all i. The resulting graph is a tree. Conversely, given a tree with two distinguished vertices x and y, there is a unique path from x to y in the tree, and the remainder of the tree consists of rooted trees attached to the vertices of the path. Now consider an object in P[T ∗]. Identify the root of each tree with a point of the set on which the permutation acts, and orient each edge of this tree towards the root. The resulting structure deﬁnes a function f on the point set, where • if v is a root, then f(v) is the image of v under the permutation; • if v is not a root, then f(v) is the unique vertex for which (v, f(v)) is a directed edge of one of the trees. Conversely, given a function f : X →X, the set Y of periodic points of f has the property that f induces a permutation on it; the pairs (v, f(v)) for which v is not a periodic point have the structure of a family of rooted trees, attached to Y at the point for which the iterated images of v under f ﬁrst enter Y. So the numbers of trees with two distinguished points is equal to the number of functions from the vertex set to itself. Thus, if there are F(n) labelled trees, we see that n2F(n) = nn, from which Cayley’s Theorem follows. Second proof As in the preceding proof, let T ∗denote the species of rooted trees. If we remove the root from a rooted tree, the result consists of an unordered collection of trees, each of which has a natural root (at the neighbour of the root of the original tree). Conversely, given a collection of rooted trees, add a new root, joined to the roots 9 of all the trees in the collection, to obtain a single rooted tree. So, if E denotes the species consisting of a single 1-vertex structure, and S the species of sets, we have T ∗∼E ×S[T ∗]. Hence, for the exponential generating functions for labelled structures, we have T ∗(x) = xexp(T ∗(x)). This is, formally, a recurrence relation for the coefﬁcients of T ∗(x), and we need to show that the nth coefﬁcient is nn−1. This can be done most easily with the technique of Lagrange inversion, which is discussed in the next section. Lagrange inversion A formal power series over a ﬁeld, with zero constant term and non-zero term in x, has an inverse with respect to composition. Indeed, the set of all such formal power series is a group, which has recently become known as the Nottingham group. However, the basic facts are much older. The associative, closure, and identity laws are obvious, and the rule for ﬁnding the inverse is known as Lagrange inversion. The basic fact can be stated as follows. Proposition 4 Let f be a formal power series over Q, with f(0) = 0 and f ′(0) ̸= 0. Then there is a unique formal power series g such that g(f(x)) = x; the coefﬁcient of yn in g(y) is dn−1 dxn−1 x f(x) n x=0 . n!. This can be expressed in a more convenient way for our purpose. Let φ(x) = x f(x). Then the inverse function g is given by the functional equation g(y) = yφ(g(y)). Then Lagrange inversion has the form g(y) = ∑ n≥1 bnyn n! , where bn = dn−1 dxn−1φ(x)n x=0 . 10 Example: Cayley’s Theorem The exponential generating function for rooted trees satisﬁes the equation T ∗(x) = xexp(T ∗(x)). With φ(x) = exp(x), we ﬁnd that the coefﬁcient of yn/n! in T ∗(y) is dn−1 dxn−1 exp(nx) x=0 = nn−1, proving Cayley’s Theorem once again. What is a species? We have proceeded this far without ever giving a precise deﬁnition of a species. The informal idea is that an object of a species is constructed from a ﬁnite set, and bijec-tions between ﬁnite sets induce isomorphisms of the objects built on them. It turns out that mathematics does provide a language to describe this, namely category theory. It would take us too far aﬁeld to give all the deﬁnitions here. In essence, a category consists of a collection of objects with a collection of morphisms between them. In the only case with which we deal, objects are sets and morphisms are set mappings. In particular, the class S whose objects are all ﬁnite sets and whose morphisms are all bijections between them satisﬁes the axioms for a category. Now a species is simply a functor F from S to itself. This means that F associates to each ﬁnite set S a set F(S), and to each bijection f : S →S′ a bijection F(f) : F(S) → F(S′), such that F respects composition and identity (that is, F(f1 f2) = F(f1)F(f2) and F(1S) = 1F(S), where 1S is the identity map on S). The standard reference on species (apart from Joyal’s original paper) is the book by Bergeron, Labelle and Leroux, Combinatorial Species and Tree-like Structures, Encyclopedia of Mathematics and its Applications 67, Cambridge University Press, Cambridge, 1998. 11
16530	https://courses.cs.washington.edu/courses/cse473/23au/notes/Search-Notes1.pdf	UW CSE 473 Notes 1 Notes by: Khushi Khandelwal, Rohan Gupta 1 Introduction 1.1 Agents An agent is an entity that perceives and acts. A rational agent, an entity that has goals or prefer-ences and tries to perform a series of actions that yield the best/optimal expected outcome given these goals. Rational agents exist in an environment, which is specific to the given instantiation of the agent. A rational agent selects actions that maximize its (expected) utility. A reflex agent chooses actions only based on the current percept. It is one that does not consider the future consequences of its actions, but rather selects an action based solely on the current state of the world. It considers how the world is. Planning agents make decisions based on (hypothesized) consequences of actions. A planning agent has a a model of how the world evolves in response to actions. 2 Search Problems Through a search problem, we try to find a new state that the agent can reach from its current state that gets it closer to its goal in the most optimal way. A search problem consists of 4 things: • A state space - The set of all possible states that are possible in your given world • A successor function - A function that takes in a state and an action and computes the cost of performing that action as well as the successor state, the state the world would be in if the given agent performed that action • A start state - The state in which an agent exists initially • A goal test - A function that takes a state as input, and determines whether it is a goal state A solution is a sequence of actions (a plan) which transforms the start state to a goal state. At the start state, the agent explores the state space using the successor function, iteratively computing successors of various states until it arrives at a goal state at which point it will have determined a path from the start state to the goal state, which is a plan. A world state contains all information about a given state, whereas a search state contains only the information about the world that’s necessary for planning. 3 State Space Graphs A graph is defined by a set of nodes and a set of edges connecting various pairs of nodes. These edges may also have weights associated with them. A state space graph is constructed with states representing nodes, with directed edges existing from a state to its successors. The arcs represent successors (action results) and the goal test is a set of goal nodes (maybe only one). In a state space graph, each state is represented exactly once. State space graph is a mathematical concept, and we do not explicitly build and store the state graph in memory as they are often huge. 1 UW CSE 473 Notes 1 4 Search Trees In a search tree, on the other hand, a state can appear multiple times. This is because though search trees are also a class of graph with states as nodes and actions as edges between states, each state/node encodes not just the state itself, but the entire path (or plan) from the start state to the given state in the state space graph. Since there often exist multiple ways to get from one state to another, states tend to show up multiple times in search trees. Search problems are solved using search trees, where we very carefully store a select few nodes to observe at a time, iteratively replacing nodes with their successors until we arrive at a goal state. There exist various methods by which to decide the order in which to conduct this iterative replacement of search tree nodes. 5 Search Algorithms and Uninformed Search For finding a plan to get from the start state to a goal state, we maintain an outer fringe of partial plans derived from the search tree. We continually expand our fringe by removing a node (which is selected using our given strategy) corresponding to a partial plan from the fringe, and replacing it on the fringe with all its children. To enable comparisons between different search algorithms, some search algorithm properties we will be discussing are as follows: Complete: Is the search algorithm guaranteed to find a solution if one exists? Optimal: Is it guaranteed to find the least cost path? Time Complexity: What is the runtime of this algorithm? Space Complexity: How much space (memory) does this algorithm need? 2 UW CSE 473 Notes 1 Since we run search algorithms on search trees, it is also important to understand features of a search tree: The branching factor b - The increase in the number of nodes on the frontier each time a frontier node is dequeued and replaced with its children is O(b). At depth k in the search tree, there exists O(bk) nodes. The maximum depth m. The depth of the shallowest solution s. We will be discussing three search algorithms in this section: Depth-First Search (DFS), Breadth-First Search (BFS), and Uniform-Cost Search (UCS). 5.1 DFS The strategy with DFS is to expand a deepest node first (dig into the search tree) and is imple-mented by treating the fringe as a Last-In-First-Out stack. Complete: No, since m could be infinite. Removing cycles in the search tree can fix this problem. With small modification to the original algorithm (ensure a node is only visited once), DFS can be complete. Optimal: No, since we could find a goal node that is deeper than s (the shallowest solution). Time Complexity: In the worst case, we go through the whole search tree. So, the runtime is O(bm) (assuming m is finite). Space Complexity: O(bm), since we only keep track of siblings on the path from the root node. 5.2 BFS The strategy with BFS is to expand a shallowest node first (sweep the search tree) and is imple-mented by treating the fringe as a First-In-First-Out queue. 3 UW CSE 473 Notes 1 Complete: Yes, since we find a solution if it exists. Optimal: Yes, if we assume all edges have equal weight. Time Complexity: O(bs), where s is the shallowest solution. In the worst case (no solution), we sweep through the whole tree. Space Complexity: O(bs), since we keep track of roughly the next layer, where s is the depth of the shallowest goal. 5.3 UCS The algorithms covered so far do not account for the weight of edges, which can likely change the lowest-cost solution. UCS helps with this problem by accounting for the weights of edges. The strategy with UCS is to expand a cheapest cost node first and is implemented by treating the fringe as a First-In-First-Out priority queue. 4 Let C be the cost of the cheapest solution and let ϵ be the lowest cost of the edges. Then, the "effective depth" is roughly C/ϵ. Complete: Yes, since we find a solution if it exists. Optimal: Yes, because we keep track of edge weights. Time Complexity: O(bC∗/ϵ). Space Complexity: O(bC∗/ϵ), since we keep track of roughly the last cost tier. What is the problem with these search algorithms? They explore options in a methodical, predetermined manner and lack information about the goal node. We will look at algorithms that overcome this limitation in the next section. 5
16531	https://ocw.mit.edu/courses/14-54-international-trade-fall-2016/resources/mit14_54f16_lecture_10/	14.54 International Trade Lecture 10: Production Functions 14.54 Week 6 Fall 2016 14.54 (Week 6) Production Functions Fall 2016 1 / 20 Today’s Plan 1 2 3 4 5 Midterm Results Properties of Production Functions (2 Factors) Isoquants Input Choice and Cost Minimization Relative Factor Demand Graphs on slides 7, 10-17, and 19 are courtesy of Marc Melitz. Used with permission. 14.54 (Week 6) Production Functions Fall 2016 2 / 20 Introduction We will now introduce another factor of production: capital Can also think about other production factors: land, skilled versus unskilled labor, ... 14.54 (Week 6) Production Functions Fall 2016 3 / 20 What issues can be addressed when production requires more than a single factor? In the short-run, some factors are more ‘flexible’ than others: how quickly and at what cost can factors move from employment in one sector to another? Example of labor and capital: After a U.S. state is hit with a regional shock, unemployment rate falls back to national average within 6 years (most inter-regional employment reallocations also involve worker reallocations across sectors) In comparison, capital depreciates over 15-20 years and structures over 30-50 years In the short-run, labor is more ”flexible” than capital across sectors Distributional consequences across factors from changes in goods prices even in the long run 14.54 (Week 6) Production Functions Fall 2016 4 / 20 Production Function Under constant returns to scale, a production function with one factor can be summarized by a single number: unit input requirement (an overall productivity index) With more than one factor, a production function also characterizes the substitutability between the factors of production (as well as an overall productivity index) We will now assume that QC = FC (KC , LC ) and QF = FF (KF , LF ) We will continue to assume constant returns to scale: F (tK , tL ) = tF (K , L) for any t > 0 And will also assume diminishing marginal returns to a single factor ... as well as factor complementarity 14.54 (Week 6) Production Functions Fall 2016 5 / 20 Properties of Production Function Marginal Products MPK = ∂F (K , L)/∂K = F K (K , L) and MPL = ∂F (K , L)/∂L = F L(K , L) For any production function, must have MPK ≥ 0 and MPL ≥ 0 Diminishing marginal returns to a single factor assumption: MPK = F K (K , L) is decreasing in K and MPL = F L(K , L) is decreasing in L Factor Complementarity: MPK = F K (K , L) is increasing in L and MPL = F L(K , L) is increasing in K Note: these assumptions are all compatible with constant returns to all factors 14.54 (Week 6) Production Functions Fall 2016 6 / 20 Production Function and Marginal Product Hold capital K constant: An increase in K will shift up both curves 14.54 (Week 6) Production Functions Fall 2016 7 / 20 Implications of Constant Returns to Scale for Marginal Products Recall the assumption that F (tK , tL ) = tF (K , L) for any t > 0 Differentiate with respect to K : F K (tK , tL )t = tF K (K , L) ⇔ F K (tK , tL ) = F K (K , L) A proportional increase in K and L leaves the MPK unchanged Similarly: F L(tK , tL ) = F L(K , L) A proportional increase in K and L leaves the MPL unchanged 14.54 (Week 6) Production Functions Fall 2016 8 / 20 Production with Competitive Output and Factor Markets Factors are paid the value of their marginal products: r = pMPK and w = pMPL Marginal cost pricing (also average cost pricing with constant returns to scale): rK + wL p = MC = AC = Q Hence, no economic profits p = pQ − rK − wL = 0 Can think of payment to capital K either as value of MPK or as revenue left over to firm owners (capital owners) after labor has been paid: rK = pQ − wL 14.54 (Week 6) Production Functions Fall 2016 9 / 20 Production Functions and Isoquants An isoquant associated with a production function FC (KC , LC ) is the set of inputs (KC , LC ) that can be used to produce a given output level QC The set of isoquants can be used to represent any given production function The curvature of the isoquant captures the substitutability of the factors in production (parallel with indifference curves) 14.54 (Week 6) Production Functions Fall 2016 10 / 20 Unit Isoquant The unit isoquant is the combination of inputs (KC , LC ) that can be used to produce one unit of output With multiple factors, firms can choose different combinations of unit input requirements aLC and aKC 14.54 (Week 6) Production Functions Fall 2016 11 / 20 Isoquants and Marginal Products The slope of any isoquant at (LC , KC ) is MPL C /MPK C Take the total derivative of FC (KC , LC ) = QC : / ∂FC ∂FC dK C ∂FC ∂LC MPL C dK C + dL C = 0 ⇔ = − / = − ∂KC ∂LC dL C ∂FC ∂KC MPK C Thus, with constant returns to scale production, slopes of isoquants do not change along any ray from the origin (why?) In addition, proportional change in output is equal to proportional change in inputs 14.54 (Week 6) Production Functions Fall 2016 12 / 20 Firm Input Choice and Cost Minimization Given factor prices w and r , firm minimizes cost of producing QC units of output rK C + wL C by choosing KC and LC such that MPL C /MPK C = w /r (see appendix to Ch. 4) Note that this is also implied by factor prices equal to the value of their marginal products: w MPL C w = pC MPL C and r = pC MPK C ⇒ = r MPK C 14.54 (Week 6) Production Functions Fall 2016 13 / 20 Firm Input Choice and Cost Minimization (Cont.) Under constant returns to scale production, a firm will always produce QC units of output with the same LC /KC as it uses to produce one unit of output At given w and r , a firm choose unit input requirements aLC and aKC such that MPL C /MPK C = w /r The minimized unit cost is wa LC + ra KC (constant AC = MC ) Firm produces QC units of output using LC = QC aLC , KC = QC aKC Thus only need to solve cost minimization for QC = 1 14.54 (Week 6) Production Functions Fall 2016 14 / 20 Relative Factor Demand The relationship between LC /KC and w /r can be represented by a relative demand curve for factors (which is independent of the amount of output QC produced) Note similarity with the relative demand curve for goods (and relationship with factor/good substitutability) Also note that textbook uses a relationship between KC /LC and w /r (which implies a positively sloped curve) 14.54 (Week 6) Production Functions Fall 2016 15 / 20 Leontief Technology: Fixed Factor Production Under Leontief technology, there is no substitutability between production factors: Producing a unit of output C requires a fixed amount of labor aLC and capital aKC The production function is QC = min {LC /aLC , KC /aKC } 14.54 (Week 6) Production Functions Fall 2016 16 / 20 Differences in Technology and Relative Factor Demands Consider the following differences between the unit isoquants: Definition of factor intensity in production: C is relatively labor intensive if, at any given relative factor price w /r LC aLC LF aLF = > = KC aKC KF aKF So C firms will always hire relatively more L (and inversely for F firms) 14.54 (Week 6) Production Functions Fall 2016 17 / 20 Factor Intensity and Factor Cost Shares If C is relatively labor intensive (relative to F ) then aLC /aKC > aLF /aKF C firms always hire relatively more labor F firms always hire relatively more capital and also wL C wL F rK C rK F and < wL C + rK C wL F + rK F wL C + rK C wL F + rK F C firms always devote a higher share of their total cost to labor F firms always devote a higher share of their total cost to capital 14.54 (Week 6) Production Functions Fall 2016 18 / 20 A Special Case What can be said about factor intensity if relative factor demand curves look like this? Then factor intensity depends on relative factor price This is called a factor intensity reversal We will not consider this special case 14.54 (Week 6) Production Functions Fall 2016 19 / 20 MIT OpenCourseWare 14.54 International Trade Fall 2016 For information about citing these materials or our Terms of Use, visit: .
16532	https://www.tsc.fl.edu/media/divisions/learning-commons/resources-by-subject/math/Slope-Intercept-Form-and-Point-Slope-Form-Updated.pdf	Slope-Intercept Form and Point-Slope Form Slope of the line m = rise run = 𝑦ଶ−𝑦ଵ 𝑥ଶ−𝑥ଵ Slope-Intercept Form 𝑦= 𝑚𝑥+ 𝑏; where m is slope and b is the y-intercept Point-Slope Form 𝑦−𝑦ଵ= 𝑚(𝑥−𝑥ଵ) or 𝑦= 𝑚(𝑥−𝑥ଵ) + 𝑦ଵ Slope of parallel lines 𝑚ଵ= 𝑚ଶ (slopes are the same) Slope of perpendicular lines 𝑚ଵ= − ଵ ௠మ (slopes are opposite & reciprocal) Equations of Horizontal and Vertical Lines Horizontal: 𝑦= 𝑏 Vertical: 𝑥= 𝑎; where a is the x intercept Example (1): Write the slope - intercept equation of a line which passes through (0,-7) whose slope is 2. Solution: Slope-intercept equation is 𝑦= 𝑚𝑥+ 𝑏 . What we need to complete this equation are slope (𝑚) & y-intercept (𝑏), and the problem provides both parts. 𝑚= 2 and 𝑏= −7 The equation of the line is 𝒚= 𝟐𝒙−𝟕 Example (2): Write the slope-intercept equation of a line which passes through (0,4) and (3, −5). Solution: Slope-intercept equation is 𝑦= 𝑚𝑥+ 𝑏 What we need to complete this equation are slope (𝑚) & y-intercept (𝑏), however, we only have y-intercept. To ﬁnd the slope, 𝑚= rise run = −5 −4 3 −0 = −9 3 = −3 𝑚= −3 and 𝑏= 4 The equation of the line is 𝒚= −𝟑𝒙+ 𝟒 Example (3): Write the slope-intercept equation of a line which passes through (−1,4) whose slope is 5. Solution: Slope-intercept equation is 𝑦= 𝑚𝑥+ 𝑏 . What we need to complete this equation are slope (𝑚) & y-intercept (𝑏), however, we only have slope. Here there are two ways to ﬁnd the equation of the line. Method I We will substitute 𝑚 and (𝑥ଵ, 𝑦ଵ) into the equation 𝑦= 𝑚𝑥+ 𝑏. 𝑚= 5 and (𝑥ଵ, 𝑦ଵ) = (−1,4) 4 = 5(−1) + 𝑏 4 = −5 + 𝑏 9 = 𝑏 𝒚= 𝟓𝒙+ 𝟗 Method II Since we are given slope m and an ordered pair (𝑥ଵ, 𝑦ଵ), we can use Point-slope form 𝑦−𝑦ଵ= 𝑚(𝑥−𝑥ଵ) to ﬁnd equation of the line. 𝑚= 5 and (𝑥ଵ, 𝑦ଵ) = (−1,4) 𝑦−4 = 5(𝑥−−1) 𝑦−4 = 5(𝑥+ 1) 𝑦−4 = 5𝑥+ 5 𝑦= 5𝑥+ 5 + 4 𝒚= 𝟓𝒙+ 𝟗 Example (4): Write the slope-intercept equation of a line which passes through (1,3) and (−5, −1) Solution: Slope-intercept equation is 𝑦= 𝑚𝑥+ 𝑏. What we need to complete this equation are slope & y-intercept. However, we are given two ordered pairs (𝑥ଵ, 𝑦ଵ) and (𝑥ଶ, 𝑦ଶ) without slope or y-intercept. Therefore, we need to ﬁnd the slope ﬁrst. Then we can use the two methods discussed on Example (3) to ﬁnd the equation of the line. m = rise run = −1 −3 −5 −1 = −4 −6 = 2 3 Method I Now we have slope, we will substitute m and (𝑥ଵ, 𝑦ଵ) in the form 𝑦= 𝑚𝑥+ 𝑏 to solve for 𝑏. 𝑦= 𝑚𝑥+ 𝑏 3 = 2 3 (1) + 𝑏 3 = 2 3 + 𝑏 3 −2 3 = 𝑏 9 3 −2 3 = 𝑏 7 3 = 𝑏 𝒚= 𝟐 𝟑𝒙+ 𝟕 𝟑 Method II We also can use Point-slope form to ﬁnd the equation of the line. 𝑦−𝑦ଵ= 𝑚(𝑥−𝑥ଵ) 𝑦−3 = 2 3 (𝑥−1) 𝑦−3 = 2 3 𝑥−2 3 𝑦= 2 3 𝑥−2 3 + 3 𝒚= 𝟐 𝟑𝒙+ 𝟕 𝟑 Example (5): Write the equations of a parallel and perpendicular line to the equation: 𝑦= 2𝑥+ 1. Parallel: Parallel lines have equal slopes, so any line with 𝑚= 2 is parallel. Perpendicular: Perpendicular lines have a negative reciprocal slope. Any line with 𝑚= − ଵ ଶ will be perpendicular. Example (6): Write the equations of a vertical line and a horizontal line passing through (3,4). Vertical: 𝑥= 3 Horizontal: 𝑦= 4 Exercises: 1. Write the slope - intercept equation of a line which passes through (1,3) whose slope is 4. 2. Write the slope-intercept equation of a line which passes through (1,2) and (5,3). 3. Write the slope-intercept equation of a line which is parallel to 𝑦= 4𝑥+ 1, passing through (8,2). 4. Write the slope-intercept equation of a line which is perpendicular to 𝑦= −𝑥+ 2, passing through (−1,1). 5. Write an equation of a horizontal line which passes through (−2, −4). 6. Write an equation of a vertical line which passes through (−5,6). Answers: 1. 𝑦= 4𝑥−1 2. 𝑦= ଵ ସ𝑥+ ଻ ସ 3. 𝑦= 4𝑥−30 4. 𝑦= 𝑥+ 2 5. 𝑦= −4 6. 𝑥= −5
16533	https://zh.wikibooks.org/wiki/%E9%AB%98%E4%B8%AD%E6%95%B0%E5%AD%A6/%E4%B8%8D%E7%AD%89%E5%BC%8F%E4%B8%8E%E6%95%B0%E5%88%97/%E4%B8%8D%E5%8A%A8%E7%82%B9%E6%B3%95	跳转到内容 \| \| \| --- \| \| 中文维基教科书即时聊天：IRC频道為Libera #wikibooks-zh 連線，Telegram群组为@wikibooks_zh，QQ群为775925391，三者通过机器人互联。 \| [關閉] \| 高中数学/不等式与数列/不动点法添加链接维基教科书，自由的教学读本 < 高中数学阅读指南 [编辑] 在前面一阶递推关系的求解章节中我们已经知道，对于这样的数列，我们可以将其改写为，从而将整体地当作一个等比数列套用公式。如果系数更麻烦一点，比如对于这样的数列，原则上还是可以拆分常数，然后在等号的左右两边分别构造出等比数列。但是此时的常数如何拆分不那么容易直接尝试出来。一个比较容易想到的办法是采用待定系数法，先提前假设构造好的数列具有的形式，初步化简后再和原来的放在一起，比较各对应项的数值，即可确定与原关系式匹配的d值。本节介绍的不动点法则是确定这个待定构造系数的更直接、简便的方法。不动点法可以解决形如的非齐次一阶常系数线性递推关系式的构造，也可以解决形如的一阶常系数分式线性递推关系式的构造。除了这2种类型的递推式之外，少数可以转换为这2种情形的特例也可以使用不动点法求解构造系数。这类不动点法的主要特点就是：适用条件容易判断用起来爽（对于不少常见情况而言）解释起来很麻烦本节花费了不少篇幅提及这些方法背后的由来，其中多半不可避免地会介绍一些来自后续数学课程中的背景知识。如果读者对于不动点法的使用完全没有经验，建议先接触一些应用不动点解题的例题。了解不动点法解题的大致流程后，再根据需要了解其背后的由来，不然容易对其中的一些推敲思路产生困惑。考试要求 [编辑] 不动点法并不被列入普通高中的考试大纲，因此在常规的高中考试中不能直接声明是在借助不动点法解题。但是利用不动点法的确可以快速求出所需的待定系数，识别适合使用不动点的问题并能快速地在草稿纸上用此方法求出所需的系数还是比较有必要掌握的。基础知识 [编辑] 不动点的概念 [编辑] 定义：对于1个给定的函数，能使成立的自变量取值叫做该函数的不动点（fixed point）。与递推式形式一致的函数也被叫做递推关系式的特征函数（characteristic function）或背景函数。递推式特征函数的不动点也叫做递推式的不动点。不动点在对应函数图象上的对应点也叫做函数图象的不动点。在解决高中数列递推问题时，我们讨论的不动点都是可逆函数的不动点。可逆函数的不动点有以下特点：如果是可逆函数的不动点，等价于是其反函数的不动点。换句话说，可逆函数的不动点与其反函数的不动点是一一对应的。函数的不动点坐标具有的形式，即不动点一定是函数图象与特殊直线的交点。如果把可逆函数与其反函数的图象画在同一个直角坐标系中，则它们的图象交点都是不动点。不动点本身是拓扑学的知识点，表示物体或空间在某种拓扑变换的作用下，仍保持固定不动的点。它的几何意义是函数图象与恒等变换直线交点的横坐标。但是它也出现在有关数列的解题方法中。利用不动点简化一阶常系数线性递推关系式的通项求解 [编辑] 在从递推式构造等比数列的过程中，使用待定系数法所求的需要分配到等式两边的数值，刚好就可以用对应函数的不动点来确定。我们先给出用不动点法解决这类数列的通用解题步骤，然后结合例题说明它的具体运用。用不动点法可以求出一阶线性递推数列 (其中p和q为常数，且)的通项的一般式。其主要步骤为： (1) 写出与对应的特征函数 (2) 求出函数的不动点，即求解方程，解得： (3) 在原始的等号两边同时减去刚才求出的不动点，即： (4) 将整体地看作一个等比数列求解，求出的表达式后，的表达式自然也就知道了。可以证明，这样求得的最终结果形式为。但是实际上，解题时并不需要记住这个最终形式，只需要记住上面的几个关键步骤即可。提示：如果求出的不动点形式比较复杂（例如算出的解带着根号跑），此时使用不动点法对于简化数列构造时的系数运算也无法提供太多帮助。求解兔子数列的通项公式问题就是一个例子。一阶常系数线性递推关系式的不动点解法由来 [编辑] 关于在一阶常系数线性递推关系式的通项求解中，为什么会与不动点的概念产生联系，我们在这里给予一个比较简单的论证。先设再假设我们期待的构造好的等比数列形式用后一式的两端同时对前一式的两端作差，可得：最后一个式子显然说明所需要的常数p就是函数的不动点。从几何上讲，从截距不为零的一次函数（对应于上述非齐次一阶线性递推关系式）变形为正比例函数（对应于构造出来的等比数列满足的关系式）的变换就是一种图象平移操作，而此类函数的不动点提供了同时沿y轴方向和x轴方向进行图象平移的距离参考值。一阶常系数分式线性递推关系式的概念及其通项的待定系数解法 [编辑] 形如的一阶常系数分式线性递推关系式的构造也可以借助不动点的求解实现。提示：我们先在本小节简述如何使用待定系数法求这类分式线性递推式通项的大体思路，再在后面的小节介绍基于不动点法的优化和解题实例，最后谈谈在这类情形下可以使用不动点法的原理。定义：具有以下形式的递推式叫做常系数分式线性递推关系：利用不动点简化一阶常系数分式线性递推关系式的通项求解 [编辑] 我们接下来从不动点的角度，指出这种常系数分式线性递推关系的通项公式求解方法。提示：本小节前半部分展示的代数变形过程比较冗长罗嗦，主要就是讲述了2种情况下最基本的分类讨论，外加上常见优化步骤的思路由来。没有耐心的读者可以先看结论和具体例题，然后有时间再回过头来大致浏览一下这些纯字母化的运算和论证。设函数满足，其不动点为。其不动点满足的方程可能只有1个解，也可能有2个不同的解（不同的实数根或不同的虚数根）。如果为以上方程的任何一个根（从而也是函数的不动点），那么可以先对原递推式两边同时减去不动点：再对上式两侧同时颠倒可得：由上式可知此时整体来说是一个一阶常系数线性递推数列，且包含有取值为常数的非齐次项。如果不动点方程的2个根是重根，则有且。此时可以继续化简：即变为以为首项、为公差的等差数列，易知其通项为：既然在这种情形下都已经得到的解析式，从而的解析式也容易知道了。如果不动点方程有2个相异的根，那么最直接的想法是只能麻烦一点，针对得到的常系数线性递推式再使用不动点法构造一次：此时构成公比为的等比数列，即：如果方程有2个不等的根和（是不同的实数根或不同的虚数根都行），此时还有一种基于不动点但是构造方向与上述思路有一些差异的解法。先在原递推式的两侧分别减去这2个根，可以得到：如果观察上述2个式子，可以发现它们的右侧都有相同的分母，且右侧的分子形式非常相似，最好能利用这个相同点去掉它们的分母。于是不妨对上述2个式子的两端同时作商（相除）可得：，其中。易知此时整体来说是一个以为首项、k为公比的等比数列，所以得到的通项公式：既然已经得到的解析式，从而的解析式也容易知道了。这种方法此时可以更快地得到同样的最终结果：综上所述，我们可以将常系数分式线性递推式的通项求解思路简化如下：当只有1个不动点时，可以对原递推式的两边同时减去不动点，然后再同时取倒数，进而转化为等差数列求解。当有2个相异的不动点时，可以先对原递推式的两边同时减去第1个不动点并取倒数，再对原递推式的两边同时减去第2个不动点并取倒数，然后将分别得到的2个式子作商，进而转化为等比数列求解。分式线性变换的概念与分式线性递推式的不动点解法由来 [编辑] 定义：具有以下形式的函数叫做分式线性变换（linear fractional transformation）：因为德国数学家奥古斯特·莫比乌斯曾对其进行过大量研究，所以分式线性变换也叫做莫比乌斯变换（Möbius transformation）。通过简单的分式分解，易知莫比乌斯函数可以分解为以下形式：由于我们只考虑自变量值取实数的情形，因此可以对于特定的一组系数可以画出函数的图像。而由上述表达式的图象性质可知，分式线性变换是可逆的。容易验证以下性质：分式线性变换的逆变换（即反函数）为：一次函数是特殊的分式线性变换。分式线性映射之间的复合结果仍然是分式线性变换。如果是分式线性变换的不动点，等价于是其反函数的不动点。这是因为可逆函数的不动点与其反函数的不动点是一一对应的。我们下面先通过一种计算量较少的思路来说明使用待定系数法时，第一步减去的常数一定是其不动点的原因。按照待定系数法的思路，假定对分式线性变换解析式的两侧同时减去某个合适的常数P再同时取倒数后，能得到形式一致的分母。先做第一步减法，得：要使上面的式子取倒数后的分母满足要求，由于分母取倒数前是位于分子的位置，基于前面的解题实例可知，我们只需要保证上式最左边的量与最右边的分子形式相似即可。即只需要使与的对应系数成相同比例即可。即：，解得：。显然，这个所需的P值是的反函数的不动点。根据前面总结的性质，可知P也是的不动点。即对两侧同时减去其不动点后，再取同时倒数，就一定能构造出一次的常系数非齐次递推式。从同胚变换的角度理解函数迭代与递推式求解 [编辑] 也可以从函数同胚变换（也就是函数相似变换）的角度出发，来理解函数的多重迭代与分式线性递推式的求解。不动点法在这两类问题中都有登场出现。首先由于一次函数属于特殊的分式线性变换，而且分式线性变换是可逆变化，所以自然想到能否将分式线性递推式还原为某种更简单的一次线性递推式的形式。求出与之相关的一次线性递推式的迭代结果后，再利用逆变换反推出原递推式的多次迭代结果。另一方面，我们已经知道使用不动点法可以简化一次线性递推式的求解，我们希望将不动点也整合到分式线性变换还原为线性递推式的系数配凑过程中。使得可以与另一个函数建立下列联系的可逆函数叫做桥函数：此时和叫做关于桥函数的一对相似函数，或者说和关于桥函数互为相似函数。用拓扑学术语来说，桥函数是一种同胚（homeomorphism）变换，它建立了和之间的拓扑共轭（topological conjugacy）。桥函数是计算函数迭代或巧妙构造数列的辅助函数。通过找到合适的桥函数，把不易得到迭代公式的变为容易得到迭代公式的形式更简单的，计算完的n次迭代结果后，再使用桥函数的逆变换即可得到的n次迭代式。这种思路把函数迭代的问题转化为桥函数的寻找与研究问题。可以通过数学归纳法验证函数的上述相似性（也就是拓扑共轭性）满足以下规律（为的桥函数，是其共轭，下标表示迭代计算的次数）：知识背景：将函数借助一个可逆的辅助函数，变换到另一个更容易求解或更容易研究的形式，在数学与物理学的许多分支中都是常见思想。矩阵对角化、傅立叶变换、Z变换、卷积等技巧都是这类思想的体现。用泛函分析的术语来说，函数可以看成抽象空间中的点，如果一个函数直接研究起来不方便，我们就可以用可逆变换将它挪到其它更简单明了的空间中再研究。变换的可逆性保证了函数本身的某些关键特性在某些恰当选取的空间中并不会改变太多，解决问题以后还可以把结果再变回到原来的空间。了解了桥函数方法的转换思想，下面使用不动点法的动机就是借助不动点套出桥函数可能满足的性质，以便确定所需桥函数中的未知待定系数。只要确定了桥函数的各个细节，就可以通过相似变换的方法巧妙地得到原函数的n次迭代式。假设，且是的不动点，那么一定也会是的不动点。这说明如果我们求出了的不动点，并且有了简单可行的，那么的不动点也就知道了。通常为了便于求解的n次迭代式，可以尝试将取为下列形式：对于这几种情况，的不动点为0或。此时如果只有唯一的不动点a，可以考虑取或；如果有2个相异的不动点a和b，则可以考虑取。就分式线性变换的例子来说，如果我们先求出了的不动点，那么如果还能通过尝试一些特例猜出合适的桥函数具有的形式，并且目标函数是一次函数，那么也一定是一次函数的不动点。即有下列关系式成立：我们先不急着移项。因为我们可以从包含无穷大的扩展实数线或扩充复平面上考虑问题，易知从方程解出的唯一不动点是。所以，即可反推：所以所求的关键待定系数k就是原函数的不动点。其它可以转换为上述情形的特例 [编辑] 对于二次函数或包含二次项的分式函数，除了恰巧可以用简单的配方法将二次项配凑进某个完全平方式的情形，一般都没有什么好办法获得n次迭代后的简洁表达式。分式线性递推的其它方法概述 [编辑] 其它可以解决分式线性递推式的方法还包括转化为二阶常系数线性递推数列和多元递推与矩阵方法。计算机求解 [编辑] Mathematica [编辑] 本节介绍的一阶常系数非齐次线性递推式和分式线性递推式都可以用Mathematica软件提供的“RSolve”命令求解出通项公式。我们在这里分别用Mathematica展示2种递推式的通项公式求解方法。它的更多递推式求解用法还会在常系数线性递推数列章节中继续补充。求解差分方程的示例： ``` RSolve[{a[n + 1] == 2 a[n] + 3, a == 1}, a[n], n]; ( 在Mathematica中表示相等，需要2个"="在一起连用 ) ``` 在新单元格（cell）输入完上述命令后，一般按组合键Shift + Enter即可得到结果。求解分式线性差分方程的示例： ``` RSolve[{a[n + 1] == (a[n] + 2) / (3 a[n] + 4), a == 1}, a[n], n]; ( 在Mathematica中表示相除，需要使用"/"符号 ) ``` 提示：如果RSolve命令没有输入错误，但是执行后出现“RSolve::deqn: Equation or list of equations expected instead of True in the first argument True.”一类的报错信息，一般是由于没有清除同名标识符中的已有信息。如果是这种情况：(1)可以先在其它地方输入独立命令“ClearAll[a]”清除标识符a中的内容，然后重新执行RSolve命令；(2)也可以改用除a以外的其它未使用过的标识符来命名要求解的数列。提示：“RSolve”命令中的“R”是递推（recurrence relation）、递归（recursion）的意思。它与“Resolve”命令拼写相似，但功能不同。Resolve是用于化简约束条件的，例如可以求解一些不等式或不等式组。相关例题1：已知在数列中，满足，使用Mathematica求它的通项公式。解答：在Mathematica软件中输入命令（因为只有一行命令，分号此时可以不写）： RSolve[{a[n + 1] == 2 a[n] + 3, a == 1}, a[n], n]; 然后按组合键Shift + Enter即可得到结果为：。答案：。相关例题2：已知数列满足，使用Mathematica求它的通项公式。解答：命令用法示例： RSolve[{a[n + 1] == (2 a[n] + 3) / (a[n] + 4), a == 1}, a[n], n]; 然后按组合键Shift + Enter即可得到结果为：。答案：。补充习题 [编辑] 参考资料 [编辑] ↑ 跳转到： 1.0 1.1 梁训果; 王永明. 第一部分数列相关问题. 高考数学最后一题. 万试无忧系列丛书 5. 重庆出版集团. 2011. ISBN 978-7-5366-8521-5 （中文（中国大陆））. ↑ 跳转到： 2.0 2.1 2.2 蔡小雄. 第4章“用竞赛策略优化高考解题”第4.1节“熟悉递推方法”第4.1.3小节“不动点法”. (编) 董德耀; 沈国明. 更高更妙的高中数学思想与方法 1. 中国杭州天目山路148号: 浙江大学出版社. 2009: 241–243. ISBN 978-7-308-06993-9 （中文（中国大陆））. ↑ 跳转到： 3.0 3.1 李世杰. 第2章“递推数列”第1节“递推数列的定义”. (编) 黄娟琴 (责任编辑); 许佳颖 (文字编辑). 递推与递推方法. 高中数学竞赛专题讲座 1. 中国杭州天目山路148号: 浙江大学出版社. 2008: 26. ISBN 978-7-308-06087-5 （中文（中国大陆））. ↑ 跳转到： 4.0 4.1 4.2 4.3 4.4 钟玉泉. 第7章“共形映射”第2节“分式线性变换”. (编) 王瑜 (策划编辑); 丁鹤龄 (责任编辑). 复变函数论. 朱慧芳 (责任校对) 1. 中国北京市西城区德外大街4号: 高等教育出版社. 2004: 283–289. ISBN 7-04-012943-4 （中文（中国大陆））. ↑ 赵嘉璐. 用不动点法求分式型递推数列的通项 (8月). 数学学习与研究: 107. 2018 （中文（中国大陆））. ↑ 跳转到： 6.0 6.1 6.2 6.3 6.4 6.5 熊斌; 朱臻; 苏勇. 第6章“函数的迭代”第6.2节“f^n (x)的求法”. (编) 孔令志 (项目编辑); 朱洪敏 (审读编辑). 函数与函数方程. 数学奥林匹克小丛书·高中卷 2. 中国上海市中山北路3663号: 华东师范大学出版社. 2012: 86–95. ISBN 978-7-5617-9170-7 （中文（中国大陆））. 外部链接 [编辑] 维基百科中的相关条目：维基百科中的相关条目：检索自“ 分类：数列高中数学
16534	https://www.reddit.com/r/AskPhysics/comments/e1zqy6/amp_volt_watt_joule_current_and_coulomb/	amp, volt, watt, joule, current and coulomb : r/AskPhysics Skip to main contentamp, volt, watt, joule, current and coulomb : r/AskPhysics Open menu Open navigationGo to Reddit Home r/AskPhysics A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to AskPhysics r/AskPhysics r/AskPhysics 1.5M Members Online •6 yr. ago MoussaAdam amp, volt, watt, joule, current and coulomb assuming I have no background physics I want to understand what's an amp, volt, watt, joule, current, charge and coulomb I tried searching, but a unit have more than one definition, idk which one is the right that acctually happen in real life, and most explainatons are circular (if that's the right way to say that) for example: an ampere is a measure of what? of one coulomb of charge per second. ok idk yet what's amp, but sure I will understand it when i know what's a "coulomb of charge". a coulomb is the charge transported by a constant current of one ampere in one second. but idk what's an ampere yet !?!?! I have to understand what's an ampere to understand what's a column to understand what's an ampere? -> simplification: to understand the definition of an ampere I have to already know what's an ampere. 😐 but this is not how definitions work AFAIK. ok let's check what is voltage, maybe this is easier. voltage aka electric potential difference aka electric pressure aka electric tension. why all those names, just to confuse me?!?!? it's just the difference in electric potential between two points points? like any two points ... in space ? what's an "electric potential" ? oh it's just the amount of work needed to move a unit of charge from a reference point to a specific point inside the field without producing an acceleration work? amount of work ? how can that be measured? charge? field? ok what's a charge? and how it relate to coulombs? Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. force? electromagnetic feild? and where is the coulombs in that? ok this will never finish let's check the joule It is equal to the energy transferred to (or work done on) an object when a force of one newton acts on that object in the direction of the force's motion through a distance of one metre(1 newton metre or N⋅m). It is also the energy dissipated as heat when an electric current of one ampere passes through a resistance of one ohm for one second Newtown, direction, forces, distance? Is that electricity anymore? what is energy in that sense? and why it should be transferred to be considered part of the definition of joule? it is also what? wait what? how can be that? what is it exactly? it can't be 2 things, I refuse to accept that Ok I give up 😭 I just want to understand in term of electrons not in terms of other units, simply bc idk the other units yet don't explain ampere with volt unless u explain volt without mentioning ampere and don't explain volt with ampere unless u explain ampere without mentioning volts Edit: edited for better readability Edit: so far i understand that charge is a fundamental property of a particule, it can be negative, like the electron's charge, or positive like the proton's charge, the unit for measuring this charge is the coulomb, the flowing of that charge throught a wire is a current, an ampere is the unit of measuring that flowing current, by measuring the amount of moved charges (in coulombs) per second, voltage is the force that make the current of charges flow in the first place, but what acctually is it? idk... what is joule and watt ? i also don't know Read more Share Related Answers Section Related Answers Unit of electric current Convert joules to amps Relationship between volts and watts Meaning of joule per coulomb Convert joules to coulombs New to Reddit? Create your account and connect with a world of communities. Continue with Google Continue with Google. Opens in new tab Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of November 26, 2019 Reddit reReddit: Top posts of November 2019 Reddit reReddit: Top posts of 2019 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
16535	https://www.ncbi.nlm.nih.gov/books/NBK1308/	Familial Lipoprotein Lipase Deficiency - GeneReviews® - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. Adam MP, Feldman J, Mirzaa GM, et al., editors. GeneReviews® [Internet]. Seattle (WA): University of Washington, Seattle; 1993-2025. GeneReviews® [Internet]. Show details Adam MP, Feldman J, Mirzaa GM, et al., editors. Seattle (WA): University of Washington, Seattle; 1993-2025. GeneReviews by Title Search term GeneReviews Advanced Search Help < PrevNext > Familial Lipoprotein Lipase Deficiency Synonym: Familial LPL Deficiency John R Burnett, MB ChB, MD, PhD, FRCPA, Amanda J Hooper, PhD, and Robert A Hegele, MD, FRCPC, FACP. Author Information and Affiliations John R Burnett, MB ChB, MD, PhD, FRCPA Department of Clinical Biochemistry Royal Perth Hospital & Fiona Stanley Hospital Network PathWest Laboratory Medicine WA; School of Medicine Faculty of Health & Medical Sciences University of Western Australia Perth, Australia Email: ua.vog.aw.htlaeh@ttenrub.nhoj Amanda J Hooper, PhD Department of Clinical Biochemistry Royal Perth Hospital & Fiona Stanley Hospital Network PathWest Laboratory Medicine WA; School of Medicine Faculty of Health & Medical Sciences University of Western Australia Perth, Australia Email: ua.vog.aw.htlaeh@repooh.adnama Robert A Hegele, MD, FRCPC, FACP Departments of Medicine and Biochemistry Schulich School of Medicine and Robarts Research Institute Western University London, Ontario, Canada Email: ac.strabor@elegeh Initial Posting: October 12, 1999; Last Update: June 22, 2017. Estimated reading time: 18 minutes Go to: Summary Clinical characteristics. Familial lipoprotein lipase (LPL) deficiency usually presents in childhood and is characterized by very severe hypertriglyceridemia with episodes of abdominal pain, recurrent acute pancreatitis, eruptive cutaneous xanthomata, and hepatosplenomegaly. Clearance of chylomicrons from the plasma is impaired, causing triglycerides to accumulate in plasma and the plasma to have a milky (lactescent or lipemic) appearance. Symptoms usually resolve with restriction of total dietary fat to ≤20 g/day. Diagnosis/testing. The diagnosis of LPL deficiency is established in a proband by the identification of biallelic pathogenic variants in LPL on molecular genetic testing. Management. Treatment of manifestations: Treatment is based on medical nutrition therapy to maintain plasma triglyceride concentration below 1000 mg/dL. Maintenance of triglyceride levels below 2000 mg/dL prevents recurrent abdominal pain. Restriction of dietary fat to ≤20 g/day or 15% of a total energy intake is usually sufficient to reduce plasma triglyceride concentration and to keep the individual with familial LPL deficiency free of symptoms. An acute pancreatitis episode is treated with standard care. Prevention of secondary complications: Prevention of recurrent acute pancreatitis decreases the risk of developing diabetes mellitus. Surveillance: Monitoring of plasma triglycerides. Agents/circumstances to avoid: Agents known to increase endogenous triglyceride concentration such as alcohol, oral estrogens, diuretics, isotretinoin, glucocorticoids, selective serotonin reuptake inhibitors, and beta-adrenergic blocking agents; fish oil supplements are contraindicated because they contribute to chylomicron levels. Pregnancy management: For pregnant women with LPL deficiency, extreme dietary fat restriction to <2 g/day during the second and third trimesters of pregnancy with close monitoring of plasma triglyceride concentration is recommended. Other: The lipid-lowering drugs that are used to treat other disorders of lipid metabolism are not effective in individuals with familial LPL deficiency. Genetic counseling. Familial LPL deficiency is inherited in an autosomal recessive manner. Each sib of an affected individual has a 25% chance of being affected, a 50% chance of being an asymptomatic carrier, and a 25% chance of being unaffected and not a carrier. Carrier testing for at-risk relatives and prenatal testing for pregnancies at increased risk are possible if the pathogenic variants in the family are known. Go to: Diagnosis Suggestive Findings Familial lipoprotein lipase (LPL) deficiency should be suspected in individuals (particularly those age <40 years) with the following clinical and supportive laboratory findings. Clinical findings Recurrent acute pancreatitis Eruptive cutaneous xanthomata Hepatosplenomegaly Supportive laboratory findings Impaired clearance of chylomicrons from plasma causing the plasma to have a milky (lactescent or lipemic) appearance Plasma triglyceride concentrations greater than 2000 mg/dL in the untreated state, regardless of fasting status Establishing the Diagnosis The diagnosis of LPL deficiency is established in a proband by the identification of biallelic pathogenic variants in LPL on molecular genetic testing (see Table 1). A consensus diagnostic algorithm has been published [Stroes et al 2017] (see Figure 3). Molecular genetic testing approaches can include single-gene testing or use of a multigene panel: Single-gene testing. Sequence analysis of LPL is performed first and followed by gene-targeted deletion/duplication analysis if only one or no pathogenic variant is found. A multigene panel that includes LPL and other genes of interest for the chylomicronemia syndrome (see Differential Diagnosis) may also be considered. Note: (1) The genes included in the panel and the diagnostic sensitivity of the testing used for each gene vary by laboratory and are likely to change over time. (2) Some multigene panels may include genes not associated with the condition discussed in this GeneReview; thus, clinicians need to determine which multigene panel is most likely to identify the genetic cause of the condition while limiting identification of variants of uncertain significance and pathogenic variants in genes that do not explain the underlying phenotype. (3) In some laboratories, panel options may include a custom laboratory-designed panel and/or custom phenotype-focused exome analysis that includes genes specified by the clinician. (4) Methods used in a panel may include sequence analysis, deletion/duplication analysis, and/or other non-sequencing-based tests. For an introduction to multigene panels click here. More detailed information for clinicians ordering genetic tests can be found here. Table 1. Molecular Genetic Testing Used in Familial Lipoprotein Lipase Deficiency View in own window \| Gene 1 \| Method \| Proportion of Probands with Pathogenic Variants 2 Detectable by Method \| --- \| LPL \| Sequence analysis 3 \| ~97%4 \| \| Gene-targeted deletion/duplication analysis5 \| ~3%6 \| See Table A. Genes and Databases for chromosomelocus and protein. See Molecular Genetics for information on allelic variants detected in this gene. Sequence analysis detects variants that are benign, likely benign, of uncertain significance, likely pathogenic, or pathogenic. Variants may include small intragenic deletions/insertions and missense, nonsense, and splice site variants; typically, exon or whole-gene deletions/duplications are not detected. For issues to consider in interpretation of sequence analysis results, click here. Brunzell & Deeb , Gilbert et al Gene-targeted deletion/duplication analysis detects intragenic deletions or duplications. Methods used may include quantitative PCR, long-range PCR, multiplex ligation-dependent probe amplification (MLPA), and a gene-targeted microarray designed to detect single-exon deletions or duplications. Brunzell & Deeb Measurement of lipoprotein lipase enzyme activity. Affected individuals have low or absent LPL enzyme activity in an assay system that contains either normal plasma or apolipoprotein C-II (apoC-II; a cofactor of LPL) and excludes hepatic lipase (HL). This assay is not routinely available and is generally only performed at specialist centers. LPL enzyme activity can be assayed in plasma taken ten minutes following intravenous administration of heparin (60 U/kg body weight). The absence of LPL enzyme activity in postheparin plasma is diagnostic of familial LPL deficiency. LPL enzyme activity may be assayed directly in biopsies of adipose tissue. Go to: Clinical Characteristics Clinical Description Familial lipoprotein lipase (LPL) deficiency usually presents in childhood with episodes of abdominal pain, recurrent acute pancreatitis, eruptive cutaneous xanthomata, and hepatosplenomegaly. Males and females are affected equally. Approximately 25% of affected children develop symptoms before age one year and the majority develop symptoms before age ten years; however, some individuals present for the first time during pregnancy. The severity of symptoms correlates with the degree of chylomicronemia. Chylomicrons are large triglyceride-rich lipoprotein particles that appear in the circulation shortly after the ingestion of dietary fat; normally, they are cleared from plasma after an overnight fast. The degree of chylomicronemia in LPL deficiency varies by dietary fat intake. The abdominal pain, which can vary from mildly bothersome to incapacitating, is usually mid-epigastric with radiation to the back. It may be diffuse and mimic an acute abdomen, often leading to unnecessary abdominal exploratory surgery. The pain probably results from chylomicronemia leading to pancreatitis. Kawashiri et al reported that individuals with LPL deficiency can lead a fairly normal life on a diet very low in total fat content. The secondary complications of pancreatitis – diabetes mellitus, steatorrhea, and pancreatic calcification – are unusual in individuals with familial LPL deficiency and rarely occur before middle age. Pancreatitis in LPL deficiency may rarely be associated with total pancreatic necrosis and death. About 50% of individuals with familial LPL deficiency have eruptive xanthomas, small yellow papules localized over the trunk, buttocks, knees, and extensor surfaces of the arms. Xanthomas are deposits of lipid in the skin that result from the extravascular phagocytosis of chylomicrons by macrophages. They can appear rapidly when plasma triglyceride concentration exceeds 2000 mg/dL and can sometimes regress if plasma triglyceride is normalized. Xanthomas may become generalized. As a single lesion, they may be several millimeters in diameter; rarely, they may coalesce into plaques. They are usually not tender unless they occur at a site susceptible to repeated abrasion. Hepatomegaly and splenomegaly often occur when plasma triglyceride concentrations are markedly increased. The organomegaly results from triglyceride uptake by macrophages, which become foam cells. When triglyceride concentrations exceed 4000 mg/dL, the retinal arterioles and venules, and often the fundus itself, develop a pale pink color ("lipemia retinalis"), caused by light scattering by large chylomicrons. This coloration is reversible and vision is not affected. Reversible neuropsychiatric findings, including mild dementia, depression, and memory loss, have also been reported with chylomicronemia [Heilman & Fisher 1974, Chait et al 1981]. Genotype-Phenotype Correlations As most pathogenic LPL variants have been reported in individual case reports or small case series, reliable systematic correlation of genotype-phenotype relationships is challenging. There are no known genotype-phenotype correlations. Nomenclature Familial LPL deficiency is the most common form of the familial chylomicronemia syndrome, which was formerly known as "type 1 hyperlipoproteinemia." Prevalence The prevalence of familial LPL deficiency is approximately one in 1,000,000 in the general US population. The disease has been described in all races. The prevalence is much higher in some areas of Quebec, Canada as a result of a founder effect. Consanguinity is observed in some families with familial LPL deficiency caused by homozygous pathogenic LPL variants. Go to: Genetically Related (Allelic) Disorders No phenotypes other than those discussed in this GeneReview are known to be associated with pathogenic variants in LPL. Mild lipid abnormalities not associated with familial LPL deficiency have been reported with common variants of LPL, such as the p.Asn291Ser allele. The p.Asn291Ser allele does not appear to have a major effect on plasma lipid concentration or on risk for coronary disease in the general population; however, it may be associated with hypertriglyceridemia in the presence of the APOE E2 allele, diabetes mellitus, familial combined hyperlipidemia, hepatic lipase (HL) deficiency, and glycogen storage disease type Ib. It is possible that the association of the p.Asn291Ser allele with these disorders is only a reflection of the relatively high frequency of this allele in the general population. Nonetheless, the allele likely predisposes to mild hypertriglyceridemia, but is insufficient in the absence of a secondary factor to cause more severe hypertriglyceridemia. Go to: Differential Diagnosis Familial lipoprotein lipase (LPL) deficiency should be considered in young persons with the chylomicronemia syndrome, defined as abdominal pain, eruptive xanthomata, plasma triglyceride concentrations greater than 2000 mg/dL, and fasting lipemic plasma. However, the majority of individuals with chylomicronemia and plasma triglyceride concentration greater than 2000 mg/dL do not have familial LPL deficiency; rather, they have one of the more common genetic disorders of triglyceride metabolism (i.e., familial combined hyperlipidemia and monogenic familial hypertriglyceridemia). Hypertriglyceridemia can also be polygenic, due to both heterozygous rare large-effect variants and accumulations of common rare small-effect variants in several genes and loci [Hegele et al 2014]. With such genetic predisposition, clinical expression of chylomicronemia typically requires the presence of secondary, non-genetic factors [Brunzell & Deeb 2001]. Secondary causes of hypertriglyceridemia include: diabetes mellitus; paraproteinemia and lymphoproliferative disorders; use of alcohol; and therapy with estrogen, glucocorticoids, selective serotonin reuptake inhibitors, atypical antipsychotic agents, isotretinoin, or certain antihypertensive agents. In one series of 123 individuals evaluated for marked hypertriglyceridemia, 110 had an acquired cause of hypertriglyceridemia combined with a common genetic form of hypertriglyceridemia, five had familial LPL deficiency, five had other rare genetic forms of hypertriglyceridemia, and three had an unknown cause [Chait & Brunzell 1983]. Other than LPL deficiency, the chylomicronemia syndrome may be caused by biallelic pathogenic variants in apolipoprotein C-II (APOC2), apolipoprotein A-V (APOA5), lipase maturation factor 1 (LMF1) or GPIHBP1 (see Table 2). Table 2. Genetic Causes of Primary Monogenic Chylomicronemia View in own window \| Gene (Gene Product) \| Homozygote Prevalence \| Gene Product Function \| Clinical Features \| Molecular Features \| % of Monogenic Variants \| References \| --- --- --- \| LPL (LPL) \| ~1 per million individuals 1 \| Hydrolysis of triglycerides & peripheral uptake of FFA \| Severe chylomicronemia in infancy or childhood \| Severely reduced or absent LPL enzyme activity \| 95.0 \| Murthy et al , Jap et al , Gotoda et al , Martín-Campos et al \| \| APOC2 (apoC-II) \| 10 families reported \| Required cofactor of LPL \| Severe chylomicronemia in childhood or adolescence \| Absent or nonfunctional apoC-II \| 2.0 \| Gotoda et al , Okubo et al \| \| GPIHBP1 (GPI-HBP1) \| 10 families reported \| Stabilizes binding of chylomicrons near LPL; supports lipolysis \| Chylomicronemia in late adulthood \| Absent or defective GPI-HBP1 \| 2.0 \| Beigneux et al 2009, Gin et al \| \| APOA5 (apoA-V) \| Three families reported \| Enhancer of LPL activity \| Chylomicronemia in late adulthood \| Absent or defective apoA-V \| 0.6 \| Calandra et al , Nilsson et al \| \| LMF1 (LMF1) \| Two families reported \| Chaperone molecule required for proper LPL folding and/or expression \| Chylomicronemia in late adulthood \| Absent or defective LMF1 \| 0.4 \| Péterfy \| From Brahm & Hegele ; reprinted by permission of Macmillan Publishers, Ltd. apoA-V = apolipoprotein A-V; apoC-II = apolipoprotein C-II; FFA = free fatty acid; GPI-HBP1 = glycosylphosphatidylinositol-anchored high density lipoprotein-binding protein 1; LMF1 = lipase maturation factor 1; LPL = lipoprotein lipase Gotoda et la Familial apolipoprotein C-II (apoC-II) deficiency (OMIM 207750). ApoC-II is a cofactor for LPL. Familial apoC-II deficiency is an extremely rare autosomal recessive disorder that differs from familial LPL deficiency in that (1) symptoms generally develop at a later age (13-60 years) and (2) individuals may develop chronic pancreatic insufficiency with steatorrhea and insulin-dependent diabetes mellitus. The diagnosis is based on the identification of biallelic pathogenic variants in APOC2, on assay of plasma apoC-II concentration or activation of a purified LPL standard, and on gel electrophoresis of VLDL apolipoproteins. Infusion of normal plasma into an individual with familial apoC-II deficiency results in dramatic reduction of the plasma triglyceride concentration. Treatment is a low-fat diet throughout life. Familial apolipoprotein A-V deficiency (OMIM 144650). It has been suggested that apoA-V facilitates the interaction of endothelial heparan sulfate with apoC-II on triglyceride-rich lipoproteins and the interaction of apoC-II with LPL on the vascular endothelium. Several families with apoA-V deficiency have been reported to have severe hypertriglyceridemia. Biallelic large-effect loss-of-function pathogenic variants in APOA5 can lead to familial chylomicronemia syndrome, while heterozygous carriers are predisposed to hypertriglyceridemia, but often demonstrate a normal clinical and biochemical phenotype in the absence of secondary non-genetic factors. Familial lipase maturation factor 1 (LMF1) deficiency (OMIM 246650). LMF1 is a transmembrane protein localized to the endoplasmic reticulum involved in the maturation of both LPL and hepatic lipase (HL). One affected individual, homozygous for a biallelicLMF1pathogenic variant, had very low LPL activity, modestly low HL activity, and chylomicronemia [Péterfy et al 2007]. Familial glycosylphosphatidylinositol-anchored high-density lipoprotein-binding protein 1 (GPIHBP1) deficiency (OMIM 615947). GPIHBP1 appears to be a binding site for LPL on the capillary endothelial surface, perhaps through binding with apoA-V [Beigneux et al 2007]. Several individuals with GPIHBP1 deficiency have been described. Biallelic large-effect loss-of-function variants of GPIHBP1 can lead to familial chylomicronemia syndrome, while heterozygous carriers are predisposed to hypertriglyceridemia, but often demonstrate a normal clinical and biochemical phenotype in the absence of secondary non-genetic factors. Recently, autoantibodies against GPIHBP1 were shown to cause chylomicronemia syndrome [Beigneux et al 2017]. Go to: Management Evaluations Following Initial Diagnosis To establish the extent of disease and needs in an individual diagnosed with familial lipoprotein lipase (LPL) deficiency, measurement of plasma triglyceride concentration is recommended. Consultation with a clinical geneticist and/or genetic counselor may also be considered. Treatment of Manifestations Medical nutrition therapy. Morbidity and mortality can be prevented by maintaining plasma triglyceride concentration at less than 2000 mg/dL; a good clinical goal is less than 1000 mg/dL [Viljoen & Wierzbicki 2012]. Restriction of dietary fat to no more than 20 g/day or 15% of total energy intake is usually sufficient to reduce plasma triglyceride concentration and to keep the individual with familial LPL deficiency free of symptoms. Medium-chain triglycerides may be used for cooking, as they are absorbed directly into the portal vein without becoming incorporated into chylomicron triglyceride. The success of therapy depends on the individual's acceptance of the fat restriction, including both unsaturated and saturated fat. Note: Fish oil supplements, which are effective in disorders of excess hepatic triglyceride production, are not effective in LPL deficiency and are contraindicated (see Agents/Circumstances to Avoid). The enlarged liver and spleen can return to normal size within one week of lowering of triglyceride concentrations. The xanthomas can clear over the course of weeks to months. Recurrent or persistent eruptive xanthomas indicate inadequate therapy. Pancreatitis associated with the chylomicronemia syndrome is treated in the manner typical for other forms of pancreatitis. Discontinuation of oral fat intake stops chylomicron triglyceride formation, and replacement with hypocaloric parenteral nutrition decreases VLDL triglyceride production. Administration of excess calories, as in hyperalimentation, is contraindicated in the acute state. The intravenous administration of lipid emulsions may lead to persistent or recurrent pancreatitis. If recurrent pancreatitis with severe hypertriglyceridemia occurs, total dietary fat intake needs to be reduced. Prevention of Primary Manifestations See Treatment of Manifestations. Prevention of Secondary Complications Prevention of recurrent acute pancreatitis decreases the risk of developing diabetes mellitus. Fat malabsorption is very rare. Surveillance Plasma triglyceride levels need to be followed over time to evaluate the affected individual’s success in following the very low-fat dietary recommendations. When the triglyceride level is above 1000 mg/dL, a fasting sample is not required for this evaluation. Other components of the lipid profile do not need to be routinely measured. Affected individuals who develop abdominal pain need to contact their physician. Agents/Circumstances to Avoid Avoidance of agents known to increase endogenous triglyceride concentration such as alcohol, oral estrogens, diuretics, isotretinoin, glucocorticoids, selective serotonin reuptake inhibitors, and beta-adrenergic blocking agents is recommended. Fish oil supplements are contraindicated as they contribute to chylomicron levels. Evaluation of Relatives at Risk It is appropriate to evaluate at-risk sibs during infancy. Early diagnosis and implementation of dietary fat intake restriction can prevent symptoms and related medical complications. Evaluations can include: Measurement of plasma triglyceride concentration; Molecular genetic testing if the pathogenic variants in the family are known. See Genetic Counseling for issues related to evaluation of at-risk relatives for genetic counseling purposes. Pregnancy Management During pregnancy in a woman with LPL deficiency, extreme dietary fat restriction to less than two grams per day during the second and third trimester with close monitoring of plasma triglyceride concentration can result in delivery of a normal infant with normal plasma concentrations of essential fatty acids [Al-Shali et al 2002]. One woman with LPL deficiency delivered a normal child following a one-gram per day fat diet and treatment with gemfibrozil (600 mg/day) [Tsai et al 2004]. Despite concerns about the possibility of essential fatty acid deficiency in the newborn, normal essential fatty acids were found in cord blood, as were normal levels of fibrate metabolites. See MotherToBaby for further information on medication use during pregnancy. Therapies Under Investigation LPLgene therapy is available in Europe and the US for the treatment of familial LPL deficiency. It consists of the LPL Ser447Ter variant in an adeno-associated virus serotype 1 (alipogene tiparvovec). Twenty percent of the general population has the Ser447Ter allelic variant, which results in a prematurely truncated LPL that is associated with increased lipolytic function and an anti-atherogenic lipid profile and can therefore be regarded as a naturally occurring gain-of-function variant (reviewed by Gaudet et al ). Note that the standard nomenclature for this variant is NP_000228.1:p.Ser474Ter (NM_000237.2:c.1421C>G). Several other targeted therapies are in development or under investigation, including inhibitors of microsomal triglyceride transfer protein (MTP) [Sacks et al 2014], diacylglycerol acyl transferase 1 (DGAT1) [Meyers et al 2015], apoC-III [Gaudet et al 2014], and angiopoietin-like protein 3 (ANGPTL3) [Gryn & Hegele 2015]. Search ClinicalTrials.gov in the US and EU Clinical Trials Register in Europe for access to information on clinical studies for a wide range of diseases and conditions. Other The lipid-lowering drugs that are used to treat other disorders of lipid metabolism are not effective in individuals with familial LPL deficiency. Although plasmapheresis and antioxidant therapy have been suggested as treatment for pancreatitis, they do not appear to be needed for either acute therapy or long-term care. Go to: Genetic Counseling Genetic counseling is the process of providing individuals and families with information on the nature, mode(s) of inheritance, and implications of genetic disorders to help them make informed medical and personal decisions. The following section deals with genetic risk assessment and the use of family history and genetic testing to clarify genetic status for family members; it is not meant to address all personal, cultural, or ethical issues that may arise or to substitute for consultation with a genetics professional. —ED. Mode of Inheritance Familial LPL deficiency is inherited in an autosomal recessive manner. Risk to Family Members Parents of a proband The parents of an affected individual are obligate heterozygotes (i.e., carriers of one LPLpathogenic variant). Heterozygotes are asymptomatic but may have moderately elevated plasma triglyceride concentrations and may be at mild risk for premature atherosclerosis. Sibs of a proband At conception, each sib of an affected individual has a 25% chance of being affected, a 50% chance of being an asymptomatic carrier, and a 25% chance of being unaffected and not a carrier. Heterozygotes are asymptomatic but may have moderately elevated plasma triglyceride concentrations and may be at mild risk for premature atherosclerosis. Offspring of a proband. The offspring of an individual with familial LPL deficiency are obligate heterozygotes (carriers) for a pathogenic variant in LPL. Other family members. Each sib of the proband's parents is at a 50% risk of being a carrier of an LPLpathogenic variant. Carrier Detection Carrier testing for at-risk relatives requires prior identification of the LPL pathogenic variants in the family. The lipid phenotype of heterozygotes varies widely from completely normal to moderate-to-severe hypertriglyceridemia and cannot be used to predict carrier status. Related Genetic Counseling Issues See Evaluation of Relatives at Risk for information on evaluating at-risk relatives for the purpose of early diagnosis and treatment. Family planning The optimal time for determination of genetic risk, clarification of carrier status, and discussion of the availability of prenatal/preimplantation genetic testing is before pregnancy. It is appropriate to offer genetic counseling (including discussion of potential risks to offspring and reproductive options) to young adults who are affected, are carriers, or are at risk of being carriers. Prenatal Testing and Preimplantation Genetic Testing Once the LPL pathogenic variants have been identified in an affected family member, prenatal testing for a pregnancy at increased risk and preimplantation genetic testing for familial LPL deficiency are possible. Differences in perspective may exist among medical professionals and in families regarding the use of prenatal testing. While most centers would consider use of prenatal testing to be a personal decision, discussion of these issues may be helpful. In practice, prenatal testing is rarely requested because of the availability of effective treatment. Go to: Resources GeneReviews staff has selected the following disease-specific and/or umbrella support organizations and/or registries for the benefit of individuals with this disorder and their families. GeneReviews is not responsible for the information provided by other organizations. For information on selection criteria, click here. Medline Plus Familial lipoprotein lipase deficiency National Library of Medicine Genetics Home Reference Familial lipoprotein lipase deficiency Go to: Molecular Genetics Information in the Molecular Genetics and OMIM tables may differ from that elsewhere in the GeneReview: tables may contain more recent information. — ED. Table A. Familial Lipoprotein Lipase Deficiency: Genes and Databases View in own window \| Gene \| Chromosome Locus \| Protein \| Locus-Specific Databases \| HGMD \| ClinVar \| --- --- --- \| \| LPL \| 8p21.3 \| Lipoprotein lipase \| LPL @ LOVD \| LPL \| LPL \| Data are compiled from the following standard references: gene from HGNC; chromosomelocus from OMIM; protein from UniProt. For a description of databases (Locus Specific, HGMD, ClinVar) to which links are provided, click here. Table B. OMIM Entries for Familial Lipoprotein Lipase Deficiency (View All in OMIM) View in own window 238600HYPERLIPOPROTEINEMIA, TYPE I 609708LIPOPROTEIN LIPASE; LPL Gene structure.LPL comprises ten exons; the last one encodes the long untranslated 3’-end of the mRNA (NM_000237.2). The sequence of LPL is highly conserved among mammalian species. Pathogenic variants. More than 220 pathogenic variants have been identified [Brunzell & Deeb 2001, Gilbert et al 2001]; see Table A, HGMD. Approximately 70% of these variants are missense, 10% nonsense, and 18% nucleotide insertions and deletions; a few splice site variants are also known. About 3% of pathogenic variants have an exon or multiexon deletion, duplication, insertion, or complex rearrangement. Normal gene product. Lipoprotein lipase (LPL) is a glycoprotein that is synthesized in adipose tissue and cardiac and skeletal muscle, but not in the postpartum liver. It is transported to the luminal surface of the capillary endothelium of extrahepatic tissues. It is essential for the hydrolysis of chylomicron and VLDL triglycerides to provide free fatty acids to tissue for energy production. LPL has two major domains: a larger NH 2-terminal domain linked by a short region to a COOH-terminal domain of approximately half its size. The globular NH 2-terminal domain specifies the catalytic properties of the lipase, whereas the COOH-terminal domain specifies substrate specificity and heparin-binding properties. LPL encodes a protein of 475 amino acids that becomes a mature protein of 448 residues after cleavage of a signal peptide (NP_000228.1). Abnormal gene product. Pathogenic deletions and nonsense and splice site variants presumably lead to absent or truncated LPL with defective catalytic activity. Most of the variants are in the highly conserved central homology region [Brunzell & Deeb 2001, Gilbert et al 2001]. Many missense variants result in LPL deficiency secondary to LPL homodimer instability. Go to: References Literature Cited Al-Shali K, Wang J, Fellows F, Huff MW, Wolfe BM, Hegele RA. Successful pregnancy outcome in a patient with severe chylomicronemia due to compound heterozygosity for mutant lipoprotein lipase. Clin Biochem. 2002;35:125–30. [PubMed: 11983347] Beigneux AP, Davies BS, Gin P, Weinstein MM, Farber E, Qiao X, Peale F, Bunting S, Walzem RL, Wong JS, Blaner WS, Ding ZM, Melford K, Wongsiriroj N, Shu X, de Sauvage F, Ryan RO, Fong LG, Bensadoun A, Young SG. Glycosylphosphatidylinositol-anchored high-density lipoprotein-binding protein 1 plays a critical role in the lipolytic processing of chylomicrons. Cell Metab. 2007;5:279–91. [PMC free article: PMC1913910] [PubMed: 17403372] Beigneux AP, Franssen R, Bensadoun A, Gin P, Melford K, Peter J, Walzem RL, Weinstein MM, Davies BS, Kuivenhoven JA, Kastelein JJ, Fong LG, Dallinga-Thie GM, Young SG. Chylomicronemia with a mutant GPIHBP1 (Q115P) that cannot bind lipoprotein lipase. Arterioscler Thromb Vasc Biol. 2009;29:956–62. [PMC free article: PMC2811263] [PubMed: 19304573] Beigneux AP, Miyashita K, Ploug M, Blom DJ, Ai M, Linton MF, Khovidhunkit W, Dufour R, Garg A, McMahon MA, Pullinger CR, Sandoval NP, Hu X, Allan CM, Larsson M, Machida T, Murakami M, Reue K, Tontonoz P, Goldberg IJ, Moulin P, Charrière S, Fong LG, Nakajima K, Young SG. Autoantibodies against GPIHBP1 as a cause of hypertriglyceridemia. N Engl J Med. 2017;376:1647–58. 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Apolipoprotein A-V; a potent triglyceride reducer. Atherosclerosis. 2011;219:15–21. [PubMed: 21831376] Okubo M, Toromanovic A, Ebara T, Murase T. Apolipoprotein C-II: a novel large deletion in APOC2 caused by Alu–Alu homologous recombination in an infant with apolipoprotein C-II deficiency. Clin Chim Acta. 2015;438:148–53. [PubMed: 25172036] Péterfy M. Lipase maturation factor 1: a lipase chaperone involved in lipid metabolism. Biochim Biophys Acta. 2012;1821:790–4. [PMC free article: PMC3288453] [PubMed: 22063272] Péterfy M, Ben-Zeev O, Mao HZ, Weissglas-Volkov D, Aouizerat BE, Pullinger CR, Frost PH, Kane JP, Malloy MJ, Reue K, Pajukanta P, Doolittle MH. Mutations in LMF1 cause combined lipase deficiency and severe hypertriglyceridemia. Nat Genet. 2007;39:1483–7. [PubMed: 17994020] Sacks FM, Stanesa M, Hegele RA. Severe hypertriglyceridemia with pancreatitis: thirteen years' treatment with lomitapide. JAMA Intern Med. 2014;174:443–7. [PubMed: 24366202] Stroes E, Moulin P, Parhofer KG, Rebours V, Lӧhr JM, Averna M. Diagnostic algorithm for familial chylomicronemia syndrome. Atheroscler Suppl. 2017;23:1–7. [PubMed: 27998715] Tsai EC, Brown JA, Veldee MY, Anderson GJ, Chait A, Brunzell JD. Potential of essential fatty acid deficiency with extremely low fat diet in lipoprotein lipase deficiency during pregnancy: A case report. BMC Pregnancy Childbirth. 2004;4:27. [PMC free article: PMC544881] [PubMed: 15610556] Viljoen A, Wierzbicki AS. Diagnosis and treatment of severe hypertriglyceridemia. Expert Rev Cardiovasc Ther. 2012;10:505–14. [PubMed: 22458582] Go to: Chapter Notes Author History John D Brunzell, MD; University of Washington (1999-2017) John R Burnett, MB ChB, MD, PhD, FRCPA (2017-present) Robert A Hegele, MD, FRCPC, FACP (2017-present) Amanda J Hooper, PhD (2017-present) Revision History 22 June 2017 (ma) Comprehensive update posted live 15 December 2011 (me) Comprehensive update posted live 28 July 2009 (me) Comprehensive update posted live 1 October 2007 (cd) Revision: deletion/duplication analysis available; prenatal testing available; mutation analysis for p.Gly188Glu done by sequence analysis 24 April 2006 (me) Comprehensive update posted live 9 April 2004 (me) Comprehensive update posted live 18 February 2002 (me) Comprehensive update posted live 12 October 1999 (me) Review posted live April 1999 (jb) Original submission Summary Diagnosis Clinical Characteristics Genetically Related (Allelic) Disorders Differential Diagnosis Management Genetic Counseling Resources Molecular Genetics References Chapter Notes Copyright © 1993-2025, University of Washington, Seattle. 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16536	https://chem-guide.blogspot.com/2010/04/periodic-discussion-and-general_02.html	CHEM-GUIDE: Periodic discussion and general characteristics of alkaline earth metals CHEM-GUIDE Tutorials for Chemistry Learning !! Periodic discussion and general characteristics of alkaline earth metals Electronic configurations Group 2 elements are called alkaline Earth metals. These elements have two electrons in the valence shell of their atoms, preceded by the noble gas configuration. Their general configuration is written as [Noble gas] ns 2 where 'n' represents the valence shell. Electronic configuration of alkaline Earth metals General Characteristics of Alkaline Earth Metals Atomic and ionic radii The atomic and ionic radii of alkaline Earth metals are smaller than the corresponding members of the alkali metals. This is due to the fact of Group 2 elements having a higher nuclear charge allow electrons to be attracted more towards the nucleus. This reduces the size of atomic and ionic radii. On moving down the group, the radii increase due to gradual increase in the number of the shells and the screening effect. Melting and boiling points The melting and boiling point of alkaline Earth metals are characteristically low and do not show regular trends down the group. However these points are higher than the corresponding alkali metals in the same period as atoms of alkaline Earth metals have smaller size compared to alkali metals. This causes them to be more closely packed in their crystal lattices. They also have two electrons per metal atom in their valence shell (as compared to one for the alkali metals) forming strong metallic bonds for binding the atoms in the crystal lattice of the metal. Hence they have higher melting and boiling points. Ionization energy Due to a fairly large size of the atoms, alkaline Earth metals have low ionization energies when compared to 'p' block elements. Down the group the ionization energy decreases as atomic size increases. This is due to new shells being added and increase in the magnitude of the screening effect of inner shell electrons. Members of group 2 have higher ionization energies values as compared to group 1 because of their smaller size, with electrons being more attracted towards the nucleus of the atoms. Correspondingly they are less electropositive than alkali members. Although IE 1 values of alkaline Earth metals are higher than that of alkali metals, the IE 2 values of alkaline Earth metals are much smaller than those of alkali metals. This occurs because in alkali metals the second electron is to be removed from a cation, which has already acquired a noble gas configuration. In the case of alkaline Earth metals the second electron is to be removed from a monovalent cation, which still has one electron in the outermost shell. Thus, the second electron can be removed more easily in the case of group 2 elements than in group 1 elements. Electropositive and metallic character Because of low ionization energies, alkaline earth metals are reasonably electropositive in nature; among the family members the electropositive character in general increases from Be to Ba. The alkaline Earth metals are not as strongly electropositive as the alkali metals of group 1 because of comparatively higher ionization energies. Density Alkaline Earth metals are denser and harder than the alkali metals of group. The comparatively smaller atomic size of alkaline Earth metals leads to more closely packed crystal lattices and hence stronger metallic bonds. This accounts for their increased hardness and high density. Characteristic flame coloration Except for Be and Mg, the alkaline Earth (metal) salts impart characteristic colors to the flame. Since their ionization energies are low, the valence electrons in their atoms can be easily excited to higher but unstable energy state, by the flame of a bunsen burner. When these excited electrons come back to ground state they emit visible radiations, giving colors to the flame. Beryllium and magnesium atoms being comparatively smaller have ionization energies that are very high. The energy of a Bunsen burner flame is not sufficient to excite their electrons to higher energy levels. These elements, therefore, do not give any color. Oxidation states Alkaline Earth metals exhibit a valency of +2 as they can lose two electrons and form bivalent ions, which are stable noble gas configurations. Thus, unlike alkali metals, alkaline Earth metals exhibit +2 oxidation states in their compound. The second ionization energy (IE 2) of alkaline Earth metals is greater than the first ionization energy (IE 1). We would expect a preference to form +1 ions (M+) rather than + 2 ions (M 2+) by alkaline Earth metals. However, they predominantly show +2 valency, e.g., Mg 2+, Ca 2+, Ba 2+ etc. This can be explained by the fact that: Divalent ions have the stable noble gas configuration. In aqueous solution, the +2 ions of alkaline Earth metals are extensively hydrated due to their smaller size and the high hydration energies of M 2+ ions, making them more stable than M+ ions. The amount of energy released when M 2+ ion is dissolved in water is much more than that for M+ion and this large amount of extra energy is more than sufficient to compensate for the high second ionization energy required for the formation of such ions. Divalent cations form stronger lattices than monovalent cations in the solid state and a lot of energy of M 2+ ion, called lattice energy, is released. It is the greater lattice energy that compensates for the high second ionization energy and is responsible for the existence and greater stability of bivalent ions in the solid state, as compared to M+ ion. Reduction properties Due to their low ionization energies all group 2 elements tend to lose their valence electrons and act as strong reducing agents. The decrease of the reduction potential values down the group, indicate an increase of strength as a reducing agent. However the members of this group are weaker reducing agents than alkali Earth metals because of the latter's higher ionization energy. No comments: Post a Comment Newer PostOlder PostHome Subscribe to: Post Comments (Atom) Contents Home General and Physical Chemistry (I) Inorganic Chemistry (I) Organic Chemistry (I) General and Physical Chemistry (II) Organic Chemistry (II) Inorganic Chemistry (II) About Me Binod SHRESTHAFRANCEView my complete profile Followers Featured Post Namaste & Welcome !! Chem-guide is a free resource for chemistry learning at school level (for 11 and 12). The resources in the blog has been compiled as pe... Awesome Inc. theme. Powered by Blogger.
16537	https://unacademy.com/content/jee/study-material/chemistry/chromyl-chloride-and-its-properties/	Chromyl Chloride and Its Properties Access free live classes and tests on the app Download + IIT JEE AFCAT AP EAMCET Bank Exam BPSC CA Foundation CAPF CAT CBSE Class 11 CBSE Class 12 CDS CLAT CSIR UGC GATE IIT JAM JEE Karnataka CET Karnataka PSC Kerala PSC MHT CET MPPSC NDA NEET PG NEET UG NTA UGC Railway Exam SSC TS EAMCET UPSC WBPSC CFA Search for: LoginJoin for Free Profile Settings Refer your friends Sign out Terms & conditions • Privacy policy About • Careers • Blog © 2023 Sorting Hat Technologies Pvt Ltd JEE Exam » JEE Study Material » Chemistry » Chromyl Chloride and its Properties Chromyl Chloride and its Properties The article explains various physical and chemical properties of this Chromyl Chloride, its methods of preparation and hazards posed by the compound. Share Chromyl chloride is an inorganic chemical compound. This chemical compound has a reddish-brown colour and can be highly volatile when placed at room temperature. It exists in a liquid form when present at room temperature. This property of Chromyl Chloride is somewhat unusual for a transition metal because most of them remain inactive at room temperature. Chromyl chloride reacts with water to release hydrochloric acid along with chromic acid. Although these acids are not particularly highly combustible, they can help other chemical substances go through combustion. In the event of accidental combustion caused by Chromyl Chloride, there should be no contact with water because instead of slowing down the fire, it will increase it. Preparation of Chromyl Chloride One of the most basic methods for preparing Chromyl Chloride includes the reaction of chemical compounds such as potassium chromate or potassium dichromate with an acid such as hydrochloric acid. This reaction usually takes place in the presence of sulphuric acid. The resultant formed by the chemical reaction goes through distillation so that it can be purified. The presence of sulphuric acid during this process is necessary because it acts as a dehydrating agent. Another method that can be used to prepare Chromyl Chloride is by exposing chromium trioxide to hydrogen chloride present in the gaseous form. This method of formation of Chromyl Chloride can also be used to detect chloride in any chemical substance. If the chemical compound contains chloride as one of its constituents, it’ll release red fumes, which is a positive sign of the presence of chloride. Physical and chemical properties of Chromyl Chloride Chromyl chloride is a chemical substance that can be found in liquid form. It is dark red in colour and possesses a pungent smell. It produces red fumes. This compound’s chemical structure is tetrahedral and is also known to be a neutral molecule in general. Another property of Chromyl Chloride includes the interactions taking place in the chemical that is occurring with the help of Van der Waal bonding. Since Van der Waal bonding is very weak in nature, this gives the chemical compound very low boiling and melting points. This property of Chromyl Chloride makes it a highly distillable liquid. Chromyl Chloride is also anionic in nature and possesses a very low electron affinity. A major property of Chromyl Chloride is that it is highly oxidising in nature. This property is shown by the fact that Chromyl Chloride rapidly hydrolyses and produces various acids. Chromyl Chloride test is also used to check the presence of chloride in any chemical compound. This chemical compound is also used for the oxidation of various alkenes and aldehydes. It is also a critical component of the Etard reaction. Chromyl chloride is a fire hazard, and it can attack multiple metals as well in the presence of water. However, it is incompatible with plastics and does not react with them. Hazardous nature of Chromyl Chloride Since Chromyl chloride is highly hazardous, and upon combustion, it leads to the formation of poisonous fumes. It is not combustible by itself, but it can enhance the combustion of other substances. If inhaled or ingested by humans, it can cause shortness of breath, severe deep burns and severe abdominal pain. If it remains in contact with the skin for a long time, it can also lead to dermatitis and prove to be cancer-causing. Ingestion of this chemical can cause severe damage to the stomach wall lining. Conclusion Hence, one can conclude that Chromyl Chloride is a chemical compound that is highly volatile when it is present in its liquid form at room temperature. It is not explosive itself, but it helps in enhancing the combustion of other chemicals. It is dark red and produces poisonous fumes. The chemical is highly oxidative in nature and has a high electron affinity. To humans, it is hazardous and can cause acute and chronic effects. It can also cause chromosomal aberrations and can even cause cancer.A Frequently Asked Questions Get answers to the most common queries related to the JEE Examination Preparation. Why is sulphuric acid required for the preparation of Chromyl Chloride? Answer- One of the most basic methods for the preparation of Chromyl Chloride includes the reaction of chemical comp...Read full What is the difference between Chromyl Chloride and other transitional metals? Answer- Chromyl chloride exists in a liquid form when present at room temperature. This property of Chromyl Chloride...Read full Why is Chromyl Chloride considered to be a highly distillable liquid? Answer- The interactions that are taking place in the chemical are occurring with the help of Van der Waal bonding. ...Read full What is the use of the Chromyl Chloride test? Answer- The Chromyl Chloride test is used to check for the presence of chloride in any chemical compound. In case ch...Read full Answer- One of the most basic methods for the preparation of Chromyl Chloride includes the reaction of chemical compounds such as potassium chromate or potassium dichromate with an acid such as hydrochloric acid. This reaction usually takes place in the presence of sulphuric acid because it acts as a dehydrating agent. Answer- Chromyl chloride exists in a liquid form when present at room temperature. This property of Chromyl Chloride is rather unusual for a transition metal because most of them remain inactive at room temperature. Answer- The interactions that are taking place in the chemical are occurring with the help of Van der Waal bonding. Since Van der Waal bonding is very weak in nature, this gives the chemical compound very low boiling and melting points. This property of Chromyl Chloride makes it a highly distillable liquid. Answer- The Chromyl Chloride test is used to check for the presence of chloride in any chemical compound. In case chloride is present, the mixture will produce red fumes. Unacademy is India’s largest online learning platform. Download our apps to start learning Starting your preparation? 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16538	https://m.acadsoc.com.cn/blog/yykyzxxx/98067/	英语打电话的情景对话，帮助你提高电话交流能力 \| 阿卡索外教网切换导航官网首页英语学习资源少儿英语英语口语商务英语托福英语雅思英语登录注册登录注册当前位置：阿卡索外教网>英语口语在线学习>英语打电话的情景对话，帮助你提高电话交流能力英语打电话的情景对话，帮助你提高电话交流能力发布时间：2024年06月05日 06:38:45 34 评论 · 2433浏览电话作为一种重要的沟通工具，在日常生活和工作中扮演着重要角色。无论是商务电话还是社交电话，有效的电话交流能力都是至关重要的。而在英语情景下的电话交流，对于许多人来说可能是一项挑战。因此，本文将通过一些英语打电话的情景对话来帮助你提高电话交流能力。顺便也分享一下我们免费的体验课给大家，感兴趣的朋友可以去试试课，不用花上万的费用也能找到效果好的英语机构了，免费的课程领取地址在这：这电话作为一种重要的沟通工具，在日常生活和工作中扮演着重要角色。无论是商务电话还是社交电话，有效的电话交流能力都是至关重要的。而在英语情景下的电话交流，对于许多人来说可能是一项挑战。因此，本文将通过一些英语打电话的情景对话来帮助你提高电话交流能力。顺便也分享一下我们免费的体验课给大家，感兴趣的朋友可以去试试课，不用花上万的费用也能找到效果好的英语机构了，免费的课程领取地址在这：这个课程都是真人外教出镜1对1直播互动教学的，和面授差不多，但上课是随时随地安排的，非常方便灵活，而且这个课程收费很便宜，才20多块钱一节，也就是大家平时喝一杯奈雪的价格吧，非常划算，我们都是聘用的100%持有TESOL等英语教学资质以及丰富教学经验的专业外教，课程涵盖了零基础，日常交流，职场商务，等级考试，雅思托福，口语专项多个板块，基本上是可以满足不同学习需求的人，在积累一定听力以及场景词汇后，和外教在真实的英语交流环境中，进一步加强输入，同时引导输出，避免哑巴英语，是最有效的英语口语练习方式！一、商务电话: 情景1: 预约会议 A: Good morning, this is John from XYZ company. May I speak to Mr. Smith? B: Speaking. How may I assist you? A: I am calling to discuss about scheduling a meeting with our team. Is Mr. Smith available on Thursday morning? B: Let me check his schedule... Yes, he's available at 10 am. How long do you anticipate the meeting will last? A: Approximately one hour. Would the conference room on the 3rd floor be available? B: I'm afraid that room is booked for another meeting at that time. However, we have a larger conference room on the 5th floor. Would that be suitable? A: Yes, that would work. Please reserve the conference room on the 5th floor for our meeting on Thursday at 10 am. B: Noted. I will send you a confirmation email shortly. A: Thank you. Looking forward to meeting then. Goodbye. B: Goodbye. 情景2: 跟进客户 A: Hello, this is Lisa from ABC company. Could I speak to Mr. Johnson, please? B: Speaking. How can I help you? A: Good morning, Mr. Johnson. I am calling to follow up on the proposal we sent you last week regarding our new product. Have you had a chance to review it? B: Yes, I have. It looks interesting, but we have a few questions regarding the pricing and delivery time. A: I understand. I would be happy to address any concerns you have. Could we schedule a time for a detailed discussion about your requirements? B: That sounds reasonable. How about tomorrow afternoon at 2 pm? Would that work for you? A: That works for me. We can set up a video conference call for our discussion. I will send you the meeting link and agenda by email. B: Great. I'll be looking forward to it. Thank you, Lisa. A: You're welcome, Mr. Johnson. Have a great day. Goodbye. B: Goodbye, Lisa. 二、社交电话: 情景1: 邀请朋友参加活动 A: Hi, Jane. This is Sarah. I was wondering if you would like to join us for a movie night this Friday? B: Hi, Sarah. Thanks for the invitation. I would love to come. What movie are you planning to watch? A: We're thinking of watching the latest action thriller. It has received great reviews. B: That sounds exciting. What time and where will the movie night take place? A: The movie will start at 7 pm at the local cinema. We can meet at the entrance at 6:45 pm. B: Sounds good. I'll make sure to be there on time. Should I bring anything? A: Just yourself. We'll take care of the snacks and drinks. Looking forward to seeing you on Friday. B: Thank you, Sarah. I'm excited for the movie night. See you then. A: See you! 情景2: 祝贺生日 A: Hello, Mike. It's Anne calling. I wanted to wish you a very happy birthday! B: Thank you, Anne. That's really kind of you. I appreciate it. A: You're welcome, Mike. I hope you have a fantastic day filled with joy and happiness. B: I'm sure I will. Do you have any plans for the day? A: Actually, I'm organizing a small get-together with a few friends in the evening. We would be honored if you could join us. B: That's very thoughtful of you, Anne. I would love to join the celebration. What time should I come? A: The party will start at 7 pm at my place. Just let me know if you need directions. B: I know the way. I'll be there at 7 pm. Once again, thank you for the birthday wishes and the invitation. A: You're welcome, Mike. See you tonight. Happy birthday! B: Thank you, Anne. See you later. 以上情景对话只是一些例子，希望能够帮助你加强英语电话交流能力。通过这些对话，你可以学习到常用的电话用语和交流技巧。记住，在打电话时保持友好、清晰和有礼貌的态度是非常重要的。不断练习，你的电话交流能力将会逐步提升。祝你成功! 顺便也分享一下我们免费的体验课给大家，感兴趣的朋友可以去试试课，不用花上万的费用也能找到效果好的英语机构了，免费的课程领取地址在这：这个课程都是真人外教出镜1对1直播互动教学的，和面授差不多，但上课是随时随地安排的，非常方便灵活，而且这个课程收费很便宜，才20多块钱一节，也就是大家平时喝一杯奈雪的价格吧，非常划算，我们都是聘用的100%持有TESOL等英语教学资质以及丰富教学经验的专业外教，课程涵盖了零基础，日常交流，职场商务，等级考试，雅思托福，口语专项多个板块，基本上是可以满足不同学习需求的人，在积累一定听力以及场景词汇后，和外教在真实的英语交流环境中，进一步加强输入，同时引导输出，避免哑巴英语，是最有效的英语口语练习方式！少儿英语培训 <上一篇探索阿卡索英语最新收费价目表：性价比高的国际与定制课程全面解析下一篇>阿卡索外教网收费多少?阿卡索外教网收费贵吗? 相关文章一对一外教网站哪个好？怎么选择一家靠谱的一对一外教网站？家长须知：孩子线上学习英语，家长不能不知道的三件事很多人都关注网上少儿英语哪家好？看了这篇你就知道！公司商务英语培训选哪家机构？快速提高商务英语水平的机构推荐线上英语一对一外教哪家好？哪家培训机构适合孩子学习英语？线上英语跟线下哪个贵？线上英语跟线下英语哪个好？评论发布评论 66 666 2022-12-07 20:47:03 游客谢谢 2021-12-25 10:16:33 游客谢谢 2021-12-25 10:16:32 游客看不到密码啊 2019-04-10 15:58:22 sally 谢谢分享 2019-04-10 15:54:06 游客谢谢 2019-04-10 15:53:08 游客谢谢 2019-04-10 08:54:52 游客谢谢 2019-01-30 19:02:15 游客看雅思词汇真经 2019-01-15 13:16:44 游客谢谢 2019-01-04 15:31:51 游客谢谢分享 2018-12-30 13:22:15 234 谢谢 2018-12-30 08:27:57 游客谢谢 2018-12-30 08:27:35 游客是多少 2018-12-28 15:26:58 233 谢谢分享 2018-12-27 07:02:32 KYL 谢谢分享 2018-12-11 12:48:17 yyy 谢谢分享 2018-12-05 21:24:27 1412345215 谢谢分享 2018-11-26 23:35:39 polarbear 谢谢分享 2018-11-16 10:37:24 youngkid 1234 2018-11-07 14:48:33 yy 回复就看到密码了吗 2018-11-05 21:19:36 卜卜谢谢啦 2018-11-01 02:25:14 哥哥不错哦哈哈 2018-10-25 09:02:53 george ggggg 2018-10-16 19:46:10 don 谢谢分享 2018-10-13 21:27:49 也许谢谢分享 2018-10-08 10:10:16 游客甲谢谢分享 2018-09-30 19:22:07 啊啊 cx 2018-09-28 14:59:21 大叔谢谢。博主。烤鸭加油 2018-09-26 14:31:13 www 希望 2018-09-22 22:24:55 去感谢楼主 2018-09-22 21:15:45 发达融入 2018-09-22 10:51:27 想学英语的人回复感谢楼主 2018-09-20 17:49:10 sdfds sdfsdf 2018-07-31 14:32:21 查看更多评论热门排行 1英语口语—你真的会用英语念电话号码和邮箱吗？ 2018年06月08日 18:39:45 82751浏览 2英语中的复合句和复杂句分别是什么，有哪些解析 2018年05月15日 11:53:40 49144浏览 3外国扑克牌怎么玩，其中有哪些常见英语词汇？ 2020年12月22日 16:23:55 47617浏览 42016最新雅思写作评分标准 2018年05月23日 12:27:37 42937浏览 5错误的雅思听力备考法你中招了没？ 2018年07月28日 11:28:06 40830浏览 6超实用的雅思备考方法 2018年05月23日 17:50:39 40763浏览最近更新成人英语培训学校那么多，阿卡索的培训效果如何？ 2025年09月28日 18:22:47 35浏览背诵英语的小技巧有哪些 2025年09月28日 18:18:15 42浏览外教英语培训哪个好？有哪些优势？ 2025年09月28日 18:13:52 44浏览阿卡索怎么样？为什么阿卡索突然流行起来？ 2025年09月28日 18:08:02 39浏览英语培训机构有哪些？分享比较流行的英语培训机构 2025年09月27日 18:09:41 78浏览学英语哪个好？培训英语机构学英语哪个好？ 2025年09月27日 17:56:35 85浏览阿卡索外教网官网价格表,优势有哪些？ 2025年09月27日 17:51:26 86浏览学英语口语哪个好？有经验的朋友给建议 2025年09月27日 17:41:36 84浏览网上英语哪个好？网上学习效果好吗？ 2025年09月27日 17:37:47 81浏览阿卡索外教网口碑怎么样？学习效果好吗？ 2025年09月27日 17:29:20 85浏览 0元领取学习大礼包获取验证码立即领取粤ICP备19156451号 · 粤公网安备 44030302001482号教育网站许可证: 粤教信息(2014)7号本站信息由深圳市阿卡索资讯股份有限公司发布。不良信息和违法举报：400-600-1166 举报邮箱：fkvip@acadsoc.com,security@acadsoc.com 地址：深圳市罗湖区笋岗街道笋西社区桃园路260号HALO广场四期九层在线客服返回顶部 X 智能验证检测中通过验证网络超时请点击此处重试由极验提供技术支持
16539	https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/parseInt	Skip to main content Skip to search Web JavaScript Reference Standard built-in objects Number parseInt() Number.parseInt() Baseline Widely available This feature is well established and works across many devices and browser versions. It’s been available across browsers since ⁨September 2015⁩. Learn more See full compatibility Report feedback The Number.parseInt() static method parses a string argument and returns an integer of the specified radix or base. Try it Syntax Parameters string : The value to parse, coerced to a string. Leading whitespace in this argument is ignored. radix Optional : An integer between 2 and 36 that represents the radix (the base in mathematical numeral systems) of the string. If `radix` is undefined or `0`, it is assumed to be `10` except when the number begins with the code unit pairs `0x` or `0X`, in which case a radix of `16` is assumed. Return value An integer parsed from the given string. If the radix is smaller than 2 or bigger than 36, or the first non-whitespace character cannot be converted to a number, NaN is returned. Examples Number.parseInt vs. parseInt This method has the same functionality as the global parseInt() function: Its purpose is modularization of globals. Please see parseInt() for more detail and examples. Specifications \| Specification \| \| ECMAScript® 2026 Language Specification # sec-number.parseint \| Browser compatibility See also Polyfill of Number.parseInt in core-js es-shims polyfill of Number.parseInt Number parseInt() Help improve MDN Learn how to contribute This page was last modified on by MDN contributors. View this page on GitHub • Report a problem with this content
16540	https://www.anec.org/en/biology/plant-cell-wall.htm	Primary/Secondary Plant Cell Wall: cellulose, pectin, lignin Anec Physics Chemistry Biology Blog Introduction▼ Development of biology 7 characteristics of life Levels of Biological Organization Material in life▼ Element in life # Free water vs bound water Water hydrogen bond Sugar (1): Monosaccharide Sugar (2): Disaccharide Sugar (3): Amylose vs Amylopectin Sugar (4): Glycogen Lipid (1): fats Lipid (2): trans fats Lipid (3): trans fat, iTFAs vs rTFAs Lipid (4): Phospholipid Lipid (5): Steroids Lipid (6): bio natural wax Protein (1): functions Protein (2): Amino acids Protein (3): Polypeptide Protein (4): 4 level structure Protein (5): 6 interactions Protein (6): denaturation Protein (7): hydrolysis, absorption Nutrients for cellular fuel Organelle▼ Plant Cell vs Animal Cell Organelle Nucleus Rough vs smooth ER Golgi Body Ribosome, Ribozyme Lysosome Plant Central Vacuole Peroxisome Plastid (1): Proplastid Plastid (2):Chromoplast Plastid (3): Chloroplast Ⅰ Plastid (4):Chloroplast Ⅱ Mitochondria (1) Mitochondria (2) Cytoskeleton: microtubule Membrane, transport▼ Plant cell wall Extracellular matrix Cell junction Membrane fluidity Membrane function Simple Diffusion Active transporter Facilitated diffusion Metabolism▼ Energy flow, negative Entropy Gibbs free energy ATP: energy currency💰 Factor for enzyme activity Enzyme, PH, temperature Enzyme: inhibitors, activators Catabolism, Respiration Aerobic Respiration Anaerobic respiration Fermentation L-lactic acid fermentation Photosynthesis (1): history Photosynthesis (2): overview Photosynthesis (4): Dark Reaction, C3 Photosynthesis (5): photorespiration Photosynthesis (6): C4 plant Photosynthesis (7): CAM plant Genetic material▼ DNA story (1): Overview DNA story (2): Friedrich Miescher DNA story (3): Albrecht Kossel DNA story (4): Phoebus Levene DNA story (5): Griffith’s Experiment DNA story (6): Oswald Avery DNA story (7): Chargaff Rules DNA story (8): Hershey-Chase experiment Evolution▼ Eve of Evolution: Creationism Eve of evolution: Catastrophism Eve of evolution: Lamarckism Eve of evolution: Voyage of Beagle Darwin Bulldog Scopes Monkey Trial Natural selection Balance selection Selective breeding Heterosis prebiotic soup Submarine white chimneys From Luca to prokaryote From procaryote to eukaryote Origin of chloroplast Multicellular algae Multicellular animal Ediacaran garden Cambrian Explosion Plant conquered land Microbes▼ Prokaryote vs Eukaryote Gram positive, negative Archaea isn't bacteria 3 common archaea VirusⅠ: Size, structure Porifera, Sponge Blue-green algae Chlamydia Mycoplasma Plant Cell Wall: Structure, Component, Function Anec>Biology>Membrane transport Table of Content ★ Primary cell wall, Middle Lamella, secondary cell wall ★ Components and structure of the cell wall The cell wall is the first structure observed by humans under a microscope. When Robert Hook observed cork slices under his self-made optical microscope in 1665, he saw neatly arranged rooms like a honeycomb. These were actually the xylem cells of plants that had already died. The walls of rooms are the plant cell walls. They are one of the significant features that distinguish plant cells from animal cells. They are a rigid structure surrounding plant plasma membrane whose main components are polysaccharides: cellulose, hemicellulose, and pectin. Woody plants also have lignin to increase hardness and resist compression. Primary cell wall, Middle Lamella, secondary cell wall The earlier synthesized components are farther from the plasma membrane. The earliest form layer is pectin-rich middle lamella that is shared by adjacent cells. When pectin binds to water, it becomes as sticky as glue, so adjacent cells are glued together. The primary cell wall is the second region to be created. Young and some mature plant cells have very thin and flexible primary cell walls. Its shape changes when cell growth. The most notable feature is that cellulose is arranged in a specific direction and regulates cell growth orientation. If plant cell is a cylinder, cellulose is like the rope that binds them. Young plant cell is hard to widen, but it is easy to elongate. The primary cell wall slowly expands to accommodate the gradually growing cells. Polysaccharides is their main component: cellulose (15%~30%), hemicellulose (20%~30%), and pectin (30%~40%). When some cells mature (such as tissues transporting water and nutrients, or providing support), they develop a secondary cell wall between primary cell wall and plasma membrane to add extra strength and prevent cell from enlarging. Its main components are more cellulose and aromatic hydrocarbon derivatives like lignin. They are divided into multiple layers. The cellulose in each layer is arranged in a different direction to hinder cell growth. Very thick secondary cell walls store considerable solar energy and carbon fixed by plant. Most of mass is concentrated here. Plants on Earth synthesize 100 billion tons of cellulose and 20 billion tons of lignin annually. Components and structure of the cell wall Cellulose No matter what kind of cell wall, cellulose is the core component and structure. Hundreds to thousands of glucose are linked together by β-1,4-glycosidic bonds to form a long, unbranched, straight chain. Each chain contains many hydroxyl groups that are key to hydrogen bond formation. About a dozen cellulose chains are connected through these hydrogen bonds and arranged in parallel to form microfibrils whose length is about several micrometers and tensile strength is comparable to steel. They are arranged so orderly and tightly that they form crystalline regions. Multiple cellulose microfibrils aggregate together to form thicker macrofibrils. Hemicellulose and pectin Hemicellulose is a polysaccharide polymerized from various monosaccharides (xylose, glucose, mannose, etc.). They are a type of linear polysaccharide, but the main chain is shorter than cellulose and has some little side chains. The gaps between cellulose microfibrils are filled by them, and adjacent microfibrils are linked by them via hydrogen bonds. The mechanical strength and rigidity of cell wall are enhanced by them. Pectin is a polysaccharide composed of galacturonic acid and other monosaccharides. They are negatively charged and branched, which is similar to the glycosaminoglycan in animal. Therefore, water is absorbed by them to form hydrated gels. Pectin is abundant in middle lamella and primary cell wall. The main function is to promote cell adhesion and fill gaps between microfibrils. Microfibrils are embedded in the hydrated gel and interact with pectin through hydrogen bonds. This not only makes cell wall resilient but also resists external pressure. Lignin is aromatic compound Lignin replaces pectin to fill the gaps between celluloses in secondary cell walls where pectin is almost absent. Benzene ring is located at the core. Their side chains linked covalently to create a three-dimensional network that greatly enhances strength and toughness. Therefore, in macroscopic world, grass is soft and wood is very hard to resist compression and bending. Even the entire tree can be supported by them.. The stable aromatic π-system makes lignin resistant to chemical and biological degradation, including pathogens and extreme climates. This is why wood takes years or even decades to degrade completely in nature. In woody plants, lignin accounts for 18%-35% of the wood mass. The proportion in softwoods (conifers) reaches to 25-35%. There is 18% to 30% lignin in hardwoods (flowering woody plants). They are harder because the cell structure is more complex, and not solely dependent on lignin. The survival strategy in herbaceous plants is rapid growth and reproduction, rather than investing in long-term substance like lignin. This strategy determines the softness of stem. That is, herbaceous plants also contain lignin, but the amount is significantly lower. The average content is about 10-15%. Bamboo is an exception. Although they categorize as herbs, they live for decades. Therefore, they have stems as hard as wood (15-25%). Frequently Asked Questions Anec>Biology>Membrane transport Prev:Cytoskeleton: Microtubule Next:Extracellular Matrix Component, Function: Adhesion, Wound Repair More
16541	https://tanujkhattar.wordpress.com/2016/01/10/the-bridge-tree-of-a-graph/	Skip to content Algorithm Tutorials by Tanuj Khattar tanujkhattar in Graph Theory 1,446 Words The “Bridge Tree” of a graph Introduction : Bridge Tree is a term coined by me that refers to the tree formed by shrinking 2-edge biconnected components of the graph . The 2-edge biconnected components shall be referred to as “bridge components” in the further post. The concept is very intuitive so let’s see more about it. Pre-requisites : Basic Graph Theory. Dfs and Bfs .Bridges in a graph. Biconnected Components and Block-Cut Tree (not really needed to understand but experience with this would be helpful in easier imagination :)) Basic Definitions : Bridge edge : A bridge edge in an undirected graph is an edge whose removal increases the number of connected components in the graph by 1. (For more info Bridges in a graph – GeeksforGeeks) Articulation Points / Cut Vertices : An articulation point in an undirected graph is a vertex whose removal (and corresponding removal of all the edges incident on that vertex) increases the no of connected components in the graph by at-least 1. (For more info Articulation Points (or Cut Vertices) in a Graph – GeeksforGeeks ).Note : Both the end points of a bridge edge are articulation vertices (given that vertex is not a leaf) . Hence, a graph having articulation vertices might not have bridges but a graph having bridges must have articulation vertices (exception : graph with only 2 nodes and 1 bridge) . eg : Biconnected Components : A biconnected component of a given graph is the maximal connected subgraph which does not contain any articulation vertices. (For more info Biconnected components) eg : In the following diagram, different colours represent different biconnected components of the graph. Block Cut Tree : If each biconnected component of a given graph is shrinked into / represented as a single node called a block, and these blocks are attached to each other at shared vertices (articulation points), then the resulting tree formed is called a Block-Cut tree. (For more info Page on inf.elte.hu ) eg : The following would be the block-cut tree of the above graph, where A,B,C are blocks attached to the articulation vertices 3 and 4. A = represents vertices 1,2,3 B = represents vertices 3,4 C = represents vertices 4,5,6 My Definitions : Bridge Component : A bridge component of a given graph is the maximal connected subgraph which does not contain any bridge edges. eg : In the following graph, different coloured vertices lie in different bridge components. The black edges are the normal edges and blue edge represents the bridge edge separating different components Bridge Tree : If each bridge component of a given graph is shrinked into/represented as a single node, and these nodes are connected to each other by the bridge edges which separated these components, then the resulting tree formed is called a Bridge Tree. eg : The following would be the bridge tree formed by shrinking the bridge components of the above given graph. Detailed Analysis : What is a Bridge Tree ? Ans: Bridges in a graph divide the graph into different components such that when we traverse a bridge, we move from one such component to another. Let’s call these components as bridge components. A “bridge tree” is a tree obtained by shrinking each of the bridge components of the graph into a single node such that an edge between two nodes in the resulting tree correspond to the bridge edge in the original graph connecting two different bridge components represented by the two nodes of the tree. Properties of the Bridge Tree Each edge in the bridge tree is the one of the bridge edges in the original graph. Since each node in the bridge tree is formed by shrinking the bridge components of original graph, therefore the bridge tree of a graph with N vertices can have at most N nodes (and N-1 edges). From the above point, it directly follows that a graph with N vertices can have at most N-1 bridges (why ?? ) How to build the Bridge Tree from a given graph ?? Ans: Now this is the most interesting part of the article and requires good understanding (and imagination) of BFS and DFS. This is an approach which I personally thought of, and you can have your own method of building the tree. (The concept understanding is more important than the code) Follow the following steps to build the bridge tree : Start a dfs from any node of the graph. In every step of the dfs, start a bfs to explore the bridge component in which this node lies. While exploring this component, as soon as a bridge edge is encountered traverse the bridge edge from the dfs step such that now the dfs step lands to other bridge component. Again launch a bfs to explore this component. Once this component has been explored, we return to the original component where we had paused the original bfs. Again this bfs continues to explore the remaining bridge component. The recursion continues. In short, each step of the dfs traverses only bridges of the original graph and once a bridge component is reached, the bfs step explores that particular component fully. (See the code for better understanding) ``` VI tree[N],graph[N];//edge list representation of graph int U[M],V[M],vis[N],arr[N],T,cmpno; bool isbridge[M]; // if i'th edge is a bridge edge or not queue Q[N]; int adj(int u,int e) { return U[e]==u?V[e]:U[e]; } int dfs0(int u,int edge) { //mark bridges vis[u]=1; arr[u]=T++; int dbe = arr[u]; for(int i=0; i Following is a shorter and more (memory) efficient implementation of bridge tree. VI tree[N],g[N];//edge list representation of graph int U[M],V[M],vis[N],arr[N],T,dsu[N]; bool isbridge[M]; // if i'th edge is a bridge edge or not int adj(int u,int e) { return U[e]^V[e]^u; } int f(int x) { return dsu[x]=(dsu[x]==x?x:f(dsu[x])); } void merge(int a,int b) { dsu[f(a)]=f(b); } int dfs0(int u,int edge) { //mark bridges vis[u]=1; arr[u]=T++; int dbe = arr[u]; for(auto e : g[u]) { int w = adj(u,e); if(!vis[w])dbe = min(dbe,dfs0(w,e)); else if(e!=edge)dbe = min(dbe,arr[w]); } if(dbe == arr[u] && edge!=-1)isbridge[edge]=true; else if(edge!=-1)merge(U[edge],V[edge]); return dbe; } void buildBridgeTree(int n,int m) { for(int i=1; i<=n; i++)dsu[i]=i; for(int i=1; i<=n; i++)if(!vis[i])dfs0(i,-1); for(int i=1; i<=m; i++) if(f(U[i])!=f(V[i])) { tree[f(U[i])].PB(f(V[i])); tree[f(V[i])].PB(f(U[i])); } } Applications of the Bridge Tree All the problems where Bridge Tree can be used can also be solved by Block-cut tree formed by Biconnected Components but many a times the bridge tree is far more intuitive than the block-cut tree eg : Q : Given an undirected connected graph with N nodes and M edges. You can add at-most 1 edge in the graph between any two nodes. Find the minimum number of bridges in the resulting graph. Ans : Build the bridge tree of the given graph and add 1 edge between the two end points of the diameter such that maximum no of bridges are destroyed. The resulting number of edges in the tree would be the answer. Also, the above code can be easily modified for maintaining some information of the compressed bridge components if required in the question.eg : Problem H – Dashboard – 2015 ACM Arabella Collegiate Programming Contest – Codeforces Other Problems (will be updated as and when found) : Sherlock and Queries on the Graph Problem – E – Codeforces Problem H : Page on codeforces.com Conclusion : Hope the explanation provided above is clear and understandable. Any comments/suggestions are most welcome. Happy Coding 🙂 Note : Link to original post Share this: Click to share on X (Opens in new window) X Click to share on Facebook (Opens in new window) Facebook Like Loading... ``` 3 thoughts on “The “Bridge Tree” of a graph” Awesome tutorial, man. LikeLike Reply 2. An easy problem : Light OJ 1291 – Real Life Traffic LikeLike Reply 3. Pingback: SnackDown Training Camp Day 3 Leave a comment Cancel reply Comment Reblog Subscribe Subscribed Algorithm Tutorials Already have a WordPress.com account? Log in now. Algorithm Tutorials Subscribe Subscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar Loading Comments... Design a site like this with WordPress.com Get started
16542	https://www.ck12.org/flexi/cbse-math/grouped-data-to-find-the-median/lesspgreaterthe-median-of-an-ungrouped-data-and-the-median-calculated-when-the-same-data-is-grouped-are-always-the-same-do-you-think-that-this-is-a-correct-statement-give-reason-less-by-pgreater/	Flexi answers - The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason. \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects CBSE Math Grouped Data to Find the Median Question The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason. Flexi Says: The statement that the median of ungrouped data and the median calculated when the same data is grouped are always the same is not correct. Here's why: Ungrouped Data: The median is the middle value when the data is ordered. If there's an odd number of data points, it's the middle one. If even, it's the average of the two middle values. Grouped Data: The median is estimated using class intervals and frequencies. It might not match the exact median of ungrouped data because it's based on an approximation. So, the medians can differ due to the approximation in grouped data. Analogy / Example Try Asking: Some teenagers collected trash for a beach clean-up. The data for the number of pounds of trash collected by each teenager are shown below: 26, 26, 21, 22, 20, 25, 35. What is the IQR of the data?How do you calculate the median of grouped data?Accountants often use the median when studying salaries for various businesses. What is the median of the following salary list: $32,019; $21,934; $27,581; $43,795; $38,871; $25,963? The median is $ enter your response here. How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
16543	https://www.baldwincpas.com/insights/construction-cost-outlook-2025-tariffs-delays-supply-risks	Published Time: Mon, 08 Sep 2025 23:16:12 GMT Construction Cost Outlook 2025: Tariffs, Delays & Supply Risks Skip to the main content. Client Login Make A Payment Who We Serve Construction Government Non-Profit Healthcare Small to Mid-size Businesses Auto Dealerships How We Serve Audit & Assurance Tax Compliance & Planning Elevate: Accounting & Payroll Payroll Services Accounting Services Advisory Services Business Valuation, Litigation & Forensic Services Employee Benefit Plan Audits HR Consulting Wealth Management About Us History Community Team Members Blog Careers Current Opportunities Contact Toggle Menu Toggle Menu Who We Serve Construction Government Non-Profit Healthcare Small to Mid-size Businesses Auto Dealerships How We Serve Audit & Assurance Tax Compliance & Planning Elevate: Accounting & Payroll Payroll Services Accounting Services Advisory Services Business Valuation, Litigation & Forensic Services Employee Benefit Plan Audits HR Consulting Wealth Management About Us History Community Team Members Blog Careers Current Opportunities Contact FacebookInstagramLinkedinXYouTubeMedium 3 min read Supply Chain Disruptions & the Construction Cost Outlook: What to Watch in Late 2025 Baldwin CPAs 8/4/25 7:45 AM Understanding the Current Landscape As we close in on the second half of 2025, construction companies are facing mounting uncertainty driven by shifting supply chain dynamics, material shortages, and evolving federal tariff policies. Contractors, developers, and builders are navigating a volatile environment in which pricing forecasts are increasingly difficult to pin down. From energy infrastructure demands to the cascading effects of global trade shifts, the cost landscape is complex and unpredictable. Recent reports indicate that even as demand in certain sectors remains steady, the cost to complete construction projects continues to rise. The pressure is not solely the result of inflation or labor shortages. Rather, it stems from a confluence of interdependent forces that include tariff policies, long lead times for critical materials, competition from large-scale industrial and energy projects, and tight global supply conditions—particularly for metals like copper and aluminum. For Kentucky-based contractors and construction companies across the U.S., these developments require careful planning and real-time financial adjustments. Material Costs and Tariff Pressures One of the biggest cost drivers heading into late 2025 is the uncertainty surrounding tariffs on essential building materials. While tariffs have not yet surged in full force, the mere anticipation of trade adjustments has already begun influencing the cost and availability of key inputs. For example, construction-grade copper—used extensively in electrical and plumbing systems—has seen price swings due to a shift in global trade routes and stockpiling activity in advance of anticipated tariff escalations. Aluminum, steel, and other metals are also showing signs of volatility. Contractors relying on imported materials may be affected first, but domestic suppliers are also adjusting their pricing models based on expectations of supply chain pressure and market realignment. These shifts are creating a climate in which estimating future material costs has become more speculative, increasing the risk of underbidding or margin compression. As these trends continue to unfold, the importance of updated procurement strategies and tax-efficient purchasing decisions will grow. Lead Times and Equipment Delays Long lead times are another factor contributing to construction cost increases in 2025. According to recent reporting, the surge in natural gas plant construction in the U.S. is placing additional strain on the availability of specialized components such as turbines, switchgear, and heavy electrical equipment. While these large energy projects are concentrated in certain states, the ripple effects are being felt across the country as manufacturers struggle to keep up with demand. Construction firms that need similar types of equipment or are working on projects with overlapping material requirements may encounter delays in procurement or be forced to pay a premium to meet scheduling obligations. These challenges extend beyond mechanical systems to include structural components and HVAC systems, particularly in commercial and industrial builds. For firms operating on tight timelines or fixed-cost contracts, these disruptions can introduce financial stress and affect project profitability. Project Backlogs and Bid Volatility Market conditions are also influencing the competitive landscape in new ways. As pricing becomes harder to predict, many contractors are adding more contingencies or cost buffers to their bids. This trend is evident in both public and private sector projects, where owners and developers are seeing greater variability in price proposals and projected timelines. Some are responding by delaying project starts or scaling down scopes, which can, in turn, affect the backlog health of mid-sized construction firms. At the same time, demand remains strong in many sectors, particularly healthcare, multifamily housing, and infrastructure. The challenge lies in aligning bid strategy with realistic projections. Overly aggressive pricing can jeopardize margins if material or labor costs spike, while overly conservative estimates risk losing out on project awards. Financial forecasting, cost tracking, and scenario modeling are becoming essential tools for maintaining operational resilience. What Construction Firms Should Do Now In light of these ongoing disruptions, construction businesses should take a proactive approach to financial management, including regular review of project cost forecasts, supplier relationships, and tax treatment of materials and inventory. Engaging in forward-looking planning can help mitigate some of the uncertainty associated with delayed shipments or price spikes. Contract terms should also be revisited, especially as they relate to escalation clauses and material substitution provisions. Firms may want to work with their legal and financial advisors to develop contract language that shares pricing risk or allows for flexibility in the face of external shocks. Additionally, companies should evaluate how current procurement practices impact their tax position and cash flow, particularly in relation to large capital purchases or stockpiled materials. Finally, firms should be tracking economic signals—such as federal trade announcements, energy sector investment trends, and lead time data from key suppliers—so they can adapt quickly to changing conditions. The goal is not to eliminate risk but to manage it with more precision and awareness. Baldwin CPAs: Supporting the Construction Industry Through Economic Uncertainty At Baldwin CPAs, we understand that rising costs and unpredictable supply chains create more than short-term headaches—they impact long-term planning, profitability, and tax exposure. Our team works with construction companies across Kentucky to help them navigate these challenges with confidence, from strategic tax planning to financial forecasting and contract advisory. Contact Baldwin CPAs today and learn how we can help you prepare for the months ahead with clarity and confidence. FAQ: Construction Costs and Supply Chain Disruptions Are tariffs already affecting material costs in the U.S. construction market? While many tariffs are still in discussion or early implementation stages, the market has already responded. Anticipation of increased duties is influencing supplier pricing, particularly for copper, steel, and aluminum. Some contractors are seeing elevated costs even before formal tariff changes take effect. What materials are experiencing the longest lead times? Electrical equipment and components—especially those used in energy projects—are experiencing extended lead times. This includes turbines, generators, switchgear, and related systems. Structural steel and HVAC systems may also face longer-than-usual delays depending on the project location and supplier network. How can construction firms manage pricing uncertainty in late 2025? Companies should revisit bidding strategies, contract terms, and supplier agreements to account for potential cost volatility. Including escalation clauses, closely monitoring supplier updates, and working with financial advisors to model different cost scenarios can help mitigate risk. Share 1 min read Should cloud computing setup costs be expensed or capitalized? Marketing : 10/5/18 12:00 AM Companies will be able to capitalize, or spread out the costs of, setting up pricey business systems that operate on cloud technology under an update... Read More 1 min read Public companies to disclose stock hedging policies and practices Marketing : 5/6/19 12:00 AM Does your company have policies in place regarding the use of hedging transactions by company insiders? Final Securities and Exchange Commission... Read More 1 min read Corporate culture: Rotten apples could spoil your financials Marketing : 3/26/18 12:00 AM Auditors often say that the tone at the top of an organization trickles down to every level of the business. Is your company’s work environment... Read More © 2025 Baldwin CPAs, PLLC Richmond Flemingsburg Lexington Louisville Maysville Sharepoint File Upload Privacy Policy © 2025 Return to top
16544	https://www.verywellhealth.com/how-long-does-sperm-live-8735193	How Long Sperm Lives Inside and Outside of the Body Skip to content Menu Health A-Z Arthritis Type 2 Diabetes Heart Disease Digestive Health Multiple Sclerosis Skin Health View All Prevention & Treatment Diet & Nutrition First Aid Surgery Health Care Health Insurance Public Health Patient Rights Disability Caregivers & Loved Ones End of Life Concerns View All News Health News Trending Tools & Resources Thyroid Test Analyzer Doctor Discussion Guides Hemoglobin A1c Test Analyzer Lipid Test Analyzer Complete Blood Count (CBC) Analyzer About Us Editorial Process Meet Our Medical Expert Board Search Search GO Health A-Z Arthritis Type 2 Diabetes Heart Disease Digestive Health Multiple Sclerosis Skin Health View All Prevention & Treatment Diet & Nutrition First Aid Surgery View All Health Care Health Insurance Public Health Patient Rights Disability Caregivers & Loved Ones End of Life Concerns View All News Health News Trending View All More in Urological Health Penis Health Peyronie’s Disease Erectile Dysfunction Bladder Health Chronic Kidney Disease Prostate Health Tools & Resources - Thyroid Test Analyzer - Doctor Discussion Guides - Hemoglobin A1c Test Analyzer - Lipid Test Analyzer - Complete Blood Count (CBC) Analyzer About Us Editorial Process Medical Review Board Verywell Health's content is for informational and educational purposes only. Our website is not intended to be a substitute for professional medical advice, diagnosis, or treatment. Ⓒ 2025 People Inc. — All rights reserved How Long Sperm Lives Inside and Outside of the Body By Patty Weasler, RN, BSN Published on January 02, 2025 Medically reviewed by Matthew Wosnitzer, MD Table of Contents View All Table of Contents Typical Sperm Lifespan Contributing Factors Pregnancy Odds When Do Sperm Die? How long sperm can live outside of the body that has produced them depends on several factors. When sperm have been deposited inside the female reproductive tract (vagina, uterus, and fallopian tubes), they can live for up to five days. Sperm that has been properly frozen can live for decades. Kathrin Ziegler / Getty Images Typical Sperm Lifespan Healthy adult testicles (testes) continuously produce sperm. On average, it takes 74 days for sperm to mature fully. Sperm production starts in the seminiferous tubules within each testicle and moves to the epididymis, where it is stored. How long it lives after it matures will depend on several factors discussed below. A Note on Gender and Sex Terminology Sperm are reproductive cells produced in the testicles, which are organs typically present in people assigned male at birth. Verywell Health acknowledges that sex and gender are related concepts, but they are not the same. To reflect our sources accurately, this article uses terms like “female,” “male,” “woman,” and “man” as the sources use them. Before Ejaculation Before ejaculation (the ejection of semen through the penis) the body stores sperm within the testes. If the sperm are not used once they are fully mature, the body will reabsorb them. After Ejaculation How long sperm can live after ejaculation will vary. Sperm that is ejaculated into the female reproductive tract can live for up to five days. However, sperm that is ejaculated and is outside of a body can live for up to an hour if it is at room temperature (68 degrees F). Sperm that a healthcare provider freezes can survive for decades. Someone might choose to do this (bank sperm) if they are undergoing treatment for certain health conditions like cancer and want to preserve the sperm for future use. Pregnancy Misconceptions Pregnancy following vaginal sex can occur when sperm travels through the vaginal canal and cervix into the uterus and into the fallopian tube, where it meets the egg. This complex movement must also occur when the female is fertile during ovulation (release of a mature egg from the ovary). Many people have misconceptions about how to become pregnant. Talking to a healthcare provider to answer specific questions is always best. Contributing Factors Many factors can contribute to the lifespan of sperm. Once sperm leaves the body, it must remain moist and warm to stay alive. Below are ways to help improve sperm health and sperm count: Avoid alcohol. Stop smoking. Do not take recreational drugs. Maintain a healthy weight with a healthy diet and regular exercise. Avoid chemical exposure (pesticides, solvents, and heavy metals). Do not wear tight-fitting underwear. Sperm Life and Pregnancy Odds With sperm's short lifespan of up to five days within the female reproductive tract, vaginal intercourse or insemination must happen at the right time for pregnancy to occur. Below are details of the likelihood of pregnancy from vaginal intercourse or artificial insemination. From Intercourse Pregnancy is more likely to occur if vaginal intercourse occurs every other day or every third day on or around a female's fertile days. A female is typically fertile between days 11 and 14 of a 28-day menstrual cycle; this is when ovulation occurs. During this fertile time, the cervical mucus changes to protect the sperm and to help it move toward the uterus, making pregnancy easier. If a female is under 40 years of age and has regular, unprotected vaginal intercourse for one year, there is an 8 out of 10 chance that they will get pregnant. Artificial Insemination Intrauterine insemination (IUI) is when a healthcare provider directly injects sperm into the uterus. This type of artificial insemination is a treatment that is used to help increase the chances of sperm fertilizing an egg. During vaginal intercourse, a few hundred sperm reach the egg, but in artificial insemination, millions of healthy sperm are placed directly inside the uterus. The success of IUI varies greatly among users. One study from 2022 found a per-cycle success rate of 10.9% and an overall success rate of 19.4%. Semen Analysis About half of fertility issues are related to sperm. If a male's fertility needs to be assessed, a healthcare provider may order a semen analysis. It is used to measure the quality and count of sperm to determine fertility. When Do Sperm Die? Sperm cannot survive in certain conditions. After ejaculation, if sperm is not in the female reproductive tract, they will die within an hour. If sperm is in the female reproductive tract, it will die after five days. Summary Sperm cannot survive for very long after ejaculation if it is not in the female reproductive tract. People who are trying to get pregnant should aim to have vaginal intercourse every two to three days around and during the female's ovulation time. Ovulation typically occurs between days 11 and 14 of a 28-day menstrual cycle. A healthcare provider can assist with determining if there are health concerns related to the sperm if infertility is suspected. Read more: Health A-Z Urological Health Penis Health 12 Sources Verywell Health uses only high-quality sources, including peer-reviewed studies, to support the facts within our articles. Read our editorial process to learn more about how we fact-check and keep our content accurate, reliable, and trustworthy. MedlinePlus. Pregnancy - identifying fertile days. Beacon Health System. Sperm: how long do they live after ejaculation? Punjani N, Wald G, Al-Hussein Alwamlh O, Feliciano M, Dudley V, Goldstein M. Optimal timing for repeat semen analysis during male infertility evaluation. F S Rep. 2021;2(2):172-175. doi:10.1016/j.xfre.2021.04.010 MedlinePlus. Sperm. University of North Carolina School of Medicine. Vasectomy. Cleveland Clinic. Sperm. University of California San Francisco Center for Reproductive Health. Conception: how it works. National Health Services. Low sperm count. National Health Services. Trying to get pregnant. 10. Penn Medicine. Intrauterine insemination (IUI). 11. Zippl AL, Wachter A, Rockenschaub P, Toth B, Seeber B. Predicting success of intrauterine insemination using a clinically based scoring system. Arch Gynecol Obstet. 2022;306(5):1777-1786. doi:10.1007/s00404-022-06758-z 12. MedlinePlus. Semen analysis. By Patty Weasler, RN, BSN Weasler is a Wisconsin-based registered nurse with over a decade of experience in pediatric critical care. See Our Editorial Process Meet Our Medical Expert Board Share Feedback Was this page helpful? Thanks for your feedback! What is your feedback? Helpful Report an Error Other Submit Related Articles 11 Causes of a Swollen Penis and What to Do About It Male Hypogonadism: What to Know What Is Erectile Dysfunction? Low Testosterone Symptoms to Recognize An Overview of an Orchiectomy Penile Fractures: What You Need to Know Balanitis: Inflammation of the Penis Head Erectile Dysfunction Injections: What You Need to Know Causes of Micropenis and Options for Treatment Low Testosterone: What Causes Symptoms in Young Males 20 and Up? How Effective Is a Vasectomy? Peyronie's Disease Statistics and Facts You Should Know Benefits and Risks of Semen Retention Causes of Bladder Infections in Children, Young, and Older Adults Top 10 Erectile Dysfunction (ED) Supplements What Are the Treatments for IgA Nephropathy? Daily Health Tips to Your Inbox Email Address Sign Up You're in! Thank you, {{form.email}}, for signing up. There was an error. Please try again. 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16545	https://chem.libretexts.org/Courses/East_Tennessee_State_University/CHEM_3110%3A_Descriptive_Inorganic_Chemistry/02%3A_Atomic_Theory/2.04%3A_Periodic_Properties_of_Atoms	2.4: Periodic Properties of Atoms - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 2: Atomic Theory CHEM 3110: Descriptive Inorganic Chemistry { } { "2.01:_Quantum_Numbers_and_Atomic_Wavefunctions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.02:_Shielding" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.03:_Aufbau_Principle" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.04:_Periodic_Properties_of_Atoms" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Introduction_to_Inorganic_Chemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Atomic_Theory" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Bonding_Theories" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Molecular_Symmetry_and_Point_Groups" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Structure_and_Energetics_of_Solids" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Acid-Base_and_Donor-Acceptor_Chemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Reduction_and_Oxidation_Chemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Coordination_Chemistry-_Structure_and_Isomers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Coordination_Chemistry-_Bonding" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Coordination_Chemistry-_Reactions_and_Mechanisms" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Organometallic_Chemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Bioinorganic_Chemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Mon, 23 Aug 2021 03:07:10 GMT 2.4: Periodic Properties of Atoms 328901 328901 Joshua Halpern { } Anonymous Anonymous User 2 false false [ "article:topic", "ionization energy", "electron affinity", "atomic radii", "periodic trends", "showtoc:no", "Ionic Radii", "license:ccbyncsa", "source-chem-151361", "source-chem-198183", "source-chem-198184", "source-chem-198185", "licenseversion:40", "citationstyle:acs" ] [ "article:topic", "ionization energy", "electron affinity", "atomic radii", "periodic trends", "showtoc:no", "Ionic Radii", "license:ccbyncsa", "source-chem-151361", "source-chem-198183", "source-chem-198184", "source-chem-198185", "licenseversion:40", "citationstyle:acs" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Campus Bookshelves 3. East Tennessee State University 4. CHEM 3110: Descriptive Inorganic Chemistry 5. 2: Atomic Theory 6. 2.4: Periodic Properties of Atoms Expand/collapse global location CHEM 3110: Descriptive Inorganic Chemistry Front Matter 1: Introduction to Inorganic Chemistry 2: Atomic Theory 3: Bonding Theories 4: Molecular Symmetry and Point Groups 5: Structure and Energetics of Solids 6: Acid-Base and Donor-Acceptor Chemistry 7: Reduction and Oxidation Chemistry 8: Coordination Chemistry- Structure and Isomers 9: Coordination Chemistry- Bonding 10: Coordination Chemistry- Reactions and Mechanisms 11: Organometallic Chemistry 12: Bioinorganic Chemistry Back Matter 2.4: Periodic Properties of Atoms Last updated Aug 23, 2021 Save as PDF 2.3: Aufbau Principle 3: Bonding Theories Page ID 328901 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Introduction 2. Ionization Energy 1. 1st, 2nd, and 3rd Ionization Energies 2. General periodic trends in ionization energy Electron Affinity EA defined as removal of an electron EA defined as addition of an electron Trends in Electron Affinity Atomic and Ionic Radii Trends in Atomic Radius Trends in Ionic Radius Cations Anions Contributors and Attributions References Introduction General periodic trends are specific patterns that are present within the periodic table; these are patterns in properties like electronegativity, ionization energy, electron affinity, atomic radius, melting point, and metallic character. General periodic trends provide chemists with an invaluable tool to quickly predict an element's properties. These trends exist because of the similar atomic structure of the elements within their respective group families or periods, and because of the periodic nature of the elements. Some of the general periodic trends are described in this section. Ionization Energy Ionization energy (IE) is the energy required to remove an electron from a neutral atom or cation in its gaseous phase. IE is also known as ionization potential. (2.4.1)A(g)n+⟶A(g)(n+1)++e−I⁢E=Δ⁢U Conceptually, ionization energy is the affinity of an element for its outermost electron (an electron it already has in its valence shell). 1 st, 2 nd, and 3 rd Ionization Energies The symbol I⁢E 1 stands for the first ionization energy(energy required to take away an electron from a neutral atom, where n=0). The symbol I⁢E 2 stands for the second ionization energy (energy required to take away an electron from an atom with a +1 charge, n=2.) First Ionization Energy, I⁢E 1 (general element, A): A(g)→A(g)1++e− Second Ionization Energy, I⁢E 2 (general element, A): A(g)1+→A(g)2++e− Third Ionization Energy, I⁢E 3 (general element, A): A(g)2+→A(g)3++e− Each succeeding ionization energy is larger than the preceding energy. This means that I⁢E 1<I⁢E 2<I⁢E 3<...<I⁢E n will always be true. For example, ionization energy increases as succeeding electrons are taken away from Mg. (2.4.2)M⁢g⁡(g)→M⁢g+⁡(g)+e−I⁢E 1=738 k⁢J/m⁢o⁢l (2.4.3)M⁢g+⁡(g)→M⁢g 2+⁡(g)+e−I⁢E 2=1451 k⁢J/m⁢o⁢l Ionization energy is correlated with the strength of attraction between the positively-charged nucleus and the negatively-charged valence electrons, and therefore the effective nuclear charge Z e⁢f⁡f. The higher the ionization energy, the stronger the attractive force between nucleus and valence electrons, and the more energy is required to remove a valence electron. The lower the ionization energy, the weaker the attractive force between nucleus and valence electrons, and the less energy required to remove a valence electron. General periodic trends in ionization energy In general, ionization energies increase from left to right and decrease down a group; however there are variations in these trends that would be expected from the effects of penetration and shielding. The trends in first ionization energy are shown in Figure 2.4.1 and are summarized below. Across a period:As Z increases across a period, the ionization energy of the elements generally increases from left to right. However there are breaks or variation in the trends in the following cases: IE is especially low when removal of an electron creates a newly empty p subshell (examples include I⁢E 1 of B, Al, Sc) IE energy is especially low where removal of an electron results in a half-filled p or d subshell (examples include I⁢E 1 of O, S) IE increases more gradually across the d- and f-subshells compared to s- and p- subshells. This is because d- and f- electrons are weakly penetrating and experience especially low Z. From one period to the next: There is an especially large decrease in IE with the start of every new period (from He to Li or from Ne to Na for example). This is consistent with the idea that IE is especially low when removal of an electron creates a newly empty s-subshell. Nobel gases:The noble gases posses very high ionization energies. Note that helium has the highest ionization energy of all the elements. Down a group: Although Z increases going down a group, there is no reliable trend in IE going down any group; in some cases IE increases going down a group, while in other cases IE decreases going down a group. Figure 2.4.1: Graph showing the First Ionization Energy of the Elements. Left: ionization energies are mapped onto the periodic table, where the magnitude of ionization energy is depicted as a bar graph. Right: the first ionization energy is plotted against atomic number. In both panels the elements of the s-block are shown in purple, p-block are shown in green, d-block are shown in red, and f-block are shown in blue. Plots of the I⁢E 1,I⁢E 2, and I⁢E 3 of elements from hydrogen to krypton (first four periods) are shown in Figure 2.4.2. Notice that I⁢E 3>I⁢E 2>I⁢E 1. Also notice that trends mentioned above for I⁢E 1 hold true for subsequent ionizations when electron configurations are considered! Figure 2.4.2. The first (I 1), second (I 2), and third (I 3) ionization energies are plotted for elements with Z = 1 to 36 (H to Kr). The position of each element in its atomic form is indicated as s- p- or d-block. (CC-BY-NC-SA; Kathryn Haas) Electron Affinity According to IUPAC, there are two different, but equivalent, definitions of electron affinity (EA).6 EA defined as removal of an electron Electron affinity can be defined as the energy required when an electron is removed from a gaseous anion. The reaction as shown in equation 2.4.4 is endothermic (positive Δ⁢U) for elements except noble gases and alkaline earth metals. Under this definition, the more positive the EA value, the higher an atom's affinity for electrons. (2.4.4)A(g)−⟶A(g)+e−E⁢A=Δ⁢U The reaction shown in equation 2.4.4 is similar those that define ionization energy. For this reason, the EA is also described as the zeroth ionization energy. EA defined as addition of an electron An alternate and more common definition is the microscopic reverse of equation 2.4.4. This more common definition states that electron affinity is the energy released when an electron is added to a gaseous atom, as shown in equation 2.4.5. The reaction as shown in equation 2.4.5 is exothermic (negative Δ⁢U) for elements except noble gases and alkaline earth metals. The more negative this EA value, the higher an atom's affinity for electrons. (2.4.5)A(g)+e−⟶A(g)−E⁢A=Δ⁢U Conceptually, this second definition is quite similar to the concept of electronegativity; but unlike electronegativity, EA is a well-defined quantitative measurement. Trends in Electron Affinity For this discussion, we will use the first definition of EA:a more positive (larger) value means that the EA is higher (meaning stronger affinity toward an electron). Across a period:Similar to ionization energy, EA generally increases across a row of the periodic table; this observation is consistent with the increase in effective nuclear charge (Z) from left to right across a period. However, there are variations across a period that are similar to variations in ionization energy and that can be explained by shielding, penetration, and electron configuration. Down a group:Like the case of ionization energy trends, EA does not consistently decrease going down a column of the periodic table despite the fact that Z∗ increases down a group. The trend in EA follows a similar zig-zag pattern as the one seen with ionization energies, except that it is displaced by one unit from the trend in I⁢E 1, two units from I⁢E 2, and so on. For example, EA peaks at F, while I⁢E 1 peaks at Ne, I⁢E 2 peaks at Na, and I⁢E 3 peaks at Mg. A plot of EA for the first 13 elements is shown overlaid on plots of I⁢E 1,I⁢E 2 and I⁢E 3 in Figure 2.4.3., where the shifts in the peaks and valleys within each zig-zag trend are indicated. Figure 2.4.3. A plot of electron affinity (the zeroth ionization energy) is overlaid on plots of the first (I 1), second (I 2), and third (I 3) ionization energies. Each plot is shown using separate y-axes. Trends in EA are similar to those in ionization energies, except the peaks and valleys of the trends are shifted by one unit, as indicated. These plots are shown in units of kJ/mole. (CC-BY-NC-SA; Kathryn Haas) Atomic and Ionic Radii There are several methods that can be used to determine radii of atoms and ions: Nonpolar atomic radii: The radius of an atom is derived from the bond lengths within nonpolar molecules; one-half the distance between the nuclei of two atoms within a covalent bond. van der Waals radius: The radius of an atom is determined by collision with other atoms. Crystal Radii: The atomic or ionic radius is determined using electron density maps from X-ray data. The measurement of atomic or ionic size will depend on a number of factors, including the covalent character of bonding in any particular molecule, coordination number, physical state (liquid, solid, gas), the identity of nearby atoms/ions, variation in crystal structure, and distortions within regular crystal structures. You should keep in mind that the size of an atom or ion is a "fuzzy" measure, and the radius under a different set of conditions will probably change slightly. Regardless, measured atomic and ionic radii reveal obvious trends across the periodic table and between atoms and ions. The relative atomic sizes shown in Figure 2.4.4 were derived from crystallographic data.9 Trends in Atomic Radius Atomic size generally decreases gradually from left to right across a period of elements. As nuclear charge (Z) increases, we expect the effective nuclear charge (Z) of the valence electrons to also increase. Increasing Z pulls electrons closer to the nucleus. However, with each additional unit of Z, there is also an additional electron. The change in size is a balance of a compression caused by increasing Z and an expansion in the number of electrons. As a result, the atomic radius decreases gradually across a period. Atomic size generally increases going down a group. As valence electrons occupying higher level shells due to the increasing quantum number (n), size increases despite the fact that Z and Z are increasing going down the group. Figure 2.4.4._Atomic Radii Calculated from Crystalographic Data. Data from Cordero, Beatriz, Veronica Gomez, Ana E. Platero-Prats, Marc Reves, Jorge Echeverria, Eduard Cremades, Flavia Barragan, and Santiago Alvarez. “Covalent Radii Revisited.” Dalton Transactions, no. 21 (2008): 2832–38. doi:10.1039/b801115j._ Trends in Ionic Radius Trends in ionic radius follow general trends in atomic radius for ions of the same charge. Ionic radius varies with charge of the ion (and number of electrons) and the electron configuration (e.g. high spin or low spin). Cations Compared to their atoms, cations have the same Z but fewer electrons. Removal of electrons from an atom to form a cation results in a significant increase in effective nuclear charge, resulting in all other electrons being more strongly attracted to the nucleus, and having a lower energy level. The result is a contraction in size from the atom to cation. Figure 2.4.5 visually illustrates the relative size of atoms and some cations of the first four periods; the data is available in tabular format in Figure 2.4.6. Anions Compared to their atoms, anions have the same Z but more electrons. Addition of electrons to an atom to form an anion results in a decrease in effective nuclear charge, which corresponds to a decrease in attractive force between the nucleus and electrons. Lower attractive force leads to expansion, where the size of the atom becomes larger in the formation of an anion. Figure 2.4.5 visually illustrates the relative size of atoms and some anions of the first four periods, while the data is available in tabular format in Figure 2.4.6. Figure 2.4.5. This figure illustrates relative size of atoms and ions of the first four periods of the periodic table. Atomic radius is indicated by grey circles. Radius of cations is shown in green (+1), lime (+2), and yellow (+3) circles. Radius of anions are shown as aqua (-1), blue (-2), and purple (-3) circles. Data from sources 1-3 Figure 2.4.6. This figure shows radii (in Angstroms) of atoms and ions of the first four periods of the periodic table. Radii from each element are listed from largest to smallest, ionic charge indicated in parentheses (); hs = high spin, ls = low spin, cn6 = coordination number is 6. Data from sources 9-11. Contributors and Attributions Swetha Ramireddy (UCD), Bingyao Zheng (UCD), Emily Nguyen (UCD) Curated or created by Kathryn Haas References Russo, Steve, and Mike Silver. Introductory Chemistry. San Francisco: Pearson, 2007. Petrucci, Ralph H, et al. General Chemistry: Principles and Modern Applications. 9th Ed. New Jersey: Pearson, 2007. Atkins, Peter et. al, Physical Chemistry, 7 th Edition, 2002, W.H Freeman and Company, New York, pg. 390. Alberty, Robert A. et. al, Physical Chemistry, 3 rd Edition, 2001, John Wiley & Sons, Inc, pg. 380. Kots, John C. et. al, Chemistry & Chemical Reactivity, 5 th Edition, 2003, Thomson Learning Inc, pg. 305-309. IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. doi.org/10.1351/goldbook. Electron Affinity (data page), Wikipedia. en.Wikipedia.org/wiki/Electron_affinity_(data_page) Accessed 12/3/19. Cordero, Beatriz, Veronica Gomez, Ana E. Platero-Prats, Marc Reves, Jorge Echeverria, Eduard Cremades, Flavia Barragan, and Santiago Alvarez. “Covalent Radii Revisited.” Dalton Transactions, no. 21 (2008): 2832–38. doi:10.1039/b801115j. R. D. Shannon (1976). "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides". Acta Crystallogr A. 32 (5): 751–767. Bibcode:1976AcCrA..32..751S. doi:10.1107/S0567739476001551. Atomic Radius, Wikipedia. Ionic Radius, Wikipedia. 2.4: Periodic Properties of Atoms is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. Back to top 2.3: Aufbau Principle 3: Bonding Theories Was this article helpful? Yes No Recommended articles 2: Atomic Theory Article typeSection or PageLicenseCC BY-NC-SALicense Version4.0Show Page TOCno on page Tags atomic radii citationstyle:acs electron affinity Ionic Radii ionization energy periodic trends source-chem-151361 source-chem-198183 source-chem-198184 source-chem-198185 © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 2.3: Aufbau Principle 3: Bonding Theories
16546	https://www.youtube.com/watch?v=WfG3nZ1WgPU	The Most Important Theorem in Finite Group Theory Bill Kinney 34700 subscribers 15 likes Description 819 views Posted: 10 Mar 2023 Lagrange's Theorem is often described as the most important theorem in finite group theory. If G is a finite group and H is a subgroup of G, then the order of H divides the order of G. A corollary says that the order of any element of G divides the order of G. The converse of Lagrange's Theorem is false. For example, the alternating group A4, which has order 12, has no subgroup of order 6 even though 6 divides 12. The proof of Lagrange's Theorem is based on properties of cosets (left cosets aH and right cosets Ha). In particular, the collection of all left cosets of a subgroup H partitions the group G into subsets with the same cardinality (number of elements). The number of distinct left cosets is called the index of H in G and equals the order of G divided by the order of H. 🔴 "Contemporary Abstract Algebra", by Joe Gallian: 🔴 Abstract Algebra Exam 1 Review Problems and Solutions: 🔴 Abstract Algebra Exam 2 Review Problems and Solutions: 🔴 Abstract Algebra Exam 3 Review Problems and Solutions: 🔴 Abstract Algebra Final Exam Review Problems and Solutions: 🔴 Abstract Algebra Problems with Solutions (including Proofs): 🔴 Abstract Algebra Lectures Playlist: 🔴 Check out my blog at: AbstractAlgebra #GroupTheory #LagrangeTheorem Links and resources 🔴 Subscribe to Bill Kinney Math: 🔴 Subscribe to my Math Blog, Infinity is Really Big: 🔴 Follow me on Twitter: 🔴 Follow me on Instagram: 🔴 You can support me by buying "Infinite Powers, How Calculus Reveals the Secrets of the Universe", by Steven Strogatz, or anything else you want to buy, starting from this link: 🔴 Check out my artist son Tyler Kinney's website: 🔴 Desiring God website: AMAZON ASSOCIATE As an Amazon Associate I earn from qualifying purchases. 3 comments Transcript: let G be not just any old group but a finite group and let h be a subgroup of G so lagrange's theorem is about subgroups of finite groups and what is the conclusion then the order of H divides the order of G I guess I don't need the word the in there order of H divides order of G that's the most basic part of the theorem it says a little more if you read in the book beyond this it says moreover the number I'll just say abbreviate number that way of what I call left cosets or right cosets I haven't said what left and right cosets are yet of H in G is the integer the order of G divided by the order of H and this number of left cosets or red cosets of H and G has a name it's called the index of H in G so it's got this last sentence which you don't really know what this means yet what is the left Cosa what is a right coset let's focus on the main conclusion the order H divides the order of G as a quick corollary of that remember Corollary corollaries uh our theorems that follow directly very easily from Court bigger theorems corollaries are important though they're not might necessarily minor facts they they could have a lot of importance and this corollary does if G is a finite group and a is an element of G the conclusion here of this corollary is then the order of a divides ah say divides the order of G the order of a as an element divides the order of G as a group the smallest positive power that a can be raised to to get back to the identity divides the number of elements in the group and such a finite power will exist if G is a finite group the order of a will not be infinite that's corollary an important Corollary that you should know for the rest of group Theory Eric Gossett Nathan's dad always called said lagrange's theorem is the most important theorem in finite group Theory you could say and as far as I'm concerned I agree with him I have no reason to disagree with him other people I've heard other people say it too okay we actually already know lagrange's theorem is true for cyclic groups Planet cyclic groups because it's the fundamental theorem of finite cyclic groups actually sayings the converse of the garnish's theorem is false for non-cyclical groups Converse of lagrange's theorem I'll just say is false in general for certain special kinds of groups it could be true cyclic groups being one special kind of group what if the converse of lagrange's theorem were true what would it say essentially it would say um well it's not really phrased in such a way that it's it's going to say exactly what I'm about to say it would say that if a number divides the order of G then there must be at least one subgroup of that order that's essentially the converse lagrange's theorem though maybe not as I've stated it here if a number divides the order of G I'll find a group then there must be a subgroup of that order that's a false statement an example illustrating that as false which you could call which you could call a counter example to the claim that it's true is that A4 has no subgroup of order 6. even though 6 divides 12 which is the order of A4 as a group right because its order is 4 factorial divided by two a four is the alternating group The group of even permutations on four objects it's a subgroup of S4 recall SN has n factorial elements a n has n factorial over two elements s ends the symmetric group on N objects the group of all permutations of a set of n objects to itself functions permutations are not necessarily isomorphisms right because the objects you're permuting are not necessarily group elements just one to one and onto is n factorial of them n is the subgroup of SN of all the even permutations that can be written as a product of an even number of two cycles two permutations are either even or odd in fact half of them are even half them are odd that's why a n has n factorial over two elements in it when n is four that's 24 24 over 2 is 12. this is an important example to realize um that A4 has no subgroup of order six excuse me is that difficult to prove I mean this is a this is a claim it would take proof how do you prove it the proof is in the book I'm forgetting which section I want it might be chapter seven I'm not I'm not remembering offhand um it's a tricky proof but it's understandable if you put the effort in and in fact I put it on exams before though as a fill in the blank kind of problem where you have to give fill in the blank when you're giving reasons now you had a full-on blank problem or two on your first exam for theorem statements like the fundamental theorem of cyclical groups I have given fill in the behind curves where that's and that's a prime example no pun intended a prime example of a proof on exam twos in the past that are given to understand it well enough that you can fill in blanks to give reasons for parts of it so it's something you should study I don't remember offhand if it's in chapter seven or if it's in a later chapter probably the chapter 7 or chapter nine or ten but I'm forgetting which one it is okay we'll come back to it I'll I'll look for it all right so this whole begs a question I hope in your mind oh by the way I should mention for your homework problem one of your homework problems due by Friday at 10 A.M uh you're gonna try to apply this Corollary and how do you apply it particular homework problems where maybe you know the order of G is say I don't know 55 [Music] and what does that mean about possible orders of elements in a group of order 55 they can only be 1 5 11 and 55. there's only four possibilities and the group doesn't have to be cyclic when you say that and maybe that restricts things that can happen maybe you can make conclusions that's one of your homework problems you got to think about applying this corollary that's a big hint for the problem the problem is going to feel difficult unless you apply this Corollary exactly what the problem is but it's one of the problems the point of the problem is to apply the grunge's theorem or this correlate for the last eight minutes here now um this begs a question how do you get to lagrange's theorem how do you prove it that's where the idea of cosets comes in and also is related to this last sentence as well what is a cosine hmm uh the book first defines this notation ah and ha okay actually people don't typically say ah and ha they say ah and h a where a is an element of G and initially H is just a subset of G not necessarily A subgroup if H is a subset of G notice I'm using subset notation here I can still Define new sets ah and H a ah and ha ah is the set of all elements that are products of the form little a times little h as little H varies over Capital H and H A ha is the set of all elements of the form little H times little a as little H varies over Capital H a is a fixed element of G H is a subset of G that could have many elements in it and in fact there's nothing here that says G or H has to be finite these sets could be infinite you can Define these subsets of G in the case where H is a subgroup of G if in fact H is a subgroup of G then we add some extra terminology we say that a h is the left coset of h n g containing a being an element of G and h a is the right coset of H in G containing a and if H is a subgroup of G the language the terminology here is consistent it really does both they really do contain a because if Capital H is a subgroup of G it's got the identity e in it e could be used in place of H in both of these sets AE equals A and E A equals a a would be elements of both sets if H is not a subgroup of G then you don't call these cosets and a is not necessarily in them because e might not be in h if H is an arbitrary subset but you can still Define these sets even if H is not a subgroup but just the subset what are the words left and right come from they come from which side of the expression a is on left coset here because a is on the left and that one right cosine here because a is on the right from that one it's where is the a h it's obviously a good thing to think about examples here we don't have time for examples so instead our last few minutes let me just give you a broad overview of what's going to happen here we mostly just focus on the left cosets we could focus on either one but we mostly just focus on left cosets because it just in general makes it easier to think about here we need to pick one of them the left one is the one we pick yay left-handers my son's left hand right left-handers are you just not appreciated by Society well you are appreciated with left cosets because we focus more on left cosets than the right core sets Venn diagram here let's say that's G what's going to happen is the distinct left cosets or this thing right cosets partition G into pieces here is one left cos that say here's another one here's another one here's another one there's another one here's another one and another one are there finitely many such cosets Maybe maybe not if G is infinite there could be infinitely many such cosets in the case where G is finite where the order of G is finite there will be finally many of these sets that partition G that are each individually cosets maybe this one is A1 H maybe this one's a2h a3h a4h A5 H where these A's are different elements of G these sets these left cosets partitions you what does that mean it means they're all non-overlapping the intersection of any two of them is the empty set I've talked about partitions before work and their Union is the entire group in fact anytime there's a partition there's an equivalence relation right equivalence relations can be used to define partitions that was the theorem in chapter zero there's some underlying equivalence relation here on G that I'm not saying yet what it is that defines this partition just as important each of these cosets has the same number of elements I could say the same order but only one of these cosets is actually a subgroup the one that contains the identity maybe it's this first one only one coset is going to be a subgroup of G when H is a subgroup of G it's the coset that contains the identity none of the others will be but they will have the same number of elements you could say the same order but keep in mind only one cos that's actually a subgroup and it's the fact that these partition G and they each have the same number of elements that is the key to the proof of lagrange's theorem and maybe you even have an intuitive feel already how you'd go about doing the proof about the number of elements in G and the number of elements in each of these cosets okay cosets have lots of properties that are hard to remember but you should work at getting used to them that's something else to focus on as you start reading chapter 7. all right have a good day
16547	https://cs.pomona.edu/~kim/CSC101S19/Lectures/Lecture27/Lecture27.pdf	Lecture 27: Unrestricted Grammars CSCI 101 Spring, 2019 Kim Bruce Unrestricted Grammars • An unrestricted, or type 0 grammar G is a quadruple (V , Σ, R, S), where: • V is an alphabet, • Σ (the set of terminals) is a subset of V , • R (the set of rules) is a ﬁnite subset of (V+ × V), • S (the start symbol) is an element of V - Σ. • The language generated by G is: {w ∈ Σ : S ⇒G w}. Example 1: • AnBnCn = {anbncn, n ≥ 0} • S → aBSc S → ε Ba → aB Bc → bc Bb → bb • Proof: • Gives only strings in AnBnCn : • All strings in AnBnCn are generated: Example 2 • {w ∈ {a, b, c} : #a(w) = #b(w) = #c(w)} • S → ABCS S → ε AB → BA BC → CB AC → CA BA → AB CA → AC CB → BC A → a B → b C → c Example 3 • WW = {ww : w ∈ {a, b}} • Idea: Generate wCwR# and then reverse last part WW = {ww : w ∈ {a, b}} • S → T# / Generate the wall exactly once. T → aTa / Generate wCwR. T → bTb ʺ T → C ʺ C → CP / Generate a pusher P Paa→ aPa / Push one character to the right to get ready to jump. Pab → bPa ʺ Pba → aPb ʺ Pbb → bPb ʺ Pa# → #a / Hop a character over the wall. Pb# → #b ʺ C# → ε Computability • Theorem: A language is generated by an unrestricted grammar if and only if it is in SD. • Proof: • (Grammar ⇒ TM): by construction of an NDTM. • (TM ⇒ grammar): by construction of a grammar that mimics the behavior of a semi-deciding TM. Proof of Equivalence • (Grammar ⇒ TM): by construction of a two-tape NDTM. • Suppose S ⇒ w. w is on tape 1. Start w/S on tape 2. • Tape 2 simulates derivation. • Non-deterministically choose production to apply to contents of tape 2. Rewrite string as appropriate. • After each step see if matches input. If yes, halt. • Semi-decides L(G). Proof of Equivalence • (TM ⇒ Grammar): Construct grammar G to simulate TM M. • Phase 1 generates a candidate string for acceptance. Phase 2 will then simulate the TM computation on the string. Phase 3 will clean up the tape, so tape only contains original candidate string. • Problem: Original string got replaced during simulation! • Solution: Duplicate it on odd cells of tape and only compute on the evens, preserving odds. Proof of Equivalence • (TM ⇒ Grammar): Construct grammar G to simulate TM M. • Phase 1: Generate a candidate string for acceptance. • Generate a string of form # ☐☐ q000 a1 a1 a2 a2 a3 a3 ☐☐ # representing input a1a2a3 and q000 is encoding of start state • Phase 2: If δ(p,a) = (q,b,→) add rule: • p’ z a → z b q‘ where p’, q’ are codes of states p,q • If δ(p,a) = (q,b,←) add x y p’ z a → q’ x y z b Note duplicates!! Proof of Equivalence • (TM ⇒ Grammar): Construct grammar G to simulate TM M. • Phase 3: If get to accept state A, clean up: • x A → A x for x ≠ #, move A to left edge of input • # A x y → x #A sweep through gathering odd -- erasing even • # A # → ε leaves original string as ﬁnal string Other formalisms • Partial recursive functions: • projection, constant, successor, closed under composition, primitive recursion, and minimization • To show equivalence with TM’s must encode conﬁgurations, conﬁguration histories, etc. as numbers. • Lambda calculus • RAM machines Undecidability • All undecidability results carry over as could use to solve corresponding TM problems using simulations. Self-Reproducing Program • Can you write a program in your favorite programming language that prints out a copy of itself? • T ry it! Virus Program • virus() = 1. For each address in address book do: 1.1. Write a copy of myself. 1.2. Mail it to the address. 2. Do something malicious like change one bit in every ﬁle on the machine. 3. Halt. • Can we implement step 1.1? • Print two copies of the following with the second in quotes: “Print two copies of the following with the second in quotes:” Fixed Points • Consider f(k) = k if k ≤ 1 = f(k-1) + f(k-2) otherwise • Function f deﬁned in terms of itself. • Think of as equation to be solved. • f = fun(k). if k≤1 then k else f(k-1)+f(k-2) • Write right side as function of g: • F(g) = fun(k). if k≤1 then k else g(k-1)+g(k-2) • Looking for f s.t. f = F(f) f is ﬁxed pt for F Recursion Theorem • Rough versions: • First Thm: If F is computable then F has a computable ﬁxed point. • Second Thm: W e can compute the program for a ﬁxed point of F from a program for F. • T rue for any formalism giving all computable fcns. • In lambda calculus, Y = λf.(λx.f(xx))(λx.f(xx)) gives ﬁxed points. I.e. for all f, if xo =Yf then f(xo) = xo Harder for Turing machines!
16548	https://www.marsja.se/sst-ssr-sse-in-r-a-complete-guide/	Mastering SST & SSE in R: A Complete Guide for Analysts Skip to content Erik Marsja Home ProgrammingMenu Toggle Python Tutorials R CV Contact Python & R resources Erik Marsja Main Menu About Contact Erik \| Curriculum Vitae Erik Marsja Privacy Python & R resources Python TutorialsMenu Toggle Data Analysis Data Visualization How to Use Python for Data Collection (e.g., in Psychology) Pandas Tutorials: Dataframe, grouping, sample, plotting, subsetting, etc. Terms of Service Mastering SST & SSE in R: A Complete Guide for Analysts Leave a Comment / By Erik Marsja / May 19, 2023 Share via: 3 Shares More Calculating the sum of squared residuals (SSR, also known as the sum of squared errors; SSE) in R provides information about the quality of our statistical models. Additionally, computing the total sum of squares (SST) is crucial for understanding the overall variability in the data. In psychology, understanding SSE in R can allow us to evaluate the accuracy of a predictive model. Moreover, it can enable us to determine how well our model fits the observed data. By quantifying the discrepancies between predicted and actual values, we can get information about the effectiveness of our theories and hypotheses. Similarly, in hearing science, we can can calculate SSR in R when analyzing auditory perception and assessing the performance of models predicting listeners’ responses. This evaluation aids in fine-tuning ourmodels and optimizing their predictive capabilities. Save Sum of Squares (Error and Total) Moreover, the computation of SST in R can play an essential role in both fields. By capturing the total variability in the data, SST allows us to differentiate between the inherent variation within the data and the variability accounted for by the model. This information can be used to examine the model’s explanatory power and to discern the significance of the predictors under investigation. In the next section, the outline of the post will be described, followed by a step-by-step guide on how to calculate SSE/SSR, and SST in R. By following this guide, you will gain a solid understanding of these calculations. Moreover, you will learn how to use them effectively in your research and data analysis. Table of Contents Outline Requirements What are These Different Sum of Squares? Synthetic Data Fit a Linear Model in R How to Calculate SST in R How to Find the SSR in R Calculating the SSR Method 1 Finding the SSE Method 2. How to Calculate the SSE, SSR, and SST from an ANOVA in R Conclusion Resources Outline I will begin the post by discussing the requirements for calculating the sum of squared errors (SSE), also known as the sum of squared residuals (SSR), and the total sum of squares (SST) in R. I will then answer the questions of what SSE, SSR, and SST are, providing explanations and highlighting their significance. Next, I will demonstrate how to fit a linear model in R and subsequently guide the readers through calculating SST and SSR/SSE. The post also explains how these calculations can be performed using an ANOVA object. Throughout the content, I will use examples and code snippets to illustrate the concepts. Lastly, I summarize the key points covered. Requirements The requirement for this post is to have a basic understanding of statistical analysis and regression models. Familiarity with R statistical programming language, including the usage of functions like lm(), %in%, and dplyr package, is essential. You should also run the latest version of R (or update R to the latest version). Additionally, a foundational understanding of the concepts (e.g., SSE/SSR, & SST) is necessary. You should, furthermore, be comfortable with data manipulation using dplyr, conditional operations with %in%, and fitting linear regression models using lm(). Prior knowledge of ANOVA (Analysis of Variance) and its use in R would be beneficial. The post assumes readers have a dataset available to perform the calculations. What are These Different Sum of Squares? This section will briefly describe the SSE/SSR, and SST. What is the sum of squared errors (SSE/SSR)? The sum of squared errors (SSE) is a statistical measure that quantifies the overall discrepancy between observed data and the predictions made by a model. It calculates the sum of the squared differences between the actual values and the corresponding predicted values. SSE is commonly used to evaluate regression or predictive models’ accuracy and goodness of fit. A lower SSE indicates a better fit, meaning the model’s predictions align more closely with the observed data. The sum of Squared Errors is sometimes called the Residuals sum of squares. What is the total sum of squares (SST)? The total sum of squares (SST) is a statistical measure representing a dataset’s total variability. It quantifies the sum of the squared differences between each data point and the overall mean of the dataset. SST provides a baseline reference for understanding the total variation in the data before any predictors or models are considered. By comparing the SST with the sum of squared residuals (SSR), one can assess how much of the total variation is accounted for by the model. SST is essential in calculating the coefficient of determination (R-squared) to evaluate the model’s explanatory power. In the next section, we will generate fake data that we will use to calculate the SSE, SSR, and SST in R. Synthetic Data Here we use the dplyr package to generate fake data: ```r library(dplyr) Creating a fake dataset set.seed(123) # Set seed for reproducibility Variables group <- rep(c("A", "B", "C"), each = 10) x <- rnorm(30, 50, 10) y <- ifelse(group %in% c("A", "B"), x 2 + rnorm(30, 0, 5), x 3 + rnorm(30, 0, 5)) Creating a data frame data <- data.frame(group, x, y) Standardizing the data data <- data %>% mutate( x_std = scale(x), y_std = scale(y) ) `` Code language:R(r) In the code chunk above, we create a fake dataset with three groups (A,B, andC) and two variables (xandy). The values ofxare randomly generated from a normal distribution, andyis calculated based on the groups andx, with added noise. Note the %in%operator in R is used in theifelsestatement to check whether the values in thegroupvariable are either “A” or “B”. Specifically, the expressiongroup %in% c("A", "B")is evaluated for each element in thegroupvariable. Next, we create a dataframe calleddataand then usemutatefromdplyrto create standardized versions ofxandyusing thescale` function. This dataset will be used to calculate SSE, SSR, and SST in the following parts of the blog post. In the next section, we will fit a regression model using the data above. Fit a Linear Model in R Here is how to fit a linear regression model in R: r model <- lm(y ~ x, data = data) Code language:R(r) In the code chunk above, we fitted a linear model using the lm() function. Of course, we used the fake data. However, if you use your own data, make sure to replace y with the name of your dependent variable. Furthermore, you need to replace x with the name of your independent variable (and add any more you have). Finally, adjust data to the name of your dataset. Save Probit Regression in R: Interpretation & Examples How to Make a Residual Plot in R & Interpret Them using ggplot2 Plot Prediction Interval in R using ggplot2 How to Standardize Data in R How to Calculate SST in R Here is a general code chunk for calculating the total sum of squares in R: php SST <- sum((data$y - mean(data$y))^2) Code language:PHP(php) In the code chunk above, twe calculate the total sum of squares (SST) using a formula. First, we calculate the mean of the dependent variable y iby using the mean function on data$y. This provides the average value of y across all observations. Next, we use each individual observation of y in the dataset data and subtract them from the mean value. This subtraction is performed for every observation and creates a vector of deviations from the mean. Next, we use these deviations are squared using the ^2 operator. Finally, we use the sum function on the squared deviations, resulting in the sum of all the squared differences between each observation of y and the mean value. Save Note that the SST can be calculated in R by summing the sum of squared differences between predicted data points and SSE. In the next section, we will calculate the residual sum of squares in R using the model (e.g., the results from the regression model we previously fitted). Save residuals sum of squares in R How to Find the SSR in R Here are two methods that we can use to calculate the sum of squared errors using R: Calculating the SSR Method 1 Here is one way to find the SSR in R: r SSR <- sum(residuals(model)^2) Code language:R(r) In the code chunk above, we calculated the sum of squared residuals in R using a formula. First, we use the residuals function to the model. This function calculates the differences between the observed values of the dependent variable and the model’s corresponding predicted values. These differences are referred to as residuals. Next, we use the ^2 operator to square each residual, element-wise. Squaring the residuals ensures that positive and negative differences contribute positively to the overall SSR calculation. Finally, the sum function is used to add up all the squared residuals, resulting in the SSR. Finding the SSE Method 2. Here is how we can find the sum of squared errors in R: SSR2 <- deviance(model) In the code chunk above, we calculate the SSE in R using a formula (the same as SSR). First, we use the deviancefunction on the model. Save As seen in the image above, we can also use the deviance function to calculate the SSE. How to Calculate the SSE, SSR, and SST from an ANOVA in R To perform the above calculations using an ANOVA object in R, you can utilize the anova() function and a fitted ANOVA model. Here is an example using the built-in dataset mtcars: ```r Fitting an ANOVA model model <- lm(mpg ~ cyl + hp, data = mtcars) anova_result <- anova(model) Extracting SSE from lmobject SSE <- sum(model$residuals^2) Calculating SST SST <- sum((mtcars$mpg - mean(mtcars$mpg))^2) `` Code language:R(r) In the code chunk above, we fit an ANOVA model to explore the relationship between the dependent variablempgand the independent variablescylandhp. We utilized thelm()function to perform the linear regression analysis, withmtcars` dataset serving as the data source. Next, we applied the anova() function to the fitted model, resulting in the anova_result object that contains the ANOVA table. Here are the results: Save ANOVA table To extract the sum of squared errors (SSE) from the fitted object, we squared the residuals obtained from modelusing the ^2 operator and summed them up using sum(). Just like in the previous examples. Notice that we already have the SSE in the ANOVA table. Finally, for calculating the total sum of squares (SST), we subtracted each observed value of mpg from the mean of mpg in mtcars, squared the differences, and summed them using sum(). Conclusion In this post, we have learned how to calculate SSE/SSR and SST in R. We discussed the importance of these measures in statistical analysis and regression modeling. By fitting a linear model in R using the lm() function, we gained hands-on experience in applying regression techniques. Throughout the post, we learned the step-by-step process of calculating SST/SSR and SSE. We learned how to find the total sum of squares by calculating the squared deviations from the mean, determining the sum of squared residuals by squaring and summing the residuals, and calculating the sum of squared errors by comparing observed and predicted values. Moreover, we demonstrated the calculations using an ANOVA object, showcasing another approach to obtaining SSE, SSR, and SST. By understanding these calculations, researchers and analysts can better assess model performance and interpret the variation in their data. I hope this post has provided you with a comprehensive understanding of SSE, SSR, and SST in R. By sharing this post on your favorite social media platforms, you can help others to learn these fundamental concepts. Furthermore, I encourage you to refer back to this post in your work, whether it be research papers, reports, or blog posts, to enhance the accuracy and clarity of your statistical analysis. Resources Here are some resources you may find helpful: How to Convert a List to a Dataframe in R – dplyr Countif function in R with Base and dplyr How to Create a Word Cloud in R Report Correlation in APA Style using R: Text & Tables How to Create a Sankey Plot in R: 4 Methods Sum Across Columns in R – dplyr & base Test for Normality in R: Three Different Methods & Interpretation Z Test in R: A Tutorial on One Sample & Two Sample Z Tests Save Share via: 3 Shares More ← Previous Post Next Post → Leave a Comment Cancel Reply Your email address will not be published.Required fields are marked Type here.. Name Email Website Δ This site uses Akismet to reduce spam. Learn how your comment data is processed. Latest Posts Understanding SSE in Regression Pandas: Drop Columns By Name in DataFrames How to Rename Columns in data.table in R (With Examples) Replace NA in data.table: Replacing with 0 and Other Values How to Use data.table to Fill NA with the Previous Value in R Privacy Terms of Service Copyright © 2025 Erik Marsja Scroll to Top Facebook Twitter Reddit LinkedIn Flipboard Telegram Email Copy Link More Networks Share via Facebook X (Twitter) LinkedIn Mix Pinterest Bluesky Tumblr Skype Buffer Pocket VKontakte Parler Xing Reddit Line Flipboard MySpace Delicious Amazon Digg Evernote Blogger LiveJournal Baidu MeWe NewsVine Yummly Yahoo WhatsApp Viber SMS Telegram Facebook Messenger Like Email Print Copy Link Copy link Copy Copied
16549	https://pubmed.ncbi.nlm.nih.gov/9647561/	Differential responses of granulosa cells from small and large follicles to follicle stimulating hormone (FSH) during the menstrual cycle and acyclicity: effects of tumour necrosis factor-alpha - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Search: Search AdvancedClipboard User Guide Save Email Send to Clipboard My Bibliography Collections Citation manager Display options Display options Format Save citation to file Format: Create file Cancel Email citation Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page. 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Report format: Send at most: [x] Send even when there aren't any new results Optional text in email: Save Cancel Create a file for external citation management software Create file Cancel Your RSS Feed Name of RSS Feed: Number of items displayed: Create RSS Cancel RSS Link Copy Actions Cite Collections Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Permalink Permalink Copy Display options Display options Format Page navigation Title & authors Abstract Similar articles Cited by Publication types MeSH terms Substances Related information Grants and funding Hum Reprod Actions Search in PubMed Search in NLM Catalog Add to Search . 1998 May;13(5):1285-91. doi: 10.1093/humrep/13.5.1285. Differential responses of granulosa cells from small and large follicles to follicle stimulating hormone (FSH) during the menstrual cycle and acyclicity: effects of tumour necrosis factor-alpha V Montgomery Rice1,S D Limback,K F Roby,P F Terranova Affiliations Expand Affiliation 1 Department of Obstetrics and Gynecology, University of Kansas Medical Center, Kansas City 66160, USA. PMID: 9647561 DOI: 10.1093/humrep/13.5.1285 Item in Clipboard Differential responses of granulosa cells from small and large follicles to follicle stimulating hormone (FSH) during the menstrual cycle and acyclicity: effects of tumour necrosis factor-alpha V Montgomery Rice et al. Hum Reprod.1998 May. Show details Display options Display options Format Hum Reprod Actions Search in PubMed Search in NLM Catalog Add to Search . 1998 May;13(5):1285-91. doi: 10.1093/humrep/13.5.1285. Authors V Montgomery Rice1,S D Limback,K F Roby,P F Terranova Affiliation 1 Department of Obstetrics and Gynecology, University of Kansas Medical Center, Kansas City 66160, USA. PMID: 9647561 DOI: 10.1093/humrep/13.5.1285 Item in Clipboard Cite Display options Display options Format Abstract This study determined effects of follicle stimulating hormone (FSH) alone and in combination with tumour necrosis factor (TNF), on granulosa cells from small (5-10 mm diameter) and large (>10-25 mm) follicles during follicular and luteal phases of the cycle and during periods of acyclicity. Granulosa cells were collected from ovaries of premenopausal women undergoing oophorectomy. The cells were cultured with human FSH (2 ng/ml) and testosterone (1 microM) in the presence or absence of human TNF-alpha (20 ng/ml). Media were removed at 48 and 96 h after culture and progesterone, oestradiol and cAMP in media were measured by radioimmunoassays. FSH stimulated the accumulation of oestradiol from granulosa cells of small follicles during the follicular and luteal phases but not during acyclicity; and TNF reduced oestradiol accumulation in the presence of FSH. Interestingly, in granulosa cells from small follicles, progesterone and cAMP secretion increased in response to FSH and neither was affected by TNF. Thus, TNF specifically inhibited the conversion of testosterone to oestradiol in granulosa cells from small follicles. FSH stimulated oestradiol production by granulosa cells of large follicles obtained only during the follicular phase of the cycle and TNF inhibited the FSH-induced oestradiol secretion. Granulosa cells obtained from large follicles during the luteal phase and during acyclicity did not accumulate oestradiol in response to FSH. However, FSH increased progesterone and cAMP secretion by granulosa cells obtained from large follicles during the follicular and luteal phases. During the luteal phase alone, TNF in combination with FSH increased progesterone accumulation above that of FSH alone. FSH did not increase progesterone, oestradiol or cAMP secretion by granulosa cells obtained from large follicles during acyclicity. Thus, FSH increases progesterone, oestradiol and cAMP secretion by granulosa cells of small follicles during the follicular and luteal phases and TNF appears to inhibit FSH-induced oestradiol secretion specifically in those cells. In large follicles, FSH-stimulated granulosa cell secretion of oestradiol is limited to the follicular phase and this effect can be inhibited by TNF. In addition, when granulosa cells of large follicles do not increase oestradiol secretion in response to FSH, TNF stimulates progesterone secretion. PubMed Disclaimer Similar articles Tumor necrosis factor alpha inhibition of follicle-stimulating hormone-induced granulosa cell estradiol secretion in the human does not involve reduction of cAMP secretion but inhibition at post-cAMP site(s).Montgomery Rice V, Limback SD, Roby KF, Terranova PF.Montgomery Rice V, et al.Endocrine. 1999 Feb;10(1):19-23. doi: 10.1385/ENDO:10:1:19.Endocrine. 1999.PMID: 10403567 Tumour necrosis factor-alpha inhibits follicle-stimulating hormone-induced granulosa cell oestradiol secretion in the human: dependence on size of follicle.Rice VM, Williams VR, Limback SD, Terranova PF.Rice VM, et al.Hum Reprod. 1996 Jun;11(6):1256-61. doi: 10.1093/oxfordjournals.humrep.a019368.Hum Reprod. 1996.PMID: 8671436 Herbimycin, a tyrosine kinase inhibitor with Src selectivity, reduces progesterone and estradiol secretion by human granulosa cells.Rice VM, Chaudhery AR, Oluola O, Limback SD, Roby KF, Terranova PF.Rice VM, et al.Endocrine. 2001 Aug;15(3):271-6. doi: 10.1385/ENDO:15:3:271.Endocrine. 2001.PMID: 11762698 Ovarian follicular and luteal physiology.Channing CP, Schaerf FW, Anderson LD, Tsafriri A.Channing CP, et al.Int Rev Physiol. 1980;22:117-201.Int Rev Physiol. 1980.PMID: 6248477 Review. Hormonal manipulations in the luteal phase to coordinate subsequent antral follicle growth during ovarian stimulation.Fanchin R, Méndez Lozano DH, Schonäuer LM, Cunha-Filho JS, Frydman R.Fanchin R, et al.Reprod Biomed Online. 2005 Jun;10(6):721-8. doi: 10.1016/s1472-6483(10)61115-7.Reprod Biomed Online. 2005.PMID: 15970000 Review. See all similar articles Cited by TNF-α haplotype association with polycystic ovary syndrome - a South Indian study.Deepika ML, Reddy KR, Yashwanth A, Rani VU, Latha KP, Jahan P.Deepika ML, et al.J Assist Reprod Genet. 2013 Nov;30(11):1493-503. doi: 10.1007/s10815-013-0080-4. Epub 2013 Aug 23.J Assist Reprod Genet. 2013.PMID: 23975191 Free PMC article. Metabolic heterogeneity of follicular amino acids in polycystic ovary syndrome is affected by obesity and related to pregnancy outcome.Zhang CM, Zhao Y, Li R, Yu Y, Yan LY, Li L, Liu NN, Liu P, Qiao J.Zhang CM, et al.BMC Pregnancy Childbirth. 2014 Jan 10;14:11. doi: 10.1186/1471-2393-14-11.BMC Pregnancy Childbirth. 2014.PMID: 24410809 Free PMC article. Obesity is associated with alterations in antral follicle dynamics in eumenorrheic women.Oldfield AL, Vanden Brink H, Carter FE, Jarrett BY, Lujan ME.Oldfield AL, et al.Hum Reprod. 2023 Mar 1;38(3):459-470. doi: 10.1093/humrep/dead007.Hum Reprod. 2023.PMID: 36708012 Free PMC article. Exploring the Association of Biochemical Characterization and Genetic Determinants of TNF-α, CXCR2, and CCR5 Delta 32 Mutation with Predisposition to Polycystic Ovary Syndrome.Almasoudi KS, Hussain E, Almotairi R, Bhat T, Mtiraoui N, Ezzidi I, Mir R.Almasoudi KS, et al.Life (Basel). 2024 Jul 28;14(8):949. doi: 10.3390/life14080949.Life (Basel). 2024.PMID: 39202691 Free PMC article. Tumor necrosis factor alpha inhibition of follicle-stimulating hormone-induced granulosa cell estradiol secretion in the human does not involve reduction of cAMP secretion but inhibition at post-cAMP site(s).Montgomery Rice V, Limback SD, Roby KF, Terranova PF.Montgomery Rice V, et al.Endocrine. 1999 Feb;10(1):19-23. doi: 10.1385/ENDO:10:1:19.Endocrine. 1999.PMID: 10403567 Publication types Research Support, U.S. Gov't, P.H.S. Actions Search in PubMed Search in MeSH Add to Search MeSH terms Adult Actions Search in PubMed Search in MeSH Add to Search Cyclic AMP / metabolism Actions Search in PubMed Search in MeSH Add to Search Drug Interactions Actions Search in PubMed Search in MeSH Add to Search Estradiol / metabolism Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Follicle Stimulating Hormone / administration & dosage Actions Search in PubMed Search in MeSH Add to Search Follicle Stimulating Hormone / pharmacology Actions Search in PubMed Search in MeSH Add to Search Follicular Phase / physiology Actions Search in PubMed Search in MeSH Add to Search Granulosa Cells / drug effects Actions Search in PubMed Search in MeSH Add to Search Granulosa Cells / physiology Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search In Vitro Techniques Actions Search in PubMed Search in MeSH Add to Search Luteal Phase / physiology Actions Search in PubMed Search in MeSH Add to Search Middle Aged Actions Search in PubMed Search in MeSH Add to Search Ovarian Follicle / cytology Actions Search in PubMed Search in MeSH Add to Search Ovarian Follicle / drug effects Actions Search in PubMed Search in MeSH Add to Search Ovarian Follicle / physiology Actions Search in PubMed Search in MeSH Add to Search Progesterone / metabolism Actions Search in PubMed Search in MeSH Add to Search Tumor Necrosis Factor-alpha / administration & dosage Actions Search in PubMed Search in MeSH Add to Search Tumor Necrosis Factor-alpha / pharmacology Actions Search in PubMed Search in MeSH Add to Search Substances Tumor Necrosis Factor-alpha Actions Search in PubMed Search in MeSH Add to Search Progesterone Actions Search in PubMed Search in MeSH Add to Search Estradiol Actions Search in PubMed Search in MeSH Add to Search Follicle Stimulating Hormone Actions Search in PubMed Search in MeSH Add to Search Cyclic AMP Actions Search in PubMed Search in MeSH Add to Search Related information PubChem Compound (MeSH Keyword) Grants and funding CA50616/CA/NCI NIH HHS/United States HD33994/HD/NICHD NIH HHS/United States [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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16550	https://baselang.com/blog/basic-grammar/spanish-imperfect-tense/	Programs REAL WORLD Unlimited one-on-one classes for a flat rate. Zero to advanced. Includes our DELE program. BOOTCAMP Zero to conversational in a month. Guaranteed. Intensive program. Not sure which program to pick? Compare programs MEDELLIN SCHOOL Our in-person school in Colombia. CORPORATE Spanish for your business. LEARN FRENCH Our sister company, LingoCulture. Blog Reviews Guide Login Sign Up Sign Up What is The Imperfect Tense in Spanish? A Beginner-Friendly Guide Adrian in Basic Grammar Get our free email course, Shortcut to Conversational. Have conversations faster, understand people when they speak fast, and other tested tips to learn faster. More info The imperfect tense in Spanish (pretérito imperfecto) is one of the most accessible verb tenses you’ll encounter. You can think of the imperfect as your tool for expressing the past in a more nuanced way. In this beginner-friendly guide, we’ll explore what the Spanish imperfect tense is, when to use it, how to conjugate it (including the few irregulars), and how it differs from the preterite. You’ll also see plenty of examples of the imperfect tense in Spanish with English translations, plus some tips to avoid common mistakes. ¡Vamos! What is the imperfect tense in Spanish? The Spanish imperfect tense is used to describe past actions that were ongoing, habitual, repeated, or not clearly completed. In simpler terms, if you want to talk about things you “used to do” or “was/were doing” in the past, you’ll likely use the imperfect tense. For example, “Cuando era niño, jugaba al baloncesto todos los sábados.” (When I was a child, I used to play basket every Saturday.) Why is it called “imperfect tense”? The name comes from the idea of an unfinished or ongoing past action, something that wasn’t a one-time event. The imperfect tense describes the background or setting in the past, instead of indicating a specific finished action. When do you use the imperfect tense in Spanish? In Spanish, the imperfect is used in several common past situations: Habitual or repeated actions (“used to …”) The imperfect tense is perfect for talking about past routines or habits: things you used to do regularly, possibly without even thinking about them. Any time you want to say “used to ___” in English, Spanish almost always reaches for the imperfect. \| \| \| --- \| \| Spanish \| English \| \| De niño, jugaba al baloncesto todos los días. \| As a child, I used to play basketball every day. \| \| Mi abuela cocinaba arroz todos los domingos. \| My grandma used to cook rice every Sunday. \| \| Nosotros íbamos al parque cada tarde después de la escuela. \| We used to go to the park every afternoon after school. \| Ongoing/interrupted actions (“was … ing”) When an action was in progress in the past, sometimes interrupted by another event, you’ll use the imperfect for the background action. The “interrupter” (the thing that suddenly happened) usually appears in the preterite. \| \| \| --- \| \| Spanish \| English \| \| Estaba estudiando cuando sonó el teléfono. \| I was studying when the phone rang. \| \| Ella leía un libro mientras yo preparaba el café. \| She was reading a book while I was making coffee. \| \| Caminábamos por la playa cuando empezó a llover. \| We were walking on the beach when it started to rain. \| Background and scene-setting (descriptions) Think of the imperfect tense as your storytelling paintbrush. Use it to describe weather, scenery, emotions, physical traits, anything that sets the stage in the past rather than pushes the plot forward. \| \| \| --- \| \| Spanish \| English \| \| La noche estaba oscura y hacía frío. \| The night was dark, and it was cold. \| \| La casa era blanca y tenía ventanas azules. \| The house was white and had blue windows. \| \| El mercado olía a pan fresco y todos parecían felices. \| The market smelled of fresh bread, and everyone seemed happy. \| Time and age in the past Whenever you mention what time it was or how old someone was, Spanish defaults to the imperfect. Use ser to tell time and tener to state age. No exceptions here! \| \| \| --- \| \| Spanish \| English \| \| Eran las ocho de la noche. \| It was 8 PM. \| \| Era medianoche y la ciudad estaba en silencio. \| It was midnight, and the city was silent. \| \| Cuando tenía doce años, aprendí a tocar la guitarra. \| When I was twelve, I learned to play the guitar. \| How to conjugate the imperfect tense (regular & irregular verbs) Luckily, conjugating verbs in the Spanish imperfect tense is straightforward. There are only two sets of endings for regular verbs, and just three irregular verbs in total (to make things more fun). Regular verb conjugations For any regular verb, take off the infinitive ending (-ar, -er, -ir) and add the imperfect endings. The endings for -ar verbs are different from those for -er and -ir verbs: -AR verbs: -aba, -abas, -aba, -ábamos, -abais, -aban Example: hablar (to speak) → yo hablaba, tú hablabas, él/ella hablaba, nosotros hablábamos, vosotros hablabais, ellos/ustedes hablaban. -ER / -IR verbs: -ía, -ías, -ía, -íamos, -íais, -ían Example: comer (to eat) → yo comía, tú comías, él/ella comía, nosotros comíamos, vosotros comíais, ellos/ustedes comían. Example: vivir (to live) → yo vivía, tú vivías, él/ella vivía, nosotros vivíamos, vosotros vivíais, ellos/ustedes vivían. 🌎 Note: In Spain, the informal second-person plural pronoun vosotros/vosotras (with endings like -abais, –íais) is standard. Throughout Latin America, speakers almost always use ustedes for both formal and informal context, and it conjugates just like the third-person plural (-aban, -ían). Spain: vosotros hablabais Latin America: ustedes hablaban Irregular verbs The imperfect tense has only three irregular verbs, and you just have to memorize these. They are ir (“to go”), ser (“to be”), and ver (“to see/watch”): \| \| \| \| \| --- --- \| \| Subject pronoun \| ir \| ser \| ver \| \| yo \| iba \| era \| veía \| \| tú \| ibas \| eras \| veías \| \| él / ella / usted \| iba \| era \| veía \| \| nosotros / nosotras \| íbamos \| éramos \| veíamos \| \| vosotros / vosotras \| ibais \| erais \| veíais \| \| ellos / ellas / ustedes \| iban \| eran \| veían \| How is the imperfect tense different from the preterite? The preterite is used for completed past events (think of it as a snapshot of a one-time moment), while the imperfect is used for ongoing, habitual, or background actions (like a continuous video or backdrop). Often, the imperfect provides the background context, and the preterite describes the specific events that occurred against that background. For example, “Caminaba por el parque cuando vi un pájaro rojo.” translates to “I was walking through the park when I saw a red bird.” Here, caminaba (“was walking”) sets the scene with the imperfect, and vi (“saw”) is the one-time event in the preterite. The theory seems pretty straightforward, but you will need practice to master it (nothing new here). In addition, there are some common mistakes you need to avoid. We’ll see them next. Practice tips & common mistakes to avoid Below are a few extra tips that will save you from the most frequent slips with the imperfect. \| \| \| \| --- \| Tip / pitfall \| Why it matters \| Quick fix + example \| \| Time & age always use the imperfect \| Learners sometimes default to the preterite (fue las ocho, tuve cinco años). Native speakers never do. \| ✅ Eran las ocho cuando llegamos. (It was eight when we arrived.) ✅ Cuando tenía cinco años, vivía en Madrid. (When I was five, I lived in Madrid.) \| \| Match markers & tenses \| Words like ayer, anoche, el año pasado point to finished actions → preterite. Pairing them with the imperfect feels off. \| ❌ Ayer estaba en la oficina todo el día. ✅ Ayer estuve en la oficina todo el día. \| \| Don’t skip the accent marks \| Missing the accent on í in -ía / íamos / íais or on é in éramos changes the meaning (or looks wrong). \| ❌ Nosotros eramos ✅ Nosotros éramos \| \| Use imperfect for “used to want/think/feel” \| Verbs that express mental or emotional states (querer, pensar, sentir, creer) typically stay imperfect when you describe how things were. \| ✅ De niño, quería ser piloto. (As a kid, I wanted to be a pilot.) ✅ En ese momento, pensaba que todo saldría bien. \| \| Combine imperfect + preterite for rich storytelling \| Use the imperfect to set the scene and the preterite to move the action forward. \| La fiesta fue increíble; todos bailaban y reían. \| \| Remember the three irregulars \| Everything else is regular, so mix-ups are obvious. \| ir – iba, ibas… / ser – era, eras… / ver – veía, veías… \| Conclusion To recap, use the imperfect tense in Spanish for past habits, ongoing or interrupted actions, and background details like time, age, and descriptions. Keep practicing these examples of imperfect tense usage, and soon choosing between “hablaba” and “hablé” will become second nature. You’ve got this! Spanish Imperfect Practice Now that you’ve gained a solid understanding of the imperfect tense in Spanish, it’s time to put your knowledge to the test! Conjugate the verbs in parentheses in the Spanish imperfect tense, being sure to choose the right conjugation to match the subject of the sentence. The answers and translations are below. Mi mamá siempre ________ (olvidar) las llaves en el auto cuando salía de compras. Claro, todos los veranos nosotros ________ (ir) allí de vacaciones. No me ________ (sentir) muy bien en ese momento, así que llamé a mi mejor amigo. En el colegio, los niños ________ (tener) clase de música los miércoles. ________ (llover) mucho cuando salí de casa. Tu siempre ________ (pagar) la cena cuando ________ (salir) con tus amigos. ________ (ser) las tres de la tarde cuando Luis salió de casa. Para ese entonces ________ (tener) ocho años, ya no recuerdo mucho. Ustedes ________ (hacer) la cena mientras ellos ________ (limpiar) la casa. Vosotras ________ (ser) muy delgadas en aquella época. Answers Mi mamá siempre olvidaba las llaves en el auto cuando salía de compras. – My mom would always forget her keys in the car when she went shopping. Claro, todos los veranos nosotros íbamos allí de vacaciones. – Sure, every summer we would go there on vacation No me sentía muy bien en ese momento, así que llamé a mi mejor amigo. – I wasn’t feeling very well at the time, so I called my best friend. En el colegio, los niños tenían clase de música los miércoles. – At school, the kids had music class on Wednesdays. 5.Llovía mucho cuando salí de casa. – It was raining a lot when I left the house. Tu siemprepagabas la cena cuando salíamos con tus amigos. – You always paid for dinner when we went out with your friends. Eran las tres de la tarde cuando Luis salió de casa. – It was three o’clock in the afternoon when Luis left the house. Para ese entonces tenía ocho años, ya no recuerdo mucho. – I was eight years old at the time, I don’t remember much anymore. Ustedes hacían la cena mientras ellos limpiaban la casa. – You made dinner while they cleaned the house. Vosotras erais muy delgadas en aquella época. – You girls were very thin at that time. Get our FREE 7-day email course, Shortcut to Conversational The exact strategies you need to become conversational in Spanish this year. Join the course now, before we come to our senses and charge for it! The Subjunctive in Spanish, Simplified For Beginners Adjectives in Spanish: When and how to use them The Complete Guide to Telling Time in Spanish This blog is presented by BaseLang: Unlimited Spanish Tutoring for $179 a Month. Learn more here. Your First Week Is Just $1. After that, it’s just $179/mo for unlimited one-on-one tutoring. Remember, the worst case scenario is you get a few free classes, don’t like it, and end up with an extra $20 in the bank. 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Biology Science Earth and space MathAddition and Subtraction: Make True Equations Addition and Subtraction: Make True Equations Grade 1 Jul 24, 2023 Introduction: Pick the equations that are true: – a. 6 = 12 – 6 b.7+ 8 = 8 + 7 c. 10 = 10 d. 7 + 3 = 3 + 7 Answers: a.6 = 12 – 6, subtract 6 from 12; the answer is 6. 6 = 6, the equation is true. b. 7 + 8 = 8 + 7; both side equations are true. c.10 = 10; both side equations are true. d. 7 + 3 = 3 + 7; both side equations are true Make the equation true. 4 + 6 = _____ + 1 Answer: Solve one side of the equation first 4 + 6 =10. to get the other side of the equation, subtract 1 from 10 10 = 9 + 1. The answer is 4 + 6 = 9 + 1 This equation is true. 3. Fill in the missing number to make this equation true: – 14 – ___ = 3 + 3 Equation True Solve one side of the equation first. 3 + 3 = 6 You can use counters to find the missing answer. 14 – ___ = 6 Take away 8 counters to get 6; the answer is 14 – 8 =6. 14 – 8 = 3 + 3. This equation is true. Solve The Following: – Observe the box. How many balls do you need to add to make the equation true? 6 + ___ = 10 To get the missing number, subtract the counters ball, i.e., 6 from 10, 10 – 6 = 4 Answer: Make the equation true: – 15 – ____ = 10 + 2 Answers: Solve one side of the equation first. 10 + 2 = 12 To get the other side of the equation, subtract 12 from 15. 15 – 12 = 3 So, the missing number is 3. 15 – 3 = 10 + 2 The equation is true Write the missing symbol {+, -, or =} and make the equation true. 17 ___ 9 ___ 4 + 4 Answers: – To get the answer, observe the equation. Solve one side of the equation [symbol inserted here] 4 + 4 = 8 Other side values are more than the next side; so, to get lesser values, you need to insert subtraction symbol, 17 – 9. Subtract 17 – 9 = 8; both the side values are the same, so insert an equal sign in the middle of both equations. The answer is 17-9=4+4. Alex has a 15-meter-long rope. He made it into two pieces. One piece is 8 m long. What is the length of the other piece? 15 – 8 = ____ Answer: Use counters and subtract the missing number. Take away 8 counters from 15; the answer is 7. 15-8=7. The length of another piece is 7 meters. The length of the other piece Peter has 14 red crayons; his brother gave him 6 more crayons. John has 12 red crayons. His father gave him 8 more crayons. Do they have an equal number of crayons? Answer: Peter has 14+6 crayons. When we add, you get the answer 20. John has 12+8 crayons. When we add, you get the answer 20. 14+6=12+8 Both of them have an equal number of crayons. Concept Summary Concept Summary Comments: Submit Related topics Addition and Multiplication Using Counters & Bar-Diagrams Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […] Read More >> Dilation: Definitions, Characteristics, and Similarities Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […] Read More >> How to Write and Interpret Numerical Expressions? Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] Read More >> System of Linear Inequalities and Equations Introduction: Systems of Linear Inequalities: A system of linear inequalities is a set of two or more linear inequalities in the same variables. The following example illustrates this, y < x + 2…………..Inequality 1 y ≥ 2x − 1…………Inequality 2 Solution of a System of Linear Inequalities: A solution of a system of linear inequalities […] Read More >> Other topics #### How to Find the Area of Rectangle? Mar 3, 2022 #### How to Solve Right Triangles? 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16552	https://gis.stackexchange.com/questions/321021/splitting-polygon-into-equal-area-polygons-using-qgis	Skip to main content Asked Modified 3 years, 11 months ago Viewed 26k times This question shows research effort; it is useful and clear 20 Save this question. Show activity on this post. There is a Plugin in QGIS 2.18 called "Polygon Divider" which divides an irregular polygon into smaller equal-area polygons of a specified/custom-sized but in QGIS 3.x version there is no "Polygon Divider" Plugin or tool to perform that task. How do I divide/split the irregular polygon into equal area/parts polygons in QGIS 3.6? As shown in the below picture: polygon qgis-3 splitting equal-area Share CC BY-SA 4.0 Improve this question Follow this question to receive notifications edited Sep 28, 2021 at 20:26 PolyGeo♦ 65.5k2929 gold badges115115 silver badges349349 bronze badges asked Apr 27, 2019 at 4:10 Kapildev adhikariKapildev adhikari 34311 gold badge22 silver badges1010 bronze badges 4 2 I don't know this plugin. But, you can try to create a grid first (processing - toolbox - create Grid - grid type : [] Polygon) , and than clip this layer with your polygon's layers. – Christophe P. Commented Apr 27, 2019 at 7:56 4 You could upgrade the 2.x plugin to work in 3.x github.com/qgis/QGIS/wiki/Plugin-migration-to-QGIS-3 – Ian Turton Commented Apr 27, 2019 at 8:19 3 Just for anyone who comes across this - I have just uploaded a version of the plugin for QGIS3, so it should be available in the Plugin Manager shortly – jonnyhuck Commented Sep 24, 2020 at 17:08 Ref.: spatialthoughts.com/2021/06/12/split-polygons-qgis – Taras ♦ Commented Jun 24, 2022 at 14:51 Add a comment \| 7 Answers 7 Reset to default This answer is useful 29 Save this answer. Show activity on this post. All of the necessary tools can be found from processing toolbox and you have to be a bit innovative to achieve this. Basically by adopting this method you could achieve pretty good results, although it doesn't give you 90 degree angles in all places, but roughly equal area: 1.Create enough random points inside a polygon. I did 10 000 Run k-means clustering for the random points. Use as many clusters as many areas you want: Run Vector Geometry --> Aggregate and use the cluster id as the aggregation field. This way you will get a multipoint layer from the clusters. Run Vector geometry --> centroids for the aggregated points Run Voronoi polygons for the centroids. Use enough buffer to cover the whole original polygon. For me this was around 40% Create lines from the voronoi polygons (Vector geometry --> polygons to lines). Currently I think you can only split a polygon layer with lines and not with another polygon layer (not 100% sure about this, but this works for sure). Finally run Split with Lines tool and use your original polygon layer as input and voronoi lines as the split layer: Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Mar 21, 2020 at 11:20 tjukanovtjukanov 63366 silver badges1010 bronze badges 7 Great solution. Though I was able to reach to the step of generating points. But when I use k-means clustering tool under Vector Analysis and set my desired clusters, I get same colored points (only one cluster) in an output layer – Waleed93 Commented Oct 6, 2021 at 12:43 Did you set the number of clusters correctly? Don't mind about the colors. Look at the attributes of the newly generated layer. – tjukanov Commented Oct 8, 2021 at 9:34 Yes, I checked. There's only one. In fact, I even did aggregation and finding centroid but there comes only one centroid (point) at the center. Strange. Btw, I'm generating 50000 random points with min distance apart of 0 meters, and creating 12 clusters – Waleed93 Commented Oct 8, 2021 at 17:08 This method worked reasonably well on all the polygons I ran it against. It works on concave polygons which the QGIS plugin Polygon Divider crashes on. Thanks for sharing! – Rob Irwin Commented Dec 8, 2021 at 16:34 Hey. this method does not work to divide an area into two or? The Part 5 The Voronoi - Polygone requires at leadt tree points and I just have two in my case gis.stackexchange.com/questions/428067/… – miraculix Commented Apr 10, 2022 at 17:20 \| Show 2 more comments This answer is useful 6 Save this answer. Show activity on this post. In this attempt, I tried to accomplish @Christophe's idea suggested in comments. Please don't be harsh, it is definitely askew and fragile nevertheless it is somehow working. The script does not include the debugging part. Proceed with Plugins > Python Console > Show Editor and paste the script below ``` defining required imports import sys defining a class class LayerIntoParts: # initiating layer's name to split and number of parts provided from user def __init__(self, layer_name, segments): self.layer_name = layer_name self.segments = segments # setting up the input layer and checking it's type def layer_geometry_check(self): # loading layer layer = QgsProject.instance().mapLayersByName(self.layer_name) if not layer.isValid(): raise ValueError("Layer failed to load!") sys.exit # checking geometry type, will work only with polygons and multipolygons if layer.wkbType() == QgsWkbTypes.Polygon: layer_valid = layer # converting multipart into single part elif layer.wkbType() == QgsWkbTypes.MultiPolygon: layer_valid = processing.run('native:multiparttosingleparts', {'INPUT': layer, 'OUTPUT': "memory:"} )["OUTPUT"] # giving an error for not polygons and multipolygons else: raise ValueError("This geometry type is not supported!\Only Polygons/Multipolygons are allowed.") sys.exit # returning the validated layer return layer_valid # defining required parameters def layer_and_parameters(self): # inheriting the output from 'layer_geometry_check' method layer = self.layer_geometry_check() # getting the layer's area area_total = [f.geometry().area() for f in layer.getFeatures()] area_total = area_total # getting the layer's extent and adjusting it for processing functions layer_extent = layer.extent() layer_extent_string = "{},{},{},{}".format(layer_extent.xMinimum(), layer_extent.xMaximum(), layer_extent.yMinimum(), layer_extent.yMaximum()) # building a dictionary to store the output parameters param = { 'layer': layer, 'layer_crs': layer.crs(), 'layer_extent': layer_extent_string, 'layer_area_total': area_total, 'area_segment': area_total / self.segments } # returning the dictionary with output parameters return param # creating grid for the input layer def layer_2_grid(self): # inheriting the output from 'layer_and_parameters' method param = self.layer_and_parameters() # creating grid for the layer with 'qgis:creategrid' algorithm layer_grid = processing.run('qgis:creategrid', {'TYPE': 2, 'CRS': param['layer_crs'], 'EXTENT': param['layer_extent'], 'HSPACING': 10000, 'VSPACING': 10000, 'HOVERLAY': 0, 'VOVERLAY': 0, 'OUTPUT': "memory:"} )["OUTPUT"] # deleting redundant attributes (appear by default when using the algorithm to create a grid) layer_grid_clean = processing.run('qgis:deletecolumn', {'INPUT': layer_grid, 'COLUMN': ['left', 'right', 'top', 'bottom'], 'OUTPUT': "memory:"} )["OUTPUT"] # returning layer as a grid return layer_grid_clean # clipping the grid with the input layer def clip_grid_and_layer(self): # inheriting the output from 'layer_2_grid' method grid = self.layer_2_grid() # inheriting the output from 'layer_and_parameters' method layer = self.layer_and_parameters()['layer'] # clipping the grid with the layer by means of 'qgis:clip' algorithm layer_clip = processing.run('qgis:clip', {'INPUT': grid, 'OVERLAY': layer, 'OUTPUT': "memory:"} )["OUTPUT"] # returning the clipped grid return layer_clip # categorizing the clipped grid layer def categorize(self): # inheriting the output from 'clip_grid_and_layer' method layer_clip = self.clip_grid_and_layer() # inheriting the output from 'layer_and_parameters' method param = self.layer_and_parameters() # transforming clipped grid layer's elements into a list with # id, calculated area, and geometry elements = [] for f in layer_clip.getFeatures(): case = {"f_id": f.id(), "area": f.geometry().area(), "geom": f.geometry()} elements.append(case) # lists for used cells' ids used_ids_i = [] used_ids_j = [] # variable for categories n = 0 # looping through all cells for i in elements: area_calc = i['area'] used_ids_i.append(i['f_id']) for j in elements: # conditions: calculated area less than a "golden" area of a segment, no duplicates of cells, and sufficient intersection of edges between cells if area_calc <= param['area_segment'] and j['f_id'] not in used_ids_i and i['f_id'] not in used_ids_j and (i['geom'].intersection(i['geom'])).length() > 0: used_ids_j.append(j['f_id']) area_calc += j['area'] n += 1 # signing a category to each cell i['cat'] = 'cat_' + str(n) i['area_calc'] = area_calc # returning categorized cells as list return elements # dissolving categorized cells and outputting the result def dissolve_result(self): # inheriting the output from 'layer_and_parameters' method param = self.layer_and_parameters() # inheriting the output from 'categorize' method elements_raw = self.categorize() # creating an empty later for output vl = QgsVectorLayer("Polygon?crs={}".format(param['layer_crs'].authid()), "output", "memory") # accessing the empty layer pr = vl.dataProvider() # embedding new field values pr.addAttributes([QgsField("id", QVariant.Int), QgsField("cat", QVariant.String), QgsField("area_calc", QVariant.Double), QgsField("area_gold", QVariant.Double) ]) # updating the empty layer vl.updateFields() # looping through a list of categorized cells and inserting them into the empty layer for elem in elements_raw: feat = QgsFeature() feat.setGeometry(elem['geom']) feat.setAttributes([elem['f_id'], elem['cat'], round(elem['area_calc'], 8), round(param['area_segment'], 8)]) pr.addFeature(feat) vl.updateExtents() # dissolving by category by means of 'qgis:dissolve' algorithm layer_dissolved = processing.run('qgis:dissolve', {'FIELD': ['cat'], 'INPUT': vl, 'OUTPUT': 'memory:'} )["OUTPUT"] # deleting 'cat' attribute using 'qgis:deletecolumn' algorithm layer_dissolved_clean = processing.run('qgis:deletecolumn', {'INPUT': layer_dissolved, 'COLUMN': ['cat'], 'OUTPUT': "memory:"} )["OUTPUT"] # updating 'id' attribute with edit(layer_dissolved_clean): for feature in layer_dissolved_clean.getFeatures(): feature.setAttribute(feature.fieldNameIndex('id'), feature.id()) layer_dissolved_clean.updateFeature(feature) # returning function output return QgsProject.instance().addMapLayer(layer_dissolved_clean) my_class = LayerIntoParts('test_area_4', 2) my_class.dissolve_result() ``` Adjust the my_class = LayerIntoParts('name of your layer', num_of_parts) Press Run script and get the output Testing. I am using a polygon of Corsica that was taken from Natural Earth \| Admin 0 – Countries Splitting into 2 parts with ``` my_class = LayerIntoParts('test_area_4', 2) my_class.dissolve_result() ``` Splitting into 3 parts with ``` my_class = LayerIntoParts('test_area_4', 3) my_class.dissolve_result() ``` Notes: (TODO) Parameters 'HSPACING' and 'VSPACING' have to be defined in advance Does not work with Multipolygons References: Docs » QGIS User Guide » Processing providers and algorithms » QGIS algorithm provider PyQGIS 101: Using expressions to compute new field values PyQGIS Developer Cookbook Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited May 9, 2020 at 19:51 answered May 9, 2020 at 19:42 Taras♦Taras 35.5k55 gold badges7777 silver badges151151 bronze badges 0 Add a comment \| This answer is useful 5 Save this answer. Show activity on this post. Since 24 September 2020 the "Polygon Divider" Plugin is again supported under the hood of the QGIS 3.x versions. P.S. It was also mentioned as a comment Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Jun 17, 2021 at 7:16 Taras♦Taras 35.5k55 gold badges7777 silver badges151151 bronze badges 0 Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. This is an approach that could be tried: 1) So far, it is most likely impossible to accurately separate a complex polygon by horizontal and vertical lines. 2) The problem lies in the combination of straight and curved lines. 3) It is possible to separate a complex polygon accurately only with curved lines. 4) Yes, with small errors it is possible to divide the polygon by horizontal and vertical lines and they will accumulate at the second (third, etc.) level, i.e. the 1st level will allow me to precisely divide the curvilinear polygon into 2 exactly equal parts, and from the second level, when we have sides with straight and curved lines, the geoinstrument will start to accumulate errors... "...until the road is overwhelmed" :-). 5) An example of the study is shown in the figures below and 6) Maybe one day I'll reconsider my point of view, but that's what came to me now :-)... Next step to think about is how to find the average value of the polygon and then bring the area of sub-plots to the average value of the polygon. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited May 11, 2020 at 22:57 PolyGeo♦ 65.5k2929 gold badges115115 silver badges349349 bronze badges answered May 11, 2020 at 11:35 Cyril MikhalchenkoCyril Mikhalchenko 4,59399 gold badges1818 silver badges4949 bronze badges 0 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. I recently "partially" updated the Split Polygon plugin to QGIS 3, created by Uroc Ilic. It is now available in the QGIS repository. But it works only for convex polygons, that is, not very irregular. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Sep 27, 2020 at 16:30 MrXsquared 36.1k2222 gold badges7474 silver badges127127 bronze badges answered Sep 27, 2020 at 16:11 Luis PerezLuis Perez 1,30466 silver badges1414 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. I have made a QGIS model, and saved it as a script, that divides the polygon in equal parts, lengthwise. I have uploaded the script. See ticket and answer here: Dividing polygon in equal parts, number of parts as attribute in QGIS Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Jun 18, 2021 at 9:13 TrajanTrajan 8111 silver badge77 bronze badges Add a comment \| This answer is useful -3 Save this answer. Show activity on this post. Now, there is a ‘Polygon Divider’ plugin for QGIS 3.x version, which is the solution. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Sep 27, 2020 at 21:50 Kadir Şahbaz 78.5k5757 gold badges260260 silver badges407407 bronze badges answered Sep 27, 2020 at 15:43 Kapil Dev AdhikariKapil Dev Adhikari 14188 bronze badges 1 1 Please edit your question to include a detailed description of the steps necessary to accomplish the tasks. Otherwise, it is likely that the link will be obsolete in the future and, as a result, the answer becomes obsolete. – Aaron ♦ Commented Sep 27, 2020 at 20:42 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions polygon qgis-3 splitting equal-area See similar questions with these tags. 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16553	http://simonrs.com/eulercircle/rtag2020/heidi-elliptic.pdf	Rational Points on Elliptic Curves Heidi Lei 1 Introduction Elliptic curves are non-singular cubic curves commonly in the form y2 = x3 + ax2 + bx + c. Figure 1.1 gives two examples of what an elliptic curve might look like when plotted in R2. It is in one connected piece if the cubic in x has one real root, and in two connected pieces if it has three real roots. x y (a) y2 = x3 −x + 1 x y (b) y2 = x3 −x Figure 1.1: Examples of elliptic curves. Elliptic curves are particularly useful in number theory since elliptic curves over the rationals are related to modular forms, and they are involved in the proof of Fermat’s last theorem through a form of the modularity theorem. Elliptic curves also have important cryptographic uses since an operation can be deﬁned on the points of an elliptic curve, and there is an analog of the discrete log problem on the points under this operation. Compared to public-key encription algorithms based on the structure of the multiplicative group Z/pZ×, the keys using elliptic curve cryptography are much smaller in size. In this paper, we prove some basic results concerning the structure of rational points on an elliptic curve mainly following the exposition of . We ﬁrst analyze rational points on conic curves to gain some intuition, then we build up the necessary tools for the main 1 results of this paper, the Nagell-Lutz theorem, which lets us compute the torsion group of rational points, and Mordell’s theorem, which states that the group of rational points is ﬁnitely generated. 2 Rational Points on Conics Before we dive into elliptic curves, we will ﬁrst look at rational points on conics, i.e., curves of degree 2 with the general formula ay2 + bxy + cx2 + dx + ey + f = 0, which gives us some insights on how rational points behave on a curve. If all the coeﬃcients of the curve a, . . . , f are rational, then we refer to the curve C as a rational conic. We are interested in the rational points on a rational conic. We have a complete characterization of rational points on conics in the sense that we can check in ﬁnite steps whether there exists a rational point, and if so, we can represent all rational points on thee conic with a closed-form parametrization. 2.1 Conics with a Known Rational Point The principal idea we develop in this section is that if we have one rational point on the conic, then in fact we have inﬁnitely many of them. The rational points on the conic can tbe parametrized by projecting the conic onto the a rational line. We assume that the conic C contains a rational point O. Let l be a rational line. Then we can establish a correspondence between rational points on the conic C and rational points on the line l. We project C onto l through the point O as shown in Figure 2.1, i.e., let P be a point on C, then its projection P ′ onto l is given by the intersection of l with the line passing through O and P. There are a couple special points that we need to take care of. The point O itself is projected onto l using the line that is tangent to C at O. The point at which the line passing through it and O is parallel to l is mapped onto their intersection at the point at inﬁnity. (We often operate in the projective space for the beneﬁt that two lines always intersect and curves have a “correct” number of intersections according to their degrees.) Conversely, if we have a point Q on the line l, then its corresponding point Q′ on C is given by the intersection of C with the line passing through O and Q, which exists since a line intersects a cubic at two points generally. Now that we have a bijective mapping between points on the conic and the line, we simply need to demonstrate that this map is also a correspondence of rational points. If P is a rational point on C, then the line passing through O and P is a rational line. It is easy to see that the intersection of two rational lines is indeed rational, so P ′ is a rational point on l. Conversely, if Q is a rational point on l, then the line passing through O and Q is a rational line. Since the coordinate of the intersection of a rational line and a rational conic is given by a quadratic with rational coeﬃcients and its roots come in conjugate pairs. Since one of the intersection O is rational, then the other intersection Q′ must also be rational. 2 C l O P P ′ Q′ Q Figure 2.1: Projecting a conic onto a line through a point O Since the rational points on the line can be easily parametrized with one of the coordi-nates, we have obtained a parametrization of rational points on a conic. We demonstrate this procedure on a circle. Example. Consider a circle given by the equation x2 + y2 = 1. We project the points on the circle from the point (−1, 0) to the y-axis parametrized by {(0, t)} as shown in Figure 2.2. x y (−1, 0) (0, t) (x, y) Figure 2.2: Rational points on the circle x2 + y2 = 1. The line passing through (−1.0) and (0, t) is given by y = t(x + 1). Since the point (x, y) lies both on the line and on the circle, we have the relationship y2 = t2(x + 1)2 = 1 −x2. 3 Since one of the intersections of the line and the circle is (−1, 0), we factor out (x + 1) from both sides, t2(x + 1) = 1 −x. Solving for x and using y = t(x + 1) to solve for y gives x = 1 −t2 1 + t2, y = 2t 1 + t2, and we have arrived at a rational parametrization of the circle. 2.2 Existence of Rational Points on Conics In the previous section, we showed how we can parametrize the inﬁnite family of rational points on a conic curve given the existence of one rational point. Now, we investigate the conditions the conic needs to satisfy to contain a rational point. The general idea is given by Hasse’s local-global principle, which states that certain families of equations have solutions in the rational numbers if and only if solutions exist in the real numbers and the p-adic numbers for each prime p. In our speciﬁc case, we can ﬁrst reduce the conic to a general form ax2 + by2 + cz2 = 0, abc ̸= 0, a, b, c pairwise relatively prime and squarefree by homogenization and a change of variables through diagonalization. (See for a complete process.) Then, the existence of a rational solution can be determined by Legendre’s theorem, and we give a proof of it using elementary number theory. Theorem 2.1 (Legendre). The homogeneous quadratic equation ax2 + by2 + cz2 = 0 with nonzero, squarefree, and pairwise relatively prime coeﬃcients a, b, c ∈Z has a nontrivial integer solution if and only if (1) a, b, c do not share the same signs. (2) −ab mod c, −bc mod a, −ac mod b are squares. Remark. Condition (1) corresponds to solutions in the real numbers in Hasse principle, while condition (2) corresponds to solutions in the p-adic numbers. Proof. ⇒(Necessity): (1) If a, b, c have the same signs, then the only solution to the equation is the trivial solution x = y = z = 0. (2) We show that −ab is a square mod c, and the other conditions can be shown similarly by symmetry. It suﬃces to show that −ab is a square mod p for some p \| c, as the rest follows by the Chinese remainder theorem. Let (x, y, z) be a nontrivial integer solution to ax2+by2+cz2 = 0. Since the equation is homogeneous, we can take gcd(x, y, z) = 1. Furthermore, x, y, z are pairwise relatively prime: Let d be a common divisor of x and y, then d2 \| ax2 + by2, which implies that 4 d2 \| cz2. Since c is squarefree, we have d \| z. Since gcd(x, y, z) = 1, we must have d = 1, so gcd(x, y) = 1, and we have shown that x, y, z are pairwise relatively prime. Reducing the equation modulo p, where p \| c, we have ax2 + by2 ≡0 mod p. Since gcd(x, y) = 1, p does not divide both x and y. Wlog we have p ∤x. Then, rearranging the equation and multiplying by −b, we have a ≡−by2 x2 mod p ⇒ −ab ≡b2y2 x2 mod p. −ab is a square mod p as claimed. ⇐(Suﬃciency): Since a, b, c do not share the same sign, assume that a > 0 and b, c < 0. If \|abc\| = 1, then the conic x2 −y2 −z2 has the nontrivial solution (1, 1, 0), so assume that \|abc\| > 1. Since −bc ≡k2 mod a, the polynomial ax2 + by2 + cz2 splits into linear factors over Z/aZ: ax2 + by2 + cz2 ≡by2 + cz2 ≡b y2 + c bz2 ≡b y2 −k2 b2 z2 ≡b y −k b z y + k b z mod a. By symmetry, ax2 + by2 + cz2 also splits into linear factors over Z/\|b\|Z and Z/\|c\|Z, so by CRT, it also factors over Z/abcZ: ax2 + by2 + cz2 ≡(αx + βy + γz)(rx + sy + tz) mod abc. Consider the set of triples of nonnegative integers S = {(x, y, z) : x, y, z ≥0, x < p \|bc\|, y < p \|ca\|, z < p \|ab\|}. Since bc, ca, ab are not perfect squares, we have \|S\| > abc. Then, by pigeonhole principle, there exists two distinct triples (x1, y1, z1) and (x2, y2, z2) such that αx1 + βy1 + γz1 ≡αx2 + βy2 + γz2 mod abc. Setting x = x1 −x2, y = y1 −y2, z = z1 −z2, we have αx + βy + γz ≡0 mod abc, which implies that ax2 + by2 + cz2 ≡(αx + βy + γz)(rx + sy + tz) ≡0 mod abc. Moreover, since x < p \|bc\|, y < p \|ca\|, z < p \|ab\|, and a > 0, b, c < 0 we have ax2 + by2 + cz2 ≤ax2 < abc, ax2 + by2 + cz2 ≥by2 + cz2 > −2abc. 5 Thus, ax2 + by2 + cz2 is either 0 or abc. In the former case, (x, y, z) is a nontrivial solution since (x1, y1, z1) and (x2, y2, z2) are distinct and we are done. In the latter case, we make the following change of variables: x′ = xz −by, y′ = yz + ax, z′ = z2 + ab and it follows that ax′2 +by′2 +cz′2 = 0. If (x′, y′, z′) is trivial, then the conic is x2 −y2 +cz2, and (1, 1, 0) is a nontrivial solution. 3 From Conics to Cubics Now with some intuition gained from analyzing the case for conics, we turn to discussing rational points on cubic curves. Similarly, we reduce the cubic to a form that is easier to deal with, namely the Weierstrass normal form: y2 = x3 + ax2 + bx + c. With a smart choice of axes in the projective space, one can show that any cubic is birationally equivalent to a cubic in the Weierstrass normal form. (See Section 1.3 of for a sketch of the transformation.) There is a bijection between rational points on the general cubic and those on the reduced cubic as a result of the birational equivalence. Note that a cubic in Weierstrass normal form is symmetric across the x-axis, i.e., if (x, y) is on the curve, then is (x, −y). An elliptic curve is a nonsingular cubic in the Weierstrass normal form. As shown in Figure 1.1, an elliptic curve could either be one or two connected pieces in R2 depending on whether they have 1 or 3 real roots. A cubic in the Weierstrass form is singular when y2 = f(x) = x3 + ax2 + bx + c has a repeated root (See Figure 3.1). If f(x) has three repeated roots, e.g., f(x) = x3, then the curve has a cusp, or a “sharp” point. If the curve has two repeated roots, e.g., f(x) = x2(x+1) then it has a pair of tangents (real or complex) at the point of singularity. Singular cubics behave quite diﬀerently from nonsingular ones. Since a line that passes through the singular point only crosses the curve at one other point, rational points on a singular cubic can be projected on to a rational line in the same way as conics, so the analysis of rational points on singular cubics can be dealt with similarly. Therefore, we will restrict our attention to singular cubics, i.e., birationally equivalent to a nonsingular cubic in the Weierstrass normal form. 3.1 Group Structure of Points on Elliptic Curves The ubiquity and usefulness of elliptic curves can be partially attributed to the fact that points on elliptic curves have a very conveninet additional structure — they form an abelian group under a natural geometric operation involving lines and intersections. Since an elliptic curve is nonsingular, a line generally intersects the curve at three points (counting multiplicities). In order to make the idea of intersections precise, we work in the 6 x y (a) y2 = x3 x y (b) y2 = x2(x + 1) Figure 3.1: Examples of singular cubics. projective space. In addition to the aﬃne part, the elliptic curve also contains a point at inﬁnity O = (0 : 1 : 0) in projective space, i.e., a point that lies on all the vertical lines parallel to the y-axis. A binary operation can be deﬁned on points on the curve by taking the line that connects the two points and ﬁnd the third intersection with the curve. More precisely, let P and Q be two points on the elliptic curve, then the point P ∗Q is the intersection of the line PQ and C. If P = Q, then the third point is the intersection of the curve with the tangent line at P, which intersects C at P “twice”. l C P Q P ∗Q l C P P ∗P Figure 3.2: A binary operation on an elliptic curve. We do not yet have an identity with the binary operation ∗, but we can turn the points on the curve into a group by buildig upon ∗. First, we need a point to be the identity. Any rational point can be chosen to be the identity, but for the ease of computation, we conventionally deﬁne the group operation such that the point at inﬁnity O be the identity. 7 The group operation, which we denote by +, is deﬁned as P + Q = O ∗(P ∗Q), i.e., we take the third intersection of the line connecting P and Q, and then connect it with O, where the third intersection between this new line and the curve is taken to be P + Q. P Q P ∗Q P + Q P −P Figure 3.3: The group operation on an elliptic curve We need to verify that the points on the curve and + do form an abelian group. First, it is easy to check that + is commutative, since ∗is commutative: P ∗Q and Q ∗P give us the same line and thus the same third intersection point. Then, we check that O is indeed the identity, i.e., P + O = O ∗(P ∗O) = P for any point P on the curve. Note that the line passing connecting O and P passes through P ∗O by deﬁnition, then the third intersection point given O and P ∗O is clearly P. Next, we check that each point P does have an inverse −P, which we show to be its reﬂection across the x-axis, i.e., if the point P = (x, y), then its inverse −P = (x, −y). Since the line passing through P and −P is vertical, its third intersection point with the curve is the point at inﬁnity O, so we have P + (−P) = O ∗(P ∗(−P)) = O ∗O. The line tangent at O is the line at inﬁnity, and its third intersection point is again at O, and we have shown that P + (−P) = O. The last thing we need to check is that + is associative. We could do so by computing the coordinates through explicit formulas, which is doable but not quite a pleasant task. Instead, we appeal to Caley-Bacharach theoreme, an important result on cubics that can be proved from Bezout’s theorem. Theorem 3.1 (Cayley-Bacharach). Let C1 and C2 be two cubic curves that intersect at exactly nine points. Then any cubic curve C that passes through eight of the points also passes through the ninth. We are interested in showing that (P + Q) + R = P + (Q + R), which is equivalent to P ∗(Q + R) = (P + Q) ∗R. We draw the points O, P, Q, R, P ∗Q, P + Q, Q ∗R, Q + R and intersection of line P(Q + R) and R(P + Q). Note that the set of three dashed lines 8 and three solid lines all pass through the above nine points, and the elliptic curve passes through the ﬁrst eight points except the intersection. Therefore, since a union of three lines is a cubic, by Cayley-Bacharach theorem, the elliptic curve C also passes through the ninth point, which implies that P(Q + R) and R(P + Q) intersects C at the same point. And we have shown P ∗(Q + R) = (P + Q) ∗R. R Q Q ∗R Q + R P ∗Q P P + Q P ∗(Q + R) = (P + Q) ∗R Figure 3.4: Associativity of the group operation. Note that the rational points on an elliptic curve are closed under + and thus form a group. If P and Q have rational coeﬃcients, then the line passing through them is a rational line. The x-coordinates of the intersections of C and l are given by a cubic with rational coeﬃcients. Since the two roots corresponding to P and Q are rational, the third root must also be rational by Vieta’s formula, so P ∗Q has rational coeﬃcients, and so does P + Q. Therefore, if we have rational points on an elliptic curve, we could potentially use the addition operation to generate more rational points on the curve. 3.2 Explicit Formulas for the Group Operation In this section, we will derive some explicit formulas for adding two points on the curve. Let P = (x1, y1) and Q = (x2, y2) be two distinct points, and we try to compute P ∗Q = (x3, y3) and P + Q = (x3, −y3). First, the equation for the line connecting P and Q is given by y = αx + β, where α = y2 −y1 x2 −x1 , β = y1 −αx1. 9 We ﬁnd the intersection of this line and the elliptic curve by substituting in y: y2 = (αx + β)2 = x3 + ax2 + bx + c. Expanding and collecting terms, we have x3 + (a −α2)x2 + (b −2αβ)x + (c −β2). Since the three intersections are P, Q, and P ∗Q, the three roots of the cubic are x1, x2, and x3. So by Vieta’s formula, we have x3 = α2 −a −x1 −x2, y3 = αx3 + β. For the case when P = Q, i.e., computing 2P, we need to ﬁnd the tangent line at P. Using the same variables, the slope α of the tangent line is given by implicit diﬀerentiation α = dy dx P = f ′(x1) 2y1 . Then, using the same formula we derived above, we have x3 = f ′(x1) 2y1 2 −a −2x1 = (3x2 1 + 2ax1 + b)2 4y2 1 −a −2x1. Simplifying and substituting in y2 1, we arrive at the formula for the x-coordinate of 2(x, y), referred to as the duplication formula: x(2(x, y)) = x4 −2bx2 −8cx + b2 −4ac 4(x3 + ax2 + bx + c) . 4 Torsion Points on Elliptic Curves As in the last section we showed how the points on an elliptic curve form a group, we can now talk about the torsion points on an elliptic curve, i.e., points with ﬁnite order. We ﬁrst characterize general points on the curve of small orders. Then we discuss speciﬁcally rational torsion points on the curve and show how they can be determined algorithmically in ﬁnite steps, as a consequence of our main result of the section, the Nagell-Lutz theorem. 4.1 Points of Order 2 and 3 Let P = (x, y) be a point of order 2, i.e., 2P = O, which is equivalent to P = −P. Since −P = (x, −y), we have y = 0, so the points of order 2 on the curve are precisely (r, 0), where r is a root of the cubic f(x) = x3 + ax2 + bx + c. Since f(x) is a nonsingular cubic, it has three distinct complex roots r1, r2, r3, so there are three points with order 2. The points satisfying 2P = O are {O, (r1, 0), (r2, 0), (r3, 0)} 10 , and they form a subgroup isomorphic to Z/2Z × Z/2Z since it is a group of four elements and all non-identity elements have order 2. Now let us move on to points of order 3. We can characterize them by noting that they are exactly the points that satisfy x(2P) = x(P) and is not the identity. By the duplication formula, we have x4 −2bx2 −8cx + b2 −4ac 4(x3 + ax2 + bx + c) = x. Simplifying and rearranging, we ﬁnd that P ̸= O is a point of order 3 if and only if x(P) is a root of the quartic φ3(x) = 3x4 + 4ax3 + 6bx2 + 12cx + 4ac −b2. We want to show that φ3(x) has four distinct roots, and we do so by relating φ3(x) to f(x), since f(x) is nonsingular. Recall the following form of the duplication formula: x(2P) = f ′(x)2 4f(x) −a −2x. Setting it to equal to x and rearranging, we have φ3(x) = f ′(x)2 −4f(x)(a + 3x) = f ′(x)2 −2f(x)f ′′(x). In order to see if φ3(x) have repeated roots, we diﬀerentiate φ′ 3(x) = 2f ′(x)f ′′(x) −(2f ′(x)f ′′(x) + 2f(x)f ′′′(x)) = −2f(x)f ′′′(x) = −12f(x). Since f(x) and f ′(x) do not have common roots, φ3(x) = f ′(x)2 −2f(x)f ′′(x) and φ′ 3(x) = −12f(x) also do not have any common roots. As we have shown that φ3(x) has four distinct roots s1, s2, s3, s4, there are eight points of order 3, and including the identity there are nine points satisfying the equation 3P = O: O, (s1, ±t1), (s2, ±t2), (s3, ±t3), (s4, ±t4). Since each non-identity point has order 3, it is easy to see that they form a group isomorphic to Z/3Z × Z/3Z. 4.2 Nagell-Lutz Theorem After a discussion of points of small orders in the ﬁeld of complex numbers, we return to the rational numbers and look at rational torsion points. If we take an elliptic curve in Weierstrass form with rational coeﬃcients, we can always make a change of variables to clear the denominators of the coeﬃcients, so we assume that the coeﬃcients a, b, c of the curve are all integers. The Nagell-Lutz theorem gives us a nice characterization of rational torsion points on the curve. 11 Theorem 4.1 (Nagell-Lutz). Let y2 = x3 + ax2 + bx + c be an elliptic curve with integer coeﬃcients a, b, c, and its discriminant be D = −4a3c + a2b2 + 18abc −4b3 −27c2. Let P = (x, y) be a rational torsion point on the curve, then its coordinates x and y are integers and either y = 0 (P has order two), or y \| D. Proof. We ﬁrst prove the second part of the statement assuming the ﬁrst part of the theorem which states that rational torsion points have integer coeﬃcient. Let P = (x, y) be a rational torsion point, then 2P must also be a rational torsion point, so P and 2P both have integer coeﬃcients. We need to show that either y = 0 or y \| D. Assume that y ̸= 0, then 2P ̸= O and the duplication formula applies: x(2P) = f ′(x) 2y 2 −a −2x. Since a, x and x(2P) are integers, we must have y \| f ′(x). Since f(x) = y2, we also have y \| f(x). By the general theorey of discriminants, we can write D as a linear combination of f(x) and f ′(x): D = r(x)f(x) + s(x)f ′(x), from which we conclude that y \| D. Next, we move onto proving the crux of the theorem: a rational torsion point on an elliptic curve has integer coordinates. We prove this rather indirectly by showing that the p-adic valuation of the coordinates is nonnegative for all prime p. To this end, let P = (x, y) be a rational torsion point and let µ = vp(x) and σ = vp(y). Then we can write the coordinates as x = mpµ, y = npσ. Plugging them into the equation of the elliptic curve, we have n2p2σ = m3p3µ + am2p2µ + bmpµ. If σ and µ are negative, then by the properties of the p-adic valuation we have 2σ = 3µ. Therefore, there exists v ∈Z+ such that µ = −2v and σ = −3v, so we can ﬁlter the rational points based on the p-adic valuation of their coordinates. Let E(pv) denote the set of rational points such that E(pv) = {O ∪(x, y) ∈E(Q) : vp(x) ≤−2v, vp(y) ≤−3v}. Then, we obtain a chain of inclusions E(Q) ⊂E(p) ⊂E(p2) ⊂. . . 12 (In p-adic topology, we have built a chain of neighborhoods around the identity O.) Our goal is to show that the only torsion point in E(p) is O. We ﬁrst make a transfor-mation to move the inﬁnite point O to the origin. Let t = x y , s = 1 y, then the curve y2 = x3 + ax2 + bx + c becomes s = t3 + at2s + bts2 + cs3. This transformation is bijective except for points of order 2, i.e., when y = 0. Let (x, y) ∈E(pv) be a point in the xy-plane, then for its corresponding point (t, s) in the ts-plane, we have vp(t) = vp x y = −2(v + k) −(−3(v + k)) = v + k ≥v, vp(s) = vp 1 y = −(−3(v + k)) ≥3v. Next, we compute explicit formula for the group law in ts-plane. Let P = (t1, s1) and Q = (t2, s2) be two distinct points, then the line passing through P and Q is given by y = αx + β, α = s2 −s1 t2 −t1 . Since P and Q are on the curve, we substitute in s = t3 + at2s + bts2 + cs3. After a bit of algebra, we arrive at α = s2 −s1 t2 −t1 = t2 2 + t1t2 + t2 1 + a (t2 + t1) s2 + bs2 2 1 −at2 1 −bt1 (s2 + s1) −c (s2 2 + s1s2 + s2 1). Similarly, we compute α when P = Q. The slope of the tangent line is given by α = ds dt P = 3t2 1 + 2at1s1 + bs2 1 1 −at2 1 −2bt1s1 −3cs2 1 . Since this is the same as the case with distinct P and Q, we simply use the former. Let us compute vp(α) for later use and recall that vp(ti) ≥v, vp(si) ≥3v. Note that for the denominator we have vp(1 −at2 1 −2bt1s1 −3cs2 1) = vp(1) = 0, so vp(α) = vp(t2 2 + t1t2 + t2 1 + a (t2 + t1) s2 + bs2 2) ≥2v. Recall that β = s1 −αt1, so we also have vp(β) ≥3v. We ﬁnd the third intersection P ∗Q = (t3, s3) of the line y = αx+β with the transformed curve by a procedure similar to when we computed explicit formulas for an elliptic curve in Weierstrass form. We substitute in the equation of the line and use Vieta’s formula to arrive at t1 + t2 + t3 = −αβ + 2bαβ + 3cα2β 1 + aα + bα2 + cα3 . 13 Since the identity in the ts-plane is (0, 0), P + Q has coordinates (−t3, −s3). We then calcualte vp(t1 + t2 + t3). Since the denominator contains 1, and recall that vp(α) ≥2v, vp(β) ≥3v, we have vp(t1 + t2 + t3) = vp(αβ + 2bαβ + 3cα2β) ≥5v. Since vp(t1), vp(t2) ≥v, we have vp(t3) ≥v as well, and we have shown that the points in E(pv) are closed under addition and hence form a subgroup of E(Q). Moreover, we obtain the relation that vp(t(P) + t(Q) −t(P + Q) ≥5v, which could be turned into a homomorphism using the ring pvZp/p5vZp. We are ready to complete the proof by showing that the only torsion point in E(p) is O. Suppose P = (t, s) is a torsion point of order n, i.e., nP = (0, 0) and thus t(nP) = 0. Let vp(t) = v, which means that P ∈E(pv) but P ̸∈E(P v+1). Then, by applying the above relation repeatedly, we have vp(nt −t(nP)) = vp(nt) ≥5v. Suppose p ∤n, then vp(nt) = vp(t) = v, a contradiction. Suppose n = kp, then we look at P ′ = kP = (t′, s′) which has order p. Let v′ = vp(P ′) So similarly we have vp(pt′) ≥5v but vp(pt′) = v + 1, a contradiction. Let us see how the Nagell-Lutz theoren can be used to compute torsion points through an simple example. Example. Consider the elliptic curve y2 = x3 + x. The only rational point of order 2 is (0, 0). Let (x, y) be a rational torsion point. Since the discriminant D = −4, then y must be ±1, ±2, or ±4. None of these cases are possible, so the only torsion points are {O, (0, 0)}, which form a group isomorphic to Z/2Z. Mazur’s theorem gives us a characterization of all the possibilities of the structure of rational torsion points of an elliptic curve. It is a beautiful and challenging theorem, and the proof is beyond the scope of this paper. Theorem 4.2 (Mazur). Let E(Q) be the group of rational points on a rational elliptic curve. Then the torsion subgroup of E(Q) is isomorphic to • either Z/nZ for 1 ≤n ≤10, or n = 12 • or Z/2Z × Z/2mZ for 1 ≤m ≤4. 5 Group of Rational Points on Elliptic Curves In this section, we will provide a sketch of proof of the celebrated Mordell’s theorem, which states that the group of rational points on an elliptic curve is ﬁnitely generated. 14 We deﬁne a measure of “complexity” of a rational number where a rational number is complex if it has a large denominator or numerator. Let the height H of a rational number be H m n = max{\|m\|, \|n\|}, and deﬁne the height of a point to be the height of its x-coordinate, i.e., H(P) = H(x). We are interested in how H(P + Q) compares to H(P) and H(Q). For the ease of notations, we use the additive counterpart of H denoted by h. Deﬁnition 5.1. The height h(P) of a point P is h(P) = log H(P). A nice property of the height function is that the set of points with height less than a real number is ﬁnite. Lemma 5.2. The set P ∈E(Q) : h(P) ≤M is ﬁnite for any M ∈R. Proof. If h(P) ≤M, then H(P) ≤eM, so there are ﬁnite choices for the numerator and denominator for x, so the set of x-coordinates is ﬁnite. Since each x-coordinate corresponds to two possible y-coordinates, the set of points is also ﬁnite. By the following lemmas, we see how the height function behave with respect to the adding and doubling points, which allows us to translate the geometric group operation to number theoretic information given by the height function. Lemma 5.3. For a ﬁxed rational point P0 and an arbitrary rational point P on an elliptic curve, h(P + P0) ≤2h(P) + κ0 for some constant κ0 speciﬁc to the curve and dependent on the choice of P0. Lemma 5.4. For a rational point P on an elliptic curve, h(2P) ≥4h(P) + κ for some constant κ speciﬁc to the curve. We note that the above two lemmas simply characterize the behavior of the height func-tion and do not rely on any fact about elliptic curves as we have explicit formulas for adding and doubling points. The proof of Lemma 5.3 simply involves direct algebraic computation with the formulas and applying triangle inequalities. The proof of Lemma 5.4 is a bit more involved as we need to show that doubling the point increases the height of the point by a lot, so there cannot be too much cancelling happening with the numerator and the denominator. We do so by writing x(2P) as a function of f(x) and f ′(x), and proving a more general bound on the height of quotients of polynomials evaluated at a rational number. The crux of Mordell’s theorem lies in the following key lemma. 15 Lemma 5.5. The index [E(Q) : 2E(Q)] is ﬁnite. The proof is quite subtle and relies on a homomorphism φ that factors the duplication map P 7→2P. φ maps C to a closely related curve C, and applying φ twice gives us C again after an appropriate scaling. Equipped with the above lemmas, we are ready to prove Mordell’s theorem. The main idea of the proof is in a similar vein to inﬁnite descent. We start with an arbitrary point and produces a smaller point with the size of points measured by the height function, and show that we arrive at a ﬁnite set, and thus the entire group is ﬁnitely generated as we can reverse the descent procedure. Theorem 5.6 (Mordell). The group of rational points E(Q) on an elliptic curve is a ﬁnitely generated abelian group. Proof. Let [E(Q) : 2E(Q)] = n, and pointe Q1, . . . , Qn be the representatives for each coset of 2E(Q). Therefore, for a point P0 ∈E(Q), we can repeatedly perform the operation Pk = Qik + 2Pk+1 to arrive at the expansion P = Qi1 + 2Qi2 + 4Qi3 + · · · + 2m−1Qim + 2mPm. Consequently, it suﬃces to show that for large enough m, we can obtain a bound for the height of Pm independent of the initial point P, which implies that the Qi’s and the ﬁnite set of points less than a certain height generate E(Q). We use the bounds in Lemma 5.3 and Lemma 5.4 to show that the height of each Pk+1 is proportionally smaller than Pk. Take −Qi to be the ﬁxed point, then for any point Pk, we have h(Pk −Qik) = h(2Pk+1) ≤2h(P) + κi. Let κ = max κi, we combine the bounds and ﬁnd that for all Qi, we have h(2Pk+1) ≤2h(Pk) + κ. Recall that h(2Pk+1) ≥4h(Pk+1) + κ′. From the upper and lower bounds on h(2Pk+1), we obtain 4h(Pk+1) + κ′ ≤2h(Pk) + κ, which we rewrite as h(Pk + 1) ≤3 4h(Pk) −1 4 (h(Pk) −(κ + κ′)) . Therefore, if h(Pk) > κ + κ′, we can always ﬁnd Pk+1 whose height is smaller by a factor of 3/4. So we are guaranteed with a positive integer m such that Pm ≤κ + κ′, which ﬁnishes our proof. 16 References Kumar, Abhinav. 18.701 Theory of Numbers. Spring 2012. Massachusetts Institute of Technology: MIT OpenCouseWare, License: Creative Commons BY-NC-SA. Silverman, Joseph H. and Tate, John T. Rational points on elliptic curve. Undergraduate Texts in Mathematics, Springer, New York 2015. 17
16554	https://pubmed.ncbi.nlm.nih.gov/505497/	Serum cardiac enzymes in stroke - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Serum cardiac enzymes in stroke J W Norris,V C Hachinski,M G Myers,J Callow,T Wong,R W Moore PMID: 505497 DOI: 10.1161/01.str.10.5.548 Item in Clipboard Serum cardiac enzymes in stroke J W Norris et al. Stroke.1979 Sep-Oct. Show details Display options Display options Format Stroke Actions Search in PubMed Search in NLM Catalog Add to Search . 1979 Sep-Oct;10(5):548-53. doi: 10.1161/01.str.10.5.548. Authors J W Norris,V C Hachinski,M G Myers,J Callow,T Wong,R W Moore PMID: 505497 DOI: 10.1161/01.str.10.5.548 Item in Clipboard Full text links Cite Display options Display options Format Abstract Serum cardiac enzyme levels (CK, LDH, SGOT) were estimated and the ECG recorded for 4 days following admission of 288 patients (Group I) to a stroke intensive care unit. Sixty-four of these patients, subsequently found not to have strokes, served as controls. Mean serum levels of all 3 cardiac enzymes were elevated in 8% of the 224 patients with stroke. The mean serum enzyme levels in patients with transient ischemic attacks (TIA) did not differ from controls. In a second group of 230 patients with stroke (Group II) serum CK levels were measured and the isoenzyems were fractionated to determine the tissue source of the enzymes. One hundred and one patients had raised total CK values and 25 of these (11%) had raised CK-MB (heart) iso-enzyme, the remainder having CK-MM (skeletal muscle) fraction. No serum CK-BB (brain) iso-enzyme was detected in any patient. Patients with positive serum levels of CK-MB has more evidence of acute myocardial ischemia on ECG (p less than 0.05), and more cardiac arrhythmias (p less than 0.001) than those with normal CK levels. Scattered areas of myocytolysis were found in the myocardium at autopsy in one patient. The acute rise in serum cardiac enzymes which we have recorded in the initial stages of stroke suggest that acute myocardial involvement is a commoner complication than is generally recognized. Also, since the CK-MB rises were modest and progressive, it is more likely that this acute myocardial dysfunction is a consequence, rather than a cause, of the acute cerebrovascular lesion. PubMed Disclaimer Similar articles Time course of serum cardiac enzymes after intracoronary thrombolytic therapy. Creatine kinase, creatine kinase MB isozyme, lactate dehydrogenase, and serum glutamic-oxaloacetic transaminase.Wei JY, Markis JE, Malagold M, Grossman W.Wei JY, et al.Arch Intern Med. 1985 Sep;145(9):1596-600.Arch Intern Med. 1985.PMID: 4026489 Direct current cardioversion. Effect on creatine kinase, lactic dehydrogenase and myocardial isoenzymes.Reiffel JA, Gambino SR, McCarthy DM, Leahey EB Jr.Reiffel JA, et al.JAMA. 1978 Jan 9;239(2):122-4. doi: 10.1001/jama.239.2.122.JAMA. 1978.PMID: 579371 Serum enzymes and isoenzymes in the diagnosis and differential diagnosis of myocardial ischemia and necrosis.Lott JA, Stang JM.Lott JA, et al.Clin Chem. 1980 Aug;26(9):1241-50.Clin Chem. 1980.PMID: 6994925 Review. Plasma MB creatine kinase activity and other conventional enzymes. Comparison in patients with chest pain and tachyarrhythmias.Strauss HD, Roberts R.Strauss HD, et al.Arch Intern Med. 1980 Mar;140(3):336-9.Arch Intern Med. 1980.PMID: 7362351 [Serum markers for ischemic myocardial damage].Fontes JP, Gonçalves M, Ribeiro VG.Fontes JP, et al.Rev Port Cardiol. 1999 Dec;18(12):1129-36.Rev Port Cardiol. 1999.PMID: 10661020 Review.Portuguese. See all similar articles Cited by The no-reflow phenomenon is a post-mortem artifact.de la Torre JC, Fortin T, Saunders JK, Butler K, Richard MT.de la Torre JC, et al.Acta Neurochir (Wien). 1992;115(1-2):37-42. doi: 10.1007/BF01400588.Acta Neurochir (Wien). 1992.PMID: 1595393 Cardiopulmonary complications of brain injury.Grunsfeld A, Fletcher JJ, Nathan BR.Grunsfeld A, et al.Curr Neurol Neurosci Rep. 2005 Nov;5(6):488-93. doi: 10.1007/s11910-005-0039-7.Curr Neurol Neurosci Rep. 2005.PMID: 16263062 Review. Recognising "painless" heart attacks.Wong CK, White HD.Wong CK, et al.Heart. 2002 Jan;87(1):3-5. doi: 10.1136/heart.87.1.3.Heart. 2002.PMID: 11751649 Free PMC article.No abstract available. Circulating leucocytes perpetuate stroke-induced aortic dysfunction.Asano S, O'Connell GC, Lemaster KC, DeVallance ER, Branyan KW, Simpkins JW, Frisbee JC, Barr TL, Chantler PD.Asano S, et al.Exp Physiol. 2017 Oct 1;102(10):1321-1331. doi: 10.1113/EP086510. Epub 2017 Sep 2.Exp Physiol. 2017.PMID: 28737253 Free PMC article. Predictive value of serum initial brain natriuretic peptide and troponin on functional prognosis in noncardiogenic patients with anterior and posterior circulation cerebral infarction.Lu WT, Du WT, Lu DS, You J, Li HY.Lu WT, et al.Arq Neuropsiquiatr. 2022 Oct;80(10):985-993. doi: 10.1055/s-0042-1755270. Epub 2022 Dec 19.Arq Neuropsiquiatr. 2022.PMID: 36535282 Free PMC article. 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16555	https://bingweb.binghamton.edu/~suzuki/ThermoStatFIles/9.8%20CP%20Chemical%20reactions%20%20van%20't%20Hoff%20equation.pdf	van ‘t Hoff equation Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: October 16, 2017) The van 't Hoff equation relates the change in the equilibrium constant, Keq, of a chemical reaction to the change in temperature, T, given the standard enthalpy change, ΔH⊖, for the process. It was proposed by Dutch chemist Jacobus Henricus van 't Hoff in 1884 in his book "Études de Dynamique chimique" (Studies in dynamic chemistry). The van 't Hoff equation has been widely utilized to explore the changes in state functions in a thermodynamic system. The van 't Hoff plot, which is derived from this equation, is especially effective in estimating the change in enthalpy, or total energy, and entropy, or amount of disorder, of a chemical reaction The chemical potential is P T k Tn k T k P B Q B B ln ) 1 ln( ) (    At the standard pressure 0 P P  = 1 bar,  0 0 ln ) 1 ln( P T k Tn k T k P B Q B B    From these relations, we have  0 0 ln ) ( P P T k P P B    For convenience, we multiply the Avogadro’s number NA. For 1 mole of the system, we have  0 0 ln ) ( P P RT P N P N A A     (for 1 mole) The chemical reaction is given by 0   i i i  or 0 )] ln( ) ( [ 0 0    i i i A i P P RT P N   or ....] ) ( ) ln[( ) ln( ) ln( ) ( 2 1 0 2 0 1 0 0 0       P P P P RT P P RT P P RT P i i i i i i i i i          The equilibrium constant is defined as .... ) ( ) ( 2 1 0 2 0 1   P P P P K  and the Gibbs free energy 0 G r  is defined as    i i i A r P N G ) ( 0 0   . Thus we have 0 ln 0    K RT G r or T G R K r 0 1 ln    Noting that the enthalpy H is expressed in terms of G as P P T G T T T G T G TS G H                       2 . (Gibbs-Helmholtz) we have 0 2 0 1 1 ln H RT T G dT d R K dT d r r      where 0 G r  is the Gibbs free energy of the reaction at 1 bar.This can be rewritten as 0 1 ln ) / 1 ( H R K T d d r    (van ’t Hoff equation) (a) Endothermic reaction (b) Exothermic reaction
16556	https://www.wallstreetmojo.com/sample-mean-vs-population-mean/	All BlogsStatistics ResourcesSample Mean vs Population Mean Sample Mean vs Population Mean Publication Date : 24 Jun, 2022 Blog Author : WallStreetMojo Team Edited by : Ashish Kumar Srivastav Reviewed by : Dheeraj Vaidya, CFA, FRM Share Table Of Contents Difference between Sample Mean vs Population Mean In statistics, there are two different averages: the sample mean and the population mean. The sample mean only considers a selected number of observations—drawn from the population data. The population mean, on the other hand, considers all the observations in the population—to compute the average value. The method of mean computation is the same in both cases. But, when the population size is large, a sample mean is determined to get an approximate value that represents the entire population. Computing the sample mean is easy, whereas computation of the population mean a tedious process. Comparative Table Sample Mean Vs Population Mean Differences are as follows: \| Basis \| Sample Mean \| Population Mean \| --- \| 1. Meaning \| The sample mean is the average of sample values picked from the population. The result resembles the population mean to a certain extent. \| Population mean is the central tendency for the entire group. \| \| 2. Number of Observations \| Compared to the population, the sample size is small. The sample size is represented by 'n.' \| The population size is large, and the sample size is denoted by 'N.' \| \| 3. Data Source \| The sample data is acquired from the population. \| The population data is collected from records, censuses, etc. \| \| 4. Calculation \| The formula used for the sample mean evaluation is: x̄ =∑xi /n \| The formula used for the population mean calculation is: μ=∑X / N \| \| 5. Level of Difficulty \| For sample mean, data collection and computation are easy. \| The collection of population data and computation is a very tedious process. \| \| 6. Computed When \| The population size is too large or infinite \| The population size is small \| \| 7. Denoted as \| The sample mean is denoted by x bar, i.e., 'x̄'. \| Population mean is represented as mu, i.e., 'μ'. \| \| 8. Standard Deviation \| The standard deviation derived from a sample mean is represented by 's.' \| The standard deviation evaluated from the population mean is represented by 'σ'. \| \| 9. Accuracy \| Since the sample mean considers random sample values taken from the whole population, it may not provide accurate results. \| The average calculated for the entire population is accurate and reliable. \| \| 10. Statistical Jargon \| A sample is a statistic. \| The population is a parameter. \| What is Sample Mean? The sample mean is a central tendency measure. The arithmetic average is computed using samples or random values taken from the population. It is evaluated as the sum of all the sample variables divided by the total number of variables. It is computed using the following formula: x̄ =∑xi /n Here, x̄ is the sample mean. ∑xi is the sum of all the sample observations. n is the sample size or the total number of observations. The simple random sampling method is considered unbiased and reliable for determining the sample mean. Simple random sampling is a process where each article or object in the population has an equal chance of getting selected. Thus, It closely represents the result an analyst would have received upon computing the average of the entire population. Sampling may not always provide the correct result. If a sample represents the population unevenly, it is referred to as a sampling error. Sample Mean Example Let us look at an example to understand sample mean calculations: A preschool authority wants to know the approximate average weight of students. Out of 360 students, 40 students are randomly selected for weighing: \| \| \| \| --- \| Number of Students \| Weight (Kg) \| Number of Students × Weight (Kg) \| \| 2 \| 14 \| 28 \| \| 3 \| 14.5 \| 43.5 \| \| 1 \| 15 \| 15 \| \| 1 \| 15.5 \| 15.5 \| \| 3 \| 16 \| 48 \| \| 2 \| 16.5 \| 33 \| \| 4 \| 17 \| 68 \| \| 3 \| 17.5 \| 52.5 \| \| 2 \| 18 \| 36 \| \| 5 \| 18.5 \| 92.5 \| \| 1 \| 19 \| 19 \| \| 4 \| 19.5 \| 78 \| \| 3 \| 20 \| 60 \| \| 1 \| 20.5 \| 20.5 \| \| 5 \| 21 \| 105 \| \| 40 714.5 \| Clearly, ∑xi = 714.5 Kgs And, n = 40 Now let us apply the values to the formula: x̄ =∑xi /n x̄ = 714.5 / 40 = 17.86 Kgs Hence, the approximate average weight of preschool students is 17.86 kgs. What is Population Mean? A population mean is the average computed from the entire group, distribution, or population. It is derived by dividing the aggregate of all the population variables by the total number of variables in the population. Population Mean is denoted by the following formula: μ=∑X / N Here, μ is the population mean. ∑X is the sum of all the observations in the population. N is the population size or the number of total variables in the data set. When the population size is too large for computation or has an infinite or uncountable number of observations, a sample mean is calculated. This is undertaken using the random sampling method. Also, the sample mean is believed to closely interpret the result of the population mean—if there are no sampling errors. Population Mean Example Let us assume that the prices of ABC stock at the end of every trading day for 20 days of a month are as follows: $15.15, $15.42, $14.91, $14.78, $15.12, $15.56, $15.36, $15.13, $15.31, $14.79, $15.11, $15.31, $15.27, $14.49, $14.77, $14.35, $14.64, $15.21, $15.38, and $15.44. Now, based on the given data, determine the population mean. Solution:μ=∑X / N μ = ($15.15 + $15.42 + $14.91 + $14.78, $15.12 + $15.56, $15.36 + $15.13 + $15.31 + $14.79 + $15.11 + $15.31 + $15.27 + $14.49 + $14.77 + $14.35 + $14.64 + $15.21 + $15.38 + $15.44) / 20 = 301.5 / 20 = $15.07 Thus, the population mean of ABC stock prices is $15.07. Sample Mean Vs Population Mean Infographics Let us look at Sample Mean Vs Population Mean infographics to understand the differences between the two averages better. Similarities Sample Mean Vs Population Mean similarities are as follows: Both are types of the arithmetic mean. Arithmetic mean is a statistical tool that determines the central tendency of a given set of data. Also, the computation methods for both the means are the same. For computing, the sum of all the observations in a given data set is divided by the total number of observations. Conclusion The sample mean is the central tendency acquired from the sample data. The sample data is drawn from the population. Statistically, the sample mean is represented by 'x̄.' On the contrary, the population mean is the average value of all the observations in a given population or group. It is represented by 'μ.' The sample mean sometimes lacks accuracy due to sampling errors and may not fairly depict the population mean. Even so, it is not always possible to determine the population mean when the population size is too large. FINANCIAL MODELING RESOURCES Create a Full Dynamic Financial Model in 2 Days (6 hours) \| Any Graduate Or Professional is eligible \| Build & Forecast IS, BS, CF from Scratch. Join WallStreetMojo YouTube INVESTMENT BANKING RESOURCES The Exact Training Used by Top Investment Banks \| FM, DCF, LBO, M&A, Accounting, Derivatives & More \| $2400+ in Exclusive Benefits \| 100+ Wall Street-Level Skills. 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16557	https://www.pnas.org/doi/10.1073/pnas.1710215115	Hamilton's inclusive fitness maintains heritable altruism polymorphism through rb = c \| PNAS ARTICLESCurrent IssueLatest ArticlesSpecial FeaturesList of IssuesPNAS Nexus Front Matter AUTHORSInformation for AuthorsEditorial and Journal PoliciesSubmission ProceduresPublication Charges Topics Physical Sciences Social Sciences Biological Sciences Featured Topics Physics Chemistry Sustainability Science Sustainable Development Goals Articles By Topic Applied Mathematics Applied Physical Sciences Astronomy Biophysics and Computational Biology Computer Sciences Earth, Atmospheric, and Planetary Sciences Engineering Environmental Sciences Mathematics Statistics Featured Topics Anthropology Sustainability Science Sustainable Development Goals Articles By Topic Demography Economic Sciences Environmental Sciences Political Sciences Psychological and Cognitive Sciences Social Sciences Featured Topic Anthropology Sustainability Science Sustainable Development Goals Articles By Topic Agricultural Sciences Applied Biological Sciences Biochemistry Biophysics and Computational Biology Cell Biology Developmental Biology Ecology Environmental Sciences Evolution Genetics Immunology and Inflammation Medical Sciences Microbiology Neuroscience Pharmacology Physiology Plant Biology Population Biology Psychological and Cognitive Sciences Systems Biology 0 Cart Sign inRegister Individual LoginInstitutional Login Submit Quick Search anywhere Enter journal, DOI, article type, keywords, authors, etc. Quick search in Citations Journal Year Volume Issue Page/eLocation ID Required field Quick Search in Journals Enter journal, DOI, article type, keywords, authors, etc. Quick Search in Journals Enter journal, DOI, article type, keywords, authors, etc. Searching: Anywhere AnywhereCitationPNAS NexusProceedings of the National Academy of Sciences Advanced SearchSearch SUGGESTED SEARCHES: Covid-19Artificial IntelligenceMpoxClimate ChangeGun Violence Sign inRegister Individual LoginInstitutional Login Quick Search anywhere Enter journal, DOI, article type, keywords, authors, etc. ARTICLES Current Issue Latest Articles Special Features List of Issues PNAS Nexus Front Matter AUTHORS Information for Authors Editorial and Journal Policies Submission Procedures Publication Charges Topics ContactSite MapTerms & Privacy PolicyAccessibility backPhysical Sciences Featured Topics Physics Chemistry Sustainability Science Sustainable Development Goals Articles By Topic Applied Mathematics Applied Physical Sciences Astronomy Biophysics and Computational Biology Computer Sciences Earth, Atmospheric, and Planetary Sciences Engineering Environmental Sciences Mathematics Statistics backSocial Sciences Featured Topics Anthropology Sustainability Science Sustainable Development Goals Articles By Topic Demography Economic Sciences Environmental Sciences Political Sciences Psychological and Cognitive Sciences Social Sciences backBiological Sciences Featured Topic Anthropology Sustainability Science Sustainable Development Goals Articles By Topic Agricultural Sciences Applied Biological Sciences Biochemistry Biophysics and Computational Biology Cell Biology Developmental Biology Ecology Environmental Sciences Evolution Genetics Immunology and Inflammation Medical Sciences Microbiology Neuroscience Pharmacology Physiology Plant Biology Population Biology Psychological and Cognitive Sciences Systems Biology backTopics Physical Sciences Social Sciences Biological Sciences Reference #1 Research Article Evolution Free access Share on Hamilton's inclusive fitness maintains heritable altruism polymorphism through rb = c Changcao Wang and Xin Lu Info & Affiliations Edited by Joan E. Strassmann, Washington University in St. Louis, St. Louis, MO, and approved December 12, 2017 (received for review June 7, 2017) January 2, 2018 115 (8) 1860-1864 View related content Commentary February 13, 2018 Cooperative breeding and the evolutionary coexistence of helper and nonhelper strategies Charlie K. Cornwallis Letter November 6, 2018 No evidence for a heritable altruism polymorphism in Tibetan ground tits Sacha C. Engelhardt, Sjouke A. Kingma, Michael Taborsky Letter March 20, 2018 Altruism or association? Owen M. Gilbert This article has a reply Letter November 6, 2018 Reply to Engelhardt et al.: Inclusive fitness does maintain a heritable altruism polymorphism in Tibetan ground tits Changcao Wang, Xin Lu Letter March 20, 2018 Reply to Gilbert: On the relationship between association and altruism Changcao Wang, Xin Lu 16,776 29 Metrics Total Views 16,776 Last 12 Months 996 Total Citations 29 Last 12 Months 3 View all metrics Track CitationsAdd to Reading List PDF Contents Vol. 115 \| No. 8 Significance Abstract Results Discussion Materials and Methods Acknowledgments Supporting Information References Information & Authors Metrics & Citations View Options References Figures Tables Media Share Significance Why should some individuals help others at the expense of their own fitness? Hamilton’s elegant formula rb>c resolves the major paradox in social evolution and has become a golden rule of sociobiology. However, rb>c only tells part of the story, namely how altruistic genotypes expand. Theoretically, altruistic genotypes can persist by coexisting stably with nonaltruistic ones relying on rb = c, which may let both genotypes have equal inclusive fitness. We present evidence for this prediction using long-term data on a species of bird. Our work suggests that altruism should be understood beyond rb>c, given that rb = c has the potential to explain widespread altruism polymorphisms in nature. Abstract How can altruism evolve or be maintained in a selfish world? Hamilton’s rule shows that the former process will occur when rb>c—the benefits to the recipients of an altruistic act b, weighted by the relatedness between the social partners r, exceed the costs to the altruists c—drives altruistic genotypes spreading against nonaltruistic ones. From this rule, we infer that altruistic genotypes will persist in a population by forming a stable heritable polymorphism with nonaltruistic genotypes if rb = c makes inclusive fitness of the two morphs equal. We test this prediction using the data of 12 years of study on a cooperatively breeding bird, the Tibetan ground tit Pseudopodoces humilis, where helping is performed by males only and kin-directed. Individual variation in ever acting as a helper was heritable (h 2 = 0.47), and the resultant altruism polymorphism remained stable as indicated by low-level annual fluctuation of the percentage of helpers among all adult males (24–28%). Helpers’ indirect fitness gains from increased lifetime reproductive success of related breeders statistically fully compensated for their lifetime direct fitness losses, suggesting that rb = c holds. While our work provides a fundamental support for Hamilton’s idea, it highlights the equivalent inclusive fitness returns to altruists and nonaltruists mediated by rb = c as a theoretically and realistically important mechanism to maintain social polymorphism. Sign up for PNAS alerts. Get alerts for new articles, or get an alert when an article is cited. Manage alerts The basis underlying Hamilton’s rule rb>c lies in the gene’s-eye view of evolution: natural selection will favor altruism over nonaltruism if the indirect fitness rewards to altruists rb, which result from copies of the genes passed indirectly via related social partners, more than offset the costs to them, c, which represent the loss of copies of the genes due to performing altruism (1). From a perspective of population genetics, c means the difference in direct fitness between altruistic and nonaltruistic genotypes. Therefore, rb>c typically describes a dynamic process during which adaptive altruistic genotypes keep increasing in frequency within a population against nonaltruistic genotypes. With its general ability and tractability to capture the essence of social evolution, Hamilton’s rule has elicited a good deal of theoretical (2–9) and empirical (10) attention. However, we emphasize that natural selection not only drives new, adaptive variants to replace old, less adaptive ones but also favors different variants to coexist stably. Thus, in addition to taking the dynamic state as specified by rb>c, altruistic genotypes may persist by forming a balanced behavioral polymorphism with nonaltruistic genotypes during the evolutionary tug-of-war between the two morphs, with the outcome depending on their relative fitness. Theoretically, the equilibrium between altruistic and nonaltruistic genotypes, or evolutionarily stable state (11), can be maintained by rb = c. This is because the equality implies that the indirect fitness benefits to individuals with altruistic genotype rb fully compensate for the direct fitness costs to them, c, permitting both morphs of the altruism polymorphism to have equal inclusive fitness so that they cannot invade each other (Fig. 1). The equivalent inclusive fitness mechanism for the maintenance of altruism polymorphism through rb = c is consistent with Fisher’s theorem of natural selection (12), which states that for a heritable trait to persist in an equilibrium population, it must cause no change in the population’s average fitness. An allele-based mathematical simulation gave a similar prediction (13). Moreover, the mechanism of fitness equivalence has been found to act in holding some morphological polymorphisms (14, 15). In the light of the ubiquity of altruism polymorphisms in nature (SI Appendix), our theoretical prediction deserves empirical testing to understand how different social morphs to coexist within a population. Fig. 1. Open in Viewer A sketch showing the testing system for the equivalent inclusive fitness mechanism mediated by rb = c to maintain altruism polymorphism. (A) Altruism enhances lifetime direct fitness of the recipients, where b is the direct fitness difference between recipients and nonrecipients. (B) The indirect fitness payoffs for altruists rb (where r is the genetic relatedness between altruists and recipients) perfectly compensate c—the lifetime direct fitness losses suffered by altruists relative to nonaltruists. Here we test the prediction using a 12-y dataset of individual relationship and lifetime reproductive success from a population of the Tibetan ground tit Pseudopodoces humilis (16). As in most avian cooperative breeders, helping in this species is kin-directed. Extra-pair paternity occurs at a very low rate (0.5% of offspring), potentially promoting kin selection (17). Cooperative breeding is facultative in that some pairs (28% on average) in the population have helpers, which aid the breeders by defending territory and feeding nestlings. A helper usually helps for one breeding season and then it breeds itself (“ever giving help”), but in contrast, some breeding individuals have never helped during their lifetime (“never giving help”). On the other side of the coin, some breeders have ever received help (“ever receiving help”), whereas the others have not (“never receiving help”). No individual has been found to adopt a floating strategy. Results To produce a robust test, we must first verify that variation in performing altruism among individuals within the study population is heritable. This is because heritability underlies the basis of defining ever and never giving help as two alternative phenotypes (the family lineages carrying either altruistic or nonaltruistic genes), which constitute a genetic polymorphism for natural selection to operate on. Specifically, an unambiguous definition of the two phenotypes makes it justified to evaluate lifetime inclusive fitness of ever-giving-help ground tits, even though they subsequently died or bred themselves. In fact, Hamilton’s theory of inclusive fitness is established on quantitative genetics (1–3, 8). Previous attempts to test this theory, including those to parameterize Hamilton’s rule, however, failed to set altruism heritability as a necessary prerequisite (10). Using the pedigree-based threshold animal model, we demonstrated a significant genetic component of individual variation in ever vs. never being a helper in the ground tit population, with the estimated narrow-sense heritability (h 2) being 0.47 (95% CI: 0.16–0.75). To our knowledge, this is the second estimate for heritability of altruistic behavior in natural populations. The first estimate was made for the Western bluebird Sialia mexicana (18), reporting a significant result (h 2 = 0.76). Nevertheless, that analysis was based on a dataset with an unbalanced sample size of helped vs. unhelped nests (22 vs. 251) and without excluding the relatively high rate of extra-pair paternity (17% of offspring). An equilibrium state of altruism polymorphism is the indicator by which we may predict that rb = c is going on. In this ground tit population, the percentages of helpers among all adult males annually varied between 24.2% and 28.3% (mean ± SD: 25.6 ± 1.0%; 3.9% for the coefficient of variation) over the 12-y study period, although the percentages of helped nests fluctuated extensively, between 13.0% and 36.1% (27.5 ± 7.4%; 26.9% for the coefficient of variation) (Fig. 2). The very low temporal variability in the relative number of helpers within the population suggests that altruistic and nonaltruistic genotypes could be at an almost stable equilibrium. Fig. 2. Open in Viewer Yearly variation in the percentages of helpers (blue line) among all adult males and of helped nests (gray line) among all nests in the Tibetan ground tit population across the study period. We then examined helping’s fitness benefits to the recipients and its fitness costs to the helpers. The general linear mixed-model analysis found that controlling for territory quality and year of birth, ever-receiving-help males had significantly higher lifetime reproductive success than never-receiving-help males (t = 10.35, df = 253.85, P< 0.001); this was also the case when comparing lifetime reproductive success between ever- and never-receiving-help females (t = 6.85, df = 159.83, P< 0.001; Fig. 3 A). On the other hand, with all else being equal, the lifetime direct reproductive success of ever-giving-help males was significantly lower than that of never-giving-help ones, an indication that helping was costly (Fig. 3 B and Table 1). Given that helpers enhanced recipients’ lifetime reproductive success at the expense of their own direct fitness, helping in the ground tits was altruistic. Fig. 3. Open in Viewer Comparisons of lifetime fitness between individual ground tits to test the prediction that equivalent inclusive fitness mediated by rb = c maintains altruism polymorphism. (A) Lifetime reproductive success of ever-receiving-help (blue columns) vs. never-receiving-help (white columns) breeders. (B) Lifetime direct reproductive success of ever-giving-help (purple columns) vs. never-giving-help (white column) males, and indirect fitness benefits of ever-giving-help males (light blue column). Values are given as mean + SE, with sample sizes above the SE bar. Table 1. Open in Viewer Results of general linear mixed-model analyses to assess whether helping behavior affected lifetime reproductive success of male ground tits (n = 217 birds), with territory quality as a covariate variable and year of birth as a random variable \| Model \| Coefficient ± SE \| t \| df \| P \| --- --- \| Without including indirect fitness \| \| \| \| \| \| Ever or never giving help \| 2.69 ± 0.80 \| 3.33 \| 208.82 \| 0.001 \| \| Territory quality \| 0.36 ± 0.32 \| 1.08 \| 209.00 \| 0.28 \| \| Including indirect fitness \| \| \| \| \| \| Ever or never giving help \| 0.24 ± 0.83 \| 0.28 \| 208.94 \| 0.78 \| \| Territory quality \| 0.55 ± 0.34 \| 1.60 \| 208.37 \| 0.12 \| EXPAND FOR MORE Alternatively, the observed significant difference in lifetime fitness between ever- and never-receiving-help breeders might be due to differential reproductive potential of the two groups of breeders rather than helpers’ contribution. However, that should not be the case. There were statistically significant differences between ever- and never-receiving-help breeders in body mass based on a t test (male: t = 1.01, df = 213, P = 0.31, BF 01 = 3.78; female: t = 0.64, df = 128, P = 0.52, BF 01 = 4.36) and tarsus length (male: t = 1.13, df = 213, P = 0.26, BF 01 = 3.36; female: t = 1.27, df = 128, P = 0.21, BF 01 = 4.84), the two commonly used indicators of individual quality (19). Furthermore, our earlier study of the ground tit population found that for a brood reared by a breeding pair plus a helper, the helper contributed to approximately one-third of the total amount of food delivered to the brood. The helped breeders of both sexes expended a significantly lower provisioning effort than the breeders that held no helper, more likely as a response to the presence of the helpers (20). This was supported by a within-individual comparison, which revealed that provisioning efforts on each nestling were 40% in males and 25% in females (n = 6 breeding pairs) lower when they were helped than when they were not. The reduced parental investment should be responsible for the higher annual survival rates and longer longevities of ever-receiving-help breeders (21), and thus the greater lifetime reproductive success as shown here. One might also argue that the lower lifetime direct reproductive success of ever-giving-help males could not be due to helping but because these males were lower quality relative to never-giving-help males. However, the lack of difference in body mass (t = 1.06, df = 175, P = 0.29, BF 01 = 5.30) and tarsus length (t = 0.68, df = 176, P = 0.50, BF 01 = 5.71) between the two groups of males does not support the assumption. Having verified that costly helping enhanced the recipients’ fitness, we next investigated whether the indirect fitness benefits gained by ever-giving-help male ground tits, rb, equaled their lifetime fitness costs, c. The general linear mixed-model analysis based on Price’s quantitative genetic idea of kin interaction (2) yielded a large P value, 0.78 (Table 1), suggesting a high probability of accepting the null hypothesis: the inclusive fitness benefits of ever-giving-help males, measured as their lifetime direct reproductive success plus their indirect fitness benefits, did not differ from the personal lifetime reproductive success of never-giving-help males (Fig. 3 B). This conclusion was reinforced by the small correlation coefficient of determination (R 2 = 0.002) for a single-predictor general linear mixed model, which indicated that whether to be a helper almost had no effect on individual variation in inclusive fitness. Actually, the difference in mean inclusive fitness between the two groups of males was slight (0.04 or 0.6%; Fig. 3 B). The very low statistical power (0.05 for a two-tailed test) of the current analysis seemed not to support the conclusion. However, this indicator has been considered useless when effect size is very small (22), such as in our case. A Bayes factor approach further provided strong evidence in favor of our prediction, with BF 01 = 24.59 indicating that the data were almost 25 times more likely under the null hypothesis than under the alternative hypothesis (23, 24). Together, these lines of evidence can lead to the conclusion that rb = c was statistically valid and the equivalent inclusive fitness mechanism may selectively maintain the heritable altruism polymorphism in the ground tit population. Discussion A major implication of our work is to offer a fundamental support for Hamilton’s theory of inclusive fitness (1), a critical concept in the textbooks of evolutionary biology and sociobiology but the focus of controversy, especially during recent years (25, 26). The evidence of altruism heritability is consistent with the core premise of Hamilton’s theory: altruistic behavior can evolve through kin selection depending on a genetic component (1–3, 8). Our empirical demonstration of rb = c, which is deduced from Hamilton’s rule, not only uncovers the role of indirect benefits in favoring altruism but also confirms the validity of rb>c, with which Hamilton formulated his theory of inclusive fitness. Importantly, our approach is robust because it follows the rigorous program of testing the theory of inclusive fitness, which demands a strict definition of altruism, the heritability of altruism as the basis for kin selection to act on, lifetime fitness to measure b and c, and quantitative genetic models to compare inclusive fitness (4–6, 27). As far as we know, our work is the first to test this theory taking such a program. Of more general importance, our work may inspire researchers to understand the evolution of altruism beyond rb>c in both theory and practice. With rb = c as a balancing state between altruistic and nonaltruistic genotypes, either of the genotypes may spread when rb>c or rb<c occurs. The framework constituted by the three equations about the evolution of altruism may be theoretically interpreted with reproductive altruism. The mechanism mediated by rb = c should operate in facultatively cooperative breeding animals (SI Appendix), given that theory suggests that extant traits including altruism should have been fixed after experiencing directional selection (28). Eusociality in which lifetime sterile workers are present in all colonies in a population persists as rb>c has pushed altruistic genotypes to go to fixation, whereas solitary breeding results when nonaltruistic genotypes have taken over the population relying on rb<c. Dynamic changes in the two genotypes in frequency may be understood by the fact that the origin and loss of solitary, facultatively cooperative, and eusocial lifestyles repeatedly occur across phylogenies (29, 30), and that these social forms shift within a single species across environments (ref. 31 and SI Appendix). Therefore, the framework based on rb vs. c relations has potential as a predictive and analytical tool to explore the evolutionary transition of social systems. Such explorations can be conducted with reproductive altruism systems by quantifying r, b, and c in association with kin structure and demography in different ecological regimes. It is also helpful for evaluating the evolution of social interactions beyond reproductive altruism, including those between nonkin (32), where r, b, and c could all be positive, negative, or zero (33). Bearing the framework for the evolution of altruism in mind is particularly important to empirical research where Hamilton’s rb>c has overwhelming influence. Indeed, previous studies attempted to test rb>c using the systems with an altruism polymorphism (10); there could have been a publication bias against the results inconsistent with the inequality (34). More unfortunately, almost all of these authors did not compare rb and c statistically but instead did so with their simple arithmetic means (10), which could hardly lead to any convincing conclusion. Employing these case studies, we may find evidence for our argument by treating the outcomes based on arithmetic means as a set of random samples from a population with an altruism polymorphism. As expected, there is no greater probability of observing rb>c than rb<c (8 vs. 5, χ 2 = 0.69, df = 1, P = 0.41), regardless of the potential publication bias favoring the former (34). In terms of alternative reproductive strategies (35), our work elaborates a kin selection-based form by demonstrating the heritability of altruism and clarifying the mechanism underlying the maintenance of individual variation in performing altruism. Specifically, while altruism polymorphism here is defined on a between-individual, lifetime basis, the altruistic behavior occurs in a within-individual, age-dependent way. In the study population, some ground tits forgo their first-year reproduction and help kin to accrue indirect fitness benefits; then they, if they survive, gain direct fitness benefits by breeding themselves. As a consequence, these ever-giving-help individuals can ultimately achieve the same lifetime inclusive fitness as those never-giving-help individuals. It is worth noting that helping could bring helpers some direct benefits, which can subsequently contribute to improving their personal reproductive success. In our study system, however, helpers seemed not to pursue such benefits through inheriting natal territory (36% of 36 vs. 40% of 131; χ 2 = 0.15, df = 1, P = 0.70). Nevertheless, other direct benefits from helping, such as acquiring breeding experience (36), remain to be investigated. There are alternative explanations for how the altruism polymorphism evolutionarily remains. One way this could occur is that altruistic alleles, if deleterious, can maintain a stable equilibrium frequency when they are eliminated by selection but are constantly reintroduced by mutation (37). The selection-mutation balance mechanism is less likely to work in the ground tit social system, where altruistic genotypes always have the same inclusive fitness as nonaltruistic genotypes so that selection against altruism should not have taken place. Also, negative frequency-dependent selection favoring rarer genotypes can allow a population to harbor both altruistic and nonaltruistic morphs (38). Almost unchangeable annual proportions of helpers relative to all adult males suggest a nearly stable altruism polymorphism, at least during the study period, excluding the mechanism of frequency-dependent selection. Even though the relative frequencies of altruistic genotypes fluctuate under frequency-dependent selection, the mean inclusive fitness of the two morphs should be statistically equal (i.e., rb = c) in the long run. Additionally, the lack of significant difference in body mass and tarsus length between ever- and never-giving-help males suggests that the maintenance of altruism polymorphism less likely depended on individual quality (19), which may indicate the competitiveness for access to territory or mate, the major ecological constraints on independent reproduction in birds (39). More indicators such as social dominance and immune level remain to be examined to explore whether the altruistic strategy is status dependent (40). Materials and Methods Study System. Data were collected during 2005–2016 from an individually banded ground tit population composed of 30–50 pairs within a 480-ha study plot (30°28′N, 91°05′E, 4,300 m above sea level) in Dangxiong County, south Tibet. Life history and social organization of this population have been described in detail elsewhere (16). Here we briefly outline the main features. Ground tits live in all-purpose territories year-round. Within territories, they construct one burrow for nesting and another for roosting during the nonbreeding period, which are a 0.8–2.9-m straight tunnel with a chamber at the end (41). During the nonbreeding period, the birds form family groups that contain male and female parents, philopatric young, and immigrants. When breeding, groups break into socially monogamous pairs, some of which have helpers. A pair usually produces one clutch per year, with a clutch size of 4–9 eggs and a brood size at fledging of 1–9 young. Incubation is undertaken by females only, lasting 15–16 d; nestlings are fed by both parents and helpers, if any, for 24–26 d. Natal dispersal is usually limited to less than five territories, and female juveniles move far more than male juveniles. Apparent annual survival rates of breeding males, breeding females, and helpers are 51%, 49%, and 43%, respectively. Pair bonds remain unless one partner dies, and only 0.5% divorce. Annual recruitment rates are 20% in male fledglings and 11% in female fledglings. Almost half (49%) of the yearling males breed independently and the other (51%) become helpers. Longevity of individuals who survive across at least one winter averages 1.9 (1–9) y in males and 1.7 (1–7) y in females. Incest and extra-pair parentage both are rare: 1% of broods in the former and 3% in the latter. A helped pair has one (85%) or two to three helpers (15%). Helpers all consist of males, the majority of which are yearlings (83%) and the minority older individuals (17%). Helping is kin-directed, toward both (62%) or one (20%) of the parents, as well as other first- or second-order relatives (18%). Most helpers help only once during their lives (90%), and a few do so for two or three breeding seasons (10%); then they breed on their own if they survive. Field and Laboratory Methods. We captured adult ground tits by placing a 1 × 1-m mist net around the entrance of nesting burrows and banded them with colored plastic and numbered metal leg rings. We checked nest contents and banded nestlings by digging a vertical shaft next to the nest chamber (the shaft was then plugged). Every winter, we banded immigrants by capturing them in roosting burrows. Social organization per nest was determined by observing behaviors of banded individuals. A blood sample of 10–30 µL was collected from each banded individual to determine its sex (supplemented by evidence from behavior and presence/absence of an incubation patch in adults) and genetic identity by means of molecular biology techniques (42, 43). Territory Quality Evaluation. We marked all nest sites located per year on a map of the study plot and determined the annual territory of each breeding pair as the areas enclosed by the midperpendicular of the connecting lines between neighboring nests. All annual territories in the plot over the study years were put together, and the maximal overlapped section around a nest was defined as a permanent territory. A total of 91 permanent territories that involved 551 nests were identified. The quality of a permanent territory was evaluated with the mean annual fledgling production by all of the pairs breeding within the territory over the study years. Heritability Estimate. A pedigree was established with 592 individuals (335 males and 257 females) who were engaged in at least one breeding event (bred independently or helped). Extra-pair parentage was taken into account when constructing the pedigree (only 1 out of the 592 individuals was sired by a male outside the study population). Birds that immigrated into the study area were assumed to be yearlings and unrelated to each other or to local individuals. Based on the pedigree, we estimated the heritability of helping behavior with a threshold animal model (44, 45), in which the binary response variable was specified as 1 for ever giving help or 0 for never giving help, with sex as a fixed factor. Analyses were conducted using the MCMCglmm package (46) implemented in R version 3.2.5 (47). A parameter-expanded prior (V = 1, ν = 1,000, α.μ = 0, α.V = 1) was set for the random effect. The binomial model with probit link was run with 16,000,000 iterations, 100,000 burn-in, and thinning of 3,000, to allow the simulations to converge adequately with weak autocorrelation (<0.03), large effective size (>5,000), and well-performed chains (46). Individual variation in acting as a helper might be a consequence that cooperative breeding depends on territory quality and high-quality territories tend to be inherited within family groups. We excluded the probability because the quality of a territory did not predict whether the territory held helpers (a generalized linear mixed-model analysis with year as the random variable: z = 1.32, df = 471.40, P = 0.19, BF 01 = 7.80). Fitness Measure. According to the definition of altruism in the context of inclusive fitness theory (4–6), b and c should be measured with lifetime direct fitness. The most appropriate proxy of lifetime fitness has been considered to be lifetime reproductive success (LRS), the total number of fledglings produced by an individual over its life, because it combines several critical components of fitness, including mating success, annual offspring productivity, survival, and longevity (48). In this study, the data were available for 422 individuals, which were banded as fledglings, bred, or helped at least once. Indirect fitness benefit (IF) gained by an ever-giving-help individual was calculated using the following equation: I F=r male(L R S helped male−mean L R S male)+r female(L R S helped female−mean LR S female) where LRS helped male and LRS helped female represent LRS of male and female breeders with whom the focal individual ever helped, respectively; mean LRS male and mean LRS female represent the average of LRS of all male and female individuals in the population, respectively; r male and r female represent the relatedness between the focal individual and the recipients, which were determined based on the pedigree. In a few cases where a breeder was helped by more than one helper, the increased direct fitness benefits from helping were equally divided to each helper. On occasion where a helper ever helped more than one nest in different years, we calculated the indirect fitness benefits to the helpers taking into account the increased direct fitness benefits to all of the involved breeders. Comparing Fitness. General linear mixed models were employed to compare the differences in lifetime reproductive success between ever- and never-receiving-help individuals for each sex. The same procedure was performed to compare the fitness differences between ever- and never-giving-help males, without and with including indirect fitness. Territory quality was assigned as a covariate variable and year of birth (the year in which an individual fledged) as a random variable to control for their potential effect on reproductive success. The inclusion of the two variables was less likely to lead to multicollinearity, because both of them, as was shown above, were unrelated to cooperative breeding. For all analyses, values of P< 0.05 were considered to be statistically significant. We also provided Bayes factors (with default mixture-of-variance priors) as an alternative approach for statistical inference when P> 0.05, with BF 01 = 3–20 indicating a positive and BF 01 = 20–150 strong support for the null hypothesis (23, 24). Acknowledgments We thank D. Ke, T. Yu, S. Tang, G. Zhang, Q. Zhao, H. Fan, W. Guo, Z. Li, and 16 other graduate students and W. Guan and five other volunteers for assistance in the field and laboratory. We thank S. Shen, Y. Tao, A. P. Møller, J. A. R. Marshall, L. Keller, S. A. West, D. C. Queller, A. F. G. Bourke, and X. H. Li for advice in preparing the manuscript; and D. W. Mock and three anonymous referees for comments on earlier versions of this manuscript. This research was supported by the National Natural Science Foundation of China (Grants 30425036, 30670337, 30830019, and 31572274). 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Available at (2016). Go to reference Google Scholar 48 GP Wagner, The measurement theory of fitness. Evolution64, 1358–1376 (2010). Go to reference PubMed Google Scholar Show all references Information & Authors Information Authors Information Published in Proceedings of the National Academy of Sciences Vol. 115 \| No. 8 February 20, 2018 PubMed: 29295937 Classifications Biological Sciences Evolution Copyright © 2018. Published under the PNAS license. Submission history Published online: January 2, 2018 Published in issue: February 20, 2018 Keywords cooperative breeding Fisher's natural selection theorem kin selection lifetime fitness quantitative genetics Acknowledgments We thank D. Ke, T. Yu, S. Tang, G. Zhang, Q. Zhao, H. Fan, W. Guo, Z. Li, and 16 other graduate students and W. Guan and five other volunteers for assistance in the field and laboratory. We thank S. Shen, Y. Tao, A. P. Møller, J. A. R. Marshall, L. Keller, S. A. West, D. C. Queller, A. F. G. Bourke, and X. H. Li for advice in preparing the manuscript; and D. W. Mock and three anonymous referees for comments on earlier versions of this manuscript. This research was supported by the National Natural Science Foundation of China (Grants 30425036, 30670337, 30830019, and 31572274). Notes This article is a PNAS Direct Submission. See Commentary on page 1684. Authors Affiliations Expand All Changcao Wang Department of Ecology, College of Life Sciences, Wuhan University, Wuhan, 430072, China; View all articles by this author Xin Lu1 Department of Ecology, College of Life Sciences, Wuhan University, Wuhan, 430072, China; The Institute for Advanced Studies, Wuhan University, Wuhan, 430072, China View all articles by this author Notes 1 To whom correspondence should be addressed. Email: luxinwh@163.com. Author contributions: X.L. designed research; C.W. and X.L. developed critical ideas; C.W. performed research; C.W. analyzed data; and X.L. wrote the paper. Competing Interests The authors declare no conflict of interest. Metrics & Citations Metrics Citations Metrics Article usage Views Citations No data available. 16,776 29 Total 6 Months 12 Months Total number of views and citations Note: The article usage is presented with a three- to four-day delay and will update daily once available. Due to ths delay, usage data will not appear immediately following publication. Citation information is sourced from Crossref Cited-by service. Altmetrics See more details Posted by 23 X users On 1 Facebook pages 127 readers on Mendeley 2 readers on CiteULike Citations Cite this article C. Wang, [...] &X. Lu, +0 authors Hamilton's inclusive fitness maintains heritable altruism polymorphism through rb = c, Proc. Natl. Acad. Sci. U.S.A. 115 (8) 1860-1864, Copy Copied! Copying failed. Export the article citation data by selecting a format from the list below and clicking Export. Format [x] Direct import Cited by Loading... View Options View options PDF Download this article as a PDF file. PDF eReader View this article with eReader. eReader Figures Open all in viewer Fig. 1. A sketch showing the testing system for the equivalent inclusive fitness mechanism mediated by rb = c to maintain altruism polymorphism. (A) Altruism enhances lifetime direct fitness of the recipients, where b is the direct fitness difference between recipients and nonrecipients. (B) The indirect fitness payoffs for altruists rb (where r is the genetic relatedness between altruists and recipients) perfectly compensate c—the lifetime direct fitness losses suffered by altruists relative to nonaltruists. Go to FigureOpen in Viewer Fig. 2. Yearly variation in the percentages of helpers (blue line) among all adult males and of helped nests (gray line) among all nests in the Tibetan ground tit population across the study period. Go to FigureOpen in Viewer Fig. 3. Comparisons of lifetime fitness between individual ground tits to test the prediction that equivalent inclusive fitness mediated by rb = c maintains altruism polymorphism. (A) Lifetime reproductive success of ever-receiving-help (blue columns) vs. never-receiving-help (white columns) breeders. (B) Lifetime direct reproductive success of ever-giving-help (purple columns) vs. never-giving-help (white column) males, and indirect fitness benefits of ever-giving-help males (light blue column). Values are given as mean + SE, with sample sizes above the SE bar. Go to FigureOpen in Viewer Tables Open all in viewer Table 1. Results of general linear mixed-model analyses to assess whether helping behavior affected lifetime reproductive success of male ground tits (n = 217 birds), with territory quality as a covariate variable and year of birth as a random variable Go to TableOpen in Viewer Media Share Share Share article link COPY LINK Copied! Copying failed. Share on social media FacebookX (formerly Twitter)LinkedInGmailemail References References 1 WD Hamilton, The genetical evolution of social behaviour. I. J Theor Biol7, 1–16 (1964). Crossref PubMed Google Scholar a [...] loss of copies of the genes due to performing altruism b [...] fitness is established on quantitative genetics c [...] support for Hamilton’s theory of inclusive fitness d [...] through kin selection depending on a genetic component 2 GR Price, Selection and covariance. Nature227, 520–521 (1970). Crossref PubMed Google Scholar a [...] rule has elicited a good deal of theoretical b [...] Price’s quantitative genetic idea of kin interaction 3 DC Queller, Expanded social fitness and Hamilton’s rule for kin, kith, and kind. Proc Natl Acad Sci USA108, 10792–10799 (2011). 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Go to reference Crossref PubMed Google Scholar 43 C Wang, X Lu, Extra-pair paternity in relation to breeding synchrony in ground tits: An individual-based approach. J Avian Biol45, 561–565 (2014). Go to reference Crossref Google Scholar 44 LEB Kruuk, Estimating genetic parameters in natural populations using the “animal model”. Philos Trans R Soc Lond B Biol Sci359, 873–890 (2004). Go to reference Crossref PubMed Google Scholar 45 AJ Wilson, et al., An ecologist’s guide to the animal model. J Anim Ecol79, 13–26 (2010). Go to reference Crossref PubMed Google Scholar 46 JD Hadfield, MCMC methods for multi-response generalised linear mixed models: The MCMCglmm R package. J Stat Softw33, 1–22 (2010). Crossref PubMed Google Scholar a [...] Analyses were conducted using the MCMCglmm package b [...] effective size (>5,000), and well-performed chains 47 ; R Core Team, R: A Language and Environment for Statistical Computing (R Foundation for Statistical Computing, Vienna), Version 3.2.5. Available at (2016). Go to reference Google Scholar 48 GP Wagner, The measurement theory of fitness. Evolution64, 1358–1376 (2010). Go to reference PubMed Google Scholar View full text\|Download PDF Figures Tables Close figure viewer Back to article Figure title goes here Change zoom level Download figure Go to figure location within the article Toggle information panel Toggle information panel All figures All tables View all material View all material xrefBack.goTo xrefBack.goTo Request permissions Expand All Collapse EXPAND FOR MORE Show all references SHOW ALL BOOKS Authors Info & Affiliations Further reading in this issue Research ArticleFebruary 7, 2018 Spatial patterns of tree yield explained by endogenous forces through a correspondence between the Ising model and ecology Andrew E. Noble, Todd S. Rosenstock, Patrick H. 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16558	https://math.stackexchange.com/questions/984660/beautiful-combinatorial-painting-problem	combinatorics - Beautiful combinatorial painting problem - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Beautiful combinatorial painting problem Ask Question Asked 10 years, 11 months ago Modified10 years, 11 months ago Viewed 425 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Mark paints squares of a white 10×10 10×10 board. He can either paints some vertical row of squares blue or some horizontal row red.(Every row is painted at most once). If blue paint is put on top of the red paint, it becomes a blue square. If red paint is put on the blue paint, the squares react and lose their color, and a white square is made. Could there be 33 33 red squares some time? I tried considering small cases n=2 k n=2 k for small k k's, and I feel it's done by induction, but I couldn't do it. How is it done? combinatorics logic Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Oct 21, 2014 at 19:14 posilon 2,283 11 11 silver badges 17 17 bronze badges asked Oct 21, 2014 at 18:42 Jackie PoehlerJackie Poehler 1,839 12 12 silver badges 23 23 bronze badges 0 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. No, it's impossible. To see this, we argue by contradiction. First, note that the order of the blue and red paint doesn't matter; in either case, an overlap of blue and red paint will result in a loss of one red square. Now suppose that Mark paints r r red horizontal rows and b b blue vertical rows. Then at the end, the total number of red squares will be: 33=10 r−r b=r(10−b)33=10 r−r b=r(10−b) Thus, since r∈{0,1,2,…,10}r∈{0,1,2,…,10} and r r is a factor of 33 33, we know that either r=1 r=1 or r=3 r=3. In the former case, we have that 10−b=33 10−b=33 so that b=−23 b=−23, which is absurd. In the latter case, we have that 10−b=11 10−b=11 so that b=−1 b=−1, which is also absurd. So it's impossible to get 33 33 red squares. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Oct 21, 2014 at 19:05 answered Oct 21, 2014 at 18:59 AdrianoAdriano 42k 4 4 gold badges 53 53 silver badges 87 87 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. This answer is pretty much the same as Adriano's, with more words and less notation. First notice the parenthetical comment that each row (horizontal row) or column (vertical row) is painted at most once. That's important, because it means that after Mark is done painting, every square has been painted in one of five ways: Not at all (so it's white) Once, with blue Once, with red Twice, first red, then blue (so it's blue) Twice, first blue, then red (so it's white) A first try at solving the problem might be to figure out how many squares there are in each of these categories, given a description of what Mark painted, say "He painted row 1, column 2, column 4, row 3, column 3, row 8, and row 7, in that order." Squares whose row and column were both never painted Squares whose column was painted, but whose row was not. Squares whose row was painted, but whose column was not. This is complicated. This is complicated. Fortunately, only #3 leaves a square red, so we don't have to figure out the complicated stuff. The number of squares in category #3 are only in rows that were painted, and there are the same number of them in each of those rows: 10 10 minus the number of columns that were painted. So the total number of red squares is (number of rows painted)××(10−10− number of columns painted). If there are 33 33 red squares, these two factors have to multiply to 33 33, but each factor is between 0 0 and 10 10, and that's impossible. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 21, 2014 at 20:45 Steve KassSteve Kass 15.4k 1 1 gold badge 24 24 silver badges 34 34 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics logic See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 12Painting chess board 17How many distinct Unruly boards are there? 4Number of ways we can paint the fence pickets 0Coloring 8x8 square board? 4Combinatorics - Painting the Unit Squares of a 3x3 Square 3Tiling a (2 n−1)×(2 n−1)(2 n−1)×(2 n−1) chessboard with one corner cut out 5Painting the board by selecting 2×2 2×2 type squares from the m×n m×n type board 1paint a board with two colors without repeating the amount painted in each row rows 2A nice combinatorics problem: show that John can implement his painting idea in at most n/2 n/2 years. 6Cuba 2003 Coloring Problem Hot Network Questions I have a lot of PTO to take, which will make the deadline impossible Lingering odor presumably from bad chicken Can you formalize the definition of infinitely divisible in FOL? 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16559	https://ourworldindata.org/data-insights/childhood-pneumonia-deaths-have-plummeted-in-nepal-and-more-lives-can-still-be-saved	Our World in Data Gdoc / Admin June 20, 2025 Childhood pneumonia deaths have plummeted in Nepal — and more lives can still be saved Veronika Samborska In the early 1980s, Nepal’s children suffered from some of the highest death rates from pneumonia in the world, with over 1,400 deaths for every 100,000 children under five. That meant around 39,000 children died from pneumonia each year, more than from any other cause . Since then, Nepal has made huge progress. The death rate has fallen almost 20-fold. This improvement is due to various measures , including pneumococcal and Hib vaccines, better access to healthcare and antibiotics, and improved nutrition. Despite this progress, pneumonia is still among the leading causes of death in children in Nepal. And we know that more progress can be made: high-income countries have achieved much lower rates , with fewer than 5 per 100,000. Explore how deaths from pneumonia among children have changed in other countries → Related topic pages: Child & Infant Mortality Pneumonia Our latest Data Insights See all Data Insights September 26 Richer countries don’t just avoid infectious disease — they also have lower rates of chronic disease deaths One of humanity’s biggest victories has been the fight against infectious diseases. This battle has led to plummeting rates of child and maternal mortality and dramatically increased life expectancy . However, there are still large differences in infectious disease rates between different parts of the world. As we might expect, deaths are much less common in high-income countries where almost everyone can access clean water , sanitation , and medication. One common misconception is that this prosperity has come at the cost of an increase in death rates from chronic diseases, such as heart disease and cancers (collectively called “non-communicable diseases”). But this is not the case. As you can see in the chart, death rates from these diseases are also lower in richer countries. It’s not the case that we’ve simply substituted one health problem for another. That means that the risk of someone of a given age dying from any cause each year tends to decrease as countries get richer. Explore more of our data on causes of death across the world → Continue reading September 24 1.5 billion people now live in countries where same-sex marriage is legal — but that’s only one in five worldwide The first nationwide law allowing same-sex couples to marry was passed in the Netherlands in 2001. Amsterdam’s mayor, Job Cohen, officiated the first couples. Twenty-five years on, these rights to same-sex marriage now cover 1.5 billion people worldwide. These people live in 39 countries with marriage equality, mainly across Western Europe and the Americas. This change in marriage laws has made a huge difference to the lives of many. But they are still in the minority globally. Four in five people still live in countries where same-sex couples are not equal under the law. Explore which countries have legalized same-sex marriage → Continue reading September 21 Fewer people work in farming in Asia’s largest countries Over the last three decades, employment has changed dramatically across Asia. In the early 1990s, almost two-thirds of the labor force in South Asia was employed in agriculture , and more than half in East Asia. Today, this is just 40% in the former, and one-quarter in the latter. The chart shows the change across several countries in these regions. The share has fallen in all of them, but some stand out as having particularly dramatic transitions away from farming. In Bangladesh and Vietnam, rates have fallen from around 70% to 38%. In China, they have dropped from 60% to 25%. In India, things have moved more slowly. This matters for people still working in agriculture and those who have moved to jobs in other sectors. Productivity gains — which can allow family members and former workers to move away from the farm — mean that the financial returns per farm worker have increased over this time. Those who have moved to jobs in industry and services often see an increase in their wages. The result is that mean incomes have increased across these countries. Read our data insight on this transition in today’s rich countries → Continue reading September 19 Honest elections matter to people everywhere Honest elections matter because they give people a say in how their country is run. They help ensure that governments reflect the people's will and that policies respond to real needs. It’s sometimes assumed that support for democracy and fair elections is limited to particular regions or cultures. However, data from the Integrated Values Surveys , which asks people across many countries how much honest elections matter to them, shows otherwise. Across all the countries surveyed, large majorities said honest elections are important in their lives. Nearly everyone said so in Indonesia, South Korea, and the UK. Even in countries where experts judge elections as unfree or unfair — like Iran, Egypt, or Russia — around 80% to 90% still said they matter. In some contexts, people may feel pressure to answer this question in a certain way, either downplaying or overstating the importance of elections. Still, the consistency of results across such different countries suggests the demand is real. Explore more data on free and fair elections for all countries → Continue reading September 17 Californians now travel millions of miles each month in driverless taxis After only two years, California’s driverless taxis now transport passengers for more than four million miles per month. Although they still make up only a fraction of taxi trips in the state, they are expanding quickly. This chart shows the monthly distance traveled in driverless trips in California. It measures the total number of passenger-miles, summing up the distance traveled by all passengers. In August 2023, California regulators fully approved self-driving taxi services in San Francisco for companies Cruise and Waymo. However, Cruise stopped operating in late 2023 due to safety and regulatory issues, so the recent growth reflects only Waymo’s service. Trips stayed under half a million miles per month until mid-2024. But since then, growth has taken off. Within a year, usage multiplied eightfold, climbing past four million miles by May 2025, the latest data available. This is a new chart on Our World in Data — we will update it every quarter based on the latest reports → Continue reading September 15 China's use of fertilizers peaked a decade ago Since the 1960s, China’s population has more than doubled . Despite having more than twice as many mouths to feed, the amount of food it produces per person has increased dramatically . Better seeds, irrigation, pest management, and improved farming techniques have all helped increase the country’s agricultural productivity. But the addition of nutrients through fertilizers has also made a huge difference. The chart shows the rapid uptake of fertilizers in China from the 1960s through the early 2000s. While fertilizers can play a crucial role in feeding more people and using less land, they also have negative environmental impacts. Excess nutrients run off into rivers and pollute coastlines, and fertilizers can emit nitrogen oxide , a powerful greenhouse gas. Using fertilizers more efficiently helps grow food while cutting pollution. China has made important progress on this in the last decade. As you can see in the chart, its fertilizer use peaked in 2014 and has fallen since then. At the same time, the country’s agricultural production has continued to increase . In 2015, China launched its “Zero-Growth Action Plan for Fertilizer”, and its government policies have played an essential role in this turnaround. Subsidies previously made fertilizers very cheap in China, which encouraged farmers to overuse them. Cutting these subsidies while offering incentives for agricultural machinery, precision technologies, farmer education , and larger farms (which tend to use less fertilizer per hectare) has made China’s farming sector much more efficient. Read more in my article “How effective are policies in reducing the environmental impacts of agriculture?” → Continue reading September 12 The latest World Bank data counts 125 million more people as living in extreme poverty — but the world has not gotten poorer To track progress towards ending extreme poverty, the United Nations relies on World Bank estimates of the number of people living below a poverty threshold called the “International Poverty Line” (IPL). In June 2025, the World Bank announced a major change to this line, raising it significantly, from $2.15 to $3 per day. As a result, 125 million people who would not have been counted as extremely poor before June are now included. The increased IPL and the higher poverty estimates are due to a mix of overlapping changes, which we explained in a recent article . Two things are particularly important to know. First, the higher estimates of extreme poverty reflect a higher poverty threshold, not that the world is poorer. In fact, the latest data shows that incomes among the world’s poorest are actually higher than previously estimated. Second, the overall message is the same whether we look at the new or previous estimates . Progress in recent decades has been enormous: well over a billion people have escaped extreme poverty since 1990. But this progress has now stalled. Incomes are stagnant in the places where most of the world’s poorest live. Unless this changes, hundreds of millions of people will be stuck in extreme poverty for years to come. Read our complete explainer on the new International Poverty Line and World Bank poverty data → Continue reading Get Data Insights delivered to your inbox Receive an email from us when we publish a Data Insight (every few days). By subscribing you are agreeing to the terms of our privacy policy. Our World in Data is free and accessible for everyone. Help us do this work by making a donation. Donate now
16560	https://courses.edx.org/assets/courseware/v1/91afa73b72b2e218b0673b0578e9d6d1/asset-v1:ImperialX+alm003+3T2019+type@asset+block/ALM003-Algebra-and-Functions-Exercise-2.pdf	Algebra and Functions Exercise 2: Composite Functions 1. Break the functions down into two or more components: a. Identify the order in which the operations “square”, “multiply by 2” and “subtract 3” are applied to 𝑥 in the function f: 𝑥↦2(𝑥−3)2, 𝑥∈ℝ b. Identify the order in which the operations “square”, “subtract 2”, “add 4” and “take the square root” are applied to 𝑥 in the function f: 𝑥↦+√(𝑥−2)2 + 4, 𝑥∈ℝ 2. The function f is defined by f: 𝑥↦3𝑥−2 for 𝑥∈ℝ and the function g is defined by g ∶𝑥↦𝑥2 + 1 for 𝑥∈ℝ . Evaluate: a. gf(2) b. fg(−1) 3. The functions f is defined by f(𝑥) = 1 𝑥−2 for 𝑥∈ℝ, 𝑥≠2 The function g is defined by g(𝑥) = 𝑥2 + 2 for 𝑥∈ℝ a. Which of the following gives the composite function fg in terms of 𝑥 along with its correct domain and range? A. fg(𝑥) = 1 𝑥2 𝑥∈ℝ fg(𝑥) > 0 B. fg(𝑥) = (𝑥−2)2 + 2 𝑥∈ℝ fg(𝑥) > 0 C. fg(𝑥) = 1 𝑥2 𝑥∈ℝ, 𝑥≠0 fg(𝑥) > 0 b. Which of the following gives the composite function gf in terms of 𝑥 along with its correct domain and range? A. gf(𝑥) = 1 (𝑥−2)2 + 2 𝑥∈ℝ, 𝑥≠2 gf(𝑥) > 0 B. gf(𝑥) = (𝑥−2)2 + 2 𝑥∈ℝ gf(𝑥) > 0 C. gf(𝑥) = 1 𝑥2 𝑥∈ℝ, 𝑥≠0 gf(𝑥) > 0 4. The function f is defined by f ∶𝑥↦𝑒 𝑥 2 −3 for domain 𝑥∈ℝ. The function g is defined by g ∶𝑥↦ln(𝑥−3) for domain 𝑥∈ℝ, 𝑥> 3. Evaluate fg(4). 5. Given the functions f and g such that f(𝑥) = 2(𝑥+ 1) for 𝑥∈ℝ and g(𝑥) = 𝑥2 −5 for 𝑥∈ℝ and that the equation gf(𝑥) −2f(𝑥) = 𝑘 (where 𝑘 is a constant) has no real roots, find the range of values that 𝑘 can take. 6. The function f is defined by f ∶𝑥↦ 𝑥+2 𝑥−1 for 𝑥∈ℝ, 𝑥≠1 and the function g is defined by g ∶𝑥↦2𝑥−3 for 𝑥∈ℝ. Solve the equation gf(𝑥) = 0. 7. If f(𝑥) = 𝑘𝑥+ 2 for 𝑥∈ℝ and (𝑥) = 𝑥−3𝑘 for 𝑥∈ℝ, find fg(𝑥). Given that fg(7) = 4, find two possible values for 𝑘. 8. If f(𝑥) = 𝑎𝑥+ 𝑏 and ff(𝑥) = 9𝑥+ 8 find possible values of 𝑎 and 𝑏.
16561	https://www.quora.com/How-do-I-use-formulas-in-Excel-to-calculate-a-remaining-balance-I-have-a-price-column-and-a-purchased-column-with-yes-and-no-marked-in-each-cell-and-would-like-a-remaining-balance-that-dis-includes-the-items-already	Something went wrong. Wait a moment and try again. Data Analysis Remaining Balance Conditional Formatting (M... Spreadsheet Programs Business Calculations Data Management Microsoft Excel Excel Functions 5 How do I use formulas in Excel to calculate a remaining balance? I have a price column and a purchased column with ‘yes’ and ‘no’ marked in each cell and would like a remaining balance that dis-includes the items already purchased. · To calculate a remaining balance in Excel based on your price column and a purchased column marked with 'yes' or 'no', you can use the SUMIF function. Here’s how to set it up step by step: Step-by-Step Guide Set Up Your Data: Assume your data is structured like this: Column A: Item Name Column B: Price Column C: Purchased (with 'yes' or 'no') \| A \| B \| C \| --------- \| Item 1 \| 10 \| yes \| \| Item 2 \| 20 \| no \| \| Item 3 \| 30 \| yes \| \| Item 4 \| 40 \| no \| Use the SUMIF Formula: To calculate the total pric To calculate a remaining balance in Excel based on your price column and a purchased column marked with 'yes' or 'no', you can use the SUMIF function. Here’s how to set it up step by step: Step-by-Step Guide Set Up Your Data: Assume your data is structured like this: Column A: Item Name Column B: Price Column C: Purchased (with 'yes' or 'no') \| A \| B \| C \| --------- \| Item 1 \| 10 \| yes \| \| Item 2 \| 20 \| no \| \| Item 3 \| 30 \| yes \| \| Item 4 \| 40 \| no \| Use the SUMIF Formula: To calculate the total price of items that have not been purchased, you can use the following formula: excel =SUMIF(C:C, "no", B:B) This formula works as follows: C:C is the range where you check for 'yes' or 'no'. "no" is the criteria; it tells Excel to sum only the prices for items that have 'no' in the purchased column. B:B is the range from which you want to sum the prices. Place the Formula: Choose a cell where you want to display the remaining balance (for example, cell D1) and enter the formula. Example Calculation Using the example data provided: Purchased Items: Item 1: 10 (yes) - Item 3: $30 (yes) - Not Purchased Items: - Item 2: $20 (no) - Item 4: $40 (no) ### Result - The formula =SUMIF(C:C, "no", B:B) will calculate: $20 (Item 2) + $40 (Item 4) = $60 So the remaining balance displayed in cell D1 will be 60. Additional Tips If you have headers in your columns, you might want to adjust the ranges to exclude them, e.g., B2:B100 and C2:C100, depending on your data size. You can also use the SUMIFS function if you want to add more criteria in the future. Feel free to ask if you need further assistance or examples! Tom Smith IT Professional (1979–present) · Author has 678 answers and 1M answer views · 5y If I understand what you are looking for, you have a list of items with associated price and a column that indicates if it was purchased. What you want is the difference between the total cost and the balance of the non-purchased items. For clarity, I am including the total of purchased items as well. So if you are looking for this: The formulas would look like this: If you want to sum the non-purchased items, just change the “=Yes” to “=No” Hope that helps. If I understand what you are looking for, you have a list of items with associated price and a column that indicates if it was purchased. What you want is the difference between the total cost and the balance of the non-purchased items. For clarity, I am including the total of purchased items as well. So if you are looking for this: The formulas would look like this: If you want to sum the non-purchased items, just change the “=Yes” to “=No” Hope that helps. Udo Boesenberg Data Analyst for many years and Excel addict. · 5y Hi, The answer sits in the use of the “=SUMIF” formula. If your amounts sit in Col A from row 1:15, and the “YES” OR “NO” sit in Col B from row 1:15, you would want to add all the numbers that have NOT purchased (“NO” values). Thus your Formula would be: =SUMIF(B1:B15,”NO”,A1:A15). This Formula says: “If B1:B15 contain “NO”, Sum the adjacent col A1:A15. Hope this does it. Related questions What formula can I use in Excel to add payments, reduce the received goods and get the balance in one row? How do I copy a formula down a column in Excel? How do I apply a formula to an entire column in Excel? How can I get the corresponding value of column B, column C and column D in Excel’s sheet 1 from sheet 2 with an Excel formula? The values in column A of both the sheets are matching but not serial-wise. How do I carry a formula down a column in Excel? M. Tithveasna Accountant (2013–present) · 5y Assume you have data layout structure:column A is Items,Column B is price and Column C contains Yes and No =SUM(B2:B100)-SUMIF(C2:C100,”Yes”,B2:B100) you can also use =SUMIF(C2:C100,”No”,B2:B100) Pat Ryan Daily MS Office user. Excel, Access, Powerpoint and others · Author has 2.1K answers and 5.8M answer views · 6y Related How do I make Excel subtract from the previous cell in a row anytime I enter a new value? How do I make Excel subtract from the previous cell in a row anytime I enter a new value? So you are entering values horizontally in a row. Each time you enter a new one, you want to subtract it from the previous entry and you want to record that result somewhere else. If I’ve understood that properly, then this is how I would do it: If you have two values in B2 and C2, the basic operation is =B2-C2 and I think it makes most sense to put that directly underneath, in C3. To copy that function across, click the tiny black square at the lower right corner of the selected cell and drag that across for How do I make Excel subtract from the previous cell in a row anytime I enter a new value? So you are entering values horizontally in a row. Each time you enter a new one, you want to subtract it from the previous entry and you want to record that result somewhere else. If I’ve understood that properly, then this is how I would do it: If you have two values in B2 and C2, the basic operation is =B2-C2 and I think it makes most sense to put that directly underneath, in C3. To copy that function across, click the tiny black square at the lower right corner of the selected cell and drag that across for as many columns as you think you’ll need (but you can extend it further later anyway.) However, if you do this, you will get negative results under the cells where you haven’t yet entered values because Excel will interpret the empty cell as zero. To get around this, place your basic function inside an IF function as shown. =IF(C2<>””,B2-C2,””) This says: If the value in C2 is not equal to empty, then give me the result of B2 minus C2. Otherwise, just give me an empty cell. When you drag that function across as many columns as you need, Excel will give you the answers that you want, but will show empty cells if you don’t yet have values above. In the example shown, cells G3 to J3 look empty, but they actually contain the same function. It’s just that in these cases, the result of the function is “empty”. So in this example 80–60=20, 60–55=5, etc. Note that if you subtract a larger number from a smaller one, you get a negative result. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. 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If the paid vs. unpaid is in the second ROW, it sounds like your data looks like this: In all the many years I’ve used excel, I’ve never built a sumifs that took criteria moving horizontally, but there’s no rule that says you can’t! The formulas would be as you can see below: And the result of the formulas are in A4 AND B4: The top answer is essentially the same. To get the original data moving vertically, copy the data and right click - “transpose”. You’ll then have the data stacked vertically. New formulas: New result: If the paid vs. unpaid is in the second ROW, it sounds like your data looks like this: In all the many years I’ve used excel, I’ve never built a sumifs that took criteria moving horizontally, but there’s no rule that says you can’t! The formulas would be as you can see below: And the result of the formulas are in A4 AND B4: The top answer is essentially the same. To get the original data moving vertically, copy the data and right click - “transpose”. You’ll then have the data stacked vertically. New formulas: New result: Related questions Can I add a column in Excel with a formula? How do I make an Excel spreadsheet with a formula deducting the capital in the first cell and the actual value of the first column and balance in the second column? What is the Excel formula to add column A to column B putting results on column C row by row? How do I create an Excel spreadsheet with formulas? How can I insert a row in Excel and keep formulas? John I use Excel extensively as a data base management tool. · Author has 16.7K answers and 70.6M answer views · 9y Related Is there a formula in Excel that returns the cell value of a table, by finding the column and row that meets certain criteria? This is a classic case for “3 dimensional” INDEX(MATCH). I’ve reproduced your sheet, here: To start, establish 3 named arrays by highlighting the cells to be included in the array, and then naming them: Yellow fill: months Blue fill: ethnics Red outline: enrollmentdata To name the arrays, after selecting the cells to be included, simply type the name in the box at left above the spreadsheet - next to the area where formulas are entered: We have also set up, on rows above your data, two drop down lists, as shown here: The Month drop down has been selected, and within the selection we have chosen Jun. This is a classic case for “3 dimensional” INDEX(MATCH). I’ve reproduced your sheet, here: To start, establish 3 named arrays by highlighting the cells to be included in the array, and then naming them: Yellow fill: months Blue fill: ethnics Red outline: enrollmentdata To name the arrays, after selecting the cells to be included, simply type the name in the box at left above the spreadsheet - next to the area where formulas are entered: We have also set up, on rows above your data, two drop down lists, as shown here: The Month drop down has been selected, and within the selection we have chosen Jun. The choices in the drop down are based on the previously created named arrays. To set it up, we click the cell where the drop down is to be placed, then, on the Data Tab, click Data Validation: A dialog box like this one appears: With the Settings selection tab chosen, enter in the Source box an = sign followed by the array whose cells you want to appear in the drop down list. You may also press F3 on the keyboard to see a list of the named arrays, from which you may select the one you want. Same thing with the drop down for Ethnicity. Using the values in the drop down list assures you won’t have any problems with typos that you might have if you simply typed in the month or ethnicity manually. Now we’re ready to find the enrollment data you want, based on selecting an ethnicity and a month. The formula is found in the cell to the right of Enrolled. Here it is: =INDEX(enrollmentdata,MATCH(E10,ethnics,0),MATCH(E9,months,0)) (0 is a zero) Let’s start from inside the formula: MATCH(E10,ethnics,0), which may be generalized to: MATCH(value,source,0). This says to find a certain value within a certain source, and return to the rest of the formula the row number on which it is found within the source (not the row number of the spreadsheet). Thus, in the actual formula of MATCH(E10,ethnics,0): E1o: is the cell where the value to be matched is found. In this case, E10 is the value you selected from the drop down list in cell E10, which is, we’ll say, “Hispanic.” ethnics (blue cells) is the name of the array in which to search for the value in E10 (“Hispanic”). 0 (zero) tells the formula to find only an exact match - not something that is just “close.” The formula returns the row number within the array on which “Hispanic” is found within the array. In this case, “Hispanic is the third item in the array, ethnics. So now, the formula has told us the vertical location of the desired value. Next, we need to find the horizontal position on that same row. This where the INDEX part takes control: INDEX(enrollmentdata,MATCH(E10,ethnics,0),MATCH(E9,months,0)) Let’s revise it to show what INDEX actually sees, so far: INDEX(enrollmentdata,3),MATCH(E9,months,0)), where: enrollmentdata: is the named array (red outline) of the enrollment numbers. 3: is the row within the array in which the desired enrollment number is to be found MATCH(E9,months,0): returns the column number (horizontal position) of the data, where: E9: is the cell containing the month to be found (from the drop down list in E9) months: is the name of the array in which the month is to be matched 0 (zero): means exact match. The MATCH formula in this case will return the number 6, because Jun is the 6th item in the array. Note that, in this case, the months array itself is horizontal, so MATCH will operate left to right, working across the array, not top to bottom. Now, here is what the formula sees: =INDEX(enrollmentdata,3,6). You could, in fact, enter the formula anywhere on the spreadsheet in exactly that way, and get the correct result What the formula says is: “In the enrollmentdata array, go down to the 3rd row, across to the 6th column, and return the value you find at that intersection. That number is 4 - the number of Hispanic enrollees in June. If you would like the actual spreadsheet to “play with,” message me and tell me where to send it. Hope that helps. Sponsored by CDW Corporation Looking for ways to leverage AI to improve customer experiences? CDW’s AI solutions include 24/7 chatbots and workshops to enhance efficiency and customer satisfaction. Brad Yundt Mechanical engineer and Excel aficionado · Author has 3.6K answers and 18.4M answer views · 2y Related How do I get a product price based on three column values in another table in Excel? Sounds like you need to match three criteria to find the right row in a lookup table. You can do that with XLOOKUP, INDEX & MATCH, LOOKUP, SUMPRODUCT, SUMIFS or FILTER. Absent other information, that’s my order of preference between them. Suppose your product price is in column D. You need column A to match a value of 10, column B to match a value of 20 and column C to match a value of 30. You can find the price corresponding to the first match using the first two formulas. below. The LOOKUP formula returns the last match, which is not a big deal if there only one such match. Note that XLOOKUP r Sounds like you need to match three criteria to find the right row in a lookup table. You can do that with XLOOKUP, INDEX & MATCH, LOOKUP, SUMPRODUCT, SUMIFS or FILTER. Absent other information, that’s my order of preference between them. Suppose your product price is in column D. You need column A to match a value of 10, column B to match a value of 20 and column C to match a value of 30. You can find the price corresponding to the first match using the first two formulas. below. The LOOKUP formula returns the last match, which is not a big deal if there only one such match. Note that XLOOKUP requires Excel 2021 or Microsoft 365. The INDEX & MATCH formula must be array-entered (Control + Shift + Enter) in Excel 2019 and earlier. =XLOOKUP(1,(A2:A1000=10)(B2:B1000=20)(C2:C1000=30),D2:D1000,"Not found") =IFNA(INDEX(D2:D1000,MATCH(1,(A2:A1000=10)(B2:B1000=20)(C2:C1000=30),0)),"Not found") =IFNA(LOOKUP(1,1/((A2:A1000=10)(B2:B1000=20)(C2:C1000=30)),D2:D1000),"Not found") SUMPRODUCT and SUMIFS only work if the value to be returned is numeric and there is only one such match. That’s because they add up the values in column D for each matching row. The formulas return 0 if there is no match. If you need a greater than or not equal criteria, put it inside quotes such as “>=” & F2, “<10” or “<>” & F2. =SUMPRODUCT((A2:A1000=10)(B2:B1000=20)(C2:C1000=30),D2:D1000) =SUMIFS(D2:D1000,A2:A1000,10,B2:B1000,20,C2:C1000,30) The FILTER function returns all matches. Preface with an at sign if you want only the first match, as shown below. Note that FILTER requires Excel 2021 or Microsoft 365. =@FILTER(D2:D1000,(A2:A1000=10)(B2:B1000=20)(C2:C1000=30),"Not found") Ted Pettit Former A/C Repair and Screen Door Sales · Author has 1.1K answers and 6M answer views · 4y Related What is the Excel formula to add totals for current month, last month, current year, and last year, on another sheet in the same workbook? One column is just a count of orders, and the other column is a sum of the orders. Here you go. My dummy data: My Results: Formulas Last Year Count: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())-1))This Year Count: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())))Last Month Count: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())),--(MONTH(Sheet1!B2:B16)=MONTH(NOW())-1))This Month Count: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())),--(MONTH(Sheet1!B2:B16)=MONTH(NOW())))Last Year Sum: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())-1),Sheet1!C2:C16)This Year Sum: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())),Sheet1!C2:C16)Last Month Sum: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())), Here you go. My dummy data: My Results: Formulas Last Year Count: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())-1))This Year Count: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())))Last Month Count: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())),--(MONTH(Sheet1!B2:B16)=MONTH(NOW())-1))This Month Count: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())),--(MONTH(Sheet1!B2:B16)=MONTH(NOW())))Last Year Sum: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())-1),Sheet1!C2:C16)This Year Sum: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())),Sheet1!C2:C16)Last Month Sum: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())),--(MONTH(Sheet1!B2:B16)=MONTH(NOW())-1),Sheet1!C2:C16)This Month Sum: =SUMPRODUCT(--(YEAR(Sheet1!B2:B16)=YEAR(NOW())),--(MONTH(Sheet1!B2:B16)=MONTH(NOW())),Sheet1!C2:C16) Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Tom Smith IT Professional (1979–present) · Author has 678 answers and 1M answer views · 4y Related What is the Excel formula to add totals for current month, last month, current year, and last year, on another sheet in the same workbook? One column is just a count of orders, and the other column is a sum of the orders. There are a lot of things that need to be asked to give you a good formula. Let me start with a few assumptions that I am drawing from your description: you have a worksheet for each month on each monthly worksheet you have: the totals from the month before the totals from the current month a running total for the current year the name of the worksheet is the month on the January worksheet, the totals from the previous year are listed Let me pause and make sure you understand the syntax to reference a cell on another worksheet. You can either reference a cell directly or, as Ryan Howe indicated, you ca There are a lot of things that need to be asked to give you a good formula. Let me start with a few assumptions that I am drawing from your description: you have a worksheet for each month on each monthly worksheet you have: the totals from the month before the totals from the current month a running total for the current year the name of the worksheet is the month on the January worksheet, the totals from the previous year are listed Let me pause and make sure you understand the syntax to reference a cell on another worksheet. You can either reference a cell directly or, as Ryan Howe indicated, you can create a lot of named cells, Generically, the syntax to reference cell A1 on a specific worksheet is: ‘worksheet name’!A1 So, to answer your question about the formula generically first. To get the sum of the monthly totals it would be, and I am using arbitrary cells: For monthly count of orders: =’last month’!D5 + ‘this month’!D100 For monthly total of orders: =’last month’!G5 + ‘this month’!G100 For the annual count of orders: =’January’!D2 + ‘this month’!D101 For the annual total of orders: =’January’!G2 + ‘this month’!G101 Now, for the thing you didn’t ask: how can you know what the last month folder is? As in most everything, there are several ways to do this. The simplest would be to have a field at the top of your form indicating the start of the period for the report/data, i.e the date of the first of the month. You can use a formula like: =EOMONTH(date,-2)+1 If the date for the page was in cell B1 then this would look like: =EOMONTH(B1,-2)+1 Then you could derive the month name, and therefore the previous worksheet name as follows: =TEXT(EOMONTH(B1,-2)+1,"mmmm") So to get this: We would use: But, out assumption was that all the monthly data would be on this month’s page, so as indicated, the formula used would be more like: =D5 + D100 But, to use the month of January as the filename, or the previous month’s name as the worksheet name we would use: =INDIRECT(TEXT(EOMONTH(B1,-2)+1,"mmmm")&"!D5") or =INDIRECT(“January”&"!D1") or =’January’!D1 And that is an elementary tutorial on cell referencing and simple date manipulation. If you are looking for something more specific, please provide some details on how your data and worksheets are set up. Brad Yundt Mechanical engineer and Excel aficionado · Author has 3.6K answers and 18.4M answer views · 6y Related How do I carry a formula down a column in Excel? Dragging a formula down a column can be so tedious when you need to go many rows. Fortunately, there are a couple of tricks. The easiest way is to select the first cell with the formula, then doubleclick the little green square at the bottom right corner of the selection marquee. The formula will be copied down until the first blank cell in the column to the left. Another easy way to copy a formula down uses the address bar, the rectangular field just above the intersection of the row numbers and column letters. Put your formula in the first cell Type the address range where you need formulas in t Dragging a formula down a column can be so tedious when you need to go many rows. Fortunately, there are a couple of tricks. The easiest way is to select the first cell with the formula, then doubleclick the little green square at the bottom right corner of the selection marquee. The formula will be copied down until the first blank cell in the column to the left. Another easy way to copy a formula down uses the address bar, the rectangular field just above the intersection of the row numbers and column letters. Put your formula in the first cell Type the address range where you need formulas in the address bar (including the one with the formula), then hit Enter. The screenshot below shows just before hitting Enter. Click in the formula bar, then CTRL + Enter Brad Yundt Mechanical engineer and Excel aficionado · Author has 3.6K answers and 18.4M answer views · 4y Related Why do formulas on my excel sheet not auto-apply to all rows and columns? Why don’t Excel formulas auto-apply to all rows and columns? While it is undeniably more convenient for you to write formulas that way, the consequence might be an annoying delay for recalculation each time you hit Enter. Excel allows 1048576 rows and 16384 columns. That’s over 17 billion cells. If Excel had to calculate a formula over all of them, you should be prepared to wait minutes after each time you hit Enter—even on a powerful workstation. If you want a formula to apply to all the cells in a column, refer to the range without row numbers such as B:B. If you want a formula to apply to all Why don’t Excel formulas auto-apply to all rows and columns? While it is undeniably more convenient for you to write formulas that way, the consequence might be an annoying delay for recalculation each time you hit Enter. Excel allows 1048576 rows and 16384 columns. That’s over 17 billion cells. If Excel had to calculate a formula over all of them, you should be prepared to wait minutes after each time you hit Enter—even on a powerful workstation. If you want a formula to apply to all the cells in a column, refer to the range without row numbers such as B:B. If you want a formula to apply to all columns, refer to the range without column letters such as 23:23. Because Excel has many more rows than columns, using B:B in a formula increases recalc time 64-fold compared to using 23:23. And for simple worksheets, the recalc delay for either way of defining ranges in formulas might be quite acceptable because modern computers are so fast and Excel is so clever. Excel became the market leading spreadsheet in large part because it recalculated much faster than Lotus 1–2–3. Lotus recalculated every cell every time a value changed. For spreadsheets with more than modest complexity, I soon learned to keep recalculation off and /PPAG to recalculate only when necessary. Excel, however, could be kept on automatic recalculation without noticeable lag on those same workbooks. It did so by maintaining lists of dependent and precedent cells so it only recalculated the cells that needed it. The trick was to use a calculation tree that told Excel which cells needed to be recalculated and in which order based on the latest change to the workbook cell values. Excel has other tricks for speeding up recalculation. Some functions (COUNTIF, SUMIFS and SUMPRODUCT for example) limit their attention to the cells in the used range of the worksheet. While Excel sometimes loses track of the extent of the used range (and goes much bigger), it is supposed to be the rectangular grid of cells that include all of your data. Users get annoyed if the recalc delay is longer than the blink of an eye, so limiting the recalc to the used range of cells makes Excel responsive. Lookup functions on the newest versions of Excel (Microsoft 365 and Excel 2019) index the lookup table so Excel know which row to go to without needing to test half the rows on average like older versions. That change can make VLOOKUP formulas over a thousand times faster when the lookup table spans an entire worksheet. PivotTables are another trick for speeding up recalculation. They eliminate the need for formulas, and let you prepare summary reports quickly as you slice your data using row and column filters. If you change the supporting data, you don’t see a recalc delay even with the gnarliest set of PivotTables. That’s because the PivotTables do not recalculate with the newest data until you rightclick and refresh them. Let’s revert to your question about making formulas auto-apply to a range of cells. Microsoft 365 has a new feature called dynamic arrays that does just that. With it, you put a formula in a single cell and it can give you a column, row or rectangular range of results automatically. UNIQUE(B:B) can return the 348 unique Chinese family names from the million names listed in column B. FILTER(A2:G1000,B2:B1000>100) returns a 7 column by n row list of values where column B in each row is more than a hundred. Gale Culbertson Author has 472 answers and 593.7K answer views · 2y Related How do you enter formulas into cells on an Excel spreadsheet? The formulas you might enter depends on what you are trying to do. That seems obvious, right? I am looking at Google Sheets right now, but the same applies to Excel, or LibreOffice, or OpenOffice. If you click on the formula icon (highlighted in green below, but in different locations with different programs), Google offers a very long list of types of formulas, of which Math is only one. The nature of your question seems to imply you are not ready for any of these, so I’ll give you some basic info. Any formula in Microsoft (MS) Excel (or Google Sheets, or LibreOffice Calc, or OpenOffice Calc …) The formulas you might enter depends on what you are trying to do. That seems obvious, right? I am looking at Google Sheets right now, but the same applies to Excel, or LibreOffice, or OpenOffice. If you click on the formula icon (highlighted in green below, but in different locations with different programs), Google offers a very long list of types of formulas, of which Math is only one. The nature of your question seems to imply you are not ready for any of these, so I’ll give you some basic info. Any formula in Microsoft (MS) Excel (or Google Sheets, or LibreOffice Calc, or OpenOffice Calc …) begins with an equal sign (=). To add two numbers you might enter the formula =3+5. When you press Enter, you see the result of your calculation: in this case, 8. You can also subtract (-), multiply (), divide (/), raise to powers (^), etc. Why bother, when even your phone has a decent calculator? If all you want to know is a single calculation, I agree — use a simple tool. But often we want to do the same calculation many times or we want to do a simple calculation on a string of numbers. You have probably begun adding a string of numbers with your calculator, then — oops! You made a mistake. How do you correct your mistake? Usually, start over 😐. With Excel or any of the others, you can simply make a list of the numbers, then add them. It’s easy to change a number or check your work. Before we begin with some calculations, let’s look at how a spreadsheet is organized. Horizontal rows are numbered, always beginning with 1. Vertical columns have letter designations: A, B, C, etc. Cells are identified by the intersection of a Column and a Row, so the first cell is A1, the first row of the second column is B1, the second row of the first column is A2, etc. In the example below Cell B2 is selected, and it contains the numeric value 5. A group of cells, such as Cells B2 through C11, comprise a ‘range’, and this range in this case contains ‘data’, in this case numeric data. Cell D2 contains the formula: =B2+C2. It adds the contents of these two cells, and displays the sum: 6. When we select Cell D2, the formula shows in the ‘formula bar’ to the right of the fx symbol. By double-clicking in D2, the formula is shown in the cell as well as in the formula bar. Going back to the first example, what do you suppose Cells D3 through D11 contain? That’s right, the formulae to add the contents of Cells B3+C3, B4+C4, etc. The interesting part is you only had to enter the formula D2, then pick the fill handle (the small blue square in the bottom right corner of the selected cell, then drag it down to Cell D11 without lifting the mouse button until you arrive. The formula is copied. Cool, huh? But wait! There’s more! Suppose you wanted to total all of the cells from B2:B11, and C2:C11, and maybe even D2:D11. There are several ways to do that. Following the example above, you could enter this formula in Cell B13: =B2+B3+B4+ … B11. If you did, you would see 57 in Cell B13. This formula used only the addition arithmetic operator (+). You could also use the SUM function, like this, entered in say, Cell B14: =sum(B2:B11) Press Enter, and see 57 in both Cells C13 and C14 — one the result of arithmetic operators and one the result of the SUM function. As when you used the fill handle to copy or fill Cells D3:D11 with the formula from D2, you could use the fill handle to fill the B13 and B14 formulas to C13:C14 and D13:D14. To see the sum of Cells B2:C11, you could use the SUM function again, like this: =sum(B2:C11), perhaps in Cell D15. See that Cells D13, D14, and D15 all have the same value: 119, done three different ways. Now, suppose you needed to change one or more of the values in Cells B2:C11. Wouldn’t it be a nuisance task to recalculate everything you just completed using a calculator? But see how easy it is with a spreadsheet! And you can quickly check both your work and your formulas. Now look through the list of functions available (even in Google Sheets!). You may find many other functions of use, such as AVERAGE. This ends this short tutorial. If I had known what your need was, we could have explored that. However, with this small bit of information, YOU can explore the vast possibilities of spreadsheets. Note that you can use text (words), too, to describe the numbers. The logical functions allow your spreadsheet to choose between things to decide how to proceed. Have fun with it. If you want to ask more questions, please include more background information so more specific answers can be offered. And DON’T FORGET TO USE YOUR SPREADSHEET PROGRAM’S BUILT-IN HELP. It’s the fastest way to an answer. Related questions What formula can I use in Excel to add payments, reduce the received goods and get the balance in one row? How do I copy a formula down a column in Excel? How do I apply a formula to an entire column in Excel? How can I get the corresponding value of column B, column C and column D in Excel’s sheet 1 from sheet 2 with an Excel formula? The values in column A of both the sheets are matching but not serial-wise. How do I carry a formula down a column in Excel? Can I add a column in Excel with a formula? How do I make an Excel spreadsheet with a formula deducting the capital in the first cell and the actual value of the first column and balance in the second column? What is the Excel formula to add column A to column B putting results on column C row by row? How do I create an Excel spreadsheet with formulas? How can I insert a row in Excel and keep formulas? How do I add rows in Excel with a formula? How do you calculate the running balance in Excel? How do you add multiple rows in Excel without messing up formulas? How do I calculate the remaining mortgage balance in Excel? 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16562	https://arxiv.org/pdf/2307.16637	arXiv:2307.16637v3 [math.NT] 23 Jul 2024 INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS ALEKSANDR TUXANIDY † AND DANIEL PANARIO Abstract. It is proven that, in any given base, there are infinitely many palindromic numbers having at most six prime divisors, each relatively large. The work involves equidistribution estimates for the palindromes in residue classes to large moduli, offering upper bounds for moments and averages of certain products closely related to exponential sums over palindromes. Contents Introduction 12. Acknowledgments 63. Paper structure 74. Notation 75. Technical lemmas 96. Exponential sums over P0 b (x) 15 7. Bounding moments of Φ N 21 8. Bounding the average 26 9. Equidistribution estimate 30 10. Equidistribution with square moduli 33 11. Palindromic almost-primes 37 References 39 1. Introduction 1.1. Tattarrattat. Generally speaking, a palindrome is a finite sequence of objects which reads the same forwards as backwards. Any such sequence is said to be palindromic . For example the sequence of the (decimal) digits of the prime number 9609457639843489367549069 is a palindrome. According to SigmaGeek (see also Iwao ) it is the largest known palindromic prime appearing in the decimal expansion of the digits of π.One of the earliest historical palindromic objects, a SATOR square, was found in the remains of the ancient city of Pompeii (see for instance Sheldon ) destroyed in the year 79 AD after the eruption of Mount Vesuvius. Palindromes appear in several facets of human endeavor, for instance in music (e.g. Joseph Haydn’s Symphony No. 47, “The Palindrome”) in literature (e.g. Georges Perec’s “Le Grand Palindrome”). In recent decades, palindromes are studied intensively in various areas of research [3, 4, 5, 8, 9, 11, 14, 15, 18, 22, 34, 35]. They also appear in nature; for example, unusually large portions of the human X and Y chromosomes † Corresponding author. 12ALEKSANDR TUXANIDY AND DANIEL PANARIO are palindromic (in a slightly different sense; see Larionov-Loskutov-Ryadchenko ). 1.2. Conjecture and previous work. From here on out, b denotes an integer larger than 1. One says that a natural number n is b-palindromic , or is a b-palindrome , if the sequence of its b-adic digits is palindromic. When the context is clear we simply say that n is palindromic (or a palindrome). Palindromic numbers belong, more broadly, in the category of sequences and functions defined in terms of digit constraints in integer bases. These have been studied abundantly in several studies. We mention for example the work of Bour-gain and Swaenepoel on primes with prescribed digits, the recent work of Dartyge et al on reversible primes, Mauduit-Rivat on the sum of digits of primes, and Maynard on primes with missing digits. We refer the reader to Dartyge et al for several other interesting works and references therein. With regards to palindromic numbers, we mention for instance the work of Banks-Hart-Sakata showing that almost all palindromes are composite (see (1) below) Banks-Shparlinski on palindromes with many prime factors, Cilleruelo-Luca-Baxter showing that if b ≥ 5, every positive integer can be written as a sum of at most three b-palindromes, Col on the distribution of palindromes in residue classes and palindromic almost-primes (see Theorems 1.2, 1.3 and (2) below) and Irving giving a lower bound for the number of b-palindromic semiprimes in the interval ( b2, b 3) with b sufficiently large. On the topic of palindromic numbers, perhaps one of the most difficult unsolved problems is the following conjecture (see for instance the concluding remarks in Banks-Hart-Sakata and Banks-Shparlinski ). First for a real number x ≥ 1, we let Pb(x) be the set of all positive b-palindromic integers at most x in size. Conjecture 1.1. Let b ≥ 2 be an integer. Then there are infinitely many prime numbers that are b-palindromic. In fact {p ∈ Pb(x)} ≍ b Pb(x) log x for x large. Conjecture 1.1 seems far out of reach with current methods. As pointed out by Banks-Shparlinski , one of the main difficulties stems from the large sparsity of b-palindromic numbers, hindering the efficacy of analytic methods. The sparsity of these is, in essence, as large as that of numbers with form n2 + 1. However Landau’s problem (posed in the year 1912) of whether or not there are infinitely many such primes, remains unsolved. On the other hand, Iwaniec proved there are infinitely many integers n such that n2 + 1 has at most two primes factors. In view of recent results (see the discussion below and the following subsection) we believe that proving the analogous result, for b-palindromes, may be feasible over the upcoming years. Some progress was made in the direction of Conjecture 1.1. In the year 2004, Banks-Hart-Sakata showed that for x large enough, (1) # {p ∈ Pb(x)} ≪ b log log log x log log x #Pb(x).INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 3 In 2009, Col improved this to the expected (2) # {p ∈ Pb(x)} ≪ b Pb(x) log x . Col also obtained (see Theorem 1.2 below) a lower bound for the amount of b-palindromic almost-primes, when the number of prime divisors is bounded above by a certain value depending only on b. In the context of the infinitude of such palindromes, Theorem 1.2 was the first, and up to now unique, result of its kind. Here Ω( n) denotes the number of prime divisors of n, counted with multiplicity. Theorem 1.2 (Col [9, Corollary 2]) . We have {n ∈ Pb(x) : Ω( n) ≤ kb} ≫ b Pb(x) log x for some value kb > 0 depending only on b and satisfying kb ∼ 24 πb as b → ∞ . In particular k2 = 60 and k10 = 372 . Theorem 1.2 was a consequence of Col’s equidistribution estimate (see Theorem 1.3 below) for b-palindromes in residue classes to large moduli, combined with well-established facts from sieve theory. Here we use the notation Pb(x, a, q ) := {n ∈ Pb(x) : n ≡ a(q)} . Theorem 1.3 (Col [9, Theorem 2]) . There exists some value β > 0, depending only on b, such that for any ǫ, A > 0, ∑ q≤xβ−ǫ (q,b 3−b)=1 sup y≤x max a∈Z ∣∣∣∣#Pb(y, a, q ) − #Pb(y) q ∣∣∣∣ ≪ǫ,A,b Pb(x) log A x . For b large, we may take β ∼ 1/6πb . Setting βb = β − ǫ, the inequality above holds for the following particular values. b 2 3 4 5 6 7 8 9 10 βb 1/30 1/94 1/74 1/122 1/114 1/158 1/150 1/194 1/186 Theorem 1.3 (see also Proposition 6.2 here) considerably improved over previous estimates of Banks-Hart-Sakata valid for β ≪ (log log x)/ log x with the implied constant small enough (see Corollary 4.5 in ). 1.3. Statement of results. In the present work we obtain the following improved versions of Theorems 1.2 and 1.3. Here P −(n) denotes the smallest prime divisor of n. Theorem 1.4. Let b ≥ 2 be an integer. Then for x ≫b 1 large enough, { n ∈ Pb(x) : Ω( n) ≤ 6, P −(n) ≥ x1/21 } ≍b Pb(x) log x . Theorem 1.4 is a consequence of Theorem 1.5 below (discussed in the following subsections) facts from sieve theory and an analogue of Theorem 1.5 with square moduli. See Proposition 10.1 for this. The latter is an application of the Baier-Zhao estimate for the large sieve with square moduli, as well as the L∞-type bound of Col (see Proposition 6.2 here). In what follows P∗ b (x) := {n ∈ Pb(x) : (n, b 3 − b) = 1 }.4 ALEKSANDR TUXANIDY AND DANIEL PANARIO Theorem 1.5. Let b ≥ 2 be an integer and let x ≥ 1. Then for any ǫ > 0, (3) ∑ q≤x1/5−ǫ (q,b 3−b)=1 sup y≤x max a∈Z ∣∣∣∣∣∣∑ n∈P∗ b(y) ( 1n≡a(q) − 1 q )∣∣∣∣∣∣ ≪b,ǫ P∗ b (x) eσ(b,ǫ )√log x , where σ(b, ǫ ) > 0 is some value depending only on b and ǫ. We now discuss some of the main ingredients in the proof of Theorem 1.5. 1.4. Setup. For the sake of brevity we denote by E the left hand side of the inequality in Theorem 1.5. Let P0 b , P∗ b be defined as in (9), (10), respectively. Set P0 b (x) := P0 b ∩ [1 , x ]. Since every b-palindrome n ≥ 1 with ⌊log b n⌋ odd (or equivalently, with an even number of digits in its b-adic expansion) is divisible by b + 1, a divisor of b3 − b, it follows P∗ b = {n ∈ P0 b : (n, b 3 − b) = 1 } . Then by the M¨ obius inversion formula 1(n,b 3−b)=1 = ∑ r\|(n,b 3−b) μ(r), the Fourier expansions 1r\|n = 1 r ∑ k(r) er (nk )and (4) 1n≡a(q) = 1 q ∑ h(q) eq (−ah )eq(nh )(with the common notation em(α) := e(α/m ) with e(α) := e2πiα ) one derives E ≪b (log x) max k∈Z ∑ 2≤q≤x1/5−ǫ (q,b 3−b)=1 1 q ∗ ∑ h(q) sup y≤x ∣∣∣∣∣∣∑ n∈P0 b(y) e ( hn q + kn b3 − b )∣∣∣∣∣∣ . The asterisk above the h-sum signifies that h runs over the invertible residue classes modulo q.The use of the decomposition in (4) is rather typical in the literature and it will shortly lead to qualitative gains in multiplicative algebraic structures, with the appearance of the products Φ N defined in (16). Unfortunately we pay the price of a quantitative loss of roughly x1/5−ǫ in the trivial bound. Thus in order for such a Fourier-type approach to succeed, we must not only be able to recover the lost x1/5−ǫ, but gain much more (in fact gain an extra exp( σ√log x)). From the works of Banks-Hart-Sakata and Col (see Lemma 6.1 here) the inner sum admits a decomposition into linear combinations of the products Φ N with 0 ≤ N ≤ 1 2 log b x. After a dyadic subdivision of the interval 2 ≤ q ≤ x1/5−ǫ,the problem of bounding E then boils down to bounding averages of the form S = ∑ 2≤q≤Q (q,b 3−b)=1 ∗ ∑ h(q) ΦN ( h q + k b3 − b ) for Q ≪ x1/5−ǫ. The trivial bound gives S ≤ Q2bN ≤ Q2√x and our job is to beat this by much more than a factor of Q.INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 5 Thanks to the work of Col (see Proposition 6.2 here) we can readily dispose of the cases with Q relatively small, say Q ≪ exp( c√log x). We are thus left with considering S, or more strongly the average (5) T = ∑ q≤Q (q,b )=1 ∗ ∑ h(q) ΦN ( h q + β ) , for exp( c√log x) ≪ Q ≪ x1/5−ǫ and β ∈ R.As a first rough thought, one could envision splitting the product Φ N into two products, one over 1 ≤ n ≤ M and the other over M < n < N , applying Cauchy-Schwarz and end up considering averages of squares (just as Bombieri-Vinogradov’s treatment of the Type II sums; see Iwaniec-Kowalski ). These could be attacked say via a direct argument (see for instance Lemma 8.2) or the large sieve inequality (see Lemma 1.6 below). The direct argument (namely that of expanding the square, switching orders of summation, using the orthogonality of the additive characters and bounding the resulting sums trivially) can at best aid in recovering a Q from the trivial bound, and is thus insufficient. This brings us to the large sieve inequality. 1.5. Large sieve. First given a real number 0 < δ ≤ 1/2, one says that a set of points α1, . . . , α R ∈ R/Z is δ-spaced if ‖αr − αs‖ ≥ δ whenever r 6 = s, where ‖α‖ := min k∈Z \|α − k\|. Lemma 1.6 (Large sieve inequality) . For any set of δ-spaced points α1, . . . , α R ∈ R/Z and any complex numbers an with M < n ≤ M + N , where 0 < δ ≤ 1/2 and N ≥ 1 is an integer, we have (6) R ∑ r=1 ∣∣∣∣∣∣∑ M<n ≤M+N ane(αrn) ∣∣∣∣∣∣ 2 ≤ (δ−1 + N − 1) ∑ M<n ≤M+N \|an\|2. Proof. See for instance Theorem 7.7 in Iwaniec-Kowalski . The large sieve inequality is a powerful and useful tool for tackling averages of exponential sums and has enjoyed several applications in the literature (see for instance [13, 24, 29, 30, 40] to name a few). Unfortunately the quality of the bounds it offers highly depends on the density η = # S/N of the set S ⊆ (M, M + N ]in which the sequence ( an) is supported on, with better bounds for denser S and worse bounds for sparser S. For example in the case when each an ≪ 1 and R ≍ Q2 ≍ δ−1 ≍ N , the large sieve inequality beats the trivial bound by a factor of # S = ηN ≍ Qη √N . In particular in order for the large sieve to beat the trivial bound by much more than Q (which is what we seek) one needs η to be much larger than 1 /√N . Unfortunately sparse sets such as, say, the b-palindromes or close relatives, have density ≪ 1/√N on ( M, M + N ]. For sequences supported on such sets, direct use of the large sieve proves insufficient. An important idea for the case of the sparse sequences is to consider instead larger moments such as R ∑ r=1 ∣∣∣∣∣∣∑ M<n ≤M+N ane(αr n) ∣∣∣∣∣∣ 2K6ALEKSANDR TUXANIDY AND DANIEL PANARIO for some integer K ≥ 2; see for example Iwaniec-Kowalski . One may arrive at these from smaller moments via H¨ older’s inequality, say. To explain the concept, note we may write  ∑ M<n ≤M+N ane(αrn)  K = ∑ KM<n ≤K(M+N) bne(αr n), where the coefficients bn are now supported on the set KS comprised of the addi-tions of any K elements from S. If the set S is sufficiently dissociated additively, then KS is much denser in ( KM, K (M + N )]. In this case we can make a more effective use of the large sieve. 1.6. Approach. Let us now return to (5). Motivated by the discussion above, we split the product Φ N into three pieces as Φ N = P1P2P3, where P1 is the product over 1 ≤ n ≤ M , P2 is the product over M < n ≤ 2M and P3 is the product over 2 M < n < N (actually this is our approach for the case when Q ≫ b( 2 5−ǫ0)N ;the case Q ≪ b( 2 5−ǫ0)N is treated in a slightly different way). The choice of M is optimized later. Applying H¨ older’s inequality with the triple ( ℓ, ℓ, 2K) (where K ≥ 2 is an integer to be chosen later and ℓ > 2 is such that 2 /ℓ + 1 /2K = 1) one obtains T ≤ ∑ q≤Q (q,b )=1 ∗ ∑ h(q) P ℓ 1  1/ℓ ∑ q≤Q (q,b )=1 ∗ ∑ h(q) P ℓ 2  1/ℓ ∑ q≤Q (q,b )=1 ∗ ∑ h(q) P 2K 3  1/2K . For K somewhat large (although bounded in terms of some parameters for our purposes) ℓ is close to 2. The first two sums corresponding to P1, P 2 are then treated by us in direct fashion (as in Lemma 8.2). Here we avoid using the large sieve which, given the sparsity of the coefficient sequences corresponding to P1, P 2,gives no significant gains over the direct approach. There are two reasons why we have chosen to raise P3 to the 2 K-th power and not P1 nor P2, namely: 1. the following algebraic identity ∑ q≤Q (q,b )=1 ∗ ∑ h(q) P 2K 3 = ∑ q≤Q (q,b )=1 ∗ ∑ h(q) Φ2KN −2M ( h q + βb 2M ) ;and 2. the sequence of coefficients corresponding to Φ N −2M is denser than those cor-responding to P1, P 2. The trivial bound shows the above is at most Q2b2K(N −2M) ≤ Q2xK(1 − 2M N) and our main goal then reduces to beating this by much more than Q for certain suitable choices of M, K . This is accomplished via Proposition 7.1 giving upper bounds for moments of Φ N for any N ≥ 2. Its proof relies partly on the large sieve. 2. Acknowledgments We would like to express our gratitude to Qiang Wang for useful advice and to Igor E. Shparlinski for conversations involving the square moduli and notify-ing us of Col’s work . We are also deeply indebted to C´ ecile Dartyge, Hirotaka Kobayashi, Yuta Suzuki and Ryota Umezawa for pointing out one important omis-sion in an earlier draft of the work and for referring us to the books by Greaves INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 7 and Halberstam-Richert . Finally we thank the anonymous referees for several helpful suggestions. D. Panario was funded by the Natural Science and Engineering Research Council of Canada (NSERC), reference number RPGIN-2018-05328. 3. Paper structure The rest of the work goes as follows. In Section 4 we describe the notation used in the work. Most of it is typical in the literature, but we also introduce notation that is particular to us. In Section 5 we compile several technical tools needed in the following two sections. In Section 6 we discuss the current (but limited) state of knowledge on exponential sums over palindromes, as well as slightly add to the literature via Proposition 6.5 and Corollary 6.6 there. In Section 7 we prove Proposition 7.1 giving an upper bound for moments of Φ N . In Section 8 we prove Proposition 8.1 bounding an average of Φ N . In Section 9 we prove Theorem 1.5 giving average equidistribution estimates for palindromes in residue classes. In Section 10 we prove Proposition 10.1 giving equidistribution estimates for square moduli. We then conclude the work in Section 11 with a proof of Theorem 1.4. 4. Notation Here we describe the notations used in the work, some of which we already encountered; these are restated here for the sake of clarity. 4.1. Literary notation. We use the asymptotic notations X ≪ Y , Y ≫ X, X = O(Y ), all signifying here that \|X\| ≤ C\|Y \| for some unspecified constant C > 0. The notation X ≍ Y is used to assert that both X ≪ Y and Y ≪ X,simultaneously. Dependence of the implied constants on parameters will be denoted by a subscript. The symbol o(X) denotes a quantity satisfying o(X)/X → 0 as X → ∞ .The letters μ, τ, ϕ denote the M¨ obius, divisor and Euler’s totient functions, re-spectively. For an integer n ≥ 1, the symbol Ω( n) denotes the total number of prime factors of n, counting their multiplicity. The symbol P −(n) denotes the smallest prime factor of n. The letter p is reserved to denote a prime number. The notation n ≡ a(q) denotes the assertion that n is congruent to a modulo q. We use d \| n to assert that d divides n, and ( m, n ) to represent the greatest common divisor (GCD) of m and n. If ( m, n ) = 1, we use the notation ord m(n) for the multiplicative order of n modulo m.The symbol ∑ a(q) denotes a summation over all the residue classes a modulo q, whereas ∗ ∑ a(q) denotes a sum over the invertible residue classes a modulo q.Given a function f : R → C, we denote by ‖f ‖1 its L1-norm and by ‖f ‖∞ its L∞-norm (whenever these exist). For a real number x and an integer q ≥ 1, we use the notation e(x) := e2πix for the complex exponential, eq(x) := e(x/q ) for the additive character modulo q and 8 ALEKSANDR TUXANIDY AND DANIEL PANARIO exp( x) := ex for the exponential. If x > 0, its natural logarithm is denoted by log x and its base-b logarithm by log b x.We use ‖x‖ to symbolize the distance from x to its nearest integer; that is, ‖x‖ := min k∈Z \|x − k\|. Moreover ⌊x⌋ denotes the floor of x and {x} := x − ⌊ x⌋ its fractional part. We use the standard trigonometric notations csc( x) = 1 / sin( x) and cot( x) = cos( x) csc( x). Given a statement E, we use 1E to symbolize the indicator function of E; that is, 1E = 1 if E is true and 1E = 0 if E is false. For example, 1n≡a(q) equals 1 if n ≡ a(q) and is zero otherwise. Given real numbers α1, . . . , α N , their discrepancy DN (α1, . . . , α N ) is defined by (7) DN (α1, . . . , α N ) := sup 0≤c≤d≤1 ∣∣∣∣ {1 ≤ n ≤ N : c ≤ { αn} < d } N − (d − c) ∣∣∣∣ and their star-discrepancy D∗ N (α1, . . . , α N ) is defined as (8) D∗ N (α1, . . . , α N ) := sup 0≤d≤1 ∣∣∣∣ {1 ≤ n ≤ N : {αn} < d } N − d ∣∣∣∣ . 4.2. Non-standard notation. The letter b is reserved to denote an integer larger than 1. The notation Pb symbolizes the set of all positive b-palindromic integers. We define the sets P0 b , P∗ b , as follows: P0 b := {n ∈ Pb : ⌊log b(n)⌋ ≡ 0(2) } ,(9) P∗ b := {n ∈ Pb : (n, b 3 − b) = 1 } .(10) From the definition above one observes that P0 b is comprised exactly of all positive b-palindromes with an odd number of digits (in their b-adic expansion). Since every b-palindrome with an even number of digits is divisible by b+1, and b+1 is a divisor of b3 − b, it follows P∗ b ⊂ P0 b . In fact P0 b = ∞ ⋃ N=0 Πb(2 N ), P∗ b = ∞ ⋃ N=0 Π∗ b (2 N ), where for an integer N ≥ 0, Πb(N ) := Pb ∩ [bN , b N +1 ) ,(11) Π∗ b (N ) := {n ∈ Πb(N ) : (n, b 3 − b) = 1 } .(12) Given a set A of integers, a real number x and integers a, q , we define the sets A (x) := A ∩ { n ∈ Z : n ≤ x} , A (x, a, q ) := {n ∈ A (x) : n ≡ a(q)} . Thus for instance P∗ b (x) := P∗ b ∩ [1 , x ],(13) P∗ b (x, a, q ) := {n ∈ P∗ b (x) : n ≡ a(q)} ,(14) with our notation. INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 9 Given a real number α and an integer N ≥ 0, we define (15) φb(α) := ∣∣∣∣∣∣∑ 0≤m<b e(αm ) ∣∣∣∣∣∣ and (16) ΦN (α) := ∏ 1≤n<N φb (α (bn + b2N −n)) with the convention Φ N (α) = 1 for N ≤ 1. The definition of Φ N also depends on b but for the sake of brevity we omit it on the left hand side above. Finally given integers n, K, b with K, b ≥ 2, we let (17) r(n; K, b ) := # { (m1, . . . , m K ) ∈ ([0 , b ) ∩ Z)K : m1 + · · · + mK = n } be the number of additive compositions of n into K integers mj , each satisfying 0 ≤ mj < b .5. Technical lemmas In this section we go over several technical tools needed in the following two sections. The first two are classical in the theory of uniform distribution of sequences modulo 1. Lemma 5.1 (Koksma-Hlawka) . Suppose f : [0 , 1] → R is a bounded integrable function of bounded variation V (f ). Then for any real numbers α1, . . . , α N with N ≥ 1 an integer, ∣∣∣∣∣ 1 N N ∑ n=1 f ({αn}) − ∫ 10 f (x)dx ∣∣∣∣∣ ≤ V (f )D∗ N (α1, . . . , α N ), where D∗ N (α1, . . . , α N ) is their star-discrepancy as defined in (8). Proof. See . Lemma 5.2 (Erd˝ os-Tur´ an) . Let α1, . . . , α N be real numbers with N ≥ 1 an integer. Then DN (α1, . . . , α N ) ≪ 1 H + ∑ 1≤h≤H 1 h ∣∣∣∣∣ 1 N N ∑ n=1 e(hα n) ∣∣∣∣∣ for any H ≥ 1, where DN (α1, . . . , α N ) is the discrepancy defined in (7). Proof. See . For the following, one says that a function f : R → C is smooth if f is infinitely differentiable; that is, f has continuous derivatives of all orders. Lemma 5.3 (Smooth exponential sum bounds) . Let f : R → C be a smooth compactly supported function and let α be a real number. Then (18) ∣∣∣∣∣∑ n∈Z f (n)e(αn ) ∣∣∣∣∣ ≤ ‖ f ‖1 + ‖f ′‖1 210 ALEKSANDR TUXANIDY AND DANIEL PANARIO and (19) ∣∣∣∣∣∑ n∈Z f (n)e(αn ) ∣∣∣∣∣ ≤ ‖f (k)‖1 \|2 sin( πα )\|k for all natural numbers k ≥ 1.Proof. See for instance Lemma 3.1 in . The following lemma appears, either implicitly or explicitly, on several works on digital functions. See for instance [7, 9, 29, 30, 31]. We give a proof for the convenience of the reader. Lemma 5.4 (Ergodic-type integral bound) . Let f : R → R+0 be a bounded, locally integrable, 1-periodic function. Then for any integer N ≥ 1, ∫ 10 ∏ 0≤n<N f (αb n)dα ≤  sup 0≤θ≤1 1 b ∑ n(b) f ( n + θ b ) N . Proof. By induction on N ≥ 1. For N = 1, we have ∫ 10 f (α)dα = 1 b ∑ 0≤h<b ∫ 10 f ( h + θ b ) dθ and the claim follows. Suppose now that the claim holds for some N ≥ 1. We have ∫ 10 ∏ 0≤n<N +1 f (αb n)dα = 1 b ∑ 0≤h<b ∫ 10 ∏ 0≤n<N +1 f (( h + θ b ) bn ) dθ = ∫ 10 1 b ∑ 0≤h<b f ( h + θ b ) ∏ 0≤n<N f (θb n) dθ. The last equality holds since f is 1-periodic by assumption. The above is ≤ sup 0≤θ≤1 1 b ∑ 0≤h<b f ( h + θ b ) ∫ 10 ∏ 0≤n<N f (αb n) dα and the claim for N + 1 now follows from the inductive hypothesis. The following lemma gives asymptotics for the number of compositions with some restrictions. We were unable to obtain a reference in the literature and we thus give a proof. If one allows the error term to depend on n, one should be able to considerably improve it with a more detailed analysis of the integrals involved. Nevertheless the error term here suffices for our purposes. Lemma 5.5 (Compositions with restrictions) . Let n, K, b be integers with K, b ≥ 2 and let r(n; K, b ) be as defined in (17). We have r(n; K, b ) bK = √ 6 π(b2 − 1) K exp ( − 6 (b2 − 1) K ( n − (b − 1) K 2 )2) O ( 1 bK 3/2 ) . Proof. An application of the identity (20) 1ℓ=0 = ∫ 1/2 −1/2 e(ℓα )dα INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 11 valid for integers ℓ, yields r(n; K, b ) = ∫ 1/2 −1/2 e(−αn )  ∑ 0≤m<b e(αm )  K dα. In what follows δ denotes a fixed real number satisfying 1 /2 < δ < 1. We allow implied constants to depend on δ. To evaluate the integral above we split it into two pieces, one over \|α\| ≤ δ/b and the other over δ/b < \|α\| ≤ 1/2. We treat each separately. When α 6 ∈ Z, the sum inside the brackets above equals e(αb ) − 1 e(α) − 1 = e ( b − 1 2 α ) sin( παb ) sin( πα ) . By Euler’s product formula, sin( παb ) sin( πα ) = b ∞ ∏ m=1 ( 1 − α2b2 m2 ) ( 1 − α2 m2 )−1 . If \|α\| < 1/b , the logarithm of the product over m ≥ 1 equals (21) − ∑ m≥1 ∑ k≥1 (b2k − 1) α2k m2kk = − ∑ k≥1 ζ(2 k) k (b2k − 1) α2k , where ζ is the Riemann zeta function. For \|α\| ≤ δ/b , the terms with k ≥ 2contribute ≪ α4b4 to the sum. Then it follows ( sin( παb ) sin( πα ) )K = bK e−ζ(2)( b2−1) Kα 2 (1 + O (α4b4K)) for 0 < \|α\| ≤ δ/b . Thus, after dividing by bK ,1 bK ∫ δ/b −δ/b e(−αn )  ∑ 0≤m<b e(αm )  K dα = ∫ δ/b −δ/b e ( b − 1 2 αK − αn ) e−ζ(2)( b2−1) Kα 2 (1 + O (α4b4K)) dα = ∫ R e ( b − 1 2 αK − αn ) e−ζ(2)( b2−1) Kα 2 dα O ( b4K ∫ R e−ζ(2)( b2−1) Kα 2 α4dα + ∫ ∞ δ/b e−ζ(2)( b2−1) Kα 2 dα ) . The quantity inside of the O-brackets can be shown to be ≪ b−1K−3/2 after sub-stitutions of variables. By the well-known formula ∫ R e−at 2 e(−θt )dt = √ π a e−π2θ2/a (θ ∈ R, a > 0) for the Fourier transform of the Gaussian e−at 2 and the fact ζ(2) = π2/6, the integral outside of the O-brackets equals √ 6 π(b2 − 1) K exp ( − 6 (b2 − 1) K ( n − (b − 1) K 2 )2) .12 ALEKSANDR TUXANIDY AND DANIEL PANARIO To conclude it suffices to show that ∫ δ/b< \|α\|≤ 1/2 e(−αn )  ∑ 0≤m<b e(αm )  K dα ≪ bK−1 K3/2 . The sum inside the brackets is at most \| csc( πα )\| ≤ 1/2\|α\| (for \|α\| ≤ 1/2) in absolute value. Then the absolute value of the integral above is ≤ 21−K ∫ 1/2 δ/b α−K dα ≤ bK−1 (2 δ)K−1(K − 1) ≪ bK−1 K3/2 for δ > 1/2 fixed. We also need some trigonometric facts involving the function φb defined as in (15). It will be useful to note that φb(α) = { b if α ∈ Z, \| sin( παb )/ sin( πα )\| otherwise. Clearly φb is 1-periodic and even; thus φb(α) = φb({α}) = φb(‖α‖) for any real α.The following fact, but with ‖α‖ in the range ‖α‖2 ≤ 6/π 2(b2 − 1), appears in (see their Lemma 3). Lemma 5.6. If ‖α‖ ≤ 1/b , then φb(α) ≤ b exp ( − π2 6 (b2 − 1) ‖α‖2 ) . Proof. We have φb(0) = b and φb(1 /b ) = 0, whence the result holds for ‖α‖ ∈ {0, 1/b }. For 0 < ‖α‖ < 1/b , the claim follows from (21). Lemma 5.7. If α and 0 ≤ δ ≤ 2/3b are real numbers with ‖α‖ ≥ δ, then φb(α) ≤ φb(δ).Proof. See Lemma 5 in . Lemma 5.8. Let α, β, γ be real numbers. Then either ∥∥α (bβ + bγ+1 )∥ ∥ ≥ ‖α(b2 − 1) bγ ‖ b + 1 or ∥∥α (bβ+1 + bγ )∥ ∥ ≥ ‖α(b2 − 1) bγ ‖ b + 1 (or both). Moreover φb (α (bβ + bγ+1 )) φb (α (bβ+1 + bγ )) ≤ bφ b ( ‖α(b2 − 1) bγ ‖ b + 1 ) . Proof. We closely follow the argument in Lemma 6 of Mauduit-Rivat . Making the substitutions u = α(bβ + bγ+1 ) and δ = ‖α(b2 − 1) bγ ‖/(b + 1), we need to show that either ‖u‖ ≥ δ or ‖bu − α(b2 − 1) bγ ‖ ≥ δ, and that (22) φb(u)φb(bu − α(b2 − 1) bγ ) ≤ bφ b(δ).INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 13 Suppose ‖bu − α(b2 − 1) bγ ‖ < δ . The triangle inequality yields b‖u‖ ≥ ‖ bu ‖ = ∥∥bu − α(b2 − 1) bγ + α(b2 − 1) bγ ∥∥ ≥ ∥∥α(b2 − 1) bγ ∥∥ − ∥∥bu − α(b2 − 1) bγ ∥∥ ≥ (b + 1) δ − δ = bδ. Thus ‖u‖ ≥ δ. Consequently either ‖u‖ ≥ δ or ‖bu − α(b2 − 1) bγ ‖ ≥ δ, while (22) now follows from Lemma 5.7. The following lemma may be useful when considering exponential sums over multisets of the form S = {{ N∑ n=1 cnαn : 0 ≤ c1, . . . , c N < b }} for real numbers α1, . . . , α N with a sufficiently small star-discrepancy. This is due to the equality ∣∣∣∣∣∣∑ γ∈S e(γ) ∣∣∣∣∣∣ = N ∏ n=1 φb (αn) . Lemma 5.9 (Weyl product bound) . There exists an absolute constant A > 0 such that, for any real numbers α1, . . . , α N with N ≥ 1 an integer, N ∏ n=1 φb(αn) ≤ ( 2 D∗ N )AbN D ∗ N , where D∗ N = D∗ N (α1, . . . , α N ) is their star-discrepancy as defined in (8). Proof. We adapt an argument of Aistleitner et al (see the proof of Theorem 1 there). Let 0 < ǫ ≤ 1/2b to be chosen later. For a real number x, we define φb,ǫ (x) := { sin( πbǫ )/\| sin( πx )\| if \|x − k b \| < ǫ for some integer k 6 ≡ 0( b) φb(x) otherwise. Clearly φb,ǫ is 1-periodic and continuous satisfying φb,ǫ (x) ≍b,ǫ 1 and φb,ǫ (x) ≥ φb(x) for every x. In particular ∏ 1≤n≤N φb(αn) ≤ ∏ 1≤n≤N φb,ǫ (αn). Denoting by PN the product on the right hand side above, taking logarithms and applying the Koksma-Hlawka inequality (Lemma 5.1) one has (23) log PN N = 1 N ∑ 1≤n≤N log φb,ǫ (αn) ≤ ∫ 10 log φb,ǫ (x)dx + V (log ◦φb,ǫ )D∗ N , where V (log ◦φb,ǫ ) is the total variation of log ◦φb,ǫ on [0 , 1]. We first bound V (log ◦φb,ǫ ). By the definition of φb,ǫ , if x ∈ S := [0 , 1] − ⋃ 0<k<b ( k b − ǫ, k b + ǫ ) = [0 , ǫ ] ∪ [1 − ǫ, 1] ∪ ⋃ 0≤k<b [ k b + ǫ, k + 1 b − ǫ ] ,14 ALEKSANDR TUXANIDY AND DANIEL PANARIO then φb,ǫ (x) = φb(x) = \| sin( πbx )\|/ sin( πx ), which we interpret as b if x ∈ { 0, 1}.Over each of the intervals on the right hand side above, φb is continuously differ-entiable and non-vanishing. Since φb(x) = φb(‖x‖) and d dt log sin( t) = cot( t), it follows ∫ S ∣∣∣∣ d dx log φb,ǫ (x) ∣∣∣∣ dx ≤ 2π ∫ ǫ 0 \|b cot( πbx ) − cot( πx )\| dx + π ∑ 0≤k<b ∫ k+1 b−ǫ k b+ǫ (b\| cot( πbx )\| + \| cot( πx )\|) dx. Since cot( x) = x−1 + O(x) for 0 < \|x\| ≤ π/ 2 and ǫ ≤ 1/2b by assumption, the integral over [0 , ǫ ] is ≪ 1. Substituting variables and using \| cot( πt )\| = cot( π‖t‖), we also have b ∫ k+1 b−ǫ k b+ǫ \| cot( πbx )\|dx = ∫ k+1 −bǫ k+bǫ \| cot( πx )\|dx = 2 ∫ 1/2 bǫ cot( πx )dx ≪ log(1 /bǫ ). Thus ∫ S ∣∣∣∣ d dx log φb,ǫ (x) ∣∣∣∣ dx ≪ b log(1 /bǫ ) + π ∑ 0≤k<b ∫ k+1 b−ǫ k b+ǫ \| cot( πx )\|dx. For x ∈ [0 , 1] − S , we have φb,ǫ (x) = sin( πbǫ )/ sin( πx ); hence \| d dx log φb,ǫ (x)\| = π\| cot( πx )\|. Thus ∫ [0 ,1] −S ∣∣∣∣ d dx log φb,ǫ (x) ∣∣∣∣ dx = π ∑ 0<k<b ∫ k b+ǫ k b−ǫ \| cot( πx )\|dx and so V (log ◦φb,ǫ ) ≪ b log(1 /bǫ ) + π ∑ 0≤k<b ∫ k+1 b−ǫ k b+ǫ \| cot( πx )\|dx + π ∑ 0<k<b ∫ k b+ǫ k b−ǫ \| cot( πx )\|dx = b log(1 /bǫ ) + π ∫ 1−ǫǫ \| cot( πx )\|dx = b log(1 /bǫ ) + 2 π ∫ 1/2 ǫ cot( πx )dx ≪ b log(1 /bǫ ).(24) With regards to the integral ∫ 10 log φb,ǫ (x)dx , splitting it similarly according to the definition of φb,ǫ , using φb(x) ≤ b, the additivity of the logarithm, and ignoring the terms log sin( πbǫ ) (these are non-positive) one derives ∫ 10 log φb,ǫ (x)dx ≤ 2ǫ log b + 2 ∫ 1/2 ǫ log csc( πx )dx + ∑ 0≤k<b ∫ k+1 b−ǫ k b+ǫ log \| sin( πbx )\|dx = 2 ǫ log b + 2 ∫ bǫ ǫ log csc( πx )dx ≪ bǫ log(1 /bǫ ).(25) The equality above follows after substituting bx with u and using ∫ k+1 −bǫ k+bǫ log \| sin( πu )\|du = ∫ 1−bǫ bǫ log \| sin( πu )\|du = −2 ∫ 1/2 bǫ log csc( πu )du. INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 15 Now the result follows from (23), (24), (25) if we let ǫ = D∗ N /2b using the fact 0 < 1/2N ≤ D∗ N ≤ 1 (see for instance Theorems 1.2 and 1.3 in Section 2 of for this). Our final tool is the following version of Vinogradov’s lemma. Lemma 5.10 (Vinogradov-type lemma) . Let A, B, θ be real numbers, A, B > 0,and let q ≥ 2 be an integer. Then ∑ n(q) min ( A, B csc 2 ( π n + θ q )) ≤ min ( A, B csc 2 ( π‖θ‖ q )) + ( 1 − 4 π2 ) Bq 2. Proof. We can write θ = m+r for some integer m and real number −1/2 < r ≤ 1/2. In fact \|r\| = ‖θ‖. Since m + n uniquely covers each residue class modulo q when so does n, the sum equals S := ∑ 0≤n<q min ( A, B csc 2 ( π n + r q )) . If r < 0, we may replace n with −n in the summand. Thus without loss of generality we may assume that r ≥ 0. Now note S ≤ min ( A, B csc 2 ( πr q )) BH q (r), where for 0 ≤ r ≤ 1/2, Hq (r) := ∑ 1≤n<q csc 2 ( π n + r q ) . Since x 7 → csc 2(πx ) is convex in 0 < x < 1, so is Hq (r) in 0 ≤ r ≤ 1/2. From the well-known identity π2 csc 2(πx ) = ∑ n∈Z 1 (n + x)2 for x 6 ∈ Z, one derives (see for instance ) Hq (r) = { (q2 − 1) /3 if r = 0 ,q2 csc 2(πr ) − csc 2(πr/q ) if 0 < r ≤ 1/2. One can show that Hq (0) ≤ Hq(1 /2) (say by using the inequality sin( πx ) ≥ 23/2x valid for 0 ≤ x ≤ 1/4) and thus Hq (r) ≤ Hq(1 /2) for 0 ≤ r ≤ 1/2 by the convexity of Hq (r) in [0 , 1/2]. We also have Hq (1 /2) = q2 − csc 2(π/ 2q) ≤ q2(1 − 4/π 2) since sin( x) ≤ x for x ≥ 0. The result now follows. Exponential sums over P0 b (x)The current state of knowledge on exponential sums over palindromes, specifi-cally on bounds for these, is rather limited. To our understanding, the literature on this topic is thus far comprised of the works of Banks-Hart-Sakata , Banks-Shparlinski and Col . Here we give a brief exposition of what is known, as well as add slightly to the literature via the results in Proposition 6.5 and Corollary 6.6 for special prime moduli. These are consequences of a result of Bourgain giving upper bounds for certain types of exponential sums over finite fields with prime order. However these are not employed by us in our proofs of Theorems 1.4 16 ALEKSANDR TUXANIDY AND DANIEL PANARIO and 1.5. As such, Proposition 6.5 and Corollary 6.6 may be regarded as addenda to the work, stated here in the interests of expanding the subject. For the sake of simplicity, and indeed, for our purposes, we focus exclusively on exponential sums over P0 b (x); that is, over b-palindromes, at most x in size, with an odd number of digits in their b-adic expansion. Sums over palindromes with an even number of digits are rather similar in nature and all results here extend naturally, albeit with minor differences, to these. The starting point, just as that in the works , , is that such sums can es-sentially be decomposed into linear combinations of objects enjoying rather useful multiplicative and algebraic structures. These properties are crucial in the upcom-ing sections. In this regard we have the following lemma, implicit in the works , . We give a proof for the convenience of the reader. Lemma 6.1. For any α, x ∈ R, x ≥ 1, ∣∣∣∣∣∣∑ n∈P0 b(x) e(αn ) ∣∣∣∣∣∣ ≤ b2 ∑ 0≤N≤1 2log bx ∑ 0≤M≤N ΦM (αb N −M ) , where ΦM is as defined in (16). Proof. From the definition of P0 b (x) and the triangle inequality, ∣∣∣∣∣∣∑ n∈P0 b(x) e(αn ) ∣∣∣∣∣∣ ≤ ∑ 0≤N≤1 2log bx ∣∣∣∣∣∣∣∣∑ n∈Πb(2 N) n≤x e(αn ) ∣∣∣∣∣∣∣∣ , where (26) Πb(2 N ) := {n ∈ Pb : ⌊log b n⌋ = 2 N } = Pb ∩ [b2N , b 2N +1 ). The sum inside the vertical brackets equals ∑ n∈Πb(2 N) n≤y e(αn ), where y is the largest palindrome in Π b(2 N ) satisfying y ≤ x. If no such palindrome y exists, the sum is trivially zero. Otherwise we may write y = yN bN + ∑ 0≤j<N yj (bj + b2N −j ) for some digits 0 ≤ yj < b with y0 > 0. We now seek to establish the inequality (27) ∣∣∣∣∣∣∣∣∑ n∈Πb(2 N) n≤y e(αn ) ∣∣∣∣∣∣∣∣ ≤ b2 ∑ 0≤M≤N ΦM (αb N −M ) . To this end, we first define for an integer λ ≥ 0, Ψλ(α) := ∑ 0≤cN<b e (αc N bN ) ∏ λ<n<N ∑ 0≤c<b e (αc (bn + b2N −n)) .INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 17 We have ∑ n∈Πb(2 N) n≤y e(αn )= Ψ 0(α) ∑ 1≤c<y 0 e (αc (1 + b2N )) Ψ 1(α)e (αy 0 (1 + b2N )) ∑ 0≤c<y 1 e (αc (b + b2N −1)) Ψ 2(α)e (αy 0 (1 + b2N ) + αy 1 (b + b2N −1)) ∑ 0≤c<y 2 e (αc (b2 + b2N −2)) · · · Ψ N −1(α)e α ∑ 0≤j<N −1 yj (bj + b2N −j ) ∑ 0≤c<y N−1 e ( αc ( bN −1 + b2N −(N −1) )) e α ∑ 0≤j<N yj (bj + b2N −j ) ∑ 0≤c≤yN e (αcb N ) . Now the result follows when we note that ∏ λ<n<N ∑ 0≤c<b e (αc (bn + b2N −n)) = ∏ 1≤n<N −λ ∑ 0≤c<b e ( αb λc ( bn + b2( N −λ)−n)) and take absolute values. From the lemma above it is clear that to study upper bounds for sums such as ∑ n∈P0 b(x) e(αn ), it is sufficient to study the products Φ N or, more generally, products such as (28) PM (α) := ∏ 1≤n≤M φb (α (bn + b2N −n)) with M ≤ 2N . Of course the product’s definition depends also on b, N , but for the sake of brevity we omit these on the left hand side above. Let us now state Col’s result in our context. This will be needed later on. We give a proof somewhat different to that of Col in its initial steps, although it ultimately boils to considering lower bounds for sums such as ∑ n≤M ‖ab n/q ‖2, just as in Col’s argument. Proposition 6.2 (Col [9, Corollary 4]) . Let h, q, k, M, N be integers with q ≥ 2, (q, h (b3 − b)) = 1 and 0 ≤ M ≤ 2N . Then for PM defined as in (28), PM ( h q + k b3 − b ) ≪b bM exp ( −σ∞(b) M log q ) uniformly in h, k, N , where σ∞(b) > 0 is some value depending only on b.Proof. We may assume that M ≥ log b q as otherwise the statement is trivial. Now for the sake of brevity let α := h q k b3−b . Grouping the factors in the product by pairs of adjacent factors, we have PM (α) ≤ b ∏ 1≤n<M √ φb (α (bn + b2N −n)) φb (α (bn+1 + b2N −n−1)) .18 ALEKSANDR TUXANIDY AND DANIEL PANARIO By Lemma 5.8 with β = 2 N − n − 1 and γ = n, φb (α (bn + b2N −n)) φb (α (bn+1 + b2N −n−1)) ≤ bφ b ( ‖α(b2 − 1) bn‖ b + 1 ) . Thus PM (α) ≤ b M −1 2+1 √√√√ ∏ 1≤n<M φb ( ‖α(b2 − 1) bn‖ b + 1 ) . By Lemma 5.6 and the 1-periodicity of ‖ · ‖ , the above is ≪ bM exp − π2(b − 1) 12( b + 1) ∑ 1≤n≤M ∥∥∥∥ h(b2 − 1) bn q ∥∥∥∥ 2  uniformly in k, N . Since q ≥ 2 and ( q, h (b3 − b)) = 1 by assumption, each term in the sum above is ≥ 1/q 2. In particular the result holds if q < b . Consider now the case when q ≥ b. Here we argue as done by Maynard in Lemma 10.1 there. Dividing the sum above into subsums over segments of length ⌊log b q⌋, we claim it is ≥ ⌊ M log b q ⌋ min (a,q )=1 ∑ 1≤n≤log bq ∥∥∥∥ ab n q ∥∥∥∥ 2 . Indeed, with the possible exception of one subsum (coming from the tail with n within a distance < ⌊log b q⌋ from M , and lower-bounded by zero) each subsum can be written as ∑ 1≤n≤log bq ∥∥∥∥ h(b2 − 1) bm+n q ∥∥∥∥ 2 for some integer m ≥ 0. There are at least ⌊M/ log b q⌋ subsums of this type. Since ( q, h (b3 − b)) = 1 by assumption, the claim follows. Now since q ≥ 2 and (q, ab ) = 1, we have ‖ab n/q ‖ ≥ 1/q for each n. Moreover if ‖ab n/q ‖ ≤ 1/2b, then ‖ab n+1 /q ‖ = b‖ab n/q ‖. Consequently there exists an integer 1 ≤ n ≤ log b q for which ‖ab n/q ‖ ≥ 1/2b2. Hence ∑ 1≤n≤log bq ∥∥∥∥ ab n q ∥∥∥∥ 2 ≥ 1 4b4 and the result follows. The bound due to Col in Proposition 6.2 is non-trivial in the range q ≪ eo(M) and is rather general in the sense of the lack of restrictions on the arithmetic shape of q. As such, it considerably improved over the corresponding general bound due to Banks-Hart-Sakata (see Lemma 3.2 there) non-trivial only in the range q = o(√M ) (strictly speaking, the bounds of Banks-Hart-Sakata were essentially given for products such as Φ N and not the product over 1 ≤ n ≤ M in Proposition 6.2, but their work easily extends to this). In what follows ord q (b) denotes the multiplicative order of b modulo q (whenever (q, b ) = 1). In the range q ≪ M 2 with P −(q) > b satisfying τ (q)√q ≪ ord q (b) ≪ M (with suitable implied constants) the arguments of Banks-Hart-Sakata (see Lemma 3.1 there) yield power-saving bounds of the form PM (h/q ) ≪ bδM with 0 < δ < 1 fixed or depending only on b. Although applicable only on the much smaller range q ≪ M 2, this is superior in strength over that in Proposition 6.2 for INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 19 such q. Indeed, the latter attains (at best) power-savings only for q ≪ 1. In few words, their argument can be explained as follows: Assume ( h, q ) = 1. Squaring PM (h/q ), applying the inequality of the arithmetic and geometric means, expanding the square and switching orders of summation, all in turn, one has P 2/M M (h/q ) ≤ ∑ 0≤c1,c 2<b 1 M ∑ 1≤n≤M eq (h(c1 − c2) (bn + b2N −n)) (29) = b + ∑ 0≤c16=c2<b 1 M ∑ 1≤n≤M eq (h(c1 − c2) (bn + b2N −n)) . If we impose the restriction P −(q) > b , then ( c1 − c2, q ) = 1 for 0 ≤ c1 6 = c2 < b .In this case (30) P 2/M M (h/q ) ≤ b + b2 M max a,k ∈(Z/q Z)× Sq(M, a, k ), where (31) Sq (M, a, k ) := ∣∣∣∣∣∣∑ 1≤n≤M eq (ab n + k¯bn)∣∣∣∣∣∣ . Here b denotes the multiplicative inverse of b modulo q. We note that sums such as these also appear for instance in Banks-Shparlinski , Bourgain (see Theorem 6.3 below) Humphries , Ostafe-Shparlinski-Voloch and Popova . Using well-known bounds for twisted Kloosterman sums, one derives (see Lemma 2.1 in ) Sq (M, a, k ) ≤ M τ (q)√q ord q(b) + ord q (b). For this to be non-trivial, it is necessary that τ (q)√q ≪ ord q (b) ≪ M ; hence their assumptions on q. One thus obtains P 2/M M (h/q ) ≤ b + b2 ( τ (q)√q ord q (b) + ord q (b) M ) . Banks-Shparlinski also studied the sum Sq(M, a, k ) after the work of Banks-Hart-Sakata and obtained a bound for PM (h/p ), with p prime, essentially compa-rable to Col’s result in Proposition 6.2 for q = p ≪ M 2/ log 4 M with no restrictions on ord p(b); see their Theorem 6. From the argument above, it is clear that improving bounds for Sq(M, a, k )and/or enlarging the range of applicability of such bounds, may yield stronger results. However due to the presence of the b term in (30) coming from the diagonal cases c1 = c2 in (29), one can never do better than PM ≪ bM/ 2 via this argument, regardless of the quality of the bounds available for Sq (M, a, k ). Soon after the works of Banks-Hart-Sakata and Banks-Shparlinski , Bour-gain obtained the following result. First given a prime p, let Fp be the finite field with p elements and let F∗ p be its subset of non-zero elements, endowed with the usual properties of Fp. Given θ ∈ F∗ p , we use the notation ord( θ) to denote its multiplicative order. 20 ALEKSANDR TUXANIDY AND DANIEL PANARIO Theorem 6.3 (Bourgain [6, Theorem 2]) . Let ǫ > 0, let p be a prime number and let θ1, . . . , θ r ∈ F∗ p satisfying ord (θj ) > p ǫ and ord (θiθ−1 j ) > p ǫ for each 1 ≤ i 6 = j ≤ r.Then for any integer N > p ǫ and any a1, . . . , a r ∈ F∗ p , ∣∣∣∣∣∣ N ∑ n=1 ep  r ∑ j=1 aj θnj ∣∣∣∣∣∣ < N pδ , where 0 < δ = δ(ǫ) < 1 is some value depending only on ǫ. We recognize that the sum, in Theorem 6.3, with r = 2 , a 1 = a, a 2 = k, θ 1 = b, θ 2 = ¯b, N = M , is precisely Sp(M, a, k ). Thus we have the following. Corollary 6.4. Let M be an integer. Then for any prime p satisfying (p, b ) = 1 and pǫ < min( M, ord p(b2)) for some ǫ > 0, we have max a,k ∈(Z/p Z)× Sp(M, a, k ) < M pδ(ǫ) , where δ(ǫ) > 0 depends only on ǫ, and Sp(M, a, k ) is defined as in (31). Proof. Suffices to note ord p(b±1) ≥ ord p(b±2) > p ǫ, the last holding by assumption. From this and (30) we may then obtain, under the same assumptions of Corollary 6.4 with p > b , PM ( h p ) ≤ bM ( 1 b + 1 pδ )M/ 2 . One may improve this considerably, in Corollary 6.6 below, by combining use of Lemma 5.9, the Erd˝ os-T´ uran inequality and Bourgain’s bound. More generally we have the following. Proposition 6.5. With the same assumptions of Theorem 6.3, ∏ 1≤n≤N φb  1 p r ∑ j=1 aj θnj  ≤ exp ( A(ǫ)b N pδ(ǫ) ) , where A(ǫ), δ (ǫ) > 0 are some values depending only on ǫ.Proof. We may view each ∑rj=1 aj θnj as a real number in the natural way. We may also assume that p ≫ǫ 1 is arbitrarily large. Now for each integer 1 ≤ n ≤ N , let αn = 1 p r ∑ j=1 aj θnj . By Lemma 5.9, ∏ 1≤n≤N φb(αn) ≤ exp ( AbN D ∗ N log ( 2 D∗ N )) for some absolute constant A > 0, where D∗ N is the star-discrepancy of ( αn)Nn=1 .By the Erd˝ os-Tur´ an inequality (Lemma 5.2) and the fact D∗ N ≤ DN , we have D∗ N ≪ 1 H + ∑ 1≤h≤H 1 h ∣∣∣∣∣∣ 1 N ∑ 1≤n≤N e(hα n) ∣∣∣∣∣∣INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 21 for any H ≥ 2. The contribution of the terms with p \| h to the series above is trivially ≪ log( H)/p . By Bourgain’s bound in Theorem 6.3, the terms with p ∤ h contribute ≪ log( H)/p δ to the sum, where 0 < δ = δ(ǫ) < 1 is some value depending only on ǫ. Thus D∗ N ≪ 1 H + log H pδ ≪ log p pδ after letting H = p. Since the function x 7 → x log(2 /x ) is increasing on (0 , 2/e ), then for p ≫ǫ 1 sufficiently large (as assumed) D∗ N log(2 /D ∗ N ) ≪ log 2 p pδ ≪δ p−δ/ 2 and the result follows. Specializing to the palindromes we have the following immediate consequence. Corollary 6.6. With the same assumptions of Corollary 6.4 and M ≤ 2N there, we have max (h,p )=1 ∏ 1≤n≤M φb ( h p (bn + b2N −n)) ≤ exp ( A(ǫ)b M pδ(ǫ) ) , where A(ǫ), δ (ǫ) > 0 are some values depending only on ǫ. Bounding moments of ΦN Here we prove Proposition 7.1 bounding moments of Φ N . Proposition 7.1 (2 K-th moment) . For any integers N, K, b ≥ 2, we have (32) ∫ 10 Φ2KN (α)dα ≤ b2( K−1) N +2 ( 1 + O ( 1 √K + b2 K )) 2N . If α1, . . . , α R ∈ R/Z are δ-spaced for some 0 < δ ≤ 1/2, then (33) R ∑ r=1 Φ2KN (αr ) ≤ (δ−1 + Kb 2N ) b2( K−1) N +2 ( 1 + O ( 1 √K + b2 K )) 2N . In particular for any Q ≥ 1 and uniformly in β ∈ R/Z, (34) ∑ q≤Q ∗ ∑ h(q) Φ2KN ( h q + β ) ≤ (Q2 + Kb 2N ) b2( K−1) N +2 ( 1 + O ( 1 √K + b2 K )) 2N . Proof. We first note that (34) follows from (33) since the points h/q + β with 1 ≤ h ≤ q ≤ Q and ( h, q ) = 1 are Q−2-spaced modulo 1. Indeed, for any two distinct such fractions h1/q 1, h 2/q 2, ∥∥∥∥ h1 q1 − h2 q2 ∥∥∥∥ ≥ 1 q1q2 ≥ 1 Q2 . Let us now set ψb(α) := ∑ 0≤m<b e(αm ). By definition, Φ2KN (α) = ∣∣∣∣∣∣∏ 1≤n<N ψKb (α (bn + b2N −n)) ∣∣∣∣∣∣ 2 .22 ALEKSANDR TUXANIDY AND DANIEL PANARIO We have ∏ 1≤n<N ψb (α (bn + b2N −n)) = ∑ 0≤c1,...,c N −1<b e α ∑ 1≤n<N cn (bn + b2N −n) and one observes ∏ 1≤n<N ψKb (α (bn + b2N −n)) = ∑ 0≤v1,...,v N −1≤(b−1) K  ∏ 1≤n<N r(vn ; K, b )  e α ∑ 1≤m<N vm (bm + b2N −m) , where r(v; K, b ) is defined as in (17). We can rewrite this as ∏ 1≤n<N ψKb (α (bn + b2N −n)) = ∑ 0≤ℓ≤Kb 2N aℓe(αℓ ), where aℓ := ∑ 0≤v1 ,...,v N −1≤(b−1) K ∑ 1≤m<N vm(bm+b2N −m)= ℓ ∏ 1≤n<N r(vn ; K, b ). Thus Φ2KN (α) = ∣∣∣∣∣∣∑ 0≤ℓ≤Kb 2N aℓe(αℓ ) ∣∣∣∣∣∣ 2 . Parseval’s identity then gives ∫ 10 Φ2KN (α) dα = ∑ 0≤ℓ≤Kb 2N a2 ℓ while the large sieve inequality (Lemma 1.6) yields R∑ r=1 Φ2KN (αr) ≤ (δ−1 + Kb 2N ) ∑ 0≤ℓ≤Kb 2N a2 ℓ for any δ-spaced points α1, . . . , α R ∈ R/Z. Thus to prove the proposition we must show that (35) ∑ 0≤ℓ≤Kb 2N a2 ℓ ≤ b2( K−1) N +2 ( 1 + O ( 1 √K + b2 K )) 2N . Expanding the square and switching orders of summation, we have ∑ 0≤ℓ≤Kb 2N a2 ℓ = ∑ 0≤u1,...,u N −1≤(b−1) K 0≤v1,...,v N −1≤(b−1) K ∑ 1≤m<N (um−vm )( bm+b2N −m)=0 ∏ 1≤n<N r(un; K, b )r(vn; K, b ).INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 23 It will be convenient later on to introduce extra variables uN , v N to the sum. To this end, we claim that ∑ 0≤u1,...,u N−1≤(b−1) K 0≤v1,...,v N−1≤(b−1) K ∑ 1≤m<N (um−vm)( bm+b2N−m)=0 ∏ 1≤n<N r(un; K, b )r(vn; K, b ) ≤ ∑ 0≤u1,...,u N≤(b−1) K 0≤v1,...,v N≤(b−1) K ∑ 1≤m≤N(um−vm)( bm+b2N−m)=0 ∏ 1≤n≤N r(un; K, b )r(vn ; K, b ). Indeed, any solution ( u′ 1 , . . . , u ′ N−1 ) × (v′ 1 , . . . , v ′ N−1 ) to the equation ∑ 1≤m<N (um − vm)( bm + b2N −m) = 0 yields the solution ( u′ 1 , . . . , u ′ N−1 , 0) × (v′ 1 , . . . , v ′ N−1 , 0) to the equation ∑ 1≤m≤N (um − vm)( bm + b2N −m) = 0 . Since r(0; K, b ) = 1 additionally, the claim follows. For any integer n, Lemma 5.5 implies r(n; K, b ) ≤ bK 10≤n≤(b−1) K √(b2 − 1) K (√ 6 π exp ( − 6 (b2 − 1) K ( n − (b − 1) K 2 )2) c K ) for some absolute constant c > 0. We fix a smooth function ν : R → [0 , 1] compactly supported on [ −2, 2] satisfying ν(t) = 1 for 0 ≤ t ≤ 1 and ‖ν(j)‖∞ ≪j 1 for each j ≥ 0. In particular 10≤n≤(b−1) K ≤ ν(n/ (b − 1) K) and r(n; K, b ) ≤ bK ν( n (b−1) K ) √(b2 − 1) K (√ 6 π exp ( − 6 (b2 − 1) K ( n − (b − 1) K 2 )2) c K ) . We can rewrite this as r(n; K, b ) ≤ bK √(b2 − 1) K ηK,b ( n √(b2 − 1) K ) , where ηK,b : R → R+0 is the smooth compactly supported function defined by (36) ηK,b (t) = √ 6 π exp −6 ( t − 1 2 √ (b − 1) K b + 1 )2 + c K  ν ( t √ b + 1 (b − 1) K ) . Then we have ∑ 0≤u1,...,u N≤(b−1) K 0≤v1,...,v N≤(b−1) K ∑ 1≤m≤N(um−vm)( bm+b2N−m)=0 ∏ 1≤n≤N r(un; K, b )r(vn; K, b ) ≤ b2KN (b2 − 1) N KN J, 24 ALEKSANDR TUXANIDY AND DANIEL PANARIO where J is the sum ∑ u1,...,u N∈Z v1,...,v N∈Z ∑ 1≤m≤N(um−vm)( bm+b2N−m)=0 ∏ 1≤n≤N ηK,b ( un √(b2 − 1) K ) ηK,b ( vn √(b2 − 1) K ) . In view of (35) we complete the proof if we show that (37) J ≤ b2 (b2 − 1) N KN b2N ( 1 + O ( 1 √K + b2 K )) 2N . To this end we proceed as follows. An application of the identity 1k=0 = ∫ 10 e(kα )dα valid for integers k, gives J = ∫ 10 ∣∣∣∣∣∣∏ 1≤n≤N ∑ v∈Z ηK,b ( v √(b2 − 1) K ) e (αv (bn + b2N −n)) ∣∣∣∣∣∣ 2 dα. Let us set (38) A := √(b2 − 1) K‖ηK,b ‖1 + ‖η′ K,b ‖1. By the symmetry of n 7 → bn + b2N −n around n = N and the bound ∣∣∣∣∣∑ v∈Z ηK,b ( v √(b2 − 1) K ) e (2αvb N )∣∣∣∣∣ ≤ A coming from (18) in Lemma 5.3, note J ≤ A ∫ 10 ∏ 1≤n< 2N ∣∣∣∣∣∑ v∈Z ηK,b ( v √(b2 − 1) K ) e (αv (bn + b2N −n)) ∣∣∣∣∣ dα. Grouping by pairs of adjacent factors and bounding square roots of two endpoint factors, we have ∏ 1≤n< 2N ∣∣∣∣∣∑ v∈Z ηK,b ( v √(b2 − 1) K ) e (αv (bn + b2N −n)) ∣∣∣∣∣ ≤ A ∏ 1≤n≤2( N−1) √√√√ ∣∣∣∣∣∑ u∈Z ηK,b ( u √(b2 − 1) K ) e (αu (bn + b2N −n)) ∣∣∣∣∣ × √√√√ ∣∣∣∣∣∑ v∈Z ηK,b ( v √(b2 − 1) K ) e (αv (bn+1 + b2N −n−1)) ∣∣∣∣∣.INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 25 By Lemma 5.3 with k = 4, the product over 1 ≤ n ≤ 2( N − 1) is at most ∏ 1≤n≤2( N−1) √√√√min ( A, ‖η(4) K,b ‖1 (b2 − 1) 3/2K3/2 sin 4(πα (bn + b2N −n)) ) × √√√√min ( A, ‖η(4) K,b ‖1 (b2 − 1) 3/2K3/2 sin 4(πα (bn+1 + b2N −n−1)) ) . By Lemma 5.8 with β = 2 N − n − 1 and γ = n, the fact that t 7 → sin( πt ) is increasing on [0 , 1/2] and \| sin( πt )\| = sin( π‖t‖), we have either ∣∣sin (πα (bn + b2N −n))∣ ∣ ≥ sin ( π ‖α(b2 − 1) bn‖ b + 1 ) or ∣∣sin (πα (bn+1 + b2N −n−1))∣ ∣ ≥ sin ( π ‖α(b2 − 1) bn‖ b + 1 ) . Then the product above is ≤ AN −1 ∏ 1≤n≤2( N−1) √√√√min ( A, ‖η(4) K,b ‖1 (b2 − 1) 3/2K3/2 sin 4(π‖α(b2 − 1) bn‖/(b + 1)) ) ≤ AN −1 ∏ 1≤n≤2( N−1) √√√√min ( A, (b + 1) 4‖η(4) K,b ‖1 (b2 − 1) 3/2K3/2 sin 4(πα (b2 − 1) bn) ) . Note in the last line we used the inequality r sin( πt/r ) ≥ sin( πt ) valid for r ≥ 1 and 0 ≤ t < 1 (which may be shown to hold, say, via Euler’s product formula). Thus ∏ 1≤n< 2N ∣∣∣∣∣∑ v∈Z ηK,b ( v √(b2 − 1) K ) e (αv (bn + b2N −n)) ∣∣∣∣∣ ≤ AN ∏ 1≤n≤2( N−1) √√√√min ( A, (b + 1) 4‖η(4) K,b ‖1 (b2 − 1) 3/2K3/2 sin 4(πα (b2 − 1) bn) ) . Inserting this in the integral, substituting αb (b2 − 1) with α and using the 1-periodicity of the integrand, we obtain J ≤ AN +1 ∫ 10 ∏ 0≤n< 2( N−1) √√√√min ( A, (b + 1) 4‖η(4) K,b ‖1 (b2 − 1) 3/2K3/2 sin 4(παb n) ) dα. (39) By Lemma 5.4 and the fact √min( A, B ) = min( √A, √B) for A, B ≥ 0, this integral (without the normalizing factor AN +1 ) is ≤ sup θ∈R 1 b ∑ n(b) min ( A1/2, (b + 1) 2‖η(4) K,b ‖1/21 (b2 − 1) 3/4K3/4 sin 2(π(n + θ)/b ) ) 2( N−1) while Lemma 5.10 implies this is ≤ ( A1/2 b + b(b + 1) 2‖η(4) K,b ‖1/21 (b2 − 1) 3/4K3/4 )2( N −1) .26 ALEKSANDR TUXANIDY AND DANIEL PANARIO Factoring 1 b (b2 − 1) 1/4K1/4 out from the inside of the brackets and factoring (b2 − 1) N +1 2 K N +1 2 out from the normalizing factor AN +1 in (39) it follows J ≤ b2 (b2 − 1) N KN b2N E(N, K, b ), where E(N, K, b ):= ( A √(b2 − 1) K )N +1 ( A1/2 (b2 − 1) 1/4K1/4 + b2(b + 1) 2‖η(4) K,b ‖1/21 (b2 − 1) K )2( N −1) ≤ ( A √(b2 − 1) K )N ( A1/2 (b2 − 1) 1/4K1/4 + 3b2‖η(4) K,b ‖1/21 K )2N = ( ‖ηK,b ‖1 + ‖η′ K,b ‖1 √(b2 − 1) K + 3b2‖η(4) K,b ‖1/21 K √ ‖ηK,b ‖1 + ‖η′ K,b ‖1 √(b2 − 1) K )2N . The inequality above follows from the facts B(√B + C)−2 ≤ 1 for any B > 0, C ≥ 0, and ( b + 1) 2 ≤ 3( b2 − 1) for b ≥ 2. The last equality holds by the definition of A in (38). From the definition of ηK,b in (36) and the assumptions on ν, we have ‖η′ K,b ‖1, ‖η(4) K,b ‖1 ≪ 1 uniformly in K, b . Moreover the fact ∫ R e−t2 dt = √π and the assumption ν ≤ 1 imply ‖ηK,b ‖1 ≤ 1 + c‖ν‖1 √K = 1 + O ( 1 √K ) . Hence E(N, K, b ) ≤ ( 1 + O ( 1 √K + b2 K )) 2N and J ≤ b2 (b2 − 1) N KN b2N ( 1 + O ( 1 √K + b2 K )) 2N . We have shown that (37) holds and the proof is thus complete. Bounding the average Here we prove Proposition 8.1 below combining our moment bounds in Propo-sition 7.1 with Col’s L∞-type bound in Proposition 6.2. As an application of Proposition 8.1, we then give a proof of Theorem 1.5 in the following section. Proposition 8.1 (L1/L 2/L 2K /L ∞ hybrid bound) . Let N ≥ 0 be an integer and let Q ≥ 1. For any 0 < ǫ ≤ 1 15 and 1 3 ≤ δ ≤ 2 5 − ǫ, (40) sup β∈R ∑ q≤Q (q,b )=1 ∗ ∑ h(q) ΦN ( h q + β ) ≪b,ǫ Q2bN (1 −δ−σ1 (b,ǫ )) + Q1− σ1 (b,ǫ ) δ bN ,INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 27 where σ1(b, ǫ ) > 0 is some value depending only on b and ǫ. Moreover (41) ∑ 2≤q≤Q (q,b 3−b)=1 ∗ ∑ h(q) ΦN ( h q + k b3 − b ) ≪b,ǫ ( Q2bN (1 −δ−σ1(b,ǫ )) + Q1− σ1(b,ǫ ) δ bN ) e− σ∞ (b)N log Q uniformly in k ∈ Z, where σ∞(b) > 0 is some value depending only on b. We will need the following bound. Lemma 8.2 (L2-bound) . Let 0 ≤ L < M < N be integers and let Q ≥ 1. Then sup β∈R ∑ q≤Q ∑ h(q) ∏ L<n ≤M φ2 b (( h q + β ) (bn + b2N −n)) ≪ǫ (Q + bM−L+ǫN ) Qb M−L for any ǫ > 0.Proof. Expanding the square, switching orders of summation, using the orthogo-nality of the additive characters modulo q and taking absolute values, we have that the left hand side above is ≤ Q ∑ 0≤uL+1 ,...,u M<b 0≤vL+1 ,...,v M<b ∑ q≤Qq\|S(u,v) 1, where S(u, v) := ∑ L<n ≤M (un − vn) (bn + b2N −n) . By the uniqueness of the representation of integers in base b, we have S(u, v) = 0 if and only if un = vn for each L < n ≤ M . The contribution to the overall sum of such (diagonal) terms is ≤ Q2bM−L. For the off-diagonal terms with un 6 = vn for some n, the inner sum is ≤ τ (S(u, v)) ≪ǫ bǫN for any ǫ > 0, where τ (m) := ∑ d\|m 1is the divisor function. Thus these contribute ≪ǫ Qb 2( M−L)+ ǫN for any ǫ > 0. Proof of Proposition 8.1. We may assume that N ≥ 100, say, as otherwise the statement is trivial. Let us treat the cases when Q ≪b,ǫ bδN and Q ≫b,ǫ bδN separately. We begin by considering the case when Q ≫b,ǫ bδN . Here one can check that both right hand sides of (40) and (41) are ≫b,ǫ Q2bN (1 −δ−σ1(b,ǫ )) (in particular the exponential factor in (41) is ≍b,ǫ 1 since δ ≥ 1/3 by assumption). Thus to prove (40) and (41) when Q ≫b,ǫ bδN it suffices to show that the left hand side of (40) is ≪b,ǫ Q2bN (1 −δ−σ1 (b,ǫ )) in this case. To this end we proceed as follows. Let M = ⌊δN ⌋ and split the product as ΦN ( h q + β ) = P1 ( h q ) P2 ( h q ) P3 ( h q ) ,28 ALEKSANDR TUXANIDY AND DANIEL PANARIO where P1 ( h q ) = ∏ 1≤n≤M φb (( h q + β ) (bn + b2N −n)) ,P2 ( h q ) = ∏ M<n ≤2M φb (( h q + β ) (bn + b2N −n)) ,P3 ( h q ) = ∏ 2M<n<N φb (( h q + β ) (bn + b2N −n)) . Let K ≥ 2 be an integer to be specified later and set ℓ := 4 K/ (2 K − 1) > 2. Note that 2 /ℓ + 1 /2K = 1. Then by H¨ older’s inequality with the triple ( ℓ, ℓ, 2K), we have ∑ q≤Q (q,b )=1 ∗ ∑ h(q) ΦN ( h q + β ) ≤ ∑ q≤Q ∑ h(q) P ℓ 1 ( h q ) 1/ℓ ∑ q≤Q ∑ h(q) P ℓ 2 ( h q ) 1/ℓ ∑ q≤Q (q,b )=1 ∗ ∑ h(q) P 2K 3 ( h q ) 1/2K . Note P ℓ 1 (h/q ) = P ℓ−21 (h/q )P 21 (h/q ) ≤ b(ℓ−2) M P 21 (h/q ) while Lemma 8.2 gives ∑ q≤Q ∑ h(q) P 21 ( h q ) ≪γ (Q + bM ) Qb M+γN for any γ > 0. Thus ∑ q≤Q ∑ h(q) P ℓ 1 ( h q ) 1/ℓ ≪γ (Q + bM )1/ℓ Q1/ℓ bM(1 −1/ℓ )+ γN for any γ > 0. Similarly we can show that ∑ q≤Q ∑ h(q) P ℓ 2 ( h q ) 1/ℓ ≪γ (Q + bM )1/ℓ Q1/ℓ bM(1 −1/ℓ )+ γN . Consider now the third sum with P3. We have ∗ ∑ h(q) P 2K 3 ( h q ) = ∗ ∑ h(q) ∏ 1≤n<N −2M φ2Kb (( h q + β ) (b2M+n + b2N −2M−n)) = ∗ ∑ h(q) ∏ 1≤n<N −2M φ2Kb (( hb 2M q + βb 2M ) ( bn + b2( N −2M)−n)) = ∗ ∑ h(q) Φ2KN −2M ( hb 2M q + βb 2M ) .INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 29 If ( q, b ) = 1, we may substitute hb 2M with h above. Thus ∑ q≤Q (q,b )=1 ∗ ∑ h(q) P 2K 3 ( h q ) = ∑ q≤Q (q,b )=1 ∗ ∑ h(q) Φ2KN −2M ( h q + βb 2M ) . Removing now the constraint ( q, b ) = 1 via positivity, Proposition 7.1 implies the above is ≤ ( Q + √Kb N −2M )2 b2( K−1)( N −2M)+2 ( 1 + c √K + cb 2 K )2( N −2M) for some absolute constant c > 0. Multiplying the three bounds together and recalling the assumptions M = ⌊δN ⌋ and 2 /ℓ + 1 /2K = 1 gives ∑ q≤Q (q,b )=1 ∗ ∑ h(q) ΦN ( h q + β ) ≪b,γ (Q + bM ) 2 ℓ (Q + bN −2M ) 1 K Q 2 ℓ bN − N K−2M ℓ+2M K+γN ( 1 + cb 2 K + c √K ) N −2M K ≪b (Q + bδN )1−1/2K ( Q + bN (1 −2δ))1/K Q1−1/2K bN (1 −δ− 1 KL(δ,b,K,γ )) , (42) where L(δ, b, K, γ ) := 1 − 5δ 2 − γK − (1 − 2δ) log b ( 1 + cb 2 K + c √K ) ≥ 5ǫ 2 − γK − 1 3 log b ( 1 + cb 2 K + c √K ) . The last line holds by the assumption 1 3 ≤ δ ≤ 2 5 − ǫ. We may choose K ≍b,ǫ 1large enough and γ ≪b,ǫ 1 small enough so that γK + 1 3 log b ( 1 + cb 2 K + c √K ) ≤ 3ǫ 2 , say. For any such choice, L(δ, b, K, γ ) ≥ ǫ. This and the assumption Q ≫b,ǫ bδN ≥ b(1 −2δ)N for δ ≥ 1/3 implies (42) is ≪b,ǫ Q2bN (1 −δ−ǫ/K ). Thus if we take σ1(b, ǫ ) = ǫ/K we see that (40) and hence (41) hold in our case of Q ≫b,ǫ bδN .Next we treat the case when Q ≪b,ǫ bδN . Here one observes that the right hand sides of (40) and (41) are ≫b,ǫ bN Q1− σ1(b,ǫ ) δ and ≫b,ǫ bN Q1− σ1(b,ǫ ) δ e−σ∞ (b)N/ log Q,respectively. It thus suffices to show that the left hand sides of (40) and (41) are ≪b,ǫ bN Q1− σ1(b,ǫ ) δ and ≪b,ǫ bN Q1− σ1(b,ǫ ) δ e−σ∞ (b)N/ log Q, respectively. We treat these two in similar fashion (but with a small difference) as follows. Since Q ≪b,ǫ bδN by assumption, we can find an integer 1 ≤ M < N − 1 such that bδ(N −M) ≍b,ǫ Q. For any such choice of M , we split the products according to whether n ≤ M or M < n < N . This yields that the left hand sides of (40) and (41) are ≤ bM Z (with Z defined below in (43)) and ≤ Z max h∈Z 2≤q≤Q (q,h (b3−b))=1 ∏ 1≤n≤M φb (( h q + k b3 − b ) (bn + b2N −n))30 ALEKSANDR TUXANIDY AND DANIEL PANARIO respectively, where (43) Z := sup β∈R ∑ q≤Q (q,b )=1 ∗ ∑ h(q) ∏ M<n<N φb (( h q + β ) (bn + b2N −n)) . With regards to the product over 1 ≤ n ≤ M above, Proposition 6.2 implies it is ≪b bM exp ( −σ∞(b) M log Q ) ≪b,ǫ bM exp ( −σ∞(b) N log Q ) . Note we used the assumption bδ(N −M) ≍b,ǫ Q implying M = N − 1 δ log b Q+ Ob,ǫ (1). Similarly as done previously, one can show that ∑ q≤Q (q,b )=1 ∗ ∑ h(q) ∏ M<n<N φb (( h q + β ) (bn + b2N −n)) = ∑ q≤Q (q,b )=1 ∗ ∑ h(q) ΦN −M ( h q + bM β ) . Recall that bδ(N −M) ≍b,ǫ Q by assumption. Since we already showed above that (40) holds for arbitrary N ≥ 1 when Q ≫b,ǫ bδN (and in particular when Q ≍b,ǫ bδN ) we may apply this to our case of Q ≍b,ǫ bδ(N −M) (substituting N there with N − M here). This shows Z ≪b,ǫ Q2b(N −M)(1 −δ−σ1(b,ǫ )) + Q1− σ1(b,ǫ ) δ bN −M ≍b,ǫ Q1− σ1(b,ǫ ) δ bN −M . It follows that (40) and (41) also hold when Q ≪b,ǫ bδN and we thus conclude the proof. Equidistribution estimate We now prove Theorem 1.5 as an application of Proposition 8.1. First we need the following fact. Lemma 9.1. For an integer N ≥ 0, let Πb(2 N ) be as defined in (26) and let Π∗ b (2 N ) := {n ∈ Πb(2 N ) : (n, b 3 − b) = 1 } . We have #Π b(2 N ) = ( b − 1) bN and (44) #Π ∗ b (2 N ) = γ2(b) ϕ(b3 − b) b3 − b #Π b(2 N ) + O (b2τ (b2 − 1) ) , where γ2(b) := { b/ (b − 1) if b is even, 1 otherwise. Moreover #P∗ b (x) ≍b √x for x ≥ 1.Proof. Every integer n ∈ Πb(2 N ) can be written uniquely as n = nN bN + ∑ 0≤j<N nj (bj + b2N −j ) = nN bN + 2 ∑ 0≤j<N nj bj + ∑ 0≤j<N nj (b2N −j − bj )INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 31 for some unique digits 0 ≤ nj < b with n0 > 0. Thus #Π b(2 N ) = ( b − 1) bN . We note that every term of the right-most sum above is divisible by b2 − 1. Then (n, b 2 − 1) = nN bN + 2 ∑ 0≤j<N nj bj , b 2 − 1  . Since ( n, b ) = ( n0, b ) and b3 − b = b(b2 − 1), it follows #Π ∗ b (2 N ) = ∑ 0≤n0,...,n N<b (n0,b )=( nNbN+2 ∑ 0≤j<N njbj,b 2−1)=1 1= ∑ 0≤n0<b (n0,b )=1 ∑ 0≤nN<b ∑ 0≤n<b N−1 (nNbN+2 n0+2 bn,b 2−1)=1 1= ∑ 0≤n0<b (n0,b )=1 ∑ 0≤nN<b (nN,2)=1 if ( b, 2)=1 ∑ 0≤n<b N−1 (nNbN+2 n0+2 bn,b 2−1)=1 1. The equality before the last follows from the unique representation of integers in base b. The last equality follows from the fact that if b is odd, whence b2 − 1 is even, then (nN bN + 2 n0 + 2 bn, b 2 − 1) = 1 implies nN is odd. The inner sum equals S(nN bN + 2 n0), where for an integer a, S(a) := ∑ 0≤n<b N−1 (a+2 bn,b 2−1)=1 1. If b odd and a is even, S(a) = 0. Thus we assume a is odd if so is b. By the M¨ obius inversion formula, S(a) = ∑ d\|(b2−1) μ(d) ∑ 0≤n<b N−1 2bn ≡− a(d) 1. If b is odd and 2 bn ≡ − a(d), then d is odd as so is a by assumption. In the case when b is even, b2 − 1 is odd and so is any divisor d of b2 − 1. Thus d runs over odd divisors of b2 − 1 regardless. These are also coprime to b as ( b2 − 1, b ) = 1. Letting 2b denote the inverse of 2 b modulo d, the above is S(a) = ∑ d\|(b2−1) (d, 2)=1 μ(d) ∑ 0≤n<b N−1 n≡− 2ba (d) 1 = ∑ d\|(b2−1) (d, 2)=1 μ(d) ( bN −1 d + O(1) ) = γ(b) ϕ(b2 − 1) b2 − 1 bN −1 + O(τ (b2 − 1)) uniformly in a, where γ(b) := { 1 if b is even 2 if b is odd . Now (44) follows after we insert this into the expression for #Π ∗ b (2 N ) above, com-pute the resulting sum and recall that #Π b(2 N ) = ( b − 1) bN .32 ALEKSANDR TUXANIDY AND DANIEL PANARIO We now establish √x ≪b #P∗ b (x) ≪b √x for x ≥ 1. Since P∗ b (x) ⊆ Pb(x)and # Pb(x) ≍b √x, the upper bound holds. To show # P∗ b (x) ≫b √x, we first note that since 1 ∈ P∗ b (x), the lower bound holds for log b x bounded and we may assume then that log b x is arbitrarily large. In this case, let N be the largest integer such that Π b(2 N ) ⊆ Pb(x). Clearly bN ≍b √x and # P∗ b (x) ≥ #Π ∗ b (2 N ) ≍b bN for N large. Proof of Theorem 1.5. Let y ≤ x. Since every b-palindromic integer n ≥ 1 with ⌊log b n⌋ odd is divisible by b + 1, and b + 1 > 1 is a divisor of b3 − b, it follows P∗ b (y) = {n ∈ P0 b (y) : (n, b 3 − b) = 1 } . Then by this and the M¨ obius inversion formula 1(b3−b,n )=1 = ∑ r\|(b3−b,n ) μ(r), we have #P∗ b (y, a, q ) − #P∗ b (y) q = ∑ r\|(b3−b) μ(r) ∑ n∈P0 b(y) r\|n ( 1n≡a(q) − 1 q ) . Note 1r\|n = 1 r ∑ k(r) er (nk )and 1n≡a(q) − 1 q = 1 q ∑ 1≤h<q eq(−ah )eq(nh ). Inserting these expressions above, switching orders of summation and taking abso-lute values, we obtain ∣∣∣∣#P∗ b (y, a, q ) − #P∗ b (y) q ∣∣∣∣ ≤ ∑ r\|(b3−b) 1 r ∑ k(r) 1 q ∑ 1≤h<q ∣∣∣∣∣∣∑ n∈P0 b(y) e (( h q + k r ) n )∣∣∣∣∣∣ ≤ b2 ∑ 0≤N≤1 2log bx ∑ 0≤M≤N ∑ r\|(b3−b) 1 r ∑ k(r) 1 q ∑ 1≤h<q ΦM ( bN −M h q + bN −M k r ) uniformly in a ∈ Z and y ≤ x, by Lemma 6.1 and a switch in the order of sum-mation. If ( q, b ) = 1, we may substitute bN −M h with h above. Thus for any Q ≥ 1, ∑ q≤Q (q,b 3−b)=1 sup y≤x max a∈Z ∣∣∣∣#P∗ b (y, a, q ) − #P∗ b (y) q ∣∣∣∣ ≤ b2 ∑ 0≤N≤1 2log bx ∑ 0≤M≤N ∑ r\|(b3−b) 1 r ∑ k(r) ∑ q≤Q (q,b 3−b)=1 1 q ∑ 1≤h<q ΦM ( h q + bN −M k r ) ≤ b2τ (b3 − b) ∑ 0≤N≤1 2log bx ∑ 0≤M≤N max r\|(b3−b) k∈Z ∑ q≤Q (q,b 3−b)=1 1 q ∑ 1≤h<q ΦM ( h q + k r ) = b2τ (b3 − b) ∑ 0≤N≤1 2log bx ∑ 0≤M≤N max k∈Z S(M, Q, k ),(45) INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 33 where S(M, Q, k ) := ∑ q≤Q (q,b 3−b)=1 1 q ∑ 1≤h<q ΦM ( h q + k b3 − b ) . Splitting the sum according to the GCD of h, q and substituting variables, we have S(M, Q, k ) ≤ ∑ d≤Q/ 2 1 d ∑ 2≤q≤Q/d (q,b 3−b)=1 1 q ∗ ∑ h(q) ΦM ( h q + k b3 − b ) . We now split the sum over 2 ≤ q ≤ Q/d into ≪ log Q/d sums over dyadic segments (R/ 2, R ] with 2 ≤ R ≤ Q/d . Let 0 < ǫ ≤ 1/15 and consider the contribution of those q in any one such interval ( R/ 2, R ]. Bounding 1 /q and applying Proposition 8.1 with δ = 2 /5 − ǫ, we get ∑ R/ 2<q ≤R (q,b 3−b)=1 1 q ∗ ∑ h(q) ΦM ( h q + k b3 − b ) ≤ 2 R ∑ 2≤q≤R (q,b 3−b)=1 ∗ ∑ h(q) ΦM ( h q + k b3 − b ) ≪b,ǫ ( Rb ( 3 5+ǫ−σ1(b,ǫ )) M bM Rσ1 (b,ǫ ) ) exp ( −σ∞(b) M log R ) ≤ Rb ( 3 5+ǫ)M bM exp ( −c(b, ǫ )√M ) for some σ1(b, ǫ ), c (b, ǫ ), σ ∞(b) > 0 depending at most on b, ǫ . Note we used e−σ1(b,ǫ ) log R−σ∞ (b)M/ log R ≤ e−c(b,ǫ )√M for R ≥ 2 and M ≥ 0. Then the com-bined contribution of the ≪ log( Q/d ) dyadic pieces to the sum over 2 ≤ q ≤ Q/d is ≪b,ǫ Q d b( 3 5+ǫ)M bM exp ( −c(b, ǫ )√M ) log Q. Dividing this by d and summing over 1 ≤ d ≤ Q/ 2 we obtain S(M, Q, k ) ≪b,ǫ Qb ( 3 5+ǫ)M bM exp ( −c(b, ǫ )√M ) log 2 Q uniformly in k ∈ Z. Finally summing over M, N on the right hand side of (45) yields, for any 1 ≤ Q ≤ x1/5 and any 0 < ǫ ≤ 1/15, ∑ q≤Q (q,b 3−b)=1 sup y≤x max a∈Z ∣∣∣∣#P∗ b (y, a, q ) − #P∗ b (y) q ∣∣∣∣ ≪b,ǫ Qx 3 10 +ǫ 2 √xe −kb,ǫ √log x , where kb,ǫ > 0 is some value depending only on b, ǫ . Since the above holds for 0 < ǫ ≤ 1/15, it also holds for ǫ > 1/15. Now the result follows when we let Q = x1/5−ǫ and use the fact √x ≪b #P∗ b (x) from Lemma 9.1. Equidistribution with square moduli Before we can prove Theorem 1.4 we also need Proposition 10.1 below. Its proof is a direct application of the estimate of Baier-Zhao for the large sieve with square moduli (see Lemma 10.2 below) and the L∞-type bound of Col in Proposition 6.2. Most definitely, it is possible to replace the log A x below with something such as exp( kb,ǫ √log x) using Col’s bound in Proposition 6.2 and our argument in the proof. Nevertheless this is more than sufficient for our purposes. 34 ALEKSANDR TUXANIDY AND DANIEL PANARIO Proposition 10.1. For any x ≥ 2 and ǫ0 > 0, ∑ q≤x1/4−ǫ0 (q,b 3−b)=1 μ2(q) sup y≤x max a∈Z ∣∣∣∣∣∣∑ n∈P∗ b(y) ( 1n≡a(q2) − 1 q2 )∣∣∣∣∣∣ ≪A,b,ǫ 0 P∗ b (x) log A x for any A > 0. We need the following two lemmas. The first is due to Baier-Zhao and Lemma 10.3 below is a direct application of it. Lemma 10.2 (Baier-Zhao [2, Theorem 1]) . For any integers M, Q, N with Q, N ≥ 1 and any sequence (γn) of complex numbers, ∑ q≤Q ∗ ∑ a(q2) ∣∣∣∣∣∣∑ M<n ≤M+N γneq2 (an ) ∣∣∣∣∣∣ 2 ≪ǫ Kǫ(Q, N ) ∑ M<n ≤M+N \|γn\|2 for any ǫ > 0, where Kǫ(Q, N ) = ( QN )ǫ ( Q3 + N + min { N √Q, √N Q 2}) . Lemma 10.3. For any Q ≥ 1 and integer N ≥ 0, (46) sup β∈R ∑ q≤Q (q,b )=1 ∗ ∑ a(q2) ΦN ( a q2 + β ) ≪b,ǫ Q3+ ǫbN/ 2 + Q15 /8+ ǫbN for any ǫ > 0.Proof. We may assume that N is arbitrarily large and set M = ⌊4N/ 9⌋ > 1. Let us first consider the case when Q ≫b bM . Here we split the product as Φ N = P1P2,where P1 is the product over 1 ≤ n ≤ M and P2 is the product over M < n < N .By the Cauchy-Schwarz inequality, the left hand side above is ≤ ∑ q≤Q ∑ a(q2) P 21 ( a q2 + β ) 1/2 ∑ q≤Q (q,b )=1 ∗ ∑ a(q2) P 22 ( a q2 + β ) 1/2 . Arguing similarly as done in the proof of Lemma 8.2 and using the assumption Q ≫b bM , one can show that ∑ q≤Q ∑ a(q2) P 21 ( a q2 + β ) ≪ǫ Q3+ ǫbM for any ǫ > 0. With regards to the second sum, as we have seen before, ∑ q≤Q (q,b )=1 ∗ ∑ a(q2) P 22 ( a q2 + β ) = ∑ q≤Q (q,b )=1 ∗ ∑ a(q2) Φ2 N−M ( a q2 + βb M ) . We may write ΦN −M ( a q2 + βb M ) = ∣∣∣∣∣∣∑ 0≤n≤b2( N−M) γneq2 (an ) ∣∣∣∣∣∣ ,INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 35 where the coefficients γn are given by γn = e (βb M n) 1n∈S with S = ∑ 1≤m<N −M cm ( bm + b2( N −M)−m) : 0 ≤ c1, . . . , c N −M−1 < b  . Then it follows from Lemma 10.2 that ∑ q≤Q (q,b )=1 ∗ ∑ a(q2) P 22 ( a q2 + β ) ≪ǫ Kǫ ( Q, b 2( N −M)) bN −M . From the definition of Kǫ in Lemma 10.2 and the assumption Q ≫b bM with M = ⌊4N/ 9⌋, one can show that Kǫ ( Q, b 2( N −M)) ≪ǫ Qǫ ( Q3 + b2( N −M)√Q ) ≪b Q3+ ǫ. Now (46) follows after multiplying the two bounds involving P1, P 2 and taking square roots. Consider now the case when Q ≪b bM . Here we may find an integer L > 1such that Q ≍b b4( N −L)/9. As done in a previous section, we split the product Φ N according to whether n ≤ L or L < n < N . Bounding the product over 1 ≤ n ≤ L trivially by bL, noting that Z := ∑ q≤Q (q,b )=1 ∗ ∑ a(q2) ∏ L<n<N φb (( a q2 + β ) (bn + b2N −n)) = ∑ q≤Q (q,b )=1 ∗ ∑ a(q2) ΦN −L ( a q2 + bLβ ) and using (46) on Z with N there substituted with N − L (we may do this since now Q ≍b b4( N −L)/9 by our assumption) we obtain ∑ q≤Q (q,b )=1 ∗ ∑ a(q2) ΦN ( a q2 + β ) ≪b,ǫ Q3+ ǫb N +L 2 Q15 /8+ ǫbN . The assumption Q ≍b b4( N −L)/9 implies bL/ 2 ≍b bN/ 2/Q 9/8, whence the left-most term, of the right hand side above, is ≪b Q15 /8+ ǫbN . It follows that (46) also holds when Q ≪b bM . Proof of Proposition 10.1. For the sake of brevity let U = x1/4−ǫ0 and let E(U )denote the left hand side of the inequality in Proposition 10.1. Now the Fourier expansion 1n≡a(q2 ) = 1 q2 ∑ h(q2) e ( −ah q2 ) e ( nh q2 )36 ALEKSANDR TUXANIDY AND DANIEL PANARIO gives E(U ) ≤ ∑ q≤U (q,b 3−b)=1 μ2(q) q2 ∑ 1≤h<q 2 sup y≤x ∣∣∣∣∣∣∑ n∈P∗ b(y) e ( hn q2 )∣∣∣∣∣∣ . Splitting the sum according to the GCD of h, q and substituting variables gives E(U ) ≤ ∑ qr ≤U (qr,b 3−b)=1 μ2(qr ) q2r2 ∑ 1≤h<q 2r (h,q )=1 sup y≤x ∣∣∣∣∣∣∑ n∈P∗ b(y) e ( hn q2r )∣∣∣∣∣∣ . We now split the last sum according to the GCD of h, r and substitute variables as before, obtaining E(U ) ≤ ∑ dqr ≤Uqr> 1(dqr,b 3−b)=1 μ2(dqr ) d2q2r2 ∗ ∑ h(q2r) sup y≤x ∣∣∣∣∣∣∑ n∈P∗ b(y) e ( hn q2r )∣∣∣∣∣∣ . By the fact P∗ b (y) = {n ∈ P0 b (y) : (n, b 3 − b) = 1 }, the M¨ obius inversion formula 1(n,b 3−b)=1 = ∑ s\|(n,b 3−b) μ(s), the Fourier expansion 1s\|n = 1 s ∑ k(s) es(nk ) and a trivial bound for the d-sum, it follows E(U ) ≪ τ (b3 − b) max k∈Z ∑ 1<qr ≤U (qr,b 3−b)=( q,r )=1 1 q2r2 ∗ ∑ h(q2r) sup y≤x ∣∣∣∣∣∣∑ n∈P0 b(y) e ( hn q2r + kn b3 − b )∣∣∣∣∣∣ . By Lemma 6.1, the Chinese remainder theorem and dyadic decompositions of the intervals, we then have E(U ) ≪b (log x)4 sup QR ≪U 0≤N≤1 2log bxk∈Z T (N, Q, R, k ), where T (N, Q, R, k ) = 1 Q2R2 ∑ q≤Qr≤Rqr> 1(qr,b 3−b)=( q,r )=1 ∗ ∑ g(q2) ∗ ∑ h(r) ΦN ( g q2 + h r + k b3 − b ) . Let B > 1000 and C > 1000 B. We bound T (N, Q, R, k ) according to the following three cases of Q, R . Case 1 (Q ≪ log B x and R ≪ log C x): Here Proposition 6.2 implies T (N, Q, R, k ) ≪b,C,D √x log −D x for any D > 0. INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 37 Case 2 (Q ≪ log B x and R ≫ log C x): Taking the worst possible value of g/q 2 and applying Proposition 8.1 with δ = 2 5 − ǫ there, one has T (N, Q, R, k ) ≤ Q R2 sup β∈R ∑ r≤R (r,b )=1 ∗ ∑ h(r) ΦN ( h r + β ) ≪b,ǫ Qx 3 10 +ǫ Q√x R ≪C √x log C−B x for ǫ > 0 small and fixed. Remark 10.4. We employed Proposition 8.1 for the sake of convenience, but it is unnecessary here. It suffices instead to proceed via a Cauchy-Schwarz argument combined with Lemma 8.2. Case 3 (Q ≫ log B x): Note (47) T (N, Q, R, k ) ≤ 1 Q2 sup β∈R ∑ q≤Q (q,b )=1 ∗ ∑ g(q2) ΦN ( g q2 + β ) while Lemma 10.3 and the assumptions 2 N ≤ log b x and log B x ≪ Q ≪ U = x1/4−ǫ0 imply this is ≪b,ǫ Q1+ ǫx1/4 + √x Q 1 8−ǫ ≪B,ǫ 0 √x log − B 8+1 x after a suitable choice of ǫ in terms of B and ǫ0. Now Proposition 10.1 follows after choosing B, C, D sufficiently large in terms of A. Palindromic almost-primes We conclude the work with a proof of Theorem 1.4. To this end we use the following version of the almost-prime sieve. Lemma 11.1 (Almost-prime linear sieve with Richert’s weights) . Let A be a finite sequence of integers in [1 , x ]. For each integer d ≥ 1, set Ad := # {n ∈ A : d \| n} . Suppose for each such d that Ad = Xg (d) + rd for some rd, real number X and some multiplicative function g satisfying 0 ≤ g(p) < 1 for each prime p and ∏ u≤p<v (1 − g(p)) −1 ≤ K log v log u for any 2 ≤ u < v ≤ x, where K > 1 is some constant. For an integer r ≥ 2, set ur := 1 + 3 −r and ∆r := r + log 3 ( 3ur 4 ) .38 ALEKSANDR TUXANIDY AND DANIEL PANARIO Suppose that ∑ d≤D \|rd\| ≪ X log 3 x and ∑ D1/4≤p≤D1/ur Ap2 ≪ X log 3 x for some D satisfying D ≥ x1/∆r +ǫ for some ǫ > 0. Then ∑ n∈A P−(n)≥D1/4 Ω( n)≤r 1 ≍r,ǫ X ∏ p<x (1 − g(p)) . Proof. For the lower bound see for instance Corollaries 1.1 and 1.2 in Section 5 of Greaves . For the upper bound see Corollary 1.1 in Section 1 of . Proof of Theorem 1.4. We may assume that x ≫b 1 is large enough so that z := x1/21 is much larger than b3 − b. In this case, S := ∑ n∈Pb(x) P−(n)≥z Ω( n)≤6 1 = ∑ n∈P∗ b(x) P−(n)≥z Ω( n)≤6 1. For any integer d ≥ 1, set Ad := ∑ n∈P∗ b(x) d\|n 1 = # P∗ b (x, 0, d ). We can write this as Ad = g(d)# P∗ b (x) + rd, where g(d) = 1(d,b 3−b)=1 d and rd = # P∗ b (x, 0, d ) − g(d)# P∗ b (x). Clearly g is multiplicative, satisfies 0 ≤ g(p) < 1 for each prime p and ∏ u≤p<v (1 − g(p)) −1 ≤ ∏ u≤p<v ( 1 − 1 p )−1 = log v log u ( 1 + O ( 1 log u )) for any 2 ≤ u < v ≤ x, by the Mertens’ theorems. Set D := z4 = x4/21 = x1/5−1/105 . Since every number in P∗ b is coprime to b3 − b, ∑ d≤D \|rd\| = ∑ d≤D (d,b 3−b)=1 ∣∣∣∣#P∗ b (x, 0, d ) − #P∗ b (x) d ∣∣∣∣ ≪b P∗ b (x) log 10 x by Theorem 1.5. It also follows from Proposition 10.1 that ∑ D1/4≤p≤D Ap2 ≪b P∗ b (x) log 10 x .INFINITUDE OF PALINDROMIC ALMOST-PRIME NUMBERS 39 For ∆ 6 defined as in Lemma 11.1, one can check that x1/∆6+1 /100 ≤ D. 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Gelfond, Sur les nombres qui ont des propri´ et´ es additives et multiplicatives donn´ ees ,Acta Arith., 13, (1968), p.259–265 G. Greaves, Sieves in number theory , Ergebnisse der Mathematik und ihrer Grenzgebiete; Folge 3, Vol. 43, Springer-Verlag Berlin Heidelberg, 2001 D. Gusfield, Algorithms on Strings, Trees, and Sequences , Cambridge Univ. Press, New York, 1997 H. Halberstam and H.-E. Richert, Sieve Methods , Lond. Math. Soc. Monographs, v.4, Aca-demic Press, London, 1974 P. Humphries, Distributing points on the torus via modular inverses , Q. J. Math. v.73 n.1 (2022), p.1–16 J. Hofbauer, A simple proof of 1 + 1 /22 + 1 /32 + · · · = π2/6 and related identities , Amer. Math. Mon., v.109, n.2 (2002), p.196–200 A.J. Irving, Diophantine Approximation with products of two primes , J. Lond. Math. Soc., 89 (2014), 581–602 H. Iwaniec, Almost-primes represented by quadratic polynomials , Invent. Math. 47 (2), (1978), 178–188 H. Iwaniec and E. Kowalski, Analytic number theory , Colloquium Publications, v.53, Amer. Math. Soc., Providence, RI, 2004 E.H. Iwao, Even more pi in the sky: Calculating 100 trillion digits of pi on Google Cloud ,Google Cloud Blog, 2022 (last visited on November 16, 2023) 40 ALEKSANDR TUXANIDY AND DANIEL PANARIO L. Kuipers and H. Niederreiter, Uniform distribution of sequences , Wiley, New York, 1974 S. Larionov, A. Loskutov, E. Ryadchenko, Chromosome evolution with naked eye: palindromic context of the life origin , Chaos 18, n.1, 013105 (2008) C. Mauduit and J. Rivat, La somme des chiffres des carr´ es , Acta Math., 203 (2009), 107–148 C. Mauduit and J. Rivat, Sur un probl` eme de Gelfond : la somme des chiffres des nombres premiers ,Ann. of Math., v.171 no.3 (2010) J. Maynard, Primes with restricted digits , Invent. Math., v.217, 127–218 (2019) J. Morgenbesser, Gelfond’s sum of digits problems , Vienna University of Tech., Diploma the-sis A. Ostafe, I.E. Shparlinski, J.F. Voloch, Equations and Character Sums with Matrix Powers, Kloosterman Sums over Small Subgroups, and Quantum Ergodicity , Int. Math. Res. Not. IMRN (2023), v.2023, n.16, 14196–14238 S.N. Popova, On sums of products in Fp × Fp, Mat. Zametki 106 (2019): 262–79 (in Russian). A.H.L. Porto, V.C. Barbosa, Finding approximate palindromes in strings , Pattern Recognit., v.35 n.11 (2002), 2581–2591 A. Rajasekaran, J. Shallit, T. Smith, Additive number theory via automata theory , Theory Comput. Syst. 64 (2020), 542–567 R.M. Sheldon, The SATOR rebus: An unsolved cryptogram? , Cryptologia, v.27 (2003), 233– 287 SigmaGeek, Find large palindromic prime numbers in the decimal expansion of π (3, 1415 . . . ),2022 (last visited on July 29, 2023) C. Swaenepoel, Prime numbers with a positive proportion of preassigned digits , Proc. Lond. Math. Soc., 121 (2020), 83–151 T. Tao, Every odd number greater than 1 is the sum of at most five primes , Math. Comp., v.83, n.286, (2014), p. 997–1038 G. Tenenbaum, Introduction to analytic and probabilistic number theory , Cambridge studies in advanced mathematics: 46, Cambridge Univ. Press, 1995 School of Mathematics and Statistics, Carleton University, 1125 Colonel By Drive, Ottawa, Ontario, K1S 5B6, Canada Email address : AleksandrTuxanidyTor@cmail.carleton.ca, daniel@math.carleton.ca
16563	https://scoop.eduncle.com/why-licl-is-more-soluble-than-cscl	Why licl is more soluble than cscl Menu Try Free Downloads Marketplace UGC NET Commerce Computer Science & Applications Economics Education English General Paper 1 Human Resource Management Law Library And Information Science Management Political Science History Geography Psychology IIT JAM Biotechnology (BT) Biological Sciences (BL) Mathematical Statistics (MS) Mathematics (MA) Physics (PH) Geology (GG) Chemistry (CY) Economics (EN) CSIR NET Chemical Sciences Earth Sciences Life Sciences Mathematical Sciences Physical Sciences General Aptitude JEE JEE Main JEE Advanced AAT BITSAT NEET NEET UG AIIMS GATE Civil Computer Science And IT Electronics And Communication Electrical Mechanical RRB RRB JE RRB SSE RRB ALP RRB NTPC RRB Group D IBPS IBPS PO IBPS SO IBPS Clerk IBPS RRB SSC SSC CGL SSC CHSL SSC CPO SSC GD Constable SSC JE/AE Others Exam Goals Choose Exam Goal GATE IIT JAM UGC NET CSIR NET IBPS SSC RRB Want FREE Counselling for Exam Preparation? 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Call Us 800-355-2211 Speak With a Friendly Mentor. OR Request a Call Back Login Sign up T My Profile Dashboard Inbox My Orders My Wishlist Subscribe Exams My Courses Library Logout IIT JAM follow IIT JAM follow Articles (77) Products (104) Reviews Top Resources Why IIT JAM? Tips by Toppers Syllabus Eligibility Question Papers Answer Keys Mock Test Download Our Results & Toppers Why Choose Eduncle? ‹› Chandan Panigrahi posted an Question October 22, 2020 • 04:00 am 30 points IIT JAM Chemistry (CY) Why licl is more soluble than cscl 3 Answer(s)Answer Now 0 Likes 3 Comments 0 Shares Likes Share Comments S Suman Kumar The solubility of the chlorides decreases as we go down the group as the hydration enthalpy of the ions decreases at a greater rate than the lattice enthalpy of these chlorides. This is because the Cl- ion is larger than a molecule of H20 meaning that the decrease in lattice enthalpy (due to the increase in atomic radius of the positive ion as you go down the group) is more significant than the decrease in hydration enthalpy of the cation (the hydration enthalpy of Cl- remains constant). For solubility hydration enthalpy should be more;Therefore, CsCl is the least soluble of these group 1 chlorides, and LiCl is most soluble. Likes(1) Reply(1) Like(0) > Priyanshu kumar Best Answer The lattice energy of LiF is very high due to small size of Li+ions and F−ions. The hydration energy of LiF is lower than lattice energy. In case of LiCl, the hydration energy is higher than the lattice energy. Hence, LiCl is water soluble. Due to higher polarization, LiCl has some covalent character. Hence, it is soluble in non polar solvents such as acetone. Likes(0) Reply(10) please ask chandan if any doubt Like(0) check this chandan Like(0) sir than can we say that Lif is more soluble than Licl in polar solvent? Like(0) Lif is insoluble in water... Like(1) In Lithium fluoride the lattice enthalpy is very high due to small size of fluoride ions. In this case the hydration enthalpy is very less. Like(1) Some exceptional case you have to keep in mind Like(1) tqq sir😇 Like(0) sir than this order is wrong? Like(1) yes Like(1) now clear chandan Like(1) > Dinesh khalmaniya 1 Solubility of chlorides of alkali metal decrease down the group.Stability of alkali metal chlorides decreases down the group. lattice enthalpy of ionic solid of cation of same valency having same anion decreases. It's due to down the group size of cation increases so attractive force between anion and cation decreases. Likes(0) Reply(2) sir than for florides your statement is not apllicable? 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16564	https://simple.wikipedia.org/wiki/Succession_to_the_British_throne	Succession to the British throne - Simple English Wikipedia, the free encyclopedia Jump to content [x] Main menu Main menu move to sidebar hide Getting around Main page Simple start Simple talk New changes Show any page Help Contact us About Wikipedia Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Give to Wikipedia Create account Log in [x] Personal tools Give to Wikipedia Create account Log in [x] Toggle the table of contents Contents move to sidebar hide Beginning 1 Line of succession 2 Notes 3 References 4 Other websites Succession to the British throne [x] 40 languages Afrikaans العربية বাংলা Bosanski Català Čeština Deutsch Eesti Ελληνικά English Español فارسی Français Gaeilge 한국어 हिन्दी Bahasa Indonesia Italiano עברית Latviešu Magyar Bahasa Melayu Nederlands 日本語 Norsk bokmål Norsk nynorsk Polski Português Română Русский Scots සිංහල Slovenčina Suomi Svenska ไทย Türkçe Українська Tiếng Việt 中文 Change links Page Talk [x] English Read Change Change source View history [x] Tools Tools move to sidebar hide Actions Read Change Change source View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Sandbox Expand all Edit interlanguage links Print/export Make a book Download as PDF Page for printing In other projects Wikidata item From Simple English Wikipedia, the free encyclopedia The line of succession to the British throne is the order in which members of the royal family would come to the throne if the reigning king or queen died. At present the first in line is William, Prince of Wales, and then Prince William's eldest child, Prince George of Wales. Traditionally, males came before females in the line of succession. However, the law changed on 26 March 2015, so at present, for people born after 28 October 2011 the succession is decided only by age: older children come before younger children. This system of inheritance is called absolute primogeniture, as opposed to male-preference primogeniture. When someone who is in line to the throne has a child, that child comes after them and their older children, but before anyone else in the line of succession. Excluded from the line of succession are Catholics and illegitimate children. Line of succession [change \| change source] No official, complete version of the line of succession is maintained. The exact number, in more remote collateral lines, of the people who would be eligible is uncertain. In 2001, American genealogist William Addams Reitwiesner compiled a list of 4,973 living descendants of the Electress Sophia in order of succession without omitting Roman Catholics. When updated in January 2011, the list included 5,753. The annotated list below covers the first part of this line of succession, being limited to descendants of the sons of King George V, King Charles III's great-grandfather. The order of the first twenty-three numbered in the list, all descendants of Queen Elizabeth II, is given on the official website of the British monarchy; other list numbers and exclusions are explained by annotations and footnotes below. People named in italics are unnumbered either because they are deceased or because sources report them to be excluded from the succession. King George V (1865–1936) King Edward VIII (1894–1972) King George VI (1895–1952) Queen Elizabeth II (1926–2022) King Charles III (born 1948) (1)William, Prince of Wales (born 1982) B D W (2)Prince George of Wales (born 2013) B D W (3)Princess Charlotte of Wales (born 2015) B D W (4)Prince Louis of Wales (born 2018) B D W (5)Prince Harry, Duke of Sussex (born 1984) B D W (6)Prince Archie of Sussex (born 2019) B D W (7)Princess Lilibet of Sussex (born 2021) B D (8)Prince Andrew, Duke of York (born 1960) B D W (9)Princess Beatrice (born 1988) B D W (10) Sienna Mapelli Mozzi (born 2021) B D (11) Athena Mapelli Mozzi (born 2025) B D (12)Princess Eugenie (born 1990) B D W (13) August Brooksbank (born 2021) B D (14) Ernest Brooksbank (born 2023) B D (15)Prince Edward, Duke of Edinburgh (born 1964) B D W (16)James Mountbatten-Windsor, Earl of Wessex (born 2007) B D W (17)Lady Louise Mountbatten-Windsor (born 2003) B D W (18)Anne, Princess Royal (born 1950) B D W (19)Peter Phillips (born 1977) B D W (20) Savannah Phillips (born 2010) B D W (21) Isla Phillips (born 2012) B D W (22)Zara Tindall (née Phillips; born 1981) B D W (23) Mia Tindall (born 2014) B D W (24) Lena Tindall (born 2018) B D W (25) Lucas Tindall (born 2021) B D Princess Margaret, Countess of Snowdon (1930–2002) (26)David Armstrong-Jones, 2nd Earl of Snowdon (born 1961) D W (27) Charles Armstrong-Jones, Viscount Linley (born 1999) D W (28)Lady Margarita Armstrong-Jones (born 2002) D W (29)Lady Sarah Chatto (née Armstrong-Jones; born 1964) D W (30) Samuel Chatto (born 1996) D W (31) Arthur Chatto (born 1999) D W Prince Henry, Duke of Gloucester (1900–1974) (32)Prince Richard, Duke of Gloucester (born 1944) D W (33)Alexander Windsor, Earl of Ulster (born 1974) D W (34) Xan Windsor, Lord Culloden (born 2007) D W (35) Lady Cosima Windsor (born 2010) D W (36)Lady Davina Windsor (born 1977) D W (37) Senna Lewis (born 2010) D W (38) Tane Lewis (born 2012) D W (39) Lady Rose Gilman (née Windsor; born 1980) D W (40) Lyla Gilman (born 2010) D W (41) Rufus Gilman (born 2012) D W Prince George, Duke of Kent (1902–1942) (42)Prince Edward, Duke of Kent (born 1935) D W (43)George Windsor, Earl of St Andrews (born 1962) M D W Edward Windsor, Lord Downpatrick (born 1988) X D W Lady Marina Windsor (born 1992) X D W (44)Lady Amelia Windsor (born 1995) D W Lord Nicholas Windsor (born 1970) X D W (45) Albert Windsor (born 2007) X? D W[note 1] (46) Leopold Windsor (born 2009) X? D W[note 1] (47) Louis Windsor (born 2014) X? D W (48)Lady Helen Taylor (née Windsor; born 1964) D W (49) Columbus Taylor (born 1994) D W (50) Cassius Taylor (born 1996) D W (51) Eloise Taylor (born 2003) D W (52) Estella Taylor (born 2004) D W (53)Prince Michael of Kent (born 1942) M D W (54)Lord Frederick Windsor (born 1979) D W (55) Maud Windsor (born 2013) D W (56) Isabella Windsor (born 2016) D W (57)Lady Gabriella Kingston (née Windsor; born 1981) D W (58)Princess Alexandra, The Honourable Lady Ogilvy (born 1936) D W (59)James Ogilvy (born 1964) D W (60) Alexander Ogilvy (born 1996) D W (61) Flora Vesterberg (née Ogilvy; born 1994) D W (62)Marina Ogilvy (born 1966) D W (63) Christian Mowatt (born 1993) D W (64) Zenouska Mowatt (born 1990) D W Notes and sources\| Mark \| Source for listing or note on exclusion from succession \| --- \| \| B \| Listed by the official website of the British monarchy, "Succession", as of February 2025[update] \| \| D \| Debrett's website (as of 29 January 2025[update]): "The Line of Succession" \| \| W \| Whitaker's Almanack 2021 \| \| M \| These people had been excluded through marriage to a Roman Catholic. This exclusion was repealed on 26 March 2015, restoring them to the line of succession, when the Perth Agreement came into effect. \| \| X \| Excluded as Roman Catholics. This exclusion is not affected by changes subsequent to the Perth Agreement. \| Notes [change \| change source] ↑ Jump up to: 1.01.1Albert and Leopold Windsor were listed on the official website of the British monarchy until 2015 and in the 2013 edition of Whitaker's Almanack as following Estella Taylor (b. 2004) and eligible to succeed; MSN News, and Whitaker's Almanack 2015 and 2021 list them after Lady Amelia Windsor and before Lady Helen Taylor. They were baptised as Catholics, and are not listed in line in editions of Whitaker's earlier than 2012. Debrett's says their place in the line of succession will be re-assessed when they reach the age of majority. References [change \| change source] ↑Reitwiesner, W. A. "Persons eligible to succeed to the British Throne as of 1 Jan 2001". wargs.com. Archived from the original on 9 June 2005. ↑Lewis, David. "Persons eligible to succeed to the British Throne as of 1 Jan 2011". wargs.com. Archived from the original on 17 May 2011. ↑"Succession". Official website of the British monarchy. Archived from the original on 13 May 2016. Retrieved 19 September 2022. ↑"Succession", retrieved 11 February 2025 ↑Debrett's, "The Line of Succession" ↑Whitaker's Almanack, 2021. Oxford: Rebellion, ISBN978-1-7810-8978-1, p.26 Other websites [change \| change source] British line of succession on any date in the past 200 years Succession at the Official web site of the British Royal Family \| Expand v t e Orders of succession \| \| Monarchies \| Bahrain·Belgium·Bhutan·Brunei·Cambodia·Denmark·Japan·Jordan·Kuwait·Lesotho·Liechtenstein·Luxembourg·Malaysia·Monaco·Morocco·Netherlands·Norway·Oman·Qatar·Saudi Arabia·Spain·Swaziland·Sweden·Thailand·Tonga·United Kingdom and other Commonwealth realms \| \| Presidencies \| Argentina·Austria·Brazil·Colombia·DR Congo·Dominican Republic·France·Finland·Germany·India·Indonesia·Ireland·Republic of Korea·Mexico·Pakistan·Peru·Philippines·Poland·Portugal·Romania·Russia·Sri Lanka·Republic of China (Taiwan)·United States \| Retrieved from " Categories: Monarchy of the United Kingdom Lines of succession of monarchies Hidden categories: Articles containing potentially dated statements from February 2025 All articles containing potentially dated statements Articles containing potentially dated statements from January 2025 This page was last changed on 11 July 2025, at 02:48. Text is available under the Creative Commons Attribution-ShareAlike License and the GFDL; additional terms may apply. See Terms of Use for details. Privacy policy About Wikipedia Disclaimers Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Succession to the British throne 40 languagesAdd topic
16565	https://people.math.osu.edu/chmutov.1/wor-gr-su24/Presentation%202.pdf	Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Boninger’s Mistake G. Black The Ohio State University July 22, 2024 Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Checkerboard Colorablitity of a Surface Definition 1 Let Σ be a closed orientable surface and D be a link diagram of a link L Ă Σ ˆ r0, 1s. We say that the diagram D is Checkerboard colorable if each connected region of ΣzD can be colored by t0, 1u such that there is no arc of D whach have 2 sides that are the same color. Figure: Checkerboard Colorable and Non-Checkerboard Colorable Embeddings of the Unknot in T2 Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Thistlethwaite’s Polynomial Definition 2 In [TH], Thistlethwaite defined the Laurent polynomial τrΓs of a Graph Γ recursively by; τrΓs “ A´1 e τrΓ ´ es AeτrΓ{es If e is not a bridge nor loop; τrΓs “ A´3 e τrΓ{es If e is a bridge; τrΓs “ A3 eτrΓ ´ es If e is a loop; τrΓ1 \ Γ2s “ dτrΓ1sτrΓ2s For disjoint union Γ1 \ Γ2; τr‚s “ 1 Where d “ ´A2 ´ A´2 and Ae “ # A if signpeq “ A´1 if signpeq “ ´ Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Boninger’s Polynomial from a Graph Definition 3 Let D be a checkerboard colorable Virtual link diagram and Γ be the associated Signed Tait Graph then we can define; νD,Γptq “ ˆ p´Aq´3wpDqτrΓs ˙ A´2“t1{2 Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix The Theorem of Boninger The following was a result given in [BO]: Theorem 4 (Boninger 23’) Let Σ be a closed orientable surface. Let D Ă Σ be a checkerboard colorable, non-split link diagram, and L Ă Σ ˆ r0, 1s the associated link. Let Γ and Γ1 be the signed Tait graphs associated to the two checkerboard colorings of D. Then, tνD,Γptq, νD,Γ1ptqu Is an isotopy invariant of L. This Theorem turns out to be false as it is currently stated. Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Counter-Example Consider the following Knot on T2 The diagram on the right, which we will call D has Tait graphs ´ We will call this graph Γ ´ We will call this graph Γ1 Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Counter-Example We can easily see that τrΓ1s “ 1. We can compute, ´ ` ´ ´ A´1 A ´A´3 ´A3 Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Counter-Example This means τrΓs “ ´A2 ´ A´2. Since wpDq “ 0 the set that Boninger’s theorem claims is an invariant is, t1, ´A2 ´ A´2u Notice that D is a diagram for the unknot, so we can campute from the other diagram given that the set Boninger’s Theorem calims is invariant is, t1, 1u This gives us a Counter-Example to Boninger’s theorem. Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Essential Annulus Definition 5 Let Σ be a closed compact orientable surface and L Ă Σ ˆ r0, 1s be a Link on Σ. We say that A Ă pΣ ˆ r0, 1sqzL is an Essential Annulus if, ‚ A is homeomorphic to the Annulus in the plane which we will denote A “ tx P R2 : 1 ď \|x\| ď 2u ‚ One boundary component of A which is denoted BA1 is contained in pΣ ˆ t1uqzL. Additionally the other boundary component of A which is denoted BA2 is contained in pΣ ˆ t0uqzL ‚ A does not bound a ball in pΣ ˆ r0, 1sqzL. Analagously this means that in pΣ ˆ r0, 1sqzL it is not possible to continuously shrink A to a line where BA1 stays contained in pΣ ˆ t1uqzL and BA2 stays contained in pΣ ˆ t0uqzL Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Example of Essential Annulus For our Counter-Example knot we have an Essential Annulus. To illustrate this, note that because of the topological nature of the definition of Essential Annulus it is a property of the link L which means it doesenn’t depend on the diagram D of the link. So consider the following diagram of the unknot of T2, Observe that the green surface is an essential Annulus. Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix A Possible Fix The following was reccomended as a possible fix to the problem in an email correspondence with Dr. Boninger. Theorem 6 Let Σ be a closed orientable surface. Let L Ă Σ ˆ r0, 1s be a Link with no Essential Annulus on Σ ˆ r0, 1s and D Ă Σ be a checkerboard colorable, non-split link diagram of L. If Γ and Γ1 are the signed Tait graphs associated to the two checkerboard colorings of D then, tνD,Γptq, νD,Γ1ptqu Is an isotopy invariant of L. Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Why Should This Fix Boninger’s Theorem Note that the counter-example we found shows that his proof for Reidemeister 2 was wrong. In fact Boninger’s proofs for Reidemeister 2 and 3 were both wrong, but it can be shown that the original statement was still true for Reidemeister 3. The Issue with the given proof is Boninger assumes that some regions of the checkerboard coloring are distinct. If we assume that all of the regions we consider under the Reidemeister moves are distinct then the proof given by Boninger holds true. Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Proving Invariance Under Reidemeister 2 Consider the following checkerboard coloring for a link L on the surface Σ described in the fixed version of the Theorem. If the upper region and lower region on the right diagram are the same region we have that there is a path from the upper region to the lower region that does not pass through any crossings in the right diagram. Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Proving Invariance Under Reidemeister 2 Since we have this path we can construct the following surface on Σ ˆ r0, 1s. We can notice that the green line drawn in this diagram represents a surface that we can see is clarly homeomorphic to an annulus because the top of the line connects back to the bottom of the line as depicted. We can note that since we have strands of the knot on both sides of the annulus it does not bound a ball in pΣ ˆ r0, 1sqzL. Thus the surface drawn is an essential annulus. Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix Proving Invariance Under Reidemeister 2 So we obtain a contradiction. This means that on the right diagram we must have that the upper region is not the same region as the lower region. Thus we can proceed to use the proof Boninger gave to show that with this additional assumption we have that the set of polynomials tνD,Γptq, νD,Γ1ptqu is invariant under the Reidemeister 2 move. Boninger’s Mistake G. Black Review Counter-Example Essential Annulus A Possible Fix References GK Greg Kuperberg,What is a virtual link?, Algebraic & Geometric Topology, Volume 3 (2003) 587–591; preprint TH Morwen B. Thistlethwaite, A spanning tree expansion of the Jones polynomial, Topology 26 (1987), no. 3, 297–309 BO Boninger, J. (2022). The Jones Polynomial from a Goeritz Matrix. Bulletin of the London Mathematical Society, 55(2), 732-755. KA L.H. Kauffman, A Tutte polynomial for signed graphs, Discrete Appl. Math. 25 (1989) 105–127.
16566	https://www.chemistrysteps.com/reduction-of-carboxylic-acids/	Skip to content Carboxylic Acids and Their DerivativesCarboxylic Acids Derivatives Reduction of Carboxylic Acids Carboxylic acids can be reduced to primary alcohols using an excess of strong reducing agents such as lithium aluminum hydride (LiAlH4 or LAH). As sodium borohydride (NaBH4) is the most common reducing agent we learn about together with LiAlH4, let’s emphasize that it is very inefficient in reducing carboxylic acids, so do not use it if you do not want to lose points on your exam. Back to the LiAlH4 reduction: Aside from the acidic workup in the last step, the reduction is overall a three-step conversation. Remember that LIAlH4 is a strong base, so the first step is the deprotonation of the carboxylic acid. In the next two steps, we have a nucleophilic addition of the hydride to the carbonyl group, thus an excess of reducing agent is needed: The formation of carboxylate ion after deprotonation is a good explanation for why NaBH4 is not effective in reducing carboxylic acids. Because of the resonance electron donation by the negatively charged oxygen, the carboxylate ion is not as electrophilic, and the addition of hydride requires a powerful reducing agent. In the case of LiAlH4, the addition of hydride ion is facilitated by the coordination of the Al to the carbonyl oxygen. This withdraws some of the electron density thus making the carbon more electron-deficient. After the hydride addition, we have an elimination step where the –OAlH2 is expelled resorting the carbonyl in a form of an aldehyde: Notice that aldehyde is another intermediate in this reaction, but we cannot stop the reduction at this stage. Remember, aldehydes are more reactive than carboxylic acids, esters, and especially carboxylate ions because there is no resonance donation of the oxygen to the carbonyl carbon which reduces the electron-deficient character of the carbonyl. The greater reactivity of the aldehyde explains why t cannot be isolated as the final product. Reduction of Carboxylic Acids with Borane We saw in the mechanism above that the carbonyl is activated by the coordination of the oxygen to Al which serves as a Lewis acid there. Now, borane is another Lewis acid, and when coordinated to an oxygen, it becomes electron-rich and thus a hydride donor. This turns out to be a good tool for reducing carboxylic acids to alcohol. It is still an addition-elimination reaction to the carbonyl, however, the intermediate is not an aldehyde by rather a boronic ester: You can read more about it in different articles including Check Also LiAlH4 and NaBH4 Carbonyl Reduction Mechanism Reduction of Carboxylic Acid Derivatives Alcohols from Carbonyl Reductions – Practice Problems Fischer Esterification Ester Hydrolysis by Acid and Base-Catalyzed Hydrolysis What is Transesterification? Esters Reaction with Amines – The Aminolysis Mechanism Ester Reactions Summary and Practice Problems Preparation of Acyl (Acid) Chlorides (ROCl) Reactions of Acid Chlorides (ROCl) with Nucleophiles Reaction of Acyl Chlorides with Grignard and Gilman (Organocuprate) Reagents Reduction of Acyl Chlorides by LiAlH4, NaBH4, and LiAl(OtBu)3H Preparation and Reaction Mechanism of Carboxylic Anhydrides Amides – Structure and Reactivity Naming Amides Amides Hydrolysis: Acid and Base-Catalyzed Mechanism Amide Dehydration Mechanism by SOCl2, POCl3, and P2O5 Amide Reduction Mechanism by LiAlH4 Amides Preparation and Reactions Summary Amides from Carboxylic Acids-DCC and EDC Coupling The Mechanism of Nitrile Hydrolysis To Carboxylic Acid Nitrile Reduction Mechanism with LiAlH4 and DIBAL to Amine or Aldehyde The Mechanism of Grignard and Organolithium Reactions with Nitriles Carboxylic Acids to Ketones Esters to Ketones Carboxylic Acids and Their Derivatives Practice Problems Carboxylic Acids and Their Derivatives Quiz Share Your Thoughts, Ask that Question! Cancel reply
16567	https://www.merriam-webster.com/word-of-the-day/hackneyed-2013-04-05	Word of the Day: Hackneyed \| Merriam-Webster Chatbot Chatbot Games Word of the Day Grammar Word Finder Slang NewNewsletters Wordplay Rhymes Thesaurus Join MWU More Games Word of the Day Grammar Wordplay Slang Rhymes Word Finder Newsletters New Thesaurus Join MWU Shop Books Merch Log In Username My Words Recents Account Log Out Est. 1828 Word of the Day : April 5, 2013 hackneyed play adjective HAK-need PrevNext What It Means : lacking in freshness or originality hackneyed in Context The new crime drama's characters are shallow stereotypes who engage one another in hackneyed dialogue. "Though it has been floating around Hollywood for a few years, and drawn some praise from insiders and would-be insiders, Noah Haidle's screenplay is pretty derivative and hackneyed." - From a review by Philip Martin in Arkansas Democrat-Gazette (Little Rock), February 1, 2013 Build your vocabulary! Get Word of the Day in your inbox every day. Test Your Vocabulary Challenging Standardized Test Words Despite the cacophony, the student tried to study. Difficult subjectLoud sounds Low lightingLate hour Test your vocabulary with our 10-question quiz! TAKE THE QUIZ Pick the best words! PLAY Did You Know? "Hackney" entered the English language in the 14th century as a noun. Some think perhaps it came from "Hakeneye" (now "Hackney"), the name of a town (now a borough) in England. Others dispute this explanation, pointing to similar forms in other European languages. The noun "hackney," in any case, refers to a horse suitable for ordinary riding or driving-as opposed to one used as a draft animal or a war charger. When "hackney" was first used as a verb in the late 16th century, it often meant "to make common or frequent use of." Later, it meant "to make trite, vulgar, or commonplace." The adjective "hackneyed" began to be used in the 18th century and now is a common synonym for "trite." Test Your Vocabulary with M-W Quizzes Previous Simplify The Convoluted Expression Quiz Play Now Spot Even More Misspelled Words Quiz Play Now Famous Novels, Last Lines Quiz Play Now Challenging Words You Should Know Play Now Commonly Confused Words Quiz Play Now Challenging Standardized Test Words Play Now Simplify The Convoluted Expression Quiz Play Now Spot Even More Misspelled Words Quiz Play Now Famous Novels, Last Lines Quiz Play Now Challenging Words You Should Know Play Now Commonly Confused Words Quiz Play Now Challenging Standardized Test Words Play Now Simplify The Convoluted Expression Quiz Play Now Spot Even More Misspelled Words Quiz Play Now Famous Novels, Last Lines Quiz Play Now Next Test Your Memory What is the meaning of "meritorious," our Word of the Day from March 6? The answer is … Podcast Theme music by Joshua Stamper ©2006 New Jerusalem Music/ASCAP More Words of the Day Sep 28 kerfuffle Sep 27 vociferous Sep 26 gesundheit Sep 25 anomaly Sep 24 brandish Sep 23 nonpareil SEE ALL WORDS OF THE DAY ### Can you solve 4 words at once? Play Play ### Can you solve 4 words at once? Play Play Love words? Need even more definitions? Subscribe to America's largest dictionary and get thousands more definitions and advanced search—ad free! Merriam-Webster unabridged Games & Quizzes See All Quordle Can you solve 4 words at once?Play Blossom Pick the best words!Play The Missing Letter A daily crossword with a twist Play Challenging Words You Should Know Not a quiz for the pusillanimous Take the quiz See All Merriam Webster Learn a new word every day. Delivered to your inbox! Help About Us Advertising Info Contact Us Privacy Policy Terms of Use Facebook Twitter YouTube Instagram © 2025 Merriam-Webster, Incorporated ✕ Do not sell or share my personal information. You have chosen to opt-out of the sale or sharing of your information from this site and any of its affiliates. To opt back in please click the "Customize my ad experience" link. This site collects information through the use of cookies and other tracking tools. Cookies and these tools do not contain any information that personally identifies a user, but personal information that would be stored about you may be linked to the information stored in and obtained from them. This information would be used and shared for Analytics, Ad Serving, Interest Based Advertising, among other purposes. For more information please visit this site's Privacy Policy. CANCEL CONTINUE Information from your device can be used to personalize your ad experience. Do not sell or share my personal information.
16568	https://es.wikipedia.org/wiki/Teor%C3%ADa_celular	Ir al contenido Teoría celular العربية Azərbaycanca Български বাংলা Català کوردی Čeština Чӑвашла Deutsch Ελληνικά English Esperanto Eesti Euskara فارسی Suomi Français Gaeilge Galego עברית हिन्दी Hrvatski Kreyòl ayisyen Magyar Bahasa Indonesia Ido Italiano 日本語 ქართული Қазақша 한국어 Кыргызча Latina Lietuvių Latviešu മലയാളം Bahasa Melayu नेपाली Nederlands Norsk bokmål Oromoo ਪੰਜਾਬੀ Papiamentu Polski پښتو Português Romnă Русский Саха тыла සිංහල Simple English Slovenčina Slovenščina Shqip Српски / srpski Svenska தமிழ் ไทย Tagalog Türkçe Українська اردو Oʻzbekcha / ўзбекча Tiếng Việt 吴语 მარგალური 中文粵語 Editar enlaces De Wikipedia, la enciclopedia libre En biología, la teoría celular es una teoría científica formulada por primera vez a mediados del siglo XIX, según la cual los organismos vivos están formados por células, que son la unidad estructural básica de todos los organismos, y que todas las células provienen de células preexistentes. Las células son la unidad básica de estructura en todos los organismos y también la unidad básica de reproducción. Los tres principios de la teoría celular son: Todos los organismos vivos están compuestos por una o más células. La célula es la unidad básica de estructura y organización en los organismos. Las células surgen de células preexistentes. La teoría alguna vez fue universalmente aceptada, pero ahora algunos biólogos consideran entidades no celulares como los virus como organismos vivos, y por lo tanto no están de acuerdo con el primer principio. A partir de 2021: "la opinión de los expertos permanece dividida aproximadamente en un tercio entre sí, no y no sé". Como no existe una definición universalmente aceptada de la vida, la discusión aún continúa. La hipótesis más aceptada para explicar el origen biológico de las células eucariotas establece que cierto tipo de procariotas necesitaron trabajar de manera grupal, de donde consecuentemente cada una fue especializándose y adquiriendo una función que más tarde estructurarían un organismo completo. De manera general se establece que el antepasado del cual surgen todas las clasificaciones y que presenta características comunes se denomina protobionte, ya que este estará dotado de los implementos necesarios para la transcripción y la traducción genética; de esta se derivan por diversas características más especializadas los modelos de procariotas (Archaea y Bacterias), las cuales permanecieron así durante un período largo de tiempo, en el cual estos organismos adaptaron su proceso metabólico a las intensas condiciones terrestres. Muchas de estas definiciones no se las pudo establecer de manera inmediata ya que se partía de que la materia se conformaba por moléculas y no se podía concluir cuales eran las unidades básicas estructurales. En cuanto a la realización de los intentos de las células por buscar su supervivencia se generaron otras etapas celulares que las describen: Heterótrofas anaerobias: Necesitadas de compuestos orgánicos disponibles en el medio, con el paso del tiempo se llegaron a limitar estas condiciones, razón por la cual cierto grupo de células tuvo que buscar otras adaptaciones, de donde se derivan. Fotosíntesis: Algunas de estas células primitivas logran fabricar sustancias orgánicas mediante la fijación y reducción de CO2, dando los primeros pasos para la fotosíntesis, medio de alimentación de carácter autótrofo, en la fotosíntesis se utiliza el agua como donante de electrones, esto nos da como origen el O2, este proceso será indispensable, mediante el cual se logra el cambio de una atmósfera reductora en la oxidante. Cianobacterias: Se especializan en el uso para su alimentación del dióxido de carbono disponible, de manera que su fuente de recursos representara un alto índice y que permitiese el desarrollo libre del otro grupo de células, y dentro de las cuales otras llegaron a adaptarlas como una respiración aerobia para su metabolismo y consecuentemente para una nutrición heterótrofa de carácter aerobio. Después de varios de estos procesos, se empezó a generar una etapa evolutiva más intensa de manera que las células que habían sobrevivido a las condiciones precarias del inicio empezaron a desarrollar mecanismos que les ayudaron a aprovechar los recursos que se iban presentando, de esa manera comenzaron a sintetizar los compuestos más básicos convirtiéndolas en sustancias más complejas catalogadas como orgánicas, de esa manera su desarrollo se empezó a dar con mayor eficiencia. De ésta síntesis, las sustancias que más se destacan por presentarse precariamente son nucleótidos y aminoácidos, las cuales se pudieron sintetizar mediante experimentos posteriores y se las consideraba como moléculas sencillas, la asociación de estas permitió la formación de moléculas de mayor complejidad como las proteínas las cuales definieron las características de los seres vivos a nivel de su composición. Antecedentes [editar] A inicios del siglo XVII Marcello Malpighi realizó sus investigaciones sobre la organización vegetal y determinó una pequeña estructura que más tarde sería denominada en referencia con la propuesta de Hooke con "cellulae" pero por los bajos fundamentos que estos presentaban se creó una controversia sobre si la célula era lo que se definía como un ser real o una cavidad. Consecuentemente se logró establecer que aquello que se observaba era lo que hoy se conoce como pared celular, con la concreción de este tema se logró establecer que la célula era completa y compleja, lo que fomentó el interés sobre la membrana celular y nuclear. : : : Otro aspecto que contribuye como promotor de planteamientos que ahora ya son teóricos y aplicables, es la consecución de mitocondrias que se basan en pruebas vestigiales muy relacionados al ADN y la construcción de proteínas, de ello se puede destacar la adopción endosimbiótica. Principios [editar] Los conceptos de materia viva y célula están estrechamente ligados a la biología ya que esta abarca todos los preceptos que la componen y la relación y función que ejercen dentro del ecosistema y, a nivel celular, en su organismo. La materia viva se distingue de la inerte (materia muerta) por su capacidad para metabolizar y autoperpetuarse, además de contar con las estructuras que hacen posible la ocurrencia de estas dos funciones; si la materia metaboliza y se autoperpetúa por sí misma, se dice que está viva, que es lo que comprende funcionalmente a los componentes para el propósito que presenta cada ser vivo. Muchos de estos organismos son capaces de cumplir con diversas funciones dependiendo de sus necesidades pero sus células presentan las mismas estructuras, ya que es a escalas macromoleculares donde se logra diferenciar lo que realmente define a cada uno. Varios científicos postularon numerosos principios para darle una estructura adecuada: Carl Woese (1928 - 2012) Robert Hooke (1635 - 1703) Anton van Leeuwenhoek (1632- 1723) Xavier Bichat (1771 - 1882) Theodor Schwann (1810 - 1882) Matthias Jacob Schleiden (1804 - 1881) Robert Hooke, observó una muestra de corcho bajo el microscopio, Hooke no vio células tal y como las conocemos actualmente, él observó que el corcho estaba formado por una serie de celdillas de color transparente, ordenadas de manera semejante a las celdas de una colmena; para referirse a cada una de estas celdas, él utiliza la palabra célula. Tiempo después este proceso sería realizado por otros científicos con evaluación microscópica de manera que se pudo visualizar correctamente el modo de funcionamiento de cada una de estas celdas formadas. Anton van Leeuwenhoek, usando unos microscopios simples, realizó observaciones sentando las bases de la morfología microscópica. Fue el primero en realizar importantes descubrimientos con microscopios fabricados por sí mismo. Desde 1674 hasta su muerte realizó numerosos descubrimientos. Introdujo mejoras en la fabricación de microscopios y fue el precursor de la biología experimental, la biología celular y la microbiología. A finales del siglo XVIII, Xavier Bichat, da la primera definición de tejido (un conjunto de células con forma y función semejantes). Más adelante, en 1819, Meyer le dará el nombre de Histología a un libro de Bichat titulado Anatomía general aplicada a la Fisiología y a la Medicina. Dos científicos alemanes, Theodor Schwann, histólogo y fisiólogo, y Jakob Schleiden, botánico, se percataron de cierta comunidad fundamental en la estructura microscópica de animales y plantas, en particular la presencia de centros o núcleos, que el botánico británico Robert Brown había descrito recientemente (1831). Publicaron juntos la obra Investigaciones microscópicas sobre la concordancia de la estructura y el crecimiento de las plantas y los animales (1839). Asentaron el primer y segundo principio de la teoría celular histórica: Todo en los seres vivos está formado por células o productos secretados por las células. Primer Principio de la teoría celular. La célula es la unidad básica de organización de la vida. Segundo principio de la teoría celular Otro alemán, el médico Rudolf Virchow, interesado en la especificidad celular de la patología (sólo algunas clases de células parecen implicadas en cada enfermedad) explicó lo que debemos considerar el tercer principio: Toda célula se ha originado a partir de otra célula, por división de esta. Tercer principio de la teoría celular Ahora estamos en condiciones de añadir que la división es por bipartición, porque a pesar de ciertas apariencias, la división es siempre, en el fondo, binaria, El principio lo popularizó Virchow en la forma de un aforismo creado por François Vincent Raspail «omnis cellula e cellula». Virchow terminó con las especulaciones que hacían descender la célula de un hipotético blastema. Su postulado, que implica la continuidad de las estirpes celulares, está en el origen de la observación por August Weismann de la existencia de una línea germinal, a través de la cual se establece en animales (incluido el hombre) la continuidad entre padres e hijos y, por lo tanto, del concepto moderno de herencia biológica. La teoría celular fue debatida a lo largo del siglo XIX, pero fue Pasteur el que, con sus experimentos sobre la multiplicación de los microorganismos unicelulares, donde se demostró que una célula partía de otra ya existente, da lugar a su aceptación rotunda y definitiva. Santiago Ramón y Cajal logró unificar todos los tejidos del cuerpo en la teoría celular, al demostrar que el tejido nervioso está formado por células. Su teoría, denominada “neuronismo” o “doctrina de la neurona”, explicaba el sistema nervioso como un conglomerado de unidades independientes. Pudo demostrarlo gracias a las técnicas de tinción de su contemporáneo Camillo Golgi, quien perfeccionó la observación de células mediante el empleo de nitrato de plata, logrando identificar una de las células nerviosas. Cajal y Golgi recibieron por ello el premio Nobel en 1906. Conjuntamente a estos postulados y con posteriores investigaciones se logró definir las características y propiedades de los seres vivos con base en ciertos parámetros que se determinan: Nivel de organización, nutrición, crecimiento. diferenciación, señalización química, respuesta a estímulos, evolución y la capacidad de autorregulación. Albert Kolliker tuvo una visión diferente y complementaria en relación con los fundamentos que se aplicaban a la explicación de tejidos y a niveles más amplios sobre el organismo en sí, de esa manera planteaba una analogía con la electricidad y explicaba como ésta a pesar de ser tan imprediscible también tenía una composición y una estructura, y por esta manera de presentarse planteaba a los tejidos como una organización de elementos básicos conocidos como células. Muchos de los análisis posteriormente junto a sus antecedentes se lograron gracias al desarrollo de microscopios de observación de células en desarrollo conjuntamente a técnicas de tratamiento que hoy se implementan en los laboratorios, este se vio precedido por la microscopia electrónica y en ese mismo siglo se intensificó los estudios sobre sustancias celulares como proteínas y enzimas, lo que permitió definir funciones y propósitos. Carl Woese, en 1980, establecía una capacidad biológica de las células primitivas para realizar actividades de funcionamiento genómico y fue de esta fundamentación que se las pudo clasificar como: Bacterias y Arqueas. llegó a obtener este conocimiento gracias a que centró sus estudios en el llamado ARN ribosómico 16s, el cual es una secuencia que está en todos los seres vivos, con evolución lenta donde se puede rastrear cambios evolutivos en los organismos. Concepto moderno [editar] El concepto moderno de teoría celular se puede resumir en los siguientes principios: Todos los seres vivos están formados por células, bacterias y otro tipo de organismos, o por sus productos de secreción. La célula es la unidad estructural de la materia viva, y dentro de los diferentes niveles de complejidad biológica, una célula puede ser suficiente para constituir un organismo. Sin embargo en la naturaleza encontramos múltiples tipos de organismos multicelulares que son parte de la naturaleza expresados a manera macroscópica manteniendo un potencial extenso en lo que respecta funciones biológicas conformativas. Las funciones vitales de los organismos ocurren dentro de las células, o en su entorno inmediato, controladas por sustancias que ellas secretan. Cada célula es un sistema abierto único e irrepetible, que intercambia materia y energía con su medio. En una célula caben todas las funciones vitales, de manera que basta una célula para tener un ser vivo (que será un ser vivo unicelular). Así pues, la célula es la unidad fisiológica de la vida. Todas las células proceden de células procariotas preexistentes, por división de éstas (Omnis cellula e cellula) o célula madre. Es la unidad de origen de todos los seres vivos. Esto determina además de la sucesión y conservación de estas unidades, las pertinencias sobre ciertas características homólogas entre las eucariotas y las procariotas de manera que fundamenta el hecho de haber establecido a las células procariotas como las primeras en este mundo y las más primitivas. Concepto actual general [editar] La célula es la unidad morfológica, fisiológica y de origen de todo ser vivo, concepto que engloba los tres principios del concepto moderno. Conjuntamente a ello se considera los postulados que hablan acerca de la manera en como se reproduce y se origina, de manera que se pueda definir exactamente las características que esta unidad estructural posee. Continuamente a ello se considera a todas las clasificaciones en el mundo de los seres vivos que se componen de ellas y también aquellos que mantienen su vida con una sola de ellas. Véase también [editar] Protobionte Procariota Eucariota Referencias [editar] ↑ Farnsworth, Keith D. (2021). «An organisational systems-biology view of viruses explains why they are not alive». Biosystems 200: 104324. ISSN 0303-2647. PMID 33307144. S2CID 228169048. doi:10.1016/j.biosystems.2020.104324. ↑ Reyes, J. (1973). «La célula: evolución y desarrollo de los sistemas celulares». Archivado desde el original el 7 de octubre de 2009. Consultado el 24 de febrero de 2017. ↑ Heras, A. (2013). «Los planetas y la vida: la aparición de las celulas y la fotosíntesis». Consultado el ↑ Saltar a: a b Peña, I. (2004). «Origen de la Vida Sobre la Tierra». Universidad Católica de Valparaíso. Consultado el 21 de enero de 2017. ↑ Barbadilla, A. (2003). «La evolución biológica». Consultado el 23 de febrero de 2017. ↑ «G Citología 1». s.f. Consultado el 20 de enero de 2017.. ↑ Saltar a: a b c d Colegio Virgen de Europa. «LA CÉLULA: Origen, organización y estructura.». Consultado el 4 de diciembre de 2016. ↑ Carrión, J. ««Los primeros organismos fotosintéticos no fueron plantas, sino bacterias»». Archivado desde el original el 2 de febrero de 2017. Consultado el 21 de enero de 2017. ↑ González, J. «Formación y evolución de la atmósfera terrestre». Nuestro planeta. Consultado el 21 de enero de 2017. ↑ Gonzales, S. (2010). «Evolución celular». Consultado el 24 de febrero de 2017. ↑ Saltar a: a b Navarra, L. (2017). «Teoría celular». Consultado el 20 de febrero de 2017. ↑ Saltar a: a b González, U. (2014). «Línea de tiempo sobre la teoría celular». Consultado el 25 de febrero de 2017. ↑ Curtis, B. (2007). «Origen de la célula». Consultado el 24 de febrero de 2017. ↑ Barcat, J. «Robert Hooke». Universidad de Buenos Aires. Consultado el 21 de enero de 2017. ↑ Saltar a: a b c Da Silva, F. (2006). «HISTORIA DE LA TEORÍA CELULAR». Consultado el 20 de enero de 2017. ↑ Teulón, A. (1982). «La teoría celular, paradigma de la biología del siglo XIX.». Consultado el 20 de enero de 2017.. ↑ Mazarello, P. (1999). «La historia de la teoría celular un concepto unificador». Historia de la Teoría Celular. Consultado el 20 de enero de 2017.. ↑ Saltar a: a b Gama, Á. BIOLOGÍA 1. p. 43. Consultado el 21 de enero de 2017. ↑ Saltar a: a b Ghershman, E. (2010). «Orígen de las células Eucariotas». Archivado desde el original el 11 de marzo de 2014. Consultado el 25 de febrero de 2017.. ↑ Martorell, E. (2010). «Unidad fundamental de la vida». Consultado el 24 de febrero de 2017. ↑ ECIR Editorial (2015). «III. Metabolismo y Autoperpetuación.». ECIR editorial. Consultado el 2 de enero de 2017. ↑ Saltar a: a b Gama, M. (2004). Enrique Quintanar Duarte, ed. BIOLOGÍA, biogénesis y microorganismos. Pearson Educación de México. p. 117. Consultado el 15 de enero de 2017. ↑ Cazares, J. (1986). «El origen de la vida y la evolución celular hoy». Consultado el 25 de febrero de 2017. ↑ Historias de la Ciencia (2010). «Historia de la ciencia - Robert Hooke». Consultado el 2 de enero de 2017. ↑ Dzul, M. (2007). «Teoría celular». Consultado el 23 de febrero de 2017. ↑ Garrido, B., Barcía, M. (20 de abril de 2013). «ANTON VAN LEEUWENHOEK, EL PRIMER CAZADOR DE MICROBIOS». II Jornada de Biologia a l'Ensenyament. Consultado el 2 de enero de 2017. ↑ Fresquet, J. (2000). «Marie François Xavier Bichat (1771-1802)». Consultado el 2 de enero de 2016. ↑ Da Silva, F. W. (2006). MESTRADO EM ENSINO DE CIÊNCIAS E MATEMÁTICA (Doctoral dissertation, Universidade Cruzeiro do Sul). ↑ Da Silva, F. W. (2006). MESTRADO DE CIÊNCIAS ↑ Puigferres, J. (1859). publicación de todas sus obras François Vincent Raspail. IMPRENTA EUTERPE. Consultado el 15 de enero de 2017. ↑ Antillón, J. (2005). Historia y Filosofía de la Medicina. Universidad de Costa Rica. p. 75. Consultado el 15 de enero de 2017. ↑ Oñate, L. (2006). Biología 1. CENGAGE LEARNING. p. 103. Consultado el 15 de enero de 2017. ↑ Campos B., et. al. (2004). «Capítulo 3». Biología. p. 75. Consultado el 15 de enero de 2017. ↑ Stanier, I. et. al. (2005). Microbiología Segunda Edición. Reverté S.A. p. 5. Consultado el 21 de enero de 2017. ↑ López, J. (2006). RAMON Y CAJAL. ESPAÑA: UNIVERSIDAD DE VALENCIA, UNIVERSIDAD DE GRANADA. p. 340. Consultado el 15 de enero de 2017. ↑ González, M. (2006). «A cien años del Premio Nobel». Archivado desde el original el 31 de octubre de 2017. Consultado el 25 de febrero de 2017. ↑ Alonso, J. (1994). Los métodos de Golgi. EDICIONES UNIVERSALES SALAMANCA. p. 35. Consultado el 15 de enero de 2017. ↑ EMP (2011). «Introducción a la Biología celular». Consultado el 25 de febrero de 2017. ↑ Teulón, A. (1982). «La teoría celular, paradigma de la biología del siglo XIX». La teoría celular, paradigma de la biología del siglo XIX. Consultado el 23 de febrero de 2017. ↑ Jaga, I. (2017). «Evolución y teoría celular». Consultado el 25 de febrero de 2017. ↑ Woese, C. (2008). «Carl Woese y las Nuevas Perspectivas en la Evolución». Carl Woese y las Nuevas Perspectivas en la Evolución. Consultado el 23 de febrero de 2017. ↑ «Biology Study Guides». sites.google.com. Consultado el 21 de octubre de 2023. ↑ «Which Cell Organelle Regulates the Movement of Molecules in and \| Biology+». StudyBiology (en inglés). Consultado el 21 de octubre de 2023. ↑ Fortoul, T. (2011). «La célula: su estructura y función». Consultado el 23 de febrero de 2017. ↑ Rilar, A. (2008). «Evolución de la teoría celular». Consultado el 22 de febrero de 2017. ↑ Martínez, G. (2016). «Biología: Teoría de la evolución celular». Consultado el 23 de febrero de 2017. ↑ Omnis cellula e cellula ↑ Saltz, E. (2005). «Estructura y función celular». Consultado el 24 de febrero de 2017. ↑ Bernedo, C. (2011). «La Evolución Celular y sus Repercusiones en la Medicina Contemporánea I». Consultado el 24 de febrero de 2017. ↑ Zamora, L. (2011). «Teorías de la evolución - biología celular». Consultado el 23 de febrero de 2017. \| \| \| --- \| \| Control de autoridades \| Proyectos Wikimedia Datos: Q177935 Diccionarios y enciclopedias Britannica: url \| Datos: Q177935 Obtenido de « Categorías: Biología celular Célula Teorías biológicas Ciencia y tecnología de Alemania del siglo XIX Ciencia de 1839 Alemania en 1839
16569	https://www.octalsteel.com/steel-pipe-dimensions-sizes/	Home Products Steel Pipe Casing and Tubing Steel Pipe Fittings Sucker Rods Drill Pipe Steel Plate Valves Equipment Resources Projects FAQ Technical Notes Skills About Octal Who we are Team & News Contact Us X Close Your Responsible Supplier Partner for Oil and Gas Products. Support & Sales +86 372 2157660 / +86 186 3727 1277 Support & Sales [email protected] Home Products Steel Pipe Welded Steel Pipe Carbon Steel Pipe Chrome Moly Pipe Stainless Steel Pipe CRA Clad & Lined Pipe Casing and Tubing Steel Casing Pipe API 5CT OCTG Tubing, API 5CT Tubing Steel Pipe Fittings Pipe Accessories & Fittings Steel Pipe Flange Sucker Rods API 11B Sucker Rod Drill Pipe API 5DP Drill Pipe Steel Plate API Carbon Steel Plate Valves API 6D Valve, Pipeline Valve Equipment Well Head X Tree Resources Projects FAQ Technical Notes Skills About Octal Who we are Team & News Contact Us Request Quote Steel Pipe Dimensions & Sizes Chart (Schedule 40, 80 Pipe) Means Home > News > Pipe Knowledge > Steel Pipe Dimensions & Sizes Chart (Schedule 40, 80 Pipe) Means Steel Pipe Dimensions & Sizes Chart (Schedule 40, 80 Pipe) Means People are often confused when it comes to accurately describing the dimensions or sizes of steel pipes, as well as effectively communicating their desired size to the seller. Any misunderstandings or the purchase of an incorrect size could lead to significant trouble. So, what are the right ways to express the dimensions of a steel pipe? For your better understanding, we will explain from the following aspects: The 3 elements of pipe dimension Dimension Standards of carbon and stainless steel pipe (ASME B36.10M & B36.19M) Pipe Size Schedule (Schedule 40 & 80 steel pipe means) Means of Nominal Pipe Size (NPS) and Nominal Diameter (DN) Steel Pipe Dimension Chart (Size chart) Pipe Weight Class Schedule (WGT) And finally, we will tells you how to describe steel pipe dimensions & sizes correctly. Steel Pipe Dimension 3 elements A complete description for steel pipe dimension includes outer diameter (OD), wall thickness (WT), pipe length (Normally 20 ft 6 meter, or 40 ft 12 meters). Through these elements we could calculate the pipe weight, how much pressure the pipe could bear, and the cost per foot or per meter. So that’s why we always need to know the correct pipe size. Steel Pipe Dimensions Chart (Size Chart) Pipe Schedule Chart unit in mm as below, view here for Pipe Schedule Chart in inch. Dimension standards for steel pipe There are different standards to describe the steel pipe size, OD and wall thickness, which mainly are ASME B 36.10, ASME B 36.19. Relevant standard specification ASME B 36.10M and B 36.19M Both ASME B36.10 and B36.19 are the standard specifications for the dimensions of the steel pipe and accessories. ASME B36.10M The standard covers the standardization of steel pipe dimensions and sizes, including seamless or welded types, which are used in high or low temperatures and pressures.The pipe is distinguished from a tube (Pipe vs Tube). In this case, the pipe is specifically used for pipeline systems and the transmission of fluids such as oil and gas, water, and slurry. The standard used is ASME B 36.10M.In this standard, when the pipe’s Outer Diameter is smaller than 12.75 in (NPS 12, DN 300), the pipe’s actual diameter is larger than NPS (Nominal Pipe Size) or DN (Nominal Diameter). As for steel tube dimensions, the actual outside diameter is the same as the pipe number for all sizes. Why pipes with sizes smaller than NPS 12 (DN 300) have different OD First, we need to acknowledge that all pipe sizes should be identified as nominal pipe size (NPS). Secondly, for NPS 1/8 (DN 6) to NPS 12 (DN 300), these sizes are based on a standard outside diameter (O.D.). Initially, the O.D. was chosen to specify the pipe dimensions, but since pipes always have a wall thickness and internal diameter (I.D.), the difference between the O.D. and I.D. varies for small pipes. However, as the pipe dimensions increase, the O.D. and I.D. become approximately equal. Since there is no direct relationship between the old standard thickness (O.D.) and Nominal Size, both of them are accepted according to the ASME B 36.10 M standard. Steel Pipe Schedule In ASME B36.10M, Pipe schedule is identified as Standard (STD), Ex-Standard (XS), or Double Extra Strong(XSS); Or with Schedule No. 5, 10, 20, 30, 40, 80, 120, 160. ASME B36.19M ASTM B36.19M is for the stainless steel pipe dimensions, including seamless and welded types (same as the previous standard.) This dimension standard is mostly the same as ASTM B36.10M. Wherever the different part is:a. For NPS 14 to NPS 22 (DN 350-550), schedule 10S;b. NPS 12 of Schedule 40Sc. NPS 10 and 12 of schedule 80S. The above-mentioned pipe thickness is different with B36.10M, so suffix “S” is used here. Ways to express pipe dimensions For Pipe wall thickness: Use Steel Pipe Schedule, like schedule 40 steel pipe, schedule 80 pipe. 2. For pipe Diameters: Nominal Pipe Size (NPS) and DN (Nominal Diameter) 3. Pipe weight Class (WGT), LB/FT (Pounds per foot), KG/M (Kg per meter) What is Steel Pipe Dimensions Schedule? Steel pipe schedule is an indicating method represented by ASME B 36.10, and used in many other standards, marked with “Sch”. Sch is the abbreviation of schedule, generally appearing in the American steel pipe standard, which is a prefix of a series number. For example, Sch 80, where “80” is a pipe number from chart/table ASME B 36.10. “Since the steel pipe’s main application is to transport the fluids under pressure, their internal diameter is their critical size. This critical size is taken as nominal bore (NB). Therefore, if steel pipe carries the fluids with pressure, it is particularly important that pipe shall have enough strength and wall thickness. So wall thickness is specified in Schedules, which means the pipe schedule, abbreviated as SCH. Here ASME is the given standard and definition for the pipe schedule.” The pipe schedule formula: Sch.=P/[ó]t×1000P is the Designed pressure, units in MPa;[ó]t is Allowable stress of materials under design temperature, Units in MPa. What does SCH mean for the steel pipe dimensions When describing the steel pipe parameter, we usually use the pipe schedule, which is a method that represents pipe wall thickness with number. Pipe schedule (sch. ) is not a wall thickness, but a wall thickness series. Different pipe schedules mean different wall thicknesses for the steel pipe in the same diameter. The most frequent indications of schedule are SCH 5, 5S, 10, 10S, 20, 20S, 30, 40, 40S, 60, 80, 80S, 100, 120, 140, 160. The larger the table number, the thicker the surface pipe wall, the higher the pressure resistance. Schedule 40, 80 steel pipe dimension means If you are new to the pipe industry, you may wonder why you always see schedule 40 or 80 steel pipes everywhere. What kind of material is used for these pipes? As you have read above articles you know that Schedule 40 and 80 represent pipe wall thickness, but why are they always searched by buyers? Here is the reason: Schedule 40 and 80 steel pipes are commonly required in various industries due to the typical pressure they can withstand. Because these pipes are designed to handle higher pressures, they are often requested in large quantities to meet the demands of different industries. The material standard for such thickness pipes has no limitations, you could ask for sch 40 stainless steel pipe, like ASTM A312 Grade 316L; Or sch 40 carbon steel pipe, such as API 5L, ASTM A53, ASTM A106B, A 179, A252, A333 etc.. What is the Nominal Pipe Size (NPS)? Nominal Pipe Size (NPS) is a North American set of standard sizes for pipes used for high or low pressures and temperatures. Pipe size is specified with two non-dimensional numbers: a nominal pipe size (NPS) based on inches, and a schedule (Sched. or Sch.). What is DN (Nominal Diameter) Nominal diameter, also known as outside diameter, is used to represent the general diameter of various pipes and pipeline accessories. Due to the thin wall of the pipe, the outside and inside diameters are almost the same. Therefore, the average value of both parameters is used as the nominal diameter of the pipe. DN (nominal diameter) ensures that pipes and pipe fittings with the same nominal diameter can be interconnected, providing interchangeability. Although the nominal diameter value is close to or equal to the inside diameter of the pipe, it does not represent the actual sense of the pipe diameter. The nominal size is denoted by a numerical symbol followed by the letter “DN” and the unit is indicated in millimeters after the symbol. For example, DN50 represents a pipe with a nominal diameter of 50 mm. DN (mm) and NPS (inch) conversion Nominal diameter DN (mm) and NPS (inch) conversion 1. Conversion DN ( mm ) that requires separate memory \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| DN (mm) \| 6 \| 8 \| 10 \| 15 \| 20 \| 25 \| 32 \| 40 \| 50 \| 65 \| 80 \| \| NPS(inch) \| 1/8 \| 1/4 \| 3/8 \| 1/2 \| 3/4 \| 1 \| 1 1/4 \| 1 1/2 \| 2 \| 2 1/2 \| 3 \| 2. Greater than or equal to DN100 diameter and NPS conversion DN=25NPS;NPS=DN/25 3. Exact conversion 1 inch =25.4 mm Pipe Weight Class Schedule The WGT class (weight class) is an indication of the pipe wall thickness that was used in the past, but is still in use today. It consists of three grades: STD (standard), XS (extra strong), and XXS (double extra strong). In the early days of pipe production, each size of pipe had only one specification, known as standard tube (STD). However, in order to handle high pressure fluids, thicker pipes known as XS were introduced. XXS pipes were then developed to handle even higher pressure fluids. As new materials processing technology emerged, there was a growing demand for more cost-effective thin-walled pipes. This eventually led to the introduction of the aforementioned pipe grades. The corresponding relationship between pipe schedules and weight classes can be found in the ASME B36.10 and ASME B36.19 specifications. How to describe steel pipe dimensions and size correctly? For example: a. Expressed as “pipe outside diameter × wall thickness”, such as Φ 88.9mm x 5.49mm (3 1/2” x 0.216” ). 114.3mm x 6.02mm (4 1/2” x 0.237”), length 6m (20ft) or 12m (40ft), Single Random Length (SRL 18-25ft), or Double Random Length (DRL 38-40ft). b. Expressed as “NPS x Schedule”, NPS 3 inch x Sch 40, NPS 4 inch x Sch 40. Same size as above specification. c. Expressed as “NPS x WGT Class”, NPS 3 inch x SCH STD, NPS 4 inch x SCH STD. Same size above. d. “Pipe Outer Diameter x lb/ft.” is commonly used in North America and South America. For example, a pipe with an outer diameter of 3 1/2” would be described as having a weight of 16.8 lb/ft. The unit “lb/ft” stands for pounds per foot. 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It includes Drill Pipe, Steel Casing Pipe and Tubing Steel Tee (Equal and Reducing Tee) – Common Use Pipe Fittings Contact Now ! Octal is located in China recognized as a leading supplier, distributor, and manufacturer union in providing piping solutions for oil and gas company. Product ranges in Steel Pipe, Casing and Tubing, Steel Plate, Sucker Rod, Steel Pipe Fittings, Valves, and Equipment for pipelines. Contact Info Unit 2544, 25/F, Tower A, Wanda Plaza, Wenfeng District, Anyang, Henan, China & Building A, No.18 Dirun Road, Zhengdong New District, Zhengzhou, Henan, China +86 372 2157660 / +86 186 3727 1277 [email protected] Home About Octal Skills Projects Products FAQ Contact Us Products Welded Pipe Casing Pipes OCTG Tubing Sucker Rod Pipe Accessories Projects Line Pipes Casing & Tubing Sucker Rods Steel Plates Pipe Fittings Copyright 2012 - 2025 OCTALSTEEL \| All Rights Reserved
16570	https://writing.wisc.edu/handbook/semicolons/	Cookie Notice We use cookies on this site. By continuing to browse without changing your browser settings to block or delete cookies, you agree to the UW–Madison Cookie Notice. Skip to main content University of Wisconsin–Madison Using Semicolons Semicolons help you connect closely related ideas when a style mark stronger than a comma is needed. Rules for Using Semicolons A semicolon is most commonly used to link (in a single sentence) two independent clauses that are closely related in thought. When a semicolon is used to join two or more ideas (parts) in a sentence, those ideas are then given equal position or rank. Some people write with a word processor; others write with a pen or pencil. Use a semicolon between two independent clauses that are connected by conjunctive adverbs or transitional phrases. However they choose to write, people are allowed to make their own decisions; as a result, many people swear by their writing methods. Use a semicolon between items in a list or series if any of the items contain commas. There are basically two ways to write: with a pen or pencil, which is inexpensive and easily accessible; or by computer and printer, which is more expensive but quick and neat. Use a semicolon between independent clauses joined by a coordinating conjunction if the clauses are already punctuated with commas or if the clauses are lengthy. Some people write with a word processor, tablet, or even a phone; but others, for different reasons, choose to write with a pen or pencil. Avoid using a comma when a semicolon is needed: Incorrect: The cow is brown, it is also old. Correct: The cow is brown; it is also old. What’s going on here? Both parts of the sentence are independent clauses, and commas should not be used to connect independent clauses if there is no coordinating conjunction. This mistake is known as a comma splice. Incorrect: I like cows, however, I hate the way they smell. Correct: I like cows; however, I hate the way they smell. What’s going on here? The conjunctive adverb however signals a connection between two independent clauses, and commas should not be used to connect independent clauses if there is no coordinating conjunction. Incorrect: I like cows: they give us milk, which tastes good, they give us beef, which also tastes good, and they give us leather, which is used for shoes and coats. Correct: I like cows: they give us milk, which tastes good; they give us beef, which also tastes good; and they give us leather, which is used for shoes and coats. What’s going on here? It’s unclear what the three list items are, since the items are separated by commas. Incorrect: Cows, though their bovine majesty has been on the wane in recent millennia, are still one of the great species of this planet, domesticated, yet proud, they ruminate silently as we humans pass tumultuously by. Correct: Cows, though their bovine majesty has been on the wane in recent millennia, are still one of the great species of this planet; domesticated, yet proud, they ruminate silently as we humans pass tumultuously by. What’s going on here? It’s unclear where the first independent clause ends and the second independent clause begins. Avoid using a semicolon when a comma is needed: Incorrect: Because cows smell; they offend me. Correct: Because cows smell, they offend me. What’s going on here? The first part is not an independent clause, so no semicolon is required. Grammar and Punctuation This is an accordion element with a series of buttons that open and close related content panels. Using Dashes Using Commas Using Semicolons Using Coordinating Conjunctions Using Conjunctive Adverbs Subject-Verb Agreement Using Gender–Neutral Pronouns in Academic Writing How to Proofread Twelve Common Errors: An Editing Checklist Clear, Concise Sentences
16571	https://map.gsfc.nasa.gov/	Wilkinson Microwave Anisotropy Probe (WMAP) Wilkinson Microwave Anisotropy Probe Search / Site Map HOME NEWS UNIVERSE MISSION RESOURCES SITE INFO WMAP Highlights Home WMAP Results WMAP Technical Papers WMAP Science Universe Tutorial Animations Teaching Resources Image Gallery Site Map History of Universe The Primordial Universe Build a Universe Facts and Questions Wilkinson Microwave Anisotropy Probe The Wilkinson Microwave Anisotropy Probe (WMAP) is a NASA Explorer mission that launched June 2001 to make fundamental measurements of cosmology -- the study of the properties of our universe as a whole. WMAP has been stunningly successful, producing our new Standard Model of Cosmology. WMAP's data stream has ended. Full analysis of the data is now complete. Publications have been submitted as of 12/20/2012. WMAP Team Receives the 2018 Breakthrough Prize "WMAP wins the Fundamental Physics Award for detailed maps of the early universe that greatly improved our knowledge of the evolution of the cosmos and the fluctuations that seeded the formation of galaxies." The prize is awarded to Charles L. Bennett, Gary Hinshaw, Norman Jarosik, Lyman Page Jr., David N. Spergel, and the 27 member WMAP Science Team. For more information about the prize visit: 2018 Breakthrough Prize - Fundamental Physics - Laureates WMAP Team Receives the 2012 Gruber Cosmology Prize "The Gruber Foundation proudly presents the 2012 Cosmology Prize to Charles Bennett and the Wilkinson Microwave Anisotropy Probe team for their exquisite measurements of anisotropies in the relic radiation from the Big Bang---the Cosmic Microwave Background. These measurements have helped to secure rigorous constraints on the origin, content, age, and geometry of the Universe, transforming our current paradigm of structure formation from appealing scenario into precise science." Other Members of the WMAP team are: Chris Barnes, Rachel Bean, Olivier Doré, Joanna Dunkley, Benjamin M. Gold, Michael Greason, Mark Halpern, Robert Hill, Gary F. Hinshaw, Norman Jarosik, Alan Kogut, Eiichiro Komatsu, David Larson, Michele Limon, Stephan S. Meyer, Michael R. Nolta, Nils Odegard, Lyman Page, Hiranya V. Peiris, Kendrick Smith, David N. Spergel, Greg S. Tucker, Licia Verde, Janet L. Weiland, Edward Wollack, and Edward L. (Ned) Wright. For more information about the prize: 2012 Gruber Cosmology Prize Press Release For more information about the WMAP Team: Charles Bennett and the WMAP Team WMAP's Top Ten The WMAP science team has� 1. ... has put the "precision" in "precision cosmology" by reducing the allowed volume of cosmological parameters by a factor in excess of 68,000. The three most highly cited physics and astronomy papers published in the new millennium are WMAP scientific papers--- reflecting WMAP's enormous impact. …mapped the pattern of tiny fluctuations in the Cosmic Microwave Background (CMB) radiation (the oldest light in the universe) and produced the first fine-resolution (0.2 degree) full-sky map of the microwave sky. …determined the universe to be 13.77 billion years old to within a half percent. …nailed down the curvature of space to within 0.4% of "flat" Euclidean. …determined that ordinary atoms (also called baryons) make up only 4.6% of the universe. …completed a census of the universe and finds that dark matter (matter not made up of atoms) is 24.0% …determined that dark energy, in the form of a cosmological constant, makes up 71.4% of the universe, causing the expansion rate of the universe to speed up. - "Lingering doubts about the existence of dark energy and the composition of the universe dissolved when the WMAP satellite took the most detailed picture ever of the cosmic microwave background (CMB)." - Science Magazine 2003, "Breakthrough of the Year" article … mapped the polarization of the microwave radiation over the full sky and discovered that the universe was reionized earlier than previously believed. - "WMAP scores on large-scale structure. By measuring the polarization in the CMB it is possible to look at the amplitude of the fluctuations of density in the universe that produced the first galaxies. That is a real breakthrough in our understanding of the origin of structure." - ScienceWatch: "What's Hot in Physics", Simon Mitton, Mar./Apr. 2008. …detected that the amplitude of the variations in the density of the universe on big scales is slightly larger than smaller scales. This, along with other results, supports "inflation", the idea is that the universe underwent a dramatic period of expansion, growing by more than a trillion trillion fold in less than a trillionth of a trillionth of a second. Tiny fluctuations were generated during this expansion that eventually grew to form galaxies. 10. … determined that the distribution of these fluctuations follows a bell curve with the same properties across the sky, and that there are equal numbers of hot and cold spots in the map. The simplest version of the inflation idea predicted these properties and remarkably, WMAP’s precision measurement of the properties of the fluctuations has confirmed these predictions, in detail.The End of the Beginning "The last word from WMAP marks the end of the beginning in our quest to understand the Universe. WMAP has brought precision to cosmology and the Universe will never be the same." -Adam Riess, recipient of the 2011 Nobel Prize in physics Steven Hawking "Stephen Hawking recently told New Scientist that WMAP's evidence for inflation was the most exciting development in physics during his career." -2013 Smart Guide,New Scientist WMAP Awarded Gruber Prize "WMAP has had a transformative impact on the field of cosmology. It provided strong confirmation of our basic picture of the universe and added unprecedented precision. It is the benchmark for almost every other cosmological measurement and sets a very high bar for future experiments." -Neil Turok, Director of the Perimeter Institute for Theoretical Physics WMAP mission scores 'world's most cited' in science pubs "WMAP results were among the most-cited scientific papers in the world across all scientific disciplines [in 2011], not just in physics and astronomy. It also happened in 2003, 2007 and 2009. This time WMAP captured the first, second and third spots in the rankings in a single year�a science trifecta." - Johns Hopkins University Gazette, 2012/04/23 THE RED-HOT RESEARCH PAPERS OF 2011 "Achieving particular distinction atop the list are three reports from the Wilkinson Microwave Anisotropy Probe (WMAP), launched in 2001. ... Paper #1 in the table, delivering the "cosmological interpretations" of the WMAP seven-year data, had already been cited more than 500 times before the end of its first year of publication." - Science Watch/ The Hottest Research of 2011 by Christopher King ----The List "Every astronomer will remember the moment he heard the results from WMAP." "Before the WMAP results, astronomers and physicists had put together a very implausible picture of our universe. It had a tiny amount of ordinary matter. It had a modest amount of dark matter, whatever that is. It had an overwhelming amount of dark energy, which is a strange beast. I have to confess I was very skeptical of this picture. But the WMAP results have convinced me." "The announcement today represents a rite of passage for cosmology from speculation to precision science." John Bahcall of the Institute for Advanced Study in Princeton, N.J. "WMAP is the instrument that finally allowed scientists to hear the celestial music and figure out what sort of instrument our cosmos is... WMAP has nearly perfect pitch." "It ends a decades-long argument about the nature of the universe and confirms that our cosmos is much, much stranger than we ever imagined." "All the arguments of the last few decades about the basic properties of the universe—its age, its expansion rate, its composition, its density—have been settled in one fell swoop." -Science Magazine 2003, "Breakthrough of the Year" article The precise and accurate WMAP result is "now the frame of reference for all cosmological investigations." It "dramatically shrinks the volume of parameter space that describes our universe." -ScienceWatch: "What's Hot in Physics", Simon Mitton, Mar./Apr. 2008 "In a sentence, the observations are spectacular and the conclusions are stunning," said Brian Greene of Columbia University in New York City. "WMAP data support the notion that galaxies are nothing but quantum mechanics writ large across the sky." "To me, this is one of the marvels of the modern scientific age." Privacy Policy and Important Notices Reproduction Guidelines / Citations Accessibility wmap.gsfc.nasa.gov Webmaster: Britt Griswold NASA Official: Dr. Edward J. Wollack Page Updated: Friday, 08-09-2024
16572	https://apps.dtic.mil/sti/tr/pdf/AD0765322.pdf	DOC1MFiNi-rT .. GVTDOC D 211. "£ 9 : 3880 -. iQ NAVAL SHIPF RESEARCH AND D EVELOPMENT CENTER -Bethesda, Maryland 20034 PROPELLER PERTURBATION PROBLEMS by Terry Brockett 0 w z 0 APPROVED FOR PUBLIC RELEASE; DISTRIBUTION UNLIMITED. D C-0.I w" uJ .J 0 SHIP PERFORMANCE DEPARTMENT RESEARCH AND DEVELOPMENT REPORT OCTOBER 1972 Report 3880 The Naval Ship Research and Development Center is a U. S. Navy center for laboratory effort directed at achieving improved sea and air vehicles. It was formed in March 1967 by merging the David Taylor Model Basin at Carderock, Maryland with the Marine Engineering Laboratory at Annapolis, Maryland. Naval Ship Research and Development Center Bethesda, Md. 20034 MAJOR NSRDC ORGANIZATIONAL COMPONENTS NSRDC COMMANDER 00 REPORT ORIGINATOR TECHNICAL DIRECTOR01 OFFICE R-IN-CHARGE OF FICE R-IN-CHA RGE CARDEROCK ANNAPOLIS 05 04 SYSTEMS DEVELOPMENT DEPARTMENT 1 I SHP PEFORMNCEAVIATION AND HIP DEPRTMEN SURFACE EFFECTS DEPARTMENT I5 DEPARTMENT 116 STRUCTURES COMPUTATION I DEPARTMENT AND MATHEMATICS DPTN 17 DEPARTMENT 18 PROPULSION AND SHIP ACOUSTICS POUSO N DEPACOUNTIS AUXILIARY SYSTEMS 19 DEPARTMENT 27 MATERIALS CENTRAL I DEPARTMENT INSTRUMENTATION 28 DEPARTMENT 29 NDW-NSRDC 3960/44 (REV. 8/71) GPO 917-872 DEPARTMENT OF THE NAVY NAVAL SHIP RESEARCH AND DEVELOPMENT CENTER BETHESDA, MD. 20034 PROPELLER PERTURBATION PROBLEMS by Terry Brockett 7 APPROVED FOR PUBLIC RELEASE; DISTRIBUTION UNLIMITED. OCTOBER 1972 Report 3880 TABLE OF CONTENTS Page ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . ADMINISTRATIVE INFORMATION ............. .................... I ACKNOWLEDGMENT ...... ... . . . INTRODUCTION . .... .... . ... ................. . . . . ••. . I A - OBJECTIVE . . . . . . . . . . . . . .. . . ... B - REGULAR- AND SINGULAR-PERTURBATION PROBLEMS ....... ........ 2 C - DESIGN AND PERFORMANCE PROBLEMS ........................... 3 D - HISTORICAL DEVELOPMENT OF PROPELLER THEORY AND STATE OF THE ART ....................... ......................... 4 E - STATEMENT OF THE PROBLEM ................. .................. 8 F - DESCRIPTION OF CONTENTS ................. .................. 9 CHAPTER 1 - PROPELLER GEOMETRY ................. .................. 10 A - COORDINATE SYSTEMS AND BLADE GEOMETRY .............. .. 10 B - SECTION GEOMETRY ......................................... 17 C - DIFFERENTIALS EXPRESSED IN PROFILE COORDINATES ........ .. 20 CHAPTER 2 - GENERAL FORMULATION ......... .................. .. 22 A - DESCRIPTION OF FLOW FIELD ......... .................. .. 22 B - GOVERNING EQUATIONS AND BOUNDARY CONDITIONS .......... .. 22 C - SOLUTION IN TERMS OF BOUNDARY VALUES .... ............ .. 25 CHAPTER 3 - REGULAR PERTURBATIONS IN PROPELLER THEORY ....... 40 A - GENERAL FIRST-ORDER SOLUTION ........ ................ .. 40 B - DESIGN PROBLEM .... ....................... 47 C - PERFORMANCE CALCULATIONS ........ ................. .. 51 D - SECOND-ORDER DESIGN THEORY ....... ................. .. 52 CHAPTER 4 - SINGULAR PERTURBATIONS IN PROPELLER THEORY ....... .. 55 A - FIRST-ORDER TERMS .............. ..................... .. 55 Outer Flow ................ ......................... .. 55 Inner Flow ................. ......................... .. 57 Matching .................. .......................... 64 B - SECOND-ORDER TERMS ............. .................... 60 Outer Flow ................ ......................... .. 60 Inner Flow ................. .............. . ........... 69 Matching .................. .......................... 71 C - PERFORMANCE ............... ........................ .. 73 CHAPTER 5 - DISCUSSION .............. ....................... .. 76 A - COMMENTS ON RESULTS ............ .................... .. 76 B - EXPRESSIONS FOR VELOCITY AS INTEGRALS OF FIRST-ORDER SINGULARITIES ............... ....................... 76 C - COMPARISON WITH OTHER FORMULATIONS ................. ... 85 ii Page D - EXAMINATION OF RADIAL COMPONENT IN BOUNDARY CONDITION . . . ................ ..................... 92 E -DESIGN PROCEDURES (HYDRODYNAMIC) ........................ 96 F - RECOMMENDATIONS ............. ...................... .. 97 SUMMARY AND CONCLUSIONS ............. ..................... .. 98 APPENDIX A - POTENTIAL FOR LIFTING LINE DERIVED FROM THE BIOT-SAVART LAW .............. ...................... .. 99 APPENDIX B - INTERMEDIATE EXPANSION OF OUTER POTENTIAL .... ...... 109 REFERENCES .................... ............................ . 115 LIST OF FIGURES Page Figure 1 - Coordinate Reference Frames for Stationary Blade .... ....... ..... 10 Figure 2 - Blade Coordinate System ........... ..................... ... 12 Figure 3 - Profile Geometry ............... ....................... .. 14 Figure 4 - Schematic of Vortex Sheet for Determining Circulation ... .......... 29 Figure 5 - Coordinate System for Describing Blade Shape .... ............. .. 40 Figure 6 - Geometry in Plane at Fixed z .......... ................... .. 58 Figure 7 - Profile Coordinates for Inner Flow ......... ................. .. 62 Figure 8 - Vortex Element ............... ........................ .. 99 Figure 9 - Path of Integration ................... ....................... 100 iii NOTATION a(:) Radius of circle which is transformed to profile in the inner flow (singular-perturbation problem) at/z) Complex coefficients in inverse mapping function - f ? Vector potential for velocity TIIR UST CT = Thrust coefficient (1/2) pU2 lrR2 E Profile shape function Unit base vector, subscript gives direction I' Inverse mapping function fl Radius of lines of constant circulation in shed vortex sheet 12' Axial position of lines of constant circulation g(eW Gage function in intermediate region i, j, k Unit base vectors in the Cartesian reference frame N Vector normal to surface, pointing into the flow field n Unit vector normal to surface, pointing into the flow field p Pressure P(z) Pitch of blade-reference surface P1(z) Pitch of shed vortex sheet at lifting line q Velocity vector q0 Velocity vector far upstream R Propeller radius r = (x, y, z) Position vector of field point s Position vector on blade surface T Thrust of propeller _ti Tangent vector to surface, subscript denotes direction U Magnitude of velocity component along propeller axis u Component of induced velocity, subscript gives direction, superscript gives order of term in expansion Vn (z) Equivalent free-stream velocity for nth term problem in inner flow (singular-perturbation problem) W(•) Propeller warp, angular position of blade-reference line x Cartesian component of position vector along axis of rotation, pointing downstream "x/e Inner variable Sx/g Intermediate variable iv Cartesian component of position vector =.v/ Inner variable y = Intermediate variable Z Number of blades z Cartesian component of position vector pointing along upward vertical rto{Z) Angle of zero lift of two-dimensional section c4/~z) Angle of attack of nth term inner problem U ta'n- Pitch of incoming flow at lifting line F = q dr Circulation enclosed by contour -B -Jump in potential across blade surface; local circulation yt' l, ) Chordwise component of velocity difference across the blade Gage function in outer flow E Thickness- or camber-to-chord ratio, small parameter in regular-perturbation problem Chord-to-diameter ratio, small parameter in singular-perturbation problem •=fl- C- + f 2e- 0 Position vector of lines of constant circulation in shed vortex sheet 0 tan"1 -Y Angular coordinate in cylindrical reference frame z 2irb 0b = Z Angular coordinate of propeller blade-reference line z A Vorticity vector /AN ) Gage function for matching to order N A = "I + 02 Radius in profile plane, inner flow Pn(f) Gage function in inner flow ti Profile coordinates, subscript denotes direction p Fluid density a Component of velocity difference across blade 0i = ie Inner variable for ti 4) Scalar potential for velocity, expressed in inner variables Scalar potential for velocity, expressed in outer variables ýoB (z) Pitch angle of profile section, measured at constant z Pp (z) Pitch angle of blade-reference surface, measured on cylinder of radius [ + + Average value of 0 across a surface v -- -"Rotational velocity of propeller, E2 in radians per unit time, E > 0 for right. hand rotation Radial coordinate in cylindrical reference frame .o=tn1 o2 Co tan Angular coordinate in profile plane, inner problem 6?( G5) Propeller rake, axial displacement of blade-reference line from propeller plane vi ABSTRACT For steady motion of a propeller operating in an inviscid fluid having an unbounded irrotational flow field, an expression for the velocity potential (in excess of the body motion) is derived in terms of the boundary values. From this expression perturbation so-lutions are determined-one for small thickness-or camber-to-chord ratio and one for small chord-to-diameter ratio. The first problem (lifting-surface theory) is a regular-perturbation problem, and the second (lifting-line theory) is a singular-perturbation problem which requires construction of matched asymptotic expansions. Two terms of each series are found. Numerical techniques are not discussed. The outer solution for the lifting line is the same as that published in the literature. The formal lifting-surface analysis differs from other developments in several ways. The most important of these is that for propellers with variable pitch, warp, and rake, the normal to the blade has a radial component which requires consideration of the radial velocity in determining the blade shape. For the case considered the sign of the con-tributions in the inner radii differ from the outer radii values; thus, this additional term might cause little effect on thrust but could be important for cavitation performance. A design procedure is discussed which involves only quantities appropriate for the lifting-surface analysis. ADMINISTRATIVE INFORMATION The research reported here is a revised version of a Ph D thesis submitted to the Naval Architecture Department of the University of California, Berkeley Campus, in March 1972. Most of the work was accomplished while participating in the U.S. Navy Integrated Advanced Training Program, and it was com-pleted with funds from the General Hydromechanics Research Program under SROZ30101, Task 00103 sponsored by the Naval Ship Systems Command, SHIPS 03412B, Work Unit 1-1544-263. ACKNOWLEDGMENT Several obscure points in the analysis were clarified during meetings with the thesis guidance committee, consisting of Profs. J.V. Wehausen, W.C. Webster, and S.A. Berger-a hearty thanks to each. INTRODUCTION A-OBJECTIVE Calculations pertaining to lifting bodies are based on linearized versions of the mathematical model used to describe the flow field about such objects. The traditional names for the two approximations used are lifting-surface and lifting-line. Each of these is associated with a different small parameter. These parameters are discussed in Section B. Until recently, developments in the linearized theory have usually been presented on an intuitive basis, rather than on a formal mathematical basis. In wing theory, the formal linearization pro-cedures have not altered the lower-order theory significantly; however, they have permitted higher-order effects to be calculated. Perhaps the best example of the formalization of the previous intuitive procedure 1 is the lifting-line theory presented by Friedrichs and Van Dyke.2 In particular, Van Dyke showed that the integral equation could be simplified to a quadrature and that higher-order terms could be found. The contribution of Friedrichs and Van Dyke will be discussed further in later paragraphs and are mentioned here as an example of the formalization. To date, a similar formal development appropriate for propeller geometry has not been presented. Accordingly, the objective of the present work is to develop the formal linearized solutions for propeller lifting-surface theory and lifting-line theory. B-REGULAR AND SINGULAR-PERTURBATION PROBLEMS A hubless propeller has three characteristic lengths: the diameter, a representative chordlength, and a representative thickness or camber. For lifting-surface theory, the ratio of thickness and/or camber to either chordlength or diameter is the small parameter, and the exact formulation is implicity expanded in a perturbation series. Such a series is herein called a regular-perturbation series, since useful information can be obtained without further analysis. However, the resulting formulation is not uniformly valid since it fails at the leading edge. The lifting-surface equations are derived in Chapter 3. Instead of the thickness and camber ratio being considered small, the chord-to-diameter ratio can be taken as the perturbation parameter. This leads to lifting-line theory which is the subject of Chapter 4. Unfortunately a vanishing chordlength leads to a singular-perturbation problem as opposed to the regular-perturbation problem in lifting-surface theory. It is often sufficient to distinguish between them by defining the regular problem as that for which the order of the governing equations and the number of boundary conditions remain fixed as the parameter goes to zero. This definition is adopted here, although an additional requirement is violated. The additional requirement is that the resulting formulation must be uniformly valid throughout the flow field. As already mentioned, the point of view adopted here is that useful information can be obtained from the regular-perturbation series without consideration of the singular region. 1Friedrichs, K.O., "Special Topics in Fluid Dynamics," New York University (1953); also published by Gordon and Breach, New York (1966). (A complete listing of references is given on page(s) 115). 2Van Dyke, M., "Lifting Line Theory as a Singular Perturbation Problem," Archiwum Mechaniki Stosowanej, Vol. 3, No. 16, pp. 601-614 (1964). A singular region can be anticipated since the formulation requires a derivative of the shape function. Although the shape relative to the reference surface is uniformly small, the derivative of the thickness function is infinite at the leading edge and small away from it. 2 For a singular-perturbation problem, the order of the governing equation is decreased, and/or one or more of the boundary conditions has to be discarded. In problems involving a lifting body of small chordlength, the limiting process as the chordlength goes to zero replaces the body by a lifting line, and, hence, the body boundary conditions cannot be applied. (The order of the equations remains unchanged.) To regain the details of the flow near the body, this region must be magnified by stretching the coordinates. The flow field near the body is calculated in these stretched variables and must be matched to the flow determined for the lifting line. The mathematical techniques involved in the analysis are explained by Van Dyke, 3 Kaplun, 4 and Cole. 5 The first application of singular-perturbation concepts to wings of finite span was made by Prandtl.6 Later, Friedrichs1 formalized the concept of inner and outer regions and derived Prandtl's inte-gral equation for circulation. Van Dyke2 obtained higher-order terms in the perturbation series and showed that the circulation was obtained as a quadrature rather than from an integral equation. A singular-perturbation solution has also been obtained for a swept wing by Thurber 7 and for a nonplanar wing by Rotta.8 The formal perturbation series assume zero thickness. Van Dyke2 compared his calculations with results from exact theory for the total lift of elliptic wings and found good agreement for aspect ratios as low as about 2.7. He was able to identify terms in the solution which gave streamline curvature and which slightly modified the angle of attack. For marine propellers the chord-to-diameter ratio is about unity, and the thickness-to-chord ratio is about one-tenth. One intuitively expects then that the lifting-surface formulation will be more accurate than the lifting-line theory for the same number of terms. C-DESIGN AND PERFORMANCE PROBLEMS In the present work a distinction is made between the design problem and the performance-evaluation problem. This distinction does not apply to the mathematical formulation but to the application 3Van Dyke, M., "Perturbation Methods in Fluid Mechanics," Academic Press, Inc., New York (1964). 4Kaplun, S., "Fluid Mechanics and Singular Perturbations," Edited by P.A. Lagerstrom, et al., Academic Press, Inc., New York (1967). 5Cole, J.D., "Perturbation Methods in Applied Mathematics," Blaisdell (1968). 6Prandtl, L.; "Application of Modern Hydrodynamics to Aeronautics," National Advisory Committee for Aeronautics Report 116 (1921). 7Thurber, J.K., "An Asymptotic Method for Determining the Lift Distribution of a Swept-Back Wing of Finite Span," Communications on Pure and Applied Mathematics, Vol. 18, pp. 733-756 (1965). 8Rotta, N.R., "The Non-Planar, Moderate Aspect Ratio, Subsonic Winds," Ph.D. Thesis, New York University, University Microfilms 69-4576 (1968). 3 of theoretical results since certain quantities are assumed known in one problem while they would be unknown in the other. In the typical design problem, one specifies the blade loading, the diameter, number of blades, blade outline and hub details. One seeks the blade-section thickness, camber and pitch angle for operation in a given flow field. Variations often occur in the specifications; in early design procedures, the camber-line shape was specified, and the radial magnitude needed to obtain a specified thrust was sought. In some modern design techniques, the spanwise and chordwise shape of the loading function as well as the thrust is specified, and the camberline shape is to be calculated. In both of these design problems, the thickness distribution is specified from strength and cavitation considerations. When the thickness distribution and loading function are specified, the design calculations become quadratures in the lifting-surface problem. This point is elaborated in Chapter 3. Recently techniques for determining the thickness distribution from a specified pressure distribution have been developed by Hille.9 This would allow a design problem for which the total pressure distribution was specified and for which the thrust and geometry would be calcu-lated. However, the combination of strength considerations and the simplicity of the design based on quadratures will probably insure continued usage of thickness specification. In contrast to the possible variations for the design problem, the performance-evaluation problem is straightforward to describe. The geometry is specified and from it one seeks to calculate some desired quantities when the propeller operates in a given upstream flow field. A formal solution of this problem is obtained as a quadrature in the lifting-line theory of Chapter 4. Unfortunately, an insufficient number of terms in the series is determined to permit accurate calculations, so investigators in this area currently attempt a solution of the integral equation from the lifting-surface analysis. D-HISTORICAL DEVELOPMENT OF PROPELLER THEORY AND STATE OF THE ART Propeller theory is based on the principles applied to conventional wings, except that the pro-peller geometry adds a significant complication. As discussed in Section B, Prandtl6 laid the foundation for lifting-line analysis of wings. In the same paper he included an approximate solution for propellers with small chord-to-diameter ratios. Several years later, Goldstein 10 obtained a more accurate solution for a propeller lifting-line with a specific circulation distribution; however, it was not until more than two decades later that MoriyaI 1 presented a derivation of the induced velocities at a lifting-line of an arbitrary 9Hille, R., "Bestimmung der Dickenlinie von Propellerfiligelprofilen bel vorgegebener Druckverteilung," Report 262 of the Institut ffir Schiffbau der Universit~t Hamburg, West Germany (1970). 10Goldstein, S., "On the Vortex Theory of Screw Propellers," Procedure Royal Society of London, Series A, Vol. 123, pp. 440-465 (1929). 11 Moriya, T., "On the Integration of Biot-Savart's Law in Propeller Theory," (in Japanese), Journal of the Society for Aeronautical Science, Japan, Vol. 9, No. 89, of 1015-1020 (1942); English translation in Selected Scientific and Technical Papers by Tomijiro Moriya, Moriya Memorial Committee, University of Tokyo, Japan, pp. 74-80 (1959). 4 propeller. This derivation was written in Japanese and went unnoticed for many years. Previous to Moriya's work, Kawada 12 had derived the potential for an infinitely long vortex filament which Lerbs13 used to develop charts useful in design. Since the work of Lerbs, only minor refinements have taken place in lifting-line theory. A derivation of the lifting-line equations and explanation of the numerical techniques used in their evaluation are given by Morgan and Wrench. 14 Even while development of the lifting-line theory was in progress, it was recognized that marine propellers were too broad for direct application of the lifting-line results. Efforts were devoted to finding correction factors which would allow use of the lifting-line results by modifying the propeller design so that they would actually deliver the required thrust. The calculation of such correction factors was quite tedious, and a considerable amount of simplification was incorporated. Ludwieg and Ginzel 1 5 derived an approximate correction factor for wide-bladed propellers which was extensively used before high-speed computers made possible more accurate calculations for the individual design. Strscheletzky 16 presented a basis and some numerical results 17 for a general lifting-surface theory; however, it was not until the use of computers became popular in the early 1960's that lifting-surface calculations became practical. Typical of 12Kawada, S., "Induced Velocity by Helical Vortices," Journal of Aeronautical Sciences, Vol. 3, pp. 86-87 (1936). 13Lerbs, H.W., "Moderately Loaded Propellers with a Finite Number of Blades and an Arbitrary Distri-bution of Circulation," Society of Naval Architects and Marine Engineers Transactions, Vol. 60, pp. 73-123 (1952). 14Morgan, W.B. and J.W. Wrench, "Some Computational Aspects of Propeller Design," Methods of Computational Physicis, No. 4, pp. 301-331, Academic Press, Inc., New York (1965). 1 5Ludwieg, H. and I. Ginzel, "Zur Theorie der Breitblattschraube," Aerodynamische Versuchsanstalt, GUttingen, Report 44/A/08 (1944); see Ginzel, G.I., "Theory of the Broad-Bladed Propeller,?" Aeronautical Research Council, Current Papers 208 (1955). 16Strscheletzky, M., "Hydrodynamische Grundlagen zur Berechnung der Schiffsschrauben," Verlag G. Braun, Karlsruhe, West Germany (1950). 17Strscheletzky, M., "Berechnungskurven f5r dreiflulgelige Schiffsschrauben," Verlag G. Braun, Karlsruhe, West Germany (1955). 5 the lifting-surface investigations are the analyses presented by Yamazaki 18-2 4 Sparenberg, 2 5 Kerwin, 2 6 ,2 7 Kerwin and Leopold, 2 8' 2 9 Pien,30 Pien and Strom-Tejsen, 3 1 Hanaoka,3 2 and Murray. 3 3 Reviews of many 1 8Yamazaki, R., "A Study on Screw Propellers," Memoirs of the Faculty of Engineering, Kyushu University, Japan, Vol. 19, No. 1, pp. 1-75 (1960). 19Yamazaki, R., "On the Theory of Screw Propellers," Fourth Symposium on Naval Hydrodynamics (1962). 2 0Yamazaki, R., "On the Theory of Screw Propellers," Memoirs of the Faculty of Engineering, Kyushu University, Japan, Vol. 23, No. 2, pp. 97-112 (1963); corrected version of 1962 paper. 2 1Yamazaki, R., "On the Theory of Screw Propellers in Non-Uniform Flows," Memoirs of the Faculty of Engineering, Kyushu University, Japan, Vol. 25, No. 2, pp. 107-174 (1966). 2 2Yamazaki, R., "On the Propulsion of Ships in Still Water (Introduction)," Memoirs of the Faculty of Engineering, Kyushu University, Japan, Vol. 27, No. 4, pp. 187-220 (1968). 2 3Yamazaki, R., "Theory of Unsteady Propeller Forces," Seventh Symposium on Naval Hydrodynamics, Rome (1968). 24Yamazaki, R., "On the Theory of Unsteady Propeller Forces," Memoirs of the Faculty of Engineering, Kyushu University, Japan, Vol. 28, No. 3, pp. 157-206 (1969). 2 5Sparenberg, J.A., "Application of Lifting Surface Theory to Ship Screws," International Shipbuilding Progress, Vol. 7, No. 67, pp. 99-106 (1960). 2 6Kerwin, J.E., "The Solution of Propeller Lifting Surface Problems by Vortex Lattice Methods," Massachusetts Institute of Technology, Naval Architecture Department Report (June 1961). 2 7Kerwin, J.E., "Linearized Theory for Propellers in Steady Flow," Massachusetts Institute of Technology, Naval Architecture Department Report (1963). 2 8Kerwin, J.E. and R. Leopold, "Propeller Incidence Correction Due to Blade Thickness," Journal of Ship Research, Vol. 7, No. 2, pp. 1-6 (1963). 2 9Kerwin, J.E. and R. Leopold, "A Design Theory for Subcavitating Propellers," Society of Naval Architects and Marine Engineers Transactions, Vol. 72, pp. 294-335 (1964). 3 0Pien, P.C., "The Calculation of Marine Propellers Based on Lifting Surface Theory," Journal of Ship Research, Vol. 5, No. 2, pp. 1-14 (1961). 3 1Pien, P.C. and J. Strom-Tejsen, "A General Theory for Marine Propellers," Seventh Symposium on Naval Hydrodynamics, Rome (1968). 3 2Hanaoka, T., "Hydrodynamics of an Oscillating Screw Propeller," Fourth Symposium on Naval Hydrodynamics (1962). 3 3Murray, M.T., "Propeller Design and Analysis by Lifting Surface Theory," International Shipbuilding "-Progress, Vol. 14, No. 160,,pp. 433-451 (1967). 6 of these investigations can be found in the paper by Wu 3 4 and the books of Isay. 3 5' 3 6 Cox 3 7 has surveyed the state of the art for subcavitating propeller design and concludes that for propellers described as lightly to moderately loaded, experimental results for propeller performance agree with the requirements set in the design procedure. (Since the design procedures are based on inviscid flow, viscous corrections must be made. Presently this is done on an empirical basis.) Cox points out that research is still needed in steady-flow design theory to extend the theory to heavily loaded propellers. The present work contributes to the heavily loaded design problem since the lifting-surface theory of Chapter 3 is carried out to second order. While these design theories were being developed, the problem of predicting the performance was also being investigated. Kerwin 3 8 calculated the performance of a series of propellers over an operating range, using approximate curved-flow corrections derived for use in design. Differences of 10 percent, com-pared with experimental results, were found, although occasionally excellent agreement was obtained. Analyses of the lifting-surface formulations to predict performance from the geometry have been attempted since Kerwin's investigation. To date most of these investigations have not compared predictions with experimental results. Yamazaki 1 9 ,20 made some comparisons with experiments, and he concluded good agreement when the pitch-to-diameter ratio was small. Murray 3 3 calculated the performance of pro-pellers he had previously designed and found discrepancies between the calculations. For one propeller the predicted and design value of the thrust differed by 8 percent. Isay and Armonat 3 9 calculated 3 4Wu, T.Y., "Some Recent Developments in Propeller Theory," Schiffstechnik, Vol. 12, No. 60, pp. 1-11 (1965). 3 5Isay, W.H., "Propellertheorie, Hydrodynamische Probleme," Springer-Verlay, Berlin (1964). 3 6Isay, W.H., "Moderne Probleme der Propellertheorie,'Springer-Verlag, Berlin (1970). 3 7Cox, G.G., "State-of-the-Art for Subcavitating Propeller Design Methods," Appendix II of the Report of the Propeller Committee, 12th International Towing Tank Conference, Rome (1969). 3 8Kerwin, J.E., "Machine Computation of Marine Propeller Characteristics," International Shipbuilding Progress, Vol. 6, No. 60, pp. 343-354 (1959). 39Isay, W.H. and R. Armonat, "Zur Berechnung der Potentialtheoretischen Druckverteilung am Fliigelblatt eines Propellers," Schiffstechnik, Vol. 13, No. 67, pp. 75-89 (1966). Lerbsl 3 defines loading ranges based on the importance of the trailing-vortex position in the calcu-lations of induced velocities at the lifting line. 7 .40 blade-pressure distribution hut made no comparisons with experiments. Sugai made extensive calculations of' blade pressure distributions, thrust coefficient, and comparative camberline studies but did not compare his results with experiments. None of the perfornance evaluation studies mentioned previously included thickness effects. All were linearized formulations, although YamazakJ18-21 formulated a theory which did not explicitly require linearization but did require zero thickness. Later papers by Yamazaki 2 2 " 2 4 included thickness effects but were based on linearized formulations. He also gave equations governing the position of the shed vortex sheet. In each case described, evaluation of the numerical formulation usually required further approximation. The linearization and approximations point out the complexity of the problem resulting from the geometry of the propeller and its shed vortex sheet. Other problems in propeller theory which are being investigated concern unsteady effects and viscous effects. Study of these two aspects of propeller theory is still in its infancy, and relatively few papers have been presented. Most of the literature on unsteady effects can be found in Yamazaki,21-24 Pien and Strom-Tejsen,31 and Tsakonas and Jacobs.4 1 Several references concerning boundary layers applicable to propeller blades can be found in Armonat. 4 2 Since these two areas are not covered in the present work, no further details of these investigations will be given. E-STATEMENT OF THE PROBLEM It is appropriate at this point to define the problem to be considered; the flow field of a hubless propeller with Z identical, equally spaced, blades operating in an unbounded, inviscid, incompressible fluid. The rectilinear and angular velocity are coaxial and constant. The exact lifting-surface theory is first formulated without further restrictions; however, the singular-perturbation problem is further restricted in geometry. Experience with wings, Van Dyke, 2 Thurber, 7 and Rotta,8 indicates that more terms can be calculated for the simplest geometry, i.e., without sweep and dihedral. Accordingly, the blade should be neither raked nor warped (these terms are defined in Chapter 1). The reason for this restriction is explained in Chapter 4. The usual assumptions (Lerbs)13 about the significant interaction effects of importance to the position of the shed vortex sheet are not made in the lifting-surface analysis. The position of the vortex sheets can be taken into account as the perturbation solutions evolve. An approximation is made in the lifting-line analysis which corresponds to moderately loaded theory. 4 0Sugai, K., "Hlydrodynamics of Screw Propellers Based on a New Lifting Surface Theory," Selected Papers from the Journal of the Society of Naval Architects of Japan, Vol. 4, Tokyo, pp. 96-106 (1970). 4 1Tsakonas, S. and W.R. Jacobs, "Propeller Loading Distributions," Journal of Ship Research, Vol. 13, No. 4, pp. 237-257 (1969). 4 2Armonat, R., "Untersuchung der Druckverteilung eines Propellers unter Berficksichtigung grenzschichtbedingter Massstabseffekte," Schiffstechnik, Vol. 16, No. 81, pp. 41-54 (1969). 8 F-DESCRIPTION OF CONTENTS Because the geometry plays such an essential role in propeller theory, the first chapter is devoted to an examination of the geometrical specification of a propeller blade. The normal to a blade is determined, and certain differentials are examined. The findings are used in later chapters. Since no derivation of the exact potential function associated with the finite dimensions of the propeller could be found in the published literature, an expression has been formulated in terms of boundary values in Chapter 2. By a straightforward linearization, the regular-perturbation problem has been obtained in Chapter 3. The applications to both design and performance calculations is discussed. The perturbation solution is carried out to second order. Although the performance calculations result in an integral equation in the regular-perturbation problem, in Chapter 4 the performance problem is considered from the viewpoint of singular perturbations, and a quadrature results. Two terms in the series have also been calculated in this analysis. In Appendix A, an integral expression for the potential of a lifting-line has been derived from the Biot-Savart Law. (Although this expression can be integrated in terms of known functions, the nested infinite series which result are not thought to be practical for computation.) In Appendix B the intermediate expansion of the outer potential is obtained. This expansion is used in Chapter 4. 9 CHAPTER 1 PROPELLER GEOMETRY The details of propeller theory are considerably more complicated than wing theory because of the geometry. In this chapter, the analytical specification of the blade shape is developed, and certain ex-pressions needed later in the analysis are obtained. First the coordinate systems are described, and the relations between them are derived. In addition, the normal to the blade is found. In another section, the blade sections are discussed, and the small parameters (thickness- and camber-to-chord ratio or chord-diameter ratio) are explicitly considered in the expressions for the derivatives appearing in the normal. In the last section, the gradients of functions expressed in these coordinate systems are obtained. A-COORDINATE SYSTEMS AND BLADE GEOMETRY To describe the flow field quantitatively, two right-handed coordinate systems are used, both attached to the moving propeller. The first of these is a Cartesian reference frame (x,y,z) with z pointing away from the axis of rotation along a reference line in the blade and with x pointing along the axis of rotation with positive displacement measured downstream from the propeller. The second system is cy-lindrical polar coordinates (C, O,x). Radial distances are measured from the axis of rotation in a (y,z] plane, and 0 is measured from the z axis in the clockwise direction looking along the positive x axis; see Figure 1. z 0 \ / \R j Figure 1 - Coordinate Reference Frames for Stationary Blade Unit base vectors in the (x,y,z) system are the usual i•j,k.) and base vectors in the (aO,x) system are denoted by (eC, f O, o), respectively, where 10 sin- si'0 j_+ cos 0 k and (1-I) e o =-cos 01-sin Ok In general, blade-section offsets for propellers are measured from a reference line lying on a cy-linder of constant radius. This viewpoint is adopted for some of the analyses (Chapters 2 and 3) but for another portion of the work (Chapter 4) a slightly different viewpoint is necessary. This second viewpoint is that the geometry is defined for sections cut by a plane at constant z on the reference blade (the reference blade is the one which lies as close as possible to the z-axis). Although a set of unit orthogonal reference vectors is assigned for each viewpoint, neither system is an orthogonal curvilinear coordinate system, except for a special case in the second system. (The orthogonal coordinate system with the cylindrical surface would be cylindrical polar coordinates.) For the sections defined on a right circular cylinder, the blade-section reference line on the surface of the cylinder is called the geometric pitch line. It makes an angle ýop(•) with the y,z plane. The geometric pitch line usually lies along the nose-tail line, which is a line on the cylindrical surface joining the leading point and the trailing point of the blade section cut by the cylinder. Another important line is the blade-reference line (also called stacking line) which is defined as the line through an approximate datum point at each radius. The datum point is usually on the nose-tail line and generally will be the midchord point. The 27rb blade is raked when the blade-reference line is given by x = R(o); 0 = 0b -= (b , 1,... Z - 1). When the reference line is given by 0 = 0b + W(&$), x = 0, the blade is said to be warped. A combination of warp and rake is called skew; usually the displacement is along a helix which passes through a straight line in the plane x = 0. On the cylindrical surface E = constant, a coordinate system (1' ý2' 13) is con-structed with ý, on the surface of the cylinder and measured from the blade-reference line along the constant pitch of the nose-tafl line. Positive t1 values point in the downstream direction. The variable t2 is normal to the t, axis and points in the upstream direction. Unit orthogonal reference vectors (el,e2 ,e.) are in the (• 1 ,•2,V} direction, respectively, but the coordinate system is only locally orthogonal, i.e., these are not the unitary base vectors described by Wills.4 3 The system is left handed to permit specification of the blade-section geometry in the conventional two-dimensional orientation (positive camber in the positive ý2 direction). A schematic of the blade-coordinate system is shown in Figure 2. Although the blade-section reference line is shown as the nose-tafl line, it is not necessary in the following development and will not always be used that way. 4 3Wills, A.P., "Vector Analysis with an Introduction to Tensor Analysis," Dover Publications, Inc., New York (1958). 11 In terms of the blade-reference system, the cylindrical polar coordinates are + cos 5 Op () + t2 sin ýOp( (-) o 0b + + (1-2) x = + tj sin s, (a) - t2 cos s1 , BLADE-REFERENCE LINE z Vx t/ / Figure 2 -Blade Coordinate System Hence the Cartesian expressions are 12 X = ('i5ý) + t/sill ý0 1 ' t2 co P v = -fsin b + W() + J (1-3) z = Z cos O b + W( ; + "P+ Pp One can express the position vector in these variables by substituting into the expression r=xi+)'j+zk and the unit base vectors in the t, and t2 direction can be found: ar el =-= sit, p-5°- cos 0 cos Op i - sin o cos L sin fOpli+ COS SOp e0 (1-4) Sr- - cos pOi - COs 0 sin p 4- sin 0 sin popk .f 2 " .s$- c =-cos OppL + sin ýOp e 0 where 0 is given in Equation (1-2). The unit base vector in the c•5 direction is the unit cylindrical polar vector in the c3 direction. It is given by Equation (1-1): j= -sin 0 + k cos 0 (The partial of r with respect to Z will not give this value since the eSP f 2 base vectors are functions of C5.) By construction e xe 2 x -The inverse of Equations (1-3) is co sot b~ 7 tn W ) + (x-6?) sin t2= sin op )/Y2 + z2 (tan,, (-) b - W) - (x-6) cos pJ ('-) VIy'2 +z2 Blade sections are given by t2 = E(t 1 , •) (1-6) 13 for c2 ( <- Cl (I• o<V<R where c2( is the distance to the leading edge, cl(iý) is the distance to the trailing edge, and R is the propeller radius. E is the sum of a single-valued (camber) and a double-valued (thickness) function which combine to describe the offsets from the tj axis on the cylinder of radius CO. The function E is considered in detail later in this chapter. Figure 3 shows the profile geometry. t2 '" BACK (+ SIDE) I QFAC E (S IDE) t,,,, Figure 3 - Profile Geometry A point on the surface of a blade is given as a function of two variables (t1, C) by s =(L + t1 sin p- E cos pi -c sin 0b + W + tl cos op + E sin °p, (1- 7) + c b+ W + cos 1 op + E sin IpP)k Two tangent vectors on the blade surface are given by 14 as (1-8) a• 12= see. e.g., Wills4 3 and a vector normal to the surface (but not a unit normal) is given by N = +t 1 X t2 (1-9) where the sign is positive on the suction or back side and negative on the pressure or face side. The components of these vectors in the (iji,k) system are lengthy but when expressed in the (f 1,c-•-2,.e_) system they are shorter. In particular, the normal is Ai,~ aE + cos ýpp+ 7 sin (1-10) +(dW (,Oj co p Esn CO ýWý + (in ap ECo P) For E = 0, the resulting expression for the normal is A? -d-ll sin Opcos dWo + cosL ýp w sin ýp which is the normal to the helical surface of pitch angle ýVp( through the blade-reference line. Components in the (i,j,k) directions can be found when Equations (1-4) and (1-1) are substituted into these expressions. Previous investigators have assumed that the only significant contribution toN 0 was e2 . Note that even for blades without rake or warp, an ew-component exists. This remaining term can be simplified by expressing ýpp(Z) in terms of the pitch, P(Z). The pitch is the distance of axial advancement in one complete revolution of a screw of pitch angle ýp (a): 15 P(2•) tan ýp,(w) = -Using this relationship, one finds d•Op sin ppCos ýPpp d dp il OCSOp _ dP _ dw dco 27r ( + p2/4 1 12) (1-12) dP cos2 ýOp dw" 27ra Thus for helical surfaces of constant pitch, this component of the normal is zero. Since few modern pro-pellers have constant pitch, this term must be retained. For the second system of blade specification the sections are cut out by a plane at a given z value. As will be explained in Chapter 4, neither rake nor warp is considered with this viewpoint. Hence the blade-reference lines are the 0b values and for the reference blade along the z axes, the nose tail line is on a plane and forms an angle zpB(Z) with the y-direction. The coordinate system is similar to the previous one, except that t1' t2 are measured in an x,y plane at a constant z value. The Cartesian components are x = t1 sin pB- t2 cos ýPB y = - (t, Cos oB + t2 sin SOB) (1-13) z=z. Similar to the previous discussion, the unit vectors are ar cos ýB efl si ýp-'BI (1-14) 3 r £2 =T2=- cos I-sin Iý ar and the third vector is the k vector such that k =- eI X f2 and here also k 9' ---. The blade surface is here assumed to be given by 2 = E( 1 , z) (1-15) where E is a sum of a double- and a single-valued function describing the thickness and camber distributions, respectively. 16 As before, a point onl the surface is found by substituting this expression in the ('artesian com-ponents given by Equation (1-13), and two tangent vectors to the surface are found by differentiation. "lhe cross product of these two tangent vectors gives a normal vector, N= L 2- + I+( -+ h'] (1-16, Sdz a at- dz k This is similar to the expression for the normal where the section shape was measured on the surface of a cylinder; however, the terms for the component normal to the template plane show marked differences. B-SECTION GEOMETRY In this section, the specific form of the shape function is examined. In general this shape function can be written t2 = E(lI t3) where t3 is either the cylindrical radius or the z variable. The blade shape is made up of a single-valued and a double-valued function. The single-valued function is the camberline, and the double-valued function is the thickness. These functions can be con-sidered separately. First the thickness can be represented by t2 =+ET (t], ts3) where =0 cI cf(• 3) and c2 = c2(t 3) are shown in Figure 3, and in general an = an(t 3). At the leading edge =c2' and at the trailing edge t, = c1. Often the same form is maintained spanwise and then an(ý3) = T(t3) an, where 7(ý3) is the thickness to chord ratio and the an are now constants. For the regular perturbation problem, )= eT( 3) (1-18) that is, the thickness is small relative to the chord. By assumption all derivatives remain of first order although by inspection this breaks down at the leading and trailing edge, where at the leading edge aET ao (1-19) '/•1"2 17 Even though this singularity is not compatible with the O() assumptions in the derivation, solutions obtained by ignoring this discrepancy have been useful. Several techniques for improvement (Friedrichs 1 and Van Dyke 3) are known but they usually break down after a few terms. In the singular-perturbation problem, the chord length is the small quantity, and in this case it is convenient to put c = ýhl 1 (t3) c2 = h2 (V3) (1-20) where 9 can be considered the ratio of maximum chordlength to diameter. In this case, the equation for thickness becomes ET=( .a2) {I Jhl } (1-21) ET 1(lv (ý-L ))' n=O an1(h, h2) The normal vector involves derivatives of this function with respect to both t1 and ý3' First, the derivative with respect to is aET 12(h +.h 2) -t , N an _ t, h2 ) n(h-h) (1-22) + -41") ( 4 •-h2)N tnan L(hl-h2)] n-i hi - h2 n=1 In Chapter 4, the form of this expression in inner or stretched variables is needed. Let the variable be a, = 1/, then converting this to the stretched coordinate, one finds aET l /2(h]+h2- 2 01 N n Ul \ --i h 2 an -1 h2 ,(h 1O) (01 -h 2 } n=O 2 2 / (1-23) +/h l.• nan h l (h+ l1-h2) n=l -This is entirely a zero-order term with respect to the chordlength parameter. 18 Second, the derivative of the thickness function with respect to the variable is T 1/2 -(eC C 2)'+ t] (cl + c2 )' N a 1 n "'C"VT ti) ( •1" c2) n=O 1c -'c2) + rcI"-ti)(•I'C 2 n a'2) (I d)2 (1-24) (c1 - c2N 1 Nen (c2 ) c2)2 i n_ n, n-2 where the prime denotes derivative with respect to •3 When this is converted to the inner variable, it be-comes aET -(-hI h 2)'+ ul(hI + h2)" N n W3 (aP t3) - 2 a'~ 7 7 nfjf~ 1 _ a]I-O) (a 1 - h2) n1=0 hI "2) N /) (1 h (1-25) +( " h1 "o 1 )(o 1 "h2 )n a' 112) (h1 - h2 " N / hJ2) eVh-r)a-2)(hI - t,2)2 n flnan h ---h2 ) n-1 Hence, this expression remains of order F For the camberline function, the appropriate expression is t2 = Ec t•' 3) where Ecj'l, t2) bn ( ) n + ýlk (1-26) (c1 - C2 ) n=O c0 2 In some current design procedures, the shape of the camberline is computed at each radius, and in general the coefficients in the summation are not a constant times the same function of t3 The derivatives of this function follow the. same patterns as the thickness function. For the regular-perturbation problem, they are 19 both the same order and for the singular-perturbation problem, the t1 derivative is of zero-order and the t3 derivative is of order T. For most camberline shapes, the t1 derivative is bounded. C-DIFFERENTIALS EXPRESSED IN PROFILE COORDINATES In some portions of later chapters, it is necessary to calculate the gradient of a function which is given in the profile coordinate system. For the hypothetical functions J1 = J1 (t1' ý2 and J2 = J2 (t1j t2' z), the gradient can be determined by straightforward application of the chain rule. Consider the function J first. atl atl ail -3 = i + -j + -- k (1-27) ax ay az where -a il at ] + ai l _L 2 ax ý1 a, a 2 ax (1-28) a= I a.,1 a,]la,1, =sin ýOp --cos ýop 02 and the expressions for2- and L are obtained similarly but are more complicated. After arrangement of terms, the expression becomes ail ai l + e+ (cos ýOp (C-Ob-W) a• ýl 2 L a• • 2d'p dW -- L all d W d6?o il 1 - 9 t . . .• •C OSp-d i op I(sin ýOp (O-Oh-W) (-9 da da d / 4 dW ddWpM - --I d. ýoP d--dJYCO dV ýPj] where (0 - 0b - W) is given in terms of (Q, ý2' Z3) in Equation (1-2). For the Q1, t2, z) variables, a similar transformation gives 20 a, a2 azJ 2 d~pB aJ 2 aj 2 1VJ2 = "el + -e2 + T + 2"1• k (1-30) at,• at2 z dz at2 This expression is used in Chapter 4 as well as one for the Laplace equation expressed in the pro-file coordinates (Q1, t2' z). From repeated application of the chain rule, one finds n2 a2 j 2 a2j 2 a2 j 2 u x2 y2 2 2 ax 2 ay 2 atl' az 2 l a2 1 d OB' 2 2 a2j 2 a 2j 2 a 22 + ýJ t2 +t 3Z 2 \dz at] 2 %2 a tj 2 /d~PB 8 2 aj 2 d2 ~p ai 2 'J 2] \dz ) at a t 2 Lz I at 2 dB/ a 2j 2 a ý 2 a 2 J 2 dz a•2az aýlaz, az 2 Obviously, the Laplace equation involves the sum of the previous two equations. It has been broken up to call attention to a feature of the expression for a2J2/1z 2. This feature is that any scale change applied to both ti and t2 cancels. This property is utilized in Chapter 4. 21 CHAPTER 2 GENERAL FORMULATION A-DESCRIPTION OF FLOW FIELD A Z-bladed propeller of finite radius R is assumed to be rotating with constant angular velocity 92 and to be advancing at a uniform rate U into an unbounded, inviscid, incompressible fluid; however, one for which the Kutta condition is satisfied at the trailing edge. The propeller produces a thrust which requires that the average pressure on one side of the blade, called either the back or suction side, is lower than on the other, called either the face or pressure side. The Kutta condition requires that no flow cross the trailing edge. The general flow pattern then is that fluid is pushed from the high-pressure side to the low-pressure side around only the leading edge and the tips. The tip flow deflects the streamlines in the vicinity of the tips toward the axis of rotation on the low-pressure side and away from the axis on the high-pressure side. This flow behavior, together with the requirement of continuity in the pressure field, results in a discontinuity in direction, but not magnitude, of the fluid velocity leaving the blades at the trailing edge. This layer of fluid slip is called a vortex sheet. The flow model thus consists of the lifting surface and the vortex sheet advancing and rotating into still fluid. B-GOVERNING EQUATIONS AND BOUNDARY CONDITIONS The unsteady flow field described in Section A can be made steady by considering the flow relative to the coordinate systems rotating with the blade. In a rotating-coordinate system, the equations of motion, e.g., Kochin, Kibel' and Roze44 Equation 2.7.8, are q .gradq+Igradp=-2_2 Xq_ A2X(_qXr) (2-1) P div q = 0 (2-2) where q(r) is the velocity vector, p is the pressure, p is the density, •-2 - Q is the angular velocity of the propeller for the assumed right-hand rotation shown in Figure 1, r is the position vector of a point in the flow field. This discussion is patterned after that given by Prandtl6 for planar wings. 44Kochin, N.E., et al., "Theoretical Hydromechanics," (Translation of Fifth Russian Edition), Interscience Publications, Inc., New York (.1964). 22 The flow field in the inertial reference frame is irrotational everywhere except on the boundaries. In the rotating coordinate system the vorticity is given by curlq =- 2 2=292i (2-3) The boundary conditions are that the upstream velocity is specified; there is no flow through the blade; the vortex sheet moves with the fluid, i.e., the normal velocity at the sheet is zero; and the velocity at the trailing edge is finite (Kutta condition). If SB" represents the pressure side of the blades, SB+ the suction side, Sv+ the continuation of SB+ onto the vortex sheet, SV- the other side of the vortex sheet, and T(z) the trailing edge, then the boundary conditions become q" q 0 = Ui + 92ae 0 as X + (2-4) q . n = OforreSb (2-5) q <+ ,,for re T (2-6) where U is the advance velocity in the inertial reference frame n is the normal pointing into the fluid Sb = S1/+. ',SB+ ,-SB- LiS - and b denotes one of the Z blades, b = 0,1, ... Z - 1 If the velocity is taken as q =7 + V x B (2-7) where B is the vector potential, and 0 is the scalar potential, then Equation (2-3) gives vX( VX B) =2 S2 and Equation (2-2) gives V2-= 0 If the boundary condition at upstream infinity is associated with B, an expression for B can be immediately stated: B = (2x- 0 U)e• (2-8) where U must be a constant for this expression to satisfy Equations (2-3) and (2-4), and VX B = q0. The symbolic vector notation is used for the differential operators in curvilinear coordinates. The velocity component U is commonly taken as a function of radius to approximate nonuniform in-flow. One would expect such shear flows to be axially variable also. Shear flows imply vorticity, and careful consideration should be given to the propagation of vorticity from upstream infinity to the propeller and on downstream. 23 The remaining unknown is now only the scalar function '1, and the problem can be reformu-lated in terms of it. The boundary conditions are: V7 - 0 as x -- o, .oo (2-9) VO.n=-qo. n for reSb (2-10) IvOPI<+-o for reT (2-11) In addition the pressure is continuous across the vortex sheet. The three simultaneous equations of motion described by Equation (2-1) have been replaced by a single equation for the scalar potential 0 which depends only on the kinematics of the flow. The dynamics of the flow enter only in the determination of pressure which can be obtained by integrating Equation (2-1). The integration is performed by using the expansion formula q .grad q =1/2V(q . q)- q X curl q = 1/2V(q . q) - 2S2 X q (2-12) and by noting that for 92-- 9i 2 X( X r) g22 S -, . 'A2 , v(2-13) 2 Hence, Equation (2-1) can be written as V(-q q+- &22 O)=0 2- -p 2 Since the expression is zero for all spatial derivatives, it must be constant. Thus q 2 1.ý, K 2-- p 2 2 where K is a constant everywhere in the fluid. Hence the pressure is p=P {K+n2a 2 -2q. q} (2-14) The condition that the pressure be continuous across the votex sheet can be expressed as (q . s .÷ " qJr_ Sq . or 24 r cSv+-(2q() + VO). o 0 (2-15) LESV-In the next section an expression for 0 is found which satisfies the boundary conditions in Equations (2-9) through (2-11) and Equation (2-15). C-SOLUTION IN TERMS OF BOUNDARY VALUES The solution of Poisson's or Laplace's equation in a three-dimensional volume can be expressed using Green's second identity; see, e.g., Tychonov and Samarski: 4 5 S 1 1 = 1 1 fff (OV (n. V 2 )d- (n" VO) dS (2-16) IrI Ir' 6 r LI I where v is the total fluid volume, 7, is the surface which bounds the volume, including the surface at 00 -, r' is the position vector, measured from a fixed point in the fluid, n is the unit normal, pointing from the bounding surface into the fluid. If 0 satisfies Laplace's equation, the volume integral is zero everywhere in the fluid, except at the point r' 0. This point can be excluded from the volume by enclosing it in a sphere. Taking the limit of the sur-face integral over the sphere as the radius goes to zero, one finds the following standard formula -4 ( V. V _- nj--n. V0 kdS (2-17) where r is an arbitrary point in the fluid, and s is a point on the boundary. 4 5Tychonov, A.N. and A.A. Samarski, "Partial Differential Equations of Mathematical Physics," Vol. 1, Holden-Day, Inc., San Francisco, Calif (1964). Tychonov and Samarski,45 for example, show that other shapes excluding the singularity are also satisfactory as long as the maximum dimension of the excluded region goes to zero. 25 It is necessary to know the value of ( on the boundary as well as in the flow field. That is, one seeks lira kr) = limy•ro+fe)-- Lr 0o) 1r ro 7740 where r is a point in the flow field ro is a point on the surface e is an arbitrary vector pointing into the fluid from.r and the + sign is used whenro is on the ± surface Two situations occur: one when.ro is a point on the boundary of a solid of finite thickness, and one when .o is a point on a boundary, both sides of which are in the flow field, across which 0 is discontinuous. In the second case, the limit r , o produces singularities in both sides of the surface. For the first case, for whichro is on a single-sided surface, one finds + OU o 1 Vck ds (2-18a) and for the second case, for whichro is on a double-sided surface, one finds 0 + -+ n -. VOI ds (2-18b) 27riro Lro~ where the improper surface integral is obtained b•€ evaluating the integral over the bounding surface, ex-cluding the region surrounding the pointro, and then by taking the limit as the maximum dimension of the excluded region tends to zero. Since the present problem is a second (or Neumann) boundary-value problem, it is important to determine the values of the gradient of 0b on the boundary. To find this limit, the pointro is excluded from the surface Z by surrounding it with different shapes s(ro, X), where X characterizes the dimensions of the region. With the same notation as used previously, one seeks lim VO7.r) = lim O (_ro + 7f) (VO)± 1 -0ro 77-+0 For a point on the boundary of a solid of finite thickness rVn) 1 li Z0 rs ( d + (2-19a) 2ir.13IL "Z-s(.ro, The ± superscripts here denote different sides of the surface as well as the limit operation when appropriate. 26 /im .L 3r -(1.•1 9a) 2 1r 3 ro -. • .. /il .K ] Cont'd ,s(ro,X) where n,) = n(r,), and 0, r When s~r 0,X) is the region of 2: interior to a right circular cylinder of radius X with axis n(, then K = 27r; when sLr 0,X) is the region of Z interior to a square of half-side Xt with axis no, then Ks = 4 2 4'2T, when S(ro,ý is the region of 2; interior to a rectangle, two sides of which are distant X from.ro and the other two sides of which are an arbitrary finite distance fromro, then Ks = 4. We call the first integral in Equation (2-19a) a Cauchy principal-value integral. For a point on a boundary in the flow field 1 lim -- (n VO) d (Vp)+ + ((Q-=2-- -[ L3 +I lim (21 _ 'ds 27r 0 -r o (ro)) 1 r °s• roS a + + + K] (2-19b) The limit operation indicated in Equations (2-19) is considered to define a singularity in the mathematical literature (Tychonov and Samarski)45 while in the aeronautical literature (Mangler, 4 6 Ashley and Landahl) 4 7 such forms are accepted. In Chapter 5, a form of the expression for velocity is obtained for which only Cauchy principal values are needed so that numerical analysis can be performed on the more convenient form. The values of the gradient on the boundary are generally to be interpreted as VF(r)lr 1= (VF)+ This expression is not equal to VF(ro) = V(F±) since the normal components are not equal. Generally the normal component is given by a separate boundary condition, and one need not explicitly consider the distinction. Throughout the text the notation for differentials of functions known on the surface V(F+) is indicated, rather than the field value evaluated on the body (VF)±. Although this notation sometimes indicates a ridiculous result if interpreted literally, it is used since it permits a shorthand-like notation to be developed. 4 6Mangler, K.W., "Improper Integrals in Theoretical Aerodynamics," Aeronautical Research Council, Current Papers 94 (1952). 4 7Ashley, H. and M. Landahl, "Aerodynamics of Wings and Bodies," Addison-Wesley Publishing Co., Inc., Reading, Mass. (1965). 27 The surface 2; consists of the Z bounding surfaces of SB %-, SV (where SB is the blade surface, and Sv the shed vortex surface) and the surface for It -+ oa. At this point it is necessary to suppose that the integral over the bounding surface at infinity gives at most a constant. (This supposition will be proven.) Hence, to within this additive constant, the potential for the propeller can be written 4S s)n-s) . Vs Lr - Vd S (2-20) where Sb is one of the M bounding surfaces sb -sv÷ - SB÷ - SB. - sv. The boundary condition in Equation (2-10) can be used to simplify Equation (2-20) since the second part of the integral becomes Sb -bs qo n d ÷ • dS sB L,• s• s(2-21) The reduction in area in Equation (2-21) follows because q0 is continuous at S, and n (on Sv+) -n (on Sv-). The normal on SV+ is directed oppositely to that on SV ., and hence Equation (2-20) can be written •-•) 4§)v -VS sB+ W, sB. (2-22) 1Z-1 i 1rb=O ffS V-8V 28 The circulation in a fluid is defined as the integral r-F , .dl1 (q0 + V d (2-23) where dl is the vector arc length along the closed curve which completely encircles the shed vorticity. Sinceqo is a continuous point function, it contributes nothing to the integral in Equation (2-23). For a curve which lies entirely within the irrotational flow except at the one point at which it cuts the shed vortex sheet P(rS) {i leSv (2-24) Further simplification is possible by considering the continuous vortex sheet to be composed of lines of constant circulation which leave the trailing edge of the blade at the point T(z). In Figure 4, from curve C2 , the circulation about the blade at T(z) is given by r(T(z)) = +(T(z)) - 0Y(T(z)) (2-25) z Figure 4 - Schematic of Vortex Sheet for Determining Circulation The word circulation will be associated with the scalar jump in potential across a surface as in Equation (2-24), and the word vorticity will be associated with the vector jump in velocity across a surface. If A is the vorticity vector then A =n X(.q+ -_q)n X V (+- •' n X VI?, where ii is the + unit normal; see Reference 48. 4 8 Milne-Thomson, L.M., "Theoretical Hydrodynamics," Fifth Edition, The Macmillan Company, New York (1968). 29 Now the circulation at r e S, is determined from the curve C1. This curve is connected to C2 by the curves C? and C4 lying along a line of constant circulation. Hence, the total curve is simply connected and 0=f JC .3dg + VA f. d4 l-+F(reSv)- (T(z)) C 3 C 4 (2-26) fC V((P+ - 0-) .dt + IQr e Sv) - 17 T(z)) (-6 C 3 By definition, C3 lies along a curve of constant circulation and from Equation (2-24), the integrand in Equation (2-26) is ar vP.t =j -= 0 (2-27) at, where t1 is a unit vector along the lines of constant circulation. Hence r(r e Sv) = F(T(zW) (2-28) Conditions are sought which describe the position vector of this line of constant circulation which leaves the trailing edge from the pointrT e T(z). To emphasize that this point is located at the blade, let it be written z,, The position vector of the line of constant circulation can be described as ý(= zOz, 0; 0bh =fl zO, 0; 0 b) Edo) + f 2(zo, 0; Ob)i (2-29) where z0 is the parameter describing the starting position of the line, 0 is the independent variable, and 2 irb 0b =- is the parameter describing the individual vortex sheets. The functions fl(zO, 0; Ob) and f2(Zo, 0; Ob) are the radius and axial position of the line, respectively. The parameter z0 ranges from zero to the propeller radius, and the variable 0 ranges from the trailing edge OTEdzo) to o. If ff = zo, then no contraction of the shed vorticity takes place. To establish differential equations" governing the position of the line, Equations (2-10) and (2-15) will be used together with the definition that the strength of the circulation is constant along the line. Since the rate of change of " is zero along the line, Alternatively, one could take " = o x; Ob). However, the usual representation given in Equation (2-29) was more convenient. The derivation here parallels that given by Ciolkowski 4 9 and Thurber. 7 4 9Ciolkowski, S.I., "The Swept-Back Wing," Ph.D. Thesis, New York University, University Microfilms 24,694 (1955). 30 V dL-r) "- = ( (2-30) Since the vortex sheet is actually continuous, the function 7(zO, 0) may be considered a function of two independent variables which describe the surface. Lines of constant circulation are obtained by holding z0 constant. Tangent vectors in the surface SV are found by constructing; tI =-•-(2-31) 12 = a (2-32) see, e.g., Wills.4 3 A vector normal to the sheet SV (but not a unit normal) is given by aý aý N =t 1 X t2 = -X -(2-33) which is assumed to be non-zero. Hence, Equation (2-10) can be expressed (go+ . N = 0 (2-34) Since p = 0 - 0.}, Equation (2-34) gives two equations: reSv vP-. N = 0 (2-35) fN. [2qo + V{@#+ 0.} (2-36) Let 4-{ + 4ý. Then Equation (2-36) is !ESv N. (2q0 + V4) = 0 (2-37) Now Equation (2-15) can be written as Vr. (qO + 1/2VVI) = 0 (2-38) It would appear that, since only two unknowns, fl and f2, are sought, the scalar Equations (2-27) or (2-30) and (2-37) would be sufficient to establish differentials of the trajectory. However, derivatives with respect to z0 enter into the equations and are not desirable. For a more elegant form of the differential equation for ', Equation (2-35) must be used. 31 Equations (2-35) and (2-30) show that VFX ( 1 XNI) = (vl .N) t 1 - (VP .tl)N= 0 (2-39) Since Equation (2-39) is zero, any multiple of it is also zero. Hence, (2q0 + V)X(V F X (1 XN)) = 0 (2-40) In another form this equation is (2q 0 + V 0) .(1XN)}I VP. [{2q+ VV) V]! Equation (2-38) simplifies this equation to be { (2qo+ vý) .Q(1X N)} VP 0 This equation holds everywhere on the sheet. Since VI1 Y 0 everywhere, it requires that (2q 0 + VO ') . (x NL) = 0 (241) Thus for the orthogonal base vector system (tl, LN, t1 X N), the average velocity at the vortex sheet, (q0 + 1/2V7t), has no N component from Equation (2-37) and not 1 XN component from Equation (2-41). Hence the average velocity is parallel to the curves of constant circulation, and the equation for " can be constructed from two of the three scalar equations which make up (2q0 + V4)Xt 1 = 0 (2-42) The vector til is a -0 aO+fif --0 i and on the shed vortex sheet (2q 0 +V=- es + a2 fl + Ž )e + U+ Hence, Equation (2-42) in component form is fU -a f a --) -=0 (2-43) 32aaoxae 32 .L l+1 0 (2-45) Thus from Equation (2-43) 8f 2 U+1/2ax 1- (2-46) 2 f,2 a 0 and from Equation (2-45) 1 a af] 2 aa a0 1 a4 (2-47) 2f12 a0 Equations (2-46) and (2-47) are nonlinear coupled integro-differential equations for the radial and axial position of the lines of constant circulation. The integral over the shed vortex sheet in the expression for 0, Equation (2-22), can be simplified using the coordinates zo, 0 used to describe the vortex sheet; see Figure 4. First, in the expression 1 r-s V -SLr-_I ir-s13 the position of the circulation element is used for s so the integration is performed for constant circulation strength. Second, the value of the vector surface area element n dS is given by ntdS = tlXIt2 dzo da _ a _ (2-48) =- X- dz0 da a TO where a is a dummy variable for 0; see, e.g., Wills.4 3 With these changes, the integral over a shed vortex sheet becomes 1k,= R x o da (2-49) o d TE(zf ) where b denotes an individual blade. Hence, 0 becomes: 33 Z-J -r-s 7-1 kf~r) - i .s)no dS+EZ Ob, V(r) S4 'r b=L-s[13 b--O 1 go - 3 b(2-50) 1 Z-1 f do' s + LI ds 4 b=O IBr-sI As described in Chapter 1, the propeller geometry is traditionally defined in terms of blade sections lying on cylinders of constant radius. The nose-tail line lies on the cylinder, and, hence, blade shape is measured from a helical surface of radially varying pitch. On the cylindrical surface aý= constant, a left-handed coordinate system (to define section geometry in the traditional orientation) is constructed with on the surface of the cylinder and is measured positively in the downstream direction along the constant pitch of the nose-tail line. The variable ½2 is normal to the ý1 axis and points in the upstream direction. Unit orthogonal vectors (el, e2 ,e) are in the (I t2' ZZ) direction, respectively, but the coordinate system is only locally orthogonal; this coordinate system is detailed in Chapter 1. Although the pitch is assumed to be that of the nose-tail line, it is not necessary in the following development and will not be used that way in later chapters. The blade shape is given by t 2 = E(t 1, aT) c2[a) <• t1 < Cl(a),O < aJ < R (2-51) = Ec( tl, "•) ± E7-(ý1, ) where E. is the camberline function and ET is the thickness function; hence, the position vector of a point on the surface is given by s (6 + 1sin Op P- E cos oP)i - asin (0b + W tICos °p P+ E sin ý° A t ICos 0Pp+ E sin ýpp S++ (ob+ a+ F k (2-52) where ýop(Z) is the pitch angle of the blade-section reference line, 63(ý) is the rake, and W(a5) is the warp. A normal is given by; see Chapter I 34 N ± i X~CS)~d -el (sin l a~ d nr7 d6I aE (3 +- os •p +- sin ýP (2-53) + + E a eas] where the + sign refers to the SB+ surface, i.e., suction or back side; and the - sign, to the SB- surface, i.e., pressure or face side. The vector element of area is nds = N A, dc (2-54) These expressions can be substituted into Equation (2-50) to give precise limits on the integration. Because of nonlinearities, no conceptual simplifications result unless the expressions are linearized. The explicit form of the normal allows the body boundary condition to be simplified. The free-stream velocity can be resolved to .0 = Iq01 cos (p p - P)eI + Iq01 sin (hp - Ne2 so that on the body, the boundary condition is N .N := " --U + 2 a2I sin -j) -cos ( Op (2-55) where = tan, U-Equations (2-49), (2-50), (2-46), (2-47) and (2-55) constitute the exact formulation. Equations (2-46) and (2-47) are nonlinear in 0. Although Equation (2-55) is linear in 0, an integral equation results in both the design and performance applications. Fortunately, however, the integral equation can be reduced to a quadrature in appropriate cases when the solution is expanded in a perturbation series. These perturbation solutions are the subject of Chapters 3 and 4. With • known, the pressure can be computed from Equation (2-14) P = P (K -[ 2 - (2q, + VO) -V and then the force on the propeller can be found from 35 Z-1 I= f p(-n)ds b=O3 SB -i [2qO + V7) -Vo]nds (2-56) b=O SB In particular, the thrust is F Zi= zff [(2qo + Vq5) • VO/ i •nds (2-57) Equation (2-54) can be used to give an explicit formula for calculating thrust. Thei component of Equation (2-53) is I ~ aE -= T cos Op + sin sp 1 •l = Cos ýP 1 + Silln ac .o 1 1 -V si 5iT (2-58) P , ,,,at Hence O~ ET-or in terms of q5: T--Z d a" O oP vc+ o --'~ ~ fe 2 ,L:: .c( "R C (cosýOp + sill nO C )-• + qo{ V (e÷ + 2-)+ V÷ - V+ SaET + V7o sin 1 s1 oa jd s 36 where 0+ and V are the values of 0 on the suction and pressure side of the blade at the points of inter-section of the line t1' ýwand the blade. By analogy with Equation (2-25), the circulation at the points of intersection of the line t1, Z and the blade is defined as S1,-, f- (VV0+ odf+ -o7- (2-59) where the integration starts at a point on the edge of the lifting surface or wake and goes to the point 1 Similar to Equation (2-37), we define aB•l,•) = g + V-til, a' f. (V0+ ° df_+ + VV- d• -(2-60) which gives 2 dZ I d 1 2q0 V1B + B oS ( Op + sin •op 0 f2 + +aET]d 1 ( 6) + (2qo - 7ýB+ 7+ •O 70 + 79P- " 7qO-) sin •o~ dt (2-61) The thrust coefficient is T Z 1 -F cr d= [ co P IU2o + n2 .92 Cos (ýOp•° ') T 12 p U 2 irR2 7r Uj2 f + P-+ 2 + g22 "•2 sin Op f cl/R J cos (sOp-3) /Iar PB aEc c2/R ar,1~ 4lPB aET\ 1ET +•1 k sin (Op -týe 2 (V('B at cot Op vrB +VFBat d+c°SOPc2 VCoBs VBd + sin-p VR (2-62) ~c/R/ aEc ' B B at, +~ (T+3(VV)2 37 Similarly the moment acting on the propeller blades is Z-I AlJ Y~ P6X(n))dS Sb=O a SB I• (2qO + 7r) Vo (s X n. dS (2-63) 2 /)= SB And tile moment about the i axis is if .°i-P Z• (2q( +7V]'V5i'(sXn)dS (2-64) This equation is not of fundamental use in the following development and is not further reduced. Before the previous formulation can be considered the complete solution, two points assumed in the development must be checked: first, that the integration over the region at infinity, which was neglected in going from Equation (2-18) to Equation (2-19), gives at most a constant; second, either that the solution is unique or that further conditions to establish uniqueness need be specified. To examine the integration over the boundary at infinity, an area consisting of a finite cylinder of radius W with ends at x + X is considered. The integration over this area as X, W -, + - is to be evaluated. By uniqueness proofs45 for nonlifting problems, the integrals of 0 and V0 are zero for that portion of the solution arising from the integration over the body. Any constant arising from integration over the wake also gives zero on the cylinder and the upstream surface. However, the downstream surface must be investigated in detail since neither 0 nor VO go to zero as x - + Co At the downstream surface the vortex sheet rolls up as it does for wings; see Cummings.50 How-ever, the total circulation remains constant and the lines of constant circulation become asymptotic to regular helical vortices of constant radius and pitch. The integral over the downstream surface requires evaluation of the two quantities J1 =f i dy' -Xy dz (2-65) X-- __0 --00_o and 5 0 Cunimings, D., "Vortex Interactions in a Propeller Wake," Massachusetts Institute of Technology, Naval Architecture Department Report 68-12 (Jun 1968). 38 X400 00-dy f xIrs dz#1 (2-66) where r = Xi +.y j + z k s =-y'j + z k and the bar through the integral sign means the vortex itself is excluded; 0 and Ox are undefined on the vortex surface, except in the limit. In Appendix A, the form of the potential for a vortex distribution on a regular helix is examined, and it is shown that both 0 and Ox have the same form far downstream. Since the velocities are known to exist in the sense of a Cauchy principal-value,11,13,14 we assume the potential does also. In particular, this means both 0 and Ox are bounded in the fluid. The region of integration can be divided into two regions; one bounded by a circle of radius A and the other one the area outside this circle and going to infinity. By taking A sufficiently large, integration over the area exterior to the circle can be made to yield as small a value as desired, since 0 and its derivatives go to zero. In the interior of the circle, the boundedness of 0 and the power of X in the deonimnator insure that the limit as X -0 gives a zero value for the integral. Thus the integral over the surface bounding the fluid does not contribute to the expression for 0 given in Equation (2-19) and the statement to that effect just previous to Equation (2-19) is justified. To examine uniqueness of the potential, two different solutions are assumed to satisfy Laplace's equation and the boundary conditions. From these a third solution is constructed by subtracting them.. This third solution also satisfies Laplace's equation and has zero normal velocity on the fixed body. However, unless the two solutions have the same circulation, the difference solution will have a shed vortex sheet with a finite value of normal velocity on it. Hence, the two solutions must have the same value of circulation in addition to satisfying the boundary conditions. Identical positions for the shed vortex sheet are assumed. Then the difference solution has no shed vortex sheet, and 0 , 0 for Irl ,-0. Green's first identity in the form (v )+ t)dr = f 0 1 (2-67) V then shows that JJJ VO-V7dT = 0 (2-68) V Since VO-VO is a positive-definite form, it follows that V7 = 0. and hence, that 0 (the difference of any two solutions with identical circulation) can be at most a constant throughout the fluid. Therefore, any two -solutions of the problem which have identical shed vortex sheets differ at most by an unessential constant; thus the solution is unique. 39 CHAPTER 3 REGULAR PERTURBATIONS IN PROPELLER THEORY A-GENERAL FIRST-ORDER SOLUTION In this chapter, the regular-perturbation problem is examined and the application of the formulation to both design and performance calculations is discussed. For the regular-perturbation problem, the propeller is assumed to deviate little from a helical reference surface of variable radial pitch angle ýp (). (This is the same description as in Chapter 2.) In design this surface might be the approximate position of the shed vortex system, and in performance calcu-lations it would be the geometrical pitch surface. On the cylindrical surface a= constant, a left-handed coordinate system (t1 ,2' ",a is constructed (as was done in Chapters 1 and 2) with ý1 along the inter-section of the cylinder with the reference surface so that ý1 increases with increasing x. The variable ý2 is on the surface, normal to the t1 axis, so that t, increases with increasing 0. The cylinder is rolled out in Figure 5. Unit vectors (f 1 C 2,2e a) are in the (t1' t2, a) direction, respectively. SAOb + W) OPp t2 tj• Figure 5 - Coordinate System for Describing Blade Shape In terms of these new independent variables, the position vector is (Chapter 1) r = (63 + t1 sin p -c2 Cos Op%)i - a sin (Ob + W ýl Cos •°p + 2si~n •°Pj + P + pcos ( Ob + W (3-1) Cos J Op + ý2 sin •OP) 40 The blade is assumed to be described as ý2 =eE(•I, 0) (3-2) = • E(tl, a) + e ET (t], "• (3-3) where E is the function describing the blade thickness and camber for the appropriate restrictions on the independent variables t1 and Ia5, e is a small parameter describing the departure from the surface t2 = 0, Ec is the camberline function, and ET is the thickness function. The description of the blade offsets differs from the presentations in Chapters 1 and 2 by explicitly making them 0(e). The position vector of a point on the blade surface is given by 2+ t1 sin •p P- e E cos sp )i - "Ssin (0 b + W Cs cos p + e E sin ýOp 4.Cos b , +/E3-4) Two tangent vectors to the blade surface are given by tj as Hence a vector normal to the blade surface can be constructed N= ±ItX2 S-2-6 e inl ýC os 4 (3-5) 41 \|i I i en', I: sin ~P , -1a M ) a/" al, d~ (3-5) + Fh- i2 (ont'd Note that even for Jp 14= = 0 and L' = L(l/) a term appears in the cadirection. This is be-cause the cross sections are measured on the surface of cylinders of different radius. For ,-, the radial component of the normal vanishes for the stated conditions. The boundary condition on the body, Equation (2-55), gives two equations 1/2 (f+°V+ -N Av- -VO-} + C? E22 Sl f,)•ýP-•O -- 1 (p3) (3-6) and CU 3ET 1/2 (N+ V-np + N-°Vp) u + w 2 S2 2 • cos (f-P-f) (3-7) The further assumption is made that (•O- O() = o(e) (3-8) (This assumption is necessary only in design, since it insures that the camber and the potential are 0(e).) As yet no approximations beyond the inviscid, incompressible fluid and the steady velocities have been made. However, to obtain the regular perturbation solution only the terms to order e are retained. Hence, a point on the surface is given by _s =s0 + E s] + O(C2) = (6 + l si?? )p)i- C0sin ( b + W) + acos b + II+ !Lý C ) k + { -E cos iO" (3-9) (o •l COS •.E sill ýpl COS ( + - wsin b+ k+ ) k}+0(62) 42 and the normal to the blade surface is NA No + eNJ + 0 (e 2) = ±E2+ dP (cos ýOp) + d6W 2 da 2,Ta +dacWs ý01p- CA) ý sin •Op P1 e E L a 7 dW(d-10) + e -"1el . at] I in f Olp F +c os POp W -d a -Ccos p ] + E -E ea + Q (C The potential function due to the blade disturbance will vanish as e -+ 0 (since then the thickness and camber vanish) and -Op -, 3. Hence, the potential function can be expanded in a perturbation series with first term 0 (e). In particular, the potential on the blade surface is expressed in a perturbation series ' (t1' e E, a) = e6 (t1 , 0') + 0 (2) (3-11) Because f0 is 0 (e), only zero-order terms in the integral over the blade in Equation (1-50) need be retained. The equation is then r-S N= 1 13s) SB-I 3 SB SSo1 0 + 0O(e 2) __O - S -0 + t dl co+ 0 (e 2) (3-12) .Actually it is the difference in velocity across the blade, Equation (3-13), which should be taken 0 (c), which is necessarily true only in design. If ýpp-0 0 (e), as would be true in performance calculations, then an integral equation must first be solved to find the pressure distribution corresponding to an uncambered, thin lifting surface at an angle of attack. The remaining problem corresponding to the camber and thickness is then 0 (e). 43 where SBO is the projectlion of tihe blade outline on the surface 2= 0 N+ (rZ'o) (6 X) cos op) Sill p {.v c1 (O + sil 0} + 1O -S il.+ C S) cos 0 -. I, sil 0 --1 sinhp +cossp Lr-s() =(x -6? -- 1si•Sil + ) 2 + + sin 0)2 + (z -Cos O)2} 3/2 ý1 Cos ýO p and 0 = 0b + W + co As in Chapter 2, the jump in potential across the blade surface can be interpreted as a local circu-lation. In terms of the velocity jump across the surface it is given by Equation (2-59) PB (ý , a){+- } Jf V0+ o d+- v-- od _-) =f L q+ - q-) di 0 + 0(e 2) (3-13) where the path starts from any point along the leading edge and TF(zo) rzo) = (q+ - q-) -d( 0 + 0 (E2) LE Similarly, the integral in Equations (2-21) and (2-50) reduces to 4---'•[r-sd~l d a "4r S 0 --aET 2U sBU + &22 / 2 +I (OP-O) dtld + 0(e 2) (3-14) 44 The position vector of the shed vortex sheet is determined by Equations (2-46) and (2-47), which to first-order in e become 0o ( ). +0( 2 f2(zo, O Tf2(/ZO, O)+ef2 (0- TZO, O 0)+ 0(C2) U -- XT(ZO0) +" 0 -(o TE (zo0)) (3-15) + e 11-20)•do; a+ 0 (e2) OTE 2 z0 flzO=f(0) l(1)2 ff, (zO, 0) +f (zO, 0) + 0 (e) (3-16) 0 =Zo.+ 1/2 2 e V4i 0) da + 0 (e 2) OTE Hence S= _1o + el + 0 (e 2) U = z0 eW+ (XTE(ZO) +"j(0 - OTE (Zo)) + e(1/2 2 n P--o) da).e+ (1/2 f { X0) (3.17) TE 0- z O) da) + 0 (e2) However, since r(zo) in Equation (2-40) is of order e the first-order solution for the potential of the shed vortex sheet requires onlySO, The zero order normal to the shed vortex sheet becomes: O -ao xazo (+ 0 zoeo +-•i•)X (feZ+--7kO XTE U OT , -Zoi+ Zo ( TE- -U (3-18) 45 and the do( p',rotduct in lhe 1c ininite integral is, +i -0 /x -•' (a O0,./) x/ + aJ/sini 0 { s .iii a " (TI: + cos a- + cos 0 z (0cm a o (3-19) "(ITv 7TI). sii a U/ j "2 0 T; I and tihe denominator is Lro[- ' (TIE +- (a 0T/j) 2 + Z-2 +- -210) -2 0COS (0 - a+OTIi 3/2 where tie position vector of an arbitrary point in space is expressed in cylindrical coordinates. This is done to facilitate the integration since past experience 11 13,14 indicates that in some cases the integration can be done analytically for such a representation. Putting together the previously described pieces, one finds the total first-order potential for the disturbance due to the blades, omitting the E I _ I? " -47r b--- ff ss d+ d • 0i (q+ - q--) df + 2+ (02 5T Lt:"E aý, (3-21) Il-.SoI-3 fR z(zc- 0 • x da 4, 1)= aa 13h This equation is the first-order solution for the regular-perturbation problem. In the next two sections, the use of this equation in design and performance predictions is discussed, and in the last section the second-order solution is outlined. The force on the propeller blade, Equation (2-56), is to first-order in E F =eff (qo VFB)NM dwd (3-22) 46 and the thrust produced by all the blades is •Z•-I •R P . ... ^U+2 ý,Zý2 T -i" F= epZf cos O V U+2 02 CQS (p -132 F(6) d a (3-23) b=O o0 The torque can be computed in a similar manner but it is not used in the following sections. A second-order theory follows by straightforward application of the previously described procedure when thickness and camber are expanded in perturbation series. The second-order effects include con-sideration of the first-order position of the shed vortex position, Equation (3-17). Second-order design theory is discussed in Section D. B-DESIGN PROBLEM For the design problem considered here the total thrust produced by the Z-blades is specified at one operating point, as well as the following geometrical data: 1. Blade outline 2. Chordwise thickness function and radial thickness ratio 3. Number of blades 4. Position of blade-reference line The designer also specifies, to within a scale factor, the chordwise difference in pressure, i.e., the pressure on the upper surface minus the pressure on the lower surface, which reduces to P- P+ N{ (q+,2 -(q)2 Since q = q0 + V0, the pressure difference is +P = q 0 o B +1 27 B ' Vr Also, since q0 has a zero-order component only along e1, the significant term of VFB is theel component. The velocity difference 7 rB can be resolved into the three orthogonal components (y, g. o), IB= -oEj+u +a (NO Xe1) From the boundary condition, Ecjuation (3-7), the component of # is known to first order ji= 2 eVU2 +,,,2 2" Cos (Op -0. NO) + 0 (e2) However, to O(e), no use is made of this knowledge. , 47 Now since V I'l = o(,), V ýj. = o(e). si, (O ) = O(c), and qo = iqOI cos (,p )_e/ + 1q,1 i I , 3)c the pressure distribution is p S= (2 + 22 "2 (e-f(I) cos (P- + O(e2)) I lence to first order, the chordwise component of the velocity difference is related to the pressurc difference across the blade by the equation p- - p+ (3-24) p g + EZ2 Los (ýp-9) For the first-order problem, no information need be specified about the other components of the velocity difference. •The component a can be found from V 17B since IB is known in terms of y(l) stated previously. An alternative is to note that when a tangential discontinuity in velocity exists across a surface, one can define the surface as a vortex sheet with a vorticity distribution given by -0 -A. nx X q+ - q-) = -. x ('eel + a (& X el)) TO+ U= ' aj uel + -(No X el) Since A must satisfy V A=O the component a can be found'from the partial differential equation governing the components. The equivalence of both these approaches is demonstrated in Chapter 5. 48 The local circulation, which is the quantity needed in the calculations, is given: by Equation (2-59) or to first-order by Equation (3-13). Since the chordwise component of velocity difference is known, d , the tangent vector in Equation (2-59), can be taken as df= tj dt, dt + (e -e2 Hence Equation (2-59) becomes FJ '{ 0 " "PB = f e 2 } ((P- d () + l C e2 L " ¢2 (Z)'e' ( { +1 ji--N+ a (f6oX el) at • 2) aET + V OB" E2 •-Cl dt S+e-(1 V +-e 2 (NE Xej))+ V OB e2e d{1 ,f2 a' When the quantities are expanded in a perturbation series, one finds 1rB ,(1) d{1 2 ,),(2) + l(!) + le f? fc2 a, -2 a EJT\ "AX efI) + ' 0b)(1) "e2-b ) -2 d + O(e 3 where dp (COS2 tOp) _ dW d6M -S 2 ( -N +X ) 1 -o sin Op '+ c os Op dw- 2 rJ5 d dco and E2 VOB(l) = 2v; v is from Equation (3-26). For the first-order problem, only thIe leading teim stated previously is needed = ( f 1(1) dt1 C 2 49 When Equation (3-24) is integrated over the chord, the total bound circulation distribution, still with the unknown scale factor, is obtained. This unknown factor can be determined from the equation for thrust. IEquation (3-23). Hlence everything needed to calculate the first-order perturbation potential is known. What is not yet known, however, is the magnitude of the camber and the blade pitch which will produce the specified thrust. To obtain these values, the boundary condition, Equation (3-6) is used. To first order, this equation is -sin -1/2 (3-25) V'U2 +(2n where i is a principal-value integral given by Equation (2-18). For conceptual consistency the blade trailing edge must lie on the reference surface so that the trailing vortices spring from the trailing edge. Thus f at', f a c f; c c2 c2 Q 1 • This camberline will consist of a shape measured from the nose-tail line and, in general, an angle of attack, This angle is added to the pitch angle of the reference surface to give the total blade pitch. This incremental angle is called the ideal angle of attack. The knowledge of the camberline completes the first approximation for the design problem. The regular perturbation solution involves an integration to find the slope of the camberline and then a straightforward integration of this slope to find the shape. However, a new term appears in Equation (3-25) which should be explicitly mentioned before proceeding with the analysis. From the equation for 4 in Equation (2-18), one sees that Equation (3-21) gives 1/2 tP when the principal value for points on the blade reference surface is taken. Hence, if the gradient of • in Equation (3-21) gives the velocity V = tueI + ve2 + w.e•- (3-26) and if for r r0 I 2 B = u(l, 0, "5)eI + v(t1, 0, ý)e 2 + w(t 1, 0, e then Equation (3-25) becomes 50 / t , 2 - -2 S i l l ( 0 , L( 3 -2 7 ) + FCos p + co-sil --(ill w(•l, 0, a() In the literature, intuitive arguments are used to derive only the first term on the right-hand side of Equation (3-27). The formal procedure presented here indicates that another term also contributes to the camberline shape. For the term containing a multiple of the distance from the reference line, the effect will be most pronounced for points near the blade edge. In the discussion section of Chapter 5, an estimate of the contribution from this term is made. In addition, other aspects of the design problem which simplify the integrals are discussed. C-PERFORMANCE CALCULATIONS To determine the performance of a propeller over a given J-range, it is assumed that the position vector of a point on the surface is given, that is, the geometry is completely specified. The unknowns of interest are the thrust and torque as a function of J and possibly the pressure distribution. The pressure e aET distribution near the leading edge will not generally be accurate because -is in general infinite and 8ý1 LE hence not of order e as assumed in the derivation. In the design problem, this point was not critical but in performance calculations it will generally give results for the pressure distribution which are not uniformly valid at the leading edge. However, it is integrable and hence the total thrust can be found. In calculating the propeller performance, the circulation is unknown, and the zero-order position of the shed vortex sheet is known. Modes for the chordwise distribution of velocity given in Equation (3-24) can be assumed which satisfy the Kutta condition and which have variable spanwise coefficients. Then, in principle, the integral equation formulation in Equation (3-25) can be solved for the spanwise coefficients. Sugai4 0 was able to successfully evaluate numerically such a procedure. Murray 3 3 used an iteration pro-cedure, starting with blade-element theory (Glauert5 ) with a reduction factor found necessary for con-vergence. As already mentioned, Murray's performance calculations did not converge to the values set in his design. Presumably the exact solution in Equation (2-50) could be used as the integral equation without much increase in complexity but no one has done this yet. 5 1Glauert, H., "Airplane Propellers," Division L, Aerodynamic Theory, Edited by W.F. Durand, J. Springer, pp. 169-360 (1935); also published by Dover Publications, Inc., New York (1963). 51 Rather than solve an integral equation it would be preferable to obtain a solution which could be constructed as quadratures as was done in the design problem. Such an expression is found in the next chapter. D-SECOND-ORDER DESIGN THEORY The second-order theory follows by retaining terms to order e 2 in the previous development. Most of the second-order effects are known from the first-order solution; however, the second-order increment in circulation must be evaluated. This is found by considering the second-order thrust T =T,(g(1) + F2 r(2)) + e2 T2 (3-28) where TI(CU(1)) is given by Equation (3-23) and ,R ~ Cl 1 ) 1 T 2 p Z d ( r(B) () cos op + 2 u+ .2 I sin (Pp-g 3) cos 0p E2 vrB(1) + cos (Pp-fg) sin iPp (3-29) (3-29+) Vi (a rB(J) a Ej) a4OB(M 8.ET All quantities in this integral are known from the first-order solution. Since in design the thrust is set and since TI(eU()) is the set value, the second-order terms must vanish: e 2 T + 2 T2 = 0 which fixes the value of r( 2) eC1 r •7B1 (1) r( 2) (•=Z -j 1/2 (3-30) c2 + 22 Z2 Cos (ýpp-(aB a)l aýB() aETe +tan ý 1 E" ) + -+ tan (sp -0) f2 vrB( d 1 52 In design the form of the chordwise circulation is assumed to remain, constant, i.e., y(• 1, •) in Equation (3-24) is a fixed chordwise variation; however, the amplitude of the function may change. The scale is fixed by the magnitude of the bound circulation.. If -yis normalized to give unity when integrated over the chord C' f c -y1, •a) d =l then the scale factor will be just the total bound circulation. In this case (q+ - q-) o e ,(I() (a) + C A-(2 (a) -Y ( 1 ' ) and FB(1) = p(1) (Z) f 1 dl (3-31) cZ where (t]' Z5) is the normalized value of Equation (3-24). Now we have previously found that •1 fl]I~y Ec B )ef e2 ' (NfX el)) "B C 2 ~ ) d , +62 •21 (2 ., 3 1 +e aET e_2 (Bo e1),1(3-32) Thus e( = r(1+e2 F 2)+ +2( l + 3E) V lPB ) + dP1)+dlje aE +e2 Vdý Hence r (2) (2) C I (aEC Ap(2) ~~ =p2c2-ff•" (P(1) + a(1) e2 (o (NX el)) aET! +42 B_1) d t, (3-33) 53 All quanttities ilppCaring under the integral sign in Equations ( 2-49) and (2-50) arc expanded in peitur•ation series and the second-order contribution found. The necessary expansions have been given previously and in Part A. In these calculations it is assumed that the thickness function is given and constant at order c. The quantity sought is the second-order correction to the camber. This is obtained from the boundary condition on the blade set in Equations (2-10) and (3-6). Because of the length of the ex-pression, it is not given explicitly but it follows directly from the previously described development. 54 CHAPTER 4 SINGULAR PERTURBATIONS IN PROPELLER THEORY As described in tile introduction, the singular perturbatiooo problem examines the solution as the chordlength goes to zero. At a fixed point r =(."' in the flow field relaJtive to the propeller, the chordlength-to-diameter ratio F" goes to zero. Since the geometry is unchanged except for chordlength, the blade forces go to zero in this limit. Hence the flow disturbance vanishes as the lifting surface vanishes. The solution in this case is just a lifting line (as the chordlength goes to zero, thickness effects give a second-order dipole distribution rather than a first-order source distribution and hence will appear in the higher-order terms of tile solution). This is the outer flow. The flow near the body is the inner flow and can be found by stretching the coordinates near the body so that the flow field is magnified sufficiently to recover the necessary details. In the inner limit the stretched variables are held constant as the chordlength goes to zero. A-FIRST-ORDER TERMS Outer Flow The starting point is the outer flow which is found by straightforward linearization of the exact solution given by Equation (2-50). Both 0 and F will be of the same order in the integrals of Equations (2-49) and (2-50). However the integral over the blade area is of higher-order than the integral over the shed vortex sheet because the area of integration goes to zero with the chordlength. The zero-order integration over the shed vortex sheet reduces to the lifting-line results. i.e., a TTFT(zO)) -b= 21rb -with the zero-order position of the vortex sheet. Formally, the circulation is expressed in a perturbation series of the form r(z 0, E) = 61 (E) r( 1) (Zo) + 62(E) r( 2) (Zo) + " (4-1) where E is the maximum chordlength-to-diameter ratio and 65 n() is an ordered set of linearly independent gage functions, with the property 6n+1 = O(6,) as F-, 0. A general form for them is a ln Oa3 n1, 9) an e- where a.m are constants. Normally the terms with a,, 4 =k 0 are considered trans-cendentally small and are ignored. The coefficients aM will be found as the solution evolves. The position vector of the curves of constant circulation in the shed vortex sheet is also expanded in a perturbation series 55 (Z0" 0; ý) =(:o 0) + 51(E) I(zO, 0) + 52(J) 2 (zo, 0) + (4-2) From Equation (2-46) and (2-47) the zero-order solution for a blade-reference line lying along a radial lirie in the x = 0 plane is U iO zOef + -(Ob)-O (4-3) and consequently the normal is given by U 30 az0 Z 2 zoi + -.. e 0 + O{S1) (4-4) Hence for the outer flow, the lowest-order solution in Equation (3-50) is 0(x, y, z; 9) = 61 01(r) + R (4-5) 01 (r] " (zO) G , o dzO) Z 1o [0 --(a-%)jZO-bsin (0-a) where G(r, zo)= d a b=O 0b U 2 a2 ÷z02 20 2 Z 1 - -H (a- 30r 1 ZOI +- --dy (4-6) b=O 0 r where r = J//x - HY)2 + Cy2 + z-2 _ 2zo -cos (0--Ob) The reason for this restriction on the reference line will be explained when the inner flow is examined. 56 This expression is also derived in Appendix A from the Biot-Savart Law applied to an elemental horseshoe vortex. Isay35 presents a similar equation which he bases on the work of Yamazaki. 19 Equation (2-50) also contains another term which appears to be of order Z; the integration of the source terms over the boundary. The first-order effect in these integrals reduces to an integration over the radius and an integration of the thickness slope over the chord. Since only closed profiles are considered, this integration is zero, and the contribution is second order. In the outer flow one expects dipoles rather than sources which is confirmed in the previous statement. Inner Flow Unfortunately, the potential is not uniquely determined since the circulation is unknown. The circulation is determined by details at the body and hence must be determined from an inner flow for which the body boundary condition is not lost as the limit of zero chordlength is taken. Because the spanwise scale is not changed as f - 0, a stretching is needed for only the chordwise variables. A suitable stretching for the reference blade along the z axis is "x = x/l and = y/g 0 <z<R (4-7) where' and T are held constant as F-, 0, The z variable remains unchanged. The previously described transformation applied to the Laplace equation gives --- + 4 2 a (4-8) a-2 a= 2 az 2 Now it can be seen why rake and warp have been excluded from the analysis: If rake were included, a displacement varying with z would have to be made in x before the stretching in Equation (4-7) could be performed,. This displacement would introduce a term of order F on the right-hand side of Equation (4-8). Rake would be equivalent to sweep in wings, which has been considered by Thurber.7 If warp were in-cluded, a displacement in y would be necessary. A displacement in either x or y prior to stretching would also lead to a transformed Laplace equation with a term of order T, unless the displacement were constant. The second-order right-hand side means that the first two nontrivial solutions are governed by a two-dimensional Laplace equation with the spanwise variable a parameter. Experience with wing perturbation problems indicates that the first term governed by Poisson's equation is a practical limit to the number of terms in the series, The potential in terms of the inner variables must also be expanded in a perturbation series. The potential in the inner region will be S(X, y, Z; r) O(-, ,z; E)=D y- , Z; -3l('x, z) + 3;2(E))2x,, z) + 0 00 (49) The inner potential is of order F because the velocities in the inner flow are required to be bounded as -, 0. 57 The inner boundary condition in Equation (2-10) SL V -go " Li forreSB has to be expressed in terms of inner variables. Instead of the T and y defined in Equation (4-9), it is convenient to use variables measured along and perpendicular to the chordline in the'xy plane. The chord-line is at an angle .pB/z) relative to the Taxis and passes through the reference line, the z axis. The variable measured along the chordline is positive in the downstream direction, and the variable t is normal to the 1 axis, positive in upstream direction. This coordinate system has been examined in Chapter I and is shown in Figure 6 below. Y Figure 6 - Geometry in Plane at Fixed z The variables I and •2 are related to x and y by x -= sin OWz) -2'cos dz)4 (4-10) y- = cos sBz)- 2 sin •, yz) Obviously if a hub existed on such a propeller, a complicated expression in (x,y,z).would be needed for its geometrical specification. For narrow blades considered here, this problem is not significant. The shape is assumed to be given relative to the reference surface bya .shape function (Chapter 1) t2 = E(t I"z) (4-11) Substituting this expression into the previous equations for x and y, and taking the appropriate differential, one can find a normal (Chapter 1) 58 x -X-l (4-12) az/ where s_= s.L , E( z),z) N-- F _ E 4B tE 1 D aE dcPB 1 L-- z a z a • 1 d z )k j (4 -1 3 ) where efI = i sinp - L cos 'pB f2 = - £ cos ýPB - L sin ýoB For the body boundary condition, the normal component of the free-stream velocity is needed: o" _ I_ 2 + -y 2 z2 (sin (iP 1 -) cos APB-P) (d(4-14) +s~y h--- -dz az a dz / where U tan j:=-The (t1, t2) variables are linear combinations of the xy variables and thus a stretching of xy is also a stretching of t1' t2. Let the stretched t1, t2 variables be Ol, u2, In Chapter 1, the transformation of the partial derivatives in the normal was discussed. It was shown that E1, z; F) = E(, = E(ol, z) aE O= 0() (4-15) 3E -0T 59 With these relations the transformation of the right-hand side of Equation (4-14) becomes _ 0 'lo U 2 +' 2 z2 (sin a- -, o g Lsn( , z) Cos (4-16 d pB 3 E aE (4.16) 2 ~+ E -z where . is given by Equation (4-7). The normal in Equation (4-13) transforms to a =+ E dl+( B2aE a E dPB\' N =+ e l.±F! e -- + E k (4-17) f2-at-(Y dz 8z a• djz ] The transformation of Equation (2-10), using inner variables (op, 02, z), then becomes (LaE BF 3E aEd I i--- -j+2 + --z e 3 a R . dz az at, dz/ L~ 4;b /P a ý L3z dz 30, a~ 02 -VU2 + E22 z2 sin (ýPB-P) --Cos }B-A) (4-18) d / PB a E aE-Te 2S br- -+ E-- - dz az a ti dz ,.. . In this section only the first-order terms are considered. Hence for 4) F'4l the equation for (1 is a24) a24,1 a + 12 0 (4-19a) aa• 2 ao 22 with the body boundary condition, on 02 = E(o1, z), 0 < z < R a% E 0% E a(2 a1 84)] V 1 (sin cr ----cos C1 ) (4-19b) 60 where Vz)2 a;(z) PWz) - OW(z) The remaining boundary condition far from the body has to be found by matching with the outer flow. This process will be formalized in the following section, but intuitively it is expected that the potential in the outer flow will match the inner flow as the outer variables 2r + y2. 0, <z <R and the inner variables cr 1 2 + o22 -, 0 <z <R. In the outer flow, as an observer approaches the reference blade, the flow should look locally like a two-dimensional line vortex with higher-order effects caused by the in-duced velocities from the trailing vortex system. Hence for the first-order inner potential, the far-field boundary condition is that the flow look like a line vortex. (Subsequent matching will confirm this hypothesis.) The inner potential could in principle be found from an inner expansion of the exact potential given in Chapter 2. Germain52 reports such procedures applied to wings lead to considerable difficulty and that it is simpler to solve the posed two-dimensional problem of Equation (4-19). In fact, using complex variables and conformal mapping of a circular cylinder, one can state the solution immediately: S_~icC 1 ei~ 4b1 (a,, 02, z) = V 1 Re f( o1 + i o2) e + a2 li a, F' (Gj"0, () 1 + i 02) " f u1 ic I ftu1 a]+') -) + i r In a -( 1 + i0 2) e -i (4-20) where 00 an(Z) f(al + i' 2 )= oa + iU 2 + Y. n=O (orl+i U 2f is the inverse mapping function an(z) are complex numbers depending on the section shape a(z) is the radius of the transformed circular cylinder w(z) = 4 7r a(z)Vl(z) sin (al(z) - ao(z)) is the value of the circulation when the Kutta condition is satisfied and cto(Z) is the angle of zero lift of the profile section 5 2Germain, P., "Recent Evolution in Problems and Methods in Aerodynamics," Journal of the Royal Aeronautical Society, Vol. 71, No. 682, pp. 673-691 (1967). 61 Several techniques exist for finding the mapping function for a given section geometry. Most of these are discussed by Thwaites. 5 3 It is not necessary at this time to examine the techniques for obtaining these coefficients, but they are considered known in terms of the geometry. Specifications of the circu-lation renders the two-dimensional problem unique. In a review of the planar wing problem, Ogilvie, 5 4 points out that the inner expansion can be taken as a Laurent series but doing so adds no new information or insight to the direct solution from conformal mapping given previously. Although the solution indicated in Equation (4-20) is the required solution, in anticipation of the matching discussed in the next section, the form of the solution at large distances from the body will be examined. Figure 7 shows complex number u1+i02, given by peiei 021 S"-------- ----v1 Figure 7 - Profile Coordinates for Inner Flow Hence pa = ¢rol2 + a22, co= tan"' 0r2 tal The expansion is straightforward, except for the log term 5 3Thwaites, B., "Incompressible Aerodynamics," Oxford University Press, England (1960). 5 4Ogilvie, T.F., "Singular Perturbation Problems in Ship Hydrodynamics," University of Michigan, Department of Naval Architecture, Report 096 (Oct 1970). 62 -in e CO ao a, In 1n -+ In + + + a a" paei(O A2e2ico n a al-1/2ao2 2 =In l- + ico + (0e-ico + -e-2ico +•° a A/2 Hence for large p, the expansion is VIl(Z 1, a2, z) -- 2 a sin (rl-co) (W + WO(z)) + Re V1(z)A a2 a2ef(al W) a 0 aO2-ai + 1 e-i(a+2c+) + -. + -. 2 [ ;eic ' p 2 e2icW "[a0 a1 -1/2ao 2 +2iasin (cl 1 -O) - + Mee W u2e2ico a2 -1/2a0 aI + 3 1U3 e3icw where coo(z) is arbitrary. By grouping terms, one finds 2a sin (a, ao) +WCOP + ~ hln(a],z) (-1 where kVra1 , z) = a cos (er-w.) + Re e-i(al+w) + 2i sin (al-aO)aoe-J 2ie2i2 k2 (al,z) +Re a e -i(,l 2r +a -- sin (al- %O) e-2ic (a1-1/2a 2) 63 Ia3 ao2-a1 k3{a 1 , z) =Re e-i(al+ 3 -)) " _aa e-(l-3'} 2i + sin ((1-oJ/) e-3ico -1/2aoaj+l/3ao2) a2 etc. The first term is a vortex; the second, a dipole. Then the higher-order terms are listed. No source terms appear, as expected, since none are in the first-order outer solution. Matching The two expressions for 0 are the leading terms of asymptotic expansions valid in different domains of the flow field. For the reference blade, the outer solution is valid everywhere in the flow field, except z ý<R, Yx2 + y2 < Co. The inner solution is valid forz <R, /y 2 + y2 < Cor C1 x2 + y2 • CO (E) where lirn -, o. This extension of the region of validity is discussed by Kaplan4 and Cole.5 In this assumed overlap region, the x and y variables are taken as x = x g(E) (4-22) y = where X and • are constants of 0(1), and g(ý) is the function which places x and y in the intermediate region. To insure x and y are in the intermediate region, g(E) must satisfy g(E) 40' E , asF O 0 g(•) ) The x and y are the outer variables, and the inner variables are obtained from Equation (4-7) as -X = g(g) x _ (4-23) X = g(j)I =-i- =Y " 6 4 64 From the definition of an asymptotic sequence, e.g., Erdelyi,55 the two solutions can be matched to order AN(") if N N g g M• 0n(7)9b(, 37, z) -n(g) T~( e' , z) n=1 n--lim =0 9,4A0 ,1(e) (4-24) (N = 1,2 Hence the dominant terms can be matched if t 1m ) 01 (g, yg, z) -'I¢flX ',y 1 6•z) A match to order F, i.e., jLl(F) = F, is possible by selecting 61(F) F rand by requiring the following equality to hold fim (91 (g, 7g, z) = 4), (oo, o, z) (4-25) g4O The matching principle given by Van Dyke3 expresses the same requirements as described previously in a more concise form. The matching principle will be used later but for now a much simpler argument can be used to find the circulation in the outer flow. The argument is that the spanwise circulation must be independent of the chordwise stretching. Thus I( 1) (z) = 4 7ra(z) Vl(z) sin (al(z) - ao(z)) (4-26) Hence the outer solution is completely determined to the first order. There are no incompatibilities; conse-quently, the first-order solution is complete. 5 5Erdelyi, A., "Asymptotic Expansions," Dover Publications, Inc.,,New York (1956). 65 A composite solution, not necessarily unique but uniformly valid throughout the flow field, can be directly written down. To within an arbitrary constant of z, which to second order is only a parameter in the inner flow, the first-order composite solution is the sum of the inner solution plus the outer solution minus the common part. The common part is either side of the equality in Equation (4-25), and from the inner solution in Equation (4-21) it is seen to be 0C.p. -- z) = 0 z>R - T V1(z) 2a(z) sin (al(z) - ar/z)) (Co + WOdz)) 0 <z R Oc j 0fl(x, Y, z) + •Abl(0 1, U 2, z) (4-27) + 2 e Vla sin(al-aO) (co)+ 0o<) Use of the matching principle will show that coo (z) is such that cw + co = tan'ly/x. This is shown in Equations (4-34) and (4-45). Such a composite solution is useful in finding information in the flow field such as field-point velocities. Since this overall view of the flow field is not required in the present analysis, it is not considered again; however, similar expressions valid to higher order can be obtained just as easily when the individual higher-order solutions are known. Composite solutions are discussed at some length by both Van Dyke3 and Cole. 5 The circulation to first order corresponds to what is known as blade-element theory. 5 1 Its use-fulness is restricted to narrow propellers with light loads. The second-order terms derived in Section B will modify the circulation distribution with higher-order effects and will extend the usefulness of the solution. B-SECOND-ORDER TERMS Outer Flow The second-order terms in the outer flow will arise from three sources. The first will be from the integration over the blade. Although this could be extracted by a Taylor expansion from the exact formulation in Equation (2-50), it is simpler to assume that a distribution of dipoles in the x and y direction will be the required singularity system. Once again, if the assumed form is not correct then the matching will not be possible. The second term in the outer series will be from the second-order term in the cir-culation expansion in Equation (4-1). This will appear in an equation similar to Equation (4-5). The last second-order contribution will appear because of the first-order modification to the position of the vortex sheet. Since this last term is too complex to treat analytically, an approximation will be made to simplify 66 it. For this term it is necessary to know the expression forl(zO, 0) in Equation (4-2). To obtain S1, the first-order effects in Equations (2-46) and (2-47) must be included. If, on the vortex sheet, 1/2 x, 1 --•-w U. and 1/2 ia-- w then the zero- and first-order terms for the position of lines of constant circu-lation can be found from fl w Z O b +Wo) da (4-28) f2 -zOf dU) (4-29) f 0 Ob 92z 0 + "'ut( 1 ) (t0) where the second term is left in the inconsistent form because of the approximation to be made. Since there is little hope of analytically integrating Equation (4-29), one seeks a reasonable approximation. The main contribution of the modified position of the shed vorticity will be from points close to the line. Hence a Taylor series expansion of the velocity is called for; however, for even greater simplicity, all but the first term is ignored, that is, the velocities are evaluated at the lifting line. Hence U + Fua(1)(Zo) f 2 0Zo (0-0b) s2z0 + 'u/l) (Zo) U + Ua(1 ) (Zo) 1 The quantity z0 is -times the pitch of the shed vortex at the lifting line. This g2z0 + Eut(P) (zO) 2r pitch is often called the hydrodynamic pitch and it is denoted by Pi(zo). Hence Pi (z 0) f2 -(0 (0-0b) (4-30) 21r It is possible to propose a slightly better model for the integral in Equation (4-29) since the induced velocity both at the lifting line and the value far downstream are known (the value for x 4 is twice that 67 at the lifting line). Numerical calculations 17 have been performed for which the information necessary to propose a variation in the induced velocities as a function of x or 0 could be obtained. However, the previous assumption is universally made and considerably simplifies the calculations. A similar universal and significantly simplifying approximation is to ignore the contraction of lines of constant circulation. This means the radial velocity in Equation (4-28) is approximated by zero. Any other constant approximation for it would lead to negative values for the radial position, clearly an unacceptable occurrence. Here too, alternatives to this crude approximation would be the numerical cal-culations or results of actuator-disk theory. These alternatives produce complications not essential to the continuation of the present investigation, and we accept the approximations as customarily used. These two approximations involved in the higher-order description of the shed vortex position are defined as the basis of moderately-loaded propeller theory.13 Hence to the first-order, the approximate position vector is Pi (z 0 ) z 0 zfea + -(0- Ob)i (4-31) 27r "ý"•0 + I _ U Pi (z 0 ) Since this approximation merely replaces the constant -in Equation (4-3) by- , no further insight is added to the problem by considering this term and from now on it will be ignored. That no essential modification to the theory results is not obvious in the formulation of the velocity potential but can be seen in the calculation of induced velocities by the Biot-Savart Law; see Appendix B. However, the cal-culation of induced velocities is usually done with expressions derived for constant Pi. Thus only the modification of the blade-element circulation distribution and the dipole distribution comprise the direct second-order outer potential. Although a general gage function 62(j) must be assumed for the circulation distribution, second-order terms from the blade integration in Equation (2-50) are easily seen to be order F2. (Change the chordwise integration to inner variables and expand the expression in a Taylor series.) Thus the two-term outer solution is Notable exceptions are Erickson et al. 56 and Cummings, 5 0 who calculate vortex interactions for lifting-line theory. 5 6Erickson, J.E. et al., "A Theory for VTOL Propeller Operation in a Static Condition," Curtiss-Wright Corporation, Caldwell, N.J. (Oct 1965). The vorticity vector A= n X VIP has no additional term from the e - component of n, since e- X v P =0. 68 j(r) = Fler) + 82(c) 012(f) + e2 d2r) (4-32) where 01 is given by Equation (4-5), and ol12(r) =--• F(2) (Z 0 ) G(r, zO) dz0 1 Z OR xul(zO) + (y + z0 sin 0b) 9 2 (zo) dz0 23 0 i2(1) JO [x2 + (y + zo sin Ob2 + (z -zo Cos O) 3/2 (4-33) where p( 2) (Zo) is the next term in the circulation expansion in Equation (4-1), and gl(z) and u2(zo) are the dipole strengths. All three of these are unknown. To match with the inner flow, one needs the asymtotic sequences which results when Equation (4-22) is substituted into the two-term expression for 0. By an indirect procedure described in Appendix B, the expansion of 0 is found to be 0(g7, g7, z; g) = Foz; g) -1--r(1) + 62 r(2), tan-1 Y + gg• u (1) (z) +÷ ut1) (z)] (4.34) Z2 x#1 (z) + Y92 (z) +• + H.O.T. 4 X +Y7 where ua(1) (z) and u/ 1 ) (z) are the axial and tangential velocities at the lifting line as induced by the first-order circulation distribution. Inner Flow For the second term in the inner flow, the governing equations must be examined. From Equation (4-18), the boundary condition on the body can be determined. One substitutes the two-term inner expansion 4i, 3, z; F) = NIP( 1 , 02; z) + v2 4) 2(0 1, (2; z) (4-35) into that equation and finds that (D2 satisfies 69 2 a1 )D o 2) (4-36) E7 Va -Y at ,aa, on 02 --E(o 2 ; z), 0 <z <R Obviously, it is most convenient to take lirn 0V2 = 0 so that the body boundary condition becomes aq 2 E a2 -° • (a]; Z) -ý ° =- 0 (4-37) on o2 = E(oI; z), 0 <z <R Similarly, in the flow field, Equation (4-8) requires 4)2 to satisfy a two-dimensional Laplace equation. As yet the boundary condition at Ol'2 + 022 -0is unknown and must be determined by matching. The simplest problem for which the two-dimensional flow cannot pass through the body is that associated with an airfoil at an angle of attack to the flow. Hence as V//o12 + U22 one assumes that the potential behaves like (D2(91, 02; z) - 01 V2(z) cos , 2(z) + 02 V2(z) sin ac2(z) (4-38) where V 2(z) and a2 (z) are unknown. This two-dimensional problem is also easily solved, using complex variables and conformal mapping. With the same notation as in Equations (4-20) and (4-21) the second-order inner potential is 42 V12 01 cos a.2 + u2 sin ct2 - 2 a sin (ag2 -riO} (Co + 0)0 (z)) a n + i-• kn(CC2, z) (4-39) Sn=1lP where wo(z) is arbitrary, and the kn(a z) are given in Equation (4-21). In order to find the necessary form of this expression for matching, one substitutes a, = a, g/F and 02 a2 91j (4-40) 70 into the expression for 'D2 and obtains V2 - g/= (ii cos a2 + U2 sin a2 ) - 2a sin (W,- ct-)) (w + w 0 (z)) V2 + k1 (Ct2, z) + O(E/g)2 (4-41) Combining this expression with the one for ID, one finds the expansion for ') has the ,form ¢(D ( g/F' FY2 g/E, z; E) = Fl(Z, F) Sk , (a,, z) "g(ý2 + 2)1/2 " P2 V2 [g/E (a, cos a a 2 + 02 sa 2) -2a sin (a2 - aO) (w + wo)+ ] (4-42) In the next section the actual matching will be performed. Matching In this section the coefficients and gage functions are to be selected so that the two expressions for 0 will be matched. In Equation (4-24) the matching condition is given for the general case. One now seeks to use this expression to match Equations (4-34) and (4-42). First, one notes that kl(al,z) in Equation (4-42) is a linear combination of cos co and sin co hence also a linear combination of and --- and this term is known from the first-order matching. By inspection this term will cancel with the dipole-distribution terms in Equation (4-34), thus implicitly giving the so-lution for their strength: P 1(z) + 92(z) V-(Z kl(1, z) =12 (4-43) Hence, the matching is reduced to the expression 71 -2F(2) tan-'Y+ eg[x Ua (Z) + tU 1 + 2 a -v2 2 I ( Cos '2 + 'U2 sin a2) (4-44) - 2a sin (a2-O) (w coo 0)+ where a2(E) is such that terms not explicitly given vanish in the limit. This matching is possible by selecting "V2= C2 62 = V2(,) = E2 (4-45) S+ 0 = tan- y/ for which x Ua(1) (z)+y u/ 1) (z) = V2 (U1 cos a2 ÷ 7 sin a 2 ) (4-46) r( 2) (z) = 41r a(z) V2(z) sin (C(2 - ao) (4-47) To determine the performance of the propeller, the circulation distribution must be known ex-plicitly. That is, both V2(z) and 4r 2 (z) must be determined in Equation (4-47). To find these terms we use Equations (4-46) and (4-10); Ui(ua(1) sin ýPB-U/1) cos 'OB) -i2uaUa) cos SOB + u/l) s OB) =oI V2 cos a2 + o2 V2 sin a2 Hence V22 = Ua(1)2 + u 1)2 (4-48) and 72 Ua(1) cos ýoB + uP) sin (oB ua(1) sin pB -iut(1) cos ýoB Thus the second-order solution is complete. In the next section, formulas for the propeller performance are given. C-PERFORMANCE Determination of the circulation from the given profile coordinates shows that the singular-perturbation problem gives a quadrature for the performance. However, the appropriate form of the inte-gral for the thrust and torque has not been determined yet. The expression for thrust is given by Equation (2-57) T= - Pz sB[2qo 0 V b) . Vol ii" nds In lifting-line theory the expression for thrust is usually obtained from the outer flow. Here we derive the appropriate expression from the inner flow since, in the preceding equation, the inner potential is the appropriate one to use. Now i n ds = i E2 T- iei) dA dz = (T cos ýPB +-sin dPB do z go = VU2 + E22 z2 [cos (oB- f] + sin (oB -)e2] + R = Vl(z)[cos a, el + sin aý e2 + F 92jT_ VO = V4) (a], 02, z; 7 + L2 + •(2D _ 42 2 +H.O.T.)+O(j)k = El auo1+e2 ( l+ see Chapter 1 for the gradient in inner variables. Thus to the first-order in F, the integrand (2 0 + V) VO involves only two-dimensional quantities, i.e., z is a parameter. By inspection, the integrand can be changed to 73 (02D+ V24) since 1qo 2D12 is a constant with respect to the chordwise integration and, hence, integrates to give zero. In two-dimensional wing theory, one defines 2 S(qo 2D + V2 cV 2 do, 2 FN (4-50) T- (q-o + 72 V Edal --• 2D where FN and F. are the forces normal to and along the chord. If one defines U + Eua(1) + •45" [Ji =tan-1(4-Si) S- ({ ut(l) + oo) then FN = L cos (SOB - Oi Fe=Lsin(ýOB- Oi where L is the lift of the blade section. Hence the thrust coefficient becomes T 2Z R CT= = ir 2 R2 p J EL [cos SoB cos (BoB -) PU2,R2 IUR f 2 + sin ýoB sin (ýoB - 0d] dz 2 Z oR r • R 2 PL cos idz (4-52) The lift is given by cU+•2a1 • .)2) L p 1(U+ uj1 )+ ) + (2z -) + 2 F(z; E) (4-53) Hence 74 fo R CT= (2()+Z ,2 +o.)(2 gu()+••.)d and by retaining terms to only second-order, one has CT 2R U2 E 1( + .g2 r(2)) Tz F2 r(I) ut(1)) dz (4-54) By a similar series of steps, one finds the power coefficient S 2 (M • ) C-cP P U 3 R2 2 2 Z 2 ((Ez ( '(1E ) + -2 r(2 )) U + E2 r(1) ua(1)) dz (4-55) (J3 R2 75 CHAPTER 5 DISCUSSION A-COMMENTS ON RESULTS The lifting-line analysis is essentially the same as that given in the literature, except for notation. One apparent difference is that the two-dimensional sections are defined on sections cut by a plane per-pendicular to the straight blade-reference line. However, a Taylor expansion of the shape defined on a cy-linder shows that to second-order they are the same since the difference z - cois 0(•2J. Of course higher-order terms would show the effect of the new section definition. The lifting-line analysis was developed for propellers with a straight lifting line. If the blade-reference line were raked or warped, the outer-flow potential could be easily found. The overall solution, however, would not be as given in Chapter 4, because the second-order problem would include solution of Poisson's rather than Laplace's equation. The appropriate solution for such cases could probably be ob-tained, but the lifting-line analysis is appropriate only for propellers with narrow blades and hence is not useful for marine propellers. For marine propellers, the most important analysis is the lifting-surface formulation, and the rest of this discussion will concern this problem. In the preceding chapters, the expressions for velocity were left in the form of the gradient operating on an integral. This is because they are principal-value integrals, and, if the differential would be taken inside some of the integrals (Equation (2-19)), they would not exist in the mathematical sense; see Tychonov and Samarski.4 5 The singularities which result when the operator is taken inside the integral are usually called Hadamard singularities, and as shown in Equation (2-19), the finite part of the integral is defined in terms of limit operations. However, if the expression for q is integrated by parts, the order of the singularity will be reduced for the limit -+ s, and the integrals will exist as Cauchy principal values. (For a principal-value integral the region surroundingr 0 must be excluded from the integral, and the limit with the maximum dimension of the excluded region going to zero must be taken; for a Cauchy principal-value integral, restrictions are placed on the shape of the region excludingz-o). In evaluating the expressions numerically, either formulation of the problem is acceptable since both expressions involve a region surrounding the singularity which requires special consideration. In fact, Sparenberg,25 Pien and Strom-Tejsen,31 and Tsakonas and Jacobs41 treat the higher-order singularity, while Pien, 3 0 Kerwin, 2 7 '2 8 Yamazki, 2 0 and Murray 3 3 work with principal-value integrals. Sparenberg 2 5 presents an analysis for both points of view. B-EXPRESSIONS FOR VELOCITY AS INTEGRALS OF FIRST-ORDER SINGULARITIES Equation (3-21) for the perturbation potential consists of three terms 0= Os + OT+ tkv (5-1) 76 In this equation Os is an integral over the surface of the blades corresponding to the lifting aspect of the propeller s = IJJ S- dl d (5-2) and kT is also an integral over the surface of the blades; however, due to the blade thickness Z-1 Tz--s01 b=O dil da (E3 SB+ Sd d 0 If-11 at,(5-3) and Ov is an integral over the shed vortex sheet OV=- f= O r(zo) dzo fN+3 da (54) 4 7r bO TE -,013 aTE where rB (c1(ay), Z) = F(•) Ns is given in Equation (3-10) and AO' ,is given in Equation (3-18) In the following work, the expressions will be considered without the effects of rake or warp since they are less complicated and permit comparison with results in the literature. The formulation with rake and warp can be obtained similarly. In design, the detailed geometry of the propeller is to be determined. Since this is so, the position of the trailing edge can be chosen to simplify the integrals. With neither rake nor warp the blade-27r reference lines are the 0 b values (0 b =T b). If the trailing edge of the blade is chosen to lie on the helical surface passing through this line, then in Equation (3-17) U •o: zo£ •(o- Ob) OTEh 0.< O ) (5-5) 77 U i.e., xTE =- 0 TE" Then Equation (3-18) reduces to Equation (4-4), etc. The expression for C reduces to 1 Z-1 R -0u a' -,2 Ifzo) dzo _ ZO--x + -ui 4 rb=O f ax 2z 0 o r aTE(zO) (5-6) where r is given in Equation (4-6). From the results of Appendix B, the velocities computed from this expression can be seen to equal those derived from the Biot-Savart Law for a bound vortex along the blade trailing edge and a shed vortex sheet trailing downstream from it. The position vector of the trailing edge is f = XTE(Z) i + ae.a(OTE(Lo) + 0 b) (5-7) Hence a vector tangent to the trailing edge is df dcXj TEdeb)+,(d dO TE t = -= • iEea)TE 0b + -d ddO OTE-+ Ob db =XTEi+ ea(OTE+ Ob)+CZeo TE (5-8) Thus the velocity is 1 Z-1 f R tx(r-X) Z Jo (z°) dz° 4 =0 0 r=-[ 3 1 Z-1 fR n d dz +- J dz ,d /d-y (5-9) 4 'r b=O dz0 0 OTE where I is given in Equation (B-10) of Appendix B. In the preceding integral, z0 is a dummy variable for CY. As shown in Appendix B this form is the same as that found in the literature, and hence the detailed explicit form used for the numerical analysis is not repeated here. When the pitch of the blade-reference surface and the shed vortex sheet are the same, Pien 3 6 found that significant savings in computing time could be obtained by rearranging the limits of the infinite integral so that it would need to be calculated only once. Naturally this left an integral over a finite 78 region to be evaluated for each new position on the blade. For points everywhere on the blade except the trailing edge, the preceding expression is a straightforward quadrature without 'singularities in the integrand. When Pien's procedure is used, Cauchy principal-value integrals occur, but apparently these are easier to handle than the repeated infinite integral. The preceding expression for the induced velocity is usually modified to include the moderately-loaded assumption discussed in Chapter 4 which consists of replacing the ý0 expression by the approximate one for So + 'El' given in Equation (4-3 1), which is derived from lifting-line theory. Although an argu-ment can be made for this approximation in lifting-line analysis, no justification is given for its use in lifting-surface theory. In any event, only a portion of the second-order effects are considered with this modification; see Section D of Chapter 3. The expression for the potential due to blade thickness can be written SZ-1 R V-c~ j E 1 ý IbO U2 + Z2 + 22 cos (dOp-9 1 ) dw " d~l (5-10) Cj2.) Lr-ioI where _r- sO1 = (x- l sin Pp) i+ y + ' sin Ob+ + S(O + -J Cos Ob + k_ Hence qT= V T is given by C aET Z Z-lfoRb~ U2 7os:0:2p-22 f~ai '-° !T fR + i•o s 2 d f -d~l (5-11) b0 0 c2 Lr-so 13 For points on the blade r -- s0, only a principal-value integral must be evaluated. Since this expression is the same as that found in the,. literature, it is convenient to make the necessary changes to demonstrate the equivalence. The changes are that the blade surface is described by Wo, 0 instead of t1, W, and that pitch P(a) instead of pitch angle •,p (a') is used. This is the description used by Kerwin. 2 7,2 8 In addition, the term cos (ýp -0 3) is ignored, and constant -pitch is assumed since the blade normal is taken to be £2, For constant pitch and neither rake nor warp, the only component of velocity in the boundary condition of Equation (3-27) is v=vipoe2 (5-12) 79 For points on the blade described by (r, 4S/, •P(r) x = r tan p Op(r) 0 =-y - r sin ( z r cos q5 the component v can be written R OTE v(r, f 0 ( 21 )2+ 4C2 d fO o(, '0) GTdO (5-13) 0 OLE where V((E27U ()2+ a2 3F-T i(zý, 0)= U -- 1(5-14) P(Za)127r Rf , , 0 and P(a•) O-P(r) oi. P( a) •i b+0-P( Z-1 27r -2ir sin (b+O GT = 2"7r ¢P ) 2 •p(b=O 02 2 312 (Ec+ a)2 + r2 + •2 -2 wr cos + 0 -(5-15) Except for notation, this agrees with Kerwin's expression. Kerwin uses 2ET for his thickness dis-tribution, so there is also a difference factor of 2. When the pitch is constant, GT is a function of( 0 -The remaining term to be examined is the integral over the lifting surface. The velocity from this component of • is given by 80 s= Os Z-1 +os'(r-so) I if S B (o) Vr 6( ) ds (5-16) b 0SB =ZIJff rB I) ___ .- I () 4- b-oJr sB+rrs°) -Soir -("-so)J ds When r - so+ 0, the singularity in this expression is second-order rather than first-order as in principal-value integrals; see Equation (2-19). To integrate by parts, the gradient of the kernel in Equation (5-16) must be expanded and rearranged S rX s 13•} dLr-1 -\r ,.-s3} But for an arbitrary vector A, nX 7 XA) X V) X A- (L7 " E) -n "A) Now, since V -- s =0 =A1 0 the preceding expression becomes a LL -11 --i 3 In (1 -s) lr_-s3 jr-s1 To actually perform the integration by parts, we determine that L1r-s r-s (nXVs)X r _ rB(nX Vs)X -+(nXVs PB)X_ (n VJX L4_s3}= Ir-sl3 Ir-sl3 Hence the velocity is 81 Is = Y if (•V( -xv7 rB) " _% 41r b=0 SB + r__s0la (5-17) S V r )I}ds +(ns X s) x FBrs3 One of Stokes' theorems is48 f ( X V)XA ds= cdX A where c is the curve bounding s, and di is along the curve in the usual orientation. Thus the expression for velocity is -1n+ X{ v( X 0 ds I ~ ~~~ OS sr) r11 I-s 4:rb= fo S++•I-so 0s Now rB is zero at the blade leading edge; and r(Z5) , at the trailing edge. The arc element along the blade outline is dl dl =- dZ' =tdW Hence qs -I(a X V -i) ds 47 W ffSB+0 OS sr) f1 f J (zO )tX dzo (5-18) 82 The line integral cancels a similar contribution from the trailing vortex sheet in Equation (5-9), and the sur-face integral is the Biot-Savart Law with a vorticity distribution: A n x vs IB EnX(q+ - q-) Since the conversion is general, it would also be possible to express velocities from the nonlinear results of Chapter 2 in terms of a vorticity distribution. To interpret the results in such a manner, it will be necessary to define an inner surface of the blade which has zero flow velocity. Then the vorticity will be simply n X V¢. Since no further insight is provided with this approach, it is not pursued further. The explicit expressions for the linearized velocity in the case of the general propeller geometry can be found in a straightforward manner. Since the results will eventually be compared with equations found in the literature, the geometrical restrictions appropriate for those cases will be made. These are that the propeller is neither raked nor warped and that the pitch is constant. For constant pitch N+ n 2 and for neither rake nor warp / 2× Hence A aRB arB tlCos2 Pp a FB Now since FR-- / -r(,7,)d,7 ac 2(W,) one finds a.••~ tjlCOS2 ýppi A= "},t- 17dr+ 'Y., e (5-19) (a W 2, Co8 83 This is the form of the vorticity vector required in the coordinate systems developed here. However, in the literature it is more common to work with cylindrical polar coordinates (ZY, 0, x). In this system the parametric representation of the helical blade-reference surface is given by P x =atan SOp 0 -y =- "•sin 0 (5-20) z = a'cos 0 ) For the surface coordinate system u1, u2 = , 0, the metric coefficients are4 3 -1 ~0 gii= (P 22 (5.21) Hence for constant pitch, the gradient of rB in this coordinate system is n X r Fe x B+ el arB Now since 0 = B f (47JI 2ý+ ýa 2 2yd 0 B 2J,, one finds /0 _ - 2 2+ a2 yd'O f flXVrB'VE2X ^tl+ C( 0 LE ( )2 2Y d) ] n X PB =2 X am OLE and hence 111A\J 0LE 'a2ydO e (5-22) This is the same expression obtained by Kerwin, 5 1 although he used the result in a different manner. 84 Explicit formulas for calculating the velocity from the Biot-Savart Law are given in the next section and hence are not repeated here. C-COMPARISON WITH OTHER FORMULATIONS In the past, linearized propeller lifting-surface formulations have been developed from three view-points-a postulated vorticity distribution (Pien,30 Kerwin,26,27 and Murray33), an acceleration potential (Sparenberg, 2 5 Hanaoka,3 2 Pien and Strom-Tejsen 3 1), or a velocity potential (Yamazaki 18" 2 4). Both velocity potential and acceleration potential formulations are based on the equations of motion, while the vortex distribution formulation is based on laws derived from the governing equations. Of course, all so-lutions should be equivalent so the development is only a matter of personal preference. The details of the formulation of the velocity-potential function are not given by Yamazaki. How-ever, examination of the potential reveals it to be compatible with the velocity potential derived from the acceleration potential. Such solutions are given in terms of the pressure difference across the blade (for constant pitch propellers, this is equivalent to the radial component of bound vorticity; see the footnote following Equation (3-24)) and an infinite integral to a variable point on the blade. In the acceleration potential method formulated by Sparenberg, the linearization is on the basis of only the axial component ofqo; other velocity components are assumed small relative to it. However, since the propeller also has a rotational velocity, it is not clear such an assumption is justified. Hanaoka linearizes about the total free-stream speed and converts his expression for the linearized acceleration po-tential to one for the velocity potential. For steady flow, his Equation (3-8) is the same as the lifting portion of Equation (3-21), when the pitch of the blade-reference surface and the shed vortex wake are identical. The formulation on the basis of the vorticity distribution is now almost classical. The vorticity vector is specified on the body and shed vortex sheet, and the velocity distribution can be calculated from the Biot-Savart Law. For a surface distribution of vorticity, the Biot-Savart equation gives the induced velocity as i r- A x(r-s) S JJS dS (5-23) qi47r so lt_s13 where A is the vorticity vector, and So is the approximate blade surfaces plus the shed vortex sheets. Across the shed vortex sheet + q- + 5a 85 and the normal to this sheet is, (from Equation (3-18)) N -n -- -cos3 i + sin I e 0 ) cos•- cos hence the vorticity vector is aF A= n X(q+ - q-) -(sin i_ + cos of 0) where tan J3=U Since s = 10, the integration over the shed vortex surface is 1 Z-1 R dF d 00 5v -Z -bO (zO) dI dy (5-24) 47b=O 0 f 0 TE where I is given by Equation (B-10) in Appendix B; this is the form given previously in Section B of this Chapter. Thus this component of velocity is the same as that given by Pien, 3 9 Morgan and Wrench,14 and others, except that this free-vorticity starts at the variable trailing edge and lies on the surface parallel to the undisturbed free-stream velocity. The bound velocity on the blade is now needed. Since most of the previous investigators have derived equations appropriate for only constant pitch, we do also. This means that the zero-order normal from Equation (3-10) is n = £2 and q+ - q- yl£ + oe-,+ #£2 where o is unknown, and y is given by Equation (3-24). Hence the bound vorticity vector is A= Yey+ ueI (5-25) The explicit form of a will be found presently but now the velocity induced by this bound-vorticity vector is desired: 1q"• z-l tR bO• cl (uel +'ye9flxr'-s 0°)d X I dR(t -fý)IX(-O b-0 0 c2 C 2 d13 86 where so is given in Equation (3-9), which for constant pitch and with neither rake nor warp be-COIliCS s, S sin +oi +e-where e -sin ob+ . +cos Ob + k Now for r xi + re/4, f, X (1-1o) =(sin ýp Pi + Cos ýP o ) X I{(x-ýi sin ýpPi+ rer - w, = sin pOp (r~e - ae 9) + cos pOp { (x-t1 sin Opp) ef' + (a- r cos ( 0 -¢)i} eaX (f-S-o) = eaX (xlsin °p)i + rer -w-e- . t, sin Op =-( sin p s . - rsin ( +. Co Hence: 1 i1R a.C•C coslsp( 'p rCosO- )-r rsinO-•) -f "a f 4dl-0 b=O 0 C2 Z-1 R Cl asin p sin (0 - 0)+ cos p (x -t, sin• p )Cos (0 --47 b=O fc2 itsO13 (5-26) 87 ",5iy (0-0) (x- 1 sin ýp )d •~ -do ]' L .Z-1 f R d -ClJ a [sin ýp (r -W-cos (0 0)+ cos ýp 1 (x-t 1 sin ýpr) sin (0 -0) q 41r~ JY.,S IL_1-b=O - -o 2 (5-26) -y (x- sin Op ) cos (0 - 0) Cont'd Scos where 0 o 0b+ r_13= { (x - t, sin ýp2+ r2+ D2 -2 r a5cos ( O- 0)} 3/2 In the literature it is common to work with 0 = rather than ýI" Since for "W= constant 0 cos pOp the preceding velocity components become I Z-1 R ad a7 OTdM Vacos Op (W-rcos(Ob+O- 0 ))-'rsin(Ob+ 0-) =- J-. J dO 7r b=O f Cos ip .r_.Soi3 (5-27) 88 -4r b=CO s OTE oLET a {N °(sinp asin 0 b + 0 -+ cos Opp (x - Jtan pp 0) sin 0 b + 0- .()) - sinOb + 0 - 0) (x -tan Op 0) + "dO Ir-so13 1 Z- 1 •R adZ q-"•¢=4-r b--° f 0 os ýOp 0 TE f.fOL a{sin ýOp(r- Wcos(Ob + 0- ))+cossOp(x - wtan op 0)sin' 0 b + -0 7y Cos O'b + 0 - 0) (x - a tan •o•0) cs +- d 0 (5-27) ] Cont'd where I-So13 =(x --tan ýOp 0)2 + r2 + a2 - 2 ir Cos ( Ob + 'O 312 In the literature it is also common to break this surface integral up into two components q=ql+q 2 (5-28) where 1 Zilrc elXir:-S'o '") SB I " 0 13's (5-29) 147 8 b=O SB 89, I e• f zX (1-10) .9 2 Tr- I~J 1 (5-30) -b -O SB,+ I-r-.13 and combine ql with the integral over the shed vortex sheet. When this is done it is customary to take the pitch angle of the shed vortex sheet and the blade-reference surface as the fOi angle computed from lifting-line theory; this corresponds to the hydrodynamic pitch introduced in Equation (4-30). When ql andqv are combined, one has the form: I z-i ~ f -Xo-Y J d5 J hdO (5-31) b=O 0 'OLE where A= ae s so OLE < 0 OTE h= -cos (3 Because of the form of this equation it is also customary in the literature to speak of both ql and qv as being induced by a trailing vortex system. In fact some authors derive their equations, postulating that the trailing vortex system starts on the blade. Because of this confusion regarding the a component of bound vorticity, it is not surprising that the literature contains several expressions for it. In order to determine a, one uses the property that ,a distribution of vorticity must satisfy the equation v. A= 0 90 For a constant-pitch helical sheet, with the vorticity vector given by the component y in thc radial direction and the component o directed in the surface perpendicular to the radial direction and tangent to a cylinder, one can determine the metric coefficients, Equation (5-21), and reciprocal unitary vectors4 3 and thus find the equation for divergence on the blade IA= (+p 2 2 ()+ (5-32) /()2 +Z•2 af 2r where P/2ir = c tan ýpp and on the shed vortex sheet VoA= cosf3 0 0 The expression in the wake is automatically satisfied since F r(a only. Now one must find the form of o on the blade. From the first expression, one obtains 0 o(a 0)=-27 OLE where f(i) is the integration "constant." The value of this constant of integration is obtained by requiring that the vorticity vector not run off the leading edge of the blade into the irrotational fluid. That is, if t is the tangent vector to the leading edge tXAI =0 0== OLE Since t=-d . . d OLE dOLE \LII dd one finds that Sd OLE f =Y "rOLE, a) cos •p d-or that 91 -y. do (5-33) af yCos 0 LE aco This is the expression obtained by Kerwin. 2 7 Pien 3 0 and Murray 3 3 state that: 0LE where OTE r'(ay) f r a(a, 0) d 0 OLE Thus the metric coefficient is absorbed in the definition of r(a) which must be w~( Oh' ly(a,0) Cos where y (W, 0) is the radial component of vorticity. Pien and Murray do calculate the velocities with an expression like that for :+ 4v in Equation (5-31). Kerwin 2 7 first integrates the value of au/a•on the blade from 0 to -o, then integrates that value over the blade surface. In a later paper, Kerwin and Leopold, 2 9 change the formulas to the form of Equation (5-3 1). D-EXAMINATION OF RADIAL COMPONENT IN BOUNDARY CONDITION Although the formal development of the lifting-surface and lifting-line equations is important academically, it does not in itself contribute to improved propeller designs. However, the lifting-surface formulation does explicitly consider the radial velocity component in the boundary condition used to derive the camber distribution in Equation (3-27). Since this term might contribute significantly to the engineering aspects of propeller design, some idea of its effect on the computed camberline must be found. In order to examine the effect of this term, a rather crude approximation will be made for the radial 92 velocity. The only radial velocity components found in the literature are those calculated by Ilough and Ordway.57 Their results are for circumferentially averaged velocity components due to a lifting-line with a circulation distribution' they call approximately optimum; for the radial component they do not actually compute values at the lifting line. Nevertheless, we use these values because no others are available. Comparison will be made with the maximum camber ratios calculated by Cheng 5 8 from the formu-lation of Pien.3 0 Comparisons of the averaged axial velocities for the optimum circulation distribution are reasonably close to those computed for the arbitrary circulation distribution actually used in the calcu-lations by Cheng. If we assume the radial velocity component is approximately constant and approximately given by the values given by Hough and Ordway at the point closest to the lifting line, then the camber due to the new term is parabolic with amplitude for symmetrical blades (C1-C(2 sin 29yp dmpp\ wv max=-_TRI + 2 U (5-34) 1 (cl-c 2 dp cos2 pp woya3) -- I- R -R 4 \2R / d 21ra U In the calculation method proposed by Pien, 3 0 Op is Pi given in Equation (4-51). Cheng tabulates these values as well as the chord-to-diameter ratio. The following table lists the maximum camber computed by Cheng for his propeller 3916A and the increment computed by using the preceding equation, both for an 8-inch-diameter propeller 5 7Hough, G.R. and D.E. Ordway, "The Generalized Actuator Diik," Developments in Theoretical and Applied Mechanics, Pergamon Press, Inc., New York, Vol. II, pp. 317-3.36 (1965). 5 8Cheng, H.M., "Hydrodynamic Aspect of Propeller Design Based on Lifting-Surface Theory-Part I-"Uniform Chordwise Load Distribution," NSRDC Report 1802 (1964). 93 Cheng, 5 8 AzEcm R Ecimax, Figs. 7 and 16 cmax 0.3 0.088 in. 0.0002 in. 0.7 0.034 in. -0.002 in. The change in sign in AEclmax reflects a change in sign of velocity component along the radius. For this propeller, the change of camber is negligible for the "'/R = 0.3 radius. This is because of the small slope of the pitch curve in this region. For the 0.7 radius, the change in camber is about 6 percent of the contribution from the conventional calculations and would be significant for propellers of larger diameter. Additional information about the importance of the radial component can be surmised by con-sidering the effect of rake or warp. The effect of warp is especially important for improved cavitation per-formance and reduced alternating thrust when operating in a wake, Miller 59 and Boswell. 6 0 To date, in-vestigations have been conducted with skewed propellers for which the blade-reference line lies in the helical surface which passes through a straight-line element in the x = 0 plane.60 For such pro-pellers the normal to the reference surface NO would contribute a radial term approximately like that described for the camber effect; however, since ý1 is measured from x = 0, an angle-of-attack term would also occur. Such propellers have blades extending considerably downstream. This overhang requires that the propeller hub and the rudder be separated more than normal. If instead of skewed propellers, only warped propellers were utilized, this problem would not occur. Hence, the effect of warp on the design will be considered now. Skew is usually given as a warp angle with the understanding that the blade-reference line lies in the helical reference surface through a 0b line. The warp angles tabulated by Boswell60 are approximately: w{~ -J.~UB)( -CDHUB) where 0s is the designated amount of "skew." 5 9Miller, M.L., 'Experimental Determination of Unsteady Propeller Forces," Seventh Symposium on Naval Hydrodynamics, Rome (1968). 6 0Boswell, R., "Design, Cavitation Performance, and Open-Water Performance of a Series of Research Skewed Propellers," NSRDC Report 3339 (1971). 94 That portion of the boundary condition of Equation (3-27) to be examined is alixc _dW W(•l, 0, ) S- Sil (5-35) If tile preceding expression for warp is used, and if the radial component is again approximated by w(a, then warp contributes an angle of attack (or pitch correction) to the camberline shape. The tangent to this pitch angle is given by the preceding expression dW -tan {Aat) = ZZý sin •Op U+ da P a2 n2 For Os =21r/Z. - tan (Aa) = F wiw) SR \ Zl /VU2 + n2 a2 'HUB 1-R For lack of other data, we approximate the ýOp values at the warped line by the 0i values for a straight blade-reference line. By applying these approximations to the five-bladed propeller considered previously, the pitch-angle corrections would be -Act in Degrees R 0.3 -0.2 0.7 0.5 These corrections are considered significant. Because of the change in sign, the preceding corrections might tend to cancel in their effect on thrust. However, they could be important for cavitation performance which is one of the main reasons for designing a propeller rather than selecting one from a series. 95 E-DESIGN PROCEDURES (HYDRODYNAMIC) In current design techniques by Morgan, Silovic, and Denny61 and Kerwin and Leopold, 2 9 a combi-nation of lifting-surface and lifting-line results is used. Briefly the mechanics of this procedure are to use the second-order, lifting-line equations for the outer flow to find the induced velocities at the blade-reference line, circulation distribution and thrust. The inner flow is ignored, and hence an integral equation must be solved. This is usually done by assuming either a tan 0i or circulation distribution and scaling the chosen quantity to eventually give the required thrust (or power). The resulting circulation distribution is used to scale the chosen radial component of vorticity y; see Equation (3-31). The lifting-line induced velocities are assumed to be approximately the average chordwise results from a lifting-surface calculation. They are used to approximate the velocities induced by the shed vortex wake and are used to give the reference surface lp/w. Hence everything is known to compute the lifting-surface results, partially taking into account second-order effects. This procedure has worked remarkably well for the performance of con-ventional propellers, Cox.3 7 On the other hand, it seems to be difficult to design a bad propeller since even the earlier, semi-empirical design procedures such as Hill's62 produced propellers which developed the desired thrust. It is in other areas, especially cavitation inception, that the more refined analysis embodied in the lengthy lifting-surface calculations are expected to have a distinct advantage. Unfortunately, few ex-perimental comparisons of the type necessary to judge the adequacy of the various hydrodynamic theories are found in the literature. However, when cavitation occurs first at the leading edge for the design condition, the design procedure has not been sufficient to produce the specified chordwise pressure distri-bution. Such is the case with a recent series of skewed propellers.60 Some differences in the numerical results of various lifting-surface formulations are given by Johnson. 6 3 Although the combination of lifting-line and lifting-surface results is quite practical, it is not entirely consistent. Specifically, the expansion for the circulation distribution in the lifting-line analysis can hardly be expected to apply for F-+ 1. However, for elliptic wings, Van Dyke 2' 3 has shown that the lift computed from the integral equation is quite close to the exact value over the entire range of aspect ratios. Thus the lifting-line results might be adequate for parametric studies since calculations can be quickly done. For final design, though, the lifting-surface equations should be used since they will give a geometry better able to meet design conditions. Another utilization of the lifting-line equations is to approximate ýpp(a) and the position vector of the shed vortex sheet. The determination of the shed vortex sheet for the lifting-surface analysis by using 6 1Morgan, W.B. et al., "Propeller Lifting-Surface Corrections," Society of Naval Architects and Marine Engineers Transactions, Vol. 76, pp. 309-347 (1965). 6 2Hill, J.G., "The Design of Propellers," Society of Naval Architects and Marine Engineers Transactions, pp. 143-192 (1949). 6 3Johnson, C.A., "Comparison of Propeller Design Techniques," Fourth Symposium on Naval Hydro-dynamics (1962). 96 (he lifting-line induced velocities is probably not accurate near the lifting surface where it is most important; in any event, the current utilization is only an approximate second-order correction for a portion of the second-order terms. Other second-order terms should be calculated to determine their magnitude also. To use the consistent first-order lifting surface calculations, the following procedure is recommended. In the first-order theory the approximate position of the shed vortex sheet is independent of the induced velocities and is known. The blade-section pitch angle ýp p is first approximated by the 0 angle. The mag-nitude of the circulation distribution needed to produce the thrust can be calculated from Equation (3-23). The propeller geometry can now be calculated. The angle of attack of the section can be used to define a new ýpp, and the calculations can be repeated. The second-order results could also be calculated if necessary. In particular the second-order thrust in Equation (3-29) could be calculated from the first-order data so that an adjustment in the circulation could be made to meet the given thrust or some other criteria used to iterate for the final solution. Obviously, these alternative procedures would be more involved than the presently-used techniques but unless an entirely consistent approach to the design problem is taken one does not know the degree of approximation involved in the present calculations or what potential exists for design based on the lifting-surface formulation. In the equations for design the chordwise component of velocity difference (or the spanwise com-ponent of vorticity) "7 is assumed to be given as well as the total bound circulation curve F(a'). This is compatible with information which a designer should supply for a propeller designed from cavitation con-siderations: Two general criteria for the avoidance of cavitation are that no suction peaks occur on the blade and that the shed vorticity be distributed as uniformly as possible in the spanwise direction. The first conditions met by selecting the functiona form of -y; the second, by the functional form of f(Z), since the gradient of r gives the shed vorticity. The number of blades and the blade-area ratio are also part of these considerations since the average pressure times the blade area determines the load which must be sufficient to produce the required thrust. F-RECOMMENDATIONS Obviously the first recommendation is to include the radial velocity component in the boundary condition for propellers designed with variable geometric pitch, warp or rake. Although the inclusion of this term is not expected to have a significant effect on the thrust, it could lead to propellers with better cavitation performance. Since cavitation considerations are one of the main reasons for designing a pro-peller rather than selecting one from a series, and since a crude analysis shows the term to have a significant effect, detailed numerical calculations should be undertaken. Since heavily-loaded propellers are becoming more common in practice, the second-order lifting-surface formulas should also be programmed for design calculations. One of the interesting aspects of such investigations would be determination of the position vector of the lines of constant circulation in the shed vortex sheet. Another interesting outcome of these calculations would be an evaluation of the magnitude of 97 the second-order terms. Presently, it is assumed that the shed vortex sheet needs to be described better than by ý0. Qualitatively this is saying that the infinite integral of 0 (e) over the shed vortex sheet produces a zero-order contribution. The proposed calculations would evaluate this assumption. The formal analysis described here should also be extended to include vorticity, i.e., an axisymmetric free-stream velocity, unsteady flow, and supercavitating propellers. Since unsteady flow with vorticity is the standard operating condition of propellers behind a ship, these two are important areas of research. SUMMARY AND CONCLUSIONS The mathematical development presented in the preceding chapters has been based on the equations of motion and certain mathematical relations which have permitted an exact formulation to be constructed. Application of formal perturbation procedures has yielded linearized solutions appropriate for numerical calculations. In Chapter 5, the velocities computed from the lifting-surface potential are shown to equal those computed by the Biot-Savart Law. These expressions are then shown to equal those found in the literature for the approximate geometrical conditions. An examination of the literature did not reveal previous explicit consideration of the radial velocity component in the first-order lifting-surface perturbation problem, Equation (3-27). Of course, other investigators realized that such a term occurred for variable pitch propellers but did not examine the consequences of neglecting it for arbitrary pitch distribution. For the example considered, the contribution of this term was as much as 6 percent of the camber ratio determined by the conventional analysis. In the lifting-line analysis, only propellers without rake and warp were considered. For these propellers, the two-dimensional sections were defined normal to the straight blade-reference line. However, to the order of the approximation considered, they could equally well have been considered as defined on cylinders as was done in the lifting-surface problem. A method of design based only on the lifting-surface formulation has been discussed. This pro-cedure makes no use of the lifting-line results and would involve greater computing effort since an iteration is appropriate. Extension of the method utilized in this work to the problem of unsteady flow and shear flow is recommended. In addition, numerical results should be obtained. The numerical results could include wake trajectories for the shed vortex sheet. 98 APPENDIX A POTENTIAL FOR LIFTING LINE DERIVED FROM THE BIOT-SAVART LAW Traditional propeller lifting-line theory (Moriya,'Il Lerbs,l 3 Morgan and Wrench, 14 ) evaluates the velocity induced by the vortex system from the law of Biot-Savart. The vortex system considered is one for which an individual vortex is an element of circulation of strength dI"/dashed at radius ('; then by the conservation of vorticity this element remains bound in the line and leaves it at the axis of rotation. It is not clear from the literature how to interpret the vortex system when the lifting line extends from a finite radius off the axis to the propeller tip. (However, see Appendix B.) Lifting-line theory for planar wings, Ashley and Landall, 4 7 uses a model which consists of an elemental horseshoe vortex of strength r. This is composed of two free vortices of equal but opposite strengh a distance Az apart, joined together at the blade by a bound vortex element of the same strength. Such a system is shown in Figure 8 for a propeller lifting line. Az Figure 8 - Vortex Element 99 A sum of these elements distributed along the radius of the lifting line then approximates the vortex system. In Ashley and Landahl, 4 7 the potential for elements an infinitesimal distance apart is derived by integrating the potential of a doublet over the length of the element. Here it will be derived from the Biot-Savart Law applied to the element. The path of integration is shown in Figure 9. / Figure 9 - Path of Integration Let this path be L. The velocity induced by the element is L df X I (Al) L-One of Stokes' theorems for vector integration is d Xd X ff(EXV)Xuds (A2) See also Lamb 64 and Kochin et al.4 4 6 4Lamb, H., "Hydrodynamics," Sixth Edition, Dover Publications, Inc., New York (1964). 100 where s is an area bounded by the curve c. Thus r-- t r--s (V'X7X ds (A3) fL Ifr-113 fs Ifr-sl3 where s is the area bounded by L. But r-s r-sr-(anX 7s)X Vrl t S V • •n -n(A4) Hence, since V (r/r?) = 0, there remains (nX Vs)X• -Vr --(A5) Ir_ -s l3 I r -sl3 since n = n(s) only. Thus the induced velocity is _{-• r 'r n)ds (A6) = -- r • n ds Hence, by definition, the potential is r r-s nds (A7) See Lamb64 Equation 4, page 212. The shed vortex sheet is envisioned as being composed of a radial distribution of these elements so that the distance between elements goes to zero and the width of an element goes to zero. In this limit the vector element of area becomes the negative of that in Equation (2-48): /ý q Sds ---dz0 dct alO az 101 and the potential for the total system becomes z R 0LI -= f ,(zo) dzo ( -da (A8) b- 0 ob ir-ýi 3 aT Z When the zero-order position vector for the trailing vortex system is substituted into Equation (A8), Equation (4-5) is obtained. As an application of Equation (A7), the potential of a circulation element for planar wings can be derived. For wings, the approximate location of the lines of constant circulation in the shed vortex sheet is on the x, y-plane at constant y values. Hence n ds - k dxAy + O(Ay 2 ) and rAy j z dxo + O(Ay 2) 41 r fp' (~o2+(o 213/2 r AY) Z~ X Z + 2 (A9) 4r(yy 0)2 + z2 L I2+'2 (,,oý+ 2] which is Equation (5-35) in Ashley and Landahl. 4 7 This can be rearranged to give pAy z (x 2 + (Y-Yo 2) + z2)1/2 + x 4r (x2 + (y-yo)2 + z2) - x2 (x2 + (y-yo)2 + z2)112 + O(Ay 2) pAy z1 47r ix 2 + (yyo) 2 + z2] 1/2 11 [x2 + (y-y 0 + 2 + 12 + O(,Ay 2) (AlO) 102 which is the form derived by von Karman and Burgers (p. 128),65 by Truckenbrodt, 6 6 and by Friedrichs (Equation 15.4).1 Returning to the propeller problem, one obtains Equation (4-5) by substituting the zero-order position vector of curves of constant circulation from Equation (4-3) into Equation (A8). Since only the kernel of Equation (4-5) is of interest, it can be considered alone: Z-1 -° [-•- (-'-a •- 11 G(r,zo) = J, f _ o da (Al1) b=O 0 Ir-f13 and the zero-order approximation to G is G(fz°) -- -° + o' (4-6) r=Vx- U-)2 + a2 +zo2- 2zO acos( 0 _ -6 b) A similar integral has been evaluated in the determination of induced velocities at a lifting line. For the induced velocities at the lifting line, Moriya1 1 was able to integrate the expression analytically to obtain an infinite series of products of modified Bessel functions. Although the first impression is that replacing an infinite integral by an infinite series is not much of an improvement, Lerbs1 3 showed that the series could be approximately summed and thus calculations could be quickly made. An excellent explanation of the integration process for the induced velocities at a lifting line and the numerical approximation to the series is given by Morgan and Wrench.1 4 Included in that article is an improved approximation for the series summation which was developed by Wrench. The rather circuitous method discovered by Moriya will be applied to the integral in Equation (4-6). Unfortunately, the kernel cannot be evaluated at an arbitrary point; however, the form far downstream is found, and it is demonstrated that velocities from this expression are twice those at the lifting line as derived from the Biot-Savart Law, as would be expected. The expression for G(r,zo) is given by Equation (4-6). From the Lipschitz integral, Watson, p.384, 6 7 one obtains 6 5von Karman, T. and J.M. Burgers, "General Aerodynamic Theory-Perfect Fluids," Vol. I'I of Aerodyraric Theory, edited by W.F. Durand, Springer; also reprinted by Dover Publications, Inc. (1963). 6 6Truckenbrodt, E., "Das Geschwindigkeitspotential der tragenden Fliche bei inkompressibler Strldmung," Zeitschrift fur angewandte Mathematik und Machanik, Berlin, Vol. 33, pp. 165-173 (1953). 6 7Watson, G.N., "Theory of Bessel Functions," Cambridge University, England (1944). 103 J= f -Ix 9 OtVW- 0 z -- eJdt z - 2z 0 acos (0 -0b) dt (A12) r 0 From one of the addition theorems, Watson,6 7 p. 128, Jo(t Y 2 + z02 - 2z0 o acos (-0 -- 0 b) 00 = I k, Jm(ta) Jm(tzo) cos (0 - a) (A13) m=0 where kmi is Neumann's factor, (km = 0, in = 0; k = 2, rn > 0). If one assumes uniform convergence for the integral, the order of integration can be switched and one obtains G(r, z0) = ktlk J7(tcw) J7(tzo)Wm(j'r,z o,t) dt (A14) nz=O 0 where zm- + Uf e- ) d7 (A15) b=W 0 Two cases occur, one for x < 0 and one for x > 0. For x < 0 the positive sign is used. For x > 0, two regions exist, one for -y <- and one when it is greater. Only the case for x > 0 will be considered since U it permits the form of the potential far downstream to be obtained. Thus for x > 0 m ,Zot) U a - xt e O-•x+ t• zo o+ 0) e f 2os m(O0 ' Ob) d7 b=Oo U 0 + Xt 00 e -"- cos m (0 -Y O b) dyf (A16) U 104 The integrals appearing in Equation (A16) can be found by integrating by parts until the same expression occurs again and then by rearranging the equation. The resulting expression for9Zm is ?m az0,t) -2z0O cosm(O- 0 b) (b)2 ft+m2 L 0 /ZO 2m Urn E_] 2(m -EUt + ---+ sin m( 0 -0 -2 +-0 U 2 sin O--0 U( 0 (A 17) Thus ) Z-1 Ut/ E2 (- -z 0 t •cosm (- 0b Gfzo= f km Jm(t Jm(tzO) eu zt ) m=O b=O 0 2 Z 2sinUm )xt (A18) + + sin(0O- Ob) - 2sinn m 0-U- 0 b) I(t ý-U-U The portion of the integral involving the exponential disappears when the form far downstream is examined. The remaining portion of the integral can be integrated using a result on page 429 of Watson. 6 7 The resulting expression far downstream is G2E2r'Zo) --- XU I ZO Ekm sin m 0 --I A (A19) b=O m=0 where Am is a combination of modified Bessel Functions 105 Im ~ n) Km z >" A- ~ " ((z U~~' Km z0 <ý (A20) The expression for • far downstream is R f-- r(zo)Go,(r,zo)dz 0 (A21) 0 which can be integrated by parts to obtain I f Rd F fo 47 0o dP lZo After some manipulation, one finds R Z-1 f z 0 -2 j2 km.sinfm ---- 0b)AmdzO (A22) 0 0 O m=O where ~2J~m)~ (om) The expression for the axial velocity,"'-, at 0 - 0, x - 0 can be easily obtained. After summing over b, the expression is SZ•/ "R d n Zzgz 0 z pzuf mF__ u'z m -d- k Ak mZ (A24) 0o dz 0 m--0• Tis is twice the value of the axial velocity at the lifting line as determined from the Biot-Savart Law, as expected. 106 An expression for the portion of the potential involving the exponential integral can be obtained by expressing the Bessel Functions as infinite series. The integral then involves a triple infinite summation of terms with an integral now of the form f "o e-at 0 t2+q2 This expression can be integrated; see Gradshteyn and Ryzhik p. 313, 3.356(1) and (2)68 in terms of sine and cosine integrals plus another finite summation. The resulting expression is quite cumbersome and is therefore not given. No use was made of it in the analysis. 6 8Gradshteyn, I.S. and I.M. Ryzhik, "Tables of Integrals Series and Products," (Translation of the Fourth Edition, Edited by A. Jeffrey), Academic Press, Inc., New York (1965). 107 APPENDIX B INTERMEDIATE EXPANSION OF OUTER POTENTIAL Equation (4-32) states the second-order potential for the outer flow. This expression is valid for a fixed point in the flow field as the parameter F, equals chord-to-diameter ratio, goes to zero. Thus the outer flow consists of a singularity system distributed along the radial lines 0 b, 0 < a< R. There are only two essential systems of singularities: one from the circulation distribution and one from the dipole distribution. The second-order contribution to the circulation will be included in the single integral involving the circulation. The two integrals will be 01 and 02' where 01 is given by Equation (4-5), and 02 is given by Equation (4-33). The 01 term will be considered first. Since considerable difficulty was encountered in an attempt to expand the potential function for the lifting-line, the velocity expression was expanded and from it the expansion for the potential was con-structed. In Appendix A, the velocity from the Biot-Savart Law was shown to be derivable from a potential. Now it is convenient to return to the Biot-Savart Law in order to find the intermediate expansion. How-ever, some rearrangement is in order before the appropriate form for expansion is obtained. The vortex system is envisioned as composed of a distribution of elements shown in Figure 8 along each of the lifting lines. The width between the free trailing portions of a vortex element is constant for all elements, and the elements are infinitesimally close together. The path ?onsists of the segment along the 0 b lines and the two free-trailing paths. Two adjacent trailing elements then have a common path of integration. Hence the total induced velocity for N of these elements is Z-1 N n r - zoe_(0b) q= -- - I ýOb)X d zo b=O f If_ - zo (O b13 b~=0 = Zn 1ze(ObI N p n -r1n+1 0o n 2" 47rj dl X n (B1) n=O Ir-0n where F0=O N+1 =0 In is the common path between the adjacent n and n+1 trailing vortices. In the limit as AZ - 0, and the distribution of circulation becomes continuous with F(0) =(R) 0, the expression for the velocity becomes 109 Z-1 ROb 1 • 7 f F(zo) r - zoe"Z•( 1 Xe"(b ~ 47 b- Ir - zo E_ 0 (O h)d z b=O 0(12 (B2) Z-1 R d ___ r- d 1 fZ dzo --Xd 4rdz 0 f- 3 h=O 0 Hence the first-order velocity distribution results when F I )F(1 (zO) + and 0 + " are sub-stituted into this equation. Since U (83) o=Zoe 0)O+-( 0 bb) one finds d e- U dO +-dO S(zo C 0 +-' i) dO and since •0 =oI -2 +- Ob + +z 0 2 ý5z 0 cos( 0 - a) one finds 0)X d 0 -- (y + zo sin a) - zo os a (x - -( Ob)) + - (z -0ocos a) +z sin a(x -- (a - 00) L {Zo.Cos(O~c)~zo2}.ida (B3) Hence 110 SZ11 fR r -wsin(0 -Ob)L-xcrxos Ob--xsi' Obk S b=O 0 o x 2 + (y + Zo sin Obý + (z - z0 cos Ob) 3/2 d O (13(B4) Z-1 R [00o-X-[ d C + I df o dr f 47b 0b [(x0--( 2- + 2 + z02- 2 z0 cos (0- a)/312 Consider first the expression from the bound vorticity. To find the intermediate expansion one substitutes the intermediate variables into the expression and rearranges it into an asymptotic sequence. By inspection of the denominator, one can see that a Taylor expansion is appropriate, except for the b = 0 lifting line. Thus R= R" xgj+gyi q-B(gx5'g-Yiz)-=--{J F(O-g f2+2+Z Z 1/dz° YB 9 93; z)= Tr7f r(zo -[g 2 (X2 +;2) -1 (z _ z oj 31 2 z Z-1 R z sin 0b d zo 7-- .f r(zo) 231 b=1 [(z0 s Ob)2 + (z -cos Ob)2131 2 + 0 (Fg) (B5) By symmetry the second term cancels, i.e., sin 01 = -snz 1, etc. The first integral can be expanded by a binomial series in each region bordered by (z - Zo) = g2 (52 + 52). The result is given by Ogilvie, Equation (2-43)54 9B(gc,g.YZ) =- + 0(Fg) (B6) 27rg !Z2 +2 Temporarily we convert to outer variables •r(z) y i - xj 9B (x, y, z) = + 0(F) (B7) and note that V tan- ly ,-i-xI hence 111 B (x, tany-y + O(F) (B8) 27r x From which one can conclude that, to within a function of z, the expansion of 0 contains a term: , , -I 2 xgyz) tan- -+ O(g ) (B9) 27r due to the bound vortex. Now the second part of the expression, the velocity due to the trailing vortices, must be determined. Letting y = a -0b, one finds Z-I R q_(, ,z)=-Y• f d ZO d 7(B 10) !T(X, Y,4z) X4dz 0 !dy B b=O 0 0 (z 0 Ocos ( 0 - 'y- Ob) - z0 2)i U U + / -•(z - zo cos + Ob) + zo sil (Y + Ob) (x - -Y)/ _ U U -f-• (y + z 0 sil (+ Ob) + zO cos (Y + 0 b) (x --y)/ k I= /(~ U 2 ~y+ Sfl(7 o))2 + (z -o cos (+ 0)) 2I-3 1 2 [(x- + +zo sill('Y + 0b 2+z-oOS Since the denominator is of positive definite form, it has a zero only for x = = 0b = Y 0, z zo. As shown by Moriya11 and Morgan and Wrench,14 the velocity vector, as x = g • and y = g 3 go to zero, is found to be + (1) (u) (1) q T(gx, gy, z)=ýua '+ cut i+•z k HOT B where Z-1 R 00 (1) f dr f Hi(z,zo,' )dy) dy "ui (z) dzo U 4r 0 0 2) 2 + z2 + z 02 - 2 z z0 cos ('y + 0 b312 112 with azzO, Y) ZO COS (Y + 0b)) ZOt U /it (z, zO, 7}-/z - z()cos (7 + 01)) -YzO Sill (7Y + 0/,)/ U zo0 Hz (z, zO, )=- [sin (7 + 0b)-- y cos (7 + 0/)) z0 and where the bar through the integral means the singular point is excluded. Except for notation, these are identical with the expressions given in Equations (11) through (13) of Morgan and Wrench, 14 Equation (7) of MoriyaI I and the equations in Appendix 2 of Lerbs.1 3 As shown by these authors, when U/l2 is con-stant, the infinite integral can be performed analytically for the axial and tangential component of velocity. From the previously described form of the velocity vector in the intermediate region, one con-cludes that the expansion for the potential due to the trailing vortices has the terms z g +j u )+• (z dz (1312) Hence the expansion for the potential due to the vortex system has the following form " (9 3F 9 g 7' Y z) = OB + 0 T -• (z) -1 =F(z;E)- Z tan I 27r -x -(1) -(1), + g(xua +y uY.t )+H.O.T. (B13) The function F(z;F) contains other unknown integration constants which are functions of z in addition to the integral of U(z). The construction of the potential expansion from the velocity components is not required since the matching could be done with the velocity components but it is simpler to work with the one potential expansion rather than the i and / velocity components. The expression for 02 valid in the intermediate region is now to be found. For the expansion of this expression, it is not necessary to examine velocity components, since a direct expansion of the potential is possible. Substitution of intermediate variables in the expression for 02 gives Z-1 R[g p (zO)+(g=+zosill b) t2z0'dz0 , z)=-- I (B 14) bW 0 [(gx) +(g9T+zosin V +(-os 113 For b : 0, the expansion is regular, and a Taylor expansion is possible. The term independent of g in the expansion is Z-1 R zO Si" 0 b P2(z0) dz 0 b=1 [(zO sin Ob) + (z -zOCos Obpi Because of the symmetry of the lifting lines, the summation of this term is zero. The term corresponding to b=O has the following expansion; see Ogilvie,54 Equation (2.43) R I ] dzo I i(Zz)+TP2(z) 47r [gxl2j 2 2 1 __32_= 0 [(g ) (g + (z - Zo) g + H.O.T. (B15) Hence -2- -Jz + Pz + H.0. T. (B16) C2 0•2 (g x, g Y, z) = 27g •2+• 21rg = .= Combining this with the expansion of the vortex terms, one finds that the expansion of the potential in the intermediate region has the form: F(z/ = ~(gg3,zF 0 Z;E-E tan-1Y 0 (g xT' g yý' z) = Fo (z"'j) --tan- x (1) -(I), + H.O.T. 21rg =2 +=2 This is the expression stated in Equation (4-34). 114 REFERENCES 1. Friedrichs, K.O., "Special Topics in Fluid Dynamics," New York University (1953); also published by Gordon and Breach, New York (1966). 2. Van Dyke, M., "Lifting Line Theory as a Singular Perturbation Problem," Archiwum Mechaniki Stosowanej, Vol. 3, No. 16, pp. 601-614 (1964). 3. Van Dyke, M., "Perturbation Methods in Fluid Mechanics," Academic Press, Inc., New York (1964). 4. Kaplun, S., "Fluid Mechanics and Singular Perturbations," Edited by P.A. Lagerstrom, et al., Academic Press, Inc., New York (1967). 5. Cole, J.D., "Perturbation Methods in Applied Mathematics," Blaisdell (1968). 6. Prandtl, L., "Application of Modern Hydrodynamics to Aeronautics," National Advisory Committee for Aeronautics Report 116 (1921). 7. Thurber, J.K., "An Asymptotic Method for Determining the Lift Distribution of a Swept-Back Wing of Finite Span," Communications on Pure and Applied Mathematics, Vol. 18, pp. 733-756 (1965). 8. Rotta, N.R., "The Non-Planar, Moderate Aspect Ratio, Subsonic Wind," Ph.D. Thesis, New York University, University Microfilms 69-4576 (1968). 9. Hille, R., "Bestimmung der Dickenlinie von PropellerflUigelprofilen bei vorgegebener Druck-verteilung," Report 262 of the Institut fifr Schiffbau der Universitiit Hamburg, West Germany (1970). 10. Goldstein, S., "On the Vortex Theory of Screw Propellers," Procedure Royal Society of London, Series A, Vol. 123, pp. 440-465 (1929). 11. Moriya, T., "On the Integration of Biot-Savart's Law in Propeller Theory," (in Japanese), Journal of the Society for Aeronautical Science, Japan, Vol. 9, No. 89, of 1015-1020 (1942); English translation in Selected Scientific and Technical Papers by Tomijiro Moriya, Moriya Memorial Committee, University of Tokyo, Japan, pp. 74-80 (1959). 12. Kawada, S., "Induced Velocity by Helical Vortices," Journal of Aeronautical Sciences, Vol. 3, pp. 86-87 (1936). 13. Lerbs, H.W., "Moderately Loaded Propellers with a Finite Number of Blades and an Arbitrary Distribution of Circulation," Society of Naval Architects and Marine Engineers Transactions, Vol. 60, pp. 73-123 (1952). 14. Morgan, W.B. and J.W. Wrench, "Some Computational Aspects of Propeller Design," Methods of Computational Physicis, No. 4, pp. 301-331, Academic Press, Inc., New York (1965). 15. Ludwieg, H. and I. Ginzel, "Zur Theorie der Breitblattschraube," Aerodynamische Versuchsanstalt, Gottingen, Report 44/A/08 (1944); see Ginzel, G.I., "Theory of the Broad-Bladed Propeller," Aeronautical Research Council, Current Papers 208 (1955). 16. Strscheletzky, M., "Hydrodynamische Grundlagen zur Berechnung der Schiffsschrauben," Verlag G. Braun, Karlsruhe, West Germany (1950). 115 17. Strscheletzky, M., "Berechnungskurwen ftir dreiflugelige Schiffsschrauben," Verlag G. Braun, Karlsruhe, West Germany (1955). 18. Yamazaki, R., "A Study on Screw Propellers," Memoirs of the Faculty of Engineering, Kyushu University, Japan, Vol. 19, No. 1, pp. 1-75 (1960). 19. Yamazaki, R., "On the Theory of Screw Propellers," Fourth Symposium on Naval Hydro-dynamics (1962). 20. Yamazaki, R., "On the Theory of Screw Propellers," Memoirs of the Faculty of Engineering, Kyushu University, Japan, Vol. 23, No. 2, pp. 97-112 (1963); corrected version of 1962 paper. 21. Yamazaki, R., "On the Theory of Screw Propellers in Non-Uniform Flows," Memoirs of the Faculty of Engineering, Kyushu University, Japan, Vol. 25, No. 2, pp. 107-174. 22. Yamazaki, R., "On the Propulsion Theory of Ships in Still Water (Introduction)," Memoirs of the Faculty of Engineering, Kyushu University, Japan, Vol. 27, No. 4, pp. 187-220 (1968). 23. Yamazaki, R., "Theory of Unsteady Propeller Forces," Seventh Symposium on Naval Hydrodynamics, Rome (1968). 24. Yamazaki, R., "On the Theory of Unsteady Propeller Forces," Memoirs of the Faculty of Engineering, Kyushu University, Japan, Vol. 28, Vol. 3, pp. 157-206 (1969). 25. Sparenberg, J.A., "Application of Lifting Surface Theory to Ship Screws," International Ship-building Progress, Vol. 7, No. 67, pp. 99-106 (1960). 26. Kerwin, J.E., "The Solution of Propeller Lifting Surface Problems by Vortex Lattice Methods," Massachusetts Institute of Technology, Naval Architecture Department Report (June 1961). 27. Kerwin, J.E., "Linearized Theory for Propellers in Steady Flow," Massachusetts Institute of Technology, Naval Architecture Department Report (1963). 28. Kerwin, J.E. and R. Leopold, "Propeller Incidence Correction Due to Blade Thickness," Journal of Ship Research, Vol. 7, No. 2, pp. 1-6 (1963). 29. Kerwin, J.E. and R. Leopold, "A Design Theory for Subcavitating Propellers," Society of Naval Architects and Marine Engineers Transactions, Vol. 72, pp. 294-335 (1964). 30. Pien, P.C., "The Calculation of Marine Propellers Based on Lifting Surface Theory," Journal of Ship Research, Vol. 5, No. 2, pp. 1-14 (1961). 31. Pien, P.C. and J. Strom-Tejsen, "A General Theory for Marine Propellers," Seventh Symposium on Naval Hydrodynamics, Rome (1968). 32. Hanaoka, T., "Hydrodynamics of an Oscillating Screw Propeller," Fourth Symposium on Naval Hydrodynamics (1962). 33. Murray, M.T., "Propeller Design and Analysis by Lifting Surface Theory," International Ship-building Progress, Vol. 14, No. 160, pp. 433-451 (1967). 34. Wu, T.Y., "Some Recent Developments in Propeller Theory," Schiffstechnik, Vol. 12, No. 60, pp. 1-11 (1965). 116 35. Isay, W.H., "Propellertheorie, Hydrodynamische Probleme," Springer-Verlag, Berlin (1964). 36. Isay, W.H., "Moderne Probleme der Propellertheorie, Springer-Verlag, Berlin (1970). 37. Cox, G.G., "State-of-the-Art for Subcavitating Propeller Design Methods," Appendix II of the Report of the Propeller Committee, 12th International Towing Tank Conference, Rome (1969). 38. Kerwin, J.E., "Machine Computation of Marine Propeller Characteristics," International Ship-building Progress, Vol. 6, No. 60, pp. 343-354 (1959). 39. Isay, W.H. and R. Armonat, "Zur Berechnung der Potentialtheoretischen Druckverteilung am Fliigelblatt eines Propellers," Schiffstechnik, Vol. 13, No. 67, pp. 75-89 (1966). 40. Sugai, K., "Hydrodynamics of Screw Propellers Based on a New Lifting Surface Theory," Selected Papers from the Joumal of the Society of Naval Architects of Japan, Vol. 4, Tokyo, pp. 96-106 (1970). 41. Tsakonas, S. and W.R. Jacobs, "Propeller Loading Distributions," Journal of Ship Research, Vol. 13, No. 4, pp. 237-257 (1969). 42. Armonat, R., "Untersuchung der Druckvertailung eines Propellers unter Bericksichtigung grenzschichthedingter Massstabseffekte," Schiffstechnik, Vol. 16, No. 81, pp. 41-54 (1969). 43. Wills, A.P., "Vector Analysis with an Introduction to Tensor Analysis," Dover Publications, Inc., New York (1958). 44. Kochin, N.E. et al., "Theoretical Hydromechanics," (Translation of Fifth Russian Edition), Interscience Publications, Inc., New York (1964). 45. Tychonov, A.N. and A.A. Samarski, "Partial Differential Equations of Mathematical Physics," Vol. 1, Holden-Day, Inc., San Francisco, Calif. (1964). 46. Mangler, K.W., "Improper Integrals in Theoretical Aerodynamics," Aeronautical Research Council, Current Papers 94 (1952). 47. Ashley, H. and M. Landahl, "Aerodynamics of Wings and Bodies," Addison-Wesley Publishing Co., Inc., Reading, Mass. (1965). 48. Milne-Thomson, L.M., "Theoretical Hydrodynamics," Fifth Edition, The Macmillan Company, New York (1968). 49. Ciolkowski, S.I., "The Swept-Back Wing," Ph.D. Thesis, New York University, University Microfilms 24,694 (1955). 50. Cummings, D., "Vortex Interactions in a Propeller Wake," Massachusetts Instutute of Technology, Naval Architecture Department .Report 68-12 (Jun 1968). 51. Glauert, H., "Airplane Propellers," Division L, Aerodynamic Theory, Edited by W.F. Durand, J. Springer, 169-360 (1935); also published by Dover Publications, Inc., New York (1963). 52. Germain, P., "Recent Evolution in Problems and Methods in Aerodynamics," Journal of the Royal Aeronautical Society, Vol. 71, No. 682, pp. 673-691 (1967). 117 53. Thwaites, B., "Incompressible Aerodynamics," Oxford University Press, England (1962). 54. Ogilvie, T.F., "Singular Perturbation Problems in Ship Hydrodynamics," University of Michigan, Department of Naval Architecture, Report 096 (Oct 1970). 55. Erdelyi, A., "Asymptotic Expansions," Dover Publications, Inc., New York (1956). 56. Erickson, J.C. et al., "A Theory for VTOL Propeller Operation in a Static Condition," Curtiss-Wright Corporation, Caldwell, N.J. (Oct 1965). 57. Hough, G.R. and D.E. Ordway, "The Generalized Actuator Disk," Developments in Theoretical and Applied Mechanics, Pergamon Press, Inc., New York, Vol. 11, pp. 317-336 (1965). 58. Cheng, H.M., "Hydrodynamic Aspect of Propeller Design Based on Lifting-Surface Theory-Part I-Uniform Chordwise Load Distribution," NSRDC Report 1802 (1964). 59. Miller, M.L., "Experimental Determination of Unsteady Propeller Forces," Seventh Symposium on Naval Hydrodynamics, Rome (1968). 60. Boswell, R., "Design, Cavitation Performance, and Open-Water Performance of a Series of Research Skewed Propellers," NSRDC Report 3339 (1971). 61. Morgan, W.B. et al., "Propeller Lifting Surface Corrections," Society of Naval Architects and Marine Engineers Transactions, Vol. 76, pp. 309-347 (1968). 62. Hill, J.C., "The Design of Propellers," Society of Naval Architects and Marine Engineers Transactions, pp. 143-192 (1949). 63. Johnsson, C.A., "Comparison of Propeller Design Techniques," Fourth Symposium on Naval Hydrodynamics (1962). 64. Lamb, H., "Hydrodynamics," Sixth Edition, Dover Publications, Inc., New York (1964). 65. von Karman, T. and J.M. Burgers, "General Aerodynamic Theory-Perfect Fluids," Vol. II of Aerodynamic Theory, Edited by W.F. Durand, Springer; also reprinted by Dover Publications, Inc (1963). 66. Truckenbrodt, E., "Das Geschwindigkeitspotential der tragenden Flache bei inkompressibler Str6mung," Zeitschrift ftir angewandte Mathematik und Mechanik, Berlin, Vol. 33, pp. 165-173 (1953). 67. Watson, G.N., "Theory of Bessel Functions," Cambridge, Mass. (1944). 68. Gradshteyn, I.S. and I.M. Ryzhik, "Tables of Integrals Series and Products," (Translation of Fourth Edition, Edited by A. Jeffrey), Academic Press, Inc., New York (1965). 118 INITIAL DISTRIBUTION Copies Copies 2 CHONR 1 NAVSHIPYD Bremerton 1 Fluid Dynamics (Code 438) 1 NAVSHIPYD Charleston 1 Sys & Res Gp (Code 492) 1 NAVSHIPYD Long Beach 1 NRL 1 NAVSHIPYD Philadelphia 1 ONR Boston 1 NAVSHIPYD Portsmouth 1 ONR Chicago 8 NAVSEC 1 ONR Pasadena 1 SEC 6100 1 ONR San Francisco 2 SEC 6110 1 SEC 6140 2 SEC 6144 1 USNPGSCOL, Monterey 2 SEC 6148 1 USNROTC 1 NAVSEC Norfolk (Code 6660) 1 NAVWARCOL 12 DDC 20 NAVSHIPS 1 Chief of R&D, OCS, Army 1 SHIPS 033 1 Army Eng R&D, Ft. Belvoir 2 SHIPS 03412B 1 SHIPS 037 1 Army Trans R&D, Ft. 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GROUP S IRF-'ORT TITLE PROPELLER PERTURBATION PROBLEMS 4 DlESCRIPlyVE NOTES (7ype ol report and inclusive dates) S AU THOR(SI (First name, middle initial, last iame) Terry F'. Brockett 6 REPORT DATE 7A. TOTAL NO. OF PAGES 7b. NO. OF REFS October 1972 128 68 SB. CONTRACT OR GRANT NO 9a. ORIGINATOR-S REPORT NUMSERIS) b. PROJECT NO. 3880 . Sbr t SO 00,Tk0 9. OTHER REPORT NOMS) (Any other numbera that may be aSSigned Subpro•jet SROZ3OIOI, Task 00103 this report) d. Work Unit 1-1544-263 10. DISTRIBUTION STATEMENT APPROVED FOR PUBLIC RELEASE: DISTRIBUTION UNLIMITED. 11. SUPPLEMENTARY NOTES 112. SPONSORING MILITARY ACTIVITY NSRDCd- GHR Program Bethesda, Md. 20034 13 ABSTRACT For steady motion of a propeller operating in an inviscid fluid having an unbounded irrotational flow field, an expression for the velocity potential (in excess of the body motion) is derived in terms of the boundary values. From this expression perturbation solutions are determined-one for small thickness-or camber-to-chord ratio and one for small chord-to-diameter ratio. The first problem (lifting-surface theory) is a regular-perturbation problem, and the second (lifting-line theory) is a singular-perturbation problem which requires construction of matched asymptotic expansions. Two terms of each series are found. Numerical techniques are not discussed. The outer solution for the lifting line is the same as that published in the literature. The formal lifting-surface differs from other developments in several ways. The most important of these is that for propellers with variable pitch, warp, and rake, the normal to the blade has a radial component which requires con-sideration of the radial velocity in determining the blade shape. For the case considered the sign of the contributions in the inner radii differ from the outer radii values; thus, this additional term might cause little effect on thrust but could be important for cavitation performance. A design procedure is discussed which involves only quantities appropriate for the lifting-surface analysis. 147M (PAGE I1) DD NOV J473 UNCLASSIFIED S/N 0101-807.6801 Security Classification UNCLASSIFIED Security Classification 14 LINK A LINK B LINK C KEY WORDS_____ ROLE WT ROLE WT ROLE WT Propellers Liflting-surfaicc theory Lifting-line theory Regular-perturbation problem Singular-pert urbation problem Subcavitating propeller Propeller design Fully wetted flow DD ID ov..1473 (BACK UNCLASSIFIED (P AGE 2) Security Classification
16573	https://www.math.stonybrook.edu/~rdhough/putnam_seminar/2020/invariants.pdf	INVARIANTS AND SEMI-INVARIANTS, ALGEBRAIC IDENTITIES ROBERT HOUGH In invariant of a system is a quantity which doesn’t change as the system evolves, like the total energy or momentum of a classical system. A semi-invariant can change as the system evolves, but only increases or decreases. Consider the following example. Example 1. Suppose n markers are given in a row, each with one white side and one black side. A move consists of selecting a white marker, not one of the outermost ones, removing it from the row, and flipping the color of its two neighbors. Prove that it is possible to reach a configuration with only two remaining markers if and only if n ´ 1 is not divisible by 3. Proof. The number of black markers is always even. Hence if two markers remain, either both are white or both are black. For each white marker, let t be the number of black markers to its left, and assign the marker the value p´1qt. Let S be the sum of the values of the white markers. The value of S modulo 3 is an invariant, as can be checked by checking the 4 possible arrangments of the adjacent markers (black, white on either side). Since the initial configuration has S of value n mod 3, and the final configuration has S ” 0 or 2 mod 3 the claim is necessary. To prove sufficiency, check that the solution is possible if there are 3 or 5 white markers. If there are n ą 5 white markers initially, successively take away the left-most available white marker 3 times. This leaves n ´ 3 white markers. □ Example 2. Given a triple of numbers, select two of them a, b and replace them with ab ? 2 , a´b ? 2 . Is it possible to change p1, ? 2, 1 ? 2q to p2, ? 2, 1 ? 2q? Proof. The sum of the squares of the numbers is an invariant. This proves that the change is impossible. □ Example 3. Let x, y, z be distinct real numbers. Prove that px ´ yq 1 3 py ´ zq 1 3 pz ´ xq 1 3 ‰ 0. Proof. Suppose the contrary. Put a “ px ´ yq 1 3, b “ py ´ zq 1 3, c “ pz ´ xq 1 3. From the identity a3 b3 c3 ´ 3abc “ pa b cqpa2 b2 c2 ´ ab ´ bc ´ caq, it follows that a3 b3 c3 “ 3abc, so abc “ 0, contradiction. □ Problem 1. Starting with an ordered quadruple of positive integers, a generalized Eu-clidean algorithm is applied successively as follows: if the numbers are x, y, u, v and x ą y then the quadruple is replaced by x ´ y, y, u v, v. Otherwise, it is replaced by x, y ´ x, u, v u. The algorithm stops when the numbers in the first pair become equal. Assume that we start with m, n, m, n. Prove that when the algorithm ends the arithmetic mean of the last two numbers is the least common multiple of m and n. Problem 2. Four congruent right triangles are given. One can cut one of them along the altitude, and repeat the operation several times with the newly obtained triangles. Prove that, no matter how the cuts are performed, we can always find among the triangles a pair that are congruent. 1 2 ROBERT HOUGH Problem 3. Starting with an ordered quadruple of integers, perform repeatedly the oper-ation pa, b, c, dq ÞÑ p\|a ´ b\|, \|b ´ c\|, \|c ´ d\|, \|d ´ a\|q. Prove that after finitely many steps, the quadruple becomes p0, 0, 0, 0q. Problem 4. Show that for no positive integer n are both n 3 and n2 3n 3 perfect cubes. Problem 5. Prove that any polynomial which takes only non-negative values can be writ-ten as the sum of the squares of two polynomials. Problem 6. Prove that for any odd integer n ě 5, ˆn 0 ˙ 5n´1 ´ ˆn 1 ˙ 5n´2 ˆn 2 ˙ 5n´2 ´ ¨ ¨ ¨ ˆ n n ´ 1 ˙ is not prime. Problem 7. Prove that, for infinitely many natural numbers a, n4a is not prime for any natural number n. Problem 8. Factor 51985 ´ 1 into a product of 3 integers, each of which is greater than 5100. Problem 9. Solve px ´ 1q 1 3 x 1 3 px 1q 1 3 “ 0. Problem 10. Suppose that n is the sum of two triangular numbers, n “ a2 a 2 b2 b 2 . Show that 4n 1 is the sum of two squares. Conversely, show that if 4n 1 is the sum of two squares, n is the sum of two triangular numbers. Problem 11. a. If a and b are consecutive integers, show that a2 b2 pabq2 is a perfect square. b. If a “ bc bc then a2 b2 c2 is the square of a rational number. c. If N differs from the two consecutive squares between which it lies by x and y, respectively, prove that N ´ xy is a square. Problem 12. Let there be given nine lattice points in three-dimensional Euclidean space. Show that there is a lattice point on the interior of one of the line segments joining two of these points. Problem 13. Remove the lower left corner square and the upper right corner square from an 8 by 8 chess board. Can the board be covered by 2 by 1 dominoes? Problem 14. a. Exploit symmetry to expand the product px2y y2z z2xqpxy2 yz2 zx2q. b. If x y z “ 0, prove that ˆx2 y2 z2 2 ˙ ˆx5 y5 z5 5 ˙ “ x7 y7 z7 7 . Problem 15. Verify that the product of four consecutive terms of an arithmetic progression plus the fourth power of the common difference is always a perfect square. INVARIANTS AND SEMI-INVARIANTS, ALGEBRAIC IDENTITIES 3 Problem 16. Given a sequence of integers x1, x2, ..., xn whose sum is 1, prove that exactly one of the cyclic shifts x1, x2, ..., xn; x2, x3, ..., xn, x1; ...; xn, x1, ..., xn´1 has all of its partial sums positive. Problem 17. Show that if a1, ..., an are non-negative numbers, then p1 a1qp1 a2q ¨ ¨ ¨ p1 anq ě p1 pa1...anq 1 nqn. Problem 18. Let x1, x2, ..., xk be real numbers such that the set A “ tcospnπx1q cospnπx2q ¨ ¨ ¨ cospnπxkq\|n ě 1u is finite. Prove that the xi are all rational. Problem 19. The positive integers are colored by two colors. Prove that there exists an infinite sequence of positive integers k1 ă k2 ă k3 ă ... with the property that 2k1 ă k1 k2 ă 2k2 ă k2 k3 ă ... all have the same color.
16574	https://math.stackexchange.com/questions/3152137/finding-the-total-probability	Finding the total probability - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Finding the total probability Ask Question Asked 6 years, 6 months ago Modified6 years, 6 months ago Viewed 60 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I know mostly how to solve probability problems, but I am stumped by the ones where it asks “What is the chance that a red marble is picked AT LEAST 1 of the 2 times?” Here’s an example, can somebody please tell me how to solve it? There is a jar of 42 marbles, 5 green and the rest blue. You pick from the jar twice, replacing the marble you picked both times. What is the chance that you pick a green marble at least once? Thanks! probability Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Mar 17, 2019 at 21:57 Isabel RobinsonIsabel Robinson 1 2 1 One minus the probability that you pick a blue marble twice.saulspatz –saulspatz 2019-03-17 22:07:21 +00:00 Commented Mar 17, 2019 at 22:07 Let X∼X∼ BINOM(2,5/52)(2,5/52) be the number of green marbles drawn in 2 draws. In this problem you can use the complement rule, so that P(X≥1)=1−P(X=0)=1−(47/52)2.P(X≥1)=1−P(X=0)=1−(47/52)2.BruceET –BruceET 2019-03-17 22:11:22 +00:00 Commented Mar 17, 2019 at 22:11 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Exactly one of the following things can occur: First pick B G B G Second pick B B G G Total number of times a green was selected 0×1✓1✓2✓First pick Second pick Total number of times a green was selected B B 0×G B 1✓B G 1✓G G 2✓ Getting at least one green corresponds to any of the events above except for the one where you selected to blues. As for approaching the problem... you may either find the probability of each outcome and add them up for those where a green was selected at least once, or what most people would do instead is find the probability that a green was not selected at least once and subtract it away from 1 1 since it is known that the probabilities should have added up to 1 1. As for actually calculating the probabilities, I will show you one of the cases and let you figure out how to repeat the process for the rest. I will show you how to calculate the probability that the first ball is green and the second is blue. The probability that the first ball selected is green is 5 42 5 42. This should be clear already... since there are five green marbles in a bag of 42 42 marbles and we selected one at random (which implies that each marble was equally likely to have been selected). Once that has happened, we then take another ball and see if it is blue. That would happen with probability 37 42 37 42. The probability that these both happen one after the other is the product of their probabilities, so the probability that the first ball is green and the second ball is blue is 5 42×37 42 5 42×37 42. Now, in this problem we are told that we replace the ball, meaning that after we pulled it out and looked at it, we put it back. In other similar problems, the ball might not have been replaced in which case depending on the outcome of the first draw, the second draw will have a different number of each ball available and a different number of total balls remaining, so take that into consideration when finding the probabilities to multiply by. Now, to continue, again, either recognize that in your problem you have "At least one green" corresponds to "green then blue" or "blue then green" or "green then green" and add the probabilities of these together. Alternatively, you find the probability that there was not at least one green and subtract this away from 1 1. They will both give the same answer. In larger, more challenging problems, you have your choice on whether you approach them directly or indirectly or a different way alltogether. Usually, you should pick the one which requires the least arithmetic or requires the least confusing argument, but that is subjective. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 17, 2019 at 22:16 JMoravitzJMoravitz 81k 5 5 gold badges 72 72 silver badges 124 124 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions probability See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 1Probability question: marbles in a jar 3Probability with changing number of marbles 0The probability of select marbles from two jars? 0conditional probability with vs. without replacement independence? 0How to find the total number of marbles? 0What is the probability that you pick 3 green, no red, 1 of each, at least 1 blue marble if a jar contains 5 blue, 3 green, and 4 red marbles. 0What is the probability of drawing at least one green and one blue marble in five draws with replacement in a bag with 100 marbles? 1Jar game and probability Hot Network Questions Discussing strategy reduces winning chances of everyone! 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16575	https://digitalcommons.uri.edu/cgi/viewcontent.cgi?article=1020&context=phy204-lecturenotes	Published Time: Mon, 01 Jul 2024 20:35:25 GMT University of Rhode Island University of Rhode Island DigitalCommons@URI DigitalCommons@URI PHY 204: Elementary Physics II -- Lecture Notes PHY 204: Elementary Physics II (2021) 10-30-2020 21. Torque acting on current loops. Magnetic dipole moment 21. Torque acting on current loops. Magnetic dipole moment Gerhard Müller University of Rhode Island , gmuller@uri.edu Robert Coyne University of Rhode Island , robcoyne@uri.edu Follow this and additional works at: Recommended Citation Recommended Citation Müller, Gerhard and Coyne, Robert, "21. Torque acting on current loops. Magnetic dipole moment" (2020). PHY 204: Elementary Physics II -- Lecture Notes. Paper 21. This Course Material is brought to you by the University of Rhode Island. It has been accepted for inclusion in PHY 204: Elementary Physics II -- Lecture Notes by an authorized administrator of DigitalCommons@URI. For more information, please contact digitalcommons-group@uri.edu . For permission to reuse copyrighted content, contact the author directly. PHY204 Lecture 21 [rln21] Dot Product Between Vectors Consider two vectors ~A=Axˆi+Ayˆj+Azˆkand ~B=Bxˆi+Byˆj+Bzˆk. •~A·~B=AB cos φ=AB A=BA B. •~A·~B=~B·~A. •~A·~B=AB if ~A‖~B. •~A·~B=0if ~A⊥~B. •~A·~B= ( Axˆi+Ayˆj+Azˆk)·(Bxˆi+Byˆj+Bzˆk)=AxBx(ˆi·ˆi) + AxBy(ˆi·ˆj) + AxBz(ˆi·ˆk)+AyBx(ˆj·ˆi) + AyBy(ˆj·ˆj) + AyBz(ˆj·ˆk)+AzBx(ˆk·ˆi) + AzBy(ˆk·ˆj) + AzBz(ˆk·ˆk). •Use ˆi·ˆi=ˆj·ˆj=ˆk·ˆk=1, ˆi·ˆj=ˆj·ˆk=ˆk·ˆi=0. •⇒~A·~B=AxBx+AyBy+AzBz. tsl229 We begin this lecture with two slides elaborating on two distinct products of vectors: the dot product and the cross product. The dot product , also named scalar product , yields a scalar i.e. a number. The slide shows how we evaluate the dot product when the product vectors are given in components. It also shows how we calculate the dot product if we know the magnitude of each factor vector and the angle between them. It is useful to recall that the dot product yields a positive number if the angle between the two factor vectors is acute ( < 90 ◦), a negative number if that angle is obtuse ( > 90 ◦), and zero if the two vectors are perpendicular to each other (at right angle, 90 ◦). The dot product is commutative, meaning that the sequencing of the factors makes no difference. We have have used the dot product for calculating electric flux Φ E = ∫ ~E ·d ~A.The dot product has also made its appearance in the calculation of electric potential from an electric field via a path integral: V = − ∫ ~E · d~ s.Moreover, we have introduced the electric current through a wire cross section as the flux of electric current density: I = ∫ ~J · d ~A.Further applications of dot products will be coming soon. 1Cross Product Between Vectors Consider two vectors ~A=Axˆi+Ayˆj+Azˆkand ~B=Bxˆi+Byˆj+Bzˆk. •~A×~B=AB sin φˆn. •~A×~B=−~B×~A. •~A×~A=0. •~A×~B=AB ˆnif ~A⊥~B. •~A×~B=0if ~A‖~B. •~A×~B= ( Axˆi+Ayˆj+Azˆk)×(Bxˆi+Byˆj+Bzˆk)=AxBx(ˆi×ˆi) + AxBy(ˆi×ˆj) + AxBz(ˆi×ˆk)+AyBx(ˆj×ˆi) + AyBy(ˆj×ˆj) + AyBz(ˆj×ˆk)+AzBx(ˆk×ˆi) + AzBy(ˆk×ˆj) + AzBz(ˆk×ˆk). •Use ˆi×ˆi=ˆj×ˆj=ˆk×ˆk=0, ˆi×ˆj=ˆk, ˆ j×ˆk=ˆi, ˆk×ˆi=ˆj. •⇒~A×~B= ( AyBz−AzBy)ˆi+ ( AzBx−AxBz)ˆj+ ( AxBy−AyBx)ˆk. tsl230 The cross product , also named vector product , yields a vector. The slide shows how we evaluate the cross product when the product vectors are given in components. If we know the magnitude of each factor vector and their orientation in space we can construct the product vector as shown and already explained in lecture 19. When the two factor vectors are parallel or anti-parallel, then the cross prod-uct vanishes. Otherwise, the orientations of the factor vectors span a plane of unique orientation. The product vector is perpendicular to that plane. There are two such orientations. The right-hand rule (see lecture 19) determines which one to pick. The cross product is non-commutative, meaning that the sequencing of the factors does matter. Interchanging the factor vectors inverts the direction of the product vector. The cross product features prominently in the expressions of magnetic force, ~F = q~ v × ~B or ~F = I ~L × ~B, as we have already seen. Further applications are just around the corner (on the next page). 2Magnetic Dipole Moment of Current Loop N: number of turns I: current through wire A: area of loop ˆn: unit vector perpendicular to plane of loop ~μ=NIA ˆn: magnetic dipole moment ~B: magnetic field ~τ=~μ×~B: torque acting on current loop tsl475 A current loop such as the one shown on the slide is a realization of a mag-netic dipole. A magnetic dipole is characterized by a vector quantity, named magnetic dipole moment ~μ.The list on the slide explains how to determine the vector ~μ from the speci-fications of a flat loop with arbitrary shape. The vector ~μ is perpendicular to the plane of the loop. That leaves two possible directions. A new rendition of the right-hand rule tells us how to pick the right one. Curl the fingers of right hand in the direction of the current around the loop. Then the thumb points in the direction of ~μ.For loops that are not flat, the construction ~μ is a bit more involved. An example will be analyzed later. There are magnetic dipoles that are not current loops. For example the electron and the proton are electric dipoles. Attempts to explain the dipole moments of the electron and the proton as manifestations of rotating charge have utterly failed. Even the neutron, which does not carry a net charge, does have a magnetic dipole moment. When a magnetic dipole moment ~μ is positioned into a magnetic field ~B it experiences a torque τ = ~μ × ~B. The torque is an agent of reorientation just as the force is an agent of relocation. The torque ~τ aims to align the vector ~μ with the vector ~B. It is strongest when ~μ is perpendicular to ~B and vanishes when ~μ is aligned with ~B.3Torque on Current Loop •magnetic field: ~B(horizontal) •area of loop: A=ab •unit vector ⊥to plane of loop: ˆn •right-hand rule: ˆnpoints up. •forces on sides a:F=IaB (vertical) •forces on sides b:F=IbB (horizontal, not shown) •torque: τ=Fb sin θ=IAB sin θ •magnetic moment: ~μ=IA ˆn •torque (vector): ~τ=~μ×~B tsl196 This slide elaborates on the construction of the torque for a rectangular loop from first principles i.e. from forces and lever arms, familiar concepts from mechanics. A rectangular loop is positioned in a uniform magnetic field as shown from two different perspectives. The forces on the two sides of length b are opposite in direction and acting along the same line. Therefore, they do not contribute to any torque. The forces on the two sides of length a are also opposite in direction but acting along different lines as shown. Therefore, they do produce a torque. The end result as worked out on the slide agrees with the more general result from the previous slide. 4Magnetic Moment of a Rotating Disk Consider a nonconducting disk of radius Rwith a uniform surface charge density σ. The disk rotates with angular velocity ~ω.Calculation of the magnetic moment ~μ: •Total charge on disk: Q=σ(πR2). •Divide the disk into concentric rings of width dr . •Period of rotation: T=2πω. •Current within ring: dI =dQ T=σ(2πrdr )ω 2π=σω rdr . •Magnetic moment of ring: dμ=dI (πr2) = πσω r3dr . •Magnetic moment of disk: μ= ∫R 0 πσω r3dr =π 4σR4ω. •Vector relation: ~μ=π 4σR4~ω=14QR 2~ω. tsl199 Here we consider a magnetic dipole that is not a simple current loop, where all current flow is on the perimeter. We are looking at a spinning disk. The disk is uniformly charged with charge density σ. The rotating charge within concentric rings of radius r and width dr is reinterpreted as the current of a loop. Each loop then produces a magnetic dipole moment dμ according to the recipe from two slides ago. Summing up the contributions from all rings amounts to an integral across the range 0 < r < R of radius. Note that the directions of the angular velocity vector ~ω and ~μ are both determined by the right-hand rule: • Curl the fingers of your right hand in the direction of the spinning motion. Then the thumb points in the direction of ~ω. • Curl the fingers of you right hand in the direction of the (effective) current. Then the thumb points in the direction of ~μ.If the charge Q on the disk is positive, then ~μ has the same direction as ~ω.Making Q negative switches the direction of ~μ only. 5Direct-Current Motor tsl408 How can we convert the torque on a current loop into an agent of continuous rotating motion? On this slide you see two primitive designs for direct-current motors. In the torque expression, τ = μB sin θ, the factor sin θ changes sign after every half turn. The trick is to undo this change of sign by changing the current direction in the loop after every half turn, which will also change the direction of the magnetic moment vector ~μ after every half turn. In both designs this switch of current direction is accomplished by brushes and commutator contacts. The consequence of this trick is that the torque maintains the same direction. The loop will keep accelerating or keep driving a load in steady rotating motion. Later in this course, we will see that alternating currents (ac) offer many advantages over direct currents (dc). One advantage of ac motors is that they do not require sliding electrical contacts. 6Galvanometer Measuring direct currents. •magnetic moment ~μ(along needle) •magnetic field ~B(toward right) •torque ~τ=~μ×~B(into plane) tsl409 The galvanometer is the prototypical analog device for measuring direct cur-rents. A current loop is mounted inside a magnet such that the plane of the loop can rotate relative to the direction of the magnetic field. The loop is held in a reference orientation by a spring as shown. When a current I flows through the loop, it produces a magnetic dipole moment ~μ in the direction of the needle. The magnetic field ~B points to the right ( →). Therefore the torque, ~τ = ~μ × ~B points into the plane ( ⊗). Use your right hand and point the thumb in the direction of ~τ . Then your fingers curl clockwise, meaning that the torque aims to reorient the needle in that direction. Its tip on the scale will move to the right. The restoring elastic force provided by the spring counteracts that angular displacement with a torque in the opposite direction. The needle will settle at a certain angle away from the reference orientation. The angular displace-ment is proportional to the current I. Once the instrument is calibrated, the current I can be read off the scale. 7Magnetic Dipole in Uniform Magnetic Field •Magnetic dipole moment: ~μ=IA ˆn •Torque exerted by magnetic field: ~τ=~μ×~B •Potential energy: U=−~μ·~BU(θ) = − ∫θπ/2 τ(θ)dθ=μB ∫θπ/2 sin θdθ=−μBcos θ Note: τ(θ)and dθhave opposite sign. I θ μ = IAn ^ n^ τ B μ . tsl198 On this page and the next, we review some basic attributes of electric and magnetic dipole moments in comparison. The importance of both kinds of dipole moments in all branches of natural science and engineering can hardly be overemphasized. Here the focus is on the magnetic dipole moment. We use the familiar real-ization of a flat current loop, here positioned perpendicular to the plane of the page. The current flows out of the page on the left and into the page on the right. The arrow shows the direction at the front of the loop. The torque aims to align the dipole moment ~μ with the magnetic field ~B. In the configuration shown, the torque ~τ is directed into the page ( ⊗). Rotating the orientation of ~μ out of alignment with ~B requires work. The dif-ferential of rotational work is τ dθ in correspondence to F dx for translational work. The orientational potential energy U of a magnetic dipole can then be ex-pressed as a dot product between the vectors ~μ and ~B as shown. It assumes the lowest value when the two vectors are aligned. Many materials contain microscopic magnetic dipoles (mostly arising from electrons). An external magnetic field tends to align these dipoles. The ma-terial thus becomes magnetized. In ferromagnets the magnetization persists when the external field is turned off. 8Electric Dipole in Uniform Electric Field •Electric dipole moment: ~p=q~L •Torque exerted by electric field: ~τ=~p×~E •Potential energy: U=−~p·~EU(θ) = − ∫θπ/2 τ(θ)dθ=pE ∫θπ/2 sin θdθ=−pE cos θ Note: τ(θ)and dθhave opposite sign. L τ θ E p−q +q p = qL tsl197 Here we switch focus to the electric dipole moment. It is an electrically neutral object. Equal amounts of positive and negative charge are slightly displaced from each other. The simplest realization of an electric dipole, as shown on the slide, are two opposite point charges at a distance L from each other. The direction of the electric dipole moment ~p points toward the positive charge. The torque expression reads “moment cross field” and the potential-energy expression reads “minus moment dot field” just as in the magnetic case. Torque aims to align the dipole moment with the field. The aligned state has the lowest potential energy. Note that here the cross product ~p × ~E produces a torque ~τ pointing out of the page ( ). Many molecules are permanent electric dipoles, the most prominent case being H 2O as illustrated and explained in an earlier lecture. External electric fields induce an electric polarization in dielectric materials as discussed earlier in the context of capacitors. On the microscopic level, this means that the (positively charged) nuclei and the surrounding shells of (negatively charged) electrons of atoms are being pulled in opposite di-rections. In consequence they are slightly displaced by the external electric field. Each atom becomes an induced electric dipole. 9Magnetic Force Application (11) If the magnetic moment of the current loop (1) is μ1=1Am 2, what are the magnetic moments μ2,μ3,μ4of the current loops (2), (3), (4), respectively? (1) (2) (3) (4) tsl206 Let us begin by establishing a coordinate system with ˆi pointing toward the front, ˆj pointing toward the right, and ˆk pointing up. Then the flat square loop (1) has magnetic dipole moment, ~μ1 = 1Am 2 ˆk. Loop (2), which is not flat, can be thought of as made of two flat square loops. The two currents along the shared side cancel. The magnetic dipole moment is ~μ2 = 1Am 2 ˆj + 1Am 2 ˆk, with magnitude μ2 = √2Am 2.In like manner, loop (3) can be interpreted as consisting of three flat square loops with two shared sides of zero current. The magnetic dipole moment is ~μ3 = 1Am 2 ˆk + 1Am 2 ˆj − 1Am 2 ˆk = 1Am 2 ˆj, with magnitude μ3 = 1Am 2.Loop (4), finally, can be interpreted as consisting of three flat square loops with three shared sides of zero current. The magnetic dipole moment is ~μ4 = 1Am 2 ˆi + 1Am 2 ˆj + 1Am 2 ˆk, with magnitude μ3 = √3Am 2.A loop positioned on an arbitrarily curved surface can thus be divided into a grid of loops small enough to be essentially flat and with shared sides of zero current. The vector sum of the magnetic moments generated by the small loops on the grid then represents the magnetic moment of the loop on the curved surface. 10 Unit Exam III: Problem #1 (Spring ’12) In a region of uniform magnetic field B=5mTˆ i, a proton (m=1.67 ×10 −27 kg, q=1.60 ×10 −19 C)is launched with velocity v0=4000m/s ˆ k.(a) Calculate the magnitude Fof the magnetic force that keeps the proton on a circular path. (b) Calculate the radius rof the circular path. (c) Calculate the time Tit takes the proton to go around that circle once. (d) Sketch the circular path of the proton in the graph. xm, q Bv0 zy Solution: (a) F=qv 0B=3.2 ×10 −18 N.(b) mv 20 r=qv 0B⇒r=mv 0 qB =8.35mm .(c) T=2πrv0 =2πmqB =13.1 μs.(d) Center of circle to the right of proton’s initial position (cw motion). tsl435 We conclude this lecture with an application of a particle moving in circular motion through a region of uniform magnetic field. The solution as given on the slide is easy to reproduce. It is a bit more challenging to come up with the solution without peeking at the solution. The best preparation for an exam (and for a professional life in science and engineering) is to use your mental power, creativity, and imagination as the only resources. The mental tools that you must be able to recall in this application are the following: • The structure of magnetic force acting on a charged particle in motion. • The concept of centripetal force, here provided magnetically. • The circumference of a circle. • The right-hand rule. 11 Magnetic Force Application (6) An electric current flows through each of the letter-shaped wires in a region of uniform magnetic field pointing into the plane. •Find the direction of the resultant magnetic force on each letter. NWESNW NE SE SW B tsl193 This is the quiz for lecture 21. Here we have another application of the vector equation ~F = I ~L × ~B.Possible answers are either no force or one of the eight compass points indi-cated. Hint: Remember what we said about the net force on a current loop in a uniform magnetic field. 12
16576	https://www.c-sharpcorner.com/article/converting-integers-to-characters-in-c-sharp/	Converting Integers to Characters in C# C# Corner Tech News Videos Forums Trainings Books Events More Interviews Jobs Live Learn Career Members Blogs Challenges Certifications Bounties Contribute Article Blog Video Ebook Interview Question Collapse Feed Dashboard Wallet Learn Achievements Network Refer Rewards SharpGPT Premium Contribute Article Blog Video Ebook Interview Question Register Login Converting Integers to Characters in C# WhatsApp Uday Dodiya 2y 44.1k 0 2 100 Article Take the challenge Introduction In C#, the need to convert integers to their corresponding character representations is a common occurrence. Whether you're working with ASCII values or dealing with character-based operations, understanding the various methods for converting integers to characters is essential for any C# developer. In this article, we will learn about the different techniques available for this conversion, providing examples and insights into each method's usage. 1.Type Casting The simplest way to change an integer into a character in C# is by typecasting. We can easily cast an integer to a character type since, at their core, characters are simply numbers (ASCII values). This approach is popular with developers since it is simple and effective. Example ```csharp using System; class Program { static void Main() { int intValue = 65; // Represents 'A' in ASCII char charValue = (char)intValue; Console.WriteLine("Convert From Int To Char In C# Using Type Casting"); Console.WriteLine("Integer Output : " + intValue); Console.WriteLine("Character Output : " + charValue); } } ``` C# Copy Output 2.Convert.ToChar() The Convert.ToChar()method in C# provides an easier and more readable way to convert an integer to its corresponding character representation compared to the direct type casting method. It is part of the System namespace and ensures a smooth conversion from an integer to a character. This method comes in handy when you want to make the conversion more explicit and easy to understand in your code. With Convert.ToChar(), you can quickly and effortlessly convert an integer to its corresponding character without any complexity. Example ```csharp using System; class Program { static void Main() { int intValue = 65; // Represents 'A' in ASCII char charValue = Convert.ToChar(intValue); Console.WriteLine("Convert From Int To Char In C# Using Convert.ToChar()"); Console.WriteLine("Integer Output : " + intValue); Console.WriteLine("Character Output : " + charValue); } } ``` C# Copy Output 3. char.ConvertFromUtf32() The char.ConvertFromUtf32() method in C# is really helpful when you're working with special characters and characters from different languages that go beyond the usual English characters. It allows you to convert a special code (a number that represents a specific character) to its actual character representation. This method is essential when you need to work with characters that are not part of the standard English alphabet or the common ASCII characters we usually deal with. It's like a magic tool that helps you handle a wider range of characters in your C# programs. Example ```csharp using System; using System.IO; class Program { static void Main() { int intValue1 = 65; // Represents 'A' in ASCII int intValue2 = 8364; // Represents the Euro sign '€' int intValue3 = 128516; // Represents the smiley face emoji '😄' int intValue4 = 960; // Represents the Greek letter 'π' (pi) string charAsString1 = char.ConvertFromUtf32(intValue1); string charAsString2 = char.ConvertFromUtf32(intValue2); string charAsString3 = char.ConvertFromUtf32(intValue3); string charAsString4 = char.ConvertFromUtf32(intValue4); // Write characters to a file File.WriteAllText("unicode_characters.txt", $"Character for ASCII 65: {charAsString1}\n" + $"Character for Unicode 8364 (Euro sign): {charAsString2}\n" + $"Character for Unicode 128516 (Smiley face): {charAsString3}\n" + $"Character for Unicode 960 (Greek letter π): {charAsString4}"); } } ``` C# Copy Output Some consoles may not support displaying special characters correctly. Therefore, we have written the output to a text file in the example. This allows us to accurately identify how the method works. The following screenshot shows the contents of the text file. Final Example In this example, we will see all three methods together and print A to Z and a to z using Type Casting and Convert.ToChar() Method and print Random special character using char.ConvertFromUtf32(). ```csharp using System; using System.IO; class Program { static void Main() { int CapitalLetterVal = 65; int SmallLetterVal = 97; Console.WriteLine("Print A to Z and a to z Using Type Casting : "); Console.Write("A to Z : "); for(int i=1;i<=26;i++, CapitalLetterVal++) { Console.Write((char)CapitalLetterVal + " "); } Console.WriteLine(); Console.Write("a to z : "); for (int i = 1; i <= 26; i++,SmallLetterVal++) { Console.Write((char)SmallLetterVal + " "); } CapitalLetterVal = 65; SmallLetterVal = 97; Console.WriteLine("\n\n"); Console.WriteLine("Print A to Z and a to z Using Convert.ToChar() : "); Console.Write("A to Z : "); for (int i = 1; i <= 26; i++, CapitalLetterVal++) { Console.Write(Convert.ToChar(CapitalLetterVal) + " "); } Console.WriteLine(); Console.Write("a to z : "); for (int i = 1; i <= 26; i++, SmallLetterVal++) { Console.Write(Convert.ToChar(SmallLetterVal) + " "); } Console.WriteLine("\n\n"); Console.WriteLine("Print Random Numbers In TextFile Using char.ConvertFromUtf32() : "); Random random = new Random(); String randomCharacters = ""; for (int i = 1; i <= 26; i++) { int utf32Value = random.Next(225, 10000); string specialCharacter = char.ConvertFromUtf32(utf32Value); randomCharacters = randomCharacters + " " + specialCharacter + " "; } // Write characters to a file File.WriteAllText("unicode_characters.txt",randomCharacters); } } ``` C# Copy Output Summary Converting integers to characters in C# is a basic task, and we have different methods to do it. Each method has its own benefits, and the choice depends on what you need for your code to be clear and easy to read. Type casting is short and straightforward. Convert.ToChar() makes the conversion explicit, and char.ConvertFromUtf32() handles special characters. People also reading (0)Comments View All Comments Press Esc key to cancel Load More Message OK Close Upcoming Events View all 29 OCT AI Agents Conference - 2025 15 NOV .NET Conf 2025 - Chennai Ebook download View all Working with Directories in C# Read by 17.8k people Download Now! 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16577	https://www.youtube.com/watch?v=XXUyZExOEFE	Partial Fractions (Comparing Coefficients Method) Houston Math Prep 51500 subscribers 564 likes Description 31169 views Posted: 24 Feb 2020 This partial fractions tutorial video explains the method of comparing coefficients, needed when doing partial fraction decomposition with repeated factors and irreducible quadratic factors. We work an example that shows how we start by using the method of setting factors equal to zero, but that method only gets us so far with this type of partial fraction decomposition. In order to finish finding the partial fractions, we introduce the method of comparing coefficients, where we distribute and compare coefficients for each power of the variable in order to solve the constants. 0:00 Introduction 0:30 Getting the equation used to solve 2:08 Why comparing coefficients is needed 3:52 Distributing and comparing coefficients 7:51 Writing our solution Houston Math Prep Partial Fractions Playlist: 17 comments Transcript: Introduction welcome back everyone to our method to video for partial fraction decomposition about comparing coefficients we have already explored our method 1 which is setting factors equal to zero that's something that is a short method and you can use it a lot but you can't use it on every problem and so we want to make sure we're familiar with the comparing coefficients method here I've set up 3x squared plus 3x minus 8 over X plus 3 times the quantity x squared plus 1 looking at our denominator that's Getting the equation used to solve already factored I know I'm going to have something over X plus 3 and I'm going to have something over x squared plus 1 okay since this is a linear factor just have an X as our highest power we will need a constant only because we have a quadratic factor x squared I need all powers below x squared which means I need an X term and I need a constant in my numerator we are going to just do a little bit of less writing here and think about what would the numerators look like once we get a common denominator so obviously the beginning fraction itself that we started with already has the common denominator it has all of the factors in the denominator to begin with so this is not going to change at all when we get a common denominator we'll have 3x squared plus 3x minus 8 what we would do to the a fraction is we would multiply in the x squared plus 1 factor to the top and the bottom so the numerator for this one would look like a times the quantity x squared + 1 and then BX plus C which is more than one term so I want to make sure that I put it in parenthesis what would we multiply the top and bottom here by to get the common denominator we would multiply the X plus 3 into this fraction so this is what the numerators would look like once we got a common denominator and remember we will be solving the numerators anyway to find a B C etc so I'm not going to write all of the denominators and then write it again hopefully you're ok with this being what we would get from the common denominator process from our method 1 video remember Why comparing coefficients is needed that was setting factors equal to 0 so if you look through here and you think about what could I set X equal to to make a factor zero you might notice on the end here that we can set X equal to negative three that would make that factor zero and some stuff would disappear here and help us solve the other factor that we have is x squared plus one and the problem becomes how do I plug in something for X so that x squared plus one equals zero and the answer is there's no real number that will work for that that happens very often when we have X square plus some number we're going to get some imaginary solution for X and we're not going to plug in imaginary values complex numbers we need to be plugging in real numbers so we'll go ahead and at least use our x equals negative three but we're not going to be able to set all of the factors equal to zero so if we go ahead and plug that in that will give us negative three squared would be 9 times three so this first term will be 27 negative three times the three that's already there would be minus nine and then we have minus eight plugging in negative three squared would be 9 plus one would give us 10 in this factor next to a and then of course plugging in negative three makes this zero which makes this whole entire thing a zero here because it's all multiplied together okay so for this one I get 27 minus nine minus eight which ends up actually being ten and that's equal to ten a so that's easy to solve divide both sides by 10 and we get that a is equal to one now we still don't have B Distributing and comparing coefficients and C and we aren't able to set this factor equal to zero so what we will need to do is we will actually need to take this stuff that we have we're trying to solve from and we need to distribute and then we will do as the title of the video says we will compare the coefficients so there's nothing to distribute on the left side we have 3x squared plus 3x minus 8 and then if I distribute a in here I get a x squared plus a if I distribute these so be x times X will give me be x squared be x times three will give me three B X C times X will give me C X and 3 times C will give us three C what we will now do is combine all of our like terms so I have 3x squared plus 3x minus 8 and then over here if I think about what do I have as far as like terms here's an x squared term and there's an x squared term so how many X Squared's do I have well I have a plus B is that's what those terms are I have ax squared plus BX squared side of a plus B X Squared's then we go through and look at X terms so I have a 3 B X and a CX there so how many X's do I have well I have 3 B of them in the one term and I have C of them in that term so I have three B plus C X's looking at the constants I have an A and I have a three C and those are my constants so I'll just put those together a plus 3c okay what we will now do is compare coefficients in other words I need to have the same amount of each respective power on both sides so if I look at my x squared terms so I have three X Squared's over here the X square coefficient over here is a plus B so that tells me right there that a plus B is equal to three if I look at how many X terms I have on both sides I have three of them here I have three B plus C of them there so I have three B plus C equal to three looking at the constant terms here I have a negative eight and here I have a plus 3 C so those must be the same so in other words a plus three C must be equal to negative eight now I have a system of equations and this might be a little bit challenging to solve and lengthy because I have three things I'm trying to solve but I have already found one of them I already know that a is 1 from up here so if I already know this is 1 then this tells me that 1 plus B is 3 and so right away I can solve that for B right if I subtract 1 from both sides then that will give me that B is 2 um I could go ahead and use either of these now I already know a and I already know B so let's just say I go with the next one if I know that B is 2 then this next 3 B + C becomes 3 times 2 which would be 6 plus C is equal to 3 and then if I subtract 6 on both sides I would get that C is negative 3 so now we've solved all our stuff you could use the last one if you want you may not have to use all of these once you know one you can just kind of pick and choose you know which equation out of these is the shortest to get the next letter that I want to get right so we have our a we Writing our solution have our B and we have our C so now we'll simply plug those into this form that we set up to begin with so our partial fraction decomposition will be u 1 over X plus 3 plus B is 2 and that's our X coefficient so that's 2 X C is negative 3 so that goes up there BX plus C is 2 X minus 3 over our factor x squared plus 1 ok so comparing coefficients that's kind of a backup if you're unable to use all of the factors and set all the factors equal to 0 maybe you have some repeated factors where you just can't get all of your A's B's and C is solved directly so this is the method that you will use to solve if you get stuck and aren't able to set all of your factors equal to 0 and get those straight away all right hopefully this helps you on comparing coefficients and partial fractions we'll see in some of our example videos
16578	https://www.pinterest.com/pin/class-height-chart-useful-for-measuring-in-cm-and-m-also-for-comparing-heights-and-ordering-etc-a8-in-2023--694750680042092598/	Skip to content Search for measuring students height length width height anchor chart teaching ruler measurement measuring kids height on wall how to teach measurement to children measuring students height When autocomplete results are available use up and down arrows to review and enter to select. Touch device users, explore by touch or with swipe gestures. Log in Sign up More about this Pin Board containing this Pin Kinder 2 1.5k Pins 23h Related interests Measuring Students Height Length Width Height Anchor Chart Teaching Ruler Measurement Measuring Kids Height On Wall How To Teach Measurement To Children Measurement Preschool Measuring Classroom Items {centimeters And Inches Kids Growth Chart Ruler Weight Conversion Chart Visit Save mungfali.com Class Height Chart - Useful For Measuring In Cm And M. Also For A8 Browse through Class Height Chart - Useful For Measuring In Cm And M. Also For , and similar images Mungfali.com Comments Add a comment More to explore More to explore More about this Pin 12 Likes 26 Shares Board containing this Pin Preescolar 1.4k Pins 2d Related interests School Wall Art Ideas School Wall Decoration Preschool Designs Giraffe Illustration Preschool Classroom Decor School Cartoon School Murals Height Chart Childrens Artwork More about this Pin 4 Likes 72 Shares Board containing this Pin Vriendjes 33 Pins 8y Related interests Elementary School Height Chart Colorful Classroom Decoration Diy Classroom Decor Ideas Classroom Setup How To Track Student Growth Math Measurement Preschool Games Reggio Inspired Classrooms Classroom Height Chart With Students' Photos Kleuters meten en vergelijken :) Meer ideetjes rond thema mijn lichaam: liestr More about this Pin 10 Likes 17 Shares Board containing this Pin Prep 271 Pins 1mo Related interests Length And Height Activities Height Preschool Activities Height Activities Kindergarten Measuring Height Kindergarten Teaching Length Measurement Activities Sciences En Maternelle Measurement Kindergarten Measurement Activities Math Measurement Measurement Activity You are signed out Sign in to get the best experience Continue with email By continuing, you agree to Pinterest's Terms of Service and acknowledge you've read our Privacy Policy. Notice at collection. Log in to see more Email Email Use 8 or more letters, numbers and symbols Password Forgot your password? Log in OR Email a login code Use QR code Facebook login is no longer available Update login method Not on Pinterest yet? Sign up Are you a business? Get started here! By continuing, you agree to Pinterest's Terms of Service and acknowledge you've read our Privacy Policy. Notice at collection. Class Height Chart - Useful For Measuring In Cm And M. Also For A8 ===============
16579	https://academichelp.net/stem/economics/compound-continuously-formula.html	Most Popular 1 day ago Best assignment service: Our top 10 1 day ago Best assignment writing service: Our top 10 – Reserve Version What Is The Continuous Compounding Formula? freepik.com Compound interest plays a pivotal role in the world of finance, and it comes in various forms. One of the most intriguing forms is continuous compounding. In this article, we will delve into the continuous compounding formula, its derivation, examples, and frequently asked questions. JOIN OUR LEARNING HUB ✅ AI Essay Writer ✅ AI Detector ✅ Plagchecker ✅ Paraphraser ✅ Summarizer ✅ Citation Generator Get Free Tools What Is Continuous Compounding Formula? Continuous compounding is a concept used in financial calculations when interest is compounded infinitely. The formula that governs this unique calculation is the continuous compounding formula, which is expressed as: In this formula: This formula is particularly useful when the problem statement explicitly mentions that the amount is compounded continuously. Deriving the Continuous Compounding Formula To understand how the continuous compounding formula is derived, let’s start with the standard compound interest formula: A = P(1 + r/n)^(nt) In this formula, n is the number of times the initial amount (P) is compounded in the time t, and A is the final amount or future value. For continuous compound interest, n approaches infinity. We can take the limit of the above formula as n approaches infinity: A = lim (P(1 + r/n)^(nt)) as n → ∞ Using one of the limit formulas, which states that lim (1 + r/n)^n = e^r, we arrive at the continuous compound interest formula: A = Pert This derivation showcases the relationship between continuous and discrete compounding, highlighting how continuous compounding becomes relevant when the compounding frequency tends towards infinity. Applying the Continuous Compounding Formula Example 1: Calculating Future Value Let’s consider an example to demonstrate the application of the continuous compounding formula. Tina invested $3000 in a bank that pays an annual interest rate of 7%, compounded continuously. We want to find out the amount she can get after 5 years. Using the continuous compounding formula: A = Pert A = 3000 × e^(0.07 × 5) ≈ 4257 Therefore, Tina can expect to have approximately $4,257 after 5 years. Example 2: Determining the Interest Rate In another scenario, we might need to find the rate of interest for a given amount to double over a specific period with continuous compounding. Let’s take an example: Suppose you have $5,300, and you want it to double in 8 years with continuous compounding. We need to find the rate of interest (r). Using the continuous compound interest formula: A = Pert 10600 = 5300 × e^(8r) After some calculations: r = (ln 2) / 8 ≈ 0.087 So, the required rate of interest is approximately 8.7%. Example 3: Long-Term Investment Jim decides to invest $5000 in a bank that pays an annual interest rate of 9%, compounded continuously. He plans to keep his money invested for 15 years and wants to know the amount he can expect to have at the end of this period. Using the continuous compounding formula: A = Pert A = 5000 × e^(0.09 × 15) ≈ 19287 Jim can anticipate having approximately $19,287 after 15 years. In conclusion, understanding the continuous compounding formula is essential for anyone involved in financial calculations. This formula provides a powerful tool for handling scenarios where interest is compounded continuously, and it can be applied to a wide range of real-world situations, as demonstrated in the examples provided above. Whether you’re calculating future values, interest rates, or long-term investments, the continuous compounding formula is a valuable asset in your financial toolkit. FAQ What is the continuous compounding formula? The continuous compounding formula is expressed as A = Pert, where A represents the final amount, P is the initial amount or principal, r is the rate of interest, t is the time, and e is the mathematical constant approximately equal to 2.7183. This formula is used for calculating compound interest when compounding occurs infinitely. How does continuous compounding differ from regular compounding? Continuous compounding differs from regular compounding in that it assumes that interest is compounded infinitely often, whereas regular compounding occurs at specific intervals, such as annually, quarterly, or monthly. Continuous compounding provides more precise calculations for scenarios where compounding happens continuously. What are the applications of continuous compounding in finance? Continuous compounding finds applications in various financial scenarios, including investments, loans, and savings accounts. It is commonly used for calculating the future value of investments, determining interest rates, and understanding the growth of investments that experience continuous compounding. What is the significance of the mathematical constant “e” in continuous compounding? The mathematical constant “e” in the continuous compounding formula is significant because it represents the base of natural logarithms. It ensures that the compounding occurs continuously and is a fundamental component in accurately calculating compound interest in continuous compounding scenarios. How does continuous compounding impact investment growth? Continuous compounding has a notable impact on investment growth. It leads to faster growth compared to compounding at discrete intervals. As compounding occurs continuously, even small changes in the interest rate or time can result in significant differences in the final amount, making it a powerful tool for maximizing investment returns over time. Follow us on Reddit for more insights and updates. Comments (0) Welcome to AHelp comments! We’re all about debate and discussion at AHelp. We value the diverse opinions of users, so you may find points of view that you don’t agree with. And that’s cool. 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16580	https://www.youtube.com/watch?v=ybyLNaQQeyk	Use this Easy Trick to Find Any Square Root - The Number Line MathBase 1130 subscribers 51 likes Description 8434 views Posted: 21 Apr 2022 How do we find the square root of a value without a calculator? Watch this video to learn how to estimate its value with a few simple steps! Square roots follow a parabolic nature in how they change from one value to another. However, when it comes to estimating their value, it is sufficient to use a linear approximation known as interpolation between two well-known square values. For example, the square root of 10 is greater than the square root of 9 but lesser than the square root of 16, both of which are well-known values; 3 and 4. So square root 10 must be something like 3.5 or 3.1! All that is left is to use some basic fractions to work out your decimal value! FRACTIONS PLAYLIST youtube.com/playlist?list=PL9c1-IKFNxAAewImx0l8GkitP_p_rgm3a LINEAR ALGEBRA PLAYLIST youtube.com/playlist?list=PL9c1-IKFNxADdbuZzzvPP3cia6IL-SXBE QUADRATICS PLAYLIST youtube.com/playlist?list=PL9c1-IKFNxADMkJDv-Ka5Haa53SfY5rqn 5 comments Transcript: Intro what is the value of square root 2 what is the value of square root 3 these are not perfect squares which means we're going to get decimal values somewhere between 1 and 2 making them fairly difficult to calculate but for the most part you don't need to get a precise value for these numbers so today i'm going to show you how to get a rough estimate using three simple tools perfect squares from 1 to 12 the number line and simple fractions Square Root 2 starting off with square root 2 prepare a list of your perfect squares on the side draw a number line marking out square root 2 at the center write out number 2 and notice that it's between 1 and 4 in your list of perfect squares these are your perfect square boundaries for two so write them out and find the square roots above them since square root one is equal to one and square root four equals two this must mean that square root two is between the values one and two so square root 2 will equal to 1 point something to find out that something take the differences between your perfect squares then take the first number and divide by the sum total of both of these numbers doing the maths gives us an estimate value of 0.3 and this is how far the value is above the lower boundary of one so one plus 0.3 is 1.3 and that is our approximate value of square root 2. the precise value for square root 2 is 1.414 meaning our estimate was only off by 0.1 units and you'll find this to be a recurring thing in our method even for larger square roots next let's check out the value for square root 3. depending on how you round your fraction you could end up with a more precise decimal Square Root 3 alright so setting up square root 3 on a number line we can see that 3 will also be between the perfect squares 1 and 4. so the value of square root 3 must also be one point something but just slightly larger than our previous answer doing our calculations we find that the answer is 1.7 when compared to the precise answer of 1.732 this is actually a pretty good estimate because we rounded two thirds as 0.7 rather than 0.6 or 0.67 and like i said this does work for larger square roots but let me prove it to you with three larger problems starting with square root 10. i'll be speeding through the rest of this video as the method will be the same but i'll focus a little more on how we handle the fractions Square Root 10 okay let's go so 10 is between 9 and 16 which means square root 10 is between 3 and 4 so it must be 3 point something taking the differences and doing the maths we find that our decimal value is 0.142857 but if you don't know this value off the top of your head that's totally fine you can either divide 1 by 7 to get the first decimal or simplify the fraction to something easier like 1 on 10 which gives you the value 0.1 either way when you add it to 3 we get the answer 3.1 technically we could have rounded 1 and 7 down to 105 or 0.2 giving us a final answer of 3.2 and that wouldn't be far off either because whether you write 3.1 or 3.2 the precise answer is 3.162 but square root 10 is still fairly small so it's no surprise that our estimate is pretty good still so what about something larger like square root of Square Root 123 123 well using the same method 123 is between 121 and 144 which means the answer should be 11 point something finding our differences we see that the fraction will be 2 on 23 which can estimate closely to 2 and 20 or 1 and 10 or 0.1 which means our answer is 11.1 the precise answer is 11.0905 which is only off by 0.01 units and when you compare this to our previous question where we converted 1 and 7 to 1 on 10 which is 3 units apart we made the same conversion from 2 on 23 to 2 and 20 because that's also 3 units apart and yet the previous answer was off by 0.06 units while this one's only off by 0.01 so clearly we can see that as the problem gets larger the estimations become more accurate let's confirm this with our final question which is way bigger square root Square Root 1354 1354 okay so obviously this is not going to be within our perfect squares list so i'm going to say that this is between the numbers 1296 and 1369 because those are the perfect squares of 36 and 37 so we expect the answer to be 36 point something carrying out our calculations 58 on 73 is very similar to 60 on 70 so we can simplify that as 6 on 7 which is 0.8 and you can find that value by doing a long division up to the first decimal place so putting that all together the answer is 36.8 the precise answer is 36.7967 meaning our estimate was only off by 0.004 units which is incredible so that is a very very accurate estimate so as long as you know how to use your perfect squares on a number line along with some fractions and long division you'll be able to estimate any square root number and with enough practice you'll be able to do it mentally so i hope you guys learned something from this video let me know what you think down in the comments below hit the like button if you enjoyed it and subscribe to the channel to learn more stuff coming in the future thanks for watching guys have a good one and i'll see you next time bye
16581	https://math.stackexchange.com/questions/4507631/maximize-the-distance-between-a-w-shaped-line-construction-and-two-points	calculus - Maximize the distance between a w shaped line construction and two points - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Maximize the distance between a w shaped line construction and two points Ask Question Asked 3 years, 1 month ago Modified3 years, 1 month ago Viewed 127 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I am trying to hang string lights on my rectangular porch in a W-shaped fashion. There are two ceiling fans interrupting the pattern and I need to maximize the distance between the ceiling fan blades and the string lights. The perpendicular distance from the center of each ceiling fan to each line should be equal. I tried to use cad to solve this and I am close but I'd like to get an exact solution if possible. I believe one way to go about it is to use the following Theorem and solve that set of equations along with the equations of the lines. There might also be a derivative involved since we are maximizing something. I began using the theorem and things got pretty nasty right away. I am hoping someone might have a more elegant solution. The unknown variables are x_1, x_2 and x_3. calculus geometry algebra-precalculus Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Aug 7, 2022 at 16:28 Mike KMike K asked Aug 7, 2022 at 8:13 Mike KMike K 143 6 6 bronze badges 2 Do you want x_3 = 292? If so, only the other two x values are unknown.bubba –bubba 2022-08-07 10:26:42 +00:00 Commented Aug 7, 2022 at 10:26 @bubba It was a mistake. The new image has now been posted.Mike K –Mike K 2022-08-07 16:26:07 +00:00 Commented Aug 7, 2022 at 16:26 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Just a few suggestions. Notation: let’s call the four diagonal lines L 1 L 1, L 2 L 2, L 3 L 3, L 4 L 4, and let’s call the circle centers C 1 C 1 and C 2 C 2. Define some distances d 1=d i s t a n c e(L 1,C 1)d 1=d i s t a n c e(L 1,C 1), d 2=d i s t a n c e(L 2,C 1)d 2=d i s t a n c e(L 2,C 1), d 3=d i s t a n c e(L 3,C 12)d 3=d i s t a n c e(L 3,C 12), and d 4=d i s t a n c e(L 4,C 2)d 4=d i s t a n c e(L 4,C 2). We want d 1=d 2=d 3=d 4 d 1=d 2=d 3=d 4. Using a CAD system, or even just PowerPoint, you can move the diagonal lines around until the four distances look roughly equal. This would probably be close enough. If your CAD system supports constraint solving, you can dimension the four distances, tell the system you want them to be equal, and let the constraint solver do its thing. You have the formula for the distance from a point to a line, so you can write three equations that express the requirements d 1=d 2 d 1=d 2, d 2=d 3 d 2=d 3 and d 3=d 4 d 3=d 4. Now we have to solve these three equations to get x 1 x 1, x 2 x 2, x 3 x 3. Solving these equations by hand is likely to be a big mess, unless someone can spot a clever algebra trick (I can’t). You can try to solve the equations using a computer algebra system like Wolfram Alpha (free) or Mathematica (far from free). If those systems can’t give you an exact algebraic solution, they will at least give you one computed by numerical methods. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 8, 2022 at 4:44 bubbabubba 44.8k 3 3 gold badges 70 70 silver badges 127 127 bronze badges 3 I think calculus can be used because If you pretend to draw the lines by hand you start in the upper left corner and Pick a D1 as a result X1 has been determined by your choice. Let D3=D2=D1, then X2 and X3 have also been determined. At this point D4 has been determined but does not necessarily equal D1. This is because the 4th line in this mental exercise is constrained by the corner point (292,110). I'm thinking you could set it up so that a derivative allows D1 to approach D4. Does this make any sense or would it be more difficult to solve then the algebraic system?Mike K –Mike K 2022-08-08 06:36:34 +00:00 Commented Aug 8, 2022 at 6:36 If you use my approach #3, this all boils down to solving a system of three equations. The numerical methods I mentioned might use derivatives, internally, but we don’t need to see that.bubba –bubba 2022-08-08 13:27:05 +00:00 Commented Aug 8, 2022 at 13:27 your cad solution worked please see above Mike K –Mike K 2022-08-10 02:44:04 +00:00 Commented Aug 10, 2022 at 2:44 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. You'll likely be forced to solve numerically no matter what, but here goes: I'm assuming the optimal solution will have a radius r r from the center C 1 C 1 to the tangent lines L 1 L 1 and L 2 L 2 and similarly for C 2 C 2. I don't know that that's necessarily the case for the optimum, but it's what I'm going with. For a given r r, we can calculate x 1 x 1 by identifying the angle the tangent line L 1 L 1 makes with the y-axis. x 1=110 tan(π 2−arctan(55 105.5)−arcsin(r 55 2+105.5 2−−−−−−−−−−√))x 1=110 tan⁡(π 2−arctan⁡(55 105.5)−arcsin⁡(r 55 2+105.5 2)) x 3 x 3 can be found the same way. x 3=292−110 tan(π 2−arctan(55 81)−arcsin(r 55 2+81 2−−−−−−−−√))x 3=292−110 tan⁡(π 2−arctan⁡(55 81)−arcsin⁡(r 55 2+81 2)) Once x 1 x 1 and x 3 x 3 are found, we can use the tangent lines L 2 L 2 and L 3 L 3 and find their intersections with the top of the rectangle. These won't necessarily agree for a given r, but will give two options x L 2 x 2 L and x R 2 x 2 R. x L 2=110 2+x 2 1−−−−−−−−√⋅sin(2 arcsin(r x 1))sin(π−2 arcsin(r x 1)−arctan(55 105.5)−arcsin(r 55 2+105.5 2√)x 2 L=110 2+x 1 2⋅sin⁡(2 arcsin⁡(r x 1))sin⁡(π−2 arcsin⁡(r x 1)−arctan⁡(55 105.5)−arcsin⁡(r 55 2+105.5 2) and x R 2=292−110 2+(292−x 3)2−−−−−−−−−−−−−−−√⋅sin(2 arcsin(r x 3))sin(π−2 arcsin(r x 3)−arctan(55 81)−arcsin(r 55 2+81 2√)x 2 R=292−110 2+(292−x 3)2⋅sin⁡(2 arcsin⁡(r x 3))sin⁡(π−2 arcsin⁡(r x 3)−arctan⁡(55 81)−arcsin⁡(r 55 2+81 2) Setting those two equal to each other and solving for r r should get you where the tangents meet at a common x 2 x 2. Barring my tangency assumptions about the optimum being wrong or me misreading the diagram, that gives me r=29.6833 r=29.6833, x 1=122.2463 x 1=122.2463, x 2=149.9779 x 2=149.9779, and x 3=205.5175 x 3=205.5175 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 8, 2022 at 12:58 Michael BiroMichael Biro 13.9k 2 2 gold badges 28 28 silver badges 40 40 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. I put it into CAD. Then checked it against the previous solution. Keep in mind that the tolerance is set to 1/64. Column got in the way! Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Aug 10, 2022 at 2:35 answered Aug 9, 2022 at 5:03 Mike KMike K 143 6 6 bronze badges 2 A numerical solution is going to be the best you can do. After doing out too much algebra: x 1=122.2463009839560914669064355869597068…x 1=122.2463009839560914669064355869597068… is the real solution of 2984 x 7−2203684 x 6+803626114 x 5−180830349625 x 4+27280292057176 x 3−2754957086055401 x 2+175453649542580508 x−5533650079159662376=0 2984 x 7−2203684 x 6+803626114 x 5−180830349625 x 4+27280292057176 x 3−2754957086055401 x 2+175453649542580508 x−5533650079159662376=0 and doesn't seem to have a solution in radicals.Michael Biro –Michael Biro 2022-08-09 12:39:39 +00:00 Commented Aug 9, 2022 at 12:39 1 @MichaelBiro your solution worked. See above.Mike K –Mike K 2022-08-10 02:42:00 +00:00 Commented Aug 10, 2022 at 2:42 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus geometry algebra-precalculus See similar questions with these tags. 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16582	https://www.cuemath.com/algebra/greatest-integer-function/	LearnPracticeDownload Greatest Integer Function The greatest integer function is also known as the step function. It rounds up the number to the nearest integer less than or equal to the given number. The graph of the greatest integer function is a step curve which we will explore in the following sections. The domain of the greatest integer function is ℝ and its range is ℤ. Therefore the greatest integer function is simply rounding off the given number to the greatest integer that is less than or equal to the given number. Here we shall learn more about the greatest integer function, its graph, and its properties. \| \| \| --- \| \| 1. \| What is Greatest Integer Function? \| \| 2. \| Domain and Range of Greatest Integer Function \| \| 3. \| Greatest Integer Function Graph \| \| 4. \| Greatest Integer Function Properties \| \| 5. \| Greatest Integer Function and Smallest Integer Function \| \| 6. \| FAQs on Greatest Integer Function \| What is Greatest Integer Function? Greatest integer function is a function that gives the greatest integer less than or equal to a given number. The greatest integer less than or equal to a number x is represented as ⌊x⌋. We will round off the given number to the nearest integer that is less than or equal to the number itself. Mathematically, the greatest integer function ⌊x⌋ can be defined as follows: ⌊x⌋ = n, where n ≤ x < n + 1 and 'n' is an integer. Here, 'x' can be any real number but 'n' is always an integer. i.e., irrespective of what number is being set as input to the function, the output is always an integer. For example, ⌊3.02⌋ = 3, as 3 ≤ 3.02 < 4. ☛Note: The greatest integer function results in the same number if the input is an integer. Examples: ⌊3⌋ = 3, ⌊-4⌋ = -4, etc. Clearly, the input variable x can take on any real value. However, the output will always be an integer. Also, all integers will occur in the output set. Domain and Range of Greatest Integer Function The greatest integer function has the domain of the function as the set of all real numbers (ℝ), while its range is the set of all integers (ℤ). Let us understand the domain and range of the function by observing the following examples of the greatest integer function in the following table: \| Values of x \| f(x)=⌊x⌋ \| --- \| \| 3.1 \| f(3.1) = ⌊3.1⌋ = 3 \| \| 2.999 \| f(2.999) = ⌊2.999⌋ = 2 \| \| −2.7 \| f(−2.7) = ⌊−2.7⌋ = −3 \| \| 4 \| f(4) = ⌊4⌋ = 4 \| \| −7 \| f(−7) = ⌊−7⌋ = −7 \| Here, the values of x can be any real number, and hence the domain of the greatest integer function is ℝ. But observe that all the values of f(x) (y-values) are integers and hence the range is ℤ. Greatest Integer Function Graph The greatest integer function graph is known as the step curve because of the step structure of the curve. Let us plot the greatest integer function graph. First, consider f(x) = ⌊x⌋, if x is an integer, then the value of f will be x itself. If x is a non-integer, then the value of x will be the integer just before x (on the left side of x). For example, For all numbers lying in the interval [0, 1), the value of f will be 0. For the entire interval [1, 2), f will give the value 1. For the interval [−1, 0), f will be the value −1 and so on. So for an integer n, all the numbers that lie in the interval [n, n+1) will have the value of the greatest integer function as n. The function has a constant value between any two integers. As soon as the next integer comes, the function value jumps by one unit. This means that the value of f at x = 1 is 1 (and not 0) hence there will be a hollow dot at (1, 0) and a solid dot at (1, 1) where a hollow dot means not including the point and solid dot represents including the point. These observations lead us to the following graph. From the graph above we can clearly see that inputs of the function can be any real number but the output will always be an integer. Thus, the domain of this function is real numbers (ℝ), while its range will be integers (ℤ). Greatest Integer Function Properties The greatest integer function has numerous properties. Some of the important properties are listed below. ⌊x+n⌋ = ⌊x⌋+n, where, (n \in \mathbb{Z}) ⌊−x⌋ = (\begin{cases} & {-\left\lfloor x\right\rfloor}, & \text{if } x \in \mathbb{Z} \ &{-\left\lfloor x \right\rfloor-1}, & \text{if } x \notin \mathbb{Z} \ \end{cases}) If ⌊f(x)⌋ ≥ L, then f(x) ≥ L If ⌊f(x)⌋ ≤ L, then f(x) < L + 1. Greatest Integer Function and Smallest Integer Function Let us understand the smallest integer function along with seeing how it is different from the greatest integer function. \| Greatest Integer Function \| Smallest Integer Function \| --- \| \| It is denoted by ⌊x⌋. \| It is denoted by ⌈x⌉. \| \| If the input is x, then the output is the immediate integer that is less than or equal to x. \| If the input is x, then the output is the immediate integer that is greater than or equal to x. \| \| It is also known as the floor function. \| It is also known as the ceiling function. \| \| Examples: ⌊5.3⌋ = 5; ⌊-5.3⌋ = -6 \| Examples: ⌈5.3⌉ = 6; ⌈-5.3⌉ = -5 \| Important Notes on Greatest Integer Function: The following points are helpful to summarize the important points of the greatest integer function. If x is a number between successive integers n and n+1, then ⌊x⌋=n. If x is an integer, then ⌊x⌋=x The domain of the greatest integer function is ℝ and its range is ℤ. The graph of the greatest integer function contains steps made of horizontal lines with one end as an open dot and the other end as a solid dot. ☛Related Topics: The following links are related to the greatest integer function Graphing Functions Types of Functions Domain and Range of a Function Floor And Ceiling Function Greatest Integer Function Examples Example 1: What is the domain of the given greatest integer function: f(x)=1/⌊x⌋? Solution: The denominator should not be 0, that is, ⌊x⌋≠0. The greatest integer part of a number is 0 if that number lies in the interval [0,1). Thus, to obtain the domain, this interval must be excluded from the set of real numbers. This means that the domain of f is R−[0,1). Answer: Domain = R−[0,1) 2. Example 2: Find the value of x such that ⌊x+1⌋ = 3. Solution: From the definition of the greatest integer function, we have 3 ≤ x+1 < 4. Subtract 1 in this inequality. We get, 2 ≤ x < 3 Answer: x can take the values greater than or equal to 2 and less than 3. 3. Example 3: Evaluate the following: (i) [13.01] (ii) [13.99] (ii) [-2.4]. Solution: The greatest integer function value for the above cases are as given below, (i) [13.01] = 13 (ii) [13.99] = 13 (iii) [-2.4] = -3 Answer: (i) 13 (ii) 13 (iii) -3. Show Answer > go to slidego to slidego to slide Breakdown tough concepts through simple visuals. Math will no longer be a tough subject, especially when you understand the concepts through visualizations. Book a Free Trial Class Practice Questions on Greatest Integer Function Check Answers > go to slidego to slide FAQs on Greatest Integer Function What is the Meaning of Greatest Integer Function? The greatest integer function of x is a function that gives the largest integer which is less than or equal to x. This function is denoted by ⌊x⌋. The given number needs to be rounded off to the nearest integer that is less than or equal to the number itself. How to Calculate the Greatest Integer Function of a Negative Number? Let us find the greatest integer function of -3.2. i.e., ⌊-3.2⌋. Don't think that this is equal to -3. By the definition, we have ⌊x⌋ = n, where n ≤ x < n + 1. So think about which integer comes on the immediate left side of -3.2, and the answer is -4. Therefore, ⌊-3.2⌋ = -4. What is the Other Name of the Greatest Integer Function? The greatest integer function is also called a floor function. Here the function takes the lowest integral value on rounding the function value and hence it is called a floor function. How to Graph Greatest Integer Function? Plotting the graph of the greatest integer function is easy. It is a step curve. Here, f(x) = ⌊x⌋, if x is an integer, then the value of f will be x itself and if x is a non-integer, then the value of x will be the integer just before x. Hence for an integer n, all the numbers of [n, n+1) will have the value of the greatest integer function as n. So for every pair of n and n + 1 on the x-axis, we get a small piece of horizontal line corresponding to y = n. But note that every horizontal line has a closed dot on the left and open dot on the right side. What is Smallest Integer Function? The smallest integer function of x is the nearest integer that is greater than or equal to the given number. It is denoted by ⌈x⌉. Here are some examples of smallest integer function's usage. ⌈4.2⌉ = 5 ⌈-2.5⌉ = -2 ⌈0⌉ = 0 To find the smallest integer function value of a number, just put the number on the number line and pick the very next integer that lies on the right side of the number. Why is the Greatest Integer Function Not Differentiable? As we check the graph of the greatest integer function, we can see that it is jumping whenever it reaches an integer. Since the curve is discontinuous at integers, it is not differentiable at those points. Therefore at each integer, the greatest integer function is not differentiable. Is the Floor Function Differentiable? The floor function or the greatest integer function is not differentiable at integers. The floor function has jumping values at integers, so its curve is known as the step curve. The curve of floor function is discontinuous at integers and hence not differentiable at integers. How to Find Greatest Integer Function of a Number? The easiest way to find the greatest integer function of a number is, just to plot it on the number line and select the first integer that comes on its left side. For example, ⌊4.15⌋ = 4, ⌊-4.15⌋ = -5. What is the Domain and Range of the Greatest Integer Function? The input of the greatest integer function can be any real number whereas the greatest integer function's output is always an integer. Thus, the domain of this function is real numbers (ℝ), and its range is integers (ℤ). 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16583	https://pmc.ncbi.nlm.nih.gov/articles/PMC6886230/	Kidney and uro-trauma: WSES-AAST guidelines - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Learn more: PMC Disclaimer \| PMC Copyright Notice World J Emerg Surg . 2019 Dec 2;14:54. doi: 10.1186/s13017-019-0274-x Search in PMC Search in PubMed View in NLM Catalog Add to search Kidney and uro-trauma: WSES-AAST guidelines Federico Coccolini Federico Coccolini 1 General, Emergency and Trauma Surgery, Pisa University Hospital, Via Paradisia, 56124 Pisa, Italy Find articles by Federico Coccolini 1,✉, Ernest E Moore Ernest E Moore 2 Trauma Surgery, Denver Health, Denver, CO USA Find articles by Ernest E Moore 2, Yoram Kluger Yoram Kluger 3 Division of General Surgery Rambam Health Care Campus, Haifa, Israel Find articles by Yoram Kluger 3, Walter Biffl Walter Biffl 4 Trauma Surgery Dept., Scripps Memorial Hospital, La Jolla, California USA Find articles by Walter Biffl 4, Ari Leppaniemi Ari Leppaniemi 5 General Surgery Dept., Mehilati Hospital, Helsinki, Finland Find articles by Ari Leppaniemi 5, Yosuke Matsumura Yosuke Matsumura 6 Department of Emergency and Critical Care Medicine, Chiba University Hospital, Chiba, Japan Find articles by Yosuke Matsumura 6, Fernando Kim Fernando Kim 7 Urology Department, University of Colorado, Denver, USA Find articles by Fernando Kim 7, Andrew B Peitzman Andrew B Peitzman 8 Surgery Department, University of Pittsburgh, Pittsburgh, PA USA Find articles by Andrew B Peitzman 8, Gustavo P Fraga Gustavo P Fraga 9 Trauma/Acute Care Surgery & Surgical Critical Care, University of Campinas, Campinas, Brazil Find articles by Gustavo P Fraga 9, Massimo Sartelli Massimo Sartelli 10 General and Emergency Surgery, Macerata Hospital, Macerata, Italy Find articles by Massimo Sartelli 10, Luca Ansaloni Luca Ansaloni 11 General, Emergency and Trauma Surgery Department, Bufalini Hospital, Cesena, Italy Find articles by Luca Ansaloni 11, Goran Augustin Goran Augustin 12 Department of Surgery, Zagreb University Hospital Centre and School of Medicine, University of Zagreb, Zagreb, Croatia Find articles by Goran Augustin 12, Andrew Kirkpatrick Andrew Kirkpatrick 13 General, Acute Care, Abdominal Wall Reconstruction, and Trauma Surgery, Foothills Medical Centre, Calgary, Alberta Canada Find articles by Andrew Kirkpatrick 13, Fikri Abu-Zidan Fikri Abu-Zidan 14 Department of Surgery, College of Medicine and Health Sciences, UAE University, Al-Ain, United Arab Emirates Find articles by Fikri Abu-Zidan 14, Imitiaz Wani Imitiaz Wani 15 Department of Surgery, DHS Hospitals, Srinagar, Kashmir India Find articles by Imitiaz Wani 15, Dieter Weber Dieter Weber 16 Department of General Surgery, Royal Perth Hospital, Perth, Australia Find articles by Dieter Weber 16, Emmanouil Pikoulis Emmanouil Pikoulis 17 3rd Department of Surgery, Attiko Hospital, National & Kapodistrian University of Athens, Athens, Greece Find articles by Emmanouil Pikoulis 17, Martha Larrea Martha Larrea 18 General Surgery, “General Calixto García”, Habana Medicine University, Havana, Cuba Find articles by Martha Larrea 18, Catherine Arvieux Catherine Arvieux 19 Clin. Univ. de Chirurgie Digestive et de l’Urgence, CHUGA-CHU Grenoble Alpes UGA-Université Grenoble Alpes, Grenoble, France Find articles by Catherine Arvieux 19, Vassil Manchev Vassil Manchev 20 General and Trauma Surgery Department, Pietermaritzburg Hospital, Pietermaritzburg, South Africa Find articles by Vassil Manchev 20, Viktor Reva Viktor Reva 21 General and Emergency Surgery, Sergei Kirov Military Academy, Saint Petersburg, Russia Find articles by Viktor Reva 21, Raul Coimbra Raul Coimbra 22 Department of General Surgery, Riverside University Health System Medical Center, Moreno Valley, CA USA Find articles by Raul Coimbra 22, Vladimir Khokha Vladimir Khokha 23 General Surgery Department, Mozir City Hospital, Mozir, Belarus Find articles by Vladimir Khokha 23, Alain Chichom Mefire Alain Chichom Mefire 24 Department of Surgery and Obstetrics and Gynecology, University of Buea, Buea, Cameroon Find articles by Alain Chichom Mefire 24, Carlos Ordonez Carlos Ordonez 25 Trauma and Acute Care Surgery, Fundacion Valle del Lili, Cali, Colombia Find articles by Carlos Ordonez 25, Massimo Chiarugi Massimo Chiarugi 1 General, Emergency and Trauma Surgery, Pisa University Hospital, Via Paradisia, 56124 Pisa, Italy Find articles by Massimo Chiarugi 1, Fernando Machado Fernando Machado 26 General and Emergency Surgery Department, Montevideo Hospital, Montevideo, Paraguay Find articles by Fernando Machado 26, Boris Sakakushev Boris Sakakushev 27 General Surgery Department, Medical University, University Hospital St George, Plovdiv, Bulgaria Find articles by Boris Sakakushev 27, Junichi Matsumoto Junichi Matsumoto 28 Department of Emergency and Critical Care Medicine, Saint-Marianna University School of Medicine, Kawasaki, Japan Find articles by Junichi Matsumoto 28, Ron Maier Ron Maier 29 Department of Surgery, Harborview Medical Centre, Seattle, USA Find articles by Ron Maier 29, Isidoro di Carlo Isidoro di Carlo 30 Department of Surgical Sciences and Advanced Technologies “GF Ingrassia”, Cannizzaro Hospital, University of Catania, Catania, Italy Find articles by Isidoro di Carlo 30, Fausto Catena Fausto Catena 31 Emergency and Trauma Surgery, Maggiore Hospital, Parma, Italy Find articles by Fausto Catena 31; WSES-AAST Expert Panel Author information Article notes Copyright and License information 1 General, Emergency and Trauma Surgery, Pisa University Hospital, Via Paradisia, 56124 Pisa, Italy 2 Trauma Surgery, Denver Health, Denver, CO USA 3 Division of General Surgery Rambam Health Care Campus, Haifa, Israel 4 Trauma Surgery Dept., Scripps Memorial Hospital, La Jolla, California USA 5 General Surgery Dept., Mehilati Hospital, Helsinki, Finland 6 Department of Emergency and Critical Care Medicine, Chiba University Hospital, Chiba, Japan 7 Urology Department, University of Colorado, Denver, USA 8 Surgery Department, University of Pittsburgh, Pittsburgh, PA USA 9 Trauma/Acute Care Surgery & Surgical Critical Care, University of Campinas, Campinas, Brazil 10 General and Emergency Surgery, Macerata Hospital, Macerata, Italy 11 General, Emergency and Trauma Surgery Department, Bufalini Hospital, Cesena, Italy 12 Department of Surgery, Zagreb University Hospital Centre and School of Medicine, University of Zagreb, Zagreb, Croatia 13 General, Acute Care, Abdominal Wall Reconstruction, and Trauma Surgery, Foothills Medical Centre, Calgary, Alberta Canada 14 Department of Surgery, College of Medicine and Health Sciences, UAE University, Al-Ain, United Arab Emirates 15 Department of Surgery, DHS Hospitals, Srinagar, Kashmir India 16 Department of General Surgery, Royal Perth Hospital, Perth, Australia 17 3rd Department of Surgery, Attiko Hospital, National & Kapodistrian University of Athens, Athens, Greece 18 General Surgery, “General Calixto García”, Habana Medicine University, Havana, Cuba 19 Clin. Univ. de Chirurgie Digestive et de l’Urgence, CHUGA-CHU Grenoble Alpes UGA-Université Grenoble Alpes, Grenoble, France 20 General and Trauma Surgery Department, Pietermaritzburg Hospital, Pietermaritzburg, South Africa 21 General and Emergency Surgery, Sergei Kirov Military Academy, Saint Petersburg, Russia 22 Department of General Surgery, Riverside University Health System Medical Center, Moreno Valley, CA USA 23 General Surgery Department, Mozir City Hospital, Mozir, Belarus 24 Department of Surgery and Obstetrics and Gynecology, University of Buea, Buea, Cameroon 25 Trauma and Acute Care Surgery, Fundacion Valle del Lili, Cali, Colombia 26 General and Emergency Surgery Department, Montevideo Hospital, Montevideo, Paraguay 27 General Surgery Department, Medical University, University Hospital St George, Plovdiv, Bulgaria 28 Department of Emergency and Critical Care Medicine, Saint-Marianna University School of Medicine, Kawasaki, Japan 29 Department of Surgery, Harborview Medical Centre, Seattle, USA 30 Department of Surgical Sciences and Advanced Technologies “GF Ingrassia”, Cannizzaro Hospital, University of Catania, Catania, Italy 31 Emergency and Trauma Surgery, Maggiore Hospital, Parma, Italy ✉ Corresponding author. Received 2019 Sep 10; Accepted 2019 Oct 23; Collection date 2019. © The Author(s). 2019 Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. The Creative Commons Public Domain Dedication waiver ( applies to the data made available in this article, unless otherwise stated. PMC Copyright notice PMCID: PMC6886230 PMID: 31827593 Abstract Renal and urogenital injuries occur in approximately 10-20% of abdominal trauma in adults and children. Optimal management should take into consideration the anatomic injury, the hemodynamic status, and the associated injuries. The management of urogenital trauma aims to restore homeostasis and normal physiology especially in pediatric patients where non-operative management is considered the gold standard. As with all traumatic conditions, the management of urogenital trauma should be multidisciplinary including urologists, interventional radiologists, and trauma surgeons, as well as emergency and ICU physicians. The aim of this paper is to present the World Society of Emergency Surgery (WSES) and the American Association for the Surgery of Trauma (AAST) kidney and urogenital trauma management guidelines. Keywords: Kidney, Urogenital, Urethra, Ureter, Bladder, Trauma, Adult, Pediatric, Classification, Guidelines, Embolization, Surgery, Operative, Non-operative, Conservative, Stenting, Urological, Endovascular trauma management, Flow chart Background In both, adult and children cohorts, urogenital trauma has a cumulative incidence of 10-20%, and the kidney is involved in 65–90% of the time [1–3]. Males are involved 3 times more than females (both in adults and children) [2, 4]. As in other abdominal injuries, the use of non-operative management (NOM) has significantly increased in last decades, particularly due to the introduction of hybrid rooms and endovascular trauma and bleeding management (EVTM) associated with modern urological mini-invasive procedures [5, 6]. Moreover, In pediatric patients, NOM should be the first option as soon as it is viable and safe. However, operative management (OM) remains the gold standard in unstable patients, after failure of NOM (fNOM), and in many injuries caused by penetrating mechanisms; in fact, in gunshot and stab wounds, OM is applied in 75% and 50% of cases, respectively . As for the other abdominopelvic lesion management, decisions should be based on physiology, anatomy, and associated injuries [6–9]. Another important consideration relates to the different management approach to kidney and urological trauma urologists and trauma surgeons . Urologic guidelines tend in general to focus more on organ preservation, whereas trauma surgeons tend to consider the stabilization of physiology more importantly than organ preservation . Despite this different point of view, an integrated approach and active collaboration between the two specialties forms the basis to achieve optimal management and the best outcomes . This is particularly true for urogenital and urinary tract injuries in which the multidisciplinary approach is the cornerstone to improve short- and long-term outcomes. Notes on the use of the guidelines The guidelines are evidence-based, with the grade of recommendation based on the evidence. The guidelines present the diagnostic and therapeutic methods for optimal management of urogenital trauma. The practice guidelines promulgated in this work do not represent a standard of practice. They are suggested plans of care, based on the best available evidence and the consensus of experts, but they do not exclude other approaches as being within the standard of practice. For example, they should not be used to compel adherence to a given method of medical management, which method should be finally determined after taking account of the conditions at the relevant medical institution (staff levels, experience, equipment, etc.) and the characteristics of the individual patient. However, responsibility for the results of treatment rests with those who are directly engaged therein, and not with the consensus group. Methods A computerized search was done by the bibliographer in different databanks (MEDLINE, Scopus, EMBASE) and citations were included for the period between January 1990 and August 2018 using the primary search strategy: kidney, injuries, trauma, urogenital, adult, pediatric, hemodynamic instability/stability, angioembolization, management, nonoperative, conservative, operative, surgery, diagnosis, follow-up, combined with AND/OR. No search restrictions were imposed. The dates were selected to allow comprehensive published abstracts of clinical trials, consensus conference, comparative studies, congresses, guidelines, government publication, multicenter studies, systematic reviews, meta-analysis, large case series, original articles, and randomized controlled trials. Case reports and small case series were excluded. Narrative review articles were also analyzed to determine if other cited studies should be included. The literature selection is reported in the flow chart (Fig.1). Fig. 1. Open in a new tab PRISMA flow chart The level of evidence (LE) was evaluated using the GRADE system (Table1). Table 1. GRADE system to evaluate the level of evidence and recommendation \| Grade of recommendation \| Clarity of risk/benefit \| Quality of supporting evidence \| Implications \| --- --- \| \| 1A \| \| Strong recommendation, high-quality evidence \| Benefits clearly outweigh risk and burdens, or vice versa \| RCTs without important limitations or overwhelming evidence from observational studies \| Strong recommendation, applies to most patients in most circumstances without reservation \| \| 1B \| \| Strong recommendation, moderate-quality evidence \| Benefits clearly outweigh risk and burdens, or vice versa \| RCTs with important limitations (inconsistent results, methodological flaws, indirect analyses or imprecise conclusions) or exceptionally strong evidence from observational studies \| Strong recommendation, applies to most patients in most circumstances without reservation \| \| 1C \| \| Strong recommendation, low-quality or very low-quality evidence \| Benefits clearly outweigh risk and burdens, or vice versa \| Observational studies or case series \| Strong recommendation but subject to change when higher quality evidence becomes available \| \| 2A \| \| Weak recommendation, high-quality evidence \| Benefits closely balanced with risks and burden \| RCTs without important limitations or overwhelming evidence from observational studies \| Weak recommendation, best action may differ depending on the patient, treatment circumstances, or social values \| \| 2B \| \| Weak recommendation, moderate-quality evidence \| Benefits closely balanced with risks and burden \| RCTs with important limitations (inconsistent results, methodological flaws, indirect or imprecise) or exceptionally strong evidence from observational studies \| Weak recommendation, best action may differ depending on the patient, treatment circumstances, or social values \| \| 2C \| \| Weak recommendation, Low-quality or very low-quality evidence \| Uncertainty in the estimates of benefits, risks, and burden; benefits, risk, and burden may be closely balanced \| Observational studies or case series \| Very weak recommendation; alternative treatments may be equally reasonable and merit consideration \| Open in a new tab A group of experts in the field coordinated by a central coordinator was contacted to express their evidence-based opinion on several issues about the pediatric (< 16 years old) and adult urogenital trauma [12, 13]. Urogenital trauma was assessed by the anatomy of the injury (kidney, urogenital tract, bladder), type of injury (blunt and penetrating injury), management (conservative and operative management), and type of patient (adults, pediatrics). Through the Delphi process, different issues were discussed in subsequent rounds. The central coordinator assembled the different answers derived from each round. Each version was then revised and improved. The definitive version was discussed during the WSES World Congress (in June 2019 in Njimengen, The Netherlands) by a combined expert group from both societies (WSES-AAST). The final version about which the agreement was reached resulted in the present manuscript. Statements are summarized in Table3. Table 3. Statements summary \| Statements \| \| Diagnostic procedures \| \| • Kidney \| - The choice of diagnostic method upon admission depends on the hemodynamic status of the patient. (GoR 1A) - E-FAST is effective and rapid to detect intra-abdominal free fluid. (GoR 1A) - E-FAST has low sensitivity and specificity in kidney trauma. (GoR 1B) - Contrast-enhanced CT scan associated with delayed urographic phase is the gold standard in hemodynamic stable or stabilized adults after blunt of penetrating trauma and in severely injured children when kidney or urinary tract injury is suspected. (GoR 1A) - In blunt trauma, contrast-enhanced CT scan associated with delayed urographic phase must be performed in cases of macro- or micro- hematuria with hypotension and after high-energy deceleration trauma regardless of the presence of hematuria. (GoR 2B) - In penetrating trauma, contrast-enhanced CT scan associated with delayed urographic phase is indicated in all hemodynamic stable or stabilized patients. (GoR 1B) - Pediatric patients with high energy/penetrating/decelerating trauma and/or in cases of drop in hematocrit associated with any degree of hematuria should undergo contrast-enhanced CT-scan with delayed urographic phase. (GoR 2A) - Ultrasound, contrast-enhanced US and eco-Doppler (E-FAST excluded) are generally not recommended as diagnostic tools during the initial evaluation of adult patients with high-energy trauma when multiple injuries and/or injury to the urinary tract and collecting system are suspected. (GoR 1C) - Ultrasound, contrast-enhanced US, and eco-doppler can be used in pregnant women and in the pediatric population as an alternative to CT-scan in the presence of hemodynamic stability during the immediate assessment and in follow-up evaluations. (GoR 1C) - In children with mild symptoms, minimal clinical findings, hematuria <50 RBCs/HPF and no other indications of CT-scanning, ultrasound and/or contrast-enhanced US and/or eco-doppler associated to blood test may be adopted for the initial evaluation. (GoR 2A) - Intravenous urography may be useful in unstable patients during surgery when a kidney injury is found intraoperatively or when CT-scanning is not available and a urinary tract injury is suspected. (GoR 2C ) \| \| • Ureter \| - Injury to the ureter should be suspected in high-energy blunt trauma, particularly in deceleration injuries with multi-system involvement and in all penetrating abdominal trauma. (GoR 1C). - Intravenous contrast-enhanced CT-scan with delayed phase should be performed in hemodynamically stable or stabilized patients if ureteral injury is suspected (GoR 1C) - Direct inspection of the ureter should be always performed during emergency laparotomy in patients with suspected ureteral injury. (GoR 1C) \| \| • Bladder \| - Retrograde cystography (conventional radiography or CT-scan) represents the diagnostic procedure of choice in bladder injuries. (GoR 1C) - Retrograde cystography should be always performed in hemodynamically stable or stabilized patients with suspected bladder injury. (GoR 1C) - Intravenous contrast-enhanced CT-scan with delayed phase is less sensitive and specific than retrograde cystography in detecting bladder injuries. (GoR 1B) - In pelvic bleeding amenable to angioembolization associated to suspected bladder injuries, cystography should be postponed until the completion of the angiographic procedure to avoid affecting the accuracy of angiography. (GoR 2A) - Direct inspection of the intraperitoneal bladder, whenever feasible, should always be performed during emergency laparotomy in patients with suspected bladder injury. Methylene blue or indigo carmine could be useful in intraoperative investigation. (GoR 1C) \| \| • Urethra \| - Patients with post-traumatic urethral hemorrhage should be investigated for urethral injuries. (GoR 1C) - During emergency laparotomy, if an urethral injury is suspected, it should be investigated directly whenever feasible. (GoR 2A) - Retrograde urethrography and selective urethroscopy represent the modalities of choice to investigate traumatic urethral injuries. (GoR 1B) - In the event of penile lesions, urethroscopy should be preferred to retrograde uretrography (GoR 2A) \| \| Management \| \| Kidney Non-operative management (NOM) \| - NOM should be the treatment of choice for all hemodynamical stable or stabilized minor (AAST I-II), moderate (AAST III) and severe (AAST IV-V) lesions. (GoR 1B) - Only in selected settings, with immediate availability of operating room, surgeons and adequate resuscitation, immediate access to blood, blood products and to high dependency / intensive care environment, and without other reasons for surgical exploration, NOM may be considered even in hemodynamically transient responder patients. (GoR 2C) - In deciding for NOM in hemodynamically stable or stabilized patients, accurate classification of the degree of injury and associated injuries with CT-scan with intravenous contrast and delayed urographic phases is mandatory. (GoR 2A) - NOM in penetrating lateral kidney injuries is feasible and effective but accurate patient selection is crucial even in the absence of other indications for laparotomy. In particular, cases without violation of the peritoneal cavity are more suitable for NOM. (GoR 2A) - Isolated urinary extravasation, in itself, is not an absolute contra-indication to NOM in absence of other indications for laparotomy. (GoR 1B) - In low resource settings, NOM could be considered in hemodynamically stable patients without evidence of associated injuries, with negative serial physical examinations and negative first level imaging and blood tests. (GoR 2C) \| \| Kidney Angiography and angioembolization \| - Angiography with eventual super-selective angioembolization is a safe and effective procedure; it may be indicated in hemodynamically stable or stabilized patients with arterial contrast extravasation, pseudoaneurysms, arteriovenous fistula, and non-self-limiting gross hematuria. (GoR 1C) - Angioembolization should be performed as selectively as possible. (GoR 1C) - Blind-angioembolization is not indicated in hemodynamically stable or stabilized patients with both kidneys when angiography is negative for active bleeding, regardless of arterial contrast extravasation on CT-scan. (GoR 1C) - In hemodynamically stable or stabilized patients with severe renal trauma with main renal artery injury, dissection or occlusion, angioembolization and/or percutaneous revascularization with stent or stentgraft is indicated in specialized centres and in patients with limited warm ischemia time (<240 min) (GoR 2C) - Endovascular selective balloon occlusion of the renal artery could be utilized as a bridge to definitive hemostasis. This procedure requires direct visualization by fluoroscopy where the balloon is advanced over a selectively placed guidewire. (GoR 2B) - In severe injury with main renal vein injury without self-limiting bleeding, angioembolization is not indicated. Patients should undergo surgical intervention. (GoR 1C) - In hemodynamically stable or stabilized patients with solitary kidney and moderate (AAST III) or severe (AAST IV-V) renal trauma with arterial contrast extravasation on CT-scan, angiography with eventual super-selective angioembolization should be considered as the first choice. (GoR 1C) - In hemodynamically stable or stabilized patients with active kidney bleeding at angiography and without other indications for surgical intervention, in case of failure of the initial angioembolization, a repeat angioembolization should be considered. (GoR 1C) - In adults, only in selected setting (immediate availability of operating room, surgeon, adequate resuscitation, immediate access to blood and blood products and to high dependency / intensive care environment) and without other reasons for surgical exploration, angioembolization might be considered in selected hemodynamically transient responder patients. (GoR 2C) - In children, angiography and eventual super-selective angioembolization should be the first choice even with active bleeding and labile hemodynamics, iof there is immediate availability of angiographic suite, immediate access to surgery and to blood and blood products, and to high dependency / intensive care environment. (GoR 2C) \| \| Kidney Operative management (OM) \| - Hemodynamically unstable and non-responder (WSES IV) patients should undergo OM. (GoR 2A) - Resuscitative Endovascular Balloon Occlusion of the Aorta (i.e., REBOA) may be used in hemodynamically unstable patients as a bridge to other more definitive procedures for hemorrhage control. (GoR 2B) - In cases of severe renal vascular injuries without self-limiting bleeding, OM is indicated. (GoR 1C) - The presence of non-viable tissue (devascularized kidney) is not an indication to OM in the acute setting in the absence of other indications for laparotomy. (GoR 2A) - Hemodynamic stable or stabilized patients having damage to the renal pelvis not amenable to endoscopic/percutaeous techniques/stent should be considered for delayed OM in absence of other indications for immediate laparotomy. (GoR 2B) \| \| Urinary tract injuries \| \| • Ureter \| - Contusions may require ureteral stenting when urine flow is impaired. (GoR 1C) - Partial lesions of the ureter should be initially treated conservatively with the use of a stent, with or without a diverting nephrostomy in the absence of other indications for laparotomy. (GoR 1C) - Partial and complete ureteral transections or avulsion not suitable for NOM may be treated with primary repair plus a double J stent or ureteral re-implant into the bladder in case of distal lesions (GoR 1C). - Ureteral injuries should be repaired operatively when discovered during laparotomy or in cases where conservative management has failed (GoR 1C) - Ureteral stenting should be attempted in cases of partial ureteral injuries diagnosed in a delayed fashion; if this approach fails, and/or in case of complete transection of the ureter, percutaneous nephrostomy with delayed surgical repair is indicated. (GoR 1C) - In any ureteral repair, stent placement is strongly recommended. (GoR 1C) \| \| • Bladder \| - Bladder contusion requires no specific treatment and might be observed clinically. (GoR 1C) - Intraperitoneal bladder rupture should be managed by surgical exploration and primary repair (GoR 1B) - Laparoscopy might be considered in repairing isolated intraperitoneal injuries in case of hemodynamic stability and no other indications for laparotomy. (GoR 2B) - In case of severe intraperitoneal bladder rupture, during damage control procedures, urinary diversion via bladder and perivesical drainage or external ureteral stenting may be used. (GoR 1C) - Uncomplicated blunt or penetrating extraperitoneal bladder injuries may be managed non-operatively, with urinary drainage via a urethral or suprapubic catheter in the absence of other indication for laparotomy. (GoR 1C) - Complex extra-peritoneal bladder ruptures—i.e., bladder neck injuries, lesions associated to pelvic ring fracture and/or vaginal or rectal injuries- should be explored and repaired. (GoR 1C) - Surgical repair of extraperitoneal bladder rupture should be considered during laparotomy for other indications and during surgical exploration of the prevesical space for orthopedic fixations. (GoR 1C) - In adult patients, urinary drainage with urethral catheter (without suprapubic catheter) after surgical management of bladder injuries is mandatory (GoR 1B); for pediatric patients suprapubic cystostomy is recommended (GoR 2C) \| \| • Urethra \| - Urinary drainage should be obtained as soon as possible in case of traumatic urethral injury. (GoR 1C) - Blunt anterior urethral injuries should be initially managed conservatively with urinary drainage (via urethral or suprapubic catheter); endoscopic treatment with realignment should be attempted before surgery. Delayed surgical repair should be considered in case of failure of conservative treatment after endoscopic approach. (GoR 1C) - Partial blunt injuries of the posterior urethra may be initially managed conservatively with urinary drainage (via urethral or suprapubic catheter) and endoscopic realignment; definitive surgical management should be delayed for 14 days if no other indications for laparotomy exist. (GoR 1C) - Injuries of the posterior urethra in cases of hemodynamic instability should be approached by immediate urinary drainage and delayed treatment. (GoR 1C) - Conservative treatment of penetrating urethral injuries is generally not recommended. (GoR 1C) - Penetrating injuries of anterior urethra should be treated with immediate direct surgical repair if the clinical conditions allow and if an experienced surgeon is available; otherwise, urinary drainage should be performed and delayed treatment planned. (GoR 1C) - Penetrating injuries of the posterior urethra should be treated with primary repair only if the clinical conditions allow. Otherwise, urinary drainage and delayed urethroplasty is recommended. (GoR 1C) - When posterior urethral injury is associated with complex pelvic fracture, definitive surgical treatment with urethroplasty should be performed after the healing of pelvic ring injury. (GoR 1C ) \| \| Short- and long-term follow-up \| \| Kidney and urinary tract \| - Follow-up imaging is not required for minor (AAST I-II) renal injuries managed non-operatively. (GoR 2B) - In moderate (AAST III) and severe (AAST IV-V) renal injuries, the need for follow-up imaging is driven by the patients’ clinical conditions. (GoR 2B) - In severe injuries (AAST IV-V), contrast-enhanced CT scan with excretory phase (in cases with possible or documented urinary extravasation) or ultrasound and contrast-enhanced US are suggested within the first 48 h after trauma in adult patients and in delayed follow-up. (GoR 2A) - Follow-up imaging in pediatric patients should be limited to moderate (AAST III) and severe (AAST IV-V) injuries. (GoR 2B) - In pediatric patients, ultrasound and contrast-enhanced US should be the first choice in the early and delayed follow-up phases. If cross-sectional imaging is required, magnetic resonance should be preferred. (GoR 2B) - CT-scan with delayed phase imaging is the method of choice for the follow-up of ureteral and bladder injuries. (GoR 2A) - Uretroscopy or uretrogram are the methods of choice for the follow-up of urethral injuries. (GoR 2A) - Return to sport activities should be allowed only after microscopic hematuria is resolved. (GoR 2B) \| Open in a new tab Definitions In adult patients, hemodynamic instability is considered the condition in which admission systolic blood pressure upon admission is < 90 mmHg with evidence of skin vasoconstriction (cool, clammy, decreased capillary refill), altered level of consciousness and/or shortness of breath, or > 90 mmHg but requiring bolus infusions/transfusions and/or vasopressor drugs and/or admission base excess (BE) > − 5 mmol/l and/or shock index > 1 and/or transfusion requirement of at least 4–6 Units of packed red blood cells within the first 24 h. Transient responder patients (adult and pediatric) are those showing an initial response to adequate fluid resuscitation, but then subsequent signs of ongoing blood loss and perfusion deficits. These patients have an initial response to therapy but do not reach sufficient stabilization to undergo interventional radiology procedures or NOM. In pediatric patients, hemodynamic stability is considered a systolic blood pressure of 90 mmHg plus twice the child’s age in years (the lower limit is inferior to 70 mmHg plus twice the child’s age in years, or inferior to 50 mmHg in some studies). An acceptable hemodynamic status in children is considered a positive response to fluid resuscitation: 3 boluses of 20 mL/kg of crystalloid replacement should be administered before blood replacement leading to heart rate reduction, cleared sensorium, return of peripheral pulses, normal skin color, increase in blood pressure and urinary output, and an increase in warmth of the skin in the extremities. Clinical judgment however is fundamental in evaluating children. WSES classification The WSES Classification (Table2) divides kidney injuries into four classes considering the AAST-OIS classification (Fig.2) and the hemodynamic status (Table3): Minor (WSES class I) Moderate (WSES class II) Severe (WSES class III and IV) Table 2. WSES kidney trauma classification \| \| WSES grade \| AAST \| Hemodynamic \| --- --- \| \| Minor \| WSES grade I \| I–II \| Stable \| \| Moderate \| WSES grade II \| III or segmental vascular injuries \| Stable \| \| Severe \| WSES grade III \| IV–V or any grade parenchymal lesion with main vessels dissection/occlusion \| Stable \| \| WSES grade IV \| Any \| Unstable \| Open in a new tab Fig. 2. Open in a new tab AAST organ injury scale for kidney trauma Minor kidney injuries: WSES class I includes hemodynamically stable AAST-OIS grade I–II blunt and penetrating lesions. Moderate kidney injuries: WSES class II includes hemodynamically stable AAST-OIS grade III blunt and penetrating lesions. Severe kidney injuries: WSES class III includes hemodynamically stable AAST-OIS grade IV–V blunt and penetrating lesions and any grade parenchymal lesion with arterial dissection/occlusion. WSES class IV includes hemodynamically unstable AAST-OIS grade I–V blunt and penetrating lesions Based on the present classification, WSES and AAST suggest a management algorithm for kidney injury shown in Fig.3 and for urogenital tract injuries in Fig.4. Fig. 3. Open in a new tab Kidney trauma management algorithm Fig. 4. Open in a new tab Uro-trauma management algorithm Patient stratification During the initial evaluation the hemodynamic status, mechanism of injury, presence of associated injuries, and anamnestic data must be considered (i.e., previous renal injuries, previous renal surgery, congenital single or pathologic kidneys or diseases), especially in children. In adults, the clinical examination in urogenital trauma should consider the presence of hematuria, flank/abdominal pain/contusion, rib fractures, and mechanism of trauma. Special attention should be given to pelvic trauma in which urethral injuries can be frequently missed but should ideally be diagnosed in the first hours . Macro or micro-hematuria is frequently present (88-94%) in cases of renal/urogenital trauma but it does not predict the grade of injury [13, 14]. Macro-hematuria is more frequently associated with major renal injuries; however, in 10–25% of high-grade kidney injury hematuria is, the same being observed in 24–50% of ureteropelvic junction and renal hilum injuries [13, 15]. In 0.1–0.5% of the patients, hemodynamic stability and micro-hematuria exist in the presence of a significant urinary tract injury [5, 16–18]. In children, the kidney is commonly injured following blunt trauma because of many anatomical reasons: less perirenal fat, thinner abdominal muscles, lack of ossification of the rib cage, larger kidney size, and fetal kidney lobulations, making them more vulnerable to injury [2, 3, 19–23]. However, even in the pediatric population, there is no clear correlation between the presence and type of hematuria and the degree of kidney injury (36–40% of renal injuries and in up to 24% of renal artery occlusions hematuria is absent) [22, 24]. However, while micro-hematuria (< 50 red blood cells (RBC) per high-power field (HPF)) is frequent in children due to the kidney anatomy and the presence of undiagnosed kidney diseases (1–36%), macro-hematuria seems to be more related to major renal injuries [22, 24]. The general suggestion is to perform imaging investigation in all those patients with blunt trauma with > 50 RBCs/HPF [13, 22]. In order to refine the use of CT scan in children, however, other factors should be considered (i.e., mechanism of injury and its energy/degree of deceleration associated with physical findings such as hypotension, flank hematoma and ecchymosis, rib fractures, cutaneous signs in the abdomen, and a drop in hematocrit associated with any degree of hematuria) [3, 13, 14, 17, 19–22, 24–34]. On the other hand, in children with minimal symptoms and/or clinical findings and < 50 RBCs/HPF, ultrasound (US), contrast-enhanced ultrasound (CEUS), Eco-Doppler, and clinical and blood test monitoring may be sufficient for the initial evaluation . In penetrating injuries, the presence of hematuria does not correlate with the grade of kidney injury. However, penetrating injuries are commonly associated with other intra-abdominal injuries [18, 26, 31, 35], therefore, independently from the degree of hematuria, all hemodynamically stable patients should be imaged following a penetrating mechanism of injury . Pathophysiology of injury Kidney The most common mechanism of injury involving the kidney is blunt trauma associated frequently to high-velocity deceleration (90% of cases); whereas penetrating trauma (gunshot and stab wounds occur in 1.4–3.3% [5, 16–18, 36]. However, these incidences depend on the geographic area of the world . The kidney is well protected in the retroperitoneum; however, it is particularly vulnerable to blunt trauma accompanied by rapid deceleration because the kidney is fixed only by the renal pelvis in the uretero-pelvic junction and by the vascular pedicle. In adults, the most frequent blunt mechanisms are falls from height, assault, skiing accidents, and road traffic–related injuries. In children, sports injuries such as skiing, snowboarding, horse riding, and bicycle and motorcycle accidents are the most frequent [2, 3, 14, 21, 23, 38–40]. In the pediatric population, isolated blunt injuries are more frequent and occur after 5 years of age, while penetrating injuries usually increase after 14 years of age [2, 4]. Penetrating trauma can affect the kidneys especially when the superior abdomen is involved . Isolated penetrating kidney injuries are rare and renal vascular injuries are more frequent than in blunt trauma [1, 22]. The majority of renal injuries (up to 90%) are minor both in adults and children and involve the parenchyma or segmental vessels [2, 16]. A unique and uncommon type of injury is the isolated renal arterial transection or intimal disruption which occurs particularly in cases of rapid deceleration . Ureter Traumatic ureteral lesions are rare (less than 1%) [41, 42]. The most common cause of ureteral injury is penetrating trauma, especially gunshot wounds [43–46]; only 1/3 of cases are caused by blunt trauma . As opposed to stab wounds, gunshot wounds can produce a blast effect even at a distance of 2 cm from the bullet path [41, 48]. In blunt trauma, ureteral injuries commonly happen at the ureteropelvic junction, especially in children and in high energy deceleration injuries [41, 44, 45, 48, 49]. Associated organ injuries are common in case of ureteral lesions [42, 45, 50]. The clinical presentation of ureteral injuries might be subtle but isolated hematuria is a common finding. Bladder Bladder injury is more frequent following blunt than penetrating trauma (65–86% vs. 14–35%) [51–53]. In particular, bladder injury is present in 3.6% of abdominal gunshot injuries and 20% of penetrating buttock injuries [41, 48, 54]. Due to the high energy necessary to damage the bladder, 60 to 90% of patients presenting with bladder injury have a pelvic bony fracture while 6–8% of patients with a pelvic fracture will have bladder injury [41, 48, 49, 54]. Pediatric patients are more susceptible to bladder injuries due to the children anatomy. However, bladder injuries in children are less associated with pelvic fractures than in adults . A Pelvic fracture with hematuria is associated to a bladder injury in 30% of cases [45, 49]. Associated prostate-urethral injuries and rupture of the bladder occur in 10–29% male patients . Bladder injuries are mainly of four types: intra-peritoneal bladder rupture (IBR), extra-peritoneal bladder rupture (EBR), bladder contusion and bladder neck avulsion. IBR occurs in 15–25% of cases [41, 45, 48, 49]. EBR is the most common and is found in 60–90% of patients, and it is more frequently associated with pelvic fractures . Combined Bladder Rupture (CBR), i.e., a combination of IBR and EBR, is found in 5–12% of cases [41, 48, 56]. EBR can be further classified into simple EBR, where the urinary leak is limited to the extra-peritoneal pelvic region, and complex injuries where extravasated urine infiltrates the anterior abdominal wall, the scrotum, and the perineum . Urethra Urethral injuries are uncommon; they mostly affect male patients and are usually diagnosed following blunt trauma [45, 57]. Urethral injuries are divided into anterior (bulbar and penile urethra) and posterior injuries (proximal to the perineal membrane, at the prostatic or membranous urethra). The main cause of anterior urethral injury is direct blunt trauma [45, 48, 50]. Penetrating injuries to the anterior urethra are rare and are mainly caused by gunshot injuries [58, 59]. Injuries to the posterior urethra usually result from pelvic trauma, Pelvic fracture urethral injury (PFUI), present in 1.5–5% of anterior pelvic fractures [60, 61]. The risk of urethral injury increases by 10% for every 1-mm increase in pubic symphysis diastasis . Posterior urethral injuries may be classified as complete (65% of lesions) or incomplete (35% of cases) . In complete injuries, a gap is present between the two injured stumps of the urethra. Penetrating injuries to the posterior urethra are extremely rare and are caused mainly by gunshot wounds; the risk of associated intra-abdominal lesions is high . The Goldman classification of urethral injuries includes five types of lesions aimed at discerning anterior from posterior and complete from incomplete and at determining whether posterior urethral injuries involve the bladder neck or the rectal wall. Associated urethral and bladder injuries are found in up to 20% of cases . Female urethral injuries are uncommon and are often caused by pelvic injuries and are usually associated with rectal and vaginal injuries [67, 68]. Diagnostic procedures There are no specific recommendations regarding the diagnosis of urogenital injuries in children. Therefore, pediatric patients should be investigated as adults considering the need to reduce, as much as possible, the exposure to ionizing radiation. Kidney The choice of diagnostic method upon admission depends on the hemodynamic status of the patient. (GoR 1A) E-FAST is effective and rapid to detect intra-abdominal free fluid. (GoR 1A) E-FAST has low sensitivity and specificity in kidney trauma. (GoR 1B) Contrast-enhanced CT scan associated with delayed urographic phase is the gold standard in hemodynamic stable or stabilized adults after blunt of penetrating trauma and in severely injured children when kidney or urinary tract injury is suspected. (GoR 1A) In blunt trauma, contrast-enhanced CT scan associated with delayed urographic phase must be performed in cases of macro- or micro-hematuria with hypotension and after high-energy deceleration trauma regardless of the presence of hematuria. (GoR 2B) In penetrating trauma, contrast-enhanced CT scan associated with delayed urographic phase is indicated in all hemodynamic stable or stabilized patients. (GoR 1B) Pediatric patients with high energy/penetrating/decelerating trauma and/or in cases of drop in hematocrit associated with any degree of hematuria should undergo contrast-enhanced CT-scan with delayed urographic phase. (GoR 2A) Ultrasound, contrast-enhanced US and eco-Doppler (E-FAST excluded) are generally not recommended as diagnostic tools during the initial evaluation of adult patients with high-energy trauma when multiple injuries and/or injury to the urinary tract and collecting system are suspected. (GoR 1C) Ultrasound, contrast-enhanced US, and eco-Doppler can be used in pregnant women and in the pediatric population as an alternative to CT scan in the presence of hemodynamic stability during the immediate assessment and in follow-up evaluations. (GoR 1C) In children with mild symptoms, minimal clinical findings, hematuria <50 RBCs/HPF and no other indications of CT-scanning, ultrasound and/or contrast-enhanced US and/or eco-doppler associated to blood test may be adopted for the initial evaluation. (GoR 2A) Intravenous urography may be useful in unstable patients during surgery when a kidney injury is found intraoperatively or when CT-scanning is not available and a urinary tract injury is suspected. (GoR 2C) Extended-focused abdominal sonography for trauma (E-FAST), Ultrasonography, and Doppler-US (DUS) are useful and reliable noninvasive methods in trauma in general [69–71], however for the assessment of the kidney, due to anatomical reasons, these modalities may underestimate injuries (up to 30%) with a sensitivity and specificity of 22–67% and 96–100%, respectively [5, 14, 16, 17, 72–76]. In particular, vascular injuries are difficult to detect even using DUS . In children, these are the methods of choice during follow-up excluding patients requiring CT-scan examination for other associated injuries [27, 77]. Usually, US/DUS can be safely used in the first 36–48 h reserving CT for selected cases or in cases of anomalies seen on US/DUS studies [22, 26, 77]. Contrast-enhanced US (CEUS) is not widely used [74, 78, 79]. Recent studies evaluated its use in abdominal trauma in the pediatric population and in fertile women as these methods seem to be effective in identifying extravasation, thrombosis, pseudoaneurysms (PSA), and post-traumautic arteriovenous fistulas [15, 80–86]. Contrast-enhanced US is thought to increase the accuracy of the E-FAST (above 80%) in stable patients in whom renal injuries are suspected but with a negative FAST or in the presence of hematuria, severe abdominal trauma, fertile women, pediatric patients, and in immediate or middle/long-term follow-up [72, 74, 76, 79–81, 86–89]. Some authors suggest using CEUS in patients with moderate and severe injuries to identify bleeding and inject a hemostatic agent percutaneously [80, 87]. Innovative US techniques with real-time 3D-enhanced imaging are promising in detecting ongoing hemorrhage [16, 90]. CEUS is not recommended in cases of suspicion of injury to the urinary tract and collecting system . In these cases, contrast-enhanced CT-scan with late urographic phase is recommended. CT scan with intravenous contrast is considered the gold standard in blunt and penetrating trauma [14, 15, 17, 75, 91–95]. In renal and urogenital trauma, the arterial and venous phases (20–30 s and 70–80 s of delay in acquiring the images, respectively) allow identification of almost all injuries and the addition of a 5-min delayed phase (excretory phase) permits the identification of urinary extravasation [5, 13, 14, 16, 75, 96–99]. This delayed phase should be added selectively in case of suspicion of urogenital injuries. CT-scanning should always be considered in patients with associated severe brain injury and in any major injuries for the high probability of occurrence of associated injuries . Three-dimensional CT reconstructions help in injury classification [95, 101, 102]. the CT cystogram is a useful and viable tool and more accurate than plain X-ray cystography . CT-scanning allows the identification of patients with high-risk criteria for NOM failure such as contrast blush, perirenal hematoma > than 3.5 cm, medial laceration with significant medial urinary extravasation (posteromedial blush/medial renal laceration) and lack of contrast in the ureter, suggesting a complete ureteropelvic junction disruption. The association of moderate or severe injuries and at least 2 of these criteria lead to a high rate to NOM failure [16, 103]. Routinely repeating CT scanning after trauma or in the follow-up phase is not recommended. A repeat CT-scan should be reserved for those cases with evident or suspected complications or significant clinical changes in moderate and severe injuries [15, 17, 75, 104, 105]. In the pediatric population, CT scanning to evaluate kidney injuries remains the gold standard in hemodynamic stable or stabilized patients with penetrating trauma or in cases where abdominal injuries are suspected independently to the grade of hematuria, when urogenital injury is suspected [10, 13, 20, 21, 24, 26, 33, 34, 106, 107]. In general, hospital CT-scan protocols should be adjusted to the ALARA (as low as reasonable achievable) principles of exposure to ionizing radiation [24, 106]. Retrograde urethrography, excretory urethrography, and intravenous urography Intravenous urography (IVU) has been almost completely replaced by CT-scanning. However, it should be used in kidney injuries discovered during surgery in unstable patients, before opening the retroperitoneal hematoma. IVU can also be used when CT is not available or in low resource settings [3, 10, 13, 14, 18, 23, 36, 105, 108]. However, IVU is frequently used by urologists, more than by trauma surgeons . The IVU false negative rate ranges between 37 and 75% . The use of excretory urethrography has been reduced during the last decade in favor of contrast-enhanced CT-scan with delayed (excretory) phase . However, in perineal trauma and/or in trauma in which pielo-uretral injuries, ureteral injuries, and bladder injuries are suspected, it might be useful [5, 109]. Another affordable tool to evaluate the urethra, especially in the operating room or in low resource settings is retrograde urethrography. Documenting a normal urethra prior to urinary catheterization in cases with a high level of suspicion for urethral lesions is advisable. Magnetic resonance image MRI can be used to diagnose renal trauma in fertile/pregnant women, in pediatric patients, in cases of iodine allergy, in some cases when CT images are equivocal, and in the follow-up phase of urinary tract injuries [15, 85, 110–112]. Ureter Injury to the ureter should be suspected in high-energy blunt trauma, particularly in deceleration injuries with multi-system involvement and in all penetrating abdominal trauma. (GoR 1C). Intravenous contrast-enhanced CT-scan with delayed phase should be performed in hemodynamically stable or stabilized patients if ureteral injury is suspected (GoR 1C) Direct inspection of the ureter should be always performed during emergency laparotomy in patients with suspected ureteral injury. (GoR 1C) Perirenal stranding or hematomas, extravasation of contrast into the perirenal space, low-density retroperitoneal fluid around the genitourinary elements at imaging are indicative of ureteral injuries [49, 113]. Macro- and microscopic hematuria [114, 115] are not reliable signs of ureteral injury because its absence occurs in up to 25% of cases. A delay in the diagnosis may have a negative impact on outcomes [41, 113]. Ultrasound plays no role in the diagnosis of ureteral injury . At Ct-scan with delayed phase peri-ureteral hematoma, partial or complete obstruction of the lumen, mild distension of the ureter, hydronephrosis, delayed pyelogram, and the lack of contrast in the ureter distal to the injury, are all signs suggestive of ureteral injury . Urinary ascites or urinoma are considered subacute/chronic findings [44, 48]. A 10-minute delayed-phase CT-scan represents a valid diagnostic tool in the diagnosis of ureteral and ureteropelvic injuries [41, 113]. In case of unclear CT-scan results, an ascending urography represents the method of choice. IVU represents an unreliable test (false negatives up to 60%) [44, 114]. In case of emergency laparotomy, direct inspection of the ureter is indicated and it can be associated with the use of renally excreted intravenous dye (i.e., indigo carmine or methylene blue) . Single-shot IVU may be indicated intraoperatively. Bladder Retrograde cystography (conventional radiography or CT-scan) represents the diagnostic procedure of choice in bladder injuries. (GoR 1C) Retrograde cystography should be always performed in hemodynamically stable or stabilized patients with suspected bladder injury. (GoR 1C) Intravenous contrast-enhanced CT-scan with delayed phase is less sensitive and specific than retrograde cystography in detecting bladder injuries. (GoR 1B) In pelvic bleeding amenable to angioembolization associated with suspected bladder injuries, cystography should be postponed until the completion of the angiographic procedure to avoid affecting the accuracy of angiography. (GoR 2A) Direct inspection of the intraperitoneal bladder, whenever feasible, should always be performed during emergency laparotomy in patients with suspected bladder injury. Methylene blue or indigo carmine could be useful in intraoperative investigation. (GoR 1C) In the presence of a pelvic fracture, macro-hematuria is associated with a bladder injury in almost one-third of cases and therefore represents an absolute indication for imaging of the bladder [48, 50]. However, micro-hematuria is not an indication for mandatory radiologic evaluation. Cystography should always be considered if other indicators of bladder injury are present such as low urine output, abdominal distension, inability to void, suprapubic tenderness, uremia or elevated creatinine level and entrance/exit wounds in the lower abdomen, perineum, or buttocks . Conventional or CT-scan cystography has similar sensitivity and specificity in identifying bladder injuries (for 95% and 100% respectively). Whenever possible CT-scan cystography would be preferred [41, 45, 48, 116–118].. If associated urethral injury is suspected, a retrograde urethrography should be obtained before bladder catheterization. Passive anterograde distension of the bladder with exclusive renal-excreted contrast by clamping of the urinary catheter during abdominopelvic CT is not an effective maneuver to diagnose bladder rupture due to the high false negative rate caused by the low intravesical urine pressure [41, 48, 49, 119]. A technical pitfall of conventional cystography is represented by the false negative results in case of injuries located in the posterior wall: the lateral view is in fact rarely feasible due to the extent of pelvic injuries. In case a bladder injury is suspected in the presence of a bleeding pelvic fracture possibly amenable to angiographic management, caution should be used as extravasated contrast in the pelvis may impair the accuracy of the angiography . Urethra Patients with post-traumatic urethral hemorrhage should be investigated for urethral injuries. (GoR 1C) During emergency laparotomy, if an urethral injury is suspected, it should be investigated directly whenever feasible. (GoR 2A) Retrograde urethrography and selective urethroscopy represent the modalities of choice to investigate traumatic urethral injuries. (GoR 1B) In the event of penile lesions, urethroscopy should be preferred to retrograde urethrography (GoR 2A) Patients with urethral trauma may present with blood at the external urethral meatus, suprapubic fullness, perineal laceration, scrotal hematoma, urinary retention, difficulty or inability to insert a urinary catheter, and superiorly displaced prostate on rectal examination [45, 50, 68, 120, 121]. If urethral injury is present or suspected, rectal and vaginal examination should be performed. Associated rectal injuries are present in up to 5% of cases [121, 122]. There are two diagnostic modalities: retrograde urethrography and flexible urethroscopy [12, 58, 68]. If urethral injury is suspected, retrograde urethrography is the procedure of choice and should be performed before attempting any other maneuvers on the genito-urinary system [45, 48, 66, 123, 124]. In case of hemodynamic instability, all the investigations on the urethra should be postponed and a urinary drainage, (i.e., suprapubic catheter) should be inserted. The placement of a urethral catheter should be postponed until urethrography is obtained. Extravasation of contrast on retrograde urography indicates an urethral injury . Pelvic MRI, although not indicated in the acute setting, represents a valuable tool for anatomic definition of the injury during the post-traumatic period . A distinction between incomplete and complete urethral lesions is difficult; in general, incomplete lesions identified on retrograde urography are often characterized by extravasation of contrast which also fills the bladder, whereas extravasation of contrast is not accompanied by bladder filling in complete lesions . In case of associated penile injuries and in women due to short urethra, urethroscopy is recommended over retrograde urethrography [67, 124–127]. Management Kidney injuries Non-operative management NOM should be the treatment of choice for all hemodynamical stable or stabilized minor (AAST I-II), moderate (AAST III) and severe (AAST IV-V) lesions. (GoR 1B) Only in selected settings, with immediate availability of operating room, surgeons and adequate resuscitation, immediate access to blood, blood products and to high dependency/intensive care environment, and without other reasons for surgical exploration, NOM may be considered even in hemodynamically transient responder patients. (GoR 2C) In deciding for NOM in hemodynamically stable or stabilized patients, accurate classification of the degree of injury and associated injuries with CT-scan with intravenous contrast and delayed urographic phases is mandatory. (GoR 2A) NOM in penetrating lateral kidney injuries is feasible and effective but accurate patient selection is crucial even in the absence of other indications for laparotomy. In particular, cases without violation of the peritoneal cavity are more suitable for NOM. (GoR 2A) Isolated urinary extravasation, in itself, is not an absolute contra-indication to NOM in absence of other indications for laparotomy. (GoR 1B) In low resource settings, NOM could be considered in hemodynamically stable patients without evidence of associated injuries, with negative serial physical examinations and negative first level imaging and blood tests. (GoR 2C) No specific recommendations exist for NOM in blunt and penetrating kidney and urogenital tract injuries in children that are different than those used for adults. Therefore, pediatric patients should be treated as adult patients keeping into account the rule that being less invasive is better. NOM in severe injuries should be considered only in those settings where close clinical observation and hemodynamic monitoring in a high dependency/intensive care environment are possible, including serial clinical examination and laboratory tests, immediate access to diagnostics, interventional radiology and surgery, and immediately available access to blood and blood products. Alternatively, NOM may be used selectively if a system for immediate transfer to a higher level of care facility exists. NOM should be considered a step-wise approach starting with conservative management, followed by the use of minimally invasive (endoscopic or angiographic) techniques [92, 116, 128].. NOM lead to a higher renal preservation rate, a shorter hospital stay and a comparable complication rate to OM [128–141]. In hemodynamically stable or stabilized patients a CT scan with contrast together with delayed images is the gold standard to select patients for NOM [1, 17, 43, 92, 108, 116, 118, 130, 131, 133, 135, 138, 139, 141–160]. Incomplete staging is a relative indication to surgical exploration [133, 156, 159–161]. Non-resolving urinomas are common complications of NOM requiring ureteric stenting or percutaneous drainage [116, 128, 145, 147, 156, 158, 161]; perirenal hematoma and renal fragmentation are not absolute indications for acute OM [108, 146, 161]. Renal pelvis injury does not contraindicate NOM; however, it may request acute or delayed, endoscopic or open repair [17, 116, 147–149], particularly when complete avulsion of the ureteropelvic junction is observed. Angioembolization of severe injuries allows continuation of NOM if after the procedure patients recovered from a hemodynamic point of view, and when no other indications for laparotomy exists [1, 17, 43, 116, 118, 135, 147, 150, 154, 161, 162]. In fact, In experienced centers with hybrid operating rooms, NOM may be attempted even in cases with a transient response to fluid resuscitation [1, 116] provided that all resources necessary for immediate operative intervention exist. Isolated penetrating injuries to the kidney are rare; they are often associated with severe injuries, multiorgan involvement, and hemodynamic instability [1, 43, 92, 137, 145, 158, 159, 163]. However, NOM may be an appropriate first-line management option in hemodynamically stable patients without other indications for open surgical exploration (peritonitis, failed embolization, persistent bleeding, expanding or pulsatile hematoma, pielo-ureteral lesions) following penetrating trauma [1, 43, 92, 116, 128, 129, 135, 136, 138, 143, 144, 146–149, 154, 156, 159, 161, 164, 165]. As for blunt trauma, in deciding the applicability of NOM, institutional factors must be considered [92, 116, 128, 130, 132, 135, 141, 143, 147, 150, 160, 161]. Moreover, a multidisciplinary approach is needed [132, 141, 143, 144, 161]. It has been demonstrated that the degree of expertise of the trauma center plays a role in the successful rate of NOM [130, 132, 136, 161, 166]. Success rate of NOM is approximately 50% in stab wounds and 40% in gunshot wounds [1, 43, 137, 146, 150, 160]. Hemodynamically unstable patients with renal trauma not responsive to fluid resuscitation should undergo OM [1, 92, 108, 116, 142, 154–156, 158, 159]. No data exist regarding the best management strategy in low resource settings, although it seems rational to use OM in those circumstances. Low resource settings, in a limited sense, could be considered similar to military settings where lack of well-equipped hospital facilities, increased distance from trauma centers, and long transport time to definitive care facilities are the norm . Other imaging modalities such as intravenous pyelography (less effective than CT in diagnosing significant renal injury) [43, 108, 116, 139, 154, 155, 158, 164], plain radiography , ultrasound (can lead to some significant false negative) [116, 139, 155, 157, 159] should be used to assess hemodynamically stable patients when CT scanning is not available. Serial physical examination is reliable in detecting significant injuries after penetrating trauma to the abdomen [130, 164, 166] if performed by experienced clinicians and preferably by the same team. Operative management Hemodynamically unstable and non-responder (WSES IV) patients should undergo OM. (GoR 2A) Resuscitative Endovascular Balloon Occlusion of the Aorta (i.e., REBOA) may be used in hemodynamically unstable patients as a bridge to other more definitive procedures for hemorrhage control. (GoR 2B) In cases of severe renal vascular injuries without self-limiting bleeding, OM is indicated. (GoR 1C) The presence of non-viable tissue (devascularized kidney) is not an indication to OM in the acute setting in the absence of other indications for laparotomy. (GoR 2A) Hemodynamic stable or stabilized patients having damage to the renal pelvis not amenable to endoscopic/percutaneous techniques/stent should be considered for delayed OM in absence of other indications for immediate laparotomy. (GoR 2B) Uncontrollable life-threatening hemorrhage with avulsion of the renal pedicle and pulsating and/or expanding retroperitoneal hematoma or renal vein lesion without self-limiting hemorrhage are indications for OM. Retroperitoneal hematoma discovered during laparotomy and not adequately studied requires exploration of the kidney if they are pulsatile or if they are the only cause of hemodynamic instability. Whenever possible, the appropriate intraoperative diagnostic study should be performed [10, 13, 15, 18, 75, 132, 136, 137, 168–179]. All penetrating injuries associated with a retroperitoneal hematoma, if not adequately studied, should be explored especially if entering the peritoneal cavity [15, 137]. A shattered kidney or avulsion of the pyelo-ureteral junction in a hemodynamically stable patient do not mandate urgent surgical intervention. Arterial injuries or severe parenchymal injuries often result in nephrectomy when discovered intraoperatively [168, 179]. The success rate of arterial repair is 25–35% [15, 18, 177]. Arterial repair should be attempted in cases of patients with only one kidney or in those with bilateral renal injuries. Urine extravasation is not by itself an indication for OM in the acute setting [18, 169, 180]. Some cases of renal injury result in significant devascularization of the organ which results in a significant renin-angiotensin-aldosterone cascade response. These patients may complain of flank pain and have unrelenting persistent hypertension not responsive to anti-hypertensives. In these rare instances, and when a contralateral kidney is functional, nephrectomy may be the only option if all other management strategies fail. Angiography and angioembolization Angiography with eventual super-selective angioembolization is a safe and effective procedure; it may be indicated in hemodynamically stable or stabilized patients with arterial contrast extravasation, pseudoaneurysms, arteriovenous fistula, and non-self-limiting gross hematuria. (GoR 1C) Angioembolization should be performed as selectively as possible. (GoR 1C) Blind-angioembolization is not indicated in hemodynamically stable or stabilized patients with both kidneys when angiography is negative for active bleeding, regardless of arterial contrast extravasation on CT-scan. (GoR 1C) In hemodynamically stable or stabilized patients with severe renal trauma with main renal artery injury, dissection or occlusion, angioembolization and/or percutaneous revascularization with stent or stentgraft is indicated in specialized centers and in patients with limited warm ischemia time (< 240 min) (GoR 2C) Endovascular selective balloon occlusion of the renal artery could be utilized as a bridge to definitive hemostasis. This procedure requires direct visualization by fluoroscopy where the balloon is advanced over a selectively placed guidewire. (GoR 2B) In severe injury with main renal vein injury without self-limiting bleeding, angioembolization is not indicated. Patients should undergo surgical intervention. (GoR 1C) In hemodynamically stable or stabilized patients with solitary kidney and moderate (AAST III) or severe (AAST IV–V) renal trauma with arterial contrast extravasation on CT-scan, angiography with eventual super-selective angioembolization should be considered as the first choice. (GoR 1C) In hemodynamically stable or stabilized patients with active kidney bleeding at angiography and without other indications for surgical intervention, in case of failure of the initial angioembolization, a repeat angioembolization should be considered. (GoR 1C) In adults, only in selected setting (immediate availability of operating room, surgeon, adequate resuscitation, immediate access to blood and blood products and to high dependency / intensive care environment) and without other reasons for surgical exploration, angioembolization might be considered in selected hemodynamically transient responder patients. (GoR 2C) In children, angiography and eventual super-selective angioembolization should be the first choice even with active bleeding and labile hemodynamics, if there is immediate availability of angiographic suite, immediate access to surgery and to blood and blood products, and to high dependency / intensive care environment. (GoR 2C) Indications to angiography and eventual selective angioembolization include arterial contrast extravasation on CT-scan in hemodynamically stable or transient responder patients [170, 181–188], gross non-self-limiting hematuria [188, 189], arteriovenous fistula [181, 188], Pseudoaneurysm (PSA) [188, 190] extended perirenal hematoma [184, 186, 191, 192] and progressive decrease in hemoglobin concentration during NOM [185, 188]. Disrupted Gerota’s fascia associated with contrast extravasation is suggested to increase the need for AE . The grade of parenchymal disruption seems not to be associated with AE need even if severe renal injuries are associated with a reduced rate of AE success [170, 183, 186, 193]. Almost 32% of blunt renal injuries with arterial contrast extravasation on CT-scan have negative angiography ; these cases can be successfully managed without AE . Overall AE success rate in blunt renal trauma ranges from 63% to 100% [135, 162, 181, 185, 188, 189, 194–200]. In case of need for a repeat AE, the success rate is similar to those seen in initial AE, so re-interventions are justified when indicated by the clinical course . Failure rates are linked to the experience of the centers . AE seems to have better results in terms of renal function and ICU length of stay compared with nephrectomy, showing similar transfusion need and re-bleeding rates . The anatomical damage to the kidney is associated with the need to repeat AE , but not with an overall AE failure . Kidney devascularisation, initial hemodynamic instability, low hemoglobin concentration, the ISS, and associated injuries did not correlate with a higher rate of AE failure [170, 193]. Age and volume of blood products given in the first 24 h, the experience of the center, and penetrating trauma are associated with a higher risk of AE failure . Renal AE has lower complication rates compared with surgery . Renal dysfunction or renovascular hypertension directly linked to AE for renal injury is rare [162, 185, 186, 189, 197, 200–203]. Long-term follow-up showed good functional and morphological results in patients with single kidney . Reported morbidity rate after AE is 25% [135, 189, 192] and includes accidental embolization of healthy arterial branches of vascularised territories, puncture-site bleeding, arterial dissection and thrombosis, contrast-induced nephropathy, post-embolization syndrome (i.e., back pain and fever), gross hematuria, renal abscess, coils migration, PSA and arteriovenous fistulae [162, 188, 189]. Shattered kidney without renal hilum avulsion could be treated with AE [185, 194], but the management of renal pedicle avulsion is still a matter of debate, with some reporting AE success rates of 80% but with the need of repeat angioembolization in almost all cases [170, 193, 204, 205], and others reporting a failure rate of 100% . Renal venous pedicle avulsion becomes the only contraindication for NOM and AE and requires immediate surgery [181, 186]. Accumulating evidence exists regarding the successful use of AE even in patients with severe trauma with liable hemodynamic parameters provided that the environment is adequate and risk is not increased [170, 186, 194, 204, 205]. In general, one in five penetrating kidney injury patients initially treated with conservatively will need either surgical or angiographic mamagement . Reported AE success rate after renal stab wounds with vascular injuries is 82-88% [203, 207]. Embolization should be performed as sub-selectively as possible to limit the associated parenchymal infarction . Agents used for AE can induce either temporary or permanent arterial occlusion. The chosen embolic agents depend on the type of vascular injury (direct bleeding, PSA, arteriovenous fistula), but the majority of procedures are performed using coils with or without gelfoam . Results of kidney artery surgical revascularization are poor, with long-term kidney function preservation rate of less than 25% [209, 210]. Conservative management of main renal artery occlusion leads to a high rate of severe hypertension, requiring subsequent nephrectomy. Percutaneous revascularization with stents showed better outcomes on renal function than surgical treatment [209, 210]. However, it must be pointed out that warm ischemia time longer than 60 min leads to significant exponential losses in kidney function [211, 212]. The placement of a peripheral stent graft may be considered for hemostasis allowing perfusion of the renal artery distal to the injury site. Selective balloon occlusion can be considered as a temporary bleeding control maneuver prior to laparotomy however fluoroscopy is required for positioning of the guidewire and balloon catheter. Selective renal artery balloon occlusion leads to less global ischemia compared with aortic balloon occlusion. Present guidelines and WSES classification consider segmental vascular injuries (SVI) as moderate lesions due to the reduced risk of organ loss and minor risk for life loss. Moreover, they have been separated from collecting system lacerations (CSL) as the overall NOM successful rate is significantly lower in SVI when compared with CLS (43% vs. 98%) . SVI may be successfully treated with AE [116, 207]. The reported success rate of AE in children with blunt renal trauma and contrast medium extravasation or PSA is 100% with a major morbidity rate of 0% [213–215]. Current indications for AE in children are not universally recognized and include moderate and severe injuries, active bleeding with contrast blush on CT-scan, ongoing hemodynamic instability and PSA [215–217] with the suggestion to proceed with NOM only in those environments allowing for it without any additional risk. AE in pediatrics fills a void between NOM in the hemodynamically stable children and OM in the highly unstable patient with severe renal injury . Urinary tract injuries Ureter Contusions may require ureteral stenting when urine flow is impaired. (GoR 1C) Partial lesions of the ureter should be initially treated conservatively with the use of a stent, with or without a diverting nephrostomy in the absence of other indications for laparotomy. (GoR 1C) Partial and complete ureteral transections or avulsion not suitable for NOM may be treated with primary repair plus a double J stent or ureteral re-implant into the bladder in case of distal lesions (GoR 1C). Ureteral injuries should be repaired operatively when discovered during laparotomy or in cases where conservative management has failed (GoR 1C) Ureteral stenting should be attempted in cases of partial ureteral injuries diagnosed in a delayed fashion; if this approach fails, and/or in case of complete transection of the ureter, percutaneous nephrostomy with delayed surgical repair is indicated. (GoR 1C) In any ureteral repair, stent placement is strongly recommended. (GoR 1C) In the absence of other indications for laparotomy, the majority of low-grade ureteral injuries (contusion or partial transection) may be managed by observation and/or ureteral stenting [43, 115]. If stenting is unsuccessful, a nephrostomy tube should be placed . If ureteral injuries are suspected during a laparotomy, direct visualization of the ureter is mandatory . Whenever possible, ureteral injuries should be repaired. Otherwise, a damage control strategy should be preferred, with ligation of the damaged ureter and urinary diversion (temporary nephrostomy), followed by delayed repair [45, 50, 115]. In cases of complete transection of the ureter, surgical repair is indicated . The two main options are primary uretero-ureterostomy or ureteral re-implant with bladder psoas hitch or a Boari flap [43, 50, 114, 115, 218]. The use of ureteral stents is recommended after all surgical repairs to reduce failures (leaks) and strictures [13, 42, 45, 50, 116, 127]. Distal injuries to the ureter (caudal to the iliac vessels) are usually treated by reimplantation of the ureter in the bladder (uretero-neocystostomy), as the traumatic insult may jeopardize the blood supply [42, 43, 45, 50, 218]. In cases of delayed diagnosis of incomplete ureteral injuries or delayed presentation, an attempt of ureteral stent placement should be done; however, retrograde stenting is often unsuccessful. In these cases, delayed surgical repair should be considered . Bladder Bladder contusion requires no specific treatment and might be observed clinically. (GoR 1C) Intraperitoneal bladder rupture should be managed by surgical exploration and primary repair (GoR 1B) Laparoscopy might be considered in repairing isolated intraperitoneal injuries in case of hemodynamic stability and no other indications for laparotomy. (GoR 2B) In case of severe intraperitoneal bladder rupture, during damage control procedures, urinary diversion via bladder and perivesical drainage or external ureteral stenting may be used. (GoR 1C) Uncomplicated blunt or penetrating extraperitoneal bladder injuries may be managed non-operatively, with urinary drainage via a urethral or suprapubic catheter in the absence of other indications for laparotomy. (GoR 1C) Complex extra-peritoneal bladder ruptures—i.e., bladder neck injuries, lesions associated with pelvic ring fracture and/or vaginal or rectal injuries—should be explored and repaired. (GoR 1C) Surgical repair of extraperitoneal bladder rupture should be considered during laparotomy for other indications and during surgical exploration of the prevesical space for orthopedic fixations. (GoR 1C) In adult patients, urinary drainage with urethral catheter (without suprapubic catheter) after surgical management of bladder injuries is mandatory (GoR 1B); for pediatric patients, suprapubic cystostomy is recommended (GoR 2C) In cases of hemodynamic instability, urethral or suprapubic catheter may be inserted as a temporary measure and the repair of the bladder injury may be postponed . All penetrating bladder injuries and Intraperitoneal bladder rupture (IBR) generally require surgical exploration and primary repair [41, 45, 53]. Laparoscopic repair of isolated IBR is a viable option . Open surgical repair of bladder injuries is in a double-layer fashion using monofilament absorbable suture . Single-layer repair is common during laparoscopic approach [12, 45, 54, 221, 222]. Uncomplicated blunt or penetrating EBR, in the absence of other indications for laparotomy, may be managed conservatively, with clinical observation, antibiotic prophylaxis and the insertion of a urethral catheter or a suprapubic percutaneous cystostomy, in case of a concomitant urethral injury . Injury healing happens within 10 days in more than 85% of cases . Surgical repair of EBR is indicated in complex injuries as bladder neck injuries or injuries associated with pelvic fractures requiring internal fixation and rectal or vaginal injuries [41, 50]. Furthermore, surgical repair of EBR may be considered in case of non-resolution of urine extravasation 4 weeks after the traumatic event . Gunshot injuries of the bladder are commonly associated to rectal injuries, which prompt fecal diversion. Commonly, these injuries are through-and-through (entry/exit site) requiring careful and complete pelvic inspection . Urethral catheterization whenever possible has the same efficacy of suprapubic cystostomy; therefore routine placement of a suprapubic tube is no longer recommended [45, 223, 224]. Suprapubic catheterization may be reserved for cases with associated perineal injuries. Suprapubic drainage is recommended in children after the surgical repair of bladder rupture . Urethra Urinary drainage should be obtained as soon as possible in case of traumatic urethral injury. (GoR 1C) Blunt anterior urethral injuries should be initially managed conservatively with urinary drainage (via urethral or suprapubic catheter); endoscopic treatment with realignment should be attempted before surgery. Delayed surgical repair should be considered in case of failure of conservative treatment after endoscopic approach. (GoR 1C) Partial blunt injuries of the posterior urethra may be initially managed conservatively with urinary drainage (via urethral or suprapubic catheter) and endoscopic realignment; definitive surgical management should be delayed for 14 days if no other indications for laparotomy exist. (GoR 1C) Injuries of the posterior urethra in cases of hemodynamic instability should be approached by immediate urinary drainage and delayed treatment. (GoR 1C) Conservative treatment of penetrating urethral injuries is generally not recommended. (GoR 1C) Penetrating injuries of anterior urethra should be treated with immediate direct surgical repair if the clinical conditions allow and if an experienced surgeon is available; otherwise, urinary drainage should be performed and delayed treatment planned. (GoR 1C) Penetrating injuries of the posterior urethra should be treated with primary repair only if the clinical conditions allow. Otherwise, urinary drainage and delayed urethroplasty are recommended. (GoR 1C) When posterior urethral injury is associated with complex pelvic fracture, definitive surgical treatment with urethroplasty should be performed after the healing of pelvic ring injury. (GoR 1C) · Bladder drainage should be obtained soon and as safe and technically feasible. In case of contrast extravasation on urethrogram, a suprapubic catheter should be considered [57, 226]. The treatment of choice in case of penetrating urethral injuries is surgical exploration and repair [227, 228]. Posterior urethral blunt injuries and selected penetrating partial injuries, in the absence of other indications for laparotomy, may be treated initially by NOM with the insertion of a suprapubic cystostomy or urethral catheter, as primary open realignment and primary open anastomosis are associated with high rates of stricture, urinary incontinence, and impotence [45, 50, 66, 123, 229]. However, the insertion of a suprapubic catheter may be difficult due to hematoma or to poor bladder filling in case of shock; an experienced provider may attempt once a careful urethral catheter placement [58, 60, 61, 67, 120, 125, 126, 226]. However, if any resistance is encountered, a suprapubic catheter should be placed under direct visualization or with ultrasound guidance . In case of anterior urethral blunt trauma, the initial treatment of choice is conservative with urinary drainage (by suprapubic or urethral catheter placement) and delayed treatment after an accurate evaluation of the extent of the injury. A trial of endoscopic realignment should be undertaken. In case of failure, surgery is recommended with urethroplasty [67, 230]. Selected cases of incomplete penetrating injuries of the anterior urethra may be managed with trans-urethral catheter placement. Urethrography should be performed every two weeks until complete healing . Unless other life-threatening injuries are present, uncomplicated penetrating lesions of the anterior urethra are best managed with prompt direct surgical repair . Cases in which damage control procedures are needed or in which anastomotic urethroplasty is not feasible due to a large anatomic defect (typically lesions > 2–3 cm in the bulbar urethra and > 1.5 cm in the penile urethra), marsupialisation of the urethra, temporary suprapubic urinary catheter placement and delayed anatomic reconstruction with graft or flap (interval urethroplasty at > 3 months) are indicated . In blunt posterior urethral injuries, initial conservative treatment is recommended with planned delayed surgical treatment, allowing multidisciplinary management involving experienced surgeons and urologists . In case of hemodynamically stable patients with complete lesions of the posterior urethra without other life-threatening injuries, immediate endoscopic realignment is preferred over immediate urethroplasty. Endoscopic realignment is associated with improved outcomes [67, 229, 231, 232]. Therefore, immediate urethroplasty is not routinely recommended. When endoscopic realignment is unsuccessful, urinary drainage with suprapubic catheter placement and delayed urethroplasty are indicated [123, 229], preferably within 14 days from the injury. In case of associated pelvic fractures, definitive surgery should be postponed until after the healing of pelvic ring injuries [50, 126, 222, 231, 233, 234]. The management of penetrating injuries to the posterior urethra depends on the presence and severity of associated injuries. In case of life-threatening associated injuries and Damage Control approach, urinary diversion and delayed urethroplasty is advised [64, 127].In hemodynamic stable patients, without associated severe injuries, immediate retropubic exploration and primary repair of the injury is recommended [64, 126]. Follow-up: Follow-up imaging is not required for minor (AAST I-II) renal injuries managed non-operatively. (GoR 2B) In moderate (AAST III) and severe (AAST IV-V) renal injuries, the need for follow-up imaging is driven by the patients’ clinical conditions. (GoR 2B) In severe injuries (AAST IV-V), contrast-enhanced CT scan with excretory phase (in cases with possible or documented urinary extravasation) or ultrasound and contrast-enhanced US are suggested within the first 48 h after trauma in adult patients and in delayed follow-up. (GoR 2A) Follow-up imaging in pediatric patients should be limited to moderate (AAST III) and severe (AAST IV-V) injuries. (GoR 2B) In pediatric patients, ultrasound and contrast-enhanced US should be the first choice in the early and delayed follow-up phases. If cross-sectional imaging is required, magnetic resonance should be preferred. (GoR 2B) CT-scan with delayed phase imaging is the method of choice for the follow-up of ureteral and bladder injuries. (GoR 2A) Ureteroscopy or urethrogram are the methods of choice for the follow-up of urethral injuries. (GoR 2A) Return to sport activities should be allowed only after microscopic hematuria is resolved. (GoR 2B) In general mild and moderate injuries have a very low complication rate [235–237]. Routine follow-up imaging may not be justified for mild injuries [236–240]. In severe injuries, CT scan with delayed excretory phase is recommended within the first 48 h after admission as urinary leak may be missed on the initial CT scan in 0.2% of all cases and in 1% of high-grade renal injuries .. Moderate injuries without urine extravasation would require follow-up imaging only in case of worsening of patient status [17, 236, 239, 241, 242]. The risk of secondary hemorrhage deserves particular mention. Secondary hemorrhage is usually caused by rupture of a PSA or arteriovenous fistula, which occurs in up to 25% of moderate/severe injuries [151, 243] within 2 weeks of the injury [151, 207, 243]. Hematuria is the most common sign suggesting these complications . It is an indication to perform contrast-enhanced CT scan or DUS or CEUS, according to the availability of the tests in the hospital. These three techniques showed to be similar in reliability regarding the detection of these complications [77, 151]. No definitive evidence exists with regard to timing of return to normal activity after renal trauma. In general, bed rest or reduced activity is recommended until gross hematuria is resolved [146, 237, 244]. Return to sport activities after a minor or moderate renal injury may occur within 2 to 6 weeks from the injury while severe injuries may require longer periods (6 to 12 months) [245, 246]. As a general rule, sports activities should be avoided until microscopic hematuria is resolved [245, 246]. Limited low-grade evidence is available with regard to the best follow-up strategy in pediatric patients with renal trauma. US or CEUS may be considered the method of choice in moderate and severe renal injuries, even if initially evaluated by CT-scan . If US or CEUS imaging is inconclusive MRI, if available, should be performed. There is no sufficient evidence regarding the relationship between renal injury severity and the rate and timing of healing or incidence of renal dysfunction [247–249]. Low-grade kidney injuries have a very low rate of late complication in pediatric patients; therefore, scheduled imaging follow-up in the potential complications is not indicated [247, 250]. The reported incidence of renal trauma-induced hypertension is 0–6.6% [244, 251–254], but in general, all those who are normotensive in the immediate post-trauma period usually do not develop signs of hypertension during follow-up . Conclusions The management of kidney and urogenital trauma is multidisciplinary. When feasible, non-operative management should always be considered as the first option. For this reason, the anatomy of the injury, its physiological effects, and the associated injuries should always be considered to define the best treatment strategy. Acknowledgements Special thanks to Ms. Franca Boschini (Bibliographer, Medical Library, Papa Giovanni XXIII Hospital, Bergamo, Italy) for the precious bibliographical work. Contributors: WSES-AAST Expert Panel Paola Fugazzola (1), Martijn Stommel (2), Mohan Rajashekar (3), Edward Tan (4), Matti Tolonen (5), Marco Ceresoli (6), Carlos Augusto Gomez (7), Niccolo Allievi (8), Mircea Chirica (9), Francesco Salvetti (10), Riccardo Bertelli (1), Offir Ben-Ishay (11), Hany Bahouth (11), Gianluca Baiocchi (12), Antonio Tarasconi (13), Stefania Cimbanassi (14), Osvaldo Chiara (14), Richard ten-Broek (2), Giulia Montori (15), Erika Picariello (1), Leonardo Solaini (16), Andreas Hecker (17), Matteo Tomasoni (1), Paola Perfetti (18), Neil Parry (19), Nicola DeAngelis (20), Bruno M Pereira (21), Joaquin Bado (22), jbado03@hotmail.com, Oreste Romeo (23), Andreas Pikoulis (24), Miklosh Bala (25), Lena Napolitano (23), Joseph Galante (26), Sandro Rizoli (27), Paula Ferrada (28), Tal Horer (29), Megan Brenner (30), m.brenner@ruhealth.org, Rao Ivatury (28) (1) General, Emergency and Trauma Surgery Dept., Bufalini Hospital, Cesena, Italy (2) Department of Surgery, Radboud University Medical Center, Nijmegen, The Netherlands (3) General Surgery, Hegde Hospital, Mangaluru, Karnataka, India (4) Emergency Med. Dept., Radboud University Medical Center, Nijmegen, The Netherlands (5) Second Department of Surgery, Meilahti Hospital, Helsinki, Finland (6) General and Emergency Surgery, Milano-Bicocca University, School of Medicine and Surgery, Monza, Italy (7) Hospital Universitário Terezinha De Jesus, Faculdade De Ciências Médicas E Da Saúde De Juiz De Fora (Suprema ) Brazil (8) General, Emergency and Trauma Surgery Dept., Papa Giovanni XXIII Hospital, Bergamo, Italy (9) Chirurgie Digestive, CHUGA-CHU Grenoble Alpes, Grenoble, France (10) General Surgery dept., Pavia University Hospital, Pavia, Italy (11) Division of General Surgery Rambam Health Care Campus Haifa, Israel (12) Department of Clinical and Experimental Sciences, Surgical Clinic, University of Brescia, Brescia, Italy. (13) Emergency and Trauma Surgery, Maggiore Hospital, Parma, Italy (14) Emergency and Trauma Surgery dept., Niguarda Hospital, Milano, Italy (15) General Surgery Dept., Aviano Hospital, Aviano, Italy (16) General Surgery Dept., Forlì Hospital, Forlì, Italy (17) Department of General and Thoracic Surgery, University Hospital of Giessen, Giessen, Germany. (18) Emergency Med. Dept., Verona Hospital, Verona, Italy (19) London Health Sciences Centre, London, ON Canada. (20) Unit of Digestive Surgery, HPB Surgery and Liver Transplant, Henri Mondor Hospital, Créteil, France (21) Trauma/ Acute Care Surgery & Surgical Critical Care, University of Campinas, Campinas, Brazil (22) General and Emergency Surgery dept., Montevideo hospital, Montevideo, Paraguay (23) Trauma and Surgical Critical Care, University of Michigan Health System, East Medical Center Drive, Ann Arbor, MI, USA (24) 3rd Department of Surgery, Attiko Hospital, National & Kapodistrian University of Athens, Greece (25) General Surgery Dept., Hadassah Medical Centre, Jerusalem, Israel (26) Trauma and Acute Care Surgery and Surgical Critical Care Trauma, Department of Surgery University of California, Davis, USA (27) Trauma Sugery, Hamad General Hospital, Doha, Qatar (28) General and Tauma Surgery, Virginia Commonwealth University, Richmond, Virginia, USA (29) Dept. of Cardiothoracic and Vascular Surgery & Dept. Of Surgery Örebro University Hospital and Örebro University, Sweden (30) Department of General Surgery, Riverside University Health System Medical Center, Moreno Valley, (California) USA Abbreviations AAST American Association for Surgery for Trauma AG/AE Angiography/angioembolization ALARA As low as reasonable achievable BE Base excess CSL Collecting system lacerations CBR Combined bladder rupture CT Computed tomography CEUS Contrast-enhanced ultrasound DUS Doppler-US EVTM Endovascular trauma and bleeding management E-FAST Extended-focused abdominal sonography for trauma EBR Extra-peritoneal bladder rupture fNOM Failure of NOM GCS Glasgow Coma Scale HPF High-power field IBR Intra-peritoneal bladder rupture ISS Injury severity score IVU Intravenous urography LE Level of evidence MRI Magnetic resonance image MTP Massive transfusion protocols NOM Non-operative management OIS Organ injury scale OM Operative management PFUI Pelvic fracture urethral injury PSA Pseudoaneurysm RBCs Red blood cells REBOA Resuscitative endovascular balloon occlusion of the aorta SVI Segmental vascular injuries US Ultrasound WSES World Society of Emergency Surgery Authors’ contributions FC, EM, YC, WB, AL, YM, FK, AP, GPF, MS, LA, GA, AK, FAZ, IW, DW, EP, ML, CA, VM, VR, RC, VK, ACM, MB, CO, MC, FM, BS, JM, RM, IDC, FC, and WSES Expert Panel: manuscript conception and draft critically revised the manuscript and contribute to important scientific knowledge giving the final approval. Funding None. Availability of data and materials Not applicable. Ethics approval and consent to participate Not applicable. Consent for publication Not applicable. Competing interests The authors declare that they have no competing interests. (: NOM should only be attempted in centers capable of a precise diagnosis of the severity of kidney injuries and capable of intensive management (close clinical observation and hemodynamic monitoring in a high dependency/intensive care environment, including serial clinical examination and laboratory assay, with immediate access to diagnostics, interventional radiology and surgery and immediately available access to blood and blood products or alternatively in presence of a rapid centralization system in those patients amenable to be transferred; @:Hemodynamic instabilityin adults is considered the condition in which patient has an admission systolic blood pressure < 90 mmHg with evidence of skin vasoconstriction (cool, clammy, decreased capillary refill), altered level of consciousness and/or shortness of breath, or > 90 mmHg but requiring bolus infusions/transfusions and/or vasopressor drugs and/or admission base excess (BE) >-5 mmol/l and/or shock index > 1 and/or transfusion requirement of at least 4-6 Units of packed red blood cells within the first 24 h; moreover transient responder patients (those showing an initial response to adequate fluid resuscitation, and then signs of ongoing loss and perfusion deficits) and more in general those responding to therapy but not amenable of sufficient stabilization to be undergone to interventional radiology treatments. In pediatric patients:Hemodynamic stability is considered systolic blood pressure of 90 mmHg plus twice the child’s age in years (the lower limit is inferior to 70 mmHg plus twice the child’s age in years, or inferior to 50 mmHg in some studies), Stabilized or acceptable hemodynamic status is considered in children with a positive response to fluids resuscitation: 3 boluses of 20 mL/kg of crystalloid replacement should be administered before blood replacement; positive response can be indicated by the heart rate reduction, the sensorium clearing, the return of peripheral pulses and normal skin color, an increase in blood pressure and urinary output, and an increase in warmth of extremity. Clinical judgment is fundamental in evaluating children. Footnotes Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Contributor Information Federico Coccolini, Phone: +39-050.992111, Email: federico.coccolini@gmail.com. Ernest E. Moore, Email: Ernest.Moore@dhha.org Yoram Kluger, Email: y_kluger@rambam.health.gov.il. Walter Biffl, Email: walt@biffl.com. Ari Leppaniemi, Email: Ari.Leppaniemi@hus.fi. Yosuke Matsumura, Email: yousuke_jpn4035@yahoo.co.jp. Fernando Kim, Email: Fernando.Kim@dhha.org. Andrew B. Peitzman, Email: peitzmanab@upmc.edu Gustavo P. Fraga, Email: fragagp2008@gmail.com Massimo Sartelli, Email: Massimo.sartelli@gmail.com. Luca Ansaloni, Email: Luca.ansaloni@auslromagna.it. Goran Augustin, Email: augustin.goran@gmail.com. Andrew Kirkpatrick, Email: Andrew.Kirkpatrick@albertahealthservices.ca. Fikri Abu-Zidan, Email: fabuzidan@uaeu.ac.ae. Imitiaz Wani, Email: imtazwani@gmail.com. Dieter Weber, Email: Dieter.Weber@health.wa.gov.au. Emmanouil Pikoulis, Email: mpikoul@med.uoa.gr. Martha Larrea, Email: larream@infomed.sld.cu. Catherine Arvieux, Email: CArvieux@chu-grenoble.fr. Vassil Manchev, Email: vassil.manchev@gmail.com. Viktor Reva, Email: vreva@mail.ru. Raul Coimbra, Email: R.Coimbra@ruhealth.org. Vladimir Khokha, Email: docvladimir@yandex.by. Alain Chichom Mefire, Email: alainchichom@yahoo.com. Carlos Ordonez, Email: ordonezcarlosa@gmail.com. Massimo Chiarugi, Email: massimo.chiarugi@med.unipi.it. Fernando Machado, Email: Fernando.Kim@dhha.org. Boris Sakakushev, Email: bsakakushev@gmail.com. Junichi Matsumoto, Email: docjun0517@gmail.com. Ron Maier, Email: ronmaier@uw.edu. Isidoro di Carlo, Email: idicarlo@unict.it. Fausto Catena, Email: faustocatena@gmail.com. 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Articles from World Journal of Emergency Surgery : WJES are provided here courtesy of BMC ACTIONS View on publisher site PDF (2.5 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Background Notes on the use of the guidelines Methods Conclusions Acknowledgements Abbreviations Authors’ contributions Funding Availability of data and materials Ethics approval and consent to participate Consent for publication Competing interests Footnotes Contributor Information References Associated Data Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
16584	https://chem.libretexts.org/Courses/University_of_Alberta_Augustana_Campus/AUCHE_252_-_Organic_Chemistry_II/06%3A_Carboxylic_Acid_Derivatives/6.02%3A_Nucleophilic_Acyl_Substitution_Reactions_of_Carboxylic_Acids	Skip to main content 6.2: Nucleophilic Acyl Substitution Reactions of Carboxylic Acids Last updated : Feb 10, 2024 Save as PDF 6.1: Nucleophilic Acyl Substitution Reactions 6.3: The Claisen Condensation Reaction Page ID : 469396 ( \newcommand{\kernel}{\mathrm{null}\,}) The poor leaving group ability of -OH makes carboxylic acids relatively unreactive towards nucleophilic acyl substitutions. The addition of a strong acid protonates the carbonyl oxygen making the carbonyl carbon more electrophilic and circumventing this problem. Also, the substitution reaction can be promoted by converting -OH into a better leaving group. Despite the lack of reactivity, under the right conditions carboxylic acids can be successfully converted into acid chlorides, acid anhydrides, esters, and amides through nucleophilic acyl substitution. Conversion of Carboxylic acids to Acid Chlorides Carboxylic acids can be converted to acid chlorides by reaction with thionyl chloride (SOCl2). During the reaction with thionyl chloride, the hydroxyl group of the carboxylic acid is converted to an acyl chlorosulfite moiety which is a better leaving group. During the reaction a nucleophilic chloride anion is produced which reacts with the acyl chlorosulfite intermediate through nucleophilic acyl substitution to produce an acid halide. General Reaction Example Mechanism The O=S-Cl bonds of thionyl chloride are analogous to the O=C-Cl bonds of an acid halide in that they can undergo a type of nuleophilicy acyl substitution. During the first step of the mechanism, the carboxylic acid reactant acts as a nucleophile and attacks the electrophilic sulfur of thionyl chloride, pushing the pi electrons of the S=O bond onto oxygen. Reforming the S=O bond and eliminating a cloride anion (Cl-) as a leaving group creates a protonated acyl chlorosulfite. The chlorosulfite group represents an excellent leaving group due to its ability to stabilize a negative charge. The carbonyl of the protonated acyl chlorosulfite intermediate is activated to nucleophilic attack, and promptly reacts with the nucleophilic chloride anion created during the second step of the mechanism to produce a tetrahedral alkoide intermediate. Subsequent reforming of the protonated carbonyl bond eliminates the chlorosulfite group as sulfurdioxide (SO2) and a chloride anion. Lastly, the chloride anion deprotonates the carbonyl to form the acid chloride product. 1) Nucleophilic attack on S=O bond 2) Removal of Cl leaving group 3) Nucleophilic attack on the carbonyl 4) Leaving group removal 5) Deprotonation Acid Anhydride Formation An acid anhydride (or just anhydride) is the product of condensation of two carboxylic acid molecules with the release of a water molecule. The most common anhydride in organic chemistry is acetic anhydride, due to the high temperatures needed to remove water. General Reaction Example Conversion of Carboxylic Acids into Esters by Alkylation Carboxylic acids can be easily converted into their conjugate bases through deprotonation with a base, such as sodium hydroxide. The resulting carboxylate can be alkylated using by an SN2 reaction with either a methyl or primary halide. If a methyl ester is required, methyl iodide (CH3I) is a commonly used reagent. Example Conversion of Carboxylic Acids to Esters: The Fischer Esterification Alcohols can be used as nucleophiles to convert carboxylic acids to esters. Due to the poor leaving group ability of -OH in carboxylic acids, an acid catalyst is required to speed up the reaction. Usually the alcohol starting material is used as the reaction solvent so that it can be present in a large excess. This helps to increase the reaction's yield by pushing reaction equilibrium to the right which can be understood via Le Chatelier's principle. General Reaction Predicting the Products of a Fischer Esterification Example Mechanism The mechanism begins with protonation of the carbonyl to increase its electrophilic character. A tetrahedral alkoxide intermediate is formed when the alcohol nucleophile adds to the protonated carbonyl, pushing the carbonyl pi electrons onto the oxygen. A different tetrahedral alkoxide intermediate is created when a proton is then transferred to one of the hydroxides (OH), turning it into a good leaving group. Reforming the protonated carbonyl bond then eliminates water as a leaving group. Finally, deprotonation of the carbonyl by water creates the ester product and regenerates the acid catalyst. All the steps of the mechanism are reversible 1) Protonation of the carbonyl 2) Nucleophilic attack on the carbonyl 3) Proton transfer 4) Removal of water as a leaving group 5) Deprotonation Isotopic Labeling Evidence to support the Fischer esterification mechanism comes from isotopic labeling experiments with oxygen-18. If the reaction is carried out with an oxygen-18 labeled alcohol, the isotope is found exclusively in the ester and not the water generated. This shows that the C-OH bond of the carboxylic acid and the H-OR bond of the alcohol that is broken during the reaction. Worked Example: Planning a Synthesis Using a Fischer Esterification How could the following molecule be made using a Fischer Esterification? Analysis The key bond formed during a Fischer Esterification is the C-O sigma bond. The target molecule can be separated into the two required starting materials by breaking this bond. The carbonyl carbon fragment gains an –OH to become a carboxylic acid and the oxygen fragment gains an H to become an alcohol. The carboxylic acid and the alcohol are reacted using a strong acid catalyst. Solution Direct Conversion of Carboxylic Acids to Amides The direct reaction of a carboxylic acid with an amine would be expected to be difficult because the basic amine would deprotonate the carboxylic acid to form a stable carboxylate salt. However when the ammonium carboxylate salt is heated to a temperature above 100 oC, water is driven off and an amide is formed. Due to the extreme conditions, this reaction is rarely used. When converting to an amide it is preferred to convert the carboxylic acid to a more reactive form such as an acid chloride first and then convert this molecule to an amide. General Reaction Conversion of Carboxylic Acids to Amides using DCC The formation of amides from carboxylic acids and amines is a reaction of great importance in biochemistry. Proteins are formed through the creating of amide bonds between amino acid residues, so a great deal of research has been performed to find efficient methods. The direct conversion of a carboxylic acids to amides is difficult because amines are basic and tend to convert carboxylic acids to their highly unreactive carboxylates. One solution to this problem is through the use of dialkylcarbodiimides (R-N=C=N-R) coupling reagents, such as dicyclohexylcarbodiimide (DCC). During a DCC amide coupling, the OH of a carboxylic acid is made into a good leaving group which can then be replaced by an amine during nucleophilic acyl substitution. Using DCC as a coupling reagent, 1o and 2o amines can be used to create 2o and 3o amides respectfully. Basic reaction Predicting the products of DCC Coupling Example Mechanism DCC deprotonates the carboxylic acid to form a carboxylate, which is a better nucleophile. The corresponding protonation of DCC increases the electrophilic character of its C=N imide bond. The caboxylate nucleophile then adds to the imide bond and pushes the C=N pi electrons onto the nitrogen. The amine nucleophile can now add itself to the carbonyl bond as part of a nuleophilic acyl substitution. The DCC coupled oxygen is eliminated as dicyclohexylurea, a good leaving group, to create the amide product. 1) Deprotonation 2) Nucleophilic attack by the carboxylate 3) Nucleophilic attack by the amine 4) Proton transfer 5) Leaving group removal Conversion of Carboxylic acids to 1o alcohols Lithium aluminum hydride (LiAlH4) Hydride nucleophiles from lithium aluminum hydride (LiAlH4) can reduce carboxylic acids to 1o alcohols. Note that NaBH4 is not a strong enough reducing agent to convert carboxylic acids or esters to alcohols. Because the incoming nucleophile is an “H” the reaction first produces an aldehyde intermediate which available for further hydride additions. The aldehyde intermediate is difficult to isolate because it is more reactive than the original carboxylic acid. This reaction represents the first example in this chapter where a carboxylic acid derivative can undergo a double nucleophilic addition. In the mechanism of this reaction, a nucleophilic acyl substitution is followed by a nucleophilic addition allowing for two hydride nucleophiles being added to the electrophilic carbonyl carbon of a carboxylic acid. General reaction Predicting the product of a hydride reduction Example Possible Mechanism Although the mechanism of this reaction is not precisely known, much of it is understood. Initially, a hydride deprotonates the carboxylic acid to form a lithium carboxylate, hydrogen gas (H2), and aluminum hydride. Then a hydride nucleophile, from aluminum hydride, adds to the carbonyl carbon as part of a nucleophile acyl substitution. The resulting high-energy dianion intermediate forms a Lewis Acid/Base complex with aluminum, making one of the oxygens a good leaving group (the negative charge on the oxygen complexed to aluminum is not shown in step 2 below). The carbonyl bond is reformed along with the elimination of OAlH2 as a leaving group to form an aldehyde. A hydride nucleophile from another molecule of LiAlH4 adds to the re-formed carbonyl carbon as part of a nucleophilic addition. The resulting alkoxide intermediate is protonated during an acidic work-up to form the 1o alcohol product. Due to the formation of a dianion intermediate the reaction requires relatively high temperature and long reaction times. 1) Deprotonation 2) Nucleopilic attack by a hydride anion 3) Leaving group removal 4) Nucleophilic attack by a hydride anion 5) Alkoxide protonation 6.1: Nucleophilic Acyl Substitution Reactions 6.3: The Claisen Condensation Reaction
16585	https://www.procoding.org/c-program-to-find-frequency-of-each-digit-in-given-number	C program to find frequency of each digit in a given number Learn how to write a C program to find the frequency of each digit in a given number. This article provides a detailed explanation and sample code for this fundamental task in C programming. Finding the frequency of each digit in a given number is a common problem in programming that helps in understanding arrays, loops, and basic arithmetic operations. This article will guide you through writing a C program to determine the frequency of each digit (0-9) in a given number, providing a detailed explanation and sample code. Steps to Find the Frequency of Each Digit To solve this problem, follow these steps: Write a C program to find frequency of each digit in a given number Here's a C program to find the frequency of each digit in a given number: #include <stdio.h> int main() { int number, digit; int frequency = {0}; // Initialize the frequency array with zeros // Input the number printf("Enter a number: "); scanf("%d", &number); // Handle negative numbers if (number < 0) { number = -number; } // Calculate the frequency of each digit while (number != 0) { digit = number % 10; // Extract the last digit frequency[digit]++; // Increment the corresponding frequency number /= 10; // Remove the last digit } // Print the frequency of each digit printf("Digit Frequency\n"); for (int i = 0; i < 10; i++) { if (frequency[i] != 0) { printf("%d %d\n", i, frequency[i]); } } return 0; } Output Enter a number: 123432 Digit Frequency 1 1 2 2 3 2 4 1 Explanation scanf number frequency while number 0 % number 12321 digit = 12321 % 10 digit = 1 frequency digit = 1 frequency++ 1 number / number = 12321 / 10 number = 1232 for frequency Detailed Steps printf scanf frequency while number % frequency / for frequency Example For an input of 12321, the program will: 12321 1 frequency 2 frequency 3 frequency 2 frequency 1 frequency Recommended Posts
16586	https://www.healthline.com/health/blood-cell-disorders/alpha-versus-beta-thalassemia	Alpha- vs. Beta-Thalassemia: Types, Symptoms, and More Health Conditions Health Conditions All Breast Cancer Cancer Care Caregiving for Alzheimer's Disease Chronic Kidney Disease Chronic Obstructive Pulmonary Disease (COPD) Digestive Health Eye Health Heart Health Menopause Mental Health Migraine Multiple Sclerosis (MS) Parkinson’s Disease Psoriasis Rheumatoid Arthritis (RA) Sleep Health Type 2 Diabetes Weight Management Condition Spotlight All Controlling Ulcerative Colitis Navigating Life with Bipolar Disorder Mastering Geographic Atrophy Managing Type 2 Diabetes Wellness Wellness Topics All CBD Fitness Healthy Aging Hearing Mental Well-Being Nutrition Parenthood Recipes Sexual Health Skin Care Sleep Health Vitamins and Supplements Women's Wellness Product Reviews All At-Home Testing Men's Health Mental Health Nutrition Sleep Vitamins and Supplements Women's Health Featured Programs All Your Guide to Glucose Health Inflammation and Aging Cold & Flu Season Survival Guide She’s Good for Real Tools Featured Video Series Pill Identifier FindCare Drugs A-Z Lessons All Crohn’s and Ulcerative Colitis Essentials Diabetes Nutrition High Cholesterol Taming Inflammation in Psoriasis Taming Inflammation in Psoriatic Arthritis Newsletters All Anxiety and Depression Digestive Health Heart Health Migraine Nutrition Edition Type 2 Diabetes Wellness Wire Lifestyle Quizzes Find a Diet Find Healthy Snacks Weight Management How Well Do You Sleep? 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Join Bezzy on the web or mobile app. All Breast Cancer Multiple Sclerosis Depression Migraine Type 2 Diabetes Psoriasis Follow us on social media Can't get enough? Connect with us for all things health. Subscribe What’s the Difference Between Alpha- and Beta-Thalassemia? Medically reviewed by Darragh O'Carroll, MD — Written by Lauren Conti, MD on February 1, 2024 Alpha-thalassemia Beta-thalassemia Risk factors Symptoms Treatment FAQ Takeaway Whether you inherit alpha- or beta-thalassemia depends on which genes are affected. The number of affected genes can determine the subtype of alpha- or beta-thalassemia and the severity of your symptoms. Thalassemias are a group of inherited blood conditions that result from problems within the genes that produce hemoglobin. Hemoglobin is the main component of your red blood cells. It’s responsible for carrying oxygen. Deletions or mutations in the genes that make hemoglobin mean that your red blood cells might form improperly, burst easily, or be unable to carry an adequate amount of oxygen. Different types of hemoglobin gene deletions and mutations result in different types of thalassemia, ranging from mild to very serious. But people with any form of thalassemia often have a low red blood cell count (anemia) and small red blood cells persistently throughout their lives. What is alpha-thalassemia? Over 95%Trusted Source of typical adult hemoglobin is made up of two alpha (α) and two beta (β) globin protein chains. Each α and β chain can bind an oxygen molecule from your lungs and deliver it to tissues throughout your body. Usually, you inherit a total of four α globin genes and two β globin genes from your parents. Together, they produce nearly all the hemoglobin in your red blood cells. People with alpha-thalassemia have a deletion of one or more of their α globin-producing genes. \| Genes (two from each parent) \| Condition \| Description \| --- \| α α / α – (1 missing) \| alpha-thalassemia carrier \| You’re a silent carrier with no significant symptoms. Your other α globin genes can still produce enough healthy hemoglobin. \| \| α α / – – α – / α – (2 missing) \| alpha-thalassemia trait \| Mild symptoms can include anemia and slightly smaller red blood cells. \| \| α – / – – (3 missing) \| hemoglobin H (HbH) disease \| You may have moderate anemia and be prone to hemolysis, especially when sick or exposed to certain medicines or chemicals. As you age, you might develop jaundice, liver and spleen enlargement, and gallbladder or bone problems. \| \| – – / – – (4 missing) \| alpha-thalassemia major \| The body’s inability to make effective hemoglobin usually causes death before birth due to severe fetal anemia and hydrops fetalis. \| What is beta-thalassemia? You have two β globin genes — one from each parent. Mutations in these genes cause beta-thalassemia. Some people have a mutation that reduces the amount of β globin they produce (β+), while others have a mutation that prevents their body from making any β globin (β 0). \| Genes (one from each parent) \| Condition \| Description \| --- \| β / β+ β / β 0 β+ / β+ \| beta-thalassemia minor \| You may have mild anemia and small red blood cells. \| \| β+ / β+ β+ / β 0 \| beta-thalassemia intermedia \| β globin production is reduced. Symptoms are less severe. \| \| β 0 / β 0 β+ / β 0 \| beta-thalassemia major \| Your body can’t produce enough β globin. Symptoms are usually more severe and tend to appear earlier, often by 2 years old. \| Symptoms of beta-thalassemia intermedia or major include: moderate to severe anemia pale skin jaundice restricted growth bone problems enlarged liver and spleen gallbladder problems pulmonary hypertension increased chance of infections You may need frequent blood transfusions, especially in beta-thalassemia major. This can cause iron overload and heart, liver, and endocrine problems. HEALTHLINE + OPTUM NOW NEWSLETTER The latest in alternative medicine, delivered Get innovative health and wellness tips delivered to your inbox with our Healthline Natural newsletter. Enter your email JOIN NOW Your privacy is important to us Who gets alpha- and beta-thalassemia? Thalassemia mutations are more common in certain populations, particularly those of African, Southeast Asian, Middle Eastern, or Mediterranean descent. These areas have historically had a high incidence of malaria, and thalassemia mutations may have been protective against this infection. If thalassemia or thalassemia trait runs in your family, consult a doctor if you begin planning a family. Since thalassemia trait doesn’t typically cause serious symptoms, you may not realize you’re a carrier. When two carriers have a baby, there’s a chance that their child will have more severe symptoms. How do the symptoms of alpha- and beta-thalassemia differ? Many symptoms of the more severe forms of alpha- and beta-thalassemia are quite similar. They stem from the complications of anemia, hemolysis, and excess iron. Alpha-thalassemia major is usually fatal before birth. Children with beta-thalassemia major appear well at birth but develop symptoms by about 2 years old. People with beta-thalassemia major generally require more frequent blood transfusions than those with HbH alpha-thalassemia, and their condition is usually more severe. How do doctors treat alpha- vs. beta-thalassemia? Treatment for either kind of thalassemia depends on the severity of your symptoms. If you’re a silent carrier, you usually won’t need any treatment. You may have mild anemia with smaller red blood cells, but you typically won’t require iron supplementation. With moderate forms of the condition, like beta-thalassemia intermedia, you may need to take medications to support red blood cell production and manage iron overload. You’ll require close monitoring by a hematologist and primary care physician. As you age, you may also need: blood transfusions iron chelation surgery to remove the spleen or gallbladder People with severe thalassemia, like beta-thalassemia major, may require treatment earlier in life. In addition to the above treatments, other treatments may include: frequent blood tests close monitoring of heart and liver function stem cell transplant Alpha-thalassemia major is usually life threatening before birth. If a doctor identifies it early, they might discuss experimental therapies like intrauterine blood transfusions followed by stem cell transplantation. Frequently asked questions The following are some commonly asked questions about alpha- and beta-thalassemia. Is alpha-thalassemia more severe? Both alpha- and beta-thalassemia minor are generally mild. Alpha-thalassemia major is more severe than the beta form, usually causing death before birth. But beta-thalassemia major can also be life threatening and is usually more severe than the HbH form of alpha-thalassemia. Are alpha- and beta-thalassemia dominant or recessive? Both alpha- and beta-thalassemia usually follow an autosomal recessive Trusted Source inheritance pattern. Which is more common, alpha- or beta-thalassemia? The prevalence of thalassemia varies by region, but experts estimate that up to 5% of the world’s population carry a gene for alpha-thalassemia, while about 1.5% carry a gene for beta-thalassemia. Can you have both alpha- and beta-thalassemia? It’s possible to inherit both α and β globin gene mutations from your parents. You can also co-inherit thalassemia mutations with other hemoglobin gene problems, such as sickle cell trait. Interestingly, people with beta-thalassemia major who also inherit an α globin gene deletion may have less severe symptoms Trusted Source. This is due to less of a mismatch between α and β globin chain production. How do I know if I have the gene for alpha- or beta-thalassemia? To diagnose thalassemia, a doctor can review blood tests such as a complete blood count and smear, iron levels, and hemoglobin electrophoresis. In some cases, they may recommend genetic testing to determine your status. Takeaway Thalassemias result from mutations in the genes responsible for making hemoglobin, a protein that allows your blood to carry oxygen. As different genes are responsible for creating α globin and β globin chains, different mutations result in either alpha- or beta-thalassemia. Several genetic combinations are possible, which determine the type of thalassemia and the severity of your symptoms. But all types typically cause some degree of anemia with other complications. Babies who inherit alpha-thalassemia major often die before birth, though screening and experimental new therapies offer hope to affected families. For other forms of alpha- or beta-thalassemia, expert monitoring and treatment can help manage the symptoms and reduce complications. How we reviewed this article: Sources History Healthline has strict sourcing guidelines and relies on peer-reviewed studies, academic research institutions, and medical journals and associations. We only use quality, credible sources to ensure content accuracy and integrity. You can learn more about how we ensure our content is accurate and current by reading our editorial policy. Alpha thalassemia. (2017). Alpha-thalassemia. (2024). Autosomal recessive disorder. (2024). Bajwa H, et al. (2023). Thalassemia. Beta-thalassemia. (2024). Harewood J, et al. (2023). Alpha thalassemia. Langer AL. (2023). Beta-thalassemia. Origa R, et al. (2017). β-thalassemia. Schwab ME, et al. (2023). The impact of in utero transfusions on perinatal outcomes in patients with alpha thalassemia major: The UCSF registry. Spotlight on alpha thalassemia major. (n.d.). Our experts continually monitor the health and wellness space, and we update our articles when new information becomes available. Current Version Feb 1, 2024 Written By Lauren Conti, MD Edited By A.L. Heywood Medically Reviewed By Darragh O'Carroll, MD Copy Edited By Jasmin Collier Share this article Medically reviewed by Darragh O'Carroll, MD — Written by Lauren Conti, MD on February 1, 2024 HEALTHLINE NEWSLETTER The best of health and wellness We do the research so you don't have to. Stay in the know with the latest in health and wellness. Enter your email JOIN NOW Your privacy is important to us related stories Are Anemia and Thalassemia the Same Thing? What to Know About Iron Supplements for Thalassemia Facial Presentation in People with Beta-Thalassemia Major Ask the Advocate: How Do I Tell My Loved Ones I Have Myelodysplastic Syndromes (MDS)? What's the Link Between Myelodysplastic Syndromes and Autoimmune Disorders? Was this article helpful? YesNo Read this next Are Anemia and Thalassemia the Same Thing? Medically reviewed by Avi Varma, MD, MPH, AAHIVS, FAAFP Thalassemia is a genetic condition that can damage the production of red blood cells, which can lead to anemia. Treatments can prevent this: READ MORE What to Know About Iron Supplements for Thalassemia Iron supplements should be avoided by anyone with thalassemia in most cases, but there are a few times when they may be used under close supervision… READ MORE Facial Presentation in People with Beta-Thalassemia Major Beta-thalassemia causes symptoms related to anemia, including weak or brittle bones, and leads to problems with how your bones develop. READ MORE Ask the Advocate: How Do I Tell My Loved Ones I Have Myelodysplastic Syndromes (MDS)? Medically reviewed by Tiffany Taft, PsyD Hematologist Dr. Mikkael Sekeres shares how people with myelodysplastic syndromes (MDS) can explain their diagnosis to their friends and family. READ MORE What's the Link Between Myelodysplastic Syndromes and Autoimmune Disorders? Myelodysplastic syndromes (MDS) and autoimmune disorders often occur together, and living with one may increase your risk of developing the other. READ MORE Is Polycythemia Vera Hereditary? Polycythemia vera results from an acquired genetic mutation rather than an inherited one. Learn more. READ MORE One Small Step: How to Conserve Your Energy with Myelodysplastic Syndromes (MDS) Medically reviewed by Angelica Balingit, MD Myelodysplastic Syndromes (MDS) can cause persistent, excessive fatigue that affects both mind and body and makes daily life challenging. Read more… READ MORE Recipe Ideas for Eating Well with Myelodysplastic Syndromes (MDS) Medically reviewed by Katherine Marengo LDN, R.D. Myelodysplastic syndromes (MDS) can cause fatigue and reduce your appetite. Here are some nutrient-dense recipes that can boost your energy, support… READ MORE Overview of Gray Platelet Syndrome Learn about gray platelet syndrome, a rare inherited disease that can cause bleeding and bruising. We cover symptoms and treatment options. READ MORE Connected by Myelodysplastic Syndromes (MDS): Five People Share Their Experiences Learn about myelodysplastic syndromes (MDS) through the experiences of five people living with the condition or in survivorship. READ MORE Healthline Related Articles Overview Leukopenia Polycythemia vera Basophils Thalassemia Thin blood RBC count What is a hematologist Jakafi Monocytes high Increase red blood cells Absolute monocytes View All Articles + Get our wellness newsletter Filter out the noise and nurture your inbox with health and wellness advice that’s inclusive and rooted in medical expertise. SIGN UP Your privacy is important to us © 2025 Healthline Media LLC. All rights reserved. Healthline Media is an RVO Health Company. Our website services, content, and products are for informational purposes only. Healthline Media does not provide medical advice, diagnosis, or treatment. See additional information. 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16587	https://studynlearn.com/what-is-a-great-circle	What is a Great Circle? Characteristics and Great Circle Route - Class 9 What is a Great Circle? Characteristics and Great Circle Route - Class 9 Geography \| January,25 2022 \| Admin What is a Great circle? A great circle is a circle that is drawn on the surface of a sphere (such as Earth) that has a radius equal to the radius of the sphere, and whose center is also the sphere's center. The Equator is the only latitude that is a great circle. All longitudes are a part of a great circle. Equator Circles that do not pass through the center of the Earth are small circles. All the parallels of latitudes other than the Equator are small circles. Arcs of great circles are the shortest route between two points on a sphere. Characteristics: A great circle is a theoretical circle formed by the intersection of the Earth's surface and an imaginary plane that passes through the center of the Earth and divides it into two equal parts. The Equator is the largest possible circle among the lines on the latitude. All such circles must pass through or touch the center of the circle. The Equator and all diagrammatically opposite longitudes touch the center of a circle and therefore are Great Circles. The infinite number of circles that touch the two opposite ends of the sphere can be drawn on a sphere. Intersecting great circles always bisect each other. Great circles are used by navigators to find the shortest distance between any two points on the Earth's surface. Do you know what is this shortest distance called? It is called the Great Circle Route. It is the shortest distance between two places on the Earth and lies on the arc of a great circle. Using this hypothesis, the shortest route from New York to Moscow would be over the North Pole. There are several that are not taken into account while finding the Great Circle Route which are: Restrictions such as weather or political restrictions are imposed by the countries concerned. ● Lack of landing and take-off facilities for airplanes. Great Circle Route (ii) Great Circle routes are especially important for places that are on opposite sides of the globe. Most globes show great circle routes between distant ports across the Atlantic, the Pacific, and the Indian Ocean. (iii) Great circles are used by meteorologists to determine climate and weather conditions in a region. Do you know who are meteorologists? Meteorologists are the scientists who study the weather. They can be weather presenters or sole researchers. Questions: How many circles touch the two opposite ends of the sphere and can be drawn on a sphere? a. Two b. Three c. Four d. Infinitely many Intersecting great circles always ____________ a. Trisect each other b. Touch each other c. Cross each other d. Bisect each other What are the scientists who study the weather called? a. Astronauts b. Paleontologists c. Archaeologists d. Meteorologists Read More: What is the International Date Line? Class 9 Geography Related post: Physical Features Of Asia: Political Division Of Asia The Largest Continent What is Human Resources? Types- Skilled and Unskilled Resources - HRD What is Ex Situ Conservation? Wildlife Conservation Projects Examples What is In Situ Conservation? Definition and Examples - Wildlife Conservation Underpopulation: Causes, Problems and Effects of Underpopulation What is Overpopulation: Causes and Effects of Overpopulation Disaster Mitigation: Strategies To Prevent Flood, Earthquakes and Fires What is Man Made Disaster? Fire Hazards, Forest Fires, Terrorism and Other Factors Affecting Location Of Industries: 12 Factors Entrepreneur Must Know Tourism in Switzerland: 3 Reasons Why Switzerland Is A Beautiful Place! Payment
16588	https://physics.stackexchange.com/questions/134975/conservative-vector-fields	homework and exercises - Conservative vector fields - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Physics helpchat Physics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Conservative vector fields Ask Question Asked 11 years ago Modified10 years, 2 months ago Viewed 3k times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. I was always told that to find whether or not a vector field is conservative, see if the curl is zero. I have now been told that just because the curl is zero does not necessarily mean it is conservative. However, if it is non-zero it is definitely not-conservative. IE: F⃗=K r 3 r^F→=K r 3 r^ (Spherical coordinates, K K is constant) To show this is conservative I would go ahead and take the curl. It will be zero - but that is not definitive proof it is conservative? How would I show it is? homework-and-exercises differential-geometry topology vector-fields Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Jul 28, 2015 at 9:53 DanielSank 25.8k 7 7 gold badges 87 87 silver badges 119 119 bronze badges asked Sep 11, 2014 at 5:44 sci-guysci-guy 817 1 1 gold badge 13 13 silver badges 23 23 bronze badges 6 mathinsight.org/path_dependent_zero_curlsci-guy –sci-guy 2014-09-11 05:46:09 +00:00 Commented Sep 11, 2014 at 5:46 In short, a conservative field is path-independent - it depends only on the initial and final points. As such, the closed line integral vanishes, and by Green's theorem the curl must also vanish. So conservativeness implies zero curl, but the reverse (∇×f=0⟹c o n s.∇×f=0⟹c o n s.) does not hold.JamalS –JamalS 2014-09-11 07:07:49 +00:00 Commented Sep 11, 2014 at 7:07 1 @JamalS perhaps that should be an answer?David Z –David Z 2014-09-11 07:17:50 +00:00 Commented Sep 11, 2014 at 7:17 Also I think perhaps this question should go to Mathematics...? (Andrew: we can migrate it if that's the case, so don't repost it there.)David Z –David Z 2014-09-11 07:18:27 +00:00 Commented Sep 11, 2014 at 7:18 @JamalS: Your comment is absolutely correct, but more can be said about when and why the reverse does not hold!DanielSank –DanielSank 2014-09-11 07:20:24 +00:00 Commented Sep 11, 2014 at 7:20 \|Show 1 more comment 1 Answer 1 Sorted by: Reset to default This answer is useful 12 Save this answer. Show activity on this post. I was always told that to find whether or not a field is conservative, see if the curl is zero. This is almost always true, but not always true. I have now been told that just because the curl is zero does not necessarily mean it is conservative. Correct! To illustrate what's going on, let's do an example. Conside the following vector field: v⃗(x,y)=−y x^+x y^x 2+y 2.v→(x,y)=−y x^+x y^x 2+y 2. Note that v⃗v→ is not defined at the origin. Is v⃗v→ conservative? Let's define "conservative" as follows A vector field v⃗v→ is conservative if for any closed path C C, the integral ∫C v⃗˙d l⃗=0.∫C v→˙d l→=0. Consider the path parametrized as x(t)=r cos(2 π t)x(t)=r cos⁡(2 π t) and y(t)=r sin(2 π t)y(t)=r sin⁡(2 π t) for t t going from 0 to 1. This path is just a circle of radius r r centered on the origin. The displacement on the path is d l⃗d t=2 π r(−x^sin(2 π t)+y^cos(2 π t)).d l→d t=2 π r(−x^sin⁡(2 π t)+y^cos⁡(2 π t)). If we integrate our example v⃗v→ on this path we get ∫C v⃗⋅d l⃗=∫1 t=0(−y x^+x y^x 2+y 2)⋅(2 π r)(−x^sin(2 π t)+y^cos(2 π t))d t=2 π∫C v→⋅d l→=∫t=0 1(−y x^+x y^x 2+y 2)⋅(2 π r)(−x^sin⁡(2 π t)+y^cos⁡(2 π t))d t=2 π which shows that v⃗v→ is definitely not conservative. Note that the integral does not depend on the radius r r of the path. Now, we compute the curl of v⃗v→. For convenience, define r≡x 2+y 2 r≡x 2+y 2, i.e. r r is the radial polar coordinate. ∇×v⃗≡d v⃗y d x−d v⃗x d y=r 2−2 x 2 r 4−−r 2+2 y 2 r 4=2 r 2−2 r 2 r 4=0.∇×v→≡d v→y d x−d v→x d y=r 2−2 x 2 r 4−−r 2+2 y 2 r 4=2 r 2−2 r 2 r 4=0. We have now shown that v⃗v→ has zero curl. A consequence of this is that if we were to integrate v⃗v→ along any little path around a point where v⃗v→ is defined, we are guaranteed to get zero. Thus, v⃗v→ has zero curl but is not conservative. What's going on? If you picture v⃗v→ you'll see that it is a swirl of vector lines circling the origin. The magnitude of the lines decreases as you move away from the origin. This decrease is just right so that if you were to integrate around a little loop which does not encircle the origin (i.e. if you check the curl) you get zero. However, because of the global circling around the origin, if you integrate along a loop which does enclose the origin, you get something non-zero. Thus, you can think of the integral as either "feeling" the presence of the origin and picking up the 2 π 2 π we calculated, or not feeling the origin and giving zero. It's like the origin is a special point worth 2 π 2 π. This is really interesting! Our field v⃗v→ is conservative everywhere locally, but if you make a path around the origin you can get a nonzero integral, so v⃗v→ is not conservative globally. Remember we pointed out that v⃗v→ is not defined at the origin? This is not an accident. Vector fields which are conservative locally but not globally must have "holes" at which they are not defined. In fact, these vector fields must be approaching infinity near their holes, which v⃗v→ most certainly is, as you can check . Those infinite points have "residues" which show up in integrals which go around them. For the experts in the audience, this is exactly the same residue you get from integrating around a simple pole in the complex plane. Let's get back to your question To show this is conservative I would go ahead and take the curl. It will be zero - but that is not definitive proof it is conservative? How would I show it is? As you have said, and we have demonstrated, having zero curl does not guarantee that a field is conservative. What does guarantee that a field is conservative is that you can express it as the gradient of a scalar function. In more general mathematical terms, if there exists a function f f such that ∇f=v⃗∇f=v→, then v⃗v→ is said to be exact. An exact vector field is absolutely 100% guaranteed to conservative. So, one answer to your question is that to show a vector field is conservative, just show that it can be written as the gradient of a function. Another answer is, calculate the general closed path integral of the vector field and show that it's identically zero in all cases. Let's go on though, because this is really super interesting stuff. Vector fields with zero curl are guaranteed to be exact, meaning that zero curl guarantees conservativeness, unless the vector field has holes (i.e. points at which it's not defined). So, the mantra you learned that zero curl indicates conservativeness is almost always true, but it fails for vector fields which have holes, like our example v⃗v→ does at the origin. Now here's the really amazing part. If I tell you that a vector field has exactly 1 hole, and that it has zero curl but is not exact, there is only one vector field it can possibly be (up to addition of other exact vector fields). In other words, if I tell you that a vector field W⃗W→ has zero curl, has one hole, and is not a gradient of any function, then you know for sure that W⃗=v⃗+λ⃗W→=v→+λ→ where λ⃗=∇f λ→=∇f for some f f. If repeat the same situation but with two holes, then you know that W⃗W→ is expressible as a linear combination of specific curl-less but not exact vector fields associated to the two holes. This whole business generalizes to high dimensional spaces. If you like it, read up on differential forms. You can try the book "Analysis on Manifolds" by Munkres, although that is a very "mathy" book. One last thing. Instead of talking about having zero curl and being the gradient of a function, you can talk about having zero divergence and being the curl of another vector field. Normally, if a vector field has zero divergence, you can write it as the curl of something else. The electric field of a point charge is conservative and has zero divergence. However, it is not the curl of any vector field. In fact, it is the only vector field in three dimensions which has zero divergence and is not a curl of something else. And of course, the electric field of a point charge goes to infinity at the charge point, so this is one of those fields where having a "hole" allows it to break the usual rules. How did Nature know to do that? The numerator of v⃗v→ goes as r r while the denominator goes as r 2 r 2. Therefore the whole thing goes as 1/r 1/r which diverges near the origin. This is a slightly incorrect statement, but it's ok for now. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Sep 11, 2014 at 15:07 answered Sep 11, 2014 at 6:59 DanielSankDanielSank 25.8k 7 7 gold badges 87 87 silver badges 119 119 bronze badges 12 what is incorrect about the "only" in ?hyportnex –hyportnex 2014-09-11 14:11:30 +00:00 Commented Sep 11, 2014 at 14:11 @user31748: In 3D, having zero divergence, one hole, and not being a curl means you must be the point charge electric field up to addition of other vector fields which are curls.DanielSank –DanielSank 2014-09-11 15:09:21 +00:00 Commented Sep 11, 2014 at 15:09 "What does guarantee that a field is conservative is that you can express it as the gradient of a scalar function. In more general mathematical terms, if there exists a function f f such that ∇f=v⃗∇f=v→, then v⃗v→ is said to be exact. An exact vector field is absolutely 100% > guaranteed to [be] conservative." What about vector fields of the form ∇(k\|r⃗\|2)∇(k\|r→\|2)? I find this very confusing: we the potential, but isn't the resulting field only locally conservative (& therefore this condition doesn't guarantee exactness always)?Rax Adaam –Rax Adaam 2016-02-19 15:41:40 +00:00 Commented Feb 19, 2016 at 15:41 @RaxAdaam what makes you think that vector field is not globally conservative?DanielSank –DanielSank 2016-02-19 17:26:45 +00:00 Commented Feb 19, 2016 at 17:26 1 @RaxAdaam Well, yeah, but that function is totally pathological. It has a branch cut. I guess I should have explained that.DanielSank –DanielSank 2016-02-20 17:27:27 +00:00 Commented Feb 20, 2016 at 17:27 \|Show 7 more comments Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. 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16589	https://en.wikipedia.org/wiki/Extreme_value_(disambiguation)	Jump to content Search Extreme value (disambiguation) Add links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Extreme values are the maximum and minimum values of a function or set. The term may also refer to: Extreme value theorem, a concept in calculus Extreme value theory, a concept in statistics Extreme value distribution, a statistical distribution Topics referred to by the same term This disambiguation page lists articles associated with the title Extreme value.If an internal link led you here, you may wish to change the link to point directly to the intended article. Retrieved from " Category: Disambiguation pages Hidden categories: Short description is different from Wikidata All article disambiguation pages All disambiguation pages Extreme value (disambiguation) Add topic
16590	https://www.frangente.com/trunk/pr_file2_2827_mf32-50-esercizi-42_d.pdf	Miriam Lettori 42/D SULLA CARTA DIDATTICA 50 ESERCIZI DI CARTEGGIO NAUTICO © 2013 Edizioni il Frangente S.a.s. Via Gaetano Trezza 12 - 37129 Verona Tel. +39 045 8012631 Fax +39 045 593881 E-mail frangente@frangente.com www.frangente.it www.frangente.com Prima edizione 2013 ISBN 978-88-98023-10-3 Testi e illustrazioni: © 2013 Miriam Lettori Stampato presso Digital book Srl - Città di Castello (PG) Printed in Italy Tutti i diritti riservati. La riproduzione e uso, anche parziale e con qualsiasi mezzo, sia esso grafico, elettronico o meccanico, non è consentita senza l’autorizzazione scritta dell’Editore. Note dell’Autore: 1. Per motivi didattici, nella carta 42/D non sono rispettati eventuali divieti di navigazione. 2. Si precisa che il lavoro svolto in carta rappresenta la teoria pura, esente da ulteriori agenti esterni, se non quelli indicati o ricavati dai testi degli esercizi. 4 Prefazione 5 Legenda acronimi 6 Le 14 regole 7 Le coordinate terrestri 8 Esercizi sulle coordinate terrestri 10 Le scale delle carte nautiche - il primo, il miglio e il nodo 11 I calcoli sessagesimali 12 Differenza di latitudine e longitudine 13 Orientamento e Rotta Vera 14 Utilizzo delle squadrette • ESERCIZI 18 Navigazione stimata • ESERCIZI 28 Angoli di rotta e di prora 29 La bussola 31 Conversione delle prore • ESERCIZI 30 - La declinazione 34 - La Deviazione 42 Correzione delle prore • ESERCIZI 50 Navigazione costiera - Rilevamenti e Punto Nave 51 Rilevamento Vero - RilV • ESERCIZI 58 Rilevamento Magnetico - RilM • ESERCIZI 66 Rilevamento Bussola - RilB • ESERCIZI 74 Rilevamenti dello stesso punto cospicuo in tempi diversi • ESERCIZI 82 Rilevamento Polare - RilP - ρ (ro) • ESERCIZI 88 Metodo del 45°/90° • ESERCIZI 94 Serie di Troub • ESERCIZI 98 Corrente e deriva (der) 100 1° problema della corrente • ESERCIZI 108 2° problema della corrente • ESERCIZI 114 4° problema della corrente • ESERCIZI 120 4° e 2° problema della corrente • ESERCIZI 130 3° problema della corrente • ESERCIZI 134 4° e 3° problema della corrente • ESERCIZI 138 2° e 3° problema della corrente • ESERCIZI 140 Vento e scarroccio (sc) • ESERCIZI 146 Deriva 4, 3 e scarroccio (sc) • ESERCIZI 148 Intercettazione • ESERCIZI 156 Intercettazione e corrente • ESERCIZI 158 Intercettazione e scarroccio • ESERCIZI CONTENUTO 1° problema della corrente 100 Nel primo problema della corrente si immagini di navigare sotto l’effetto di una corrente nota mantenendo un determinato orientamento di prora, PV. N.B. Nella risoluzione grafica dei problemi in presenza di corrente si costruisce un parallelogramma tenendo sempre ben presente che in ognuno dei 2 triangoli che lo costituiscono, ogni linea rappresenta 2 valori che sono sempre TASSATI VAMENTE INSIEME: DC, PV e RV sono tutte direzioni espresse in GRADI - si riportano con le squadrette. VC, VP e VE sono tutte velocità espresse in NODI (sempre ORARIE) - si riportano con il compasso. DC con VC PV con VP RV con VE Si intende calcolare quale RV e quale VE si stanno effettivamente navigando. Al timone di una unità in navigazione, senza ausilio del GPS, è possibile mante nere una PB (quindi conoscere con quale PV si naviga) e una VP (indicata dal LOG, contamiglia). Se non si contrasta la corrente, la RV (reale percorso rispetto al fondo del mare) sarà sicuramente diversa dalla PV tenuta e la VE (reale velocità ri spetto al fondo del mare) potrebbe es sere diversa dalla VP prodotta. DATI: INCOGNITE : DC VC RV VE PV VP ESEMPIO DI RISOLUZIONE DEL 1° PROBLEMA Partenza dal punto alle ore 08:00 con elementi della corrente noti, DC e VC. Navigando con PV e VP note si intende determinare su quale RV e con quale VE si sta navigando, visto che non si contrasta ma si subisce la deriva. Ricorda che sotto l’effetto di una corrente, VP e PV, con relativa PB, sono gli unici elementi della navigazione controllabili da bordo. DATI: INCOGNITE: DC RV VC VE PV VP Regole 9, 10, 11 1° problema della corrente 101 1. Quali sono i dati a disposizione? DC/VC PV/VP 2. Chiudere il parallelogramma. 3. Qual è la RV, cioè qual è il reale percorso rispetto al fondo del mare, e a quale VE si naviga rispetto al fondo del mare? Dalla fine del vettore DC/VC riportare con le squadrette la PV e su di essa definire con il compasso la VP. Dal punto di partenza, con le squadrette si traccia il vettore della DC e con il com passo si definisce sullo stesso la VC. Riportare parallelamente il vettore DC/ VC fino alla fine del vettore PV/VP. Chiudere ora il parallelogramma tra slando parallelamente a se stesso anche il vettore PV/VP alla fine del vettore DC/VC. Tracciare la RV in diagonale nel paral lelogramma dal punto di partenza fino all’angolo opposto che chiameremo punto chiave (punto stimato dopo 1h di navigazione). N.B. Al timoniere viene data la direzione in gradi da seguire dalla conversione della PV in PB. N.B. La Rotta Vera è sempre la diago nale. Il segmento diagonale misurato con il compasso, tra il punto di partenza e il punto chiave, corrisponde alla VE che è la risultante del calcolo vettoriale. 1° problema della corrente 102 esercizio n. 29 Un’unità alle ore 08:00 si trova nel PNSat φ 41°20’·0N λ 008°47’·2E. Naviga con PB 035° e VP 7,2 nd in presenza di corrente con DC 298° VC 2,6 nd. Determinare RV e VE. Calcolare inoltre il valore dell’angolo di deriva (der) e il PS dopo 1h 15m di navigazione. Descrizione • Dal PN Satellitare tracciare il vettore DC/VC. • Alla fine del vettore DC/VC tracciare il vettore PV/VP e identificare il punto chiave. • Non è indispensabile, ma per comprendere meglio la costruzione, chiudere il parallelogramma trasportando parallelamente a se stessi i vettori DC/VC e PV/VP. • La RV, che è sempre la diagonale del parallelogramma, si traccia tra il punto di partenza e il punto chiave. • Per determinare il valore dell’angolo di deriva calcolare la differenza tra il valore di Prora e quello di Rotta. Per attribuire il segno al valore angolare osservare l’effetto della corrente sulla direzione dell’imbarcazione. In questo esercizio la corrente fa derivare l’unità verso la propria sinistra, quindi l’angolo di deriva ha segno negativo: 036° − 015° = − 021°. Esecuzione PV = 036° RV = 015° VE = 7,3 nd der = − 021° PS h 09:15 = φ 41°28’·8N λ 008°50’·3E Soluzione d = 2°40’W Regole 9, 10, 11 esercizio n. 29 103 1° problema della corrente 104 esercizio n. 30 Alle ore 22:10 una unità si trova a 5 miglia da Cap de Feno esattamente sull’alli neamento tra il faro di Cap de Feno e il faro di Capo Pertusato. Naviga con PV 163° e con VP 5,2 nd in zona con corrente DC 098° e VC 3 nd. Determinare: ΔV (differenza velocità) e angolo di deriva, der. Descrizione • Per determinare il PN identificare i fari in carta e riportare la distanza con il compasso, dal faro di Cap de Feno, sull’allineamento stesso. • Tracciare il vettore DC/VC e di seguito il vettore PV/VP. • Chiudere il parallelogramma e tracciare la RV sulla quale si legge la VE. • Calcolare la differenza di velocità tra VP e VE. Poiché la corrente è a favore, proviene più verso poppa rispetto al traverso, la differenza di velocità ha segno positivo. • Calcolare il valore dell’angolo di deriva con la differenza tra il valore di RV e quello di PV. L’angolo di deriva ha segno negativo perché l’unità ha subìto la corrente derivando verso sinistra. Esecuzione (PN = φ 41°25’·2N λ 008°59’·6E RV = 140° VE = 7 nd) ΔV = + 1,8 nd der = − 023° Soluzione esercizio n. 30 105 1° problema della corrente 106 esercizio n. 31 Si parte da Punta Sperone (φ 41°22’·0N λ 009°13’·2E) alle ore 20:40 con VP 8 nd e con l’intento di tenere PV 198°, in presenza di corrente SW con velocità di 2,5 nodi. Determinare l’ora e il punto in cui si rileva per RilP 300° il settore rosso del faro di Santa Teresa di Gallura. Determinare inoltre l’ora e la località d’arrivo in costa. Descrizione • Dal punto di partenza tracciare il vettore DC/VC e di seguito il vettore PV/VP. Non ha importanza se costruendo in carta il calcolo vettoriale si tracciano le linee sulla terra, è solo una costruzione grafica. • Dal punto di partenza al punto chiave tracciare la RV e misurare con il com passo la VE. • Tracciare il RilV del fanale di pericolo isolato corretto dal RilP fino a incrociare il prolungamento della RV. ATTENZIONE Il rilevamento polare si considera orientato a 300° rispetto alla prora, ma lo si legge comunque sulla rotta. In navigazione con corrente (e come poi si vedrà anche con vento) è ancora più importante ricordare la formula di cor rezione del RilP in RilV: RilV = PV + (±ρ) ora PV e RV non coincidono più. • Una volta tracciato il rilevamento, calcolare l’ora in funzione delle miglia navi gate dal punto di partenza al luogo del rilevamento, e della VE tenuta. • Calcolare ora, con la solita formula, anche il tempo di arrivo in costa utilizzando sempre la VE ottenuta dal calcolo vettoriale. Esecuzione (RV = 204° VE = 10,3 nd) RilV = 138° h 21:16 PS = φ 41°16’·4N λ 009°09’·9E Arrivo a Cala Spinosa alle ore 21:28 Soluzione esercizio n. 31 107 2° problema della corrente 108 Nel primo problema della corrente l’effetto della deriva non viene contrastato e l’unità naviga su una RV diversa dalla PV tenuta. Il secondo problema della cor rente invece permette di calcolare graficamente quale PV tenere per contrastare la deriva e quindi navigare effettivamente su una predeterminata RV. N.B. LA CORRENTE MI FA CAMBIARE ROTTA SE MANTENGO UNA PRORA QUINDI CAMBIO PRORA PER MANTENERE LA ROTTA Si intende calcolare quale PV si deve tenere per contrastare la deriva mante nendosi in rotta, e quale sarà la velocità reale rispetto al fondo del mare, cioè la risultante del calcolo vettoriale, la VE. Nel GOVERNO della barca, RV e VE non hanno importanza! Ne hanno invece PB, PV, e VP, poiché sono gli unici elementi che si possono controllare da bordo. Se i calcoli sono giusti navigheremo co munque sulla RV voluta e con la VE cal colata. Nel caso dell’esempio a lato la VE risulta essere maggiore della VP; la corrente “arriva” da oltre il traverso più verso poppa, quindi “spinge” l’unità. DATI: INCOGNITE: DC VC PV VE VP RV ESEMPIO DI RISOLUZIONE DEL 2° PROBLEMA DATI: INCOGNITE: DC PV VC VE RV E.T.A. VP Partenza alle ore 08:00 con note RV, VP e DC, VC. Determinare PV da mantenere, VE con cui si navigherà e quindi si calcolerà il tempo d’arrivo, E.T.A. 2° problema della corrente 109 N.B. Al timoniere viene data la direzione in gradi da seguire dalla conversione della PV in PB. Dal punto di partenza, dopo aver trac ciato la RV, tracciare con la squadretta il vettore DC e con il compasso aperto, con la velocità in nodi della corrente, de finire la VC sullo stesso vettore. Con il compasso, prendere la misura rela tiva alla VP. Con la misura relativa a VP, puntare il compasso sulla fine del vettore DC/VC e con l’altra punta del compasso chiudere sulla RV per determinare il punto chiave. Segnare il punto chiave (Punto Sti mato dopo 1h di navigazione). Unire con una linea la fine del vettore DC/VC e il punto chiave, e su quella linea scrivere VP. DOVE SCRIVO VP LEGGO con la squa dretta il “contrario”, cioè PV. Sulla RV tra il punto di partenza e il punto chiave tracciato precedentemente si mi sura la VE, risultante del calcolo vetto riale, con la quale si effettuano i calcoli per l’E.T.A. Solo per completezza teorica, nell’esempio sopra è stato chiuso il parallelogramma. Nel 2° problema della corrente è sufficiente costruire graficamente il triangolo per de terminare gli elementi della navigazione, PV e VE. 2° problema della corrente 110 esercizio n. 32 Partiamo alle ore 10:00 a SE dell’Isola Rossa (φ 41°00’·6N λ 008°52’·6E) con VP 6,5 nd in direzione Castelsardo (φ 40°54’·9N λ 008°42’·3E). Sapendo che navigheremo con corrente W e VC 2,2 nd, determinare: PV, VE, E.T.A. per arrivare a Castelsardo. Descrizione • Dal punto a SE dell’isola tracciare il vettore DC/VC. • Dallo stesso punto di partenza tracciare la rotta per Castelsardo. • Aprire il compasso del valore corrispondente alla VP; posizionare una punta del compasso alla fine del vettore DC/VC e far tornare l’altra punta sulla RV per deter minare il punto chiave. • Unendo il punto chiave con la fine del vettore DC/VC si ottiene il vettore corrispon dente alla VP (riportata) e alla PV (da leggere con le squadrette = 222°) da tenere per contrastare la forza della corrente. • Sulla RV dal punto di partenza al punto chiave si misura la VE (8,1 nd), diversa dalla VP, perché risultante della somma vettoriale delle due forze (VP e VC). • Per determinare l’E.T.A. misurare le miglia da percorrere e utilizzare la VE. Tm = M/VE x 60 Esecuzione PV = 222° VE = 8,1 nd E.T.A. = h 11:11 Soluzione esercizio n. 32 111 2° problema della corrente 112 esercizio n. 33 Partiamo da Punta de Rondinara (φ 41°27’·8N λ 009°16’·7E) alle ore 09:13 con DC 035° e VC 1,5 nd. Navigando con VP 7° nd, determinare PV, PB, VE e E.T.A. per rag giungere il punto a Nord dell’Isola La Presa (φ 41°18’·7N λ 009°22’·6E). Descrizione • Da Punta de Sponsaglia tracciare il vettore DC/VC. • Unire il punto di partenza con il punto a Nord di La Presa per tracciare la RV. • Aprire il compasso del valore corrispondente alla VP; posizionare una punta del compasso alla fine del vettore DC/VC e far tornare l’altra punta sulla RV per deter minare il punto chiave. • Unendo il punto chiave con la fine del vettore DC/VC si ottiene il vettore corrispon dente alla VP (riportata) e alla PV (da leggere con le squadrette = 165°) da tenere per contrastare la forza della corrente. • Sulla RV, dal punto di partenza al punto chiave, si misura la VE (6,2 nd), diversa dalla VP, perché risultante della somma vettoriale delle due forze (VP e VC). • Per determinare l’E.T.A. misurare le miglia da percorrere e utilizzare la VE. Esecuzione PV = 165° PB = 164° VE = 6,1 nd E.T.A. = h 10:52 Soluzione d = 2°55’E esercizio n. 33 113
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16592	https://www.mathsgenie.co.uk/error-intervals.php	Error Intervals When we use error intervals we are looking at a number that has been rounded. An error interval tells us the range of numbers that we could have had before the number was rounded. Example 1: A number x has been rounded to one decimal placeThe result is 4.8 Write down the error interval for x The number has been rounded to 4.8 to one decimal place.The number below is 4.7 and the number above is 4.9 (to one decimal place) We need to look half way between 4.7 and 4.8 for the lower bound and half way between 4.8 and 4.9 for the upper bound. The number must have been between 4.75 and 4.85 to round to 4.8 to one decimal place. This can be written as an inequality.4.75 would round to 4.8, so we use an or equal symbol4.85 would not round to 4.8 we just use the less than symbol. 4.75 ≤ x < 4.85 Example 2: A number y has been rounded to one significant figureThe result is 7000 Write down the error interval for y The number below is 6000 and the number above is 8000 (to one significant figure) We need to look half way between 6000 and 7000 for the lower bound and half way between 7000 and 8000 for the upper bound. The number must have been between 6500 and 7500 to round to 7000 to one significant figure. We write this as an inequality. The number could have been equal to 6500, but not equal to 7500. 6500 ≤ y < 7500 Try these: A number x has been rounded to 1 decimal place.The result is 18.3Write an error interval for x. A number y has been rounded to 2 decimal places.The result is 8.03Write an error interval for y. A number z has been rounded to 3 significant figures.The result is 71.2Write an error interval for z. Truncating numbers When a number has been truncated it has been cut off. If we truncate a number to one decimal place we delete all digits after one decimal place. If we truncate a number to two decimal places we delete all digits after two decimal places. Example 3: Truncate 4.5891 to one decimal place We delete all the digits after one decimal place:4.5~~891~~4.5891 truncated to one decimal place is 4.5 Example 4: Truncate 19.68572 to two decimal places We remove all the digits after two decimal places:19.68~~572~~19.68572 truncated to two decimal place sis 19.68 Example 5: A number x has been truncated to one decimal placeThe result is 1.4Write an error interval for x The number must have began with 1.4 in order for it to be truncated to 1.4 All of the numbers between 1.4 and 1.5 begin with 1.4 The number could have been equal to 1.4 but not equal to 1.5 (It could have been anything up to 1.5, but not 1.5) 1.4 ≤ x < 1.5 Try these: A number x has been truncated to 1 decimal place.The result is 15.6Write an error interval for x. A number y has been truncated to 2 decimal places.The result is 1.32Write an error interval for y. Learn GCSE Maths Edexcel Exam Papers OCR Exam Papers AQA Exam Papers Eduqas Exam Papers Edexcel IGCSE Maths GCSE Statistics A Level Learn A Level Maths Edexcel A Level Papers AQA A Level Papers OCR A Level Papers OCR MEI A Level Papers Old Spec A Level Other Links Primary School Maths Free Maths Course More Resources Privacy and Cookie Policy This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License. Copyright © Maths Genie. Maths Genie Limited is a company registered in England and Wales with company number 14341280. Registered Office: 39 Church Street, Wye, Kent, TN25 5BL.
16593	https://www.palomar.edu/math/wp-content/uploads/sites/134/2019/01/Complex-numbers-for-trigonometric-identities.pdf	Complex numbers and Trigonometric Identities The shortest path between two truths in the real domain passes through the complex domain. Jacques Hadamard Simplicity in linearity • In Mathematics, we know that the distributive property states: • a(b + c) = ab + ac • But why is this even true to begin with? • Here is a visual proof where we can think of the real number values representing the lengths of rectangles and their products the area of their associated rectangles. • Even the proof for natural numbers takes effort. • Since 𝑚𝑚⋅𝑛𝑛is just 𝑛𝑛+ 𝑛𝑛+ ⋯+ 𝑛𝑛, repeated 𝑚𝑚times. Then by using the commutative property: 𝑎𝑎𝑏𝑏+ 𝑐𝑐= 𝑏𝑏+ 𝑐𝑐𝑎𝑎= 𝑎𝑎+ 𝑎𝑎+ ⋯+ 𝑎𝑎 𝑏𝑏+𝑐𝑐𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 (𝑎𝑎+ 𝑎𝑎+ ⋯+ 𝑎𝑎) 𝑏𝑏𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 + (𝑎𝑎+ 𝑎𝑎+ ⋯+ 𝑎𝑎) 𝑐𝑐𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑏𝑏⋅𝑎𝑎+ 𝑐𝑐⋅𝑎𝑎= 𝑎𝑎𝑎𝑎+ 𝑏𝑏𝑏𝑏 Simplicity in linearity Its definitively ALIVE!!! • Previous theorem show how you will see in Calculus 1 how the derivative of two functions does behave linearly. • What other mathematical objects have this nice linear property? • Lets take another result from calc 1, the definite integral Simplicity in Linearity • Linearity property of the definite integral As expected… 𝐸𝐸[𝑋𝑋] = ෍ 𝑖𝑖=1 𝑘𝑘 𝑥𝑥𝑖𝑖𝑝𝑝𝑥𝑥𝑖𝑖= 𝑥𝑥1𝑝𝑝𝑥𝑥1 + 𝑥𝑥2𝑝𝑝𝑥𝑥2 + ⋯+ 𝑥𝑥𝑘𝑘𝑝𝑝(𝑥𝑥𝑘𝑘) Discrete case: Continuous case: 𝐸𝐸𝑋𝑋= න ℝ 𝑥𝑥𝑥𝑥(𝑥𝑥) 𝑑𝑑𝑑𝑑 Lets go beyond calculus, and go into probability theory. Expect to see this in a statistics class (Math 120). The expected value of a random variable, intuitively, is the long-run average value of repetitions of the experiment it represents. As expected… a proof • For discrete random variables X and Y. By the basic definition of expected value, • Note that we have never used any properties of independence in this proof, and thus linearity of expectation holds for all random variables! • For continuous random variables, the proof is essentially the same except that the summations are replaced by integrals. Linear Transformations • A Linear Transformation (or linear map) is a special type of function where: F(u + v) = F(u) + F(v) and F(cv) = cF(v) for a constant/scalar c. These functions are extensively studied in Linear Algebra (Math 200) and get their name by always mapping a line into a line. The function f: ℝ2 →ℝ2 with f(x,y)=(2x,y) is a linear map. This function scales the x component of a vector by the factor 2. The function is additive: It doesn't matter whether first vectors are added and then mapped or whether they are mapped and finally added: f(u+v)=f(u)+f(v) The function is homogeneous: It doesn't matter whether a vector is first scaled and then mapped or first mapped and then scaled: f(cv)=cf(v) Is everything linear? • Can we apply this procedure universally, just because its easy? • Lets consider a “freshman sum“ from math students, when requested to add fractions • 2 7 + 3 5= 5 12 Right!! • In reality this “mistake” leads us to interesting area and study of numbers called the Mediant 𝑎𝑎 𝑏𝑏+ 𝑐𝑐 𝑑𝑑= 𝑎𝑎+𝑐𝑐 𝑏𝑏+𝑑𝑑. • IF we do allow this new interesting relationships can occur  More on mediant • Not todays topic, but want to entice your math curiosity, look at: • Farey sequences • Stern-Brocot tree Fundamental Tragedy of algebra • That sinister bug has raised his head, • And like a germ he is starting to spread. • Distributing exponents is the sign, • That this bug is on the climb. • Look at that student over there, • Distributing exponents without a care. • Please listen to your maker, • Distributing exponents will bring the undertaker. • Dear Lord please open your gates. • Being a math student was not his fate. • Distributing exponents was his only sin. • But that’s enough to do an algebra student in. • An example, his demise should serve, • For other students who haven’t heard, • Distributing exponents is a sin. • It’s enough to do an algebra student in. • • Donald E. Brook • Mt. San Antonio College • 𝑎𝑎+ 𝑏𝑏2 ≠𝑎𝑎2 + 𝑏𝑏2 • • 𝑎𝑎2 + 𝑏𝑏2 ≠𝑎𝑎+ 𝑏𝑏 • • 𝑎𝑎2 + 𝑏𝑏2 1/2 ≠𝑎𝑎+ 𝑏𝑏 • • 𝑎𝑎−1 + 𝑏𝑏−1 ≠𝑎𝑎+ 𝑏𝑏−1 • • 𝑎𝑎−1 + 𝑏𝑏−1 ≠− 1 𝑎𝑎+𝑏𝑏 • • 3 𝑎𝑎3 + 𝑏𝑏3 ≠𝑎𝑎+ 𝑏𝑏 • • 𝑎𝑎+ 𝑏𝑏3 ≠𝑎𝑎3 + 𝑏𝑏3 • • 𝑎𝑎+ 𝑏𝑏4 ≠𝑎𝑎4 + 𝑏𝑏4 • • 𝑎𝑎+ 𝑏𝑏 2 ≠𝑎𝑎+ 𝑏𝑏 Fundamental Tragedy of linearity • Linear breakdowns in other areas of mathematics: Algebra: logb(x) + logb(y) ≠ logb(x + y) Recall logb x=y ↔by=x Correct identity logb(x) + logb(y) = logb(xy), Proof: Let bA=x and bB=y, then logb(xy)=logb(bAbB)=logb(bA+B)=A+B= logb(x) + logb(y) Logic: ¬(p∧q)≠ ¬p∧¬q, likewise ¬(p∨q)≠ ¬p∨¬q Rather by DeMorgans law ¬(p∧q)= ¬p∨¬q likewise ¬(p∨q)= ¬p∧¬q p q p∧q ¬(p∧q) ¬p∧¬q ¬p∨¬q F F F T T T F T F T F T T F F T F T T T T F F F p q p∧q ¬(p∨q) ¬p∨¬q ¬p∧¬q F F F T T T F T F F T F T F F F T F T T T F F F Fundamental Tragedy of linearity • Linear breakdowns in other areas of mathematics: Sets: (𝐴𝐴∩𝐵𝐵)′ ≠𝐴𝐴𝐴∩𝐵𝐵𝐵likewise (𝐴𝐴∪𝐵𝐵)′ ≠𝐴𝐴𝐴∪𝐵𝐵𝐵 Rather by DeMorgans law (𝐴𝐴∩𝐵𝐵)′ ≠𝐴𝐴𝐴∪𝐵𝐵𝐵, proof below. Likewise 𝐴𝐴∪𝐵𝐵′ = 𝐴𝐴𝐴∩𝐵𝐵𝐵. Try proving this second version. Attempt at Linearity What should cos 𝑥𝑥+ 𝑦𝑦and sin 𝑥𝑥+ 𝑦𝑦be? Do these trigonometric functions behave linearly? Is cos 𝑥𝑥+ 𝑦𝑦= cos 𝑥𝑥+ cos 𝑦𝑦and sin 𝑥𝑥+ 𝑦𝑦= sin 𝑦𝑦+ sin 𝑦𝑦? Try with some known values: cos 𝜋𝜋 6 + 𝜋𝜋 3 = cos 𝜋𝜋 6 + cos 𝜋𝜋 3 cos 3𝜋𝜋 6 = cos 𝜋𝜋 6 + cos 𝜋𝜋 3 cos 𝜋𝜋 2 = 3 2 + 1 2 0 = 1+ 3 2 ? sin 𝜋𝜋 6 + 𝜋𝜋 3 = sin 𝜋𝜋 6 + sin 𝜋𝜋 3 sin 3𝜋𝜋 6 = sin 𝜋𝜋 6 + sin 𝜋𝜋 3 sin 𝜋𝜋 2 = 1 2 + 3 2 1 = 1+ 3 2 ? So that failed… let’s try distance! Find cos 𝑥𝑥−𝑦𝑦based on the unit circle. So that failed… let’s try distance! Find cos 𝑥𝑥−𝑦𝑦based on the unit circle. Label the coordinates of each point. x y x – y (cos x, sin x) (cos y, sin y) (cos (x – y), sin (x – y)) (1, 0) Figure 1 Figure 2 Find cos 𝑥𝑥−𝑦𝑦based on the unit circle. Distance between the two labeled points in Figure 1. 𝑑𝑑= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐2 + 𝑠𝑠𝑖𝑖𝑖𝑖𝑥𝑥−𝑠𝑠𝑠𝑠𝑠𝑠𝑦𝑦2 𝑑𝑑= 𝑐𝑐𝑐𝑐𝑐𝑐2𝑥𝑥−2𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐+ 𝑐𝑐𝑐𝑐𝑐𝑐2𝑦𝑦+ 𝑠𝑠𝑠𝑠𝑠𝑠2𝑥𝑥−2𝑠𝑠𝑠𝑠𝑠𝑠𝑥𝑥𝑠𝑠𝑠𝑠𝑠𝑠𝑦𝑦+ 𝑠𝑠𝑠𝑠𝑠𝑠2𝑦𝑦 𝑑𝑑= 𝑐𝑐𝑐𝑐𝑐𝑐2𝑥𝑥+ 𝑠𝑠𝑠𝑠𝑠𝑠2𝑥𝑥+ 𝑐𝑐𝑐𝑐𝑐𝑐2𝑦𝑦+ 𝑠𝑠𝑠𝑠𝑠𝑠2𝑦𝑦−2𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐−2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑= 1 + 1 −2𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐−2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑= 2 −2𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐−2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑= 2 −2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐+ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 Keep this in mind as we move to the next part. Find cos 𝑥𝑥−𝑦𝑦based on the unit circle. Distance between the two labeled points in Figure 2. 𝑑𝑑= 𝑐𝑐𝑐𝑐𝑐𝑐𝑥𝑥−𝑦𝑦−1 2 + 𝑠𝑠𝑖𝑖𝑖𝑖𝑥𝑥−𝑦𝑦−0 2 𝑑𝑑= 𝑐𝑐𝑐𝑐𝑐𝑐2 𝑥𝑥−𝑦𝑦−2𝑐𝑐𝑐𝑐𝑐𝑐𝑥𝑥−𝑦𝑦+ 1 + 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥−𝑦𝑦 𝑑𝑑= 𝑐𝑐𝑐𝑐𝑐𝑐2 𝑥𝑥−𝑦𝑦+ 𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥−𝑦𝑦 + 1 −2𝑐𝑐𝑐𝑐𝑐𝑐𝑥𝑥−𝑦𝑦 𝑑𝑑= 1 + 1 −2𝑐𝑐𝑐𝑐𝑐𝑐𝑥𝑥−𝑦𝑦= 2 −2𝑐𝑐𝑐𝑐𝑐𝑐𝑥𝑥−𝑦𝑦 Now compare this to the previous distance: 𝑑𝑑= 2 −2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐+ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 Since the distances must be the same, we conclude that: 𝒄𝒄𝒄𝒄𝒄𝒄𝒙𝒙−𝒚𝒚= 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄+ 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 Want 𝑠𝑠𝑖𝑖𝑖𝑖𝑥𝑥−𝑦𝑦, then just check the complement (cofunction Identity): 𝒔𝒔𝒔𝒔𝒔𝒔𝒙𝒙−𝒚𝒚= 𝑐𝑐𝑐𝑐𝑐𝑐𝜋𝜋 2 −𝑥𝑥−𝑦𝑦 = 𝑐𝑐𝑐𝑐𝑐𝑐 𝜋𝜋 2 −𝑥𝑥−−𝑦𝑦 = cos 𝜋𝜋 2 −𝑥𝑥cos −𝑦𝑦+ sin 𝜋𝜋 2 −𝑥𝑥sin(−𝑦𝑦) = 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔−𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 What about the other ones? You can use a similar picture to graph addition, but in this case, you’ll need to think of a clockwise rotation of an angle, so one angle will be negative. Summary: 𝑐𝑐𝑐𝑐𝑐𝑐𝑥𝑥± 𝑦𝑦= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐∓𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑥𝑥± 𝑦𝑦= 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠± 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 Deriving sum identity using SOHCAHTOA, and without the Unit circle. Deriving sum identity using SOHCAHTOA, and without the Unit circle. Deriving sum identity using SOHCAHTOA, and without the Unit circle. Deriving sum identity using SOHCAHTOA, and without the Unit circle. Deriving sum identity using SOHCAHTOA, and without the Unit circle. Deriving sum identity using SOHCAHTOA, and without the Unit circle. Deriving sum identity using SOHCAHTOA, and without the Unit circle. • Consider the small right triangle in the figure above, which gives Deriving sum identity using SOHCAHTOA, and without the Unit circle. • Now, the usual trigonometric definitions applied to the large right triangle give Deriving sum identity using SOHCAHTOA, and without the Unit circle. • Solving these two equations simultaneously for the variables sin(α+β) and cos(α +β) then immediately gives Deriving sum identity using SOHCAHTOA, and without the Unit circle. • These can be put into the familiar forms with the aid of the trigonometric identities • which can be verified by direct multiplication. • Plugging these back to the equations in the previous slide gives: Other pictures My eyes hurt • Is there another way, that does not involve the unit circle with distance formula or ‘weird’ pictures. • Is there a way that doesn’t require long arithmetic, memorizing or remembering some picture, or remembering some method to create a picture that would work? • Lets enter an imaginary word of opportunities. 33 Leonhard Euler [1707-1783] • Euler is considered the most prolific mathematician in history. (What about Erdős?) • His contemporaries called him “analysis incarnate.” • “He calculated without effort, just as men breathe or as eagles sustain themselves in the air.” The Saga of Mathematics: A Brief History 34 Leonhard Euler [1707-1783] •Euler was born in Basel, Switzerland, on April 15, 1707. •He received his first schooling from his father Paul, a Calvinist minister, who had studied mathematics under Jacob Bernoulli. •Euler's father wanted his son to follow in his footsteps and, in 1720 at the age of 14, sent him to the University of Basel to prepare for the ministry. The Saga of Mathematics: A Brief History 35 Leonhard Euler [1707-1783] •At the age of 15, he received his Bachelor’s degree. •In 1723 at the age of 16, Euler completed his Master's degree in philosophy having compared and contrasted the philosophical ideas of Descartes and Newton. •His father demanded he study theology and he did, but eventually through the persuading of Johann Bernoulli, Jacob’s brother, Euler switched to mathematics. The Saga of Mathematics: A Brief History 36 Leonhard Euler [1707-1783] • Euler completed his studies at the University of Basel in 1726. • He had studied many mathematical works including those by Varignon, Descartes, Newton, Galileo, von Schooten, Jacob Bernoulli, Hermann, Taylor and Wallis. • By 1727, he had already published a couple of articles on isochronous curves and submitted an entry for the 1727 Grand Prize of the French Academy on the optimum placement of masts on a ship. The Saga of Mathematics: A Brief History 37 Leonhard Euler [1707-1783] •Euler did not win but instead received an honorable mention. •He eventually would recoup from this loss by winning the prize 12 times. •What is interesting is that Euler had never been on a ship having come from landlocked Switzerland. •The strength of his work was in the analysis. The Saga of Mathematics: A Brief History 38 Leonhard Euler [1707-1783] • Euler was in a sense the creator of modern mathematical expression. • In terms of mathematical notation, Euler was the person who gave us: • π for pi •i for √−1 •∆y for the change in y •f(x) for a function • Σ for summation The Saga of Mathematics: A Brief History 39 Leonhard Euler [1707-1783] •To get an idea of the magnitude of Euler’s work it is worth noting that: •Euler wrote more than 500 books and papers during his lifetime – about 800 pages per year. •After Euler’s death, it took over forty years for the backlog of his work to appear in print. • Approximately 400 more publications. The Saga of Mathematics: A Brief History 40 Leonhard Euler [1707-1783] •He published so many mathematics articles that his collected works Opera Omnia already fill 73 large volumes – tens of thousands of pages – with more volumes still to come. •More than half of the volumes of Opera Omnia deal with applications of mathematics – acoustics, engineering, mechanics, astronomy, and optical devices (telescopes and microscopes). The Saga of Mathematics: A Brief History The number e and compound interest • Invest $1 • Interest rate 100% Interest applied each Sum at end of the year Year $2.00000 Half-year $2.25000 Quarter $2.44141 Month $2.61304 Week $2.69260 Day $2.71457 Hour $2.71813 Minute $2.71828 Second $2.71828 𝐴𝐴= 𝑃𝑃1 + 𝑟𝑟 𝑛𝑛 𝑛𝑛𝑡𝑡 “Compound interest is the most powerful force in the universe.” Albert Einstein. What about ‘continuous’ compounding? The number e as a limit 𝑒𝑒= lim 𝑛𝑛→∞1 + 1 𝑛𝑛 𝑛𝑛 = 2.7182818284590452353602874... So for continuous compounding, let x=n/r we would get 𝐴𝐴= lim 𝑛𝑛→∞𝑃𝑃1 + 𝑟𝑟 𝑛𝑛 𝑛𝑛𝑡𝑡 = 𝑃𝑃⋅lim 𝑥𝑥→∞1 + 1 𝑥𝑥 𝑥𝑥𝑥𝑥𝑡𝑡 = 𝑃𝑃lim 𝑥𝑥→∞1 + 1 𝑥𝑥 𝑥𝑥𝑟𝑟𝑟𝑟 = 𝑃𝑃𝑒𝑒𝑟𝑟𝑟𝑟 Euler on complex numbers Of such numbers we may truly assert that they are neither nothing, nor greater than nothing, nor less than nothing, which necessarily constitutes them imaginary or impossible. Complex Numbers William Rowan Hamilton 1805 - 1865 We define a complex number as a pair (a, b) of real numbers. Complex Numbers William Rowan Hamilton 1805 - 1865 We define a complex number as a pair (a, b) of real numbers. They are added as follows: (a, b) + (c, d) = (a + c, b + d); (1, 2) + (3, 4) = (4, 6) Complex Numbers William Rowan Hamilton 1805 - 1865 We define a complex number as a pair (a, b) of real numbers. They are added as follows: (a, b) + (c, d) = (a + c, b + d); They are multiplied as follows: (a, b) x (c, d) = (ac - bd, ad + bc); (1, 2) × (3, 4) = (3 – 8, 4 + 6) = (-5, 10) Complex Numbers William Rowan Hamilton 1805 - 1865 We define a complex number as a pair (a, b) of real numbers. They are added as follows: (a, b) + (c, d) = (a + c, b + d); They are multiplied as follows: (a, b) x (c, d) = (ac - bd, ad + bc); The pair (a, 0) then corresponds to the real number a the pair (0, 1) corresponds to the imaginary number i Complex Numbers William Rowan Hamilton 1805 - 1865 We define a complex number as a pair (a, b) of real numbers. They are added as follows: (a, b) + (c, d) = (a + c, b + d); They are multiplied as follows: (a, b) x (c, d) = (ac - bd, ad + bc); The pair (a, 0) then corresponds to the real number a the pair (0, 1) corresponds to the imaginary number i Then (0, 1) x (0, 1) = (-1, 0), which corresponds to the relation i x i = i 2= - 1. Representing Complex numbers geometrically Caspar Wessel in 1799 In this representation, called the complex plane, two axes are drawn at right angles – the real axis and the imaginary axis – and the complex number a + b −1 is represented by the point at a distance a in the direction of the real axis and at height b in the direction of the imaginary axis. Complicated Extensions Powers of the complex expression (1+iπ/n)n • We can approximate numerically these complex powers without evaluating trigonometric functions The exponential function ez can be defined as the limit of (1 + z/n)n, as n approaches infinity, and thus eiπ is the limit of (1 + iπ/n)n. In this animation n takes various increasing values from 1 to 100. The computation of (1 + iπ/n)n is displayed as the combined effect of n repeated multiplications in the complex plane, with the final point being the actual value of (1 + iπ/n)n. As n gets larger (1 + iπ/n)n approaches a limit of −1. This animation depicts points moving along the graphs of the sine function (in blue) and the cosine function (in green) corresponding to a point moving around the unit circle Source: 𝑒𝑒𝑖𝑖𝑖𝑖= cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥 Three-dimensional visualization of Euler's formula. See also circular polarization. Tailor made for math Borrowing results from calculus we can find the Taylor series expansions for the transcendental functions: cos 𝑥𝑥= 1 -𝑥𝑥2 2! + 𝑥𝑥4 4! -𝑥𝑥6 6! + 𝑥𝑥8 8! -𝑥𝑥10 10! + 𝑥𝑥12 12! + ⋅⋅⋅ sin 𝑥𝑥= 𝑥𝑥-𝑥𝑥3 3! + 𝑥𝑥5 5! -𝑥𝑥7 7! + 𝑥𝑥9 9! -𝑥𝑥11 11! + 𝑥𝑥13 13! + ⋅⋅⋅ 𝑒𝑒𝑥𝑥=1 + 𝑥𝑥+ 𝑥𝑥2 2! + 𝑥𝑥3 3! + 𝑥𝑥4 4! + 𝑥𝑥5 5! + 𝑥𝑥6 6! + 𝑥𝑥7 7! + 𝑥𝑥8 8! + ⋅⋅⋅ 𝑥𝑥is measured in radians The exponential function ex (in blue), and the sum of the first n + 1 terms of its Taylor series at 0 (in red). 𝑒𝑒𝑖𝑖𝑖𝑖= cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥 cos 𝑥𝑥= 1 - 𝑥𝑥2 2! + 𝑥𝑥4 4! - 𝑥𝑥6 6! + 𝑥𝑥8 8! - 𝑥𝑥10 10! + 𝑥𝑥12 12! + ⋅⋅⋅ 𝑖𝑖sin 𝑥𝑥=𝑖𝑖𝑥𝑥- 𝑖𝑖𝑥𝑥3 3! + 𝑖𝑖𝑥𝑥5 5! - 𝑖𝑖𝑥𝑥7 7! +𝑖𝑖𝑥𝑥9 9! - 𝑖𝑖𝑥𝑥11 11! + 𝑖𝑖𝑥𝑥13 13! + ⋅⋅⋅ 𝑒𝑒𝑖𝑖𝑖𝑖= cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥 cos 𝑥𝑥= 1 - 𝑥𝑥2 2! + 𝑥𝑥4 4! - 𝑥𝑥6 6! + 𝑥𝑥8 8! - 𝑥𝑥10 10! + 𝑥𝑥12 12! + ⋅⋅⋅ 𝑖𝑖sin 𝑥𝑥=𝑖𝑖𝑥𝑥- 𝑖𝑖𝑥𝑥3 3! + 𝑖𝑖𝑥𝑥5 5! - 𝑖𝑖𝑥𝑥7 7! +𝑖𝑖𝑥𝑥9 9! - 𝑖𝑖𝑥𝑥11 11! + 𝑖𝑖𝑥𝑥13 13! + ⋅⋅⋅ Add to get 1 + 𝑖𝑖𝑥𝑥-𝑥𝑥2 2! - 𝑖𝑖 𝑥𝑥3 3! + 𝑥𝑥4 4! + 𝑖𝑖 𝑥𝑥5 5! -𝑥𝑥6 6! - 𝑖𝑖 𝑥𝑥7 7! + 𝑥𝑥8 8! + 𝑖𝑖 𝑥𝑥9 9! ⋅⋅⋅ 𝑒𝑒𝑖𝑖𝑖𝑖= cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥 cos 𝑥𝑥= 1 - 𝑥𝑥2 2! + 𝑥𝑥4 4! - 𝑥𝑥6 6! + 𝑥𝑥8 8! - 𝑥𝑥10 10! + 𝑥𝑥12 12! + ⋅⋅⋅ 𝑖𝑖sin 𝑥𝑥=𝑖𝑖𝑥𝑥- 𝑖𝑖𝑥𝑥3 3! + 𝑖𝑖𝑥𝑥5 5! - 𝑖𝑖𝑥𝑥7 7! +𝑖𝑖𝑥𝑥9 9! - 𝑖𝑖𝑥𝑥11 11! + 𝑖𝑖𝑥𝑥13 13! + ⋅⋅⋅ Add to get 1 + 𝑖𝑖𝑥𝑥-𝑥𝑥2 2! - 𝑖𝑖 𝑥𝑥3 3! + 𝑥𝑥4 4! + 𝑖𝑖 𝑥𝑥5 5! -𝑥𝑥6 6! - 𝑖𝑖 𝑥𝑥7 7! + 𝑥𝑥8 8! + 𝑖𝑖 𝑥𝑥9 9! ⋅⋅⋅ Lets compare it with 𝑒𝑒𝑖𝑖𝑥𝑥= 1+ 𝑖𝑖𝑥𝑥 1! + (𝑖𝑖𝑖𝑖)2 2! + (𝑖𝑖𝑖𝑖)3 3! + (𝑖𝑖𝑖𝑖)4 4! + (𝑖𝑖𝑖𝑖)5 5! + (𝑖𝑖𝑖𝑖)6 6! + (𝑖𝑖𝑖𝑖)7 7! + ⋅⋅⋅ 𝑒𝑒𝑖𝑖𝑖𝑖= cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥 Add to get 1 + 𝑖𝑖𝑥𝑥- 𝑥𝑥2 2! - 𝑖𝑖𝑥𝑥3 3! + 𝑥𝑥4 4! + 𝑖𝑖𝑥𝑥5 5! - 𝑥𝑥6 6! - 𝑖𝑖𝑥𝑥7 7! + 𝑥𝑥8 8! + 𝑖𝑖𝑥𝑥9 9! ⋅⋅⋅ which is 𝑒𝑒𝑖𝑖𝑥𝑥= 1+ 𝑖𝑖𝑥𝑥 1! + (𝑖𝑖𝑖𝑖)2 2! + (𝑖𝑖𝑖𝑖)3 3! + (𝑖𝑖𝑖𝑖)4 4! + (𝑖𝑖𝑖𝑖)5 5! + (𝑖𝑖𝑖𝑖)6 6! + (𝑖𝑖𝑖𝑖)7 7! + ⋅⋅⋅ Note: 𝑖𝑖2 = -1 𝑖𝑖3 = - 𝑖𝑖 𝑖𝑖4 = 1 𝑖𝑖5 = 𝑖𝑖and so on Euler’s formula in Introductio, 1748 From which it can be worked out in what way the exponentials of imaginary quantities can be reduced to the sines and cosines of real arcs 𝑒𝑒𝑖𝑖𝑖𝑖= cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥 Now that we have the basics, let’s check how it works: • 𝑒𝑒𝑖𝑖𝜋𝜋 3 = cos 𝜋𝜋 3 + 𝑖𝑖sin 𝜋𝜋 3 = 1 2 + 3 2 𝑖𝑖. • This corresponds to the point 1 2 , 3 2 on the unit circle. Now if we set 𝑥𝑥equal to π 𝑒𝑒𝑖𝑖π = cos π+ 𝑖𝑖sin π and use cos π = -1 and sin π = 0 giving 𝑒𝑒𝑖𝑖π = -1 or 𝒆𝒆𝒊𝒊𝝅𝝅+ 1 = 0 𝑒𝑒𝑖𝑖𝜋𝜋+ 1 = 0 This links five of the most important constants in mathematics: • 0 is the additive identity, which when added to any number leaves the number unchanged • 1 is the multiplicative identity, which multiplied by any number leaves the number unchanged • e of the exponential function which we have defined before and base of the natural logarithms • 𝜋𝜋which is the ratio of a circle’s circumference to its diameter • i is the imaginary unit, which is the square root of -1 • Richard Feynman, an incredibly famous physicist, claimed this was the jewel of mathematics. Some have written that because of the innate beauty and simplicity of this equation, that Euler used it as a proof that god must exist. 𝑒𝑒𝑖𝑖𝑖𝑖= cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥 Futher unexpected results: if we raise the imaginary unit to itself as a power, the answer is a REAL number. Set 𝑥𝑥equal to π/2 and use cos π/2 = 0 and sin π/2 = 1 Then raise both sides to the power of i. 𝑖𝑖𝑖𝑖= (𝑒𝑒𝑖𝑖(𝜋𝜋/2))𝑖𝑖= 𝑒𝑒𝑖𝑖∗𝑖𝑖𝜋𝜋/2 = 𝑒𝑒−𝜋𝜋/2 = 𝑒𝑒𝜋𝜋−1/2 = 1 𝑒𝑒𝜋𝜋=0.207879576350… 𝑒𝑒𝑖𝑖𝑖𝑖= cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥 Set 𝑥𝑥equal to π/2 and use cos π/2 = 0 and sin π/2 = 1 Then raise both sides to the power of i. 𝑖𝑖𝑖𝑖= 1 𝑒𝑒𝜋𝜋=0.207879576350… “… we have not the slightest idea of what this equation means , but we may be certain that it means something very important” Benjamin Peirce Why is this formula so powerful? 𝑒𝑒𝑖𝑖𝑥𝑥+𝑦𝑦= cos 𝑥𝑥+ 𝑦𝑦+ 𝑖𝑖sin 𝑥𝑥+ 𝑦𝑦 (1) 𝑒𝑒𝑖𝑖𝑥𝑥+𝑦𝑦= 𝑒𝑒𝑖𝑖𝑥𝑥𝑒𝑒𝑖𝑖𝑖𝑖 𝑒𝑒𝑖𝑖𝑥𝑥+𝑦𝑦= cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥 cos 𝑦𝑦+ 𝑖𝑖sin 𝑦𝑦 𝑒𝑒𝑖𝑖𝑥𝑥+𝑦𝑦= cos 𝑥𝑥cos 𝑦𝑦+ 𝑖𝑖sin 𝑥𝑥cos 𝑦𝑦+ 𝑖𝑖sin 𝑦𝑦cos 𝑥𝑥+ 𝑖𝑖2 sin 𝑥𝑥sin 𝑦𝑦 𝑒𝑒𝑖𝑖𝑥𝑥+𝑦𝑦= cos 𝑥𝑥cos 𝑦𝑦+ 𝑖𝑖sin 𝑥𝑥cos 𝑦𝑦+ 𝑖𝑖sin 𝑦𝑦cos 𝑥𝑥−sin 𝑥𝑥sin 𝑦𝑦 𝑒𝑒𝑖𝑖𝑥𝑥+𝑦𝑦= cos 𝑥𝑥cos 𝑦𝑦−sin 𝑥𝑥sin 𝑦𝑦+ 𝑖𝑖sin 𝑥𝑥cos 𝑦𝑦+ sin 𝑦𝑦cos 𝑥𝑥 (2) At this point, we will need to equate the real parts and the imaginary parts of the two equations. Why is this formula so powerful? 𝑒𝑒𝑖𝑖𝑥𝑥+𝑦𝑦= cos 𝑥𝑥+ 𝑦𝑦 + 𝑖𝑖sin 𝑥𝑥+ 𝑦𝑦 𝑒𝑒𝑖𝑖𝑥𝑥+𝑦𝑦= cos 𝑥𝑥cos 𝑦𝑦−sin 𝑥𝑥sin 𝑦𝑦 + 𝑖𝑖sin 𝑥𝑥cos 𝑦𝑦+ sin 𝑦𝑦cos 𝑥𝑥 Once you do this, you can see two identities with just one quick expansion of the distributive property: • Red (Real): cos 𝑥𝑥+ 𝑦𝑦= cos 𝑥𝑥cos 𝑦𝑦−sin 𝑥𝑥sin 𝑦𝑦 • Green (Imaginary): sin 𝑥𝑥+ 𝑦𝑦= sin 𝑥𝑥cos 𝑦𝑦+ sin 𝑦𝑦cos 𝑥𝑥 You could continue doing this with 𝑒𝑒𝑖𝑖𝑥𝑥−𝑦𝑦as well; no pictures needed, and very little algebra. As a bonus, the derivation is extremely quick, and you get two formulas each time! Why is this formula so powerful? What about other identities? Maybe Even/Odd? 𝑒𝑒𝑖𝑖−𝑥𝑥= cos −𝑥𝑥+ 𝑖𝑖sin −𝑥𝑥 𝑒𝑒𝑖𝑖−𝑥𝑥= 𝑒𝑒−𝑖𝑖𝑖𝑖= 1 𝑒𝑒𝑖𝑖𝑖𝑖= 1 cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥 Multiply by the complex conjugate. 1 cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥 cos 𝑥𝑥−𝑖𝑖sin 𝑥𝑥 cos 𝑥𝑥−𝑖𝑖sin 𝑥𝑥 = cos 𝑥𝑥−𝑖𝑖sin 𝑥𝑥 cos2 𝑥𝑥+ sin2 𝑥𝑥= cos 𝑥𝑥+ 𝑖𝑖−sin 𝑥𝑥 So this means that cos −𝑥𝑥= cos 𝑥𝑥and sin −𝑥𝑥= −sin 𝑥𝑥 No pictures… just a little algebra! Application to Trigonometric Identities What about other identities? 𝑒𝑒𝑖𝑖𝑖𝑖𝑖= cos 2 𝜃𝜃+ 𝑖𝑖sin 2 𝜃𝜃 On the other hand, 𝑒𝑒𝑖𝑖𝑖𝑖𝑖= 𝑒𝑒𝑖𝑖𝑖𝑖2 = cos 𝜃𝜃+ 𝑖𝑖sin 𝜃𝜃2 = cos2 𝜃𝜃−sin2 𝜃𝜃+ 𝑖𝑖2 cos 𝜃𝜃sin 𝜃𝜃 Equating real and imaginary parts of the two expressions yield 𝑡𝑡ℎ𝑒𝑒𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎identities: cos 2 𝜃𝜃= cos2 𝜃𝜃−sin2 𝜃𝜃 sin 2 𝜃𝜃= 2 cos 𝜃𝜃sin 𝜃𝜃 69 Application to Trigonometric Identities (cont.) In general, 𝑒𝑒𝑖𝑖𝑖𝑖𝑖𝑖= cos 𝑛𝑛𝜃𝜃+ 𝑖𝑖sin 𝑛𝑛𝜃𝜃= 𝑒𝑒𝑖𝑖𝑖𝑖𝑛𝑛= cos 𝜃𝜃+ 𝑖𝑖sin 𝜃𝜃𝑛𝑛 Expand using binomial theorem, then equate real and imaginary parts to obtain new identities. 𝑒𝑒𝑖𝑖𝑖𝑖= cos 𝑧𝑧+ 𝑖𝑖sin 𝑧𝑧 𝑒𝑒−𝑖𝑖𝑖𝑖= cos 𝑧𝑧−𝑖𝑖sin 𝑧𝑧 ⇒cos 𝑧𝑧= 𝑒𝑒𝑖𝑖𝑖𝑖+ 𝑒𝑒−𝑖𝑖𝑖𝑖 2 , sin 𝑧𝑧= 𝑒𝑒𝑖𝑖𝑖𝑖−𝑒𝑒−𝑖𝑖𝑖𝑖 2𝑖𝑖 ⇒cos 𝑖𝑖𝑖𝑖= 𝑒𝑒𝑧𝑧+ 𝑒𝑒−𝑧𝑧 2 = cosh 𝑧𝑧, sin 𝑖𝑖𝑖𝑖= −𝑒𝑒𝑧𝑧−𝑒𝑒−𝑧𝑧 2𝑖𝑖 = 𝑖𝑖sinh 𝑧𝑧 Connecting them to the hyperbolic cosine and hyperbolic sine 70 DeMoivre’s Theorem 71 θ 2 k θ π − 2 k θ π + x y z z 𝑧𝑧𝑛𝑛= 𝑟𝑟𝑒𝑒𝑖𝑖𝑖𝑖𝑛𝑛= 𝑟𝑟𝑛𝑛𝑒𝑒𝑖𝑖𝑖𝑖𝑖𝑖= 𝑟𝑟𝑛𝑛cos 𝑛𝑛𝜃𝜃+ 𝑖𝑖sin 𝑛𝑛𝜃𝜃= 𝑟𝑟𝑛𝑛∠𝑛𝑛𝑛𝑛 Roots of a Complex Number 𝑧𝑧𝑛𝑛= 𝑟𝑟𝑒𝑒𝑖𝑖𝑖𝑖𝑛𝑛= 𝑟𝑟𝑛𝑛𝑒𝑒𝑖𝑖𝑖𝑖𝑖𝑖= 𝑟𝑟𝑛𝑛cos 𝑛𝑛𝜃𝜃+ 𝑖𝑖sin 𝑛𝑛𝜃𝜃= 𝑟𝑟𝑛𝑛∠𝑛𝑛𝑛𝑛(DeMoivre′s Theorem) Applies also for 𝑛𝑛not an integer, but in this case, the result Example: 𝑛𝑛th root of a complex number: 72 Roots of a Complex Number (cont.) −8𝑖𝑖 1 3 = ൞ 3 −𝑖𝑖, 2𝑖𝑖, −3 −𝑖𝑖 Note that the 𝑛𝑛th root of 𝑧𝑧can also be expressed in terms z x y 8i − u v w 𝑤𝑤= 𝑧𝑧1/3 = −8𝑖𝑖1/3 Re Im 1 0 ∠° 1 120 ∠ ° 1 240 ∠ ° Cube root of unity 73 𝑤𝑤= 𝑢𝑢+ 𝑖𝑖𝑖𝑖 Could we do more? Heck yes! We could use this to find the sum to product formulas, but they require the ability to remember a substitution in the middle of the problem. These types of substitutions do come up in calculus, but we’ll leave them off for now. What kind of substitutions? When you look at 𝑥𝑥, you probably don’t immediately think 𝑥𝑥= 𝑥𝑥+ 0. Further, you probably don’t think 𝑥𝑥+ 0 = 𝑥𝑥+ 𝑦𝑦 2 −𝑦𝑦 2, right? Oh, and once you do think about these, do you immediately think: 𝑥𝑥+ 𝑦𝑦 2 −𝑦𝑦 2 = 𝑥𝑥 2 + 𝑥𝑥 2 + 𝑦𝑦 2 −𝑦𝑦 2 = 𝑥𝑥 2 + 𝑦𝑦 2 + 𝑥𝑥 2 −𝑦𝑦 2 = 𝑥𝑥+𝑦𝑦 2 + 𝑥𝑥−𝑦𝑦 2 .  Well, those are the types of substitutions needed for the sum-to-product formulas. Why use it? Euler’s formula is a tool you can use, but don’t have to. Like many things in math, it is extremely useful at times and not as useful in other situations. Try it out. As you get comfortable with it, you may find some very interesting results! Bonus, this opens up a whole new world. Indeed, if 𝑒𝑒𝑖𝑖𝜋𝜋= −1, which we confirmed that it did on an earlier slide, then we could rewrite this using logarithms: 𝑒𝑒𝑖𝑖𝜋𝜋= −1 ↔𝑙𝑙𝑙𝑙−1 = 𝑖𝑖𝑖𝑖 So logarithms could be defined over negative and even complex numbers if we allowed complex number outputs. Test this on your calculator in complex mode to see the result. Back when covering logarithms, the domain was restricted to all non-negative real numbers… but that was needed to get the result to be a real number. Expanding our definition/domain will allow us to do MORE, not less. Now, go explore for yourselves! How Important are the Addition Formulas? Back in the 1950s, Professor Hans Rademacher showed that all of trigonometry could be developed with just two functions that he called “C” and “S” where: 1. 𝐶𝐶𝑥𝑥−𝑦𝑦= 𝐶𝐶𝑥𝑥𝐶𝐶𝑦𝑦+ 𝑆𝑆𝑥𝑥𝑆𝑆𝑦𝑦 2. 𝑆𝑆𝑥𝑥−𝑦𝑦= 𝑆𝑆𝑥𝑥𝐶𝐶𝑦𝑦−𝑆𝑆𝑥𝑥𝐶𝐶𝑦𝑦 3. lim 𝑥𝑥→0+ 𝑆𝑆𝑥𝑥 𝑥𝑥 = 1 Those of you who have taken calculus may recognize (3) as one of the most important limits in calculus, necessary for the formulation of trigonometric derivatives. Mathematics Teacher Vol L (January 1957) pp. 45-48. Calc link (power reduction) න𝑐𝑐𝑐𝑐𝑐𝑐2 𝑥𝑥𝑑𝑑𝑑𝑑= න𝑐𝑐𝑐𝑐𝑐𝑐2𝑥𝑥+ 1 2 𝑑𝑑𝑑𝑑 𝑒𝑒𝑖𝑖𝑖𝑖= cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥 𝑒𝑒−𝑖𝑖𝑖𝑖= cos 𝑥𝑥−𝑖𝑖sin 𝑥𝑥 So 𝑒𝑒𝑖𝑖𝑖𝑖+ 𝑒𝑒−𝑖𝑖𝑖𝑖= 2cos 𝑥𝑥which means cos 𝑥𝑥= 𝑒𝑒𝑖𝑖𝑖𝑖+𝑒𝑒−𝑖𝑖𝑖𝑖 2 This means 𝑐𝑐𝑐𝑐𝑐𝑐2 𝑥𝑥= 𝑒𝑒𝑖𝑖𝑖𝑖+𝑒𝑒−𝑖𝑖𝑖𝑖 2 2 = 𝑒𝑒𝑖𝑖𝑖𝑖𝑖+2+𝑒𝑒−𝑖𝑖𝑖𝑖𝑖 4 = 2𝑐𝑐𝑐𝑐𝑐𝑐2𝑥𝑥+2 4 = 𝑐𝑐𝑐𝑐𝑐𝑐2𝑥𝑥+1 2 Calc link (power reduction) Expanding the technique from the previous page, we can even do something like this: න𝑐𝑐𝑐𝑐𝑐𝑐6 𝑥𝑥𝑑𝑑𝑑𝑑= න2𝑐𝑐𝑐𝑐𝑐𝑐6𝑥𝑥+ 12𝑐𝑐𝑐𝑐𝑐𝑐4𝑥𝑥+ 30𝑐𝑐𝑐𝑐𝑐𝑐2𝑥𝑥+ 20 64 𝑑𝑑𝑑𝑑 It helps to remember Pascal’s triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 Calc link (power reduction) cos 𝑥𝑥= 𝑒𝑒𝑖𝑖𝑖𝑖+𝑒𝑒−𝑖𝑖𝑖𝑖 2 , which means 𝑒𝑒𝑖𝑖𝑖𝑖+ 𝑒𝑒−𝑖𝑖𝑖𝑖= 2cos 𝑥𝑥. cos6 𝑥𝑥= 𝑒𝑒𝑖𝑖𝑖𝑖+ 𝑒𝑒−𝑖𝑖𝑖𝑖 2 6 = 𝑒𝑒𝑖𝑖𝑖𝑖+ 𝑒𝑒−𝑖𝑖𝑖𝑖6 26 Using Pascal’s Triangle for coefficients: 𝑎𝑎+ 𝑏𝑏6 = 𝑎𝑎6 + 6𝑎𝑎5𝑏𝑏+ 15𝑎𝑎4𝑏𝑏2 + 20𝑎𝑎3𝑏𝑏3 +15𝑎𝑎2 𝑏𝑏4 + 6𝑎𝑎𝑏𝑏5 + 𝑏𝑏6 Expanding the binomial: 𝑒𝑒𝑖𝑖𝑖𝑖+ 𝑒𝑒−𝑖𝑖𝑖𝑖6 = 𝑒𝑒𝑖𝑖𝑖𝑖6 + 6 𝑒𝑒𝑖𝑖𝑖𝑖5 𝑒𝑒−𝑖𝑖𝑖𝑖+ 15 𝑒𝑒𝑖𝑖𝑖𝑖4 𝑒𝑒−𝑖𝑖𝑖𝑖2 + 20 𝑒𝑒𝑖𝑖𝑖𝑖3 𝑒𝑒−𝑖𝑖𝑖𝑖3 + 15 𝑒𝑒𝑖𝑖𝑖𝑖2 𝑒𝑒−𝑖𝑖𝑖𝑖4 + 6 𝑒𝑒𝑖𝑖𝑖𝑖 𝑒𝑒−𝑖𝑖𝑖𝑖5 + 𝑒𝑒−𝑖𝑖𝑖𝑖6 Simplifying exponents: 𝑒𝑒𝑖𝑖𝑖𝑖+ 𝑒𝑒−𝑖𝑖𝑖𝑖6 = 𝑒𝑒𝑖𝑖6𝑥𝑥+ 6𝑒𝑒𝑖𝑖5𝑥𝑥𝑒𝑒−𝑖𝑖𝑖𝑖+ 15𝑒𝑒𝑖𝑖4𝑥𝑥𝑒𝑒−𝑖𝑖2𝑥𝑥+ 20𝑒𝑒𝑖𝑖3𝑥𝑥𝑒𝑒−𝑖𝑖3𝑥𝑥+ 15𝑒𝑒𝑖𝑖2𝑥𝑥𝑒𝑒−𝑖𝑖4𝑥𝑥+ 6𝑒𝑒𝑖𝑖𝑖𝑖𝑒𝑒−𝑖𝑖5𝑥𝑥+ 𝑒𝑒−𝑖𝑖6𝑥𝑥 Calc link (power reduction) Simplifying exponents: 𝑒𝑒𝑖𝑖𝑖𝑖+ 𝑒𝑒−𝑖𝑖𝑖𝑖6 = 𝑒𝑒𝑖𝑖6𝑥𝑥+ 𝑒𝑒−𝑖𝑖6𝑥𝑥+ 6𝑒𝑒𝑖𝑖4𝑥𝑥+ 6𝑒𝑒−𝑖𝑖4𝑥𝑥+ 15𝑒𝑒𝑖𝑖2𝑥𝑥+ 15𝑒𝑒−𝑖𝑖2𝑥𝑥+ 20𝑒𝑒𝑖𝑖0𝑥𝑥 Grouping like objects: 𝑒𝑒𝑖𝑖𝑖𝑖+ 𝑒𝑒−𝑖𝑖𝑖𝑖6 = 𝑒𝑒𝑖𝑖6𝑥𝑥+ 𝑒𝑒−𝑖𝑖6𝑥𝑥+ 6 𝑒𝑒𝑖𝑖4𝑥𝑥+ 𝑒𝑒−𝑖𝑖4𝑥𝑥+ 15 𝑒𝑒𝑖𝑖2𝑥𝑥+ 15𝑒𝑒−𝑖𝑖2𝑥𝑥+ 20 Euler’s formula then substitutes since 𝑒𝑒𝑖𝑖𝑖𝑖+ 𝑒𝑒−𝑖𝑖𝑖𝑖= 2cos 𝑥𝑥: 𝑒𝑒𝑖𝑖𝑖𝑖+ 𝑒𝑒−𝑖𝑖𝑖𝑖6 = 2𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐+ 6 2𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐+ 15 2𝑐𝑐𝑐𝑐𝑐𝑐2𝑥𝑥+ 20 This shows why 𝑐𝑐𝑐𝑐𝑐𝑐6 𝑥𝑥= 2𝑐𝑐𝑐𝑐𝑐𝑐6𝑥𝑥+12𝑐𝑐𝑐𝑐𝑐𝑐4𝑥𝑥+30𝑐𝑐𝑐𝑐𝑐𝑐2𝑥𝑥+20 64 . And now you can do the integration in calculus II much quicker: න𝑐𝑐𝑐𝑐𝑐𝑐6 𝑥𝑥𝑑𝑑𝑑𝑑= න2𝑐𝑐𝑐𝑐𝑐𝑐6𝑥𝑥+ 12𝑐𝑐𝑐𝑐𝑐𝑐4𝑥𝑥+ 30𝑐𝑐𝑐𝑐𝑐𝑐2𝑥𝑥+ 20 64 𝑑𝑑𝑑𝑑 Lets raise the roof! Recall: 𝑒𝑒𝑥𝑥= 1 + 𝑥𝑥+ 𝑥𝑥2 2! + 𝑥𝑥3 3! + ⋯= ෍ 𝑛𝑛=0 ∞𝑥𝑥𝑛𝑛 𝑛𝑛! Define extension to complex variable(𝑥𝑥→𝑧𝑧= 𝑥𝑥+ 𝑖𝑖𝑖𝑖): 𝑒𝑒𝑧𝑧≡1 + 𝑧𝑧+ 𝑧𝑧2 2! + 𝑧𝑧3 3! + ⋯= ෍ 𝑛𝑛=0 ∞𝑧𝑧𝑛𝑛 𝑛𝑛! (converges for all 𝑧𝑧) ⇒𝑒𝑒𝑖𝑖𝑖𝑖= ෍ 𝑛𝑛=0 ∞ 𝑖𝑖𝑖𝑖𝑛𝑛 𝑛𝑛! = 1 −𝜃𝜃2 2! + 𝜃𝜃4 4! −⋯+ 𝑖𝑖𝜃𝜃−𝜃𝜃3 3! + 𝜃𝜃5 5! −⋯ = cos 𝜃𝜃+ 𝑖𝑖sin 𝜃𝜃 ⇒𝑒𝑒𝑖𝑖𝑖𝑖= cos 𝜃𝜃+ 𝑖𝑖sin 𝜃𝜃𝑒𝑒−𝑖𝑖𝑖𝑖= cos 𝜃𝜃−𝑖𝑖sin 𝜃𝜃 81 Further connections In Calculus 2 (Math 141) you will see Taylor and McLaurin series. Used to confirm Euler’s formula 𝑒𝑒𝑖𝑖𝑖𝑖= cos 𝑥𝑥+ 𝑖𝑖sin 𝑥𝑥. In Discrete Math (Math 245) you can prove that the previous formulas are true for all values of n, using formal mathematical induction. Calc I and II (Math 140 and 141) often must substitute in power reduction formulas for powers of trigonometric functions.
16594	https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14?srsltid=AfmBOoq2REHfBiMZ8aDusL5cTl5_IgkEZjHE3SbQyBbrVI7E6_20B1gL	Art of Problem Solving 2004 AMC 12A Problems/Problem 14 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2004 AMC 12A Problems/Problem 14 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2004 AMC 12A Problems/Problem 14 The following problem is from both the 2004 AMC 12A #14 and 2004 AMC 10A #18, so both problems redirect to this page. Contents 1 Problem 2 Solution 1 3 Solution 2 4 Solution 3 5 Solution 4 6 Solution 5 6.1 Solution 6(Partially Brute Force) 7 Video Solution 8 Video Solution by OmegaLearn 9 See also Problem A sequence of three real numbers forms an arithmetic progression with a first term of . If is added to the second term and is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression? Solution 1 Let be the common difference. Then , , are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, . The smallest possible value occurs when , and the third term is . Solution 2 Let be the common difference and be the common ratio. Then the arithmetic sequence is , , and . The geometric sequence (when expressed in terms of ) has the terms , , and . Thus, we get the following equations: Plugging in the first equation into the second, our equation becomes . By the quadratic formula, can either be or . If is , the third term (of the geometric sequence) would be , and if is , the third term would be . Clearly the minimum possible value for the third term of the geometric sequence is . Solution 3 Let the three numbers be, in increasing order, Hence, we have that . Also, from the second part of information given, we get that Plugging back in... Simplifying, we get that Applying the quadratic formula, we get that Obviously, in order to minimize the value of , we have to subtract. Hence, However, the problem asks for the minimum value of the third term in a geometric progression. Hence, the answer is Solution 4 Let the arithmetic sequence be and let the geometric sequence be . Now, we just try all the solutions. If the last term is , then . This gives the geometric sequence which indeed works. The answer is Solution by franzliszt Solution 5 The terms of the arithmetic progression are 9, , and for some real number . The terms of the geometric progression are 9, , and . ThereforeThus or . The corresponding geometric progressions are and so the smallest possible value for the third term of the geometric progression is . Solution 6(Partially Brute Force) List out the first few terms arithmetic progressions and their corresponding geometric progressions. List both the positive and negative. Then you will find that the answer is . Video Solution Education, the Study of Everything Video Solution by OmegaLearn ~ pi_is_3.14 See also 2004 AMC 12A (Problems • Answer Key • Resources) Preceded by Problem 13Followed by Problem 15 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 All AMC 12 Problems and Solutions 2004 AMC 10A (Problems • Answer Key • Resources) Preceded by Problem 17Followed by Problem 19 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 All AMC 10 Problems and Solutions Retrieved from " Category: Introductory Algebra Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16595	https://www.nice.org.uk/guidance/ng235/chapter/Recommendations	Recommendations \| Intrapartum care \| Guidance \| NICE Cookies on the NICE website and services Cookies are files saved on your phone, tablet or computer when you visit a website. We use cookies to store information about how you use the NICE website and services, such as the pages you visit. For more information, view our cookie statement.(Opens in a new window) Accept all cookies Reject cookies Essential cookies These cookies enable basic functions like page navigation and access to secure areas of the website. Our website cannot function properly without these cookies and they can only be deactivated by changing your browser preferences. Preference cookies These cookies remember information that changes the way our website behaves or looks, like your preferred layout or previously viewed pages. 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Show vendors within this category [x] Marketing and advertising cookies On Off Skip to content Accessibility help Search NICE… Skip to content Menu Sign in Guidance Standards and indicators Clinical Knowledge Summaries (CKS) British National Formulary(BNF) British National Formulary for Children (BNFC) Life sciences More from NICE Skip More from NICE submenu Explore more from NICE What NICE does About us Reusing our content Implementing NICE guidance News, blogs and podcasts Contact us Get involved Careers Join a committee Comment on a consultation Close menu My account Consultation responses Sign in You are here: Home NICE Guidance Conditions and diseases Fertility, pregnancy and childbirth Intrapartum care Intrapartum care NICE guideline Reference number: NG235 Published: 29 September 2023 Last updated: 18 June 2025 Guidance Tools and resources Information for the public Evidence History Download guidance (PDF) Overview Recommendations Recommendations for research Rationale and impact Context Appendix A: Adverse outcomes for different places of birth Appendix B: Outcomes for different places of birth – by BMI at booking Appendix C: Outcomes for intravenous remifentanil patient-controlled analgesia (PCA) compared with intramuscular pethidine Finding more information and committee details Update information On this page 1.1 Antenatal education about labour 1.2 Service organisation 1.3 Planning place of birth 1.4 Care throughout labour in all birth settings 1.5 Transfer of care and changing place of birth 1.6 Pain relief during labour 1.7 Prelabour rupture of membranes at term 1.8 First stage of labour 1.9 Second stage of labour 1.10 Third stage of labour 1.11 Care of the newborn baby 1.12 Care of the woman after birth Terms used in this guideline Recommendations People have the right to be involved in discussions and make informed decisions about their care, as described in NICE's information on making decisions about your care. Making decisions using NICE guidelines explains how we use words to show the strength (or certainty) of our recommendations, and has information about prescribing medicines (including off-label use), professional guidelines, standards and laws (including on consent and mental capacity), and safeguarding. 1.1 Antenatal education about labour 1.1.1 Give all women information antenatally about: what to expect in the latent first stage of labour how to work with any pain they experience how to contact their midwifery care team and what to do in an emergency. [2014, amended 2023] 1.1.2 Offer all women antenatal education about the signs of labour, consisting of: how to differentiate between Braxton Hicks contractions and active labour contractions the expected frequency of contractions and how long they last recognition of amniotic fluid ('waters breaking') description of normal vaginal loss. [2014, amended 2023] 1.1.3 For all women, discuss their preferences and choices for care during labour and birth as early as possible in their pregnancy, and record these choices. Emphasise that: making and recording care choices in advance will mean they will have more time to think about all their options they are free to make their decisions and change their mind at any time, including during labour or while giving birth choices and decisions may need to be discussed again if problems or changes occur during pregnancy or labour. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on antenatal education about labour. Loading. Please wait. Hide rationale and impact section on antenatal education about labour 1.2 Service organisation 1.2.1 Commissioners and providers, including networks of providers, should ensure that all 4 birth settings (home, freestanding midwifery unit, alongside midwifery unit and obstetric unit) are available to all women (in the local area or in a neighbouring area). [2014, amended 2023] 1.2.2 Ensure that all women giving birth have timely access to an obstetric unit if they need transfer of care for medical reasons or because they request regional analgesia. Audit and publish transfer times and reasons for delay in transfers so women can be informed of local service availability. [2014, amended 2023] 1.2.3 Commissioners and providers, including networks of providers, should ensure that there are: robust protocols in place for transfer of care between settings (see also section 1.5) clear local pathways for the continued care of women who are transferred from one setting to another, including: when crossing provider boundaries if the nearest obstetric or neonatal unit is closed to admissions or the local midwifery-led unit is full. 1.2.4 Commissioners and providers, including networks of providers, should ensure that there are multidisciplinary clinical governance structures in place to enable the oversight of all birth settings. These structures should include, as a minimum, midwifery, obstetric, anaesthetic and neonatal expertise, and adequately supported user representation. [2014, amended 2023] 1.3 Planning place of birth All women at low risk of complications 1.3.1 Explain to both multiparous and nulliparous women who are at low risk of complications that giving birth is generally very safe for both the woman and her baby. 1.3.2 Advise women that additional resources to help them plan their place of birth are available on the tools and resources page for this guideline and the NHS website. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on planning place of birth. Loading. Please wait. Hide rationale and impact section on planning place of birth 1.3.3 Explain to both multiparous and nulliparous women that they may choose any birth setting (home, freestanding midwifery unit, alongside midwifery unit or obstetric unit), and support them in their choice of setting wherever they choose to give birth: Advise low-risk multiparous women that planning to give birth at home or in a midwifery-led unit (freestanding or alongside) is associated with a lower rate of interventions and the outcome for the baby is no different compared with an obstetric unit. Advise low-risk nulliparous women that planning to give birth in a midwifery-led unit (freestanding or alongside) is associated with a lower rate of interventions and the outcome for the baby is no different compared with an obstetric unit. Explain that if they plan birth at home, there is a small increase in the risk of an adverse outcome for the baby. [2014, amended 2023] 1.3.4 Ensure that all healthcare professionals involved in the care of pregnant women are familiar with the types and frequencies of serious medical problems that can affect babies (see appendix A), in order to be able to provide this information to women if they request it. 1.3.5 Discuss with women the following information, including local statistics, about all local birth settings, and update them if this changes during their pregnancy: access to midwives, including: the likelihood of being cared for in labour by a familiar midwife the likelihood of receiving one-to-one care throughout labour (not necessarily being cared for by the same midwife for the whole of labour) access to medical staff (obstetric, anaesthetic and neonatal) availability of birthing pools access to pain relief, including Entonox (a 50:50 mixture of oxygen and nitrous oxide) and medicines (for example, pethidine, diamorphine, patient-controlled analgesia and regional analgesia) the likelihood of being transferred to an obstetric unit (if this is not the woman's chosen place of birth), the reasons why this might happen, the time it may take, the delay in obstetric or neonatal care this may cause, and how her birth companion will travel; refer to table 1 if no local data is available. More information on transfer to an obstetric unit for different groups of women is included in table 2 and table 4. [2014, amended 2023] Table 1 Primary reasons for transfer to an obstetric unit by number of women transferred (% of total transferred from each setting)\| Main reason for transfer to an obstetric unit for each woman (there may be more than 1 reason) \| From home (n=3,529) \| From a freestanding midwifery unit (n=2,457) \| From an alongside midwifery unit (n=4,401) \| --- --- \| \| Delay during first or second stage of labour \| 1,144 (32.4%) \| 912 (37.1%) \| 1,548 (35.2%) \| \| Abnormal fetal heart rate \| 246 (7.0%) \| 259 (10.5%) \| 477 (10.8%) \| \| Request for regional analgesia \| 180 (5.1%) \| 163 (6.6%) \| 585 (13.3%) \| \| Meconium staining \| 432 (12.2%) \| 301 (12.2%) \| 538 (12.2%) \| \| Retained placenta \| 250 (7.0%) \| 179 (7.3%) \| 203 (4.6%) \| \| Repair of perineal trauma \| 386 (10.9%) \| 184 (7.5%) \| 369 (8.4%) \| \| Postpartum neonatal concerns \| 180 (5.1%) \| 63 (2.6%) \| 5 (0.11%) \| \| Other \| 711 (20.1%) \| 396 (16.2%) \| 676 (16.3%) \| Source: Birthplace in England research study, 2011. Impact of BMI on choice of place of birth 1.3.6 Advise women that, in general, the higher their body mass index (BMI) at booking (and particularly with a BMI above 35 kg/m 2), the greater the likelihood of complications, so this may be something they wish to think about when planning their place of birth. Discuss with them that: these complications include unplanned caesarean birth, postpartum haemorrhage, transfer from home to an obstetric unit, stillbirth, neonatal death or the baby needing neonatal care the risk of complications may depend on whether the woman is nulliparous or multiparous, but in general the risks of complications are higher for nulliparous women with an increased BMI compared with multiparous women with an increased BMI in the event of complications arising, advanced care can generally be given more quickly in an obstetric unit or an alongside midwifery unit than at home or in a freestanding midwifery unit. For more detail on the risks associated with increased BMI in different places of birth, see appendix B. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on impact of BMI on choice of place of birth. Full details of the evidence and the committee's discussion are in evidence review A: impact of BMI on choice of place of birth. Loading. Please wait. Hide rationale and impact section on impact of BMI on choice of place of birth Low-risk multiparous women 1.3.7 Using tables 2 and 3, explain to low-risk multiparous women that: planning birth at home or in a freestanding midwifery unit is associated with a higher rate of spontaneous vaginal birth than planning birth in an alongside midwifery unit, and these 3 settings are associated with higher rates of spontaneous vaginal birth than planning birth in an obstetric unit planning birth in an obstetric unit is associated with a higher rate of interventions, such as vaginal birth with forceps or ventouse, unplanned caesarean birth and episiotomy, compared with planning birth in other settings there are no differences in outcomes for the baby associated with planning birth in any setting. [2014, amended 2023] Table 2 Low-risk multiparous women: Rates of different modes of birth, transfer to an obstetric unit and obstetric interventions for each planned place of birth (number of incidences per 1,000 women giving birth by location)\| Type of birth \| Home \| Freestanding midwifery unit \| Alongside midwifery unit \| Obstetric unit \| --- --- \| Spontaneous vaginal birth \| 984 \| 980 \| 967 \| 927 \| \| Birth with forceps or ventouse \| 9 \| 12 \| 23 \| 38 \| \| Unplanned caesarean birth \| 7 \| 8 \| 10 \| 35 \| \| Transfer to an obstetric unit \| 115 \| 94 \| 125 \| 10 \| \| Regional analgesia (epidural and/or spinal) \| 28 \| 40 \| 60 \| 121 \| \| Episiotomy \| 15 \| 23 \| 35 \| 56 \| \| Blood transfusion \| 4 \| 4 \| 5 \| 8 \| Sources: Birthplace in England research study, 2011; Outcomes of planned home births and planned hospital births in low-risk women in Norway between 1990 and 2007 (Blix et al. 2012). For obstetric unit transfer to an obstetric unit, the 10 cases noted in table 2 are the estimated transfer rate from one obstetric unit to a different obstetric unit owing to lack of capacity or expertise. For regional anaesthesia, Blix et al. reported epidural analgesia and the Birthplace in England study reported spinal or epidural analgesia. Table 3 Low-risk multiparous women: Outcomes for the baby for each planned place of birth (by number of babies per 1,000 births)\| Population \| Home \| Freestanding midwifery unit \| Alongside midwifery unit \| Obstetric unit \| --- --- \| Babies without serious medical problems \| 997 \| 997 \| 998 \| 997 \| \| Babies with serious medical problems \| 3 \| 3 \| 2 \| 3 \| Source: Birthplace in England research study, 2011. Serious medical problems were combined in table 3: neonatal encephalopathy and meconium aspiration syndrome were the most common adverse events, together accounting for 75% of the total. Stillbirths after the start of care in labour and death of the baby in the first week of life accounted for 13% of the events. Fractured humerus and clavicle were uncommon outcomes (less than 4% of adverse events). For the frequency of these events (how often any of them actually occurred), see appendix A. Low-risk nulliparous women 1.3.8 Using tables 4 and 5, explain to low-risk nulliparous women that: planning birth at home or in a freestanding midwifery unit is associated with a higher rate of spontaneous vaginal birth than planning birth in an alongside midwifery unit, and these 3 settings are associated with higher rates of spontaneous vaginal birth than planning birth in an obstetric unit planning birth in an obstetric unit is associated with a higher rate of interventions, such as vaginal birth with forceps or ventouse, unplanned caesarean birth and episiotomy, compared with planning birth in other settings there are no differences in outcomes for the baby associated with planning birth in an alongside midwifery unit, a freestanding midwifery unit or an obstetric unit planning birth at home is associated with an overall small increase in the risk of a baby having a serious medical problem (about 4 more per 1,000 births) compared with planning birth in other settings. [2014, amended 2023] Table 4 Low-risk nulliparous women: Rates of different modes of birth, transfer to an obstetric unit and obstetric interventions for each planned place of birth (number of incidences per 1,000 women giving birth by location)\| Type of birth \| Home \| Freestanding midwifery unit \| Alongside midwifery unit \| Obstetric unit \| --- --- \| Spontaneous vaginal birth \| 794 \| 813 \| 765 \| 688 \| \| Birth with (forceps or ventouse) \| 126 \| 118 \| 159 \| 191 \| \| Unplanned caesarean birth \| 80 \| 69 \| 76 \| 121 \| \| Transfer to an obstetric unit \| 450 \| 363 \| 402 \| 10 \| \| Regional analgesia (epidural and/or spinal) \| 218 \| 200 \| 240 \| 349 \| \| Episiotomy \| 165 \| 165 \| 216 \| 242 \| \| Blood transfusion \| 12 \| 8 \| 11 \| 16 \| Sources: Birthplace in England research study, 2011; Outcomes of planned home births and planned hospital births in low-risk women in Norway between 1990 and 2007 (Blix et al. 2012). For obstetric unit transfer to an obstetric unit, the 10 cases noted in table 4 are the estimated transfer rate from one obstetric unit to a different obstetric unit owing to lack of capacity or expertise. For regional anaesthesia, Blix et al. reported epidural analgesia and the Birthplace in England study reported spinal or epidural analgesia. Table 5 Low-risk nulliparous women: Outcomes for the baby for each planned place of birth (by number of babies per 1,000 births)\| Population \| Home \| Freestanding midwifery unit \| Alongside midwifery unit \| Obstetric unit \| --- --- \| Babies without serious medical problems \| 991 \| 995 \| 995 \| 995 \| \| Babies with serious medical problems \| 9 \| 5 \| 5 \| 5 \| Source: Birthplace in England research study, 2011. Serious medical problems were combined in table 5: neonatal encephalopathy and meconium aspiration syndrome were the most common adverse events, together accounting for 75% of the total. Stillbirths after the start of care in labour and death of the baby in the first week of life accounted for 13% of the events. Fractured humerus and clavicle were uncommon outcomes (less than 4% of adverse events). For the frequency of these events (how often any of them actually occurred), see appendix A. Medical conditions and other factors that may affect planned place of birth 1.3.9 Use tables 6 to 9 as part of an assessment for a woman choosing her planned place of birth: tables 6 and 7 show medical conditions or other situations in which there is increased risk for the woman or baby during or shortly after labour, where care in an obstetric unit would be expected to reduce this risk the factors listed in tables 8 and 9 are not reasons in themselves for advising birth within an obstetric unit, but indicate that further consideration of birth setting may be needed discuss these risks and the additional care that can be provided in the obstetric unit with the woman so that she can make an informed choice about planned place of birth. [2007, amended 2014] Table 6 Medical conditions indicating increased risk and suggesting planned birth at an obstetric unit\| Disease area \| Medical condition \| --- \| \| Cardiovascular \| Confirmed cardiac disease Hypertensive disorders \| \| Respiratory \| Asthma requiring an increase in treatment or hospital treatment Cystic fibrosis \| \| Haematological \| Haemoglobinopathies, such as sickle-cell disease, beta-thalassaemia major History of thromboembolic disorders Immune thrombocytopenia purpura or other platelet disorder or platelet count below 100×10 9/litre Von Willebrand's disease Bleeding disorder in the woman or unborn baby Atypical antibodies which carry a risk of haemolytic disease of the newborn \| \| Endocrine \| Hyperthyroidism Diabetes needing medication \| \| Infective \| Hepatitis B or C with abnormal liver function tests Toxoplasmosis – women receiving treatment Current active infection of chicken pox, rubella or genital herpes in the woman or baby Tuberculosis under treatment \| \| Immune \| Systemic lupus erythematosus Scleroderma \| \| Renal \| Abnormal renal function Renal disease requiring supervision by a renal specialist \| \| Neurological \| Epilepsy Myasthenia gravis Previous cerebrovascular accident \| \| Gastrointestinal \| Liver disease associated with current abnormal liver function tests \| \| Psychiatric \| Psychiatric disorder requiring current inpatient care \| Table 7 Other factors indicating increased risk and suggesting planned birth at an obstetric unit\| Factor \| Additional information \| --- \| \| Previous complications \| Unexplained stillbirth or neonatal death, or previous death related to intrapartum difficulty Previous baby with neonatal encephalopathy Pre-eclampsia requiring preterm birth Placental abruption with adverse outcome Eclampsia Uterine rupture Primary postpartum haemorrhage requiring additional treatment or blood transfusion Caesarean birth Shoulder dystocia \| \| Current pregnancy \| Multiple birth Placenta praevia Pre-eclampsia or pregnancy-induced hypertension Preterm labour or preterm prelabour rupture of membranes Placental abruption Anaemia – haemoglobin less than 85 g/litre at onset of labour Confirmed intrauterine death Substance misuse Alcohol dependency requiring assessment or treatment Gestational diabetes needing medication Malpresentation – breech or transverse lie Recurrent antepartum haemorrhage Small for gestational age in this pregnancy (less than third centile or reduced growth velocity on ultrasound as defined in the NHS Saving babies lives version 3) Abnormal fetal heart rate, umbilical or fetal doppler studies Ultrasound diagnosis of oligo- or polyhydramnios \| \| Previous gynaecological history \| Myomectomy Hysterotomy \| Table 8 Medical conditions indicating individual assessment is needed when planning place of birth\| Disease area \| Medical condition \| --- \| \| Cardiovascular \| Cardiac disease without intrapartum implications \| \| Haematological \| Atypical antibodies not putting the baby at risk of haemolytic disease Sickle-cell trait Thalassaemia trait Anaemia – haemoglobin 85 to 105 g/litre at onset of labour \| \| Endocrine \| Unstable hypothyroidism such that a change in treatment is needed \| \| Infective \| Group B streptococcus where intrapartum intravenous antibiotics are recommended Hepatitis B or C with normal liver function tests (as baby will need paediatric review after birth) Carrier of or infected with HIV \| \| Immune \| Non-specific connective tissue disorders \| \| Skeletal/Neurological \| Spinal abnormalities Previous fractured pelvis Neurological deficits \| \| Gastrointestinal \| Liver disease without current abnormal liver function Crohn's disease Ulcerative colitis \| Table 9 Other factors to take into account when planning place of birth\| Factor \| Additional information \| --- \| \| Previous complications \| Stillbirth or neonatal death with a known non-recurrent cause Pre-eclampsia developing at term Placental abruption with good outcome History of previous baby more than 4.5 kg Extensive vaginal, cervical, or third- or fourth-degree perineal trauma Retained placenta needing manual removal in theatre Previous term baby with jaundice requiring exchange transfusion Major gynaecological surgery \| \| Current pregnancy \| Antepartum bleeding of unknown origin (single episode after 24 weeks of pregnancy) Body mass index (BMI) at booking (see the section on impact of BMI on choice of place of birth) Blood pressure of 140 mmHg systolic or 90 mmHg diastolic or more on 2 occasions Clinical or ultrasound suspicion of macrosomia Induction of labour Grand multiparity (parity 4 or more) Recreational drug use Under current outpatient psychiatric care Age 40 or over at booking Fibroids Fetal abnormality \| 1.3.10 If further discussion is wanted by either the midwife or the woman about the choice of planned place of birth, arrange this with an appropriately trained senior or consultant midwife and/or a senior or consultant obstetrician if there are obstetric issues. [2014, amended 2023] 1.3.11 When discussing the woman's choice of place of birth with her, do not disclose personal views or judgements about her choices. 1.4 Care throughout labour in all birth settings 1.4.1 For all women giving birth in all birth settings, follow the principles in the NICE guideline on patient experience in adult NHS services and the NICE guideline on shared decision making, and support the woman's choices. [2014, amended 2023] 1.4.2 When providing information on the benefits and risks of care options or suggested interventions: encourage the woman to ask questions if possible, give her time to think about the options and help her make a supported decision. Obtain consent before carrying out the chosen care option or intervention. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on care throughout labour in all birth settings. Loading. Please wait. Hide rationale and impact section on care throughout labour in all birth settings 1.4.3 All staff and organisations should ensure that all birth settings have a culture of respect for each woman as an individual undergoing a significant and emotionally intense life experience, so that the woman is in control, is listened to, her choices are supported, and she is cared for with compassion. [2014, amended 2023] 1.4.4 All staff should demonstrate, through their own words and behaviour, appropriate ways of relating to and talking about women and their birth companion(s), and of talking about birth and the choices to be made when giving birth. [2014, amended 2023] One-to-one care in all birth settings 1.4.5 Maternity services should: provide a model of care that supports one-to-one care in labour for all women and benchmark services and identify overstaffing or understaffing by using workforce planning models and/or woman-to-midwife ratios. 1.4.6 Do not leave a woman in established labour on her own except for short periods or at the woman's request. 1.4.7 For guidance on ensuring continuity of care, see recommendation 1.4.1 in the NICE guideline on patient experience in adult NHS services. Communication 1.4.8 When giving a woman (and her birth companions) information about care during labour: use clear, plain language and confirm with the woman that they have understood the information tailor the content and delivery of information to the needs and preferences of the woman ensure that the woman is empowered to make a supported decision with her healthcare team, which may include: using reliable interpreting services when needed (for example, for languages other than English, British Sign Language, or Makaton) using interpreters who are independent of the woman (rather than, for example, a family member or friend) using culturally sensitive language adapting communication when necessary, for example, by using healthcare passports for people with learning disabilities or autism. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on communication. Loading. Please wait. Hide rationale and impact section on communication 1.4.9 Treat all women in labour with kindness, dignity and respect. [2007, amended 2023] 1.4.10 Ensure that the woman is empowered, informed and central to making decisions about her care, and recognise that the way in which care is given is key to this. Support the woman so she: can continue to make decisions about her care feels confident that her care team is there to assist her understands that she can accept or decline care that is offered, can change her mind, and that decisions she makes will not affect how care is provided to her. To ensure this happens, establish a rapport with the woman, ask her about her wants and expectations for labour, and be aware of the importance of both tone and demeanour and the actual words used. Use this information to support her and guide her care through her labour. [2007, amended 2023] 1.4.11 To establish communication with the woman: greet her and her birth companion(s) with a smile and a personal welcome, introduce yourself and explain your role in her care ask her what she wants to be called maintain a calm, confident and professional approach respect the woman's personal space, privacy and dignity, and ask others to do the same (for example, knock and wait before entering the woman's room) ask how the woman is feeling and whether there is anything in particular she would like to discuss or if she has any concerns discuss the woman's labour and birth preferences, and review and discuss any written birth plan ensure the woman is aware of pain relief options, and provide both the opportunity to discuss these options and give information if she requests it to establish what her choices are encourage the woman to adapt the environment to meet her individual needs explain all procedures and observations before they take place and ask for consent for them, focusing on the woman rather than the technology or the documentation show the woman and her birth companion(s) how to summon help, and reassure her that she can do so whenever and as often as she needs to when leaving the room, let her know when you will return involve the woman in any handover of care to another professional, either when additional expertise has been brought in or at the end of a shift – this should occur in the room, when appropriate, with the woman at the centre of the handover discussion. [2007, amended 2023] Position and mobilisation 1.4.12 Encourage and help the woman to move and adopt whatever positions she finds most comfortable throughout labour, except lying flat on her back. [2007, amended 2023] Support 1.4.13 Encourage the woman to have support from birth companion(s) of her choice. Hygiene measures 1.4.14 Tap water may be used if cleansing is needed before vaginal examination. 1.4.15 Routine hygiene measures taken by staff caring for women in labour, including standard hand hygiene and single-use, non-sterile gloves, are appropriate to reduce cross-contamination between women, babies and healthcare professionals. 1.4.16 Selection of personal protective equipment for healthcare professionals must be based on an assessment of the risk of exposure to blood and/or bodily fluids, non-intact skin or mucous membranes. Standard infection control procedures to prevent transmission of recognised and unrecognised infections must be followed. See NHS England's national infection prevention and control manual. [2007, amended 2023] 1.5 Transfer of care and changing place of birth Transfer of care refers to the transfer between midwifery-led care and obstetric-led care. This may or may not involve transport from one location to another. Women who are receiving midwifery-led care in an obstetric unit can have responsibility for their care transferred to being obstetric-led without being moved. 1.5.1 Base any decisions about transfer of care on clinical findings and discuss the options with the woman and her birth companion(s). 1.5.2 If contemplating transfer of care: talk with the woman and her birth companion(s) about the reasons for this and what they can expect, including the time needed for transfer address any concerns she has and try to allay any anxiety about the transfer ensure that her wishes are respected and her informed consent is obtained. 1.5.3 When arranging transfer of care, the midwife attending the labour should contact the ambulance service (if appropriate) and the coordinating midwife in the obstetric unit. The coordinating midwife should then alert the relevant healthcare professionals (obstetric, anaesthetic and neonatal). 1.5.4 Carry out transfer of care of women in labour as soon as possible after the decision to transfer has been made. Categorise transfers as: life-threatening emergency (ambulance service category 1) urgent, for example for pain relief (ambulance service category 2). For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on transfer of care and changing place of birth. Loading. Please wait. Hide rationale and impact section on transfer of care and changing place of birth 1.5.5 When arranging transfer from one location to another, ensure the following: before transfer, the woman is dressed, wrapped in a blanket or otherwise covered in a way that she feels is comfortable and appropriate the woman is made to feel as comfortable as possible before and during transfer any ambulance staff or other personnel involved are aware that some positions may make the woman uncomfortable or afraid and could affect her labour, so she should be encouraged to choose how to move and what position to adopt if possible, in accordance with ambulance service protocols communication and companionship are maintained: explain the arrangements for transfer to the woman and her birth companion(s) ensure a midwife who has been involved in the woman's care up to that point travels with her carry out a handover of care that involves the woman the woman is monitored throughout the transfer, as appropriate for her stage of labour, including intermittent auscultation of the fetal heart if possible and safe to do enable the woman's birth companion(s) to travel with her in the ambulance if that is what she wants and this is agreed by her care team and the ambulance crew. [2014, amended 2023] 1.5.6 If a woman is transferred to an obstetric unit after the birth (see the section on care of the woman after birth), ensure that her baby goes with her. 1.6 Pain relief during labour Attitudes to pain and pain relief in childbirth 1.6.1 Healthcare professionals should think about how their own values and beliefs inform their attitude to coping with pain in labour and ensure their care supports the woman's choice. 1.6.2 Take into account that every woman's experience of pain is unique and may be expressed in different ways, both verbally and non-verbally. In particular, this may vary because of: their cultural background and beliefs their socioeconomic status any neurodiverse conditions they may have. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on attitudes to pain and pain relief in childbirth. Loading. Please wait. Hide rationale and impact section on attitudes to pain and pain relief in childbirth Non-pharmacological pain-relieving strategies 1.6.3 Advise women that breathing exercises, having a shower or bath, and massage may reduce pain during the latent first stage of labour. [2014, amended 2023] 1.6.4 Do not offer or advise aromatherapy, yoga or acupressure for pain relief during the latent first stage of labour. If a woman wants to use any of these techniques, support her choice. [2014, amended 2023] 1.6.5 If a woman chooses to use breathing and relaxation techniques in labour, support her choice. 1.6.6 If a woman chooses to use massage techniques in labour that have been taught to birth companions, support her choice. 1.6.7 Advise women who wish to use transcutaneous electrical nerve stimulation (TENS) that: TENS devices are not provided by the NHS, but if a woman wants to use TENS to manage her comfort during labour, support her choice there is very little evidence of its effectiveness in established labour, but no evidence of harm other forms of pain relief can be used alongside TENS if needed by the woman.[2007, amended 2023] 1.6.8 Do not offer acupuncture, acupressure or hypnosis during labour. If a woman wants to use any of these techniques, support her choice. [2007, amended 2023] 1.6.9 Support the playing of music of the woman's choice in labour. 1.6.10 Offer the woman the opportunity to labour in water for pain relief. 1.6.11 For women labouring in water, monitor the temperature of the woman and the water hourly to ensure that the woman is comfortable and not becoming pyrexial. The temperature of the water should not be above 37.5°C. 1.6.12 Keep baths and birthing pools clean using a protocol agreed with the local microbiology department or infection control guidance and, in the case of birthing pools, in accordance with the manufacturer's guidelines. [2007, amended 2023] Sterile water injections 1.6.13 Consider intracutaneous or subcutaneous sterile water injections as a pain relief option for women in labour with back pain. These injections can be given by a midwife trained in the use of sterile water injections. 1.6.14 Explain to the woman that sterile water injections can provide relief of back pain from 10 minutes after the injection for up to 3 hours, but there can be an initial stinging sensation. 1.6.15 If the woman chooses to have sterile water injections, give these at 4 different injection points around the Rhombus of Michaelis, using doses of 0.1 ml intracutaneously or 0.5 ml subcutaneously at each injection point. For a short explanation of why the committee made the 2023 recommendations and how they might affect practice, see the rationale and impact section on sterile water injections. Full details of the evidence and the committee's discussion are in evidence review C: sterile water injections. Loading. Please wait. Hide rationale and impact section on sterile water injections Inhalational analgesia 1.6.16 Ensure that Entonox (a 50:50 mixture of oxygen and nitrous oxide) is available in all birth settings as it may reduce pain in labour, but inform the woman that it may make her feel nauseous and light-headed. Pharmacological analgesia 1.6.17 Ensure that pethidine, diamorphine or other opioids are available in all birth settings. Inform the woman that these will provide limited pain relief during labour and may have significant side effects for both her (for example, drowsiness, nausea and vomiting) and her baby (for example, short-term respiratory depression and drowsiness, which may last several days and may make it more difficult to breastfeed). [2007, amended 2023] 1.6.18 If an intravenous or intramuscular opioid is used, also administer an antiemetic. 1.6.19 Women should not enter water (a birthing pool or bath) within 2 hours of opioid administration or if they feel drowsy. 1.6.20 Consider intravenous remifentanil patient-controlled analgesia (PCA), at 40 micrograms per bolus with a 2‑minute lockout period, as an option for women who want ongoing pain relief during labour and birth. In September 2023, this was an off-label use of remifentanil. See NICE's information on prescribing medicines. 1.6.21 Only use remifentanil PCA in obstetric units because of the risk of respiratory depression in women that may need anaesthetic support. 1.6.22 Discuss the risks and benefits of remifentanil PCA with the woman and help them make a supported decision about its use. Explain that, with 40 micrograms of remifentanil compared with intramuscular pethidine: they are less likely to need an epidural or have a birth using forceps or ventouse if using remifentanil PCA they are more likely to have a spontaneous vaginal birth or need supplemental oxygen when using remifentanil PCA. For more detail on the risks and benefits associated with the use of remifentanil, see appendix C. 1.6.23 When using remifentanil PCA, ensure that: there is the continuous presence of a midwife trained in the care of women receiving remifentanil PCA (one-to-one care) there is continuous cardiotocography monitoring if there are other risk factors there is continuous monitoring of respiratory function (observation of breathing and pulse oximetry) units have clear guidelines on responding to respiratory depression supplemental oxygen is immediately available immediate anaesthetic support is available in case of respiratory depression. For a short explanation of why the committee made the 2023 recommendations and how they might affect practice, see the rationale and impact section on remifentanil patient-controlled analgesia. Full details of the evidence and the committee's discussion are in evidence review D: remifentanil patient-controlled analgesia. Loading. Please wait. Hide rationale and impact section on remifentanil patient-controlled analgesia Regional analgesia Information about regional analgesia 1.6.24 If a woman requests regional analgesia, talk with her about the benefits and risks and the effect it may have on her pain and her labour. [2007, amended 2023] 1.6.25 Provide information to women about epidural analgesia, including the following: it is available only in obstetric units so transfer will be necessary if she is in another setting it provides more effective pain relief than opioids it may not always be fully effective and may need to be adjusted or replaced complications during insertion of the epidural may cause a severe postnatal headache it is not associated with long-term backache it is not associated with a longer first stage of labour or an increased chance of an unplanned caesarean birth it is associated with a longer second stage of labour and an increased chance of birth with forceps or ventouse it will be accompanied by a more intensive level of monitoring and intravenous access, so mobility may be reduced. [2007, amended 2023] 1.6.26 If, after a discussion of the benefits and risks, a woman in labour chooses regional analgesia, support her decision. [2007, amended 2023] Care and observations for women with regional analgesia 1.6.27 Always secure intravenous access before starting regional analgesia. 1.6.28 Preloading and maintenance fluid infusion do not need to be administered routinely before establishing low-dose epidural analgesia and combined spinal–epidural analgesia. 1.6.29 Undertake the following additional observations for women with regional analgesia: during establishment of regional analgesia or after further boluses (10 ml or more of low-dose solutions), measure blood pressure every 5 minutes for 15 minutes if the woman is not pain-free 30 minutes after each administration of local anaesthetic or opioid solution, ask the anaesthetist to review assess the level of the sensory block hourly if the woman is not mobilising, assess the level of motor block hourly by asking the woman to do a straight leg raise. If she is unable to do this, ask the anaesthetist to review. [2007, amended 2023] 1.6.30 Encourage women with regional analgesia to adopt whatever positions, including upright, they find comfortable throughout labour, except lying flat on their back. [2007, amended 2023] 1.6.31 Advise women with an epidural in situ that if they have sufficient leg strength and sensation, as checked and confirmed by a midwife trained in caring for women with epidurals, they can mobilise with assistance, but their legs may feel heavier than usual. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on care and observations for women with regional analgesia. Loading. Please wait. Hide rationale and impact section on care and observations for women with regional analgesia 1.6.32 Once established, continue regional analgesia until after completion of the third stage of labour and any necessary perineal repair. 1.6.33 On confirmation of full cervical dilatation in a woman with regional analgesia, unless the woman has an urge to push or the baby's head is visible, pushing may be delayed by 1 hour for multiparous women and up to 2 hours for nulliparous women, after which actively encourage her to push during contractions. [2007, amended 2023] 1.6.34 Do not routinely use oxytocin in the second stage of labour for women with regional analgesia. 1.6.35 Perform continuous cardiotocography for at least 30 minutes during establishment of regional analgesia and after administration of each further bolus of 10 ml or more. [2007, amended 2014] Establishing and maintaining regional analgesia 1.6.36 Use either epidural or combined spinal–epidural analgesia for establishing regional analgesia in labour. 1.6.37 If rapid analgesia is needed, use combined spinal–epidural analgesia. 1.6.38 Establish combined spinal–epidural analgesia with bupivacaine (or an equivalent local anaesthetic) and fentanyl. [2007, amended 2023] 1.6.39 Establish epidural analgesia with a low-concentration local anaesthetic and fentanyl solution. The initial dose is essentially a test dose, so administer it cautiously to ensure that inadvertent intrathecal or intravascular placement of the epidural catheter has not occurred. [2007, amended 2023] 1.6.40 Use low-concentration local anaesthetic and opioid solutions (0.0625% to 0.1% bupivacaine or equivalent combined with 2.0 micrograms per ml fentanyl) for maintaining epidural analgesia in labour. 1.6.41 Use patient-controlled epidural analgesia, programmed intermittent epidural bolus or intermittent bolus given by a trained healthcare professional for maintaining epidural analgesia. 1.6.42 Do not use high concentrations of local anaesthetic solutions (0.25% or above of bupivacaine or equivalent) routinely for either establishing or maintaining epidural analgesia. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on programmed intermittent epidural bolus. Full details of the evidence and the committee's discussion are in evidence review E: programmed intermittent epidural bolus. Loading. Please wait. Hide rationale and impact section on programmed intermittent epidural bolus 1.7 Prelabour rupture of membranes at term 1.7.1 Advise women with suspected rupture of membranes after 37+0 weeks to contact their midwife or maternity unit to have an initial triage assessment over the phone with a midwife. This should include when the membranes ruptured and an assessment of any risk factors, such as: meconium-stained liquor vaginal bleeding blood-stained liquor reduced fetal movements continuous abdominal pain unpleasant smelling liquor, or any change in the colour or smell of her vaginal loss the woman feeling unwell group B streptococcus carriage or infection in this or a previous pregnancy where a plan has been made for intrapartum antibiotic prophylaxis in this pregnancy the baby has abnormal lie or presentation (for example, transverse lie or breech) fetal growth restriction low-lying placenta. If any of these factors are present or if there is any uncertainty, the woman should be advised to immediately attend the maternity unit for an urgent in-person review (see the recommendations on the assessment of a woman in the first stage of labour, and recommendations 1.7.3 and 1.7.4). 1.7.2 For women after 37+0 weeks with suspected rupture of the membranes but no risk factors on initial phone triage assessment (see recommendation 1.7.1): see the woman in person as soon as possible if she has any concerns or wishes to be induced immediately or within 12 hours and if anything changes or the woman has any concerns, advise her to call her midwife or maternity unit back sooner than the planned review. Offer to carry out the review at the woman's home, in a midwifery-led unit, or an assessment centre at an obstetric unit. 1.7.3 Do not carry out a speculum examination if it is certain that the membranes have ruptured. 1.7.4 If it is uncertain whether prelabour rupture of the membranes has occurred, offer the woman a speculum examination to determine whether the membranes have ruptured. Avoid digital vaginal examination in the absence of contractions. 1.7.5 Advise women presenting with prelabour rupture of the membranes at term that: the risk of serious neonatal infection is 1%, rather than 0.5% for women with intact membranes, and may increase over time intrapartum antibiotics are recommended in some situations (see the section on intrapartum antibiotics in NICE's guideline on neonatal infection) 60% of women with prelabour rupture of the membranes will go into labour within 24 hours. [2007, amended 2023] 1.7.6 Offer women with prelabour rupture of membranes at term (at or after 37+0 weeks) a choice of: expectant management for up to 24 hours or induction of labour as soon as possible (see the section on induction of labour in specific circumstances in NICE's guideline on inducing labour). Discuss the benefits and risks of these options with the woman, and take into account her individual circumstances and preferences. 1.7.7 For women who choose expectant management after prelabour rupture of the membranes at term (at or after 37+0 weeks), offer induction of labour if labour has not started naturally after approximately 24 hours. (see the section on induction of labour in specific circumstances in NICE's guideline on inducing labour). 1.7.8 Until the induction is started or if expectant management beyond 24 hours is chosen by the woman: do not offer lower vaginal swabs and measurement of maternal C‑reactive protein to detect any infection that may be developing, advise the woman to record her temperature every 4 hours during waking hours and to report immediately any change in the colour or smell of her vaginal loss inform the woman that bathing or showering is not associated with an increase in infection, but that having sexual intercourse may be. 1.7.9 Assess fetal movement and heart rate at initial contact and then every 24 hours after rupture of the membranes while the woman is not in labour, and advise the woman to report immediately any decrease in fetal movements. 1.7.10 If labour has not started 24 hours after rupture of the membranes, advise the woman to give birth where there is access to neonatal services (this may be in an obstetric unit or an alongside midwifery unit) and to stay in hospital for at least 12 hours after the birth. [2007, amended 2023] 1.7.11 If a woman has prelabour rupture of membranes at term (at or after 37+0 weeks) and has had a positive group B streptococcus test at any time in their current pregnancy, or a previous pregnancy where the baby developed group B streptococcus infection, offer immediate induction of labour, or caesarean birth if it had been planned. For advice on intrapartum antibiotics, see the section on intrapartum antibiotics in NICE's guideline on neonatal infection. For a short explanation of why the committee made the 2023 recommendations and how they might affect practice, see the rationale and impact section on initial assessment of women reporting prelabour rupture of membranes. Full details of the evidence and the committee's discussion are in evidence review B: initial assessment of women reporting prelabour rupture of membranes. Loading. Please wait. Hide rationale and impact section on initial assessment of women reporting prelabour rupture of membranes 1.8 First stage of labour Definitions of the latent and established first stages of labour and early triage 1.8.1 For the purposes of this guideline, use the following definitions of labour: latent first stage of labour is a period of time, not necessarily continuous, when: there are contractions and there is some cervical change, including cervical position, consistency, effacement and dilatation up to 4 cm established first stage of labour is when: there are regular contractions and there is progressive cervical dilatation from 4 cm. [2007, amended 2023] 1.8.2 If a woman in labour contacts her midwife or maternity unit for advice, carry out an assessment of labour by telephone triage and determine whether a face-to-face assessment is needed. [2014, amended 2023] 1.8.3 Carry out a face-to-face early assessment of labour either: at home (regardless of planned place of birth) or in her planned place of birth (midwifery-led unit or obstetric unit), comprising one-to-one midwifery care for at least 1 hour. [2014, amended 2023] 1.8.4 Include the following in any early or triage assessment of labour: ask the woman how she is, and about her wishes, expectations and any concerns she has ask the woman about the baby's movements, including any changes give information about what the woman can expect in the latent first stage of labour and how to work with any pain she experiences give information about what to expect when she accesses care agree a plan of care with the woman, including guidance about who she should contact next and when provide guidance and support to the woman's birth companion(s). 1.8.5 The triage midwife should document the guidance that she gives to the woman. 1.8.6 If a woman seeks advice or attends a midwifery-led unit or obstetric unit with painful contractions, but is not in established labour: recognise that a woman may experience painful contractions without cervical change and offer her individualised support and analgesia if needed encourage her to remain at or return home, unless doing so leads to a significant risk that she could give birth without a midwife present or become distressed. [2014, amended 2023] Assessment of women in the first stage of labour 1.8.7 When performing an initial assessment of a woman in labour, listen to her story and support her preferences and her emotional and psychological needs. [2014, amended 2023] 1.8.8 Carry out an initial assessment to determine if midwifery-led care in any setting is suitable for the woman, irrespective of any previous plan including consultant-led care. This assessment should comprise the following: maternal factors: review and discuss the antenatal notes (including all antenatal screening results) review the personalised care plan review if there are any antenatal or intrapartum risk factors for fetal hypoxia (see the NICE guideline on fetal monitoring in labour) ask her about the length, strength and frequency of her contractions ask her about any pain she is experiencing and discuss her options for pain relief record her pulse, blood pressure, temperature and respiratory rate, and carry out urinalysis record if she has had any vaginal loss check if she needs intrapartum antibiotics for group B streptococcus prophylaxis and, if so, that these are available in her chosen place of birth if needed (see the section on intrapartum antibiotics in NICE's guideline on neonatal infection) observations of the unborn baby: ask the woman about the baby's movements in the last 24 hours palpate the woman's abdomen to determine the fundal height, the baby's lie, presentation, position, engagement of the presenting part, and frequency and duration of contractions auscultate the fetal heart rate for a minimum of 1 minute immediately after a contraction; palpate the woman's pulse to differentiate between the heartbeats of the woman and the baby if there is uncertainty about whether the woman is in established labour, a vaginal examination may be helpful after a period of assessment, but is not always necessary if the woman appears to be in established labour, offer a vaginal examination. [2014, amended 2023] 1.8.9 When conducting a vaginal examination: be sure that the examination is necessary and will add important information to the decision-making process recognise that a vaginal examination can be very distressing for a woman, especially if she is already in pain, highly anxious and in an unfamiliar environment explain the reason for the examination and what will be involved ensure the woman's informed consent, privacy, dignity and comfort explain sensitively the findings of the examination and any impact on the birth plan to the woman and her birth companion(s) advise the woman that she can decline the examination before it starts, or ask to stop at any stage during the examination. [2014, amended 2023] 1.8.10 When performing a vaginal examination, determine: the station of the presenting part the position of the presenting part the presence or absence of caput or moulding cervical effacement cervical dilatation presence or absence of membranes. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on assessment of women in the first stage of labour. Loading. Please wait. Hide rationale and impact section on assessment of women in the first stage of labour 1.8.11 Transfer the woman to obstetric-led care, following the general principles for transfer of care described in section 1.5, if any of the following are observed on initial assessment, taking into account that multiple risk factors may increase the urgency of the transfer, particularly if they have a cumulative effect: observations of the woman: pulse over 120 beats a minute on 2 occasions 15 to 30 minutes apart a single reading of either raised diastolic blood pressure of 110 mmHg or more or raised systolic blood pressure of 160 mmHg or more either raised diastolic blood pressure of 90 mmHg or more or raised systolic blood pressure of 140 mmHg or more on 2 consecutive readings taken 15 to 30 minutes apart respiratory rate of less than 9 or more than 21 breaths per minute on 2 occasions 15 to 30 minutes apart a reading of 2+ of protein on urinalysis and a single reading of either raised diastolic blood pressure (90 mmHg or more) or raised systolic blood pressure (140 mmHg or more) temperature of 38°C or above on a single reading, or 37.5°C or above on 2 consecutive readings 1 hour apart; for advice about intrapartum antibiotics, see the section on intrapartum antibiotics in NICE's guideline on neonatal infection fresh red bleeding or blood-stained liquor rupture of membranes more than 24 hours before the onset of established labour the presence of meconium (see the section on the presence of meconium) pain reported by the woman that differs from the pain normally associated with contractions any risk factors recorded in the woman's notes that indicate the need for obstetric-led care observations of the unborn baby: non-cephalic fetal presentation high (4/5 to 5/5 palpable) or free-floating head in a nulliparous woman suspected or diagnosed small for gestational age diagnosed fetal growth restriction diagnosis of oligohydramnios or anhydramnios on ultrasound concerns about fetal monitoring, as described in the NICE guideline on fetal monitoring in labour reduced fetal movements in the last 24 hours reported by the woman cord presentation. If none of these are observed, continue with midwifery-led care unless the woman requests transfer. [2014, amended 2023] 1.8.12 If any of the factors in recommendation 1.8.11 are observed but birth is imminent, assess whether birth in the current location is preferable to transferring the woman to an obstetric unit and discuss this with the woman and the coordinating midwife. [2014, amended 2023] Measuring fetal heart rate as part of initial assessment 1.8.13 Offer auscultation of the fetal heart rate at first contact with a woman in suspected or established labour, and at each further assessment. [2017, amended 2023] 1.8.14 For advice on the choice and method of fetal monitoring during labour, including risk assessment and indications for continuous cardiotocography, see the NICE guideline on fetal monitoring in labour. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on measuring fetal heart rate as part of the initial assessment. Loading. Please wait. Hide rationale and impact section on measuring fetal heart rate as part of the initial assessment Controlling gastric acidity 1.8.15 Do not routinely offer proton pump inhibitors to low-risk women. [2007, amended 2023] 1.8.16 Consider proton pump inhibitors (for example, omeprazole) for women who receive opioids, or who have or develop risk factors that make a caesarean birth more likely. [2007, amended 2023] Eating and drinking 1.8.17 Explain to the woman or pregnant person that they should drink during labour when they are thirsty, and that isotonic drinks may be more beneficial than water. Also explain that there is no benefit to drinking any more than normal, and overconsumption may be harmful (see sections on fluid intake and output and confirmed peripartum hyponatraemia. [2007, amended 2025] 1.8.18 Inform the woman or pregnant person that they can eat a light diet in established labour if they wish, unless they have received opioids or they develop risk factors that make a caesarean birth more likely. [2007, amended 2025] Ongoing assessment during the first stage of labour 1.8.19 Record the following observations during the first stage of labour: half-hourly documentation of frequency of contractions hourly pulse 4‑hourly temperature, blood pressure and respiratory rate offer a 4‑hourly vaginal examination (see recommendations 1.8.9 and 1.8.10), or in response to the woman's wishes if there is concern about progress (after abdominal palpation and assessment of vaginal loss). 1.8.20 Carry out an hourly risk assessment of the woman and her baby, and if any of the following risks have developed, transfer the woman to obstetric-led care (following the general principles for transfer of care described in section 1.5), unless the risks of transfer outweigh the benefits. Take into account that multiple risk factors may increase the urgency of the transfer, particularly if they have a cumulative effect: observations of the woman: pulse over 120 beats/minute on 2 occasions 15 to 30 minutes apart a single reading of either raised diastolic blood pressure of 110 mmHg or more or raised systolic blood pressure of 160 mmHg or more either raised diastolic blood pressure of 90 mmHg or more or raised systolic blood pressure of 140 mmHg or more on 2 consecutive readings taken 15 to 30 minutes apart a reading of 2+ of protein on urinalysis and a single reading of either raised diastolic blood pressure (90 mmHg or more) or raised systolic blood pressure (140 mmHg or more) respiratory rate of less than 9 or more than 21 breaths per minute on 2 occasions 15 to 30 minutes apart temperature of 38°C or above on a single reading, or 37.5°C or above on 2 consecutive occasions 1 hour apart; for advice on intrapartum antibiotics, see the section on intrapartum antibiotics in NICE's guideline on neonatal infection fresh red bleeding or blood-stained liquor the new appearance of meconium (see the section on the presence of meconium) pain reported by the woman that differs from the pain normally associated with contractions confirmed delay in the first stage of labour request by the woman for additional pain relief using regional analgesia obstetric emergency, including antepartum haemorrhage, cord prolapse, maternal seizure or collapse, or a need for advanced neonatal resuscitation observations of the unborn baby: any non-cephalic presentation, including cord presentation high (4/5 to 5/5 palpable) or free-floating head in a nulliparous woman suspected fetal growth restriction or macrosomia suspected anhydramnios or polyhydramnios any changes in the fetal heart rate pattern (see the NICE guideline on fetal monitoring in labour) If none of these are observed, continue with midwifery-led care unless the woman requests transfer. [2014, amended 2023] 1.8.21 Do not routinely use verbal assessment using a numerical pain score. 1.8.22 Use a pictorial record of labour (partogram) once labour is established. [2007, amended 2023] 1.8.23 Give ongoing consideration to the woman's emotional and psychological needs, including her desire for pain relief. 1.8.24 Encourage the woman to say if she needs more analgesia at any point during labour. Bladder care 1.8.25 Review bladder care for women at least every 4 hours. This should include: frequency of passing urine and bladder sensation fluid balance monitoring if sensation is abnormal or absent, if there is an inability to pass urine, or the woman is receiving intravenous fluids (including oxytocin) offering to insert a catheter if there are any ongoing concerns over the woman's ability to pass urine. [2023, amended 2025] For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on bladder care. Loading. Please wait. Hide rationale and impact section on bladder care Fluid intake and output 1.8.26 Discuss with the woman or pregnant person that: it is important to drink during labour when thirsty (see recommendation 1.8.17) it is important to regularly empty the bladder (see recommendation 1.8.18) excessive intake of oral or intravenous fluids may be harmful as this can cause hyponatraemia (a sodium level of less than 130 mmol/L in a pregnant woman or pregnant person) and lead to maternal and neonatal seizures or death their midwife will ask about and check up on their fluid intake and output throughout labour fluid balance monitoring may be advised during labour to reduce the risk of hyponatraemia or dehydration. 1.8.27 Monitor and record fluid balance, if: there are any concerns about fluid intake, for example the woman or pregnant person is drinking too much (also take into account fluid intake before labour care began) the woman or pregnant person is receiving intravenous fluids the woman or pregnant person is receiving an oxytocin infusion there are any concerns about fluid output, for example there is an inability to pass urine, nausea, vomiting or diarrhoea there are certain medical conditions, such as haemorrhage or pre-eclampsia (also see NICE's guideline on hypertension in pregnancy). 1.8.28 If there is a positive fluid balance of 1500 ml or more, or there are clinical concerns (for example, signs and symptoms of hyponatraemia): explain to the woman or pregnant person that it is possible they are developing, or have developed, hyponatraemia request an obstetric review offer a blood test to check their sodium level advise that they will need to be transferred to an obstetric setting if they are currently in a midwifery-led setting (see the section on transfer of care and changing place of birth). 1.8.29 Do not routinely advise oral fluids or give intravenous fluids for the treatment of ketonuria in women and pregnant people who are not diabetic. Confirmed peripartum hyponatraemia 1.8.30 If the woman or pregnant person has peripartum hyponatraemia, follow local protocols for the management of hyponatraemia. 1.8.31 If the woman or pregnant person is hyponatraemic inform the following care staff as soon as possible, or immediately if the woman or pregnant person is symptomatic or their sodium level is less than 125 mmol/L: the consultant obstetrician the consultant anaesthetist the neonatal team the delivery suite coordinator. For a short explanation of why the committee made the 2025 recommendations, see the rationale and impact section on fluid balance and hyponatraemia. Full details about the committee's discussion are in the supporting document for the recommendations on fluid balance and peripartum hyponatraemia. Loading. Please wait. Hide rationale and impact section on fluid balance and hyponatraemia Presence of meconium 1.8.32 As part of ongoing assessment, document the presence or absence of meconium. [2014, amended 2023] 1.8.33 If meconium is present, consider the character of the meconium and discuss the option of transfer to obstetric-led care with the woman. Explain that meconium: may increase the risk to the baby means that continuous cardiotocography monitoring may be advised (see the NICE guideline on fetal monitoring in labour) may mean that healthcare professionals trained in advanced neonatal life support are needed as soon as the baby is born. [2014, amended 2023] 1.8.34 If the woman wishes to be transferred, provided that it is safe to do so and the birth is unlikely to occur before transfer is completed, follow the general principles for transfer of care described in section 1.5. Take into account that the presence of other risk factors (in addition to meconium) may increase the urgency of the transfer. [2014, amended 2023] 1.8.35 Be aware that meconium is more common after full term but should still trigger a full risk assessment and discussion with the woman about the option of transfer to obstetric-led care. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on presence of meconium. Loading. Please wait. Hide rationale and impact section on presence of meconium Duration of the first stage 1.8.36 Inform women that, while the length of established first stage of labour varies between women: first labours last on average 8 hours and are unlikely to last over 18 hours second and subsequent labours last on average 5 hours and are unlikely to last over 12 hours. 1.8.37 Do not offer or advise clinical intervention if labour is progressing with no complications or concerns and the woman or pregnant person and baby are well. [2007, amended 2025] 1.8.38 In all stages of labour, women and pregnant people who have moved to specialist care pathways because of the development of complications can return to standard care if or when the complication is resolved. [2007, amended 2025] Interventions in the first stage 1.8.39 Do not routinely perform amniotomy if labour is progressing with no complications or concerns. [2007, amended 2025] 1.8.40 Do not routinely use combined early amniotomy with use of oxytocin. Delay in the first stage 1.8.41 If delay in the established first stage is suspected, take the following into account: parity cervical dilatation and rate of change uterine contractions station and position of presenting part. Offer the woman support, hydration, and appropriate and effective pain relief. [2007, amended 2023] 1.8.42 If delay in the established first stage is suspected, assess all aspects of progress in labour when diagnosing delay, including: cervical dilatation of less than 2 cm in 4 hours for first labours cervical dilatation of less than 2 cm in 4 hours or a slowing in the progress of labour for second or subsequent labours descent and rotation of the baby's head changes in the strength, duration and frequency of uterine contractions. 1.8.43 If delay in the established first stage of labour is suspected, discuss the findings (see recommendation 1.8.42) and the options available with the woman, and support her decision. [2007, amended 2023] 1.8.44 Offer all women with delay in the established first stage of labour support and effective pain relief. 1.8.45 Advise all women with suspected delay in the established first stage of labour to have a vaginal examination 2 hours later, and diagnose delay if progress is less than 1 cm. 1.8.46 If delay in the established first stage of labour is diagnosed, consider amniotomy for all women with intact membranes, after explanation of the procedure and advice that it will shorten labour by about an hour and may increase the strength and pain of contractions. [2007, amended 2023] 1.8.47 Do not advise transfer to obstetric-led care for amniotomy alone. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on delay in the first stage. Loading. Please wait. Hide rationale and impact section on delay in the first stage 1.8.48 After amniotomy, advise the woman to have a repeat vaginal examination 2 hours later. [2007, amended 2023] 1.8.49 If there is no progress 2 hours after the amniotomy, diagnose delay and transfer the woman to obstetric-led care. Follow the general principles for transfer of care described in section 1.5. Take into account that the presence of other risk factors (in addition to delay) may increase the urgency of the transfer. [2014, amended 2023] 1.8.50 For all women with confirmed delay in the established first stage of labour, an obstetrician should offer a full assessment. The obstetric review should include abdominal palpation and vaginal examination and consideration of oxytocin. [2007, amended 2023] 1.8.51 Discuss the use of oxytocin with the woman and make a decision with her about its use. Explain that: her choice to start, stop or restart the oxytocin will be supported using oxytocin after spontaneous or artificial rupture of the membranes will bring forward the time of birth but will not influence the mode of birth or neonatal outcomes oxytocin will increase the frequency and strength of contractions and that its use will mean that her contractions and her baby's heartbeat will be monitored continuously using cardiotocography; see the NICE guideline on fetal monitoring in labour oxytocin can cause hyperstimulation, which may increase the chance of transient fetal hypoxia, and if hyperstimulation occurs the dose will be reduced or stopped. [2007, amended 2023] 1.8.52 Offer the woman an epidural before oxytocin is started or if she requests it later. [2007, amended 2023] 1.8.53 When starting intravenous oxytocin in the first stage of labour: do not start separate intravenous fluids without a clinical indication (for example, the woman is not drinking, is dehydrated, or is hypotensive) monitor fluid balance. 1.8.54 If oxytocin is used in the first stage of labour, ensure that the time between increments of the dose is no more frequent than every 30 minutes. Increase oxytocin until there are 3 to 4 contractions in 10 minutes. 1.8.55 Use oxytocin in labour with caution. If the woman has contractions more frequently than 4 in 10 minutes, reduce or stop the oxytocin until the woman is having 4 or fewer contractions in 10 minutes. 1.8.56 Oxytocin must be discontinued immediately if the cardiotocography is pathological, and urgent obstetrician or senior midwife review sought. See the NICE guideline on fetal monitoring in labour. 1.8.57 Consider restarting oxytocin in the first stage of labour if: obstetric review has been carried out and the cardiotocography is no longer pathological the woman agrees that it can be restarted. Base the dose when restarting on a full clinical assessment, taking into consideration the previous dose. 1.8.58 Advise the woman to have a vaginal examination 4 hours after the oxytocin infusion has led to regular contractions in established labour: if cervical dilatation has increased by less than 2 cm after 4 hours of oxytocin, further obstetric review is needed to assess whether a caesarean birth is advisable if cervical dilatation has increased by 2 cm or more, advise 4‑hourly vaginal examinations. [2007, amended 2023] 1.8.59 If oxytocin is restarted in the first stage of labour, base the timing of the next vaginal examination on a clinical assessment of the woman and her individual circumstances. For a short explanation of why the committee made the 2023 recommendations and how they might affect practice, see the rationale and impact section on the use of oxytocin in the first or second stage of labour. Full details of the evidence and the committee's discussion are in evidence review F: oxytocin in the first or second stage of labour. Loading. Please wait. Hide rationale and impact section on the use of oxytocin in the first or second stage of labour 1.9 Second stage of labour Definition of the second stage 1.9.1 For the purposes of this guideline, use the following definitions of labour: passive second stage of labour: when there is full dilatation of the cervix (determined by either vaginal examination or noting other external signs of full dilatation) before or in the absence of involuntary or active pushing the passive second stage of labour may be up to 2 hours when a woman with an epidural in place has been advised to delay pushing (see recommendations 1.9.7 to 1.9.10) onset of the active second stage of labour is when: the baby is visible or there is involuntary or active pushing with full dilatation of the cervix. [2007, amended 2023] Assessment of women during the second stage of labour 1.9.2 Continue with observations of the woman and baby, and assessment of risk as described for the first stage of labour (see recommendations 1.8.19 and 1.8.20), but be aware that the frequency of fetal monitoring should increase. See the NICE guideline on fetal monitoring in labour. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on assessment of women during the second stage of labour. Loading. Please wait. Hide rationale and impact section on assessment of women during the second stage of labour 1.9.3 Offer a vaginal examination (see recommendations 1.8.9 and 1.8.10) hourly in the active second stage, or in response to the woman's wishes (after abdominal palpation and assessment of vaginal loss). To assess progress, the vaginal examination should include: position of the head descent caput and moulding. [2007, amended 2023] 1.9.4 During the second stage of labour: continue to take the woman's emotional and psychological needs into account assess progress, which should include the woman's behaviour, the effectiveness of pushing and the baby's wellbeing, taking into account the baby's position and station at the onset of the second stage; these factors will assist in deciding the timing of further vaginal examinations and any need for transfer to obstetric-led care assess the frequency, strength and duration of contractions perform intermittent auscultation of the fetal heart rate immediately after a contraction for at least 1 minute, at least every 5 minutes. Palpate the woman's pulse every 5 minutes to differentiate between the 2 heartbeats; see the NICE guideline on fetal monitoring in labour ongoing consideration should be given to the woman's position, hydration, coping strategies and pain relief throughout the second stage. [2007, amended 2023] The woman's position and pushing in the second stage 1.9.5 Advise a woman with an epidural in place during the second stage of labour that: lying flat on her back can lead to a decrease in blood pressure and may reduce placental blood flow lying on her side may increase the chance of a spontaneous vaginal birth, but she can use any other position she finds comfortable to give birth, including upright positions. 1.9.6 Advise a woman without an epidural in place during the second stage of labour that: lying flat on her back can lead to a decrease in blood pressure and may reduce placental blood flow she can use any other position she finds comfortable to give birth upright positions and keeping mobile may be beneficial (as they may reduce fetal heart rate abnormalities, episiotomy rates and improve her birthing experience). For a short explanation of why the committee made the 2023 recommendations and how they might affect practice, see the rationale and impact section on position for birth. Full details of the evidence and the committee's discussion are in evidence review G: position for birth. Loading. Please wait. Hide rationale and impact section on position for birth 1.9.7 Advise women without an epidural in place that: spontaneous pushing may shorten the second stage of labour compared with directed pushing if directed pushing is used, pushing while exhaling may shorten the active second stage of labour for multiparous women. 1.9.8 If full dilatation of the cervix has been confirmed in a woman without an epidural in place, but she does not get an urge to push, offer to carry out further assessment after 1 hour. 1.9.9 Advise nulliparous women with an epidural that: directed pushing rather than spontaneous pushing may reduce the likelihood of having an unplanned caesarean birth delayed directed pushing (up to 2 hours after full dilatation) may shorten the active second stage of labour. 1.9.10 Advise multiparous women with an epidural that: delayed directed pushing (by 1 hour after full dilatation) may reduce the likelihood of needing birth with forceps or ventouse delayed directed pushing (by 1 hour after full dilatation) may shorten the active second stage of labour. 1.9.11 If pushing is ineffective or if requested by the woman, offer strategies to assist birth, such as support, change of position, emptying of the bladder and encouragement. For a short explanation of why the committee made the 2023 recommendations and how they might affect practice, see the rationale and impact section on pushing techniques. Full details of the evidence and the committee's discussion are in evidence review H: pushing techniques. Loading. Please wait. Hide rationale and impact section on pushing techniques Intrapartum interventions to reduce perineal trauma 1.9.12 Discuss the woman's preferences for techniques to reduce perineal trauma during birth and support her choices. 1.9.13 Once the presenting part distends the perineum in the second stage of labour, offer to apply a warm wet compress to the perineum and continue this until birth. Check the temperature of the compress is comfortable for the woman. 1.9.14 Consider massage of the perineum with a water-soluble lubricant in the second stage of labour, if perineal massage is acceptable to the woman and she prefers this to a warm compress. 1.9.15 Do not offer lidocaine spray to reduce pain in the second stage of labour. 1.9.16 Do not carry out a routine episiotomy during spontaneous vaginal birth. 1.9.17 Inform any woman with a history of severe perineal trauma that her risk of repeat severe perineal trauma is not increased in a subsequent birth, compared with women having their first baby. 1.9.18 Do not offer episiotomy routinely at vaginal birth after previous third- or fourth-degree trauma. 1.9.19 In order for a woman who has had previous third- or fourth-degree trauma to make an informed choice, talk with her about the future mode of birth, encompassing: current urgency or incontinence symptoms the degree of previous trauma risk of recurrence the success of the repair undertaken the psychological effect of the previous trauma management of her labour. 1.9.20 Inform any woman with infibulated genital mutilation of the risks of difficulty with vaginal examination, catheterisation and application of fetal scalp electrodes. Inform her of the risks of delay in the second stage and spontaneous laceration together with the possible need for defibulation in labour. [2007, amended 2023] 1.9.21 If an episiotomy is performed, the recommended technique is a mediolateral episiotomy originating at the vaginal fourchette and usually directed to the right side. The angle to the vertical axis should be between 45 and 60 degrees at the time of the episiotomy. 1.9.22 Perform an episiotomy if there is a clinical need, such as birth with forceps or ventouse or suspected fetal compromise. 1.9.23 Provide tested, effective analgesia before carrying out an episiotomy, except in an emergency because of acute fetal compromise. For a short explanation of why the committee made the 2023 recommendations and how they might affect practice, see the rationale and impact section on interventions to reduce perineal trauma. Full details of the evidence and the committee's discussion are in evidence review I: interventions to reduce perineal trauma. Loading. Please wait. Hide rationale and impact section on interventions to reduce perineal trauma Water birth 1.9.24 Inform women that there is insufficient high-quality evidence to either support or discourage giving birth in water. Duration of the active second stage and definition of delay 1.9.25 For a nulliparous woman without an epidural: birth would be expected to take place within 3 hours of the start of the active second stage in most women after 1 hour of active pushing, reassess the clinical picture, including progress, contractions, and maternal and fetal wellbeing: if there are signs of progress (in terms of rotation or descent of the presenting part), encourage the woman to continue pushing if there are no signs of progress, offer vaginal examination and consider amniotomy if the membranes are intact; if there is still no progress, diagnose delay and escalate for senior review if birth is not imminent after 2 hours of pushing, refer the woman for a senior review and a decision on place and mode of birth. [2007, amended 2023] 1.9.26 For a multiparous woman without an epidural: birth would be expected to take place within 2 hours of the start of the active second stage in most women after 30 minutes of active pushing, reassess clinical picture, including progress, contractions, and maternal and fetal wellbeing: if there are signs of progress (in terms of rotation or descent of the presenting part), encourage the woman to continue pushing if there are no signs of progress, offer vaginal examination and consider amniotomy if the membranes are intact; if there is still no progress, diagnose delay and escalate for senior review if birth is not imminent after 1 hour of pushing, refer the woman for senior review and decision on place and mode of birth. [2007, amended 2023] 1.9.27 For a nulliparous woman with an epidural: birth would be expected to take place within 3 hours of the start of the active second stage in most women, but be aware that these women may have had a passive stage of up to 2 hours after full dilatation before commencing active pushing (see recommendation 1.9.9) after 1 hour of active pushing, reassess the clinical picture, including progress, contractions, and maternal and fetal wellbeing: if there are signs of progress (in terms of rotation or descent of the presenting part), encourage the woman to continue pushing if there are no signs of progress, offer vaginal examination and consider amniotomy if the membranes are intact; if there is still no progress, diagnose delay and escalate for senior review if birth is not imminent after 2 hours of pushing, refer the woman for a senior review and decision on place and mode of birth. [2007, amended 2023] 1.9.28 For a multiparous woman with an epidural: birth would be expected to take place within 2 hours of the start of the active second stage in most women, but be aware that these women may have had a passive stage of up to 1 hour after full dilatation before commencing active pushing (see recommendation 1.9.10) after 30 minutes of active pushing, reassess clinical picture, including progress, contractions, and maternal and fetal wellbeing: if there are signs of progress (in terms of rotation or descent of the presenting part), encourage the woman to continue pushing if there are no signs of progress, offer vaginal examination and consider amniotomy if the membranes are intact; if there is still no progress, diagnose delay and escalate for senior review if birth is not imminent after 1 hour of pushing, refer the woman for a senior review and decision on place and mode of birth. [2007, amended 2023] Delay in the second stage 1.9.29 If there is delay in the second stage of labour (see the section on the duration of the active second stage), or if the woman is excessively distressed, provide support and sensitive encouragement and ask her if she needs analgesia or anaesthesia. [2007, amended 2023] 1.9.30 If there is delay in the second stage of labour and the decision is made to transfer the woman to obstetric-led care, follow the general principles for transfer of care described in section 1.5. Take into account that the presence of other risk factors (in addition to delay) may increase the urgency of the transfer. [2014, amended 2023] 1.9.31 An obstetrician should carry out an in-person assessment of a woman with confirmed delay in the second stage after transfer to obstetric-led care before contemplating the use of oxytocin. This should include: assessment and confirmation of fetal wellbeing (including presentation, position and heart rate) differentiation between the fetal and maternal heart rates confirmation that there are no signs of obstructed labour confirmation that contractions are infrequent or ineffective. [2014, amended 2023] 1.9.32 If the decision is made to start oxytocin in the second stage of labour, ensure that the time between increments of the dose is no more frequent than every 30 minutes. Increase oxytocin until there are 3 to 4 contractions in 10 minutes (see recommendation 1.8.54). 1.9.33 After initial obstetric assessment of a woman with delay in the second stage, maintain ongoing obstetric review every 15 to 30 minutes. Expediting birth 1.9.34 If the birth needs to be expedited for maternal or fetal reasons, assess both the risk to the baby and the safety of the woman. The assessment should include: the degree of urgency clinical findings on abdominal and vaginal examination the mode of birth (and whether to use forceps or ventouse if indicated) anticipated degree of difficulty, including the likelihood of success if birth with forceps or ventouse is attempted location any time that may be needed for transfer to obstetric-led care the need for additional analgesia or anaesthesia the woman's preferences. 1.9.35 Talk with the woman and her birth companion(s) about why the birth needs to be expedited and what the options are. 1.9.36 Inform the team about the degree of urgency. 1.9.37 Record the time at which the decision to expedite the birth is made. Birth with forceps or ventouse in delayed second stage 1.9.38 Offer birth with forceps or ventouse if there is concern about the baby's wellbeing, there is a prolonged second stage or the woman requests assistance. [2007, amended 2023] 1.9.39 If a woman declines a birth with forceps or ventouse: discuss her remaining options (vaginal birth, caesarean birth or reconsidering her decision about forceps or ventouse) advise her that her choices may be limited by clinical safety or degree of urgency (for example, if a caesarean birth is no longer an option because the baby's head is too low in the pelvis) support her decision. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on birth with forceps or ventouse in delayed second stage. Loading. Please wait. Hide rationale and impact section on birth with forceps or ventouse in delayed second stage 1.9.40 Base the choice of instrument on a balance of clinical circumstance and practitioner experience. [2007, amended 2023] 1.9.41 Discuss pain relief options for birth with forceps or ventouse. The option used should be based on the woman's preference and the clinical situation. [2007, amended 2023] 1.9.42 Ensure the level of pain relief is acceptable to the woman before using forceps or ventouse during birth. [2007, amended 2023] 1.9.43 Offer women who have had a birth with forceps or ventouse a single dose of intravenous co-amoxiclav (or a locally agreed alternative for women who are allergic to penicillin) within 6 hours after cord clamping. 1.9.44 Advise the woman to have a caesarean birth if vaginal birth is not possible. See the NICE guideline on caesarean birth. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on prophylactic antibiotics for birth with forceps or ventouse. Full details of the evidence and the committee's discussion are in evidence review J: prophylactic antibiotics for birth with forceps or ventouse. Loading. Please wait. Hide rationale and impact section on prophylactic antibiotics for birth with forceps or ventouse 1.10 Third stage of labour 1.10.1 Recognise that the time immediately after the birth is when the woman and her birth companion(s) are meeting and getting to know the baby. Ensure that any care or interventions are sensitive to this and minimise separation or disruption of the mother and baby. Definition of the third stage 1.10.2 For the purposes of this guideline, use the following definitions: the third stage of labour is the time from the birth of the baby to the expulsion of the placenta and membranes active management of the third stage involves a package of care comprising the following components: routine use of uterotonic drugs cord clamping and cutting of the cord (see recommendation 1.10.14) controlled cord traction after signs of separation of the placenta physiological management of the third stage involves a package of care that includes the following components: no routine use of uterotonic drugs no clamping of the cord until pulsation has stopped, or after delivery of the placenta delivery of the placenta spontaneously or by maternal effort. [2014, amended 2023] Observations in the third stage 1.10.3 Record the following observations for a woman in the third stage of labour: her general physical condition, as shown by her colour, respiration and her own report of how she feels vaginal blood loss. 1.10.4 If there is postpartum haemorrhage, a retained placenta or maternal collapse, or any other concerns about the woman's wellbeing: carry out frequent observations to assess whether resuscitation is needed transfer her to obstetric-led care; follow the general principles for transfer of care described in section 1.5, taking into account that multiple risk factors may increase the urgency of the transfer, particularly if they have a cumulative effect. [2014, amended 2023] Management of the third stage 1.10.5 Discuss with the woman antenatally, during her initial assessment and in labour: the different options for managing the third stage of labour, and what to expect with each option the benefits and risks associated with active and physiological management of the third stage (see tables 10 and 11). Table 10 Outcomes that were more or less likely for women having active management of the third stage compared with physiological management\| Outcome \| Active management of the third stage of labour \| Physiological management of the third stage of labour \| Risk difference \| --- --- \| \| Haemorrhage of more than 500 mL \| About 68 women per 1,000 would be expected to have a haemorrhage of more than 500 mL (so 932 per 1,000 would not) \| About 188 women per 1,000 would be expected to have a haemorrhage of more than 500 mL (so 812 per 1,000 would not) \| About 120 per 1,000 fewer women would be expected to have a haemorrhage of more than 500 mL with active management, so for 880 there would be no difference \| \| Haemorrhage of more than 1 litre \| About 13 women per 1,000 would be expected to have a haemorrhage of more than 1 litre (so 987 per 1,000 would not) \| About 29 women per 1,000 would be expected to have a haemorrhage of more than 1 litre (so 971 per 1,000 would not) \| About 16 per 1,000 fewer women would be expected to have a haemorrhage of more than 1 litre with active management, so for 984 there would be no difference \| \| Need for blood transfusion \| About 13 women per 1,000 would be expected to need a blood transfusion (so 987 per 1,000 would not) \| About 35 women per 1,000 would be expected to need a blood transfusion (so 965 per 1,000 would not) \| About 22 per 1,000 fewer women would be expected to need a blood transfusion with active management, so for 978 there would be no difference \| \| Postpartum anaemia (haemoglobin less than 9 g/dL) \| About 30 women per 1,000 would be expected to have anaemia (so 970 per 1,000 would not) \| About 60 women per 1,000 would be expected to have anaemia (so 940 per 1,000 would not) \| About 30 per 1,000 fewer women would be expected to have anaemia with active management, so for 970 there would be no difference \| \| Need for further uterotonics \| About 47 women per 1,000 would be expected to need further uterotonics (so 953 per 1,000 would not) \| About 247 women per 1,000 would be expected to need further uterotonics (so 753 per 1,000 would not) \| About 200 per 1,000 fewer women would be expected to need further uterotonics with active management, so for 800 there would be no difference \| \| Side effects (nausea and vomiting, headache, hypertension, readmission for bleeding) \| About 186 women per 1,000 would be expected to have these side effects (so 814 per 1,000 would not) \| About 90 women per 1,000 would be expected to have these side effects (so 910 per 1,000 would not) \| About 96 per 1,000 more women would be expected to have these side effects with active management, so for 904 there would be no difference \| Table 11 Outcomes that were similar for both active and physiological management of the third stage\| Outcome \| \| Retained placenta beyond 1 hour or need for manual removal \| \| Antibiotics for bleeding (discharge to 6 weeks) \| \| Satisfied with third-stage management \| \| Felt in control during labour \| 1.10.6 Advise women that active management of the third stage of labour is associated with a lower risk of a postpartum haemorrhage or blood transfusion. 1.10.7 If a woman requests physiological management of the third stage: discuss her level of risk so she can make an informed decision and support her in her choice. [2014, amended 2023] 1.10.8 Document in her records the decision that is agreed with the woman about management of the third stage. 1.10.9 For a woman who is having a vaginal birth and has chosen to have an active third stage, discuss the choice of uterotonic for active management. Include that: oxytocin plus ergometrine may be more effective than oxytocin alone at reducing the risk of postpartum haemorrhage oxytocin plus ergometrine is advised if there are risk factors which could increase the risk of postpartum haemorrhage oxytocin plus ergometrine is more likely to lead to nausea and vomiting compared with oxytocin alone oxytocin plus ergometrine is contraindicated in women with severe hypertension, pre-eclampsia, eclampsia, or severe cardiac, hepatic or renal disease. 1.10.10 Offer antiemetics (for example, cyclizine) to women having oxytocin plus ergometrine. 1.10.11 For active management after vaginal birth, administer 10 units of oxytocin (by intramuscular injection), 5 units of oxytocin (by intravenous injection, see recommendation 1.10.12) or 5 units of oxytocin plus 500 micrograms of ergometrine (by intramuscular injection) immediately after the birth of the baby and before the cord is clamped and cut. 1.10.12 If oxytocin is used, administer it by: intramuscular injection or slow intravenous injection over 3 to 5 minutes for women who have received oxytocin during labour. 1.10.13 For women who have had a caesarean birth, offer carbetocin by slow intravenous injection for the prevention of postpartum haemorrhage. For a short explanation of why the committee made the 2023 recommendations and how they might affect practice, see the rationale and impact section on management of the third stage of labour. Full details of the evidence and the committee's discussion are in: evidence review K: active and physiological management of the third stage evidence review L: route of administration of oxytocin in the third stage of labour evidence review M: uterotonics for prevention of postpartum haemorrhage. Loading. Please wait. Hide rationale and impact section on management of the third stage of labour 1.10.14 After administering the uterotonic, clamp and cut the cord, but: do not clamp the cord earlier than 1 minute from the birth of the baby unless there is concern about the integrity of the cord or the baby has a heart rate below 60 beats a minute that is not getting faster clamp the cord before 5 minutes in order to perform controlled cord traction as part of active management if the woman requests that the cord is clamped and cut later than 5 minutes, support her choice. [2014, amended 2023] For a short explanation of why the committee did not make any recommendations about the position of the baby during cord clamping, see the rationale and impact section on position of the baby during cord clamping. Full details of the evidence and the committee's discussion are in evidence review N: position of the baby during cord clamping. Loading. Please wait. Hide rationale and impact section on position of the baby during cord clamping 1.10.15 After cutting the cord, perform controlled cord traction as part of active management only after administration of oxytocin and signs of separation of the placenta. [2014, amended 2023] 1.10.16 Record the timing of cord clamping in both active and physiological management. 1.10.17 Advise a change from physiological management to active management if either of the following occur: haemorrhage the placenta is not delivered within 1 hour of the birth of the baby. 1.10.18 Offer a change from physiological management to active management if the woman wants to shorten the third stage. 1.10.19 Do not use either umbilical oxytocin infusion or prostaglandin routinely in the third stage of labour. Prolonged third stage 1.10.20 Diagnose a prolonged third stage of labour if it is not completed within 30 minutes of the birth with active management or within 60 minutes of the birth with physiological management. Follow the recommendations on managing a retained placenta. Retained placenta 1.10.21 Secure intravenous access if the placenta is retained, and explain to the woman why this is needed. 1.10.22 Do not use umbilical vein agents if the placenta is retained. 1.10.23 Do not use intravenous oxytocic agents routinely to deliver a retained placenta. 1.10.24 Give intravenous oxytocic agents if the placenta is retained and the woman is bleeding excessively. 1.10.25 If the placenta is retained and there is concern about the woman's condition: offer a vaginal examination to assess the need to undertake manual removal of the placenta explain that this assessment can be painful and advise her to have analgesia. 1.10.26 If the woman reports inadequate analgesia during the assessment, stop the examination and address this immediately. 1.10.27 If the placenta is retained and the woman is not already in an obstetric unit, arrange transfer. Follow the general principles for transfer of care described in section 1.5, taking into account that multiple risk factors may increase the urgency of the transfer, particularly if they have a cumulative effect. [2014, amended 2023] 1.10.28 Do not carry out uterine exploration or manual removal of the placenta without an anaesthetic. Postpartum haemorrhage Risk factors for postpartum haemorrhage 1.10.29 Advise women with antenatal risk factors for postpartum haemorrhage to give birth in an obstetric unit, where more emergency treatment options are available. Risk factors include: previous postpartum haemorrhage over 1,000 mL or requiring blood transfusion placenta accreta spectrum pre-eclampsia maternal haemoglobin level below 85 g/litre at onset of labour BMI greater than 35 kg/m 2 grand multiparity (parity 4 or more) antepartum haemorrhage or placental abruption overdistention of the uterus (for example, multiple pregnancy, polyhydramnios) existing uterine abnormalities (for example, fibroids) low-lying placenta. [2007, amended 2023] 1.10.30 Continue to assess risk factors for postpartum haemorrhage during labour, taking into account antenatal risk factors and any risk factors that have arisen during labour. These can include: induction or augmentation of labour with oxytocin or prostaglandins prolonged first or second stage of labour sepsis oxytocin use during labour precipitate labour birth with forceps or ventouse caesarean birth shoulder dystocia delay in delivery of the placenta. [2007, amended 2023] 1.10.31 Be aware that taking selective serotonin reuptake inhibitor (SSRI) or serotonin-noradrenaline reuptake inhibitor (SNRI) antidepressants in the month before birth may result in a small increased risk of postpartum haemorrhage, and that this should be taken into account as part of the bleeding and thrombotic risk assessment. See the Medicines and Healthcare products Regulatory Agency (MHRA) advice on the use of SSRI and SNRI antidepressants in the month before birth. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on risk factors for postpartum haemorrhage. Loading. Please wait. Hide rationale and impact section on risk factors for postpartum haemorrhage 1.10.32 If a woman has risk factors for postpartum haemorrhage, highlight these in her notes, and agree with her a care plan covering the third stage of labour. Management of postpartum haemorrhage 1.10.33 If a woman has a postpartum haemorrhage: call for help give immediate clinical treatment: emptying of the bladder and uterine massage and uterotonic drugs and intravenous fluids and controlled cord traction if the placenta has not yet been delivered continuously assess blood loss and the woman's condition, and identify the source of the bleeding consider giving supplementary oxygen (starting at 15 L/minute to obtain a target oxygen saturation of 94% to 98%, using a non-rebreathing mask with a reservoir bag) arrange for transfer of the woman to obstetric-led care (following the general principles for transfer of care described in section 1.5). [2014, amended 2023] 1.10.34 Administer 1 of the following (see table 12) as first-line treatment for postpartum haemorrhage, taking into account which uterotonics have already been administered as part of active management of the third stage of labour. Offer further treatment for postpartum haemorrhage if needed. Table 12 Choice of uterotonics for the treatment of postpartum haemorrhage because of uterine atony [2023, amended 2024]\| Uterotonic used in the third stage of labour as prophylaxis \| Suggested first-line treatment of postpartum haemorrhage \| Suggested second-line treatment of postpartum haemorrhage \| Additional treatments that can be offered, depending on clinical need \| --- --- \| \| No uterotonic used – physiological management \| Oxytocin plus ergometrine by intramuscular injection (if contraindicated, give carboprost), or Oxytocin infusion as soon as intravenous access is available \| Carboprost intramuscular injection \| Carboprost intramuscular injection (can be repeated at intervals not less than 15 minutes up to a maximum of 8 doses), or Misoprostol 800 micrograms sublingually or rectally (may be used earlier if intravenous route not available) \| \| Oxytocin alone \| Ergometrine intramuscular injection (if contraindicated give carboprost), or Oxytocin infusion as soon as intravenous access is available \| Carboprost intramuscular injection \| Carboprost intramuscular injection (can be repeated at intervals not less than 15 minutes up to a maximum of 8 doses), or Misoprostol 800 micrograms sublingually or rectally (may be used earlier if intravenous route not available) \| \| Oxytocin plus ergometrine \| Carboprost intramuscular injection, or Oxytocin infusion as soon as intravenous access is available \| Repeat carboprost after 15 minutes \| Carboprost intramuscular injection (can be repeated at intervals not less than 15 minutes up to a maximum of 8 doses), or Misoprostol 800 micrograms sublingually or rectally (may be used earlier if intravenous route not available) \| \| Carbetocin \| Ergometrine intramuscular injection \| Carboprost intramuscular injection \| Carboprost intramuscular injection (can be repeated at intervals not less than 15 minutes up to a maximum of 8 doses), or Misoprostol 800 micrograms sublingually or rectally \| Please note that: Not all medicines in table 12 will be available in all settings, and this may impact on choice and order of use. A repeat dose of ergometrine or the combination of oxytocin and ergometrine may be given if other medicines are not available, for example in a home birth setting. Note that after intramuscular administration, oxytocin acts in about 2.5 minutes and the effects last about 30 minutes to 1 hour. Ergometrine acts in about 7 minutes and the effects last about 3 hours. In December 2024, this was an off-label use of misoprostol, so the dosage is included in table 12. Consult the BNF for dosages of other drugs listed. See NICE's information on prescribing medicines. In September 2024, NHS England published a National Patient Safety Alert on the risks of oxytocin overdose during labour and childbirth which contains advice on actions needed. 1.10.35 In addition to uterotonic drugs, give tranexamic acid (1 g by intravenous injection over 10 minutes). Repeat if necessary after at least 30 minutes for managing continuing postpartum haemorrhage. For a short explanation of why the committee made the 2023 recommendations and how they might affect practice, see the rationale and impact section on the pharmacological management of postpartum haemorrhage. Full details of the evidence and the committee's discussion are in evidence review O: pharmacological management of postpartum haemorrhage. Loading. Please wait. Hide rationale and impact section on the pharmacological management of postpartum haemorrhage For a short explanation of why the committee made the 2024 update to table 12 and how it might affect practice, see the rationale and impact section on uterotonics for postpartum haemorrhage. Full details of the evidence and the committee's discussion are in evidence review O: pharmacological management of postpartum haemorrhage. Loading. Please wait. Hide rationale and impact section on uterotonics for postpartum haemorrhage 1.10.36 Allocate a member of the healthcare team to stay with the woman and her birth companion(s), explain what is happening, answer any questions and offer support throughout the emergency situation. 1.10.37 If the haemorrhage continues: consider near-patient coagulation testing, if available consider administration of blood products (for example, packed red cells and clotting products) perform examination under anaesthetic ensure that the uterus is empty and repair any trauma consider balloon tamponade before surgical options. [2014, amended 2023] 1.10.38 Be aware that no particular surgical procedure can be recommended over any other for treating postpartum haemorrhage. 1.10.39 Ensure the maternity service and ambulance service have strategies in place to respond quickly and appropriately if a woman has a postpartum haemorrhage in any setting. 1.11 Care of the newborn baby Initial assessment of the newborn baby and mother–baby bonding 1.11.1 Record the Apgar score routinely at 1 minute and 5 minutes for all births. 1.11.2 When assessing the colour element of the Apgar score: assess central oxygenation by looking inside the mouth at the mucous membranes and tongue assess peripheral oxygenation by looking at the colour of the nail beds. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on initial assessment of the newborn baby. Loading. Please wait. Hide rationale and impact section on initial assessment of the newborn baby 1.11.3 Record the time from birth to the onset of regular respirations. 1.11.4 If the baby is born in poor condition (for example, with abnormal breathing, heart rate or tone): follow recommendations on neonatal resuscitationand take paired cord-blood samples for blood gas analysis, after double-clamping the cord using 2 clamps. Continue to evaluate and record the baby's condition until it is improved and stable. 1.11.5 Do not take paired cord-blood samples (for blood gas analysis) routinely. 1.11.6 Ensure that a second clamp to allow double-clamping of the cord is available in all birth settings. 1.11.7 Encourage women to have skin-to-skin contact with their babies as soon as possible after the birth. If the woman is not well enough, encourage her birth companion to have skin-to-skin contact instead. [2007, amended 2023] 1.11.8 In order to keep the baby warm, dry and cover them with a warm, dry blanket or towel while maintaining skin-to-skin contact with the woman. 1.11.9 Prioritise optimal baby airway positioning, ensuring the head is supported so the airway does not become obstructed during skin-to-skin contact and explain to the woman and her birth companion(s) how to maintain the baby's airway. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on optimal positioning during skin-to-skin contact. Loading. Please wait. Hide rationale and impact section on optimal positioning during skin-to-skin contact 1.11.10 Avoid separating the woman and her baby within the first hour of the birth for routine postnatal procedures, for example, weighing, measuring and bathing, unless these measures are requested by the woman or are necessary for the immediate care of the baby. 1.11.11 Encourage initiation of breastfeeding as soon as possible after the birth, ideally within 1 hour. 1.11.12 Record head circumference, and birth weight soon after the first hour following birth and plot on the centile chart. [2007, amended 2023] 1.11.13 Check the baby's body temperature is in the normal range. [2007, amended 2023] 1.11.14 Undertake an initial examination to detect any major physical abnormality and to identify any problems that need referral. 1.11.15 Undertake additional monitoring of the baby for women who have taken SSRI or SNRI antidepressants during pregnancy as these may result in a small increased risk of persistent pulmonary hypertension of the newborn or neonatal withdrawal symptoms. See the NICE guideline on antenatal and postnatal mental health and the MHRA advice on the use of SSRI and SNRI antidepressants in pregnancy. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on additional monitoring for babies exposed to antidepressants in utero. Loading. Please wait. Hide rationale and impact section on additional monitoring for babies exposed to antidepressants in utero 1.11.16 Ensure that any examination or treatment of the baby is undertaken with the consent of the parents and either in their presence or, if this is not possible, with their knowledge. Neonatal resuscitation 1.11.17 In the first minutes after birth, evaluate the condition of the baby (specifically respiration, heart rate and tone) in order to determine whether resuscitation is needed according to nationally accredited guidelines on neonatal resuscitation. 1.11.18 All relevant healthcare professionals caring for women during birth should attend annually a course in neonatal resuscitation that is consistent with nationally accredited guidelines on neonatal resuscitation. 1.11.19 In all birth settings: bear in mind that it will be necessary to call for help if the baby needs resuscitation, and plan accordingly ensure that there are facilities for resuscitation, and for transferring the baby to another location if necessary develop emergency referral pathways for both the woman and the baby, and implement these if necessary. 1.11.20 If a newborn baby needs basic resuscitation, start with air. 1.11.21 Minimise separation of the baby and mother, taking into account the clinical circumstances. 1.11.22 Throughout an emergency situation in which the baby needs resuscitation, allocate a member of the healthcare team to talk with, and offer support to, the woman and any birth companion(s). Care of babies in the presence of meconium 1.11.23 In the presence of any degree of meconium: do not suction the baby's upper airways (nasopharynx and oropharynx) before birth of the shoulders and trunk do not suction the baby's upper airways (nasopharynx and oropharynx) if the baby has normal respiration, heart rate and tone do not intubate if the baby has normal respiration, heart rate and tone. 1.11.24 If there has been any degree of meconium and the baby does not have normal respiration, heart rate and tone, follow nationally accredited guidelines on neonatal resuscitation. [2014, amended 2023] 1.11.25 If there has been significant meconium and the baby is healthy, closely observe the baby within a unit with immediate access to a neonatologist. Perform these observations at 1 and 2 hours old and then 2‑hourly until 12 hours old. 1.11.26 If there has been non-significant meconium, observe the baby at 1 and 2 hours old in all birth settings. 1.11.27 If any of the following are observed after any degree of meconium, ask a neonatologist to assess the baby. Transfer both the woman and baby if they are at home or in a freestanding midwifery unit. Follow the general principles for transfer of care described in section 1.5, taking into account that multiple risk factors may increase the urgency of the transfer, particularly if they have a cumulative effect: respiratory rate above 60 breaths per minute the presence of grunting heart rate below 100 or above 160 beats per minute capillary refill time above 3 seconds body temperature of 38°C or above, or 37.5°C on 2 occasions 15 to 30 minutes apart oxygen saturation below 95% (measuring oxygen saturation is optional after non-significant meconium) presence of central cyanosis, confirmed by pulse oximetry if available. [2014, amended 2023] 1.11.28 Explain the findings to the woman, and inform her about what to look out for and who to talk to if she has any concerns. Babies born to women with prelabour rupture of the membranes at term 1.11.29 Closely observe any baby born to a woman with prelabour rupture of the membranes (more than 24 hours before the onset of established labour) at term for the first 12 hours of life (at 1, 2, 6 and 12 hours) in all settings. Include assessment of: temperature heart rate respiratory rate presence of respiratory grunting significant subcostal recession presence of nasal flare presence of central cyanosis, confirmed by pulse oximetry if available skin perfusion assessed by capillary refill floppiness concerns about general wellbeing and feeding. If any of these are observed, ask a neonatologist to assess the baby. Transfer both the woman and baby if they are at home or in a freestanding midwifery unit. Follow the general principles for transfer of care described in section 1.5 and take into account that multiple risk factors may increase the urgency of the transfer, particularly if they have a cumulative effect. [2014, amended 2023] 1.11.30 If there are no signs of infection in the woman or in the baby, do not give antibiotics to either the woman or the baby, even if the membranes have been ruptured for over 24 hours. [2007, amended 2023] 1.11.31 If there is evidence of infection in the woman or in the baby, see the section on antibiotics for suspected early-onset infection in NICE's guideline on neonatal infection for advice on when to consider antibiotics. [2007, amended 2023] 1.11.32 Advise women with prelabour rupture of the membranes to inform their healthcare professionals immediately of any concerns they have about their baby's wellbeing in the first 5 days after birth, particularly in the first 12 hours when the risk of infection is greatest. 1.11.33 Do not perform blood, cerebrospinal fluid and/or surface culture tests in an asymptomatic baby. 1.11.34 Refer a baby with any symptom of possible sepsis, or born to a woman who has evidence of chorioamnionitis, to a neonatal care specialist immediately. 1.12 Care of the woman after birth Initial assessment 1.12.1 Carry out the following observations of the woman after birth: record her temperature, pulse, blood pressure and respiratory rate. Transfer the woman (with her baby) to obstetric-led care if any of the relevant indications listed in recommendation 1.8.20 are met check uterine contraction and lochia examine the placenta and membranes: assess their condition, structure, cord vessels and completeness; transfer the woman (with her baby) to obstetric-led care if the placenta is incomplete make an early assessment of the woman's emotional and psychological condition in response to labour and birth; see the NICE guideline on postnatal care check for successful voiding of the bladder; if, after 6 hours, her bladder is palpable and she is unable to pass urine, advise catheterisation and consider transferring the woman (with her baby) to obstetric-led care. If transferring the woman to obstetric-led care, follow the general principles for transfer of care described in section 1.5 and take into account that multiple risk factors may increase the urgency of the transfer, particularly if they have a cumulative effect. [2014, amended 2023] 1.12.2 Check that women who have had regional analgesia or anaesthesia can perform a straight leg raise by 4 hours after the last anaesthetic dose. If not, contact the obstetric anaesthetist for urgent review. For a short explanation of why the committee made the 2023 recommendation and how it might affect practice, see the rationale and impact section on care of the woman after birth. Loading. Please wait. Hide rationale and impact section on care of the woman after birth Perineal care 1.12.3 Define perineal or genital trauma caused by either tearing or episiotomy as follows: first degree – injury to skin only second degree – injury to the perineal muscles but not the anal sphincter third degree – injury to the perineum involving the anal sphincter complex: 3a – less than 50% of external anal sphincter thickness torn 3b – more than 50% of external anal sphincter thickness torn 3c – internal anal sphincter torn fourth degree – injury to the perineum involving the anal sphincter complex (external and internal anal sphincter) and anal epithelium. 1.12.4 Before assessing for genital trauma: explain to the woman what is planned and why offer inhalational analgesia ensure good lighting position the woman so that she is comfortable and so that the genital structures can be seen clearly. 1.12.5 Offer all women systematic assessment, including a rectal examination, to exclude all genital tract trauma, including buttonhole tears. [2007, amended 2023] 1.12.6 Perform the initial examination gently and with sensitivity. It may be done in the immediate period after birth. 1.12.7 Include the following in a systematic assessment of genital trauma: further explanation of what is planned and why confirmation by the woman that tested effective local or regional analgesia is in place visual assessment of the extent of perineal trauma to include the structures involved, the apex of the injury and assessment of bleeding a rectal examination to assess whether there has been any damage to the external or internal anal sphincter if there is any suspicion that the perineal muscles are damaged. 1.12.8 Ensure that the timing of this systematic assessment does not interfere with mother–baby bonding unless the woman has bleeding that requires urgent attention. 1.12.9 Assist the woman to adopt a position that allows adequate visual assessment of the degree of trauma and for repair. Only maintain this position for as long as necessary for systematic assessment and repair. If it is not possible to adequately assess the trauma, transfer the woman (with her baby) to obstetric-led care, following the general principles for transfer of care described in section 1.5. [2007, amended 2014] 1.12.10 Seek advice from a more experienced midwife or obstetrician if there is uncertainty about the nature or extent of the trauma. Transfer the woman (with her baby) to obstetric-led care (following the general principles for transfer of care described in section 1.5) if the repair needs further surgical or anaesthetic expertise. [2007, amended 2014] 1.12.11 Document the systematic assessment and its results fully, possibly pictorially. 1.12.12 All relevant healthcare professionals should attend training in perineal/genital assessment and repair and ensure that they maintain these skills. 1.12.13 Undertake repair of the perineum as soon as possible to minimise the risk of infection and blood loss. 1.12.14 When carrying out perineal repair: ensure that tested effective analgesia is in place, using infiltration with up to 20 ml of 1% lidocaine or equivalent top up the epidural or insert a spinal anaesthetic if necessary. 1.12.15 If the woman reports inadequate pain relief at any point, address this immediately. 1.12.16 Advise the woman that in the case of first-degree trauma, the wound should be sutured in order to improve healing, unless the skin edges are well opposed. 1.12.17 Advise the woman that in the case of second-degree trauma, the muscle should be sutured in order to improve healing. 1.12.18 If the skin is opposed after suturing of the muscle in second-degree trauma, there is no need to suture it. 1.12.19 If the skin does need suturing, use a continuous subcuticular technique. 1.12.20 Undertake perineal repair using a continuous non-locked suturing technique for the vaginal wall and muscle layer. 1.12.21 Use an absorbable synthetic suture material to suture the perineum. 1.12.22 Offer rectal non-steroidal anti-inflammatory drugs routinely after perineal repair of first- and second-degree trauma provided these drugs are not contraindicated. 1.12.23 Observe the following basic principles when performing perineal repairs: repair perineal trauma using aseptic techniques check equipment and count swabs and needles before and after the procedure ensure good lighting is available to see and identify the structures involved ensure that trauma is repaired in the appropriate place with appropriate anaesthesia, according to the clinician's judgement ensure that good anatomical alignment of the wound is achieved and that consideration is given to the cosmetic results ensure that suture material has not been accidentally inserted through the rectal mucosa by carrying out a rectal examination after completing the repair after completion of the repair, document an accurate detailed account covering the extent of the trauma, the method of repair and the materials used give the woman information about the extent of the trauma, pain relief, diet, hygiene and the importance of learning to do pelvic floor exercises, what to expect as they recover, and where and when to seek advice or psychological support if needed. [2007, amended 2023] Terms used in this guideline This section defines terms that have been used in a particular way for this guideline. For other definitions, see the NICE glossary and the Think Local, Act Personal Care and Support Jargon Buster. Cephalic The baby is positioned head-down, ready to be born head-first through the vagina. Hyperstimulation This is overactivity of the uterus. It is variously defined as uterine tachysystole (more than 5 contractions per 10 minutes for at least 20 minutes), and uterine hypersystole or hypertonicity (a contraction lasting at least 2 minutes). These may or may not be associated with changes in the fetal heart rate pattern (persistent decelerations, tachycardia, or increased or decreased short-term variability). Hyponatraemia Peripartum hyponatraemia occurs during labour and birth when the sodium concentration in the blood drops below 130 mmol/L and can have life-threatening consequences. Women or pregnant people in labour are predisposed to developing hyponatraemia because of lower baseline sodium level during pregnancy, lowered ability to excrete water in the third trimester and exposure to the anti-diuretic effect of oxytocin (both natural and synthetic). Maternal hyponatraemia also impacts the unborn baby because water crosses the placenta freely, which can lead to hyponatraemia in the newborn. Early symptoms and signs include headache, loss of appetite, nausea and lethargy, while more advanced symptoms and signs include agitation, confusion, seizures, depressed reflexes and coma. Significant meconium This is defined as dark green or black amniotic fluid that is thick or tenacious, or any meconium-stained amniotic fluid containing lumps of meconium. 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16596	https://link.springer.com/article/10.1007/s12045-016-0423-4	Log in The arithmetic mean - Geometric mean - Harmonic mean: Inequalities and a spectrum of applications Classroom Published: Volume 21, pages 1119–1133, (2016) Cite this article Resonance Aims and scope Submit manuscript Prithwijit De1 306 Accesses 4 Citations 2 Altmetric 1 Mention Explore all metrics Abstract The Arithmetic Mean–Geometric Mean–Harmonic Mean inequality, AM–GM–HM inequality in short, is one of the fundamental inequalities in Algebra, and it is used extensively in olympiad mathematics to solve many problems. The aim of this article is to acquaint students with the inequality, its proof and various applications. This is a preview of subscription content, log in via an institution to check access. Access this article Log in via an institution Subscribe and save Springer+ from $39.99 /Month Starting from 10 chapters or articles per month Access and download chapters and articles from more than 300k books and 2,500 journals Cancel anytime View plans Buy Now Price excludes VAT (USA) Tax calculation will be finalised during checkout. Instant access to the full article PDF. Institutional subscriptions Similar content being viewed by others An Arithmetic-Geometric Mean of a Third Kind! Chapter © 2019 Explicit forms of invariant means: complementary results to Gauss (\left( {\mathcal {A}},{\mathcal {G}}\right) )-theorem and some applications Article Open access 14 July 2023 Notes on the arithmetic–geometric mean inequality Article 17 January 2024 Explore related subjects Discover the latest articles, books and news in related subjects, suggested using machine learning. Abstract Harmonic Analysis Algebra Geometry Mathematics Mathematics in the Humanities and Social Sciences Social Inequality Suggested Reading B J Venkatachala, Inequalities - An Approach Through Problems, Hindustan Book Agency. Download references Author information Authors and Affiliations Homi Bhabha Centre for Science Education, Tata Institute of Fundamental Research, Mumbai, India Prithwijit De Authors Prithwijit De View author publications Search author on:PubMed Google Scholar Corresponding author Correspondence to Prithwijit De. Rights and permissions Reprints and permissions About this article Cite this article De, P. The arithmetic mean - Geometric mean - Harmonic mean: Inequalities and a spectrum of applications. Reson 21, 1119–1133 (2016). Download citation Published: Issue Date: DOI: Share this article Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Provided by the Springer Nature SharedIt content-sharing initiative Keywords Arithmetic mean geometric mean harmonic mean Nesbit’s inequality Euler’s inequality
16597	https://brainly.com/question/13222945	[FREE] Calculate the densities of a hydrogen nucleus and a hydrogen atom using the information in this chapter on - brainly.com 3 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +59,9k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +45,8k Ace exams faster, with practice that adapts to you Practice Worksheets +7k Guided help for every grade, topic or textbook Complete See more / Chemistry Textbook & Expert-Verified Textbook & Expert-Verified Calculate the densities of a hydrogen nucleus and a hydrogen atom using the information in this chapter on the mass of the proton, the mass of the electron, and the sizes of the nucleus and the atom. Radius of a hydrogen atom: 1×1 0−8 cm Density of hydrogen nucleus: Density of hydrogen atom: 1 See answer Explain with Learning Companion NEW Asked by jjtavory6921 • 09/10/2019 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 4587084 people 4M 0.0 0 Upload your school material for a more relevant answer The density of a hydrogen nucleus is approximately 1.24 x 10^17 kg/m^3, and the density of a hydrogen atom is approximately 8.38 x 10^16 kg/m^3. Explanation To calculate the density of a hydrogen nucleus, we can use the formula for density, which is mass divided by volume. The mass of a proton is approximately 1.67 x 10^-27 kg, and the volume of a sphere can be calculated using the formula V = (4/3)πr^3. Let's assume the radius of a hydrogen nucleus is equal to the diameter of a proton, which is approximately 1.2 fm. Plugging these values into the formula, we can calculate the density of a hydrogen nucleus. density = mass / volume = 1.67 x 10^-27 kg / ((4/3)π(0.6 x 10^-15 m)^3) = 1.24 x 10^17 kg/m^3 So, the density of a hydrogen nucleus is approximately 1.24 x 10^17 kg/m^3. To calculate the density of a hydrogen atom, we need to know its mass and volume. The mass of an electron is approximately 9.11 x 10^-31 kg, and the volume of a sphere can be calculated using the formula V = (4/3)πr^3. Let's assume the radius of a hydrogen atom is equal to the given value, which is 1 x 10^-8 cm. Plugging these values into the formula, we can calculate the density of a hydrogen atom. density = mass / volume = 9.11 x 10^-31 kg / ((4/3)π(0.5 x 10^-10 m)^3) = 8.38 x 10^16 kg/m^3 So, the density of a hydrogen atom is approximately 8.38 x 10^16 kg/m^3. Answered by JoanFontaine •10.5K answers•4.6M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 4587084 people 4M 0.0 0 The AP Physics Collection - Gregg Wolfe, Erika Gasper, John Stoke, Julie Kretchman, David Anderson, Nathan Czuba, Sudhi Oberoi, Liza Pujji, Irina Lyublinskaya, Douglas Ingram Physics - Boundless Chemistry - Paul Flowers Upload your school material for a more relevant answer The density of a hydrogen nucleus is approximately 2.31 x 10^17 kg/m³, while the density of a hydrogen atom is about 4.00 x 10² kg/m³. The calculations show that the nucleus is significantly denser than the atom as a whole, illustrating the compact nature of nuclear matter compared to atomic matter. Explanation To calculate the densities of a hydrogen nucleus and a hydrogen atom, we will use the formula for density: Density=Volume Mass 1. Density of Hydrogen Nucleus: For the hydrogen nucleus, we consider just the proton since it is the only nucleon in a hydrogen atom. The mass of a proton is approximately: m p=1.67×1 0−27 kg. The radius of a hydrogen nucleus (approximately the size of a proton) is about r n≈1.2 femtometers (fm), which is 1.2×1 0−15 m. The volume of the nucleus, assuming it's spherical, can be calculated using the formula for the volume of a sphere: V=3 4π r 3 Substituting for radius: V n=3 4π(1.2×1 0−15 m)3 Calculating the volume: V n≈3 4π(1.728×1 0−45)≈7.24×1 0−45 m 3 Now, calculating the density: Density n=V nm p≈7.24×1 0−45 m 3 1.67×1 0−27 kg≈2.31×1 0 17 kg/m 3 2. Density of Hydrogen Atom: For the hydrogen atom: The total mass of a hydrogen atom is approximately the mass of a proton plus the mass of an electron: Mass of an electron: m e=9.11×1 0−31 kg. Total mass of hydrogen atom: m H≈1.67×1 0−27+9.11×1 0−31≈1.68×1 0−27 kg The radius of the hydrogen atom is given as r a=1×1 0−8 cm = 1×1 0−10 m. The volume of the hydrogen atom is: V a=3 4π(1×1 0−10 m)3 Calculating the volume: V a≈3 4π(1×1 0−30)≈4.19×1 0−30 m 3 Now calculating the density: Density a=V am H≈4.19×1 0−30 m 3 1.68×1 0−27 kg≈4.00×1 0 2 kg/m 3 Conclusion: The density of the hydrogen nucleus is approximately 2.31×1 0 17 kg/m 3. The density of the hydrogen atom is approximately 4.00×1 0 2 kg/m 3. Examples & Evidence For example, nuclear densities are incredibly high compared to everyday materials. The density of water is 1 g/cm³, which is about 1,000 kg/m³, whereas the nucleus's density is several orders of magnitude higher (about 2.31 x 10^17 kg/m³). This demonstrates how tightly packed the protons and neutrons are within the nucleus compared to the electrons that occupy the larger space of the atom. The calculations are based on established physics regarding the mass and volume of subatomic particles, supporting the conclusion that the densities of the nucleons are much higher than the densities attributed to the entire atom. Thanks 0 0.0 (0 votes) Advertisement jjtavory6921 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Chemistry solutions and answers Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __ Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction. Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power. Community Answer 5 Dimensional Analysis 1. I have 470 milligrams of table salt, which is the chemical compound NaCl. How many liters of NaCl solution can I make if I want the solution to be 0.90% NaCl? (9 grams of salt per 1000 grams of solution). The density of the NaCl solution is 1.0 g solution/mL solution. Community Answer 8 For which pair of functions is the exponential consistently growing at a faster rate than the quadratic over the interval mc015-1. Jpg? mc015-2. Jpg mc015-3. Jpg mc015-4. Jpg mc015-5. Jpg. Community Answer 4.8 268 How did Mendeleev come up with the first periodic table of the elements? (1 point) A He determined the mass of atoms of each element. B He estimated the number of electrons in atoms of each element. C He arranged the elements by different properties to find a pattern. D He organized the elements by their atomic number. New questions in Chemistry Consider the chemical equations shown here. C H 4(g)+2 O 2(g)→C O 2(g)+2 H 2O(g)Δ H 1=−802 k J 2 H 2O(g)→2 H 2O(l)Δ H 2=−88 k J Which equation shows how to calculate Δ H r x n for the equation below? C H 4(g)+2 O 2(g)→C O 2(g)+2 H 2O(l) What is Δ H r x n for the overall reaction? Consider the chemical equations shown here. 2 H 2(g)+O 2(g)→2 H 2O(g)Δ H 1=−483.6 k J+2=−241.8 k J/m o l 3 O 2(g)→2 O 3(g)Δ H 2=284.6 k J+2=142.3 k J/m o l What is the overall enthalpy of reaction for the equation shown below? 3 H 2(g)+O 3(g)→3 H 2O(g)□ kJ Consider the chemical equations shown here. P 4(s)+3 O 2(g)→P 4O 6(s)Δ H 1=−1,640.1 k J P 4O 10(s)→P 4(s)+5 O 2(g)Δ H 2=2,940.1 k J What is the overall enthalpy of reaction for the equation shown below? Round the answer to the nearest whole number. P 4O 6(s)+2 O 2(g)→P 4O 10(s) Prepare a "Coal and Petroleum" fact file. What is a double bond? A. A covalent bond between 2 atoms where each atom contributes 2 electrons. B. A covalent bond between 2 atoms where each atom contributes 4 electrons. C. An ionic bond between 2 atoms where 2 electrons have been transferred. D. A covalent bond between 2 atoms where each atom contributes 1 electron. 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16598	https://simple.wikipedia.org/wiki/Tangent_(geometry)	Jump to content Search Contents Beginning 1 Uses 2 Related pages Tangent (geometry) العربية Asturianu বাংলা Български Català Чӑвашла Čeština Cymraeg Dansk Deutsch Eesti English Español Esperanto Euskara فارسی Français Gaeilge Galego 한국어 Հայերեն हिन्दी Hrvatski Bahasa Indonesia Íslenska Italiano עברית Қазақша Кыргызча Lietuvių Lombard Magyar Македонски മലയാളം Bahasa Melayu Nederlands नेपाली 日本語 Nordfriisk Norsk bokmål Norsk nynorsk Piemontèis Polski Português Romnă Runa Simi Русский Shqip Slovenščina Soomaaliga کوردی Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska Tagalog தமிழ் ไทย Türkçe Українська Tiếng Việt 文言吴语粵語中文 Change links Page Talk Read Change Change source View history Tools Actions Read Change Change source View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Make a book Download as PDF Page for printing In other projects Wikimedia Commons Wikidata item Appearance From Simple English Wikipedia, the free encyclopedia In geometry, a tangent is a straight line that touches a curve at one point. At the place where they touch, the line and the curve both have the same slope (they are both "going in the same direction"). For this reason, a tangent line is a good approximation of the curve near that point. The curve and the tangent line are almost exactly the same near the intersection point. A straight line which touches a circle at a single point is perpendicular to the circle's radius and is considered tangent to the circle. Uses [change \| change source] Tangent lines are useful in calculus because they can magnify the slope of a curve at a single point. The slope of some curve at the point P is often called the instantaneous rate of change of the curve at P. The instantaneous rate of change at a specific point on the curve can be found by evaluating the derivative function for the curve at that point. Related pages [change \| change source] Derivative (mathematics) Calculus Trigonometric function This short article about mathematics can be made longer. You can help Wikipedia by adding to it. Retrieved from " Category: Geometry Hidden category: Math stubs Tangent (geometry) Add topic
16599	https://en.wikipedia.org/wiki/African_leopard	Jump to content African leopard العربية Azərbaycanca Български Čeština Dansk Ελληνικά Español Esperanto Euskara فارسی Français Gagauz 한국어 Հայերեն Bahasa Indonesia Italiano עברית Kotava Magyar Македонски മലയാളം مصرى Монгол Nederlands Norsk bokmål ଓଡ଼ିଆ Oʻzbekcha / ўзбекча پنجابی Polski Português Русский Shqip Slovenščina Suomi Taqbaylit ไทย Türkçe Українська Tiếng Việt 中文 Edit links From Wikipedia, the free encyclopedia Leopard subspecies \| African leopard \| \| --- \| \| \| \| \| Leopard in Kruger National Park \| \| \| Conservation status \| \| \| Vulnerable (IUCN 3.1) \| \| \| Scientific classification \| \| \| Kingdom: \| Animalia \| \| Phylum: \| Chordata \| \| Class: \| Mammalia \| \| Order: \| Carnivora \| \| Suborder: \| Feliformia \| \| Family: \| Felidae \| \| Subfamily: \| Pantherinae \| \| Genus: \| Panthera \| \| Species: \| P. pardus \| \| Subspecies: \| P. p. pardus \| \| Trinomial name \| \| \| Panthera pardus pardus (Linnaeus, 1758) \| \| \| Synonyms \| \| \| List + P. p. panthera (Schreber, 1777) + P. p. leopardus (Schreber, 1777) + P. p. melanotica (Günther, 1885) + P. p. suahelicus (Neumann, 1900) + P. p. nanopardus (Thomas, 1904) + P. p. ruwenzorii (Camerano, 1906) + P. p. chui (Heller, 1913) + P. p. reichenowi (Cabrera, 1918) + P. p. antinorii (de Beaux, 1923) + P. p. ituriensis (Allen, 1924) + P. p. adusta Pocock, 1927 + P. p. shortridgei (Pocock, 1932) + P. p. brockmani (Pocock, 1932) + P. p. puella (Pocock, 1932) + P. p. adersi Pocock, 1932 \| \| The African leopard (Panthera pardus pardus) is the nominate subspecies of the leopard, native to many countries in Africa. It is widely distributed in most of sub-Saharan Africa, but the historical range has been fragmented in the course of habitat conversion. Leopards have also been recorded in North Africa as well. Taxonomy [edit] Felis pardus was the scientific name used by Carl Linnaeus in the 10th edition of Systema Naturae in 1758. His description was based on descriptions by earlier naturalists such as Conrad Gessner. He assumed that the leopard occurred in India. In the 18th and 19th centuries, several naturalists described various leopard skins and skulls from Africa, including: Felis pardus panthera proposed by Johann Christian Daniel von Schreber in 1778 based on descriptions by earlier naturalists Felis leopardus var. melanotica by Albert Günther in 1885 from the Cape of Good Hope, Southern Africa Felis leopardus suahelicus by Oscar Neumann in 1900 from the Tanganyika territory Felis leopardus nanopardus by Oldfield Thomas in 1904 from Italian Somaliland Felis pardus ruwenzorii by Lorenzo Camerano in 1906 from the Ruwenzori and Virunga Mountains Felis pardus chui by Edmund Heller in 1913 from Uganda Felis pardus iturensis by Joel Asaph Allen in 1924 from the Belgian Congo Felis pardus reichenowi by Ángel Cabrera in 1927 from Cameroon Panthera pardus adusta by Reginald Innes Pocock in 1927 from the Ethiopian Highlands Panthera pardus adersi by Pocock in 1932 from Unguja Island, Zanzibar Panthera pardus brockmani by Pocock in 1932 from Somaliland Results of genetic analyses indicate that all African leopard populations are generally closely related and represent only one subspecies, namely P. p. pardus. However, results of an analysis of molecular variance and the pairwise fixation index of African leopard museum specimens shows differences in the ND-5 locus spanning five major haplogroups, namely in Central–Southern Africa, Southern Africa, West Africa, coastal West–Central Africa, and Central–East Africa. In some cases, fixation indices showed higher diversity than for Panthera pardus nimr and Panthera pardus tulliana in Asia. Characteristics [edit] The African leopard exhibits great variation in coat color, depending on location and habitat. Coat colour varies from pale yellow to deep gold or tawny, and sometimes black, and is patterned with black rosettes while the head, lower limbs and belly are spotted with solid black. Male leopards are larger, averaging 58 kg (128 lb) with 90 kg (200 lb) being the maximum weight attained by a male. Females weigh about 37.5 kg (83 lb) on average. The African leopard is sexually dimorphic; males are larger and heavier than females. Between 1996 and 2000, 11 adult leopards were radio-collared on Namibian farmlands. Males weighed 37.5 to 52.3 kg (83 to 115 lb) only, and females 24 to 33.5 kg (53 to 74 lb). The heaviest known leopard weighed about 96 kg (212 lb), and was recorded in South West Africa. According to Alfred Edward Pease, black leopards in North Africa were similar in size to lions. An Algerian leopard killed in 1913 was reported to have measured approximately 8 ft 10 in (2.69 m), before being skinned. Leopards inhabiting the mountains of the Cape Provinces appear smaller and less heavy than leopards further north. Leopards in Somalia and Ethiopia are also said to be smaller. The skull of a West African leopard specimen measured 11.25 in (286 mm) in basal length, and 7.125 in (181.0 mm) in breadth, and weighed 1 lb 12 oz (0.79 kg). To compare, that of an Indian leopard measured 11.2 in (280 mm) in basal length, and 7.9 in (20 cm) in breadth, and weighed 2 lb 4 oz (1.0 kg). Distribution and habitat [edit] The African leopards inhabited a wide range of habitats within Africa, from mountainous forests to grasslands and savannahs, excluding only extremely sandy desert. It is most at risk in areas of semi-desert, where scarce resources often result in conflict with nomadic farmers and their livestock. It used to occur in most of sub-Saharan Africa, occupying both rainforest and arid desert habitats. It lived in all habitats with annual rainfall above 50 mm (2.0 in), and can penetrate areas with less than this amount of rainfall along river courses. It ranges up to 5,700 m (18,700 ft), has been sighted on high slopes of the Ruwenzori and Virunga volcanoes, and observed when drinking thermal water 37 °C (99 °F) in the Virunga National Park. The African leopard appears to be successful at adapting to altered natural habitat and settled environments in the absence of intense persecution. It has often been recorded close to major cities. But already in the 1980s, it has become rare throughout much of West Africa. Now, it remains patchily distributed within historical limits. During surveys in 2013, it was recorded in Gbarpolu County and Bong County in the Upper Guinean forests of Liberia. Leopards are extremely rare in North Africa. A relict population persisted in the Atlas Mountains of Morocco until the 1990s, in forest and mountain steppe in elevations of 300 to 2,500 m (980 to 8,200 ft), where the climate is temperate to cold. The last individual in the area was recorded in 1996. In 2005, leopard DNA was detected in the Hoggar Mountains of southern Algeria, in the central Sahara. In 2014, a leopard was killed in the Elba Protected Area in southeastern Egypt. This was the first sighting of a leopard in the country since the 1950s. In 2016, a leopard was recorded for the first time in a semi-arid area of Yechilay in northern Ethiopia. Behavior and ecology [edit] In Kruger National Park, male leopards and female leopards with cubs were more active at night than solitary females. The highest rates of daytime activity were recorded for leopards using thorn thickets during the wet season, when impala also used them. Leopards are generally most active between sunset and sunrise, and kill more prey at this time. Diet and hunting [edit] The leopard has an exceptional ability to adapt to changes in prey availability, and has a very broad diet. It takes small prey where large ungulates are less common. The known prey of leopards ranges from dung beetles to adult elands, which can reach 900 kg (2,000 lb). In sub-Saharan Africa, at least 92 prey species have been documented in leopard scat, including rodents, birds, small and large antelopes, hyraxes, hares, and arthropods. Leopards generally focus their hunting activity on locally abundant medium-sized ungulates in the 20 to 80 kg (44 to 176 lb) range, while opportunistically taking other prey. Average intervals between ungulate kills range from seven to 12–13 days. Leopards often hide large kills in trees, a behavior for which great strength is required. There have been several observations of leopards hoisting carcasses of young giraffes, estimated to weigh up to 125 kg (276 lb), i.e. 2–3 times the weight of the leopard, up to 5.7 m (19 ft) into trees. In Serengeti National Park, leopards were radio-collared for the first time in the early 1970s. Their hunting at night was difficult to watch; the best time for observing them was after dawn. Of their 64 daytime hunts, only three were successful. In this woodland area, they preyed mostly on impalas, both adult and young, and caught some Thomson's gazelles in the dry season. Occasionally, they successfully hunted warthogs, dik-diks, reedbucks, duikers, steenboks, blue wildebeest and topi calves, jackals, Cape hares, guineafowl and starlings. They were less successful in hunting plains zebras, Coke's hartebeests, giraffes, mongooses, genets, hyraxes and small birds. Scavenging from the carcasses of large animals made up a small proportion of their food. In the tropical rainforests of Central Africa, their diet consists of duikers and primates. Some individual leopards have shown a strong preference for pangolins and porcupines. In North Africa, the leopard preys on Barbary macaque (Macaca sylvanus). Analysis of leopard scat in Taï National Park revealed that primates are primary leopard prey during the day. In Gabon's Lope National Park, the most important prey species was found to be the red river hog (Potamochoerus porcus), African buffaloes (Syncerus caffer) and greater cane rats (Thryonomys swinderianus), comprised 13% each of the consumed biomass. In the Central African Republic's Dzanga-Sangha Complex of Protected Areas, a leopard reportedly attacked and pursued a large western lowland gorilla, but did not catch it. Gorilla parts found in leopard scat indicates that the leopard either scavenged on gorilla remains or killed it. African leopards were observed preying on adult eastern gorillas in the Kisoro area near Uganda's borders with Rwanda and the Democratic Republic of the Congo. Threats [edit] Throughout Africa, the major threats to leopards are habitat conversion and intense persecution, especially in retribution for real and perceived livestock loss. The Upper Guinean forests in Liberia are considered a biodiversity hotspot, but have already been fragmented into two blocks. Large tracts are affected by commercial logging and mining activities, and are converted for agricultural use including large-scale oil palm plantations in concessions obtained by a foreign company. The impact of trophy hunting on populations is unclear, but may have impacts at the demographic and population level, especially when females are shot. In Tanzania, only males are allowed to be hunted, but females comprised 28.6% of 77 trophies shot between 1995 and 1998. Removing an excessively high number of males may produce a cascade of deleterious effects on the population. Although male leopards provide no parental care to cubs, the presence of the sire allows females to raise cubs with a reduced risk of infanticide by other males. There are few reliable observations of infanticide in leopards, but new males entering the population are likely to kill existing cubs. Analysis of leopard scats and camera trapping surveys in contiguous forest landscapes in the Congo Basin revealed a high dietary niche overlap and an exploitative competition between leopards and bushmeat hunters. With increasing proximity to settlements and concomitant human hunting pressure, leopards exploit smaller prey and occur at considerably reduced population densities. In the presence of intensive bushmeat hunting surrounding human settlements, leopards appear entirely absent. Transhumant pastoralists from the border area between Sudan and the Central African Republic take their livestock to the Chinko area. They are accompanied by armed merchants who engage in poaching large herbivores, sale of bushmeat and trading leopard skins in Am Dafok. Surveys in the area revealed that the leopard population decreased from 97 individuals in 2012 to 50 individuals in 2017. Rangers confiscated large amounts of poison in the camps of livestock herders, who admitted that they use it for poisoning predators. Conservation [edit] The leopard is listed in CITES Appendix I. Hunting is banned in Zambia and Botswana, and was suspended in South Africa for 2016. 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"Calibrating a camera trap-based biased mark-recapture sampling design to survey the leopard population in the N'wanetsi concession, Kruger National Park, South Africa" (PDF). African Journal of Ecology. 51 (3): 422–430. Bibcode:2013AfJEc..51..422M. doi:10.1111/aje.12047. hdl:2263/31024. External links [edit] Wikimedia Commons has media related to Panthera pardus pardus. Species portrait African leopard; IUCN/SSC Cat Specialist Group[permanent dead link] Leopards .:. wild-cat.org — Information about research and conservation of leopards The Cape Leopard Trust, South Africa Safarinow.com: African Leopard » Panthera pardus » 'Luiperd' South Africa Wildlife: Leopard {Panthera Pardus} Image of a leopard from the Central African forests of Gabon (archived 14 January 2014) Video of the Barbary Leopard in the wild (broken link) (archived 20 February 2010) Predation on a child Archived 2020-09-26 at the Wayback Machine at Queen Elizabeth National Park, Uganda \| v t e Extant Carnivora species \| \| --- \| \| Kingdom: Animalia Phylum: Chordata Class: Mammalia Infraclass: Eutheria Superorder: Laurasiatheria \| \| \| \| Feliformia ("cat-like" carnivorans) \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| Feliformia \| \| \| \| --- \| \| African palm civet (Nandinia binotata) \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| Feloidea \| \| \| \| --- \| \| Prionodon (Asiatic linsangs) \| Banded linsang (P. linsang) Spotted linsang (P. pardicolor) \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| Felidae (cats) \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| \| \| \| \| \| \| \| --- --- --- \| \| Pantherinae \| \| \| \| --- \| \| Neofelis \| Sunda clouded leopard (N. diardi) Clouded leopard (N. nebulosa) \| \| Panthera \| Snow leopard (P. uncia) Tiger (P. tigris) Jaguar (P. onca) Lion (P. leo) Leopard (P. pardus) \| \| \| \| \| Felinae sensu stricto \| \| \| \| \| \| \| \| --- --- --- \| \| Bay cat lineage \| \| \| \| --- \| \| Pardofelis \| Marbled cat (P. marmorata) \| \| Catopuma \| Bay cat (C. badia) Asian golden cat (C. temminckii) \| \| \| Caracal lineage \| \| \| \| --- \| \| Serval (Leptailurus serval) \| \| \| Caracal \| African golden cat (C. aurata) Caracal (C. caracal) \| \| \| Leopardus \| Pampas cat (L. colocola) Geoffroy's cat (L. geoffroyi) Kodkod (L. guigna) Southern tiger cat (L. guttulus) Andean mountain cat (L. jacobita) Ocelot (L. pardalis) Oncilla (L. tigrinus) Margay (L. wiedii) \| \| Lynx \| Canada lynx (L. canadensis) Eurasian lynx (L. lynx) Iberian lynx (L. pardinus) Bobcat (L. rufus) \| \| Puma lineage \| \| \| \| --- \| \| Acinonyx \| Cheetah (A. jubatus) \| \| Jaguarundi (Herpailurus yagouaroundi) \| \| \| Puma \| Cougar (P. concolor) \| \| \| Leopard cat lineage \| \| \| \| --- \| \| Pallas's cat (Otocolobus manul) \| \| \| Prionailurus \| Leopard cat (P. bengalensis) Sunda leopard cat (P. javanensis) Flat-headed cat (P. planiceps) Rusty-spotted cat (P. rubiginosus) Fishing cat (P. viverrinus) \| \| \| Felis \| Chinese mountain cat (F. bieti) Domestic cat (F. catus) Jungle cat (F. chaus) African wildcat (F. lybica) Sand cat (F. margarita) Black-footed cat (F. nigripes) European wildcat (F. silvestris) \| \| \| \| \| \| \| \| \| Viverroidea \| + see below↓ \| \| \| \| \| \| \| \| Viverroidea \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| Viverridae \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| Palm civets \| \| \| \| --- \| \| Hemigalinae \| Owston's palm civet (Chrotogale owstoni) Otter civet (Cynogale bennettii) Hose's palm civet (Diplogale hosei) Banded palm civet (Hemigalus derbyanus) \| \| Paradoxurinae \| \| \| \| --- \| \| Binturong (Arctictis binturong) Small-toothed palm civet (Arctogalidia trivirgata) Sulawesi palm civet (Macrogalidia musschenbroekii) Masked palm civet (Paguma larvata) \| \| \| Paradoxurus \| Asian palm civet (P. hermaphroditus) Brown palm civet (P. jerdoni) Golden palm civet (P. zeylonensis) \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| Viverrinae sensu lato \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| \| \| \| \| \| \| \| --- --- --- \| \| Viverrinae sensu stricto \| \| \| \| --- \| \| Small Indian civet (Viverricula indica) African civet (Civettictis civetta) \| \| \| Viverra \| Malabar large-spotted civet (V. civettina) Large-spotted civet (V. megaspila) Malayan civet (V. tangalunga) Large Indian civet (V. zibetha) \| \| \| \| \| Genettinae \| \| \| \| \| \| --- --- \| \| \| \| \| --- \| \| Poiana (African linsangs) \| Central African oyan (P. richardsonii) West African oyan (P. leightoni) \| \| \| \| Genetta (genets) \| \| \| \| --- \| \| subgenus Genetta (paraphyletic) \| Abyssinian genet (G. abyssinica) Common genet (G. genetta) South African small-spotted genet (G. felina) \| \| subgenus Eugenetta (paraphyletic) \| Angolan genet (G. angolensis) Cape genet (G. tigrina) \| \| subgenus Herpailuropoda (paraphyletic) \| Bourlon's genet (G. bourloni) Johnston's genet (G. johnstoni) Hausa genet (G. thierryi) Giant forest genet (G. victoriae) \| \| subgenus Pardogale (paraphyletic) \| Pardine genet (G. pardina) King genet (G. poensis) \| \| subgenus Prionailuropoda \| Letaba genet (G. letabae) Rusty-spotted genet (G. maculata) Schouteden's genet (G. schoutedeni) \| \| subgenus Leptailuropoda (paraphyletic) \| Crested servaline genet (G. cristata) Servaline genet (G. servalina) \| \| subgenus Osbornictis \| Aquatic genet (G. piscivora) \| \| \| \| \| \| \| \| \| \| \| --- \| \| Herpestoidea \| + see below↓ \| \| \| \| \| \| \| \| \| Herpestoidea \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| Hyaenidae (hyenas) \| \| \| \| --- \| \| Proteles \| Aardwolf (P. cristata) \| \| Hyaeninae (bone-crushing hyenas) \| \| \| \| --- \| \| Striped hyena (Hyaena hyaena) Brown hyena (Parahyaena brunnea) \| \| \| Crocuta \| Spotted hyena (C. crocuta) \| \| \| \| \| \| Herpestidae sensu lato \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- \| \| Eupleridae (Malagasy carnivorans) \| \| \| \| \| \| \| \| --- --- --- \| \| Euplerinae (Malagasy civets) \| \| \| \| --- \| \| Fossa (Cryptoprocta ferox) Malagasy civet (Fossa fossana) \| \| \| Eupleres (falanoucs) \| Eastern falanouc (E. goudotii) Western falanouc (E. major) \| \| \| Galidiinae (vontsira) \| \| \| \| --- \| \| Ring-tailed vontsira (Galidia elegans) \| \| \| Galidictis \| Broad-striped Malagasy mongoose (G. fasciata) Grandidier's mongoose (G. grandidieri) \| \| Narrow-striped mongoose (Mungotictis decemlineata) \| \| \| Salanoia \| Brown-tailed mongoose (S. concolor) Durrell's vontsira (S. durrelli) \| \| \| \| \| \| Herpestidae sensu stricto (mongooses) \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- \| \| Mungotinae \| \| \| \| --- \| \| Suricata \| Meerkat (S. suricatta) \| \| Liberian mongoose (Liberiictus kuhni) \| \| \| Mungos \| Gambian mongoose (M. gambianus) Banded mongoose (M. mungo) \| \| Pousargues's mongoose (Dologale dybowskii) \| \| \| Helogale \| Ethiopian dwarf mongoose (H. hirtula) Common dwarf mongoose (H. parvula) \| \| Crossarchus (kusimanses) \| Alexander's kusimanse (C. alexandri) Angolan kusimanse (C. ansorgei) Common kusimanse (C. obscurus) Flat-headed kusimanse (C. platycephalus) \| \| \| Herpestinae \| \| \| \| --- \| \| Marsh mongoose (Atilax paludinosus) Long-nosed mongoose (Xenogale naso) \| \| \| Urva (Asian mongooses) \| Small Indian mongoose (U. auropunctata) Short-tailed mongoose (U. brachyura) Indian grey mongoose (U. edwardsii) Indian brown mongoose (U. fusca) Javan mongoose (U. javanica) Collared mongoose (U. semitorquata) Ruddy mongoose (U. smithii) Crab-eating mongoose (U. urva) Stripe-necked mongoose (U. vitticolla) \| \| White-tailed mongoose (Ichneumia albicauda) Yellow mongoose (Cynictis penicillata) Selous's mongoose (Paracynictis selousi) Meller's mongoose (Rhynchogale melleri) \| \| \| Bdeogale \| Bushy-tailed mongoose (B. crassicauda) Jackson's mongoose (B. jacksoni) Black-footed mongoose (B. nigripes) \| \| Herpestes (slender mongooses) \| Angolan slender mongoose (H. flavescens) Egyptian mongoose (H. ichneumon) Somalian slender mongoose (H. ochracea) Cape gray mongoose (H. pulverulenta) Common slender mongoose (H. sanguinea) \| \| \| \| \| \| \| \| \| \| Caniformia ("dog-like" carnivorans) \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| Canidae (dogs) \| \| \| \| \| \| --- --- \| \| \| \| \| --- \| \| Urocyon \| Gray fox (U. cinereoargenteus) Island fox (U. littoralis) \| \| \| \| Vulpini \| \| \| \| --- \| \| Bat-eared fox (Otocyon megalotis) \| \| \| Nyctereutes (raccoon dogs) \| Common raccoon dog (N. procyonoides) Japanese raccoon dog (N. viverrinus) \| \| Vulpes (true foxes) \| Bengal fox (V. bengalensis) Blanford's fox (V. cana) Cape fox (V. chama) Corsac fox (V. corsac) Tibetan fox (V. ferrilata) Arctic fox (V. lagopus) Kit fox (V. macrotis) Pale fox (V. pallida) Rüppell's fox (V. rueppelli) Swift fox (V. velox) Red fox (V. vulpes) Fennec fox (V. zerda) \| \| \| Canini (true dogs) \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| Cerdocyonina (zorro) \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| \| \| \| --- \| \| Maned wolf (Chrysocyon brachyurus) \| \| \| Speothos \| Bush dog (S. venaticus) \| \| Short-eared dog (Atelocynus microtis) Crab-eating fox (Cerdocyon thous) \| \| \| \| \| Lycalopex (South American foxes) \| Culpeo (L. culpaeus) Darwin's fox (L. fulvipes) South American gray fox (L. griseus) Pampas fox (L. gymnocercus) Sechuran fox (L. sechurae) Hoary fox (L. vetulus) \| \| \| Canina (wolf-like canids) \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| \| \| \| --- \| \| Lupulella \| Side-striped jackal (L. adustus) Black-backed jackal (L. mesomelas) \| \| Lycaon \| African wild dog (L. pictus) \| \| Dhole (Cuon alpinus) \| \| \| \| \| Canis \| Golden jackal (C. aureus) Domestic dog (C. familiaris) Coyote (C. latrans) African wolf (C. lupaster) Wolf (C. lupus) Eastern wolf (C. lycaon) Red wolf (C. rufus) Ethiopian wolf (C. simensis) \| \| \| \| \| Arctoidea \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| Ursidae (bears) \| \| \| \| --- \| \| Ailuropoda \| Giant panda (A. melanoleuca) \| \| Tremarctos \| Spectacled bear (T. ornatus) \| \| Ursinae \| \| \| \| --- \| \| Sloth bear (Melursus ursinus) Sun bear (Helarctos malayanus) \| \| \| Ursus \| American black bear (U. americanus) Asian black bear (U. thibetanus) Brown bear (U. arctos) Polar bear (U. maritimus) \| \| \| \| \| \| \| \| \| \| \| --- --- --- \| \| Mustelida \| \| \| \| --- \| \| Pinnipedia (seals) \| + see below↓ \| \| Musteloidea \| + see below↓ \| \| \| \| \| \| \| \| \| \| \| Pinnipedia (seals) \| \| --- \| \| \| \| \| \| \| --- --- \| \| \| \| \| --- \| \| Odobenidae \| Walrus (Odobenus rosmarus) \| \| \| \| Otariidae (eared seals) \| \| \| \| --- \| \| Callorhinus (northern fur seals) \| Northern fur seal (C. ursinus) \| \| Otariinae (sea lions) \| \| \| \| --- \| \| Steller sea lion (Eumetopias jubatus) \| \| \| Zalophus \| California sea lion (Z. californianus) Galápagos sea lion (Z. wollebaeki) \| \| \| South American sea lion (Otaria flavescens) \| \| \| Neophoca \| Australian sea lion (N. cinerea) \| \| New Zealand sea lion (Phocarctos hookeri) \| \| \| Arctocephalus (southern fur seals) \| South American fur seal (A. australis) Australasian fur seal (A. forsteri) Galápagos fur seal (A. galapagoensis) Antarctic fur seal (A. gazella) Juan Fernández fur seal (A. philippii) Brown fur seal (A. pusillus) Guadalupe fur seal (A. townsendi) Subantarctic fur seal (A. tropicalis) \| \| \| Phocidae (earless seals or true seals) \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| Phocinae ("northern seals") \| \| \| \| --- \| \| Bearded seal (Erignathus barbatus) Hooded seal (Cystophora cristata) \| \| \| Phocini \| \| \| \| --- \| \| Harp seal (Pagophilus groenlandicus) Ribbon seal (Histriophoca fasciata) Grey seal (Halichoerus grypus) \| \| \| Phoca \| Spotted seal (P. largha) Harbor seal (P. vitulina) \| \| Pusa \| Caspian seal (P. caspica) Ringed seal (P. hispida) Baikal seal (P. sibirica) \| \| \| \| Monachinae ("southern seals") \| \| \| \| \| \| \| \| --- --- --- \| \| Monachini (monk seals) \| \| \| \| --- \| \| Mediterranean monk seal (Monachus monachus) \| \| \| Neomonachus \| Hawaiian monk seal (N. schauinslandi) \| \| \| Mirounga (elephant seals) \| Northern elephant seal (M. angustirostris) Southern elephant seal (M. leonina) \| \| Lobodontini (Antarctic seals) \| \| \| \| --- \| \| Leopard seal (Hydrurga leptonyx) Weddell seal (Leptonychotes weddellii) Crabeater seal (Lobodon carcinophagus) Ross seal (Ommatophoca rossi) \| \| \| \| \| \| \| \| \| \| \| Musteloidea \| \| --- \| \| \| \| \| \| \| --- --- \| \| \| \| \| --- \| \| Ailuridae \| Red panda (Ailurus fulgens) \| \| \| \| Mephitidae (skunks) \| \| \| \| --- \| \| Conepatus (hog-nosed skunks) \| Molina's hog-nosed skunk (C. chinga) Humboldt's hog-nosed skunk (C. humboldtii) American hog-nosed skunk (C. leuconotus) Striped hog-nosed skunk (C. semistriatus) \| \| Mephitis \| Hooded skunk (M. macroura) Striped skunk (M. mephitis) \| \| Mydaus (stink badgers) \| Sunda stink badger (M. javanensis) Palawan stink badger (M. marchei) \| \| Spilogale (spotted skunks) \| Southern spotted skunk (S. angustifrons) Western spotted skunk (S. gracilis) Eastern spotted skunk (S. putorius) Pygmy spotted skunk (S. pygmaea) \| \| \| Procyonidae \| \| \| \| --- \| \| Kinkajou (Potos flavus) \| \| \| Bassariscus \| Ringtail (B. astutus) Cacomistle (B. sumichrasti) \| \| Procyon (raccoons) \| Crab-eating raccoon (P. cancrivorus) Raccoon (P. lotor) Cozumel raccoon (P. pygmaeus) \| \| Bassaricyon (olingos) \| Eastern lowland olingo (B. alleni) Northern olingo (B. gabbii) Western lowland olingo (B. medius) Olinguito (B. neblina) \| \| Nasuina (coatis) \| \| \| \| --- \| \| Nasua \| White-nosed coati (N. narica) South American coati (N. nasua) \| \| Nasuella (mountain coatis) \| Eastern mountain coati (N. meridensis) Western mountain coati (N. olivacea) \| \| \| \| \| \| \| --- \| \| Mustelidae \| + see below↓ \| \| \| \| \| \| \| \| \| \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| Mustelidae \| \| \| \| --- \| \| American badger (Taxidea taxus) \| \| \| Mellivora \| Honey badger (M. capensis) \| \| \| \| \| \| \| \| \| --- --- --- \| \| \| \| \| \| --- \| \| Arctonyx (hog badgers) \| Northern hog badger (A. albogularis) Greater hog badger (A. collaris) Sumatran hog badger (A. hoevenii) \| \| Meles (Eurasian badgers) \| Japanese badger (M. anakuma) Caucasian badger (M. canescens) Asian badger (M. leucurus) European badger (M. meles) \| \| \| \| \| Melogale (ferret-badgers) \| Vietnam ferret-badger (M. cucphuongensis) Bornean ferret badger (M. everetti) Chinese ferret-badger (M. moschata) Javan ferret-badger (M. orientalis) Burmese ferret-badger (M. personata) Formosan ferret-badger (M. subaurantiaca) \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| \| \| --- \| \| Tayra (Eira barbara) \| \| \| Pekania \| Fisher (P. pennanti) \| \| Gulo \| Wolverine (G. gulo) \| \| Martes (martens) \| American marten (M. americana) Pacific marten (M. caurina) Yellow-throated marten (M. flavigula) Beech marten (M. foina) Nilgiri marten (M. gwatkinsii) European pine marten (M. martes) Japanese marten (M. melampus) Sable (M. zibellina) \| \| \| \| \| \| \| \| \| \| \| --- --- --- \| \| Lyncodontini \| \| \| \| --- \| \| Patagonian weasel (Lyncodon patagonicus) \| \| \| Galictis (grisons) \| Lesser grison (G. cuja) Greater grison (G. vittata) \| \| \| Ictonychini (African polecats) \| \| \| \| --- \| \| Vormela \| Marbled polecat (V. peregusna) \| \| African striped weasel (Poecilogale albinucha) \| \| \| Ictonyx \| Saharan striped polecat (I. libycus) Striped polecat (I. striatus) \| \| \| \| Lutrinae (otters) \| \| \| \| --- \| \| Giant otter (Pteronura brasiliensis) \| \| \| Lontra \| North American river otter (L. canadensis) Marine otter (L. felina) Neotropical otter (L. longicaudis) Southern river otter (L. provocax) \| \| Enhydra \| Sea otter (E. lutris) \| \| Spotted-necked otter (Hydrictis maculicollis) \| \| \| Lutra \| Eurasian otter (L. lutra) Hairy-nosed otter (L. sumatrana) \| \| Lutrogale \| Smooth-coated otter (L. perspicillata) \| \| Aonyx \| African clawless otter (A. capensis) Asian small-clawed otter (A. cinereus) Congo clawless otter (A. congicus) \| \| \| \| \| \| \| \| \| --- --- \| \| \| \| \| --- \| \| Neogale (New World weasels) \| Amazon weasel (N. africana) Colombian weasel (N. felipei) Long-tailed weasel (N. frenata) American mink (N. vison) \| \| \| \| Mustela (weasels) \| \| \| \| --- \| \| subgenus Mustela (paraphyletic) \| Sichuan weasel (M. aistoodonnivalis) Mountain weasel (M. altaica) Stoat/Beringian ermine (M. erminea) Haida ermine (M. haidarum) Yellow-bellied weasel (M. kathiah) Least weasel (M. nivalis) American ermine (M. richardsonii) \| \| subgenus Lutreola (paraphyletic) \| Japanese weasel (M. itatsi) European mink (M. lutreola) Indonesian mountain weasel (M. lutreolina) Malayan weasel (M. nudipes) Siberian weasel (M. sibirica) Back-striped weasel (M. strigidorsa) \| \| subgenus Putorius \| Steppe polecat (M. eversmannii) Ferret (M. furo) Black-footed ferret (M. nigripes) European polecat (M. putorius) \| \| \| \| \| \| \| See also: Mink Polecat \| \| \| \| \| \| \| Taxon identifiers \| \| --- \| \| Panthera pardus pardus \| Wikidata: Q39153 Wikispecies: Panthera pardus pardus CoL: 5K5LY EoL: 1270484 GBIF: 7193915 iNaturalist: 319574 ITIS: 726464 MSW: 14000251 \| Retrieved from " Categories: IUCN Red List vulnerable species Leopards Mammals described in 1758 Animal taxa named by Carl Linnaeus Mammals of Sub-Saharan Africa Hidden categories: CS1: long volume value CS1 Latin-language sources (la) CS1 maint: article number as page number Webarchive template wayback links Articles with short description Short description is different from Wikidata Articles with 'species' microformats Commons link from Wikidata All articles with dead external links Articles with dead external links from August 2025 Articles with permanently dead external links

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