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14200	https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_3?srsltid=AfmBOoq72mJr_vZbv2qKG2uP5VvGXuBMtPpBBHA-PCwNdVvIR52OL9l1	Art of Problem Solving 1991 IMO Problems/Problem 3 - 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 1991 IMO Problems/Problem 3 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 1991 IMO Problems/Problem 3 Contents 1 Problem 2 Solution 1 3 Solution 2 4 See Also Problem Let . Find the smallest integer such that each -element subset of contains five numbers which are pairwise relatively prime. Solution 1 Let's look at this another way. We aim to find the least number of integers we can remove from S to leave a set which does not contain 5 pairwise coprime integers. Let And more generally And finally It is clear that these sets taken over all collections of primes and 1 forms a partition of S. If there are representatives from five sets , there are five mutually coprime integers, so all but four of the sets must be completely removed. It is clear that (with the exception of 1, which can clearly be removed first) if so it is worth removing before removing (if we are striving for a minimum). Furthermore, if we remove two sets , we must also (to stop there from being 5 mutually coprime integers) remove and so on, and these have similar ordering relations. So once we have removed everything we need to keeping sizes of sets removed to a minimum, we are left with and the multi-index sets not coprime to all of these. In other words, we have the set of multiples of 2,3,5 and 7, which can be calculated (by lots of inclusion-exclusion) to have cardinality . Therefore, a subset of of size must contain 5 coprime integers. This solution was posted and copyrighted by Ilthigore. The original thread for this problem can be found here: Firstly note that . Therefore, we only need to check primes to check if a number from to is prime. Solution 2 I claim the answer is . First, let's look at a subset of that consists only of multiples of . It is easy to calculate by PIE that the total number of elements in this set isNow, note that no matter what element subset we consider of this subset we just made, consisting of multiples of , we will always have at least of those elements that exist in the subset that share a factor greater than by the Pigeonhole Principle since . Thus, we know from this. We also know that there are composite numbers and prime numbers (obviously just ) in this subset of . Now we do casework on the numbers (we don't have to check higher numbers because remember !) to find the other composite numbers. We find that all of numbers, fills in the rest of the composite numbers from to . This gives a total of numbers when counted. So we can now count up the total as composite numbers and the remaining must be prime so prime numbers ( not included as prime or composite). Now suppose is a subset of our set such that ( represents its cardinality or the number of elements in the subset ). Also suppose that there is no element subset of such that all elements are relatively prime to each other. This means we would have to have at most prime numbers and at least composite numbers to make this subset . This means the set (everything that is in but not in ) has at most composite numbers. Now consider the following sets and notice that at least one of these must be a subset of !:And obviously each of these sets has elements that are all relatively prime numbers, as desired. This solution was posted and copyrighted by Wave-Particle. The original thread for this problem can be found here: See Also 1991 IMO (Problems) • Resources Preceded by Problem 21•2•3•4•5•6Followed by Problem 4 All IMO Problems and Solutions 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.
14201	https://www.webassign.net/labsgraceperiod/ucscgencheml1/lab_6/manual.html	Lab 6 - Quantum States for the Visible Hydrogen Atomic Emission Spectrum Contents> Lab 6 - Quantum States for the Visible Hydrogen Atomic Emission Spectrum Lab 6 - Quantum States for the Visible Hydrogen Atomic Emission Spectrum Goal and Overview The relationship between color, wavelength, and frequency of visible light will be determined using a Spec 20 spectrometer. The visible emission spectrum of atomic hydrogen will be analyzed in a spectrometer that has been calibrated based on the visible emission spectrum of helium. Based on the hydrogen atomic emission, the principal quantum numbers (electronic energy levels) of the initial and final states for the atoms (before and after emission) will be determined. A numerical value of the Rydberg constant will also be extracted from a graphical analysis of the emission wavelengths. Objectives and Science Skills • Explain and use the relationship between photon wavelength and energy, both qualitatively and quantitatively. • Understand and explain atomic absorption and emission in relation to allowed energy levels (states) in an atom as well as their relationship to photon wavelength and energy. • Evaluate the precision of a spectrometer by comparing measured He emission lines to literature values. • Observe atomic H line spectra, fit data to the Rydberg equation, and compare experimental results to theoretical values. • Analyze and discuss factors that limit the precision of the results. Suggested Review and External Reading • data analysis and spectroscopy information; relevant textbook information on quantum theory Background The study of the interaction of light with matter is called spectroscopy. Spectroscopy had been a fundamental part in the development of chemical molecular theory, and it can be used for both qualitative and quantitative analysis of matter. Most modern chemistry, biology, geology, astronomy, and physics labs use some type of spectroscopy. Wave-Particle Duality of Light and Matter Modern wave theory says that light is composed of oscillating perpendicular electric and magnetic fields, as shown. These oscillating magnetic and electric fields are responsible for interactions of photons with the electric and magnetic properties of matter. Figure 1 The speed of light, c, through a vacuum, is 2.998 × 10 8 m/s. Light travels at different speeds through materials because of the light's interaction with the material. 1 The peak-to-peak distance is called the wavelength, λ. 2 The number of oscillations per second is the oscillation frequency, ν. Two cases are illustrated: one wave has a wavelength three times that of the other. Wavelength and frequency are inversely proportional; as one increases, the other decreases. The two are related through the equation: λ ν = c. This relationship allows you to convert between wavelength and frequency. Figure 2 Light can also be thought of as individual particles or packets of energy called photons. A photon has an energy proportional to its frequency, ν : E photon = h ν = hc/λ. The proportionality constant is Planck's constant, h = 6.626 × 10–34 J · s.Notice that wavelength is inversely proportional to frequency. Low energy photons, like radio waves, have wavelengths of thousands of meters or more. High energy photons, like x-rays, have wavelengths of a billionth of a meter or less. Visible photons correspond to a very small part of the electromagnetic spectrum, lying between 400 (violet) and 700 (red) nanometers (1 nm = 10–9 m, or 1 m = 10 9 nm).In the mid-1920s, Louis DeBroglie proposed a new notion of a wave-particle duality for matter. Wave-particle duality for light and matter quickly led to the creation and development of quantum theory. Atomic Absorption and Emission When an atom absorbs a photon of light, its final energy, E final, is greater than its initial energy, E initial; the atom experiences a positive energy change. When an atom emits a photon, the atom's initial energy, E initial, is greater than the final energy, E final; the atom experiences a negative energy change. Absorption and emission strictly follows the law of conservation of energy: E photon = \|Δ E atom\|. The change in the atom's energy is equivalent to energy of the photon, h ν. ( 1 ) Δ E atom = Δ E levels n i → n f = E f − E i = E h ν = h ν = hc λ Figure 3 This experiment focuses on the emission of visible light from excited hydrogen atoms.The wave description of matter explains why electrons may only exist in very specific states, or orbitals (e.g., 1 s, 2 s, 2 p, etc.). The orbitals have very specific energies. Atoms can be "excited" to higher energy states in several ways, including strongly heating the atom or putting it in an electric discharge or colliding it with energetic particles. The excited atoms will eventually spontaneously lose that excitation energy and "fall" to lower energy states, emitting light at wavelengths given by Eq. 1 Δ E atom = Δ E levels n i → n f = E f − E i = E h ν = h ν = hc λ. Atomic emission spectra are called line spectra because they appear as sets of discrete lines. This reflects the fact that atoms can only emit photons with energies corresponding to the energy difference between two discrete electronic states.Each type of atom shows a characteristic emission spectrum, owing to its own unique orbital energies. This enables its identification by emission spectroscopy. For example, before helium was discovered on the earth, it was found to exist in the sun because He's emission spectrum was observed in sunlight.The hydrogen atom is the simplest atom, with the most abundant isotope consisting of one proton and one electron. The observed emission lines of excited hydrogen atoms can be empirically fitted to a simple equation. ( 2 ) Rydberg Equation:E h ν = R H 1 n f 2 − 1 n i 2 , for n i>n f where n f and n i are the principal quantum numbers specifying the final and initial energy states of the H atom. R H is Rydberg's constant, and it has been shown to contain fundamental quantities like π, electron charge and mass, Planck's constant and the speed of light. ( 3 ) Rydberg's constant:R H = 2 π e 4 m h 3 c = 2.180 × 10−18 J = 1.096776 × 10 7 m−1 Because the H atom is relatively simple, the Schrödinger equation can be solved for the quantum states and energy levels of its electron. The math supports the experimentally-derived Rydberg equation. The allowed set of energy levels for the electron in the H atom follows the equation below. ( 4 ) E n = − R H n 2 ,where n is an integer = 1, 2, 3, Note the following: i The energy of a stable H atom must obey this equation, where n, the principal quantum number, corresponds to electron orbitals. For example, n = 1 gives the energy of the 1 s orbital; n = 2, the 2 s and 2 p orbitals; etc. ii All energy levels of H atoms are negative, which means that they are lower than the value chosen to be zero. For H atoms, the zero of energy is that of a proton and an electron separated such that the charges no longer feel each other. iii The energy level depends on 1/n 2, just as the Rydberg equation shows. When n = 1, the H atom has the lowest possible energy: ( 5a ) −R H/1 2 = −2.18 × 10−18 J; this is the ground state. When n = 2, the H atom is in the first excited state with energy of: ( 5b ) −R H/2 2 = −2.18 × 10−18 J/4 = −5.45 × 10−19 J. When n = 3, the H atom is in the second excited state with energy of: ( 5c ) −R H/3 2 = −2.18 × 10−18 J/9 = −2.42 × 10−19 J. And, so on.The wavelengths of light expected to appear in the H atomic emission spectrum can be calculated using the equation below. ( 6 ) E h ν = R H 1 n f 2 − 1 n i 2 E f − E i atom For example, the photon energy predicted for the transition, n i = 3 to n f = 2, is below. ( 7 ) E h ν, expected = E f − E i = −R H 1 2 2 − 1 3 2 −5.45 × 10−19 + 2.42 × 10−19 J = 3.03 × 10−19 J Eq. 6 E h ν = R H 1 n f 2 − 1 n i 2 = E f − E i atomgives the positive value, which would be appropriate for an emitted photon. You should understand that the energy change of the atom is positive for absorption (n f>n i) and is negative for emission (n f<n i).Note the following: i Each photon wavelength (or energy) in the observed spectrum is associated with two quantum numbers (for the initial and final states). ii For emission, the initial energy is greater than the final energy, and the initial quantum number, n i, is greater than the final quantum number, n f. The absolute value is included so that the photon energy is positive even though the energy change of the atom is actually negative for emission. iii Evidence of quantized energy levels in matter comes from atomic emission spectra, in which photons are observed as electrons change from one orbital to another. If electrons could have any energy, you would observe all wavelengths. However, only discrete wavelengths are emitted, and they correspond to the energy difference between two orbitals of specific energy, i.e., 1 s, 2 s, 2 p, 3 s... and so on. A special quality of the hydrogen atom emission spectrum is that the emitted photon energies come in separate wavelength sets, called series. In each series, all electronic transitions share a common final state (n f). Figure 4 The lowest few energy levels for the H atom are shown (Eq. 4 E n = − R H n 2 ,where n is an integer = 1, 2, 3,), with transitions that would give rise to three series in the H emission spectrum indicated by vertical lines (Eq. 6 E h ν = R H 1 n f 2 − 1 n i 2 = E f − E i atom).The emission series shown are: Lyman series, with n f = 1 (UV); Balmer series, with n f = 2 (visible); and, Paschen series, with n f = 3 (IR). This lab provides you with experimental evidence that the quantum theory of matter is actually true (neat!).You will characterize the emission of excited hydrogen atoms and verify that you are observing the Balmer series (n final = 2). Using your measured wavelengths (energies), you will determine the quantum numbers that fit the data and the Rydberg equation. Helium atomic emission is used to calibrate the spectrometer. Procedure Part 1: Visible light - wavelength vs. color This part of the experiment is not designed to take a lot of time. Caution: A small Lucite rod is taped in to your Spec 20. The rod's orientation is critical, so please do not remove the tape and please do not touch the rod. The Lucite rod transmits the light up from within so you can see the color. 1 Write the wavelengths from 400 to 760 nm at 20 nm intervals (uniformly spaced down a page in your notebook). 2 Turn the wavelength knob and note the colors you see while looking at the rod. Get a quick sense of how the colors change as you change wavelength. 3 Look carefully at the light of each wavelength at 20 nm intervals. 4 Broadly classify the color as violet, blue, green, yellow, orange, or red. Also describe the color in the most precise words that you can. Figure 5 5 Record the boundary wavelength between different colors. 6 Record the wavelength where each color has its maximum intensity, λ max. 7 Note how broad the wavelength range is for some colors and how narrow it is for others. Part 2: Spectrometer calibration (He spectra) and usage (H spectra) Caution: Do not touch the high voltage discharge tubes or power supply connections. Helium gas, He, and hydrogen gas, H 2, are sealed in glass discharge tubes. These tubes can be connected to a high voltage source. The high voltage across the tube of gas creates a discharge in the gas. When a spark passes through the gas, its energy is raised greatly. These lamps are like fluorescent tubes or neon lights. In the case of He, the atoms are excited and give off radiation (some visible) as they fall back to lower energy levels. You will study this radiation in Part 2b. In the H 2 tube, some H 2 molecules are broken into H atoms, and many of these are in excited states. The excited H atoms emit photons as they transition to lower energy states. Some of these photons lie at visible wavelengths. You will study this radiation in Part 2c. Excited atoms give rise to line spectra. The emission spectrum of excited H 2 molecules would contain 'bands' due to the allowed vibrational levels within each electronic level. Figure 6: Schematic of Emission Spectrometer Light enters a slit in the spectrometer and is dispersed (spread out in the way a prism does). The spectrometer has an eyepiece through which you see the colored spectral lines superimposed on a wavelength scale (e.g., 6.60 × 10–5 cm or 660 nm). When observing the emission lines, it is important to be consistent in how you position your head and where you hold your eye. The position of the lines on the scale will appear to move if you move your line of sight through the eyepiece. Part 2a: Handheld Spectrometers (if available) 1 Using a handheld spectrometer, point the slit at the fluorescent lights in the ceiling. Note the spectral lines in the eyepiece. The bright blue and green lines you see are mercury atom emission. 2 Adjust the side shutter on the spectrometer to let more or less light in to illuminate the wavelength scale. The more light you get on the scale, the harder it is to see your spectral lines. Part 2b: Helium Emission Lines The types of spectrometers we use are not highly precise instruments, but their accuracy should be relatively good. You will use the emission wavelengths of excited helium atoms to assess the accuracy of the instrument. Figure 7 1 Point the table-top spectrometer towards the He discharge tube and position the slit so that you observe colored lines on the scale. Write down the wavelength and color of each observed spectral line. You may not see all of the lines. Some are too close to each other to distinguish individual lines; the lowest- and highest-wavelength lines can also be hard to observe. 2 Even though the theoretical wavelength values are to four significant figures, the difficulty in reading the spectrometer's scale limits the precision to which you can read wavelength. An uncertainty of ±10 nm around the wavelength value you report is typical. Note: Since you can only estimate the position of the emission lines within ±10 nm precision, your uncertainty is ±10 nm, and your reported wavelengths for the He and H emission should be reported to the ±10 nm (e.g., 660 nm, not 656 nm). 3 Compare theoretical wavelength values to your experimental wavelengths. Calculate the percent error in each that you observed to two significant figures. How accurate is your spectrometer? Part 2c: Hydrogen Atomic Emission 1 The lines of the hydrogen atomic emission spectrum in the visible are known to be at 410.1, 434.0, 486.1, and 656.2 nm. Calculate the energy of each line to four significant figures in Joules using Eq. 1 Δ E atom = Δ E levels n i → n f = E f − E i = E h ν = h ν = hc λ E h ν = hc λ . 2 View the hydrogen gas tube with the spectrometer that you used in part 2b. Note the multi-colored bands due to excited H 2 molecules. These are not present in the He emission because helium exists naturally as atoms. 3 Position your head so that the bright blue-green emission line shows on the same about 490 nm. Do not move your head from this position. Write down the wavelength (to ±10 nm) and the color of each observed H emission line. You should have 3 or 4 values. 4 Calculate the energy of each emitted photon to two significant figures in Joules using Eq. 1 Δ E atom = Δ E levels n i → n f = E f − E i = E h ν = h ν = hc λ E h ν = hc λ . The calculated energy equals the change in atom's energy, E final − E initial = Δ E atom. The atom's energy change is negative even though photon energy is usually given as positive. The emission wavelengths that are in the visible range for the H atom are known to be those of the Balmer series (n f = 2). Are your values similar? 5 Calculate the expected transition energies to two significant figures for the Balmer series using Eq. 6 E h ν = R H 1 n f 2 − 1 n i 2 = E f − E i atom E h ν = R H 1 n f 2 − 1 n i 2 E f − E i . Note n f = 2; n i = 3, 4, 5, and (maybe) 6. 6 Calculate the percent error in each experimental energy relative to each known literature value (to two significant figures). A set of example data is shown. Calculations were performed in Excel®, so scientific notation appears as #E# instead of # × 10#. You must use your own values. Figure 8 Hopefully, you will see only a small percent error. This experiment supports the quantum mechanical treatment of the atom (and you saw it in lab). Your data also allow you to determine an experimental value for the Rydberg constant—how does yours compare to the true value? 7 To determine your experimental value of the Rydberg constant, R H, you need to rearrange the Rydberg equation into the form of a straight line: ( 8 ) E h ν = R H 1 n f 2 − 1 n i 2 ⇒Δ E atom=−R H 1 n i 2 +R H n f 2 y=m x+b where slope = −R H, y-intercept = R H n f 2 , x values: 1 n i 2 , and y values: E h ν. 8 Plot E h ν for each emission line vs. 1/n i 2 (recall n f = 2 for the Balmer series). The points should lie on a straight line. a You should have 3 or 4 data points for the graph, one for each photon energy and its associated initial state, n i. b The lowest energy photon (at the red end of the spectrum) corresponds to the lowest energy transition. For the Balmer series this would be n i = 3 → n f = 2. The energies increase as the wavelengths decrease. c Draw a best-fit line through the points on your graph. d Determine the slope of the best-fit line to three significant figures. Pick two points on the line, one near the top and one near the bottom. The slope is Δ E h ν over Δ(1/n i 2) values (this is rise over run, or Δ y/Δ x). e Confirm you have selected the correct series (value for n f) by checking three values from your graph. i The slope of your best-fit line should be approximately –R H. Find the percent error relative to the true value (R H = 2.18 × 10–18 J). ii The y-intercept should be R H/n f 2 (2.18 × 10–18 J/2 2); find your value to three significant figures. iii The x-intercept (the point where the line crosses the x-axis) should be the point at which Δ E is zero and 1/n i 2 = 1/n f 2 = 1/2 2. Three significant figures are fine for this value. For the Balmer series, n f = 2. The line on which your data points lie should have a y-intercept = R H/4 and a x-intercept of 1/4 (1/n f 2 = 0.250). The slope should be negative. An example plot is shown using the data from above. Figure 9 Figure 10 Experimental:slope (R H, exp)~2 × 10−18 J x−intercept~0.245 Literature:R H, lit 2.18 × 10−18 J 1/n f 2 Balmer 0.250 = 1/2 2 NOTE: YOU MUST USE YOUR OWN EXPERIMENTAL DATA. IT WILL NOT BE EXACTLY THE SAME AS THE EXAMPLES GIVEN. Reporting Results Complete your lab summary or write a report (as instructed). Abstract Results Wavelength versus color observations He emission calibration data and plot H emission wavelength (measured and corrected) data, colors, and error Calculated value table – theoretical Δ E, measured Δ E, etc. (see example table) Rydberg plot, experimental R H and error (relative to literature value) Sample Calculations Photon energy Energy level calculation and Δ E calculation Rydberg slope and error Discussion What you did, how you did it, and what you determined Spec 20 color to wavelength correlation Spectrometer Calibration – purpose, etc. H atom emission spectra and the Rydberg plot Experimental Rydberg constant? What is n f for observed emission? Which series is this? What is n i for each observed photon? What else can you conclude from this study? Review Copyright © 2011 Advanced Instructional Systems, Inc. and the University of California, Santa Cruz \| Credits
14202	https://eprints.whiterose.ac.uk/id/eprint/174625/1/Accepted_CJC-2021-0064-R1.pdf	This is a repository copy of Solidification transformations in liquid phase separated metastable monotectic Cu – 50 at.% Co alloy. White Rose Research Online URL for this paper: Version: Accepted Version Article: Jegede, OE, Haque, N, Mullis, AM orcid.org/0000-0002-5215-9959 et al. (1 more author) (2021) Solidification transformations in liquid phase separated metastable monotectic Cu – 50 at.% Co alloy. Canadian Journal of Chemistry. ISSN 0008-4042 © 2021. This is an author produced version of a journal article published in Canadian Journal of Chemistry. Uploaded in accordance with the publisher's self-archiving policy. eprints@whiterose.ac.uk Reuse Items deposited in White Rose Research Online are protected by copyright, with all rights reserved unless indicated otherwise. They may be downloaded and/or printed for private study, or other acts as permitted by national copyright laws. The publisher or other rights holders may allow further reproduction and re-use of the full text version. This is indicated by the licence information on the White Rose Research Online record for the item. Takedown If you consider content in White Rose Research Online to be in breach of UK law, please notify us by emailing eprints@whiterose.ac.uk including the URL of the record and the reason for the withdrawal request. 1 Solidification transformations in liquid phase separated metastable monotectic Cu – 50 at. % Co alloy. Oluwatoyin E. Jegede a, Nafisul Haque a,b, , Andrew M. Mullis a and Robert F. Cochrane a a School of Chemical & Process Engineering, University of Leeds, Leeds LS2 9JT, UK. b Department of Metallurgical Engineering, NEDUET, University Road, Karachi 75270, Pakistan Corresponding author: Oluwatoyin E. Jegede (email: pmoej@leeds.ac.uk) 2 Metastable monotectic Cu – 50 at. % Co alloy produced by arc melting has been processed under micro gravity condition using a drop tube and subjected to differential thermal analysis (DTA). Microstructural evidence from the as solidified sample revealed that rapid cooling of the arc melt process was enough to incite liquid phase separation in the alloy. In the drop tube samples, the melting temperature of the β- phase (Cu – rich) was determined to be 1294.8 K while that of the α- phase (Co – rich) was found to vary with cobalt content. Keywords: Liquid phase separation, Monotectic solidification, Differential thermal analysis (DTA), Drop tube, metastable alloys. 3 1 INTRODUCTION There has been significant research on the equi – atomic Cu – Co alloy (Cu50Co50) due to the symmetrical metastable miscibility gap (MG) said to exist in the alloy system [1–3]. Nakagawa , Robinson et al. and Elder et al. all observed a metastable MG whose highest point (critical point) corresponds to the equi – atomic composition. Different values have been cited for the critical undercooling necessary to bring about liquid phase separation (LPS) in the alloy; the value was placed at 96 K below its equilibrium liquidus by Yamauchi et al. , 90 K was reported by Nakagawa while Robinson et al. reported 80 K. The peritectic reaction of the system has been said to occur at the temperature (Tp) of 1385 K. However, the accepted MG of the system is that determined by Cao et al. using DTA and glass fluxing on alloys of composition 16 – 87.2 at. % Co. This slightly symmetrical MG had critical composition of 53 at. % Cu and a binodal boundary with composition range of 16 – 89.3 at. % Cu. Liquid phase separation was found to occur at this critical composition at temperature of 1547 K which was 108 K below its equilibrium liquidus. Palumbo et al. however stated that the critical point of the MG occurred at 58.5 at. % Cu with a temperature of 1556 K. Generally, LPS occurs in the Cu – Co alloy system into L1 (Co – rich) and L2 (Cu – rich) phases when the parent melt is undercooled into the MG beyond a certain temperature limit (Tsep) which is composition dependant [4,7,8]. The solidification path of the demixed liquids differ to each other and to the initial homogeneous melt when traced on the metastable phase diagram due to each liquid having different undercooling. Also, secondary liquid phase separation of one liquid phase inside another has been said to occur at high undercooling due to truncated diffusion . 4 Different microstructures have been reported in this alloy [2,4,7–11]. Liquid phase separated structures have been reported at varying degree of undercooling. Yamauchi et al. observed liquid phase separated features at undercooling of 123 K while in the long drop tube experiment (105 m), LPS was said to occur at undercooling of 300 K a value which exceeds the metastable phase diagram estimate for an alloy of such bulk composition . In electro magnetic levitation (EML) studies carried out by Zhang et al. and Davidoff et al. on the alloy, liquid phase separated features were also reported. Davidoff et al. using a combination of EML and splat quenching, observed that in splat quenched samples at estimated cooling rates of 107 Ks-1 an undercooling of 220 K is achieved and the microstructural morphology consisted of large coagulated Co – rich phase that showed evidence of secondary liquid phase separation a conclusion based on the presence of dispersed L2 – rich particles in it. At higher cooling rates (magnitude of 106 to 107 Ks-1); spinodal decomposition was observed . In their EML sample on the other hand, the microstructural morphology at undercooling above 120 K consisted of large distorted / spherical particles with very high copper content (96 at. % Cu) in a L1 - rich matrix (84 at. % Co). The interface between the two liquids had dendrites extending from the L1 – rich phase into the L2 – rich phase an indication they were likely formed post solidification of the Co – rich phase . However, in the EML study of Munitz and Abbaschian they undercooled the alloy to above 110 K and did not observe any evidence of LPS but rather zones of fine and coarse dendrites whose distinctive boundaries were observed to fade with increasing undercooling. They also observed that at 200 K, the elements of the coarse dendrites were found within the fine dendritic zones. The existence of distorted spherulites in other compositions was cited as evidence of LPS in those alloys. This is a probability as the distorted nature of the spherulites is thought to be as a result of the convective flow induced by the electromagnetic 5 stirring during the EML process. This is however not expected to be observed in drop tube experiments one of which reported the presence of core shell microstructures in the alloy . The core shell microstructures resulted from the interplay of interfacial energies and temperature and or composition gradient leading to Marangoni motion of dispersed particles of the minority phase formed close to the surface of the parent droplets to the center of the parent droplet where they converge to form a core. Mechanism of formation and characteristics of these microstructures in drop tube processed 50 at. % Co alloy is available in literature [12,13]. Other researchers have also reported that the alloy is dendritic and that LPS is only feasible in the alloy at higher degree of undercooling [2,8,9]. Even though the alloy has been extensively researched there has been little report of phase separation in the alloy which is contrary to the calculations of the metastable phase diagram reported in which states that the alloy should be able to phase separate both spinodally and by binodal phase separation after the critical alloy because of the minimal undercooling estimated to get the alloy into the miscibility gap. Undercooling is difficult to determine in drop tube experiments however, it can be inferred from the cooling rate which can be estimated from the diameter of droplets obtained in drop tube experiments using balance of heat fluxes within the falling droplet as described in . In this article, differential thermal analysis (DTA) is used to gain insights into the accuracy of the calculated metastable phase diagram of the Co – Cu alloy system described in as well as to understand the solidification transformations in drop tube processed Cu – 50 at. % Co alloy. 2 EXPERIMENTAL METHODS 6 The Cu – 50 at. % Co alloy was produced by alloying pure Co (99.998 %) and Cu (99.999 %) in an arc melting furnace with a protective argon atmosphere. The arc melting process was repeated nine times to ensure homogeneity of the ingot. Slices from the arc melt ingot were then processed in a 6.5 m drop tube under a nitrogen environment. The drop tube was evacuated to 10-4 Pa and back filled with nitrogen gas to 40 kPa. The alloy sample inductively heated was super-heated to 200 K above its equilibrium liquidus and injected into the drop tube and a dispersion of droplets of varying size solidified during their free fall down the tube. The collected droplets were sieved into standard sieve size fractions. Due to the limited amount of sample retrieved from the drop tube, samples from the sieve size fraction with the most amount of powder available were selected for DTA measurements (i.e. 850+ and < 38 μm). 31.4 mg and 9.7 mg of powder from the 850+ and < 38 μm size fractions respectively were measured out and melted in Al2O3 crucibles using a PerkinElmerTM STA 8000 simultaneous thermal analyser at heating and cooling rates of 15 Kmin-1. Only the first heating and cooling cycle curves were considered in this study as the samples showed signs of Cu oxide contamination (samples had black appearance) and this is thought to be the origin of some of the peaks observed in the second process cycle. Also, that DTA curves from the second process cycle upwards are not likely to be representative of the starting alloy composition. Baseline artefacts identified on the curves were not discussed as they are not part of the transformation details of the alloy but of the reference sample hence they were subtracted from all DTA curves. All the samples were then metallographic processed and their microstructure examined with a BX51 Olympus optical microscope fitted with a Zeiss AxioCam™ MRc5 camera and a Carl Zeiss Evo MA 15 SEM fitted with energy dispersive X-ray spectrometer (EDS). Backscattered electron imaging (BSE) mode was employed. 7 3 RESULTS AND DISCUSSION The metastable phase diagram in (figure 1) had critical composition of 58.7 at. % Cu and liquidus temperature of 1636 K. The estimated undercooling for LPS to occur in this critical alloy either by nucleation in the binode region or spinodally within the spinode was placed at 13 K. The Cu – 50 at. % Co alloy was estimated to undergo binodal LPS at undercooling of 41 K and spinodal decomposition at 52 K. The liquidus temperature of the alloy was determined to be 1639 K. Figure 1: Metastable phase diagram of the Cu – Co system with super imposed miscibility gap . Black vertical line shows the solidification path of the Cu – 50 at. % Co alloy. Slices from the arc melt ingot showed two regions, one having a homogeneous appearance and the other which appeared unmixed contained a huge region of un-melted cobalt surrounded by 8 Co – rich coarse dendrites which decreased in size away from it towards the bottom of the sample. In the homogeneous section, shown by the optical micrograph in figure 2, dark structures dispersed in a light matrix in regions close to the copper hearth of the arc melt furnace are observed (figure 2a). These particles appear clustered in some areas. The appearance of the un-clustered particles looked like columnar structures and these were initially thought to be dendrites similar to the zones of fine and coarse dendrites observed by Munitz and Abbaschian in the alloy but these were revealed to be spherical and spheroidised at higher magnification (figure 2b). EDS analysis (figure 3) gave the average composition of the dark particles as 87.6 at. % Co while the light matrix was observed to be Cu – rich. The microstructure also revealed considerate particle free zones indicating the volume fraction of the dispersed Co – rich particles in the matrix was low. These images seem to suggest the arc melted Cu – 50 at. % Co alloy had undergone LPS. Figure 2: Optical microscopy images of slices of arc melt ingot of Cu – 50 at. % Co alloy, (a) shows dark cobalt rich structures in a copper rich matrix, (b) magnified view of the dispersed spherical particles. 9 Figure 3: EDS readings of the matrix and of the dark inclusions respectively. Extensive classification of structures observed in undercooled drop tube processed Cu- Co alloys have been given in literature . These structures are basically of two categories; liquid phase separated structures and non – liquid phase separated structures. Figure 4a – d shows some of the reported microstructures. 10 Figure 4: Back scattered SEM images of reported microstructures in undercooled drop-tube processed Co-Cu alloys, dark phase is Co – rich while the lighter phase is Cu – rich. (a) and (b) are evolving core shell structures of diameters 250 µm and 410 µm respectively, (c) 60 µm fully evolved core shell structure and (d) 72 µm dendritic structure . The microstructure of the arc melt sample of the alloy presented in figure 2 establishes the occurrence of LPS in the alloy. This is evidenced by the dispersion of minority Co – rich spherical particles in a majority Cu – rich matrix. The appearance of these nucleated spherical particles indicates the alloy was successfully undercooled into the binodal region of the MG since nucleation is not known to occur under the spinodal curve. The migration and growth of 11 these particles in the binodal region occurs by diffusion and this could explain the clustering nature of the spherical particles which could be an indication of high rate of transport and this in turn is expected to lead to coarsening effect but the particles appeared to be of uniform size. No microstructural evidence was found to suggest concentration gradient in any of the arc melt slices; it is therefore likely the migration of these particles was due to temperature variation. At close proximity to the copper hearth, thermal gradient is higher and as such Marangoni convection move the particles away from the cooler end however, this may have been overwhelmed by flow induced by the arc’s electric current which would also explain the spheroidised / rod – like appearance of some of the Co – rich particles. At the extreme end away from the copper hearth, the temperature difference is not as high but due to minute density difference between cobalt and copper, Stokes effect which should be prominent is rather negligible. This would explain why most of the particles are clustered at this side of the particle. The arc melt sample was not used for DTA due to its majorly inhomogeneous appearance and apparent difficulty in extracting the homogeneous section. However, in the 850+ drop tube powder the first cycle heating curve (figure 5) had two endothermic peaks at 1385.5 K and 1645.6 K with onset temperatures of 1294.8 K and 1522 K respectively. Due to the fact that the microstructure in figure 6a revealed no evidence of metastable phase transformation these temperatures are analyzed using the equilibrium phase diagram. The first onset (1294.8 K) falls within α + β region with the α – phase (i.e. Co – rich) having volume fraction of 50.7 at. %. Since the alloy has equal atomic volume at this temperature, the observed peak is assumed to be due to release of heat of fusion on melting of the β – phase which is Cu – rich. Hence melting temperature (Tm) of the Cu – rich β – phase is taken as 1294.8 K (melting point of pure copper is 12 1358 K). The second endothermic onset falls within L + α region. There is however, more α – phase in this region than the liquid phase with α – phase having a volume fraction of 53 %. The α - Co rich phase contained 83.8 at. % Co while the composition of the liquid phase is shown to contain 12.5 at. % Co. The melting temperature of α – phase is taken as 1522 K (melting point of pure cobalt is 1768 K). On the cooling curve two prominent exothermic events can also be observed. The first event had an onset temperature of 1589 K which is a shift of 67 K to its corresponding endothermic event. This temperature is also traced to the L + α region on the equilibrium phase diagram however, the composition of the liquid had increased to 19.4 at. % Co. The compositional increase is thought to be the reason for the shift between the endothermic and exothermic peaks as during solidification the composition is constantly being adjusted. The volume fraction of α – Co phase was determined to be 48.9 %. The nucleation temperature (TN) of α – Co phase is therefore taken as 1589 K. The second exothermic onset corresponded to the solidification temperature (Ts) of the Cu – rich phase. 13 Figure 5: DTA plots of the 850+ μm drop tube powder of the Cu – 50 at. % Co alloy. Figure 6: SEM images of DTA processed drop tube powder of the Cu – 50 at. % Co alloy: (a) 850+ μm powder size, (b) < 38 μm powder size. 14 In the < 38 μm drop tube powder size, two endothermic events were also observed on its heating curve (figure 7). The microstructure of the sample shown in figure 6b also displayed no signs of metastable transformation hence the equilibrium phase diagram is also used in analysing its DTA plot. The first onset temperature (1374 K) corresponds to the Tm of the Cu – rich phase having a composition of 94.3 at. % Cu. A second endothermic event signified by a broad peak had an onset temperature of 1621 K which is also traced to the L + α region however, the liquid composition (71.6 at. % Cu) is lesser than that of the 850+ μm powder. Volume fraction of α – Co phase is calculated to be 40.6 %. On the cooling curve, alloy at the onset temperature of the first exothermic peak (1613 K) has composition of 74.8 at. % Cu which is observed lower than that in the larger powder size. This is thought to be the temperature at which α – Co starts to nucleate (TN). Volume fraction of the Co – rich phase had however reduced to 49.9 %. Onset temperature of the second exothermic peak (1383 K) which corresponds to the Ts of the Cu – rich phase is higher than that observed for the 850+ powder size. The volume fraction of α – Co phase was also observed to have increased to 53 %. 15 Figure 7: DTA plots of the < 38 μm drop tube powder of the Cu – 50 at. % Co alloy. The value quoted for the liquidus temperature in the phase diagram in figure 1 is lower than that quoted by Davidoff et al. who quoted 1655 K as the liquidus of the alloy and an estimated undercooling of 106 K to get into the binode. However, their values seem not supported by their phase diagram. According to their calculations, the critical composition of their MG was 52.7 at. % Cu with corresponding temperature of 1547 K, they also stated that their miscibility gap was symmetrical. If that is the case, the Cu – 50 at. % Co alloy is very close to their critical composition and as such expected to contact the binode curve very close to the temperature of their critical alloy. 16 4 CONCLUSION The occurrence of LPS in the arc melt sample is contrary to previous cited works which concludes that the alloy is dendritic unless when subjected to high undercooling into the MG. Although the temperature and cooling rate of the arc melt processing is unknown, the fact that rapid cooling of the process was able to cause LPS to occur in the alloy is an indication that the undercooling required to cool into the MG is not as large as indicated in those literature but in line with the predictions of the calculated metastable phase diagram in figure 1. Dendritic structures consistent with equilibrium phase diagram predictions were observed in the DTA samples of the drop tube processed powders. The melting temperature of the Cu – rich and Co – rich liquids was found to vary with droplet size which is also in line with phase diagram prediction. There is increased undercooling in smaller droplets which in turn favours the enrichment with the Co – rich phase. The higher the cobalt content, the higher the melting point of the two phases. ACKNOWLEDGEMENT Oluwatoyin Jegede is a commonwealth scholar, sponsored by the UK government. COMPETING INTEREST STATEMENT The authors declare there are no competing interests. 17 AUTHOR CONTRIBUTION STATEMENT All authors actively contributed towards the writing of the manuscript and have given approval to its final version. REFERENCES Y. Nakagawa, Liquid immiscibility in copper-iron and copper-cobalt systems in the supercooled state, Acta Metall. 6 (1958) 704–711. M.B. Robinson, D. Li, T.J. Rathz, G. Williams, Undercooling, liquid separation and solidification of Cu-Co alloys, J. Mater. Sci. 34 (1999) 3747–3753. G. Elder, SP and Munitz, A and Abbaschian, Metastable Liquid Immiscibility in Fe-Cu and Co-Cu Alloys, Mater. Sci. Forum. 50 (1989) 137–150. I. Yamauchi, N. and Ueno, M. and Shimaoka, I. and Ohnaka, Undercooling in Co – Cu alloys and its effect on solidification structure, J. Mater. Sci. 33 (1998) 371–378. C.. Cao, G.. Görler, D.. Herlach, B. Wei, Liquid–liquid phase separation in undercooled Co–Cu alloys, Mater. Sci. Eng. A. 325 (2002) 503–510. M. Palumbo, S. Curiotto, L. Battezzati, Thermodynamic analysis of the stable and metastable Co–Cu and Co–Cu–Fe phase diagrams, Calphad. 30 (2006) 171–178. 18 A. Munitz, R. Abbaschian, Microstructure of Cu-Co Alloys Solidified at Various Supercoolings, Metall. Mater. Trans. A. 27 (1996) 4049–4059. A. Munitz, R. Abbaschian, Two-melt separation in supercooled Cu-Co alloys solidifying in a drop-tube, J. Mater. Sci. 26 (1991) 6458–6466. A. Munitz, R. Abbaschian, Liquid separation in Cu – Co and Cu – Co – Fe alloys solidified at high cooling rates, Mater. Sci. 33 (1998) 3639–3649. Y.K. Zhang, J. Gao, D. Nagamatsu, T. Fukuda, H. Yasuda, M. Kolbe, J.C. He, Reduced droplet coarsening in electromagnetically levitated and phase-separated Cu-Co alloys by imposition of a static magnetic field, Scr. Mater. 59 (2008) 1002–1005. E. Davidoff, P.. Galenko, D.. Herlach, M. Kolbe, N. Wanderka, Spinodally decomposed patterns in rapidly quenched Co–Cu melts, Acta Mater. 61 (2013) 1078–1092. O.E. Jegede, R.F. Cochrane, A.M. Mullis, Metastable monotectic phase separation in Co– Cu alloys, J. Mater. Sci. 53 (2018) 11749–11764. A.M. Mullis, O.E. Jegede, T.D. Bigg, R.F. Cochrane, Dynamics of core--shell particle formation in drop-tube processed metastable monotectic alloys, Acta Mater. 188 (2020) 591–598. 19 M. Erol, U. Böyük, T. Volkmann, D.. Herlach, Containerless solidification of Ag–Al and Ag–Cu eutectic alloys in a drop tube, J. Alloys Compd. 575 (2013) 96–103. LIST OF FIGURES Figure 1: Metastable phase diagram of the Cu – Co system with super imposed miscibility gap . Black vertical line shows the solidification path of the Cu – 50 at. % Co alloy. ................... 7 Figure 2: Optical microscopy images of slices of arc melt ingot of Cu – 50 at. % Co alloy, (a) shows dark cobalt rich structures in a copper rich matrix, (b) magnified view of the dispersed spherical particles............................................................................................................................ 8 Figure 3: EDS readings of the matrix and of the dark inclusions respectively ............................. 9 Figure 4: Back scattered SEM images of reported microstructures in undercooled drop-tube processed Co-Cu alloys, dark phase is Co – rich while the lighter phase is Cu – rich. (a) and (b) are evolving core shell structures of diameters 250 µm and 410 µm respectively, (c) 60 µm fully evolved core shell structure and (d) 55 µm dual structure droplet showing fragmented dendrites and spherical particles . .......................................................................................................... 10 Figure 5: DTA plots of the 850+ μm drop tube powder of the Cu – 50 at. % Co alloy. ............ 13 Figure 6: SEM images of DTA processed drop tube powder of the Cu – 50 at. % Co alloy: (a) 850+ μm powder size, (b) < 38 μm powder size. ......................................................................... 13 Figure 7: DTA plots of the < 38 μm drop tube powder of the Cu – 50 at. % Co alloy. .............. 15
14203	https://www.reddit.com/r/learnmath/comments/ag3o54/geometry_triangle_projections/	Geometry Triangle Projections : r/learnmath Skip to main contentGeometry Triangle Projections : r/learnmath Open menu Open navigationGo to Reddit Home r/learnmath A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to learnmath r/learnmath r/learnmath Post all of your math-learning resources here. Questions, no matter how basic, will be answered (to the best ability of the online subscribers). 403K Members Online •7 yr. ago Butterlemons11 Geometry Triangle Projections I have a question that has a projection in it and I don’t know where the projection is (there’s no pictures). The question is “In a right triangle whose hypotenuse measures 10, the projection of the longer leg on the hypotenuse measures 6.4. Find the measure of the longer leg.” I’m not sure if if there’s just one triangle with the hypotenuse as 10 and the shorter side as 6.4 and I should use the Pythagorean theorem or if the original triangle is split into two with the hypotenuse of the original triangle as 10 and the altitude or projection as 6.4. Or I could be completely wrong because I don’t know what a projection is. The definition my teacher gave was confusing since I don’t understand what a geometric mean is. :( Any help is welcome! Read more Share Related Answers Section Related Answers Effective strategies for mastering algebra Mastering algebra can be challenging, but with the right strategies and resources, it is definitely achievable. Here are some effective strategies and resources recommended by Redditors: General Study Tips Consistent Practice: Doing more problems and revisiting them regularly helps solidify understanding. 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YouTube Channels: Many Redditors suggest using YouTube for visual explanations and step-by-step problem-solving. "Google 'organic chemistry tutor' on YouTube. He walks through a variety of problems for different subjects, but they’re easy to follow and he explains his solutions." Textbooks and Workbooks: Using textbooks and workbooks can provide a structured approach to learning and ample practice problems. "Get a used copy of beginning algebra by Lial or Elayn Martin-Gay or any other big college textbook." Teaching Others: Explaining concepts to someone else can help identify gaps in your own understanding. "Try teaching math to someone else... This is called the 'Feynman Technique'." Dealing with Learning Disabilities Accommodations: If you have a learning disability like dyscalculia, it's crucial to seek accommodations from your school or university. "Accommodations are how I got through my engineering degree." Alternative Methods: Using visual aids, real-world examples, and different learning styles can help. "Manipulate variables not numbers until the very end." Additional Tips Focus on Fundamentals: Make sure you have a strong grasp of the basics before moving on to more complex topics. "Know the fundamentals. Problem modelling and solving, logic and proofs." Growth Mindset: Believe that you can improve with effort and persistence. "The most important qualities are curiosity, persistence, and humility." Subreddits for Further Help r/learnmath r/mathematics r/matheducation r/dyscalculia These strategies and resources should help you on your journey to mastering algebra. Good luck! See Answer Tips for improving mental math skills Exploring real-world uses of number theory Comparing different methods of integration Best practices for preparing for math exams New to Reddit? Create your account and connect with a world of communities. Continue with Google Continue with Google. 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14204	https://www.youtube.com/playlist?list=PLUPEBWbAHUsyJkm0CwwrOdOTpT0avzW9w	Fractions - YouTube Back Skip navigation Search Search with your voice Sign in Home HomeShorts ShortsSubscriptions SubscriptionsYou YouHistory History Play all Fractions by mathantics • Playlist•8 videos•1,359,373 views Play all PLAY ALL Fractions 8 videos 1,359,373 views Last updated on Apr 1, 2020 Save playlist Shuffle play Share Show more mathantics mathantics Subscribe Play all Fractions by mathantics Playlist•8 videos•1,359,373 views Play all 1 5:16 5:16 Now playing Math Antics - Fractions Are Parts mathantics mathantics • 2.4M views • 13 years ago • 2 5:56 5:56 Now playing Math Antics - Working With Parts mathantics mathantics • 845K views • 13 years ago • 3 3:03 3:03 Now playing Math Antics - Fractions Are Division mathantics mathantics • 824K views • 13 years ago • 4 6:42 6:42 Now playing Math Antics - Types of Fractions mathantics mathantics • 1.1M views • 13 years ago • 5 9:15 9:15 Now playing Math Antics - Fractions and Decimals mathantics mathantics • 2M views • 13 years ago • 6 6:59 6:59 Now playing Math Antics - Converting Base-10 Fractions mathantics mathantics • 2.4M views • 13 years ago • 7 6:36 6:36 Now playing Math Antics - Converting Any Fraction mathantics mathantics • 1.1M views • 13 years ago • 8 6:31 6:31 Now playing Math Antics - Comparing Fractions mathantics mathantics • 2.8M views • 13 years ago • Search Info Shopping Tap to unmute 2x If playback doesn't begin shortly, try restarting your device. • You're signed out Videos you watch may be added to the TV's watch history and influence TV recommendations. To avoid this, cancel and sign in to YouTube on your computer. Cancel Confirm Share - [x] Include playlist An error occurred while retrieving sharing information. Please try again later. Watch later Share Copy link 0:00 / •Watch full video Live • • NaN / NaN [](
14205	https://ams.confex.com/ams/pdfpapers/108410.pdf	9A.2 Tropical Cyclone Satellite Tutorial Online Through The COMET Program Thomas F. Lee Steven D. Miller F. Joseph Turk Jeffrey D. Hawkins Naval Research Laboratory, Monterey CA Patrick Dills Sherwood Wang Cooperative Program for Operational Meteorology, Education, and Training (UCAR/COMET®) Boulder CO 27th Conference on Hurricanes and Tropical Meteorology 24-28 April 2006 Monterey CA 1. Introduction The Cooperative Program for Meteorology, Education and Training (COMET) has produced an online educational module that introduces forecasters to observation and analysis of tropical cyclones. Sponsored by the National Polar-orbiting Operational Environmental Satellite System (NPOESS) Integrated Program Office (IPO), this module foresees forecaster applications when the NPOESS Preparatory Project (NPP) and NPOESS satellites are launched starting later this decade. The training, however, can be used now. Even as it previews modernized applications, the module reviews important capabilities from existing polar-orbiting satellite sensors including Defense Meteorological Satellite Program (DMSP) Special Sensor Microwave Imager (SSM/I) (and SSMIS, final “S” standing for sounder), Tropical Rainfall Measuring Mission (TRMM) Microwave Imager (TMI), NASA Aqua Advanced Microwave Scanning Radiometer (AMSR-E), Coriolis WindSat, and NOAA Advanced Microwave Sounding Unit (AMSU-B). The module is freely available to the public and takes about one hour to complete. While the content is designed for personnel with basic training in meteorology or oceanography, the general public may also find it useful. The module is voice narrated and interactive, ending with a quiz to test comprehension of key concepts. Feedback received from surveys indicates that a wide variety of users worldwide are benefiting. Corresponding author: Thomas F. Lee, 7 Grace Hopper Ave Monterey CA 93943, 831 656 4883, thomas.lee@nrlmry.navy.mil 2. The COMET Program The COMET Program has provided distance-learning education to a wide spectrum of users in the atmospheric science community since 1990. Distance-learning technologies applied by COMET focus on Web-based training modules, teletraining offerings developed in conjunction with the NOAA/NWS VISIT (Virtual Institute for Satellite Integration Training) program, and course that blend distance-learning with residence attendence. COMET also has the capacity to offer a variety of residence courses and workshops in its classroom facility in Boulder, Colorado. The full exploitation of improved information resulting from rapid advances in global observing systems like NPOESS is contingent upon strong education and training processes. For the NPOESS IPO, the COMET program’s focus is on highlighting and demonstrating the future capabilities and applications of the NPOESS system for operational forecasters and other user communities. COMET works closely with these user communities to stimulate greater utilization of both the training materials and current operational and research polar-orbiting satellite data observations and products. To meet these goals, the NPOESS training effort generates web-modules, teletraining sessions, webcasts, workshops, and maintains a Web-based information resource portal, the NPOESS Userport. The Userport Website provides links to polar-orbiting satellite multimedia learning resources and real-time data for forecasters, scientists, and the general public interested in learning more about the mission, spacecraft, instrumentation, data processing, products, and applications: 3. Satellite Training on Tropical Cyclones at the Naval Research Laboratory The Naval Research Laboratory (NRL) Monterey is charged with providing appropriate examples and content to COMET developers to demonstrate future NPOESS capabilities. With respect to tropical cyclones, NRL is uniquely positioned for this role because of its web site which shows live satellite images, especially from microwave sensors, for every storm on a global, year-round basis: ( using existing satellite data streams from polar and geostationary satellites (Hawkins et al. 2001; Lee et al. 2002). The Tropical Cyclone Analysis module which focuses on the use of passive microwave images for the analysis and fixing of tropical cyclones (front page shown in Fig. 1) is one of the most popular COMET modules on satellite meteorology: The module is divided into ten sections (Table 1) with illustrations and example products appearing throughout. There are also tables to summarize key information (e.g., Table 2). The module emphasizes the difference between infrared images, the standard used for analyzing tropical cyclones, and passive microwave images that have been increasingly used by forecasters since the launch of the first SSM/I in 1987. Fig. 2 shows an example of an infrared image of Hurricane Frances in 2004, showing a cirrus canopy with low temperatures in red near the center. Fig. 3 is the corresponding composite based on observations from the U.S. Navy research and development microwave sensor, WindSat, aboard the Coriolis satellite. The microwave data reveals a double eye wall structure that is hidden beneath the cirrus canopy on Fig. 2. Fig. 4 also demonstrates concentric eyewall characteristics, but this time for Hurricane Juliette in the Eastern Pacific based on TRMM TMI data. Figs. 5 and 6 present a quiz question, the former asking the question and the latter showing the answer. The interspersed questions throughout the module help reinforce key elements of the material and accelerate learning. Fig. 7 shows the difference between cross track and conical scan strategies. Both strategies are used in the passive microwave imaging of tropical cyclones, but there are important ramifications to the use of each. In Fig. 8 we see the distortion arising at the edge of scan with the cross-scanning arising from the AMSU-B sensor. Fig. 9 illustrates how satellite parallax creates displacement of elevated features, such as clouds, with respect to the earth. Fig. 10 illustrates the effect of parallax on microwave images. 4. Other NRL Training NRL also maintains training on its NexSat satellite site (Miller et al 2006): This site is a realtime display of a number of experimental satellite products to illustrate advances in polar satellite capabilities in anticipation of NPOESS. Products which blend geostationary and polar capabilities are also emphasized, for example “geostationary color” as shown in Fig. 11. This example of Hurricane Katrina shows visible data overlaid upon the NASA blue marble background during the daytime. At night infrared data are overlaid upon a background that incorporates nighttime city lights from the National Geophysical Data Center (NGDC). The NexSat page contains a detailed training tutorial in pdf form (green button shown in Fig. 12). Similar training product tutorials are available for all of the experimental applications listed on the left side (“Products” list) of the interface. 5. Acknowledgements The support of the research sponsor, the National Polar-orbiting Operational Environmental Satellite System’s (NPOESS) Integrated Program Office (IPO) located in Silver Spring, MD, is gratefully acknowledged. The support of the research sponsors, the Oceanographer of the Navy through the program office at the PEO C4I&Space/PMW-180, under program element PE-0603207N and the Office of Naval Research under program element PE-0602435N is gratefully acknowledged. 6. References Hawkins, J.D., T. Lee, J. Turk, C. Sampson, J. Kent, and K. Richardson, 2001: Real-time Internet distribution of satellite products for tropical cyclone reconnaissance. Bull. Amer. Met. Soc., 82, 567-578. Lee, T.F., F.J. Turk, J.D. Hawkins, and K.A. Richardson, 2002: Interpretation of TRMM TMI images of tropical cyclones. Earth Interactions E-Journal 6:3. Miller, S. D., J. D. Hawkins, J. Kent, F. J. Turk, T. F. Lee, A. P. Kuciauskas, K. Richardson, R. Wade, and C. Hoffman, 2006: NEXSAT: Previewing NPOESS/VIIRS Imagery Capabilities, Bull. Amer. Met. Soc., in press (April issue). Table 1 Table of Contents Using Microwave Observations for Tropical Cyclone Analysis 1.0 Overview 2.0 Tropical Cyclone Examples 3.0 Current Passive Microwave Sensors 4.0 Improvements with NPOESS CMIS 5.0 Characteristics of Microwave Imagery 6.0 85 to 91 GHz Imagery -- Interpretation 7.0 Multispectral Examples and Exercises 8.0 Concentric Eyewall Characteristics 9.0 Summary 10.0 Exercises Table 2 Comparing Active and Passive Microwave Sensors
14206	https://bmcbioinformatics.biomedcentral.com/articles/10.1186/s12859-014-0354-6	MLGO: phylogeny reconstruction and ancestral inference from gene-order data \| BMC Bioinformatics \| Full Text Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Advertisement Search Explore journals Get published About BMC Login Menu Explore journals Get published About BMC Login Search all BMC articles Search BMC Bioinformatics Home About Articles Submission Guidelines Collections join the board Submit manuscript MLGO: phylogeny reconstruction and ancestral inference from gene-order data Download PDF Download PDF Software Open access Published: 08 November 2014 MLGO: phylogeny reconstruction and ancestral inference from gene-order data Fei Hu1,2, Yu Lin3& Jijun Tang1,2 BMC Bioinformaticsvolume 15, Article number:354 (2014) Cite this article 8585 Accesses 103 Citations 9 Altmetric Metrics details Abstract Background The rapid accumulation of whole-genome data has renewed interest in the study of using gene-order data for phylogenetic analyses and ancestral reconstruction. Current software and web servers typically do not support duplication and loss events along with rearrangements. Results MLGO MLGO (Maximum Likelihood for Gene-Order Analysis) is a web tool for the reconstruction of phylogeny and/or ancestral genomes from gene-order data. MLGO MLGO is based on likelihood computation and shows advantages over existing methods in terms of accuracy, scalability and flexibility. Conclusions To the best of our knowledge, it is the first web tool for analysis of large-scale genomic changes including not only rearrangements but also gene insertions, deletions and duplications. The web tool is available from Background As whole genomes are sequenced at increasing rates, using gene-order data a for phylogenetic analyses and ancestral reconstruction is attracting increasing interest. Comparative genomics, evolutionary biology, and cancer research all require tools to elucidate the history and consequences of the large-scale genomic changes, such as rearrangements, duplications, losses. However, using gene-order data has proved far more challenging than using sequence data and numerous problems plague existing methods: oversimplified models, poor accuracy, poor scaling, lack of robustness, lack of statistical assessment, etc. Genome rearrangement operations change the ordering of genes on chromosomes. An inversion operation (also called reversal) reverses both the order and orientation of a segment of a chromosome. A transposition is an operation that swaps two adjacent segments of a chromosome. In case of multiple chromosomes, a translocation breaks a chromosome and reattaches a part to another chromosome, while a fusion joins two chromosomes and a fission splits one chromosome into two. Yancopoulos et al. proposed a universal double-cut-and-join (DCJ) operation that accounts for all rearrangements used to date. None of these operations alter the gene content of genomes, whereas deletions (or losses) delete segments of (one or more) contiguous genes from a chromosome, while insertions introduce a segment of (one or more) contiguous genes from external sources into a chromosome. and duplications copies an existing segment within the genome and inserts into a chromosome. Finally, whole genome duplication (WGD) creates an additional copy of the entire genome of a species. As phylogenies play a central role in biological research, over the past decade many methods were developed to reconstruct phylogenies from gene-order data. The first algorithm for phylogeny inference from gene-order data was BPAnalysis based on breakpoint distances . Moret et al. later extended this approach with GRAPPA GRAPPA by using inversion distances. While these methods were limited to unichromosomal genomes, Bourque and Pevzner developed MGR MGR to handle multichromosomal genomes. These approaches are parsimony-based: they solve the so-called Big Parsimony Problem (BPP) and all suffer from serious scalability issues. In contrast with parsimony-based methods, distance-based methods run in time polynomial in the number and size of genomes. Lin et al. have demonstrated the accuracy and scalability of a distance-based method that uses NJ and FastME with an accurate distance estimator . Instead of working directly with the evolutionary events of the model, one can also transform the problem into the familiar sequence-based reconstruction problem. Wang et al. first proposed a parsimony-based approach, MPBE MPBE (Maximum Parsimony on Binary Encoding). Recently Hu et al. developed MLBE MLBE, later refined by Lin et al. with MLWD MLWD, both of which demonstrate that using maximum-likelihood approaches is the decisive factor in improving the modest accuracy of MPBE. If the tree is fixed, then computing its parsimony score is known as the Small Parsimony Problem (SPP). Ancestral reconstruction has been studied through several optimization schemes for SPP on gene-order data—using adjacencies -, using conserved intervals (Roci Roci—Reconstruction of Conserved Intervals ), using multiple breakpoint graphs (MGRA MGRA) and supporting whole-genome duplications ,, where continuous regions or complete ancestral genomes have been inferred. Relatively few of these tools are offered through web servers. Lin et al. had developed a web-server version of MGR MGR with new heuristics to speed up the original MGR MGR algorithm, but the site is no longer accessible. Both Roci Roci and MGRA MGRA (for ancestral reconstruction only) are offered through web servers, but none can handle complex events such as gene insertions, deletions and duplications. We present a new tool MLGO MLGO for the reconstruction of phylogeny and/or ancestral genomes from gene-order data. MLGO MLGO relies on two methods we have developed: MLWD MLWD for phylogenetic reconstruction and PMAG+PMAG+ for ancestral genome reconstruction. Our tool takes the advantage of binary encoding on gene-order data, supports a fairly general model of genomic evolution (rearrangements plus duplications, insertions, and losses of genomic regions), and successfully accommodates itself into the framework of maximized likelihood. The results of extensive testing on both simulated and real data show that both MLWD MLWD and PMAG+PMAG+ can achieve great performance, scalability and flexibility, suggesting MLGO MLGO a suitable tool for large-scale analysis of high-resolution data. Furthermore, MLGO MLGO is deployed as a web service, providing the first web tool that is suitable for large scale genomic analysis with a general model of evolution. Implementation MLGO MLGO preprocesses the gene-order data, configures the transition model, reconstructs a phylogeny, and finally solves the SPP on that phylogeny. Terminology Given a set of n genes labeled as {1,2,.,n}, gene-order data for a genome consists of lists of genes in the order in which they are placed along one or more chromosomes. Each gene is assigned with an orientation that is either positive, written i, or negative, written −i. Two genes i and j form an adjacency (i,j) if i is immediately followed by j, or, equivalently, −j is immediately followed by −i. If gene k lies at one end of a linear chromosome, we let k be adjacent to an extremity o to mark the beginning or ending of the chromosome, written as (o,k) or (k,o), and called telomere. Phylogeny reconstruction The data preprocessing and the configuration of the transition model follow the approach of MLWD MLWD. Each adjacency that appears at least once in the collection of input genomes corresponds to a unique character position in the sequence and the presence or absence of any of these adjacencies in a given genomes is coded by a 1 (presence) or a 0 (absence). Since our encodings are binary sequences, the parameters of the model are simply the transition probability from presence (1) to absence (0) and that from absence (0) to presence (1). Lin et al. gave the following derivation for these parameters. A DCJ operation selects uniformly at random two adjacencies (or telomeres) and replaces them by two new adjacencies (or telomeres). Since a genome with n genes and O(1) chromosomes has n+O(1) adjacencies and telomeres, the transition probability from 1 to 0 is 2 n+O(1) under one DCJ operation; and since there are up to 2 n+2 2 possible adjacencies and telomeres, the transition probability from 0 to 1 is 2 2 n 2+O(n). Thus the transition from 0 to 1 is roughly 2 n times less likely than that from 1 to 0. Despite the restrictive assumption that all DCJ operations are equally likely, this result is in line with the observed bias in transitions of adjacencies given by Sankoff and Blanchette : the probability of breaking a given ancestral adjacency is high while that of creating a particular adjacency along several lineages is low (a version of homoplasy for adjacencies). Finally, the encoding adds characters and a transition probability for the presence or absence of each unique gene. Due to duplicated genes, there is no one-to-one correspondence between genomes and the final encodings of multisets of genes, adjacencies, and telomeres. Once we have the binary sequences and transition parameters, we can reconstruct a phylogeny using maximum likelihood. Of the many implementations of this method, we chose RAxML for its speed and its dedicated handling of binary sequences. Bootstrap support A distinct advantage of using sequence encoding is the ability to use the bootstrap method to assess the robustness of the inferred phylogeny. Doing so with gene-order data is not possible, because a chromosome with n distinct genes presents a single character (the ordering) with 2 n×n! possible states (the first term is for the strandedness of each gene and the second for the possible permutations in the ordering). This single character is equivalent to an alignment with a single column, albeit one where each character can take any of a huge number of states—we cannot meaningfully resample a single character. The binary encoding effectively maps this single character into a high-dimensional binary vector, so that the standard phylogenetic bootstrap can be used. While the evolution of a specific adjacency depends directly on several others, independence can be assumed if, once an adjacency is broken during evolution, it is not formed again—an analog of Dollo parsimony, but one that is very likely in rearrangement data due to the enormous state space . Ancestral inference Using the phylogeny thus computed, we then proceed to solve the SPP, now following the approach of Hu et al.. The first step involves the estimation of ancestral gene contents from the contents of the input genomes. Our inference of ancestral contents relies on viewing genes and adjacencies as independent binary characters, as described for the encoding. Whether or not an ancestral genome contains a gene or an adjacency is determined by the conditional probability of the presence state of the gene or the adjacency, computed by the marginal probabilistic reconstruction method suggested by Yang et al.. If such probability is larger than 50%, we conclude that the gene belongs to the genome. We extend this approach to compute the probability of observing each adjacency. We then reduce the adjacency assembly problem for any given ancestral genome to an instance of the Travelling Salesperson Problem (TSP), by representing genes as vertices and adjacencies as edges, and finally solve the TSP by using Concorde Concorde. Results and discussion MLGO MLGO is written in C++ and Perl as a web tool. Figure 1 shows the screen shot of the web interface for MLGO MLGO. The input format of the dataset is that used by GRAPPA GRAPPA and MGR MGR: FASTA-like headers for the names of the genomes (> followed by an alphanumeric sequence followed by a newline), each chromosome represented by a signed permutation of integers ending with a $ symbol and a newline character. Phylogenies are output as trees in Newick format. Figure 1 The screen shot of the web interface forMLGO MLGO. Full size image We used the genomes of 12 fully sequenced drosophila species to demonstrate the performance of MLGO MLGO. Figure 2 shows the consensus phylogeny reconstructed by MLGO MLGO with the bootstrap support values obtained using 100 replicates. Compared to the study using sequence data published by Clark et al., all major groups in those 12 drosophila genomes were correctly identified with strong support (bootstrap value >90), except for one median support at the bipartition between D. simulans, D. sechellia and the rest. The total running time for reconstructing the phylogeny of 12 drosophila species is less than 1 minute, while ancestral reconstruction adds less than 30 minutes. We also tested the performance of MLGO MLGO on 15 Metazoan genomes from the eGOB (Eukaryotic Gene Order Browser) database , and the reconstructed phylogeny tree shown in Figure 3 is perfectly supported from existing studies ,. Figure 2 The consensus phylogeny of 12 drosophila genomes with bootstrap support values from 100 replicates. Full size image Figure 3 The reconstructed phylogeny of 15 Metazoan genomes. Full size image Conclusion As whole genomes are sequenced at increasing rates, using gene-order data for phylogenetic analyses and ancestral reconstruction is attracting increasing interest, especially coupled with the recent advances in identifying conserved synteny blocks among multiple species -. MLGO (Maximum Likelihood for Gene-Order Analysis) is the first web tool for likelihood-based inference of both the phylogeny and ancestral genomes. It provides fast and scalable analyses with bootstrap support of large-scale genomic changes including not only rearrangements but also gene insertions, deletions and duplications. Availability and requirements The web tool is available from name: MLGOProject home page: system(s): Platform independentProgramming language: PerlOther requirements: NoneLicense: GNURestrictions for use by non-academics: None Endnote a We use the term "gene" as this is in fact a common form of syntenic blocks, but other kinds of markers could be used. Authors' contributions FH implemented the web server. YL contributed to the phylogeny reconstruction part with the help of FH and JT. FH and JT contributed to the ancestral inference part. JT provided advice and oversight of the project. All authors drafted, read and approved the final manuscript. References Yancopoulos S, Attie O, Friedberg R: Efficient sorting of genomic permutations by translocation, inversion and block interchange. Bioinformatics. 2005, 21 (16): 3340-3346. 10.1093/bioinformatics/bti535. ArticlePubMedCASGoogle Scholar Blanchette M, Bourque G, Sankoff D: Breakpoint phylogenies. Genome Inform. 1997, 1997: 25-34. 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Proc. 4th Int'l Workshop Algs. in Bioinformatics (WABI'04). 2004, Springer, Germany, 14-25. Google Scholar Alekseyev MA, Pevzner PA: Breakpoint graphs and ancestral genome reconstructions. Genome Res. 2009, 19 (5): 943-957. 10.1101/gr.082784.108. ArticlePubMed CentralPubMedCASGoogle Scholar Murat F, Xu J-H, Tannier E, Abrouk M, Guilhot N, Pont C, Messing J, Salse J: Ancestral grass karyotype reconstruction unravels new mechanisms of genome shuffling as a source of plant evolution. Genome Res. 2010, 20 (11): 1545-1557. 10.1101/gr.109744.110. ArticlePubMed CentralPubMedCASGoogle Scholar Ouangraoua A, Tannier E, Chauve C: Reconstructing the architecture of the ancestral amniote genome. Bioinformatics. 2011, 27 (19): 2664-2671. 10.1093/bioinformatics/btr461. ArticlePubMedCASGoogle Scholar Lin CH, Zhao H, Lowcay SH, Shahab A, Bourque G: webmgr: an online tool for the multiple genome rearrangement problem. Bioinformatics. 2010, 26 (3): 408-410. 10.1093/bioinformatics/btp689. ArticlePubMedCASGoogle Scholar Hu F, Zhou J, Zhou L, Tang J: Probabilistic reconstruction of ancestral genomes with gene insertions and deletions. IEEE/ACM Trans Comput Biol Bioinformatics. 2014, 11 (4): 667-672. 10.1109/TCBB.2014.2309602. ArticleGoogle Scholar Sankoff D, Blanchette M: Probability models for genome rearrangement and linear invariants for phylogenetic inference. Proc. 3rd Int'l Conf. Comput. Mol. Biol. (RECOMB'99). 1999, ACM, USA, 302-309. Google Scholar Stamatakis A: Raxml-vi-hpc: maximum likelihood-based phylogenetic analyses with thousands of taxa and mixed models. Bioinformatics. 2006, 22 (21): 2688-2690. 10.1093/bioinformatics/btl446. ArticlePubMedCASGoogle Scholar Felsenstein J: Confidence limits on phylogenies: an approach using the bootstrap. Evol. 1985, 39: 783-791. 10.2307/2408678. ArticleGoogle Scholar Lin Y, Rajan V, Moret BME: Bootstrapping phylogenies inferred from rearrangement data. Proc. 11th Workshop Algs. in Bioinf. (WABI'11), Lecture Notes in Computer Science, Vol. 6833. 2011, Springer, Germany, 175-187. Google Scholar Yang Z, Kumar S, Nei M: A new method of inference of ancestral nucleotide and amino acid sequences. Genetics. 1995, 141 (4): 1641-1650. PubMed CentralPubMedCASGoogle Scholar Applegate D, Bixby R, Chvatal V, Cook W: Concorde tsp solver2006. [], [ Clark AG, Eisen MB, Smith DR, Bergman CM, Oliver B, Markow TA, Kaufman TC, Kellis M, Gelbart W, Iyer VN, Pollard DA, Sackton TB, Larracuente AM, Singh ND, Abad JP, Abt DN, Adryan B, Aguade M, Akashi H, Anderson WW, Aquadro CF, Ardell DH, Arguello R, Artieri CG, Barbash DA, Barker D, Barsanti P, Batterham P, Batzoglou S, et al: Evolution of genes and genomes on the drosophila phylogeny. Nature. 2007, 450 (7167): 203-218. 10.1038/nature06341. ArticlePubMedGoogle Scholar López MD, Samuelsson T: eGOB: eukaryotic gene order browser. Bioinformatics. 2011, 27 (8): 1150-1151. 10.1093/bioinformatics/btr075. ArticlePubMedGoogle Scholar Ponting CP: The functional repertoires of metazoan genomes. Nat Rev Genet. 2008, 9 (9): 689-698. 10.1038/nrg2413. ArticlePubMedCASGoogle Scholar Srivastava M, Begovic E, Chapman J, Putnam NH, Hellsten U, Kawashima T, Kuo A, Mitros T, Salamov A, Carpenter ML, Signorovitch AY, Moreno MA, Kamm K, Grimwood J, Schmutz J, Shapiro H, Grigoriev IV, Buss LW, Schierwater B, Dellaporta SL, Rokhsar DS: The trichoplax genome and the nature of placozoans. Nature. 2008, 454 (7207): 955-960. 10.1038/nature07191. ArticlePubMedCASGoogle Scholar Simillion C, Janssens K, Sterck L, Van de Peer Y: i-adhore 2.0: an improved tool to detect degenerated genomic homology using genomic profiles. Bioinformatics. 2008, 24 (1): 127-128. 10.1093/bioinformatics/btm449. ArticlePubMedCASGoogle Scholar Pham SK, Pevzner PA: Drimm-synteny: decomposing genomes into evolutionary conserved segments. Bioinformatics. 2010, 26 (20): 2509-2516. 10.1093/bioinformatics/btq465. ArticlePubMedCASGoogle Scholar Rödelsperger C, Dieterich C: Cyntenator: progressive gene order alignment of 17 vertebrate genomes. PloS one. 2010, 5 (1): 8861-10.1371/journal.pone.0008861. ArticleGoogle Scholar Download references Acknowledgements We thank Bernard Moret for helpful discussions. FH and JT were funded by NSF IIS 1161586 and an internal grant from Tianjin University, China. YL was supported by a fellowship of the Swiss National Science Foundation (grant no. 146708). The funders had no role in study design, data collection and analysis, decision to publish, or preparation of the manuscript. Author information Authors and Affiliations Tianjin Key Laboratory of Cognitive Computing and Application, Tianjin University, Tianjin, 300072, China Fei Hu&Jijun Tang Department of Computer Science and Engineering, University of South Carolina, Columbia, 29208, SC, USA Fei Hu&Jijun Tang Department of Computer Science and Engineering, University of California, San Diego, 92093, La Jolla, CA, USA Yu Lin Authors 1. Fei HuView author publications Search author on:PubMedGoogle Scholar 2. Yu LinView author publications Search author on:PubMedGoogle Scholar 3. Jijun TangView author publications Search author on:PubMedGoogle Scholar Corresponding author Correspondence to Jijun Tang. Additional information Competing interests The authors declare that they have no competing interests. Authors’ original submitted files for images Below are the links to the authors’ original submitted files for images. 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To view a copy of this licence, visit The Creative Commons Public Domain Dedication waiver ( applies to the data made available in this article, unless otherwise stated in a credit line to the data. Reprints and permissions About this article Cite this article Hu, F., Lin, Y. & Tang, J. MLGO: phylogeny reconstruction and ancestral inference from gene-order data. BMC Bioinformatics15, 354 (2014). Download citation Received: 16 July 2014 Accepted: 16 October 2014 Published: 08 November 2014 DOI: Share this article Anyone you share the following link with will be able to read this content: Get shareable link Sorry, a shareable link is not currently available for this article. Copy shareable link to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Keywords Phylogeny reconstruction Ancestral inference Genome rearrangement Maximum likelihood Download PDF Sections Figures References Abstract Background Implementation Results and discussion Conclusion Availability and requirements Endnote Authors' contributions References Acknowledgements Author information Additional information Authors’ original submitted files for images Rights and permissions About this article Advertisement Figure 1 View in articleFull size image Figure 2 View in articleFull size image Figure 3 View in articleFull size image Yancopoulos S, Attie O, Friedberg R: Efficient sorting of genomic permutations by translocation, inversion and block interchange. Bioinformatics. 2005, 21 (16): 3340-3346. 10.1093/bioinformatics/bti535. ArticlePubMedCASGoogle Scholar Blanchette M, Bourque G, Sankoff D: Breakpoint phylogenies. Genome Inform. 1997, 1997: 25-34. Google Scholar Moret B, Wang L, Warnow T, Wyman S: New approaches for reconstructing phylogenies from gene order data. Bioinformatics. 2001, 17 (suppl 1): 165-173. 10.1093/bioinformatics/17.suppl_1.S165. ArticleGoogle Scholar Bourque G, Pevzner P: Genome-scale evolution: reconstructing gene orders in the ancestral species. Genome Res. 2002, 12 (1): 26-36. PubMed CentralPubMedCASGoogle Scholar Lin Y, Rajan V, Moret BME: TIBA: a tool for phylogeny inference from rearrangement data with bootstrap analysis. Bioinformatics. 2012, 28 (24): 3324-3325. 10.1093/bioinformatics/bts603. ArticlePubMedCASGoogle Scholar Saitou N, Nei M: The neighbor-joining method: a new method for reconstructing phylogenetic trees. Mol Biol Evol. 1987, 4 (4): 406-425. PubMedCASGoogle Scholar Desper R, Gascuel O: Fast and accurate phylogeny reconstruction algorithms based on the minimum-evolution principle. J Comput Biol. 2002, 9 (5): 687-705. 10.1089/106652702761034136. ArticlePubMedCASGoogle Scholar Lin Y, Moret BME: Estimating true evolutionary distances under the DCJ model. Bioinformatics. 2008, 24 (13): i114-i122. 10.1093/bioinformatics/btn148. ArticlePubMed CentralPubMedCASGoogle Scholar Wang L-S, Jansen R, Moret BME, Raubeson L, Warnow T: Fast phylogenetic methods for the analysis of genome rearrangement data: an empirical study. Pacific Symposium on Biocomputing. Pacific Symposium on Biocomputing (PSB). 2001, World Scientific, Singapore, 524-535. Google Scholar Hu F, Gao N, Zhang M, Tang J: Maximum likelihood phylogenetic reconstruction using gene order encodings. Computational Intelligence in Bioinformatics and Computational Biology (CIBCB), 2011 IEEE Symposium On. 2011, IEEE, USA, 1-6. Google Scholar Lin Y, Hu F, Tang J, Moret BME: Maximum likelihood phylogenetic reconstruction from high-resolution whole-genome data and a tree of 68 eukaryotes. Proc. 18th Pacific Symp. on Biocomputing, (PSB). 2013, World Scientific, Singapore, 285-296. Google Scholar Ma J, Zhang L, Suh BB, Raney BJ, Burhans RC, Kent WJ, Blanchette M, Haussler D, Miller W: Reconstructing contiguous regions of an ancestral genome. Genome Res. 2006, 16 (12): 1557-1565. 10.1101/gr.5383506. ArticlePubMed CentralPubMedCASGoogle Scholar Ma J, Ratan A, Raney BJ, Suh BB, Zhang L, Miller W, Haussler D: Dupcar: reconstructing contiguous ancestral regions with duplications. J Comput Biol. 2008, 15 (8): 1007-1027. 10.1089/cmb.2008.0069. ArticlePubMed CentralPubMedCASGoogle Scholar Ma J: A probabilistic framework for inferring ancestral genomic orders. Bioinformatics and Biomedicine (BIBM), 2010 IEEE International Conference On. 2010, IEEE, USA, 179-184. ChapterGoogle Scholar Gagnon Y, Blanchette M, El-Mabrouk N: A flexible ancestral genome reconstruction method based on gapped adjacencies. BMC Bioinformatics. 2012, 13 (Suppl 19): 4- Google Scholar Bergeron A, Blanchette M, Chateau A, Chauve C: Reconstructing ancestral gene orders using conserved intervals. Proc. 4th Int'l Workshop Algs. in Bioinformatics (WABI'04). 2004, Springer, Germany, 14-25. Google Scholar Alekseyev MA, Pevzner PA: Breakpoint graphs and ancestral genome reconstructions. Genome Res. 2009, 19 (5): 943-957. 10.1101/gr.082784.108. ArticlePubMed CentralPubMedCASGoogle Scholar Murat F, Xu J-H, Tannier E, Abrouk M, Guilhot N, Pont C, Messing J, Salse J: Ancestral grass karyotype reconstruction unravels new mechanisms of genome shuffling as a source of plant evolution. Genome Res. 2010, 20 (11): 1545-1557. 10.1101/gr.109744.110. ArticlePubMed CentralPubMedCASGoogle Scholar Ouangraoua A, Tannier E, Chauve C: Reconstructing the architecture of the ancestral amniote genome. Bioinformatics. 2011, 27 (19): 2664-2671. 10.1093/bioinformatics/btr461. ArticlePubMedCASGoogle Scholar Lin CH, Zhao H, Lowcay SH, Shahab A, Bourque G: webmgr: an online tool for the multiple genome rearrangement problem. Bioinformatics. 2010, 26 (3): 408-410. 10.1093/bioinformatics/btp689. ArticlePubMedCASGoogle Scholar Hu F, Zhou J, Zhou L, Tang J: Probabilistic reconstruction of ancestral genomes with gene insertions and deletions. IEEE/ACM Trans Comput Biol Bioinformatics. 2014, 11 (4): 667-672. 10.1109/TCBB.2014.2309602. ArticleGoogle Scholar Sankoff D, Blanchette M: Probability models for genome rearrangement and linear invariants for phylogenetic inference. Proc. 3rd Int'l Conf. Comput. Mol. Biol. (RECOMB'99). 1999, ACM, USA, 302-309. Google Scholar Stamatakis A: Raxml-vi-hpc: maximum likelihood-based phylogenetic analyses with thousands of taxa and mixed models. Bioinformatics. 2006, 22 (21): 2688-2690. 10.1093/bioinformatics/btl446. ArticlePubMedCASGoogle Scholar Felsenstein J: Confidence limits on phylogenies: an approach using the bootstrap. Evol. 1985, 39: 783-791. 10.2307/2408678. ArticleGoogle Scholar Lin Y, Rajan V, Moret BME: Bootstrapping phylogenies inferred from rearrangement data. Proc. 11th Workshop Algs. in Bioinf. (WABI'11), Lecture Notes in Computer Science, Vol. 6833. 2011, Springer, Germany, 175-187. Google Scholar Yang Z, Kumar S, Nei M: A new method of inference of ancestral nucleotide and amino acid sequences. Genetics. 1995, 141 (4): 1641-1650. PubMed CentralPubMedCASGoogle Scholar Applegate D, Bixby R, Chvatal V, Cook W: Concorde tsp solver2006. [], [ Clark AG, Eisen MB, Smith DR, Bergman CM, Oliver B, Markow TA, Kaufman TC, Kellis M, Gelbart W, Iyer VN, Pollard DA, Sackton TB, Larracuente AM, Singh ND, Abad JP, Abt DN, Adryan B, Aguade M, Akashi H, Anderson WW, Aquadro CF, Ardell DH, Arguello R, Artieri CG, Barbash DA, Barker D, Barsanti P, Batterham P, Batzoglou S, et al: Evolution of genes and genomes on the drosophila phylogeny. Nature. 2007, 450 (7167): 203-218. 10.1038/nature06341. ArticlePubMedGoogle Scholar López MD, Samuelsson T: eGOB: eukaryotic gene order browser. Bioinformatics. 2011, 27 (8): 1150-1151. 10.1093/bioinformatics/btr075. ArticlePubMedGoogle Scholar Ponting CP: The functional repertoires of metazoan genomes. Nat Rev Genet. 2008, 9 (9): 689-698. 10.1038/nrg2413. ArticlePubMedCASGoogle Scholar Srivastava M, Begovic E, Chapman J, Putnam NH, Hellsten U, Kawashima T, Kuo A, Mitros T, Salamov A, Carpenter ML, Signorovitch AY, Moreno MA, Kamm K, Grimwood J, Schmutz J, Shapiro H, Grigoriev IV, Buss LW, Schierwater B, Dellaporta SL, Rokhsar DS: The trichoplax genome and the nature of placozoans. Nature. 2008, 454 (7207): 955-960. 10.1038/nature07191. ArticlePubMedCASGoogle Scholar Simillion C, Janssens K, Sterck L, Van de Peer Y: i-adhore 2.0: an improved tool to detect degenerated genomic homology using genomic profiles. Bioinformatics. 2008, 24 (1): 127-128. 10.1093/bioinformatics/btm449. ArticlePubMedCASGoogle Scholar Pham SK, Pevzner PA: Drimm-synteny: decomposing genomes into evolutionary conserved segments. Bioinformatics. 2010, 26 (20): 2509-2516. 10.1093/bioinformatics/btq465. ArticlePubMedCASGoogle Scholar Rödelsperger C, Dieterich C: Cyntenator: progressive gene order alignment of 17 vertebrate genomes. PloS one. 2010, 5 (1): 8861-10.1371/journal.pone.0008861. ArticleGoogle Scholar BMC Bioinformatics ISSN: 1471-2105 Contact us General enquiries: journalsubmissions@springernature.com Read more on our blogs Receive BMC newsletters Manage article alerts Language editing for authors Scientific editing for authors Policies Accessibility Press center Support and Contact Leave feedback Careers Follow BMC BMC Twitter page BMC Facebook page BMC Weibo page By using this website, you agree to our Terms and Conditions, Your US state privacy rights, Privacy statement and Cookies policy. Your privacy choices/Manage cookies we use in the preference centre. © 2025 BioMed Central Ltd unless otherwise stated. Part of Springer Nature.
14207	https://www.geeksforgeeks.org/chemistry/why-do-all-the-isotopes-of-an-element-have-similar-chemical-properties/	Why do all the Isotopes of an Element have similar Chemical Properties? Last Updated : 04 Mar, 2022 Suggest changes 1 Like The fundamental building units of matter are atoms and molecules. The existence of many types of matter is due to the various atoms that contain them. The atom was no longer seen as a simple, indivisible unit by 1900. It did, however, contain at least one subatomic particle, the electron. J.J. Thomson was the one who discovered it. E Goldstein discovered the presence of electrons in 1886. An electron is usually denoted by the letter e–, while a proton is denoted by the letter p+. The number of protons that a chemical element has in its nucleus is called its atomic number. For example- The atomic number of Hydrogen is 1. The total number of protons and neutrons in an atom is called its mass number. For example- the mass number of Carbon is 12 (6 protons and 6 neutrons.) What are Isotopes? Isotope is a variant of an element where it has an equal number of protons but a different number of neutrons of the atom due to which, the isotopes have different masses. Elements with an odd atomic numbers like Hydrogen, Lithium, Sodium or Nitrogen have one or two stable isotopes. Elements with even atomic number like Sulphur or Oxygen is more likely to have at least 3 stable isotopes. The exceptions here are carbon, helium, and beryllium. Properties of isotopes: Chemical Properties: The chemical properties are almost identical as different isotopes show almost identical chemical behaviours. Physical Properties: Physical properties like mass, melting or boiling point, density, and freezing point are different and depends on the mass of each isotope. Isotopes are either stable or radioactive. Radioactive isotopes are called radioisotopes or radionuclides. Isotopes that do not decay radioactively are called stable isotopes or stable nuclides. Isotopes of Hydrogen and Carbon Hydrogen has three stable isotopes called protium, deuterium, and tritium. All three of them have a same number of protons but a different number of neutrons. Deuterium has one neutron and tritium has two. Protium has zero neutrons while deuterium has one and tritium has two. Carbon has three isotopes namely Carbon-12 (stable isotope), Carbon-13, and Carbon-14 (radioactive isotope) where 12, 13, and 14 are the isotopes’ atomic masses. Isotopes and Nuclides A nuclide is a species of an atom with a specific number of protons and neutrons in the nucleus. E.g. carbon-13 with 6 protons and 7 neutrons. Nuclear qualities are prioritized over chemical properties in the nuclide idea. Chemical isotopes are prioritized over nuclear isotopes in the isotope idea. Even for the lightest elements, it usually has only a minor impact, though it can be significant in some cases. All the Isotopes of an Element Have Identical Chemical Properties All the Isotopes of an element have identical chemical properties because they have the same number of electrons as an atom of that element but they have different numbers of neutrons. The different number of neutrons affects the mass number. The mass number determines the physical properties while the atomic number determines the chemical properties. Therefore, isotopes have similar chemical properties but different physical properties. Radioisotopes, radionuclides, and radioactive nuclides are all terms used to describe radioactive isotopes. At least one radioactive isotope exists for each chemical element. Hydrogen, for example, has three isotopes with mass numbers 1, 2, and 3, with tritium being radioactive and the other two being stable. There are about 1,000 radioactive isotopes of the elements known, with 50 of them occurring naturally and the rest being created artificially as direct products of nuclear processes or as radioactive descendants of these products. The half-life of a substance is the time it takes for its radioactivity to decline to half of its original value. Applications of Isotopes Medical Field: Cobalt-60 is a radiation source used in medicine to slow the progression of cancer. Other radioactive isotopes are employed as tracers in metabolic research and for diagnostic purposes. In a breath test, carbon-14 is utilised to detect the ulcer-causing bacteria Helicobacter pylori. Industry: Radioactive isotopes are employed in industry to measure the thickness of metal or plastic sheets, with the strength of the radiations that penetrate the substance being investigated indicating the precise thickness. They can also be utilised as small electrical power sources. Plutonium-238 in spacecraft is an example. Sample Questions Question 1: Define mass number and atomic number. Answer: The number of protons that a chemical element has in its nucleus is called its atomic number. For example- The atomic number of Hydrogen is 1. The total number of protons and neutrons in an atom is called its mass number. For example- the mass number of Carbon is 12 (6 protons and 6 neutrons.) Question 2: What are isotopes and isobars? Answer: Isotope is a variant of an element where the variant will have equal number of protons but would differ in the number of neutrons of the atom. Isobars are atoms that have same number of nucleons. Isobars of different chemical elements have different atomic number but have the same mass number. Question 3: What are the isotopes of Hydrogen and Carbon? Answer: Hydrogen has three stable isotopes called protium, deuterium, and tritium. All three of them have same number of protons but a different number of neutrons. Deuterium has one neutron and tritium has two. Protium has zero neutrons while deuterium has one and tritium has two. Carbon has three isotopes namely Carbon-12 (stable isotope), Carbon-13, and Carbon-14 (radioactive isotope) where 12, 13, and 14 are the isotopes’ atomic masses. Question 4: What are the chemical and physical properties of isotopes? Answer: The chemical properties of isotopes of a given element are almost identical as different isotopes show almost identical chemical behaviors. Physical properties of an isotopes like mass, melting or boiling point, density, and freezing point are different and depends on the mass of each isotope. Question 5: Why do all the Isotopes of an Element Have Identical Chemical Properties? Answer: All the Isotopes of an element have identical chemical properties. The reason for this is because isotopes of an element have the same number of electrons as an atom of that element but they have different number of neutrons which affects the mass number. Mass number determines the physical properties while atomic number determines the chemical properties. Therefore, isotopes have similar chemical properties but different physical properties. P prateek sharma 7 Improve Article Tags : School Learning Class 9 Chemistry Chemistry-Class-9 Explore Basic Concepts Importance of Chemistry in Everyday Life 10 min readWhat is Matter ? 9 min readProperties of Matter 9 min readMeasurement Uncertainty 9 min readLaws of Chemical Combination 7 min readDalton's Atomic Theory 8 min readGram Atomic and Gram Molecular Mass 7 min readMole Concept 10 min readPercentage Composition - Definition, Formula, Examples 5 min readStoichiometry and Stoichiometric Calculations 7 min read Structure of Atom Composition of an Atom 8 min readAtomic Structure 15+ min readDevelopments Leading to Bohr's Model of Atom 6 min readBohr's Model of the Hydrogen Atom 9 min readQuantum Mechanical Atomic Model 8 min read Classification of Periodicity Classification of Elements 8 min readPeriodic Classification of Elements 10 min readModern Periodic Law 6 min read118 Elements and Their Symbols 9 min readElectronic Configuration in Periods and Groups 9 min readElectron Configuration 8 min readS Block Elements 9 min readPeriodic Table Trends 13 min read Bonding and Molecular Structure Chemical Bonding 12 min readIonic Bond 8 min readBond Parameters - Definition, Order, Angle, Length 7 min readVSEPR Theory 9 min readValence Bond Theory 7 min readHybridization 7 min readMolecular Orbital Theory 7 min readHydrogen Bonding 13 min read Thermodynamics Basics Concepts of Thermodynamics 12 min readApplications of First Law of Thermodynamics 8 min readInternal Energy as a State of System 8 min readEnthalpy Change of a Reaction 9 min readEnthalpies for Different Types of Reactions 10 min readWhat is Spontaneity? - Definition, Types, Gibbs Energy 7 min readGibbs Energy Change and Equilibrium 10 min read Equilibrium Equilibrium in Physical Processes 11 min readEquilibrium in Chemical Processes 7 min readLaw of Chemical Equilibrium and Equilibrium Constant 8 min readDifference between Homogeneous and Heterogeneous Equilibria 7 min readApplications of Equilibrium Constants 6 min readWhat is the Relation between Equilibrium Constant, Reaction Quotient and Gibbs Energy? 8 min readFactors Affecting Chemical Equilibrium 8 min readIonic Equilibrium 5 min readAcids, Bases and Salts 15+ min readIonization of Acids and Bases 6 min readBuffer Solution 10 min readSolubility Equilibria 5 min read Redox Reactions Redox Reactions 14 min readRedox Reactions in terms of Electron Transfer 4 min readOxidation Number \| Definition, How To Find, Examples 13 min readRedox Reactions and Electrode Processes 8 min read Basic Principles and Techniques Organic Chemistry - Some Basic Principles and Techniques 10 min readWhat is Catenation and Tetravalency? 6 min readStructural Representations of Organic Compounds 5 min readClassification of Organic Compounds 12 min readIUPAC Nomenclature of Organic Compounds 13 min readIsomerism 6 min readFundamental Concepts in Organic Reaction Mechanism 15+ min readPurification of Organic Compounds 5 min readQualitative Analysis of Organic Compounds 10 min readWhat is Quantitative Analysis? 9 min read Hydrocarbons What are Hydrocarbons? 11 min readClassification of Hydrocarbons 10 min readAlkanes - Definition, Nomenclature, Preparation, Properties 7 min readAlkenes - Definition, Nomenclature, Preparation, Properties 6 min readAlkynes - Definition, Structure, Preparation, Properties 8 min readAromatic Compounds 9 min read We use cookies to ensure you have the best browsing experience on our website. 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14208	https://artofproblemsolving.com/wiki/index.php/Sequence?srsltid=AfmBOopImfSHbesKdUI-TBqHFmfi1tkZi_IkGGI8fx5zYNcLwkinEmDG	Art of Problem Solving Sequence - 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 Sequence Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Sequence A sequence is an ordered list of terms. Sequences may be either finite or infinite. Contents 1 Definition 2 Convergence 3 Monotone Sequences 4 Resources 5 See Also Definition A sequence of real numbers is simply a function . For instance, the function defined on corresponds to the sequence . Convergence Intuitively, a sequence converges if its terms approach a particular number. Formally, a sequence of reals converges to if and only if for all positive reals , there exists a positive integer such that for all integers , we have . If converges to , is called the limit of and is written . The statement that converges to can be written as . A classic example of convergence would be to show that as . Claim: . Proof: Let be arbitrary and choose . Then for we see that which proves that , so as Monotone Sequences Many significant sequences have their terms continually increasing, such as , or continually decreasing, such as . This motivates the following definitions: A sequence of reals is said to be increasing if for all and strictly increasing if for all , decreasing if for all and strictly decreasing if for all , monotone if it is either decreasing or increasing. Resources Online Encyclopedia of Integer Sequences See Also Arithmetic sequence Geometric sequence Bolzano-Weierstrass Theorem This article is a stub. Help us out by expanding it. Retrieved from " Category: Stubs 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.
14209	https://flexbooks.ck12.org/cbook/ck-12-middle-school-math-concepts-grade-8/section/6.14/primary/lesson/recognizing-translation-transformations-msm8/	Skip to content Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 6.14 Recognizing Translation Transformations Written by:Brenda Meery \| Jen Kershaw, M.ed Fact-checked by:The CK-12 Editorial Team Last Modified: Aug 01, 2025 Have you ever seen an image like this? This diagram represents a transformation. Do you know which one? In this concept, you will learn to recognize translation transformations. Translation A transformation is the movement of a geometric figure on the coordinate plane. There are several different types of transformations. One of these is called a translation. A translation is when a geometric figure slides up, down, left or right on the coordinate plane. The figure moves its location, but doesn’t change its orientation. It also doesn’t change its size or shape. When you perform translations, you slide a figure left or right, up or down. This means that, in the coordinate plane, the coordinates for the vertices of the figure will change. You can represent this triangle by using coordinate notation. Coordinate notation is when you write ordered pairs to represent each of the vertices of a geometric figure that has been graphed on the coordinate plane. These are the coordinates of the vertices of the triangle. If you slide this triangle 3 places down, all of its vertices will shift 3 places down the -axis. That means that the ordered pairs for the new vertices will change. Specifically, the -coordinate in each pair will decrease by 3. Now the -coordinate of each ordered pair decreased by three units. You can see how the ordered pairs changed from the first image to the next image. The -coordinate changed from 5 to 2, from -1 to 2 and from 2 to -1. As you move down, the value of the coordinate also moved down. Similarly, if you were to move the image up three units on the -axis, then you would increase the value of the -coordinate by three units. Also if you were to move the image to the right then you would increase the -coordinate. If you were to move it to the left, then you would decrease the -coordinate. You can translate figures in other ways too. You can move figures diagonally by changing both their - and -coordinates. One way to recognize translations, then, is to compare their points. The -coordinates will all change the same way, and the -coordinates will all change the same way. To graph a translation, you perform the same change for each point. Now let’s try graphing a translation. Take a look at this one. Graph the following translation five places to the right. First, you can see by looking at this square that there are four vertices, so there are four sets of ordered pairs to represent these vertices. List the ordered pairs. Next, the translation is to move the square five places to the right. That means that you are going to change the -coordinate and not the -coordinate. List the new ordered pairs. Then, let’s look at the graph of the translation. This is the answer. Notice that while it is helpful to graph the square both first and then as a translation, it isn’t necessary to do so to figure out the coordinate notation. If you know the vertices of the figure that you are translating and you know how you are moving it, then you can figure out the new coordinates of the vertices. Examples Example 1 Looking at this diagram, you can see that the figure, a quadrilateral has been shifted to the right and then up. It has not been flipped or turned. It has been moved, so this is a slide. Another name for a slide is a translation. The answer is that this transformation is a slide. Example 2 A triangle with the coordinates (0, 2), (2, 2) and (2, 5) is graphed on the coordinate grid. Find the coordinates of a translation moved three units down. Then graph the translation. First, notice the direction of the translation. It is to move the triangle three units down. Down means that you will be subtracting three and down also means that you will be changing the -coordinate since up and down involves the -axis. List the ordered pairs for the translated triangle. Next, graph the translation. This is the answer. Use this diagram to answer the following questions. Example 3 Is this figure a translation? Look at and . What is happening to the triangles? The vertices of have moved to the right 6 units and down 4 units. is the translation of . The answer is that yes, this is a translation. Remember a translation is a transformation that is informally called a slide or a glide. Example 4 How many units up or down has the figure been moved? First, look at the vertices in the triangles to compare. Point has coordinates (-2, 8) and Point has coordinates (4, 4). All other vertices change in the same way. Next, look at the change in the coordinates. Asking if the figure means up or down is the same as asking what the change in the -coordinate is The answer is that the figure moves 4 units down or -4. Example 5 How many units to the right or left? First, look at the vertices in the triangles to compare. Point has coordinates (-2, 8) and Point has coordinates (4, 4). All other vertices change in the same way. Next, look at the change in the coordinates. Asking if the figure means left or right is the same as asking what the change in the -coordinate is The answer is that the figure moves 6 units to the right or +6. Review Use the following diagram to answer each question. What kind of transformation is shown in the diagram? What are the coordinates of the original blue triangle? What are the coordinates of the translated red triangle? What direction was the triangle first moved? By how many units horizontally? In what vertical direction was the triangle translated? By how many units vertically? What are the coordinates of the original blue triangle? What are the coordinates of the translated red triangle? In what horizontal direction was the triangle translated? By how many units horizontally? In what vertical direction was the triangle translated? By how many units vertically? True or false. Another name for a slide is a translation. True or false. A rotation and a translation have the same characteristics. Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. 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14210	https://mathworld.wolfram.com/ComplexNumber.html	Complex Number Download Wolfram Notebook The complex numbers are the field of numbers of the form , where and are real numbers and i is the imaginary unit equal to the square root of , . When a single letter is used to denote a complex number, it is sometimes called an "affix." In component notation, can be written . The field of complex numbers includes the field of real numbers as a subfield. The set of complex numbers is implemented in the Wolfram Language as Complexes. A number can then be tested to see if it is complex using the command Element[x, Complexes], and expressions that are complex numbers have the Head of Complex. Complex numbers are useful abstract quantities that can be used in calculations and result in physically meaningful solutions. However, recognition of this fact is one that took a long time for mathematicians to accept. For example, John Wallis wrote, "These Imaginary Quantities (as they are commonly called) arising from the Supposed Root of a Negative Square (when they happen) are reputed to imply that the Case proposed is Impossible" (Wells 1986, p. 22). Through the Euler formula, a complex number \| \| \| (1) \| may be written in "phasor" form \| \| \| (2) \| Here, is known as the complex modulus (or sometimes the complex norm) and is known as the complex argument or phase. The plot above shows what is known as an Argand diagram of the point , where the dashed circle represents the complex modulus of and the angle represents its complex argument. Historically, the geometric representation of a complex number as simply a point in the plane was important because it made the whole idea of a complex number more acceptable. In particular, "imaginary" numbers became accepted partly through their visualization. Unlike real numbers, complex numbers do not have a natural ordering, so there is no analog of complex-valued inequalities. This property is not so surprising however when they are viewed as being elements in the complex plane, since points in a plane also lack a natural ordering. The absolute square of is defined by , with the complex conjugate, and the argument may be computed from \| \| \| (3) \| The real and imaginary parts are given by \| \| \| (4) \| \| (5) \| \| (6) \| \| (7) \| de Moivre's identity relates powers of complex numbers for real by \| \| \| (8) \| A power of complex number to a positive integer exponent can be written in closed form as \| \| \| (9) \| The first few are explicitly \| \| \| (10) \| \| (11) \| \| (12) \| \| (13) \| (Abramowitz and Stegun 1972). Complex addition \| \| \| (14) \| complex subtraction \| \| \| (15) \| complex multiplication \| \| \| (16) \| and complex division \| \| \| (17) \| can also be defined for complex numbers. Complex numbers may also be taken to complex powers. For example, complex exponentiation obeys where is the complex argument. See also Absolute Square, Argand Diagram, Complex Argument, Complex Division, Complex Exponentiation, Complex Modulus, Complex Multiplication, Complex Plane, Complex Subtraction, i, Imaginary Number, Phase, Phasor, Real Number, Surreal Number Explore this topic in the MathWorld classroom Explore with Wolfram\|Alpha More things to try: complex number i complex number i References Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 16-17, 1972.Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 353-357, 1985.Bold, B. "Complex Numbers." Ch. 3 in Famous Problems of Geometry and How to Solve Them. New York: Dover, pp. 19-27, 1982.Courant, R. and Robbins, H. "Complex Numbers." §2.5 in What Is Mathematics?: An Elementary Approach to Ideas and Methods, 2nd ed. Oxford, England: Oxford University Press, pp. 88-103, 1996.Ebbinghaus, H. D.; Hirzebruch, F.; Hermes, H.; Prestel, A; Koecher, M.; Mainzer, M.; and Remmert, R. Numbers. New York: Springer-Verlag, 1990.Krantz, S. G. "Complex Arithmetic." §1.1 in Handbook of Complex Variables. Boston, MA: Birkhäuser, pp. 1-7, 1999.Mazur, B. Imagining Numbers (Particularly the Square Root of Minus Fifteen). Farrar, Straus and Giroux, 2003.Morse, P. M. and Feshbach, H. "Complex Numbers and Variables." §4.1 in Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 349-356, 1953.Nahin, P. J. An Imaginary Tale: The Story of -1. Princeton, NJ: Princeton University Press, 2007.Press, W. H.; Flannery, B. P.; Teukolsky, S. A.; and Vetterling, W. T. "Complex Arithmetic." §5.4 in Numerical Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed. Cambridge, England: Cambridge University Press, pp. 171-172, 1992.Wells, D. The Penguin Dictionary of Curious and Interesting Numbers. Middlesex, England: Penguin Books, pp. 21-23, 1986.Wolfram, S. A New Kind of Science. Champaign, IL: Wolfram Media, p. 1168, 2002. Referenced on Wolfram\|Alpha Complex Number Cite this as: Weisstein, Eric W. "Complex Number." From MathWorld--A Wolfram Resource. Subject classifications
14211	https://math-physics-problems.fandom.com/wiki/Electric_Field_of_a_Charged_Disk	Skip to content Math & Physics Problems Wiki 531 pages in: Electromagnetism and Optics, Electricity and Magnetism, Electromagnetism, Physics Electric Field of a Charged Disk Sign in to edit History Purge Talk (0) Problem[] Find the electric ﬁeld a distance along the axis from a disc of radius and uniform charge density Hint[] (1) Think of this disk as a bunch of concentric rings. Begin by ﬁnding the electric field a distance along the axis up from a thin ring of charge and radius (2) You may use the integral Solution[] Consider the ring problem first. Due to the symmetry of this geometry, there is a cancellation effect in the y-direction. Therefore, all contributions to the electric field in the x-direction. The distance of a point on the x-axis from the ring is Hence, the differential element of electric field (along the x-axis) is or Integrating the differential yields: For the disk problem, replace the differential charge with because the areal charge density is defined as and for a disk. Since we are integrating with respect to the variable now, the variable radii is replaced with . Therefore, the electric field is modified to Using the given integral yields, Therefore, the electric ﬁeld a distance along the axis from a disc of radius and uniform charge density is Categories Community content is available under CC-BY-SA unless otherwise noted. Comments Start a conversation Sign in to share your thoughts and get the conversation going. SIGN IN Don't have account? Register now Search this wiki Search all wikis Do Not Sell My Personal Information When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings to find out more and change our default settings according to your preference. You cannot opt-out of our First Party Strictly Necessary Cookies as they are deployed in order to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more information about the First and Third Party Cookies used please follow this link. Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Targeting Cookies These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising. Social Media Cookies These cookies are set by a range of social media services that we have added to the site to enable you to share our content with your friends and networks. They are capable of tracking your browser across other sites and building up a profile of your interests. This may impact the content and messages you see on other websites you visit. If you do not allow these cookies you may not be able to use or see these sharing tools.
14212	https://macaulay2.com/doc/Macaulay2/share/doc/Macaulay2/Parametrization/html/index.html	Parametrization -- Rational parametrization of rational curves and related computations Macaulay2 » Documentation Packages » Parametrization :: Parametrization next \| previous \| forward \| backward \| up \| index \| toc Parametrization -- Rational parametrization of rational curves and related computations Description Overview: Parametrization is a package to compute rational parametrizations of rational curves defined over \mathbb{Q}. Suppose C is a rational plane curve C of degree n defined over \mathbb{Q}. We use the package AdjointIdeal to compute the adjoint ideal of C. (The package exports also all functions available in AdjointIdeal, e.g., geometricGenus.) The corresponding linear system maps the curve birationally to a rational normal curve in \mathbb{P}^{n-2}. Iterating the anticanonical map we give a projection of the rational normal curve to \mathbb{P}^{1} for n odd or to a conic C_2 in \mathbb{P}^{2} for n even. In the case that n is even we test for the existence of a rational point on the conic and if so give a rational parametrization of the conic. By inverting the birational map of C to \mathbb{P}^{1} or the conic we obtain a rational parametrization of C. If n is odd or C_2 has a rational point C is parametrized by \mathbb{P}^{1} otherwise by C_2. The main focus of the algorithm is to avoid unnecessary choices to obtain a parametrization of small height. For more theoretical details see J. Boehm: Rational parametrization of rational curves, The package is work in progress, so there will be future improvements and more testing is necessary. Key user functions: parametrize -- This is the universal rational parametrization function, it works for plane rational curves, in particular conics, and rational normal curves. testParametrization -- Test a parametrization. rationalPointOnConic -- Test for a rational point on a conic and find it if it exists. mapToRNC -- Map a plane rational curve to a rational normal curve. Setup: This package uses the package AdjointIdeal, so set up this first. Place the file Parametrization.m2 somewhere into the M2 search path and install the package by doing installPackage("Parametrization") Author Janko Boehm<boehm@mathematik.uni-kl.de> Version This documentation describes version 0.6 of Parametrization. Citation If you have used this package in your research, please cite it as follows: bib @misc{ParametrizationSource, title = {{Parametrization: rational parametrization of rational plane curves and related computations. Version~0.6}}, author = {Janko Boehm}, howpublished = {A \emph{Macaulay2} package available at \url{ } Exports Functions and commands chineseRemainder -- Solve simultaneous congruences. invertBirationalMap -- Computes the inverse of a birational map. isomorphicProjectionOfRNC -- Parametrizing linear system of a rational normal curve. legendreSymbol -- Compute the Legendresymbol. mapToRNC -- Map plane rational curve to rational normal curve. modularInverse -- Compute the inverse in Z/nZ. modularPower -- Compute the modular power. modularSquareRoot -- Compute the modular square root. parametrize -- Rational parametrization of rational curves. rationalPointOnConic -- Rational point on a conic. rParametrizeConic -- Compute a rational parametrization of a conic. rParametrizePlaneCurve -- Rational parametrization of rational plane curves. rParametrizeRNC -- Compute a rational parametrization of a rational normal curve. testParametrization -- Test if parametrization. Methods chineseRemainder(List,List) -- see chineseRemainder -- Solve simultaneous congruences. invertBirationalMap(Ideal,Matrix) -- see invertBirationalMap -- Computes the inverse of a birational map. isomorphicProjectionOfRNC(Ideal) -- see isomorphicProjectionOfRNC -- Parametrizing linear system of a rational normal curve. legendreSymbol(ZZ,ZZ) -- see legendreSymbol -- Compute the Legendresymbol. mapToRNC(Ideal) -- see mapToRNC -- Map plane rational curve to rational normal curve. mapToRNC(Ideal,Ideal) -- see mapToRNC -- Map plane rational curve to rational normal curve. mapToRNC(Ideal,Matrix) -- see mapToRNC -- Map plane rational curve to rational normal curve. modularInverse(ZZ,ZZ) -- see modularInverse -- Compute the inverse in Z/nZ. modularPower(ZZ,ZZ,ZZ) -- see modularPower -- Compute the modular power. modularSquareRoot(ZZ,ZZ) -- see modularSquareRoot -- Compute the modular square root. parametrize(Ideal) -- see parametrize -- Rational parametrization of rational curves. parametrize(Ideal,Ideal) -- see parametrize -- Rational parametrization of rational curves. rationalPointOnConic(Ideal) -- see rationalPointOnConic -- Rational point on a conic. rParametrizeConic(Ideal) -- see rParametrizeConic -- Compute a rational parametrization of a conic. rParametrizePlaneCurve(Ideal,Ideal) -- see rParametrizePlaneCurve -- Rational parametrization of rational plane curves. rParametrizeRNC(Ideal) -- see rParametrizeRNC -- Compute a rational parametrization of a rational normal curve. testParametrization(Ideal,Matrix) -- see testParametrization -- Test if parametrization. Symbols parametrizeConic -- Option whether to rationally parametrize conics. vb -- Option whether to print intermediate results. For the programmer The object Parametrization is a package, defined in Parametrization.m2, with auxiliary files in Parametrization/. The source of this document is in Parametrization.m2:1167:0.
14213	https://devguide.tech/docs/data-structure-and-algorithm/algorithmic-complexity/horner-method	数据结构与算法-算法复杂度分析-多项式与霍纳法则 \| 开发指南跳到主要内容开发指南博客Java数据结构与算法概述算法复杂度分析大O符号与渐近分析什么是算法？算法复杂度多项式与霍纳法则数据结构算法算法复杂度分析多项式与霍纳法则本页总览多项式与霍纳法则多项式求值 P(x)=a n x n+a n−1 x n−1+a n−2 x n−2+...+a 3 x 3+a 2 x 2+a 1 x P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{3}x^{3}+a_{2}x^{2}+a_{1}x P(x)=a nx n+a n−1x n−1+a n−2x n−2+...+a 3x 3+a 2x 2+a 1x，在多项式的计算中通常乘法比加法慢。因此，乘法是主导的基本操作。假设有一个次数为 n n n 的多项式，并且各项系数均不为 0。常规解法使用常规解法就必须求下列各幂的值。 x n,x n−1,x n+2,x n−3,...,x 3,x 2,x 1 x^n,\;x^{n-1},\;x^{n+2},\;x^{n-3},\;...\;,\;x^3,\;x^2,\;x^1 x n,x n−1,x n+2,x n−3,...,x 3,x 2,x 1 第一项执行 n−1 n-1 n−1 次乘法，第二项是 n−2 n-2 n−2 以此类推，执行乘法的总数是 (n−1)+(n−2)+(n−3)+...+3+2+1=n(n−1)2=n 2 2−n 2(n-1)+ (n-2) + (n-3) + ... + 3 + 2 + 1 = \frac{n(n-1)}{2} = \frac{n^2}{2} - \frac{n}{2} (n−1)+(n−2)+(n−3)+...+3+2+1=2 n(n−1)=2 n 2−2 n 除常数外，每项需要与系数的一次乘法，总共是 n n n 次，因此总的乘法次数是 n 2 2−n 2+n=n 2 2+n 2\frac{n^2}{2} - \frac{n}{2} + n = \frac{n^2}{2} + \frac{n}{2} 2 n 2−2 n+n=2 n 2+2 n 根据上面的步骤推导出计算多项式的算法的复杂度为 O(n 2)O(n^2)O(n 2) 使用霍纳法则（Horner's method）霍纳法则（Horner's method）是一种高效的多项式计算方法，用于在给定点快速计算多项式的值。它将多项式计算的时间复杂度从 O(n 2)O(n^2)O(n 2) 降低到 O(n)O(n)O(n)。霍纳法则将其重写为嵌套形式： P(x)=a 0+x(a 1+x(a 2+⋯+x(a n−1+x⋅a n)…))P(x) = a_0 + x(a_1 + x(a_2 + \dots + x(a_{n-1} + x\cdot a_n) \dots ))P(x)=a 0+x(a 1+x(a 2+⋯+x(a n−1+x⋅a n)…)) 信息该算法最早由南宋数学家秦九韶创造，所以又称秦九韶算法。准确来说就应该称为秦九韶算法南宋数学家秦九韶将贾宪的增乘开方术推广，以求解任意高次方程的实数根的数值解。秦九韶的《数书九章》详细叙述用秦九韶算法求解二十六个二次到十次方程的的实数根的数值解，其中包含二十个二次方程，一个三次方程，四个四次方程和一个十次方程。其中有些得到精确解；多数得近似解。 19世纪初，英国数学家威廉·乔治·霍纳重新发现并证明，后世称作霍纳算法（Horner's method、Horner scheme）。但是，19世纪英国传教士伟烈亚力 Alexander Wylie. (1815–1887) 最早对霍纳的发明权提出质疑。此后，日本数学史家三上义夫在《中日数学史》一书中在详述秦九韶的正负开方术后写道：“谁能否认，霍纳的辉煌方法，至少在早于欧洲六百年之前，已经在中国运用了。” 霍纳在1819年发表的《解所有次方程》论文中的算例，其算法程序和数字处理都远不及五百多年前的秦九韶有条理；秦九韶算法不仅在时间上早于霍纳，也比较成熟。霍纳法则的步骤从最高次项开始，将多项式的系数依次乘以 x x x 并累加。重复上述步骤，直到处理完所有系数。示例假设我们要计算多项式 P(x)=2 x 3+3 x 2+4 x+5 P(x) = 2x^3 + 3x^2 + 4x + 5 P(x)=2 x 3+3 x 2+4 x+5 在 x=2 x=2 x=2 处的值。可以将多项式写为嵌套形式：P(x)=5+x(4+x(3+x⋅2))P(x) = 5 + x(4 + x(3 + x\cdot 2))P(x)=5+x(4+x(3+x⋅2)) 按照霍纳法则计算：步骤 1: 结果 = 2 × 2 + 3 = 7步骤 2: 结果 = 7 × 2 + 4 = 18步骤 3: 结果 = 18 × 2 + 5 = 41 所以 P(2)=41 P(2) = 41 P(2)=41 使用霍纳法则重写多项式后，只需要进行 3 次乘法运算。参考资料 Mark Allen Weiss. 《数据结构与算法分析（Java语言描述符）》. 机械工业出版社. 2008: P22-P42 维基百科. 秦九韶算法. 维基百科. 增乘开立方术. 标签：数据结构算法上一页算法复杂度下一页数据结构常规解法使用霍纳法则（Horner's method）参考资料 Docs Java Community Stack Overflow Discord Twitter More Blog GitHub Copyright © 2024 DEVGUIDE.TECH, Inc. Built with Docusaurus. 本站所有作品均采用CC BY-NC-ND 4.0许可协议，未经允许，禁止用于商业用途，转载需注明出处。
14214	https://en.wikipedia.org/wiki/List_of_date_formats_by_country	Jump to content Search Contents (Top) 1 Usage map 2 Listing 2.1 Table coding 3 See also 4 References 5 External links List of date formats by country 한국어 Italiano 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 \| \| \| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "List of date formats by country" – news · newspapers · books · scholar · JSTOR (February 2017) (Learn how and when to remove this message) \| The legal and cultural expectations for date and time representation vary between countries, and it is important to be aware of the forms of all-numeric calendar dates used in a particular country to know what date is intended. Writers have traditionally written abbreviated dates according to their local custom, creating all-numeric equivalents to day–month formats such as "26 September 2025" (26/09/25, 26/09/2025, 26-09-2025 or 26.09.2025) and month–day formats such as "September 26, 2025" (09/26/25 or 09/26/2025). This can result in dates that are impossible to understand correctly without knowing the context. For instance, depending on the order style, the abbreviated date "01/11/06" can be interpreted as "1 November 2006" for DMY, "January 11, 2006" for MDY, and "2001 November 6" for YMD. The ISO 8601 format YYYY-MM-DD (2025-09-26) is intended to harmonize these formats and ensure accuracy in all situations. Many countries have adopted it as their sole official date format, though even in these areas writers may adopt abbreviated formats that are no longer recommended. The Unicode CLDR (Common Locale Data Repository) Project is the world's largest repository documenting a wide variety of time and date representations for different countries and language groups. Usage map [edit] \| Colour \| Order styles \| Main regions and countries \| --- \| \| DMY \| Europe: Italy, Netherlands, Turkey, Ireland, etc.North America: Mexico, various Caribbean islandsCentral America: Guatemala, Honduras, Panama, etc.South America: Brazil, Colombia, Chile, Argentina, Peru, Venezuela, etc.North Africa: Egypt, Algeria, Morocco, Tunisia, etc.East Africa: SomaliaWest, Central, and Southern Africa: Nigeria, Ethiopia, DRC, Tanzania, Sudan, Uganda, South Africa, etc.West Asia: Iraq, Saudi Arabia, Yemen, etc.Central Asia: Tajikistan, Kyrgyzstan, TurkmenistanEast and Southeast Asia: Indonesia, Thailand, Cambodia, etc.South Asia: Pakistan, BangladeshOceania: Papua New Guinea, New Zealand, etc.Middle East: United Arab Emirates, Qatar, Oman, Yemen, Saudi Arabia, Kuwait, Iraq, Egypt, Syria, Lebanon, Israel, Jordan \| \| \| YMD \| China, Japan, South Korea, Taiwan, Hungary, Mongolia, Lithuania, Bhutan \| \| \| MDY \| Some US island territories \| \| \| DMY, YMD \| India, Russia, Vietnam, Germany, Iran, United Kingdom, France, Myanmar, Spain, Poland, Uzbekistan, Afghanistan, Nepal, Australia, Cameroon, Sri Lanka, Malaysia, Singapore, etc. \| \| \| DMY, MDY \| Philippines, Togo, Puerto Rico, Cayman Islands, Greenland \| \| \| MDY, YMD \| United States \| \| \| MDY, DMY, YMD \| Kenya, Canada, Ghana \| Listing [edit] Table coding [edit] : All examples use example date 2021-03-31 / 2021 March 31 / 31 March 2021 / March 31, 2021 – except where a single-digit day is illustrated. Basic components of a calendar date for the most common calendar systems: D – day M – month Y – year Specific formats for the basic components: yy – two-digit year, e.g. 24 yyyy – four-digit year, e.g. 2024 m – one-digit month for months below 10, e.g. 3 mm – two-digit month, e.g. 03 mmm – three-letter abbreviation for month, e.g. Mar mmmm – month spelled out in full, e.g. March d – one-digit day of the month for days below 10, e.g. 2 dd – two-digit day of the month, e.g. 02 ddd – three-letter abbreviation for day of the week, e.g. Fri dddd – day of the week spelled out in full, e.g. Friday Separators of the components: / – oblique stroke (slash) . – full stop, dot or point (period) – hyphen (dash) – space \| Country \| All-numeric date format \| Details \| Official standard \| --- --- \| \| YMD \| DMY \| MDY \| \| Afghanistan \| Yes \| Yes \| No \| Short format: d/m/yyyy (Year first, month, and day in right-to-left writing direction)Long format: yyyy mmmm d (Day first, full month name, and year in right-to-left writing direction \| \| \| Åland \| Yes \| Yes \| No \| Short format: yyyy-mm-ddLong format: d mmmm yyyy \| \| \| Albania \| Yes \| Yes \| No \| dd/mm/yyyySome YMD \| \| \| Algeria \| No \| Yes \| No \| (dd/mm/yyyy) \| \| \| American Samoa \| No \| No \| Yes \| (mm/dd/yy) \| \| \| Andorra \| No \| Yes \| No \| \| \| \| Angola \| No \| Yes \| No \| \| \| \| Anguilla \| No \| Yes \| No \| \| \| \| Antigua and Barbuda \| No \| Yes \| No \| \| \| \| Argentina \| Sometimes \| Yes \| No \| Numeric format: yyyyMMdd (Example: 20030613)Short format: dd/mm/yy (Example: 13/06/03)Medium format: dd/mm/yyyy (Example: 13/06/2003)Long format: d' de 'mmmm' de 'yyyy (Example: 13 de junio de 2003)Full format: dddd d' de 'mmmm' de 'yyyy (Example: viernes 13 de junio de 2003). \| \| \| Armenia \| No \| Yes \| No \| (dd.mm.yyyy) \| \| \| Aruba \| No \| Yes \| No \| \| \| \| Australia \| Yes \| Yes \| No \| mmmm d, yyyy is sometimes used, usually informally in the mastheads of magazines and newspapers, and in advertisements, video games, news, and TV shows, especially those emanating from the United States. MDY in numeric-only form is never used.The ISO 8601 date format (2025-09-26) is the recommended short date format for government publications. \| AS/NZS ISO 8601.1:2021 \| \| Austria \| Yes \| Yes \| No \| (Using dots (which denote ordinal numbering) as in d.m.(yy)yy or sometimes d. month (yy)yy). \| ÖNORM ISO 8601 \| \| Azerbaijan \| No \| Yes \| No \| (dd.mm.yyyy) \| \| \| The Bahamas \| No \| Yes \| No \| [citation needed] \| \| \| Bahrain \| No \| Yes \| No \| \| \| \| Bangladesh \| No \| Yes \| No \| Not officially standardized. Bengali calendar dates are also used: দদ-মম-বববব \| \| \| Barbados \| No \| Yes \| No \| \| BNS 50:2000 \| \| Belarus \| No \| Yes \| No \| (dd.mm.yyyy) \| \| \| Belgium \| No \| Yes \| No \| (dd/mm/yyyy) or (dd.mm.yyyy) \| NBN Z 01-002 \| \| Belize \| No \| Yes \| No \| \| \| \| Benin \| No \| Yes \| No \| \| \| \| Bermuda \| No \| Yes \| No \| \| \| \| Bhutan \| Yes \| No \| No \| \| \| \| Bolivia \| No \| Yes \| No \| \| \| \| Bonaire \| No \| Yes \| No \| \| \| \| Bosnia and Herzegovina \| No \| Yes \| No \| (d. m. yyyy. or d. mmmm yyyy.) \| \| \| Botswana \| Yes \| Yes \| No \| yyyy-mm-dd for Setswana and dd/mm/yyyy for English \| \| \| Brazil \| No \| Yes \| No \| (dd/mm/yyyy) or (dd.mm.yyyy) \| NBR 5892:2019 \| \| British Indian Ocean Territory \| No \| Yes \| No \| \| \| \| British Virgin Islands \| No \| Yes \| No \| \| \| \| Brunei \| No \| Yes \| No \| \| \| \| Bulgaria \| No \| Yes \| No \| (dd.mm.yyyy) \| \| \| Burkina Faso \| No \| Yes \| No \| \| \| \| Burundi \| No \| Yes \| No \| \| \| \| Cambodia \| No \| Yes \| No \| Short format: dd/mm/yyLong format: d mmmm yyyy \| \| \| Cameroon \| Yes \| Yes \| No \| (d)d/(m)m/yyyy or d mmmm yyyy for Aghem, Bafia, Basaa, Duala, English, Ewondo, French, Fula, Kako, Kwasio, Mundang, Ngiemboon and Yangbenyyyy-mm-dd for Meta' and Ngomba \| \| \| Canada \| Yes \| Yes \| Yes \| ISO 8601 is the only format that the Government of Canada and Standards Council of Canada officially recommend for all-numeric dates. However, usage differs with context.All three long forms are used in Canada.For English speakers, MDY (mmmm-dd-yyyy) (example: April 9, 2019) is used by many English-language publications and media company products as well as the majority of government documents written in English.For French and English speakers, DMY (dd-mmmm-yyyy) is used (example: 9 April 2019/le 9 avril 2019). This form is used in formal letters, academic papers, military, many media companies and some government documents, particularly in French-language ones.Federal regulations for shelf life dates on perishable goods mandate a year/month/day format, but allow the month to be written in full, in both official languages, or with a set of standardized two-letter bilingual codes such as 2019 AL 09 or 19 AL 09. \| CAN/CSA-Z234.4-89 (R2007) \| \| Cape Verde \| No \| Yes \| No \| \| \| \| Cayman Islands \| No \| Yes \| Yes \| DMY and MDY are used interchangeably. Official forms generally tend towards DMY. Month is often spelled out to avoid confusion.[citation needed] \| \| \| Central African Republic \| No \| Yes \| No \| \| \| \| Chad \| No \| Yes \| No \| \| \| \| Chile \| No \| Yes \| No \| In Chile the format dd/mm/yyyy is used only, or you can also say "3 June 2023" or in Spanish "3 de junio del 2023"You can also use the short format, example "03/06/23". \| \| \| China \| Yes \| No \| No \| National standard format is yyyy-mm-dd (with leading zeroes) and (yy)yy年(m)m月(d)d日 (with or without leading zeroes)Uyghur languages in Xinjiang usually give date examples in the form 2017-يىل 18-ئاۋغۇست or 2017-8-18 (i.e. yyyy-d-mmm) but this form is never used when writing in Chinese; casually many people use (yy)yy/(m)m/(d)d or (yy)yy.(m)m.(d)d (with or without leading zeroes). See Dates in Chinese. \| GB/T 7408.1-2023 \| \| Christmas Island \| Yes \| Yes \| No \| \| \| \| Cocos (Keeling) Islands \| Yes \| Yes \| No \| \| \| \| Colombia \| No \| Yes \| No \| \| \| \| Comoros \| No \| Yes \| No \| \| \| \| Congo (East and West) \| No \| Yes \| No \| \| \| \| Cook Islands \| No \| Yes \| No \| \| \| \| Costa Rica \| No \| Yes \| No \| \| \| \| Croatia \| No \| Yes \| No \| (d. m. yyyy. or d. mmmm yyyy.) See Date and time notation in Croatia for details on cases used. \| \| \| Cuba \| Yes \| Yes \| No \| \| \| \| Curaçao \| No \| Yes \| No \| \| \| \| Cyprus \| No \| Yes \| No \| dd/mm/yyyy \| \| \| Czech Republic \| Yes \| Yes \| No \| (d. m. yyyy or d. month yyyy) \| ČSN ISO 8601 \| \| Denmark \| Yes \| Yes \| No \| Examples: Long date: 7. juni 1994. Long date with weekday: onsdag(,) den 21. december 1994. Numeric date: 1994-06-0749yy is the traditional Danish date format. The international format yyyy-mm-dd or yyyymmdd is also accepted, though this format is not commonly used. The formats d. 'month name' yyyy and in handwriting d/m-yy or d/m yyyy are also acceptable.) \| DS/ISO 8601:2005 \| \| Djibouti \| Yes \| Yes \| No \| Short format: dd/mm/yyyy (Day first, month number and year in left-to-right writing direction) in Afar, French and Somali ("d/m/yy" is a common alternative). Gregorian dates follow the same rules but tend to be written in the yyyy/m/d format (Day first, month number, and year in right-to-left writing direction) in Arabic language.Long format: d mmmm yyyy or mmmm dd, yyyy (Day first, full month name, and year or first full month name, day, and year, in left-to-right writing direction) in Afar, French and Somali and yyyy ،mmmm d (Day first, full month name, and year in right-to-left writing direction) in Arabic \| \| \| Dominica \| No \| Yes \| No \| \| \| \| Dominican Republic \| No \| Yes \| No \| \| \| \| East Timor \| No \| Yes \| No \| \| \| \| Ecuador \| No \| Yes \| No \| \| \| \| Egypt \| No \| Yes \| No \| \| \| \| El Salvador \| No \| Yes \| No \| \| \| \| Equatorial Guinea \| No \| Yes \| No \| (dd/mm/yyyy or d mmmm yyyy) for French and Spanish \| \| \| Eritrea \| Yes \| Yes \| Sometimes \| Short format: dd/mm/yyyy for Afar, Bilen, English, Saho, Tigre and Tigrinya. Gregorian dates follow the same rules but tend to be written in the yyyy/m/d (Day first, month number and year in right-to-left writing direction) format in Arabic language.Long format: D MMMM YYYY (Day first, full month name, and year in left-to-right writing direction) for Bilen, English, Tigre and Tigrinya, YYYY ،MMMM D (Day first, full month name, and year in right-to-left writing direction) for Arabic and MMMM DD, YYYY (First full month name, day and year in left-to-right writing direction) for Afar and Saho \| \| \| Estonia \| Sometimes \| Yes \| No \| dd.mm.yyyy, d.m.(yy)yy or d. mmmm yyyy (mmmm may be substituted by Roman numerals). In more formal, international contexts yyyy-mm-dd is the preferred allowed format. \| \| \| Eswatini \| Yes \| Yes \| No \| YMD (in Swati), DMY (in English) \| \| \| Ethiopia \| No \| Yes \| Sometimes \| (dd/mm/yyyy or dd mmmm yyyy) for Amharic, Tigrinya and Wolaytta(dd/mm/yyyy or mmmm dd, yyyy) for Afar, Oromo and Somali \| \| \| Falkland Islands \| No \| Yes \| No \| \| \| \| Faroe Islands \| No \| Yes \| No \| \| \| \| Federated States of Micronesia \| No \| No \| Yes \| \| \| \| Fiji \| No \| Yes \| No \| \| \| \| Finland \| No \| Yes \| Sometimes \| Finnish: d.m.yyyy or in long format d. mmmm yyyyInari Sami: mmmm d. p. yyyyNorthern Sami: mmmm d. b. yyyySkolt Sami: mmmm d. p. yyyySwedish: d mmmm yyyy(Note: Month and year can be shortened) \| \| \| France \| Yes \| Yes \| No \| (dd/mm/yyyy) for Alsatian, Catalan, Corsican, French and Occitan65 for Breton, Basque and Interlingua \| NF Z69-200 \| \| French Guiana \| No \| Yes \| No \| \| \| \| French Polynesia \| No \| Yes \| No \| \| \| \| Gabon \| No \| Yes \| No \| \| \| \| The Gambia \| No \| Yes \| No \| \| \| \| Georgia \| No \| Yes \| No \| (dd.mm.yyyy) (In Georgian calendar dates, century digits may be omitted, e.g., dd-mm-yy.) \| \| \| Germany \| Yes \| Yes \| No \| The format dd.mm.yyyy using dots (which denote ordinal numbering) is the traditional German date format, and continues to be the most commonly used. In 1996, the international format yyyy-mm-dd was made the official date format in standardized contexts such as government, education, engineering and sciences. However, as it failed to establish itself, the traditional format (d)d.(m)m.(yy)yy was allowed again as an alternative in 2006 (except in areas where there is risk of ambiguity). The handwritten form d. mmmm yyyy is also accepted (compare DIN 5008).[citation needed] See Date and time notation in Europe. \| DIN ISO 8601:2006-09, used in DIN 5008:2011-04 (see Datumsformat) \| \| Ghana \| Yes \| Yes \| Yes \| (yyyy/mm/dd) for Akan(dd/mm/yyyy)(m/d/yyyy) for Ewe[citation needed] \| \| \| Gibraltar \| No \| Yes \| No \| \| \| \| Greece \| No \| Yes \| No \| Short format: d/m/yyyy or rarely d-m-yyyyLong format: dddd, d mmmm, yyyy (month in genitive) \| ELOT EN 28601 \| \| Greenland \| No \| Yes \| Yes \| Danish: d. mmmm yyyyGreenlandic: mmmm d.-at, yyyy[citation needed] \| \| \| Grenada \| No \| Yes \| No \| \| \| \| Guadeloupe \| No \| Yes \| No \| \| \| \| Guam \| No \| No \| Yes \| [citation needed] \| \| \| Guatemala \| No \| Yes \| No \| Short format: dd/mm/yyyyLong format: d de mmmm de yyyy or dddd, d de mmmm de yyyy \| \| \| Guernsey \| No \| Yes \| No \| \| \| \| Guinea \| Yes \| Yes \| Sometimes \| Short format: dd/mm/yyyy (Day first, month and year in left-to-right writing direction) in French and Fulah. Gregorian dates follow the same rules but tend to be written in yyyy/mm/dd (Day first, month number, and year in right-to-left writing direction) format in N'ko language.Long format: D MMMM YYYY (Day first, month and year in left-to-right writing direction) for French and Fulah and YYYY, DD MMMM (First full month name, day, and year in right-to-left writing direction) for N'ko \| \| \| Guinea-Bissau \| No \| Yes \| No \| \| \| \| Guyana \| No \| Yes \| No \| \| \| \| Haiti \| No \| Yes \| No \| \| \| \| Hong Kong \| Yes \| Yes \| Rarely \| (yy)yy年(m)m月(d)d日 (if without leading zeros) for Chinese and in British English, (d)d/(m)m/(yy)yy in short format. d mmmm yyyy (Casually many people use with commas: d mmmm, yyyy) in long format.Both expanded forms dd-mmmm-yyyy and mmmm-dd-yyyy are used interchangeably in Hong Kong, except the latter was more frequently used in media publications and commercial purpose, such as The Standard. \| \| \| Honduras \| No \| Yes \| No \| \| \| \| Hungary \| Yes \| Sometimes \| No \| yyyy. mm. (d)d.The year is written in Arabic numerals. The name of the month can be written out in full or abbreviated, or it can be indicated by Roman numerals or Arabic numerals. The day is written in Arabic numerals. \| MSZ ISO 8601:2003 \| \| Iceland \| No \| Yes \| No \| (dd.mm.yyyy) \| IST EN 28601:1992 \| \| India \| Yes \| Yes \| Sometimes \| In India, the dd-mm-yyyy is the predominant short form of the numeric date usage. Almost all government documents need to be filled up in the dd-mm-yyyy format. An example of dd-mm-yyyy usage is the passport application form. Though not yet a common practice, the BIS (Bureau of Indian Standards) of the Government of India introduced the standard named "IS 7900:2001 (Revised in 2006) Data Elements And Interchange Formats – Information Interchange – Representation Of Dates And Times" which officially recommends use of the date format yyyy-mm-dd;[citation needed] for example, 2013-09-10, 20130910, or 2013 09 10 for the date 10 September 2013. Dates in the Bodo language are in mm/dd/yyyy.The majority of English-language newspapers and media publications in India use mmmm dd, yyyy.[citation needed] \| IS 7900:2001 \| \| Indonesia \| No \| Yes \| Rarely \| On English-written materials, Indonesians tend to use the M-D-Y but was more widely used in non-governmental contexts.[citation needed]English-language governmental and academic documents use DMY. \| \| \| Iran \| Yes \| Yes \| No \| Short format: yyyy/mm/dd in Persian Calendar system ("yy/m/d" is a common alternative). Gregorian dates follow the same rules in Persian literature but tend to be written in the dd/mm/yyyy format in official English documents.Long format: YYYY MMMM D (Day first, full month name, and year in right-to-left writing direction) \| \| \| Iraq \| No \| Yes \| No \| Short format: (dd/mm/yyyy) \| \| \| Ireland \| No \| Yes \| No \| (dd-mm-yyyy). dd/mm/yyyy is also in common use \| IS/EN 28601:1993 \| \| Isle of Man \| No \| Yes \| No \| \| \| \| Israel \| No \| Yes \| No \| The format dd.mm.yyyy using dots is the common format. dd/mm/yyyy is also in common use. The Jewish calendar is in limited use, mainly for Jewish holidays, and follows the DMY format. \| \| \| Italy \| No \| Yes \| No \| (dd/mm/yyyy) \| UNI EN 28601 \| \| Ivory Coast \| No \| Yes \| No \| \| \| \| Jamaica \| Yes \| Yes \| No \| \| \| \| Jan Mayen \| No \| Yes \| No \| \| \| \| Japan \| Yes \| No \| No \| Often in the form yyyy年mm月dd日; sometimes Japanese era year is used, e.g. 平成18年12月30日. \| JIS X 0301:2002 \| \| Jersey \| No \| Yes \| No \| \| \| \| Jordan \| No \| Yes \| No \| \| \| \| Kazakhstan \| Sometimes \| Yes \| No \| Short format: (yyyy.dd.mm) in Kazakh[obsolete source] and (dd.mm.(yy)yy) in Russian[obsolete source]Long format: yyyy 'ж'. d mmmm in Kazakh; d MMMM yyyy in RussianFull format in Kazakh: yyyy 'ж'. dd mmmm \| \| \| Kenya \| Yes \| Yes \| Yes \| (yy/mm/dd)98(m/d/yyyy) for Swahili \| \| \| Kiribati \| No \| Yes \| No \| \| \| \| North Korea \| Yes \| No \| No \| \| \| \| South Korea \| Yes \| No \| No \| National standard format is yyyy-mm-dd (with leading zeroes) and (yy)yy년 (m)m월 (d)d일 (with or without leading zeroes)casually many people use (yy)yy.(m)m.(d)d(.) (with or without leading zeroes, with or without the last full stop). \| KS X ISO 8601 \| \| Kosovo \| No \| Yes \| No \| \| \| \| Kuwait \| No \| Yes \| No \| \| \| \| Kyrgyzstan \| No \| Yes \| No \| (dd.mm.yyyy) \| \| \| Laos \| No \| Yes \| No \| \| \| \| Latvia \| No \| Yes \| No \| Short format: dd.mm.yyyy.Long format: yyyy. gada d. mmmm \| \| \| Lebanon \| No \| Yes \| No \| \| \| \| Lesotho \| Yes \| Yes \| No \| yyyy-mm-dd for Sesotho and dd/mm/yyyy for English \| \| \| Liberia \| No \| Yes \| No \| \| \| \| Libya \| No \| Yes \| No \| \| \| \| Liechtenstein \| No \| Yes \| No \| (dd.mm.yyyy) \| \| \| Lithuania \| Yes \| Sometimes \| No \| (yyyy-mm-dd)yyyy d \| LST ISO 8601:1997 (obsolete)LST ISO 8601:2006 (current) \| \| Luxembourg \| No \| Yes \| No \| (dd.mm.yyyy) \| ITM-EN 28601 \| \| Macau \| Yes \| Yes \| No \| YMD（年月日）(Same as Hong Kong)DMY (in Portuguese and British English) \| \| \| Madagascar \| No \| Yes \| No \| \| \| \| Malawi \| No \| Yes \| No \| \| \| \| Malaysia \| Yes \| Yes \| Sometimes \| dd-mm-yyyy or dd/mm/yyyy is more commonly used, especially in English and Malay in both short and long format. yyyy-mm-dd is used in other instances particularly in documentation and organizing and also in Chinese (yyyy年m月d日), in short and long format. MMDDYYYY in long format is sometimes used in media, especially written English media, but less frequently compared to the others.There is no 'official' date format used but they are used interchangeably based on the situation. \| \| \| Maldives \| Yes \| Yes \| No \| Short format: yy/mm/dd (Day first, month next and year last in right-to-left writing direction)Long format: dd mmmm yyyy (Year first, full month name and day last in right-to-left writing direction) \| \| \| Mali \| No \| Yes \| No \| \| \| \| Malta \| No \| Yes \| No \| \| \| \| Marshall Islands \| No \| No \| Yes \| [citation needed] \| \| \| Martinique \| No \| Yes \| No \| \| \| \| Mauritania \| No \| Yes \| No \| \| \| \| Mauritius \| No \| Yes \| No \| \| \| \| Mayotte \| No \| Yes \| No \| \| \| \| Mexico \| No \| Yes \| No \| \| NOM-008-SCFI-2002 \| \| Moldova \| No \| Yes \| No \| \| \| \| Monaco \| No \| Yes \| No \| \| \| \| Mongolia \| Yes \| No \| No \| National standard format is yyyy-mm-dd (with leading zeroes) and yyyy оны (m)m сарын (d)d (with or without leading zeroes)Traditional Mongolian languages in Mongolia usually give date examples in the form 2017ᠣᠨ ᠵᠢᠷᠭᠤᠳᠤᠭᠠᠷ ᠰᠠᠷ᠎ᠠ 2ᠡᠳᠦᠷ but this form is never used when writing in Mongolian Cyrillic; casually many people use yyyy/(m)m/(d)d or yyyy.(m)m.(d)d (with or without leading zeroes). \| MNS-ISO 8601 \| \| Montenegro \| No \| Yes \| No \| Both d.m.yyyy. and dd.mm.yyyy. are accepted. A period is used as a separator and after the year because the Montenegrin language writes these numbers as ordinal numbers that are written as the corresponding cardinal number, with a period at the end. \| \| \| Montserrat \| No \| Yes \| No \| \| \| \| Morocco \| No \| Yes \| No \| \| \| \| Mozambique \| No \| Yes \| No \| \| \| \| Myanmar \| Yes \| Yes \| No \| YMD for Burmese calendar. DMY for Gregorian calendar. \| \| \| Namibia \| Yes \| Yes \| No \| DMY \| \| \| Nauru \| No \| Yes \| No \| \| \| \| Nepal \| Yes \| Yes \| Sometimes \| DMY,[citation needed] YMD in official Nepali Vikram Samvat calendar (also see Nepal Sambat which is also in use); MDY in Gregorian dates are used for newspapers (English language) and PCs \| \| \| Netherlands \| No \| Yes \| No \| Using hyphens as in "dd-mm-yyyy". \| NEN ISO 8601, NEN EN 28601, NEN 2772 \| \| New Caledonia \| No \| Yes \| No \| \| \| \| New Zealand \| Yes \| Yes \| No \| \| AS/NZS ISO 8601.1:2021 \| \| Nicaragua \| No \| Yes \| No \| \| \| \| Niger \| No \| Yes \| No \| \| \| \| Nigeria \| No \| Yes \| Sometimes \| Short format: (d)d/(m)m/(yy)yy for Edo, English, Fulani, Hausa, Ibibio, Igbo, Kanuri and Yoruba languageLong format: d mmmm yyyy for English, Hausa, Igbo and Yoruba, and mmmm dd, yyyy for Edo, Fulani, Ibibio and Kanuri. \| \| \| Niue \| No \| Yes \| No \| dd/mm/yyyy \| \| \| Norfolk Island \| No \| Yes \| No \| \| \| \| North Macedonia \| No \| Yes \| No \| (dd.mm.yyyy) \| \| \| Northern Mariana Islands \| No \| No \| Yes \| [citation needed] \| \| \| Norway \| Yes \| Yes \| Rarely \| dd.mm.yyyy; leading zeroes and century digits may be omitted, e.g., 10.02.16; ddmmyy (six figures, no century digits, no delimiters) allowed in tables. ISO dates yyyy-mm-dd can be used for "technical" purposes. The fraction form d/m-y is incorrect, but is common and considered passable in handwriting. Lule Sami and Southern Sami dates mmmm d. b. yyyy. \| NS-ISO 8601 \| \| Oman \| No \| Yes \| No \| \| \| \| Pakistan \| No \| Yes \| No \| \| \| \| Palestine (Palestinian Authority, West Bank and Gaza Strip) \| No \| Yes \| No \| (dd/mm/yyyy) \| \| \| Palau \| No \| Yes \| Rarely \| Formerly including: (m)m/(d)d/(yy)yy in English and (yy)yy/m(m)/(d)d in Japanese \| \| \| Panama \| No \| Yes \| No \| Short format: dd/mm/yyyyLong format: d de mmmm de yyyy \| \| \| Papua New Guinea \| No \| Yes \| No \| \| \| \| Paraguay \| No \| Yes \| No \| \| \| \| Peru \| No \| Yes \| No \| \| \| \| Philippines \| No \| Yes \| Yes \| Long formats: English: mmmm d, yyyyDMY dates are also used occasionally, primarily by, but not limited to, government institutions such as on the data page of passports, and immigration and customs forms.Filipino (Tagalog): ika-d ng mmmm(,) yyyy136 may also be written in mmmm d, yyyy format in civil use but still pronounced as above.)Short/numerical format: mm/dd/yyyy for both languages. \| \| \| Pitcairn Islands \| No \| Yes \| No \| \| \| \| Poland \| Sometimes \| Yes \| No \| Traditional format (DMY): (dd.mm.yyyy, often with dots as separators; more official is d yyyy, or, less frequently, d yyyy)Official format (YMD): The ISO 8601 YYYY-MM-DD format is used in official documents, banks, computer systems[citation needed] and the internet[citation needed] in Poland. \| PN-90/N-01204 \| \| Portugal \| Yes \| Yes \| No \| Mostly (dd/mm/yyyy) and (dd-mm-yyyy); some newer documents use (yyyy-mm-dd). \| NP EN 28601 \| \| Puerto Rico \| No \| Yes \| Yes \| English: mmmm d, yyyySpanish: d de mmmm de yyyy \| \| \| Qatar \| No \| Yes \| No \| \| \| \| Réunion \| No \| Yes \| No \| \| \| \| Romania \| No \| Yes \| No \| (dd.mm.yyyy) Also widely used: (d)d-mmm-yyyy (3 letters of month name with the notable exception of Nov for November, which would otherwise be noiembrie) and (d)d-XII-yyyy (month number as a Roman numeral with lines above AND below, slowly deprecating) \| \| \| Russia \| Yes \| Yes \| No \| yyyy-mm-dddd.mm.yyyy(dd.mm.(yy)yy); more official is d yyyy г. (= g., short for goda, i.e. year in genitive) Bashkir, Ossetian, Sakha and Tatar languages in Russia usually give date examples in the form 22 май 2017 й, 22 майы, 2017 аз, ыам ыйын 22 күнэ 2017 с., 22 май 2017 ел but this form is never used when writing in Russian. \| GOST R 7.0.64-2018GOST R 7.0.97-2016 \| \| Rwanda \| Yes \| Yes \| No \| (yyyy/mm/dd or yyyy mmmm dd) for Kinyarwanda(dd/mm/yyyy or d mmmm yyyy) for English and French \| \| \| Saba \| No \| Yes \| No \| \| \| \| Saint Barthélemy \| No \| Yes \| No \| \| \| \| Saint Helena, Ascension and Tristan da Cunha \| No \| Yes \| No \| \| \| \| Saint Kitts and Nevis \| No \| Yes \| No \| \| \| \| Saint Lucia \| No \| Yes \| No \| \| \| \| Saint Martin \| No \| Yes \| No \| \| \| \| Saint Pierre and Miquelon \| No \| Yes \| No \| \| \| \| Saint Vincent and the Grenadines \| No \| Yes \| No \| \| \| \| Samoa \| No \| Yes \| No \| \| \| \| San Marino \| No \| Yes \| No \| \| \| \| São Tomé and Príncipe \| No \| Yes \| No \| \| \| \| Saudi Arabia \| No \| Yes \| No \| (dd/mm/yyyy in Islamic and Gregorian calendar systems, \| \| \| Senegal \| No \| Yes \| No \| \| \| \| Serbia \| No \| Yes \| No \| (d.m.yyyy. or d. mmmm yyyy.) \| \| \| Seychelles \| No \| Yes \| No \| \| \| \| Sierra Leone \| No \| Yes \| No \| \| \| \| Singapore \| Yes \| Yes \| Sometimes \| (Chinese representation: yyyy年m月d日, no leading zeroes)DMY in English, Malay and Tamil languagesMDY (in long format) also sometimes used, especially in media publications, commercial usage, and some governmental websites.[citation needed] \| \| \| Sint Eustatius \| No \| Yes \| No \| \| \| \| Sint Maarten \| No \| Yes \| No \| \| \| \| Slovakia \| No \| Yes \| No \| (d. m. yyyy) \| \| \| Slovenia \| No \| Yes \| No \| (d. m. yyyy or d. mmmm yyyy) \| \| \| Solomon Islands \| No \| Yes \| No \| \| \| \| Somalia \| No \| Yes \| No \| Short format: dd/mm/yyyy \| \| \| South Africa \| Yes \| Yes \| Sometimes \| (yyyy/mm/dd and dd mmmm yyyy) in English(yyyy-mm-dd and dd mmmm yyyy) in Afrikaans156 in XhosaMDY in Zulu \| SANS 8601:2009 \| \| Spain \| Yes \| Yes \| No \| (dd/mm/yyyy) for Asturian, Catalan, Galician, Spanish and Valencian160 for Basque \| UNE EN 28601 \| \| Sri Lanka \| Yes \| Yes \| Rarely \| (yyyy-mm-dd) for Sinhala and (d-m-yyyy) for TamilEnglish-language media and commercial publications use Month-day-year in long format, but only Day-month-year format (both long and short numeric) are used in governmental and other English documents of official contexts. \| \| \| Sudan \| No \| Yes \| No \| \| \| \| South Sudan \| No \| Yes \| No \| \| \| \| Suriname \| No \| Yes \| No \| \| \| \| Svalbard \| No \| Yes \| No \| \| \| \| Sweden \| Yes \| Yes \| No \| National standard format is yyyy-mm-dd. dd.mm.yyyy format is used in some places where it is required by EU regulations, for example for best-before dates on food and on driver's licenses. d/m format is used casually, when the year is obvious from the context, and for date ranges, e.g. 28-31/8 for 28–31 August.The textual format is "d mmmm yyyy" or "den d mmmm yyyy". \| SS-ISO 8601 \| \| Switzerland \| No \| Yes \| No \| (dd.mm.yyyy or d. mmmm yyyy) for French, German, Italian and Romansh[failed verification] \| SN ISO 8601:2005-08 \| \| Syria \| No \| Yes \| No \| \| \| \| Taiwan \| Yes \| No \| No \| Short format: yyyy/(m)m/(d)d or yyyy-mm-dd Long format: yyyy年m月d日, in most context year is represented using ROC era system: 民國95年12月30日. \| CNS 7648 \| \| Tajikistan \| No \| Yes \| No \| (dd.mm.yyyy) \| \| \| Tanzania \| No \| Yes \| No \| \| \| \| Thailand \| No \| Yes \| No \| dd/mm/yyyy (in governmental sector with Buddhist Era years instead of Common Era) \| TIS 1111:2535 in 1992 \| \| Togo \| No \| Yes \| Yes \| (dd/mm/yyyy) in French and (mm/dd/(yy)yy) in Ewe \| \| \| Tokelau \| No \| Yes \| No \| \| \| \| Tonga \| No \| Yes \| No \| \| \| \| Trinidad and Tobago \| No \| Yes \| No \| \| \| \| Tunisia \| No \| Yes \| No \| \| \| \| Turkey \| No \| Yes \| No \| Short format: dd.mm.yyyyLong format: d mmmm yyyy Full format: d mmmm yyyy dddd \| \| \| Turkmenistan \| No \| Yes \| No \| (dd.mm.(yy)yy ý.), yyyy-nji ýylyň d-nji mmmm \| \| \| Turks and Caicos Islands \| No \| Yes \| No \| \| \| \| Tuvalu \| No \| Yes \| No \| \| \| \| Uganda \| No \| Yes \| No \| \| \| \| Ukraine \| No \| Yes \| No \| (dd.mm.(yy)yy; some cases of dd/mm/yyyy) \| \| \| United Arab Emirates \| No \| Yes \| No \| \| \| \| United Kingdom \| Yes \| Yes \| No \| Most style guides follow the DMY convention by recommending d mmmm yyyy (sometimes written dd/mm/yyyy) format in articles (e.g. The Guardian's, and the Oxford Style Manual).Some newspapers use dddd mmmm d, yyyy for both the banner and articles, while others stick to DMY for both.In addition, YMD with four-digit year is used increasingly especially in applications associated with computers, and as per British standard BS ISO 8601:2019+A1:2022, avoiding the ambiguity of the numerical versions of the DMY/MDY formats. \| 8601:2019+A1:2022 \| \| United States Minor Outlying Islands \| No \| No \| Yes \| Same as the US \| \| \| United States \| Yes \| Rarely \| Yes \| (Civilian vernacular: m/d/yy or m/d/yyyy; other formats, especially d mmm(m) yyyy (but no short DMY formats) and yyyy-mm-dd (but rarely any other short YMD formats and rarely any long YMD formats), are sometimes prescribed or used—particularly in military, academic, scientific, computing, industrial, or governmental contexts. See Date and time notation in the United States.) \| ANSI INCITS 30-1997 (R2008) \| \| United States Virgin Islands \| No \| No \| Yes \| \| \| \| Uruguay \| No \| Yes \| No \| \| \| \| Uzbekistan \| Yes \| Yes \| No \| (dd.mm.yyyy Cyrillic, dd/mm yyyy Latin) \| \| \| Vanuatu \| No \| Yes \| No \| \| \| \| Vatican City \| Rarely \| Yes \| No \| (dd m yyyy), with p.C.n. following (post Christum natum) if CE, and a.C.n. (ante Christum natum) if BCE. Likely from similar phrases used in ecclesiastical latin. \| \| \| Venezuela \| No \| Yes \| No \| \| \| \| Vietnam \| Yes \| Yes \| Sometimes \| Long format: "Ngày (d)d tháng (m)m năm yyyy" (leading zeros required by Circular No. 01/2011/TT-BNV by the Ministry of Home Affairs) or ngày (d)d tháng (month in textform) năm yyyy.Short format (interchangeably): (d)d/(m)m/yyyy or (d)d-(m)m-yyyy; (d)d.(m)m.yyyy is also in use.In English documents: Short format: yyyy-mm-dd Long format: mmmm d, yyyy In historical documents: era names năm thứ _ tháng [m]m (or in textform) ngày(mồng) [d]d (or in textform). \| \| \| Wallis and Futuna \| No \| Yes \| No \| \| \| \| Yemen \| No \| Yes \| No \| \| \| \| Zambia \| No \| Yes \| No \| \| \| \| Zimbabwe \| No \| Yes \| No \| \| \| See also [edit] Date and time representation by country Common Locale Data Repository, a database that covers national date and time notations ISO 8601 References [edit] ^ "Date/Time Patterns". 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14215	https://www.chemicalforums.com/index.php?topic=15281.0	Chemical Forums: chromate/dichromate equilibrium Chemical Forums September 29, 2025, 05:51:33 AM Forum Rules: Read This Before Posting Home Help Search Login Register Discover more Forums forum Science Simple Machines Forum science SMF Forum Iron(III) chloride sulfur Bookshelves Chemical Forums Chemistry Forums for Students High School Chemistry Forum chromate/dichromate equilibrium « previousnext » Print Pages: Go Down Topic: chromate/dichromate equilibrium (Read 13890 times) 0 Members and 1 Guest are viewing this topic. jennielynn_1980 Full Member Posts: 167 Mole Snacks: +8/-4 Gender: chromate/dichromate equilibrium « on: December 20, 2006, 02:11:19 PM » I am doing a lab about the chromate/dichromate equilibrium represented by the following equation: H+(aq) + 2CrO 4 2- (aq) <--> Cr 2 O 7 2- (aq) + OH- (aq) I am trying to figure out why 1) NH 3 acts as a base when added to chromate and dichromate solution 2) why Ca(OH)2 only has a small effect on the equilibrium 3) why C 2 H 5 OH has on effect on the equilibrium For 1) I know it has something to do with the two unpaired electrons in the ammonia but other than that, I don't know specifically what is happening For 2) I think it has to do with the low solubility of Ca(OH)2. If it has low solubiility, it is unable to react with the other species in an aqueous solution For 3) I am pretty stumped. The only thing I really know about ethanol is that it react strangely with water. Thanks Logged Discover more Sciences Groceries sulfur Forum Iron(III) chloride Bookshelves SMF Forums forum Internet forum DevaDevil Chemist Full Member Posts: 690 Mole Snacks: +55/-9 Gender: postdoc at ANL Re: chromate/dichromate equilibrium « Reply #1 on: December 20, 2006, 02:31:12 PM » First of all the equilibrium is generally noted as 2H+(aq) + 2CrO 4 2- (aq) <--> Cr 2 O 7 2- (aq) + H 2 O (l) As you can never have both OH- and H+ in abundance in solution; think of the water equilibrium. 1) Are protons (hydronium ions) in abundance in the solution? So what does that tell you about the acid / base equilibrium of ammonia-ammonium? 2) true low solubility means the acidity [H+] doesn't change much, hence according to Le Chatelier's principle the equilibrium will not change much. 3) Dichromate combined with an acidic solution is used as an oxidiser. It will Oxidise Ethanol to Ethanal; so you can write down the formula? Logged jennielynn_1980 Full Member Posts: 167 Mole Snacks: +8/-4 Gender: Re: chromate/dichromate equilibrium « Reply #2 on: December 20, 2006, 02:46:44 PM » So 1) for the ammonia - ammonium because there is lots of H+, the ammonia reacts with water to give hydroxide ions and ammonium. So because one of the products is hydroxide ions, the ammonia added will act as a base. The equation being: NH 3 + H 2 O --> NH 4+ + OH- For 3) I don't know what would happen here. So if ethanol is oxidized it loses electrons meaning the hydrogen? My only guess is that it dissociates kind of so that there is both OH- and H+ and they cancel each other out.I haven't studied oxidation reactions yet. Logged Borek Mr. pH Administrator Deity Member Posts: 27959 Mole Snacks: +1827/-412 Gender: I am known to be occasionally wrong. Re: chromate/dichromate equilibrium « Reply #3 on: December 20, 2006, 03:26:29 PM » Quote from: jennielynn_1980 on December 20, 2006, 02:46:44 PM So 1) for the ammonia - ammonium because there is lots of H+, the ammonia reacts with water to give hydroxide ions and ammonium. I don't get this 'because there is lots of H+' part. Quote So because one of the products is hydroxide ions, the ammonia added will act as a base.The equation being: NH3+ H2O -->NH4+ + OH- That's nothing else but Bronsted-Lowry base definition at work Logged Chembuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info Discover more sulfur science forum School supplies Internet forum SMF Forums Science Groceries Iron(III) chloride jennielynn_1980 Full Member Posts: 167 Mole Snacks: +8/-4 Gender: Re: chromate/dichromate equilibrium « Reply #4 on: December 20, 2006, 03:29:03 PM » I meant that "there is lots of H+" in the equilibrium equation. The H+ would react with the NH3. But the equation is correct? Logged DevaDevil Chemist Full Member Posts: 690 Mole Snacks: +55/-9 Gender: postdoc at ANL Re: chromate/dichromate equilibrium « Reply #5 on: December 20, 2006, 03:30:52 PM » 1) like I said in the first reply; if [H+] is big, [OH-] will be minimal because of the water equilibrium. You might want to change your equilibrium with ammonia to include H+, not OH-, else it is fine. Like borek said; acid-base chemistry 3) can you write a reaction with Cr 2 O 7 2-, H+ and ethanol, with final products (a.o.) Cr 3+ and ethanal (which is the final product of ethanol, it will not dissociate)? Also, what will this do to the original chromate/dichromate equilibrium? Logged Borek Mr. pH Administrator Deity Member Posts: 27959 Mole Snacks: +1827/-412 Gender: I am known to be occasionally wrong. Re: chromate/dichromate equilibrium « Reply #6 on: December 20, 2006, 03:49:40 PM » Quote from: jennielynn_1980 on December 20, 2006, 03:29:03 PM I meant that "there is lots of H+" in the equilibrium equation.The H+ would react with the NH3. But the equation is correct? If there is 'lots of H+' ammonia doesn't have to react with water - it can get protonated directly Logged Chembuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info Discover more Sciences Forum School supplies Bookshelves science Science SMF Internet forum forum Forums jennielynn_1980 Full Member Posts: 167 Mole Snacks: +8/-4 Gender: Re: chromate/dichromate equilibrium « Reply #7 on: January 10, 2007, 03:31:10 PM » So for the NH 3, the equation would be: NH 3 + H+ --> NH 4+ And there would be H+ ions in the solution already from the chromate/dichromate equilibrium. For the ethanol, the equation would be: Cr 2 O 7 2- + 8H+ + 3C 2 H 5 OH --> 2Cr 3+ + 3C 2 H 4 O + H 2 O That would give us ethanal that will not further dissociate. I don't know what it will do the to the equilbrium though. According to the experiment, it has no effect whatsoever. Why, I still don't know. Logged DevaDevil Chemist Full Member Posts: 690 Mole Snacks: +55/-9 Gender: postdoc at ANL Re: chromate/dichromate equilibrium « Reply #8 on: January 10, 2007, 06:52:21 PM » Quote from: jennielynn_1980 on January 10, 2007, 03:31:10 PM For the ethanol, the equation would be: Cr2O72-+ 8H+ + 3C2H5OH -->2Cr3++ 3C2H4O + H2O That would give us ethanal that will not further dissociate.I don't know what it will do the to the equilbrium though.According to the experiment, it has no effect whatsoever.Why, I still don't know. If you'd balance it completely (H 2 O) you have it right Well, the effect should be that you lower the Cr 2 O 7 2- content, thereby pushing the equilibrium a little to this side (Le Chatelier's principle). Meaning the concentration of CrO 4 2- will be reduced as well to counterbalance the loss in dichromate concentration form the reaction. Logged jennielynn_1980 Full Member Posts: 167 Mole Snacks: +8/-4 Gender: Re: chromate/dichromate equilibrium « Reply #9 on: January 10, 2007, 09:36:46 PM » So, because in effect the concentration of both chromate and dichromate are lowered, there APPEARS to be no change when the ethanol is added? Or there should be a somewhat visible change (in the colour of the solution I mean)? I am asking because the experiment says there should be no change in colour. I didn't actually do the experiment by the way. It was a simulation in which I was given the results and asked to analyze the data. Thanks for all your help Logged DevaDevil Chemist Full Member Posts: 690 Mole Snacks: +55/-9 Gender: postdoc at ANL Re: chromate/dichromate equilibrium « Reply #10 on: January 11, 2007, 02:25:59 PM » well; any reaction in equilibrium has an equilibruim constand as you know. K = conc products / conc starting in this case K = [OH-][dichromate] / [H+][chromate], so K = Kw[dichromate] / [H+]2[chromate] In this case you remove more H+ than dichromate. (8:1) so what should happen to the equilibrium? I must recall my words before, the conc. chromate may even increase a little. It mainly depends on the change of [H+] as compared to the change of [dichromate]. But in amounts; I don't know If your acid was very strong to start with (high [H+], aka low change in conc) and your alcohol volume low, the change may be negligable, if your acid was weak, or your dichromate concentration very low, the change will be more pronounced. Since you influence both sides of the equilibrium, it is not a clear thing to say unless you know in what proportions the products are there Logged Print Pages: Go Up « previousnext » Sponsored Links Discover more sulfur Iron(III) chloride Science science forum Groceries Forum School supplies Simple Machines Forum Forums Chemical Forums Chemistry Forums for Students High School Chemistry Forum chromate/dichromate equilibrium Jump to: Site Friends: Chembuddy \| Chem Reddit \| Powered by SMF 2.0.17 \| SMF © 2011, Simple Machines\| Design by Webtiryaki XHTML RSS WAP2 Page created in 0.429 seconds with 20 queries.
14216	https://www.ebsco.com/research-starters/health-and-medicine/lesch-nyhan-syndrome	Menu Lesch-Nyhan syndrome Lesch-Nyhan syndrome is a rare genetic disorder primarily affecting males, stemming from a mutation in the HPRT1 gene located on the X chromosome. This condition disrupts purine metabolism, leading to an accumulation of uric acid in the body, which can result in symptoms such as kidney stones, joint pain, and a distinctive orange crystalline deposit in urine. Individuals with the syndrome often exhibit neurological and behavioral challenges, including self-injury, which emerges as compulsive behaviors like biting and head banging. These behaviors are believed to be linked to neurotransmitter abnormalities, and it is important to note that affected individuals do not wish to harm themselves; they are unable to control these actions. Diagnosis involves a physical examination and specific tests to measure HPRT enzyme activity. While there is currently no cure for Lesch-Nyhan syndrome, treatments can help manage symptoms, such as medications to control uric acid levels and behavioral interventions. The prognosis typically allows for an average life expectancy into the early to mid-twenties, with many patients living longer with comprehensive medical care. Families with a history of the disorder may benefit from genetic counseling to understand potential risks for future children. Lesch-Nyhan syndrome ALSO KNOWN AS: Hypoxanthine-guanine phosphoribosyltransferase deficiency or HPRT deficiency; Lesch-Nyhan disease DEFINITION Lesch-Nyhan syndrome is a genetic disorder that affects the metabolism of purines in the body. Purines are protein molecules that are important for the metabolism of ribonucleic acid (RNA) and deoxyribonucleic acid (DNA), which make up genetic codes. Lesch-Nyhan syndrome is characterized by uric acid buildup and self-injury. This disease, which mainly affects men, is rare, occurring in 1 of every 100,000 males. Risk Factors Males and individuals who have male family members on their mother’s side of the family with Lesch-Nyhan syndrome are at risk for developing the disorder. Etiology and Genetics Lesch-Nyhan syndrome results from a mutation in the HPRT1 gene, found on the long arm of the X chromosome at position Xq26.1. This gene encodes the hypoxanthine phosphoribosyltransferase 1, which is an essential enzyme in the salvage pathway. Proper functioning of this pathway allows cells to recycle purines, one of the building blocks of DNA and RNA, rather than having to synthesize them from scratch. In the absence of the HPRT1 enzyme, the pathway is blocked and there is an accumulation in the body of uric acid, a waste product of purine decomposition. Excess uric acid can cause arthritis, kidney stones, and bladder stones It is unclear how the enzyme deficiency causes the behavioral and neurological problems associated with this disease. Inheritance of Lesch-Nyhan syndrome follows a strict sex-linked recessive pattern. Only males are affected, and they inherit the defective gene from their mothers. Mothers who carry the mutated gene on one of their two X chromosomes are unaffected, but they face a 50 percent chance of transmitting this disorder to each of their male children. Female children have a 50 percent chance of inheriting the gene and becoming carriers like their mothers. Affected males rarely live to reproduce, but in that unlikely event they would pass the mutation on to all of their daughters but to none of their sons. Symptoms The first symptom of Lesch-Nyhan syndrome is an orange-colored crystal-like deposit in the diaper. This may occur in children as young as three months. These deposits are caused by increased uric acid in the urine. Other symptoms include irritability and nervous system impairment. Symptoms of nervous system impairment for an infant who is from four to six months old include a lack of muscle tone and an inability to lift the head. Symptoms in infants who are six months old include an unusual arching of the back; symptoms in a nine-month-old child include the inability to crawl or stand. At twelve months, a child’s symptoms include an inability to walk. Symptoms in children who are older than twelve months include spasms of the limbs and facial muscles. Additional symptoms include kidney stones, blood in the urine, pain and swelling of joints, difficulty swallowing (dysphagia), impaired kidney function, self-injury, and uric acid deposits in the joints. Self-mutilating behavior is the hallmark of this disease. Children begin to bite their fingers, their lips, and the insides of their mouths as early as two years old. As children grow, self-injury becomes increasingly compulsive and severe. Eventually, mechanical physical restraints will be necessary to prevent head and leg banging, nose gouging, loss of fingers and lips from biting, and loss of vision from eye rubbing, among other behaviors. In addition to self-injury, older children and teens will become physically and verbally aggressive. The cause of these behaviors is not entirely understood. However, some experts believe they are related to abnormalities in brain chemicals called neurotransmitters. It should be stressed that these children do not want to hurt themselves or others, but they are incapable of preventing these behaviors. Individuals with Lesch-Nyhan syndrome have been described as “doing the opposite” of what they really want. Screening and Diagnosis The doctor will ask about symptoms, behavior traits, and medical history and will perform a physical exam. Tests may include a measurement of HPRT enzyme activity to confirm the diagnosis. Molecular of the HPRT1 gene may be done to confirm the diagnosis and to detect if an unaffected female is a carrier of the gene mutation. Treatment and Therapy There is no treatment to cure Lesch-Nyhan. However, certain medications may help to alleviate some of its symptoms. For example, allopurinol (Aloprim, Zyloprim) may be prescribed to control excessive levels of uric acid in the body. Such treatment should begin as early as possible to prevent complications from developing. Diazepam (Diastat, Valium), haloperidol (Haldol), and phenobarbital (Luminal) can help reduce some of the problem behaviors. A 2006 report suggests that administration of s-adenosylmethionine, a food supplement, may reduce self-mutilating behaviors in adults with Lesch-Nyhan syndrome. This supplement, which is available in health food stores, is naturally synthesized by the human body and is important for many bodily processes. Patients should talk to their health care providers before taking any supplements. Subsequent studies found additional compounds that show promise and can be used as food supplements. With treatment, the average life expectancy for Lesch-Nyhan patients is early to mid-twenties. There may be an increased risk of sudden death due to respiratory causes. However, many patients live longer with good medical and psychological care. Prevention and Outcomes There are no guidelines to prevent Lesch-Nyhan syndrome. Individuals with a family history of this condition can talk to a genetic counselor when deciding whether to have children. Bibliography EBSCO Publishing. Health Library: Lesch-Nyhan Syndrome. Ipswich, Mass.: Author, 2009. Available through Glick, N. “Dramatic Reduction in Self-Injury in Lesch-Nyhan Disease Following S-Adenosylmethionine Administration.” Journal of Inherited Metabolic Disease 29, no. 5 (October 2006): 687. Morales, Pamilla C. “Lesch-Nyhan Syndrome.” In Handbook of Neurodevelopmental and Genetic Disorders in Children, edited by Sam Goldstein and Cecil R. Reynolds. New York: Guilford Press, 1999. Neychev, V. K., and H. A. Jinnah. “Sudden Death in Lesch-Nyhan Disease.” Developmental Medicine and Child Neurology 48, no. 11 (November 2006): 923-926. Ruillier, Valentin, et al. "Rescuing Compounds for Lesch-Nyhan Disease Identified Using Stem Cell-Based Phenotypic Screening." JCI Insight, vol. 5, no. 4, 2020, doi.org/10.1172/jci.insight.132094. Accessed 4 Sept. 2024. Schroeder, Stephan R., Mary Lou Oster-Granite, and Travis Thompson, eds. Self-Injurious Behavior: Gene-Brain-Behavior Relationships. Washington, D.C.: American Psychological Association, 2002. Visser, Jasper E. “Lesch-Nyhan Syndrome.” In Handbook of Neurodevelopmental and Genetic Disorders in Adults, edited by Sam Goldstein and Cecil R. Reynolds. New York: Guilford Press, 2005. 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14218	https://optimization.cbe.cornell.edu/index.php?title=Sequential_quadratic_programming	Sequential quadratic programming From Cornell University Computational Optimization Open Textbook - Optimization Wiki Jump to navigation Jump to search This web page is a duplicate of Authored by: Ben Goodman (ChE 345 Spring 2016) Steward: Dajun Yue and Fenqi You Contents 1 Introduction 2 Background: Prerequisite Methods 2.1 Karush-Kuhn-Tucker (KKT) Conditions and the Lagrangian Function 2.2 The Active Set Method and its Limitations 2.3 Newton's Method 3 The SQP Algorithm 3.1 Example Problem 4 Conclusion 5 Sources Introduction Sequential quadratic programming (SQP) is a class of algorithms for solving non-linear optimization problems (NLP) in the real world. It is powerful enough for real problems because it can handle any degree of non-linearity including non-linearity in the constraints. The main disadvantage is that the method incorporates several derivatives, which likely need to be worked analytically in advance of iterating to a solution, so SQP becomes quite cumbersome for large problems with many variables or constraints. The method dates back to 1963 and was developed and refined in the 1970's . SQP combines two fundamental algorithms for solving non-linear optimization problems: an active set method and Newton’s method, both of which are explained briefly below. Previous exposure to the component methods as well as to Lagrangian multipliers and Karush-Kuhn-Tucker (KKT) conditions is helpful in understanding SQP. The abstracted, general problem below will be used for the remainder of this page to explain and discuss SQP: with f(x), h(x), and g(x) each potentially non-linear. is potentially a vector of many variables for the optimization, in which case h(x) and g(x) are systems. Background: Prerequisite Methods Karush-Kuhn-Tucker (KKT) Conditions and the Lagrangian Function The Lagrangian function combines all the information about the problem into one function using Lagrangian multipliers for equality constraints and for inequality constraints: A single function can be optimized by finding critical points where the gradient is zero. This procedure now includes and as variables (which are vectors for multi-constraint NLP). The system formed from this gradient is given the label KKT conditions: The second KKT condition is merely feasibility; h(x) were constrained to zero in the original NLP. The third KKT condition is a bit trickier in that only the set of active inequality constraints need satisfy this equality, the active set being denoted by . Inequality constraints that are nowhere near the optimal solution are inconsequential, but constraints that actively participate in determining the optimal solution will be at their limit of zero, and thus the third KKT condition holds. Ultimately, the Lagrangian multipliers describe the change in the objective function with respect to a change in a constraint, so is zero for inactive constraints, so those inactive constraints can be considered removed from the Lagrangian function before the gradient is even taken. The Active Set Method and its Limitations The active set method solves the KKT conditions using guess and check to find critical points. Guessing that every inequality constraints is inactive is conventionally the first step. After solving the remaining system for , feasibility can be checked. If any constraints are violated, they should be considered active in the next iteration, and if any multipliers are found to be negative, their constraints should be considered inactive in the next iteration. Efficient convergence and potentially large systems of equations are of some concern, but the main limitation of the active set method is that many of the derivative expressions in the KKT conditions could still be highly non-linear and thus difficult to solve. Indeed, only quadratic problems seem reasonable to tackle with the active set method because the KKT conditions are linear. Sequential Quadratic Programming addresses this key limitation by incorporating a means of handling highly non-linear functions: Newton's Method. Newton's Method The main idea behind Newton's Method is to improve a guess in proportion to how quickly the function is changing at the guess and inversely proportional to how the function is accelerating at the guess. Walking through a few extreme scenarios makes this approach more intuitive: a long, steep incline in a function will not be close to a critical point, so the improvement should be large, and a shallow incline that is rapidly expiring is likely to be near a critical point, so the improvement should be small. The iterations converge to critical values of any function with improvement steps that follow the form below: The negative sign is important. Near minimums, a positive gradient should decrease the guess and vice versa, and the divergence is positive. Near maximums, a positive gradient should increase the guess and vice versa, but the divergence is negative. This sign convention also prevents the algorithm from escaping a single convex or concave region; the improvement will reverse direction if it overshoots. This is an important consideration in non-convex problems with multiple local maximums and minimums. Newton's method will find the critical point closest to the original guess. Incorporating Newton's Method into the active set method will transform the iteration above into a matrix equation. The SQP Algorithm Critical points of the objective function will also be critical points of the Lagrangian function and vice versa because the Lagrangian function is equal to the objective function at a KKT point; all constraints are either equal to zero or inactive. The algorithm is thus simply iterating Newton's method to find critical points of the Lagrangian function. Since the Lagrangian multipliers are additional variables, the iteration forms a system: Recall: Then Unlike the active set method, the need to ever solve a system of non-linear equations has been entirely eliminated in SQP, no matter how non-linear the objective and constraints. Theoretically, If the derivative expressions above can be formulated analytically then coded, software could iterate very quickly because the system doesn't change. In practice, however, it is likely that the divergence will not be an invertible matrix because variables are likely to be linearly bound from above and below. The improvement direction "p" for the Newton's Method iterations is thus typically found in a more indirect fashion: with a quadratic minimization sub-problem that is solved using quadratic algorithms. The subproblem is derived as follows: Since p is an incremental change to the objective function, this equation then resembles a two-term Taylor Series for the derivative of the objective function, which shows that a Taylor expansion with the increment p as a variable is equivalent to a Newton iteration. Decomposing the different equations within this system and cutting the second order term in half to match Taylor Series concepts, a minimization sub-problem can be obtained. This problem is quadratic and thus must be solved with non-linear methods, which once again introduces the need to solve a non-linear problem into the algorithm, but this predictable sub-problem with one variable is much easier to tackle than the parent problem. Example Problem This example problem was chosen for being highly non linear but also easy to solve by inspection as a reference. The objective function Z is a trigonometric identity: The first constraint then just restricts the feasible zone to the first half of a period of the sine function, making the problem convex. The maximum of the sine function within this region occurs at , as shown in Figure 1. The last constraint then makes the problem easy to solve algebraically: and Now, the problem will be solved using the sequential quadratic programming algorithm. The Lagrangian funtion with its gradient and divergence are as follows: with The first important limitation in using SQP is now apparent: the divergence matrix is not invertible because it is not full rank. We will switch to the quadratic minimization sub-problem above, but even with this alternate framework, the gradient of the constraints must be full rank. This can be handled for now by artificially constraining the problem a bit further so that the derivatives of the inequality constraints are not linearly dependent. This can be accomplished through a small modification to the constraint. If , then it is also true that The addition to the left-hand-side is relatively close to zero for the range of possible values of , so the feasible region has not changed much. This complication is certainly annoying, however, for a problem that's easily solved by inspection. It illustrates that SQP is truly best for problems with highly non-linear objectives and constraints. Now, the problem is ready to be solved. The MATLAB code in figure two was implemented, using the function fmincon to solve the minimization subproblems. fmincon is itself an SQP piece of software. In each step, the incumbent guess is plugged into the gradient, hessian, and constraint arrays, which then become parameters for the minimization problem. Conclusion SQP is powerful enough to be used in commercial software but also burdened by some intricacy. In addition to the complication from needing full-rank constraint gradients, the divergence matrix can be very difficult or laborious to assemble analytically. Commercial SQP packages include checks for the feasibility of the sub-problem in order to account for rank deficiencies. In addition to fmincon, SNOPT and FILTERSQP are two other commercial SQP packages, and each uses a different non-linear method to solve the quadratic subproblem. Line search methods and trust-region methods are trusted options for this step, and sub-gradient methods have also been proposed. The other common modification to SQP (ubiquitous to commercial packages) is to invoke quasi-Newton methods in order to avoid computing the Hessian entirely. SQP is thus very much a family of algorithms rather than a stand-alone tool for optimization. At its core, it is a method for turning large, very non-linear problems into a sequence of small quadratic problems to reduce the computational expense of the problem. Sources Nocedal, J. and Wright, S. Numerical Optimization, 2nd. ed., Ch. 18. Springer, 2006. You, Fengqi. Lecture Notes, Chemical Engineering 345 Optimization. Northwestern University, 2015. Retrieved from " Navigation menu
14219	https://www.ijfs.ir/article_45454.html	Pregnancy in Non-Communicating Rudimentary Horn of A Unicornuate Uterus This website uses cookies to ensure you get the best experience on our website. Got it! Toggle navigation Home Search by Contents By Issue By Author By Subject Author Index Keyword Index Indexed in Journal Info Aims and Scope About Journal Editorial Policy Peer Review Process Related Links Journal Metrics Editors Editorial Board Section Editors Reviewers Reviewers List Reviewers Guideline Review Checklist Reviewers invitation Authors Guide for Authors Checklist Submit Manuscript Fee Iranian Authors International Authors Contact Us Login Register Pregnancy in Non-Communicating Rudimentary Horn of A Unicornuate Uterus Document Type : Case Report Authors Mania Kaveh1 Abolfazl Mehdizadeh Kashi2 Kambiz Sadegi3 Forough Forghani4 1 Endometriosis and Gynecological Disorder Research Center, Iran University of Medical Science, Tehran, Iran;Department of Obstetrics and Gynecology, Zabol University of Medical Science, Zabol, Iran 2 Endometriosis and Gynecological Disorder Research Center, Iran University of Medical Science, Tehran, Iran 3 Pain Research Center, Iran University of Medical Science, Tehran, Iran;4Department of Anesthesiology, Zabol University of Medical Science, Zabol, Iran 4 Department of Obstetrics and Gynecology, Zabol University of Medical Science, Zabol, Iran 10.22074/ijfs.2018.5022 Abstract Diagnosis and management of pre-rupture stage of the pregnant horn are difficult and usually missed on a routine ul- trasound scan. Also most cases are detected after rupture of pregnant horn. We presented a 28-year-oldG2 L1 woman with diagnosis of rudimentary horn pregnancy (RHP) at 14 weeks of gestation. We diagnosed her with a normal intrauterine pregnancy, whereas a pregnancy in a right-sided non-communicating rudimentary horn with massive he- moperitoneum was later discovered on laparotomy. RHP has a high risk of death for mother, so there must be a strong clinical suspicion for the diagnosis of RHP. Although there is a major advancement in field of diagnostic ultrasound and other imaging modalities, prenatal diagnosis has remained elusive and a laparotomy surgery is considered as a definitive diagnosis. Keywords Pregnancy Rudimentary Uterus 20.1001.1.2008076.2018.11.4.14.0 Full Text Introduction Rudimentary horn pregnancy (RHP) as a rare incidence, has been estimated at 1:76,000-1:160,000 pregnancies (1). It has also been reported that 75-83% of cases are the pregnancy in non-communicating rudimentary horn that is caused by transmigration of peritoneal sperm or fertilized ovum (2). Gynecological and obstetrical complications of pregnancy in unicornuate uterus with a rudimentary horn are as following: i. Spontaneous abortion, ii. Preterm labor, iii. Infertility, iv. Endometriosis, v. Hematometra, vi. Intrauterine growth restriction (IUGR), vii. Intra peritoneal bleeding, and viii.. Uterine rupture. Kidney abnormalities have also been reported in 31% of cases, while the patients were diagnosed after reaching their stable condition (3). Rupture of RHP is considered as a life threatening condition for mothers. We used timely laparotomy, excision of the horn and blood transfusion to save a 28-year-old G2 L1 woman who was initially diagnosed with a normal intrauterine pregnancy, but a 14-week pregnancy in a right-sided non-communicating rudimentary horn with massive hemoperitoneum was later discovered on laparotomy. Case report A 28-year-old G2L1 woman who was 14 weeks pregnant was admitted at Amiralmomenin Hospital, Zabol, Iran, in November 2015, with generalized abdominal pain, nausea and vomiting. The patient who had a previous cesarean section received early prenatal care two years ago and an ultrasound exam at 14 weeks of gestation (a day before she was admitted at the hospital). The patient suffering from hypovolemic shock, was extremely pale, and had a weak pulse of 120-130 beats per minute and a blood pressure of 80/60 mm Hg. Physical exam revealed impaired consciousness and agitation, generalized abdominal tenderness with sharp right lower quadrant (RLQ) pain, no vaginal bleeding, and a closed cervix. A portable ultrasound detected more than 2 liters of free fluid in the abdomen and pelvis that confirmed the presence of unicorn ate uterus and a 14-week pregnancy in right-sided rudimentary horn. After fluid resuscitation, the patient was transferred to the operating room for an emergency laparotomy. Her blood pressure was 90/60 mm Hg at the time of laparotomy. During laparotomy, we founded that right-sided noncommunicating rudimentary horn was already ruptured and the fetus with amniotic sac extruded into the peritoneal cavity with presence of about a 3-liter hemoperitoneum (Figs .1-3). The rudimentary horn was then excised (Fig .4) and the abdomen closed following hemostasis. Furthermore, patient received 3 unites of pack cell and recovered well after surgery. She was discharged with satisfactory condition on fifth post-operative day after the kidney anomalies ruled out. This study was approved by the Ethics Committee of Iran University of Medical Sciences, Tehran, Iran. Written informed consent was obtained from case. Fig.1 Rudimentary horn pregnancy still attached to the main horn. Fig.2 A 4-cm ruptured rudimentary horn with placenta partially protruding from it. Fig.3 Fetus with placenta. Fig.4 Excised rudimentary horn. Discussion Pregnancy in a non-communicating rudimentary horn is the results of developmental defect of one Müllerian duct or incomplete connection with Müllerian ducts on the opposite site that has been estimated at 1:76,000-1:160,000 pregnancies (1,4). The first case of uterine rupture following RHP was reported by Kanagal and Hanumanalu (5). It has been reported that the timing of a ruptured rudimentary horn that is mainly associated with horn musculature and its ability to hypertrophy is estimated between 5 and 35 weeks. The early diagnosis of RHP is likely to prevent maternal morbidity and mortality. The best management strategies for early diagnosis of RHP are as: i. Ultrasound, ii. Hysterosalpingography, hysteroscopy, laparoscopy, as well as iii. Magnetic resonance imaging (MRI) (5). It is noted that the sensitivity of ultrasound is 26%, although its sensitivity decreases when the maternal age increases (6). However, RHP is likely to be missed by the most experienced radiologist. The most common ultrasound reports that leads to misdiagnose RHP are as follows: i. Tubal pregnancy, ii. Cornal pregnancy, iii. Intrauterine pregnancy, and iv. Abdominal pregnancy (7). It is difficult to confirm a rudimentary horn with thin myometrium diagnosis because of obscuring adjacent anatomical structures. The following diagnostic criteria for RHP were indicated by Tsafrir et al. (8) using ultrasonography: i. Pseudo pattern of asymmetrical bicornuate uterus, ii. Non-continuity between tissue surrounding the gestational sac and the uterine cervical canal, and iii. The presence of myometrial tissue surrounding the gestational sac. Furthermore, a hyper vascularization pattern like placenta accrete is considered as an indication for the diagnosis of RHP that is detected by both Color Doppler ultrasound and Doppler ultrasound. Samuels and Awonuga (9) have reported a uterine rupture after labor induction with misoprostol. The application of different methods of labor induction for termination of RHP was unsuccessful and led to a uterine rupture. The surgical approach is considered as the first management strategy. There are several reports of early diagnosis and laparoscopic excision of rudimentary horn (10,12). Edelman has reported a successful strategy treatment of RHP including the use of methotrexate (MTX) and laparoscopic excision of rudimentary horn in the first weeks of pregnancy (13). Emergency surgery after diagnosis even in cases of un-ruptured rudimentary horn has been recommended (3). Also, prophylactic removal of the rudimentary horn has been suggested (14). There is a report of RHP reaching a full-term delivery that led to live birth using cesarean section (15). The reproductive outcome of a unicornuate uterus is discussed in some articles. However, afew of them have discussed the reproductive outcome after resection of rudimentary horn. Those patients who have ever undergone resection of a rudimentary horn should be considered as a high-risk group in the fallowing pregnancy (16,17). This case report had no ethical consideration for patient. Conclusion RHP has high risk of death for mother, so there must be a strong clinical suspicion for the diagnosis of RHP. Although there is a major advancement in field of diagnostic ultrasound and other imaging modalities, prenatal diagnosis has remained elusive and a laparotomy surgery is considered as a definitive diagnosis. Early diagnosis, timely resuscitation, laparotomy, and blood transfusion are the necessary management steps to save a patient. ) Volume 11, Issue 4 January 2018 Pages 318-320 Files XML PDF 621.07 K Cited by 14 Share "Email") "Print") How to cite RIS EndNote Mendeley BibTeX APA MLA HARVARD CHICAGO VANCOUVER Statistics Article View: 873 PDF Download: 347 × APA Kaveh, M. , Mehdizadeh Kashi, A. , Sadegi, K. and Forghani, F. (2018). Pregnancy in Non-Communicating Rudimentary Horn of A Unicornuate Uterus. International Journal of Fertility and Sterility, 11(4), 318-320. doi: 10.22074/ijfs.2018.5022 × MLA Kaveh, M. , , Mehdizadeh Kashi, A. , , Sadegi, K. , and Forghani, F. . "Pregnancy in Non-Communicating Rudimentary Horn of A Unicornuate Uterus", International Journal of Fertility and Sterility, 11, 4, 2018, 318-320. doi: 10.22074/ijfs.2018.5022 × HARVARD Kaveh, M., Mehdizadeh Kashi, A., Sadegi, K., Forghani, F. (2018). 'Pregnancy in Non-Communicating Rudimentary Horn of A Unicornuate Uterus', International Journal of Fertility and Sterility, 11(4), pp. 318-320. doi: 10.22074/ijfs.2018.5022 × CHICAGO M. Kaveh , A. Mehdizadeh Kashi , K. Sadegi and F. Forghani, "Pregnancy in Non-Communicating Rudimentary Horn of A Unicornuate Uterus," International Journal of Fertility and Sterility, 11 4 (2018): 318-320, doi: 10.22074/ijfs.2018.5022 × VANCOUVER Kaveh, M., Mehdizadeh Kashi, A., Sadegi, K., Forghani, F. Pregnancy in Non-Communicating Rudimentary Horn of A Unicornuate Uterus. International Journal of Fertility and Sterility, 2018; 11(4): 318-320. doi: 10.22074/ijfs.2018.5022 Explore Journal Home About Journal Editorial Board Submit Manuscript Contact Us Sitemap Latest News Royan International E-Summer School (2020)2020-12-01 Related Links Royan Education Royan Institute The Int J Fertil Steril site and its metadata are licensed under a Creative Commons Attribution Non-Commercial 3.0 (CC BY-NC 3.0). Newsletter Subscription Subscribe to the journal newsletter and receive the latest news and updates Subscribe © Journal management system. designed by sinaweb ×
14220	https://www.khanacademy.org/test-prep/lsat-prep/xdf35b2883be7178a:lsat-prep-lessons/xdf35b2883be7178a:lsat-prep-logic-toolbox/a/logic-toolbox--article--if-x-then-y--sufficiency-and-necessity	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
14221	https://www.investopedia.com/articles/optioninvestor/03/073003.asp	Getting Acquainted With Options Trading Skip to content News Markets Companies Earnings CD Rates Mortgage Rates Economy Government Crypto Live Markets News Personal Finance View All Investing Stocks Cryptocurrency Bonds ETFs Options and Derivatives Commodities Trading Automated Investing Brokers Fundamental Analysis Markets View All Simulator Login / Portfolio Trade Research My Games Leaderboard Banking Savings Accounts Certificates of Deposit (CDs) Money Market Accounts Checking Accounts View All Personal Finance Budgeting and Saving Personal Loans Insurance Mortgages Credit and Debt Student Loans Taxes Credit Cards Financial Literacy Retirement View All Economy Government and Policy Monetary Policy Fiscal Policy Economics View All Reviews Best Online Brokers Best Crypto Exchanges Best Savings Rates Best CD Rates Best Life Insurance Best Mortgage Rates Best Robo-Advisors Best Personal Loans Best Debt Relief Companies View All Trade Search Search Please fill out this field. 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News News Markets Companies Earnings CD Rates Mortgage Rates Economy Government Crypto Live Markets News Personal Finance View All Investing Investing Stocks Cryptocurrency Bonds ETFs Options and Derivatives Commodities Trading Automated Investing Brokers Fundamental Analysis Markets View All Simulator Simulator Login / Portfolio Trade Research My Games Leaderboard Banking Banking Savings Accounts Certificates of Deposit (CDs) Money Market Accounts Checking Accounts View All Personal Finance Personal Finance Budgeting and Saving Personal Loans Insurance Mortgages Credit and Debt Student Loans Taxes Credit Cards Financial Literacy Retirement View All Economy Economy Government and Policy Monetary Policy Fiscal Policy Economics View All Reviews Reviews Best Online Brokers Best Crypto Exchanges Best Savings Rates Best CD Rates Best Life Insurance Best Mortgage Rates Best Robo-Advisors Best Personal Loans Best Debt Relief Companies View All Follow Us Table of Contents Expand Table of Contents How Options Work Types of Options Option Pricing Options Profitability Expiration Dates FAQs The Bottom Line Getting Acquainted With Options Trading By Barclay Palmer Full Bio Barclay Palmer is a creative executive with 10+ years of creating or managing premium programming and brands/businesses across various platforms. Learn about our editorial policies Updated October 11, 2024 Reviewed by Thomas Brock Reviewed by Thomas Brock Full Bio Thomas J. Brock is a CFA and CPA with more than 20 years of experience in various areas including investing, insurance portfolio management, finance and accounting, personal investment and financial planning advice, and development of educational materials about life insurance and annuities. Learn about our Financial Review Board Fact checked by Kimberly Overcast Fact checked by Kimberly Overcast Full Bio Kimberly Overcast is an award-winning writer and fact-checker. She has ghostwritten political, health, and Christian nonfiction books for several authors, including several New York Times bestsellers. Kimberly also holds a Class C private investigator license. Learn about our editorial policies Part of the Series Day Trading Introduction Day Trading: The Basics and How To Get Started Day Trading Basics Day Trader Definition Day Trading Rate of Return Day Trading Instruments Stock Definition Options TradingCURRENT ARTICLE Forex (FX) Trading Trading Platforms, Tools, Brokers Trading Platform Definition Choosing Day Trading Software Choosing an Online Broker Paper Day Trading Trading Order Types Market Order Definition Limit Order Definition Stop Order Definition Strategy & Analysis Beginner Day Trading Strategies Choosing Stocks for Day Trading 10 Rules For Successful Trading Short Selling Risk & Money Management Guide to Day Trading on Margin Market Risk Risk Management Techniques Risk/Reward Ratio Definition Day Trading Psychology Trading Psychology Definition Importance of Trading Psychology 0 seconds of 0 seconds Volume 0% Press shift question mark to access a list of keyboard shortcuts Keyboard Shortcuts Enabled Disabled Shortcuts Open/Close/ or ? Play/Pause SPACE Increase Volume↑ Decrease Volume↓ Seek Forward→ Seek Backward← Captions On/Off c Fullscreen/Exit Fullscreen f Mute/Unmute m Decrease Caption Size- Increase Caption Size+ or = Seek %0-9 Next Up 10 Biggest Companies in the World 00:34 Subtitle Settings Off English Font Color White Font Opacity 100% Font Size 100% Font Family Arial Character Edge None Edge Color Black Background Color Black Background Opacity 65 Window Color Black Window Opacity 0% Reset White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25% 200%175%150%125%100%75%50% Arial Courier Georgia Impact Lucida Console Tahoma Times New Roman Trebuchet MS Verdana None Raised Depressed Uniform Drop Shadow White Black Red Green Blue Yellow Magenta Cyan White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25%0% White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25%0% Loading ad Live 00:25 00:00 02:04 More Videos 02:04 How to Get Acquainted with Options Trading 00:34 10 Biggest Companies in the World 00:36 25 Highest Paid Occupations in the US 00:32 8 Most Successful Products From "Shark Tank" 00:32 6 Strategies to Protect Income From Taxes 00:35 9 States With No Income Tax Close Trading options is very different from trading stocks because options have distinct characteristics from stocks. Investors need to take the time to understand the terminology and concepts involved with options before trading them. Options are financial derivatives, meaning that they derive their value from the underlying security or stock. Options give the buyer the right, but not the obligation, to buy or sell the underlying stock at a pre-determined price. Key Takeaways Options give a buyer the right, but not the obligation, to buy (call) or sell (put) the underlying stock at a pre-set price called the strike price. Options have a cost associated with them, called a premium, and expiration date. A call option is profitable when the strike price is below the stock's market price since the trader can buy the stock at a lower price. A put option is profitable when the strike is higher than the stock's market price since the trader can sell the stock at a higher price. Alison Czinkota {Copyright} Investopedia, 2019. How Stock Options Trading Works Trading options is more like betting on horses at the racetrack: Each person bets against all the other people there. The track simply takes a small cut for the facilities. So trading options, like betting at the horse track, is a zero-sum game. The option buyer's gain is the option seller's loss and vice versa. One important difference between stocks and options is that stocks give you a small piece of ownership in a company, while options are just contracts that give you the right to buy or sell the stock at a specific price by a specific date. It's important to remember that there are always two sides to every option transaction: a buyer and a seller. In other words, for every option purchased, there's always someone else selling it. Important Options are not meant for novice traders. They are complicated and sophisticated financial assets that should only be traded once you develop a lot of experience. Types of Options The two types of options are calls and puts: When you buy a call option, you have the right, but not the obligation, to purchase a stock at a set price, called the strike price, any time before the option expires. When you buy a put option, you have the right, but not the obligation, to sell a stock at the strike price any time before the expiration date. When individuals sell options, they effectively create a security that didn't exist before. This is known as writing an option, and it explains one of the main sources of options since neither the associated company nor the options exchange issues the options. \| \| Call Options \| Put Options \| --- \| Description \| Buyer has the right (not the obligation) to buy the underlying security \| Buyer has the right (not the obligation) to sell the underlying security \| \| Buy When? \| The price of the underlying asset will rise \| The price of the underlying asset will drop \| \| Sell When? \| The price of the underlying asset will drop \| The price of the underlying asset will rise \| When you write a call, you may be obligated to sell shares at the strike price any time before the expiration date. When you write a put, you may be obligated to buy shares at the strike price any time before expiration. There are also two basic styles of options: American and European. An American-style option can be exercised at any time between the date of purchase and the expiration date. A European-style option can only be exercised on the expiration date.1 Most exchange-traded options are American style, and all stock options are American style. Many index options are European style. Option Pricing The price of an option is called the premium. The buyer of an option can't lose more than the initial premium paid for the contract, no matter what happens to the underlying security. So the risk to the buyer is never more than the amount paid for the option. The profit potential, on the other hand, is theoretically unlimited. In return for the premium received from the buyer, the seller of an option assumes the risk of having to deliver (if a call option) or taking delivery (if a put option) of the shares of the stock. Unless that option is covered by another option or a position in the underlying stock, the seller's loss can be open-ended, meaning the seller can lose much more than the original premium received. Please note that options are not available at just any price. Stock options are generally traded with strike prices in intervals of $0.50 or $1, but can also be in intervals of $2.50 and $5 for higher-priced stocks.2 Also, only strike prices within a reasonable range around the current stock price are generally traded. Far in- or out-of-the-money options might not be available. Options Profitability When the strike price of a call option is above the current price of the stock, the call is not profitable or out-of-the-money. In other words, an investor is not going to buy a stock at a higher price (the strike) than the current market price of the stock. When the call option strike price is below the stock's price, it's considered in-the-money since the investor can buy the stock for a lower price than in the current market. Put options are the exact opposite. They're considered out-of-the-money when the strike price is below the stock price since an investor wouldn't sell the stock at a lower price (the strike) than in the market. Put options are in the money when the strike price is above the stock price since investors can sell the stock at a higher (strike) price than the market price of the stock. Expiration Dates All stock options expire on a certain date, called the expiration date. For normal listed options, this can be up to nine months from the date the options are first listed for trading. Longer-term option contracts, called long-term equity anticipation securities (LEAPS), are also available on many stocks. These can have expiration dates up to three years from the listing date.3 Options expire at market close on Friday, unless it falls on a market holiday, in which case expiration is moved back one business day.Monthly options expire on the third Friday of the expiration month, while weekly options expire on each of the other Fridays in a month.4 Like shares of stock, options settle under the T+1 rule. This means they settle the next day. To settle on the expiration date, you have to exercise or trade the option by the end of the day on Friday.5 What Is a Stock Options Contract? A stock option contract entitles the owner of the contract to 100 shares of the underlying stock upon expiration.6 So, if you purchase seven call option contracts, you are acquiring the right to purchase 700 shares.And, if the owner of a call option decides to exercise their right to buy the stock at a particular price, the option writer must deliver the stock at that price. What Do Stock Options Cost? Options contracts usually represent 100 shares of the underlying security, and the buyer will pay a premium fee for each contract.6 For example, if an option has a premium of $0.55 per contract, buying one option would cost$55 ($0.55 x 100 = $55). How Do You Make Money Trading Options? You can make money by being anoptionbuyer or an option writer.If you are a call option buyer, you can make a profit if the underlying stock rises above the strike price before the expiration date. If you are a put option buyer, you can make a profit if the price falls below the strike price before the expiration date. Is Options Trading Better Than Stocks? Options trading can be riskier than trading stocks. However, when it is done properly, it can be more profitable for the investor than traditional stock market investing. The Bottom Line Options are complex financial contracts that are based on the price of an underlying security like stocks. bonds, and ETFs among others. 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14222	https://www.khanacademy.org/math/math3/x5549cc1686316ba5:rationals/x5549cc1686316ba5:common-factors/v/simplifying-rational-expressions-w-higher-degree-terms	Simplifying rational expressions: higher degree terms (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Math: Florida B.E.S.T. Grade 3 math (FL B.E.S.T.) Grade 4 math (FL B.E.S.T.) Grade 5 math (FL B.E.S.T.) Grade 6 math (FL B.E.S.T.) Grade 7 math (FL B.E.S.T.) Grade 8 math (FL B.E.S.T.) Algebra 1 (FL B.E.S.T.) Geometry (FL B.E.S.T.) Algebra 2 (FL B.E.S.T.) 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Skip to lesson content Integrated math 3 Course: Integrated math 3>Unit 13 Lesson 1: Cancelling common factors Reducing rational expressions to lowest terms Intro to rational expressions Reducing rational expressions to lowest terms Simplifying rational expressions: common monomial factors Reduce rational expressions to lowest terms: Error analysis Simplifying rational expressions: common binomial factors Simplifying rational expressions: opposite common binomial factors Simplifying rational expressions (advanced) Reduce rational expressions to lowest terms Simplifying rational expressions: grouping Simplifying rational expressions: higher degree terms Simplifying rational expressions: two variables Simplify rational expressions (advanced) Math> Integrated math 3> Rational functions> Cancelling common factors © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Simplifying rational expressions: higher degree terms FL.BEST.Math: MA.912.AR.1.9 Google Classroom Microsoft Teams About About this video Transcript Sal simplifies & states the domain of (x⁴+8x²+7)/(3x⁵-3x). Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Manuel Del Río Rodríguez 9 years ago Posted 9 years ago. Direct link to Manuel Del Río Rodríguez's post “As the exercise is about ...” more As the exercise is about 'simplifying' the rational expression, would it not have been better to put in the denominator 3x(x+1)(x-1)? It is the equivalent of Sal's two examples, but more 'factored out' if you will. Answer Button navigates to signup page •Comment Button navigates to signup page (16 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Clement 9 years ago Posted 9 years ago. Direct link to Clement's post “You are right. But it mig...” more You are right. But it might be more simplified in the last form. I did notice Sal missed a step. Sal might have wanted to avoid the long process of multiplying 3x(x+1)(x-1) which equals to 3x(x^2-1). See, it would lead us to the same step. In conclusion, we simplify the denominator to lead us no factors left behind. Hopefully it helps! Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Walter Green 8 years ago Posted 8 years ago. Direct link to Walter Green's post “In the case of test quest...” more In the case of test questions in general, do you expect answers in an expanded form or in a factored form after cancellations? That has confused me and caused some of my answers, which were technically correct, to be counted as incorrect. I do note that on some questions you do specify "expanded" form for answers. Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer R. E. Banks 8 years ago Posted 8 years ago. Direct link to R. E. Banks's post “In my experience, Khan Ac...” more In my experience, Khan Academy Practice Problems software will accept answers whether they're in expanded form or not; for example, they list both '(x - 2)/4(x-3)' and '(x - 2)/(4x-12)' as answers to the question. So maybe you're not simplifying down to those two basic equations? One thing that kept marking me as wrong (and that I incorrectly assumed was due to the fact of my simplification in the wrong form) was that I did not choose a second option for what x cannot equal; I thought it was not needed to say that x cannot -3 in the equation (x+6)/(x+3), since in previous videos Mr. Khan said it was unnecessary since it was obvious if you just look at the equation; but the Practice Problems want you to select ALL the correct options that x cannot be equal to. Hope this solves some of your problems, Walter. Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more antonio marcos 4 years ago Posted 4 years ago. Direct link to antonio marcos's post “Can I simplify x⁴+8x²+7 t...” more Can I simplify x⁴+8x²+7 to x⁴+5x+3x+7? Then I could regroup x⁴+5x+3x+7 to 3x⁵+5x+7? I'm doing this in order to simplify (x⁴+8x²+7)/(3x⁵-3x). By regrouping the numerator I'm trying to eliminate the 3x⁵ from both expressions. Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 4 years ago Posted 4 years ago. Direct link to Kim Seidel's post “You have multiple errors....” more You have multiple errors. 1) 5x+3x is not the same as 8x². Addition does not change exponents. 2) Similarly, x⁴+3x does not equal 3x⁵. Addition is not the same as multiplication which is what you would have done to get to 3x⁵ 3) Making the first term into 3x⁵ will not allow to you eliminate the 3x⁵ in the denominator. When we reduce fractions, we can only cancel out common factors (items being multiplied). You are trying to cancel out terms (items being added/subtracted). To do your problem, you need to completely factor both the numerator and denominator. And then only cancel out common factors. Hope this helps. Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Alissa 5 years ago Posted 5 years ago. Direct link to Alissa's post “I need help factoring x³-...” more I need help factoring x³-1. The answer is (x-1)(x²+x-1). I see that the answer checks out when I multiply it together, but I don't understand how to arrive at the answer from the problem. Can anyone explain or point me to a video that will explain? Thought this one might, but I still don't understand my problem, hence my asking this. Thanks! Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 5 years ago Posted 5 years ago. Direct link to Kim Seidel's post “You have a difference of ...” more You have a difference of 2 cubes. It is factored using a pattern. Search on KA for "factoring difference of cubes" and you should find the video. FYI - You have a sign error on the last 1, it should be +1. x³-1 = (x-1)(x²+x+1) Hope this helps. 1 comment Comment on Kim Seidel's post “You have a difference of ...” (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Arbaaz Ibrahim 6 years ago Posted 6 years ago. Direct link to Arbaaz Ibrahim's post “At about 5 minutes into t...” more At about 5 minutes into the video, Sal cancelled out x^2+1. Can someone please explain why this can never be equal to zero? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer kubleeka 6 years ago Posted 6 years ago. Direct link to kubleeka's post “If x²+1=0, then x²=-1. So...” more If x²+1=0, then x²=-1. So multiplying a number by itself yields a negative number, which is impossible in the real numbers. So x²+1 cannot be 0. 1 comment Comment on kubleeka's post “If x²+1=0, then x²=-1. So...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Federico 6 years ago Posted 6 years ago. Direct link to Federico's post “In the exercises later th...” more In the exercises later they ask: "Simplify the following rational expression and express in expanded form." What does mean "express in expanded form"? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Beaniebopbunyip 6 years ago Posted 6 years ago. Direct link to Beaniebopbunyip's post “Don’t leave something in ...” more Don’t leave something in factored form, like 2(x+3). You want to multiply everything out like this: 2x+6. Hope this helps! 1 comment Comment on Beaniebopbunyip's post “Don’t leave something in ...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more troy0bush881 8 years ago Posted 8 years ago. Direct link to troy0bush881's post “At 0:37 where did the oth...” more At 0:37 where did the other x^2 go? because i thought it was an x^4... ?? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Evan Kim 25 days ago Posted 25 days ago. Direct link to Evan Kim's post “wait so if it cant work f...” more wait so if it cant work for -1 bc its squared (x^2 +1) then why is -1 included in the list? (+_ 1) Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Video transcript [Voiceover] Let's see if we can simplify this expression. So pause the video and have a try at it, and then we're gonna do it together right now. All right, so when you look at this, it looks like both the numerator and denominator, they might, you might be able to factor them, and maybe they have some common factors that you can divide the numerator and the denominator by to simplify it. So let's first try to factor the numerator. X to the fourth, plus eight x squared, plus seven. At first it might be a little intimidating, because you have an x to the fourth here. It's not a quadratic; it's a fourth degree polynomial, but like any, if you, like a lot of quadratics that we've seen in the past, it does seem to have a pattern. For example, if this said x squared plus eight x plus seven, you'd say, oh, well, this is pretty straightforward to factor. What two numbers add up to eight and when I take their product I get seven? Well, there's only two numbers where you take their product and you get positive seven that are going to be positive, and they need to be positive, if they're going to add up to positive eight, and that's one and seven. So this would be x plus seven times x plus one. Well, if you just think of, instead of thinking in terms of x and x squared, if you just think in terms of x squared and x to the fourth, it's going to be the exact same thing. So this thing can be written as x squared plus seven times x squared plus one. If you want, you can do some type of a substitution saying, saying that a is equaled to x squared, in which case, so if you said that a is equal to x squared, then this thing would become a squared plus eight a plus seven, and then you would factor this into a plus seven and a plus one, and then you would undo the substitution, and that's x squared plus seven and x squared plus one. But hopefully you see what's going on here; this is the higher order term, and then this is half the degree of that, so it fits this mold. And so you could do a substitution, or you could just recognize, oh okay, instead of dealing with x squared, I'm dealing with x to the fourth. All right, so that's the numerator. Now let's think about, let's think about the denominator. So the denominator, both of these terms are divisible by three x. So let's factor out a three x. So it's three x times, three x times, if you factor out a three x here, three divided by three is one, x to the fifth divided by x is x to the fourth, and then if you factor out a three x here, you're just gonna get one. And so far this doesn't seem too helpful. I don't see an x to the fourth minus one, or a three x in the numerator, but maybe I can factor this out further, x to the fourth minus one. And that's because it is a difference of squares. And you might say, wait, I'm always used to recognizing a difference of squares as something like a squared minus one, which you could write as a plus one times a minus one. Well, this would be a squared minus one if you say that a is equal to x squared. Then this would be a squared minus one. So let's rewrite all of this. So let's rewrite. So this is all going to be equal to, same numerator, let's see, let's do it in green. Same numerator: x squared plus seven, can't factor that out any more, times x squared plus one, can't factor that out any more, all of that over three x, but this I can view as a difference of squares. So this is x squared squared, and this is obviously one squared, so this is going to be x squared plus one times x squared minus, times x squared minus one. Now clearly have an x squared in the numerator, x squared minus one in the numerator, x squared, sorry, x squared plus one in the numerator, x squared plus one in the denominator, and so I could cancel them out, and I'm going to be left with, in the numerator, x squared plus seven, over three x times x squared minus one. Now, this looks pretty simple, and we want to be a little careful, because whenever we do this cancelling out we don't want, we want to make sure that we restrict the xes for which the expression's defined, if we want them to be algebraically equivalent. So this one, would this be, this would obviously be undefined, so x cannot be equal to zero, x cannot be equal to plus or minus one. Positive or negative one would make this expression right over here equal zero, so it cannot be equal to zero, x cannot be equal to, I'll write plus or minus one; that would make this part zero. But this right over here, this one, unless, we're assuming we're dealing only with real numbers, this one can't ever equal zero if you're dealing with real numbers, because x squared is always going to be non-negative and you're adding it to a positive value, and so this part, this factor, would have never made the entire thing undefined. So we can actually just factor it out, or cancel it out, without worrying much about it. And so this is actually algebraically equivalent to what we had originally. Now we could write these constraints on it, if we want. If someone were to ask me, you know, for what x is this expression not defined, well, it's clear it's not defined for x, that would make the denominator zero, dividing by zero not defined, or if x is plus or minus one, it would make the denominator equal zero. But that is, that comes straight of this expression, so this expression and our original expression are algebraically equivalent. Now if you wanted to, you could expand the bottom out a little bit, you could multiply it out if you like. That's equivalent to, so x squared plus seven over three x times x squared is three x to the third minus three x. So these are all, these are all equivalent expressions, and we are done. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: video Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. 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14223	https://arxiv.org/abs/2211.06320	[2211.06320] Fast Evaluation of Real and Complex Polynomials Skip to main content We gratefully acknowledge support from the Simons Foundation, member institutions, and all contributors.Donate >math> arXiv:2211.06320 Help \| Advanced Search Search GO quick links Login Help Pages About Mathematics > Numerical Analysis arXiv:2211.06320 (math) [Submitted on 20 Oct 2022] Title:Fast Evaluation of Real and Complex Polynomials Authors:Ramona Anton (IMJ-PRG (UMR_7586)), Nicolae Mihalache (LAMA), François Vigneron (LMR) View a PDF of the paper titled Fast Evaluation of Real and Complex Polynomials, by Ramona Anton (IMJ-PRG (UMR_7586)) and 2 other authors View PDF Abstract:We propose an algorithm for quickly evaluating polynomials. It pre-conditions a complex polynomial Pof degree din time O(d\log d), with a low multiplicative constant independent of the precision. Subsequent evaluations of Pcomputed with a fixed precision of pbits are performed in average arithmetic complexity O\big(\sqrt{d(p+\log d)}\big)and memory O(dp). The average complexity is computed with respect to points z \in \mathbb{C}, weighted by the spherical area of \overline{\mathbb{C}}. The worst case does not exceed the complexity of H{ö}rner's scheme. In particular, our algorithm performs asymptotically as O(\sqrt{d\log d})per evaluation. For many classes of polynomials, in particular those with random coefficients in a bounded region of \mathbb{C}, or for sparse polynomials, our algorithm performs much better than this upper bound, without any modification or this http URLarticle contains a detailed analysis of the complexity and a full error analysis, which guarantees that the algorithm performs as well as H\''orner's scheme, only faster. Our algorithm is implemented in a companion library, written in standard C and released as an open-source project [MV22].Our claims regarding complexity and accuracy are confirmed in practice by a set of comprehensive benchmarks. Subjects:Numerical Analysis (math.NA); Mathematical Software (cs.MS); Dynamical Systems (math.DS) Cite as:arXiv:2211.06320 [math.NA] (or arXiv:2211.06320v1 [math.NA] for this version) Focus to learn more arXiv-issued DOI via DataCite Submission history From: Francois Vigneron [view email][via CCSD proxy] [v1] Thu, 20 Oct 2022 06:35:25 UTC (7,703 KB) Full-text links: Access Paper: View a PDF of the paper titled Fast Evaluation of Real and Complex Polynomials, by Ramona Anton (IMJ-PRG (UMR_7586)) and 2 other authors View PDF TeX Source Other Formats view license Current browse context: math.NA <prev \| next> new \| recent \| 2022-11 Change to browse by: cs cs.MS cs.NA math math.DS References & Citations NASA ADS Google Scholar Semantic Scholar aexport BibTeX citation Loading... BibTeX formatted citation × Data provided by: Bookmark Bibliographic Tools Bibliographic and Citation Tools [x] Bibliographic Explorer Toggle Bibliographic Explorer (What is the Explorer?) [x] Connected Papers Toggle Connected Papers (What is Connected Papers?) [x] Litmaps Toggle Litmaps (What is Litmaps?) [x] scite.ai Toggle scite Smart Citations (What are Smart Citations?) Code, Data, Media Code, Data and Media Associated with this Article [x] alphaXiv Toggle alphaXiv (What is alphaXiv?) [x] Links to Code Toggle CatalyzeX Code Finder for Papers (What is CatalyzeX?) [x] DagsHub Toggle DagsHub (What is DagsHub?) [x] GotitPub Toggle Gotit.pub (What is GotitPub?) [x] Huggingface Toggle Hugging Face (What is Huggingface?) [x] Links to Code Toggle Papers with Code (What is Papers with Code?) [x] ScienceCast Toggle ScienceCast (What is ScienceCast?) Demos Demos [x] Replicate Toggle Replicate (What is Replicate?) [x] Spaces Toggle Hugging Face Spaces (What is Spaces?) [x] Spaces Toggle TXYZ.AI (What is TXYZ.AI?) Related Papers Recommenders and Search Tools [x] Link to Influence Flower Influence Flower (What are Influence Flowers?) [x] Core recommender toggle CORE Recommender (What is CORE?) Author Venue Institution Topic About arXivLabs arXivLabs: experimental projects with community collaborators arXivLabs is a framework that allows collaborators to develop and share new arXiv features directly on our website. Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. arXiv is committed to these values and only works with partners that adhere to them. Have an idea for a project that will add value for arXiv's community? Learn more about arXivLabs. Which authors of this paper are endorsers? \| Disable MathJax (What is MathJax?) About Help Contact Subscribe Copyright Privacy Policy Web Accessibility Assistance arXiv Operational Status Get status notifications via email or slack
14224	https://support.google.com/docs/answer/9083932?hl=en	AMORLINC function - Google Docs Editors Help Skip to main content Google Docs Editors Help Sign in Send feedback on... This help content & information General Help Center experience Next Help Center Community Gemini in Docs Editors Google Docs Editors AMORLINC function The AMORLINC function returns the depreciation for an accounting period, or the prorated depreciation if the asset was purchased in the middle of a period.This function is available for users of the French accounting system. Parts of an AMORLINC function AMORLINC(cost, purchase_date, first_period_end, salvage, period, rate, [basis]) PartDescriptionNotes costThe asset's purchase cost purchase_dateThe date the asset was purchased The purchase date should be before the first period end date. first_period_endThe end date of the first period salvageThe asset's value at the end of its life (i.e. its salvage value) periodThe period for which to calculate depreciation The period should be a non-negative value. Fractional values less than 1 automatically round up to 1, and fractional values greater than 1 round down. rateThe annual depreciation rate. The depreciation rate may be specified as either a decimal or a percentage. day_count_convention(Optional)An indicator of what day count method to use, marked 0 by default 0indicates US (NASD) 30/360. This assumes 30-day months and 360-day years, per the National Association of Securities Dealers (NASD) standard, and performs specific adjustments to entered dates that fall at the ends of months. 1indicates Actual/Actual. This calculates based on the actual number of days between the specified dates and the actual number of days in the intervening years. 2indicates Actual/360. This calculates based on the actual number of days between the specified dates, but assumes a 360-day year. 3indicates Actual/365. This calculates based on the actual number of days between the specified dates, but assumes a 365-day year. 4indicates European 30/360. Similar to0, this calculates on a 30-day month and a 360-day year, but adjusts end-of-month dates according to European financial conventions. basis(Optional)The year basis to use Sample formulas AMORLINC(1000, "7/20/1969", "8/20/1969", 100, 6, 15%) AMORLINC(1234.56, DATE(1969, 7, 20), DATE(1969, 8, 20), 123.45, 6.5, 0.15, 1) AMORLINC(A1, A2, A3, A4, 6, 15%) Examples This example shows the sixth period depreciation of an asset with a purchase cost of $1,000, a purchase date of 7/20/1969, a first period end date of 8/20/1969, a salvage value of $100, and a depreciation rate of 15% using the default 30-day month and 360-day year counting convention: \| \| A \| B \| --- \| 1 \| Cost \| $1,000 \| \| 2 \| Purchase date \| 7/20/1969 \| \| 3 \| First period end date \| 8/20/1969 \| \| 4 \| Salvage value \| $100 \| \| 5 \| Period \| 6 \| \| 6 \| Depreciation rate \| 15% \| \| 7 \| Result \| 137.5 \| \| 8 \| Formula \| =AMORLINC(B1, B2, B3, B4, B5, B6) \| This example shows the sixth period depreciation of an asset with a purchase cost of $1,000, a purchase date of 7/20/1969, a first period end date of 8/20/1969, a salvage value of $100, and a depreciation rate of 15%using the actual days-per-month and actual days-per-year day counting convention: \| \| A \| B \| --- \| 1 \| Cost \| $1,000 \| \| 2 \| Purchase date \| 7/20/1969 \| \| 3 \| First period end date \| 8/20/1969 \| \| 4 \| Salvage value \| $100 \| \| 5 \| Period \| 6 \| \| 6 \| Depreciation rate \| 15% \| \| 7 \| Day count convention \| 1 \| \| 8 \| Formula \| =AMORLINC(B1, B2, B3, B4, B5, B6, B7) \| \| 9 \| Result \| 137.26 \| Related functions DDB:The DDB function calculates the depreciation of an asset for a specified period using the double-declining balance method. VDB:The VDB function returns the depreciation of an asset for a particular period (or partial period). DB:The DB function calculates the depreciation of an asset for a specified period using the arithmetic declining balance method. SLN:The SLN function calculates the depreciation of an asset for one period using the straight-line method. SYD:The SYD function calculates the depreciation of an asset for a specified period using the sum of years digits method. Give feedback about this article Choose a section to give feedback on Need more help? Try these next steps: Post to the help community Get answers from community members true Financial 1 of 51 Google Sheets function list 2 of 51 ACCRINT 3 of 51 ACCRINTM 4 of 51 AMORLINC function 5 of 51 COUPDAYBS 6 of 51 COUPDAYS 7 of 51 COUPDAYSNC 8 of 51 COUPNCD 9 of 51 COUPNUM 10 of 51 COUPPCD 11 of 51 CUMIPMT 12 of 51 CUMPRINC 13 of 51 DB 14 of 51 DDB 15 of 51 DISC 16 of 51 DOLLARDE 17 of 51 DOLLARFR 18 of 51 DURATION 19 of 51 EFFECT 20 of 51 FV 21 of 51 FVSCHEDULE 22 of 51 INTRATE 23 of 51 IPMT 24 of 51 IRR 25 of 51 ISPMT function 26 of 51 MDURATION 27 of 51 MIRR 28 of 51 NOMINAL 29 of 51 NPER 30 of 51 NPV function 31 of 51 PDURATION function 32 of 51 PMT 33 of 51 PPMT 34 of 51 PRICE 35 of 51 PRICEDISC 36 of 51 PRICEMAT 37 of 51 PV 38 of 51 RATE 39 of 51 RECEIVED 40 of 51 RRI function 41 of 51 SLN 42 of 51 SYD 43 of 51 TBILLEQ 44 of 51 TBILLPRICE 45 of 51 TBILLYIELD 46 of 51 VDB function 47 of 51 XIRR 48 of 51 XNPV 49 of 51 YIELD 50 of 51 YIELDDISC 51 of 51 YIELDMAT function Visit the Learning Center Using Google products, like Google Docs, at work or school? 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14225	http://www.ext.msstate.edu/sites/default/files/topic-files/drying/dry-kiln-operators-manual/chapter10.pdf	Chapter 10 Log and Lumber Storage Log storage 220 Kiln drying is only one step in the harvesting, handling, Dry storage 220 and processing of wood products. The best results can Logs with bark 221 be obtained in kiln drying, therefore, when adequate Debarked logs 222 attention is paid to related phases of wood processing. Transpiration drying 222 Although a dry kiln operator may have no responsi-Wet storage 223 bility for these related phases, knowledge of them is Pond storage 223 required to understand how they interact with dry-Water sprinkling 224 ing. Problems that occur in drying, or that are erro-Effects of climate on lumber storage 225 neously blamed on drying, are sometimes related to the Relative humidity 225 methods used to store logs and lumber before drying Temperature 225 and those used to store kiln-dried lumber and finished Rainfall 225 products. Average equilibrium moisture content conditions by region and season 225 Lumber storage 225 Outdoor storage 225 Green lumber 226 Partly dried lumber 226 Kiln-dried lumber 226 Pile covers 228 Open shed storage 228 Green lumber 228 Partly dried lumber 228 Kiln-dried lumber 228 Closed, unheated shed storage 229 Green lumber 229 Partly dried lumber 229 Kiln-dried lumber 229 Closed, heated shed storage 230 Green lumber 230 Partly dried lumber 230 Kiln-dried lumber 230 Conditioned storage sheds 230 Treating stored lumber 230 When is chemical treatment needed? 231 When and where to apply treatment 232 How to apply treatment 232 Treating area and equipment 232 Dipping operation 233 Treating for insect control 233 Precautions for handling chemicals 233 Lumber handling and storage in transit 234 Truck transport 234 Rail transport 235 Ship transport 235 Literature cited 236 Sources of additional information 236 Tables 237 Logs and lumber go through various storage and trans-port periods while moving through the processing se-quence. Log storage and transit really begin when the tree is felled and continue until the log is sawed into lumber. Similarly, lumber storage and transit include the time between sawing and drying and the time be-tween drying and end use. The moisture content of lumber should be controlled in storage and transit. Large increases in moisture content during storage may make lumber unsuitable or out of specifications for many uses, cause lumber to warp, or cause the devel-opment of stain or decay. Large decreases in moisture content may cause checks and warp to occur or make machining and fastening difficult. Chapter 10 was revised by William T. Simpson, Supervisory Research Forest Products Technologist, and James C. Ward, Research Forest Products Technologist. 219 Figure 10-1—Splits in black cherry millwork from lumber that was sawn from a wind-damaged tree. (M88 0170) Log Storage The source of some lumber drying problems can be traced to changes in the wood that began in the tree just before or during the timber harvesting operation. Logs that have been salvaged from forests that were damaged by hurricanes or tornadoes may yield lum-ber that is likely to split during drying and subsequent machining (fig. 10-1). Felling the tree with a clipping or shearing machine can initiate radial splits and ring failures in the end of the log, which may lengthen con-siderably with lumber drying. After felling, the main stem of the tree is detached from the crown, except when transpiration drying is desired. At most commercial logging operations in North Amer-ica, the main stem of the felled tree is either left full length (tree-length log) or cut (bucked) into shorter logs with lengths that correspond to lengths of the in-tended lumber. Tree-length and standard-length logs should be sawed into lumber as soon as possible af-ter felling, especially during warm weather. However, prompt sawing of logs is not always possible because of log transportation difficulties or the economic need to stockpile logs at the sawmill. This section suggests methods for reducing drying defects that result from prolonged storage of logs. Logs need to be stored under conditions that will min-imize defects associated with shrinkage, mainly end checking, and attacks by fungi, bacteria, and insects. Defects associated with shrinkage are minimal during periods of cloudy, wet weather and low temperatures. Fungi and insects are inactive at temperatures below 32 °F or under conditions of wet storage with low levels of oxygen. On the other hand, many types of bacteria can grow in wood under wet, anaerobic conditions, but not at subfreezing temperatures. There are two general methods for storing logs: dry storage and wet storage. Precautions must be taken with each storage method to ensure defect-free lumber. Dry Storage Most sawlogs in North America are stored under dry conditions with the bark intact. Occasionally, kiln 220 Figure 10-2—Splits in the end of a red oak log resulting from ruptures caused by an imbalance in tree growth stresses after felling. (M88 0169) operators may encounter logs from diseased or insect-damaged trees where most or all of the bark has fallen off. Because lumber from logs subjected to transpiration drying may show up in the drying operation, this sub-ject will be discussed as a part of dry storage. Logs With Bark Most lumber that needs to be kiln dried will be sawed from logs that were stored on land with the bark intact. If the logs do not contain wetwood, then any lumber drying problems will usually be associated with serious end checking of the logs, insect attack, and sapwood stains. End checks can occur in all species of logs and are more pronounced in the denser hardwoods. Deep end splits can sometimes occur in the log ends, but these are the result of residual tree growth stresses that be-come unbalanced after the log is bucked, and they can-not be prevented by measures for reducing end check-ing (fig. 10-2). End checks are minimized by keeping the log ends in cool, moist, and shaded locations. If the logs are valuable and cannot be sawed into lum-ber within a short time, then the ends should be coated with a suitable end-sealing compound (fig. 10-3). The end coating should be thick enough to cover all wood pores, cracks, and irregularities on the surface, yet viscous enough so that it neither cracks nor “sags” excessively. It is good practice to treat the log ends with chemical fungicide before end coating to prevent sapwood staining. Figure 10-3—Oak logs 8 months after they were cut and the ends treated with preservatives. All but two logs were also end sealed; no end checking developed in these logs. The preservative treatment of the unsealed logs (topmost and lower left) was of little value once the barrier of the surface-treated wood was ruptured by seasoning checks. (M 81288). 221 Figure 10-4—Sweetgum logs with heavy sapwood stain at the ends. Under conditions favorable for staining, end stain may appear within 2 weeks and the discol-oration may penetrate into the log as rapidly as 1 ft per month. (M 38236) Fungal blue stain will develop in the sapwood of exposed log ends and debarked surfaces during warm weather within 2 weeks after the tree is felled (fig. 10-4). Applying or spraying chemical fungicides on all exposed log surfaces will provide adequate pro-tection if the wood does not check or split. These chemically treated areas should then be coated with a log end-sealing compound to prevent checking and the opening of untreated inner wood to fungal attack. Since wood-boring insects can carry spores and hyphae of sapwood-staining fungi into the logs, even through areas with attached bark, logs may need to be sprayed with a mixture of chemicals that control both insects and fungi. Figure 10-5—Fungal blue stain and chemical brown stain in sapwood and wetwood of a kiln-dried eastern white pine board sawed from a log stored during early spring on a log deck in the forest. (M88 0168) 222 Chemical changes will occur in moist sapwood during log storage that may cause chemical discolorations dur-ing subsequent drying. These discolorations can vary from gray, yellow, and pinkish to deep brown. Chem-ical stains are likely to occur in lumber from logs that were stored under moist, shaded conditions for the pur-pose of preventing end checking. Brown stain and blue stain will develop together in lumber from logs that were stored in the forest or similarly shaded locations (fig. 10-5). Prompt sawing of freshly cut logs is the eas-iest way to control chemical sapwood stains because treating the logs with fungicides that prevent blue stain will not be effective. The most effective method for controlling chemical stains is to freeze freshly cut sapwood. This can be done economically only by sawing winter-cut logs in northern climates during cold weather and using proper kiln-drying conditions. At some northern mills, short logs of birch, maple, and pine that are to be sawed into specialty products are frozen to ensure white color. The short logs from winter-cut timber are placed in ground depressions and sprayed with water to form a coating of ice. The frozen log decks are covered with sawdust, wood shavings, or other available insulating material so that the wood remains frozen well into the summer months. Debarked Logs Most logs intended for lumber are debarked on the day of sawing; the problems associated with the storing of debarked logs are thus not carried over into the dry-ing operation. Logs intended for poles and pulpwood are debarked soon after felling to reduce losses from in-sect borers and decay and to lower sapwood moisture content. The disadvantages of early debarking are ex-tensive surface checking and end splitting. Sapwood staining can also be quite substantial. Transpiration Drying After a tree is cut or girdled, the main stem will lose more moisture if the crown is left attached than if the stem is bucked into logs. This method of drying is called transpiration drying. Teak trees in the forests of southeast Asia are girdled and left standing for at least a year before felling so that the logs will be light enough to float to the sawmills via the river systems. In North America, there is some interest in transpira-tion drying because of its potential application to wood energy production. If tree-length logs are stored in the woods for a short time, leaving the crown attached also seems to provide some protection from ambrosia beetle attack. The amount of moisture lost depends upon viable fo-liage in the crown and the amount of sapwood in the stem. Softwoods will undergo transpiration drying throughout the year if winter temperatures are not below freezing, but deciduous hardwoods can only be dried during the summer when the leaves are present. During transpiration drying, oak logs with narrow rings of sapwood will not lose much more than 10 per-cent moisture content whereas sweetgum and yellow-poplar, species with wide bands of sapwood, can lose over 30 percent moisture content. The maximum mois-ture loss from hardwoods will occur in 1 to 2 weeks, but this period will usually be longer for conifers. On the west coast of Washington, Douglas-fir will undergo a maximum moisture loss of 30 to 50 percent in about 90 days. In South Africa and Holland, Visser and Vermaas (1986) found that transpiration drying of both hard-wood and softwood trees resulted in total energy sav-ings because of easier handling of green lumber and reduced kiln-drying times. A mass loss of 30 percent after 1 month of transpirational drying resulted in an energy saving of approximately 55 percent in the kiln, while a maas loss of 10 percent after 1 week of drying resulted in an energy saving of approximately 25 per-cent. These authors also noted that the sudden drop in moisture content with transpirational drying helps to suppress the development of blue stain in South African timber. Wet Storage When logs must be stored for a long time at temper-atures above freezing, it is desirable (when possible) to keep them soaking wet. This prevents drying and checking and inhibits attacks by insects and sapwood-stain fungi. However, some types of bacteria are not inhibited, and the wood may become predisposed to developing chemical stains. Pond Storage Pond storage includes logs that are stored in lakes, rivers, and salt water estuaries as well as mill ponds. Although pond storage was once a regular practice, it is now rare in North American mills. Nevertheless, a dry kiln operator may receive lumber from logs that have been submerged in water. Considerable volumes of logs are rafted from woods to mills along the coast of the Pacific Northwest. Foreign lumber is frequently sawed from pond-stored logs, and some lumber is salvaged from old submerged logs and timber. Pond-stored logs are usually banded together to in-crease the log-holding capacity of the pond and to pre-vent wetwood (sinker) logs from sinking to the bottom (fig. 10-6). Some logs in the bundle will be above wa-ter and are subject to insect attack, stain, and decay. Until recent Environmental Protection Agency (EPA) Figure 10-6—Logs banded together in a log pond in merged while others are entirely out of the water. northern California. Some logs are completely sub- (M88 0167) 223 Figure 10-7—Water sprinkling of decked hardwood logs. A fine mist effectively covers log surfaces and ends. (M 144876). prohibitions, these types of damage were controlled for several weeks by spraying the exposed parts of the log bundles with insecticides and fungicides. Logs rafted and stored in ocean water are also subject to attack by marine borers and salt water micro-organisms. Most damage to submerged logs can be traced to growth of bacteria in the sapwood. In softwoods, pit membranes in the sapwood are destroyed so that the wood becomes more permeable, and the wood will dry somewhat faster. However, the lumber will also overab-sorb chemicals used to stabilize and preserve the wood, and finishing can be a problem. Honeycomb, ring fail-ure, and collapse are likely to develop in lumber from logs and timber that have been submerged for over a year. Chemical brown stain has been a frequent prob-lem with the drying of ponderosa pine and sugar pine lumber from pond-stored logs. In rare situations, the iron content of the water is unusually high, and woods gradually acquire a grayish color because of an iron-tannate reaction. Water Sprinkling Where log decking is a preferred manner of storage, sprinkling the decks with water provides an effective method for reducing checking, sapwood stains, and de-cay when temperatures are above freezing (fig. 10-7). Sprinkling will not provide certain protection from in-sect attack although it tends to be more effective than dry log storage in some localities. Nevertheless, the beneficial effects of using water sprays during warm weather have been reported for western softwoods and eastern hardwoods, especially in the South. For sprinkling to be effective, the log ends and ex-posed, debarked wood surfaces must be kept contin-uously wet during the entire period of storage. This prevents shrinkage and checking of the exposed wood. Water sprays reduce temperatures in and around the log decks, but the reduction of oxygen from continuous soaking of the wood is the major deterrent to sapwood-staining and decay fungi. Bacteria and slime molds, less common in dry-stored logs, may develop extensively in sprayed logs. Bac-teria can be responsible for chemical stains and in-creased porosity in lumber from wet-stored logs, but these problems are greater in pond-stored logs than in logs stored on sprinkled decks. Bacterial problems with sprinkling can be prevented by not drawing the water from stagnant reservoirs where drainage from the wetted logs is returned and recycled. Under wa-ter sprays, bacteria from wetwood zones in the log may extend their growth into the sapwood, which will then develop brown stains during drying. 224 Water sprinkling requires constant maintenance to guard against clogging of hoses and spray nozzles from debris and slime in the water. Adequate drainage must be provided in the log yard to prevent handling prob-lems with forklift vehicles. Effects of Climate On Lumber Storage Relative humidity, air temperature, and rainfall of the storage region are the main factors that determine the rate and amount of moisture content change in the lumber and the procedures necessary to protect lum-ber stored outdoors or in unheated sheds. Relative Humidity Relative humidity has a much greater effect on wood equilibrium moisture content (EMC) than does temper-ature (table 1-6). The more humid a region, the more moisture the lumber will absorb and the more rapid the rate of absorption. Seasonal estimates of the average wood EMC for a region can be helpful when trying to control moisture change in lumber stored outdoors. Storage methods to retain low moisture content in kiln-dried lumber will differ between humid regions like the gulf coast and dry regions like the Southwest. Likewise, storage requirements may differ from month to month in regions where average relative humidity varies con-siderably with the season, such as inland California. Temperature Air temperature has a minor effect on EMC (table 1-6), but its main effect is on the rate of moisture content change. Moisture content changes occur faster at warm temperatures than at cool temperatures. Therefore, if lumber has to be stored at EMC conditions different than the moisture content of the lumber, the tempera-ture should be taken into consideration. Some moisture equalization can be effected in storage; the warmer the temperature, the faster the rate of equalization. Warm temperatures also increase the hazard of fungal infection in stored lumber. All lumber is practically immune to fungal infection below 30 °F. When green lumber is solid piled, mold, stain, and decay fungi will grow at temperatures from 40°F to 100 °F with the rate of attack increasing rapidly at higher temperatures in this range. Dipping or spraying freshly sawed lumber with an approved fungicide reduces the chance of fungal growth. Rainfall When lumber is protected while stored outdoors, rain-fall does not greatly affect its moisture content. Solid-piled green lumber is often unprotected while temporar-ily stored outdoors before stacking for air or kiln dry-ing. Some wetting of green lumber is not considered hazardous. If, however, green lumber has been treated with a fungicide for extended green storage or ship-ment, protection from rain is needed to prevent leach-ing of the chemicals. Solid-piled dry lumber should be protected from rain, preferably in storage sheds. Redrying solid-piled lum-ber that has been wetted by rain is difficult. Solid-piled lumber that has been thoroughly soaked requires stick-ering before it is redried, and redrying may result in drying losses. Also, if rain increases the moisture con-tent of the lumber to 20 percent or more, fungi may grow and cause stain and decay. Average Equilibrium Moisture Content Conditions by Region and Season Estimated monthly wood EMC conditions at various locations throughout the United States are given in ta-ble 10-1. They represent average values from climato-logical data and thus may vary from year to year. Also, EMC conditions are often influenced by microclimates within regions, so more localized values can be deter-mined from local weather stations. The Southwestern States are generally the driest re-gions, and the coastal regions, the wettest. During summer months, the states west of the Mississippi River are much drier than during the spring months. East of the Mississippi, the summer months are slightly more humid than the spring. Fall is usually more hu-mid than spring or summer in most of the United States, and winter is generally even more humid. Lumber Storage Lumber storage can be classified into five major types: outdoors, open shed, closed and unheated ‘shed, closed and heated shed, and conditioned shed. The desirable type of storage depends on the moisture content of the lumber and the weather conditions during storage. Outdoor Storage Lumber is often stored outdoors because shed or ware-house facilities are not available. Unprotected outdoor storage is satisfactory for small timbers and lumber for less exacting end uses, although precautions to prevent stain, decay, and insect infestation may be necessary. 225 Kiln-dried lumber stored outdoors without protection will have a rapid increase in moisture content. Protection against rain is more important for solid-piled lumber than for stickered lumber because rain-water cannot evaporate readily from solid piles. Fur-thermore, rain that penetrates solid-piled lumber may in time increase the moisture content enough that stain and decay can grow. Storage areas should be open, well drained, and kept free of weeds and debris that restrict air movement along the surface of the ground, harbor fungi and insects, and create a hazard when dry. The ground, particularly along runways for lumber-handling equipment, should be surfaced with gravel, crushed rock, asphalt, or concrete. Surfacing or paving per-mits vehicles to operate efficiently in all weather and restricts weed growth. The method of piling for out-door storage depends on the species involved, its mois-ture content, and the degree of drying desired during the storage period. Green Lumber Green lumber dries during storage. To reduce drying defects and kiln-drying time as much as possible, the principles of good air-drying practice should be fol-lowed (Reitz and Page 1971). Briefly, these include (1) stacking the lumber properly with dry stickers spaced correctly so as to minimize warp, (2) provid-ing good pile foundations, (3) laying out the yard with adequate spacing between piles and rows of piles, and (4) providing good pile roofs. If green lumber must be stored in solid piles for more than 24 h in warm weather, it should be dipped in an approved antistain solution. Green lumber properly stacked and protected on a good site will lose moisture rapidly with a minimum of defects and can remain out-doors indefinitely without excessive deterioration. Figure 10-8—High-grade Douglas-fir stored temporarily under water spray while the mill accumulates enough for a full kiln load. (M88 0165) Sometimes high-quality green lumber is stored tem-porarily under water spray (fig. 10-8), while lumber is being accumulated for a kiln load. Partly Dried Lumber If the moisture content of lumber is above 20 percent or if further drying is desired, the lumber can be stored like green lumber. Lumber that is below 20 percent moisture content can be solid piled if no additional drying is desired. The piles should be fully protected against infiltration of rainwater. Water that penetrates a solid lumber pile is not readily evaporated and is likely to cause stain or decay. Lumber surfaces that are alternately wetted and dried are likely to check. Kiln-Dried Lumber Lumber kiln dried to a moisture content of 12 percent or less can be stored outdoors in dry weather in stick-ered or solid piles for a short time. Extended storage will result in excessive moisture regain. Figure 10-9 shows the change in moisture content of southern pine during yard and shed storage in solid piles in inland Louisiana. If the lumber had been piled on stickers, its moisture content would have risen to the maximum of about 13-1/2 percent in a much shorter time. During the warm, dry season in areas such as the arid South-west and in parts of Idaho, Montana, Nevada, Oregon, and Washington, the outside storage period can be ex-tended considerably without serious effects if pile covers are used. Kiln-dried lumber can and often is afforded temporary protection, particularly in transit, by wrapping in vari-ous types of coated paper. Such wrap for unit packages of lumber (fig. 10-10) will adequately protect kiln-dried softwood lumber under short-term storage conditions such as long-haul transport on flatcars, interim stor-age at distribution centers, and short-term outdoor storage at construction sites. However, coated paper wrappings should not be considered a substitute for storage sheds when long-term storage of dried lumber is involved. The lumber could deteriorate during storage and is susceptible to tearing during handling. If such storage is unavoidable, the protective wrap should be inspected periodically for tears or other deterioration. Water that enters packages through tears in the pro-tective wrapping can collect and cause more regain of moisture than if no wrap were used. To avoid trapping water in torn packages, the bottom is often left open. However, moisture from ground water can enter pack-ages if not enough ground clearance is provided by the pile foundations. 226 Figure 10-9—Change in average moisture content of kiln-dried southern pine 1- by 4-in flooring and 1- by 8-in boards during storage in solid piles within sheds and in a yard with a protective roof over each pile. (ML88 5557) Figure10-10—Covering packages of lumber with wa- the package bottom, and thus will not be damaged by terproof kraft paper wrap. The wrap does not cover forklift handling nor will it trap rainwater. (M 120954) 227 Figure 10-11—Stickered lumber yarded for air drying. The well-braced pile foundations of stringers and cross-beams prevent tipping. Most piles are covered with a prefabricated board and batten roof (M 134963) Pile Covers High-grade lumber stored in a yard, whether solid piled or stickered, green or dried, should be protected from the weather. Lumber surfaces exposed to alter-nate wetting and drying will check, warp, and discolor. Stacks of lumber in storage yards can be provided with pile covers the same as are used in air-drying yards (fig. 10-11). Open Shed Storage Open sheds provide excellent protection for green and partially dried lumber. Lumber that has been kiln dried to a low moisture content can also be stored in open sheds for varying periods, depending on weather conditions. An open shed is a roofed lumber storage yard. Lum-ber dried to moisture contents as low as 12 to 14 per-cent can be stored in open sheds without significant re-gain of moisture. The atmospheric conditions within an open shed are the same as those outdoors except that lumber is protected from direct contact with rain and sun. A shed may be open on all sides or on one side only (fig. 10-12). Often the side facing the prevailing winds can be closed to keep out driving rain. The shed should be located on an open, well-drained area. It should be large enough to permit rapid han-dling of the lumber and have a floor of gravel, crushed rock, blacktop, or concrete firm enough to support the piles of lumber and the weight of lumber-handling equipment. The roof should overhang far enough be-yond the piles of lumber to protect them from driving rain and snow. Figure 16-12—Open storage for packages of dry, sur-faced lumber. (M88 0164) Green Lumber Green lumber can be stored for long times in open sheds without danger of serious deterioration, provided it is stickered. Such sheds protect the lumber from the sun, rain, and snow, thereby keeping end and surface checks and splits to a minimum. To obtain good air drying in open sheds, adequate spaces should be pro-vided between the sides and ends of the stacks. By al-lowing this free circulation of outdoor air, lumber will dry to as low a moisture content as it does in the open air. The drying time in an open shed is usually shorter and the lumber brighter than if stored outdoors be-cause rewetting is avoided. Partly Dried Lumber Open sheds afford excellent protection to partly dried lumber. If the moisture content is above 20 percent, the lumber should be stacked on dry stickers. If it is below 20 percent, it can be solid piled unless further drying is desired, in which case it should be stickered. Kiln-Dried Lumber Kiln-dried lumber can be well protected from sun, rain, and melting snow when stored in open sheds. An open shed will not, however, prevent regain of moisture dur-ing periods of high humidity, especially if temperatures are also high. Therefore, storage time should be lim-ited during warm, humid weather. Lumber piles can be either solid or stickered. Solid-piled lumber will re-gain moisture more slowly than stickered lumber. In-crease in moisture content will be greatest at the ends and in the outer tiers of a solid pile, as illustrated in figure 10-13. The effect of long-term storage in an open shed on moisture content of solid-piled, kiln-dried lum-ber is also shown in figure 10-9. 228 Figure 10-13—Change in average moisture content of solid-piled, surfaced 1- by 8-in Doughs-fir boards stored in an open shed. (ML88 5556) Figure 10-14—Closed, unheated storage shed at a dis-tributing yard. (M88 0166) Closed, Unheated Shed Storage Closed, unheated sheds (fig. 10-14) are generally used for storing kiln-dried lumber, although they also can be used for storing green or partly dried lumber. This type of shed should be provided with reasonably tight-fitting doors. Ventilators are sometimes provided, and their need depends on the moisture content of the stored lumber and the tightness of the building. Green Lumber Green lumber is sometimes stored in closed sheds, al-though this type of storage will retard drying. The dry-ing can be retarded enough that the growth of mold be-comes a problem. Some drying capability can be added to closed-shed storage by exhaust vents and circulation fans. The solar heat that is absorbed through the roof and walls of a shed will provide some energy for drying. Care should be exercised for species that are suscepti-ble to surface checking. If air circulation is inadequate, the temperature near the roof will rise and could cause surface checking. Partly Dried Lumber Partly dried lumber that is properly piled can be stored in a closed shed without developing drying defects. Lumber should be stickered if it has a moisture con-tent greater than 20 percent. If below 20 percent and no further drying is desired, lumber can be solid piled. If further drying is desired, the lumber should be stick-ered, and it may be advantageous to add fans to circu-late air through the lumber. High shed temperatures from solar energy generally will not cause checking or splitting in partly dried lumber because these defects usually occur when moisture contents are higher. Kiln-Dried Lumber The object of storing kiln-dried lumber in closed sheds is to minimize pickup of moisture. Thus, lumber should be solid piled. Although kiln-dried lumber will regain some moisture during periods of high relative humidity, the percentage regained will be less than if the lumber were stored outdoors. In dry regions, kiln-dried lumber can be stored indefinitely during hot, dry weather. The ultimate moisture content lumber will reach in a closed shed depends on the local weather. If sunny weather prevails, the roof and walls of the shed will ab-sorb solar radiation and heat the air inside. This lowers the relative humidity in the shed and thus the EMC conditions. Prolonged periods of sunshine can thus result in low moisture contents. Conversely, if cloudy weather prevails, moisture contents will not be much lower than in an open shed. Lumber dried to a moisture content of 10 percent or less, and items manufactured from it, will regain mois-ture if stored for extended periods under conditions of high relative humidity. Excessive regain of moisture fre-quently results in (1) swelling of whole pieces or of cer-tain parts, such as the ends of the pieces, (2) warping of items such as glued panels, and (3) wood or glueline failures in solid-piled items where the moisture regain is confined to the ends. During fabrication and use, lumber and items that have adsorbed excessive moisture during storage may (1) end check and split when the high-moisture-content sur-faces are exposed to low relative humidities in heated buildings, (2) shrink excessively, (3) warp, (4) suffer ex-tension of end splits, and (5) open at glue joints. 229 Closed, Heated Shed Storage If air in a shed is heated, the relative humidity and EMC are lowered as long as no additional moisture is added to the air. Thus, storage in closed, heated sheds provides excellent protection in preventing kiln-dried lumber from regaining moisture. Lumber for use in fi-nal products such as furniture and millwork that will be used in a heated environment should be stored in heated sheds. A heated shed should be reasonably tight and can be insulated or uninsulated. Heat can be sup-plied by any convenient means as long as the system can maintain up to 30 °F above outside temperatures. Circulation is desirable to maintain uniform tempera-ture. Ventilators are generally not necessary but should be provided if any drying is anticipated. Temperature can be controlled by a simple thermostat that regulates the heating system. The shed should be located on a well-drained site. Its floor should be of gravel, crushed rock, asphalt, or con-crete, and it should be sufficiently firm to support piles of lumber. Green Lumber Green lumber is not ordinarily stored in heated sheds because the higher temperatures within the shed may cause end and surface checks or splits. If drying in a heated shed is considered, predryers should be used, as described in chapter 2. Partly Dried Lumber Partly dried lumber can be stored in heated sheds for further drying. Stickering and ventilating are necessary. If further drying is not desired, lumber should be stored in open or unheated sheds because it will dry further in a heated shed. Kiln-Dried Lumber Closed, heated sheds are ideal for storing lumber kiln dried to 12 percent moisture content or less. The de-sired EMC of the lumber can be regulated simply by increasing the temperature in the shed by a certain amount over the outside temperature. This can be done with thermostats that measure temperature dif-ferentials. When outside air is heated without adding moisture, even though the absolute humidity remains the same, the relative humidity decreases and thus the EMC decreases. The outside temperature and relative humidity must be known to determine the amount by which the temperature in the shed must be increased to attain a certain EMC. For example, if the outside air is at a temperature of 50 °F and is at 80 percent relative humidity, how much must the temperature in the shed be raised to attain an EMC of 6 percent? The 230 answer can be determined by using figure 10-15. Enter the graph along the arrows that lead from 50 °F and 80 percent relative humidity to the point where they in-tersect. Note that this is at an EMC of about 16.5 per-cent and an absolute humidity of about 0.0625 pound of water per pound of dry air (at a barometric pressure of 29.92 in Hg). Since no moisture is being added to the air in the shed, the absolute humidity will remain the same as we raise the temperature. Therefore, follow the arrowed line down parallel to the absolute humid-ity lines to the point where it intersects the 6 percent EMC line. From this point drop straight down to the temperature axis and read the required temperature in the shed as 80 °F or a 30 °F temperature rise. An alternative way to control conditions in a heated shed is to control the heater with a humidistat. When the relative humidity is above the set point of the hu-midistat, the heater will be on until the relative hu-midity falls to set point. For example, we know from table 1-6 of chapter 1 and figure 10-15 that to maintain an EMC of 6 percent, the relative humidity should be controlled at about 30 percent. Conditioned Storage Sheds Kiln-dried lumber and finished products can also be held at any desired moisture content in storage by con-trolling both relative humidity and temperature. This is the most costly method of controlling EMC because of the equipment involved. However, when it is desir-able or necessary to maintain temperature within cer-tain limits, then it may not be possible to maintain rel-ative humidity simply by manipulating temperature. For example, to attain 6 percent EMC when the out-side air is at a temperature of 85 °F and a relative hu-midity of 80 percent, the temperature must be raised to 114 °F. This temperature is unreasonable in a work area where people must spend any length of time. In this case, refrigeration equipment is required to attain 6 percent EMC at a comfortable temperature. Treating Stored Lumber Fungal infection and insect attack both pose serious hazards to stored lumber. Fungal infection was found to be the principal cause of degrade in a study of grade loss in l-in southern pine lumber. Insect infestation also causes serious losses in stored lumber, particularly in the warmer parts of the United States. For protec-tion from fungi and insects, lumber may require a dip or spray treatment in a chemical solution at the storage installation. In some cases, this treatment will supple-ment an earlier dip or spray at the sawmill. Figure 10-15—Psychrometric chart showing the rela-tionship between temperature, relative humidity, abso-lute humidity, and equilibrium moisture content (EMC) of wood at a barometric pressure of 29.92 in Hg. The chart and arrowed lines illustrate the temperature rise required to attain 6 percent EMC by heating outside air originally at 50 °F and 80 percent relative humidity. (ML88 5558) To minimize fungal and insect attacks on stored lum-ber, air-drying yards should be kept sanitary and as open as possible to air circulation. Recommended prac-tice includes locating yards and sheds on well-drained ground. Remove debris, which is a source of infection, and weeds, which reduce air circulation. Piling meth-ods should permit rapid drying of the lumber and also protect against wetting. Open sheds should be well maintained, with an ample roof overhang to prevent wetting from rain. In areas where termites or water-conducting fungi may be trou-blesome, stock to be held for long periods should be set on foundations high enough to be inspected from beneath. When Is Chemical Treatment Needed? Prompt drying will often protect untreated lumber from attack by stain, decay, and some insects. For in-stance, untreated lumber uniformly below 20 percent moisture content is immune to attack by fungi. With protective storage it will keep that immunity. However, dried lumber that regains moisture to a level of more than 20 percent again becomes susceptible to stain and decay. The sapwood of all wood species is more susceptible than heartwood to decay, stain, or insects. Therefore, the hazards are highest for woods that usually con-tain a high percentage of sapwood. The heartwood of such species as redwood, the cedars, and some white oaks has high natural resistance to fungi and most in-sects. But few products--even from these woods-are of heartwood only. 231 Damp weather can increase the damage from stain and decay fungi. Rainfall and humid conditions increase the hazard to unprotected wood in both open and solid piles. Air temperature is highly important. The stain and de-cay fungi grow most rapidly at 70 to 90 °F, grow no more than one-fifth as rapidly at 50 to 60 °F, and cease growth at about 32 °F. As a result, wood at about 25 to 30 percent moisture content, stored in solid piles in warm weather, may show evidences of stain within a week and early decay infection within a month. The initial infections, which are not visible, probably started shortly after the wood was sawed. With temperatures of 50 to 60 °F, similar deterioration requires five or more times as long. At 32 °F or below, the lumber can remain in solid piles indefinitely without adverse effects. High humidity favors subterranean termites but does not affect drywood termites or powder-post beetles. The influence of temperature on insect activity, how-ever, is pronounced. Insects are inactive at temper-atures of 50 °F or below but increase their activity rapidly as the temperature rises above this level. In-sects will approximately double their activity with each increase of 10 ° above 50 °F, reaching maximum activity levels at about 80 °F. When and Where to Apply Treatment Stain and decay in lumber are normally controlled at sawmills, collection points, and drying yards by drying the wood as rapidly as possible below 20 percent mois-ture content. Lumber to be air dried may be treated with fungicidal solution by dip or spray before the dry-ing period begins. Sometimes an insecticide is mixed into the solution if insects are likely to be a problem. The layer of wood chemically protected by a dip or spray is only “skin deep” and will not stop fungi or in-sects that have already entered the wood. This is why stock is dipped as soon as possible after it is sawed. To illustrate how quickly the dipping must be accom-plished, the safe times are estimated as follows: 1 day at temperatures of 80 °F or above; 2 days at 70 °F; 1 week at 60 °F; and 1 month at 50 °F. Longer delays at these temperatures progressively lower the benefit from surface treatments. Generally, dip or spray treatments immediately after cutting are designed to protect green stock only when it is drying. If treated green lumber is not air dried to below 20 percent moisture content, prolonged storage may require redipping or respraying of the lumber. Lumber properly dipped in an antistain solution at the sawmill can be stored in solid piles for up to 1 month in warm weather if further drying is not required. If longer bulk storage is anticipated, dip-treated stock should be redipped. Additional dipping can protect pines and hardwoods from stain and decay for 6 to 8 weeks in warm weather and western softwoods other than pines for 4 to 6 months. If the lumber was not dipped at the sawmill, dipping at the storage yard may still protect it from fungi dur-ing bulk storage provided the stock is not already in-fected. Infection would not occur if daytime temper-atures in the interval between sawing and receipt at the yard did not exceed about 40 °F. If temperatures were higher, however, fungus infection may have al-ready taken place, and solid piling should be avoided. Instead, lumber may be dipped in a fungicidal solution and open piled. Because a number of factors affect safe storage time, all dipped bundles should be labeled with the date on which they were treated. Representative bundles should be opened from time to time to determine the condi-tion of the stock. Any lumber that shows signs of being inadequately protected should be designated for early use, redipped, or stickered for air drying. How to Apply Treatment Lumber to be dipped at the storage installation will probably be in unit packages. Thus, the dipping pro-cedures explained here are for unit packages. When lumber is dipped, the amount of solution absorbed will be about 4 to 8 percent of the wood weight, depending on type of wood and moisture content at the time of treatment. Treating Area and Equipment Location of the treating plant affects the costs and ef-ficiency of the treating operation. Ready access of the plant to packaging and storage areas-and to railroad spurs or shipping docks-will keep costs to a minimum and ensure an efficient handling operation. Equipment for treating lumber often includes an elec-tric hoist that runs on a monorail attached to the ridge of a long, open shed. The treating vat can be installed in or ab1llres Rail Transport Some years ago, the Forest Products Laboratory stud-ied the changes in moisture content of softwood lum-ber shipped in tight railroad boxcars from West Coast sawmills to midwestern U.S. markets. These stud-ies involved five boxcar loads of l-in clear Douglas-fir shipped from a West Coast sawmill to the Chicago, IL, area during late winter and spring. The time in rail transit averaged 18.5 days; the shortest period was 14 days and the longest, 22 days. Average mois-ture content of the five carloads of kiln-dried boards at the time of loading was 8 percent, and the aver-age gain in moisture content was 0.2 percent. These values were baaed on an average of 18 teat boards dis-tributed throughout the boxcar load in each shipment. In another study, test boards in a carload of Douglas-fir quarter-round and crown molding, which were at 8 per-cent moisture content when loaded, regained 0.8 per-cent in moisture in a 20-day transit period from the West Coast to the vicinity of Chicago. Thus, no signif-icant change in moisture content of dry lumber need be expected during the usual haul in tight boxcars. A study of moisture changes in rail shipments of kiln-dried hardwood lumber was conducted by the Forest Products Laboratory. These shipments were of kiln-dried pecan lumber, transported in wide-door boxcars from midsouth Mississippi to a furniture company in North Carolina, a distance of about 900 mi. Each load of unitized lumber packages contained four test boards for moisture analysis. Test shipments were made from June through November, and the increase in moisture content was less than 0.5 percent moisture content. Conventional flatcars have become widely used for the transport of dried lumber because they can (1) save handling time and shipping cost, (2) hold twice the load of conventional boxcars, and (3) be loaded by lift trucks to save handling time. Improvements in unitized package wrapping have made it possible to obtain these advantages without much increase in moisture content, even on long hauls. Unitized packages on flatcars are usually protected, ei-ther partially by tarpaulins or entirely by flexible, wa-terproof packaging that completely “tailor-wraps” each package. One common type of waterproof packaging uses composite kraft paper that is reinforced with glass fiber coated with polymer. The packaging is frequently supplied with additional reinforcement at stress points such as edges and corners. Improvements in packaging materials have made possible the shipment of kiln-dried lumber with little change in moisture content and a good retention of brightness. Wrapping for unitized packages of lumber should be free from rips to be effective. Rain that enters through rips is held by the sheeting, and the package may act as a humidifier. If so, moisture regain may be higher than if the lumber were unprotected. Ship Transport Lumber is often transported overseas in ships while it is either green, partly dried, or kiln dried. A study con-ducted in Canada, which involved 33 shipments of l-in lumber from the Canadian west coast to five different ports, concluded that seasoned lumber stored below decks, either by itself or together with green lumber, will not undergo moisture regain of serious propor-tions (table 10-2). This study also indicates that well-dried lumber may undergo significant moisture regain if stored on deck, although it is not commonly stored in this way. Similar tests were made with 2-in Douglas-fir lumber. The kiln-dried lumber had a moisture content of 9 to 10 percent when stowed. The overall average moisture gains for the seasoned 2-in lumber were as follows: Lumber stowed below decks with dry lumber . . . . . . . 1.3 percent Lumber stowed below decks with green lumber . . . . . . 2.4 percent Lumber stowed on deck with green lumber . . . . . . . . . 4.2 percent 235 Literature Cited Nielson, R. W.; MacKay, J. F. K. 1985. Sorting of dry and green lodgepole pine before kiln drying. In: Proceedings, joint meeting Western Dry Kiln Clubs; 1985 May 8-10; Vancouver, BC. Corvallis, OR 97331: Oregon State University School of Forestry: 63-69. Reitz, Raymond C.; Page, Rufus H. 1971. Air dry-ing of lumber: A guide to industry practice. Agric. Handb. 402. Washington, DC: U.S. Department of Agriculture. 110 p. Visser, J. J.; Vermaas, H. F. 1986. Biological drying of Pinus radiata and Eucalyptus cladocalyx trees. Journal of the Institute of Wood Science. 10(5): 197-201. Sources of Additional Information Cech, M. Y.; Pfaff, F. 1977. Kiln operators manual for eastern Canada. Report OPX 192E. Ottawa, ON: East-ern Forest Products Laboratory. 189 p. Ellwood, E. L.; Ecklund, B. A. 1959. Bacterial attack of pine logs in pond storage. Forest Products Journal. 9(9): 283-292. Findlay, W. P. K. 1967. Timber pests and diseases. Pergamon Series of Monographs on Furniture and Timber, Vol. 5. Pergamon Press. 280 p. Johnson, N. E.; Zingg, J. G. 1969. Transpirational dry-ing of Douglas-fir: Effect on log moisture content and insect attack. Journal of Forestry. 67(11): 816-819. McMillen, J. M. 1956. Coatings for the prevention of end checks in logs and lumber. Forest Products Lab-oratory Report No. 1435. Madison, WI: U.S. Depart-ment of Agriculture, Forest Service, Forest Products Laboratory. (Out of print) McMinn, J. W. 1986. Transpirational drying of red oaks, sweetgum, and yellow-poplar in the upper Pied-mont of Georgia. Forest Products Journal. 36(3): 25-27. Reitz, Raymond C. 1978. Storage of lumber. Agric. Handb. 531. Washington, DC: U.S. Department of Agriculture. 63 p. Salamon, M. 1973. Drying of lodgepole pine and spruce studs cut from flooded timber: A progress report. In: Proceedings, 24th annual meeting Western Dry Kiln Clubs, Oregon State University, Corvallis; 51-56. Scheffer, T. C. 1958. Control of decay and sap stain in logs and green lumber. Forest Products Laboratory Re-port No. 2107. Madison, WI: U.S. Department of Agri-culture, Forest Service, Forest Products Laboratory. 13 p. (Out of print) Scheffer, T. C. 1961. Protecting stored logs and pulp-wood in North America. Material und Organismen. 4(3): 167-199. Wagner, F. J. Jr. 1978. Preventing degrade in stored southern logs. Forest Products Utilization Bulletin. Atlanta, GA 30309: U.S. Department of Agriculture, Forest Service, State and Private Forestry. 4 p. Williams, L. H.; Mauldin, J. K. 1985. Laboratory dip diffusion treatment of unseasoned banak (Virola spp.) lumber with boron compounds. Res. Note SO-313. New Orleans, LA: U.S. Department of Agriculture, Forest Service, Southern Station. 8 p. 236 Table 10-1—Equilibrium moisture content of wood, exposed to outdoor atmosphere, in the United States Equilibrium moisture content in different months (percent) 1 Location Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec. Portland, ME Concord, NH Boston, MA Providence, RI Bridgeport, CT New York, NY Newark, NJ Wilmington, DE Philadelphia, PA Baltimore, MD Norfolk, VA Wilmington, NC Charleston, SC Savannah, GA Key West, FL Burlington, VT Cleveland, OH South Bend, IN Charleston, WV Louisville, KY Nashville, TN Mobile, AL Jackson, MS Detroit, Ml Milwaukee, WI Chicago, IL Des Moines, IA Kansas City, MO Little Rock, AK New Orleans, LA Duluth, MN Bismark, ND Huron, SD Omaha, NE Wichita, KS Tulsa, OK Galveston, TX Missoula, MT Casper, WY Denver, CO Salt Lake City, UT Albuquerque, NM Tuscon, AZ Boise, ID Reno, NV Seattle-Tacoma, WA Portland, OR San Francisco, CA Juneau, AK San Juan, PR Honolulu, HI 1The values were calculated by means of average monthly temperatures and relative humidities given in Climatological Data monthly reports of the Weather Bureau and the wood equilibrium moisture content to relative humidity relationship. 237 Table 10-2—Average gain in lumber moisture content during ocean shipment 1 Lumber moisture content increase (percent) Number of shipments Shipment destination Time in transit (days) Stowed with dry lumber below decks Stowed with green lumber below decks Stowed on deck with green lumber 11 10 6 3 3 (Average) England Australia South Africa Eastern Canada Trinidad 1Lumber used was 1-in kiln-dried Douglas-fir. 238
14226	https://cubeforteachers.com/post/TUh3suFyRtjAM3U4igVC8vrPhhRx7uvt	Base Ten Blocks \| Teaching Tools \| Toy Theater Educational Games - CubeForTeachers - Cube For Teachers Toggle Navigation Login Register Filters Teacher Finds 26 Educational Posters for Kids This set of 26 Educational Posters for Kids features fun, colorful visuals that teach letters, numbers, shapes, and moreâperfect for early learning spaces. As an Amazon Associate, we may earn a commission on some purchases. Base Ten Blocks \| Teaching Tools \| Toy Theater Educational Games Cubed by Math Monkey on 0000-00-00 00:00:00 Use these blocks of one and ten to model place value and help kids visualize the concept of 10. Teacher Finds 26 Educational Posters for Kids This set of 26 Educational Posters for Kids features fun, colorful visuals that teach letters, numbers, shapes, and moreâperfect for early learning spaces. As an Amazon Associate, we may earn a commission on some purchases.
14227	https://webstersdictionary1828.com/Dictionary/Allure	Websters 1828 - Webster's Dictionary 1828 - Allure Menu Home Menu Home Preface History Quotations Constitution The Bible Literature Grammar Education Science Mathematics Medical American Dictionary of the English Language Dictionary Search Home Preface History Quotations Noah Webster Topics Bible Constitution Literature Grammar Education Science Mathematics Medical Allure ALLU'RE, verb transitive To attempt to draw to; to tempt by the offer of some good, real or apparent; to invite by something flattering or acceptable; as, rewards allure men to brave danger. Sometimes used in a bad sense, to allure to evil; but in this sense entice is more common. In Hosea 2:14, allure is used in its genuine sense; 2 Peter 2:18, in the sense of entice. Websters Dictionary 1828 SITEMAP Home Preface History Quotations INFORMATION About Us Statement of Accessibility Terms of Use Privacy Policy Copyright Notice OUR WEBSITES The Kings Bible King James Bible 1611 King James Bible Dictionary Websters Dictionary 1828 Textus Receptus Bibles Treasury of Scripture Knowledge ©Copyright 2025 MasonSoft Technology Ltd v4 (7.10.2024) X Cookies & Privacy We use cookies to ensure you get the best experience on our website. By using our site you acknowledge that you have read and understand our Privacy Policy. Accept Reject
14228	https://www.ck12.org/flexi/cbse-math/area-of-regular-and-irregular-polygons/	Area of Regular and Irregular Polygons \| Flexi Homework help & answers \| CK-12 Foundation What are you looking for? Search Math Grade 6 Grade 7 Grade 8 Algebra 1 Geometry Algebra 2 PreCalculus Science Earth Science Life Science Physical Science Biology Chemistry Physics Social Studies Economics Geography Government Philosophy Sociology Subject Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up All Subjects CBSE Math Area of Regular and Irregular Polygons Area of Regular and Irregular Polygons Concept Summary: A regular polygon is a polygon with all equal sides and all equal angles. The formula for the area of a regular polygon is: Area=1 2⋅Perimeter⋅apothem or Area=1 2⋅side length⋅number of sides⋅apothem An irregular polygon is a polygon that does not have all equal sides and all equal angles. The area of an irregular polygon can be found by decomposing the shape into shapes that you know how to find the area of (i.e., triangles, parallelograms, trapezoids). The approximate area of irregular shapes can also be found using squared paper. We find the approximate area of irregular shapes in the following way:- We count the number of squares that are completely enclosed or whose more than half the parts are enclosed by the figure as one.- We count the squares whose exactly half parts are enclosed as 1 2- We leave out all the squares which are less than half in the figure Ask your own question Find the area of a regular hexagon with an apothem 11.3 centimeters long and a side 13 centimeters long. Round your answer to the nearest tenth. Find the area of a regular hexagon with an apothem 9.5 centimeters long and a side 11 centimeters long. Round your answer to the nearest tenth. A rectangular pan has a length that is double the width. The total area of the pan is 432 in². What is the width of the cake pan? A rectangular pan has a length that is 4/3 the width. The total area of the pan is 432 in^2. What is the width of the cake pan? A square piece of construction paper has sides that are 6 inches long. What is the piece of paper's area? (Answer in square inches.) How many sides does a regular polygon have if each exterior angle measures 24°? And what is the sum of the interior angles? A regular hexagon has a perimeter of 60 m. Find its area. Leave your answer in simplest radical form. For the hexagon with 6 dots, there are 2 dots on each side. For the hexagon with 12 dots, there are 3 dots on each side. How many dots are there on the hexagon if there are 23 dots on each side? The total area of two square windows is 1,025 in². Each side of the larger window is 5 in. longer than the sides of the smaller window. How long are the sides of the smaller window? By joining all the vertices of a dodecagon (12 sides), how many diagonals are formed? No. of diagonals = ? The ratio of an interior angle to an exterior angle of a regular polygon is 5:1. How many sides does this polygon have? A parking garage has 9 levels and there are 678 parking spots on each level. How many parking spots are there in total? Find the area of a regular hexagon with a side length of 10 m. Round your answer to the nearest tenth. Find the area of a regular hexagon with side length of 8 m. Round your answer to the nearest tenth. Find the area of a regular decagon, with a side length of 10 ft. and the apothem equal to 6 ft. Find the area of a regular octagon, with a side length of 8 ft. and the apothem equal to 12 ft. Find the area of a regular nonagon, with a side length of 7 ft. and the apothem equal to 6 ft. Find the area of a regular nonagon, with a side length of 6 ft. and the apothem equal to 4 ft. If an interior angle of a regular polygon measures 60°, how many sides does the polygon have? A certain polygon has 54 diagonals. How many sides does it have? How to calculate the area of a regular octagon? How to calculate the area of a regular hexagon? How to calculate the area of a regular polygon? Define a convex polygon. What is the area of the polygon? How do I calculate the area of a polygon? How do I calculate the area of a polygon in math? What is the method for solving a regular polygon? What is the definition of the area of a polygon? How do you find the area of an irregular shape? How do you calculate the area of an irregular shape? How do you find the area of a 10-sided polygon? How to calculate the area of a polygon? What is the procedure to find the area of a polygon? What are some examples of polygons? Can a polygon be concave? What are the properties of similar polygons? How do you calculate the area of an irregular polygon? State the properties of similar polygons. What is a regular polygon with an exterior angle measuring 120 degrees called? Which pairs of polygons are similar? 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14229	https://www.scirp.org/html/26485.html	Counting Runs of Ones and Ones in Runs of Ones in Binary Strings Home Journals Books Conferences News About Us Jobs Paper Menu >> ●Abstract ●Full-Text PDF ●Full-Text HTML ●Full-Text ePUB ●Linked References ●How to Cite this Paper Journal Menu >> ●Indexing ●View Papers ●Aims & Scope ●Editorial Board ●Guideline ●Article Processing Charges ●Paper Submission ●OJAppS Subscription ●Free Newsletter Subscription ●Most popular papers in OJAppS ●Publication Ethics Statement ●About OJAppS News ●Frequently Asked Questions ●Open Special Issues ●Published Special Issues ●Special Issues Guideline Counting Runs of Ones and Ones in Runs of Ones in Binary Strings Counting runs in binary strings Frosso S. Makri, Zaharias M. Psillakis, Nikolaos Kollas Departments of Mathematics and Physics University of Patras Patras, Greece makri@math.upatras.gr; psillaki@physics.upatras.gr Abstract—Consider a binary string (a symmetric Bernoulli sequence) of length . For a positive integer , we exactly enumerate, in all  possible binary strings of length , the number of all runs of 1s of length (equal, at least)  and the number of 1s in all runs of 1s of length at least . To solve these counting problems, we use probability theory and we obtain simple and easy to compute explicit formulae as well as recursive schemes, for these potential useful in engineering numbers. Keywords-runs; symmetric Bernoulli trials; probability theory; combinatorial problems Introduction and Preliminaries Nowadays, the increasing use of the computer science in diverse applications including encoding, compression and transmission of digital information calls for understanding the distribution of runs of 1s or 0s. For instance, such knowledge would help in analyzing, and comparing also, several techniques used in communication networks (wired or wireless). In such networks binary data, ranging from a few bytes (e.g. e-mails) to many gigabytes of greedy multimedia applications (e.g. video on demand), are highly processed. For details, see [1-2] and the references therein. Another area where the study of the distribution of runs of 1s and 0s has become increasingly useful is the field of bioinformatics or computational biology. In particular, molecular biologists examine tandem repeats among DNA (Deoxyribonucleic acid) segments trying to specify how probable are runs of matches, denoted as 1s, in adjacent segments of a DNA sequence. See, e.g. [3-5]. In such applications, as the indicative ones mentioned above, a key point is the understanding how 1s and 0s are distributed and combined among the elements of a binary sequence (finite or infinite, memoryless or not) and eventually forming runs of 1s and 0s according to certain enumeration rules (counting schemes). Each enumeration rule defines how runs of same symbols (i.e. 1s or 0s) are formed and consequently counted. A rule may depend on, among other considerations, whether overlapping counting is allowed or not as well as if the counting starts or not from scratch when a run of a certain size has been so far enumerated. For extensive reviews of the runs literature we refer to [6-8]. The topic is still active and attractive too, because of the wide range of its application in many areas of applied probability and engineering including hypothesis testing, quality control, system reliability and financial engineering. Some recent contributions on the subject, among others, are the works of – . Let be a sequence of binary (two-state) random variables (RVs) taking on the values zero (0) or one (1) ordered on a line. According to Mood’s enumeration scheme a run of 1s (1-run) is defined to be a sequence of consecutive 1s preceded and succeeded by 0s or by nothing. The number of 1s in a 1-run is referred to as its length (or size). For a positive integer  let  denote the number of 1-runs of length exactly  in the first  , binary trials. Following Makri and Psillakis , we use the indicator functions   ending at    with the convention . Consequently, the statistic  can be expressed as     . The RV , which is a fundamental one in the run literature, besides its independent merit, may be used for the representation of other interesting statistics, too. Among them, the following two have been frequently discussed in the literature and in particular in financial engineering and bioinformatics [4, 17-18]. They refer to 1-runs of length exceeding a positive integer , , in  binary trials, and they are the number  of 1-runs of length at least ;    , and the number  of 1s in all 1-runs of length at least ;      An alternative interpretation of  is that it denotes the sum of the lengths of the 1-runs of length greater than or equal to  The statistics  have been studied on binary sequences of several internal structures by many researches who used various methods. See, e.g. [1, 4, 9, 11-14, 17-19, 23-30]. In this brief note we show how someone can easily enumerate explicitly the (total) number of occurrences of all 1- runs associated with the first two mentioned statistics, as well as, the (total) number of 1s according to the third one, in all possible  binary strings of length . Our approach is relied on simple and efficient probabilistic arguments. It provides an Open Journal of Applied Sciences Supplement：2012 world Congress on Engineering and Technology 44 Copyright © 2012 SciRes. alternative way to recapture explicit formulae for numbers associated with 1-runs and it also establishes a new explicit expression for the number of 1s in certain 1-runs. A unified recursive scheme for these numbers is provided, too. Main Results Let   stand for the RVs ,,  for  , respectively. The support (range set) of   is        . (1) Next, we consider a sequence   of length of independent (i.e. derived by a memoryless source) and identically distributed 0-1 RVs with a common probability of 1s  i.e.  ,    Such a sequence, called a finite Bernoulli sequence, is of particular importance in studies of applied probability because of its simplicity, and also since it may be considered as a special case of a sequence with dependent elements; e.g. a Markovian or an exchangeable one. For a Bernoulli sequence of length , let    denote the probability mass function (PMF) of the RV  ; i.e.    ,    ,  . (2) Then, the expected value of  ,          (3) is given by (see Makri et al. )                       In the sequel, we consider a symmetric () finite Bernoulli sequence (i.e. a finite binary string) of length  for which we obtain our main results. Since the cardinality of a proper sample space is  (i.e. there are  binary strings that are equally likely to occur) the classical definition of probability implies that        . (5) The numbers   ,   , for  admit the following interpretation: (i)   ,   is the number of all binary strings of length  with exactly     , 1-runs of length [exactly (), at least ()] , among all the  possible binary strings of length , . (ii)   is the number of all binary strings of length  with exactly  n, 1s contained in all 1-runs of length at least , among all the  possible binary strings of length , . Simple explicit expressions of   (not repeated here) are given in Makri and Psillakis . The authors provided an explicit formula of   , for  in terms of binomial coefficients. Their method is based on the solution of a combinatorial problem; specifically, the allocation of balls into cells under certain constrains (see Lemma 2.2 of ). An explicit expression of   , in terms of binomial coefficients too, is given by Sinha and Sinha who used a generating function approach. The latter expression contains an additional sum; therefore it may be evaluated slower computationally than that provided in . The numbers   allow us to establish (we do not actually need their specific expressions according to the new proposed approach) respective numbers referring to all possible  binary strings of length . They are defined as     , (6) i.e.   is the total number of occurrences of all 1-runs of length [exactly (), at least ()] , and the total number of 1s in all 1-runs of length at least  [], in all possible  binary strings of length , . Since      , hence by (5) and (6),    . Therefore (4) implies                      (7) Readily, by symmetry,   provides the respective numbers associated with 0-runs and 0s in 0-runs. By (7) it is noted that for a fixed ,  ,   decreases exponentially as  increases, and for a fixed  , as , it holds                (8) Furthermore,      , since   , . Table I presents the three numbers     in binary strings of length  bits,  , and for . The entries of the table confirm the previously noted behavior of the depicted numbers. Sinha was the first who addressed the usefulness of the number   and also provided its formula. Then, Sinha and Sinha obtained an explicit expression of   whereas Copyright © 2012 SciRes. 45 Makri and Psillakis derived the same formula for it and they also established an explicit expression of  . In both papers ( and ) the approach was relied on the definition of   via   , . Recently, Sinha and Sinha reestablished   solving explicitly a recursive generation scheme for it. The proposed, in the present note, approach is a new one and treats under the same frame all the numbers   in a simple, unified and systematic way. Accordingly, by (7) we effortless get a recursive scheme for  ,  . It gives a way to generate   from   and it offers further insight in understanding the interdependencies among the studied numbers. Specifically, it holds                 (9) with     . We note that for the particular case  we capture by the relevant entries of (9) and (7), Theorems 2 and 3 of , respectively. TABLE I. NUMBERS OF OCCURRENCES OF 1-RUNS,   AND NUMBER OF 1 S, , IN BINARY STRINGS OF LENGTH                         4 1 12 20 32 2 5 8 20 3 2 3 10 4 1 1 4 16 1 147456 278528 524288 2 69632 131072 376832 3 32768 61440 237568 4 15360 28672 139264 5 7168 13312 77824 6 3328 6144 41984 7 1536 2816 22016 8 704 1280 11264 9 320 576 5632 10 144 256 2752 11 64 112 1312 12 28 48 608 13 12 20 272 14 5 8 116 15 2 3 46 16 1 1 16 Conlusions In this note we stated three run statistics which are important in many areas of applied probability. We defined them on a binary (0-1) sequence, and we then provided explicitly their mean values for a Bernoulli sequence. After that, we considered binary strings (symmetric Bernoulli sequences) and we showed how the analytic expressions of the means of these RVs provide eventually the respective explicit expressions of three numbers studied recently by different methods. Finally, as a byproduct of our approach, we proposed a unified recursive scheme which clarifies further the interdependencies among these numbers. The examined numbers are potential useful in many engineering applications like the ones mentioned briefly in the Introduction. Early results are encouraging in this direction. R EFERENCES K. Sinha and B. P. Sinha, “On the distribution of runs of ones in binary strings,” Comput. Math. Appl., vol. 58, pp. 1816-1829, 2009. K. Sinha and B. P. Sinha, “Energy-efficient communication: understanding the distribution of runs in binary strings,” 1 st International Conference on Recent Advances in Information Technology (RAIT-2012), pp. 177-181, 2012. G. Benson, “Tandem repeats finder: a program to analyze DNA sequences,” Nucleic Acids Res., vol. 27, pp. 573- 580, 1999. W.Y. W. Lou, “The exact distribution of the “-tuple statistic for sequence homoloy,” Statist. Probab. Lett., vol. 61, pp. 51-59, 2003. G. Nuel, L. Regad, J. Martin and A.C. Camproux, “Exact distribution of a pattern in a set of random sequences generated by a Markov source: applications to biological data,” Alg. Mol. Biol., vol. 5, pp. 1-18, 2010. N. Balakrishnan and M. V. Koutras, Runs and Scans with Applications, New York: Wiley, 2002. J. C. Fu and W. Y. W. Lou, Distribution Theory of Runs and Patterns and its Applications: A Finite Markov Imbedding Approach, New Jersey: World Scientific, 2003. M. V. Koutras, “Applications of Markov chains to the distribution theory of runs and patterns,” in Handbook of Statistics, vol. 21, D. N. Shanbhag, C. R. Rao, Eds. North Holland: Elsevier, 2003, pp. 431-472. S. Eryilmaz, “Success runs in a sequence of exchangeable binary trials,” J. Statist. Plann. Inference, vol. 137, pp. 2954-2963, 2007. F. S. Makri, A.N. Philippou and Z. M. Psillakis, “Polya, inverse Polya, and circular Polya distributions of order  for -overlapping success runs,” Commun. Statist. Theory Methods, vol. 36, pp. 657-668, 2007. F. S. Makri, A. N. Philippou and Z. M. Psillakis, “Success run statistics defined on an urn model,” Adv. Appl. Probab., vol. 39, pp. 991-1019, 2007. S. Eryilmaz, “Run statistics defined on the multicolor urn model,” J. Appl. Probab., vol. 45, pp. 1007-1023, 2008. S. Demir and S. Eryilmaz, “Run statistics in a sequence of arbitrarily dependent binary trials,” Stat. Papers, vol. 51, pp. 959-973, 2010. K. Inoue and S. Aki, “On the conditional and unconditional distributions of the number of success runs on a circle with applications,” Statist. Probab. Lett., vol. 80, pp. 874-885, 2010. 46 Copyright © 2012 SciRes. F. S. Makri, “On occurences of  strings in linearly and circularly ordered binary sequences,” J. Appl. Probab., vol. 47, pp. 157-178, 2010. F. S. Makri, “Minimum and maximum distances between failures in binary sequences,” Statist. Probab. Lett., vol. 81, pp. 402-410, 2011. F. S. Makri and Z. M. Psillakis, “On runs of length exceeding a threshold: normal approximation,” Stat. Papers, vol. 52, pp. 531-551, 2011. F. S. Makri and Z. M. Psillakis, “On success runs of length exceeded a threshold,” Methodol. Comput. Appl. Probab., vol. 13, pp. 269-305, 2011. F. S. Makri and Z. M. Psillakis, “On success runs of a fixed length in Bernoull sequences: exact and aymptotic results,” Comput. Math. Appl., vol. 61, pp. 761-772, 2011. S. D. Dafnis, A. N. Philippou and D. L. Anztoulakos, “Distributions of patterns of two successes separeted by a string of  failures,” Stat. Papers, vol. 53, pp. 323- 344, 2012. M. V. Koutras and F. S. Milienos, “Exact and asymptotic results for pattern waiting times,” J. Statist. Plann. Inference, vol. 142, pp. 1464-1479, 2012. F. S. Makri and Z. M. Psillakis, “Counting certain binary strings,” J. Statist. Plann. Inference, vol. 142, pp. 908-924, A. M. Mood, “The distribution theory of runs,” Ann. Math. Stat., vol. 11, pp. 367-392, 1940. J. C. Fu and M. V. Koutras, “Distribution theory of runs: a Markov chain approach,” J. Amer. Statist. Assoc., vol. 89, pp. 1050-1058, 1994. D. L. Antzoulakos, “On waiting time problems associated with runs in Markov dependent trials,” Ann. Inst. Statist. Math., vol. 51, pp. 323-330, 1999. Q. Han and S. Aki, “Joint distributions of runs in a sequence of multi-trials,” Ann. Inst. Statist. Math., vol. 51, pp. 419-447, 1999. J. C. Fu, W. Y. W. Lou, Z. Bai and G. Li, “The exact and limiting distributions for the number of successes in success runs within a sequence of Markov-dependent two-state trials,” Ann. Inst. Statist. Math., vol. 54, pp. 719-730, 2002. K. Sen, M. L. Agarwal and S. Chakraborty, “Lenths of runs and waiting time distributions by using Polya- Eggenberger sampling scheme,” Studia Sci. Math. Hungar., vol. 2, pp. 309-332, 2002. D. L. Antzoulakos, S. Bersimis and M. V. Koutras, “On the distribution of the total number of run lengths,” Ann. Inst. Statist. Math., vol. 55, pp. 865-884, 2003. D. E. Martin, “Distribution of the number of successes in success runs of length at least  in higher-order Markovian sequences,” Methodol. Comput. Appl. Probab., vol. 7, pp. 543-554, 2005. K. Sinha, Location and communication issues in mobile networks. Ph. D. Dissertaion, Department of Computer Science and Engineering, Jadavpur University, Calcutta, India , 2007. Copyright © 2012 SciRes. 47 Home \| About SCIRP \| Sitemap \| Contact Us Copyright ? 2006-2013 Scientific Research Publishing Inc. All rights reserved.
14230	https://brainly.com/question/24850861	[FREE] The product –2 • (–5) is shown on the number line. Use the drop-down menus to explain how the model on the - brainly.com 4 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 +56,7k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +45,2k Ace exams faster, with practice that adapts to you Practice Worksheets +6,2k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified The product –2 • (–5) is shown on the number line. Use the drop-down menus to explain how the model on the number line could be used to find the answer. Number line from -10 to 10. Start at -10. Count up by 1. 2 black arrows are shown above the number line. First from -5 to 10 above which from -5 to 0 Starting at 0, move twice to the Choose... 5 spaces. 2 • –5= Choose... . Since –2 • (–5) is Choose... 2 • (–5), the number line shows the same distance to the Choose... of 0. 1 See answer Explain with Learning Companion NEW Asked by Sydnee08 • 09/29/2021 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 10684415 people 10M 0.0 0 Upload your school material for a more relevant answer Visually representing (-2) · (-5) on a number line involves drawing a leftward segment for -2, extending it left for -5, and mirroring the result to the right, confirming the positive product of 10. To represent the product (-2) · (-5) on a number line, follow these steps: Begin by drawing the distance of the first negative number, which is -2. Since negative numbers represent leftward distances on the number line, draw a line segment to the left with a length corresponding to the magnitude of -2. Extend this initial distance by the magnitude of the second negative number, which is -5. Since -5 is a larger magnitude, extend the line segment to the left even further, maintaining its negative direction. Now, draw the distance of the reflection of the result found in step 2. To do this, imagine the line segment as a mirror and draw an equal length to the right of the original line. Due to the algebraic property that the product of two negative numbers is positive, the reflection results in a rightward distance. Visually, you now have a line segment to the left representing the product of (-2) and (-5), and a mirrored, equal-length segment to the right, indicating the positive result of the multiplication, which is 10. This representation aligns with the concept that the product of two negative numbers is positive. Answered by bamkinbose •15.7K answers•10.7M people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 10684415 people 10M 0.0 0 Upload your school material for a more relevant answer To find −2⋅(−5) using a number line, you start at 0 and move 2 spaces left, considering the multiplication as repeated movement to reflect positively due to both numbers being negative. This results in a conclusion that the product is positive, specifically 10. The number line effectively demonstrates this concept through its representation of distance and direction. Explanation To understand the product −2⋅(−5) using a number line, we follow these steps: Start at the Origin: Begin at the point 0 on the number line. Understanding -2: Since we are multiplying by -2, imagine moving 2 spaces to the left from 0. This respectively represents the negative direction along the number line. Understanding -5: Next, you take the next step by recognizing we are still working with -5; this indicates that you focus on the distance moving away from zero as if you were to mirror it. Making the First Move: From the starting point at -2, you can think of how many times that number (or -5) fits in the entire distance. Count back 5 spaces two times (because we have -2 which indicates multiplying two times) reflecting towards a positive direction. Final Movement and Reflection: This means that moving twice the distance of -5 virtually leads to a positive result because you're moving in the opposite direction of the negative; thus you get a final result that appears as a positive value. Concluding Result: Thus, we conclude that −2⋅(−5)=10. The number line effectively demonstrates that two negatives yield a positive through this logical reflection. Hence, the number line allows us to find this value visually by extending leftward for the negative numbers and then reflecting positively, confirming the product is indeed positive. This is consistent with the multiplication rules which convey that the product of two negative numbers will always result in a positive outcome. Examples & Evidence A helpful example is moving from 0: when you move left by 2 (to -2) and then visualize reflecting that to the right by the distance of 5; this spatial movement reinforces that two negatives create a positive result. This follows the mathematical principle that states when multiplying two negative numbers, the product is positive, as confirmed by the rules of signed numbers. Thanks 0 0.0 (0 votes) Advertisement Sydnee08 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics A farmer has a total of 7805 beet plants. He wants to plant 40 beet plants in each row. He does the following work in long division: $\begin{array}{c} 195 R 5 \ 4 0 \longdiv { 7 8 0 5 } \ \frac{-40}{380} \ \frac{-360}{205} \ \frac{-200}{5} \end{array}$ Which statement is correct? A. The farmer plants 5 rows of beets and has 40 beet plants left over. B. The farmer plants 40 rows of beets and has 195 beet plants left over. C. The farmer plants 195 rows of beets and has 5 beet plants left over. D. The farmer plants 5 rows of beets and has 195 beet plants left over. Write an equation of the line that passes through (−2,−3) and is parallel to the line defined by 5 x+y=−5. Write the answer in slope-intercept form and in standard form (A x+B y=C) with smallest integer coefficients. What are the x-intercepts of the function f(x)=−2 x 2−3 x+20? Given the function f(x)=−5 x 2−x+20, find f(3). A circular garden with a radius of 8 feet is surrounded by a circular path with a width of 3 feet. What is the approximate area of the path alone? Use 3.14 for π. A. 172.70 f t 2 B. 178.98 f t 2 C. 200.96 f t 2 D. 379.94 f t 2 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
14231	https://digital.library.unt.edu/ark:/67531/metadc177270/m2/1/high_res_d/AcceptMan-Pub-599.pdf	Accepted Manuscript Title: A Group Contribution Model for Determining the Sublimation Enthalpy of Organic Compounds at the Standard Reference Temperature of 298 K Author: Farhad Gharagheizi Poorandokht Ilani-Kashkouli William E. Acree Jr. Amir H. Mohammadi Deresh RamjugernathTel.: + 27 312603128; fax: + 27 312601118. PII: S0378-3812(13)00321-X DOI: Reference: FLUID 9636 To appear in: Fluid Phase Equilibria Received date: 26-3-2013 Revised date: 18-6-2013 Accepted date: 21-6-2013 Please cite this article as: F. Gharagheizi, P. Ilani-Kashkouli, W.E. Acree Jr., A.H. Mohammadi, D. Ramjugernath, A Group Contribution Model for Determining the Sublimation Enthalpy of Organic Compounds at the Standard Reference Temperature of 298 K, Fluid Phase Equilibria (2013), This is a PDF ﬁle of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its ﬁnal form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain. Page 1 of 79 Accepted Manuscript 1 A Group Contribution Model for Determining the Sublimation Enthalpy of Organic Compounds at the Standard Reference Temperature of 298 K Farhad Gharagheizi,a Poorandokht Ilani-Kashkouli,a William E. Acree Jr.,b Amir H. Mohammadi,,a,c Deresh Ramjugernath,a a Thermodynamics Research Unit, School of Engineering, University of KwaZulu-Natal, Howard College Campus, King George V Avenue, Durban 4041, South Africa bDepartment of Chemistry, 1155 Union Circle Drive #305070, University of North Texas, Denton, TX 76203-5017, USA cInstitut de Recherche en Génie Chimique et Pétrolier (IRGCP), Paris Cedex, France Abstract  The sublimation enthalpy provides a measure of molecular interactions in the solid phase. Practical applications involving sublimation enthalpies include the estimation of the crystal lattice energy of molecular crystals, estimation of the enthalpy of solvation of crystalline organic compounds, and prediction of the environmental fate and vapor pressures of solid compounds. Recently an extensive compilation of phase change enthalpies, including sublimation enthalpies of pure organic and organometallic compounds, was published . This collection of sublimation enthalpies for 1269 compounds at the standard temperature of 298.15 K was used in this study for the development of a predictive model. The compounds in the collection are composed of carbon, hydrogen, nitrogen, oxygen, phosphorous, sulfur, fluorine, chlorine, bromine, and iodine. This paper presents a reliable group contribution model for the estimation of the sublimation enthalpies of organic compounds. The group contribution model developed is able to predict the standard molar enthalpies of sublimation to within an average absolute relative deviation of 6.4%, which is of sufficient accuracy for many practical applications. Keywords: Sublimation enthalpy, Group Contribution, Organic Compounds; Chemical Structure, Reliable model. Corresponding author, Email: a.h.m@irgcp.fr Tel.: + (33) 1 64 69 49 70. Fax: + (33) 1 64 69 49 68. : Corresponding author, Email: ramjuger@ukzn.ac.za Tel.: + (27) 312603128. Fax: + (27) 312601118. Page 2 of 79 Accepted Manuscript 2 1. Introduction The enthalpy change associated with a change in phase is of great importance in various disciplines such as chemical and environmental engineering, chemistry, and physics due to the fact that it provides a measure of intra- and intermolecular interactions . As one of the phase change enthalpies, the sublimation enthalpy is considered as a measure of intermolecular forces of substances in the solid phase . The crystal lattice energy which is the energy that constitutes a crystal from the isolated gas phase molecules is directly calculated from the sublimation enthalpy. Therefore, this parameter is used to determine the specific packing of solid state crystalline substances . As a consequence of such capability, sublimation enthalpy is used to describe the solvation of molecules, particularly drug molecules [4-9]. The sublimation enthalpy is also used to evaluate the transport of contaminants in the atmosphere; in environmental fate modeling; to determine discoloration of materials; and determine dispersion of dyes . Furthermore, it can be used to calculate the standard molar enthalpy of formation of crystalline compounds in the gas phase from measured enthalpy of combustion data. Moreover, it can be used to estimate other physical properties such as vapor pressure through the well know Clausius-Clapeyron equation . Several methods have so far been proposed for the estimation of sublimation enthalpies of pure compounds at standard temperature, viz. 298.15 K. Rice et al. used the properties associated with quantum mechanically determined electrostatic potentials of isolated molecules to correlate the sublimation enthalpies of a dataset of 35 pure organic compounds. The root mean Page 3 of 79 Accepted Manuscript 3 square error (RMSE) and the maximum deviation of the model from experimental data were reported as 15 and 52 kJ.mole-1, respectively. Several other proposed models, viz. Politzer et al. , Matheieu and Simonetti , and Kim et al. independently modified the van der Waals electrostatic surface potentials. Implementing these modifications, they developed several parameters to calculate the sublimation enthalpy. Their models showed low deviation from experimental data for a small dataset of 34 organic compounds. In another proposed model, Ouvrard and Mitchell used the number of occurrences of various atom types as descriptors to correlate the sublimation enthalpy. The authors employed a training set comprised of 226 compound for developing the basic model and another dataset of 35 compounds as a test set for assessing the model’s predictive capability for which they reported squared correlation coefficients (R2) of 0.925 and 0.937, respectively. Politzer et al. suggested a three-term expression for correlating the sublimation enthalpies that employed as input parameters the molecular surface area and information based on surface electrostatic potential. The input parameters were computed at the B3PW91/6-31G level. The correlation model derived predicted the sublimation enthalpies of 105 amino acids and small organic compounds to within an average absolute deviation of 11.7 kJ mol-1. Byrd and Rice used quantum mechanical data to predict the sublimation enthalpy. They stated that their model can estimate the sublimation enthalpies of 35 organic compounds with a RMSE and maximum deviation of 12.5 and 217.7 kJ.mole-1, respectively. To date, a few models have been proposed for the estimation of sublimation enthalpy at the triple point. As a first attempt, Gaharagheizi proposed a 5-parameter quantitative structure property relationship (QSPR) for the estimation of the sublimation enthalpies of 1348 pure chemical compounds. The R2, RMSE, and maximum absolute relative deviation of the Page 4 of 79 Accepted Manuscript 4 model from DIPPR 801 data were 0.9746, 5.46, and 27.56 kJ.mole-1, respectively. In a follow-up study, Gharagheizi et al. used an artificial neural network-group contribution approach to correlate the sublimation enthalpies of 1384 pure chemical compounds. The model demonstrated good descriptive ability as evidenced by the R2, average absolute relative deviation (AARD%), and root-mean square error of 0.986, 3.54% and 4.21 kJ.mole-1, respectively. Mathieu used a subset of the dataset used by Gharagheizi (1300 out of 1348 data) to develop a 31 parameter-model based on the fragment contributions. The R2, RMSE, and AARD% of the model compared with DIPPR 801 are 0.986, 4 kJ.mole-1, and 3.1%, respectively. More recently, Salahinejd et al. employed another subset of the dataset used by Gharagheizi (1304 out of 1348 data) to obtain a 4 parameter-QSPR model for the prediction of the sublimation enthalpy. The authors reported values of 0.96 and 7.9 kJ.mole-1 for the R2 and the average absolute error of their model, respectively. The results show that the latter model predicts the sublimation enthalpy with a lower accuracy than the one proposed by Gharagheizi . Neither Mathieu , nor Salahinejd et al. mentioned why they eliminated 48 and 44 compounds, respectively, from the complete dataset implemented by Gharagheizi 20. A thorough comparison among the previous models proposed for the estimation of the sublimation enthalpy of pure chemical compounds reveals that: 1- Most of the previous models for the estimation the sublimation enthalpy at the standard temperature of 298.15 K have been developed/evaluated for small chemical groups/families of compounds. Furthermore, the largest data set was used by Ouvrard (261 compounds). Page 5 of 79 Accepted Manuscript 5 2- Among the various models for the estimation of the sublimation enthalpy at the triple point, the model proposed by Gharagheizi et al. shows better results and is more comprehensive that the others. Recently, Acree and Chickos reviewed the literature for published phase change enthalpies at the standard temperature of 298.15 K over the period of 1880-2010 and presented their results as a massive compilation. The main aim of this study is to develop a group contribution method using the data compilation, along with published enthalpy of sublimation [23-119] data over the past three years. 2. Experimental enthalpy of sublimation database As mentioned earlier, the data compilation presented by Acree and Chickos together with recently published data over the past three years [23-119] was implemented to provide the dataset of sublimation enthalpies of compounds used in this study. The enthalpies of sublimation were determined by well-established experimental methodologies, including “vacuum sublimation” drop microcalorimetry, Knudsen mass-loss effusion, transpiration, and correlation gas chromatography combined with differential scanning calorimetric measurement of the enthalpy of fusion. Correlation gas chromatographic measurements which pertain to enthalpies of vaporization, ΔvaporizationHm o, and the enthalpy of fusion, ΔfusionHm o, are needed to convert the measured ΔvaporizationHm o to ΔsublimationHm o , e.g., ΔsublimationHm o (T = 298.15 K) = ΔvaporizationHm o (T = 298.15 K) + ΔfusionHm o (T = 298.15 K). For many of the compounds the measurements were performed at mean temperatures, Tmean, higher than 298.15 K, in which case the measured Page 6 of 79 Accepted Manuscript 6 enthalpy of sublimation was corrected back to 298.15 K using the standard thermodynamic relationship ) 15 . 298 ( ) ( ) 15 . 298 ( lim lim mean o p mean o m ation sub o m ation sub T C T T H T H         (1) where ΔCp o is the molar heat capacity difference between the crystalline and gaseous forms of the organic compound. Researchers reporting ΔsublimationHm o data have estimated the required ΔCp o values in different ways. Some used a generic value of -8.314 J mol-1 K-1 , while others used group contribution methods , or measured heat capacity data for the crystalline compound combined with estimated gas phase heat capacities from statistical thermodynamics using the vibrational frequencies from quantum mechanical B3LYP/6–31G(d) calculations . We have used the enthalpies of sublimation as reported by the authors as there was often insufficient experimental data given in the published papers for us to make corrections in a consistent manner. Some papers simply gave the the enthalpy of sublimation corrected to 298 K with no additional information. The reported experimental uncertainty given by the reporting authors rarely included the uncertainty associated with extrapolating the measured values to 298.15 K. Roux et al. compiled and critically evaluated published thermodynamic data for polycyclic aromatic hydrocarbons (PAHs). As part of their evaluations the authors did recommend numerical values for ΔsublimationHm o (T = 298.15 K) for the compounds they studied. Our database includes includes many of (though not all of) the PAH compounds considered by Roux et al. We considered only those compounds where the reporting authors had given a ΔsublimationHm o (T = 298.15 K) value. For most of the PAH compounds common to both databases, the average values that we have used in developing our group contribution method Page 7 of 79 Accepted Manuscript 7 were within 1 to 2 kJ.mol-1 of the recommended values of Roux et al. . The notable exceptions were for chrysene, dibenz[a,c]anthracene, dibenz[a,h]anthracene, naphthacene, and pentacene where we have elected to use the ΔsublimationHm o (T = 298.15 K) values given by the reporting author . Roux et al. stated that they believed the correction used to extrapolate the measured ΔsublimationHm (T = Tmean) back to 298 K was too large, and they used a different set of ΔCp o values. The database used in the present study is comprised of 1645 experimental data points for 1269 compounds. A single experimental value was reported for 1018 compounds. Multiple values were reported for 251 of the 1269 compounds, in which case the arithmetic averages were used. No attempt was made to select between the independently determined values, which for the most part differed by less than 6 kJ mol-1. Several of the more notable exceptions to this were: 2-imadazolinone where the observed enthalpy of sublimation ranged from 83.7 kJ to 96.6 kJ mol-1; 3,4-dihydroxy-3-cyclobutene-1,2-dione where the enthalpy of sublimation ranged from 83.7 kJ mol-1 to 154.3 kJ mol-1; cytosine where the enthalpy of sublimation ranged from 155 to 176 kJ mol-1; tetrahydro-2-pyrimidone where the enthalpy of sublimation ranged from 89.3 kJ to 113.4 kJ mol-1; 1,3-dithiane where the enthalpy of sublimation ranged from 52.3 to 69.9 kJ mol-1; 4-hydroxypyridine where the enthalpy of sublimation ranged from 103.8 to 118.6 kJ mol-1; 5-methyluracil from 131.3 to 138 kJ mol-1; 3-pyridinecarboxylic acid where the enthalpy of sublimation ranged from 105.2 to 123.9 kJ mol-1; 1,2,3-trihydroxybenzene where the enthalpy of sublimation ranged from 104 to 116.9 kJ mol-1; hexanamide where the enthalpy of sublimation ranged from 85 to 98.7 kJ mol-1; 2-brombenzoic acid where the enthalpy of sublimation ranged from 95.9 to 108.5 kJ mol-1; 2-iodobenzoic acid where the enthalpy of sublimation ranged from 92.6 to 112.8 kJ mol-1; 2-iodobenzoic acid where the enthalpy of sublimation ranged from 96.4 Page 8 of 79 Accepted Manuscript 8 to 111.1 kJ mol-1; 4-iodobenzoic acid where the enthalpy of sublimation ranged from 99.3 to 112.9 kJ mol-1; 2,4,6-trinitrotoluene where the enthalpy of sublimation ranged from 104.6 to 113.2 kJ mol-1; benzimidazole where the enthalpy of sublimation ranged from 94.3 to 102.2 kJ mole-1; 3-hydroxybenzoic acid where the enthalpy of sublimation ranged from 118.3 to 125 kJ mol-1; 4-hydroxybenzamide where the enthalpy of sublimation ranged from 117.8 to 129.7 kJ mol-1; 1,3-dimethylxanthine where the enthalpy of sublimation ranged from 135 to 144 kJ mol-1; (2,4-dichlorophenoxy)acetic acid where the enthalpy of sublimation ranged from 115 to 125 kJ mol-1; for coumarin where the enthalpy of sublimation ranged from 83.1 to 95.4 kJ mol-1; 2-(2,4-dichlorophenoxy)propanoic acid where the enthalpy of sublimation ranged from 116 to 130 kJ mol-1; for 2,4,6-trimethylphenol where the enthalpy of sublimation ranged from 82.8 to 95 kJ mol-1; 2-adamantone where the enthalpy of sublimation ranged from 66.3 to 80.3 kJ mol-1; and tetraphenylmethane where the enthalpy of sublimation ranged from 140 to 150.6 kJ mol-1. In total a collection of standard molar enthalpies of sublimation was obtained for 1270 unique pure chemical compounds at 298.15 K. A careful analysis of the compounds within the dataset shows that the sublimation enthalpies range between 34 and 240 kJ.mole-1. The compounds are composed of carbon (1 to 34 atoms per compound), hydrogen (1 to 48 atoms per compound), nitrogen (1 to 7 atoms per compound), oxygen (from 1 to 14 atoms per compound), phosphorus (only 1 atom per compound), sulfur (1 to 6 atoms per compound), fluorine (1 to 34 atoms per compound), chlorine (1 to 6 atoms per compound), bromine (1 to 4 atoms per compound), and iodine (1 to 2 atoms per compound) atoms. There are 117 hydrocarbons (C and H compounds) in the dataset whose sublimation enthalpies range from 37 to 182 kJ.mole-1. The dataset includes 113 nitrogen compounds whose sublimation enthalpies range from 34 to 199 kJ.mole-1. The elemental Page 9 of 79 Accepted Manuscript 9 composition analysis of the dataset further shows that there are 918 oxygen compounds whose sublimation enthalpies range from 45 to 239 kJ.mole-1. There are 112 sulfur compounds in the dataset having sublimation enthalpies that range from 54 to184 kJ.mole-1. There are a significant number of halogen compounds within the dataset: 55 fluorine-containing compounds with sublimation enthalpies between 62 and 134 kJ mol-1; 116 chlorine-containing compounds with sublimation enthalpies between 60 and 183 kJ mol-1; 33 bromine-containing compounds having sublimation enthalpies that range from 54 to 152 kJ.mole-1; and 16 iodine-containing compounds whose sublimation enthalpies range from 70 to 127 kJ.mole-1. The number of phosphorous compounds in the dataset is much smaller (5 compounds) and their sublimation enthalpies range from 75 to 143 kJ.mole-1. The chemical diversity of the dataset considered in the present study is significantly greater than datasets used in earlier studies [13-23] involving the prediction of sublimation enthalpies. In order to obtain a predictive model, the data set was split into three sub-data sets; the first set for developing the model (called the “training set”), the second set for assessing the internal validity of the model (called the ”validation set”), and the final set for evaluating the predictive capability of the derived model (called the “test set”). The division of the data can be performed randomly; however, this may lead to an inappropriate allocation of compounds to each sub-dataset; in other words all of the larger enthalpies of sublimation might end up in the test set. In order to avoid this potential problem, one can use the K-means clustering technique [125, 126]. This method partitions a dataset into n sub-datasets in which each data point belongs to the subset with the closest mean. This procedure resolves the issue of inappropriate allocation of datasets. Another point is the quota of each sub-dataset from the main dataset. It has been shown that if training set is too small, the produced model doesn’t have predictive power. Page 10 of 79 Accepted Manuscript 10 Moreover, if the dataset is too large, the model may produce significantly better results for the training set rather than for the validation and test sets . In order to prevent these issues, nearly 80% of the data set was allocated to the training set (1015 data points) and the remaining data points were allocated evenly between the respective validation and test sets (127 data points each). 3. Model development To develop a reliable correlation model, one must use parameters which enable one to distinguish each compound from the others. In other words, one needs a unique set of parameters for each compound that can adequately describe the sublimation enthalpy. Based on past experience [20, 21] it was decided to generate the parameters from the molecular structures. As a result, a collection of 294 chemical substructures were gathered which have previously been implemented by the authors to correlate other important physical properties [20, 128-131]. In the next step, the frequency of appearance of each of these 294 chemical substructures was counted in each compound. The pair correlation between each pair of the 294 chemical substructures was then evaluated to avoid entering irrelevant parameters into the final model. In the next step, if the pair correlation of a pair of chemical substructures was more than the threshold value of 0.95, one of them was eliminated and the other kept for the next step. Performing this procedure, the collection of the chemical substructures was reduced to 251 chemical substructures. In order to determine the final model and to choose the optimal subset of chemical substructures affecting the sublimation enthalpy, the sequential search method was applied . The major target of a sequential search is to find an optimal subset of chemical substructures for a specified model Page 11 of 79 Accepted Manuscript 11 size. The basic idea of the method is to replace each chemical substructure, one at a time, with all the remaining ones and see whether a better model is obtained. To accomplish this, both R2 and AARD% were used to evaluate the improvement by adding a new chemical substructure to the model. The statistical parameters used in this article are defined in Appendix A. 4. Result and discussion In order to obtain a reliable model, the collection of 251 chemical substructures prepared in the previous step, was introduced to the sequential search algorithm. The gradual changes in R2 and AARD% as a function of an incremental increase in the number of chemical substructures is depicted in Figure 1. Figure 1 In order to find the optimal model in terms of both the number of chemical substructures and accuracy, a threshold value of 0.01 was considered for the decrease in AARD% as a stopping criterion. It means that when the improvement of the model AARD% was less than 0.01, the sequential search algorithm was automatically stopped and reported the final model. The optimal model was obtained using 147 chemical substructures. This point is depicted as a green pentagram sign in Figure 1. The model obtained is as follows: (2) Page 12 of 79 Accepted Manuscript 12 where , and are the intercept of the equation, the contribution of the ith chemical substructure to the sublimation enthalpy, and the number of occurrences of the ith chemical substructure in every chemical structure of pure compounds, respectively. The subset of 147 chemical substructures and their contribution to the sublimation enthalpy are tabulated in Table 1. Table 1 The predicted sublimation enthalpies and their absolute relative deviation from the experimental values are presented as a supplementary table. The model results show that it can successfully predict the standard molar enthalpies of sublimation of pure organic compounds at 298 K. The average absolute relative deviation, standard deviation error, and root mean square error of the model are 6.3%, 10.5, and,10.5 for the training set; 6.3%, 10.7, and 10.7 for the validation set; and 6.3%, 12.7, and 10.8 for the test set, respectively. The values are based on the model predictions and their corresponding experimental values. For graphical presentation of the applicability domain of the model and the outliers, the Williams plot is depicted in Figure 2. Figure 2 Page 13 of 79 Accepted Manuscript 13 This plot shows the correlation of hat values and standardized residuals. It should be noted that the hat values and standardized residuals values are presented in a supplementary table for all the compounds. A warning leverage (h=0.35) - blue vertical line – is generally fixed at 3n/p, where n is number of training chemicals and p the number of model variables plus one. The leverage of 3 is considered as a cut-off value to accept the points that lie ± 3 (two horizontal red lines) standard deviations from the mean (to cover 99% normally distributed data). The applicability domain is located in the region of 0≤h≤0.35 and -3≤R≤+3. Existence of the majority of data points in this domain shows that both model development and prediction are performed within the applicability domain which results in a valid model. The points depicted with red circles (3<R or R<-3) are "bad high leverage" points and represent outliers of the model. This erroneous prediction could probably be attributed to incorrect experimental data rather than to the molecular structure . These points are highlighted in the supplementary table. According to the results, the model can predict the sublimation enthalpies of 117 hydrocarbons with an AARD% of 6.7%. There are 24 hydrocarbons for which the model shows a deviation of more than 10%. A careful consideration of the hydrocarbons demonstrates that they are multi-ring complicated compounds. The chemical structures of these 24 hydrocarbons are shown in Table 2. We do note that chrysene, dibenz[a,c]anthracene, dibenz[a,h]anthracene, naphthacene, and pentacene are among the 24 hydrocarbons showing the larger deviations. Our predicted values of 114.4 kJ.mol-1 (123.4 kJ mol-1) for chrysene, of 148.9 kJ.mol-1 (135.4 kJ.mol-1) for diben[a,h]anthracene, 129.9 kJ.mol-1 (145.9 kJ.mol-1) for dibenz[a,c]anthracene, 116.3 kJ.mol-1 (135.9 kJ.mol-1) for naphthacene, and 137.3 kJ.mol-1 (165.5 kJ.mol-1) for pentacene are in better agreement with the recommended values of Roux et al. (which are given in parentheses) Page 14 of 79 Accepted Manuscript 14 than the reported ΔsublimationHm o (T = 298.15 K) values of DeKruif . Figure 3 depicts the predicted sublimation enthalpies of hydrocarbons versus the corresponding experimental values. Figure 3 Table 2 The model predictions for nitrogen compounds versus. the corresponding experimental sublimation enthalpies are presented in Figure 4. As demonstrated, the AARD% of the model from experimental data is 6.8%. According to the Williams plot depicted in Figure 2, many outliers of the model are for nitrogen compounds. This may be the major cause of high deviation in the prediction of the sublimation enthalpy for these nitrogen compounds. Figure 4 The AARD% of the model results from experimental sublimation enthalpy for oxygen compounds is 6.2%. The predicted values versus the corresponding experimental data are shown in Figure 5. Like nitrogen compounds, the majority of highly deviating oxygen compounds are outliers. Therefore, their experimental data may be erroneous. Figure 5 There are just 5 phosphorous compounds within the data set for which the model gives a promising AARD% of 0.1%. Sulfur compounds are another class of compounds for which the model shows an AARD% of 4.7%. There are 18 compounds for which the model gives an ARD% of higher than 10%. The Page 15 of 79 Accepted Manuscript 15 compounds are presented in Table 3. The predicted versus experimental sublimation are shown in Figure 6. Table 3 Figure 6 Fluorine compounds are one of the important classes of compounds whose sublimation enthalpies are predicted by the model with an AARD% of 5.9%. Their predicted versus experimental sublimation data are shown in Figure 7. Figure 7 The model predicts the sublimation enthalpies of chlorine compounds better than fluorine compounds in terms of AARD% (5% vs. 5.9%). Their predicted versus experimental sublimation data are shown in Figure 8. Figure 8 Based on the model analysis, the sublimation enthalpies of bromine compounds are successfully predicted by the model. The model AARD% for this class of compounds is 3.1% which is less than those of halogen compounds mentioned above. Figure 9 depicts the predicted sublimation enthalpies of bromine compounds versus their corresponding experimental values. Figure 9 Page 16 of 79 Accepted Manuscript 16 Iodine compounds are another class of halogen compounds whose sublimation enthalpies are successfully predicted by the model. The model shows a low AARD% of 2.6% which is the minimum deviation among all the lighter halogen compounds that have been studied. Figure 10 depicts the predicted sublimation enthalpies of iodine compounds versus their corresponding experimental values. Figure 10 The AARD% of the model from experimental sublimation enthalpies of various classes of compounds are shown in Table 4. Table 4 Unfortunately, a comprehensive comparison between the presented model and the previous models is not possible because they have mostly developed for small groups/classes of compounds. Even the largest dataset used by Ouvrard and Mitchell which comprised of sublimation enthalpies of 261 organic compounds, when compared with the data used in this study is very small. In order to compare the performance of the presented model developed in this study with that proposed by Ouvrard and Mitchell , a comparison was made based on the chemical families of compounds that were used by Ouvrard and Mitchell in their studies. They categorized the compounds within their data set as aliphatic hydrocarbons, aromatic hydrocarbons, and non-hydrogen bonding compounds. We used the same classification for our main dataset in order to make a comparison. The results are presented in Table 5. As can be seen, the model presented by Ouvrard and Mitchell predicts the sublimation enthalpies of the Page 17 of 79 Accepted Manuscript 17 aliphatic and aromatic hydrocarbons slightly better than the model presented in our study. A similar behavior can be observed for non- hydrogen bonding compounds. It should be noted that the number of hydrocarbons in our dataset is significantly larger than that of Ouvrard and Mitchell . Another point to consider is that majority of the compounds for which the experimental sublimation enthalpies have been reported are capable of forming hydrogen bonding. However, most of the compounds used by Ouvrard and Mitchell to develop their model are non-hydrogen bonding. This latter detail may be considered as a drawback of thir model. Table 5 4. Conclusions A group contribution model was developed for the prediction of the standard molar enthalpies of sublimation at 298.15 K, ΔsublimationHm o (T = 298.15 K), for organic compounds. The validity and the predictive capability of the model were assessed using a validation set and a test set, respectively. The model is capable of predicting ΔsublimationHm o (T = 298.15 K) values of organic compounds with an acceptable average absolute relative deviation between predicted and experimental values of 6.4%. The dataset used in this study is comprised of 1269 organic compounds containing carbon, hydrogen, nitrogen, oxygen, phosphorous, sulfur, fluorine, chlorine, bromine and iodine atoms. Analysis of the model shows that the model can estimate the sublimation enthalpies of hydrocarbons, and compounds containing nitrogen, oxygen, Page 18 of 79 Accepted Manuscript 18 phosphorous, sulfur, fluorine, chlorine, bromine and iodine atoms to within acceptable average relative deviations of 6.9%, 6.8%, 6.2%, 0.1%, 5.7%, 5.9%, 5%, 3.1% and 2.6% from the corresponding experimental values, respectively. The parameters needed to predict the sublimation enthalpies are the number of occurrences of 147 simple chemical substructures in the compound under consideration and the numerical values of each substructure contribution to ΔsublimationHm o (T = 298.15 K) given in Table 1. Acknowledgements This work is based upon research supported by the South African Research Chairs Initiative of the Department of Science and Technology and National Research Foundation. Appendix A The mathematical definition of the relative deviation (RD%), average absolute relative deviation (AARD%), root mean square error (RMSE), standard deviation error (Std), and squared correlation coefficient (R2) are presented as follows: (A1) Page 19 of 79 Accepted Manuscript 19 lit lit pred RD   100 % (A2)    N i i lit i lit i pred N AARD ) ( \| ) ( ) ( \| 100 % (A3) N i lit i pred RMSE N i 2 1 )) ( ) ( (    (A4)      N 2 pred pred(i) N 1 i Std (A5)       N i N i lit i pred i lit i pred R 2 2 2 ) ) ( ( )) ( ) ( ( 1 where pred and lit denote the predicted value by model and its corresponding value reported by literature, respectively. The terms pred and lit refer to the mean values over the predicted values by the model and the mean value over the literature reported data. N is the number of data point in each data set or subset. Page 20 of 79 Accepted Manuscript 20 Appendix B Example 1: (1R,2R,3R,5S)-(-)-isopinocamheol HO Experimental value: 80.5±1.1 kJ.mole-1 Predicted value: 82.74 kJ.mole-1 (2.8% deviation) ID ΔSublimationHm i Number Value Number X VALUE ΔsublimationHm 0 17.22158 17.22157658 1 ΔsublimationHm 1 3 1.839585 5.518755471 2 ΔsublimationHm 2 3 4.324475 12.97342471 3 ΔsublimationHm 3 3 -3.17657 -9.529707593 4 ΔsublimationHm 4 1 5.180637 5.18063742 23 ΔsublimationHm 23 1 2.102287 2.102286687 24 ΔsublimationHm 24 1 -10.1085 -10.10848286 39 ΔsublimationHm 39 1 59.12675 59.12674626 49 ΔsublimationHm 49 13 -0.11698 -1.520792582 50 ΔsublimationHm 50 1 1.524469 1.524468656 51 ΔsublimationHm 51 1 -41.7694 -41.76940137 Page 21 of 79 Accepted Manuscript 21 66 ΔsublimationHm 66 1 -5.34759 -5.347590443 68 ΔsublimationHm 68 1 1.167072 1.167072277 69 ΔsublimationHm 69 1 13.44807 13.44806904 73 ΔsublimationHm 73 1 -6.3383 -6.338298652 75 ΔsublimationHm 75 1 -2.32134 -2.321344256 78 ΔsublimationHm 78 1 1.067491 1.067491379 80 ΔsublimationHm 80 1 -1.53055 -1.530549801 111 ΔsublimationHm 111 11 5.538397 60.92236595 113 ΔsublimationHm 113 1 7.904142 7.904142211 119 ΔsublimationHm 119 16 -2.11715 -33.87436543 121 ΔsublimationHm 121 2 -1.40174 -2.803484879 123 ΔsublimationHm 123 13 1.344881 17.48345031 126 ΔsublimationHm 126 5 -1.13923 -5.696139159 128 ΔsublimationHm 128 2 -1.02864 -2.057274921 Sum 82.74 Example 2: 2-phenyl-4H-1-benzopyran-4-one (flavone) O O Experimental value: 108.2±1.7 kJ.mole-1 Predicted value: 110.75 kJ.mole-1 (2.4% deviation) ID ΔSublimationHm i Number Value Number X VALUE ΔsublimationHm 0 17.22157658 17.22157658 5 ΔsublimationHm 5 12 2.327303476 27.92764171 6 ΔsublimationHm 6 3 -2.84795394 -8.54386182 7 ΔsublimationHm 7 3 1.719792347 5.159377041 26 ΔsublimationHm 26 1 5.235888943 5.235888943 42 ΔsublimationHm 42 1 -1.786805532 -1.786805532 Page 22 of 79 Accepted Manuscript 22 43 ΔsublimationHm 43 9 -1.949630378 -17.5466734 44 ΔsublimationHm 44 2 3.237334284 6.474668567 50 ΔsublimationHm 50 10 1.524468656 15.24468656 53 ΔsublimationHm 53 1 4.370113154 4.370113154 54 ΔsublimationHm 54 1 -4.089886384 -4.089886384 66 ΔsublimationHm 66 1 -5.347590443 -5.347590443 68 ΔsublimationHm 68 1 1.167072277 1.167072277 69 ΔsublimationHm 69 1 13.44806904 13.44806904 73 ΔsublimationHm 73 1 -6.338298652 -6.338298652 75 ΔsublimationHm 75 1 -2.321344256 -2.321344256 78 ΔsublimationHm 78 1 1.067491379 1.067491379 79 ΔsublimationHm 79 1 0.360241352 0.360241352 80 ΔsublimationHm 80 1 -1.530549801 -1.530549801 94 ΔsublimationHm 94 1 1.26499858 1.26499858 95 ΔsublimationHm 95 1 -1.389579357 -1.389579357 102 ΔsublimationHm 102 1 0.778841644 0.778841644 111 ΔsublimationHm 111 16 5.538396905 88.61435047 113 ΔsublimationHm 113 3 7.904142211 23.71242663 119 ΔsublimationHm 119 20 -2.117147839 -42.34295679 121 ΔsublimationHm 121 6 -1.401742439 -8.410454636 123 ΔsublimationHm 123 18 1.344880793 24.20785428 126 ΔsublimationHm 126 14 -1.139227832 -15.94918964 128 ΔsublimationHm 128 6 -1.028637461 -6.171824764 133 ΔsublimationHm 133 13 -0.268166117 -3.486159527 145 ΔsublimationHm 145 1 -0.251418535 -0.251418535 Sum 110.75 Example 3: pentacyclo[18.2.2.2(9,12).0(4,15).0(4,15).0(6,17)]hexacos-4,6(17),9,11,-15,20,22,23,25-nonane (triple layered [2.2]paracyclophane) Page 23 of 79 Accepted Manuscript 23 Experimental value: 125.9±2.5 kJ.mole-1 Predicted value: 132.77 kJ.mole-1 (5.5% deviation) ID ΔSublimationHm i Number Value Number X VALUE ΔsublimationHm 0 17.22157658 17.22157658 2 ΔsublimationHm 2 8 4.324474902 34.59579921 3 ΔsublimationHm 3 8 -3.176569198 -25.41255358 5 ΔsublimationHm 5 18 2.327303476 41.89146256 6 ΔsublimationHm 6 8 -2.84795394 -22.78363152 43 ΔsublimationHm 43 10 -1.949630378 -19.49630378 44 ΔsublimationHm 44 8 3.237334284 25.89867427 49 ΔsublimationHm 49 16 -0.116984045 -1.871744717 50 ΔsublimationHm 50 10 1.524468656 15.24468656 68 ΔsublimationHm 68 1 1.167072277 1.167072277 73 ΔsublimationHm 73 1 -6.338298652 -6.338298652 79 ΔsublimationHm 79 1 0.360241352 0.360241352 94 ΔsublimationHm 94 1 1.26499858 1.26499858 102 ΔsublimationHm 102 1 0.778841644 0.778841644 108 ΔsublimationHm 108 1 2.124815779 2.124815779 111 ΔsublimationHm 111 30 5.538396905 166.1519071 119 ΔsublimationHm 119 42 -2.117147839 -88.92020925 123 ΔsublimationHm 123 47 1.344880793 63.20939728 126 ΔsublimationHm 126 47 -1.139227832 -53.54370809 133 ΔsublimationHm 133 55 -0.268166117 -14.74913646 145 ΔsublimationHm 145 16 -0.251418535 -4.022696558 Sum 132.77 Page 24 of 79 Accepted Manuscript 24 References W. 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Hare, Predicting heats of formation of energetic materials using quantum mechanical calculations, Combustion and Flame, 118 (1999) 445-458. Page 26 of 79 Accepted Manuscript 26 P. Politzer, J.S. Murray, M.E. Grice, M. Desalvo, E. Miller, Calculation of heats of sublimation and solid phase heats of formation, Molecular Physics, 91 (1997) 923-928. D. Mathieu, P. Simonetti, Evaluation of solid-state formation enthalpies for energetic materials and related compounds, Thermochimica Acta, 384 (2002) 369-375. C.K. Kim, K.A. Lee, K.H. Hyun, H.J. Park, I.Y. Kwack, C.K. Kim, H.W. Lee, B.S.U. Lee, Prediction of physicochemical properties of organic molecules using van der waals surface electrostatic potentials, Journal of Computational Chemistry, 25 (2004) 2073-2079. C. Ouvrard, J.B.O. Mitchell, Can we predict lattice energy from molecular structure?, Acta Crystallographica Section B: Structural Science, 59 (2003) 676-685. P. Politzer, Y. Ma, P. Lane, M.C. Concha, Computational prediction of standard gas, liquid, and solid-phase heats of formation and heats of vaporization and sublimation, International Journal of Quantum Chemistry, 105 (2005) 341-347. E.F.C. Byrd, B.M. Rice, Improved prediction of heats of formation of energetic materials using quantum mechanical calculations, Journal of Physical Chemistry A, 110 (2006) 1005-1013. F. Gharagheizi, A new molecular-based model for prediction of enthalpy of sublimation of pure components, Thermochimica Acta, 469 (2008) 8-11. F. Gharagheizi, M. Sattari, B. Tirandazi, Prediction of crystal lattice energy using enthalpy of sublimation: A group contribution-based model, Industrial and Engineering Chemistry Research, 50 (2011) 2482-2486. Page 27 of 79 Accepted Manuscript 27 D. Mathieu, Simple alternative to neural networks for predicting sublimation enthalpies from fragment contributions, Industrial and Engineering Chemistry Research, 51 (2012) 2814-2819. M. Salahinejad, T.C. Le, D.A. Winkler, Capturing the Crystal: Prediction of Enthalpy of Sublimation, Crystal Lattice Energy, and Melting Points of Organic Compounds, Journal of Chemical Information and Modeling, 53 (2013) 223-229. A.R.R.P. Almeida, M.A.R. Matos, M.J.S. Monte, V.M.F. Morais, Experimental and computational thermodynamic study of ortho-, meta-, and para-methylbenzamide, Journal of Chemical Thermodynamics, 47 (2012) 81-89. A.R.R.P. Almeida, M.J.S. Monte, Thermodynamic study of benzamide, N-methylbenzamide, and N, N -dimethylbenzamide: Vapor pressures, phase diagrams, and hydrogen bond enthalpy, Journal of Chemical and Engineering Data, 55 (2010) 3507-3512. A.R.R.P. Almeida, M.J.S. Monte, Thermodynamic study of the three fluorobenzamides: Vapor pressures, phase diagrams, and hydrogen bonds, Journal of Chemical and Engineering Data, 55 (2010) 5230-5236. A.R.R.P. Almeida, M.J.S. Monte, Vapor pressures and phase diagrams of two methyl esters of substituted benzoic acids, Journal of Chemical and Engineering Data, 56 (2011) 4862-4867. A.R.R.P. Almeida, M.J.S. Monte, Thermodynamic study of phase transitions of imidazoles and 1-methylimidazoles, Journal of Chemical Thermodynamics, 44 (2012) 163-168. Page 28 of 79 Accepted Manuscript 28 A.R.R.P. Almeida, M.J.S. Monte, Thermodynamic study of phase transitions in methyl esters of ortho- meta- and para-aminobenzoic acids, Journal of Chemical Thermodynamics, 53 (2012) 100-107. A.R.R.P. Almeida, M.J.S. Monte, The influence of the halogen size in the volatility and melting of methyl p-halobenzoic esters and of their parent acids, Journal of Chemical Thermodynamics, 57 (2013) 160-168. R.D. Chirico, A.F. Kazakov, W.V. Steele, Thermodynamic properties of three-ring aza-aromatics. 1. Experimental results for phenazine and acridine, and mutual validation of experiments and computational methods, Journal of Chemical Thermodynamics, 42 (2010) 571-580. J.C.S. Costa, C.F.R.A.C. Lima, M.A.A. Rocha, L.R. Gomes, L.M.N.B.F. Santos, Phase transition equilibrium of terthiophene isomers, Journal of Chemical Thermodynamics, 43 (2011) 133-139. J.Z. Dàvalos, A. Guerrero, R. Herrero, P. Jimenez, A. Chana, J.L.M. Abboud, C.F.R.A.C. Lima, L.M.N.B.F. Santos, A.F. Lago, Neutral, ion gas-phase energetics and structural properties of hydroxybenzophenones, Journal of Organic Chemistry, 75 (2010) 2564-2571. J.Z. Dávalos, R. Herrero, A. Chana, A. Guerrero, P. Jiménez, J.M. Santiuste, Energetics and structural properties, in the gas phase, of trans -hydroxycinnamic acids, Journal of Physical Chemistry A, 116 (2012) 2261-2267. Page 29 of 79 Accepted Manuscript 29 V.N. Emel'Yanenko, E.N. Stepurko, S.P. Verevkin, G.N. Roganov, The thermodynamic properties of 1,4-dioxane-2,6-dione, Russian Journal of Physical Chemistry A, 85 (2011) 179-185. V.N. Emel'yanenko, S.P. Verevkin, A.A. Pimerzin, The thermodynamic properties of DL-and L-lactides, Russian Journal of Physical Chemistry A, 83 (2009) 2013-2021. V.N. Emel'Yanenko, S.P. Verevkin, R.V. Ralys, V.V. Turovtsev, V.Y. Orlov, Enthalpy of phase transitions of lactams, Russian Journal of Physical Chemistry A, 86 (2012) 1493-1499. V.N. Emel'Yanenko, S.P. Verevkin, E.N. Stepurko, G.N. Roganov, M.K. Georgieva, Thermodynamic properties of glycolic acid and glycolide, Russian Journal of Physical Chemistry A, 84 (2010) 1301-1308. A.I.M.C.L. Ferreira, M.A.V. Ribeiro Da Silva, Thermochemical study of three dibromophenol isomers, Journal of Chemical Thermodynamics, 43 (2011) 227-234. J.M.S. Fonseca, O. Pfohl, R. Dohrn, Development and test of a new Knudsen effusion apparatus for the measurement of low vapour pressures, Journal of Chemical Thermodynamics, 43 (2011) 1942-1949. J.M.S. Fonseca, L.M.N.B.F. Santos, M.J.S. Monte, Thermodynamic study of 4- n -alkyloxybenzoic acids, Journal of Chemical and Engineering Data, 55 (2010) 2238-2245. V.L.S. Freitas, J.R.B. Gomes, L. Gales, A.M. Damas, M.D.M.C. Ribeiro Da Silva, Experimental and computational studies on the structural and thermodynamic properties of two Page 30 of 79 Accepted Manuscript 30 sulfur heterocyclic keto compounds, Journal of Chemical and Engineering Data, 55 (2010) 5009-5017. V.L.S. Freitas, J.R.B. Gomes, M.D.M.C. Ribeiro da Silva, Energetic effects of ether and ketone functional groups in 9,10-dihydroanthracene compound, Journal of Chemical Thermodynamics, 42 (2010) 1248-1254. V.L.S. Freitas, J.R.B. Gomes, M.D.M.C. Ribeiro Da Silva, Experimental and computational thermochemical studies of 9-R-xanthene derivatives (ROH, COOH, CONH 2), Journal of Chemical Thermodynamics, 54 (2012) 108-117. I.V. Garist, S.P. Verevkin, J.E. Bara, M.S. Hindman, S.P.O. Danielsen, Building blocks for ionic liquids: Vapor pressures and vaporization enthalpies of 1-(n-alkyl)-benzimidazoles, Journal of Chemical and Engineering Data, 57 (2012) 1803-1809. E.M. Gonçalves, C.E.S. Bernardes, H.P. Diogo, M.E. Minas Da Piedade, Energetics and structure of nicotinic acid (Niacin), Journal of Physical Chemistry B, 114 (2010) 5475-5485. D. Hasty, T. Subramanian, T.C. Winter, J.S. Chickos, A.A. Samarov, A.V. Yermalayeu, S.P. Verevkin, Applications of correlation gas chromatography and transpiration studies for the evaluation of the vaporization and sublimation enthalpies of some perfluorinated hydrocarbons, Journal of Chemical and Engineering Data, 57 (2012) 2350-2359. C.F.R.A.C. Lima, J.C.S. Costa, L.M.N.B.F. Santos, Thermodynamic insights on the structure and energetics of s-triphenyltriazine, Journal of Physical Chemistry A, 115 (2011) 9249-9258. Page 31 of 79 Accepted Manuscript 31 C.F.R.A.C. Lima, M.A.A. Rocha, A. Melo, L.R. Gomes, J.N. Low, L.M.N.B.F. Santos, Structural and thermodynamic characterization of polyphenylbenzenes, Journal of Physical Chemistry A, 115 (2011) 11876-11888. C.F.R.A.C. Lima, C.A.D. Sousa, J.E. Rodriguez-Borges, A. Melo, L.R. Gomes, J.N. Low, L.M.N.B.F. Santos, The role of aromatic interactions in the structure and energetics of benzyl ketones, Physical Chemistry Chemical Physics, 12 (2010) 11228-11237. D. Lipkind, N. Rath, J.S. Chickos, V.A. Pozdeev, S.P. Verevkin, The vaporization enthalpies of 2- and 4-(N,N-dimethylamino)pyridine, 1,5-diazabicyclo[4.3.0]non-5-ene, 1,8-diazabicyclo[5.4.0]undec-7-ene, imidazo[1,2-a]pyridine and 1,2,4-triazolo[1,5-a]pyrimidine by correlation-gas chromatography, Journal of Physical Chemistry B, 115 (2011) 8785-8796. A.I.M.C. Lobo Ferreira, M.A.V. Ribeiro Da Silva, Experimental and computational thermochemical study of the three monoiodophenol isomers, Journal of Chemical and Engineering Data, 56 (2011) 4881-4890. A.I.M.C. Lobo Ferreira, M.A.V. Ribeiro Da Silva, Experimental and computational study of the molecular energetics of the monoiodoanisole isomers, Journal of Chemical Thermodynamics, 48 (2012) 84-92. M.A.R. Matos, C.C.S. Sousa, V.M.F. Morais, Thermochemistry of chromone- and coumarin-3-carboxylic acid, Journal of Thermal Analysis and Calorimetry, 100 (2010) 519-526. Page 32 of 79 Accepted Manuscript 32 R. Maxwell, J. Chickos, An examination of the thermodynamics of fusion, vaporization, and sublimation of ibuprofen and naproxen by correlation gas chromatography, Journal of Pharmaceutical Sciences, 101 (2012) 805-814. M.S. Miranda, M.A.R. Matos, V.M.F. Morais, J.F. Liebman, Combined experimental and computational study on the energetics of 1,2-benzisothiazol-3(2H)-one and 1,4-benzothiazin-3(2H, 4H)-one, Journal of Chemical Thermodynamics, 43 (2011) 635-644. M.S. Miranda, M.A.R. Matos, V.M.F. Morais, J.F. Liebman, Study of energetics and structure of 1,2,3-benzotriazin-4(3H)-one and its 1H and Enol tautomers, Journal of Physical Chemistry B, 115 (2011) 6616-6622. M.S. Miranda, M.A.R. Matos, V.M.F. Morais, J.F. Liebman, Energetics of quinazoline-2,4(1 H,3 H)-dione: An experimental and computational study, Journal of Chemical and Engineering Data, 56 (2011) 4516-4523. M.S. Miranda, M.A.R. Matos, V.M.F. Morais, J.F. Liebman, 2,1,3-Benzothiadiazole: Study of its structure, energetics and aromaticity, Journal of Chemical Thermodynamics, 50 (2012) 30-36. E.A. Miroshnichenko, T.S. Kon'Kova, Y.N. Matyushin, Y.O. Inozemtsev, Bond dissociation energies in nitramines, Russian Chemical Bulletin, 58 (2009) 2015-2019. M.J.S. Monte, R. Notario, M.M.G. Calvinho, A.R.R.P. Almeida, L.M.P.F. Amaral, A.I.M.C. Lobo Ferreira, M.D.M.C. Ribeiro Da Silva, Experimental and computational study of Page 33 of 79 Accepted Manuscript 33 the thermodynamic properties of 9-fluorenone and 9-fluorenol, Journal of Chemical and Engineering Data, 57 (2012) 2486-2496. M.J.S. Monte, R. Notario, S.P. Pinto, A.I.M.C. Lobo Ferreira, M.D.M.C. Ribeiro Da Silva, Thermodynamic properties of fluoranthene: An experimental and computational study, Journal of Chemical Thermodynamics, 49 (2012) 159-164. M.J.S. Monte, S.P. Pinto, A.I.M.C. Lobo Ferreira, L.M.P.F. Amaral, V.L.S. Freitas, M.D.M.C. Ribeiro Da Silva, Fluorene: An extended experimental thermodynamic study, Journal of Chemical Thermodynamics, 45 (2012) 53-58. M.J.S. Monte, L.M.N.B.F. Santos, J.M.S. Fonseca, C.A.D. Sousa, Vapour pressures, enthalpies and entropies of sublimation of para substituted benzoic acids, Journal of Thermal Analysis and Calorimetry, 100 (2010) 465-474. M.J.S. Monte, C.A.D. Sousa, Thermodynamic study on the sublimation of diphenyl and triphenyl substituted acetic and propanoic acids, Journal of Thermal Analysis and Calorimetry, 106 (2011) 913-920. V.M.F. Morais, C.C.S. Sousa, M.A.R. Matos, Experimental and computational study of the energetics of methoxycoumarins, Journal of Molecular Structure: THEOCHEM, 946 (2010) 13-19. A.G. Nazmutdinov, I.A. Nesterov, T.A. Nazmutdinov, T.N. Nesterova, S.V. Tarazanov, S.V. Vostrikov, L.L. Pashchenko, E.A. Miroshnichenko, S.P. Verevkin, Vapour pressures and Page 34 of 79 Accepted Manuscript 34 enthalpies of vaporization of a series of the alkylbiphenyls, Fluid Phase Equilibria, 335 (2012) 88-98. R. Notario, M.V. Roux, C. Foces-Foces, M.A.V. Ribeiro Da Silva, M.D.D.M.C. Ribeiro Da Silva, A.F.L.O.M. Santos, R. Guzmán-Mejía, E. Juaristi, Experimental and computational thermochemical study of N-benzylalanines, Journal of Physical Chemistry B, 115 (2011) 9401-9409. G.L. Perlovich, A.M. Ryzhakov, N.N. Strakhova, V.P. Kazachenko, K.J. Schaper, O.A. Raevsky, Thermodynamic aspects of solubility, solvation and partitioning processes of some sulfonamides, Journal of Chemical Thermodynamics, 43 (2011) 683-689. G.L. Perlovich, T.V. Volkova, A.N. Proshin, D.Y. Sergeev, C.T. Bui, L.N. Petrova, S.O. Bachurin, Synthesis, pharmacology, crystal properties, and quantitative solvation studies from a drug transport perspective for three new 1,2,4-thiadiazoles, Journal of Pharmaceutical Sciences, 99 (2010) 3754-3768. R. Picciochi, H.P. Diogo, M.E. Minas Da Piedade, Thermochemistry of paracetamol, Journal of Thermal Analysis and Calorimetry, 100 (2010) 391-401. R. Picciochi, H.P. Diogo, M.E. Minas Da Piedade, Thermodynamic characterization of three polymorphic forms of piracetam, Journal of Pharmaceutical Sciences, 100 (2011) 594-603. S.V. Portnova, E.L. Krasnykh, S.P. Verevkin, Vapour pressure and enthalpy of vaporization of di-iso-propyl and di-tert-butyl esters of dicarboxylic acids, Fluid Phase Equilibria, 309 (2011) 114-120. Page 35 of 79 Accepted Manuscript 35 M.A.V. Ribeiro Da Silva, L.M.P.F. Amaral, Standard molar enthalpies of formation of monochloroacetophenone isomers, Journal of Chemical Thermodynamics, 42 (2010) 1473-1477. M.A.V. Ribeiro Da Silva, L.M.P.F. Amaral, Thermochemical study of 2,5-dimethyl-3-furancarboxylic acid, 4,5-dimethyl-2-furaldehyde, and 3-acetyl-2,5-dimethylfuran, Journal of Chemical Thermodynamics, 43 (2011) 1-8. M.A.V. Ribeiro Da Silva, L.M.P.F. Amaral, Thermochemical study of some dichloroacetophenone isomers, Journal of Chemical Thermodynamics, 43 (2011) 255-261. M.A.V. Ribeiro Da Silva, L.M.P.F. Amaral, Standard molar enthalpies of formation of 3′-and 4′-nitroacetophenones, Journal of Chemical Thermodynamics, 43 (2011) 876-881. M.A.V. Ribeiro Da Silva, L.M.P.F. Amaral, R.V. Ortiz, Experimental study on the thermochemistry of 3-nitrobenzophenone, 4-nitrobenzophenone and 3,3′-dinitrobenzophenone, Journal of Chemical Thermodynamics, 43 (2011) 546-551. M.A.V. Ribeiro Da Silva, L.M.P.F. Amaral, P. Szterner, Experimental study on the thermochemistry of 2-thiouracil, 5-methyl-2-thiouracil and 6-methyl-2-thiouracil, Journal of Chemical Thermodynamics, 57 (2013) 380-386. M.A.V. Ribeiro Da Silva, J.I.T.A. Cabral, Standard molar enthalpies of formation of 5- and 6-nitroindazole, Journal of Thermal Analysis and Calorimetry, 100 (2010) 457-464. M.A.V. Ribeiro Da Silva, J.I.T.A. Cabral, Standard molar enthalpies of formation of three methyl-pyrazole derivatives, Journal of Chemical Thermodynamics, 47 (2012) 138-143. Page 36 of 79 Accepted Manuscript 36 M.A.V. Ribeiro Da Silva, A.I.M.C. Lobo Ferreira, Thermochemistry of hydroxymethylphenol isomers, Journal of Thermal Analysis and Calorimetry, 100 (2010) 447-455. M.A.V. Ribeiro Da Silva, A.I.M.C. Lobo Ferreira, A.L.M. Barros, A.R.C. Bessa, B.C.S.A. Brito, J.A.S. Vieira, S.A.P. Martins, Standard molar enthalpies of formation of 1- and 2-cyanonaphthalene, Journal of Chemical Thermodynamics, 43 (2011) 1306-1314. M.A.V. Ribeiro Da Silva, A.I.M.C. Lobo Ferreira, A. Cimas, Experimental and computational study on the molecular energetics of benzyloxyphenol isomers, Journal of Chemical Thermodynamics, 43 (2011) 1857-1864. M.A.V. Ribeiro Da Silva, A.I.M.C. Lobo Ferreira, Á. Cimas, Calorimetric and computational study of the thermochemistry of phenoxyphenols, Journal of Organic Chemistry, 76 (2011) 3754-3764. M.A.V. Ribeiro Da Silva, M.J.S. Monte, A.I.M.C. Lobo Ferreira, J.A.S.A. Oliveira, A. Cimas, Experimental and Computational Thermodynamic Study of Three Monofluoronitrobenzene Isomers, Journal of Physical Chemistry B, 114 (2010) 7909-7919. M.A.V. Ribeiro Da Silva, M.J.S. Monte, A.I.M.C. Lobo Ferreira, J.A.S.A. Oliveira, Á. Cimas, A combined experimental and computational thermodynamic study of difluoronitrobenzene isomers, Journal of Physical Chemistry B, 114 (2010) 12914-12925. Page 37 of 79 Accepted Manuscript 37 M.A.V. Ribeiro Da Silva, M.J.S. Monte, I.M. Rocha, A. Cimas, Energetic study applied to the knowledge of the structural and electronic properties of monofluorobenzonitriles, Journal of Organic Chemistry, 77 (2012) 4312-4322. M.A.V. Ribeiro da Silva, M.D.M.C. Ribeiro da Silva, A.F.L.O.M. Santos, A.I.M.C.L. Ferreira, T.L.P. Galvão, Experimental thermochemical study of two chlorodinitroaniline isomers, Journal of Chemical Thermodynamics, 42 (2010) 496-501. M.A.V. Ribeiro Da Silva, A.F.L.O.M. Santos, Thermochemical properties of two nitrothiophene derivatives : 222-acetyl-5-nitrothiophene and 5-nitro-2-thiophenecarboxaldehyde, Journal of Thermal Analysis and Calorimetry, 100 (2010) 403-411. M.A.V. Ribeiro Da Silva, A.F.L.O.M. Santos, Energetics and molecular structure of 2,5-dimethyl-1-phenylpyrrole and 2,5-dimethyl-1-(4-nitrophenyl)pyrrole, Journal of Physical Chemistry B, 114 (2010) 16214-16222. M.D.M.C. Ribeiro Da Silva, V.L.S. Freitas, M.A.A. Vieira, M.J. Sottomayor, W.E. Acree Jr, Energetic and structural properties of 4-nitro-2,1,3-benzothiadiazole, Journal of Chemical Thermodynamics, 49 (2012) 146-153. M.D.M.C. Ribeiro da Silva, M.S. Miranda, C.M.V. Vaz, M.A.R. Matos, W.E. Acree Jr, Experimental thermochemical study of three monosubstituted pyrazines, The Journal of Chemical Thermodynamics, 37 (2005) 49-53. M.V. Roux, C. Foces-Foces, R. Notario, M.A.V. Ribeiro Da Silva, M.D.D.M.C. Ribeiro Da Silva, A.F.L.O.M. Santos, E. Juaristi, Experimental and computational thermochemical study of Page 38 of 79 Accepted Manuscript 38 sulfur-containing amino acids: L -cysteine, l -cystine, and l -cysteine-derived radicals. S-S, S-H, and C-S bond dissociation enthalpies, Journal of Physical Chemistry B, 114 (2010) 10530-10540. M.V. Roux, R. Notario, C. Foces-Foces, M. Temprado, F. Ros, V.N. Emel'Yanenko, S.P. Verevkin, Experimental and computational thermochemical study and solid-phase structure of 5,5-dimethylbarbituric acid, Journal of Physical Chemistry A, 114 (2010) 3583-3590. M.V. Roux, R. Notario, C. Foces-Foces, M. Temprado, F. Ros, V.N. Emel'Yanenko, S.P. Verevkin, Experimental and computational thermochemical study of barbituric acids: Structure-energy relationship in 1,3-dimethylbarbituric acid, Journal of Physical Chemistry A, 115 (2011) 3167-3173. M.V. Roux, R. Notario, M. Segura, J.S. Chickos, Thermophysical study of 2-thiobarbituric acids by differential scanning calorimetry, Journal of Chemical and Engineering Data, 57 (2012) 249-255. M.V. Roux, R. Notario, D.H. Zaitsau, V.N. Emel'Yanenko, S.P. Verevkin, Experimental and computational thermochemical study of 2-thiobarbituric acid: Structure-energy relationship, Journal of Physical Chemistry A, 116 (2012) 4639-4645. M.V. Roux, M. Temprado, P. Jiménez, C. Foces-Foces, R. Notario, A.R. Parameswar, A.V. Demchenko, J.S. Chickos, C.A. Deakyne, J.F. Liebman, Experimental and theoretical study of the structures and enthalpies of formation of 3 H -1,3-benzoxazole-2-thione, 3 H -1,3-benzothiazole-2-thione, and their tautomers, Journal of Physical Chemistry A, 114 (2010) 6336-6341. Page 39 of 79 Accepted Manuscript 39 M.V. Roux, M. Temprado, P. Jiménez, R. Notario, A.R. Parameswar, A.V. Demchenko, J.S. Chickos, C.A. Deakyne, J.F. Liebman, Knowledge of a molecule: An experimental and theoretical study of the structure and enthalpy of formation of tetrahydro-2 H -1,3-oxazine-2-thione, Journal of Chemical and Engineering Data, 56 (2011) 4725-4732. A.F.L.O.M. Santos, M.A.V.R. Da Silva, Energetics of 1-(aminophenyl)pyrroles: A joint calorimetric and computational study, Journal of Chemical Thermodynamics, 43 (2011) 1480-1487. A.F.L.O.M. Santos, A.R. Monteiro, J.M. Gonalves, W.E. Acree, M.D.M.C. Ribeiro Da Silva, Thermochemistry of 2,2′-dipyridil N-oxide and 2,2′-dipyridil N,N′-dioxide. the dissociation enthalpies of the N-O bonds, Journal of Chemical Thermodynamics, 43 (2011) 1044-1049. A.F.L.O.M. Santos, M.A.V. Ribeiro da Silva, Experimental and computational thermochemistry of 1-phenylpyrrole and 1-(4-methylphenyl)pyrrole, Journal of Chemical Thermodynamics, 42 (2010) 734-741. A.F.L.O.M. Santos, M.A.V. Ribeiro Da Silva, Experimental and computational energetic study of two halogenated 2-acetylpyrrole derivatives: 2-Trichloroacetylpyrrole and 2-trifluoroacetylpyrrole, Journal of Chemical Thermodynamics, 42 (2010) 1079-1086. A.F.L.O.M. Santos, M.A.V. Ribeiro Da Silva, A calorimetric and computational study of the thermochemistry of halogenated 1-phenylpyrrole derivatives, Journal of Chemical Thermodynamics, 42 (2010) 1441-1450. Page 40 of 79 Accepted Manuscript 40 A.F.L.O.M. Santos, M.A.V. Ribeiro Da Silva, Diaminobenzenes: An experimental and computational study, Journal of Physical Chemistry B, 115 (2011) 4939-4948. A.F.L.O.M. Santos, M.A.V. Ribeiro Da Silva, Molecular energetics of pyrrolecarbonitriles and derivatives: A combined calorimetric and computational study, Journal of Chemical Thermodynamics, 48 (2012) 194-200. A.F.L.O.M. Santos, M.A.V. Ribeiro Da Silva, The enthalpies of formation of alkyl carbamates: Experimental and computational redetermination, Journal of Chemical Thermodynamics, 57 (2013) 454-460. L.M.N.B.F. Santos, L.M.S.S. Lima, C.F.R.A.C. Lima, F.D. Magalhães, M.C. Torres, B. Schröder, M.A.V. Ribeiro Da Silva, New Knudsen effusion apparatus with simultaneous gravimetric and quartz crystal microbalance mass loss detection, Journal of Chemical Thermodynamics, 43 (2011) 834-843. A.L.R. Silva, Á. Cimas, M.D.M.C. Ribeiro Da Silva, Experimental and computational thermochemical studies of benzoxazole and two chlorobenzoxadole derivatives, Journal of Chemical Thermodynamics, 57 (2013) 212-219. C.C.S. Sousa, M.A.R. Matos, V.M.F. Morais, When theory and experiment hold hands: The thermochemistry of γ-pyrone derivatives, Journal of Chemical Thermodynamics, 43 (2011) 1159-1163. C.C.S. Sousa, M.A.R. Matos, V.M.F. Morais, Calorimetric and computational study of 7-hydroxycoumarin, Journal of Chemical Thermodynamics, 43 (2011) 1435-1440. Page 41 of 79 Accepted Manuscript 41 C.C.S. Sousa, V.M.F. Morais, M.A.R. Matos, Energetics of the isomers: 3- and 4-hydroxycoumarin, Journal of Chemical Thermodynamics, 42 (2010) 1372-1378. P. Umnahanant, D. Hasty, J. Chickos, An examination of the thermodynamics of fusion, vaporization, and sublimation of (R,S)- and (R)-flurbiprofen by correlation gas chromatography, Journal of Pharmaceutical Sciences, 101 (2012) 2045-2054. M.A. Varfolomeev, D.I. Abaidullina, B.N. Solomonov, S.P. Verevkin, V.N. Emel'Yanenko, Pairwise substitution effects, inter- and intramolecular hydrogen bonds in methoxyphenols and dimethoxybenzenes. Thermochemistry, calorimetry, and first-principles calculations, Journal of Physical Chemistry B, 114 (2010) 16503-16516. R.M. Varushchenko, A.A. Efimova, A.I. Druzhinina, E.S. Tkachenko, I.A. Nesterov, T.N. Nesterova, S.P. Verevkin, The heat capacities and thermodynamic functions of 4-methylbiphenyl and 4-tert-butylbiphenyl, Journal of Chemical Thermodynamics, 42 (2010) 1265-1272. S.P. Verevkin, V.N. Emel'yanenko, A.A. Pimerzin, E.E. Vishnevskaya, Thermodynamic analysis of strain in the five-membered oxygen and nitrogen heterocyclic compounds, Journal of Physical Chemistry A, 115 (2011) 1992-2004. S.P. Verevkin, V.N. Emel'Yanenko, A.A. Pimerzin, E.E. Vishnevskaya, Thermodynamic analysis of strain in heteroatom derivatives of indene, Journal of Physical Chemistry A, 115 (2011) 12271-12279. Page 42 of 79 Accepted Manuscript 42 M.L.F. Viveiros, V.L.S. Freitas, N. Vale, J.R.B. Gomes, P. Gomes, M.D.M.C.R. Da Silva, Synthesis and thermochemical study of quinoxaline-N-oxides: Enthalpies of dissociation of the N-O bond, Journal of Physical Organic Chemistry, 25 (2012) 420-426. J.A. Widegrenand, T.J. Bruno, Vapor pressure measurements on low-volatility terpenoid compounds by the concatenated gas saturation method, Environmental Science and Technology, 44 (2010) 388-393. W.E. Acree, V.V. Simirskii, A.A. Kozyro, A.P. Krasulin, G.Y. Kabo, M.L. Frenkel, Thermodynamic properties of organic compounds. 2. Combustion and sublimation enthalpies of 2,4,6-trimethylbenzonitrile N-oxide, Journal of Chemical & Engineering Data, 37 (1992) 131-133. W.E. Acree, J.R. Powell, S.A. Tucker, M.D.M.C. Ribeiro da Silva, M.A.R. Matos, J.M. Gonçalves, L.M.N.B.F. Santos, V.M.F. Morais, G. Pilcher, Thermochemical and Theoretical Study of Some Quinoxaline 1,4-Dioxides and of Pyrazine 1,4-Dioxide, The Journal of Organic Chemistry, 62 (1997) 3722-3726. L.M.P.F. Amaral, A.F.L.O.M. Santos, M.D.D.M.C. Ribeiro Da Silva, R. Notario, Thermochemistry of sarcosine and sarcosine anhydride: Theoretical and experimental studies, Journal of Chemical Thermodynamics, 58 (2013) 315-321. M.V. Roux, M. Temprado, J.S. Chickos, Y. Nagano, Critically evaluated thermochemical properties of polycyclic aromatic hydrocarbons, Journal of Physical and Chemical Reference Data, 37 (2008) 1855-1996. Page 43 of 79 Accepted Manuscript 43 C.G. De Kruif, Enthalpies of sublimation and vapour pressures of 11 polycyclic hydrocarbons, The Journal of Chemical Thermodynamics, 12 (1980) 243-248. G.A.F. Seber, Multivariate Observations, John Wiley & Sons, Inc., Hoboken, NJ, 1984. H. Spath, Cluster Dissection and Analysis: Theory, FORTRAN Programs, Examples. (Translated by J. Goldschmidt.), Halsted Press, New York, 1985. F. Gharagheizi, QSPR analysis for intrinsic viscosity of polymer solutions by means of GA-MLR and RBFNN, Computational Materials Science, 40 (2007) 159-167. F. Gharagheizi, An accurate model for prediction of autoignition temperature of pure compounds, Journal of Hazardous Materials, 189 (2011) 211-221. F. Gharagheizi, A. Eslamimanesh, P. Ilani-Kashkouli, A.H. Mohammadi, D. Richon, Determination of vapor pressure of chemical compounds: A group contribution model for an extremely large database, Industrial and Engineering Chemistry Research, 51 (2012) 7119-7125. F. Gharagheizi, A. Eslamimanesh, A.H. Mohammadi, D. Richon, Group contribution model for determination of molecular diffusivity of non-electrolyte organic compounds in air at ambient conditions, Chemical Engineering Science, 68 (2012) 290-304. F. Gharagheizi, S.A. Mirkhani, A.R. Tofangchi Mahyari, Prediction of standard enthalpy of combustion of pure compounds using a very accurate group-contribution-based method, Energy and Fuels, 25 (2011) 2651-2654. P. Gramatica, Principles of QSAR models validation: internal and external, QSAR & Combinatorial Science, 26 (2007) 694-701. Page 44 of 79 Accepted Manuscript 44 Figure Captions Figure 1- The gradual change of R2 and ARD% as function of number of chemical substructures. Figure 2- Williams plot – red circles shows the outliers of the model for which the experimental values may be erroneous. Figure 3- The sublimation enthalpies of hydrocarbons predicted by the model vs. experimental data. The left side (o) should be read for R2 and the right side (+) should be read for RD% Figure 4- The sublimation enthalpies of nitrogen compounds predicted by the model vs. experimental data. The left side (o) should be read for R2 and the right side (+) should be read for RD% Figure 5- The sublimation enthalpies of oxygen compounds predicted by the model vs. experimental data. The left side (o) should be read for R2 and the right side (+) should be read for RD% Figure 6- The sublimation enthalpies of sulfur compounds predicted by the model vs. experimental data. The left side (o) should be read for R2 and the right side (+) should be read for RD% Figure 7- The sublimation enthalpies of fluorine compounds predicted by the model vs. experimental data. The left side (o) should be read for R2 and the right side (+) should be read for RD% Page 45 of 79 Accepted Manuscript 45 Figure 8- The sublimation enthalpies of chlorine compounds predicted by the model vs. experimental data. The left side (o) should be read for R2 and the right side (+) should be read for RD% Figure 9- The sublimation enthalpies of bromine compounds predicted by the model vs. experimental data. The left side (o) should be read for R2 and the right side (+) should be read for RD% Figure 10- The sublimation enthalpies of iodine compounds predicted by the model vs. experimental data. The left side (o) should be read for R2 and the right side (+) should be read for RD% Page 46 of 79 Accepted Manuscript 46 Figure 1. Page 47 of 79 Accepted Manuscript 47 Figure 2. Page 48 of 79 Accepted Manuscript 48 Figure 3. Page 49 of 79 Accepted Manuscript 49 Figure 4. Page 50 of 79 Accepted Manuscript 50 Figure 5. Page 51 of 79 Accepted Manuscript 51 Figure 6. Page 52 of 79 Accepted Manuscript 52 Figure 7. Page 53 of 79 Accepted Manuscript 53 Figure 8. Page 54 of 79 Accepted Manuscript 54 Figure 9. Page 55 of 79 Accepted Manuscript 55 Figure 10. Page 56 of 79 Accepted Manuscript 56 Table 1- The contribution of each chemical substructure to the sublimation enthalpy of organic compounds (parameters of eq. (2)). ID ΔSublimationHm i Chemical substructure Comment Value ΔsublimationHm 0 17.22158 1 ΔsublimationHm 1 number of terminal primary C(sp3) Y = any terminal atom or heteroaromatic group (i.e. H, X, OH, NH2, etc.) 1.839585 2 ΔsublimationHm 2 number of total secondary C(sp3) Y = H or any heteroatom 4.324475 3 ΔsublimationHm 3 number of ring secondary C(sp3) Y = H or any heteroatom -3.17657 4 ΔsublimationHm 4 number of ring quaternary C(sp3) 5.180637 C C Y Y Y C C Y Y C Y Y Y Y Page 57 of 79 Accepted Manuscript 57 5 ΔsublimationHm 5 Sum of all the carbons belonging to any aromatic and heteroaromatic structure number of aromatic C(sp2) 2.327303 6 ΔsublimationHm 6 number of substituted benzene C(sp2) Y = carbon or any heteroatom -2.84795 7 ΔsublimationHm 7 number of non-aromatic conjugated C(sp2) 1.719792 8 ΔsublimationHm 8 number of terminal primary C(sp2) Y = any terminal atom or heteroaromatic group (i.e. H, X, OH, NH2, etc.) -3.80592 9 ΔsublimationHm 9 number of allenes groups 11.52741 10 ΔsublimationHm 10 number of non-terminal C(sp) Y = C or any non-terminal heteroatom 5.037913 Y O Y Y Y Page 58 of 79 Accepted Manuscript 58 11 ΔsublimationHm 11 number of esters (aromatic) Y = Al or Ar -2.10042 12 ΔsublimationHm 12 number of secondary amides (aliphatic) Y = Ar or Al (not H, not C = O) Al = H or aliphatic group linked through C 7.354295 13 ΔsublimationHm 13 number of tertiary amides (aliphatic) Y = Ar or Al (not H, not C = O) Al = H or aliphatic group linked through C 12.00637 14 ΔsublimationHm 14 number of (thio-) carbamates (aliphatic) Y = O or S Al = H or aliphatic group linked through any atom -10.6307 15 ΔsublimationHm 15 number of ketones (aliphatic) 3.564898 O O Y Ar O NH Y Al O N Y Al Y Y N Al Y Al Al O Al Al Page 59 of 79 Accepted Manuscript 59 16 ΔsublimationHm 16 number of carbonate (-thio) derivatives Y = O or S -10.5264 17 ΔsublimationHm 17 Ar-NH2 number of primary amines (aromatic) -16.3392 18 ΔsublimationHm 18 number of secondary amines (aliphatic) Al = aliphatic group linked through C (not C = O) -32.4857 19 ΔsublimationHm 19 number of secondary amines (aromatic) Y = Ar or Al (not C = O) -17.7938 20 ΔsublimationHm 20 number of tertiary amines (aliphatic) Al = aliphatic group linked through C (not C = O) -10.6959 21 ΔsublimationHm 21 Ar-CN number of nitriles (aromatic) 3.199378 22 ΔsublimationHm 22 >N< number of quaternary N -19.5828 23 ΔsublimationHm 23 Al-OH number of hydroxyl groups Al = aliphatic group linked through any atom 2.102287 24 ΔsublimationHm 24 number of secondary alcohols -10.1085 Y Y Y H N Al Al H N Y Ar Al N Al Al OH Page 60 of 79 Accepted Manuscript 60 25 ΔsublimationHm 25 Al-O-Al number of ethers (aliphatic) Al = aliphatic group linked through C (not C = O, not C # N) -4.02326 26 ΔsublimationHm 26 Ar-O-Y number of ethers (aromatic) Y = Ar or Al (not C = O, not C # N) 5.235889 27 ΔsublimationHm 27 number of anhydrides (thio-) Y = O or S 12.35995 28 ΔsublimationHm 28 S=C< number of thioketones -29.7795 29 ΔsublimationHm 29 -S-number of sulfides -11.2922 30 ΔsublimationHm 30 -S-S-number of disulfides 5.389454 31 ΔsublimationHm 31 number of sulfones 9.439112 32 ΔsublimationHm 32 number of phosphoranes / thiophosphoranes Y = O or S 17.70081 33 ΔsublimationHm 33 number of CHR2X -6.79778 Y O Y O S O O S O O S O S Y P Y1 Y1 Y1 Y1 P Y1 Y1 Y1 Y1 C C X H C Page 61 of 79 Accepted Manuscript 61 34 ΔsublimationHm 34 number of CRX3 -9.81816 35 ΔsublimationHm 35 Ar-X number of X on aromatic ring 3.208322 36 ΔsublimationHm 36 number of X on ring C(sp2) 22.24842 37 ΔsublimationHm 37 number of Furanes -1.45294 38 ΔsublimationHm 38 number of Thiophenes -16.1342 39 ΔsublimationHm 39 Sum of the hydrogens linked to all of the Os and Ns in the molecule number of donor atoms for H-bonds (N and O) 59.12675 X C X X C X O S Page 62 of 79 Accepted Manuscript 62 40 ΔsublimationHm 40 number of intramolecular H-bonds Y1 = B, N, O, Al, P, S Y2 = N, O, F The geometric distance between H and Y2 must be in the range 1 - 2,7 A. -5.99035 41 ΔsublimationHm 41 CH2X2 5.988704 42 ΔsublimationHm 42 =CHR -1.78681 43 ΔsublimationHm 43 R--CH--R -1.94963 44 ΔsublimationHm 44 R--CR--R 3.237334 45 ΔsublimationHm 45 X--CH--X -6.81541 46 ΔsublimationHm 46 X--CX--X -8.72027 47 ΔsublimationHm 47 R-C(=X)-X or R-C#X or X=C=X 0.085745 48 ΔsublimationHm 48 X--CX..X -13.0023 49 ΔsublimationHm 49 Ha attached to C0(sp3) no X attached to next C -0.11698 50 ΔsublimationHm 50 Ha attached to C1(sp3) or C0(sp2) 1.524469 51 ΔsublimationHm 51 H attached to heteroatom -41.7694 52 ΔsublimationHm 52 Ha attached to C0(sp3) with 2X attached to next C -1.16842 53 ΔsublimationHm 53 =O 4.370113 54 ΔsublimationHm 54 Al-O-Ar or Ar-O-Ar or R..O..R or R-O-C=X -4.08989 55 ΔsublimationHm 55 O-- b 2.422218 56 ΔsublimationHm 56 O- (negatively charged) 23.85644 57 ΔsublimationHm 57 R-O-O-R 11.10866 58 ΔsublimationHm 58 Al2-NH 17.85208 59 ΔsublimationHm 59 Ar-NH-Al 7.191607 60 ΔsublimationHm 60 RCO-N< or >N-X=X -7.10496 61 ΔsublimationHm 61 R#N or R=N-3.820022 62 ΔsublimationHm 62 Ar-NO2 or R--N(--R)--Oc or RO-NO 10.98812 63 ΔsublimationHm 63 Fa attached to C1(sp2) -0.59666 H Y1 Y2 H Y1 Y2 Page 63 of 79 Accepted Manuscript 63 64 ΔsublimationHm 64 Bra attached to C2(sp2)-C4(sp2) or C1(sp) or C4(sp3) or X 3.598674 65 ΔsublimationHm 65 R-SH 62.64204 66 ΔsublimationHm 66 presence (0) or absence (1) of C-O -5.34759 67 ΔsublimationHm 67 presence (0) or absence (1) of N-P -65.8101 68 ΔsublimationHm 68 presence (0) or absence (1) of C-A-C A means any atom 1.167072 69 ΔsublimationHm 69 presence (0) or absence (1) of C-A-O A means any atom 13.44807 70 ΔsublimationHm 70 presence (0) or absence (1) of C-A-Br A means any atom 3.07181 71 ΔsublimationHm 71 presence (0) or absence (1) of N-A-N A means any atom 11.17699 72 ΔsublimationHm 72 presence (0) or absence (1) of N-A-N A means any atom 2.047634 73 ΔsublimationHm 73 presence (0) or absence (1) of C-A1-A2-C A1 and A2 means any atom -6.3383 74 ΔsublimationHm 74 presence (0) or absence (1) of C-A1-A2-N A1 and A2 means any atom 5.717817 75 ΔsublimationHm 75 presence (0) or absence (1) of C-A1-A2-O A1 and A2 means any atom -2.32134 76 ΔsublimationHm 76 presence (0) or absence (1) of C-A1-A2-S A1 and A2 means any atom 6.836506 77 ΔsublimationHm 77 presence (0) or absence (1) of C-A1-A2-A3-N A1 , A2 and A3, means any atom -1.9211 78 ΔsublimationHm 78 presence (0) or absence (1) of C-A1-A2-A3-O A1 , A2 and A3, means any atom 1.067491 79 ΔsublimationHm 79 presence (0) or absence (1) of C-A1-A2-A3-A4-C A1 , A2 , A3, and A4 means any atom 0.360241 80 ΔsublimationHm 80 presence (0) or absence (1) of C-A1-A2-A3-A4-O A1 , A2 , A3, and A4 means any atom -1.53055 81 ΔsublimationHm 81 presence (0) or absence (1) of C-A1-A2-A3-A4-S A1 , A2 , A3, and A4 means any atom -6.38755 82 ΔsublimationHm 82 presence (0) or absence (1) of N-A1-A2-A3-A4-O A1 , A2 , A3, and A4 means any atom 0.519688 83 ΔsublimationHm 83 presence (0) or absence (1) of N-A1-A2-A3-A4-F A1 , A2 , A3, and A4 means any atom 7.439548 84 ΔsublimationHm 84 presence (0) or absence (1) of O-A1-A2-A3-A4-O A1 , A2 , A3, and A4 means any atom 5.36676 85 ΔsublimationHm 85 presence (0) or absence (1) of Cl-A1-A2-A3-A4-Br A1 , A2 , A3, and A4 means any atom 8.366331 86 ΔsublimationHm 86 presence (0) or absence (1) of Cl-A1-A2-A3-A4-I A1 , A2 , A3, and A4 means any atom 7.186903 87 ΔsublimationHm 87 presence (0) or absence (1) of Br-A1-A2-A3-A4-I A1 , A2 , A3, and A4 means any atom 8.570828 88 ΔsublimationHm 88 presence (0) or absence (1) of C-A1-A2-A3-A4-A5-S A1 , A2 , A3, A4 and A5 means any atom -5.59339 89 ΔsublimationHm 89 presence (0) or absence (1) of C-A1-A2-A3-A4-A5-Br A1 , A2 , A3, A4 and A5 means any atom -0.37507 90 ΔsublimationHm 90 presence (0) or absence (1) of N-A1-A2-A3-A4-A5-Cl A1 , A2 , A3, A4 and A5 means any atom 27.02776 91 ΔsublimationHm 91 presence (0) or absence (1) of O-A1-A2-A3-A4-A5-O A1 , A2 , A3, A4 and A5 means any atom 5.795304 Page 64 of 79 Accepted Manuscript 64 92 ΔsublimationHm 92 presence (0) or absence (1) of O-A1-A2-A3-A4-A5-S A1 , A2 , A3, A4 and A5 means any atom -25.9605 93 ΔsublimationHm 93 presence (0) or absence (1) of S-A1-A2-A3-A4-A5-F A1 , A2 , A3, A4 and A5 means any atom -16.4801 94 ΔsublimationHm 94 presence (0) or absence (1) of C-A1-A2-A3-A4-A5-A6-C A1 , A2 , A3, A4 ,A5 and A6 means any atom 1.264999 95 ΔsublimationHm 95 presence (0) or absence (1) of C-A1-A2-A3-A4-A5-A6-O A1 , A2 , A3, A4 ,A5 and A6 means any atom -1.38958 96 ΔsublimationHm 96 presence (0) or absence (1) of C-A1-A2-A3-A4-A5-A6-F A1 , A2 , A3, A4 ,A5 and A6 means any atom 4.043397 97 ΔsublimationHm 97 presence (0) or absence (1) of C-A1-A2-A3-A4-A5-A6-I A1 , A2 , A3, A4 ,A5 and A6 means any atom 6.879248 98 ΔsublimationHm 98 presence (0) or absence (1) of N-A1-A2-A3-A4-A5-A6-N A1 , A2 , A3, A4 ,A5 and A6 means any atom 6.246321 99 ΔsublimationHm 99 presence (0) or absence (1) of N-A1-A2-A3-A4-A5-A6-S A1 , A2 , A3, A4 ,A5 and A6 means any atom 82.38669 100 ΔsublimationHm 100 presence (0) or absence (1) of N-A1-A2-A3-A4-A5-A6-F A1 , A2 , A3, A4 ,A5 and A6 means any atom -36.297 101 ΔsublimationHm 101 presence (0) or absence (1) of O-A1-A2-A3-A4-A5-A6-O A1 , A2 , A3, A4 ,A5 and A6 means any atom 6.121689 102 ΔsublimationHm 102 presence (0) or absence (1) of C-A1-A2-A3-A4-A5-A6-A7-C A1 , A2 , A3, A4 ,A5 ,A6 and A7 means any atom 0.778842 103 ΔsublimationHm 103 presence (0) or absence (1) of C-A1-A2-A3-A4-A5-A6-A7-O A1 , A2 , A3, A4 ,A5 ,A6 and A7 means any atom -1.67133 104 ΔsublimationHm 104 presence (0) or absence (1) of N-A1-A2-A3-A4-A5-A6-A7-N A1 , A2 , A3, A4 ,A5 ,A6 and A7 means any atom 36.78154 105 ΔsublimationHm 105 presence (0) or absence (1) of O-A1-A2-A3-A4-A5-A6-A7-O A1 , A2 , A3, A4 ,A5 ,A6 and A7 means any atom 7.876963 106 ΔsublimationHm 106 presence (0) or absence (1) of C-A1-A2-A3-A4-A5-A6-A7-A8-S A1 , A2 , A3, A4 ,A5 ,A6 ,A7 and A8 means any atom 30.18888 107 ΔsublimationHm 107 presence (0) or absence (1) of Cl-A1-A2-A3-A4-A5-A6-A7-A8-Cl A1 , A2 , A3, A4 ,A5 ,A6 ,A7 and A8 means any atom 12.66579 108 ΔsublimationHm 108 presence (0) or absence (1) of C-A1-A2-A3-A4-A5-A6-A7-A8-A9-C A1 , A2 , A3, A4 ,A5 ,A6 ,A7, A8 and A9 means any atom 2.124816 109 ΔsublimationHm 109 presence (0) or absence (1) of C-A1-A2-A3-A4-A5-A6-A7-A8-A9-S A1 , A2 , A3, A4 ,A5 ,A6 ,A7, A8 and A9 means any atom 11.74033 110 ΔsublimationHm 110 presence (0) or absence (1) of O-A1-A2-A3-A4-A5-A6-A7-A8-A9-Cl A1 , A2 , A3, A4 ,A5 ,A6 ,A7, A8 and A9 means any atom 15.2838 111 ΔsublimationHm 111 number of C-C 5.538397 112 ΔsublimationHm 112 number of C-N 7.531674 113 ΔsublimationHm 113 number of C-O 7.904142 114 ΔsublimationHm 114 number of C-S 14.03693 115 ΔsublimationHm 115 number of C-P 7.820161 116 ΔsublimationHm 116 number of C-Br 5.199863 Page 65 of 79 Accepted Manuscript 65 117 ΔsublimationHm 117 number of C-I 12.0811 118 ΔsublimationHm 118 number of N-N 8.231438 119 ΔsublimationHm 119 number of C-A-C A means any atom -2.11715 120 ΔsublimationHm 120 number of C-A-N A means any atom -0.41774 121 ΔsublimationHm 121 number of C-A-O A means any atom -1.40174 122 ΔsublimationHm 122 number of N-A-N A means any atom -1.92971 123 ΔsublimationHm 123 number of C-A1-A2-C A1 and A2 means any atom 1.344881 124 ΔsublimationHm 124 number of C-A1-A2-Cl A1 and A2 means any atom 0.696158 125 ΔsublimationHm 125 number of Cl-A1-A2-Cl A1 and A2 means any atom 3.321792 126 ΔsublimationHm 126 number of C-A1-A2-A3-C A1 , A2 and A3, means any atom -1.13923 127 ΔsublimationHm 127 number of C-A1-A2-A3-N A1 , A2 and A3, means any atom -1.49467 128 ΔsublimationHm 128 number of C-A1-A2-A3-O A1 , A2 and A3, means any atom -1.02864 129 ΔsublimationHm 129 number of C-A1-A2-A3-Cl A1 , A2 and A3, means any atom 1.623283 130 ΔsublimationHm 130 number of N-A1-A2-A3-S A1 , A2 and A3, means any atom -6.56591 131 ΔsublimationHm 131 number of N-A1-A2-A3-Br A1 , A2 and A3, means any atom 3.740106 132 ΔsublimationHm 132 number of O-A1-A2-A3-S A1 , A2 and A3, means any atom -3.61868 133 ΔsublimationHm 133 number of C-A1-A2-A3-A4-C A1 , A2 , A3, and A4 means any atom -0.26817 134 ΔsublimationHm 134 number of C-A1-A2-A3-A4-S A1 , A2 , A3, and A4 means any atom 3.892686 135 ΔsublimationHm 135 number of N-A1-A2-A3-A4-N A1 , A2 , A3, and A4 means any atom 5.043057 136 ΔsublimationHm 136 number of N-A1-A2-A3-A4-F A1 , A2 , A3, and A4 means any atom -3.94056 137 ΔsublimationHm 137 number of O-A1-A2-A3-A4-O A1 , A2 , A3, and A4 means any atom -0.64996 138 ΔsublimationHm 138 number of N-A1-A2-A3-A4-A5-S A1 , A2 , A3, A4 and A5 means any atom -7.00242 139 ΔsublimationHm 139 number of O-A1-A2-A3-A4-A5-F A1 , A2 , A3, A4 and A5 means any atom -3.71243 140 ΔsublimationHm 140 number of C-A1-A2-A3-A4-A5-A6-Cl A1 , A2 , A3, A4 ,A5 and A6 means any atom -2.10959 141 ΔsublimationHm 141 number of O-A1-A2-A3-A4-A5-A6-Cl A1 , A2 , A3, A4 ,A5 and A6 means any atom -4.48158 142 ΔsublimationHm 142 number of C-A1-A2-A3-A4-A5-A6-A7-Cl A1 , A2 , A3, A4 ,A5 ,A6 and A7 means any atom 5.212202 143 ΔsublimationHm 143 number of N-A1-A2-A3-A4-A5-A6-A7-N A1 , A2 , A3, A4 ,A5 ,A6 and A7 means any atom -11.5283 144 ΔsublimationHm 144 number of N-A1-A2-A3-A4-A5-A6-A7-O A1 , A2 , A3, A4 ,A5 ,A6 and A7 means any atom -10.9808 Page 66 of 79 Accepted Manuscript 66 145 ΔsublimationHm 145 number of C-A1-A2-A3-A4-A5-A6-A7-A8-C A1 , A2 , A3, A4 ,A5 ,A6 ,A7 and A8 means any atom -0.25142 146 ΔsublimationHm 146 number of O-A1-A2-A3-A4-A5-A6-A7-A8-O A1 , A2 , A3, A4 ,A5 ,A6 ,A7 and A8 means any atom 3.57817 147 ΔsublimationHm 147 number of C-A1-A2-A3-A4-A5-A6-A7-A8-A9-N A1 , A2 , A3, A4 ,A5 ,A6 ,A7, A8 and A9 means any atom -19.235 R represents any group linked through carbon; X represents any electronegative atom (O, N, S, P, Se, halogens); Al and Ar represent aliphatic and aromatic groups, respectively; = represents a double bond; # represents a triple bond; -- represents an aromatic bond as in benzene or delocalized bonds such as the N-O bond in a nitro group .. represents aromatic single bonds as the C-N bond in pyrrole a The superscript represents the formal oxidation number. The formal oxidation number of a carbon atom equals the sum of the conventional bond orders with electronegative atoms; the C--N bond order in pyridine may be considered as 2 while we have one such bond and 1.5 when we have two such bonds; the C..X bond order in pyrrole or furan may be considered as 1. Page 67 of 79 Accepted Manuscript 67 b As in nitro, N-oxides c Pyridine N-oxide type structure. Page 68 of 79 Accepted Manuscript 68 Table 2- Highly deviant hydrocarbons. No. Structure Δsublimation Hm exp/ kJ.mole-1 Δsublimation Hm pred/ kJ.mole-1 %ARD 1 1,1,2,2-tetra-tert-butylethane 74.3 40.8 45 2 bicyclo[3.2.2]non-6-ene 48 61.2 27.5 3 1,1,2-triphenylethane 92.2 117.2 27.1 4 pentacene 184 137.3 25.4 5 119.7 144.5 20.8 Page 69 of 79 Accepted Manuscript 69 9,9’-dimethyl-9,9’-bifluorenyl 6 bicyclo[3.3.2]decane 58.2 69.9 20 7 naphthacene 143.7 116.3 19 8 dibenz[a,c]anthracene 159 129.9 18.3 9 Pentacyclo[5.4.0.02,6.03,10.05,9]undecane 54.9 64.2 16.9 Page 70 of 79 Accepted Manuscript 70 10 1,1’-biadamantane 113.8 94.9 16.6 11 dibenz[a,h]anthracene 162 135.4 16.4 12 cyclododecane 76.2 63.7 16.4 13 tetraphenylmethane 145.3 121.8 16.2 14 acenaphthylene 71.5 82.8 15.8 Page 71 of 79 Accepted Manuscript 71 15 1,4,5,8-tetramethylnaphthalene 99.8 85.6 14.2 16 triphenylene 126.5 109.4 13.5 17 bicyclo[4.2.1]non-3-ene 49.7 56.2 13.2 18 pentacyclo[4.4.0.02,503,804,7]dec-9-ene (basketene) 55.3 62.5 13.1 19 chrysene 131 114.4 12.6 Page 72 of 79 Accepted Manuscript 72 20 trans-heptacyclene 149 131.8 11.6 21 bicyclo[3.3.1]non-2-ene 48.2 53.8 11.5 22 9,10-dimethylanthracene 113 100.5 11.1 23 bicyclo[3.3.1]nonane 50.6 56.1 10.9 24 1,4-di-tert-butylbenzene 82.8 74.3 10.2 Page 73 of 79 Accepted Manuscript 73 Table 3- Highly deviant sulfur compounds. No. Structure Δsublimation Hm exp/ kJ.mole-1 Δsublimation Hm pred/ kJ.mole-1 %ARD 1 tetrahydro-2H-1,3-oxazine-2-thione 108.9 80 26.6 2 cis-5a,6,11a,12-tetrahdro[1,4]benzothiazino[3,2-b]-[1,4]-benzothiazine 123.3 149.9 21.6 3 N-theonylthiocarbamic-O-propyl ester 136.5 110.4 19.1 4 4-amino-N-(5-chloro-2-methylphenyl)benzenesulfonamide 130 151.7 16.7 Page 74 of 79 Accepted Manuscript 74 5 N-(diethylaminothiocarbonyl)benzamideine 126 145.8 15.7 6 3,6-diphenyl-1,2-dithiin 183.1 155.1 15.3 7 1,3-dithiole-2-thione 75.4 86.2 14.3 8 4-cyanothiazole 73.9 84.2 13.9 9 benzo[b]thiophene 65.7 74 12.6 10 61.7 69.4 12.5 Page 75 of 79 Accepted Manuscript 75 1,3-dithiane 11 2-thiouracil 138.5 121.8 12.1 12 5-methyl-2-thiouracil 137.3 121.2 11.7 13 thioacetamide 83.05 73.4 11.6 14 monthiodibenzoylmethane 125.5 139.9 11.5 15 6-methyl-2-thiouracil 140.7 125.9 10.5 Page 76 of 79 Accepted Manuscript 76 16 phenoxathiin 95.6 105.5 10.3 17 L-(d)-methionine 164 147.4 10.2 18 phenothiazine 114.5 103.1 10 Page 77 of 79 Accepted Manuscript 77 Table 4- The deviation of the model results from experimental data for some important chemical families of the compounds. ID Chemical groups/families AARD% N 1 aliphatic esters 6.3 31 2 aromatic esters 6.4 30 3 aliphatic carboxylic acids 6.4 86 4 aromatic carboxylic acids 6.4 149 5 primary aliphatic amides 6.3 23 6 primary aromatic amides 6.5 14 7 secondary aliphatic amides 6.3 18 8 secondary aromatic amides 5.7 4 9 tertiary aliphatic amides 6.5 4 10 tertiary aromatic amides 5.5 2 11 aliphatic aldehydes 5.5 1 12 aromatic aldehydes 5 6 13 aliphatic ketones 6.3 40 14 aromatic ketones 6.3 68 15 primary aliphatic amines 6.2 20 16 primary aromatic amines 6.3 86 17 secondary aliphatic amines 6.8 17 18 secondary aromatic amines 6.5 21 19 tertiary aliphatic amines 6.5 5 20 tertiary aromatic amines 6.4 8 21 aliphatic alcohols 6.4 298 22 aromatic alcohols 6.4 156 Page 78 of 79 Accepted Manuscript 78 Table 5- The comparison between the presented model and the one proposed by Ouvrard and Mitchell Chemical family Statistical parameter Ouvrard and Mitchell The presented model Aliphatic hydrocarbons n 33 38 R2 0.968 0.932 RMSE 7.42 8.61 Aromatic hydrocarbons n 50 79 R2 0.965 0.836 RMSE 7 10.85 non-hydrogen bonding compounds n 156 164 R2 0.896 0.885 RMSE 9.98 10.56 Various compounds n 226 1269 R2 0.925 0.826 RMSE 9.58 10.79 Page 79 of 79 Accepted Manuscript 79 Research Highlights  A new group contribution model is presented for the estimation of sublimation enthalpy.  A compendium of experimental data for 1271 compounds is used to develop and validate the model.  The model shows low deviation from experimental data.
14232	https://www.heartrhythmcasereports.com/article/S2214-0271(16)30018-5/fulltext	Concomitant pulsus and pseudoelectrical alternans in severe systolic dysfunction - HeartRhythm Case Reports Skip to Main ContentSkip to Main Menu Login to your account Email/Username Your email address is a required field. E.g., j.smith@mail.com Password Show Your password is a required field. Forgot password? [x] Remember me Don’t have an account? Create a Free Account If you don't remember your password, you can reset it by entering your email address and clicking the Reset Password button. 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Ok Short communicationVolume 2, Issue 3p277-279 May 2016 Open access Download Full Issue Download started Ok Concomitant pulsus and pseudoelectrical alternans in severe systolic dysfunction Dun-Hui Yang, MD Dun-Hui Yang, MD Affiliations Department of Radiology, Tainan Municipal Hospital, Tainan, Taiwan Search for articles by this author ∙ Ching-Yu Julius Chen, MD Ching-Yu Julius Chen, MD Correspondence Address reprints and correspondence: Dr Ching-Yu Chen, Department of Internal Medicine, National Taiwan University Hospital, No. 7, Chun-Shan S. Rd, Taipei, Taiwan Juliuschen1984@gmail.com Affiliations Department of Internal Medicine, National Taiwan University Hospital, Taipei, Taiwan Search for articles by this author †Juliuschen1984@gmail.com Affiliations & Notes Article Info Department of Radiology, Tainan Municipal Hospital, Tainan, Taiwan †Department of Internal Medicine, National Taiwan University Hospital, Taipei, Taiwan Publication History: Published online March 11, 2016 DOI: 10.1016/j.hrcr.2016.03.002 External LinkAlso available on ScienceDirect External Link Copyright: © 2016 Heart Rhythm Society. Published by Elsevier Inc. User License: Creative Commons Attribution – NonCommercial – NoDerivs (CC BY-NC-ND 4.0) \| Elsevier's open access license policy Download PDF Download PDF Outline Outline Keywords Introduction Case report Discussion References Article metrics Related Articles Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley bluesky Add to my reading list More More Download PDF Download PDF Cite Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley Bluesky Add to my reading list Set Alert Get Rights Reprints Download Full Issue Download started Ok Previous articleNext article Show Outline Hide Outline Keywords Introduction Case report Discussion References Article metrics Related Articles Keywords Pulsus alternans Pseudoelectrical alternans Systolic dysfunction Calcium cycling Electromechanical coupling Introduction • The pathophysiology of pulsus alternans in systolic dysfunction is attributed to 2 major mechanisms: Frank-Starling relationship and impaired calcium cycling. • The difference between peak pressure of the first Korotkoff sound and the pressure when the Korotkoff sounds double indicates the pressure gap of the pulsus alternans. • Pseudoelectrical alternans is the phenomenon describing beat-to-beat variation in axis or amplitude owing to alternation in conduction rather than excessive cardiac motion, and may occur in severe systolic dysfunction. KEY TEACHING POINTS Open table in a new tab Pulsus alternans and electrical alternans are important and useful clinical signs, indicating several conditions of cardiac dysfunction. Pulsus alternans is associated with severe systolic dysfunction, which is involved in calcium overload and impaired calcium cycling. To date, the proposed mechanisms of pulsus alternans include the alternative change in left ventricular end-diastolic pressure and stroke volume mediated by Frank-Starling law, as well as alternation of calcium transient coupling to myocardial contractility. On the other hand, electrical alternans is usually caused by massive pericardial effusion owing to cardiac motion, and occasionally found in sustained tachycardia due to alternans of conduction or refractoriness. Regarding to heart failure, T wave alternans has been well studied, whereas QRS alternans was less discussed. Here we demonstrated a case with both of these phenomena, which may share a common underlying mechanism, such as impaired calcium release and reuptake resulting in rhythmic variation of calcium transient. Case report A 41-year-old man complained of progressive dyspnea on exertion for 6 months. At presentation, tachycardia with S3 gallop was noted, along with bilateral basal crackles in chest auscultation, implying the condition of heart failure. During blood pressure measurement, Korotkoff sounds were heard with the rate of 60 beats per minute when the cuff pressure was fixed at 140 mm Hg, while the rate doubled at 130 mm Hg, suggesting pulsus alternans with pressure gap of 10 mm Hg. Echocardiography revealed left ventricular ejection fraction around 10%, as well as the rhythmic variation of the peak flow velocity at the ascending aorta (Figure 1), indicating the alternative change of stroke volume. Electrocardiography showed a beat-to-beat shift in the QRS axis and amplitude (Figure 2). Typically, electrical alternans was seen in patients with massive pericardial effusion owing to cardiac motion, but there was no effusion in this case. Therefore, this case is better described as “pseudoelectrical alternans,” which is the phenomenon where beat-to-beat variation in axis or amplitude is due to alternation in conduction rather than mechanical shifting. However, there was no evidence of PR interval and QRS duration between beats, which excluded the possibilities of intermittent ventricular pre-excitation, aberrancy, and ventricular bigeminy with fusion beats, all of which can be causes of pseudoalternans. Figure viewer Figure 1 The rhythmic variation of the peak flow velocity at ascending aorta demonstrated by pulse-wave doppler. Figure viewer Figure 2 Electrical alternans. The QRS axis and amplitude varied alternatively, along with a premature ventricular complex. Discussion The pathophysiology of pulsus alternans in systolic dysfunction is attributed to 2 major mechanisms: Frank-Starling relationship and impaired calcium cycling. According to the former theory, congestive heart failure with elevated end-diastolic pressure generates higher myocardial contractility, which empties more effectively and lowers the end-diastolic pressure in the same cardiac cycle; the contractility of the next cycle then becomes weaker, thus the end-diastolic pressure re-elevates again and the cyclic beat-to-beat alternation is maintained. The theory of impaired calcium cycling is composed by the positive relationship between the end-diastolic calcium content in sarcoplasmic reticulum and its release, along with spontaneous calcium leak from sarcoplasmic reticulum and reduced calcium reuptake in heart failure, all of which contribute to the alternative change in cytosolic calcium concentration and its coupling to membrane voltage as well as contractility.1 1. Edwards, J.N. ∙ Blatter, L.A. Cardiac alternans and intracellular calcium cycling Clin Exp Pharmacol Physiol. 2014; 41:524-532 Crossref Scopus (75) PubMed Google Scholar Regarding electrical alternans, it includes the alternans in QRS-T amplitude and QT interval, and the microvolt T-wave alternans provided an accurate means to predict ventricular tachyarrhythmias and sudden cardiac death in patients with systolic heart failure.2 2. Hohnloser, S.H. ∙ Ikeda, T. ∙ Cohen, R.J. Evidence regarding clinical use of microvolt T-wave alternans Heart Rhythm. 2009; 6:S36-44 Full Text Full Text (PDF) Scopus (71) PubMed Google Scholar QRS alternans is mostly caused by massive pericardial effusion owing to cardiac motion, and sometimes develops in supraventricular tachycardia because of the intermittent refractoriness of the conduction system.3 3. Surawicz, B. ∙ Fisch, C. Cardiac alternans: diverse mechanisms and clinical manifestations J Am Coll Cardiol. 1992; 20:483-499 Abstract Full Text (PDF) Scopus (184) PubMed Google Scholar The latter, involving alternans in conduction, is called “pseudoelectrical alternans,” which was first described by Klein et al4 4. Klein, H.O. ∙ Di Segni, E. ∙ Kaplinsky, E. Procainamide-induced left anterior hemiblock of the 2:1 type (pseudoelectrical alternans) Chest. 1978; 74:230-233 Full Text Full Text (PDF) Scopus (9) PubMed Google Scholar regarding a case with procainamide-induced left anterior hemiblock of the 2:1 type. The beat-to-beat alternation of cytosolic calcium regulation is an important reason for both mechanical and pseudoeletrical alternans,5 5. Kanaporis, G. ∙ Blatter, L.A. The mechanisms of calcium cycling and action potential dynamics in cardiac alternans Circ Res. 2015; 116:846-856 Crossref Scopus (69) PubMed Google Scholar and the concomitant existence in this case implied the major contribution of impaired calcium cycling in the underlying mechanisms of both phenomena. References 1. Edwards, J.N. ∙ Blatter, L.A. Cardiac alternans and intracellular calcium cycling Clin Exp Pharmacol Physiol. 2014; 41:524-532 Crossref Scopus (75) PubMed Google Scholar 2. Hohnloser, S.H. ∙ Ikeda, T. ∙ Cohen, R.J. Evidence regarding clinical use of microvolt T-wave alternans Heart Rhythm. 2009; 6:S36-44 Full Text Full Text (PDF) Scopus (71) PubMed Google Scholar 3. Surawicz, B. ∙ Fisch, C. Cardiac alternans: diverse mechanisms and clinical manifestations J Am Coll Cardiol. 1992; 20:483-499 Abstract Full Text (PDF) Scopus (184) PubMed Google Scholar 4. Klein, H.O. ∙ Di Segni, E. ∙ Kaplinsky, E. Procainamide-induced left anterior hemiblock of the 2:1 type (pseudoelectrical alternans) Chest. 1978; 74:230-233 Full Text Full Text (PDF) Scopus (9) PubMed Google Scholar 5. Kanaporis, G. ∙ Blatter, L.A. The mechanisms of calcium cycling and action potential dynamics in cardiac alternans Circ Res. 2015; 116:846-856 Crossref Scopus (69) PubMed Google Scholar Figures (2)Figure Viewer Article metrics Related Articles Open in viewer Concomitant pulsus and pseudoelectrical alternans in severe systolic dysfunction Hide CaptionDownloadSee figure in Article Toggle Thumbstrip Figure 1 Figure 2 Download .PPT Go to Go to Show all references Expand All Collapse Expand Table Authors Info & Affiliations Home Access for Developing Countries Articles and Issues Articles in Press Current Issue List of Issues HRCR Rare Diseases Article Collection A Case for Education Quiz Current Quiz Quiz Archive For Authors About Open Access Author Information Permissions Researcher Academy Submit a Manuscript Journal Info About Open Access Info for Advertisers Contact Information Editorial Board Society Info Heart Rhythm Society Heart Rhythm Society Journals Heart Rhythm Heart Rhythm O 2 Cardiovascular Digital Health Journal Follow Us Facebook Twitter RSS Feed The content on this site is intended for healthcare professionals. 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14233	https://www.youtube.com/watch?v=hhvCsY1Qqxo	Process Flow Chart with Microsoft Excel Hands-On-Data 1980 subscribers 1577 likes Description 266304 views Posted: 11 May 2020 In this video, I use Microsoft Excel to show how you can create a flow chart. This can also be used for organization charts and the likes. flowchart #projectmanagement #exceltutorial 📈 Get Personalized Training and Financial Modelling Expertise! 📊 Are you looking to enhance your financial planning and Excel skills? Need a professional to handle your financial modeling tasks? I'm here to help! Contact info below 👨‍🏫 One-on-One Training: I offer personalized, one-on-one hourly training sessions to help you master Microsoft Excel and financial modeling. Contact info below 📊 Financial Modelling Services: Need assistance with financial modeling? With over 15 years of experience in financial planning and analysis, I can provide you with top-notch financial models to support your business decisions. 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I look forward to working with you! 111 comments Transcript: Introduction hello everyone and welcome to this tutorial in this tutorial I want to show you how to draw process chats in Microsoft Excel so in the internet and on this side net mind dotnet for component effective process map I saw this and I want to replicate it in Microsoft Excel so for those of us who work Michaels of the excel and would want to present visuals of these nature or process charts of this nature we can also just use Excel to do something like this and then use it for our presentation let me show you how to do that so here we have Excel and the first Remove Gridlines thing I want to do is I want to go to my View tab and take out the gridlines so I have justice so the next thing I want to do is to look out for things that I would use here and bring them in in other words I want to look out for the objects that we use here so here I know I have a database stat I never have rectangle I have SEC holes I have a decision box and all of that so that's Insert Shapes exactly what I would be doing here and to do that I will go to my insert tab I'll go to illustrations and then shapes so here is where I would have the shapes I would work with okay so these I recently used shapes and you can see that much of what we'll be using a here in the recently used shape so let me pick the rectangle and the first thing I want to do is what will be in the header you know what will be in the header so here have technology acquisition process so that's what I will do take technology acquisition process so haven't done that I can always select it and in my home tab I can change color to whatever I can also in my format tab I can also take out the shape outline so I always like taking it out because like you can see looks smooth and fine here you can also select it and then either increase your text here let's put it at the middle so that's it for us so we can use it in this way so the next thing I would like Swim Lanes to do is to come up with this the swim lanes so I'm going to replicate this replicate it here okay I can use my arrow buttons to just adjust where you should be and I think it's fine we do not need this text now so I'm taking it out and what I'm going to do is just extend it this way okay and this time around I need an outline so I'm going in here pick an outline take a line and give it a color and once I do that I take out the few colors here so so I have something I can use yet so I'm going to replicate this one and just place it side-by-side with this you know okay like this and then we do a third one that third one is just going to stay below here okay so looks like we're fine yeah so your arrow buttons can help you position it the way you want so it looks like we're fine here so one two and three the first one here is empty so I'm not going to be doing anything with that so we have executive team business unit and project teams okay so to do that what I will be doing is I still can use a rectangle in fact it's all has to do with how creative you could be with these tools so what I'm going to do is just I'm just going to create something with this and while it is you can see this yeah I'm just going funny so there it goes in and just size it appropriately okay so that is what it is and and then having it this way I just replicate and using control d selected and then control D like this puts it right there for us so we're fine with this okay so the next thing we do is we go and look for text box so I have a text box here I'm going to bring in that exports put in executive team okay so once again you find out that you have this here so it allows us itani in whichever way we want this time we just want it this way okay so I'm gonna move it here the killer team let's make it a little bigger and make it white and the font color white and we can make it bold if we want to so that's for that so again control D I want to replicate it and we have it here and also there we have the other one so here we have the project team so I'm just in the project team and here we have the business unit business unit so you can see that it's coming into shape like this so what I would just do is let me just minimize this so that we can see it this way so you can always bring it by by double clicking and it comes up this way okay so okay so we have it this Add Shapes way so we now see what things we need to bring in into it so I need so I go to my insert and inside we go to shapes and then we'll be bringing in tact okay so here we should have it okay that's it here so I'm just going to use it here so it's this touch so we have a start here and I like to just make it middle bring it to this point increase it a little bit just like that okay so that's how's that I want to replicate it because I know that I have somewhere where it's going to be an end and it's the same thing so we just put it here and then that does for that and then go back and Add Rectangles we have rectangles one two three so I'm going back to my insert I'm going to get rectangles and now we can pick those rectangles from anywhere but let me just have this there is one in each of the lanes from what we see there so I'm going to just replicate this I think there's one here and there is one somewhere here so we'll position them where they should be okay actually I have two here somewhere around here too and then we have the third one here okay one two three okay then the next thing we need a decision box so I have a decision box here okay so we have this one here and I'm going to replicate it because there is another one here but this time is smaller one okay so we'll have down one just here and okay so we have those two dishes on boxes there and then we have this other one here it's called off page connector so we're going to just pick it and it's somewhere here so it's somewhere here and we're going to take out its outline we don't want the outline there same for each of this okay so there they are so the next thing we will be doing it's Add Circles we will have this still go back to our shapes and bring in circles what would the circles do they would help us for the numbers and so I'm going to use a field color they stick out the outline and so it has a value of 1 now you can see where the one is it's not ideal so you just come here and then use this for it so to get into the way you want them okay so we need to just position them they should be we have one here and we have one so replicating it is easy white is selected just do your ctrl D and then you'll be able to have it copy so here I'm just going to change this to okay and the next one it's replicate this and we just move it here change it to three and here we have it four and five so we replicate move it to your desired position and this is four and here I just move it down here I think and that's fine okay so that's about what we have here that's about Add Arrows what we have here and then the next thing would do is bring in the arrows okay so that's one interesting thing when you're working with shapes next cell so the arrows are here and so we have the straight one so once you select an arrow and bring it closer a shape you're going to do these circles around it you see that there are points that you can use to to stick your your arrows you see so I'm going to choose this one hold down my mouse button and I'm going to draw it and I'm coming to this shape I'm seeing another one and I just aim at the center of it and then leave it so what does that do you'll see here so you see as I move these they are joined together and I can always make it straight and then we have this one here okay okay so that's how you're going to do for all of the others except that the kind of arrows you choose might be different so for this one I'm going to pick this and I think we have like this coming in here okay let me confirm that okay that's exactly what we have so still using this we use it to build to this one and then we have that that's this other one going we have one coming to the end here so we're going to pick it again and then it's going to come from this end and to this other end now because we have it like this and it's joined I can always adjust and you can see it follows with it you get it and also with the arrows I can do some adjustments in the line and then then we have this one so it doesn't it look nice yes it does we can do all of this in Excel and then we have this other one coming from here okay so I need go to my shapes and take this and so this one goes like this okay so I'm I want to shift some space here okay so there we have it you can see that it's coming to what it is here get another one and so you can see how it is done it's quite easy just get whatever objects you feel is appropriate and then implements them as it is done as is done right here okay so we're going to have another one here that's coming in here okay right about that and just two more this goes straight down to this so straight down to this we may need to do Add Connector so you see the green buttons that come up to tell us that yes okay it's we've made a connection between two objects of that sort so you get it so we're able to build it that way so for this other one all we need to do is use same this time an arrow so most of the time you're going to find yourself using the elbow arrow or the elbow connector so we just pick that and we come to this end here so you notice that they are different just pick this knob here pick this point here and try to make them fit in so you can see that it does just what we want here so you can see that we've done all of these things bring them in here and that's what we have the next thing you need to do is bring in all whatever you want to write within it now I'll show you something like this one that says that a proof project okay so I approve project so if you see you find out that Format Shapes it doesn't fit in the way we won so how do we what do we do to make it fit in so what I'm going to do is that I'm going to right-click on it and format shapes so once you do that you'll see this size and property icon here so we're going to click on that icon clicking on that icon we are going to come to text box if it's not opened if it's like this just click on text box and you will see this so what I need to do is come to the margins come to the magic these are they they what controls the margin so what I'm going to do is I'm going to reduce all of them to zero I'm going to reduce all of them to zero so that gives me some space within here so once I've done that I can I can you see so we now have that there so we can increase it we can bring it to the middle like this and we can increase it to as much as we can take so that is what we have here so most of the times you're going to find out that it may not fit in what you want to write just make sure that all the margins are zero and then you you are just the size and then certainly you're going to get it something you want inside it so I'm going to just take some time to put in the things that are inside and come back to you all right so we're back again and this is what we have we've been able to bring in each of these things inside the box in it so I just wanted you to see how we could bring in these things that we have here you see approve reject requirement need more work and all of those things in it so let's look at how we bring in this approve into it so what we do is that we we go to our inserts we go to look for a text box and here it is so we put it there and what we want to bring in is approved approved so we have this approved in here so what we just need to do is bring Approve it in here and I think that's where it is okay so it's in there and one is in there all you need to do is you know give it a white field so it has a white field there and you just resize it as it should be let me move this with you and so that is it so you can see there and it is just as if it is what it is here so you could use any color for it to to to get it what color can I use here let me just use green for approve so you have it Rejected just there so you can really keep this one because you have another place for example rejected so I'm going to just copy this one bring it somewhere here and what we do is just use rejected and rejected yes we can use red for that so we just use red for that let's use a data rate for it here so you can see how Data Rate it fits in fine with it so you could use it to do all all the others you know like these are the one just country D move it here and we see that this one is approved project our project approved whatever is there so all I need to do is just extend it so that it shows you get it so that's the way you can you can bring it in and have something really really fine and presentable and you can use it to do a lot of your presentation and then have them in such a way that makes it interesting for you to use in presenting things that are related to processes you know so we're gonna put this one here like this so it just looks like it's there okay and I think that's that's quite it okay just requirement need more work okay let's have another one here with things this time I just want to put rework you know so we have this one here you can just adjust it you know you can just adjust it there and that means rework so you can see how this came out and you already know how you could use colors for all of this those are things that we already know how to use colors in such a way that it brings out the information just the way you can present it so it makes it nice sound okay so because you're using a background like this you might have to change one or two things about it so this is how you can come up with a process chart in Select Objects Microsoft Excel it's quite presentable if you want to select everything maybe you won't take it PowerPoint or do any of such things just come to this find and select and let me just try it's in the Home tab find and select just come down select objects so that makes it possible for you to select all of them as one so you see everything is selected here and so you could do anything one you can copy it that way or you can go to your format and then group them group all of them to become one and that's it so you can move so when you when you select it you can move just anywhere as one object so I hope you learned something from this tutorial on how to view the process charts with Microsoft Excel so I hope to see in another tutorial if you enjoy what you've seen why not subscribe so that you can enjoy more from my tutorials I would appreciate if you have questions or if you have things you feel Excel can do you wanna see done please feel free to ask me in the comments below and I will give it a shot have a nice time too - video
14234	https://math.okstate.edu/people/binegar/4023/4023-l01.pdf	LECTURE 1 Proofs and Logic The primary purpose of this course is to introduce you, most of whom are mathematics majors, to the most fundamental skills of a mathematician; the ability to read, write, and understand proofs. Indeed, this is a course where proofs matter more than results. That said, I should also stress that this is not supposed to be a killer course. Yes, we are going to be rigorous and meticulous; but we will take our time to cover the material. And while we will be often dealing in abstractions; our purpose shall be to develop concrete ways of handling far reaching concepts. 1. Logic In order to get our bearings, let us begin with a discussion of logic and proof. Much of this discussion will appear as common sense. However, not all common sense is logical, nor does every common sensical argument constitute a proof. For this reason, we must delineate from the start, exactly what constitutes a logical argument. 1.1. Statements. Definition 1.1. A statement is a declarative sentence that is either true or false. Each of the following sentences is a statement: Every square has four sides. π is a rational number. Orange is the best color. Note that the ﬁrst statement is true, the second is false and the third is merely an opinion. Let us now lay out the means by which we manipulate statements in a logical manner. 1.2. Compound Statements. Suppose P and Q are statements which are either true or false. Then there are several ways we can create new statements which are also either true or false. 1.2.1. Logical Connectives. Definition 1.2. If P and Q are statements, then “P and Q” is a true statement only if P and Q are both true; otherwise “P and Q” is false. Thus, P is true and Q is true.} ⇒ “P and Q” is true P is true and Q is false. Q is true and P is false. Q is false and P is false.    ⇒ “P and Q” is false 1 1. LOGIC 2 In standard English the conjunction “or” can be used in two distinct ways depending on the context: ﬁrst of all it can be used to exclude one or the other of two possibilities: The result of a coin toss is either head or tails. We refer to this usage as the exclusive or. The word “or” can also be used to include two possibilities: I need 6 or 7 dollars. In mathematics, one always uses the logical connective “or” in the inclusive sense. Definition 1.3. If P and Q are statements, then “P or Q” is a true statement only if P, Q is true, or P and Q are both true; otherwise “P or Q” is false. So there are always three possibilities for “P or Q” to be a true mathematical statement and only one possibility for it to be false: P is true and Q is true. P is true and Q is false. Q is true and P is false.    ⇒ “P or Q” is true P is false and Q is false.} ⇒ “P or Q” is false 1.2.2. Universal Quantiﬁers. The following statements contain universal quantiﬁers. For all real numbers x, x2 ̸= −1. All triangles have three sides. For each real number a, a2 ≥0. Notice that in each of the statements above, a property is attributed to all members of a set; this is what we mean by a universal quantiﬁer. Notation 1.4. As a short hand for the phrase “for all” we shall use the symbol ∀(an up-side-down A). 1.2.3. Existential Quantiﬁers. The following statements contain existential quantiﬁers Some integers are prime. There exists a integer between 7.5 and 9.1. The exists an irrational real number. Notice that in each of these statements a property is attributed to at least one element of a set; this is what one means by a existential quantiﬁer. Notation 1.5. As a short hand for the phrase “there exists”, we shall often use the symbol ∃(a backwards E). Notation 1.6. As a short hand for the phrase “such that”, we shall often use the abbreviation “s.t.”. As an example of our notational short-hand we note that ∀x, ∃y s.t. y = x2 translates as “for all x, there exists a y such that y equals x2”. 1. LOGIC 3 1.2.4. Negation. The negation, not-P, of a statement P is the statement such that not-P is true exactly when P is false, and not-P is false exactly when P is true. In most cases you can transform a statement into its negation by inserting a “not” in the appropriate place. A is B. − →A is not B. The negation of compound statements works as follows: The negation of “P and Q” is “not-P or not-Q”. The negation of “P or Q” is “not-P and not-Q”. The negation of universal and existential quantiﬁers works as follows: The negation of a statement with a universal quantiﬁer is a statement with an existential quantiﬁer. The negation of a statement with an existential quantiﬁer is a statement with a universal quantiﬁer. For example, the negation of the statement “All crayons are blue′′, which has a universal quantiﬁer is “Not all crayons are blue” which if true, would of course imply that at least one crayon was not blue; i.e. a statement with an existential quantiﬁer. In summary A is B negation − − − − − − − − − → A is not B A and B negation − − − − − − − − − → not-A or not-B A or B negation − − − − − − − − − → not-A and not-B All A are B negation − − − − − − − − − → at least one A is not-B At least one A is B negation − − − − − − − − − → all A are not-B Notation 1.7. If P is a statement we shall sometimes employ the notation ˜P to indeicate its negation “not-P”. 1.3. Conditional Statements. In mathematics one deals primarily with conditional statements; that is to say statements of the form If P, then Q. which is written symbolically as P ⇒Q . Such a statement means that the truth of P guarantees the truth of Q. More explicitly • P ⇒Q is true if Q is true whenever P is true. • P ⇒Q is false if Q can be false when P is true. 1. LOGIC 4 The statement P is called the hypothesis, or premise, and the statement Q is called the conclusion. Here are some examples: If x and y are integers, then x + y is an integer. x ̸= 0 ⇒x2 > 0. There are several ways of phrasing a conditional statement, all of which mean the same thing: If P, then Q. P implies Q. P is suﬃcient for Q. Q provided that P. Q whenever P.            ∼ P ⇒Q 1.4. The Contrapositive of a Conditional Statement. The contrapositive of a conditional state-ment If P, then Q. is the conditional statement If not-Q , then not-P For example, the contrapositive of If x < 6, then x < 8 is If x is not less than 8, then x is not less than 6 or, equivalently, If x ≥8, then x ≥6. In this example, the truth of the original conditional statement seems to guarantee the truth of its contra-positive. In fact, The conditional statement “P ⇒Q” is equivalent to its contrapositive “not-Q ⇒not-P”. Let’s prove “ P ⇒Q” implies “not-Q ⇒not-P” . By hypothesis, if P is true, then Q is true. Suppose not-Q is true. Then Q is false. But then P can not be true, since that would contradict our hypothesis. So not-P must be true. Example 1.8. Prove that “not-Q ⇒not-P” implies “P ⇒Q” . 1. LOGIC 5 1.5. The Converse of a Conditional Statement. The converse of the conditional statement P ⇒Q is the conditional statement Q ⇒P . It is important to note that the truth of a conditional statement does not imply the truth of its converse. For example, it is true that If x is an integer, then x is a real number; but the converse of this statement If x is a real number, then x is an integer, is certainly not true. However, there are some situations in which both a conditional statement and its converse are true. For example, both If the integer x is even, then the integer x + 1 is odd and its converse If integer x + 1 is odd, then the integer x is even are true. We can state this fact more succinctly by saying The integer x is even if and only if the integer x + 1 is odd . More generally, the statement P if and only if Q which may be abbreviated P iﬀQ or P ⇔Q means “P ⇒Q” and “Q ⇒P” . “P if and only if Q” is called a biconditional statement. When “P ⇔ Q” is a true biconditional statement, P is true exactly when Q is true, and so the statements P and Q can be regarded as equivalent statements (when inserted in other statements).
14235	https://www.pcepurnia.org/wp-content/uploads/2020/03/Notes-Engineering-Graphics-and-Design-2.pdf	Engineering Drawing Lecture 10 Projection of Solids 1 Solids A 3-D object having length, breadth and thickness and bounded by surfaces which may be either plane or curved, or combination of the two. ▪ Classified under two main headings ▪ Polyhedron ▪ Solids of revolution ▪ Regular polyhedron – solid bounded only by plane surfaces (faces). Its faces are formed by regular polygons of same size and all dihedral angles are equal to one another. ▪ Other polyhedra – when faces of a polyhedron are not formed by equal identical faces, they may be classified into prisms and pyramids. Five regular polyhedra Tetrahedron – four equal equilateral triangular faces Cube/hexahedron – six equal square faces Octahedron– eight equal equilateral triangular faces Dodecahedron – twelve equal regular pentagonal faces Icosahedron– twenty equal equilateral triangular faces Prism – a polyhedron formed by two equal parallel regular polygon, end faces connected by side faces which are either rectangles or parallelograms. Different types of prisms Pyramids – a polyhedron formed by a plane surface as its base and a number of triangles as its side faces, all meeting at a point, called vertex or apex. Axis – the imaginary line connecting the apex and the center of the base. Inclined/slant faces – inclined triangular side faces. the apex and the base corners. Right pyramid – when the axis of the pyramid is perpendicular to its base. Oblique pyramid – when the axis of the pyramid is inclined to its base. Triangular pyramid Square pyramid Rectangular pyramid Pentagonal pyramid Hexagonal pyramid Oblique pyramid Solids of revolution – when some of the plane figures are revolved about one of their sides – solids of revolution is generated. Cylinder – when a rectangle Cone – Sphere – when a semi-circle is revolved about one of its sides, the other parallel side generates a cylinder. revolved about one of its sides, the hypotenuse of the right triangle generates a cone. is revolved about one of its diameter, a sphere is generated.. Oblique cylinder – when a parallelogram is revolved about one of its sides, the other parallel side generates a cylinder. Oblique cone Truncated and frustums of solids – when prisms, pyramids, cylinders are cut by cutting planes, the lower portion of the solids (without their top portions) are called, either truncated or frustum of these solids. Visibility – when drawing the orthographic views of an object, it will be required to show some of the hidden details as invisible and are shown on the orthographic views by dashed lines Rules of visibility All outlines of every view are visible – the outlines of all the views are shown by full lines. In the top view, the highest portions of the object are visible. Frustum of a pentagonal pyramid – the top face ABCDE is the highest, it is completely visible in the top view. In the top view, edges ab, bc, cd, de and ea are shown as full lines. The bottom pentagonal faces A1B1C1D1E1 is smaller than the top face, hence invisible. The slant edges AA1, BB1, CC1, DD1 and EE1 are invisible in the top view, hence they are shown as lines of dashes. The line connecting a visible point and an invisible point is shown as an invisible line of dashes unless they are outlines. In the front view - the front faces of the object are visible. In the front view – the faces ABB1A1 and BCC1B1 are the front faces, hence are visible. In the front view, the corners a, b, c and a1, b1, c1 are visible to the observer. Hence in the front view, the lines a’a’1, b’b’1 and c’c’1 are shown as full lines. The corners d, e, d1 and e1 are invisible in the front view. The lines, e’e’1, d’d’1 are invisible, hence shown as dashed lines. The top rear edges a’e’, e’d’ and d’c’ coincide with the top front visible edges a’b’ and b’c’. Object In the side view - the face lying on that side are visible. As sheen in the left side view, the corners e, a, b and e1, a1, b1 lie on left side and are visible in the left view. 1 a”a1” and b”b1” are shown as full lines. The edges d”d1”, c”c1” coincide with the visible edges e”e1” and a”a1” respectively. Projections of solids placed in different positions The solids may be placed on HP in various positions (1) The way the axis of the solid is held with respect to HP or VP or both - ▪ Perpendicular to HP or VP ▪ Parallel to either HP or VP and inclined to the other ▪ Inclined to both HP and VP Axis of the solid perpendicular to HP A solid when placed on HP with its axis perpendicular to it, then it will have its base on HP. This is the simplest position in which a solid can be placed. When the solid is placed with the base on HP position, in the top view, the base will be projected in its true shape. Hence, when the base of the solid is on HP, the top view is drawn first and then the front view and the side views are projected from it. Only one position in which a cylinder or a cone may be placed with its base on HP. Four positions of a prism placed with its base on HP. Four positions of a triangular pyramid placed with its base on HP Axis of the solid perpendicular to VP When a solid is placed with its axis perpendicular to VP, the base of the solid will always be perpendicular to HP and parallel to VP. Hence in the front view, base will be projected in true shape Therefore, when the axis of the solid is perpendicular to VP, the front view is drawn first and then the top and side views are drawn from it. When a cylinder rests on HP with its axis perpendicular to VP, one of its generators will be on HP. When a cone rests on HP with its axis perpendicular to VP, one of the points on the circumference of the base will be on HP. Prism placed with their axis perpendicular to VP in three different positions. Pyramid placed with their axis perpendicular to VP in three different positions. Axis of the solid inclined to HP and parallel to VP When a solid is placed on HP with its axis inclined to HP, the elemental portion of the solid that lies on HP depends upon the type of the solid. When a prism is placed on HP with its axis inclined to it, then it will lie either on one of its base edges or on one of its corners on HP. When a pyramid is placed on HP with its axis inclined to HP, then we will have one of its base edges on HP or one of its base corners on HP or one of its slant edges on HP or one of its triangular faces on HP or an apex on HP. Case 1. When the solid lies with an edge of base on HP If the solid is required to be placed with an edge of the base on HP, then initially the solid has to be placed with its base on HP such that an edge of the base is perpendicular to VP, i.e., to XY line in top view preferably to lie on the right side. When the solid lies with an edge of base on HP When a pentagonal prism has to be placed with an edge of base on HP such that the base or axis is inclined to HP, then initially, the prism is placed with its base on HP with an edge of the base perpendicular to VP and the lying on the right side. In this position, the first set of top and front views are drawn with the base edges (c1)(d1) perpendicular to XY line in the top view. In the front view, this edge c1’(d1’) appears as a point. Since the prism has to lie with an edge of the base on HP, the front view of the prism is tilted on the edge c1’(d1’) such that the axis is inclined at  to HP. Redraw the first front view in the tilted position Whenever the inclination of axis  with HP is given, first the base is drawn at (90- ) in the front view, otherwise improper selection of the position of the axis may result in the base edge c1’(d1’) lying above or below the XY line. The second top view is projected by drawing the vertical projectors from the corners of the second front view and the horizontal projectors from the first top view. . Top and the front views of a hexagonal pyramid when it lies on HP on one of its base edges with its axis or the base inclined to HP. Case.2 : When the solid lies on one of its corners of the base on HP When a solid lies on one of its corners of the base on HP, then the two edges of the base containing the corner on which it lies make either equal inclinations or different inclination with HP. Corner of the base on HP with two base edges containing the corner on which it rests make equal inclinations with HP Initially the solid should be placed with its base on HP such that an imaginary line connecting the center of the base and one of its corners is parallel to VP, i.e. to XY line in the top view, and preferably to lie on the right side. For example, when a hexagonal prism has to be placed with a corner of the base on HP such that the base or the axis is inclined to HP, then initially the the prism is placed with its base o HP such that an imaginary line connecting the center of the base and a corner is parallel to VP and it lies on the right side. In this position, the first set of top and front views are drawn – the line (o1)(d1) is parallel to the XY line in the top view. n Since the prism has to lie on one of its corners of the base on HP, the front view of the prism is tilted on the corner d1’ such that the axis is inclined at  to HP. Redraw the front view in the tilted position. The base edge is drawn at (90- ) in the front view. The second top view is projected by drawing the vertical projectors from the corners of the second front view and horizontal projectors from the first top view. s with HP. Case.2 for Pyramid: The top and front views of the pyramid when it rests on HP on one of its base corners such that the two base edges containing the corner on which it rests make equal inclination Case-3 When a pyramid lies on one of its triangular faces on HP If a pyramid has to be placed on one of its triangular faces on HP, then initially let the pyramid be placed with its base on HP. In the first front view, the right side inclined line, i.e., o’c’(d’) represents a triangular face. Redraw the front view such that the triangular face o’c’(d’) lies on HP. Project the top view in this position. CASE-4: When a pyramid lies on one of its slant edges on HP When a pyramid lies with one of its slant edges on HP, then two triangular faces containing the slant edge on which it rests make either equal inclinations or different inclinations with HP. SOLID WITH AXIS INCLINED TO BOTH THE RPs Methods of drawing the projections of solids Two methods 1. Change of position method - the solids are placed first in the simple position and then tilted successively in two or three stages to obtain the final position. 2. Auxiliary plane method (Change of reference-line method) – the solids are placed initially in the simple position and then one or two auxiliary planes are setup to obtain the views in the required position. Problem.1 A cube of 30 mm side rests with one of its edges on HP such that one of the square faces containing that edge is inclined at 300 to HP and the edge on which it rests being inclined to 600 to VP. Draw its projections. Problem2. Draw the top and front views of a rectangular pyramid of sides of base 40x 50 mm and height 70 mm when it lies on one of its larger triangular faces on HP. The longer edge of the base of the triangular face lying on HP is inclined at 600 to VP in the top view with the apex of the pyramid being nearer to VP. Problem-3: A Hexagonal Pyramid, base 25 mm side and axis 55 mm long, has one of its slant edges on the ground. A plane containing that edge and the axis is perpendicular to the HP and inclined at 450 to VP. Draw the projections when the apex is nearer to the VP than the base. Draw the TV of the pyramid with a side of base parallel to XY. The slant edges AO and DO will also be parallel to XY. Draw FV also. Tilt the FV so that d’o’ is in XY. Project the second TV. Draw a new reference line X1Y1 making 45o angle with o1p1 (the top view of the axis) and project the final FV. Problems on cones Problem4. A cone of base 80 mm diameter and height 100 mm lies with one of its generators on HP and the axis appears to be inclined to VP at an angle of 400 in the top view. Draw its top and front views. Problem5. ▪ A cone of base 60 mm diameter and the axis 80 mm long lies on HP with its axis inclined at 450 and 300 to HP and VP, respectively. Draw the top and front views of the cone. SOLID WITH AXIS INCLINED TO BOTH THE RPs If the axis of a solid is inclined to both the RPs then the problem is solved in three stages. Example: A triangular pyramid of edge of base ‘s’ mm and length of axis ‘h’ mm is resting on a side of base on the HP. The axis of the pyramid is inclined at 8° to the HP and ø° to the VP. Draw its projections. Engineering Drawing Sections of solids 1 Section Views ➢ Sectional drawings are multiview technical drawings t h a t contain special views of a part or parts, views that reveal interior features. ➢ Used to improve clarity and reveal interior features o f parts. ➢ interior features of complicated assemblies. ➢ A primary reason for creating a section v i e w i s the elimination of hidden lines, so that a drawing can be more easily understood or visualized. Section Views • Traditional section views are based on the use of an imaginary cutting plane that cuts through the object to reveal interior features. • This imaginary cutting plane is controlled by the designer and can (a) go completely through the object (full section); (b) go half-way through the object (half section); (c) be bent to go through features that are not aligned (offset section); or (d) go through part of the object (broken-out section). CUTTING PLANE LINES – which show where the cutting plane passes through the object, represent the edge view of the cutting plane and are drawn in the view(s) adjacent to the section view. In the figure the cutting plane line is drawn in the top view, which is adjacent to the sectioned front view. Cutting plane lines are thick (0.7 mm) dashed lines, that extend past the edge of the object 6 mm and have line segments at each end drawn at 90 degrees and terminated with arrows. The arrows represent the direction of the line of sight for the section view and they point away from the sectioned view. Two types of lines are acceptable for cutting plane lines in multi-view drawings Line B-B is composed of alternating long and two short dashes, which is one of the two standard methods. The length of the long dashes varies according to the size of the drawing, and is approximately 20 to 40 mm. For a very large section view drawing, the long dashes are made very long to save drawing time. The short dashes are approximately 3 mm long. The open space between the lines is approximately 1.5 mm. Capital letters are placed at each end of the cutting plane line, for clarity or when more than one cutting plane is used on a drawing. The second method used for cutting plane lines is shown by line C-C, which is composed of equal-length dashed lines. Each dash is approximately 6 mm long, with a 1.5 mm space between. If the cutting plane line is in the same position as a center line, the cutting plane line has precedence. Types of Cutting Planes and Their Representation • Frontal or Vertical Cutting/ Section Plane • Horizontal Cutting/ Section Planes • Profile Cutting / Section Planes • Auxiliary Section Plane – Auxiliary Inclined Plane (AIP) – Auxiliary Inclined Plane (AVP) • Oblique Section Plane In this figure, the cutting plane appears as an edge in the top view and is normal in the front view; therefore, it is a frontal cutting plane or Vertical Section Plane. The front half of the object is "removed" and the front view is drawn in section. If the cutting plane appears as an edge in the front view and is normal in the top view, it is a horizontal cutting/section plane. The top half of the object is "removed" and the top view is drawn in section. If the cutting plane appears as an edge in the top and front views and is normal in the profile view, it is a profile cutting/section plane. The left (or righ) half of the object is "removed" and the left (or right) side view is drawn in section. Multiple sections can be done on a single object, as shown in the figure. In this example, two cutting planes are used: one a horizontal and the other a profile cutting plane. Both cutting planes appear on edge in the front view, and are represented by cutting plane lines A-A and B-B, respectively. Each cutting plane will create a section view, and each section view is drawn as if the other cutting plane did not exist. Section Line Practices Section lines or cross-hatch lines are added to a section view to indicate the surfaces that are cut by the imaginary cutting plane. Different section line symbols can be used to represent various types of materials. However, there are so many different materials used in engineering design that the general symbol (i.e., the one used for cast iron) may be used for most purposes on engineering drawings. The actual type of material required is then noted in the title block or parts list or as a note on the drawing. The angle at which lines are drawn is usually 45 degrees to the horizontal, but this can be changed for adjacent parts shown in the same section. Also the spacing between section lines is uniform on a section view. Material Symbols The type of section line used to represent a surface varies according to the type of material. However, the general purpose section line symbol used in most section view drawings is that of cast iron. The specific type of steel to be used will be indicated in the title block or parts list. Occasionally, with assembly section views, material symbols are used to identify different parts of the assembly. Drawing Techniques The general purpose cast iron section line is drawn at a 45- degree angle and spaced 1.5 mm to 3 mm or more, depending on the size of the drawing. As a general rule, use 3mm spacing. Section lines are drawn as thin (.35 mm) black lines, using an H or 2H pencil. The section lines should be evenly spaced and of equal thickness, and should be thinner than visible lines Also, do not run section lines beyond the visible outlines or stop them too short Section lines should not run parallel or perpendicular to the visible outline. If the visible outline to be sectioned is drawn at a 45- degree angle, the section lines are drawn at a different angle, such as 30 degrees. Avoid placing dimensions or notes within the section lined areas. If the dimension or note must be placed within the sectioned area, omit the section lines in the area of the note Outline Sections An outline section view is created by drawing partial section outlines adjacent to all object lines in the section view. For large parts, outline sectioning may be used to save time. Thin Wall Sections Very thin parts such as washers and gaskets are not easily represented with section lines, so conventional practice calls for representing the thin part in solid black. Gasket is drawn solid black to show that it is sectioned Section lined areas are bounded by visible lines, never by hidden lines, because the bounding lines are visible in the section view Points of Intersection (POI) • Whenever a section plane cuts a solid, it intersects (and or coincides with) the edges of solids. The point at which the section plane intersects an edge of the solid is called the point of intersection (POI). A section view is created by passing an imaginary cutting plane vertically through the center of the part. This figure is a 3D representation of the part after it is sectioned. This section view more clearly shows the interior features of the part. The corners of the section view are numbered so that they can be compared with the orthographic section view. The line of sight for the section view is perpendicular to the cut surfaces, which means they are drawn true size and shape in the section view. Also, no hidden lines are drawn and all visible surfaces and edges behind the cutting plane are drawn as object lines. All the surfaces touched by the cutting plane are marked with section lines. Because all the surfaces are the same part, the section lines are identical and are drawn in the same direction. The center line is added to the counter bored hole to complete the section view. Types of Section Views • Full sections • Half sections • Offset sections • Broken-out sections • Revolved sections • Removed sections Full Section View • In a full section view, the cutting plane cuts across the entire object • Note that hidden lines become visible in a section view Full Section View • Show cutting plane in the top view – New line type – • Make a full section in the front view • Note how the cutting plane is drawn and how the crosshatching lines mark the surfaces of material cut by the cutting plane. Half Section View • The cutting planes do not cut all the way through to the object. • They cut only half way and intersect at the centerline. Half Section View Half Section is used mainly for symmetric objects Offset Sections Offset sections are features that do not lie along a straight line Offset Sections Offset Sections Broken Out Sections A broken-out section view is off part of the object to reveal interior features Broken Out Sections Hidden lines are used only when needed for clarity. Revolved Sections Revolved sections show the shape of an object's cross-section superimposed on a longitudinal view Beam Any part with an odd number of spokes or ribs will give an unsymmetrical and misleading section if the principle of true projections are strictly adhered to. 1) The spoke is rotated to the path of the vertical cutting plane and then projected on the side view. 2) Neither of the spokes should be sectioned (hatched). Section of solids • Section plane parallel to VP ( cube) • Section plane parallel to HP ( prism, pyramid) • Section plane inclined to VP ( Pyramid, cylinder) • Section plane for which its true shape is given • Sectional views for a complex object Section plane parallel to VP Draw the projection of the solid without section plane. (i.e. top view and front view according to the given conditions). Then introduce the section plane in the top view. As it is parallel to the VP, is seen as a line in top view. Carry it to the front view. Section plane parallel to HP A triangular prism, side of base 30 mm and axis 50 mm long is lying on the HP on one of its rectangular faces with its axis inclined at 30 to the VP. It is cut by a horizontal section plane at a distance of 12 mm above the ground. Draw its front view, side view and sectional top view. Draw the projections of the un-cut prism. As the section plane is parallel to HP, it will be seen as a straight line parallel to XY in the front view. Project the section to the top view. Section plane parallel to HP….. A pentagonal pyramid, side of base 30 mm and axis 65 mm long, has its base horizontal and an edge of the base parallel to the VP. A horizontal section plane cuts it at a distance of 25 mm above the base. Draw its front view and sectional top view. Section plane Inclined to VP A pentagonal pyramid has its base on the HP. Base of the pyramid is 30 mm in side, axis 50 mm long. The edge of the base nearer to VP is parallel to it. A vertical section plane, inclined at 45 to the VP, cuts the pyramid at a distance of 6 mm from the axis. Draw the top view, sectional front view and the auxiliary front view on an AVP parallel to the section plane. Sections of Cylinders: Section plan inclined to the base Problem.1 A cylinder of 40 mm diameter, 60 mm height and having its axis vertical is cut by a section plane, perpendicular to the VP, inclined at 45 to the HP and intersecting the axis 32 mm above the base. Draw its front view, sectional top view, sectional side view and the true shape of the section Practice Example-1: A cube of 70 mm long edges has its vertical faces equally inclined to the VP. It is cut by an AIP in such a way that the true shape of the cut part is a regular hexagon. Determine the inclination of the cutting plane with the HP. Draw FV, sectional TV and true shape of the section. Step-1 Draw TV and FV of the cube as shown. Step-2 As the true shape of the section is a hexagon, the cutting plane must cut the prism at 6 points. plane will cut two edges of the top, two edges of the base and two vertical edges. The POIs at two vertical edges will be farthest from each other. These points will represent the two opposite corners of the hexagon and the distance between them will be equal to b( b1)– d( d1). Practice Example 2: Example for a complex object: Draw the sectional FV, TV and SV of the object shown in Figure below A A Engineering Drawing Intersections of Solids 1 Whenever two or more solids combine, a definite curve is seen at their intersection. This curve is called the curve of intersection (COI). CASES OF INTERSECTION The cases of intersection depend on the type of intersecting solids and the manner in which they intersect. Two intersecting solids may be of the same type (e.g., prism and prism) or of different types (e.g., prism and pyramid). The possible combinations are shown in Table below. The two solids may intersect in different ways. The axes of the solids may be parallel, inclined or perpendicular to each other. The axes may be intersecting, offset or coinciding. Therefore, the following sub-cases exist: (i) Axes perpendicular and intersecting (ii) Axes perpendicular and offset (iii) Axes inclined and intersecting (iv) Axes inclined and offset (v) Axes parallel and coinciding (vi) Axes parallel and offset Intersection The type of intersection created depends on the types of geometric forms, which can be two- or three- dimensional. Intersections must be represented on multiview drawings correctly and clearly. For example, when a conical and a cylindrical shape intersect, the type of intersection that occurs depends on their sizes and on the angle of intersection relative to their axes. The line of intersection is determined using auxiliary views and cutting planes Methods – (1) Line and (2) Cutting-plane methods Line method: A number of lines are drawn on the lateral surface of one of the solids and in the region of the line of intersection. Points of intersection of these lines with the surface of the other solid are then located. These points will lie on the required line of intersection. They are more easily located from the view in which the lateral surface of the second solid appears edgewise (i.e. as a line). The curve drawn through these points will be the line of intersection. Cutting-plane method: The two solids are assumed to be cut by a series of cutting planes. The cutting planes may be vertical (i.e. perpendicular to the H.P.), edgewise (i.e. perpendicular to the V.P.) or oblique. The cutting planes are so selected as to cut the surface of one of the solids in straight lines and that of the other in straight lines or circles. Intersection of two prisms Prisms have plane surfaces as their faces. The line of intersection between two plane surfaces is obtaine by locating the positions of points at which the edges of one surface and then joining the points by a straight line. These points are called vertices The line of intersection between two prisms is therefore a closed figure composed of a number of such lines meeting at the vertices d Intersection of two prisms A vertical square prism, base 50 mm side, is completely penetrated by a horizontal square prism, base 35 mm side, so that their axes intersect. The axis of the horizontal prism is parallel to the prism., while the faces of the two prisms are equally inclined to the prism. Draw the projections of the solids, showing lines of intersection. (Assume suitable lengths for the prisms.) Steps: Draw the projections of the prisms in the required position. The faces of the vertical prism are seen as lines in the top view. Hence, let us first locate the points of intersection in that view. Steps: Lines 1-1 and 3-3 intersect the edge of the vertical prism at points p1 and p3 (coinciding with a). Lines 2-2 and 4-4 intersect the faces at p2 and p4 respectively. The exact positions of these points along the length of the prism may now be determined by projecting them on corresponding lines in the front view. For example, p2 is projected to p2' on the line 2'2'. Note that p4' coincides with p2'. Intersection of two prisms Draw lines p1’p2' and p2‘p3'. Lines p1‘p4' and p3‘p4' coincide with the front lines. Thes lines show the line of intersection. Lines q1'q2' and q2‘q3' on the other side are obtained in the same manner Note that the lines for the hidden portion of the edges are shown as dashed lines. The portions p1’p3' and q1’q3' of vertical edges a'a' and c'c' do not exist and hence, must be removed or kept fainter. e Intersection of Cylinder and Cylinder Intersection of Cylinder and Cylinder As cylinders have their lateral surfaces curved – the line of intersection between them will also be curved. Points on this line may be located by any of the methods. For plotting an accurate curve, certain critical or key points, at which the curve changes direction, must also be located. These are the points at which outermost or extreme lines of each cylinder pierce the surface of the other cylinder. Intersection of Cylinder and Cylinder Example - A vertical cylinder of 80 mm diameter is completely penetrated by another cylinder of 60 mm diameter, their axes bisecting each other at right angles. Draw their projections showing curves of penetration, assuming the axis of the penetrating cylinder to be parallel to the VP. Intersection of Cylinder and Cylinder Assume a series of horizontal cutting planes passing through the the horizontal cylinder and cutting both cylinders. Sections of the horizontal cylinde will be rectangles, while those of the vertical cylinder will always be circles of the same diameter as its own. Points at which sides of the rectangles intersect the circle will be the curve of intersection. For example, let a horizontal section pass through points 2 and 12 r In the front view, it will be seen as a line coinciding with line 2' 2'. The section of the horizontal cylinder will be a rectangle of width (i.e. the line 2-12). The section of the vertical cylinder will be a circle. Points p2 and p12 at which the sides (2-2 and 12-12) of the rectangle cuts the circle, lie on the curve. These points are first marked in the top view and then projected to points p2' and p12' on lines 2’2’ and 12’12’ in the front view. Points on the other side of the axis are located in the same manner. Intersection of Cone and Cylinder r Intersection of Cone and Cylinder Example - A vertical cone, diameter of base 75 mm and axis 100 mm long, is completely penetrated by a cylinde of 45 mm diameter. The axis of the cylinder is parallel to HP and the VP and intersects the axis of the cone at a point 22 mm above the base. Draw the projections of the solids showing curves of intersection. Cutting-Plane Method Draw lines dividing the surface of the cylinder into twelve equal parts. Assume a horizontal cutting plane passing through say, point 2. The section of the cylinder will be a rectangle of width w (i.e. the line 2- 12), while that of the cone will be a circle of diameter ee. These two sections intersect at points p2 and p12. These sections are clearly indicated in the top view by the rectangle 2- 2-12-12 and the circle of diameter ee . In the front view, the cutting plane is seen as a line coinciding with 2' 2’. Points p2 and p12 when projected on the line 2' 2’ (with which the line 12'- 12' coincides) will give a point p2' (with which p12' will coincide). Then p2' and p12' are the points on the curve of intersection. To obtain the points systematically, draw circles with centre 0 and diameters dd, ee, ff, etc. cutting lines through 1, 2 and 12, 3 and 11 etc. at points p1, p2 and p12, p3 and pll etc . Project these points to the corresponding lines in the front view. Final views Engineering Drawing Intersections of Solids 2 1 Questions for Practice INTERSECTION OF PRISM AND PRISM (with axis perpendicular and offset) Prob. 1) A vertical square prism, base 50 mm side, is completely penetrated by a horizontal square prism, base 35 mm side, so that their axis are 6 mm apart. The axis of the horizontal prism is parallel to the VP, while the faces of both prisms are equally inclined to the VP. Draw the projections of the prisms showing lines of intersection. (Assume that the length of both the prisms is 100 mm). (Book: N. D. Bhatt) INTERSECTION OF PRISM AND PRISM Prob. 2) A vertical square prism, base 50 mm side and height 90 mm has a face inclined at 30O to the VP. It is completely penetrated by another horizontal square prism, base 40 mm side and axis 100 mm long, faces of which are equally inclined to the VP. The axis of the two prisms are parallel to the VP and bisect each other at right angles. Draw the projections showing lines of intersection. (Book: N. D. Bhatt) Prob. 3) A square prism of 40 mm edge of base and 90 mm high rests vertically with its base on HP such that the front right vertical rectangular face is inclined at 60° to VP. This prism is penetrated by another horizontal square prism whose rectangular faces make equal inclination with both HP and VP. The axis of the horizontal prism is passing at the mid height at a distance of 10 mm infront of the vertical prism. The horizontal square prism is of the same dimensions as that of the vertical square prism. Draw the lines of intersection (Taken from K.R. Gopslakrishna, Engg. Drawing, subhas store book center) (Taken from K.R. Gopslakrishna, Engg. Drawing, subhas store book center) INTERSECTION OF CYLINDER AND CONE (with axis perpendicular and offset) Prob. 4) A cone with a base diameter of 64 mm and an axis length of 70 mm is kept on its base on the HP. A cylinder of diameter 30 mm and length 90 mm penetrates the cone horizontally. The axis of the cylinder is 20 mm above the base of the cone and 5 mm away from the axis of the latter. Draw the three views of the solids showing curve of intersection. (Taken from Dhananjay A Jolhe, Engg. Drawing, MGH) INTERSECTION OF PRISM AND CYLINDER (with axis perpendicular and offset) Prob. 5) A vertical cylinder with a 60 mm diameter is penetrated by a horizontal square prism with a 40 mm base side, the axis of which is parallel to the VP and 10 mm away from the axis of the cylinder. A face of the prism makes an angle of 30° with the HP. Draw their projections showing curves of intersection. (Taken from Dhananjay A Jolhe, Engg. Drawing, MGH) Prob. 6) A vertical pentagonal prism 30 mm edge of base and height 100 mm has one of its rectangular faces parallel to VP and nearer to it. It is penetrated by a rectangular prism of side 40mm x 20 mm and 100 mm high, with its front largest lower front rectangular face inclined at 60° to HP. The axis of the rectangular prism is inclined at 30 ° to HP and parallel to VP, 5 mm infront of the axis of the pentagonal prism and appears to bisect it in the front view. Draw the interpenetration line. (Taken from K.R. Gopslakrishna, Engg. Drawing, subhas store book center) (Taken from K.R. Gopslakrishna, Engg. Drawing, subhas store book center) (Taken from K.R. Gopslakrishna, Engg. Drawing, subhas store book center) Prob. 7) A vertical cylinder of 40 mm diameter and 80 mm high is intersected by another cylinder of 35 mm diameter and 80 mm long. The axis of the penetrating cylinder is inclined at 30° to HP, parallel to VP, 6 mm infront of the vertical cylinder and appears to bisect it in front view. Draw the intersection curve. (Taken from K.R. Gopslakrishna, Engg. Drawing, subhas store book center) (Taken from K.R. Gopslakrishna, Engg. Drawing, subhas store book center) INTERSECTION OF PRISM AND PYRAMID Prob. 8) A square pyramid with a base side of 55 mm and an axis length of 80 mm stands on its base on the HP with the sides of base equally inclined to the VP. A triangular prism with a base side of 34 mm and length of axis 100 mm, penetrates the pyramid completely. The axis of the prism is perpendicular to the VP and intersects the axis of pyramid at 24 mm from the HP. One of the lateral faces of the prism is perpendicular to the HP. Draw the three views of the solids showing LOI. (Taken from Dhananjay A Jolhe, Engg. Drawing, MGH) Engineering Drawing Development of Surfaces 1 Development of surfaces A development is the unfold/unrolled flat / plane figure of a 3-D object. Called also a pattern, the plane may show the true size of each area of the object. When the pattern is cut, it can be rolled or folded back into the original object. Methods of development of surfaces are: ➢ Parallel line development ➢ Radial line development ➢ Triangulation development ➢ Approximate development ➢ Parallel line development uses parallel lines to construct t h e expanded pattern of each three-dimensional shape. The method divides the surface into a series of parallel lines to determine the shape of a pattern. Example: Prism, Cylinder. ➢ Radial line development uses lines radiating from a central p o i n t to construct the expanded pattern of each three-dimensional shape. Example: Cone, Pyramid. ➢ Triangulation developments a r e made from polyhedrons, single- curved surfaces, and wrapped surfaces. Example: Tetrahedron and other polyhedrons. ➢ In approximate development, t h e shape obtained is only approximate. After joining, the part is stretched or distorted to obtain the final shape. Example: Sphere. Examples of Developments A true development is one in which no stretching or distortion of the surfaces occurs and every surface of the development is the same size and shape as the corresponding surface on the 3-D object. e.g. polyhedrons and single curved surfaces Polyhedrons are composed entirely of plane surfaces that can be flattened true size onto a plane in a connected sequence. Single curved surfaces are composed of consecutive pairs of straight-line elements in the same plane. An approximate development is one in which stretching or distortion occurs in the process of creating the development. The resulting flat surfaces are not the same size and shape as the corresponding surfaces on the 3-D object. Wrapped surfaces do not produce true developments, because pairs of consecutive straight-line elements do not form a plane. Also double-curved surfaces, such as a sphere do not produce true developments, because they do not contain any straight lines. 1. Parallel-line developments are made from common solids that are composed of parallel lateral edges or elements. e.g. Prisms and cylinders The cylinder is positioned such that one element lies on the development plane. The cylinder is then unrolled until it is flat on the development plane. The base and top of the cylinder are circles, with a circumference equal to the length of the development. All elements of the cylinder are parallel and are perpendicular to the base and the top. When cylinders are developed, all elements are parallel and any perpendicular section appears as a stretch-out line that is perpendicular to the elements. 2. Radial-line development Radial-line developments are made from figures such as cones and pyramids. In the development, all the elements of the figure become radial lines that have the vertex as their origin. The cone is positioned such that one element lies on the development plane. The cone is then unrolled until it is flat on the development plane. One end of all the elements is at the vertex of the cone. The other ends describe a curved line. The base of the cone is a circle, with a circumference equal to the length of the curved line. 3. Triangulation developments: Made from polyhedrons, single- curved surfaces, and wrapped surfaces. The development involve subdividing any ruled surface into a series of triangular areas. If each side of every triangle is true length, any number of triangles can be connected into a flat plane to form a development Triangulation for single curved surfaces increases in accuracy through the use of smaller and more numerous triangles. Triangulation developments of wrapped surfaces produces only approximate of those surfaces. 4. Approximate developments Approximate developments are used for double curved surfaces, such as spheres. Approximate developments are constructed through the use of conical sections of the object. The material of the object is then stretched through various machine applications to produce the development of the object. Development of a right rectangular prism Parallel-line developments Developments of objects with parallel elements or parallel lateral edges begins by constructing a stretch-out line that is parallel to a right section of the object and is therefore, perpendicular to the elements or lateral edges. In the front view, all lateral edges of the prism appear parallel to each other and are true length. The lateral edges are also true length in the development. The length, or the stretch-out, of the development is equal to the true distance around a right section of the object. Step 1. To start the development, draw the stretch-out line in the front view, along the base of the prism and equal in length to the perimeter of the prism. Draw another line in the front view along the top of the prism and equal in length to the stretch-out line. Draw vertical lines between the ends of the two lines, to create the rectangular pattern of the prism. Step 2. Locate the fold line on the pattern by transferring distances along the stretch-out line in length to the sides of the prism, 1-2, 2-3, 3-4, 4-1. Draw thin, dashed vertical lines from points 2, 3, and 4 to represent the fold lines. Add the bottom and top surfaces of the prism to the development, taking measurements from the top view. Add the seam to one end of the development and the bottom and top. Development of a truncated prism Step 1: Draw the stretch-out line in the front view, along the base of the prism and equal in length to the perimeter of the prism. Locate the fold lines on the pattern along the stretch-out line equal in length to the sides of the prism, 1-2, 2-3, 3-4, and 4-1. Draw perpendicular construction lines at each of these points. Project the points 1, 2, 3, and 4 from the front view Step 2: Darken lines 1-2-3 and 4-1. Construct the bottom and top, as shown and add the seam to one end of the development and the top and bottom Development of a right circular cylinder Step 1. In the front view, draw the stretch-out line aligned with the base of the cylinder and equal in length to the circumference of the base circle. At each end of this line, construct vertical lines equal in length to the height of the cylinder. Step 2. Add the seam to the right end of the development, and add the bottom and top circles. Development of a truncated right circular cylinder The top circular view of the cylinder is divided into a number of equal parts, e.g 12. The stretch-out line, equal in length to the circumference of the circle, is aligned with the base in the F.V. view and is divided into 12 equal parts from which vertical lines are constructed. The intersection points in the T.V. are projected into the F.V. , where the projected lines intersect the angled edge view of the truncated surface of the cylinder. These intersection points are in turn projected into the development. The intersections between these projections and the vertical lines constructed from the stretch-out line are points along the curve representing the top line of the truncated cylinder. Development of a truncated right circular cylinder Development of a right circular cone To begin this development, use a true-length element of the cone as the radius for an arc and as one side of the development. A true- length element of a right circular cone is the limiting element of the cone in the front view. Draw an arc whose length is equal to the circumference of the base of the cone. Draw another line from the end of the arc to the apex and draw the circular base to complete the development. Question: A cone of base diameter 40 mm and slant height 60 mm is kept on the ground on its base. An AIP inclined at 45° to the HP cuts the cone through the midpoint of the axis. Draw the development. Solution Refer Fig. 16.10. 1. Draw FV and TV as shown. Locate the AIP. 2. Divide the TV into 12 equal parts and draw the corresponding lateral lines (i.e., generators) in FV. Mark points p1’, p2’, p3’, …, p12’ at the points of intersections of the AIP with generators of the cone. 3. Obtain the included angle of the sector. 8 = (20/60) 360 = 120°. 4. Draw O–1 parallel and equal to o’–7. Then draw sector O–1–1– O with O as a centre and included angle 120°. 5. Divide the sector into 12 equal parts (i.e., 10° each). Draw lines O–2, O–3, O– 4, …, O–12. 6. Project points p1’, p2’, p3’, …, p12’ from FV to corresponding lines in development and mark points P1, P2, P3, …, P12 respectively. Join all these points by a smooth freehand curve. Development of Transition pieces used in industry Source : Internet Triangulation development Employed to obtain the development of Transition Pieces Transition pieces are the sheet metal objects used for connecting pipes or openings either of different shapes of cross sections or of same cross sections but not arranged in identical positions. 1. Transition pieces joining a curved cross section to a non curved cross section (e,g, Square to round, hexagon to round , square to ellipse, etc.) 2. Joining two non-curved cross sections (e.g. square to hexagon, square to rectangle, square to square in un- identical positions) 3. Joining only two curve sections (e.g. Circle to oval, circle to an ellipse, etc) In this method, the lateral surfaces of the transition pieces are divided in to a number of triangles. By finding the true lengths of the sides of each triangle, the development is drawn by laying each one of the triangles in their true shapes adjoining each other. Transition pieces joining curved to Non-curved cross sections The lateral surface must be divided into curved and non-curved triangles. Divide the curved cross section into a number of equal parts equal to the number of sides of non-curved cross-section. Division points on the curved cross section are obtained by drawing bisectors of each side of the non-curved cross section. The division points thus obtained when connected to the ends of the respective sides of the non-curved cross-section produces plane triangles In between two plane triangles there lies a curved triangle After dividing in to a number of triangles, the development is drawn by triangulation method. The transition piece consists of 4 plane and 4 curved triangles 1da, 5ab, 9bc, and 13cd are plane triangles and 1a5, 5b9, 9c13 and 13d1 are curved triangles. Since the transition piece is symmetrical about the horizontal axis pq in the top view, the development is drawn only for one half of the transition piece. The front semicircle in the top view is divided into eight equal parts 1,2,3,4, etc. Connect points 1,2,3,4 and 5 to point a. Project points 1,2,3,etc to the front view to 1’,2’,3’. etc. Connect 1’, 2’, 3’ etc to a’ and 5’, 6’, 7’, 8’ 9’ to b’. True length di agD rr a a m w va er n ti d cad l le in v ee X lo Y.pT m he en fir tst triangle to be drawn is 1pa The true length of sides 1p and 1a are found from the true length diagram. To obtain true length of sides 1p and 1a, step off the distances 1p and 1a on the horizontal drawn through X to get the point 1P’ and 1A’. Connect these two points to Y. The length Y-1P’ and Y-1A’ are the true lengths of the sides 1p and 1a respectively. DEVELOPMENT Draw a line 11P = Y-1P’. Draw another line with center 11 and radius Y-1A’. With P as center and radius pa, as measured from the top view, draw an arc to cut the line 11-A to meet at A. With A as center and radius equal to true length of the line 2a (i.e Y-2A’), draw an arc. With 11 as center and radius equal to 1-2 (T.V), draw another arc intersecting the pervious arc at 21. Similarly determine the points 31, 41 and 51. A -11-21-31- 41- 51 is the development of the curved triangle 1-a-5. AB is the true length of the plain triangle a-5-b. Similar procedure is repeated for the other three curved triangles and plain triangles. Square to hexagon transition The transition piece is assumed to cut along PQ. Triangles 1pa and 1a2 and trapezium a23b are obtained. To develop the lateral surface a23b, it is divided into two triangles by connecting either a3 or 2b and completed by triangulation method. True length diagram is drawn and development obtained by the previous method. Transition pieces joining two curved surfaces Draw TV and FV of conical reducing pieces Divide the two circles into twelve equal parts. Connect point 1a, 2b, 3c, etc in the TV and 1’a’, 2’b’,etc in the FV. These lines are called radial lines The radial lines divide the lateral surface into a number of equal quadrilaterals. Their diagonals are connected (dashed lines) forming a number of triangles. The true length diagram are drawn separately for radial and diagonal lines. Conical reducing piece to connect two circular holes of diameters 80 mm and 50 mm. The holes are 90 mm apart and center offset by 15 mm. True length diagram for radial lines: For the radial line 7-g. Draw XX equal to vertical height (90mm). With X as center and radius = 7g (from the top view), draw a horizontal offset line from X (in the true length diagram) to obtain point 71. Join 71 and X, which is the true length of radial line 7g. Similarly we can obtain true lengths for all the radial lines. For drawing convenience, the offset points are drawn on both sides of the line XX Similarly true length diagram for the diagonal lines can be obtained. PRACTICE 2 Engineering Drawing Lecture 15 Isometric Projections The axonometric projection is produced by multiple parallel lines of sight perpendicular to the plane of projection, with the observer at infinity and the object rotated about an axis to produce a pictorial view Axonometric projection - is a parallel projection technique used to create a pictorial drawing of an object by rotating the object on an axis relative to a projection or picture plane. The differences between a multiview drawing and an axonometric drawing are that, in a multiview, only two dimensions of an object are visible in each view and more than one view is required to define the object; whereas, in an axonometric drawing, the object is rotated about an axis to display all three dimensions, and only one view is required. Isometric axes can be positioned in a number of ways to create different views of the same object. Figure A is a regular isometric, in which the viewpoint is looking down on the top of the object. In a regular isometric, the axes at 30° to the horizontal are drawn upward from the horizontal. For the reversed axis isometric, the viewpoint is looking up on the bottom of the object, and the 30° axes are drawn downward from the horizontal. For the long axis isometric, the viewpoint is looking from the right or from the left of the object, and one axis is drawn at 60 ° to the horizontal. ISOMETRIC PROJECTION and ISOMETRIC DRAWING Isometric drawings are almost always preferred over isometric projection for engineering drawings, because they are easier to produce. An isometric drawing is an axonometric pictorial drawing for which the angle between each axis equals 120° and the scale used is full scale. Size comparison of Isometric Drawing and True Isometric Projection Isometric Axonometric Projections An isometric projection is a true representation of the isometric view of an object. An isometric view of an object is created by rotating the object 45o about a vertical axis, then tilting the object (see figure - in this case, a cube) forward until the body diagonal (AB) appears as a point in the front view The angle the cube is tilted forward is 35° 16’. The 3 axes that meet at A, B form equal angles of 120° and are called the isometric axes. Each edge of the cube is parallel to one of the isometric axes. Line parallel to one of the legs of the isometric axis is an isometric line. Planes of the cube faces & all planes parallel to them are isometric planes The forward tilt of the cube causes the edges and planes of the cube to become shortened as it is projected onto the picture plane. The lengths of the projected lines are equal to the cosine of 35° 16’, or 0.81647 times the true length. In other words, the projected lengths are approximately 82% of the true lengths. A drawing produced using a scale of 0.816 is called an isometric projection and is a true representation of the object. However, if the drawing is produced using full scale, it is called an isometric drawing, which is the same proportion as an isometric projection, but is larger by a factor of 1.23 to 1. Isometric scale is produced by positioning a regular scale at 45 ° to the horizontal and projecting lines vertically to a 30° line. In an isometric drawing, true length distances can only be measured along isometric lines, that is, lines that run parallel to any of the isometric axes. Any line that does not run parallel to an isometric axis is called a non-isometric line. Non-isometric lines include inclined and oblique lines and can not be measured directly. Instead they must be created by locating two end points. adjacent isometric axes. relative to isometric axes Planes that are not parallel to any isometric plane are called non-isometric planes (Figure B) Figure B: Non-isometric plane Figure A is an isometric drawing of a cube. The three faces of the isometric cube are isometric planes, because they are parallel to the isometric surfaces formed by any two Figure A: Isometric planes Standards for Hidden Lines, Center Lines and Dimensions In isometric drawings, hidden lines are omitted unless they are absolutely necessary to completely describe the object. Most isometric drawings will not have hidden lines. To avoid using hidden lines, choose the most descriptive viewpoint. However, if an isometric viewpoint cannot be found that clearly depicts all the major features, hidden lines may be used. Centerlines are drawn only for showing symmetry or for dimensioning. Normally, centerlines are not shown, because many isometric drawings are used to communicate to non- technical people and not for engineering purposes. As per the Standards: Dimension lines, extension lines, and lines being dimensioned shall lie in the same plane. All dimensions and notes should be unidirectional, reading from the bottom of the drawing upward and should be located outside the view whenever possible. The texts is read from the bottom, using horizontal guidelines. ISOMETRIC VIEWS OF STANDARD SHAPES Square Consider a square ABCD with a 30 mm side shown in Fig. If the square lies in the vertical plane, it will appear as a rhombus with a 30 mm side in isometric view as shown in Fig. (a) or (b), depending on its orientation, i.e., right-hand vertical face or left-hand vertical face. If the square lies in the horizontal plane (like the top face of a cube), it will appear as in Fig.(c). The sides AB and AD, both, are inclined to the horizontal reference line at 30°. Taken from Dhananjay A Jolhe, Engg. Drawing, MGH Rectangle A rectangle appears as a parallelogram in isometric view. Three versions are possible depending on the orientation of the rectangle, i.e., right-hand vertical face, left-hand vertical face or horizontal face. Taken from Dhananjay A Jolhe, Engg. Drawing, MGH Triangle A triangle of any type can be easily obtained in isometric view as explained below. First enclose the triangle in rectangle ABCD. Obtain parallelogram ABCD for the rectangle as shown in Fig. (a) or (b) or (c). Then locate point 1 in the parallelogram such that C–1 in the parallelogram is equal to C–1 in the rectangle. A–B–1 represents the isometric view of the triangle. Taken from Dhananjay A Jolhe, Engg. Drawing, MGH Pentagon Enclose the given pentagon in a rectangle and obtain the parallelogram as in Fig. 18.9(a) or (b) or (c). Locate points 1, 2, 3, 4 and 5 on the rectangle and mark them on the parallelogram. The distances A–1, B–2, C–3, C–4 and D–5 in isometric drawing are same as the corresponding distances on the pentagon enclosed in the rectangle. Taken from Dhananjay A Jolhe, Engg. Drawing, MGH Circle The isometric view or isometric projection of a circle is an ellipse. It is obtained by using four-centre method explained below. Four-Centre Method : First, enclose the given circle into a square ABCD. Draw rhombus ABCD as an isometric view of the square. Join the farthest corners of the rhombus, i.e., A and C. Obtain midpoints 3 and 4 of sides CD and AD respectively. Locate points 1 and 2 at the intersection of AC with B–3 and B– 4 respectively. Now with 1 as a centre and radius 1–3, draw a small arc 3–5. Draw another arc 4–6 with same radius but 2 as a centre. With B as a centre and radius B–3, draw an arc 3–4. Draw another arc 5–6 with same radius but with D as a centre. Taken from Dhananjay A Jolhe, Engg. Drawing, MGH Any irregular Shape Any irregular shape 1–2–3–4–5–6–7 can be drawn in isometric view as follows: The figure is enclosed in a rectangle first. The parallelogram is obtained in isometric for the rectangle as shown. The isolines B–2, D–2, C–3, E–3, G–4, F–4, H–5, H–6 and A–7 has the same length as in original shape, e.g., B–2 in isometric = B–2 in irregular shape. Taken from Dhananjay A Jolhe, Engg. Drawing, MGH Isometric views for solids The Boxing-in Method The four basic steps for creating an isometric drawing are: Determine the isometric viewpoint that clearly depicts the features of the object, then draw the isometric axes which will produce that view-point. height (H), and depth (D) of the object, such that the object will be totally enclosed in a box. Locate details on the isometric planes. Darken all visible lines, and eliminate hidden lines unless absolutely necessary to describe the object. Sketch from an actual object STEPS 1. Positioning object. 2. Select isometric axis. 3. Sketch enclosing box. 4. Add details. 5. Darken visible lines. Note In isometric sketch/drawing), hidden lines are omitted unless they are absolutely necessary to completely describe the object.Sketch from an actual object
14236	https://brainly.com/question/60246147	[FREE] Using Lagrange multipliers, find the endpoints of the major and minor axes of the ellipse 3x^2 - 2xy + 3y^2 - 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 +63,8k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +14,6k Ace exams faster, with practice that adapts to you Practice Worksheets +8k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified Using Lagrange multipliers, find the endpoints of the major and minor axes of the ellipse 3 x 2−2 x y+3 y 2=4. 1 See answer Explain with Learning Companion NEW Asked by manuel347163 • 04/11/2025 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 1182974143 people 1182M 0.0 0 Upload your school material for a more relevant answer We wish to find the endpoints of the major and minor axes of the ellipse 3 x 2−2 x y+3 y 2=4. One effective method is to maximize and minimize the distance from the origin subject to the ellipse constraint. Since the distance from the origin is given by d=x 2+y 2, it is equivalent (and simpler) to work with the squared distance f(x,y)=x 2+y 2. We then introduce the constraint function g(x,y)=3 x 2−2 x y+3 y 2−4=0. The method of Lagrange multipliers tells us that any extreme point must satisfy ∇f(x,y)=λ∇g(x,y), together with the constraint g(x,y)=0, where λ is the Lagrange multiplier. ────────────────────────────── Step 1. Compute the Gradients The gradient of f(x,y) is ∇f(x,y)=(∂x∂f,∂y∂f)=(2 x,2 y). The gradient of g(x,y) is ∇g(x,y)=(∂x∂g,∂y∂g)=(6 x−2 y,−2 x+6 y). ────────────────────────────── Step 2. Set Up the Equations The Lagrange multiplier conditions are: 2 x=λ(6 x−2 y)and 2 y=λ(−2 x+6 y). Along with the constraint 3 x 2−2 x y+3 y 2=4. ────────────────────────────── Step 3. Solve the Equations A direct way to seek a relation between x and y is to consider nontrivial solutions (i.e. not both x=0 and y=0). Notice that the two equations are symmetric, and it is natural to test the possibilities y=x and y=−x. ––––––– Case 1. Assume y=x. Substitute y=x into the first equation: 2 x=λ(6 x−2 x)=λ(4 x). For a nonzero x, we can cancel x to get: 2=4 λ⟹λ=2 1. Now, substitute y=x into the constraint: 3 x 2−2 x 2+3 x 2=4 x 2=4. Thus, x 2=1, or x=±1. Since y=x, the corresponding points are (1,1)and(−1,−1). The squared distance for these points is f(x,y)=1 2+1 2=2. ––––––– Case 2. Assume y=−x. Substitute y=−x into the first equation: 2 x=λ(6 x−2(−x))=λ(6 x+2 x)=λ(8 x). Again, cancelling x (with x=0) we find: 2=8 λ⟹λ=4 1. Now, substitute y=−x into the constraint: 3 x 2−2 x(−x)+3(−x)2=3 x 2+2 x 2+3 x 2=8 x 2=4. Thus, x 2=2 1, or x=±21. Since y=−x, the points are (21,−21)and(−21,21). The squared distance for these points is f(x,y)=(21)2+(−21)2=2 1+2 1=1. ────────────────────────────── Step 4. Determine Major and Minor Axes Since the distance from the origin (or f(x,y)) is larger for the points where y=x, these points are the endpoints of the major axis. Therefore, the endpoints of the major axis are (1,1)and(−1,−1). Likewise, the endpoints of the minor axis are the points where y=−x, which are (21,−21)and(−21,21). ────────────────────────────── Conclusion The ellipse 3 x 2−2 x y+3 y 2=4 has: Major axis endpoints: (1,1)and(−1,−1), Minor axis endpoints: (21,−21)and(−21,21). These are the required ends of the major and minor axes of the ellipse. Answered by GinnyAnswer •8M answers•1.2B people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) 0.0 0 Upload your school material for a more relevant answer Using Lagrange multipliers, we found the endpoints of the major and minor axes of the ellipse defined by the equation 3 x 2−2 x y+3 y 2=4. The major axis endpoints are (1,1) and (−1,−1), while the minor axis endpoints are (21,−21) and (−21,21). Explanation To find the endpoints of the major and minor axes of the ellipse given by the equation 3 x 2−2 x y+3 y 2=4 we will use the method of Lagrange multipliers. Step 1: Define the Objective Function and Constraint The first step is to establish our functions. We want to maximize and minimize the distance from the origin, given by the formula: d=x 2+y 2 However, to simplify our calculations, we will work with the squared distance: f(x,y)=x 2+y 2 This is our objective function that we want to maximize/minimize. Next, we define the constraint function based on our ellipse equation: g(x,y)=3 x 2−2 x y+3 y 2−4=0 Step 2: Calculate the Gradients To apply Lagrange multipliers, we compute the gradients of both functions: For the function f: ∇f(x,y)=(∂x∂f,∂y∂f)=(2 x,2 y) For the function g: ∇g(x,y)=(∂x∂g,∂y∂g)=(6 x−2 y,−2 x+6 y) Step 3: Set Up the Lagrange Equations The method of Lagrange multipliers states that we have: ∇f(x,y)=λ∇g(x,y) This leads to the following system of equations: 2 x=λ(6 x−2 y)2 y=λ(−2 x+6 y) Along with the constraint: 3 x 2−2 x y+3 y 2=4 Step 4: Solve the Equations These equations allow us to relate x and y. To explore possible solutions, we can consider cases based on symmetry in the equations: Case 1: Assume y=x Substituting y=x leads us to: 2 x=λ(6 x−2 x)=4 λ x Assuming x=0, we can cancel x and obtain: 2=4 λ⇒λ=2 1 Now substituting this back into the constraint gives: 3 x 2−2 x 2+3 x 2=4⇒4 x 2=4⇒x 2=1⇒x=±1 Thus, for this case, the points are: (1,1)and(−1,−1) Case 2: Assume y=−x Substituting y=−x yields: 2 x=λ(6 x−2(−x))=λ(8 x) Again assuming x=0 leads us to: 2=8 λ⇒λ=4 1 Substituting back gives: 3 x 2+2 x 2+3 x 2=4⇒8 x 2=4⇒x 2=2 1⇒x=±21 Thus, the corresponding points are: (21,−21)and(−21,21) Step 5: Identify Major and Minor Axes To determine which points belong to the major and minor axes, we compare the squared distances from the origin. The squared distances for the points we found are: For (1,1) and (−1,−1): f(1,1)=2 For (21,−21) and (−21,21): f(21,−21)=1 Since the maximum distance corresponds to the points (1,1) and (-1,-1), these points are the endpoints of the major axis. The endpoints of the minor axis are the points (21,−21) and (−21,21). Conclusion The endpoints of the major and minor axes of the ellipse 3 x 2−2 x y+3 y 2=4 are: Major axis endpoints: (1,1)and(−1,−1) Minor axis endpoints: (21,−21)and(−21,21) Examples & Evidence For example, you can visualize the major axis points (1, 1) and (-1, -1) lying diagonally on the graph, indicating that they are further from the origin compared to the minor axis points, which are closer and positioned along the axes. The calculations and processes used conform to the principles of finding extrema using Lagrange multipliers and are supported by standard mathematical methods for analyzing conic sections. Thanks 0 0.0 (0 votes) Advertisement manuel347163 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer Use Lagrange Multipliers to find the largest area of rectangle with sides parallel to the axes which can be inscribed in the ellipse (the corners are on the ellipse) x² + 4y² = 4. Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. New questions in Mathematics Maggie solved a math problem and came up with the answer 28. Which of the following expressions did she solve? A. 12+3⋅10÷2 B. 15(5−3)−12 C. 2 2(6)+8−6 D. 22+(6⋅3 3)÷27 Write your answer as a simplified improper fraction. 5 4 3+2 5 2= Factorise fully the following expressions. (a) x 2+2 x (b) 10 x 2+2 x (c) 3 a 2−5 a (d) 14 x y−21 x You want to buy 12 apples for $0.45 each and a box of cereal that costs $3.35. What is the total cost of your purchase? Consider the function f(x)={sin(x π)0x=0 x=0 on the interval [−2,2].i. Does the premise of the IVT hold? In other words, is f continuous on [−2,2]?ii. Does the conclusion of the IVT hold? In other words, for every real number M between f(−2) and f(2) can you find c∈(−2,2) such that f(c)=M? 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14237	https://www.floridabar.org/the-florida-bar-journal/time-to-end-lets-pretend/	The Florida Bar Rules, Ethics & Professionalism Login Find a Lawyer Florida Bar Journal Home Journal & News Time to End Let’s Pretend Vol. 71, No. 5 May 1997 Pg 97 Douglas C. Kaplan Real Property, Probate and Trust Law Several short years ago, the Florida Legislature designed and enacted “transaction brokerage” as a legal relationship between the real estate licensee and the client.1 The concept of transaction brokerage was created to liberate the real estate licensee and the general public from the licensee’s conflicting fiduciary relationships and responsibilities and the parties’ resulting exposure to litigation. The transaction broker functioned in the nature of an independent contractor representing neither the buyer nor the seller.2 At this writing, a proposed amendment to F.S. Chapter 475 entitled “an Act Relating to Real Estate Transactions” was being submitted at the 1997 session of the Florida Legislature. The proposed act grafts on to the traditional definition of transaction brokerage a provision that makes the transaction broker a “limited representative” for a buyer or a seller, or both. The proposed act provides, however, that the transaction broker represents no one in a “fiduciary capacity.”3 The proposed act flies in the face of the traditional role of transaction brokerage and, in reality, creates the following two kinds of agency: 1) a conventional single agency of a buyer or a seller, and 2) an agency-clone without fiduciary responsibility, created out of transaction brokerage. Ironically, the word “represent,” the very word used in the proposed act to polarize the formerly neutral concept of transaction broker, converts the transaction broker to an agent look-alike. Indeed, the word “represent” is the effective equivalent of agency. One who represents is a “representative.”4 Thus, a single word changes the entire concept of transaction brokerage from independence to a quasi-principal and agent relationship. Since a transaction broker has no fiduciary obligations to a client under the proposed act, will prospective buyers who retain transaction brokers to “represent” them understand the ramifications? That is, will prospective buyers know that such representation does not include (in fact, it expressly excludes) the following items defined within the act as fiduciary obligations: trust and confidence, loyalty, complete confidentiality, obedience, full disclosure, and diligence? Will clients understand that, in hiring transaction brokers to “represent” them in a limited capacity, that inadvertently they may be exposing themselves to the traditional dangers involved in a conventional agency relationship? Those dangers include: 1) liability for the brokers’ false or erroneous representations to third parties; 2) the clients’ tort liability arising from brokers’ negligence; 3) the clients’ contractual commitments from brokers’ agreements; and 4) the clients’ responsibility for brokers’ receipt of notice and delivery. In exchange for that exposure, the clients would receive: “limited representation”— whatever that means. Curiously, the proposed act defines “limited representation” by what it is not, i.e. : It is not fiduciary obligations; it is not full confidentiality; it is not full disclosure. What the proposed act fails to tell us is what limited representation is. In an unsuccessful effort to fill that void, the act merely lists a litany of obligations: honesty; accounting; use of skill, care, and diligence; and Johnson v. Davis disclosures—all of which are universal obligations of all licensees. One can conclude only that there is no clear definition of a limited representative’s duties and responsibilities. In the context of the transaction broker, limited representation appears to be little more than mere empty words to suggest a representative service that does not exist, and can never be rendered. Origin of Agency Brokerage In only 10 years, the real estate brokerage industry has experienced a revelation that might have taken a lifetime. It did not go willingly in search of its soul but, like Lot’s wife, was tantalized by curiosity to look, thus to be changed forever. When the mist cleared, what was exposed was a fundamental conflict in the DNA of agency brokerage, i.e. , fiduciary duties to clients conflicted with transaction-generated commissions. Agency and subagency ruled mightily in the post-World War II real estate brokerage industry. The notion was that real estate broker agency relationships came to this country with the Magna Carta or, at least, resided in some cherished historical document in the Washington Bureau of Archives. Or, so one would think by the refrain, “Haven’t we always been agents?” Certainly, the law of agency had sound legal history in the English common law. When third parties were involved, subagency appeared as a natural and logical extension of the agency relationship. Although agency is still used today in many areas of commerce, the justified demise of the bewildering concept of subagency is imminent. How did agency become the accepted legal relationship for real estate brokers? One view urges that agency was the device used by the civil courts to impose a fiduciary obligation on the broker in the absence of rules or constraints. Another suggests that the agency relationship with its fiduciary obligation engendered a better quality professional fee. Or simply, sellers may have believed that making the broker his or her agent would place the broker on the seller’s team. Whatever the source of the agency connection for the real estate broker, the system seemed to work. Decades of real estate brokers earned their livelihoods using the ill-fitting law of agency with all its fiduciary responsibilities as the legal mucilage that bound them to their clients. Agency and Transaction-Generated Fees Other professionals, such as doctors and architects, get paid for their time, effort, and service whether or not the clients utilize or are ultimately benefitted by such services. Imagine doctors, who only get paid for cures; or architects, who get paid only when and if their plans result in actual construction. How the agency concept of “I will be your representative, your advisor, your expert, your confidante” became mingled—no mangled—with “pay me only when, and if, I get you to sign a contract” (commission-based compensation) is puzzling. The two together make an explosive compound, indeed, for the concept foolishly and unnecessarily pits the professional broker against the broker’s own instinct for economic survival. What happened to provoke change was subtle, sophisticated, but nevertheless significant. The real estate industry began to view itself seriously as a profession. A number of factors distinguish an occupation from a profession. far, the most important are an enforced code of ethics and standards of quality, service, and honesty. Although nonprofessional businesses also must function honestly to survive, generally no structure exists within the business world to enforce codes of ethics or standards of honesty. Discovery of a Conflict The medical profession publishes a Physicians Desk Reference, a text which identifies the ingredients of modern medicines and describes how they work. The Physicians Desk Reference identifies what ingredients or medications should not be taken with others since their interaction can be harmful, or even fatal. Physicians are charged with the knowledge and responsibility of what medications are dangerous when administered together. The real estate brokerage industry has long needed to look at its two most important elements of the service it offers to see if they can both interact and co-exist in a professional setting. They are agency and transaction-based compensation for real estate brokerage services. • Agency Agency is a term (which by definition includes a “ fiduciary ” obligation) involving trust, fidelity, full disclosure, and confidentiality. A fiduciary relationship requires that fiduciaries may not take selfish advantage of their trust relationship or deal with the subject matter of that trust in such a way as to benefit themselves or prejudice their principals. • Transaction-Based Compensation for Real Estate Brokerage Services Transaction-based compensation for real estate brokerage services, in contrast, means that payment occurs only if the transaction or deal is produced, i.e., no compensation is due for time, effort, or facility without a signed contract. Skinning the Agency Cat Sadly, there is no lack of misdirected ingenuity within the real estate brokerage industry. So intent are certain factions on preserving the fiction of fiduciary representation that they have moved some state legislatures to enact potentially misdirected laws. These laws can be grouped as 1) abrogation of common law agency for real estate agents; 2) designated agency; and 3) limited agency, limited disclosure, limited representation, and limited honesty. • Abrogation of Common Law Agency for Real Estate Agents For the most part, this design is a transparent effort to change the rules of agency by removing most of the fiduciary obligations that make agents vulnerable yet, still allowing them to call themselves agents. It is the special icon of those who would like to practice dual agency without legal exposure. How are the consumers in such a state going to distinguish between the diluted responsibilities of real estate agents and the obligations of all other kinds of agents with whom they deal? They can not, unless they are accompanied by a lawyer and carry with them a copy of the state statutes. And how will such a scenario play to unsuspecting consumers accustomed to traditional fiduciary roles when they buy or sell in the growing interstate, Internet marketplace? • Designated Agency Some legislatures have enacted a unique program sometimes called “designated agency.” Under that scenario, such as in Illinois, a single real estate broker can identify a person in his or her office who can act as a representative of the buyer and a different person who can act as a representative of the seller in the same transaction. Presumptively, in some way, this separates the confidentiality of each salesperson; the obligations to disclose of each; and the fiduciary duties of each. Designated agency has been enacted in states where salespeople do not have their own independent identity sufficient to advertise in their own names; list property on the multiple listing service; create their own depository trust accounts; collect real estate commissions for themselves; take listings for themselves; and maintain their own offices. In all of the foregoing instances, the salespeople function under the direction and supervision of a single real estate broker. Does that broker have two heads, neither of which talks to the other, or shares thoughts with the other? Only in a world of fantasy would salespeople within one office not share communications and disclosures with each other and with the principal broker. This agency design appears to be little more than a house of cards built upon a foundation of dual agency. • Limited Agency; Limited Disclosure; Limited Representation; Limited Honesty At no time in the annals of Anglo-American jurisprudence has the word “limited” received such unlimited usage as in the real estate industry during this decade. In recent years, we have seen the advent of “limited agency,” “limited disclosure,” and “limited representation.” Limited to what? Limited to how much? Why limited? Essentially, the word “limited” has been used in an effort to suggest the existence of an agency fiduciary relationship, while at the same time attempting to limit the exposure of the fiduciary. The word “limited” is the current weasel-word displaying, merely, unlimited imagination and limited candor. If an agent is going to serve as someone’s fiduciary and valued representative, how can the agent quantify or limit the sincerity, commitment, and dedication of such representation? Who weighs the amount of representation, disclosure, and fiduciary obligation? This posturing is commercially unnecessary, dangerous, and counterproductive. It exists for one purpose, i.e., to try and justify a commercial licensee’s claim that he or she represents the buyer and the seller and, in certain cases, to collect a commission from both. Effect of Proposed Florida Act Is it mere coincidence that the same proposed act (that has been before the 1997 Florida Legislature), which abrogates dual agency, also changes the genetic code of independent transaction broker to create an agent-like limited representative? Why would anyone tamper with an existing and working model of an independent transaction broker when apparently there is little to gain and much to lose? The answer may lie in the new transaction broker’s capacity to offer, in the same real estate transaction, limited representation to both buyer and seller without having to comply with standards of fiduciary obligations. What really is created by making the transaction broker a limited representative is dual agency in transaction-broker clothing. Under existing law, even a dual agent is bound by significant fiduciary obligations to the clients. A transaction broker representing both buyer and seller will be little more than a dual agent functioning with no common law agency protections to the principals. Given the prior opposition of the Department of Professional Regulation and the Florida Real Estate Commission to the abrogation of common law agency, it is hard to understand how this ill-conceived and destructive proposal could have gotten this far. The proposed change to transaction brokerage is a minefield for Florida real estate licensees. The rights of irate clients “represented” by transaction brokers will not be asserted before a legislative committee, state agencies, or a voluntary association of brokers. They will be pressed in county courthouses throughout the state. Judges will not measure with a micrometer the anguished cries of the real estate licensee who defends by saying, “but I was only a ‘limited’ representative.” Plaintiffs will cite the licensee’s assurances that he or she could and would represent the plaintiff and examine whether the licensee did, in fact, do so. Plaintiff’s counsel will point to the fact that by all historical standards, a representative is another expression for agent and that, if it “walks like a duck, then. . . . ” Instead of insulating transaction brokers from litigation and liability, the proposed change blurs their roles, denies their independence, and assures their involvement in the controversy. Florida was, and still is, on the cutting-edge of developments in the real estate brokerage industry. Some states—including Alabama, Colorado, Georgia, Idaho, Kentucky, Michigan, Montana, New Jersey, New Mexico, Tennessee, Virginia, and Wyoming—now offer transaction brokerage, or other nonagency options. Other states—Alaska, Kansas, Maryland, Mississippi, Missouri, Oklahoma, and Pennsylvania—all have new transaction brokerage legislation under consideration. The scuttlebutt is that the proposed act is a compromise. One would do well to keep in mind the sage words of Tryon Edwards, the American theologian who stated that “Compromise is but the sacrifice of one right or good in the hope of retaining another—too often ending in the loss of both.” If the state of Florida contaminates transaction brokerage by mixing it with the authority to represent clients in the same transaction, it will be a step backward in what has previously been Florida’s progressive move toward the future in real estate brokerage.q 1 F la. Stat. §475.01(1)(k). 2 See the excellent article Real Estate Brokerage and Agency: Florida Contemplates Change , by Professor Glenn H. Boggs of Florida State University College of Business, 71 Fla. B.J. 46 (Feb. 1997). 3 Florida Senate Bill No. SB 0082 and Florida House of Representatives Bill No. HB 0339, 1997 Legislative Session 4 Definitions: 1) Fla. Stat. §671.201(35), part of the Uniform Commercial Code, expressly recites that the word “‘representative’ includes an agent. . . . ” 2) Black’s Law Dictionary, Sixth Edition, page 1302, cites Sunset Mill & Grain Co. v. Anderson , 39 Cal.2d 733, 249 P.2d 24, 27, for the proposition that “one who represents others. . . is interchangeable with ‘agent’” and its definition of “Representative” directs the reader to: “See also Agent.” 3) The Random House Dictionary of the English Language, Second Edition, page 1635, defines “representative” as “an agent or deputy: a legal repre sentative.” Douglas C. Kaplan is the board attorney for the Realtor Association of Hollywood-South Broward, Inc., sponsor of the Independent Real Estate Broker initiative. He is a partner in the law firm of Kaplan, Jaffe and Gates, P.A., in Hollywood. This column is submitted on behalf of the Real Property, Probate and Trust Law Section, Robert W. Goldman, chair, and David H. Simmons and Brian Felcoski, editors. Real Property, Probate and Trust Law Search Journal Archives Browse by Issue Latest Digital Edition Book Reviews Theodore Boone: Kid Lawyer More Book Reviews Columns Administrative Law Animal Law Appellate Practice Aviation Law Business Law City, County and Local Government Criminal Law Education Law Elder Law Entertainment, Arts and Sports Law Environmental & Land Use Law Family Law Government Lawyer Health Law International Law Labor and Employment Law Out of State Division Prepaid Legal Services Public Interest Law Real Property, Probate and Trust Law Solo and Small Firm Trial Lawyers Workers' Compensation Young Lawyers Division
14238	https://byjus.com/maths/pie-chart/	A pie chart is a type of graph that represents the data in the circular graph. The slices of pie show the relative size of the data, and it is a type of pictorial representation of data. A pie chart requires a list of categorical variables and numerical variables. Here, the term “pie” represents the whole, and the “slices” represent the parts of the whole. \| \| \| Table of Contents: Definition Formula How to Create Pie Chart Pie Chart Maker How to Solve Pie Chart Examples Uses Advantages Disadvantages Practice Problem FAQs \| What is a Pie Chart? The “pie chart” is also known as a “circle chart”, dividing the circular statistical graphic into sectors or sections to illustrate the numerical problems. Each sector denotes a proportionate part of the whole. To find out the composition of something, Pie-chart works the best at that time. In most cases, pie charts replace other graphs like the bar graph, line plots, histograms, etc. Formula The pie chart is an important type of data representation. It contains different segments and sectors in which each segment and sector of a pie chart forms a specific portion of the total(percentage). The sum of all the data is equal to 360°. The total value of the pie is always 100%. To work out with the percentage for a pie chart, follow the steps given below: Categorize the data Calculate the total Divide the categories Convert into percentages Finally, calculate the degrees Therefore, the pie chart formula is given as (Given Data/Total value of Data) × 360° Note: It is not mandatory to convert the given data into percentages until it is specified. We can directly calculate the degrees for given data values and draw the pie chart accordingly. How to Create a Pie Chart? Imagine a teacher surveys her class on the basis of favourite Sports of students: \| \| \| \| \| \| --- --- \| Football \| Hockey \| Cricket \| Basketball \| Badminton \| \| 10 \| 5 \| 5 \| 10 \| 10 \| The data above can be represented by a pie chart as following and by using the circle graph formula, i.e. the pie chart formula given below. It makes the size of the portion easy to understand. Step 1: First, Enter the data into the table. \| \| \| \| \| \| --- --- \| Football \| Hockey \| Cricket \| Basketball \| Badminton \| \| 10 \| 5 \| 5 \| 10 \| 10 \| Step 2: Add all the values in the table to get the total. I.e. Total students are 40 in this case. Step 3: Next, divide each value by the total and multiply by 100 to get a per cent: \| \| \| \| \| \| --- --- \| Football \| Hockey \| Cricket \| Basketball \| Badminton \| \| (10/40) × 100 \| (5/ 40) × 100 \| (5/40) ×100 =12.5% \| (10/ 40) ×100 =25% \| (10/40)× 100 =25% \| Step 4: Next to know how many degrees for each “pie sector” we need, we will take a full circle of 360° and follow the calculations below: The central angle of each component = (Value of each component/sum of values of all the components)✕360° \| \| \| \| \| \| --- --- \| Football \| Hockey \| Cricket \| Basketball \| Badminton \| \| (10/ 40)× 360° \| (5 / 40) × 360° \| (5/40) × 360° =45° \| (10/ 40)× 360° =90° \| (10/ 40) × 360° =90° \| Now you can draw a pie chart. Step 5: Draw a circle and use the protractor to measure the degree of each sector. Let us take an example for a pie chart with an explanation here to understand the concept in a better way. Question: The percentages of various cops cultivated in a village of particular distinct are given in the following table. \| \| \| \| \| \| \| \| --- --- --- \| Items \| Wheat \| Pulses \| Jowar \| Groundnuts \| Vegetables \| Total \| \| Percentage of cops \| 125/3 \| 125/6 \| 25/2 \| 50/3 \| 25/3 \| 100 \| Represent this information using a pie-chart. Solution: The central angle = (component value/100) × 360° The central angle for each category is calculated as follows \| \| \| \| --- \| Items \| Percentage of cops \| Central angle \| \| Wheat \| 125/3 \| [(125/3)/100] × 360° = 150° \| \| Pulses \| 125/6 \| [(125/6)/100] × 360° = 75° \| \| Jowar \| 25/2 \| [(25/2)/100] × 360° = 45° \| \| Groundnuts \| 50/3 \| [(50/3)/100] × 360° = 60° \| \| Vegetables \| 25/3 \| [(25/3)/100] × 360° = 30° \| \| Total \| 100 \| 360° \| Now, the pie-chart can be constructed by using the given data. Steps to construct: Step 1: Draw the circle of an appropriate radius. Step 2: Draw a vertical radius anywhere inside the circle. Step 3: Choose the largest central angle. Construct a sector of a central angle, whose one radius coincides with the radius drawn in step 2, and the other radius is in the clockwise direction to the vertical radius. Step 4: Construct other sectors representing other values in the clockwise direction in descending order of magnitudes of their central angles. Step 5: Shade the sectors obtained by different colours and label them as shown in the figure below. Pie Chart Maker Till now you understood how to draw a pie chart for the given data using geometric tools. In this section, you will know how to make the pie chart using an online tool. People often use a graphing feature in excel sheets to get the desired pie chart. However, we have provided an online pie chart maker. Click here to get the pie chart calculator. How to Solve Pie Chart Questions? In this section, you will learn how to solve or interpret the pie chart to get the original values. For this, we need to check whether the given chart is given in percentages, degrees or without any value. Based on this information, we can solve the questions related to pie charts. Let’s have a look at the solved example to understand this thoroughly. Question: The pie-chart shows the marks obtained by a student in an examination. If the student secures 440 marks in all, calculate the marks in each of the given subjects. Solution: The given pie chart shows the marks obtained in the form of degrees. Given, total marks obtained = 440 i.e. 360 degrees = 440 marks Now, we can calculate the marks obtained in each subject as follows. Marks secured in mathematics = (central angle of maths/ 360°) × Total score secured = (108°/ 360°) × 440 = 132 marks Marks secured in science = (central angle of science / 360°) × Total score secured = (81°/ 360°) × 440 = 99 marks Marks secured in English = (central angle of English/ 360°) × Total score secured = (72°/ 360°) × 440 = 88 marks Marks secured in Hindi = (central angle of Hindi / 360°) × Total score secured = (54°/ 360°) × 440 = 66 marks Marks secured in social science = (central angle of social science / 360°) × Total score secured = (45°/ 360°) × 440 = 55 marks This can be tabulated as: \| \| \| \| \| \| \| \| --- --- --- \| Subject \| Mathematics \| Science \| English \| Hindi \| Social science \| Total \| \| Marks \| 132 \| 99 \| 88 \| 66 \| 55 \| 440 \| Examples A pie chart can be used to represent the relative size of a variety of data such as: The type of houses (1bhk, 2bhk, 3bhk, etc.) people have Types of 2 wheelers or 4 wheelers people have Number of customers a retail market has in all weekdays Weights of students in a class Types of cuisine liked by different people in an event Monthly expenditure of a family, etc. Uses of Pie Chart Within a business, it is used to compare areas of growth, such as turnover, profit and exposure. To represent categorical data. To show the performance of a student in a test, etc. Also, check some important topics here: \| \| \| Graphical Representation Types of Graphs Bar Graph Linear Graph Histogram Box and Whisker Plot \| Advantages The picture is simple and easy-to-understand Data can be represented visually as a fractional part of a whole It helps in providing an effective communication tool for the even uninformed audience Provides a data comparison for the audience at a glance to give an immediate analysis or to quickly understand information No need for readers to examine or measure underlying numbers themselves, which can be removed by using this chart To emphasize a few points you want to make, you can manipulate pieces of data in the pie chart Disadvantages It becomes less effective if there are too many pieces of data to use If there are too many pieces of data. Even if you add data labels and numbers may not help here, they themselves may become crowded and hard to read As this chart only represents one data set, you need a series to compare multiple sets This may make it more difficult for readers when it comes to analyze and assimilate information quickly You can practice another pie chart question for Class 8, given below: Practice Problem Question: Priya lists down her monthly expenditure as follows: \| \| \| --- \| \| Expenditure \| Amount \| \| Rent \| 4000 \| \| Food \| 5400 \| \| Clothing \| 2800 \| \| Savings \| 400 \| Draw a pie chart for her monthly expenses. You might also like to practice some more Mathematical topics. Download BYJU’S- The Learning App for a variety of topics and concepts to learn. Frequently Asked Questions – FAQs Q1 What is a pie chart? A pie chart is a pictorial representation of data. The slices of pie here shows the relative sizes of data. The same data is represented in different sizes with the help of pie charts. Q2 Why do we use pie charts? Pie charts are used to represent the proportional data or relative data in a single chart. The concept of pie slices is used to show the percentage of a particular data from the whole pie. Q3 How to calculate the percentage of data in the pie chart? Measure the angle of each slice of the pie chart and divide by 360 degrees. Now multiply the value by 100. The percentage of particular data will be calculated. Q4 How to find the total number of pieces of data in a slice of a pie chart? To find the total number of pieces of data in a slice of a pie chart, multiply the slice percentage with the total number of data set and then divide by 100.For example, a slice of the pie chart is equal to 60% and the pie chart contains a total data set of 150. Then, the value of 60% of pie slice is: (60×150)/100 = 90. Q5 What are the examples of a pie chart? There are many real-life examples of pie charts, such as:Representation of marks obtained by students in a classRepresentation of kinds of cars sold in a monthTo show the type of food liked by people in a room Test your Knowledge on Pie Chart Q5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” buttonCheck your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Maths related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted Login To View Results Did not receive OTP? Request OTP on Login To View Results Comments Leave a Comment Cancel reply Riya January 14, 2020 at 12:31 pm very helpful example. Reply - Ola April 26, 2020 at 8:51 pm Thank you so much for your support Reply - Gigie December 10, 2020 at 10:49 pm Wow one of the clearest online math examples Reply Register with BYJU'S & Download Free PDFs Register with BYJU'S & Watch Live Videos
14239	https://quillbot.com/blog/commonly-confused-words/cannot-or-can-not/	Cannot or Can Not \| Difference, Meaning & Spelling Cannot (one word) is the negative form of the modal verb “can.” Spelling it as “can not” (two words) is incorrect. Occasionally, however, you need to use the verb “can” followed by “not” as part of a subsequent phrase (e.g., when using the construction “not only… but also”). So, it’s sometimes correct to write “can” directly followed by “not.” Can vs cannot examples \| Cannot in a sentence \| Can not in a sentence \| \| I cannot play the piano. I can not play the piano. \| The talented 10-year-old can not only sing but also play the piano. The talented 10-year-old cannot only sing but also play the piano \| \| You cannot force me to go to the party. You can not force me to go to the party. \| Nobody’s forcing you; you can not go if you want to. Nobody’s forcing you; you cannot go if you want to. \| Free Grammar Checker Table of contents Can’t or cannot Frequently asked questions about can not vs cannot Can’t or cannot Can’t is the contraction (short form) of cannot. Contractions are shortened forms of word combinations frequently used in speech. For instance, “it’s” is the contraction of “it is,” “wanna” is the contraction of “want to,” and “can’t” is the contraction of “cannot.” It’s generally best to avoid contractions in formal writing (e.g., academic writing). Can’t vs cannot examples \| Can’t in a sentence \| Cannot in a sentence \| \| I can’t find my keys anywhere. \| The management cannot accept responsibility for loss or damage of personal items. \| \| Can’t you hurry up? We’re going to be late! \| However, this theory cannot fully explain the phenomenon in all its complexity. \| Frequently asked questions about can not vs cannot How do you spell can’t? : Can’t (the contraction of “cannot”) is spelled with an apostrophe between the “n” and the “t” (i.e., “cant” is a common misspelling of “can’t”). Along the same lines, confusion over can not or cannot leads to the common misspelling of “cannot” as “can not” (two words). The word “cant” (with no apostrophe, and pronounced so it rhymes with “ant”) is an uncountable noun referring to statements that are not genuinely believed by the person who makes them but said because they are the normal, accepted thing to say in the given situation (e.g., “His speech was filled with empty platitudes and cant”). You can use QuillBot’s free Grammar Checker to help you avoid misspellings and typos such as “cant” for “can’t.” Is can’t a word? : Yes, can’t is the contraction of “cannot” (i.e., the negative form of the modal verb “can”). It’s generally best to use “cannot” instead of “can’t” in formal writing. Along the same lines, confusion over cannot vs can not sometimes leads to the misspelling of “cannot” as “can not” (two words instead of one). “Cant” (with no apostrophe) is a noun referring to statements that the speaker doesn’t really believe in, often made because they are considered the usual, accepted thing to say in the given situation (e.g., “It was nothing more than cant; stock phrases and jargon that told us nothing new”). You can use QuillBot’s free Grammar Checker to help you avoid missing out apostrophes in contractions like “can’t.” What type of word is can’t? : Can’t is a contraction. It is the contraction of “cannot” (i.e., the negative form of the modal verb “can”). Other commonly used contractions include “I’m,” “we’ve,” and “mustn’t.” These short forms are very common in everyday speech, but they are generally avoided in formal writing (e.g., academic writing). It’s important to note that “cannot” is not a contraction, and cannot and can not are not interchangeable. The correct negative form of “can” is “cannot” (written as one word). Use QuillBot’s free Grammar Checker to help you use contractions correctly in your writing. Is cannot a contraction? : Cannot is not a contraction. It is the full negative form of the modal verb “can” (e.g., “I cannot attend the meeting today”). It’s important to note that cannot and can not are not interchangeable. A contraction is a short form of word combinations we commonly use in speech. For instance, “I’m” is the contraction of “I am,” “let’s” is the contraction of “let us,” and “can’t” is the contraction of “cannot.” Most contractions have an apostrophe where a part of the full form is left out, but some don’t (e.g., “gonna” for “going to). We don’t generally use contractions in formal writing (e.g., academic writing). You can use QuillBot’s free Grammar Checker to help you use contractions appropriately in your writing. Cite this Quillbot article We encourage the use of reliable sources in all types of writing. You can copy and paste the citation or click the "Cite this article" button to automatically add it to our free Citation Generator. Is this article helpful? 0 You have already voted. Thanks :-) Your vote is saved :-) Processing your vote... Tom Challenger, BA Tom holds a teaching diploma and is an experienced English language teacher, teacher trainer, and translator. He has taught university courses and worked as a teacher trainer on Cambridge CELTA courses. Do it all with QuillBot Writing Tools AI Generator Tools PDF Tools Other Tools AI Humanizer Paraphraser Grammar Checker AI Detector Plagiarism Checker Translate Summarizer Sentence Rewriter Rewording Tool Word Counter Paragraph Rewriter AI Proofreader Spell Checker Punctuation Checker Essay Checker Citation Generator APA Citation Generator MLA Citation Generator Chicago Citation Generator AI Chat Ask AI AI Paragraph Generator AI Story Generator AI Email Writer AI Text Generator AI Writer Cover Letter Generator AI Acronym Generator AI Title Generator Hashtag Generator AI Lyric Generator AI Product Name Generator AI Business Name Generator AI Meta Description Generator AI Instagram Caption Generator AI Product Title Generator AI Caption Generator AI Job Description Generator AI Article Rewriter Chat PDF PDF Editor Speech To Text QR Code Generator Invisible Character Text to Speech View all Others also liked #### Everyone vs Every One \| Difference & Examples Everyone (one word) means “everybody.” It does not mean the same as every one (two words). The phrase “every one of” means “all of.” #### Into vs In To \| Examples, Definition & Differences “Into” is a preposition used to describe insertion, collision, transformation, or entry. “In to” is a combination of the two prepositions “in” and “to… #### There, Their, They’re \| Difference, Meaning & Examples “There” is often used with “is”/“are” to state that something exists; “their” is a possessive adjective; “they’re” is short for “they are.”
14240	https://www.youtube.com/watch?v=ixAJrLxuynk	Atomic Number for Protium, Deuterium, and Tritium Wayne Breslyn (Dr. B.) 889000 subscribers 195 likes Description 15316 views Posted: 15 Jul 2022 The Atomic Number for an element is equal to the number of protons for that element. Hydrogen, and all of its isotopes, have only one proton. Therefore, the atomic number for Protium, Deuterium, and Tritium (which are isotopes of Hydrogen) will be 1. The only difference between Protium, Deuterium, and Tritium is the number of neutrons in the nucleus. Protium does not have any, Deuterium has one, and Tritium has two. So while they will have different mass numbers, their atomic number is the same. ---Learning Resources--- Isotopes of Hydrogen: Protons, Neutrons, and Electrons for Hydrogen: Introduction to Isotopes: Image from: [url= Deuterium Tritium Nuclei Schmatic-en[/url] Dirk Hünniger; Derivative work in english - Balajijagadesh, CC BY-SA 3.0 via Wikimedia Commons 15 comments Transcript: i was asked how to find the atomic number for protium deuterium and tritium these are all isotopes of hydrogen so if you look at each atomic model here for each isotope you'll notice something that's similar all of them have one proton so the atomic number that's the number of protons so for protium the atomic number is one because it has one proton for deuterium the atomic number is one one proton and finally tritium one proton you guessed it the atomic number number of protons is one when you look at the periodic table the atomic number for hydrogen that's just one because it has one proton what's different for these isotopes of hydrogen it's the mass number so the mass number that's the number of protons plus neutrons so for protium we just have a proton so the mass number is one and we write that here above the atomic number the number of protons for deuterium we have a proton and a neutron one plus one that gives us a mass number of two the atomic number still stays the same though and then for tritium here we have one plus one plus one we have three that's the mass number and the atomic number for tritium that's one we only have one proton what you'll note the periodic table has an average atomic mass and it's really close to one the reason is is that 99 more than 99 of hydrogen atoms are protium and the mass number for protein that's just one so this number is really close to the mass number for protium since it's more common we do have a little bit of deuterium and tritium and that's what makes this number a little bit bigger than one but in answer to our question the atomic number for protium deuterium and tritium that's the same as the number of protons which is one for each isotope of hydrogen this is dr b thanks for watching
14241	https://pmc.ncbi.nlm.nih.gov/articles/PMC6374094/	Treatment of Adult Obstructive Sleep Apnea with Positive Airway Pressure: An American Academy of Sleep Medicine Clinical Practice Guideline - 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. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice J Clin Sleep Med . 2019 Feb 15;15(2):335–343. doi: 10.5664/jcsm.7640 Search in PMC Search in PubMed View in NLM Catalog Add to search Treatment of Adult Obstructive Sleep Apnea with Positive Airway Pressure: An American Academy of Sleep Medicine Clinical Practice Guideline Susheel P Patil Susheel P Patil, MD, PhD 1 Johns Hopkins University, Baltimore, Maryland Find articles by Susheel P Patil 1,✉, Indu A Ayappa Indu A Ayappa, PhD 2 Icahn School of Medicine at Mount Sinai, New York, New York Find articles by Indu A Ayappa 2, Sean M Caples Sean M Caples, DO 3 Mayo Clinic, Rochester, Minnesota Find articles by Sean M Caples 3, R John Kimoff R John Kimoff, MD 4 McGill University Health Centre, Montreal, Quebec, Canada Find articles by R John Kimoff 4, Sanjay R Patel Sanjay R Patel, MD 5 University of Pittsburgh, Pittsburgh, Pennsylvania Find articles by Sanjay R Patel 5, Christopher G Harrod Christopher G Harrod, MS 6 American Academy of Sleep Medicine, Darien, Illinois Find articles by Christopher G Harrod 6 Author information Article notes Copyright and License information 1 Johns Hopkins University, Baltimore, Maryland 2 Icahn School of Medicine at Mount Sinai, New York, New York 3 Mayo Clinic, Rochester, Minnesota 4 McGill University Health Centre, Montreal, Quebec, Canada 5 University of Pittsburgh, Pittsburgh, Pennsylvania 6 American Academy of Sleep Medicine, Darien, Illinois ✉ Address correspondence to: Susheel P. Patil, MD, PhD; 2510 N. Frontage Road, Darien, IL, Phone: (630) 737-9700, FAX: (630) 737-9790, Email: research@aasm.org Received 2018 Dec 18; Revised 2019 Jan 14; Accepted 2019 Jan 14. © 2019 American Academy of Sleep Medicine PMC Copyright notice PMCID: PMC6374094 PMID: 30736887 Abstract Introduction: This guideline establishes clinical practice recommendations for positive airway pressure (PAP) treatment of obstructive sleep apnea (OSA) in adults and is intended for use in conjunction with other American Academy of Sleep Medicine (AASM) guidelines in the evaluation and treatment of sleep-disordered breathing in adults. Methods: The AASM commissioned a task force of experts in sleep medicine. A systematic review was conducted to identify studies, and the Grading of Recommendations Assessment, Development and Evaluation (GRADE) process was used to assess the evidence. The task force developed recommendations and assigned strengths based on the quality of evidence, the balance of clinically significant benefits and harms, patient values and preferences, and resource use. In addition, the task force adopted recommendations from prior guidelines as “good practice statements” that establish the basis for appropriate and effective treatment of OSA. The AASM Board of Directors approved the final recommendations. Good Practice Statements: The following good practice statements are based on expert consensus, and their implementation is necessary for appropriate and effective management of patients with OSA treated with positive airway pressure: Treatment of OSA with PAP therapy should be based on a diagnosis of OSA established using objective sleep apnea testing. Adequate follow-up, including troubleshooting and monitoring of objective efficacy and usage data to ensure adequate treatment and adherence, should occur following PAP therapy initiation and during treatment of OSA. Recommendations: The following recommendations are intended as a guide for clinicians using PAP to treat OSA in adults. A STRONG (ie, “We recommend…”) recommendation is one that clinicians should follow under most circumstances. A CONDITIONAL recommendation (ie, “We suggest…”) reflects a lower degree of certainty regarding the outcome and appropriateness of the patient-care strategy for all patients. The ultimate judgment regarding any specific care must be made by the treating clinician and the patient, taking into consideration the individual circumstances of the patient, available treatment options, and resources. We recommend that clinicians use PAP, compared to no therapy, to treat OSA in adults with excessive sleepiness. (STRONG) We suggest that clinicians use PAP, compared to no therapy, to treat OSA in adults with impaired sleep-related quality of life. (CONDITIONAL) We suggest that clinicians use PAP, compared to no therapy, to treat OSA in adults with comorbid hypertension. (CONDITIONAL) We recommend that PAP therapy be initiated using either APAP at home or in-laboratory PAP titration in adults with OSA and no significant comorbidities. (STRONG) We recommend that clinicians use either CPAP or APAP for ongoing treatment of OSA in adults. (STRONG) We suggest that clinicians use CPAP or APAP over BPAP in the routine treatment of OSA in adults. (CONDITIONAL) We recommend that educational interventions be given with initiation of PAP therapy in adults with OSA. (STRONG) We suggest that behavioral and/or troubleshooting interventions be given during the initial period of PAP therapy in adults with OSA. (CONDITIONAL) We suggest that clinicians use telemonitoring-guided interventions during the initial period of PAP therapy in adults with OSA. (CONDITIONAL) Citation: Patil SP, Ayappa IA, Caples SM, Kimoff RJ, Patel SR, Harrod CG. Treatment of adult obstructive sleep apnea with positive airway pressure: an American Academy of Sleep Medicine clinical practice guideline. J Clin Sleep Med. 2019;15(2):335–343. Keywords: obstructive sleep apnea, OSA, positive airway pressure, PAP INTRODUCTION Since the publication of the previous American Academy of Sleep Medicine (AASM) positive airway pressure (PAP) practice parameters,1–3 the scientific literature has continued to expand regarding the effects of PAP on clinical outcomes in adults with obstructive sleep apnea (OSA). In addition, research on improving PAP adherence and advancements in device technology have continued to evolve. Given these advancements, updating the prior practice parameters was considered timely. The AASM commissioned a task force (TF) of content experts to update and consolidate previous AASM PAP practice parameters and reviews relevant to the treatment of adult OSA with PAP modalities.1,3,4 This guideline does not address the initiation and management of PAP in patients with obesity hypoventilation syndrome, sleep-related hypoventilation, or those with concurrent forms of obstructive and central sleep apnea. The efficacy of continuous PAP (CPAP), auto-adjusting PAP (APAP), bilevel PAP (BPAP), and other advanced PAP modalities for central sleep apnea and hypoventilation are addressed in other active AASM guidelines.5,6 Furthermore, several prior recommendations on the management of OSA with PAP were not readdressed in the present guideline. Nevertheless, the TF adopted and modified two prior statements as good practice statements, as they were considered essential to providing high quality care to patients with OSA who are treated with PAP. This guideline, in conjunction with the accompanying systematic review,7 provides a comprehensive update of the available evidence and a synthesis of clinical practice recommendations. METHODS The AASM commissioned a TF of both board-certified sleep medicine specialists and experts with proficiency in the use of PAP in adults with OSA to develop this guideline. The TF was required to disclose all potential conflicts of interest (COI) per the AASM's COI policy prior to being appointed to the TF, and throughout the research and writing of this paper. In accordance with the AASM's conflicts of interest policy, TF members with a Level 1 conflict were not allowed to participate. TF members with a Level 2 conflict were required to recuse themselves from any related discussion or writing responsibilities. All relevant conflicts of interest are listed in the disclosure statement. The TF conducted a systematic review of the scientific literature to answer 11 PICO (Patient, Population or Problem, Intervention, Comparison, and Outcomes) questions regarding the initiation of PAP therapy in patients with OSA that focus on patient-oriented, clinically relevant outcomes (see systematic review, Table 1).7 The purpose of the review was to determine the effectiveness of PAP, alternative PAP modes (ie, APAP, BPAP), and concurrent strategies designed to improve outcomes by enhancing acceptance and use of PAP for OSA treatment (eg, patient education and telemonitoring). The TF did not compare PAP against other treatment options (eg, oral appliance therapy, surgical therapy). Assessment of the evidence was performed according to the Grading of Recommendations Assessment, Development and Evaluation (GRADE) process.8 The TF assessed the following four components to determine the direction and strength of a recommendation: quality of evidence, balance of beneficial and harmful effects, patient values and preferences, and resource use. Details of these assessments can be found in the accompanying systematic review.7 Taking these major factors into consideration, each recommendation statement was assigned a strength (STRONG or CONDITIONAL). Additional information is provided in the form of remarks immediately following the recommendation statements, when deemed necessary by the TF. Remarks are based on the evidence evaluated during the systematic review and are intended to provide context for the recommendations to guide clinicians in the implementation of the recommendations in daily practice. Table 1. Implications of STRONG and CONDITIONAL recommendations for users of AASM clinical practice guidelines. Open in a new tab As this guideline focuses on the indications for PAP therapy for OSA in adults, rather than the use of specific components or accessories of the PAP device, recommendations for three of the PICO questions were not included. A summary of the systematic review and meta-analyses of the evidence for these PICO questions can be found in the additional considerations section, as these factors are still important for clinicians to consider in the context of their individual patient's circumstances, when initiating PAP therapy. The AASM expects this guideline to have an impact on professional behavior, patient outcomes and, possibly, health care costs. This clinical practice guideline reflects the state of knowledge at the time of publication and will be reviewed and updated as new information becomes available. GOOD PRACTICE STATEMENTS The following are good practice statements, the implementation of which is necessary for appropriate and effective management of patients with OSA who are treated with PAP. Treatment of OSA with PAP therapy should be based on a diagnosis of OSA established using objective testing.9 This good practice statement applies specifically to a new diagnosis of OSA, which should be established by either a home sleep apnea test or in-laboratory sleep testing prior to initiation of treatment for OSA. Patients with a previously established diagnosis of OSA who are currently on PAP therapy and have good symptom control should continue PAP therapy, even when prior testing results are not readily available. Adequate follow-up, including troubleshooting and monitoring of objective efficacy and usage data to ensure adequate treatment and adherence, should occur following PAP therapy initiation and during treatment of OSA. OSA is a chronic disease that rarely resolves except with substantial weight loss or successful corrective surgery. As with other chronic diseases, periodic follow-up by a qualified clinician (eg, physician or advanced practice provider) is necessary to confirm adequate treatment, assess symptom resolution, and promote continued adherence to treatment. Initial treatment of OSA requires close monitoring and early identification of difficulties with PAP use, as adherence over the first few days to weeks has been shown to predict long-term adherence.10,11 Objective monitoring of PAP therapy should be performed to complement patient reporting of difficulties with PAP use, as patients often overestimate their use of PAP treatment.12 The timing of adequate follow-up after treatment is initiated will vary depending on patient circumstances. However, patients should be followed in the initial weeks to months after PAP initiation to promote adherence and assess response to treatment. Subsequently, yearly evaluation by a trained health care provider is reasonable, although longer periods of follow-up may be appropriate for selected patients who are highly adherent to PAP therapy, have sustained resolution of OSA-related symptoms, and have no concerns regarding their PAP therapy. In contrast, patients with persistent or recurrent sleep-related complaints or persistent difficulties with PAP use should receive more frequent follow-up to address their issues. Routine sleep testing to re-evaluate OSA status in patients on PAP therapy with good symptom control and no change in clinical status (eg, significant weight loss or upper airway surgery) is considered low value care. CLINICAL PRACTICE RECOMMENDATIONS The following clinical practice recommendations are based on a companion systematic review, which evaluated the evidence using the GRADE methodology and should be read concurrently with this clinical practice guideline.7 The recommendations in this guideline define principles of practice that should meet the needs of most patients in most situations. A STRONG recommendation is one that clinicians should follow for almost all patients (ie, something that might qualify as a quality measure). A CONDITIONAL recommendation reflects a lower degree of certainty in the appropriateness of the patient-care strategy for all patients. It requires that the clinician use clinical knowledge and experience, while strongly considering the individual patient's values and preferences to determine the best course of action. The ultimate judgment regarding any specific care must be made by the treating clinician and the patient, taking into consideration the individual circumstances of the patient, available treatment options, and resources. The implications of the strength of recommendations for clinicians, patients, and policymakers are summarized in Table 1. Remarks are provided to guide clinicians in the implementation of these recommendations. A flowchart for the implementation of the recommendations is presented in Figure 1. Figure 1. Flow chart for implementation of clinical practice guideline. Open in a new tab a = Kapur et al., 2017.9 b = symptoms that can impair sleep-related QOL include but are not limited to snoring, sleep-related choking, insomnia, disruption of bedpartner's sleep, morning headaches, nocturia, impairments in productivity or social functioning, and daytime fatigue. c = comorbidities may include: congestive heart failure, chronic opiate use, significant lung disease such as chronic obstructive pulmonary disease, neuromuscular disease, history of uvulopalatopharyngoplasty, those with known sleep-related oxygen requirements or expected to have nocturnal arterial oxyhemoglobin desaturation due to conditions other than OSA including hypoventilation syndromes and central sleep apnea syndromes. d = alternative therapies may include, but are not limited to, weight loss, positional therapy, oral appliance therapy or surgical interventions. e = BPAP is defined as a respiratory assist device that delivers inspiratory and expiratory positive airway pressure. f = BPAP devices may need to be used for patients with therapeutic pressure requirements greater than can be provided with CPAP or APAP; the decision to use BPAP should be based on the clinician's clinical judgement and needs of the individual patient. g = PAP therapy should be performed in conjunction with adequate follow-up to ensure adequate treatment and adherence. h = recommendations included within these boxes should be considered concurrently. i = educational interventions include those focused primarily on providing information about what OSA is, downstream consequences of untreated OSA, what PAP therapy is, how to use it, and the potential benefits of PAP therapy. j = behavioral interventions include those focused on behavior change related to use of PAP therapy using strategies such as cognitive behavioral therapy or motivational enhancement. Troubleshooting interventions include those focused on close patient communication to identify PAP-related problems and to initiate potential solutions. k = telemonitoring interventions include those that remotely monitor data obtained from a PAP device to identify PAP-related problems and to initiate potential solutions. l = when implementing the above recommendations, providers should consider additional strategies that will maximize the individual patient's comfort and adherence. APAP = auto-adjusting positive airway pressure, BPAP = bilevel positive airway pressure, CPAP = continuous positive airway pressure, OSA = obstructive sleep apnea, PAP = positive airway pressure, QOL = quality of life. PAP Therapy Recommendation 1: We recommend that clinicians use positive airway pressure, compared to no therapy, to treat OSA in adults with excessive sleepiness. (STRONG) The TF assessed whether PAP should be offered to adult patients with OSA, based on improvements in the critical outcome of sleepiness, compared to no therapy. The TF identified 38 randomized controlled trials (RCTs) that assessed the efficacy of PAP to reduce excessive sleepiness. Meta-analyses demonstrated a clinically significant improvement in sleepiness with the use of PAP to treat adults with OSA as compared with no treatment. The overall quality of evidence, based on the critical outcome of sleepiness, was high. While a benefit of PAP use includes a reduction in daytime sleepiness, the potential harms include side effects such as nasal dryness or irritation, dry mouth, sore throat, and sinus infection as well as loss of intimacy,4,9 all of which can be mitigated with appropriate interventions or are reversible with discontinuation of PAP. The potential burdens to the patient may include the costs of treatment and inconvenience such as maintaining the equipment and attending follow-up visits with the sleep clinician. The TF concluded that in adult patients with OSA and excessive sleepiness, the benefits of PAP therapy compared to no PAP therapy likely outweigh the potential harms and burdens, and that the majority of well-informed patients would choose the intervention over no treatment. Recommendation 2: We suggest that clinicians use positive airway pressure, compared to no therapy, to treat OSA in adults with impaired sleep-related quality of life. (CONDITIONAL) Remarks: Sleep-related quality of life (QOL) in adult patients with OSA may be adversely affected by OSA-related symptoms. Examples of such symptoms include: snoring, sleep-related choking, insomnia, disruption of bedpartner's sleep, morning headaches, nocturia, impairments in productivity or social functioning, and daytime fatigue. The TF assessed whether PAP compared to no therapy should be offered to adult patients with OSA to improve the critical outcome of sleep-related QOL. OSA-related symptoms that can reduce sleep-related QOL include, but are not limited to; snoring, nocturnal choking, insomnia, disruption of their partner's sleep, morning headaches, nocturia, impairments in productivity or social functioning, and daytime fatigue. The TF identified 19 RCTs that assessed the efficacy of PAP to improve sleep-related QOL. Meta-analyses of sleep-related QOL, as assessed by the Calgary Sleep Apnea QOL Index (SAQLI) and the Functional Outcomes of Sleep Questionnaire (FOSQ), demonstrated a clinically significant improvement. However, meta-analyses of global QOL, as assessed by the SF-36 component scores, demonstrated no clinically significant improvement. The overall quality of evidence, based on the critical outcome of sleep-related QOL, was moderate due to imprecision. The benefits and potential harms of PAP for patients with impaired sleep-related QOL are the same as for patients with sleepiness.4,9 The TF concluded that in adult patients with OSA and impaired sleep-related QOL, the benefits of PAP therapy compared to no PAP therapy likely outweigh the potential harms and burdens, and that the majority of well-informed patients would choose the intervention over no treatment. Recommendation 3: We suggest that clinicians use positive airway pressure, compared to no therapy, to treat OSA in adults with comorbid hypertension. (CONDITIONAL) The TF assessed whether PAP should be offered to adult patients with hypertension and OSA, compared to no therapy to reduce the critical outcome of blood pressure (BP). The TF identified 5 RCTs that reported on the efficacy of PAP therapy on BP in this patient population. Meta-analyses demonstrated clinically significant BP reductions in nocturnal, daytime, and 24-hour systolic and diastolic BP when all patients in the studies were considered, with the largest effects seen for nocturnal measurements. When stratified by resistant hypertensive, hypertensive, and normotensive status, BP reduction was clinically significant in the meta-analyses for the group with hypertension and most BP measures in the group with resistant hypertension. The overall quality of evidence, based on the critical outcome of mean arterial BP, was moderate due to imprecision. The TF notes that the majority of studies evaluating the impact of PAP on BP recruited patients with predominantly moderate to severe OSA. (Note: studies did not systematically report BP based on OSA severity, which limited the ability to make recommendations specific to OSA severity.) The benefits and potential harms of PAP for patients with comorbid hyper-tension are the same as for patients with excessive sleepiness.4,9 The TF recognized that patients experiencing symptoms of OSA (eg, excessive sleepiness) may be more accepting of PAP therapy, with the possibility of secondary benefits related to blood pressure reduction. Non-sleepy patients with OSA, however, may have a more nuanced view of whether to pursue treatment of OSA, particularly given the efficacy of standard antihypertensive treatments. The TF recognized that some non-sleepy patients will place a high value on any intervention that potentially reduces blood pressure, including PAP therapy. Nevertheless, the TF concluded that in adult patients with OSA and comorbid hypertension, the benefits of PAP therapy compared to no PAP therapy likely outweigh the potential harms and burdens, and that the majority of well-informed patients would choose the intervention over no treatment. There is insufficient and inconclusive evidence to either recommend or withhold PAP to treat non-sleepy adults with OSA as a means to reduce cardiovascular events or mortality. The TF assessed whether PAP compared to no therapy should be offered to adult patients with OSA to improve the critical outcomes of cardiovascular (CV) event and mortality risk. The TF identified 17 studies (11 observational studies and 6 RCTs) that assessed the impact of PAP therapy on cardiovascular events, and 13 studies (9 observational studies and 4 RCTs) that assessed the impact of PAP therapy on all-cause mortality. Meta-analyses of observational studies suggested a reduction in CV events and mortality with PAP therapy. In contrast, meta-analyses of randomized controlled trials demonstrated no clinically significant improvements in CV events or mortality. The quality of evidence for incident CV events and mortality ranged from very low to moderate due to study type and imprecision. Thus, the TF judged that the meta-analyses demonstrated insufficient and inconclusive findings regarding the impact of PAP therapy on incident CV events and mortality. Therefore, no recommendation is made regarding the use of PAP based on reduced incident CV events and mortality. Some patients may, however; place a high value on any intervention that potentially reduces CV risk even when they are non-sleepy. In this situation, the patient and clinician should have a balanced discussion about the current state of the evidence about CV risk reduction with PAP therapy for OSA and the potential harms of PAP therapy when there are no other indications to treat the patient's OSA. Conversely, the uncertainty of any CV benefit, may lead some non-sleepy patients with OSA to decline treatment of their OSA. In these patients, conservative management of OSA, with monitoring for development of OSA symptoms over time, may be appropriate. Initiation of PAP Therapy Recommendation 4: We recommend positive airway pressure therapy be initiated using either APAP at home or in-laboratory PAP titration in adults with OSA and no significant comorbidities. (STRONG) Remarks: When APAP is initiated in the home setting, therapy is maintained over the long-term by either using a fixed, continuous pressure setting determined from PAP monitoring data or continuing in the auto-adjusting mode. In-laboratory titration refers to both full-night and split-night titration. The choice of PAP initiation in the home or lab should be based on access, cost-effectiveness, patient preference, sleep clinician judgement, and other factors. This recommendation is based on studies that excluded patients with the following comorbidities or conditions: congestive heart failure, chronic opiate use, significant lung disease such as chronic obstructive pulmonary disease, neuromuscular disease, history of uvulopalatopharyngoplasty, sleep-related oxygen requirements, or expectation for nocturnal arterial oxyhemoglobin desaturation due to conditions other than OSA, including hypoventilation syndromes and central sleep apnea syndromes. This recommendation is based on the clinical trials reviewed, in which mask fittings and education on PAP use at a sleep center and/or close follow-up by trained staff during the treatment period were provided to the home APAP group. In some studies, daytime acclimatization to PAP was included. The TF examined whether initiation of PAP using APAP at home (ie, without an in-laboratory titration) versus in-laboratory titration improved the critical outcomes of adherence, sleepiness, and QOL. The TF did not specifically assess patient outcomes with split-night titration testing as this was recently assessed by an AASM clinical practice guideline on diagnostic sleep testing and was deemed to be a reasonable approach under certain circumstances.9 The guideline assumes that a diagnosis of OSA has already been established. The TF identified 10 RCTs that compared initiation of PAP using home APAP versus an in-laboratory PAP titration. Meta-analyses demonstrated no clinically significant differences in adherence, sleepiness, or QOL between APAP at home and in-laboratory PAP titration. The overall quality of evidence for this recommendation was high. Potential benefits of using APAP in the home setting include lower cost, reduced time away from home, faster initiation of treatment, and greater access to care, whereas the potential harms may include inadequate patient education and the inability to identify and rectify problems related to mask fit, leak, or other PAP-related issues on the night of APAP use. The potential benefits of an in-laboratory PAP titration include providing education by a trained sleep technologist, real-time visual identification of efficacy of therapy, and the ability to provide immediate interventions to make PAP treatment more comfortable for the patient, whereas the potential harms include the need for an overnight stay away from home at the testing facility with the associated costs and patient time and the potential delay in initiation of therapy. The TF concluded that the majority of well-informed adult patients with OSA and without significant comorbidities would prefer initiation of PAP using the most rapid, convenient and cost-effective strategy. This recommendation assumes that adequate education on PAP use and mask fittings with or without daytime acclimatization by trained staff are available. Furthermore, when APAP is implemented, the clinician is strongly encouraged to monitor the clinical response and PAP usage and therapy data within the first few weeks to make necessary PAP adjustments when indicated. Independent of payor restrictions, home APAP will be more rapid and convenient for most patients and has been shown to be more cost-effective.13–15 Nevertheless, final determination of which strategy is ideal for an individual patient should be based on patient preferences and abilities, the sleep clinician's judgment, anticipated or known previous difficulty with PAP treatment, and availability of resources and cost of each strategy in a particular region. PAP Modalities Recommendation 5: We recommend that clinicians use either APAP or CPAP for ongoing treatment of OSA in adults. (STRONG) Remarks: This recommendation is based on studies that mostly excluded patients with the following comorbidities or conditions: congestive heart failure, chronic opiate use, significant lung disease such as chronic obstructive pulmonary disease, neuromuscular disease, history of uvulopalatopharyngoplasty, sleep-related oxygen requirements or expectation of nocturnal arterial oxyhemoglobin desaturation due to conditions other than OSA, including hypoventilation syndromes and central sleep apnea syndromes. The TF examined whether APAP versus CPAP improved the critical outcomes of adherence, sleepiness, and QOL. The TF identified 26 RCTs that investigated the effects of ongoing treatment with APAP compared with fixed CPAP. Meta-analyses demonstrated no clinically significant differences between APAP versus CPAP in adherence, self-reported and objective sleepiness, or QOL. The overall quality of evidence for this recommendation was moderate due to imprecision. The TF judged that the benefits and harms of APAP and CPAP are similar, and the balance of effects does not favor either intervention. The main potential benefit of APAP to patients is the ability to automatically adjust pressure requirements over time in response to acute and chronic changes (eg, alcohol consumption, body position, or weight changes). No substantial differences in harms were identified for APAP versus CPAP use. Although meta-analyses demonstrated a lack of clinically significant differences in treatment adherence and outcomes, and patient preference varied between studies, the TF determined that individual patient tolerance of PAP, adherence, and symptom response may differ for one form of PAP or the other. The TF concluded that either APAP or CPAP should be used for ongoing treatment of adult OSA, with the choice of therapy being tailored to patient tolerance and symptom response. Recommendation 6: We suggest that clinicians use CPAP or APAP over BPAP in the routine treatment of OSA in adults. (CONDITIONAL) Remarks: This recommendation is based on BPAP defined as a respiratory assist device that delivers inspiratory and expiratory positive airway pressure. This recommendation applies to all BPAP devices including flexible, modified, and auto-adjusting BPAP. BPAP devices may need to be used for patients with higher therapeutic pressure requirements than can be provided by CPAP or APAP devices. The decision to use BPAP should be based on the clinician's judgement and needs of the individual patient. Furthermore, this recommendation is for the initial treatment of OSA and does not address management of patients who have previously failed CPAP or APAP. In addition, treatment of other forms of sleep-related breathing disorders associated with hypercapnia, which may require the use of BPAP, are covered in other AASM guidelines.5,16 To improve PAP adherence, BPAP has been used as an alternative to CPAP, in part due to issues of patient intolerance of high CPAP settings.17 The TF examined whether BPAP versus CPAP improves the critical outcomes of adherence, sleepiness, and QOL (Note: while no direct evidence was available for the comparison of APAP to BPAP, the TF considers APAP to be equivalent to CPAP for the implementation of this recommendation. See recommendation 5 and Figure 1. The TF identified 5 RCTs that compared the use of BPAP to CPAP. Meta-analyses demonstrated no clinically significant differences in adherence, self-reported sleepiness, and residual OSA with BPAP compared to CPAP. Studies reporting on sleep-related QOL and sleep quality also demonstrated no clinically significant differences. The overall quality of evidence for this recommendation was very low due to publication bias from industry funding and imprecision associated with small sample size. The main potential benefit of BPAP over CPAP or APAP is improved comfort by lowering the pressure during exhalation, which may then increase adherence. The potential harms of BPAP over CPAP or APAP are a sub-optimally low expiratory pressure level that fails to prevent the occurrence of obstructive breathing events and the higher cost of BPAP. Furthermore, the historically perceived benefits of BPAP are less likely to be relevant since modified pressure profile technology, which also lowers expiratory pressures, has been integrated into modern PAP devices. The TF judged that although the benefits of treatment with BPAP versus CPAP or APAP are similar, the low quality of evidence and potential harms or burdens did not favor the regular use of BPAP for the routine treatment of OSA. Therefore, the TF concluded that the majority of well-informed adult patients with OSA would prefer initiation of treatment with CPAP or APAP over BPAP. However, there is a small subset of patients that require PAP treatment with pressures higher than 20 cm H 2 O, which CPAP units are not typically capable of delivering. In these situations, BPAP devices may be needed for optimal treatment and can be utilized during an initial or subsequent in-laboratory PAP titration study. In addition, for specific patients who are unable to tolerate CPAP or APAP due to high pressure requirements and despite the use of modified pressure profiles, a trial of BPAP may be offered either during the initial in-laboratory titration or following a period of demonstrated non-acceptance. Educational and Behavioral Interventions With PAP Recommendation 7: We recommend that educational interventions be given prior to initiation of PAP therapy in adults with OSA. (STRONG) Recommendation 8: We suggest that behavioral and/ or troubleshooting interventions be given during the initial period of PAP therapy in adults with OSA. (CONDITIONAL) Remarks: These recommendations are based on interventions defined as follows: Educational interventions: Interventions focused primarily on providing information prior to initiation of PAP about what OSA is, its downstream consequences, what PAP therapy is, and the potential benefits of PAP therapy. Behavioral interventions: Interventions focused on behavior change prior to and during the initiation and subsequent use of PAP therapy using strategies such as cognitive behavioral therapy or motivational enhancement. Troubleshooting interventions: Interventions focused on close patient communication to identify PAP-related problems and to initiate potential solutions during the initial period of PAP therapy. The intervention period may include interactions prior to, during and after PAP titration and follow-up. The TF examined whether an educational, behavioral, or troubleshooting intervention versus no intervention prior to or during PAP treatment improves the critical outcome of adherence. QOL was initially considered a critical outcome; however, none of the accepted studies reported on this outcome. The TF identified 18 RCTs that evaluated the use of some combination of an educational, behavioral, or troubleshooting intervention as an adjunct to initiation of PAP therapy compared to PAP therapy with usual care. Meta-analyses demonstrated a clinically significant improvement in PAP adherence with all three types of interventions. The overall quality of evidence was moderate due to imprecision. The potential harms of each of these interventions are minimal. The potential burdens to the patient are negligible for educational interventions but include the time required to receive the intervention and the cost of the additional care for the more intensive behavioral and troubleshooting interventions. The TF judged that the benefits of educational interventions outweigh potential harms and burdens, while the benefits of behavioral and troubleshooting interventions likely outweigh the potential harms and burdens in most patients. As such, the TF concluded that the vast majority of well-informed adult patients with OSA would prefer that an educational intervention be provided with initiation of PAP therapy over initiation of PAP without education. In addition, the TF concluded that the majority of well-informed adult patients with OSA would likely prefer that a behavioral and/or troubleshooting intervention be given during the initial period of PAP therapy over no such intervention. Monitoring During Treatment Recommendation 9: We suggest that clinicians use telemonitoring-guided interventions during the initial period of PAP therapy in adults with OSA. (CONDITIONAL) Remarks: This recommendation is based on interventions defined as follows: Telemonitoring includes the remote monitoring of PAP parameters such as PAP use, residual OSA severity, unintentional mask leaks, and PAP settings during treatment initiation and follow-up. The TF examined whether behavioral, educational and troubleshooting interventions guided by remote monitoring of PAP therapy versus no remote monitoring improved the critical outcomes of adherence, sleepiness, and side effects. QOL was initially considered a critical outcome; however, none of the accepted studies reported on this outcome. The TF identified 5 RCTs that evaluated the use of remote monitoring of PAP variables to trigger early interventions versus no such system as an adjunct to PAP therapy. Meta-analyses demonstrated a clinically significant improvement in adherence, but not in sleepiness, with use of telemonitoring. PAP-associated side effect severity scores tended to be lower with a telemonitoring-guided intervention; however, these differences were not clinically significant. The overall quality of evidence was moderate due to imprecision. The potential harms of remote monitoring interventions are small, primarily related to potential loss of privacy. How tele-monitoring guided interventions are implemented could lead to substantial increases in costs to health-care systems, or conversely, may reduce costs if reductions in healthcare utilization substantially offset the investment in tele-monitoring systems. Nevertheless, the TF determined that the benefits of telemonitoring-guided adherence interventions likely outweigh the potential harms and burdens in most patients. Based on clinical experience, the TF concluded that the improvement in adherence would be valued by most patients. The TF concluded that the majority of well-informed adult patients with OSA would prefer enrollment in such a system compared to treatment without such an intervention. ADDITIONAL CONSIDERATIONS The AASM supports patient-centered care in which an individual patient's specific health needs and desired health outcomes are the driving force behind all health care decisions. When implementing the recommendations in this guideline, clinicians should consider strategies that will maximize the individual patient's comfort and adherence. The TF performed a systematic review of several interventions that aim to improve comfort and adherence including modified pressure profile, mask interfaces, and humidification. These are summarized below. A detailed evidence review can be found in the accompanying systematic review.7 Modified Pressure Profile PAP Many PAP devices have now integrated modified pressure profiles, which lower the treatment pressure used during expiration. The potential benefit of using modified pressure profile PAP is increased patient comfort, while the potential harms of modified pressure profile PAP are similar to standard PAP. The TF identified 7 RCTs that investigated the use of modified pressure profile PAP to improve clinical outcomes and reduce side effects. Meta-analyses demonstrated no clinically significant differences in adherence, sleepiness, and QOL with modified pressure profile PAP versus standard PAP. Insufficient, standardized data were available to perform a meta-analysis on side effects; however, the reported data demonstrated no clinically significant differences. The data suggest that there are no systematic benefits to the routine initiation of treatment with modified pressure profile PAP, compared to standard PAP for OSA, despite perceived minimal potential harms and burdens. However, modified pressure profile PAP may have value in some patients in other contexts (eg, poorly adherent patients or patients with difficulty tolerating CPAP during in-laboratory titration studies) that have not yet been well studied. Mask Selection Appropriate mask selection will benefit patients by reducing side effects such as air leak and discomfort, which may then potentially improve adherence and subsequently patient outcomes. The TF identified 11 studies (3 observational studies and 8 RCTs) that evaluated the effects of different PAP interfaces on reducing the apnea-hypopnea index (AHI); improving adherence, sleepiness and QOL; and reducing side effects. Meta-analyses demonstrated a clinically significant improvement in adherence with nasal PAP versus oronasal interfaces, but there was no clinically significant difference in adherence between nasal or intra-nasal interfaces. With regards to self-reported sleepiness, meta-analyses demonstrated no clinically significant differences between interfaces. Studies reporting on QOL demonstrated no clinically significant differences between intra-nasal versus nasal interfaces. Finally, the studies analyzed indicated that there were fewer side effects with nasal compared with oronasal and oral interfaces. These data suggest that, for the routine initiation of PAP therapy in adults with OSA, clinicians should generally use nasal or intranasal mask interfaces over oronasal or oral interfaces. However, individual patient factors or preferences vary; therefore, the mask interface that minimizes side effects and optimizes efficacy and adherence should be used. Humidified PAP The use of humidification could also potentially reduce side effects from PAP therapy. The TF identified 9 RCTs that evaluated the use of PAP with humidification versus PAP without humidification to improve adherence, sleepiness, QOL, or PAP-related side effects. Meta-analyses demonstrated a clinically significant reduction in several side effects associated with the use of PAP, including dry mouth/throat, nasal discharge, nasal congestion, dry nose, bleeding nose, sinus pain or headache, sore throat, hoarse voice, and reduced smell. However, meta-analyses demonstrated no clinically significant improvement in PAP adherence, sleepiness, or QOL with the use of humidification compared with no humidification. The possibility of humidification causing “rain out” (ie, condensation of water into the PAP circuit, face, and nose or mouth of the patient) may deter its use. While the use of humidification may increase the ongoing costs (eg, purchasing distilled water, heated hoses) and maintenance requirements, these data suggest clinicians should generally use heated humidification with PAP devices to reduce side effects that may occur while treating adults with OSA. SUMMARY This clinical practice guideline provides recommendations for the use of PAP, approaches to the initiation of PAP treatment, and interventions to promote PAP adherence in adults with OSA. The recommendations are the result of the task force interpretation of evidence collected for the systematic review and application to the clinical care of adults with OSA. Four recommendations are strongly suggested and include: (1) using PAP to treat excessive sleepiness, (2) initiating PAP therapy with either APAP at home or an in-laboratory CPAP titration, (3) continuing PAP therapy for OSA with either CPAP or APAP, and (4) using educational interventions to initiate PAP therapy in adults with OSA. All other recommendations were conditional and include using PAP to treat impaired sleep-related QOL or concomitant hypertension; implementing CPAP or APAP over BPAP in the routine treatment of OSA; and utilizing behavioral, troubleshooting, and telemonitoring interventions during the initial period of PAP therapy. The TF determined that a recommendation could not be made for the use or withholding of PAP therapy to treat non-sleepy patients to reduce incident cardiovascular disease. When implementing the recommendations, providers should consider additional strategies that will maximize the individual patient's comfort and adherence such as nasal/intranasal over oronasal mask interface and heated humidification, as discussed in the additional considerations section. Readers are strongly encouraged to read the companion systematic review7 for a more detailed presentation and evaluation of the evidence. This clinical practice guideline reflects the state of knowledge at the time of publication and will be reviewed and updated as new information becomes available. DISCLOSURE STATEMENT Dr. Indu Ayappa disclosed that she receives royalties from patents held on some PAP devices. Dr. Ayappa's potential conflict was managed by requesting that she refrain from participation on any discussions pertaining to devices for which she may hold a patent. She was also asked to refrain from voting on recommendations pertaining to those devices. Dr. Sanjay Patel disclosed that he receives compensation from Bayer Pharmaceuticals to assess the impact of OSA and its treatment on pulmonary hypertension and other cardiovascular outcomes. He was asked to refrain from participation on any discussion pertaining to the use of PAP to treat hypertension. He was also asked to refrain from voting on recommendations pertaining to the use of PAP to treat hypertension. Mr. Harrod is employed by the American Academy of Sleep Medicine. No other task force members had any relevant conflicts of interest to disclose. ACKNOWLEDGMENTS The task force thanks Dr. Romola Bucks (University of Western Australia) and Dr. Gerry Taylor (Case Western Reserve University) for lending their expertise in determining how neurocognitive outcomes should be addressed in this guideline. REFERENCES 1.Kushida CA, Littner MR, Hirshkowitz M, et al. Practice parameters for the use of continuous and bilevel positive airway pressure devices to treat adult patients with sleep-related breathing disorders. Sleep. 2006;29(3):375–380. doi: 10.1093/sleep/29.3.375. [DOI] [PubMed] [Google Scholar] 2.Kushida CA, Chediak A, Berry RB, et al. Clinical guidelines for the manual titration of positive airway pressure in patients with obstructive sleep apnea. J Clin Sleep Med. 2008;4(2):157–171. [PMC free article] [PubMed] [Google Scholar] 3.Morgenthaler TI, Aurora RN, Brown T, et al. Practice parameters for the use of autotitrating continuous positive airway pressure devices for titrating pressures and treating adult patients with obstructive sleep apnea syndrome: an update for 2007. An American Academy of Sleep Medicine report. Sleep. 2008;31(1):141–147. doi: 10.1093/sleep/31.1.141. [DOI] [PMC free article] [PubMed] [Google Scholar] 4.Gay P, Weaver T, Loube D, et al. 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[DOI] [PubMed] [Google Scholar] Articles from Journal of Clinical Sleep Medicine : JCSM : Official Publication of the American Academy of Sleep Medicine are provided here courtesy of American Academy of Sleep Medicine ACTIONS View on publisher site PDF (1.0 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract INTRODUCTION METHODS GOOD PRACTICE STATEMENTS CLINICAL PRACTICE RECOMMENDATIONS ADDITIONAL CONSIDERATIONS SUMMARY DISCLOSURE STATEMENT ACKNOWLEDGMENTS 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
14242	https://iuuk.mff.cuni.cz/~rakdver/linalg/lesson15-1.pdf	Dot product and inner product Zdenˇ ek Dvoˇ r´ ak February 24, 2015 1 Dot (scalar) product of real vectors Deﬁnition 1. Let u = (α1, . . . , αn) and v = (β1, . . . , βn) be vectors from Rn. The dot product of u and v is u · v = α1β1 + α2β2 + . . . + αnβn. Deﬁnition 2. The Euclidean norm of v = (α1, . . . , αn) ∈Rn is \|v\| = q α2 1 + α2 2 + . . . + α2 n = √v · v. u v θ Lemma 1 (Geometric interpretation). For any u, v ∈Rn such that the angle between u and v is θ, u · v = \|u\|\|v\| cos θ. Proof. Note that the dot product is commutative and linear in both argu-ments, and thus (u −v) · (u −v) = u · (u −v) −v · (u −v) = (u · u −u · v) −(v · u −v · v) = u · u + v · v −2u · v Recall that in a triangle 1 a b c θ u v u −v θ we have c2 = a2 + b2 −2ab cos θ, and thus \|u −v\|2 = \|u\|2 + \|v\|2 −2\|u\|\|v\| cos θ. It follows that \|u\|\|v\| cos θ = \|u\|2 + \|v\|2 −\|u −v\|2 2 = u · u + v · v −(u −v) · (u −v) 2 = u · u + v · v −(u · u + v · v −2u · v) 2 = 2u · v 2 = u · v. Uses of dot product: • Determining the angle between two vectors: u v θ θ = arccos u · v \|u\|\|v\| • Two vectors are perpendicular iﬀtheir dot product is 0. 2 • Orthogonal projection: u v p θ The projection p of u on v (where θ is the angle between u and v) has norm \|u\| cos θ = u · v \|v\| and the same direction as v, hence p = v \|v\| · u · v \|v\| = u · v v · v v. • Determining coordinates in an orthogonal basis (projections to basis vectors). Example 1. Let u1 = ( √ 2/2, √ 2/2) and u2 = (− √ 2/2, √ 2/2). Determine the coordinates of (3, 5) with respect to the basis B = u1, u2. u1 u2 (3, 5) Note that u1 · u2 = 0 (the vectors u1 and u2 are perpendicular) and \|u1\| = \|u2\| = 1. Hence, the coordinates are (3, 5) · u1 = 4 √ 2 3 and (3, 5) · u2 = √ 2 2 Inner product spaces Recall: • R: the ﬁeld of real numbers • C: the ﬁeld of complex numbers • complex conjugation: – α + βi = α −βi – x + y = x + y – xy = x y – xx = \|x\|2, where \|α + βi\| = p α2 + β2 Deﬁnition 3. Let F be either R or C. Inner product space is a vector space V over F, together with an inner product ⟨·, ·⟩: V2 →F satisfying the following axioms: positive deﬁniteness For all v ∈V, ⟨v, v⟩is a non-negative real number, and ⟨v, v⟩= 0 if and only if v = o. linearity in the ﬁrst argument For all u, v, w ∈V and α ∈F, ⟨u + v, w⟩= ⟨u, w⟩+ ⟨v, w⟩ ⟨αu, w⟩= α ⟨u, w⟩ conjugate commutativity For all u, v ∈V, ⟨u, v⟩= ⟨v, u⟩. Remark: • ⟨o, v⟩= 0 = ⟨v, o⟩for every v ∈V. • If F = R, then 4 – the last axiom states commutativity ⟨u, v⟩= ⟨v, u⟩, and – ⟨·, ·⟩is linear in the second argument as well ⟨w, u + v⟩= ⟨w, u⟩+ ⟨w, v⟩ ⟨w, αu⟩= α ⟨w, u⟩ • If F = C, then ⟨w, u + v⟩= ⟨u + v, w⟩ = ⟨u, w⟩+ ⟨v, w⟩ = ⟨u, w⟩+ ⟨v, w⟩ = ⟨w, u⟩+ ⟨w, v⟩ ⟨w, αu⟩= ⟨αu, w⟩ = α ⟨u, w⟩ = α⟨u, w⟩ = α ⟨w, u⟩. – ⟨·, ·⟩is not linear in the second argument, because of the conjuga-tion in scalar multiplication. Example 2. • Dot product gives an inner product on Rn. • Another example of possible inner product on R2: ⟨(α1, α2), (β1, β2)⟩= 2α1β1 + α2β2 −α1β2 −α2β1 – positive deﬁniteness: ⟨(α1, α2), (α1, α2)⟩= α2 1 + (α1 −α2)2 ≥0, and equal to 0 if and only if α1 = 0 and α1 −α2 = 0 ⇒α2 = 0. • Complex dot product on Cn: (α1, . . . , αn) · (β1, . . . , βn) = α1β1 + . . . + αnβn. • Standard inner product on the space of continuous functions f : [α, β] → R: ⟨f, g⟩= Z β α f(x)g(x)dx 5 Deﬁnition 4. Let V be an inner product space. Vectors u, v ∈V are orthogonal if ⟨u, v⟩= 0. We write u ⊥v. Example 3. • (1, 0) and (0, 1) are orthogonal with respect to the dot product, since (1, 0) · (0, 1) = 0. • (1, 0) and (0, 1) are not orthogonal with respect to the inner product ⟨(α1, α2), (β1, β2)⟩= 2α1β1 + α2β2 −α1β2 −α2β1, since ⟨(1, 0), (0, 1)⟩= −1. • f(x) = sin x and g(x) = 1 are orthogonal with respect to the standard inner product on the space of continuous functions from [0, 2π]: Z 2π 0 f(x)g(x)dx = Z 2π 0 sin x dx = [−cos x]2π 0 = cos 0 −cos(2π) = 0. Theorem 2 (Pythagoras theorem). Let V be an inner product space and let u, v ∈V. If u ⊥v, then ⟨u, u⟩+ ⟨v, v⟩= ⟨u + v, u + v⟩. Proof. ⟨u + v, u + v⟩= ⟨u, u⟩+ ⟨v, v⟩+ ⟨u, v⟩+ ⟨v, u⟩ = ⟨u, u⟩+ ⟨v, v⟩, since ⟨u, v⟩= 0 and ⟨v, u⟩= ⟨u, v⟩= 0. Theorem 3 (Cauchy-Schwarz inequality). Let V be an inner product space. Then for all u, v ∈V, \| ⟨u, v⟩\|2 ≤⟨u, u⟩⟨v, v⟩, and if u and v are linearly independent, then the inequality is sharp. Proof. The claim is clearly true if v = o, hence assume that ⟨v, v⟩> 0. 6 u v w z Let w = ⟨u,v⟩ ⟨v,v⟩v and z = u −w. Then ⟨z, v⟩= u −⟨u, v⟩ ⟨v, v⟩v, v = ⟨u, v⟩− ⟨u, v⟩ ⟨v, v⟩v, v = ⟨u, v⟩−⟨u, v⟩ ⟨v, v⟩⟨v, v⟩ = 0, and thus v ⊥z and w ⊥z. Since u = w + z, Pythagoras theorem implies ⟨u, u⟩= ⟨w, w⟩+ ⟨z, z⟩ ≥⟨w, w⟩ = ⟨u, v⟩ ⟨v, v⟩ ⟨u, v⟩ ⟨v, v⟩⟨v, v⟩ = ⟨u, v⟩ ⟨v, v⟩ 2 ⟨v, v⟩ = \| ⟨u, v⟩\|2 ⟨v, v⟩, and thus ⟨u, u⟩⟨v, v⟩≥\| ⟨u, v⟩\|2. The equality holds only if z = o, i.e., if u = ⟨u,v⟩ ⟨v,v⟩v, which implies that u and v are linearly dependent. Example 4. Let x1, . . . , xn be positive real numbers. Prove that x2 1 + . . . + x2 n ≥(x1 + . . . + xn)2 n , 7 where the equality holds if and only if x1 = x2 = . . . = xn. Proof. We apply the Cauchy-Schwarz inequality for the dot product of u = (x1, . . . , xn) and v = (1, . . . , 1): (x2 1 + . . . + x2 n)n = (u · u)(v · v) ≥(u · v)2 = (x1 + . . . + xn)2, where the equality only holds if u and v are linearly dependent, i.e., x1 = . . . = xn. Deﬁnition 5. Let V be a vector space over a ﬁeld F ∈{R, C}. A function s : V →R is a norm if • s(v) ≥0 for every v ∈V, and s(v) = 0 if and only if v = o. • s(αv) = \|α\|s(v) for every v ∈V and α ∈F. • s(u + v) ≤s(u) + s(v) for every u, v ∈V (triangle inequality). Deﬁnition 6. The norm induced by an inner product is ∥v∥= p ⟨v, v⟩. • If ⟨·, ·⟩is the dot product, then ∥· ∥is the Euclidean norm. • Pythagoras theorem reformulated using the norm: if u ⊥v, then ∥u∥2 + ∥v∥2 = ∥u + v∥2 • Cauchy-Schwarz inequality reformulated using the norm: \| ⟨u, v⟩\| ≤∥u∥∥v∥ • The triangle inequality holds because of Cauchy-Schwarz: ∥u + v∥2 = ⟨u + v, u + v⟩ = ∥u∥2 + ∥v∥2 + ⟨u, v⟩+ ⟨v, u⟩ ≤∥u∥2 + ∥v∥2 + 2\| ⟨u, v⟩\| ≤∥u∥2 + ∥v∥2 + 2∥u∥∥v∥ = (∥u∥+ ∥v∥)2 8
14243	https://www.teachoo.com/subjects/cbse-maths/class-11th/ch10-11th-straight-lines/	Class 6 Maths (Ganita Prakash) Class 6 Maths (old NCERT) Class 6 Science (Curiosity) Class 6 Science (old NCERT) Class 6 Social Science Class 6 English Class 7 Maths (Ganita Prakash) Class 7 Maths (old NCERT) Class 7 Science (Curiosity) Class 7 Science (old NCERT) Class 7 Social Science Class 7 English Class 8 Maths (Ganita Prakash) Class 8 Maths (old NCERT) Class 8 Science (Curiosity) Class 8 Science (old NCERT) Class 8 Social Science Class 8 English Class 9 Maths Class 9 Science Class 9 Social Science Class 9 English Class 10 Maths Class 10 Science Class 10 Social Science Class 10 English Class 11 Maths Class 11 Computer Science (Python) Class 11 English Class 12 Maths Class 12 English Class 12 Economics Class 12 Accountancy Class 12 Physics Class 12 Chemistry Class 12 Biology Class 12 Computer Science (Python) Class 12 Physical Education GST and Accounting Course Excel Course Tally Course Finance and CMA Data Course Payroll Course Learn English Learn Excel Learn Tally Learn GST (Goods and Services Tax) Learn Accounting and Finance GST Demo GST Tax Invoice Format Accounts Tax Practical Tally Ledger List Remove Ads Login Chapter 9 Class 11 Straight Lines Master Chapter 9 Class 11 Straight Lines with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons. Start Learning Now Updated for new NCERT 2026-2026. Get solutions of Chapter 9 Class 11 Straight Lines of the NCERT Book. Answers of all exercise questions, examples and miscellaneous are provided for your reference. Let's see what this chapter is all about. We learned about Coordinate Geometry in Class 10. In this chapter, we will Revise our Coordinate Geometry concepts (Distance between points, Section Formula, Area of triangle) and solve some questions What is slope of a line, and different methods to find it (angle, 2 points) Finding angle between two lines using slope Checking if lines are parallel or perpendicular, using slope Proving three points collinear using slope (Slope of AB = Slope of BC) Then, we study equation of lines Equation of lines parallel to x and y axis Slope of lines parallel to x and y axis Equation of line when 1 point and slope is given Equation of line when 2 points are given Equation of line when slope and intercept is given(both x intercept and y intercept) Equation of line when both x and y intercepts are given (Equation of line in intercept form) Equation of line in normal form General Equation of line, and converting any equation into General Form Finding Distance of a point from the line Finding Distance between two parallel lines Distance of a point from a line along a line Mix Questions using all the concepts You can check out the solutions by clicking the exercise links below, or learn the concepts (with questions), the NCERT way. Serial order wise Ex 9.1 Start Learning/category/Ex-10.1/ "Master Ex 9.1 step-by-step") Ex 9.2 Start Learning Ex 9.3 Start Learning Examples Start Learning Miscellaneous Start Learning-x---(4---k2)y---k2/category/Miscellaneous/ "Master Miscellaneous step-by-step") Concept wise Cordinate geometry questions Start Learning/category/Cordinate-geometry-questions/ "Master Cordinate geometry questions step-by-step") Slope - Finding slope Start Learning Slope - Prependicularity Start Learning--(4--8)-is-perpendicular/category/Slope---Prependicularity/ "Master Slope - Prependicularity step-by-step") Angle between two lines by Slope Start Learning Collinearity of 3 points by sliope Start Learning/category/Collinearity-of-3-points-by-sliope/ "Master Collinearity of 3 points by sliope step-by-step") Point Slope form Start Learning--slope--4/category/Point-Slope-form/ "Master Point Slope form step-by-step") Slope-Intercept form Start Learning Intercept form Start Learning Normal form Start Learning Given Statement Start Learning Two lines // or/and prependicular Start Learning-and-perpendicular/category/Two-lines----or-and-prependicular/ "Master Two lines // or/and prependicular step-by-step") Angle between two lines Start Learning Distance - Between two parallel lines Start Learning Distance of a point from a line Start Learning-from-line-3x---4y---26--0/category/Distance-of-a-point-from-a-line/ "Master Distance of a point from a line step-by-step") Distance of a point from a line along a line Start Learning/category/Distance-of-a-point-from-a-line-along-a-line/ "Master Distance of a point from a line along a line step-by-step") Other Type of questions - Area of Triangle formed Start Learning Other Type of questions - 3 lines Concurrent Start Learning Other Type of questions - Image Start Learning Other Type of questions - Mix Start Learning-reflects-on-x-axis/category/Other-Type-of-questions---Mix/ "Master Other Type of questions - Mix step-by-step") Why Learn This With Teachoo? Updated for new NCERT 2026-2026. Get solutions of Chapter 9 Class 11 Straight Lines of the NCERT Book. Answers of all exercise questions, examples and miscellaneous are provided for your reference. Let's see what this chapter is all about. We learned about Coordinate Geometry in Class 10. In this chapter, we will Revise our Coordinate Geometry concepts (Distance between points, Section Formula, Area of triangle) and solve some questions What is slope of a line, and different methods to find it (angle, 2 points) Finding angle between two lines using slope Checking if lines are parallel or perpendicular, using slope Proving three points collinear using slope (Slope of AB = Slope of BC) Then, we study equation of lines Equation of lines parallel to x and y axis Slope of lines parallel to x and y axis Equation of line when 1 point and slope is given Equation of line when 2 points are given Equation of line when slope and intercept is given(both x intercept and y intercept) Equation of line when both x and y intercepts are given (Equation of line in intercept form) Equation of line in normal form General Equation of line, and converting any equation into General Form Finding Distance of a point from the line Finding Distance between two parallel lines Distance of a point from a line along a line Mix Questions using all the concepts You can check out the solutions by clicking the exercise links below, or learn the concepts (with questions), the NCERT way. 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14244	https://faculty.ucmerced.edu/sim3/teaching/spring15/lecture_note/lecture14-1.pdf	CSE 135: Introduction to Theory of Computation Rice’s Theorem and Closure Properties Sungjin Im University of California, Merced 04-21-2015 Mapping Reductions Deﬁnition A function f : Σ∗→Σ∗is computable if there is some Turing Machine M that on every input w halts with f (w) on the tape. Deﬁnition A reduction (a.k.a. mapping reduction/many-one reduction) from a language A to a language B is a computable function f : Σ∗→Σ∗such that w ∈A if and only if f (w) ∈B In this case, we say A is reducible to B, and we denote it by A ≤m B. Reductions and Recursive Enumerability Proposition If A ≤m B and B is r.e., then A is r.e. Proof. Let f be a reduction from A to B and let MB be a Turing Machine recognizing B. Then the Turing machine recognizing A is On input w Compute f (w) Run MB on f (w) Accept if MB accepts, and reject if MB rejects □ Corollary If A ≤m B and A is not r.e., then B is not r.e. Reductions and Decidability Proposition If A ≤m B and B is decidable, then A is decidable. Proof. Let f be a reduction from A to B and let MB be a Turing Machine deciding B. Then a Turing machine that decides A is On input w Compute f (w) Run MB on f (w) Accept if MB accepts, and reject if MB rejects □ Corollary If A ≤m B and A is undecidable, then B is undecidable. The Halting Problem Proposition The language HALT = {⟨M, w⟩\| M halts on input w} is undecidable. Proof. Recall Atm = {⟨M, w⟩\| w ∈L(M)} is undecidable. Will give reduction f to show Atm ≤m HALT = ⇒HALT undecidable. Let f (⟨M, w⟩) = ⟨N, w⟩where N is a TM that behaves as follows: On input x Run M on x If M accepts then halt and accept If M rejects then go into an infinite loop N halts on input w if and only if M accepts w. The Halting Problem Proposition The language HALT = {⟨M, w⟩\| M halts on input w} is undecidable. Proof. Recall Atm = {⟨M, w⟩\| w ∈L(M)} is undecidable. Will give reduction f to show Atm ≤m HALT = ⇒HALT undecidable. Let f (⟨M, w⟩) = ⟨N, w⟩where N is a TM that behaves as follows: On input x Run M on x If M accepts then halt and accept If M rejects then go into an infinite loop N halts on input w if and only if M accepts w. i.e., ⟨M, w⟩∈Atm iﬀf (⟨M, w⟩) ∈HALT □ Emptiness of Turing Machines Proposition The language Etm = {M \| L(M) = ∅} is not decidable. Note: in fact, Etm is not recognizable. Emptiness of Turing Machines Proposition The language Etm = {M \| L(M) = ∅} is not decidable. Note: in fact, Etm is not recognizable. Proof. Recall Atm = {⟨M, w⟩\| w ∈L(M)} is undecidable. Emptiness of Turing Machines Proposition The language Etm = {M \| L(M) = ∅} is not decidable. Note: in fact, Etm is not recognizable. Proof. Recall Atm = {⟨M, w⟩\| w ∈L(M)} is undecidable. For the sake of contradiction, suppose there is a decider B for Etm. Emptiness of Turing Machines Proposition The language Etm = {M \| L(M) = ∅} is not decidable. Note: in fact, Etm is not recognizable. Proof. Recall Atm = {⟨M, w⟩\| w ∈L(M)} is undecidable. For the sake of contradiction, suppose there is a decider B for Etm. Then we ﬁrst transform ⟨M, w⟩to ⟨M1⟩which is the following: On input x If x ̸= w, reject else run M on w , and accept if M accepts w , and accept if B rejects ⟨M1⟩, and rejects if B accepts ⟨M1⟩. Emptiness of Turing Machines Proposition The language Etm = {M \| L(M) = ∅} is not decidable. Note: in fact, Etm is not recognizable. Proof. Recall Atm = {⟨M, w⟩\| w ∈L(M)} is undecidable. For the sake of contradiction, suppose there is a decider B for Etm. Then we ﬁrst transform ⟨M, w⟩to ⟨M1⟩which is the following: On input x If x ̸= w, reject else run M on w , and accept if M accepts w , and accept if B rejects ⟨M1⟩, and rejects if B accepts ⟨M1⟩. Then we show that (1) if ⟨M, w⟩∈Atm, then accept, and (2) ⟨M, w⟩∈Atm, then reject. (how?) Emptiness of Turing Machines Proposition The language Etm = {M \| L(M) = ∅} is not decidable. Note: in fact, Etm is not recognizable. Proof. Recall Atm = {⟨M, w⟩\| w ∈L(M)} is undecidable. For the sake of contradiction, suppose there is a decider B for Etm. Then we ﬁrst transform ⟨M, w⟩to ⟨M1⟩which is the following: On input x If x ̸= w, reject else run M on w , and accept if M accepts w , and accept if B rejects ⟨M1⟩, and rejects if B accepts ⟨M1⟩. Then we show that (1) if ⟨M, w⟩∈Atm, then accept, and (2) ⟨M, w⟩∈Atm, then reject. (how?) This implies Atm is decidable, which is a contradiction. □ Checking Regularity Proposition The language REGULAR = {M \| L(M) is regular} is undecidable. Checking Regularity Proposition The language REGULAR = {M \| L(M) is regular} is undecidable. Proof. We give a reduction f from Atm to REGULAR. Checking Regularity Proposition The language REGULAR = {M \| L(M) is regular} is undecidable. Proof. We give a reduction f from Atm to REGULAR. Let f (⟨M, w⟩) = N, where N is a TM that works as follows: On input x If x is of the form 0n1n then accept x else run M on w and accept x only if M does Checking Regularity Proposition The language REGULAR = {M \| L(M) is regular} is undecidable. Proof. We give a reduction f from Atm to REGULAR. Let f (⟨M, w⟩) = N, where N is a TM that works as follows: On input x If x is of the form 0n1n then accept x else run M on w and accept x only if M does If w ∈L(M) then L(N) = Checking Regularity Proposition The language REGULAR = {M \| L(M) is regular} is undecidable. Proof. We give a reduction f from Atm to REGULAR. Let f (⟨M, w⟩) = N, where N is a TM that works as follows: On input x If x is of the form 0n1n then accept x else run M on w and accept x only if M does If w ∈L(M) then L(N) = Σ∗. Checking Regularity Proposition The language REGULAR = {M \| L(M) is regular} is undecidable. Proof. We give a reduction f from Atm to REGULAR. Let f (⟨M, w⟩) = N, where N is a TM that works as follows: On input x If x is of the form 0n1n then accept x else run M on w and accept x only if M does If w ∈L(M) then L(N) = Σ∗. If w ̸∈L(M) then L(N) = Checking Regularity Proposition The language REGULAR = {M \| L(M) is regular} is undecidable. Proof. We give a reduction f from Atm to REGULAR. Let f (⟨M, w⟩) = N, where N is a TM that works as follows: On input x If x is of the form 0n1n then accept x else run M on w and accept x only if M does If w ∈L(M) then L(N) = Σ∗. If w ̸∈L(M) then L(N) = {0n1n \| n ≥0}. Checking Regularity Proposition The language REGULAR = {M \| L(M) is regular} is undecidable. Proof. We give a reduction f from Atm to REGULAR. Let f (⟨M, w⟩) = N, where N is a TM that works as follows: On input x If x is of the form 0n1n then accept x else run M on w and accept x only if M does If w ∈L(M) then L(N) = Σ∗. If w ̸∈L(M) then L(N) = {0n1n \| n ≥0}. Thus, ⟨N⟩∈REGULAR if and only if ⟨M, w⟩∈Atm □ Checking Equality Proposition EQtm = {⟨M1, M2⟩\| L(M1) = L(M2)} is not r.e. Checking Equality Proposition EQtm = {⟨M1, M2⟩\| L(M1) = L(M2)} is not r.e. Proof. We will give a reduction f from Etm (assume that we know Etm is R.E.) to EQtm. Checking Equality Proposition EQtm = {⟨M1, M2⟩\| L(M1) = L(M2)} is not r.e. Proof. We will give a reduction f from Etm (assume that we know Etm is R.E.) to EQtm. Let M1 be the Turing machine that on any input, halts and rejects Checking Equality Proposition EQtm = {⟨M1, M2⟩\| L(M1) = L(M2)} is not r.e. Proof. We will give a reduction f from Etm (assume that we know Etm is R.E.) to EQtm. Let M1 be the Turing machine that on any input, halts and rejects i.e., L(M1) = ∅. Take f (M) = ⟨M, M1⟩. Checking Equality Proposition EQtm = {⟨M1, M2⟩\| L(M1) = L(M2)} is not r.e. Proof. We will give a reduction f from Etm (assume that we know Etm is R.E.) to EQtm. Let M1 be the Turing machine that on any input, halts and rejects i.e., L(M1) = ∅. Take f (M) = ⟨M, M1⟩. Observe M ∈Etm iﬀL(M) = ∅iﬀL(M) = L(M1) iﬀ ⟨M, M1⟩∈EQtm. □ Checking Properties Given M Does L(M) contain M? Is L(M) non-empty? Is L(M) empty?   Undecidable Is L(M) inﬁnite? Is L(M) ﬁnite? Is L(M) co-ﬁnite (i.e., is L(M) ﬁnite)? Is L(M) = Σ∗? Which of these properties can be decided? Checking Properties Given M Does L(M) contain M? Is L(M) non-empty? Is L(M) empty?   Undecidable Is L(M) inﬁnite? Is L(M) ﬁnite? Is L(M) co-ﬁnite (i.e., is L(M) ﬁnite)? Is L(M) = Σ∗?        Undecidable Which of these properties can be decided? None! Checking Properties Given M Does L(M) contain M? Is L(M) non-empty? Is L(M) empty?   Undecidable Is L(M) inﬁnite? Is L(M) ﬁnite? Is L(M) co-ﬁnite (i.e., is L(M) ﬁnite)? Is L(M) = Σ∗?        Undecidable Which of these properties can be decided? None! By Rice’s Theorem Properties Deﬁnition A property of languages is simply a set of languages. Properties Deﬁnition A property of languages is simply a set of languages. We say L satisﬁes the property P if L ∈P. Properties Deﬁnition A property of languages is simply a set of languages. We say L satisﬁes the property P if L ∈P. Deﬁnition For any property P, deﬁne language LP to consist of Turing Machines which accept a language in P: LP = {M \| L(M) ∈P} Properties Deﬁnition A property of languages is simply a set of languages. We say L satisﬁes the property P if L ∈P. Deﬁnition For any property P, deﬁne language LP to consist of Turing Machines which accept a language in P: LP = {M \| L(M) ∈P} Deciding LP: deciding if a language represented as a TM satisﬁes the property P. ▶Example: {M \| L(M) is inﬁnite} Properties Deﬁnition A property of languages is simply a set of languages. We say L satisﬁes the property P if L ∈P. Deﬁnition For any property P, deﬁne language LP to consist of Turing Machines which accept a language in P: LP = {M \| L(M) ∈P} Deciding LP: deciding if a language represented as a TM satisﬁes the property P. ▶Example: {M \| L(M) is inﬁnite}; Etm = {M \| L(M) = ∅} Properties Deﬁnition A property of languages is simply a set of languages. We say L satisﬁes the property P if L ∈P. Deﬁnition For any property P, deﬁne language LP to consist of Turing Machines which accept a language in P: LP = {M \| L(M) ∈P} Deciding LP: deciding if a language represented as a TM satisﬁes the property P. ▶Example: {M \| L(M) is inﬁnite}; Etm = {M \| L(M) = ∅} ▶Non-example: {M \| M has 15 states} Properties Deﬁnition A property of languages is simply a set of languages. We say L satisﬁes the property P if L ∈P. Deﬁnition For any property P, deﬁne language LP to consist of Turing Machines which accept a language in P: LP = {M \| L(M) ∈P} Deciding LP: deciding if a language represented as a TM satisﬁes the property P. ▶Example: {M \| L(M) is inﬁnite}; Etm = {M \| L(M) = ∅} ▶Non-example: {M \| M has 15 states} ← −This is a property of TMs, and not languages! Trivial Properties Deﬁnition A property is trivial if either it is not satisﬁed by any r.e. language, or if it is satisﬁed by all r.e. languages. Trivial Properties Deﬁnition A property is trivial if either it is not satisﬁed by any r.e. language, or if it is satisﬁed by all r.e. languages. Otherwise it is non-trivial. Trivial Properties Deﬁnition A property is trivial if either it is not satisﬁed by any r.e. language, or if it is satisﬁed by all r.e. languages. Otherwise it is non-trivial. Example Some trivial properties: ▶Pall = set of all languages ▶Pr.e. = set of all r.e. languages ▶P where P is trivial Trivial Properties Deﬁnition A property is trivial if either it is not satisﬁed by any r.e. language, or if it is satisﬁed by all r.e. languages. Otherwise it is non-trivial. Example Some trivial properties: ▶Pall = set of all languages ▶Pr.e. = set of all r.e. languages ▶P where P is trivial ▶P = {L \| L is recognized by a TM with an even number of states} Trivial Properties Deﬁnition A property is trivial if either it is not satisﬁed by any r.e. language, or if it is satisﬁed by all r.e. languages. Otherwise it is non-trivial. Example Some trivial properties: ▶Pall = set of all languages ▶Pr.e. = set of all r.e. languages ▶P where P is trivial ▶P = {L \| L is recognized by a TM with an even number of states} = Pr.e. Trivial Properties Deﬁnition A property is trivial if either it is not satisﬁed by any r.e. language, or if it is satisﬁed by all r.e. languages. Otherwise it is non-trivial. Example Some trivial properties: ▶Pall = set of all languages ▶Pr.e. = set of all r.e. languages ▶P where P is trivial ▶P = {L \| L is recognized by a TM with an even number of states} = Pr.e. Observation. For any trivial property P, LP is decidable. (Why?) Trivial Properties Deﬁnition A property is trivial if either it is not satisﬁed by any r.e. language, or if it is satisﬁed by all r.e. languages. Otherwise it is non-trivial. Example Some trivial properties: ▶Pall = set of all languages ▶Pr.e. = set of all r.e. languages ▶P where P is trivial ▶P = {L \| L is recognized by a TM with an even number of states} = Pr.e. Observation. For any trivial property P, LP is decidable. (Why?) Then LP = Σ∗or LP = ∅. Rice’s Theorem Proposition If P is a non-trivial property, then LP is undecidable. Rice’s Theorem Proposition If P is a non-trivial property, then LP is undecidable. ▶Thus {M \| L(M) ∈P} is not decidable (unless P is trivial) Rice’s Theorem Proposition If P is a non-trivial property, then LP is undecidable. ▶Thus {M \| L(M) ∈P} is not decidable (unless P is trivial) We cannot algorithmically determine any interesting property of languages represented as Turing Machines! Properties of TMs Note. Properties of TMs, as opposed to those of languages they accept, may or may not be decidable. Properties of TMs Note. Properties of TMs, as opposed to those of languages they accept, may or may not be decidable. Example {⟨M⟩\| M has 193 states} {⟨M⟩\| M uses at most 32 tape cells on blank input} Decidable {⟨M⟩\| M halts on blank input} {⟨M⟩\| on input 0011 M at some point writes the symbol $ on its tape}   Undecidable Proof of Rice’s Theorem Rice’s Theorem If P is a non-trivial property, then LP is undecidable. Proof. Proof of Rice’s Theorem Rice’s Theorem If P is a non-trivial property, then LP is undecidable. Proof. ▶Suppose P non-trivial and ∅̸∈P. Proof of Rice’s Theorem Rice’s Theorem If P is a non-trivial property, then LP is undecidable. Proof. ▶Suppose P non-trivial and ∅̸∈P. ▶(If ∅∈P, then in the following we will be showing LP is undecidable. Then LP = LP is also undecidable.) Proof of Rice’s Theorem Rice’s Theorem If P is a non-trivial property, then LP is undecidable. Proof. ▶Suppose P non-trivial and ∅̸∈P. ▶(If ∅∈P, then in the following we will be showing LP is undecidable. Then LP = LP is also undecidable.) ▶Recall LP = {⟨M⟩\| L(M) satisﬁes P}. We’ll reduce Atm to LP. Proof of Rice’s Theorem Rice’s Theorem If P is a non-trivial property, then LP is undecidable. Proof. ▶Suppose P non-trivial and ∅̸∈P. ▶(If ∅∈P, then in the following we will be showing LP is undecidable. Then LP = LP is also undecidable.) ▶Recall LP = {⟨M⟩\| L(M) satisﬁes P}. We’ll reduce Atm to LP. ▶Then, since Atm is undecidable, LP is also undecidable. Proof of Rice’s Theorem Rice’s Theorem If P is a non-trivial property, then LP is undecidable. Proof. ▶Suppose P non-trivial and ∅̸∈P. ▶(If ∅∈P, then in the following we will be showing LP is undecidable. Then LP = LP is also undecidable.) ▶Recall LP = {⟨M⟩\| L(M) satisﬁes P}. We’ll reduce Atm to LP. ▶Then, since Atm is undecidable, LP is also undecidable. ··→ Proof of Rice’s Theorem Proof (contd). Since P is non-trivial, at least one r.e. language satisﬁes P. Proof of Rice’s Theorem Proof (contd). Since P is non-trivial, at least one r.e. language satisﬁes P. i.e., L(M0) ∈P for some TM M0. Proof of Rice’s Theorem Proof (contd). Since P is non-trivial, at least one r.e. language satisﬁes P. i.e., L(M0) ∈P for some TM M0. Will show a reduction f that maps an instance ⟨M, w⟩for Atm, to N such that ▶If M accepts w then N accepts the same language as M0. ▶Then L(N) = L(M0) ∈P ▶If M does not accept w then N accepts ∅. ▶Then L(N) = ∅̸∈P Proof of Rice’s Theorem Proof (contd). Since P is non-trivial, at least one r.e. language satisﬁes P. i.e., L(M0) ∈P for some TM M0. Will show a reduction f that maps an instance ⟨M, w⟩for Atm, to N such that ▶If M accepts w then N accepts the same language as M0. ▶Then L(N) = L(M0) ∈P ▶If M does not accept w then N accepts ∅. ▶Then L(N) = ∅̸∈P Thus, ⟨M, w⟩∈Atm iﬀN ∈LP. Proof of Rice’s Theorem Proof (contd). Since P is non-trivial, at least one r.e. language satisﬁes P. i.e., L(M0) ∈P for some TM M0. Will show a reduction f that maps an instance ⟨M, w⟩for Atm, to N such that ▶If M accepts w then N accepts the same language as M0. ▶Then L(N) = L(M0) ∈P ▶If M does not accept w then N accepts ∅. ▶Then L(N) = ∅̸∈P Thus, ⟨M, w⟩∈Atm iﬀN ∈LP. ··→ Proof of Rice’s Theorem Proof (contd). The reduction f maps ⟨M, w⟩to N, where N is a TM that behaves as follows: On input x Ignore the input and run M on w If M does not accept (or doesn’t halt) then do not accept x (or do not halt) If M does accept w then run M0 on x and accept x iff M0 does. Notice that indeed if M accepts w then L(N) = L(M0). Otherwise L(N) = ∅. □ Rice’s Theorem Recap Every non-trivial property of r.e. languages is undecidable Rice’s Theorem Recap Every non-trivial property of r.e. languages is undecidable ▶Rice’s theorem says nothing about properties of Turing machines Rice’s Theorem Recap Every non-trivial property of r.e. languages is undecidable ▶Rice’s theorem says nothing about properties of Turing machines ▶Rice’s theorem says nothing about whether a property of languages is recurisvely enumerable or not. Big Picture . . . again Regular CFL L0n1n Decidable Lanbncn Recursively Enumerable Languages Ld, Atm, Etm Atm, Etm, HALT Big Picture . . . again Regular CFL L0n1n Decidable Lanbncn Recursively Enumerable Languages Ld, Atm, Etm “almost all” properties! Atm, Etm, HALT Boolean Operators Boolean Operators Proposition Decidable languages are closed under union, intersection, and complementation. Boolean Operators Proposition Decidable languages are closed under union, intersection, and complementation. Proof. Given TMs M1, M2 that decide languages L1, and L2 ▶A TM that decides L1 ∪L2: on input x, run M1 and M2 on x, and accept iﬀeither accepts. Boolean Operators Proposition Decidable languages are closed under union, intersection, and complementation. Proof. Given TMs M1, M2 that decide languages L1, and L2 ▶A TM that decides L1 ∪L2: on input x, run M1 and M2 on x, and accept iﬀeither accepts. (Similarly for intersection.) Boolean Operators Proposition Decidable languages are closed under union, intersection, and complementation. Proof. Given TMs M1, M2 that decide languages L1, and L2 ▶A TM that decides L1 ∪L2: on input x, run M1 and M2 on x, and accept iﬀeither accepts. (Similarly for intersection.) ▶A TM that decides L1: On input x, run M1 on x, and accept if M1 rejects, and reject if M1 accepts. □ Regular Operators Proposition Decidable languages are closed under concatenation and Kleene Closure. Proof. Given TMs M1 and M2 that decide languages L1 and L2. ▶A TM to decide L1L2: Regular Operators Proposition Decidable languages are closed under concatenation and Kleene Closure. Proof. Given TMs M1 and M2 that decide languages L1 and L2. ▶A TM to decide L1L2: On input x, for each of the \|x\| + 1 ways to divide x as yz: run M1 on y and M2 on z, and accept if both accept. Else reject. Regular Operators Proposition Decidable languages are closed under concatenation and Kleene Closure. Proof. Given TMs M1 and M2 that decide languages L1 and L2. ▶A TM to decide L1L2: On input x, for each of the \|x\| + 1 ways to divide x as yz: run M1 on y and M2 on z, and accept if both accept. Else reject. ▶A TM to decide L∗ 1: Regular Operators Proposition Decidable languages are closed under concatenation and Kleene Closure. Proof. Given TMs M1 and M2 that decide languages L1 and L2. ▶A TM to decide L1L2: On input x, for each of the \|x\| + 1 ways to divide x as yz: run M1 on y and M2 on z, and accept if both accept. Else reject. ▶A TM to decide L∗ 1: On input x, if x = ϵ accept. Else, for each of the 2\|x\|−1 ways to divide x as w1 . . . wk (wi ̸= ϵ): run M1 on each wi and accept if M1 accepts all. Else reject. □ Boolean Operators Proposition R.E. languages are closed under union, and intersection. Boolean Operators Proposition R.E. languages are closed under union, and intersection. Proof. Given TMs M1, M2 that recognize languages L1, L2 Boolean Operators Proposition R.E. languages are closed under union, and intersection. Proof. Given TMs M1, M2 that recognize languages L1, L2 ▶A TM that recognizes L1 ∪L2: on input x, run M1 and M2 on x in parallel, and accept iﬀeither accepts. Boolean Operators Proposition R.E. languages are closed under union, and intersection. Proof. Given TMs M1, M2 that recognize languages L1, L2 ▶A TM that recognizes L1 ∪L2: on input x, run M1 and M2 on x in parallel, and accept iﬀeither accepts. (Similarly for intersection; but no need for parallel simulation) □ Complementation Proposition R.E. languages are not closed under complementation. Proof. Atm is r.e. but Atm is not. □ Regular Operations Proposition R.E languages are closed under concatenation and Kleene closure. Proof. Given TMs M1 and M2 recognizing L1 and L2 ▶A TM to recognize L1L2: Regular Operations Proposition R.E languages are closed under concatenation and Kleene closure. Proof. Given TMs M1 and M2 recognizing L1 and L2 ▶A TM to recognize L1L2: On input x, do in parallel, for each of the \|x\| + 1 ways to divide x as yz: run M1 on y and M2 on z, and accept if both accept. Else reject. Regular Operations Proposition R.E languages are closed under concatenation and Kleene closure. Proof. Given TMs M1 and M2 recognizing L1 and L2 ▶A TM to recognize L1L2: On input x, do in parallel, for each of the \|x\| + 1 ways to divide x as yz: run M1 on y and M2 on z, and accept if both accept. Else reject. ▶A TM to recognize L∗ 1: Regular Operations Proposition R.E languages are closed under concatenation and Kleene closure. Proof. Given TMs M1 and M2 recognizing L1 and L2 ▶A TM to recognize L1L2: On input x, do in parallel, for each of the \|x\| + 1 ways to divide x as yz: run M1 on y and M2 on z, and accept if both accept. Else reject. ▶A TM to recognize L∗ 1: On input x, if x = ϵ accept. Else, do in parallel, for each of the 2\|x\|−1 ways to divide x as w1 . . . wk (wi ̸= ϵ): run M1 on each wi and accept if M1 accepts all. Else reject. □
14245	https://www.studypug.com/basic-math-help/understand-integer-multiplication	Home Basic Math Multiplying and Dividing Integers Understanding integer multiplication Get the most by viewing this topic in your current grade. Pick your course now. Now Playing:Understand integer multiplication– Example 0 Introduction to integer multiplication Examples Write the multiplication statement for each diagram. Solve. (+2)×(+5) (-4)×(+6) (+7)×(+3) (-2)×(-9) Dave can do 15 sit-ups in one minute. How many sit-ups can he do in 8 minutes? A building has 8 stories above ground and 4 stories below ground. Each story has a height of 5m. What is the total height of the building above ground? What is the total depth of the building below ground? Mary spends $30 per week on meals for a year. Sally spends $125 per month on meals for a year. Who spends more money on meal, and by what amount? View All Practice Build your skill!Try your hand with these practice questions. Sign Up Free to Join! StudyPug is a learning help platform covering math and science from grade 4 all the way to second year university. Our video tutorials, unlimited practice problems, and step-by-step explanations provide you or your child with all the help you need to master concepts. On top of that, it's fun — with achievements, customizable avatars, and awards to keep you motivated. Students Parents Try Free Easily See Your Progress We track the progress you've made on a topic so you know what you've done. From the course view you can easily see what topics have what and the progress you've made on them. Fill the rings to completely master that section or mouse over the icon to see more details. #### Make Use of Our Learning Aids ###### Last Viewed ###### Practice Accuracy ###### Suggested Tasks Get quick access to the topic you're currently learning. See how well your practice sessions are going over time. Stay on track with our daily recommendations. Try Free #### Earn Achievements as You Learn Make the most of your time as you use StudyPug to help you achieve your goals. Earn fun little badges the more you watch, practice, and use our service. #### Create and Customize Your Avatar Play with our fun little avatar builder to create and customize your own avatar on StudyPug. Choose your face, eye colour, hair colour and style, and background. Unlock more options the more you use StudyPug. Try Free Understanding integer multiplication Jump to:NotesConceptExampleFAQsPrerequisitesRelated Notes In this section, we will learn the multiplication of positive and negative integers. At the beginning, diagrams will be used to help us getting familiar with the concept. We will then practice more with actual numbers. Concept Introduction: Understanding Integer Multiplication Integer multiplication is a fundamental mathematical operation that extends beyond simple whole number calculations. Our lesson begins with an essential introduction video, which lays the groundwork for comprehending this crucial concept. This video serves as a visual aid, making the abstract notion of integer multiplication more tangible and accessible. As we delve deeper, we'll explore the intricacies of multiplying both positive and negative integers. Through the use of clear diagrams and practical examples, we'll demystify the process, enabling you to confidently tackle various multiplication scenarios. You'll learn how the signs of integers affect the outcome and discover the patterns that emerge when working with positive and negative numbers. By the end of this lesson, you'll have a solid grasp of integer multiplication, a skill that forms the basis for more advanced mathematical concepts and real-world applications. Example Write the multiplication statement for each diagram. Step 1: Introduction to the Problem Hi, welcome to this question right here. This might seem challenging at first, but it's actually quite simple once you understand the concept. We are trying to figure out the result when two integers are multiplied together. To make this easier, we will use a diagram to illustrate the process. This visual representation will help you understand the multiplication of integers more clearly. Step 2: Understanding the Diagram In the diagram, you will see red circles and blue circles. For the purpose of this explanation, treat the red circles as positive integers and the blue circles as negative integers. Initially, we have six positive circles and six negative circles, which balance each other out to zero. This is because six positives and six negatives cancel each other out. Step 3: Identifying the Elements Next, observe the green arrows in the diagram. These arrows indicate that we are taking away some circles. The act of taking away is represented by a negative sign. Now, count the number of circles being taken away. In this case, we are removing three green circles. This is represented as negative three (-3). Step 4: Counting the Positive Circles Within each of the green circles that we are taking away, there are two red circles. Remember, the red circles are positive. Therefore, each green circle contains two positive integers. This is represented as plus two (+2). Step 5: Formulating the Multiplication Statement Now, we combine these elements to form the multiplication statement. We are taking away three sets of two positive circles. This can be written as -3 (the number of sets being taken away) multiplied by +2 (the number of positive circles in each set). The multiplication statement is thus -3 +2. Step 6: Calculating the Result To find the result, multiply the numbers together. Three times two equals six. Since we are multiplying a negative number by a positive number, the result is negative. Therefore, -3 +2 equals -6. This means that when you take away three sets of two positive circles, you are left with six negative circles. Step 7: Conclusion In conclusion, the diagram helps us visualize the multiplication of integers. By identifying the elements and understanding their significance, we can easily formulate and solve the multiplication statement. The final result of -3 +2 is -6, which is represented by the six negative circles left in the diagram. FAQs What is the rule for multiplying integers with different signs? When multiplying integers with different signs (one positive and one negative), the result is always negative. For example, 5 × (-3) = -15 and (-4) × 2 = -8. This rule is often summarized as "unlike signs make negative." 2. How do integer chips help in understanding multiplication? Integer chips provide a visual representation of multiplication. Red chips represent positive integers, while blue chips represent negative integers. When multiplying, you either "insert" or "remove" chips based on the factors. This method helps students visualize the process and understand why multiplying two negative numbers results in a positive product. 3. Why does multiplying two negative numbers result in a positive product? Multiplying two negative numbers results in a positive product because it's equivalent to "removing" a negative amount a negative number of times. For example, (-3) × (-4) can be thought of as removing 3 negative chips 4 times, which results in adding 12 positive chips, giving a final product of 12. 4. How can I quickly determine the sign of a product when multiplying multiple integers? To quickly determine the sign of a product when multiplying multiple integers, count the number of negative factors. If there's an even number of negative factors, the product will be positive. If there's an odd number of negative factors, the product will be negative. For example, in (-2) × 3 × (-4) × 5, there are two negative factors, so the product will be positive. 5. What are some real-world applications of integer multiplication? Integer multiplication has numerous real-world applications, including: Financial calculations (e.g., calculating profits and losses) Physics problems (e.g., force calculations) Everyday scenarios like party planning or grocery shopping Area calculations for home improvement projects Estimating fuel costs for travelUnderstanding integer multiplication is crucial for solving problems in various fields and everyday situations. Prerequisites Understanding integer multiplication is a crucial skill in mathematics, but to master it, students must first grasp several fundamental concepts. One of the most important prerequisites is comparing and ordering numbers, particularly when dealing with positive and negative numbers. This skill forms the foundation for understanding how integers behave in multiplication. Before diving into multiplication, students should be comfortable with adding integers. Techniques such as adding integers with chips can help visualize the process, which is essential when transitioning to multiplication. The application of integer operations in real-world scenarios further reinforces these concepts, making integer multiplication more relatable and easier to grasp. A solid understanding of the distributive property of multiplication is crucial when working with integer multiplication. This property helps students break down complex multiplication problems into simpler parts, making the process more manageable and less intimidating. While it may seem unrelated at first, proficiency in multiplying decimals can greatly benefit students when learning integer multiplication. The principles of place value and the mechanics of multiplication remain consistent, whether dealing with decimals or integers. Lastly, comparing and ordering rational numbers is an essential skill that ties directly into integer multiplication. Understanding how rational numbers behave in multiplication scenarios helps students develop a more comprehensive grasp of number systems and their operations. By mastering these prerequisite topics, students build a strong foundation for understanding integer multiplication. Each concept contributes to a deeper comprehension of how integers interact when multiplied, making the learning process smoother and more intuitive. As students progress, they'll find that these fundamental skills not only aid in integer multiplication but also serve as building blocks for more advanced mathematical concepts. It's important to note that while some students may be tempted to skip over these prerequisites, doing so can lead to gaps in understanding and difficulties down the line. Taking the time to thoroughly explore and practice these foundational concepts will pay dividends in the long run, not just for integer multiplication, but for mathematics as a whole. By approaching the subject with a solid grounding in these prerequisite topics, students set themselves up for success and a more enjoyable learning experience in mathematics. Place value Determining Common Factors Adding integers
14246	https://ws.engr.illinois.edu/sitemanager/getfile.asp?id=196	106 C H A P T E R 4 Wave Propagation in Free Space In Chapters 2 and 3, we learned Maxwell’s equations in integral form and in differential form. We now have the knowledge of the fundamental laws of electromagnetics that enable us to embark upon the study of their applications. Many of these applications are based on electromagnetic wave phenomena, and hence it is necessary to gain an understanding of the basic principles of wave propagation, which is our goal in this chapter. In particular, we shall consider wave propagation in free space. We shall then in the next chapter consider the interaction of the wave fields with materials to extend the application of Maxwell’s equations to material media and discuss wave propagation in material media. We shall employ an approach in this chapter that will enable us not only to learn how the coupling between space-variations and time-variations of the electric and magnetic fields, as indicated by Maxwell’s equations, results in wave motion, but also to illustrate the basic principle of radiation of waves from an antenna, which will be treated in detail in Chapter 9. In this process, we will also learn several techniques of analysis pertinent to field problems. We shall augment our discussion of radiation and propagation of waves by considering such examples as the principle of an antenna array and polar-ization. Finally, we shall discuss power flow and energy storage associated with the wave motion and introduce the Poynting vector. 4.1 THE INFINITE PLANE CURRENT SHEET In Chapter 3, we learned that the space-variations of the electric and magnetic field components are related to the time-variations of the magnetic and electric field com-ponents, respectively, through Maxwell’s equations. This interdependence gives rise to the phenomenon of electromagnetic wave propagation. In the general case, electro-magnetic wave propagation involves electric and magnetic fields having more than one component, each dependent on all three coordinates, in addition to time. However, a simple and very useful type of wave that serves as a building block in the study of elec-tromagnetic waves consists of electric and magnetic fields that are perpendicular to each other and to the direction of propagation and are uniform in planes perpendicular to the direction of propagation. These waves are known as uniform plane waves . By orienting the coordinate axes such that the electric field is in the -direction, the magnetic field is in the -direction, and the direction of propagation is in the -direction, as shown in Figure 4.1, we have (4.1) (4.2) Uniform plane waves do not exist in practice because they cannot be produced by finite-sized antennas. At large distances from physical antennas and ground, however, the waves can be approximated as uniform plane waves. Furthermore, the principles of guiding of electromagnetic waves along transmission lines and waveguides and the principles of many other wave phenomena can be studied basically in terms of uniform plane waves. Hence, it is very important that we understand the principles of uniform plane wave propagation. H = Hy(z, t)ay E = Ex(z, t)axzyx 4.1 The Infinite Plane Current Sheet 107 E z y x H Direction of Propa gation FIGURE 4.1 Directions of electric and magnetic fields and direction of propagation for a simple case of uniform plane wave. In order to illustrate the phenomenon of interaction of electric and magnetic fields giving rise to uniform plane electromagnetic wave propagation, and the principle of radiation of electromagnetic waves from an antenna, we shall consider a simple, idealized, hypothetical source. This source consists of an infinite sheet lying in the -plane, as shown in Figure 4.2. On this infinite plane sheet a uniformly distributed current varying sinusoidally with time flows in the negative -direction. Since the cur-rent is distributed on a surface, we talk of surface current density in order to express the current distribution mathematically. The surface current density, denoted by the symbol , is a vector quantity having the magnitude equal to the current per unit width (A/m) crossing an infinitesimally long line, on the surface, oriented so as to maximize the current. The direction of is then normal to the line and toward the side of the current flow. In the present case, the surface current density is given by (4.3) where is a constant and is the radian frequency of the sinusoidal time-variation of the current density. Because of the uniformity of the surface current density on the infinite sheet, if we consider any line of width w parallel to the -axis, as shown in Figure 4.2, the y vJS0 JS = -JS0 cos vt ax for z = 0 JS JSxxy 108 Chapter 4 Wave Propagation in Free Space current crossing that line is simply given by times the current density, that is, . If the current density is nonuniform, we have to perform an integration along the width of the line in order to find the current crossing the line. In view of the sinusoidal time-variation of the current density, the current crossing the width w actually alternates between negative - and positive -directions, that is, downward and upward. The time history of the current flow for one period of the sinusoidal variation is illustrated in Figure 4.3, with the lengths of the lines indicating the mag-nitudes of the current. xx wJS0 cos vt w 0p2p wJ S0wJ S0 vtvt FIGURE 4.3 Time history of current flow across a line of width wparallel to the y-axis for the current sheet of Figure 4.2. 4.2 MAGNETIC FIELD ADJACENT TO THE CURRENT SHEET In the previous section, we introduced the infinite current sheet lying in the -plane and upon which a surface current flows with density given by (4.4) Our goal is to find the electromagnetic field due to this time-varying current distribu-tion. In order to do this, we have to solve Faraday’s and Ampere’s circuital laws simul-taneously. Since we have here only an -component of the current density independent x JS = -JS0 cos vt axxy w z y x JS FIGURE 4.2 Infinite plane sheet in the -plane carrying surface current of uniform density. xy 4.2 Magnetic Field Adjacent to the Current Sheet 109 of x and y, the equations of interest are (4.5) (4.6) The quantity on the right side of (4.6) represents volume current density, whereas we now have a surface current density. Furthermore, in the free space on either side of the current sheet the current density is zero and the differential equations reduce to (4.7) (4.8) To obtain the solutions for and on either side of the current sheet, we therefore have to solve these two differential equations simultaneously. To obtain a start on the solution, however, we need to consider the surface cur-rent distribution and find the magnetic field immediately adjacent to the current sheet. This is done by making use of Ampere’s circuital law in integral form given by (4.9) and applying it to a rectangular closed path abcda , as shown in Figure 4.4, with the sides and lying immediately adjacent to the current sheet, that is, touching the current sheet, and on either side of it. This choice of the rectangular path is not arbi-trary but is intentionally chosen to achieve the task of finding the required magnetic field. First, we note from (4.6) that an -directed current density gives rise to a xcd ab CC H # dl = LS J# dS + ddt LS D # dS HyEx 0Hy 0z = - 0Dx 0t 0Ex 0z = - 0By 0tJx 0Hy 0z = - aJx + 0Dx 0t b 0Ex 0z = - 0By 0t z y d bc a x JS FIGURE 4.4 Rectangular path enclosing a portion of the current on the infinite plane current sheet. 110 Chapter 4 Wave Propagation in Free Space magnetic field in the -direction. At the source of the current, this magnetic field must also have a differential in the third direction, namely, the -direction. In fact, from sym-metry considerations, we can say that on and must be equal in magnitude and opposite in direction. If we now consider the line integral of H around the rectangular path abcda , we have (4.10) The second and the fourth integrals on the right side of (4.10) are, however, equal to zero, since H is normal to the sides bc and da and furthermore and are infinitesi-mally small. The first and third integrals on the right side of (4.10) are given by Thus, (4.11) since .We have just evaluated the left side of (4.9) for the particular problem under consideration here. To complete the task of finding the magnetic field adjacent to the current sheet, we now evaluate the right side of (4.9), which consists of two terms. The second term is, however, zero, since the area enclosed by the rectangular path is zero in view of the infinitesimally small thickness of the current sheet. The first term is not zero, since there is a current flowing on the sheet. Thus, the first term is simply equal to the current enclosed by the path abcda in the right-hand sense, that is, the current crossing the width ab toward the negative -direction. This is equal to the surface cur-rent density multiplied by the width , that is, . Thus, substituting for the quantities on either side of (4.9), we have or (4.12) It then follows that (4.13) [Hy]cd = - JS02 cos vt [Hy]ab = JS02 cos vt 2[ Hy]ab (ab ) = JS0 cos vt (ab ) JS0 cos vt (ab )ab x [Hy]cd = -[Hy]ab Cabcda H # dl = [Hy]ab (ab ) - [Hy]cd (cd ) = 2[ Hy]ab (ab ) L dc H # dl = -[Hy]cd (cd ) L ba H # dl = [Hy]ab (ab ) da bc Labcda H # dl = L ba H # dl + L cb H # dl + L dc H # dl + L ad H # dl cd ab Hyzy4.3 Successive Solution of Maxwell’s Equations 111 Thus, immediately adjacent to the current sheet the magnetic field intensity has a magnitude and is directed in the positive -direction on the side and in the negative -direction on the side . This is illustrated in Figure 4.5. It is cau-tioned that this result is true only for points right next to the current sheet, since if we consider points at some distance from the current sheet, the second term on the right side of (4.9) will no longer be zero. z 6 0yz 7 0yJS02 cos vt z z!0z"0 y HH x JS FIGURE 4.5 Magnetic field adjacent to and on either side of the infinite plane current sheet. This section may be omitted without loss of continuity. The technique we have used here for finding the magnetic field adjacent to the time-varying current sheet by using Ampere’s circuital law in integral form is a stan-dard procedure for finding the static electric and magnetic fields due to static charge and current distributions, possessing certain symmetries, by using Gauss’ law for the electric field and Ampere’s circuital law in integral forms, respectively, as we have al-ready demonstrated in Chapter 2. Since for the static field case the terms involving time derivatives are zero, Ampere’s circuital law simplifies to Hence, if the current distribution were not varying with time, then in order to compute the magnetic field we can choose a rectangular path of any width bc and it would still enclose the same current, namely, the current on the sheet. Thus, the magnetic field would be independent of the distance away from the sheet on either side of it. There are several problems in static fields that can be solved in this manner. We shall not dis-cuss these here; instead, we shall include a few cases in the problems for the interested reader and shall continue with the derivation of the electromagnetic field due to our time-varying current sheet in the following section. 4.3 SUCCESSIVE SOLUTION OF MAXWELL’S EQUATIONS In the preceding section, we found the magnetic field adjacent to the infinite plane sheet of current introduced in Section 4.1. Now, to find the solutions for the fields everywhere on either side of the current sheet, let us first consider the region .z 7 0 CC H # dl = LS J# dS112 Chapter 4 Wave Propagation in Free Space In this region, the fields simultaneously satisfy the two differential equations (4.7) and (4.8) and with the constraint that the magnetic field at is given by (4.12). To find the solutions for these differential equations, we have a choice of starting with the solution for given by (4.12) and solving them successively and repeatedly in a step-by-step manner until the solutions satisfy both differential equations or of combining the two differential equations into one and then solving the single equation subject to the constraint at . Although it is somewhat longer and tedious, we shall use the first approach in this section in order to obtain a feeling for the mechanism of inter-action between the electric and magnetic fields. We shall consider the second and more conventional approach in the following section. To simplify the task of the repetitive solution of the two differential equations (4.7) and (4.8), we shall employ the phasor technique. Thus, by letting (4.14) (4.15) where Re stands for real part of and and are the phasors corresponding to the time functions and , respectively, and replacing the time functions in (4.7) and (4.8) by the corresponding phasor functions and by , we obtain the dif-ferential equations for the phasor functions as (4.16) (4.17) We also note that since (4.12) can be written as the solution for the phasor at is given by (4.18) We start with (4.18) and solve (4.16) and (4.17) successively and repeatedly, and after obtaining the final solutions for and , we put them in (4.14) and (4.15), respec-tively, to obtain the solutions for the real fields. Thus, starting with (4.18) and substituting it in (4.16), we get Integrating both sides of this equation with respect to z, we have E – x = -jvm 0 JS0z 2 + C – 0E – x 0z = -jvm 0 JS02 H – y E – x [H – y]z = 0 = JS02 z = 0H – y [Hy]ab = Re a JS02 ejvt b 0H – y 0z = -jvD – x = -jvP0E – x 0E – x 0z = -jvB – y = -jvm 0H – y jv0>0tHy(z, t)Ex(z, t) H – y (z)E – x (z) Hy(z, t) = Re [H – y (z)ejvt ] Ex(z, t) = Re [E – x (z)ejvt ] z = 0 Hyz = 04.3 Successive Solution of Maxwell’s Equations 113 where is the constant of integration. This constant of integration must, however, be equal to since the first term on the right side tends to zero as Thus, (4.19) Now, substituting (4.19) into (4.17), we obtain (4.20) We have thus obtained a second-order solution for , which, however, does not satisfy (4.16) together with the solution for given by (4.19). Hence, we must continue the step-by-step solution by substituting (4.20) into (4.16) and finding a higher-order solution for , and so on. Thus, by substituting (4.20) into (4.16), we get (4.21) From (4.17), we then have (4.22) + JS02 a1 - v 2m0P0z22 + v 4m20P20z424 b = -jvP0[E – x]z = 0 az - v2m0P0z36 b v 2m0P0JS02 a z22 - v2m0P0z424 b + [H – y ]z = 0 H – y = -jvP0[E – x]z = 0 az - v2m0P0z36 b 0H – y 0z = -jvP0[E – x]z = 0 a1 - v2m0P0z22 b - v2m0P0JS02 az - v2m0P0z36 b = [E – x]z = 0 a1 - v2m0P0z22 b - jvm 0JS02 az - v2m0P0z36 b E – x = -v2m0P0 z22 [E – x]z = 0 - jvm 0 JS02 az - v2m0P0z36 b + [E – x]z = 0 = -v2m0P0z[E – x]z = 0 - jvm 0 JS02 a1 - v2m0P0z22 b 0E – x 0z = -jvm 0 e -jvP0z[E – x]z = 0 + JS02 a1 - v2m0P0z22 b f E – xE – xH – y = -jvP0z[E – x]z = 0 + JS02 a1 - v2m0P0z22 b = -jvP0z[E – x]z = 0 - v2m0P0 JS0z24 + JS02 H – y = -jvP0z[E – x]z = 0 - v2m0P0 JS0z24 + [H – y]z = 0 = -jvP0[E – x]z = 0 - v 2m0P0 JS0z 2 0H – y 0z = -jvP0 e -jvm 0 JS0z 2 + [E – x]z = 0 f E – x = -jvm 0 JS0z 2 + [E – x]z = 0 z : 0. [E – x ]z = 0 , C –114 Chapter 4 Wave Propagation in Free Space Continuing in this manner, we will get infinite series expressions for and as follows: (4.23) (4.24) where we have introduced the notations (4.25) (4.26) It is left to the student to verify that the two expressions (4.23) and (4.24) simultane-ously satisfy the two differential equations (4.16) and (4.17). Now, noting that and substituting into (4.23) and (4.24), we have (4.27) (4.28) We now obtain the expressions for the real fields by putting (4.27) and (4.28) into (4.14) and (4.15), respectively. Thus, (4.29) = cos bz (C cos vt + D sin vt) + h0JS02 sin bz sin vt = cos bz Re 5[E – x]z = 0 ejvt 6 + h0JS02 sin bz Re [ej(vt - p>2) ] Ex(z, t) = Re e [E – x]z = 0 cos bz ejvt - jh0JS02 sin bz ejvt f H – y = -j 1 h0[E – x]z = 0 sin bz + JS02 cos bz E – x = [E – x]z = 0 cos bz - jh0JS02 sin bz sin bz = bz - (bz)33! + (bz)55! + Á cos bz = 1 - (bz)22! + (bz)44! - Á h0 = A m0 P0 b = v1m0P0 JS02 c1 - (bz)22! + (bz)44! - Á d H – y = -j 1 h0[E – x]z = 0 cbz - (bz)33! + (bz)55! - Á d jh0JS02 cbz - (bz)33! + (bz)55! - Á d E – x = [E – x]z = 0 c1 - (bz)22! + (bz)44! - Á d H – y E – x4.4 Solution by Wave Equation 115 (4.30) where we have replaced the quantity by , in which C and D are arbitrary constants to be determined. Making use of trigonometric identities and proceeding further, we write (4.29) and (4.30) as (4.31) (4.32) Equation (4.32) is the solution for that together with the solution for given by (4.31) satisfies the two differential equations (4.7) and (4.8) and that reduces to (4.12) for . Likewise, we can obtain the solutions for and for the region by starting with given by (4.13) and proceeding in a similar manner. We shall, however, proceed with the evaluation of the constants C and D in (4.31) and (4.32). In order to do this, we first have to understand the meanings of the functions and . We shall do this in Section 4.5. 4.4 SOLUTION BY WAVE EQUATION In Section 4.3, we found the solutions to the two simultaneous differential equations (4.7) and (4.8) by solving them successively and repeatedly in a step-by-step manner. In this section, we shall consider an alternative and more conventional method by com-bining the two equations into a single equation and then solving it. We recall that the two simultaneous differential equations to be satisfied in the free space on either side of the current sheet are (4.33) (4.34) 0Hy 0z = - 0Dx 0t = -P 0 0Ex 0t 0Ex 0z = - 0By 0t = -m0 0Hy 0t sin ( vt < bz)cos (vt < bz)[Hy]z = 0 -z 6 0 ExHyz = 0 ExHy D 2h0sin (vt - bz) - D 2h0sin (vt + bz) Hy(z, t) = 2C + hJS04h0cos (vt - bz) - 2C - h0JS04h0cos (vt + bz) D 2 sin (vt - bz) + D 2 sin (vt + bz) Ex(z, t) = 2C + h0JS04 cos (vt - bz) + 2C - h0JS04 cos (vt + bz)(C cos vt + D sin vt)Re 5[E – x ]z = 0 ejvt 6 = 1 h0 sin bz (C sin vt - D cos vt) + JS02 cos bz cos vt = 1 h0 sin bz Re 5[E – x ]z = 0 ej(vt - p>2) 6 + JS02 cos bz Re [ ejvt ] Hy(z, t) = Re e -j 1 h0 [E – x ]z = 0 sin bz ejvt + JS02 cos bz ejvt f116 Chapter 4 Wave Propagation in Free Space Differentiating (4.33) with respect to z and then substituting for from (4.34), we obtain or (4.35) We have thus eliminated from (4.33) and (4.34) and obtained a single second-order partial differential equation involving only. Equation (4.35) is known as the wave equation . A technique of solving this equa-tion is the separation of variables technique. Since it is a differential equation involving two variables, z and t, the technique consists of assuming that the required solution is the product of two functions, one of which is a function of z only and the second is a function of t only. Denoting these functions to be Z and T, respectively, we have (4.36) Substituting (4.36) into (4.35) and dividing throughout by , we obtain (4.37) In (4.37), the left side is a function of z only and the right side is a function of t only. In order for this to be satisfied, they both must be equal to a constant. Hence, setting them equal to a constant, say , we have (4.38a) (4.38b) We have thus obtained two ordinary differential equations involving separately the two variables z and t; hence, the technique is known as the separation of variables technique. The constant in (4.38a) and (4.38b) is not arbitrary, since for the case of the sinusoidally time-varying current source the fields must also be sinusoidally time-varying with the same frequency, although not necessarily in phase with the source. Thus, the solution for T(t) must be of the form (4.39) where A and B are arbitrary constants to be determined. Substitution of (4.39) into (4.38b) gives us . The solution for (4.38a) is then given by (4.40) = A¿ cos bz + B¿ sin bz Z(z) = A¿ cos v1m 0P0z + B¿ sin v1m0P0z a2 = -v2 T(t) = A cos vt + B sin vt a2 d2Tdt 2 = a2T d2Zdz 2 = a2m0P0Z a21 m0P0Z d2Zdz 2 = 1 T d2Tdt 2 m0P0Z(z)T(t) Ex(z, t) = Z(z)T(t) ExHy 0 2Ex 0z2 = m0P0 0 2Ex 0t2 0 2Ex 0z2 = -m0 00z a 0Hy 0t b = -m0 00t a 0Hy 0z b = -m 0 00t a -P 0 0Ex 0t b 0Hy>0z4.4 Solution by Wave Equation 117 where and are arbitrary constants to be determined and we have defined (4.41) The solution for is then given by (4.42) The corresponding solution for can be obtained by substituting (4.42) into one of the two equations (4.33) and (4.34). Thus, using (4.34), we get Defining (4.43) we have (4.44) Equation (4.44) is the general solution for valid on both sides of the current sheet. In order to deduce the arbitrary constants, we first recall that the magnetic field adjacent to the current sheet is given by (4.45) Thus, for ,or C¿ = 0 and D¿ = h0JS021 h0[-C¿ sin vt + D¿cos vt] = JS02 cos vtz 7 0 Hy = d JS02 cos vt for z = 0+- JS02 cos vt for z = 0- Hy C¿ cos bz sin vt + D¿ cos bz cos vt] Hy = 1 h0[C sin bz sin vt - D sin bz cos vt h0 = bvP0 = v1m0P0 vP0 = A m0 P0 C¿ cos bz sin vt + D¿cos bz cos vt ] Hy = vP0 b [C sin bz sin vt - D sin bz cos vt v C¿ sin bz sin vt + vD¿sin bz cos vt ] 0Hy 0z = -P 0[-v C cos bz sin vt + vD cos bz cos vt Hy C¿ sin bz cos vt + D¿ sin bz sin vt = C cos bz cos vt + D cos bz sin vt Ex = (A¿ cos bz + B¿ sin bz)( A cos vt + B sin vt) Ex b = v1m0P0 B¿A¿118 Chapter 4 Wave Propagation in Free Space giving us (4.46) (4.47) Making use of trigonometric identities and proceeding further, we write (4.47) and (4.46) as (4.48) (4.49) Equation (4.49) is the solution for that together with the solution for given by (4.48) satisfies the two differential equations (4.7) and (4.8) and that reduces to (4.12) for . Similarly, we can obtain the solutions for and for the region by using the value of to evaluate and in (4.44). We shall, however, proceed with the evaluation of the constants C and D in (4.48) and (4.49). In order to do this, we first have to understand the meanings of the functions and . We shall do this in the following section. 4.5 UNIFORM PLANE WAVES In the previous two sections, we derived the solutions for and , due to the infinite plane sheet of sinusoidally time-varying uniform current density, for the region .These solutions consist of the functions and , which are de-pendent on both time and distance. Let us first consider the function . To understand the behavior of this function, we note that for a fixed value of time it varies in a cosinusoidal manner with the distance z. Let us therefore consider three values of time, and , and examine the sketches of this function versus z for these three times. By noting that for t = p 2v, cos (vt - bz) = cos a p 2 - bzb = sin bz for t = p 4v, cos (vt - bz) = cos a p 4 - bzb for t = 0, cos (vt - bz) = cos (-bz) = cos bzt = p>2vt = 0, t = p>4v,cos (vt - bz)sin (vt < bz)cos (vt < bz) z 7 0 HyEx sin ( vt < bz)cos (vt < bz) D¿C¿[Hy]z = 0 -z 6 0 ExHyz = 0 ExHy D 2h0sin (vt - bz) - D 2h0sin (vt + bz) Hy (z, t) = 2C + h0JS04h0cos (vt - bz) - 2C - h0JS04h0cos (vt + bz) D 2 sin (vt - bz) + D 2 sin (vt + bz) Ex (z, t) = 2C + h0JS04 cos (vt - bz) + 2C - h0JS04 cos (vt + bz) Ex = h0JS02 sin bz sin vt + cos bz (C cos vt + D sin vt ) Hy = JS02 cos bz cos vt + 1 h0sin bz (C sin vt - D cos vt )4.5 Uniform Plane Waves 119 we draw the sketches of the three functions as shown in Figure 4.6. co s(vt#bz) b pp 2b 3p 2b 5p 2b 2p b p 4v t$p 2v t$t$0 1 0z FIGURE 4.6 Sketches of the function versus zfor three values of . tcos (vt-bz) It is evident from Figure 4.6 that the sketch of the function for is a replica of the function for except that it is shifted by a distance of toward the positive z-direction. Similarly, the sketch of the function for is a replica of the function for except that it is shifted by a distance of toward the positive z-direction. Thus as time progresses, the function shifts bodily to the right, that is, toward increasing values of z. In fact, we can even find the velocity with which the function is traveling by dividing the distance moved by the time elapsed. This gives which is the velocity of light in free space, denoted c. Thus, the function represents a traveling wave moving with a velocity toward the direction of increasing z. The wave is also known as the positive going wave , or wave. Similarly, by considering three values of time, , , and ,for the function , we obtain the sketches shown in Figure 4.7. An examination of these sketches reveals that represents a traveling wave moving with a velocity toward the direction of decreasing values of z. The wave is also known as the negative going wave , or wave. Since the sine functions are cosine functions shifted in phase by , it follows that and represent traveling waves moving in the positive and negative z-directions, respectively. sin (vt + bz)sin (vt - bz)p>2 (-) v>b cos (vt + bz)cos (vt + bz) t = p>2vt = p>4vt = 0 (+) v>b cos (vt - bz) = 3 10 8 m/s = 1 1m0P0 = 1 24p 10 - 7 10 - 9 >36 p velocity = p>b - p>2bp>2v - 0 = vb = vv1m0P0 p>2bt = 0 t = p>2vp>4bt = 0 t = p>4v120 Chapter 4 Wave Propagation in Free Space Returning to the solutions for and given by (4.31) and (4.32) or (4.48) and (4.49), we now know that these solutions consist of superpositions of traveling waves propagating away from and toward the current sheet. In the region , however, we have to rule out traveling waves propagating toward the current sheet, because such a situation requires a source of waves to the right of the sheet or an object that reflects the wave back toward the sheet. Thus, we have which give us finally (4.50) Having found the solutions for the fields in the region , we can now consider the solutions for the fields in the region . From our discussion of the functions , we know that these solutions must be of the form , since this function represents a traveling wave progressing in the negative z-direction, that is, away from the sheet in the region . Recalling that the magnetic field adjacent to the current sheet and to the left of it is given by we get (4.51a) Hy = - JS02 cos (vt + bz) for z 6 0[Hy]z = 0 - = - JS02 cos vtz 6 0cos (vt + bz)cos (vt < bz) z 6 0 z 7 0 Ex = h0JS02 cos (vt - bz) Hy = JS02 cos (vt - bz) s for z 7 02C - h0JS0 = 0 or C = h0JS02 D = 0 z 7 0 HyEx co s(vt%bz) b pp 2b 3p 2b 5p 2b 2p b p 4v t$ p 2v t$t$0 1 0z FIGURE 4.7 Sketches of the function cos versus zfor three values of t.(vt+bz)4.5 Uniform Plane Waves 121 The corresponding can be obtained by simply substituting the result just obtained for into one of the two differential equations (4.7) and (4.8). Thus using (4.7), we obtain (4.51b) Combining (4.50) and (4.51), we find that the solution for the electromagnetic field due to the infinite plane current sheet in the xy -plane characterized by is given by (4.52a) (4.52b) These results are illustrated in Figure 4.8, which shows sketches of the current density on the sheet and the distance-variation of the electric and magnetic fields on either side of the current sheet for a few values of t. It can be seen from these sketches that the phenomenon is one of electromagnetic waves radiating away from the current sheet to either side of it, in step with the time-variation of the current density on the sheet. The solutions that we have just obtained for the fields due to the time-varying infinite plane current sheet are said to correspond to uniform plane electromagnetic waves propagating away from the current sheet to either side of it. The terminology arises from the fact that the fields are uniform (i.e., they do not vary with position) over the . Thus, the phase of the fields, that is, the quantity , as well as the amplitudes of the fields, is uniform over the . The mag-nitude of the rate of change of phase with distance z for any fixed time is . The quan-tity is therefore known as the phase constant . Since the velocity of propagation of the wave, that is, , is the velocity with which a given constant phase progresses along the z-direction, that is, along the direction of propagation, it is known as the phase velocity and is denoted by the symbol . Thus, (4.53) vp = vb vp v>bbb planes z = constant (vt ; bz)planes z = constant H = ; JS02 cos (vt < bz) ay for z ! 0 E = h0JS02 cos (vt < bz) ax for z ! 0 JS = -JS0 cos vt ax = h0JS02 cos (vt + bz) for z 6 0 Ex = m0JS02 vb cos (vt + bz) 0Ex 0z = - 0By 0t = - m0JS02 v sin (vt + bz) HyEx122 Chapter 4 Wave Propagation in Free Space z y x JS z y x x JS z y H E E H E H H H H E E E JS $ # JS 0 co s vt ax t $ 0, JS $ # JS 0ax JS $ # ax JS 0 t $ , p 4v 2 JS $ 0t $ , p 2v FIGURE 4.8 Time history of uniform plane electromagnetic wave radiating away from an infinite plane current sheet in free space. 4.5 Uniform Plane Waves 123 The distance in which the phase changes by radians for a fixed time is . This quantity is known as the wavelength and is denoted by the symbol . Thus, (4.54) Substituting (4.53) into (4.54), we obtain or (4.55) Equation (4.55) is a simple relationship between the wavelength , which is a parameter governing the variation of the field with distance for a fixed time, and the frequency f,which is a parameter governing the variation of the field with time for a fixed value of z.Since for free space , we have (4.56) Other properties of uniform plane waves evident from (4.52) are that the electric and magnetic fields have components lying in the planes of constant phase and per-pendicular to each other and to the direction of propagation. In fact, the cross product of E and H results in a vector that is directed along the direction of propagation, as can be seen by noting that (4.57) Finally, we note that the ratio of to is given by (4.58) The quantity , which is equal to , is known as the intrinsic impedance of free space. Its value is given by (4.59) = 120 p Æ = 377 Æ h0 = B 14p 10 - 7 2 H/m 110 - 9 >36 p2 F/m = 21144 p2 10 22 H/F 1m0>P0h0 ExHy = e h0 for z 7 0, that is, for the 1+2 wave -h0 for z 6 0, that is, for the 1-2 wave HyEx = ; h0J2 S0 4 cos 2(vt < bz) az for z ! 0 E : H = Exax : Hyay l in meters f in MHz = 300 l in meters f in Hz = 3 10 8 vp = 3 10 8 m/s llf = vp l = 2pv>vp = vpf l = 2pbl 2p>b2p124 Chapter 4 Wave Propagation in Free Space Example 4.1 The electric field of a uniform plane wave is given by . Let us identify the various parameters associated with the uniform plane wave. We recognize that Also, . From (4.58), and since the given field represents a ( ) wave, Example 4.2 An antenna array consists of two or more antenna elements spaced appropriately and excited with currents having the appropriate amplitudes and phases in order to obtain a desired radia-tion characteristic. To illustrate the principle of an antenna array, let us consider two infinite plane parallel current sheets, spaced apart and carrying currents of equal amplitudes but out of phase by , as given by the densities and find the electric field due to the array of the two current sheets. We apply the result given by (4.52) to each current sheet separately and then use superpo-sition to find the required total electric field due to the array of the two current sheets. Thus, for the current sheet in the plane, we have E1 = μ h0JS02 cos 1vt - bz2 ax for z 7 0 h0JS02 cos 1vt + bz2 ax for z 6 0 z = 0 JS2 = -JS0 sin vt ax z = l 4 JS1 = -JS0 cos vt ax z = 0 p>2 l>4 H = Ex h0 ay = 10 377 cos 13p 10 8 t - pz2 ay A/m + lf = vp = 2 1.5 10 8 = 3 10 8 m/s vp = vb = 3p 10 8 p = 3 10 8 m/s l = 2pb = 2 m b = p rad/m f = v 2p = 1.5 10 8 Hz = 150 MHz v = 3p 10 8 rad/s E = 10 cos 13p 10 8 t - pz2 ax V/m 4.5 Uniform Plane Waves 125 For the current sheet in the plane, we have Now, using superposition, we find the total electric field due to the two current sheets to be Thus, the total field is zero in the region , and hence there is no radiation toward that side of the array. In the region the total field is twice that of the field due to a single sheet. The phenomenon is illustrated in Figure 4.9, which shows sketches of the individual fields and and the total field for a few values of t. The result that we have obtained here for the total field due to the array of two current sheets, spaced apart and fed with currents of equal amplitudes but out of phase by , is said to correspond to an endfire radiation pattern. In Section 1.4, we introduced polarization of sinusoidally time-varying fields, which is of relevance here in wave propagation. To extend the discussion, in the case of circular and elliptical polarizations, since the circle or the ellipse can be traversed in one of two opposite senses relative to the direction of the wave propagation, we talk of right-handed or clockwise polarization and left-handed or counterclockwise polarization. The conven-tion is that if in a given constant phase plane, the tip of the field vector of a circularly polarized wave rotates with time in the clockwise sense as seen looking along the direc-tion of propagation of the wave, the wave is said to be right circularly polarized. If the tip of the field vector rotates in the counterclockwise sense, the wave is said to be left circu-larly polarized. Similar considerations hold for elliptically polarized waves, which arise due to the superposition of two linearly polarized waves in the general case. For example, for a uniform plane wave propagating in the z-direction and having the electric field, (4.60) E = 10 sin 13p 10 8t - pz2 ax + 10 cos 13p 10 8t - pz2 ay V/m + p>2 l>4 Ex = Ex1 + Ex2Ex2 Ex1 z 7 l>4 z 6 0 = e h0JS0 cos 1vt - bz2 ax for z 7 l 4 h0JS0 sin vt sin bz ax for 0 6 z 6 l 40 for z 6 0 E = E1 + E2 = μ h0JS02 cos 1vt - bz2 ax for z 7 l 4 h0JS02 cos 1vt + bz2 ax for z 6 l 4 = μ h0JS02 sin avt - bz + p 2 b ax for z 7 l 4 h0JS02 sin avt + bz - p 2 b ax for z 6 l 4 E2 = μ h0JS02 sin cvt - b az - l 4 b d ax for z 7 l 4 h0JS02 sin cvt + b az - l 4 b d ax for z 6 l 4 z = l>4126 Chapter 4 Wave Propagation in Free Space z z$0 t$0 Ex1 Ex2 Ex z$l 4 p 4v z z$0 Ex1 Ex2 Ex z$l 4 z t$ p 2v t$ Ex1 Ex2 Ex FIGURE 4.9 Time history of individual fields and the total field due to an array of two infinite plane parallel current sheets. the two components of E are equal in amplitude, perpendicular, and out of phase by 90°. Therefore, the wave is circularly polarized. To determine if the polarization is right-handed or left-handed, we look at the electric field vectors in the plane for two values of time, and . These are shown in Figure 4.10. As time progresses, the tip of the vector rotates in the counterclockwise sense, as seen looking in the z-direction. Hence, the wave is left circularly polarized. + t = 16 10 - 8 s 13p 10 8t = p>22t = 0 z = 04.5 Uniform Plane Waves 127 Thus far, we have considered a source of single frequency. We found that wave propagation in free space is characterized by a phase velocity equal to c and intrinsic impedance , independent of frequency. Let us now consider a nonsinusoidal excitation for the current sheet. Then, since the propagation characteristics are the same for each frequency component of the nonsinu-soidal excitation, the resulting fields at any given value of z will have the same shape as that of the source with time, that is, they propagate without change in shape with time. Thus, for an infinite plane current sheet of surface current density given by (4.61) the solution for the electromagnetic field is given by (4.62a) (4.62b) The time variation of the electric field component in a given constant plane is the same as the current density variation delayed by the time and multiplied by . The time variation of the magnetic field component in a given constant plane is the same as the current density variation delayed by and multiplied by ,depending on . Using these properties, one can construct plots of the field com-ponents versus time for fixed values of z and versus z for fixed values of t. Example 4.3 Let us consider the function in (4.61) to be that given in Figure 4.11. We wish to find and sketch (a) versus t for , (b) versus t for , (c) versus z for ,and (d) versus z for .t = 2.5 msHyt = 1 msExz = -450 m Hyz = 300 m ExJS1t2 z ! 0 ; 12ƒzƒ>vpz =h0>2 ƒzƒ>vpz =Ex H1z, t2 = ; 12JS at < zvp bay for z ! 0 E1z, t2 = h02 JS at < zvp bax for z ! 0 JS1t2 = -JS1t2ax for z = 0 h0 1= 377 Æ21= 3 10 8 m/s 2 vp y z x [E]t$0 [E]t$&10 –8 s 1 6 FIGURE 4.10 For the determination of the sense of circular polarization for the field of equation (4.60). 102 0.1 t,ms JS, A/m A B CDE FIGURE 4.11 Plot of versus tfor Example 4.3. JS 128 Chapter 4 Wave Propagation in Free Space 1 0 (a) 2 3 4 5 18.85 t, ms 18.85 A B C D E [Ex]z $ 300 m , V/m 1 0 (b) 2 3 4 5 0.05 t, ms 0.05 A B C D E [Hy]z $ #450 m , A/m 900 #600 #300 (d) 0 300 600 900 0.05 0.05 [Hy]t $ 2.5 ms, A/m A A B B C D E CDE 900 #600 #300 (c) 0 300 600 900 z, m z, m 18.85 [Ex]t $ 1 ms, V/m AA BB C FIGURE 4.12 Plots of field components versus t for fixed values of z and versus z for fixed values of t for Example 4.3. 4.6 Poynting Vector and Energy Storage 129 (a) Since , the time delay corresponding to 300 m is . Thus, the plot of versus t for is the same as that of multiplied by , or 188.5, and delayed by , as shown in Figure 4.12(a). (b) The time delay corresponding to 450 m is . Thus, the plot of versus t for is the same as that of multiplied by and delayed by , as shown in Figure 4.12(b). (c) To sketch versus z for a fixed value of t, say, , we use the argument that a given value of existing at the source at an earlier value of time, say, , travels away from the source by the distance equal to times . Thus, at , the values of correspond-ing to points A and B in Figure 4.11 move to the locations and ,respectively, and the value of corresponding to point C exists right at the source. Hence, the plot of versus z for is as shown in Figure 4.12(c). Note that points beyond C in Figure 4.11 correspond to , and therefore they do not appear in the plot of Figure 4.12(c). (d) Using arguments as in part (c), we see that at , the values of corresponding to points A, B, C, D, and E in Figure 4.11 move to the locations ,, and , respectively, as shown in Figure 4.12(d). Note that the plot is an odd function of z, since the factor by which is multiplied to obtain is , depending on . 4.6 POYNTING VECTOR AND ENERGY STORAGE In the preceding section, we found the solution for the electromagnetic field due to an infinite plane current sheet situated in the plane. For a surface current flowing in the negative x-direction, we found the electric field on the sheet to be directed in the positive x-direction. Since the current is flowing against the force due to the electric field, a certain amount of work must be done by the source of the current in order to maintain the current flow on the sheet. Let us consider a rectan-gular area of length and width on the current sheet, as shown in Figure 4.13. Since the current density is , the charge crossing the width in time is . The force exerted on this charge by the electric field is given by (4.63) The amount of work required to be done against the electric field in displacing this charge by the distance is (4.64) Thus, the power supplied by the source of the current in maintaining the surface current over the area is (4.65) dw dt = JS0 Ex cos vt ¢x ¢y ¢x ¢ydw = Fx ¢x = JS0 Ex cos vt dt ¢x ¢y ¢x F = dq E = JS0 ¢y cos vt dt Exaxdq = JS0 ¢y cos vt dt dt ¢yJS0 cos vt ¢y¢xz = 0 z " 0 ; 12HyJS0 ;150 m ;300 m ;450 m z = ; 750 m, ;600 m, Hyt = 2.5 ms t 7 1 ms t = 1 msEx Exz = ; 150 m z = ; 300 m Ext = 1 msvp1t1 - t22 t 2Ext1Ex 1.5 ms-1>2JS1t2z = -450 m Hy1.5 ms1 ms h0 >2JS(t)z = 300 m Ex 1 msvp = c = 3 10 8 m/s 130 Chapter 4 Wave Propagation in Free Space Recalling that on the sheet is , we obtain (4.66) We would expect the power given by (4.66) to be carried by the electromagnetic wave, half of it to either side of the current sheet. To investigate this, we note that the quantity has the units of which represents power density. Let us then consider the rectangular box enclosing the area on the current sheet and with its sides almost touching the current sheet on either side of it, as shown in Figure 4.13. Recalling that is given by (4.57) and evaluating the surface integral of over the surface of the rectangular box, we obtain the power flow out of the box as (4.67) = h0 J2 S0 2 cos 2 vt ¢x ¢y a -h0 J2 S0 4 cos 2 vt azb # (- ¢ x ¢y az2 CE : H # dS = h0 J2 S0 4 cos 2 vt az # ¢x ¢y az E : HE : H ¢x ¢y = newton-meters second 1 1meter 22 = watts 1meter 22newtons coulomb amperes meter = newtons coulomb coulomb second-meter meter meter E : H dw dt = h0 J2 S0 2 cos 2 vt ¢x ¢y h0 JS02 cos vtEx y x 'x 'y z FIGURE 4.13 For the determination of power flow density associated with the electromagnetic field. 4.6 Poynting Vector and Energy Storage 131 This result is exactly equal to the power supplied by the current source as given by (4.66). We now interpret the quantity as the power flow density vector associated with the electromagnetic field. It is known as the Poynting vector , after J. H. Poynting, and is denoted by the symbol P. Although we have here introduced the Poynting vec-tor by considering the specific case of the electromagnetic field due to the infinite plane current sheet, the interpretation that is equal to the power flow out of the closed surface S is applicable in the general case. Example 4.4 Far from a physical antenna, that is, at a distance of several wavelengths from the antenna, the radiated electromagnetic waves are approximately uniform plane waves with their constant phase surfaces lying normal to the radial directions away from the antenna, as shown for two directions in Figure 4.14. We wish to show from the Poynting vector and physical considerations that the electric and magnetic fields due to the antenna vary inversely proportional to the radial distance away from the antenna. AS E : H # dSE : H Con stant Pha se Surface s Antenna ra ra rb rb FIGURE 4.14 Radiation of electromagnetic waves far from a physical antenna. From considerations of electric and magnetic fields of a uniform plane wave, the Poynting vector is directed everywhere in the radial direction indicating power flow radially away from the antenna and is proportional to the square of the magnitude of the electric field intensity. Let us now consider two spherical surfaces of radii and and centered at the antenna and insert a cone through these two surfaces such that the vertex is at the antenna, as shown in Figure 4.14. Then the power crossing the portion of the spherical surface of radius inside the cone must be the same as the power crossing the portion of the spherical surface of radius inside the cone. Since these surface areas are proportional to the square of the radius and since the surface inte-gral of the Poynting vector gives the power, the Poynting vector must be inversely proportional to the square of the radius. This in turn means that the electric field intensity and hence the mag-netic field intensity must be inversely proportional to the radius. rarbrbra132 Chapter 4 Wave Propagation in Free Space Thus, from these simple considerations we have established that far from a radiating antenna the electromagnetic field is inversely proportional to the radial distance away from the antenna. This reduction of the field intensity inversely proportional to the distance is known as the free space reduction . For example, let us consider communication from earth to the moon. The distance from the earth to the moon is approximately , or . Hence, the free space reduction factor for the field intensity is or, in terms of decibels, the reduction is , or 171.6 db. Returning to the electromagnetic field due to the infinite plane current sheet, let us consider the region . The magnitude of the Poynting vector in this region is given by (4.68) The variation of with z for is shown in Figure 4.15. If we now consider a rec-tangluar box lying between and planes and having dimensions and in the x- and y-directions, respectively, we would in general obtain a nonzero result for the power flowing out of the box, since is not everywhere zero. Thus, there is some energy stored in the volume of the box. We then ask ourselves the ques-tion, “Where does this energy reside?” A convenient way of interpretation is to at-tribute the energy storage to the electric and magnetic fields. 0Pz>0z ¢y ¢xz = z + ¢ zz = zt = 0PzPz = ExHy = h0 J2 S0 4 cos 2 1vt - bz2 z 7 020 log 10 38 10 710 - 7 >38 38 10 7 m38 10 4 km z 0 [Pz]t$0 p b 2p b zz%'z 4 S0h0J2 FIGURE 4.15 For the discussion of energy storage in electric and magnetic fields. To discuss the energy storage in the electric and magnetic fields further, we evaluate the power flow out of the rectangular box. Thus, (4.69) = 0Pz 0z ¢v = [Pz]z + ¢ z - [Pz]z ¢z ¢x ¢y ¢z CS P # dS = [Pz]z + ¢ z ¢x ¢y - [Pz]z ¢x ¢ySummary 133 where is the volume of the box. Letting equal and using (4.7) and (4.8), we obtain (4.70) Equation (4.70), which is known as Poynting’s theorem, tells us that the power flow out of the box is equal to the sum of the time rates of decrease of the quantities and . These quantities are obviously the energies stored in the electric and magnetic fields, respectively, in the volume of the box. It then follows that the energy densities associated with the electric and magnetic fields are and , respec-tively. It is left to the student to verify that the quantities and do indeed have the units . Once again, although we have obtained these results by consider-ing the particular case of the uniform plane wave, they hold in general. Summarizing our discussion in this section, we have introduced the Poynting vec-tor as the power flow density associated with the electromagnetic field characterized by the electric and magnetic fields, E and H, respectively. The surface in-tegral of P over a closed surface always gives the correct result for the power flow out of that surface. There is energy storage associated with the electric and magnetic fields with the energy densities given by (4.71) and (4.72) respectively. SUMMARY In this chapter, we studied the principles of uniform plane wave propagation in free space. Uniform plane waves are a building block in the study of electromagnetic wave propagation. They are the simplest type of solutions resulting from the coupling of the electric and magnetic fields in Maxwell’s curl equations. We learned that uniform plane waves have their electric and magnetic fields perpendicular to each other and to wm = 12 m0H2 we = 12 P0E2 P = E : H J/m 312 m0H212 P0E212 m0H2 y 12 P0E2 x 12 m0H2 y ¢v 12 P0E2 x ¢v = - 00t a 12 m0H2 y ¢vb - 00t a 12 P0E2 x ¢vb = -m0Hy 0Hy 0t ¢v - P 0Ex 0Ex 0t ¢v = a -Hy 0By 0t - Ex 0Dx 0t b ¢v = aHy 0Ex 0z + Ex 0Hy 0z b ¢v CS P # dS = 00z [ExHy]¢vExHyPz¢v134 Chapter 4 Wave Propagation in Free Space the direction of propagation. The fields are uniform in the planes perpendicular to the direction of propagation. We obtained the uniform plane wave solution to Maxwell’s equations by consider-ing an infinite plane current sheet in the xy -plane with uniform surface current den-sity given by (4.73) and deriving the electromagnetic field due to the current sheet to be given by (4.74a) (4.74b) In (4.74a) and (4.74b), cos represents wave motion in the positive z-direction, whereas cos represents wave motion in the negative z-direction. Thus, (4.74a) and (4.74b) correspond to waves propagating away from the current sheet to either side of it. Since the fields are independent of x and y, they represent uniform plane waves. The quantity is the phase constant, that is, the magnitude of the rate of change of phase with distance along the direction of propagation, for a fixed time. The phase velocity that is, the velocity with which a particular constant phase progresses along the direction of propagation, is given by (4.75) The wavelength , that is, the distance along the direction of propagation in which the phase changes by radians, for a fixed time, is given by (4.76) The wavelength is related to the frequency f in a simple manner as given by (4.77) which follows from (4.75) and (4.76). The quantity is the intrinsic impe-dance of free space. It is the ratio of the magnitude of E to the magnitude of H and has a value of .In the process of deriving the electromagnetic field due to the infinite plane cur-rent sheet, we used two approaches and learned several useful techniques. These are discussed in the following: The determination of the magnetic field adjacent to the current sheet by employing Ampere’s circuital law in integral form: This is a common procedure used in the computa-tion of static fields due to charge and current distributions possessing certain symmetries. In Chapter 5 we shall derive the boundary conditions , that is, the relationships between the fields on either side of an interface between two different media, by applying Maxwell’s equations in integral form to closed paths and surfaces straddling the boundary as we have done here in the case of the current sheet. 120 p Æ h0 1= 2m0>P02 vp = lf l = 2pb 2pl vp = vb vp, b 1= v1m0P021vt + bz21vt - bz2 H = ; JS 02 cos 1vt < bz2 ay for z ! 0 E = h0JS02 cos 1vt < bz2 ax for z ! 0 JS = -JS0 cos vt ax A/m Review Questions 135 The successive, step-by-step solution of the two Maxwell’s curl equations, to obtain the final solution consistent with the two equations, starting with the solution obtained for the field adjacent to the current sheet: This technique provided us a feel for the phenomenon of radiation of electromagnetic waves resulting from the time-varying current distribution and the interaction between the electric and magnetic fields. We shall use this kind of approach and the knowledge gained on wave propagation to obtain in Chapter 9 the complete electromagnetic field due to an elemental antenna, which forms the basis for the study of physical antennas. The solution of wave equation by the separation of variables technique: This is the stan-dard technique employed in the solution of partial differential equations involving multiple variables. The application of phasor technique for the solution of the differential equations: The phasor technique is a convenient tool for analyzing sinusoidal steady-state problems as we learned in Chapter 1. We discussed (a) polarization of sinusoidally time-varying fields, as it pertains to uni-form plane wave propagation, and (b) nonsinusoidal excitation giving rise to nonsinu-soidal waves propagating in free space without change in shape, in view of phase velocity independent of frequency. We also learned that there is power flow and energy storage associated with the wave propagation that accounts for the work done in maintaining the current flow on the sheet. The power flow density is given by the Poynting vector and the energy densities associated with the electric and magnetic fields are given, respectively, by The surface integral of the Poynting vector over a given closed surface gives the total power flow out of the volume bounded by that surface. Finally, we have augmented our study of uniform plane wave propagation in free space by illustrating (a) the principle of an antenna array, and (b) the inverse distance dependence of the fields far from a physical antenna. wm = 12 m0H2 we = 12 P0E2 P = E : H REVIEW QUESTIONS 4.1. What is a uniform plane wave? 4.2. Why is the study of uniform plane waves important? 4.3. How is the surface current density vector defined? Distinguish it from the volume cur-rent density vector. 4.4. How do you find the current crossing a given line on a sheet of surface current? 4.5. Why is it that Ampere’s circuital law in integral form is used to find the magnetic field adjacent to the current sheet of Figure 4.2? 136 Chapter 4 Wave Propagation in Free Space 4.6. Why is the path chosen to evaluate the magnetic field in Figure 4.4 rectangular? 4.7. Outline the application of Ampere’s circuital law in integral form to find the magnetic field adjacent to the current sheet of Figure 4.2. 4.8. Why is the displacement current enclosed by the rectangular path abcda in Figure 4.4 equal to zero? 4.9. How would you use Ampere’s circuital law in differential form to find the magnetic field adjacent to the current sheet? 4.10. If the current density on the infinite plane current sheet of Figure 4.2 were directed in the positive y-direction, what would be the directions of the magnetic field adjacent to the current sheet and on either side of it? 4.11. Why are the results given by (4.12) and (4.13) for the magnetic field not valid for points at some distance from the current sheet? 4.12. Under what conditions would a result obtained for the magnetic field adjacent to the infi-nite plane current sheet of Figure 4.2 be valid at points distant from the current sheet? 4.13. Briefly outline the procedure involved in the successive solution of Maxwell’s equations. 4.14. How does the technique of successive solution of Maxwell’s equations reveal the inter-action between the electric and magnetic fields giving rise to wave propagation? 4.15. State the wave equation for the case of How is it derived? 4.16. Briefly outline the separation of variables technique of solving the wave equation. 4.17. Discuss how the function represents a traveling wave propagating in the positive z-direction. 4.18. Discuss how the function represents a traveling wave propagating in the negative z-direction. 4.19. Discuss how the solution for the electromagnetic field given by (4.52) corresponds to that of a uniform plane wave. 4.20. Why is the quantity in known as the phase constant? 4.21. What is phase velocity? How is it related to the radian frequency and the phase con-stant of the wave? 4.22. Define wavelength. How is it related to the phase constant? 4.23. What is the relationship between frequency, wavelength, and phase velocity? What is the wavelength in free space for a frequency of 15 MHz? 4.24. What is the direction of propagation for a uniform plane wave having its electric field in the negative y-direction and its magnetic field in the positive z-direction? 4.25. What is the direction of the magnetic field for a uniform plane wave having its electric field in the positive z-direction and propagating in the positive x-direction? 4.26. What is intrinsic impedance? What is its value for free space? 4.27. Discuss the principle of an antenna array. 4.28. What should be the spacing and the relative phase angle of the current densities for an array of two infinite, plane, parallel current sheets of uniform densities, equal in magni-tude, to confine their radiation to the region between the two sheets? 4.29. Discuss polarization of sinusoidally time-varying fields, as it is relevant to propagation of uniform plane waves. 4.30. Discuss the propagation of uniform plane waves arising from an infinite plane current sheet of nonsinusoidally time-varying surface current density. 4.31. Why is a certain amount of work involved in maintaining current flow on the sheet of Figure 4.2? How is this work accounted for? cos 1vt - bz2b cos 1vt + bz2 cos 1vt - bz2 E = Ex1z, t2ax.Problems 137 4.32. What is a Poynting vector? What is its physical significance? 4.33. What is the physical interpretation of the surface integral of the Poynting vector over a closed surface? 4.34. Discuss how the fields far from a physical antenna vary inversely proportional to the distance from the antenna. 4.35. Discuss the interpretation of energy storage in the electric and magnetic fields of a uni-form plane wave. 4.36. What are the energy densities associated with the electric and magnetic fields? PROBLEMS 4.1. An infinite plane sheet lying in the plane carries a current of uniform density A/m. Find the currents crossing the following straight lines: (a) from (0, 0, 0) to (0, 2, 0); (b) from (0, 0, 0) to (2, 0, 0); (c) from (0, 0, 0) to (2, 2, 0). 4.2. An infinite plane sheet lying in the plane carries a current of nonuniform density A/m. Find the currents crossing the following straight lines: (a) from (0, 0, 0) to (0, 1, 0); (b) from (0, 0, 0) to (0, , 0); (c) from (0, 0, 0) to (1, 1, 0). 4.3. An infinite plane sheet lying in the plane carries a current of uniform density Find the currents crossing the following straight lines: (a) from (0, 0, 0) to (0, 2, 0); (b) from (0, 0, 0) to (2, 0, 0); (c) from (0, 0, 0) to (2, 2, 0). 4.4. An infinite plane sheet lying in the plane carries a current of uniform density Find the magnetic field intensities adjacent to the sheet and on either side of it. What is the polarization of the field? 4.5. An infinite plane sheet lying in the plane carries a current of nonuniform density . Find the magnetic field intensities adjacent to the current sheet and on either side of it at (a) the point (0, 1, 0) and (b) the point (2, 2, 0). 4.6. Current flows with uniform density in the region . Using Ampere’s circuital law in integral form and symmetry considerations, find H everywhere. 4.7. Current flows with nonuniform density in the region where is a constant. Using Ampere’s circuital law in integral form and symmetry con-siderations, find H everywhere. 4.8. For an infinite plane sheet of charge lying in the -plane with uniform surface charge density , find the electric field intensity on both sides of the sheet by using Gauss’ law for the electric field in integral form and symmetry considerations. 4.9. Charge is distributed with uniform density in the region Using Gauss’ law for the electric field in integral form and symmetry considerations, find E everywhere. 4.10. Charge is distributed with nonuniform density in the region where is a constant. Using Gauss’ law for the electric field in integral form and symmetry considerations, find E everywhere. 4.11. Verify that expressions (4.23) and (4.24) simultaneously satisfy the differential equations (4.16) and (4.17). r0\| x \| 6 a, r = r011 - \| x \|>a2 C/m 3\| x \| 6 a.r = r0 C/m 3 rS0 C/m 2 xy J0\| z \| 6 a,J = J0(1 - \| z \|>a2ax A/m 2\| z \| 6 aJ = J0 ax A/m 2 JS = -0.2 e -\|y\| cos vt ax A/m z = 0 JS = (-0.2 cos vt ax + 0.2 sin vt ay2 A/m z = 0 JS = 1-0.1 cos vt ax + 0.1 sin vt ay2 A/m z = 0 q JS = -0.1 e -\|y\| axz = 0 JS = -0.1 axz = 0138 Chapter 4 Wave Propagation in Free Space 4.12. For the infinite plane current sheet in the plane carrying surface current of density where is a constant, find the magnetic field adjacent to the cur-rent sheet . T hen use the method of successive solution of Maxwell’s equations to show that for ,where C is a constant. 4.13. For the infinite plane current sheet in the plane carrying surface current of density where is a constant, find the magnetic field adjacent to the cur-rent sheet . T hen use the method of successive solution of Maxwell’s equations to show that for where C is a constant. 4.14. Verify that expressions (4.48) and (4.49) simultaneously satisfy the differential equations (4.7) and (4.8), and that (4.49) reduces to (4.12) for . 4.15. Show that and are solutions of the wave equation. With the aid of sketches, discuss the nature of these functions. 4.16. For arbitrary time-variation of the fields, show that the solutions for the differential equations (4.33) and (4.34) are where A and B are arbitrary constants. Discuss the nature of the functions . 4.17. In Problems 4.12 and 4.13, evaluate the constant C and obtain the solutions for and in the region Then write the solutions for and in the region 4.18. The electric field intensity of a uniform plane wave is given by Find (a) the frequency, (b) the wavelength, (c) the phase velocity, (d) the direction of propagation of the wave, and (e) the associated magnetic field intensity vector H. 4.19. An infinite plane sheet lying in the plane carries a surface current of density Find the expressions for the electric and magnetic fields on either side of the sheet. JS = 1-0.2 cos 6 p 10 8t ax - 0.1 cos 12 p 10 8t ax2 A/m z = 0 E = 37.7 cos 16p 10 8 t + 2pz2 ay V/m. z 6 0. HyExz 7 0. HyEx and g1t + z1m0P02 f1t - z1m0P02 Hy = 1 h0[Af 1t - z1m0P0 2 - Bg 1t + z1m0P0 2] Ex = Af 1t - z1m0 P02 + Bg 1t + z1m0 P021t + z1m0P022 1t - z1m0P022 z = 0+ Hy = a 2C + h0JS04h0 b1 t - z1m0P0 22 - a 2C - h0JS04h0 b1 t + z1m0 P022 Ex = a 2C + h0 JS04 b1 t - z1m0P0 22 + a 2C - h0 JS04 b1 t + z1m0 P022 z 7 0, JS0JS = -JS0 t2 ax A/m, z = 0 Hy = a 2C + h0JS04h0 b1 t - z1m0 P02 - a 2C - h0JS04h0 b1 t + z1m0 P02 Ex = a 2C + h0JS04 b1 t - z1m0 P02 + a 2C - h0JS04 b1 t + z1m0 P02 z 7 0 JS0JS = -JS0 t ax A/m, z = 0Problems 139 4.20. An array is formed by two infinite plane parallel current sheets with the current densi-ties given by where is a constant. Find the electric field intensity in all three regions: (a) (b) (c) . 4.21. Determine the spacing, relative amplitudes, and phase angles of current densities for an array of two infinite plane parallel current sheets required to obtain a radiation charac-teristic such that the field radiated to one side of the array is twice that of the field radi-ated to the other side of the array. 4.22. For two infinite plane parallel current sheets with the current densities given by where is a constant, find the electric field in all three regions: (a) (b) (c) . Discuss the polarization of the field in all three regions. 4.23. For each of the following fields, determine if the polarization is right- or left-circular. (a) (b) 4.24. For each of the following fields, determine if the polarization is right- or left-elliptical. (a) (b) 4.25. Express the following uniform plane wave electric field as a superposition of right- and left-circularly polarized fields: 4.26. Repeat Problem 4.25 for the following electric field: 4.27. Write the expression for the electric field intensity of a sinusoidally time-varying uni-form plane wave propagating in free space and having the following characteristics: (a) ; (b) direction of propagation is the -direction; and (c) polarization is right circular with the electric field in the plane at having an -component equal to and a -component equal to 4.28. An infinite plane sheet lying in the plane carries a surface current of density where is the periodic function shown in Figure 4.16. Find and sketch (a) versus for (b) versus for and (c) versus for t = 1 ms. zExz = 150 m, tExz = 0+,tHyJS 1t2JS = -JS1t2ax, z = 00.75 E0 .yE0 xt = 0z = 0 +zf = 100 MHz E0ay cos 1vt - bz + p>62 E0 ax cos 1vt - bz + p>32 - E0ax cos 1vt + bz2 E0 cos 1vt - bx2 az - E0 sin 1vt - bx + p>42 ayE0 cos 1vt + by2 ax - 2E0 sin 1vt + by2 azE0 cos 1vt + bx2 ay + E0 sin 1vt + bx2 azE0 cos 1vt - by2 az + E0 sin 1vt - by2 axz 7 l>20 6 z 6 l>2; z 6 0; JS0 JS2 = -JS0 cos vt ay z = l 2 JS1 = -JS0 cos vt ax z = 0 z 7 l>20 6 z 6 l>2; z 6 0; JS0 JS2 = -JS0 cos vt ax z = l 2 JS1 = -JS0 cos vt ax z = 0 FIGURE 4.16 For Problem 4.28. t,ms 0.2 3210#1#2 JS, A/m 140 Chapter 4 Wave Propagation in Free Space 4.29. The time-variation of the electric field intensity in the plane of a uniform plane wave propagating away from an infinite plane current sheet lying in the plane is given by the periodic function shown in Figure 4.17. Find and sketch (a) versus for (b) versus for and (c) versus for t = 13 ms. zHyt = 0, zExz = 200 m, tExz = 0 z = 600 m Ex 4.30. The time-variation of the electric field intensity in the plane of a uniform plane wave propagating away from an infinite plane current sheet lying in the plane is given by the aperiodic function shown in Figure 4.18. Find and sketch (a) versus for (b) versus for and (c) versus for t = 2 ms. zHyt = 1 ms, zExz = 600 m, tExz = 0 z = 300 m Ex FIGURE 4.17 For Problem 4.29. FIGURE 4.18 For Problem 4.30. Ex, V/m 75.4 37.7 t,ms 2#10125 3 2 3 1 3 4 3 7 3 4.31. Show that the time-average value of the magnitude of the Poynting vector given by (4.68) is one-half its peak value. For an antenna radiating a time-average power of 150 kW, find the peak value of the electric field intensity at a distance of 100 km from the antenna. Assume the antenna to be radiating equally in all directions. 4.32. The electric field of a uniform plane wave propagating in the positive -direction is given by where is a constant. (a) Find the corresponding magnetic field H. (b) Find the Poynting vector. 4.33. Show that the quantities and have the units J/ . 4.34. Show that the energy is stored equally in the electric and magnetic fields of a traveling wave. m312 m0H212 P0E2 E0 E = E0 cos 1vt - bz2 ax + E0 sin 1vt - bz2 ayz
14247	https://pubmed.ncbi.nlm.nih.gov/3049365/	Odynophagia/dysphagia in AIDS - 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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PMID: 3049365 Item in Clipboard Review Odynophagia/dysphagia in AIDS J P Raufman. Gastroenterol Clin North Am.1988 Sep. Show details Display options Display options Format Gastroenterol Clin North Am Actions Search in PubMed Search in NLM Catalog Add to Search . 1988 Sep;17(3):599-614. Author J P Raufman1 Affiliation 1 Department of Medicine, State University of New York-Health Science Center, Brooklyn. PMID: 3049365 Item in Clipboard Cite Display options Display options Format Abstract Odynophagia and dysphagia are common symptoms of treatable disorders of the esophagus in patients with AIDS. Esophageal candidiasis is the most frequent cause of these symptoms. In patients with AIDS or AIDS-related complex, thrush in combination with odynophagia or dysphagia almost certainly indicates the presence of esophageal candidiasis. Other causes of swallowing disorders in AIDS include opportunistic infection of the esophagus with herpes simplex virus, cytomegalovirus, or, rarely, cryptosporidiosis. Recently, ulcerative esophagitis in AIDS associated with unidentified viral-like particles has been described. Infrequently, Kaposi's sarcoma or lymphoma may involve the posterior pharynx or esophagus, respectively. Because Candida esophagitis is so frequently the cause of odynophagia and/or dysphagia in AIDS, it is suggested that in most cases, a therapeutic trial with an antifungal agent, like ketoconazole, may be appropriate before radiologic or endoscopic examination. Further investigation can be reserved for patients who do not respond to this trial or who have clinical evidence suggesting another esophageal disorder. Herpes simplex and cytomegalovirus esophagitis can be treated with antiviral agents, such as acyclovir and ganciclovir, respectively. Maintenance therapy with antifungal agents to prevent recurrent esophageal candidiasis may be beneficial, but the efficacy and cost effectiveness of this approach remain to be determined. Because of the increasing numbers of patients with AIDS, frequency of esophageal disorders, such as candidiasis, in these patients and the morbidity of these disorders, an expansion of clinical research efforts to determine effective treatment and prophylaxis for these disorders is warranted. PubMed Disclaimer Similar articles Opportunistic infections of the esophagus not responding to oral systemic antifungals in patients with AIDS: their frequency and treatment.Parente F, Cernuschi M, Rizzardini G, Lazzarin A, Valsecchi L, Bianchi Porro G.Parente F, et al.Am J Gastroenterol. 1991 Dec;86(12):1729-34.Am J Gastroenterol. 1991.PMID: 1962617 [Esophageal pathology in patients with the AIDS virus. Etiology and diagnosis].Varsky CG, Yahni VD, Freire MC, Patrizio E, Balbo V, Benetucci J, Boffi A, Mattoni RA, Luis, Alicia M.Varsky CG, et al.Acta Gastroenterol Latinoam. 1991;21(2):67-83.Acta Gastroenterol Latinoam. 1991.PMID: 1820692 Spanish. Timing and necessity of endoscopy in AIDS patients with dysphagia or odynophagia.Lai YP, Wu MS, Chen MY, Chuang CY, Shun CT, Lin JT.Lai YP, et al.Hepatogastroenterology. 1998 Nov-Dec;45(24):2186-9.Hepatogastroenterology. 1998.PMID: 9951891 HIV infection and AIDS.Lloyd A.Lloyd A.P N G Med J. 1996 Sep;39(3):174-80.P N G Med J. 1996.PMID: 9795558 Review. [The compromise of esophagus in HIV/AIDS diseases].Corti M, Villafañe MF.Corti M, et al.Acta Gastroenterol Latinoam. 2003;33(4):211-20.Acta Gastroenterol Latinoam. 2003.PMID: 14708474 Review.Spanish. See all similar articles Cited by Mycobacterial infections in AIDS.Hill AR.Hill AR.Can J Infect Dis. 1991 Spring;2(1):19-29. doi: 10.1155/1991/476503.Can J Infect Dis. 1991.PMID: 22451748 Free PMC article. Clinical management of the complications of HIV infection.Cooney TG.Cooney TG.J Gen Intern Med. 1991 Jan-Feb;6(1 Suppl):S12-8. doi: 10.1007/BF02599252.J Gen Intern Med. 1991.PMID: 2005472 Review. Gastrointestinal symptoms in patients infected with human immunodeficiency virus: relevance of infective agents isolated from gastrointestinal tract.Ullrich R, Heise W, Bergs C, L'age M, Riecken EO, Zeitz M.Ullrich R, et al.Gut. 1992 Aug;33(8):1080-4. doi: 10.1136/gut.33.8.1080.Gut. 1992.PMID: 1327982 Free PMC article. Gastrointestinal opportunistic infections in human immunodeficiency virus disease.Al Anazi AR.Al Anazi AR.Saudi J Gastroenterol. 2009 Apr;15(2):95-9. doi: 10.4103/1319-3767.48965.Saudi J Gastroenterol. 2009.PMID: 19568572 Free PMC article. Imaging the gastrointestinal tract of children with AIDS.Haller JO.Haller JO.Pediatr Radiol. 1995;25(2):94-6. doi: 10.1007/BF02010314.Pediatr Radiol. 1995.PMID: 7596673 Review.No abstract available. See all "Cited by" articles Publication types Review Actions Search in PubMed Search in MeSH Add to Search MeSH terms Acquired Immunodeficiency Syndrome / complications Actions Search in PubMed Search in MeSH Add to Search Candidiasis, Oral / complications Actions Search in PubMed Search in MeSH Add to Search Deglutition Disorders / complications Actions Search in PubMed Search in MeSH Add to Search Esophageal Diseases / complications Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Opportunistic Infections / complications Actions Search in PubMed Search in MeSH Add to Search Parasitic Diseases / complications Actions Search in PubMed Search in MeSH Add to Search Virus Diseases / complications Actions Search in PubMed Search in MeSH Add to Search Related information MedGen [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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14248	https://puzzleaday.wordpress.com/2025/07/13/a-clock-question/	A Clock Question \| Puzzle a Day Puzzle a Day A new puzzle, riddle, game or interesting tidbit posted every day MenuSkip to content Home About Search Search for: A Clock Question July 13, 2025 June 19, 2025 / Puzzling You have a 12-hour digital clock. If you look at it at a random time, what is the probability that the time will contain the digit 1? Note: Try to intuitively guess the answer before doing the maths. Scroll down for a clue and further down for the answer. Clue:All the times starting with 1:XY, 10:XY, 11:XY, and 12:XY contain a 1. Now, focus on the minute values for the remaining hours. Scroll down for the answer. Answer:There is a ½ probability the clock will contain the digit 1. All the times starting with 1:XY, 10:XY, 11:XY, and 12:XY contain a 1. This accounts for 4 hours. For the remaining 8 hours, the digit 1 appears in the minutes 1, 10-19, 21, 31, 41, and 51. This adds up to 15 minutes per hour of the remaining 8 hours, which is 2 hours. Summing these values, we get the digit 1 appearing for 6 hours, which is 50% of the time on the 12-hour clock—a nice, round answer! Share this: Click to share on X (Opens in new window)X Click to share on Facebook (Opens in new window)Facebook Like Loading... Related General Puzzle, maths, Number Puzzle puzzle, puzzles Post navigation ← A Puzzle From Only Connect Three Rebus Puzzles → Leave a comment Cancel reply Δ My new book, The Encyclopedia of Science Oddities,is available on Amazon. It features the most mind-blowing facts in all of science—in an encyclopedic format. Check it out here. My book,The World's First Book of TRUE Lateral Thinking Puzzles, is available on Amazon. Every puzzle in the book actually happened! The puzzles are all new & each answer contains a fascinating, factual backstory. Check it out here. My book,The Ultimate Book of Geography Oddities, is now available on Amazon. Prepare to be taken on a whirlwind journey through more than 1,000 of the world’s most surprising, unbelievable and downright bizarre geographical curiosities. Check it out here. Follow Blog via Email Enter your email address to follow this blog and receive notifications of new posts by email. Email Address: Follow Recent Posts A Matchstick Puzzle A Curious Animal Sequence A Quick Geometry Puzzle Three Rebus Puzzles A Cube Puzzle Categories Chess Codes Do It Yourself games General Knowledge General Puzzle Geography How many… Illusion Interactive Job Interview Questions Language Lateral thinking Letter Puzzle Logic magic Matchstick maths Murder mystery My Creations Number Puzzle Number Tricks Only Connect Paradox Picture puzzle Raven's Matrices Rebus Riddles Science Spatial Tests Uncategorized Video wordle Blog at WordPress.com. Comment Reblog SubscribeSubscribed Puzzle a Day Join 316 other subscribers Sign me up Already have a WordPress.com account? Log in now. Puzzle a Day SubscribeSubscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar Loading Comments... Write a Comment... Email (Required) Name (Required) Website %d Design a site like this with WordPress.com Get started
14249	https://math.stackexchange.com/questions/2637690/is-there-a-formula-to-calculate-the-area-of-a-trapezoid-knowing-the-length-of-al	Skip to main content Asked Modified 6 years, 9 months ago Viewed 24k times This question shows research effort; it is useful and clear Save this question. Show activity on this post. 1 3 What do you mean by "trapezoid"? In North America, it's a quadrilateral with at least one pair of parallel sides; elsewhere, it's a quadrilateral with no parallel sides. – David Richerby Commented Feb 6, 2018 at 22:27 Add a comment \| 11 Answers 11 Reset to default This answer is useful 38 Save this answer. Show activity on this post. This problem is more subtle than some of the other answers here let on. A great deal hinges on whether "trapezoid" is defined inclusively (i.e. as a quadrilateral with at least one pair of parallel sides) or exclusively (i.e. as a quadrilateral with exactly one pair of parallel sides). The former definition is widely considered more mathematically sophisticated, but the latter definition is more traditional, is still extensively used in K-12 education in the United States, and has some advantages. As the other responses have pointed out, if one defines "trapezoid" inclusively, then any parallelogram is automatically a trapezoid, and as the side-lengths of a parallelogram do not determine its area, it is not possible (even conceptually) that there could be a formula for the area of a trapezoid in terms of its side lengths. However, if "trapezoid" is defined exclusively, then things are quite different. Consider a trapezoid with parallel bases of length a and b with b>a. Let θ and ϕ respectively denote the angles formed by the legs c and d with the base b. Then we have the following relationships: ccosθ+dcosϕ=b−a csinθ=dsinϕ These conditions uniquely determine θ and ϕ, and therefore among non-parallelogram trapezoids, choosing the lengths of the parallel sides and the lengths of the bases uniquely determines the figure. In particular we would have cosθ=(b−a)2+c2−d22c(b−a) . The height of the trapezoid would then be h=csinθ (or if you prefer h=dsinϕ, which is equal to it), so the area of the trapezoid can (in principal) be computed. If you really want to carry it out, you would have sinθ=1−((b−a)2+c2−d22c(b−a))2−−−−−−−−−−−−−−−−−−−−−−√ so the area would be A=a+b2c1−((b−a)2+c2−d22c(b−a))2−−−−−−−−−−−−−−−−−−−−−−√ I am not sure if there is a simpler expression, however. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Feb 5, 2018 at 22:15 answered Feb 5, 2018 at 22:07 mweissmweiss 24.6k33 gold badges5050 silver badges9393 bronze badges 8 2 You're right. I was thinking this through in response to your comment. One way to see it geometrically rather than algebraically is to imagine the two parallel edges joined at one pair of ends by a third edge. As you rotate that edge around a semicircle, changing the angles it makes with the parallel sides, the distance between the other ends of the parallel edges varies continuously with no repeat values. – Ethan Bolker Commented Feb 5, 2018 at 22:12 1 This is the address: 1728.org/quadtrap.htm – Seyed Commented Feb 5, 2018 at 22:22 9 On the other hand, the question does not even specify which two of the four lengths correspond to the pair of parallel sides. If you are only given four lengths, do they determine a unique trapezoid? – user856 Commented Feb 6, 2018 at 6:41 3 @Rahul For some "lucky" values of the four length, you can deduce which two are parallel, and there is a unique area. But in general, there is more than one way of choosing the parallel sides. Given lengths 1,2,2,4, you can either take 4 and 1 as the parallel sides, or you can take 4 and 2 as the parallel sides. This leads to different trapezoids with different areas (547–√ and 3415−−√, respectively, if I do it correctly). – Jeppe Stig Nielsen Commented Feb 6, 2018 at 15:48 3 "A great deal hinges on whether 'trapezoid' is defined inclusively (i.e. as a quadrilateral with at least one pair of parallel sides) or exclusively (i.e. as a quadrilateral with exactly one pair of parallel sides)." Stuff hinges on more than that! Outside North America, a trapezoid is a quadrilateral with no parallel sides (at least one pair would be a "trapezium" in that nomenclature). – David Richerby Commented Feb 6, 2018 at 22:29 \| Show 3 more comments This answer is useful 17 Save this answer. Show activity on this post. To add a derivation that puts the square root factor in a Heronian context ... For c:=\|CD¯¯¯¯¯¯¯¯\|≠\|AB¯¯¯¯¯¯¯¯\|=:a, \|□ABCD\|=12(a+c)⋅h=12(a+c)⋅2\|△AC′D\|\|C′D¯¯¯¯¯¯¯¯¯¯\|=a+c\|a−c\|\|△AC′D\| Then, applying Heron's Formula to a triangle with side-lengths b, d, c−a, we have \|△AC′D\|=14((c−a)+b+d)(−(c−a)+b+d)((c−a)−b+d)((c−a)+b−d)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√ Share CC BY-SA 3.0 Follow this answer to receive notifications edited Feb 6, 2018 at 9:48 answered Feb 6, 2018 at 7:15 BlueBlue 84.2k1414 gold badges128128 silver badges265265 bronze badges 1 1 This derivation is far better than my own! – mweiss Commented Feb 6, 2018 at 15:05 Add a comment \| This answer is useful 8 Save this answer. Show activity on this post. A note for the case when only two sides are parallel, just the set of four side lengths do not determine the area. An additional information is needed to define, which pair of sides are parallel. An illustrative example for side lengths 19,23,29,31: Share CC BY-SA 3.0 Follow this answer to receive notifications answered Feb 6, 2018 at 15:22 g.kovg.kov 13.9k11 gold badge1414 silver badges3939 bronze badges 5 Yeah, I was just writing a comment to mweiss's answer about the same thing. – Jeppe Stig Nielsen Commented Feb 6, 2018 at 15:53 I was expecting a slightly different diagram, like your first in terms of which sides have which length, but with 29 parallel to 31 instead of 19 parallel to 23 – Henry Commented Feb 7, 2018 at 12:35 1 @Henry: Well, have you tried to construct it with side of length 29 parallel to side of length 31 with the other two 19, 23? – g.kov Commented Feb 7, 2018 at 13:33 @g.kov - good point, as in effect I would need to construct a 19,23,2 triangle. So does this suggest that given the order of the sides and them all being different, there would only be one possibility (and its reflection) for a pair of parallel sides, and so only one area? – Henry Commented Feb 7, 2018 at 15:05 @Henry: This is probably an interesting new question. – g.kov Commented Feb 7, 2018 at 15:17 Add a comment \| This answer is useful 6 Save this answer. Show activity on this post. This is how to calculate the area of a trapezoid when the four sides are known: Share CC BY-SA 3.0 Follow this answer to receive notifications edited Feb 5, 2018 at 23:50 answered Feb 5, 2018 at 22:09 SeyedSeyed 9,06144 gold badges2323 silver badges3131 bronze badges 5 2 Could you explain the derivation of that formula for h, or cite a source? It looks like it comes from a method similar to the one I derived in my answer, but has a nice symmetric form to it. – mweiss Commented Feb 5, 2018 at 22:16 5 You need to assume that a is not c since you have (a−c)2 in a denominator. – Somos Commented Feb 5, 2018 at 23:22 6 @Somos : Which is precisely the condition that the trapezium is not a parallelogram. – Martin Bonner supports Monica Commented Feb 6, 2018 at 12:53 1 Wikipedia Link – Steven Alexis Gregory Commented Feb 6, 2018 at 13:18 1 Don't put plain English text in italics in formulae! I see that the formula is made in Microsoft Word. Write your English text within double quotes to tell Word to treat it as normal text. – Andreas Rejbrand Commented Feb 7, 2018 at 8:30 Add a comment \| This answer is useful 6 Save this answer. Show activity on this post. Hint (if we know the parallel sides): From The picture: take: a=AB,b=BC,c=CD,d=DA,x=AE so we have: h=ED=d2−x2−−−−−−√=b2−(a−c−x)2−−−−−−−−−−−−−−√=CF solve for x and find h=d2−x2−−−−−−√ Find the area A=a+b2h Share CC BY-SA 3.0 Follow this answer to receive notifications edited Feb 6, 2018 at 17:01 answered Feb 5, 2018 at 22:46 Emilio NovatiEmilio Novati 64.4k66 gold badges4949 silver badges128128 bronze badges 2 Shouldn't the second line be b2−( a−c−x)2−−−−−−−−−−−−−−√=CF – crb233 Commented Feb 6, 2018 at 16:36 Yes! Thank you. I edit the typo...:) – Emilio Novati Commented Feb 6, 2018 at 17:01 Add a comment \| This answer is useful 4 Save this answer. Show activity on this post. There can't be such a formula. The side lengths do not determine the area. Think about all the rhombi with four sides of length 1. They are all trapezoids (even parallelograms) with the same side lengths but different areas, which can be anything between 0 and 1. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Feb 5, 2018 at 21:56 answered Feb 5, 2018 at 21:52 Ethan BolkerEthan Bolker 104k77 gold badges127127 silver badges221221 bronze badges 4 I dont understood.How? – Newuser Commented Feb 5, 2018 at 21:54 3 If "trapezoid" is defined exclusively (i.e. as a quadrilateral with exactly one pair of parallel sides, so that parallelograms are not trapezoids) then do the side lengths determine the figure? – mweiss Commented Feb 5, 2018 at 21:55 @mweiss You're right. See my comment on your answer. – Ethan Bolker Commented Feb 5, 2018 at 22:14 1 if these conditions are met a+c+d>b , a+c<b , a+d<b , there should be a unique trapzoid in universe with these parameters. and of course with a calculable area. – Abr001am Commented Feb 6, 2018 at 16:08 Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. I derived the formula for the area using the same procedure as is used to show Heron's Formula. Looking at the end formula, I realized that the formula follows quite simply from Heron's Formula. Given a trapezoid with unequal parallel bases a and b, consider the triangle with base \|b−a\| and sides c and d: Using s=\|b−a\|+c+d2, the area of the triangle is Area of Triangle=s(s−c)(s−d)(s−\|b−a\|)−−−−−−−−−−−−−−−−−−−−−−√ The altitude of the trapezoid and the triangle are the same, so the area is proportional to the average of the lengths of the bases. That is, Area of Trapezoid=b+a\|b−a\|s(s−c)(s−d)(s−\|b−a\|)−−−−−−−−−−−−−−−−−−−−−−√ Share CC BY-SA 4.0 Follow this answer to receive notifications edited Nov 22, 2018 at 17:33 answered Nov 20, 2018 at 12:42 robjohn♦robjohn 354k3838 gold badges497497 silver badges890890 bronze badges 1 I get it now. I had some difficulty at first ... but then I visualised it as two seperate triangles made by drawing a diagonal across the trapezium: one with a as base (or lid, rather) & the other with b as base: the former has area a/\|b-a\| × that of your Heron triange - the latter b/\|b-a\| × it; so the sum of them, the area of the trapezium, is (b+a)/\|b-a\| × it! Neat! – AmbretteOrrisey Commented Nov 23, 2018 at 21:34 Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. -For the trapzoid abcd to have parallel sides it requires all these conditions to be set for uniqueness of area: a+b+c>d, b+a<d, c+a<d. -Neverthless, a trapzoid with non parallel sides can not be defined just by his side lengths, but a fifth coordinate added should. In this experience we show how come multiple trapzoid shapes can be formed with same lengths of edges. Imagine we bring a fork and a knife to start dining on some digestable geometrical concepts : A thread and two pins: We instill the pins on some flat table: Then take the fork and the knife, choose two fixed points in the string, then pull it from these points with those tools without changing the fulcrum points. Distance of the chord from the pinpoints to the coordinates of fulcrums dosn't change, while the shape of the trapzoid changes infinitely! now envisage that h1 is figured by the fork, h2 symbolised by the knife, h1 is directly relative to h2 always regarding the same side lengths. We will show in the following trigonometric relations: cosθ=b‘/a, sinθ=(h2−h1)/a⟹1−(b‘/a)2−−−−−−−−−√=(h2−h1)/a b‘‘/d=1−(h2/d)2−−−−−−−−−√ (b−b‘−b‘‘)/c=1−(h1/c)2−−−−−−−−−√ Since there is 4 unknowns b‘,b‘‘,h1,h2 we can formulate h1 in function of h2 and 4 side constants. Credits for the images goes to canstockphoto.com Share CC BY-SA 3.0 Follow this answer to receive notifications answered Feb 7, 2018 at 2:02 Abr001amAbr001am 76666 silver badges1515 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. I think you mean Brahmagupta's formula, not Heron's formula. There is no formula for the area of trapezoid given the lengths of the sides, because the sides alone do not determine the area. This is true even for a parallelogram. Imagine a parallelogram made of four sticks, joined together by pins at the corners. Then you can slide it closed by moved the top side parallel to the bottom side. You'll get zero area when the top and bottom coincide, and maximum are when you have a rectangle. In the case of Brahmagupta's formula, the quadrilateral is circumscribable, and you can't change the sides like that. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Feb 5, 2018 at 21:56 saulspatzsaulspatz 53.8k77 gold badges3737 silver badges7979 bronze badges 3 2 "This is true even for a parallelogram" - in fact, as other answers show, it is true only for a parallelogram. For a trapezium which is not a parallelogram the size lengths do determine the area (provided you are told which are the lengths of the parallel sides). – Martin Bonner supports Monica Commented Feb 6, 2018 at 12:55 @MartinBonner True, but I've never heard the definition of a trapezoid that excludes a parallelogram. – saulspatz Commented Feb 6, 2018 at 14:31 @saulspatz In the United States, at least, that is the norm at the pre-college level, and has been for more than a century. See my answer on MESE at matheducators.stackexchange.com/a/13766/29 for a historical survey. – mweiss Commented Nov 27, 2018 at 1:29 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. It depends on your definition of trapezoid. If it includes parallelograms, then No: here are two trapezoids, same side-lengths, different area: Share CC BY-SA 4.0 Follow this answer to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 answered Feb 7, 2018 at 12:23 M. WinterM. Winter 30.9k88 gold badges5151 silver badges112112 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. M Weiss put A=a+b2c1−((b−a)2+c2−d22c(b−a))2−−−−−−−−−−−−−−−−−−−−−−√. I would rather have that answer symmetrical in c & d, which would be A=a+b4(b−a)(b−a)4+2(b−a)2(c2+d2)+(c2−d2)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√. But it's just that to my taste it looks odd that it's not symmetrical in c & d ... my mind just protests that it ought to be! Or A=a+b4(b−a)(b−a)2((b−a)2+2(c2+d2))+(c2−d2)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√, even. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Nov 23, 2018 at 8:14 answered Nov 23, 2018 at 8:03 AmbretteOrriseyAmbretteOrrisey 1,00166 silver badges1212 bronze badges 3 This is definitely a better form than mine. Thank you! – mweiss Commented Nov 23, 2018 at 16:28 Glad you like it! That kind of symmetry is a bit of a thing with me ... I kind of - need it! – AmbretteOrrisey Commented Nov 23, 2018 at 21:14 Have you seen the one below though, that adapts Heron's formula? – AmbretteOrrisey Commented Nov 23, 2018 at 21:35 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 geometry euclidean-geometry area quadrilateral See similar questions with these tags. Featured on Meta Upcoming initiatives on Stack Overflow and across the Stack Exchange network... 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14250	https://en.wikipedia.org/wiki/Weinan	Jump to content Search Contents 1 History 1.1 Ancient 1.2 Modern 2 Administration 3 Climate 4 Demographics 5 Transportation 5.1 Public Transportation 5.2 Roads 5.3 Rail 5.4 Air 6 Tourism 7 Education 7.1 Schools 7.2 Universities 8 Notable people 9 International relations 10 References 11 External links Weinan Afrikaans العربية 閩南語 / Bn-lm-gí Català Cebuano Čeština Dansk Deutsch Español Euskara فارسی Français Gĩkũyũ 한국어 Italiano ქართული Kurdî Magyar Македонски Malagasy 閩東語 / Mìng-dĕ̤ng-ngṳ̄ Nederlands 日本語 Norsk bokmål Norsk nynorsk Polski Русский Suomi Svenska Türkçe Українська اردو Vepsän kel’ Tiếng Việt 文言 Winaray 吴语粵語中文 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 Wikimedia Commons Wikidata item Appearance Coordinates: 34°31′14″N 109°28′16″E / 34.5206°N 109.4710°E / 34.5206; 109.4710 From Wikipedia, the free encyclopedia For other uses, see Weinan (disambiguation). Prefecture-level city in Shaanxi, People's Republic of China \| Weinan 渭南市 \| \| Prefecture-level city \| \| Location of Weinan City jurisdiction in Shaanxi \| \| Coordinates (Weinan municipal government): 34°31′14″N 109°28′16″E / 34.5206°N 109.4710°E / 34.5206; 109.4710 \| \| Country \| People's Republic of China \| \| Province \| Shaanxi \| \| Settled \| 668 BC \| \| Municipal seat \| Linwei District \| \| DivisionsCounty-level:Township-level: \| 2 district2 county-level cities7 counties143 towns and villages \| \| Government \| \| • CPC Ctte Secretary \| Li Mingyuan \| \| • Mayor \| Li Yi \| \| Area \| \| • Land \| 13,134 km2 (5,071 sq mi) \| \| • Urban (2018) \| 78 km2 (30 sq mi) \| \| Population (2010) \| \| • Prefecture-level city \| 5,520,772 \| \| • Urban (2018) \| 660,000 \| \| • Urban density \| 8,500/km2 (22,000/sq mi) \| \| Demonym \| Guanzhong dialect \| \| GDP \| \| • Prefecture-level city \| CN¥ 143 billionUS$ 23 billion \| \| • Per capita \| CN¥ 26,729US$ 4,292 \| \| Time zone \| UTC+8 (China Standard) \| \| Postal code \| 714000 \| \| Area code \| 0913 \| \| ISO 3166 code \| CN-SN-05 \| \| Licence plate prefixes \| 陕E \| \| Website \| (in Chinese) Weinan.gov.cn(in English) EN.Weinan.gov.cn \| Weinan (Chinese: 渭南; pinyin: Wèinán) is a prefecture-level city in east-central Shaanxi province, northwest China. The city lies on the lower section of the Wei River confluence into the Yellow River, about 60 km (37 mi) east of the provincial capital Xi'an, and borders the provinces of Shanxi and Henan to the east. The name "Weinan", literally meaning "south of the Wei River", describes the location of the city's urban districts being mostly south of the Wei River, although majority of its metropolitan area actually lies on the north side of the river. History [edit] As a significant area between the ancient Chinese capital Xi'an and Luoyang, Weinan has a long history. Ancient [edit] The ancient Dali Man lived in the modern area of Weinan. The Xiagui county was settled in the year of 668 BC by the state of Qin. Weinan got its name in the year of 360 by the Former Qin state. In the Tang dynasty, 10 emperors were buried in Weinan after their death. On the morning of 23 January 1556, the deadliest earthquake on record with its epicenter in Huaxian killed approximately 830,000 people, destroying an 840 kilometre-wide (520 mi) area. Modern [edit] The Weinan prefecture-level city was established in 1995, in replacement of the Weinan prefecture. Due to the construction of the Sanmenxia Dam, the economy of the city was restricted to agricultural sections, and therefore the development level of the city is much lower than other cities in the province. The city developed rapidly after economic reform in China. The east part of Guanzhong Plain now belongs to the city, making Weinan the second most populated city in the Shaanxi province, after the capital Xi'an. Administration [edit] The municipal executive, legislature and judiciary are in Linwei District, together with the CPC and Public Security bureaux. \| Administrative divisions of Weinan City \| \| Linwei Huazhou TongguanCounty DaliCounty HeyangCounty ChengchengCounty PuchengCounty BaishuiCounty FupingCounty Hancheng(city) Huayin(city) \| \| Division code \| English name \| Chinese name \| Pinyin \| Area (km) \| Seat \| Postal code \| Subdivisions \| \| Subdistricts \| Towns \| Residential communities \| Villages \| \| 610500 \| Weinan City \| 渭南市 \| Wèinán Shì \| 13030.56 \| Linwei District \| 714000 \| 12 \| 130 \| 215 \| 3218 \| \| 610502 \| Linwei District \| 临渭区 \| Línwèi Qū \| 1263.76 \| Duqiao Subdistrict (杜桥街道) \| 714000 \| 8 \| 16 \| 56 \| 514 \| \| 610503 \| Huazhou District \| 华州区 \| Huàzhōu Qū \| 1132.46 \| Huazhou Subdistrict (华州街道) \| 714100 \| 1 \| 9 \| 15 \| 242 \| \| 610522 \| Tongguan County \| 潼关县 \| Tóngguān Xiàn \| 427.35 \| Chengguan Subdistrict (城关街道) \| 714300 \| 1 \| 4 \| 15 \| 78 \| \| 610523 \| Dali County \| 大荔县 \| Dàlì Xiàn \| 1690.60 \| Chengguan Subdistrict (城关街道) \| 715100 \| 1 \| 15 \| 26 \| 400 \| \| 610524 \| Heyang County \| 合阳县 \| Héyáng Xiàn \| 1317.15 \| Chengguan Subdistrict (城关街道) \| 715300 \| 1 \| 11 \| 19 \| 353 \| \| 610525 \| Chengcheng County \| 澄城县 \| Chéngchéng Xiàn \| 1121.64 \| Chengguan Subdistrict (城关街道) \| 715200 \| 1 \| 9 \| 14 \| 266 \| \| 610526 \| Pucheng County \| 蒲城县 \| Púchéng Xiàn \| 1579.81 \| Chengguan Subdistrict (城关街道) \| 715500 \| 1 \| 15 \| 8 \| 373 \| \| 610527 \| Baishui County \| 白水县 \| Báishuǐ Xiàn \| 983.95 \| Chengguan Subdistrict (城关街道) \| 715600 \| 1 \| 7 \| 11 \| 194 \| \| 610528 \| Fuping County \| 富平县 \| Fùpíng Xiàn \| 1245.99 \| Chengguan Subdistrict (城关街道) \| 711700 \| 1 \| 14 \| 9 \| 337 \| \| 610581 \| Hancheng City \| 韩城市 \| Hánchéng Shì \| 1591.60 \| Xincheng Subdistrict (新城街道) \| 715400 \| 2 \| 10 \| 26 \| 275 \| \| 610582 \| Huayin City \| 华阴市 \| Huàyīn Shì \| 676.26 \| Taihua Road Subdistrict (太华路街道) \| 714200 \| 2 \| 4 \| 16 \| 186 \| \| Note: The statistic of Linwei District includes the 2 subdistricts of Weinan High-Tech Industrial Development Zone. \| Climate [edit] \| Climate data for Weinan, elevation 350 m (1,150 ft), (1991–2020 normals, extremes 1981–2010) \| \| Month \| Jan \| Feb \| Mar \| Apr \| May \| Jun \| Jul \| Aug \| Sep \| Oct \| Nov \| Dec \| Year \| \| Record high °C (°F) \| 17.3(63.1) \| 24.2(75.6) \| 30.9(87.6) \| 36.2(97.2) \| 37.5(99.5) \| 42.8(109.0) \| 40.4(104.7) \| 39.0(102.2) \| 37.9(100.2) \| 32.0(89.6) \| 25.6(78.1) \| 22.5(72.5) \| 42.8(109.0) \| \| Mean daily maximum °C (°F) \| 5.3(41.5) \| 9.9(49.8) \| 16.0(60.8) \| 22.7(72.9) \| 27.7(81.9) \| 32.5(90.5) \| 33.2(91.8) \| 30.9(87.6) \| 26.1(79.0) \| 20.3(68.5) \| 12.9(55.2) \| 6.6(43.9) \| 20.3(68.6) \| \| Daily mean °C (°F) \| −0.1(31.8) \| 3.9(39.0) \| 9.5(49.1) \| 15.8(60.4) \| 20.9(69.6) \| 25.9(78.6) \| 27.4(81.3) \| 25.3(77.5) \| 20.4(68.7) \| 14.5(58.1) \| 7.3(45.1) \| 1.4(34.5) \| 14.4(57.8) \| \| Mean daily minimum °C (°F) \| −4.3(24.3) \| −0.8(30.6) \| 4.1(39.4) \| 9.7(49.5) \| 14.5(58.1) \| 19.5(67.1) \| 22.3(72.1) \| 20.8(69.4) \| 16.0(60.8) \| 10.0(50.0) \| 2.9(37.2) \| −2.8(27.0) \| 9.3(48.8) \| \| Record low °C (°F) \| −13.7(7.3) \| −14.5(5.9) \| −7.5(18.5) \| −1.2(29.8) \| 2.4(36.3) \| 10.3(50.5) \| 15.0(59.0) \| 13.0(55.4) \| 6.6(43.9) \| −3.3(26.1) \| −7.7(18.1) \| −16.7(1.9) \| −16.7(1.9) \| \| Average precipitation mm (inches) \| 5.4(0.21) \| 10.1(0.40) \| 21.9(0.86) \| 42.9(1.69) \| 55.3(2.18) \| 56.4(2.22) \| 82.8(3.26) \| 82.0(3.23) \| 96.2(3.79) \| 54.5(2.15) \| 25.9(1.02) \| 5.1(0.20) \| 538.5(21.21) \| \| Average precipitation days (≥ 0.1 mm) \| 3.3 \| 4.0 \| 5.6 \| 7.1 \| 8.4 \| 7.2 \| 8.9 \| 9.2 \| 11.0 \| 8.5 \| 5.6 \| 2.6 \| 81.4 \| \| Average snowy days \| 4.0 \| 2.8 \| 1.1 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| 1.1 \| 3.2 \| 12.2 \| \| Average relative humidity (%) \| 64 \| 63 \| 62 \| 64 \| 64 \| 61 \| 72 \| 78 \| 80 \| 77 \| 74 \| 67 \| 69 \| \| Mean monthly sunshine hours \| 130.8 \| 129.3 \| 167.8 \| 194.8 \| 217.5 \| 218.0 \| 228.5 \| 203.7 \| 146.3 \| 138.6 \| 126.7 \| 128.8 \| 2,030.8 \| \| Percentage possible sunshine \| 42 \| 42 \| 45 \| 49 \| 50 \| 50 \| 52 \| 49 \| 40 \| 40 \| 41 \| 42 \| 45 \| \| Source: China Meteorological Administration \| Demographics [edit] Breakdown of Weinan population by district and county (2010 census) \| Division \| Permanent residents8 \| Hukou residents9 \| \| Total \| Percentage \| Population density (persons/km2) \| \| Weinan City \| 5286077 \| 100 \| 405.67 \| 5600599 \| \| Linwei District \| 877142 \| 16.59 \| 694.07 \| 976822 \| \| Huazhou District \| 322148 \| 6.09 \| 284.47 \| 347913 \| \| Tongguan County \| 155463 \| 2.94 \| 363.79 \| 165860 \| \| Dali County \| 693392 \| 13.12 \| 410.15 \| 718340 \| \| Heyang County \| 436441 \| 8.26 \| 331.35 \| 451983 \| \| Chengcheng County \| 386150 \| 7.31 \| 344.27 \| 404663 \| \| Pucheng County \| 743000 \| 14.06 \| 470.31 \| 782571 \| \| Baishui County \| 279679 \| 5.29 \| 284.24 \| 294823 \| \| Fuping County \| 743385 \| 14.06 \| 596.62 \| 791246 \| \| Hancheng City \| 391164 \| 7.40 \| 245.77 \| 399696 \| \| Huayin City \| 258113 \| 4.88 \| 381.68 \| 266682 \| \| Note:The permanent residents of Linwei district include the 49209 permanent residents of Weinan High-Tech Industrial Development Zone. \| According to the sixth National Population Census of the People's Republic of China, there are 5,286,077 people resident in Weinan. Compared to the previous census data from 2000, the population has decreased by 108,729 persons. Among the residents, 50.75% are males and 49.25% are females. The number of male and female residents are separately 2,682,710 and 2,603,367. The sex ratio is 103.05. Transportation [edit] Due to its location on the plain, Weinan is well connected, especially to the provincial capital Xi'an. Public Transportation [edit] As of March 2015, there are 20 bus routes operated in the urban area of the city. Apart from line 1 and 2 with air-conditioning carrying a fare of 2 RMB, all other bus route fares are 1 RMB. There are also 1000 taxis in the urban city, The base fare is currently ¥5 which covers the first 2.5 km. Additional kilometers cost ¥1.4 or ¥1.5 each. Roads [edit] China National Highway 108 and China National Highway 310 pass through the city. Major expressways in the city are G5 Beijing–Kunming Expressway, G30 Lianyungang–Khorgas Expressway and G65E Yulin–Lantian Expressway. 5 bridges connect north and south part of the city on the Wei River, there are also 3 road bridges connecting the city to the neighboring Shanxi Province on the Yellow River. Rail [edit] The urban city of Weinan owns 4 railway stations. Weinan railway station was open in 1934. Located on the important Longhai railway, almost all Chinese major cities can be reached through the train. Weinan North railway station is served by high-speed trains on the Xuzhou–Lanzhou and Datong–Xi'an lines. The station is part of the Weinan Weihe Grand Bridge, one of the longest bridges in the world. Weinan South railway station and Weinan West railway station are on the Nanjing–Xi'an Railway, they are both only served by the slowest trains. In the suburban area, there are also 2 stations for high-speed railway: Huashan North Railway Station and Dali Railway Station, there are as well also stations for conventional rail like the Huashan railway station and Hancheng railway station. Air [edit] There is no commercial passagenger airport in Weinan, the nearest airport is Xi'an Xianyang International Airport. A shuttle bus connects the Weinan Railway Station with the airport frequently throughout the day. A smaller airport is under construction in Huazhou District to serve the tourism of Hua Mountain. Tourism [edit] Mount Hua, one of China's Five Great Mountains, which has a long history of religious significance. Xiyue Temple, Temple for the God of Mount Hua. Tong Pass, an important strategic pass with historical significance. Education [edit] Schools [edit] As of the end of 2012, there are 2859 schools of all kinds; 75 high schools, 70 vocational schools, 310 junior high schools, 1266 primary schools, 1002 kindergartens and 7 special education schools. Universities [edit] Weinan has 3 higher educational institutions. The Weinan Normal University is a comprehensive multi-discipline research university. Weinan has 2 other tertiary institutions, they are Shaanxi Railway Institute and Weinan Vocational and Technical College. Notable people [edit] Sima Qian, historian of the Han dynasty, widely considered the father of Chinese historiography for his work, the Records of the Grand Historian. Emperor Wen of Sui, founder of the Sui dynasty. Guo Ziyi, general during the Tang dynasty, ended the An Shi Rebellion. Bai Juyi, renowned Chinese poet and Tang dynasty government official. Kou Zhun, a much-praised official of Song dynasty. Yang Hucheng, general who conducted the famous Xi'an Incident in 1936. Xi Zhongxun, communist revolutionary and a political leader of PRC, father of Xi Jinping. Bai Yulu, snooker player, 2 times Woman Snooker World Champion. International relations [edit] See also: List of twin towns and sister cities in China Weinan's twin towns and sister cities are: \| \| \| Szeged, Hungary (since 1999) Gumi, South Korea (since 2014) North Las Vegas, United States (since 2015) Changwon, South Korea (since 2015) Komsomolsk-on-Amur, Russia (since 2016) \| References [edit] ^ a b Cox, W (2018). Demographia World Urban Areas. 14th Annual Edition (PDF). St. Louis: Demographia. p. 22. ^ 陕西省统计局、国家统计局陕西调查总队 (August 2016). 《陕西统计年鉴－2016》. China Statistics Press. ISBN 978-7-5037-7918-3. Archived from the original on 2017-05-29. ^ "China's History of Massive Earthquakes". 12 May 2008. Archived from the original on 14 May 2008. Retrieved 16 June 2008. ^ 国家统计局统计用区划代码 (in Chinese). National Bureau of Statistics of China. Retrieved 2015-06-26. ^ Ministry of Civil Affairs of the People's Republic of China (2014). 《中国民政统计年鉴2014》 (in Simplified Chinese). China Statistics Print. ISBN 978-7-5037-7130-9. ^ 中国气象数据网 – WeatherBk Data (in Simplified Chinese). China Meteorological Administration. Retrieved 26 August 2023. ^ "Experience Template" 中国气象数据网 (in Simplified Chinese). China Meteorological Administration. Retrieved 26 August 2023. ^ National Bureau of Statistics of China (2012). 《中国2010年人口普查分县资料》 (in Simplified Chinese). China Statistics Print. ISBN 978-7-5037-6659-6. ^ Ministry of Public Security of China (1986). 《中华人民共和国全国分县市人口统计资料2010》 (in Simplified Chinese). 群众出版社. ISBN 978-7-5014-4917-0. ^ a b 渭南市2010年第六次全国人口普查主要数据公报. Bureau of Health and Family Planning of Weinan City, in Chinese. Archived from the original on 2015-06-26. Retrieved 2015-06-26. ^ 我市与俄罗斯阿穆尔共青城结为友好城市. Weinan People's Government. Archived from the original on 2017-02-19. Retrieved 2016-05-14. External links [edit] Wikimedia Commons has media related to Weinan. Look up Weinan in Wiktionary, the free dictionary. Weinan Government official website Archived 2018-04-24 at the Wayback Machine (in English and Chinese) Official website of Weinan Normal University (in English, Chinese, and Russian) \| Links to related articles \| \| \| v t e Prefecture-level city of Weinan \| \| Districts \| Linwei Huazhou \| \| Cities \| Hancheng Huayin \| \| Counties \| Baishui Chengcheng Dali Fuping Heyang Pucheng Tongguan \| \| Landmarks \| Mount Hua Tong Pass Heyang Qiachuan National Park Grave and Ancestral Temple of Sima Qian Xiyue Temple Fawang Temple Weinan Sports Center Stadium \| \| Culture & demographics \| People Qinqiang Opera Crystal cake \| \| Education \| Weinan Normal University \| \| Economy \| Weinan Hi-Tech Industries Development Zone \| \| Transport \| Weinan railway station Weinan North railway station Huashan North railway station Dali railway station Weinan Weihe Grand Bridge China National Highway 108 China National Highway 310 G5 Expressway G30 Expressway \| \| v t e County-level divisions of Shaanxi Province \| \| Xi'an (capital) \| \| Sub-provincial city \| \| \| \| --- \| \| Xi'an \| Xincheng District Lianhu District Beilin District Baqiao District Weiyang District Yanta District Yanliang District Lintong District Chang'an District Gaoling District Huyi District Lantian County Zhouzhi County \| \| \| Prefecture-level cities \| \| \| \| --- \| \| Tongchuan \| Yaozhou District Wangyi District Yintai District Yijun County \| \| Baoji \| Weibin District Jintai District Chencang District Fengxiang District Qishan County Fufeng County Mei County Long County Qianyang County Linyou County Feng County Taibai County \| \| Xianyang \| Qindu District Weicheng District Yangling District Xingping city Binzhou city Sanyuan County Jingyang County Qian County Liquan County Yongshou County Changwu County Xunyi County Chunhua County Wugong County \| \| Weinan \| Linwei District Huazhou District Huayin city Hancheng city Tongguan County Dali County Pucheng County Chengcheng County Baishui County Heyang County Fuping County \| \| Yan'an \| Baota District Ansai District Zichang city Yanchang County Yanchuan County Zhidan County Wuqi County Ganquan County Fu County Luochuan County Yichuan County Huanglong County Huangling County \| \| Hanzhong \| Hantai District Nanzheng District Chenggu County Yang County Xixiang County Mian County Ningqiang County Lüeyang County Zhenba County Liuba County Foping County \| \| Yulin \| Yuyang District Hengshan District Shenmu city Fugu County Jingbian County Dingbian County Suide County Mizhi County Jia County Wubu County Qingjian County Zizhou County \| \| Ankang \| Hanbin District Xunyang city Hanyin County Shiquan County Ningshan County Ziyang County Langao County Pingli County Zhenping County Baihe County \| \| Shangluo \| Shangzhou District Luonan County Danfeng County Shangnan County Shanyang County Zhen'an County Zhashui County \| \| \| v t e Prefecture-level divisions of China \| \| Notes: Provincial capitals, ★Sub-provincial cities, ☆Sub-provincial autonomous prefecture Sub prefectural-level divisions, ✧"Comparatively larger city [zh]" (较大的市) as approved by the State Council \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| Provinces \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| --- \| \| Anhui \| Hefei Wuhu Bengbu Huainan Ma'anshan Huaibei Tongling Anqing Huangshan Chuzhou Fuyang Suzhou Lu'an Bozhou Chizhou Xuancheng \| \| Fujian \| Fuzhou ★Xiamen Putian Sanming Quanzhou Zhangzhou Nanping Longyan Ningde \| \| Gansu \| Lanzhou Jiayuguan Jinchang Baiyin Tianshui Wuwei Zhangye Pingliang Jiuquan Qingyang Dingxi Longnan Linxia (Hui) Gannan (Tibetan) \| \| Guangdong \| ★Guangzhou ★Shenzhen Shaoguan Zhuhai Shantou Foshan Jiangmen Zhanjiang Maoming Zhaoqing Huizhou Meizhou Yangjiang Shanwei Heyuan Qingyuan Dongguan Zhongshan Jieyang Yunfu \| \| Guizhou \| Guiyang Liupanshui Zunyi Anshun Bijie Tongren Qianxinan (Buyei and Miao) Qiannan (Buyei and Miao) Qiandongnan (Miao and Dong) \| \| Hainan \| Haikou Sanya Danzhou Sansha Wuzhishan Qionghai Wenchang Wanning Dongfang Ding'an County Tunchang County Chengmai County Lingao County Baisha County (Li) Changjiang County (Li) Ledong County (Li) Lingshui County (Li) Baoting County (Li and Miao) Qiongzhong County (Li and Miao) \| \| Hebei \| Shijiazhuang ✧Tangshan Qinhuangdao Handan Xingtai Baoding Zhangjiakou "Kalgan" Chengde Cangzhou Langfang Hengshui \| \| Henan \| Zhengzhou Kaifeng Luoyang Pingdingshan Anyang Hebi Xinxiang Jiaozuo Puyang Xuchang Luohe Sanmenxia Nanyang Shangqiu Xinyang Zhoukou Zhumadian Jiyuan \| \| Hubei \| ★Wuhan Huangshi Shiyan Yichang Xiangyang Ezhou Jingmen Xiaogan Jingzhou Huanggang Xianning Suizhou Enshi (Tujia and Miao) Xiantao Qianjiang Tianmen Shennongjia Forestry District \| \| Heilongjiang \| ★Harbin Qiqihar Jixi Hegang Shuangyashan Daqing Yichun Jiamusi Qitaihe Mudanjiang Heihe Suihua Daxing'anling Prefecture \| \| Hunan \| Changsha Zhuzhou Xiangtan Hengyang Shaoyang Yueyang Changde Zhangjiajie Yiyang Chenzhou Yongzhou Huaihua Loudi Xiangxi (Tujia and Miao) \| \| Jilin \| ★Changchun Jilin Siping Liaoyuan Tonghua Baishan Songyuan Baicheng Yanbian (Korean) \| \| Jiangsu \| ★Nanjing Wuxi Xuzhou Changzhou ✧Suzhou Nantong Lianyungang Huai'an Yancheng Yangzhou Zhenjiang Taizhou Suqian \| \| Jiangxi \| Nanchang Jingdezhen Pingxiang Jiujiang Xinyu Yingtan Ganzhou Ji'an Yichun Fuzhou Shangrao \| \| Liaoning \| ★Shenyang ★Dalian Anshan Fushun Benxi Dandong Jinzhou Yingkou Fuxin Liaoyang Panjin Tieling Chaoyang Huludao \| \| Qinghai \| Xining Haidong Haibei (Tibetan) Huangnan (Tibetan) Hainan (Tibetan) Golog "Guolog" (Tibetan) Yushu (Tibetan) Haixi "Qaidam" (Mongol and Tibetan) \| \| Sichuan \| ★Chengdu Zigong Panzhihua Luzhou Deyang Mianyang Guangyuan Suining Neijiang Leshan Nanchong Meishan Yibin Guang'an Dazhou Bazhong Ziyang Ya'an Ngawa "Aba" (Tibetan and Qiang) Garzê "Ganzi" (Tibetan) Liangshan (Yi) \| \| Shaanxi \| ★Xi'an Tongchuan Baoji Xianyang Weinan Yan'an Hanzhong Yulin Ankang Shangluo \| \| Shandong \| ★Jinan ★Qingdao Zibo Zaozhuang Dongying Yantai Weifang Jining Tai'an Weihai Rizhao Linyi Dezhou Liaocheng Binzhou Heze \| \| Shanxi \| Taiyuan Datong Yangquan Changzhi Jincheng Shuozhou Jinzhong Yuncheng Xinzhou Linfen Lüliang \| \| Taiwan \| Taibei Gaoxiong Xinbei Taizhong Tainan Taoyuan \| \| Yunnan \| Kunming Qujing Yuxi Baoshan Zhaotong Lijiang Pu'er Lincang Chuxiong (Yi) Honghe (Hani and Yi) Wenshan (Zhuang and Miao) Xishuangbanna (Dai) Dali (Bai) Dehong (Dai and Jingpo) Nujiang (Lisu) Dêqên (Tibetan) \| \| Zhejiang \| ★Hangzhou ★Ningbo ✧Wenzhou Jiaxing Huzhou Shaoxing Jinhua Quzhou Zhoushan Taizhou Lishui \| \| \| \| Autonomous regions \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| --- \| \| Guangxi \| Nanning Liuzhou Guilin Wuzhou Beihai Fangchenggang Qinzhou Guigang Yulin Baise Hezhou Hechi Laibin Chongzuo \| \| Ningxia \| Yinchuan Shizuishan Wuzhong Guyuan Zhongwei \| \| Inner Mongolia \| Hohhot ✧Baotou Wuhai Chifeng "Ulankhad" Tongliao Ordos Hulunbuir Bayannur "Bayannao'er" Ulanqab Hinggan League Xilingol League Alxa League "Ālāshàn League" \| \| Xinjiang \| Ürümqi Karamay Turpan Hami Changji (Hui) Bortala (Mongol) Bayingolin (Mongol) Kizilsu (Kyrgyz) ( ☆Ili (Kazakh) Tacheng Prefecture Altay Prefecture ) Aksu Prefecture Kashgar "Kashi" Prefecture Hotan Prefecture Shihezi Aral Tumxuk Wujiaqu Beitun Tiemenguan Shuanghe Kokdala Kunyu Huyanghe Xinxing \| \| Tibet \| Lhasa Shigatse "Xigazê" Chamdo "Qamdo" Nyingchi "Linzhi" Shannan Nagqu Ngari Prefecture \| \| \| \| \| \| \| \| --- \| \| Direct-administered municipalities \| Beijing Tianjin Shanghai Chongqing \| \| Special administrative regions \| Hong Kong Macau \| \| \| See also: List of prefectures in China, List of cities in China \| \| v t e Largest cities in Shaanxi Source: China Urban Construction Statistical Yearbook 2018 Urban Population and Urban Temporary Population \| \| \| Rank \| \| Pop. \| Rank \| \| Pop. \| --- --- --- \| \| 1 \| Xi'an \| 5,866,100 \| 11 \| Xingping \| 225,800 \| \| 2 \| Xianyang \| 1,023,100 \| 12 \| Shenmu \| 211,800 \| \| 3 \| Baoji \| 893,200 \| 13 \| Hancheng \| 173,200 \| \| 4 \| Yulin \| 631,100 \| 14 \| Yangling \| 154,900 \| \| 5 \| Hanzhong \| 569,500 \| 15 \| Huayin \| 115,500 \| \| 6 \| Weinan \| 551,600 \| 16 \| Binzhou \| 114,100 \| \| 7 \| Yan'an \| 416,700 \| \| 8 \| Tongchuan \| 413,300 \| \| 9 \| Ankang \| 350,700 \| \| 10 \| Shangluo \| 262,500 \| \| \| v t e Major cities along the Yellow River \| \| Province-levelsubdivisions \| Cities (from upper reaches to lower reaches) \| \| Gansu \| Lanzhou Baiyin \| \| Ningxia \| Zhongwei Wuzhong Yinchuan Shizuishan \| \| Inner Mongolia \| Wuhai Ordos Bayan Nur Baotou Hohhot (boundaries of upper and middle reaches) \| \| Shanxi \| Pinglu County \| \| Shaanxi \| Fugu County Jia County Wubu County Hancheng \| \| Henan \| Sanmenxia Luoyang Jiyuan Jiaozuo Zhengzhou(boundaries of middle and lower reaches) Xinxiang Kaifeng Puyang \| \| Shandong \| Liaocheng Tai'an Jinan Dezhou Binzhou Zibo Dongying \| \| Major cities along the Pearl River · Major cities along the Yangtze River \| \| v t e Metropolitan cities of China \| \| \| Major metropolitan regions \| \| \| \| \| Jing-Jin-Ji (BJ-TJ-HE) Greater Bay Area Yangtze Delta (SH-JS-ZJ) Zhongyuan Chengyu Cross-Strait Western Coast Guanzhong Mid-Southern Liaoning Shandong Peninsula Yangtze River Midstream (Yangtze River Valley) \| \| \| \| \| Major cities \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- \| \| \| \| --- \| \| National Central Cities \| Beijinga Chongqinga Guangzhoub2 Shanghaia2 Tianjina2 \| \| Special administrative regions \| Hong Kong Macau \| \| Regional Central Cities \| Chengdub Nanjingb Shenyangb Shenzhenc1 Wuhanb Xi'anb \| \| Sub-provincial cities \| Changchunb Chengdub Dalianc2 Guangzhoub2 Hangzhoub Harbinb Jinanb Nanjingb Ningboc2 Qingdaoc2 Shenyangb Shenzhenc1 Wuhanb Xiamenc1 Xi'anb \| \| Provincial capitals(Prefecture-level) \| Changsha Fuzhou2 Guiyang Haikou Hefei Kunming Lanzhou Nanchang Shijiazhuang Taiyuan Xining Zhengzhou Taibei5 \| \| Autonomous regional capitals \| Hohhot Lhasa Nanning Ürümqi Yinchuan \| \| Comparatively large cities \| Anshan Baotou Benxi Datong Fushun Jiujiang Handan Huainan Jilin Luoyang Qiqihar Suzhou Tangshan Wuxi Xuzhou Zibo \| \| \| \| \| \| Prefecture-level cities by province \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| --- \| \| Hebei \| Shijiazhuang Tangshan Qinhuangdao2 Handan Xingtai Baoding Zhangjiakou Chengde Cangzhou Langfang Hengshui \| \| Shanxi \| Taiyuan Datong Yangquan Changzhi Jincheng Shuozhou Jinzhong Yuncheng Xinzhou Linfen Lüliang \| \| Inner Mongolia \| Hohhot Baotou Wuhai Chifeng Tongliao Ordos Hulunbuir Bayannur Ulanqab \| \| Liaoning \| Shenyang Dalian Anshan Fushun Benxi Dandong Jinzhou Yingkou Fuxin Liaoyang Panjin Tieling Chaoyang Huludao \| \| Jilin \| Changchun Jilin Siping Liaoyuan Tonghua Baishan Songyuan Baicheng \| \| Heilongjiang \| Harbin Qiqihar Jixi Hegang Shuangyashan Daqing Yīchun Jiamusi Qitaihe Mudanjiang Heihe Suihua \| \| Jiangsu \| Nanjing Wuxi Xuzhou Changzhou Suzhou Nantong Lianyungang2 Huai'an Yancheng Yangzhou Zhenjiang Tàizhou Suqian \| \| Zhejiang \| Hangzhou Ningbo Wenzhou2 Jiaxing Huzhou Shaoxing Jinhua Quzhou Zhoushan Tāizhou Lishui \| \| Anhui \| Hefei Wuhu Bengbu Huainan Ma'anshan Huaibei Tongling Anqing Huangshan Chuzhou Fuyang Sùzhou Lu'an Bozhou Chizhou Xuancheng \| \| Fujian \| Fuzhou Xiamen Putian Sanming Quanzhou Zhangzhou Nanping Longyan Ningde \| \| Jiangxi \| Nanchang Jingdezhen Pingxiang Jiujiang Xinyu Yingtan Ganzhou Ji'an Yíchun Fǔzhou Shangrao \| \| Shandong \| Jinan Qingdao Zibo Zaozhuang Dongying Yantai2 Weifang Jining Tai'an Weihai Rizhao Laiwu Linyi Dezhou Liaocheng Binzhou Heze \| \| Henan \| Zhengzhou Kaifeng Luoyang Pingdingshan Anyang Hebi Xinxiang Jiaozuo Puyang Xuchang Luohe Sanmenxia Nanyang Shangqiu Xinyang Zhoukou Zhumadian \| \| Hubei \| Wuhan Huangshi Shiyan Yichang Xiangyang Ezhou Jingmen Xiaogan Jinzhou Huanggang Xianning Suizhou \| \| Hunan \| Changsha Zhuzhou Xiangtan Hengyang Shaoyang Yueyang Changde Zhangjiajie Yiyang Chenzhou Yongzhou Huaihua Loudi \| \| Guangdong \| Guangzhou Shaoguan Shenzhen Zhuhai1 Shantou1 Foshan Jiangmen Zhanjiang2 Maoming Zhaoqing Huizhou Meizhou Shanwei Heyuan Yangjiang Qingyuan Dongguan Zhongshan Chaozhou Jieyang Yunfu \| \| Guangxi \| Nanning Liuzhou Guilin Wuzhou Beihai2 Fangchenggang Qinzhou Guigang Yùlin Baise Hezhou Hechi Laibin Chongzuo \| \| Hainan1 \| Haikou Sanya Sansha4 Danzhou \| \| Sichuan \| Chengdu Zigong Panzhihua Luzhou Deyang Mianyang Guangyuan Suining Neijiang Leshan Nanchong Meishan Yibin Guang'an Dazhou Ya'an Bazhong Ziyang \| \| Guizhou \| Guiyang Liupanshui Zunyi Anshun Bijie Tongren \| \| Yunnan \| Kunming Qujing Yuxi Baoshan Zhaotong Lijiang Pu'er Lincang \| \| Tibet \| Lhasa Shigatse Chamdo Nyingchi Shannan \| \| Shaanxi \| Xi'an Tongchuan Baoji Xianyang Weinan Yan'an Hanzhong Yúlin Ankang Shangluo \| \| Gansu \| Lanzhou Jiayuguan Jinchang Baiyin Tianshui Wuwei Zhangye Pingliang Jiuquan Qingyang Dingxi Longnan \| \| Qinghai \| Xining Haidong \| \| Ningxia \| Yinchuan Shizuishan Wuzhong Guyuan Zhongwei \| \| Xinjiang \| Ürümqi Karamay Turpan Hami \| \| Taiwan5 \| (none) \| \| \| \| \| \| Other cities (partly shown below) \| \| \| \| \| --- \| \| Prefecture-level capitals(County-level) \| (Inner Mongolia: Ulanhot Xilinhot) Jiagedaqi3, Heilongjiang Enshi, Hubei Jishou, Hunan (Sichuan：Xichang Kangding Barkam) (Guizhou: Xingyi Kaili Duyun) (Yunnan: Chuxiong Mengzi Wenshan Jinghong Dali Mangshi Shangri-La Lushui) (Gansu: Linxia Hezuo) (Qinghai: Yushu Delingha) (Xinjiang: Changji Bole Korla Yining Artush Aksu Kashgar1 Hotan Tacheng Altay) \| \| Province-governed cities(Sub-prefecture-level) \| Jiyuan, Henan (Hubei: Xiantao Qiánjiang Tianmen Shennongjia) (Hainan1: Wuzhishan Qionghai Wenchang Wanning Dongfang) (Xinjiang - XPCC(Bingtuan) cities: Shihezi Aral Tumxuk Wujiaqu Beitun Tiemenguan Shuanghe Kokdala Kunyu) \| \| Former Prefecture-level cities \| Chaohu, Anhui Yumen，Gansu Dongchuan, Yunnan Shashi, Hubei (Sichuan: Fuling Wanxian) (Jilin: Meihekou Gongzhuling) \| \| Sub-prefecture-level cities(Prefecture-governed) \| Qian'an, Hebei Manzhouli, Inner Mongolia Erenhot, Inner Mongolia Golmud, Qinghai \| \| \| \| \| County-level cities by province \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| --- \| \| Hebei \| Xinji Jinzhou Xinle Zunhua Qian'an Wu'an Nangong Shahe Zhuozhou Dingzhou Anguo Gaobeidian Botou Renqiu Huanghua Hejian Bazhou Sanhe Shenzhou \| \| Shanxi \| Gujiao Lucheng Gaoping Jiexiu Yongji Hejin Yuanping Houma Huozhou Xiaoyi Fenyang \| \| Inner Mongolia \| Holingol Manzhouli Yakeshi Zhalantun Ergun Genhe Fengzhen Ulanhot Arxan Erenhot Xilinhot \| \| Liaoning \| Xinmin Wafangdian Zhuanghe Haicheng Donggang Fengcheng Linghai Beizhen Gaizhou Dashiqiao Dengta Diaobingshan Kaiyuan Beipiao Lingyuan Xingcheng \| \| Jilin \| Yushu Dehui Jiaohe Huadian Shulan Panshi Gongzhuling Shuangliao Meihekou Ji'an Linjiang Fuyu Taonan Da'an Yanji Tumen Dunhua Hunchun Longjing Helong \| \| Heilongjiang \| Shangzhi Wuchang Nehe Hulin Mishan Tieli Tongjiang Fujin Fuyuan Suifenhe Hailin Ning'an Muling Dongning Bei'an Wudalianchi Anda Zhaodong Hailun \| \| Jiangsu \| Jiangyin Yixing Xinyi Pizhou Liyang Changshu Zhangjiagang Kunshan Taicang Qidong Rugao Haimen Dongtai Yizheng Gaoyou Danyang Yangzhong Jurong Jingjiang Taixing Xinghua \| \| Zhejiang \| Jiande Lin'an Yuyao Cixi Fenghua Rui'an Yueqing Haining Pinghu Tongxiang Zhuji Shengzhou Lanxi Yiwu Dongyang Yongkang Jiangshan Wenling Linhai Longquan \| \| Anhui \| Chaohu Jieshou Tongcheng Tianchang Mingguang Ningguo \| \| Fujian \| Fuqing Changle Yong'an Shishi Jinjiang Nan'an Longhai Shaowu Wuyishan Jian'ou Zhangping Fu'an Fuding \| \| Jiangxi \| Leping Ruichang Gongqingcheng Lushan Guixi Ruijin Jinggangshan Fengcheng Zhangshu Gao'an Dexing \| \| Shandong \| Zhangqiu Jiaozhou Jimo Pingdu Laixi Tengzhou Longkou Laiyang Laizhou Penglai Zhaoyuan Qixia Haiyang Qingzhou Zhucheng Shouguang Anqiu Gaomi Changyi Qufu Zoucheng Xintai Feicheng Rongcheng Rushan Laoling Yucheng Linqing \| \| Henan \| Gongyi Xingyang Xinmi Xinzheng Dengfeng Yanshi Wugang Ruzhou Linzhou Weihui Huixian Qinyang Mengzhou Yuzhou Changge Yima Lingbao Dengzhou Yongcheng Xiangcheng Jiyuan \| \| Hubei \| Daye Danjiangkou Yidu Dangyang Zhijiang Laohekou Zaoyang Yicheng Zhongxiang Yingcheng Anlu Hanchuan Shishou Honghu Songzi Macheng Wuxue Chibi Guangshui Enshi Lichuan Xiantao Qianjiang Tianmen \| \| Hunan \| Liuyang Liling Xiangxiang Shaoshan Leiyang Changning Wugang Miluo Linxiang Jinshi Yuanjiang Zixing Hongjiang Lengshuijiang Lianyuan Jishou \| \| Guangdong \| Lechang Nanxiong Taishan Kaiping Heshan Enping Lianjiang Leizhou Wuchuan Gaozhou Huazhou Xinyi Sihui Xingning Lufeng Yangchun Yingde Lianzhou Puning Luoding \| \| Guangxi \| Cenxi Dongxing Guiping Beiliu Jingxi Yizhou Heshan Pingxiang \| \| Hainan \| Wuzhishan Qionghai Wenchang Wanning Dongfang \| \| Sichuan \| Dujiangyan Pengzhou Qionglai Chongzhou Jianyang Guanghan Shifang Mianzhu Jiangyou Emeishan Langzhong Huaying Wanyuan Barkam Kangding Xichang \| \| Guizhou \| Qingzhen Chishui Renhuai Xingyi Kaili Duyun Fuquan \| \| Yunnan \| Anning Xuanwei Tengchong Chuxiong Mengzi Gejiu Kaiyuan Mile Wenshan Jinghong Dali Ruili Mangshi Lushui Shangri-La \| \| Tibet \| (none) \| \| Shaanxi \| Xingping Hancheng Huayin \| \| Gansu \| Yumen Dunhuang Linxia Hezuo \| \| Qinghai \| Yushu Golmud Delingha \| \| Ningxia \| Lingwu Qingtongxia \| \| Xinjiang \| Changji Fukang Bole Alashankou Korla Aksu Artush Kashgar Hotan Yining Kuytun Korgas Tacheng Wusu Altay Shihezi Aral Tumxuk Wujiaqu Beitun Tiemenguan Shuanghe Kokdala Kunyu \| \| Taiwan5 \| (none) \| \| \| \| \| \| \| \| \| \| \| Indicates this city has already occurred above. aDirect-administered municipalities. bSub-provincial cities as provincial capitals. cSeparate state-planning cities. 1Special economic-zone cities. 2Open coastal cities. 3Prefecture capital status established by Heilongjiang Province and not recognized by Ministry of Civil Affairs. Disputed by Oroqen Autonomous Banner, Hulunbuir, Inner Mongolia as part of it. 4Only administers islands and waters in South China Sea and have no urban core comparable to typical cities in China. 5The claimed province of Taiwan no longer have any internal division announced by Ministry of Civil Affairs of PRC, due to lack of actual jurisdiction. See Administrative divisions of Taiwan instead. All provincial capitals are listed first in prefecture-level cities by province. \| \| \| \| v t e Major regions and cities of China \| \| Nationalmegalopolises \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| Jing-Jin-Ji(Inner) Bohai Economic Rim \| \| \| \| --- \| \| Beijing \| Beijing + Changping + Daxing + Fangshan + Mentougou + Shunyi + Tongzhou \| \| Tianjin \| Tianjin + Binhai + Dongli + Jinnan + Wuqing \| \| Hebei \| Baoding + Xiong'an Cangzhou Chengde Langfang Shijiazhuang Tangshan + Caofeidian + Qian'an Zhangjiakou \| \| \| Yangtze Delta(Economic Zone) \| \| \| \| --- \| \| Jiangsu \| Changzhou Lianyungang Nanjing Nantong Suzhou Taizhou Wuxi Xuzhou Yangzhou Zhenjiang \| \| Shanghai \| Shanghai + Baoshan + Jiading + Minhang + Pudong + Qingpu + Songjiang \| \| Zhejiang \| Hangzhou Huzhou Jiaxing Jinhua Lishui Ningbo Quzhou Shaoxing Taizhou Wenzhou Zhoushan \| \| Anhui \| Chuzhou Hefei Huainan Ma'anshan Wuhu \| \| \| Pearl River Deltaa.k.a. Greater Bay Area(Economic Zone) \| \| \| \| --- \| \| Guangdong \| Dongguan Foshan Guangzhou + Huadu + Nansha + Panyu Huizhou Jiangmen Shenzhen + Bao'an Zhaoqing Zhongshan Zhuhai + Hengqin \| \| SARs \| Hong Kong + Kowloon Macau + Cotai \| \| \| West TriangleEconomic Zone \| \| \| \| --- \| \| Chongqing \| Chongqing + Fuling + Liangjiang + Qianjiang + Wanzhou + Xinbei \| \| Sichuan \| Chengdu Dazhou Deyang Guang'an Leshan Luzhou Meishan Mianyang Nanchong Neijiang Suining Yibin Zigong Ziyang \| \| Shaanxi \| Baoji Tongchuan Weinan Xi'an Xianyang + Yangling \| \| \| Central Plain Megalopolis \| \| \| \| --- \| \| Henan \| Jiaozuo Jiyuan Kaifeng Luohe Luoyang Pingdingshan Xinxiang Xuchang Zhengzhou \| \| \| Harbin-Changchun Megalopolis(Northeastern Cities) \| \| \| \| --- \| \| Heilongjiang \| Harbin Daqing Mudanjiang Qiqihar \| \| Jilin \| Changchun Jilin Siping Yanji \| \| \| Middle Reaches of the Yangtze River/Yangtze River Valley(Central Triangle Economic Zone) \| \| \| \| --- \| \| Hubei \| Ezhou Huanggang Huangshi Qianjiang Tianmen Wuhan Xianning Xiantao Xiaogan \| \| Hunan \| Changde Changsha Hengyang Loudi Xiangtan Yiyang Yueyang Zhuzhou \| \| Jiangxi \| Fuzhou Ji'an Jingdezhen Jiujiang Nanchang Shangrao Xinyu Yichun Yingtan \| \| Anhui \| Anqing Bengbu Chizhou Chuzhou Hefei Huainan Lu'an Ma'anshan Tongling Wuhu Xuancheng \| \| \| (North) Bohai Economic Rim \| \| \| \| --- \| \| Liaoning \| Anshan Dalian + Lüshun Huludao Jinzhou Panjin Shenyang Yingkou \| \| \| (South) Bohai Economic Rim \| \| \| \| --- \| \| Shandong \| Binzhou Dongying Jinan Qingdao Weifang Weihai Yantai Zibo \| \| \| \| Regions \| East + Northeast North South Central + Central + South + Huizhou Western + Northwestern + Southwestern \| \| Administrative divisions \| By GDP By GDP per capita By Human Development Index Prefecture-level divisions County-level divisions \| \| Cities \| Direct-administered municipality Prefecture-level city Sub-provincial city County-level city List of cities in China + by GDP per capita + by population \| \| Capitals \| Historical capitals Current and former capitals \| \| Categories: Subdivisions Regions Cities \| \| \| Authority control databases \| \| International \| \| \| Other \| Yale LUX \| Retrieved from " Categories: Weinan Cities in Shaanxi Hidden categories: Pages using gadget WikiMiniAtlas CS1: unfit URL CS1 uses Chinese-language script (zh) CS1 Chinese-language sources (zh) CS1 Simplified Chinese-language sources (zh-hans) Articles with short description Short description is different from Wikidata Coordinates on Wikidata Articles containing Chinese-language text Articles with Chinese-language sources (zh) Articles containing simplified Chinese-language text Commons category link is on Wikidata Webarchive template wayback links Articles with Russian-language sources (ru) Pages using largest cities with nav class Pages using largest cities with unknown parameters Weinan Add topic
14251	https://math.stackexchange.com/questions/3602888/does-stewarts-calculus-consider-endpoints-of-the-domain-of-a-nice-function-to	education - Does Stewart's calculus consider endpoints of the domain of a (nice) function to be critical numbers? - 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 Does Stewart's calculus consider endpoints of the domain of a (nice) function to be critical numbers? Ask Question Asked 5 years, 6 months ago Modified3 years, 6 months ago Viewed 206 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. This is a question of convention - specifically the convention used in Stewart's Calculus. In Stewart's calculus (the latest version), in chapter 4.1, definition 6 defines a critical number of a function f f to be a number c c in the domain of f f such that either f′(c)=0 f′(c)=0 or f′(c)f′(c) does not exist. Suppose f f is obtained from a nice function by restricting to a closed interval [a,b][a,b]. Say, g(x)=x g(x)=x and f=g\|[0,1]f=g\|[0,1]. Does Stewart consider 0 and 1 to be critical points of f f? In other words, does f′(0)f′(0) and f′(1)f′(1)exist? The book seems to be extremely elusive above this, going so far as to not including any exercises that might elucidate which convention he uses. I'm asking this because I'm an instructor trying to decide what convention to take. I'm hoping that by choosing one convention over the other, I do not inadvertently subtly contradict Stewart somewhere down the road. In Stewart's description of "The closed interval method" (Section 4.1), he first asks you to find the values of f f at critical numbers in (a,b)(a,b), and then to compute the values of f f at the endpoints a a and b b (and doesn't simply say compute values of f f at all critical numbers). The fact that he does this would suggest that he does not consider the endpoints to be critical points. On the other hand, taking this convention would seem to imply that f′(a)f′(a) does exist, but he defines f′(a):=lim h→0 f(a+h)−f(a)h f′(a):=lim h→0 f(a+h)−f(a)h, and he only ever defines the limit of a function at points on the interior of its domain, so technically according to Stewart this limit is "undefined", so "does not exist", but then it is very puzzling why he phrases the closed interval method in the way that he does. calculus education Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Mar 31, 2020 at 7:08 stupid_question_botstupid_question_bot asked Mar 31, 2020 at 6:38 stupid_question_botstupid_question_bot 1,253 7 7 silver badges 8 8 bronze badges 9 Stewart is not claiming that f′(a)f′(a) or f′(b)f′(b) exist or don't exist. He is not calling the endpoints critical numbers. Just because something is not a critical point does not mean its derivative has to exist - that is not logically equivalent to its converse Ninad Munshi –Ninad Munshi 2020-03-31 06:42:44 +00:00 Commented Mar 31, 2020 at 6:42 @NinadMunshi His definition of a critical point is a point where either the derivative vanishes or it does not exist. Thus, this means that not being a critical point means the derivative exists and is nonvanishing. A definition is an "if and only if".stupid_question_bot –stupid_question_bot 2020-03-31 06:45:44 +00:00 Commented Mar 31, 2020 at 6:45 A definition is not always an "if and only if"Ninad Munshi –Ninad Munshi 2020-03-31 06:46:49 +00:00 Commented Mar 31, 2020 at 6:46 @NinadMunshi Give me one example of a peer-reviewed math paper where a definition is not an if and only if.stupid_question_bot –stupid_question_bot 2020-03-31 06:47:56 +00:00 Commented Mar 31, 2020 at 6:47 You're giving me a textbook written for first year calculus students, it seems a little unfair to expect me to provide you a paper no? Focusing on the purpose of a definition is more important than the pathologies of an imprecise definition. When Stewart is mentioning limits not existing, we both know intuitively he is referring to cusps and vertical tangents (as he is assuming continuity). End points don't fit in that picture. Logically, it would not be a good idea to focus on that pathology as it is not the point of the concept - because Stewart did not aim for a real analysis audience Ninad Munshi –Ninad Munshi 2020-03-31 06:56:51 +00:00 Commented Mar 31, 2020 at 6:56 \|Show 4 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. You are correct. The derivative does not exist at the endpoints based on Stewart's definition, and therefore by Stewarts' definition endpoints are critical points. However, as Ninad Munshi, suggests, do not focus on this. If a student asks, you can have the same discussion we are having here. Whether you test endpoints to find a global max or min or you include them as "critical points" is only a matter of naming. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 23, 2022 at 4:36 Applied Squared MathematicianApplied Squared Mathematician 835 1 1 gold badge 6 6 silver badges 23 23 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 calculus education 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 6Does a limit at infinity exist? 0Single variable calculus beginner - is this interval contained in the domain of the function? 1Closed Interval Method 0Relationship between local max/min and absolute max/min 1Can a function be differentiable at its endpoints? 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14252	https://artofproblemsolving.com/wiki/index.php/Change_of_base_formula?srsltid=AfmBOoqE7vbPVJzWmAEPXWjUwEK6NmyNF_341Mo097o7AIYkpjaVMpyi	Art of Problem Solving Change of base formula - 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 Change of base formula Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Change of base formula The change of base formula is a formula for expressing a logarithm in one base in terms of logarithms in other bases. For any positivereal numbers such that neither nor are , we have This allows us to rewrite a logarithm in base in terms of logarithms in any base . This formula can also be written Proof Let . Then . And, taking the of both sides, we get By the properties of logarithms, Substituting for y, Use for computations The change of base formula is useful for simplifying certain computations involving logarithms. For example, we have by the change of base formula that The formula can also be useful when calculating logarithms on a calculator. Many calculators have only functions for calculating base-10 and base-e logarithms. But you can still calculate logs in other bases, you just need to use the change of base formula to put in in base 10. For example, if you wanted to calculate , you would first convert it to the form . Then you would evaluate it using the base-10 log function on the calculator. Special cases and consequences Many other logarithm rules can be written in terms of the change of base formula. For example, we have that . Using the second form of the change of base formula gives . One consequence of the change of base formula is that for positive constants , the functions and differ by a constant factor, for all . This article is a stub. Help us out by expanding it. Retrieved from " Category: Stubs 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.
14253	https://study.com/learn/lesson/boolean-expressions-statements-logic-operators-examples.html	Boolean Expression & Operators \| Definition & Application - Lesson \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Business Courses Business 109: Intro to Computing Boolean Expression & Operators \| Definition & Application Contributors: Aruna AG, Paul Zandbergen Author Author: Aruna AG Show more Instructor Instructor: Paul Zandbergen Show more Explore what a Boolean expression is and learn how to write a Boolean expression. Read the definition of Boolean operators and see what Boolean logic operators are. Updated: 11/21/2023 Table of Contents What is a Boolean Expression? Boolean Operators: Definition Applications: Boolean Coding or Boolean Programming Lesson Summary Show FAQ What is the meaning of Boolean expression? Boolean expressions are the expressions that evaluate a condition and result in a Boolean value i.e true or false. Ex: (a>b && a> c) is a Boolean expression. It evaluates the condition by comparing if 'a' is greater than 'b' and also if 'a' is greater than 'c'. If both the conditions are true only then does the Boolean expression result is true. If any one condition is not true then the Boolean expression will result in false. What is an example of a Boolean operator? Boolean operators are used to execute Boolean expressions. Boolean operators compare the conditional expressions and return a Boolean value. The most common Boolean operators are: AND, OR and NOT operators. (condition1 AND condition2) AND operator - results true if both the conditions are true (condition1 OR condition2) OR operator - results true if any one of the conditions is true NOT (condition) NOT operator - returns true when the condition is false. How do you write Boolean expressions? Boolean expressions are written using Boolean operators (AND) &&, (OR)\|\| and (NOT) !. Example: (x>1) && (x<5) - returns true if both the conditions are true, i.e if the value of 'x' is between 1 and 5. (x%x) \|\| (x%1) - returns true if any one condition is true, i.e: either 'x' is divisible by itself OR if 'x' is divisible by 1. !(x==0) - returns true if the value of 'x' is other than 0. If the value of 'x' is 0 then it returns false. Create an account LessonTranscript VideoQuizCourse Click for sound 6:48 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? 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Your next lesson will play in 10 seconds 0:06 Boolean Data Type 1:14 Boolean Expression 2:51 Boolean Operators 6:07 Lesson Summary QuizCourseView Video Only Save Timeline 57K views Recommended lessons and courses for you Related LessonsRelated Courses ##### Prolog in AI: Definition & Uses 5:06 ##### What Is Boolean Logic? - Definition, Diagram & Examples 5:54 ##### Data Types in Programming: Numbers, Strings and Others 8:00 ##### Programming Language \| Levels, Interpreters & Compilers 8:55 ##### Programming Language \| Definition, Types & List 5:43 ##### How to Set up a Coding Environment in Python 6:27 ##### Simple Reflex Agents: Definition, Uses & Examples 4:22 ##### First-Order Logic in AI: Identification, Uses & Calculations 7:07 ##### Utility-based Agents: Definition, Interactions & Decision Making 4:24 ##### Types of Artificial Intelligence 6:23 ##### Learning Agents: Definition, Components & Examples 3:13 ##### Agent-Based Modeling \| Process & Examples 5:58 ##### Goal-based Agents: Definition & Examples 4:41 ##### Knowledge Engineering \| Process , Role & Examples 6:01 ##### Game Theory in Artificial Intelligence 8:06 ##### Heuristic Methods in AI: Definition, Uses & Examples 5:37 ##### Using Artificial Intelligence (AI) and Expert Systems to Solve Complex Problems 7:44 ##### Probabilistic Reasoning & Artificial Intelligence 5:02 ##### Memory Segmentation in an Operating System \| Definition & Purpose 5:23 ##### How to Write a Program: Coding, Testing & Debugging 7:36 ##### Computer Science 303: Database Management ##### Computer Science 107: Database Fundamentals ##### Business 104: Information Systems and Computer Applications ##### Workplace Communications with Computers ##### Intro to Excel: Essential Training & Tutorials ##### CLEP Information Systems Study Guide and Exam Prep ##### DSST Computing and Information Technology Study Guide and Test Prep ##### Intermediate Excel Training: Help & Tutorials ##### Microsoft Excel Certification: Practice & Study Guide ##### UExcel Business Information Systems: Study Guide & Test Prep ##### DSST Management Information Systems Study Guide and Test Prep ##### Business 303: Management Information Systems ##### System Analysis & Design: Help & Review ##### Computer Science 102: Fundamentals of Information Technology ##### Computer Science 111: Programming in C ##### Computer Science 115: Programming in Java ##### Computer Science 305: Operating Systems ##### Computer Science 306: Computer Architecture ##### Computer Science 302: Systems Analysis & Design ##### Computer Science 201: Data Structures & Algorithms What is a Boolean Expression? ----------------------------- In computer programming, data is a set of instructions or facts which is processed as information and stored in a computer system. The data can be in the form of a document, audio or video. Computers interpret organized data in the form of binary numbers (0 and 1). Data type determines the type of values which can be stored or used to perform operations on them. In computer programming, the most commonly used data types are integer, float, double, character, string, Boolean, date and time. Let us learn more about Boolean data. The term Booleanis named after a mathematician called George Boole who invented mathematical logic and defined Boolean algebra. The Boolean data type can store either 1 or 0, which represents true or false. The primary use of Boolean data type is in conditional statements to check whether a condition is true or false. The Boolean data type can be used to write a well-structured program that allows the programmer to control the execution of a set of instructions by choosing between yes or no options. Boolean expressions are used to make decisions. Boolean data is used in Boolean expressions which produce a result as true or false. Boolean expressions help decide whether to execute a set of instructions or not depending on the result of Boolean data. Example: if (condition) statement1 else statement2 In this example, the condition is a Boolean expression that results in either true or false. If the Boolean expression results true, a statement1 will be executed. If the Boolean expression results false, then statement2 will be executed. This way programs can be controlled with logical decisions. The next set of instructions will be executed depending on the result of a Boolean expression. How to Write a Boolean Expression For example, consider the code in which it accepts 2 input values a and b with a value of 5 and 10 respectively. Compare which is greater by using a condition (a>b). a=5 b=10 a>b statement1 In this, (a>b) is the Boolean expression. This expression compares whether a is having a value greater than b. The result of this code will be either true or false depending on the input values. In this example, the value of a is less than the value of b, hence the Boolean expression will give a result as false. Depending on the result, the next set of instructions will be executed. In this example, since the Boolean expression (a>b) results in false, statement1 will not be executed. Statement1 will be executed only when the Boolean expression (a>b) results true. Boolean Expression Examples if(age>18) - age>18 is a Boolean expression that returns true if the input value for age is greater than 18. If the input value is less than 18, then the Boolean expression will return false. while(n!=0) - n!=0 is a Boolean expression. In this example, the Boolean expression returns true when the value of n is other than 0. If the value of n=0, the Boolean expression returns false. Boolean Operators: Definition ----------------------------- The Boolean operators are used to perform Boolean logic operations using Boolean expressions to make a logical decision in a programming language. The most common Boolean operators used are: AND (&), OR (\|) and NOT (!). These Boolean operators, when used in a Boolean expression, control the program flow based on the Boolean expression result. Boolean operators are used to making choices in a program. It is used in searching for a particular topic. For example, if we want to search for an article about the Benefits of Yoga, we also require information about Simple Yoga Poses. The Boolean operator AND will narrow the search by including all the content or articles related to both yoga poses and the benefits of yoga. AND operator helps to connect various information and search the actual required content we are looking for. OR operator is used to choosing between 2 topics. Using the NOT operator, we can negotiate the content we do not require. AND operator represented using Venn diagram OR operator represented using Venn diagram NOT operator represented using Venn diagram Boolean Logic Operators \| Operator \| Operator name \| Function \| --- \| & \| Bitwise logical AND \| Bitwise AND operator returns 1 only when both the bits are 1 otherwise it will return 0. It is used to compare the data at the bit level. \| \| \| \| Bitwise logical OR \| Bitwise OR operator returns 1 if one of the integers has 1 otherwise it returns 0. It is used to compare the data at the bit level. \| \| ^ \| Bitwise logical exclusive OR (XOR) \| Bitwise OR operator returns 1 if only one of the integers has 1 otherwise it returns 0. It is used to compare the data at the bit level. \| \| && \| Conditional AND \| It is used to compare 2 conditional statements and return 1 if both the conditional statements are true otherwise return false. It is used effectively in searching. It helps in narrowing the search and finding the specific content required. \| \| \|\| \| Conditional OR \| It is used to compare 2 conditional statements and return 1 if any one of the conditional statements is true otherwise return false. OR operator helps to broaden the searching, by choosing any of the terms mentioned in the search. \| \| ! \| Boolean logical NOT \| NOT operator is used to reverse the logical state. NOT operator helps to exclude the Not required content from the search. \| \| &= \| AND assignment \| It is used to perform a bitwise AND operation and assign the result to a variable. It is used to reduce the memory storage by assigning the result to the same variable. \| \| \|= \| OR assignment \| It is used to perform a bitwise OR operation and assign the result to a variable. It is used to reduce the memory storage by assigning the result to the same variable. \| \| ^= \| XOR assignment \| It is used to perform a bitwise XOR operation and assign the result to a variable. It is used to reduce the memory storage by assigning the result to the same variable. \| \| == \| Equal to \| This operator compares 2 operands and returns true if the operands are equal otherwise will return false.It is used to compare 2 operands or expressions of equality. \| \| != \| Not equal to \| This operator compares 2 operands and returns true if the operands are not equal otherwise will return false. It is used to compare 2 operands or expressions of in-equality. \| Boolean Operators Examples \| Operator \| Operator name \| Example \| --- \| & \| Bitwise logical AND \| Returns 1 only when both the bits are 1 otherwise it will return 0. Ex: int a=2, b=3 In binary 2 is represented by 010 and 3 is represented by 011.a & b i.e: 010 & 011 which results in 010. \| \| \| \| Bitwise logical OR \| Returns 1 if one of the integers has 1 otherwise it returns 0.Ex: int a=2, b=3 In binary 2 is represented by 010 and 3 is represented by 011.a \| b i.e: 010 \| 011 which results in 011. \| \| ^ \| Bitwise logical exclusive OR (XOR) \| Returns 1 if only one of the integers has 1 otherwise it returns 0.Ex: int a=4, b=5 In binary, 4 is represented by 110 and 5 is represented by 101.a ^ b i.e: 110 ^ 101 which results in 011. \| \| && \| Conditional AND \| Compare 2 conditional statements and return 1 if both the conditional statements are true otherwise return false.Ex: (a>b) && (a>c)a should be greater than b and c only then the AND operator results in 1 otherwise 0. \| \| \|\| \| Conditional OR \| Compare 2 conditional statements and return 1 if any one of the conditional statements is true otherwise return false.Ex: (a>b) \|\| (a>c)either a should be greater than b or a should be greater than c only then the OR operator results in 1 otherwise 0. \| \| ! \| Boolean logical NOT \| Reverse the logical state. Ex: !(a>b)if a is greater than b, the NOT operator returns 0 otherwise it will return 1. \| \| &= \| AND assignment \| Perform a bitwise AND operation and assign the result to a variable.Ex: int n=3 n &= 5 In binary, 3 is represented by 011 and 5 is represented by 101.n&=5 performs bitwise AND operation on 3 and 5 then assigns the result to n. \| \| \|= \| OR assignment \| Perform a bitwise OR operation and assign the result to a variable.Ex: int n=3 n \|= 5 In binary, 3 is represented by 011 and 5 is represented by 101.n\|=5 performs bitwise OR operation on 3 and 5 then assigns the result to n. \| \| ^= \| XOR assignment \| Perform a bitwise XOR operation and assign the result to a variable.Ex: int n=3 n ^= 5 In binary, 3 is represented by 011 and 5 is represented by 101.n^=5 performs bitwise XOR operation on 3 and 5 then assigns the result to n. \| \| == \| Equal to \| Compares 2 operands and returns true if the operands are equal otherwise will return false.Ex:int n=2(n==4)In this example, the value of n is not equal to 2 hence it results in false. If (n==2) only then it results true. \| \| != \| Not equal to \| Compares 2 operands and returns true if the operands are not equal otherwise will return false.Ex:int n=2(n!=4)In this example, the value of n is not equal to 4 hence it results in true. If the value of n is 4 only then it results true. \| Applications: Boolean Coding or Boolean Programming --------------------------------------------------- Boolean data types, Boolean expression and Boolean operators are used in a programming language to make choices and decisions. Boolean coding is used to control the flow of the program based on the conditions. Example 1:For every application, we set a password. When we type the password, the application compares the password given by the user with the saved password. If the entered password matches with the password stored in the application, it will allow the user to login into the application otherwise it will give an error message. In this case, the AND operator returns true, when both the password matches, otherwise returns false. Example 2:In any search engine, when you search for any information like what is Boolean programming. The search engine automatically adds AND in between each word like what AND is AND Boolean AND programming. The search results will contain the most similar content. Example 3: If we want to get information about global warming, we can search for global warming OR temperature change OR overheating. Any of these words are valid and help in searching a wide range of content related to a similar topic. Lesson Summary -------------- The Boolean data type can result in only 1 value either true or false based on the condition. Boolean expressions are the conditional statements that return true or false by comparing the given condition. Boolean expression helps to control the flow of the program structure. Boolean operators are used to execute Boolean expressions. AND, OR and NOT are the basic Boolean operators used commonly in most programming languages. These operators are used in coding to control the flow, search engines to extract the appropriate information and database queries to get the required record of information. Video Transcript Boolean Data Type Programming uses a number of different data types. The data type of an object determines what type of values an object can have and what operations can be performed on the object. Commonly used data types include strings, numbers, lists and arrays. This lesson will look more closely at one data type that is widely used: Boolean data. The Boolean data type can only represent two values: true or false. Typically, a 1 is used to represent true, and a 0 is used to represent false. Boolean data is widely used when working with conditions. If you want to ask a basic question and the answer can be only yes or no, you need a Boolean. The term 'Boolean' comes from a 19th century mathematician called George Boole who came up with the original idea of what we now call Boolean logic in his book The Laws of Thought. Boolean Expression Boolean data are used in Boolean expressions, which are expressions in a programming language that produce a Boolean value. An expression in programming is any combination of values, variables and operators that produce a new value. For example, 2 + 3 is an expression, and the result is the new value 5. When you use a Boolean expression, the only logical result can be true or false. Consider the following example where a user inputs two values, and a computer program determines whether the first one is smaller than the second one or not. x = 8 y = 7 x< y In this example, the part 'x<y' is the Boolean expression. You are asking whether x is less than y, and the answer can only be a yes or a no - which means true or false in programming. In the example, the value of x is in fact not smaller than the value of y, and the program therefore results in a Boolean value of false. In programming language, we say that the expression is evaluated and returns a value of false. The Boolean type is the primary result of conditional statements, which are used to control workflow in program. For example, if a particular condition is true, then do this; if the condition is false, then do something else. Boolean Operators In addition to Boolean data, there are Boolean operators, which are used to carry out Boolean algebra. There are three main Boolean operators: AND, OR and NOT. The first two are used to combine two expressions; the third is used as a negation operator. Let's look at each of these in more detail. The simplest Boolean operator is the NOT operator. It simply turns true into false, and vice versa. Consider the following example. x = 8 y = 7 NOT (x < y) This returns a value of true. We know that x is greater than y, so the expression 'x<y' returns a value of false. The NOT operator turns this into a value of true. Now, let's look at the AND and OR operators. The AND operator compares two expressions. It only returns a value of true if both expressions are true; otherwise, it returns a value of false. Consider the following example: x = 8 y = 7 z = 6 (x<y) AND (z<y) The first expression is false, and the second expression is true. The AND operator combines both expressions, and since one of them is false, the final result is false. Now, let's look at the OR operator. The OR operator also compares two expressions. It returns a value of true if one of the expressions is true or if both expressions are true. If both expressions are false, it returns a value of false. x = 8 y = 7 z = 6 (x<y) OR (z<y) The first expression is false, and the second expression is true. The OR operator combines both expressions, and since one of them is true, the final result is true. These Boolean operators illustrate the use of Boolean logic. Boolean logic is widely used when writing programs. It is also widely used as part of search algorithms and database queries. For example, when you use a search engine, you can go into advanced settings to give you more control. One option is to search for all these words; you want to find pages where all these words occur together. This is similar to using a Boolean AND operator, since you want the results for finding those words on the same page to be true. This is typically the default for any search engine. Another option is to search for any of these words; you want to find the page where one or more of these words occur, but they don't all have to occur together. This is similar to using a Boolean OR operator, since you don't need the results to contain all of these words together; it just needs to be true for one of them. So, you are using Boolean operators every time you do an online search using two or more words. So, now you know a bit more about how those search engines actually work. Lesson Summary The Boolean data type can only represent two values: true or false. Boolean expressions are expressions in a programming language that produce a Boolean value. This is like asking a question where the logical answers can only be true or false. Boolean operators are used to carry out Boolean algebra. The three main Boolean operators are AND, OR and NOT. The first two are used to combine two expressions, and the third is used to return the opposite value. Boolean logic is used in programming, search engines and database queries. Learning Outcomes After you've reviewed this video lesson, you should be able to: Define Boolean data, Boolean expressions and Boolean operators Explain the uses of the three main Boolean operators Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a member Already a member? Log in Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. Teacher Try it now Go back Create an account to start this course today Start today. Try it now Business 109: Intro to Computing 10 chapters 99 lessons 9 flashcard sets Chapter 1 Application Software Application Software for Personal, Enterprise & Workgroup Objectives 8:21 min Microsoft Office and Open Office: Office Suite Applications 5:32 min What Is Word Processing Software? - Definition, Types & Examples 5:29 min Automating Tasks Using Macros 8:48 min Presentation Software & Graphic Suites: Purpose & Examples 4:30 min Types of Images: Vector & Raster Graphics 6:05 min Database and Spreadsheet Software: Tools & Management 8:06 min Communication Software: Purpose & Examples 4:45 min Software Licensing: Proprietary and Free and Open-Source Licenses 10:29 min Multimedia Software: Working with Audio and Video 8:30 min Chapter 2 Systems Software Systems Software: Utility Software, Device Drivers and Firmware 7:10 min Command Line Interface: Commands, Parameters & Options 6:08 min Computer Operating Systems: Managing Hardware and Software Resources 7:56 min Enterprise, Workgroup & Personal Operating Systems 8:10 min Computer Security \| Definition, Components & Threats 6:29 min Systems Security: Firewalls, Encryption, Passwords & Biometrics 8:22 min Computer Security Best Practices & Tips What Is Computer Management? - Maintenance & Tools 10:29 min File Systems: FAT, NTFS, and HFS+ 8:50 min File Extensions and File Types: MP3, GIF, JPG, DOCX, XLSX, EXE, & More 9:02 min Chapter 3 Computer Hardware Types of Computer Systems in Business 5:56 min Computer System Components: Computer Parts & Functions 7:59 min Computer Input Devices \| Types, Function & Examples 6:48 min Computer Output Devices: Monitors, Speakers, & Printers 6:33 min Units of Measurement: Megapixels, Kilobytes & Gigahertz 12:12 min ASCII and Unicode to Represent Characters in Binary Code 3:56 min Computer Memory \| Definition, Types & Functions 5:40 min External and Internal Storage Devices: Optical, Magnetic & Semiconductor Storage 6:52 min RAID Arrays & Data Redundancy 9:00 min Chapter 4 Social Impacts and History of Computing History of Computers \| Definition & Types 9:55 min How Has Information Technology Impacted the Economy 7:40 min Globalization, Outsourcing & Insourcing: Impact of Technology on Careers 7:06 min Ergonomics, Telecommuting, Virtual Teams & Job Design: Technology Effects on Careers 7:56 min Personal Privacy Issue in Information Technology \| Types & Ethics 10:43 min Management Information Systems \| MIS Definition, Scope & Careers 7:17 min Information Systems Jobs & Career Options 9:13 min Information Systems Resources: Networks, Hardware, Software, Data & People 3:51 min Computer Security & Threat Prevention for Individuals & Organizations 10:01 min Intellectual Property and Open Source Software: Issues and Concerns 12:54 min Chapter 5 Data Communications Types of Networks: LAN, WAN, WLAN, MAN, SAN, PAN, EPN & VPN 9:17 min Client/Server and Mainframe Systems Used in Telecommunication Systems 7:56 min What is a Database Management System? - Purpose and Function 10:03 min Models of Database Management Systems (DBMS) 10:03 min What is a Relational Database? - Elements, Design & Advantages 10:19 min What is a Distributed Database? - Architecture, Principles & Advantages 6:10 min SQL vs. NoSQL Databases NoSQL Databases: Design & Types Types of Data: Text, Numbers & Multimedia 11:43 min What is Database Management? - How Databases Help Organizations 7:09 min Database Concepts and Structures: The Elements That Make Up a Database 6:24 min Database Administration and Security: Definition and Purpose 6:23 min Cloud Computing and Databases: Technology to Improve Database Management 5:55 min Chapter 6 World Wide Web History of the Internet \| Overview & Timeline 8:13 min Internet Connectivity and Communication Standards 4:55 min Hypertext Markup Language: Software to Create Web Pages 6:50 min Web Page Design and Programming Languages: HTML, XML, CSS & JavaScript 11:58 min Web Scripting: Client-Side and Server-Side 7:53 min Scripting Languages: Perl, JavaScript, VBScript & AppleScript 7:52 min What is JavaScript? \| Overview, Functions & Examples 4:31 min Search Engines, Keywords & Web Portals 6:16 min World Wide Web: How the Web Works 6:36 min The Internet: IP Addresses, URLs, ISPs, DNS & ARPANET 8:47 min What are Newsgroups and Discussion Forums? 7:37 min Chapter 7 Computer & Information Security What is Information Security? - Definition & Best Practices 4:47 min Definition of Threat in Information Security What is an Information Security Policy? - Definition & Types Hacking Definition, Techniques & Historical Context 6:04 min Malware \| Meaning, Types & Examples 4:55 min What is a Trojan Horse Virus? - Definition, Examples & Removal Options 4:52 min What is a Worm Virus? - Definition, Examples & Removal Tools 5:10 min Spyware Definition & Examples 5:48 min What is Cybercrime? - Definition, History, Types & Laws 3:36 min Social Media Threats, Attacks & Security 5:23 min What is Internet Security? - Privacy, Protection & Essentials 11:46 min Chapter 8 Software Development Systems Development Methods and Tools 6:37 min Systems Development Life Cycles: Software Development Process 6:12 min Agile Methodology Overview & Steps 5:45 min Agile Requirements Management: Tools & Process Configuration Management \| Definition, Plan & Process 6:04 min Application Development Processes: Internal & External 5:02 min Developing a Graphical User Interface (GUI) 6:39 min Chapter 9 Programming Methodology 5 Basic Elements Of Programming 9:16 min Programming Definition, Software & Languages 3:47 min Object-Oriented Programming vs. Procedural Programming 7:17 min Functional Programming and Procedural Programming 6:58 min Using Pseudocode to Map Code 6:16 min Pseudocode in Programming \| Definition, Examples & Advantages 5:52 min Viewing now Boolean Expression & Operators \| Definition & Application 6:56 min Up next What is a Computer Algorithm? 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14254	https://www.healthline.com/health/humerus-bone	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 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 ### 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. The Humerus Bone: Anatomy, Breaks, and Function Medically reviewed by Kevin Martinez, M.D. — Written by Jill Seladi-Schulman, Ph.D. — Updated on February 28, 2022 The humerus bone is located in the upper arm between the elbow and shoulder. It’s the longest bone in the arm, and supports movement in the arm and shoulder. What to know about the humerus bone: anatomy, function, and fractures The humerus is the bone in your upper arm that’s located between your elbow and your shoulder. Its main function is to provide support for your shoulder and a wide variety of movements for your arm. Fractures are the most common injury to the humerus, and often occur because of a direct blow to the bone. Keep reading to learn more about your humerus, its different parts, why it’s important, and what types of injuries it can sustain. Fast facts about the humerus Your humerus is classifiedTrusted Source as a long bone. Other types of long bones include the radius and ulna in your forearm and the femur in your upper leg. Speaking of long, the humerus is the longest boneTrusted Source in your arm. Despite its name, when you hit your “funny bone,” you’re not hitting your humerus. You’re actually hitting your ulnar nerve, which is located behind your elbow joint. Bones in your arms or hands are more likely Trusted Sourceto be broken. This is because we commonly use these parts of our body to break a fall or reduce the impact of some kind of trauma or blow. Humerus anatomy Your humerus is the only bone in your upper arm. It can be found between your elbow and your shoulder. There are some important terms to know with regard to the anatomy of your humerus: Proximal: This is the upper part of your humerus closest to your shoulder. Body or shaft: This is the long, middle portion of your humerus. Distal: This is the lower area of your humerus that’s closest to your elbow. In addition to the areas mentioned above, the humerus is made up of other parts, too. Let’s explore some of these. Parts of the humerus Head: This rounded area of the proximal humerus interacts with your shoulder blade (scapula) to form your shoulder joint. Tubercles: These bony areas of the proximal humerus serve as attachment points for the muscles of your shoulder joint. You have two tubercles — the greater and lesser tubercle. Surgical neck: This part is located at the base of the proximal humerus and is a common siteTrusted Source for fractures. Epicondyles: These are bony protrusions at the distal (lower end) of your humerus. You have two — the medial and lateral epicondyle. They serve as attachment points for the muscles of your lower arm, wrist, and hand. Trochlea: This part of your distal humerus interacts with the ulna bone in your lower arm. Capitulum: The capitulum is the part of the humerus that interacts with the radius bone of your lower arm. Fossae: You have three fossae, which are depressions that help to accommodate the bones of your lower arm when your elbow joint is moved. Humerus fractures Fractures are one of the most common injuries to the humerus. Humerus fractures are classified by their location: Proximal: A break that occurs at the end of your humerus closest to your shoulder. Mid-shaft or middle: A break that occurs in the shaft or body of your humerus. Distal: A break that happens at the end of your humerus that’s closest to your elbow. Humerus fracture causes A humerus fracture most often occurs due to a direct blow. This type of injury often happens in contact sports or car accidents. You can also break your humerus if you fall with your arm outstretched. Sometimes a humerus fracture can happen due to an underlying health condition. This is called a pathologic fracture and can be caused by conditions such as osteoporosis or cancer. Humerus fracture symptoms Some of the most common signs that you may have fractured your humerus include: arm pain, which can be severe and often gets worse with movement a cracking or snapping sound that happens at the time of the injury swelling bruising a visible lump or bump in your upper arm decreased range of motion Treatment for proximal humerus fractures Fractures to the proximal humerus occur near the shoulder joint. Most fractures to the proximal humerus can be treated without surgery, as long as the bones haven’t shifted out of their original position. In cases where the bones have moved, surgery may be recommended. The most common surgery in these cases involves realigning the bone fragments using plates, pins, or screws. In cases where surgery is not required, your doctor may recommend going through physical therapy to help you regain strength and flexibility in the area. Treatment for mid-shaft humerus fractures A mid-shaft humerus fracture occurs when there is a direct blow to the upper arm. In most cases, there is a high likelihood that bones can be realigned by using a splint or sling in order to keep the bone in place and reduce movement. Your doctor will likely prescribe medication to relieve any initial pain or swelling, and they will follow up with you in a week order to evaluate whether it’s healing correctly. These fractures may take up to 4 months to fully heal. Treatment for distal humerus fractures A distal fracture refers to when you break the lower part of the humerus, and is most commonly caused by directly hitting or falling on the bone. Although a fracture in this area can be painful, there are ways to successfully treat it and relieve any pain or swelling. The type of treatment will depend on the severity of the fracture. Some treatment options may include one or a combination of the following: applying ice to reduce pain and swelling medications to help ease pain and swelling immobilization using a sling or splint to prevent movement of your upper arm and help with healing surgery for severe fractures, which can include plates and screws to help the bone to mend physical therapy to help you maintain or regain strength, flexibility, and movement in the affected arm The most common surgical procedures to treat a distal fracture are determined by the severity of your fracture. For more severe breaks, your doctor may recommend external fixation, where the surgeon will apply a frame to hold the bone in place during the operation. The most common surgical treatment for distal fractures is open reduction and internal fixation. During the operation, your surgeon will reposition bone fragments into their original alignment and then hold them in place using plates and screws that are attached to the outside of the bones. Recovery time can also depend on the severity of your injury as well as your overall health. Depending on the type and severity of the fracture, it could take up to a year for the fracture to fully heal. Keep in mind that you may need to do physical therapy for several months afterward. Humerus function Your humerus has two important functions. These are movement and support. Let’s explore them in a little more detail. The connections that the humerus makes at your shoulder and elbow allow for a wide variety of arm movements, such as: rotation at the shoulder joint raising your arms away from your body (abduction) lowering your arms back toward your body (adduction) moving your arm behind your torso (extension) moving your arm in front of your torso (flexion) straightening your elbow (extension) bending your elbow (flexion) In addition to being crucial for various arm movements, your humerus is also important for support. For example, parts of the humerus serve as a connection point for muscles in your shoulder and arm. Other humerus issues Other potential issues associated with the humerus include: Radial nerve injury: The radial nerve runs through your arm. Radial nerve injury can be a complication of a humerus fracture, particularly middle or distal fractures. Metastatic bone disease: Metastatic bone disease is when cancer that developed in one area of the body, such as the lungs or breast, spreads to the bones. According to the American Academy of Orthopaedic Surgeons, the humerus is commonly impacted. Osteochondroses: This is a group of disorders in which bone growth is affected. One type, called Panner’s disease, can impact the distal area of the humerus, leading to pain in the elbow. Summary Your humerus is the long bone in your upper arm. The connections it makes at the shoulder and elbow enable you to make many different arm movements. The humerus is also a connection point for arm and shoulder muscles. Humerus fractures are a common injury that are often caused by falls, car accidents, or contact sports. These fractures are typically treated with medications, immobilization, and physical therapy. Surgery may be needed in more severe cases. See a doctor if you have upper arm pain that is severe, unexplained by another health condition, or affects your range of motion. Your doctor can help diagnose what may be causing your pain and develop a treatment plan. ADVERTISEMENT Erectile dysfunction treatment at a more affordable cost Hims offers doctor-trusted generic medications around 95% cheaper than brand name alternatives. Visit online to find the right treatment for you with ease. Hard Mints - starting at less than $2/dose GET STARTED No insurance needed 200K+ people treated Discreet delivery How we reviewed this article: 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. 8.2: Bones of the upper arm. (n.d.). Bounds EJ, et al. (2021). Humeral shaft fractures. Classification of bones. (n.d.). Distal humerus fractures of the elbow. (2016). Elbow (olecranon) fractures (2016). Fractures (broken bones). (2021). Gragossian A, et al. (2020). Radial nerve injury. Jo MJ, et al. Proximal humerus fractures. (2012). Metastatic bone disease. (2021). Mostafa E, et al. (2021). Anatomy, shoulder and upper limb, humerus. Suraj A, et al. (2019). Apophysitis and osteochondrosis: Common causes of pain in growing bones. Our experts continually monitor the health and wellness space, and we update our articles when new information becomes available. Current Version Feb 28, 2022 Written By Jill Seladi-Schulman, PhD Edited By Amy Boshnack Medically Reviewed By Kevin Martinez, MD Copy Edited By Megan McMorris May 27, 2020 Written By Jill Seladi-Schulman, PhD Edited By Claire Brocato Medically Reviewed By Alana Biggers, MD, MPH Copy Edited By Amy Whitley Share this article Medically reviewed by Kevin Martinez, M.D. — Written by Jill Seladi-Schulman, Ph.D. — Updated on February 28, 2022 Was this article helpful? Yes No Read this next Humerus Fracture: How Long Will It Take to Heal? Medically reviewed by William Morrison, M.D. A humerus fracture is a break in the large bone of your upper arm. There are several types of humerus fractures, depending on the location of the… READ MORE What Is a Spiral Fracture? Medically reviewed by Daniel Murrell, M.D. A spiral fracture, also known as torsion fracture, is a type of complete fracture that occurs due to a rotational, or twisting, force. READ MORE All About Bone Spurs in Shoulders Medically reviewed by Timothy Gossett, M.D. Learn what can cause bone spurs in your shoulders, see images of bone spurs, how to recognize common symptoms, and how to seek treatment. READ MORE Tennis Elbow Medically reviewed by William Morrison, M.D. Tennis elbow often occurs when a specific muscle in the forearm, the extensor carpi radialis brevis (ECRB) muscle, is damaged. The ECRB helps raise… READ MORE How to Make Bone Broth, Plus Health Benefits Written by Kayla McDonnell, RD Bone broth is easy to make and may provide many health benefits. Here are 6 reasons to drink bone broth, as well as a recipe to get you started. READ MORE Bone Pain Medically reviewed by William Morrison, M.D. Bone pain is an extreme tenderness or aching in one or more bones. It’s commonly linked to diseases that affect normal bone function or structure. READ MORE What to Know About More Common Pediatric Musculoskeletal Disorders Medically reviewed by Karen Gill, M.D. Pediatric musculoskeletal disorders affect 1.7 billion people worldwide. Here's information about the main types that affect children. READ MORE What Can Cause Musculoskeletal Chest Pain? Musculoskeletal chest pain can have many causes, such as a pulled muscle or arthritis. Here’s what you need to know about the symptoms, causes, and… READ MORE Overview of Work-Related Musculoskeletal Disorders Musculoskeletal injuries are common in the workplace. They can affect those who are often on their feet or those sitting for extended periods. READ MORE
14255	https://www.youtube.com/watch?v=2H1xz64_lzo	Work To Stretch A Spring Calculus Using Hooke's Law + Force and Spring Constant Angie Teaches Math 6800 subscribers 13 likes Description 2208 views Posted: 14 Feb 2023 All the steps to find the work needed to stretch a spring past its natural length using Hooke’s Law and calculus. We’ll first calculate the spring constant using Hooke’s Law, then set up the integral to compute the work needed to stretch a spring beyond the natural length at equilibrium and stretch a spring to a given length. Calculate the work needed using an integral for a spring. ▶ Watch Work to Lift Rope/Cable: Calc 2 Playlist: ⏰ Time Stamps 0:00 Work to stretch spring beyond equilibrium in foot pounds 5:30 Work to stretch spring from 15 cm to 20 cm long in Joules Looking for more great videos like this one? Check out these: My Math Channel Transcript: Work to stretch spring beyond equilibrium in foot pounds let's talk about work and springs work is equal to force times distance for the spring the distance is going to be How Far We've either stretched or compressed that spring from its natural length which you could also call equilibrium force is going to be based on hooke's law so let me go ahead and make some room down here my Force based on hooke's law says that the force is directly proportional so equals some K times the length that we have either stretched or compressed our Spring by so we have this K which is our spring constant and X which is the distance stretched or compressed I can add K to my table here with all my units it's okay if you're using English units is going to be pounds per foot and if you are using metric units is going to be Newtons per meter for this particular example we have pounds and feet so we are going to be using the English units that means that I want that force in pounds and my final answer the work needed is going to be in foot pound but the very first thing that we've got to do with these problems whenever we're doing a spring and work problem is to figure out that spring constant K I'm going to use this first statement to figure that out it says a force of one pound is required to hold a spring 0.4 feet Beyond its natural length that 0.4 is the amount stretched so I can go ahead and say that 0.4 in this case is equal to X as I put this together then I've got a force of one pound is equal to I'm looking for K times that distance stretched which is 0.4 feet if I do my Division I have 1 divided by 0.4 which is going to equal 1 divided by four tenths also known as 10 4 or 2.5 if you were to follow those units through we had pounds divide divided by feet and this does turn out to be pounds per foot this is my spring constant it's attached to the spring and I can use it to calculate the force for any given length stretched now we can take a look at the second part of this question it says how much work is required to stretch the spring one foot Beyond its natural length we've already established that we can come up with the force based on that spring constant so my force is given to me by this formula and thus the spring constant 2.5 times x 2.5 is in pounds per foot and X is the distance stretched in feet so this is going to be times feet so we end up with the unit that we needed for Force let me move some more things out of the room here so we can continue working so I've got my force and it is a variable Force so instead of using work is equal to force times distance just as that formula I need to use the integral instead so work is equal to the integral of force times DX which is equal to let's figure out those limits of integration I am now stretching that spring one foot beyond the natural length so as I write this spring here stretched one foot Beyond I am going from 0 to 1 foot those are my limits of integration as I'm stretching along there my force needed is going to change right the further away I am from the natural length the harder I'm going to need to pull on the string so my limits of integration is directly related to the distance stretched so I've got zero to one my force is 2.5 x and then I've got my d x let's give us some more room we are ready to integrate that is going to give me 2.5 integrating the X gives me an x squared divided by 2 and we're going to evaluate that from 0 to 1. let's go ahead and divide 2.5 by 2 that gives me 1.25 so the antiderivative that I'm going to be evaluating is 1.25 x squared from 0 to 1. this gives me 1.25 times 1 squared minus 0 squared and I end up with 1.25 now we can do a really quick units check the unit for my Force right here we had already checked that one to be in pounds DX is related to the distance stretched and I'm going from 0 to 1 foot so this one is in feet so I'm feeling really good because I've got the unit of work that I needed which is in foot pounds Work to stretch spring from 15 cm to 20 cm long in Joules in this next example we are given this comparison it takes 35 Newtons to hold a spring stretch to a length of 15 centimeters from a natural length of 10 centimeters let's go ahead and label some of these things here so 10 centimeters is my natural length and it takes 35 Newtons to get us stretched to the total length here of 15 centimeters so 15 centimeters the distance that we're looking for is that difference so how far has it been stretched from the natural length that's our x value which in this case 15 minus 10 is going to equal 5. this distance of 5 is 5 Centimeters now I've got centimeters and Newtons so I am in the metric system let's make sure we've got exactly what we need for the metric system in the metric system I have meters per second squared and then for the unit of work joules I need meters square squared per second squared so I need to do a unit conversion here from centimeters into meters so I'm going to do that first so 5 centimeters times I want to get rid of centimeters there are a hundred centimeters in one meter so I can cancel my centimeters and I end up with .05 which is 5 divided by 100 meters so I can replace my five centimeters with .05 meters now I'm ready to find that spring constant K so step number one we've got this force of 35 Newtons so force is equal to K times the distance traveled I'm going to go ahead and keep that distance as meters so my force of 35 Newtons is equal to K times .05 and if I go ahead and divide both sides by .05 this gives me 700 is equal to K so our Force function is going to be 700 times x so 700 times x and we want to make sure that X is in meters now we're ready to compute the work done so it says how much work is required to stretch the spring from a total of 15 centimeters long where I am with that second Spring right now two 20 centimeters long so if I were to draw a third spring in here that is even more stretched now I'm going from a total length of 15 centimeters to a total length of 20 centimeters but remember our value X is displacement how far away am I from that natural length I'm going to draw these in in pink instead so how far am I from the natural length at 15 x is equal to 5 Centimeters I am five beyond the natural length of 10 and at 20 I am compared to my natural length of 10. I am 10 centimeters Beyond now I do know that I want everything in terms of meters instead so I need to divide both of these by 100 both of these by 100 and that's going to put me on the interval 5 divided by 100 is .05 and 10 divided by 100 is 0.10 these are my limits of integration I'm going to stretch that starting at a distance of 0.05 meters beyond the natural length to a distance of 0.10 beyond the natural length let's put this together in our work integral so in our integral this is Step number two now I've got I think just about everything that we need work is equal to the integral force times DX so F DX in the case of our spring my force is 700 times x and I was really careful to make sure that the units here are Newtons so 700 times x and DX my limits of integration are going to start me off at .05 meters beyond the natural length 2.10 so I've got .05 2.10 the hard work is over I am ready to integrate rate as I integrate this one it turns out really pretty nice I'm going to keep the 700 and then x squared divided by that new power 2 and I'll be evaluating this from .05 to 0.10 2 goes into 700 and that leaves me with 350 so I'm going to take this antiderivative of 350 x squared evaluated from .05 to 0.10 so in my calculator I'm going to go ahead and take 350 times 0.10 squared minus 350 times .05 squared and I end up with 2.625 so this is equal to 2.625 I made sure that all of my units were in meters so this turns out to be 6.25 Newton meters or we could say the unit is simply a unit of joules I hope this was helpful do take a look at the next one you guys are doing great
14256	https://artofproblemsolving.com/wiki/index.php/Circumradius?srsltid=AfmBOoqf_6HqltleQdK6_FrEjx1NGHFiC9j5s9b59FxuPBo63ScT9Sw6	Art of Problem Solving Circumradius - 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 Circumradius Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Circumradius The circumradius of a cyclicpolygon is the radius of the circumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle. Contents [hide] 1 Formula for a Triangle 2 Proof 3 Formula for Circumradius 4 Circumradius, bisector and altitude 5 Euler's Theorem for a Triangle 6 Proof 7 Right triangles 7.1 Theorem 8 Equilateral triangles 9 If all three sides are known 10 If you know just one side and its opposite angle 11 See also Formula for a Triangle Let and denote the triangle's three sides and let denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply . This can be rewritten as . Proof We let , , , , and . We know that is a right angle because is the diameter. Also, because they both subtend arc . Therefore, by AA similarity, so we have or However, remember that . Substituting this in gives us and then simplifying to get and we are done. Formula for Circumradius Where is the circumradius, is the inradius, and , , and are the respective sides of the triangle and is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that . But, if you don't know the inradius, you can find the area of the triangle by Heron’s Formula: Circumradius, bisector and altitude Circumradius and altitude are isogonals with respect bisector and vertex of triangle. Euler's Theorem for a Triangle Let have circumcenter and incenter .Then Proof See Right triangles The hypotenuse of the triangle is the diameter of its circumcircle, and the circumcenter is its midpoint, so the circumradius is equal to half of the hypotenuse of the right triangle. This results in a well-known theorem: Theorem The midpoint of the hypotenuse is equidistant from the vertices of the right triangle. The midpoint of the hypotenuse is the circumcenter of a right triangle. Equilateral triangles where is the length of a side of the triangle. If all three sides are known Which follows from the Heron's Formula and . If you know just one side and its opposite angle by the Law of Sines. (Extended Law of Sines) See also Inradius Semiperimeter Retrieved from " Category: Geometry 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.
14257	https://kconrad.math.uconn.edu/math3240s20/handouts/pelleqn1.pdf	PELL’S EQUATION, I KEITH CONRAD 1. Introduction For a positive integer d that is not a square, an equation of the form x2 −dy2 = 1 is called Pell’s equation. We are interested in x and y that are both integers, and the term “solution” will always mean an integral solution. The obvious solutions (x, y) = (±1, 0), are called the trivial solutions. They are the only solutions where x = ±1 or y = 0 (separately). Solutions where x > 0 and y > 0 will be called positive solutions. Every nontrivial solution can be made into a positive solution by changing the sign of x or y. We don’t consider the case when d is a square, since if d = c2 with c ∈Z then x2 −dy2 = x2 −(cy)2 and the only squares that diﬀer by 1 are 0 and 1, so x2 −(cy)2 = 1 = ⇒x = ±1 and y = 0. Thus Pell’s equation for square d only has trivial solutions. In Section 2 we’ll show how solutions to Pell’s equation can be found. In Section 3 we’ll discuss an elementary problem about polygonal numbers that is equivalent to a speciﬁc Pell equation. Section 4 describes how to create new solutions of Pell’s equation if we know one nontrivial solution and in Section 5 we will see how all solutions can be generated from a minimal nontrivial solution. In the ﬁnal Section 6 a generalized Pell equation is introduced, where the right side is not 1. 2. Examples of Solutions To ﬁnd a nontrivial solution of x2 −dy2 = 1 by elementary methods, rewrite the equation as x2 = dy2 +1 and then set y = 1, 2, 3, . . . until you reach a value where dy2 +1 is a perfect square. Call that value x2 and then we have a solution (x, y). Example 2.1. Two positive solutions of x2 −2y2 = 1 are (3, 2) and (17, 12), since 2y2 + 1 is a square when y = 2 and 12, where it has values 9 = 32 and 289 = 172. See below. y 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 2y2 + 1 3 9 19 33 51 73 99 129 163 201 243 289 339 393 451 Square? X ✓ X X X X X X X X X ✓ X X X Example 2.2. Three positive solutions of x2 −3y2 = 1 are (2, 1) and (7, 4), and (26, 15), as shown by the table below. y 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 3y2 + 1 4 13 28 49 76 109 148 193 244 301 364 433 508 589 676 Square? ✓ X X ✓ X X X X X X X X X X ✓ 1 2 KEITH CONRAD The table below gives a positive solution to x2 −dy2 = 1 for nonsquare d from 2 to 24 where x and y are as small as possible. d 2 3 5 6 7 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 x 3 2 9 5 8 3 19 10 7 649 15 4 33 17 170 9 55 197 24 5 y 2 1 4 2 3 1 6 3 2 180 4 1 8 4 39 2 12 42 5 1 The theorem suggested by such data is a hard result of Lagrange. Theorem 2.3 (Lagrange, 1768). For any positive integer d that is not a square, the equation x2 −dy2 = 1 has a nontrivial solution. This theorem, which will be proved in Part II, is our hunting license to search for solutions by tabulating dy2 + 1 until it takes a square value. We are guaranteed this search will eventually terminate, but we are not assured how long it will take. In fact, the smallest positive solution of x2 −dy2 = 1 can be unusually large compared to the size of d. The table above illustrates this if we compare the smallest positive solutions when d = 12, 13, and 14. As more extreme examples, see the smallest positive solutions below when d = 61 or 109 compared with nearby values of d. d 60 61 62 108 109 110 x 31 1766319049 63 1351 158070671986249 21 y 4 226153980 8 130 15140424455100 2 While Lagrange was the ﬁrst person to give a proof that Pell’s equation for general (nonsquare) d has a nontrivial solution, 100 years earlier Fermat claimed to have a proof and challenged other mathematicians in Europe to prove it. In one letter he wrote that anyone failing this task should at least try to ﬁnd solutions to x2 −61y2 = 1 and x2 −109y2 = 1, where he said he chose small coeﬃcients “pour ne vous donner pas trop de peine” (so you don’t have too much work). He clearly was being mischievous. If Fermat had posed his challenge to mathematicians in India then he may have gotten a positive response: a nontrivial solution to x2 −61y2 = 1 had already been known there for 500 years. 3. Triangular–Square Numbers A positive integer n is called triangular if n dots can be arranged to look like an equilateral triangle. The ﬁrst four triangular numbers are 1 (a degenerate case), 3, 6, and 10. In the pictures below, the shading shows how each triangular number is built from the previous one by adding a new side. T1 = 1 T2 = 3 T3 = 6 T4 = 10 For any k ≥3, a k-gonal number is a positive integer n for which n dots can be arranged to look like a regular k-gon. The ﬁrst four square and pentagonal numbers, corresponding to k = 4 and k = 5, are shown below. Both sequences start with 1 as a degenerate case. PELL’S EQUATION, I 3 S1 = 1 S2 = 4 S3 = 9 S4 = 16 P1 = 1 P2 = 5 P3 = 12 P4 = 22 A formula for the nth square number Sn is obvious: Sn = n2. To get a formula for the nth triangular and pentagonal numbers, Tn and Pn, the ﬁrst few values suggest how to write them as a sum of terms in an arithmetic progression (which are their real deﬁnitions): Tn = 1 + 2 + · · · + n = n X k=1 k, Pn = 1 + 4 + · · · + (3n −2) = n X k=1 (3k −2). This works for square numbers too: Sn = 1+3+· · ·+(2n−1) = Pn k=1(2k −1) is n2. Using the formula for the sum of terms in an arithmetic progression, Tn = n(n + 1) 2 and Pn = n(3n −1) 2 . With these formulas we ﬁll in the table below of the ﬁrst 10 triangular, square, and pen-tagonal numbers. n 1 2 3 4 5 6 7 8 9 10 Tn 1 3 6 10 15 21 28 36 45 55 Sn 1 4 9 16 25 36 49 64 81 100 Pn 1 5 12 22 35 51 70 92 117 145 Besides the common value 1, we see 36 is both triangular and square: 36 = T8 = S6. Call a positive integer a triangular–square number if it is both Tm for some m and Sn for some n. Finding these numbers is the same as solving a particular Pell equation. Theorem 3.1. Triangular–square numbers correspond to solutions of x2 −2y2 = 1 in positive integers x and y. 4 KEITH CONRAD Proof. Using the formulas for Tm and Sn, Tm = Sn ⇐ ⇒ m(m + 1) 2 = n2 ⇐ ⇒ m2 + m = 2n2 ⇐ ⇒ m + 1 2 2 −1 4 = 2n2 ⇐ ⇒ (2m + 1)2 −1 = 2(2n)2 ⇐ ⇒ (2m + 1)2 −2(2n)2 = 1. Because every step is reversible, ﬁnding triangular–square numbers is equivalent to solving x2 −2y2 = 1 in positive integers x and y where x = 2m + 1 is odd and y = 2n is even: T(x−1)/2 = Sy/2. (While we want x = 2m+1 with m ≥1, we can say x > 0 instead of x ≥3 because the only solution of x2 −2y2 = 1 with x = 1 has y = 0, which is not positive.) Including the constraints that x is odd and y is even in the correspondence between triangular–square numbers and positive solutions of x2 −2y2 = 1 is unnecessary because they are forced by the equation x2 −2y2 = 1. Indeed, writing the equation as x2 = 2y2 + 1 shows x2 is odd, so x is odd. Then x = 2m + 1 for some integer m, and feeding that into the Pell equation makes 4m2 + 4m + 1 −2y2 = 1, so y2 = 2m2 + 2m. Thus y2 is even, so y is even. □ Example 3.2. From the solutions (x, y) = (3, 2) and (17, 12) of x2 −2y2 = 1 we get the triangular–square numbers T1 = S1 = 1 and T8 = S6 = 36 by writing x = 2m + 1 and y = 2n in each case to ﬁnd m and n. As practice with the ideas in the proof of Theorem 3.1, show that ﬁnding square– pentagonal numbers, which are numbers of the form Sm and Pn for some positive integers m and n, is the same as solving x2 −6y2 = 1 in positive integers where x is one less than a multiple of 6. The ﬁrst three positive solutions of x2 −6y2 = 1 are (5, 2), (49, 20), (485, 198), and only the ﬁrst and third have x one less than a multiple of 6; they lead to the square– pentagonal numbers 1 = S1 = P1 and 9801 = S99 = P81 if you work out the details. 4. New Solutions from Old Solutions We found in Section 2, by making a table, that two solutions of x2−2y2 = 1 are (3, 2) and (17, 12). They are closely related when we convert the pair (x, y) into the number x + y √ 2: (4.1) 17 + 12 √ 2 = (3 + 2 √ 2)2. Let’s raise 3 + 2 √ 2 to a few powers beyond the second: (4.2) (3+2 √ 2)3 = 99+70 √ 2, (3+2 √ 2)4 = 577+408 √ 2, (3+2 √ 2)5 = 3363+2378 √ 2. The coeﬃcient pairs (99,70), (577,408), and (3363,2378) are all solutions to x2 −2y2 = 1. Similarly, we previously found three solutions of x2 −3y2 = 1: (2, 1), (7, 4), and (26, 15). When we convert the pair (x, y) into the number x + y √ 3 we have (4.3) 7 + 4 √ 3 = (2 + √ 3)2 and 26 + 15 √ 3 = (2 + √ 3)3. The key to solving x2 −dy2 = 1 in Z is to study numbers of the form x + y √ d where x, y ∈Z. Such numbers are closed under multiplication: (4.4) (x + y √ d)(x′ + y′√ d) = (xx′ + dyy′) + (xy′ + yx′) √ d PELL’S EQUATION, I 5 and xx′ + dyy′ and xy′ + yx′ are both integers.1 This formula is similar to the rule for multiplying complex numbers: (x + yi)(x′ + y′i) = (xx′ −yy′) + (xy′ + yx′)i, which is the case d = −1 (for Pell’s equation we are taking d > 0). Just as a complex number x + yi has a real part x and an imaginary part y, a number x + y √ d with x, y ∈Z has coeﬃcients x and y. The coeﬃcients of such a number are unique: if x+y √ d = x′ +y′√ d with x, y, x′, y′ ∈Z then x = x′ and y = y′. Indeed, if y ̸= y′ then √ d = (x −x′)/(y′ −y) is rational, which is a contradiction (nonsquare integers have irrational square roots). Thus y = y′, so x + y √ d = x′ + y √ d, which implies x = x′. Theorem 4.1. If X2 −dY 2 = 1 has solutions (x, y) and (x′, y′) then the coeﬃcients of (x + y √ d)(x′ + y′√ d) are also a solution. Proof. Using the coeﬃcients from (4.4) we compute (xx + dyy′)2 −d(xy′ + yx′)2 = (x2x′2 + 2dxx′yy′ + d2y2y′2) −d(x2y′2 + 2xx′yy′ + y2x′2) = x2x′2 + d2y2y′2 −dx2y′2 −dy2x′2 = x2(x′2 −dy′2) −dy2(x′2 −dy′2) = x2 −dy2 = 1. □ Corollary 4.2. If X2−dY 2 = 1 has a solution (x, y) then the coeﬃcients of (x+y √ d)n are also a solution for all n ∈Z. In particular, this Pell equation has inﬁnitely many solutions if it has a nontrivial solution. Proof. The coeﬃcients of (x + y √ d)n are solutions for n ≥1 by repeated multiplication using Theorem 4.1. If (x, y) ̸= (±1, 0) then x + y √ d ̸= ±1, so the powers (x + y √ d)n for n ≥1 are distinct and give us inﬁnitely many solutions of X2 −dY 2 = 1. To show the coeﬃcients of (x + y √ d)n are solutions for n < 0, write n = −N and set (x + y √ d)N = xN + yN √ d with xN, yN ∈Z. Then x2 N −dy2 N = 1, so (x + y √ d)−N = 1 (x + y √ d)N = 1 xN + yN √ d = xN −yN √ d (xN + yN √ d)(xN −yN √ d) = xN −yN √ d x2 N −dy2 N = xN −yN √ d and (xN, −yN) is a solution. Finally, the coeﬃcients of (x + y √ d)0 are (1, 0). □ Example 4.3. Since (3 + 2 √ 2)4 = 577 + 408 √ 2, we have (3 + 2 √ 2)−4 = 577 −408 √ 2. If we were not dealing with solutions of Pell’s equation, negative powers would not have integer coeﬃcients. e.g., (5 + 2 √ 2)−1 = 5/17 −(2/17) √ 2. 1If we use a cube root instead of a square root, such sums would not be closed under multiplication, e.g., (1 + 3 √ 2)(1 − 3 √ 2) = 1 − 3 √ 4 ̸= x + y 3 √ 2 for any x and y in Z. 6 KEITH CONRAD 5. All Solutions to a Pell Equation We will describe all solutions to x2 −dy2 = 1 using inequalities on numbers x + y √ d. Comparing the size of such numbers is not generally the same as comparing coeﬃcients: x + y √ d < x′ + y′√ d is not the same as x < x′ and y < y′. Consider 1 + 2 √ 2 < 7 − √ 2. But for Pell solutions, under a mild condition it is the same! Lemma 5.1. If x2 −dy2 = 1 and x + y √ d > 1 then x > 1 and y > 0. Proof. The crucial point is that 1/(x + y √ d) = x −y √ d when x2 −dy2 = 1. Therefore x + y √ d > 1 > x −y √ d > 0. From x + y √ d > x −y √ d we get 2y √ d > 0, so y > 0. Then y ≥1 since y is an integer, so x > y √ d ≥ √ d > 1. □ Without a hypothesis like x2 −dy2 = 1 in Lemma 5.1 there are counterexamples. For instance, 5 − √ 2 > 1 and −2 + 3 √ 2 > 1. Lemma 5.2. Suppose x2 −dy2 = 1 and a2 −db2 = 1 where a, b ≥0. Then a + b √ d < x + y √ d ⇐ ⇒a < x and b < y. Proof. The implication (⇐) is obvious. To prove (⇒), we have a ≥1 since a ≥0 by hypothesis and from a2 −db2 = 1 we can’t have a = 0 (why?), so x+y √ d > 1 and therefore x and y are positive by Lemma 5.1. Reciprocating the inequality a + b √ d < x + y √ d we get x −y √ d < a −b √ d and adding these inequalities gives (a + x) + (b −y) √ d < (a + x) + (y −b) √ d. Subtracting a + x from both sides and dividing by √ d we get b −y < y −b, so 2b < 2y and thus b < y. Then a2 = 1 + db2 < 1 + dy2 = x2, so a < x from positivity of a and x. □ Theorem 5.3. Assume x2 −dy2 = 1 has a solution in positive integers and let (x1, y1) be such a solution where y1 is minimal. Then all solutions to x2 −dy2 = 1 in integers are, up to sign, generated from (x1, y1) by taking powers of x1 + y1 √ d: x + y √ d = ±(x1 + y1 √ d)n for some n ∈Z and some sign. Proof. By Corollary 4.2, for each n ∈Z the coeﬃcients of (x1 +y1 √ d)n satisfy x2 −dy2 = 1, and clearly this is also true for coeﬃcients of −(x + y √ d)n. Conversely, suppose integers x and y satisfy x2 −dy2 = 1. If x and y are positive we’ll show x + y √ d = (x1 + y1 √ d)n for some n ≥1. Since x + y √ d > 1 and the numbers (x1 + y1 √ d)n for n = 0, 1, 2, . . . are an increasing sequence that starts at (x1 + y1 √ d)0 = 1 and tends to ∞, x + y √ d equals or lies between two powers of x1 + y1 √ d: (5.1) (x1 + y1 √ d)n ≤x + y √ d < (x1 + y1 √ d)n+1 for some integer n ≥0. Dividing through (5.1) by (x1 + y1 √ d)n, 1 ≤(x + y √ d)(x1 + y1 √ d)−n < x1 + y1 √ d. PELL’S EQUATION, I 7 The number (x + y √ d)(x1 −y1 √ d)−n has coeﬃcients that are a Pell solution since Pell solutions are closed under multiplication and under raising to integer powers (Theorem 4.1, Corollary 4.2). Therefore (x + y √ d)(x1 −y1 √ d)−n = a + b √ d for some a, b ∈Z and (5.2) 1 ≤a + b √ d < x1 + y1 √ d. If 1 < a + b √ d then a and b are positive by Lemma 5.1, so b < y1 by Lemma 5.2. This contradicts the minimality of y1 among positive Pell solutions, so we must have 1 = a+b √ d. That implies x + y √ d = (x1 + y1 √ d)n. Since x ≥1 and y ≥1, n is not 0, so n ≥1. What if x and y are not both positive? Then α := x + y √ d is not in (1, ∞). If α ̸= ±1 then α lies in one of the intervals (0, 1), (−1, 0), and (−∞, −1), which makes (exactly) one of the numbers 1/α, −1/α, or −α belong to (1, ∞). Each of these is a Pell solution too: 1 α = 1 x + y √ d = x −y √ d, −1 α = −(x −y √ d) = −x + y √ d −α = −x −y √ d. The number among these in (1, ∞) has positive coeﬃcients, so by our previous reasoning ±α±1 = (x1 +y1 √ d)N for some N ≥1 and some signs on the left side. Thus α = x+y √ d = ±(x1 + y1 √ d)±N. If α = ±1 then it also arises in this way using N = 0. □ Example 5.4. The positive solution of x2 −2y2 = 1 with least y-value is (3, 2), so every positive solution comes from coeﬃcients of (3 + 2 √ 2)n for n ≥1. Example 5.5. The positive solution of x2 −3y2 = 1 with least y-value is (2, 1), so the positive solutions are the coeﬃcients of (2 + √ 3)n for n ≥1. Example 5.6. The positive solution of x2 −5y2 = 1 with least y-value is (9, 4), so every positive solution comes from coeﬃcients of (9 + 4 √ 5)n for n ≥1. 6. Generalized Pell Equations The equation x2 −dy2 = n where n ∈Z −{0} is a generalized Pell equation. The special case x2 −dy2 = −1 is called a negative Pell equation. To ﬁnd a solution (in Z) rewrite the equation as x2 = dy2 + n and compute the right side for y = 1, 2, . . . until it’s a square. In the tables below we try to solve x2 −2y2 = −1 and x2 −3y2 = −1. For 1 ≤y ≤15 two solutions are found for the ﬁrst equation, (1, 1) and (7, 5), and none for the second. y 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 2y2 −1 1 7 17 31 49 71 97 127 161 199 241 287 337 391 449 Square? ✓ X X X ✓ X X X X X X X X X X y 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 3y2 −1 2 11 26 47 74 107 146 191 242 299 362 431 506 587 674 Square? X X X X X X X X X X X X X X X A longer search in the second case would be fruitless: x2 −3y2 = −1 has no integral solutions. To prove some generalized Pell equation like x2 −3y2 = −1 has no solution, an argument by contradiction using modular arithmetic often works. 8 KEITH CONRAD Example 6.1. If integers x and y satisfy x2 −3y2 = −1 then reduce both sides mod 3 to get x2 ≡−1 mod 3. This congruence has no solution: the only squares mod 3 are 0 and 1. Thus x2 −3y2 = −1 has no solution (in Z). Example 6.2. The equation x2 −5y2 = 2 has no integral solution because reducing the equation mod 5 makes it x2 ≡2 mod 5, which has no solution. By the same idea, the generalized Pell equations x2 −5y2 = 3 and x2 −5y2 = 7 have no solutions. Example 6.3. The equation x2 −5y2 = 6 has no solution, but we can’t prove this by reducing both sides mod 5 to get x2 ≡6 mod 5, since that congruence has a solution so there is no contradiction. Reduce mod 3 instead: the equation becomes x2−5y2 ≡0 mod 3, or x2 ≡5y2 ≡2y2 mod 3. This too has a solution, namely (0, 0), so it doesn’t seem like progress has been made. But this is progress because (0, 0) is the only solution mod 3, since the squares mod 3 are 0 and 1, and the only way one of these is twice the other mod 3 is when they’re both 0. Therefore if x2 −5y2 = 6 in Z then x and y are both multiples of 3. That makes the left side a multiple of 9, which contradicts the right side being 6! We used modular arithmetic here (reducing mod 3), but in a more subtle way than in the previous two examples. Here is a problem about sums of squares whose solution is equivalent to solving a partic-ular generalized Pell equation. Theorem 6.4. Finding positive integers a and b satisfying (6.1) a2 + (a + 1)2 = b2 + (b + 1)2 + (b + 2)2 is the same as solving x2 −6y2 = 3 in positive integers x and y other than (3, 1). Proof. Expanding the squares and combining like terms, a2 + (a + 1)2 = b2 + (b + 1)2 + (b + 2)2 ⇐ ⇒ 2a2 + 2a + 1 = 3b2 + 6b + 5 ⇐ ⇒ 2(a2 + a) = 3(b2 + 2b) + 4 ⇐ ⇒ 2 a + 1 2 2 −1 4 ! = 3((b + 1)2 −1) + 4 ⇐ ⇒ ((2a + 1)2 −1) = 6((b + 1)2 −1) + 8 ⇐ ⇒ (2a + 1)2 −6(b + 1)2 = 3. All steps are reversible, so solving the original equation with a, b ≥1 is equivalent to solving x2 −6y2 = 3 with odd x ≥3 and any y ≥2. Requiring x to be odd can be dropped since it is forced by the equation x2 −6y2 = 3: the number x2 = 6y2 + 3 must be odd, so x must be odd. □ For a Pell equation x2 −dy2 = 1, multiplying two known solutions as numbers x + y √ d leads to a third solution (Theorem 4.1). For a generalized Pell equation x2 −dy2 = n, multiplying a known solution with a solution of the Pell equation x2 −dy2 = 1 leads to a new solution of x2 −dy2 = n. The proof is like that of Theorem 4.1. Details are left to the reader. Example 6.5. One solution of x2 −6y2 = 3 is (3, 1). A nontrivial solution of x2 −6y2 = 1 is (5, 2). Therefore a second solution of x2 −6y2 = 3 comes from the coeﬃcients of (3 + √ 6)(5 + 2 √ 6) = 27 + 11 √ 6. PELL’S EQUATION, I 9 Check 272 −6 · 112 = 3. In the context of Theorem 6.4 the solution (27, 11) of x2 −6y2 = 3 leads to a solution of (6.1): 2a + 1 = 27 ⇒a = 13 and b + 1 = 11 ⇒b = 10, so 132 + 142 = 102+112+122. This is more attractive if we swap the two sides: 102+112+122 = 132+142. As an application of math to art, consider the painting in Figure 1 by Bogdanov-Belsky, where the children have to calculate (102 +112 +122 +132 +142)/365 in their heads. If they knew 102 + 112 + 122 = 132 + 142, they could ﬁnd the numerator as 2(132 + 142), so only two squares would have to be computed instead of ﬁve. Since 2(132 +142) = 2(169+196) = 2(365), dividing by 365 shows the fraction is 2. Figure 1. Bogdanov-Belsky’s Mental Calculation (1895) The bold artistic details on the blackboard and knowledge of the period when the painting was produced help us ﬁnd the deeper meaning of this work of art: it is advocating for the inclusion of generalized Pell equations in the math curriculum of 19th century peasants.
14258	https://english.stackexchange.com/questions/355752/commas-in-plurals-of-numbers-1000s-vs-1-000s-in-web-writing	Skip to main content Commas in plurals of numbers (1000s vs. 1,000s) in web writing [duplicate] Ask Question Asked Modified 8 years, 9 months ago Viewed 9k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Would I use a comma within the number 1,000 when expressing it as a plural? For example: The laboratory contained 100s or even 1,000s of rats. vs. The laboratory contained 100s or even 1000s of rats. Chicago Manual of Style dictates that these kinds of numbers should be spelled out (hundreds or even thousands of rats), but in web style, numerals are almost always preferred. I'm also working with a very small space, so spelling out the number isn't an option. Note that this question differs from previously answered questions in that it refers to plurals of numbers. grammatical-number numbers Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Oct 29, 2016 at 21:24 Nicole L asked Oct 27, 2016 at 22:21 Nicole LNicole L 5911 silver badge88 bronze badges 6 Why wouldn't you use a comma? As someone who trained as an accountant 55 years ago, one of the first things I learned was the importance of writing figures clearly, lining them up properly when tabulating, and ALWAYS USING COMMAS. And if there is one OCD disorder that I express it is about numbers which are sloppily written without commas. It causes error and confusion. – WS2 Commented Oct 27, 2016 at 22:30 1 What is web writing? – tchrist ♦ Commented Oct 27, 2016 at 22:50 @tchrist Spidery scrawl. – Edwin Ashworth Commented Oct 27, 2016 at 22:51 1 "in web style, numerals are almost always preferred." What is web style? Does it have a manual? If not, you can do whatever you feel is best. Personally, I find the 2nd version (without a comma) more readable. – michael.hor257k Commented Oct 28, 2016 at 6:23 1 No style guide I have ever seen would recommend using numerals to write these numbers. And for good reason—it looks terrible, and it's misleading. Just write them out. You're saving a total of eight characters. How tight can your space possibly be? – Janus Bahs Jacquet Commented Oct 29, 2016 at 21:29 \| Show 1 more comment 1 Answer 1 Reset to default This answer is useful 4 Save this answer. Show activity on this post. Chapter 12 (“Numerals”) of the U.S. Government Printing Office (GPO) Style Manual (a 467-page PDF document) has a dozen pages on how to write numbers, and yet I can’t find anything in it explicitly answering this question. The closest I could find was paragraph 12.14, “Punctuation” (on page 289) which says, The comma is used in a number containing four or more digits, except in serial numbers, common and decimal fractions, astronomical and military time, and kilocycles and meters of not more than four figures pertaining to radio. and, since your case isn’t one of the listed exceptions, it looks like they specify that you should use the comma. However, I noticed that paragraph 12.9(l), “Measurement and time” / “Percentage”⁠ (on page 287) says you can use either “a 1,100-percent increase” or “an 1100-percent increase.” They don’t explain this; I guess it’s because “1,100” is pronounced “one thousand one hundred” and “1100” is pronounced “eleven hundred.” ___________ Chapter 12 begins on page 283. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Oct 28, 2016 at 0:34 Scott - Слава УкраїніScott - Слава Україні 7,3861616 gold badges3737 silver badges5454 bronze badges 2 My company typically uses the Chicago Manual of Style since it tends to be more pertinent for web writing (no spiders, only Internets!), but the GPO style manual is also officially on the books as a reference. Neither of the two seems to have a definitive answer on this one. My inclination is to put the comma in, but the stakeholder wants to leave the comma out. Since there isn't a rule anywhere I can find, I guess I'll defer to the stakeholder. Thanks for your answer. The part about "a" vs. "an" does seem to imply that either could be correct—it is a very interesting example. – Nicole L Commented Oct 29, 2016 at 21:06 1 @NicoleL You can always write to the CMoS and ask them—in my experience, they're usually quite good at getting back to you with their views on edge cases that aren't mentioned outright in the guide. (Though here my bet would be that they would say spell it out.) – Janus Bahs Jacquet Commented Oct 29, 2016 at 21:31 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions grammatical-number numbers See similar questions with these tags. Featured on Meta Upcoming initiatives on Stack Overflow and across the Stack Exchange network... Will you help build our new visual identity? Linked 19 When writing large numbers, should a comma be inserted? 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14259	https://link.springer.com/article/10.1007/s11845-024-03634-4	Advertisement Vitamin B12 deficiency in diabetic patients treated with metformin: A narrative review You have full access to this open access article 10k Accesses 7 Citations 1 Altmetric Explore all metrics Abstract Metformin is the most prescribed oral hypoglycemic drug and is considered by many health practitioners as the first-line treatment for non-insulin-dependent diabetes mellitus (T2DM). It is used either as a monotherapy or adjuvant to other anti-hyperglycemic agents. Most of its side effects are usually mild and self-limiting. However, several studies have shown an association between the use of metformin and low vitamin B12 levels in diabetic patients. The current review aimed to provide a literature review of the current published reports on the association, the possible mechanisms, and the related individualized risk factors that might lead to this incidence. The most accepted mechanism of the effect of metformin on vitamin B12 level is related to the absorption process where metformin antagonism of the calcium cation and interference with the calcium-dependent IF-vitamin B12 complex binding to the ileal cubilin receptor. In addition, many risk factors have been associated with the impact of metformin on vitamin B12 levels in diabetic patients such as dose and duration where longer durations showed a greater prevalence of developing vitamin B12 deficiency. Male patients showed lower levels of vitamin B12 compared to females. Black race showed a lower prevalence of vitamin B12 deficiency in metformin-treated patients. Moreover, chronic diseases including T2DM, hyperlipidemia, coronary artery disease, polycystic ovary disease (PCOD), obesity, and metformin therapy were significantly associated with increased risk of vitamin B12 deficiency. Graphical abstract Metformin impacts vitamin B12 by (A) inhibiting calcium-dependent IF-B12 binding. (B) Prolonged use raises deficiency risk. (C) Males have lower B12 levels than females. (D) Black individuals show lower deficiency rates. (E) Conditions like T2DM, hyperlipidemia, coronary artery disease, PCOD, obesity, and metformin use heighten deficiency risk. Similar content being viewed by others Vitamin B12 screening in metformin-treated diabetics in primary care: were elderly patients less likely to be tested? Association between vitamin B12 level and clinical peripheral neuropathy in type 2 diabetic patients on metformin therapy Metformin Usage Index and assessment of vitamin B12 deficiency among metformin and non-metformin users with type 2 diabetes mellitus Explore related subjects Avoid common mistakes on your manuscript. Introduction Diabetes mellitus is a complex metabolic disorder. The major clinical manifestation is chronic hyperglycemia which results from impaired insulin secretion or/and impaired insulin action. Diabetes mellitus classifications include type 1 insulin-dependent, type 2 insulin-independent, gestational diabetes, and other less common types (e.g., MODY) . In addition, certain criteria for the diagnosis of diabetes mellitus are shown in Table 1 . Generally, type 2 diabetes mellitus (T2DM) is a global health concern that is steadily rising . For instance, an estimated 422 million adults with diabetes worldwide was reported in 2014 . Diabetes prevalence expanded from 4.7% in 1980 to 8.5% in 2014 in adults, with the greatest increase in low and middle-income countries compared to high-income nations . Additionally, 1.1 million children and adolescents aged 14–19 years have T1DM, as estimated by the International Diabetes Federation (IDF), and without interventions to stop the rise in diabetes, by 2045, there will be at least 629 million diabetic patients . Type 2 diabetes mellitus is one of the leading causes of morbidity and mortality worldwide, and it is associated with many systemic vascular complications, which can reduce the quality of life and result in social and economic burdens [7,8,9,10]. Moreover, the financial cost of the healthcare of diabetes mellitus is another economic burden. For. Instance, in many countries, around 5–10% of the healthcare budget is assigned for the treatment of diabetes mellitus . Vitamin B12 deficiency is a significant concern in diabetic patients, particularly those treated with metformin. Several studies have highlighted the association between metformin use and vitamin B12 deficiency in individuals with type 2 diabetes mellitus [13,14,15,16,17,18,19,20,21,22]. The prevalence of vitamin B12 deficiency in diabetic patients on metformin has been reported to be as high as 93% . Furthermore, the impact of vitamin B12 deficiency on peripheral neuropathy in diabetic patients has been a subject of investigation, with studies demonstrating an association between vitamin B12 deficiency and peripheral neuropathy in individuals with type 2 diabetes mellitus [14, 18, 19, 23]. Additionally, the prevalence of vitamin B12 deficiency has been found to be higher in diabetic patients compared to the general population . This deficiency has also been linked to gastroparesis in patients with type 2 diabetes . Moreover, the prevalence of vitamin B12 deficiency has been reported to be higher in individuals with pre-diabetes and diabetes compared to those without these conditions . These findings underscore the importance of routine screening for vitamin B12 deficiency and the potential need for supplementation among diabetic patients, especially those on metformin therapy [27, 28]. In recent years, there has been increasing interest in the association between metformin use and vitamin B12 deficiency in patients with type 2 diabetes mellitus (T2DM)[15, 18]. Several studies have investigated the prevalence of vitamin B12 deficiency and its associated factors among patients with T2DM who are on metformin [20, 29,30,31]. Some studies have indicated a correlation between longer duration of metformin use and increased risk of vitamin B12 deficiency [15, 32, 33]. Additionally, there is evidence suggesting a link between metformin use and diabetic neuropathy [18, 19, 21, 30]. The American Diabetes Association guidelines now recommend routine evaluation for vitamin B12 deficiency in patients taking metformin [19, 29]. It has been suggested that physicians should consider screening for vitamin B12 deficiency in diabetic patients before starting metformin therapy, and periodic monitoring of vitamin B12 levels has been recommended for all patients using metformin, particularly for those using the medication long-term [21, 34]. Moreover, the potential role of vitamin B12 deficiency in exacerbating conditions such as diabetic neuropathy and gastroparesis in patients with T2DM has been highlighted, emphasizing the importance of addressing this issue in clinical management . The research on vitamin B12 deficiency in diabetic patients treated with metformin underscores the need for increased awareness and monitoring of vitamin B12 levels in this patient population. The evidence suggests a potential association between metformin use and vitamin B12 deficiency, with implications for the management of diabetic patients. Hence, this review provides additional insights into the mechanisms underlying this association and guides the development of targeted interventions to mitigate the risk of vitamin B12 deficiency in diabetic patients using metformin. Methods We searched Google Scholar and PubMed using the keywords metformin, diabetes, vitamin B12, deficiency, metabolism, mechanism, risk factors, and side effects. We excluded articles that were not related to our research and the ones that did not have sufficient data. In the end, 25 articles were included from the years 2009 to 2021. Out of the 25 articles, 21 showed an association between the use of metformin and vitamin B12 deficiency, while 4 did not. All clinical trials included the use of metformin. However, not always the participants were diabetics. Vitamin B12 deficiency percentage was calculated in some of the studies and others were not. Table 2 summarizes the main data. Result Metformin Metformin belongs to a group of oral hypoglycemic drugs called biguanides. Some aspects of metformin’s mechanism of action are still not fully understood . It is suggested that metformin primarily achieves its glucose-lowering effects through the inhibition of gluconeogenesis in the liver, therefore lowering the production of glucose. Other mechanisms include increased insulin sensitivity and inhibiting lipogenesis as well as increasing the uptake of glucose in the intestine and muscles. In addition, metformin delays gastric emptying which contributes to a decrease in appetite . Figure 1 illustrates metformin’s mechanism of action. A summary of metformin’s mechanism of action is shown in Fig. 1. Illustration of metformin’s mechanism of action Metformin is routinely prescribed to 120 million patients around the world, as it is the first-line treatment for individuals with diabetes mellitus (T2DM) and normal kidney function [62, 63]. It is currently the most prescribed oral anti-diabetic agent and recommended as first-line therapy for type 2 diabetes because of its safety, effectiveness, and possibility for use in combination with other anti-diabetic medications . Both the American Diabetes Association and the European Association for the Study of Diabetes recommend the use of metformin as the first therapeutic choice in the management of type 2 diabetes mellitus (T2DM). Metformin enhances peripheral insulin sensitivity and cardiovascular mortality risk and contributes to weight loss. Medical situations that may require the prescription of metformin, as well as the precautions of its use, are shown in Table 2. The majority of the side effects associated with metformin are mild. One or more episodes of nausea, vomiting, or diarrhea are experienced in 30–45% of patients . Other less common side effects include headache, diaphoresis, weakness, and rhinitis. Vitamin B12 Vitamin B12 also called cobalamin is the largest and most complex vitamin known. It is a water-soluble vitamin with a molecular weight of 1355.4 [65, 66]. It is found mainly in animal sources such as meat, milk, egg, fish, and shellfish, explaining why strict vegetarians are highly prone to developing vitamin B12 deficiency . However, it can also be found in large quantities in some plants like edible algae or blue-green algae . Cobalamin is introduced to the body through the oral cavity, where it binds to its first carrier protein known as transcobalamin I or R-protein. R-protein protects cobalamin from gastric acidity and low pH in the stomach. Gastric parietal cells secret the intrinsic factor IF, the second carrier of cobalamin. In the small intestine, the R-protein is hydrolyzed releasing cobalamin which then binds to intrinsic factors. The cobalamin-IF complex is then absorbed in the distal ileum [65, 69]. An illustration of vitamin B12 metabolism is shown in Fig. 2. There are many ways to diagnose vitamin B12 deficiency such as identification of macrocytic anemia and findings of hyper-segmented (more than 5 lobes) neutrophils on a blood smear. Nevertheless, the most sensitive and diagnostic criteria are low levels of serum cobalamin (< 148 pmol/L). Mild vitamin B12 deficiency can result in fatigue, weakness, and memory loss. Severe deficiency can result in macrocytic anemia, peripheral neuropathy, and mental psychiatric changes [71,72,73]. The prevalence of deficiency varies by age group. In the US, 3% of those aged 20–39, 4% of those aged 40–59, and 6% of those aged 70 have vitamin B12 deficiency . On the other hand, it is estimated that the subclinical deficiency of vitamin B12 in the US ranges from 10–15% among those > 60 years old to 23–35% in > 80 years old . In Asian and African countries, the prevalence is much higher; for example, in India, 70% of adults are deficient . Illustration of absorption pathway of vitamin B12 in the human body The deficiency of vitamin B12 can arise from its lack in one’s diet or a defect in gastrointestinal absorption . Some genetic studies reported an association between certain genetic variants and the deficiency of vitamin B12 [77, 78]. Other risk factors that have been suggested to be related to the development of vitamin B12 deficiency include pernicious anemia and the long-term use of certain drugs such as acid-suppressing medications and metformin . Metformin and vitamin B12 deficiency Metformin-induced vitamin B12 deficiency was reported as early as 1971 when Tomkin et al. recommended that all patients on long-term metformin therapy should be tested for serum B12 deficiency annually . In a randomized placebo-controlled trial, metformin treatment was associated with a mean decrease in vitamin B12 concentration by 19% and an increase in homocysteine concentration by 5% . Former studies have reported that the prevalence of vitamin B12 deficiency among metformin-treated patients varied greatly and ranged between 5.8% and 52% [43, 46,47,48,49, 52, 81,82,83]. A recent meta-analysis that included thirty-one studies reported that patients who received metformin had a significantly higher risk of vitamin B12 deficiency in comparison with diabetic patients not taking metformin, and significantly lower serum vitamin B12 concentrations which depended on dose and duration of treatment . In a retrospective study, it was observed that subjects receiving doses of metformin higher than 2000 mg/day or for more than 4 years had low levels of vitamin B12 . In another review of patients with type 2 diabetes taking metformin after up to 4 months, they showed a decrease in B12 level by 57 pmol/L, which would be predicted to lead to a frank deficiency in a significant proportion of patients based on European data for B12 status . Contrary to the increase in homocysteine levels that is stated above, a cross-sectional study that concluded 1111 patients with type 2 diabetes who took metformin for at least 6 months reported that homocysteine levels were negatively correlated with vitamin B12 levels, and suggested a hypothesis that B12 deficiency due to the use of metformin occurred at the tissue level . While numerous studies reported an association between vitamin B12 deficiency and metformin treatment, in 2017, Rodríguez-Gutiérrez et al., found no variation in vitamin B12 levels between participants receiving metformin and those naive to therapy . Suggested mechanisms Absorption Different mechanisms have been suggested clarifying how metformin interposes with vitamin B12 absorption. In 1977, Caspary and Creutzfeldt proposed a mechanism that was how bacterial overgrowth in the intestine resulted in bacterial binding with IF-B12 complex instead of the latter getting absorbed . Another mechanism that was suggested is the alteration of metformin on intestinal motility, thereby reducing the absorption of vitamin B12 . The process of B12-intrinsic factor complex uptake is known to be dependent on calcium availability. Therefore, out of the many mechanisms that were suggested of how metformin interferes with the absorption of vitamin B12, metformin antagonism of the calcium cation and interference with the calcium-dependent IF-vitamin B12 complex binding to the ileal cubilin receptor was the most accepted one. Bauman et al. suggested that the protonated metformin molecule directs itself towards the hydrocarbon core of the ileal cell membrane, displacing the divalent calcium cations by giving a positive charge to the membrane surface. An effect that can be reversed by increasing calcium intake, consequently greatly supporting the mechanism . Dose and duration Many studies stated that serum vitamin B12 concentrations are inversely related to long-term therapy and/or higher doses of metformin use [50, 52, 80, 89]. In addition, a meta-analysis of four clinical trials demonstrated that after three to 6 months of metformin use it significantly reduced vitamin B12 levels . Several studies found an association between metformin dose and B12 deficiency, while there was no correlation with its duration. Higher doses of metformin were associated with lower levels of vitamin B12. Hence, it is important to consider metformin dose in recommendations for screening for cobalamin deficiency [19, 35, 40, 43]. Nevertheless, other research demonstrated a relationship between metformin usage, with higher doses, and longer durations, showing a greater prevalence of developing vitamin B12 deficiency [18, 41, 84]. Other risk factors Alvarez et al. found that male patients had lower levels of vitamin B12 in comparison to females . Black race was found to be a protective factor for vitamin B12 deficiency in metformin-treated patients . A meta-analysis by Niafar et al. found that patients with T2DM, hyperlipidemia, coronary artery disease, polycystic ovary disease (PCOD), or obesity, and on metformin therapy were significantly associated with increased risk of vitamin B12 deficiency and lower serum vitamin B12 concentrations . Increased metformin exposure was hypothesized to be associated with lower levels of B12 and more severe peripheral neuropathy . Impaired vibration sensation and proprioception and paresthesia are unfortunately similar in both diabetic neuropathy and vitamin B12 deficiency . Consequently, it has been suggested that serum B12 levels should be screened routinely in long-term metformin users . In a comparison between T2DM patients having neuropathy, and those who do not, it was found that the first group had higher levels of B12 deficiency than the other group . Moreover, in a dose-dependent manner, both borderline and low levels of vitamin B12 occurred to be associated with the presence of distinct neuropathies and macrocytic anemia . T2DM patients with neuropathy treated with MET 1000 mg/d manifested lower levels of vitamin B12 . Diabetic neuropathy relationship with vitamin B12 deficiency has prominent importance considering vitamin B12 deficiency is profoundly common, especially among patients with diabetic neuropathy. Furthermore, diabetic or pre-diabetic patients diagnosed with diabetic neuropathy may have neuropathy due to vitamin B12 deficiency. Therefore, before initiating the treatment of diabetic neuropathy, the other condition should be excluded . A cross-sectional study stated that patients possibly get diagnosed with diabetic neuropathy instead of vitamin B12 deficiency induced by metformin which leads to neurologic damage with symptoms of peripheral neuropathy . On the other hand, a cross-sectional study by Ahmed et al. found that there was no difference among those with normal and decreased vitamin B12 levels and the presence of neuropathy . Smoking was reported to be associated with lower vitamin B12 levels than in non-smokers (Table 3) . Cognitive impairment A meta-analysis reported that cognitive impairment prevalence happened to be less significant in people with diabetes treated with metformin. Additionally, six studies showed that dementia incidence also had a reduced risk. Campbell et al. reported that there is no available evidence supporting the use of metformin by non-diabetic individuals in an attempt to prevent dementia. Nevertheless, in patients at risk of developing dementia or Alzheimer’s disease, metformin should continue to be used as first-line therapy for diabetes . A contradicted study, by Moore et al., reported a significant finding of impaired cognitive performance in diabetic patients treated with metformin, which might be alleviated by vitamin B12 and calcium supplements . Multivitamins Individuals who are receiving supplementation of multivitamins may potentially have protection against B12 deficiency in comparison to those not receiving any [40, 42]. PPIs and/or H2RAs In consideration of the expanding prevalence of obesity, T2DM, and GORD, there is now more potential for the use of acid-suppressing medications and anti-diabetics concomitantly. Considering that the solitary use of either metformin, PPIs, or H2RAs, has been shown to considerably deplete vitamin B12, co-prescription of metformin with either PPIs or H2RAs can have additional adverse effects on vitamin B12 status . The production of stomach acid by the gastric parietal cells is needed for the conversion of pepsinogen to pepsin, which releases vitamin B12 from ingested proteins. PPIs and H2RAs inhibit this acid production. PPIs block gastric H+K+-ATPase, which is responsible for pumping H+ ions from within gastric parietal cells into the gastric lumen, where they interact with Cl− ions to form HCl. On the other hand, H2RAs inhibit the interaction of histamine with the parietal cell histamine H2 receptor. This blocks the cAMP-dependent pathway that promotes H+K+-ATPase function, thus reducing gastric acid production. A reduction or lack of gastric acid and pepsin diminishes the release of vitamin B12 from food and hence decreases its availability for absorption in the ileum . Long et al. observed that the association of vitamin B12 deficiency along with the concomitant use of metformin and proton pump inhibitors was significantly greater than those on monotherapy. 34.15% of patients with co-prescription were vitamin B12 deficient; in contrast, those on metformin (21.91%) or PPIs (25.58%) monotherapy had lesser deficiency suggesting an additional impact . Nevertheless, there is no clear indication that biochemical or functional vitamin B12 deficiency would occur due to decreased serum vitamin B12 that is caused by these medications, as indicated by circulating homocysteine and methylmalonic acid concentrations, or to the hematologic and neurological manifestations of clinical deficiency . Until other studies are done, Miller recommends those who are co-prescribed to these drugs to observe vitamin B12 status and take vitamin B12 supplements if needed . However, Romero and Lozano found that there was no notable variation in plasma vitamin B12 levels among those receiving and not receiving PPIs . Sulfonylurea and/or insulin In a cross-sectional study by Kang et al., it was demonstrated that T2DM patients need to monitor their vitamin B12 deficiency and keep an ordinary regulation of their vitamin B12 levels especially those who were prescribed metformin in combination with sulfonylurea. in contrast to insulin metformin and sulfonylurea, co-prescription has been shown to decrease the mean blood vitamin B12 level and the prevalence of vitamin B12 deficiency was significantly increased. Moreover, even after modifications for the daily dosage and duration of metformin among the patients taking the maximal dosage of sulfonylurea, this finding persisted to be significant . Rosiglitazone In a 6-week study to find the impacts of treatment with metformin or rosiglitazone on serum concentrations of homocysteine, folate, and vitamin B12 in patients with recently diagnosed T2DM where 165 patients have been tested, Sahin et al. observed that metformin use was associated with an increase in homocysteine levels, but vitamin B12 did not vary significantly. Whereas management with rosiglitazone showed a decrease in homocysteine levels, with no significant change in vitamin B1 levels . Conclusion Until this day, some of the aspects of metformin’s mechanism of action are still not fully understood. We found in our review that there is an undeniable association between the use of metformin and the progression of vitamin B12 deficiency in some diabetic patients. However, the benefits outweigh the risks. When spotted early, vitamin B12 deficiency is easily treated. We recommend further studies to achieve a better understanding of the possible mechanisms, risk factors, and relation of vitamin B12 deficiency to the dose and duration of metformin use. In addition, we advise physicians and health practitioners to always be aware of these side effects. Routine monitoring of vitamin B12 for patients on long-term metformin was also suggested by other studies. These recommendations were associated with higher dosages and longer durations of usage. Some of these studies recommended the regular check for doses ranging more than 1000 to 2000 mg/day, while others did not specify a certain dose or duration. Screening was mostly advised to be annual [14, 18, 19, 36, 39, 41, 43,44,45, 49, 52, 83, 93, 96]. On the contrary, a recommendation for screening by Rodríguez-Gutiérrez et al. could not be made . Data availability The data that supports the findings in this study are available from the corresponding authors upon reasonable request. 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Author information Authors and Affiliations Department of Basic Medical Sciences, Faculty of Medicine, Yarmouk University, Irbid, 211-63, Jordan Rasha Al Kreasha & Sarah Aqel Department of Basic Medical Sciences, Faculty of Medicine, Yarmouk University, Irbid, 211-63, Jordan Mazhar Salim Al Zoubi & Ahmad Saeed Surgical Research Section, Department of Surgery, Hamad Medical Corporation, 3050, Doha, Qatar Ahmad R. Al-Qudimat & Raed M. Al-Zoubi Department of Biomedical Sciences, College of Health Sciences, QU-Health, Qatar University, Doha, 2713, Qatar Raed M. Al-Zoubi Department of Chemistry, Jordan University of Science and Technology, P.O.Box 3030, Irbid, 22110, Jordan Raed M. Al-Zoubi Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Contributions All authors contributed toward writing the initial draft and reviewing the manuscript. All agreed on the published version of the article. Corresponding author Correspondence to Raed M. Al-Zoubi. Ethics declarations Conflict of interest The authors declare no competing interests. Additional information Publisher's Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Rights and permissions Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit Reprints and permissions About this article Cite this article Al Zoubi, M.S., Al Kreasha, R., Aqel, S. et al. Vitamin B12 deficiency in diabetic patients treated with metformin: A narrative review. Ir J Med Sci 193, 1827–1835 (2024). Download citation Received: 25 January 2024 Accepted: 12 February 2024 Published: 21 February 2024 Issue Date: August 2024 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 Profiles Avoid common mistakes on your manuscript. Advertisement Search Navigation Discover content Publish with us Products and services Our brands 3.88.92.158 Not affiliated © 2025 Springer Nature
14260	https://www.youtube.com/watch?v=-3LTXSEJ4qs	Problem 4.35 (Nilsson Riedel) Electric Circuits 12th Edition - Mesh-Current Method Ardi Satriawan 32800 subscribers 12 likes Description 814 views Posted: 30 May 2024 4.35 Solve Problem 4.26 using the mesh-current method 4.26 a) Use the node-voltage method to find the power dissipated in the 5 Ω resistor in the circuit in Fig. P4.26. b) Find the power supplied by the 500 V source Playlists: Alexander Sadiku 5th Ed: Fundamental of Electric Circuits Chapter 3: Alexander Sadiku 5th Ed: Fundamental of Electric Circuits Chapter 4: Alexander Sadiku 5th Ed: Fundamental of Electric Circuits Chapter 5: Alexander Sadiku 5th Ed: Fundamental of Electric Circuits Chapter 6: Alexander Sadiku 5th Ed: Fundamental of Electric Circuits Chapter 15: Basic integrals: E-mail: ardiantosatriawan@gmail.com Twitter: twitter.com/ardisatriawan 3 comments Transcript: hello my name is Rd and we are going to solve problem 4.35 from Nelson and rle book so the question is solve problem 4.26 using m method but the 4.26 has this circuit here and previously we need to use note voltage but this time we need to use the so let's do it by drawing the loop first okay let's name this as Loop i1 and then this Loop here as Loop I2 and then this Loop here as Loop I3 and this Loop here s Loop I4 so we will have four equation with four variables here by using k l for its Loops okay let's start let's do k l F Loop i1 okay cap L say that the sum of the voltage in a loop will equal to zero okay we start from this voltage source and the current is Flowing from negative polarity to positive polarity so the sign will be negative so we have - 500 and then plus we have 4 oh here multiplied by i1 okay however this I2 is Flowing to the opposite direction so I have minus I2 here and then we have three here so we have plus three and then again multipli by i1 this I3 is Flowing to the opposite direction so I will have minus I3 okay once more we will have six oh here and again we multiply by I all for this time we will have I4 flowing to the opposite direction so we will have minus I4 and all of that will equal to zero okay but I'm space so let's move that slightly to the left so I can write this will equal to zero on the right hand side okay and now we will have 4 + 3+ plus six that will be 7 + 6 is 13 yeah I think we'll have 13 i1 and then for I2 less only from here so we'll have - 4 I2 and I3 only came from here so we have minus 3 I3 and I4 is only came from this bracket here so we'll have - 6 I and let's move this minus 500 to right hand side so that will equal to 500 and let's save this s equation okay one equation done and now let's do another equation let's do at Loop I2 l f I2 again the sum of the voltage will equal to zero okay now what can we do from here we will have 5 I2 or we will start from here maybe first so we will have 4 I2 but i1 is Flowing to the opposite direction so I'll have minus i1 and then plus we'll have 5 and six here in series so we'll have 11 I2 two and then we will have this two here so we will have plus two then multip by I2 however this I3 is Flowing to the opposite so we will have this one and all of that will equal to okay so we will have Min - 4 i - 4 I and then for I2 we have 4 + 11 which is 15 and then 15 + 2 that will be 7 so we have 17 I2 and then we will have - 2 I3 so we have - 2 I3 but we also have Z for I4 so we have plus 0 I4 maybe do not WR that okay so that's all that's all equal to zero and let's sa this as equation okay two equation to go so let's do K but this time at Lo I3 k at Lo three again the sum of the voltage in a loop will equal to zero and let's start from this three Ohm resistor so 3 MTI by I3 but then i1 is Flowing to the oppos so minus one and then plus two again multili by I3 but then I2 is Flowing to the opposite okay and then four and the only current that pass through it is just I and then what else this two okay 2 MTI by I3 this is I4 flowing to the opp side all of that will equal to zero that is it slightly okay now this one for i1 I only came from this bracket so we'll have - 3 I okay and then I2 only came from here so we have- 2 then for I3 we will have 3 + 2 that will be 5 5 + 4 that will be 9 9 + 2 that will be 11 so we have 11 I3 and then I4 only came from here so we'll have Min - 2 I all of that will equal to zero and let's save this as equation number okay we still need to do one more equation okay now let's do K at Loop I4 k l F four okay again the sum of the voltage will equal to zero and let's start from this six oh resistor here so we have 6 by I4 but i1 flowing to the opposite direction so minus i1 and then + 2 again multiplied by I4 but then I3 to the opposite direction so minus I3 okay and then plus we have three and one in series so we will have four four and the only current that pass through it is I4 all of that will equal to Z and this will be - 6 i1 and then plus I don't have I2 here so 0 I2 uh and then we will have - 2 I3 - 2 I3 and then we will have 6 + 2 is 8 8 + 4 is 12 so I have + 12 all of that will equal to Z and let's set this as equation so we have four equation with four variables okay let's clean up the bo and solve it using okay we have four equation and four variables from the previous calculation and we should be able to solve it but how I think the easiest way is to use W from alpha okay maybe let's put the so minus 13 I'm using a for i1 and then B for I2 and and then C for I3 and then D for i a c right again so we have 3 A- + 11 C and then is = and then the last equation we will have - 6 a and then you need to be careful here because that is - 2 I3 which is C and then plus 12 I4 which is T and z and let's SL from our Bo there okay and that should be a b c in the approxim so we have deal okay so i1 is - [Music] 26 is 6 is I2 I2 is - 7576 is and then I3 what is I3 I3 is - 11 36 364 and then what is I4 I4 is just - 15515 and the unit is is and now what is the question used to find the power disip in the five store one okay so the question eight we ask about the power so the power is i² mtip by the resistance right so we will have this is I2 right so we will have - 7576 squ MTI 5 so we will have the power what is it 7 okay and then square and then multip by 5 so we will have 28 7 maybe 2 286 98 and the unit here is what okay and then the power supply okay we have this one so we have the power is the voltage multipli by the current the voltage is 500 then the current here okay the direction is not important so the sign we can we can ignore the sign so we will have 6.5 5ti 500 okay so we will have 100 or maybe 13.26 1326 and the unit will be K yeah that is all for this problem hopefully this help you see you in the next video bye-bye
14261	https://www.facebook.com/NeilDoesMaths/videos/quick-maths-are-these-divisible-by-7-divisibility-test-divisible-by-7-divide-by-/1044260821017124/	Facebook Log In Log In Forgot Account? Quick Maths! Are These Divisible By 7? ➗ [Divisibility Test, Divisible By 7, Divide By 7, How To Tell If A Number Is Divisible By 7, Maths Tricks, Co… See more Neil Does Maths · Original audio Neil Does Maths · Original audio Neil Does Maths · Original audio Neil Does Maths Public 2.7K 176 192 Video See more on Facebook See more on Facebook Email or phone number Password Log In Forgot password? or Create new account
14262	http://www.360doc.com/content/24/1130/11/83544406_1140789139.shtml	已知对数型复合函数单调性，如何求参数范围？搜索我的图书馆查看信箱系统消息官方通知设置开始对话有 11 人和你对话，查看忽略历史对话记录通知设置发文章发文工具撰写网文摘手文档视频思维导图随笔相册原创同步助手其他工具图片转文字文件清理AI助手留言交流搜索分享 QQ空间QQ好友新浪微博微信生成长图转Word打印朗读全屏修改转藏+1 × 微信扫一扫关注作者 × 微信扫一扫将文章发送给好友了解公众号宣传计划【原】已知对数型复合函数单调性，如何求参数范围？数海碎片 2024-11-30 发布于湖南来源\|70阅读\|1 转藏大中小转藏全屏朗读打印转Word生成长图分享 QQ空间QQ好友新浪微博微信展开全文刚过去不久，很多学校的高一已经学完对数函数，进度慢的也快学完。单独考查对数函数不难，若是结合其它考点，则难度会增加不少。八月份，聊过一个类似的问题：今天，继续聊这个话题，来谈谈其与复合函数单调性的问题。求复合函数单调性，遵循同增异减法则，即：复合函数外层函数与内层函数单调性相同，复合之后为增函数；单调性相异，复合之后为减函数。常规求单调区间的题型，较为直接，暂且不提。已知复合函数单调性，求参数范围，在考试中更为常见。不管是月考，还是期末考试，出现的频率可能会更高。下面就以外层函数为对数函数，内层函数为熟悉的一次函数、二次函数为例，具体分析一二。对数函数的单调性，与底数的范围有关，但不管如何，且是毋庸置疑的。这就使得内层的一次函数单调性，即可确定下来，单调递减。但函数在区间上递减，根据同增异减法则，外层的对数函数，需单调递增，才能满足题意。因此，底数。不要忘记，对数函数定义域（真数部分）需大于零，即在区间上大于零恒成立，才能满足题意。上述为分析过程，具体解析如下：对数函数的单调性，与底数的范围有关，这句话在这题当中同样适用，但由于二次函数的单调性，与对称轴、区间的相对位置有关。因此，需分类讨论。整体思路是：先确定内层函数的单调性，再去确定外层函数的单调性，由内往外。因为底数，所以里层的二次函数开口向上。再根据对称轴分布在区间的左右两侧，可得出二次函数在区间上的单调性。同时，结合同增异减，倒推对数函数的单调性，是底数，还是。最后，考虑定义域大于零，只需保证区间上的最小值大于零即可。具体解析如下：当然，此题也可以以底数还是作为分类依据，即先确定外层函数的单调性，再去确定内层函数的单调性。整体思路相差不大，只是在细节上稍有不同。以上两题，难度中等，其中穿插着分类讨论、数形结合等思维，弄懂它们，对函数的理解也更为深刻。好了，就说这么多。那啥，最近创建了个读者交流群，基于数学与教育为主的聊天群，主要有全国各地的家长与教师，感兴趣的读者，可以通过下方留言或后台，拉你进群。都看到这里了，点个右下方的在看与朋友圈再走吧。 +关注数海碎片欢迎关注同名微信公众号:数海碎片，从高中数学开始，整理一条学习之路共 123 篇原创微信公众号：微信扫一扫关注赞赏转藏分享 QQ空间QQ好友新浪微博微信献花（0） +1 来自：数海碎片>《待分类》举报/认领上一篇：高中数学成绩好，其实跟老师关系不大下一篇：这一道零点题，有点难度猜你喜欢 0 条评论发表请遵守用户评论公约查看更多评论类似文章更多函数的单调性中同增异减怎么理解？所谓“同增异减性”即是在复合函数中，内层函数与外层函数的单调性相同，则复合函数为增函数，内层函数与外层函数的单调性相反，则复合... 以对数为载体的复合函数的单调性把一个复杂的函数看成由两个简单函数复合而成（本题对数函数和二次函数都是我们熟悉的，方便研究它们的性质），体现了数学的转化与化归... 基本初等函数二次函数在区间上最值的求解要注意利用二次函数在该区间上的图象.二次函数的对称轴与区间的位置通常有三种情况：（1）定义域区间在对称轴的右侧；点评：发掘、提炼多变元问题中变元间的相互依存、相互... 高中数学 \| 高中学习阶段三种重要的函数模型高中数学 \| 高中学习阶段三种重要的函数模型。解析：在同一坐标系中作出四个指数函数的图象，并作出直线的图象，且它与指数函数图象有交... 高考数学：指数函数专题训练第一部分，熟记三条性质高一上学期所学的函数：二次函数、幂函数、指数函数和对数函数中，指数函数算是比较简单的函数之一。外层函数是一个指数函数。虽然函数f... 【必修1】函数的概念·基础知识5条在函数关系式的表述中,函数的定义域有时可以省略,这时就约定这个函数的定义域就是使得这个函数关系式有意义的实数的全体构成的集合.在实际问题中,函数的定义域还要受到自变量实际意义的制约.函数值域是... 数学概念的通俗理解之二复合函数若y=f(t)，t=u(x)，则y是x的关于对应法则h与f的复合函数，h(x)为内层函数，f(t)为外层函数。t也有两重身份，内层函数h的函数值，也是外层函数f的自变量；y即是内层函数h的函数值，也是复合函数f(h)的函... 中级数学10-指数、对数函数中级数学10-指数、对数函数。复合函数、反函数指数函数对数函数对数的性质。复合函数是一种嵌套函数形式，即一个函数的输出是另一个函数... 数海拾贝函数的零点问题在高中函数知识模块中占有极其重要的地位，它把函数与方程紧密地联系在一起，而复合函数的零点问题是函数零点问题中典型的一类，也是高考和模考中的热点问题．复合函数涉及内层函数与外... 个图VIP年卡，限时优惠价168元>>x 数海碎片教育领域优质作者关注对话 TA的最新馆藏关于计算，再多说几点谈一谈暑假初升高数学的上课思路初高衔接：三次多项式的因式分解如何判断高中数学老师的水平？可以试试这几题为何小学不教方程，非要用算术法解题？小学生学高中数学，这事靠谱吗？喜欢该文的人也喜欢更多普通人怎样用好Deepseek?5000字长文，最全史诗级Deepseek玩法！阅129 如果你想挣大钱，想翻身，狠狠记住这10句话阅123 心理测试：测测你是苦命打工人还是上天宠爱的幸运鹅？阅109 结节最怕一种“草”，对淋巴、乳腺结节最有效，还能治肿瘤、调血...阅987 如何在一分钟内讲清楚一个复杂的问题！阅81 热门阅读换一换幼儿园新教师上岗培训内容阅32383 销售员绩效考核方案阅27232 发挥党员先锋模范作用方面的不足三篇阅61491 【最新湘教版】小学美术一年级上册教案阅34487 中考名著《朝花夕拾》知识点汇总（附常考习题）阅17016 复制打印文章发送到手机微信扫码，在手机上查看选中内容全屏阅读朗读全文分享文章QQ空间QQ好友新浪微博微信 AI解释复制打印文章 adsbygoogle.js 发送到手机微信扫码，在手机上查看选中内容全屏阅读朗读全文 AI助手阅读时有疑惑？点击向AI助手提问吧联系客服在线客服： 360doc小助手2 客服QQ： 1732698931 联系电话：4000-999-276 客服工作时间9:00-18:00，晚上非工作时间，请在QQ留言，第二天客服上班后会立即联系您。
14263	http://hcmswkiser.weebly.com/uploads/2/2/7/9/22796448/_tempconversion.pdf	Main Page \| Full Conversion Table \| Full Converter \| Partial Conversion Table&Converter Prefixes \| Bibliography Temperature Conversion Table From To Fahrenheit To Celsius To Kelvin Fahrenheit (F) F (F - 32) 5/9 (F - 32) 5/9 + 273.15 Celsius (C or o) (C 9/5) + 32 C C + 273.15 Kelvin (K) (K - 273.15) 9/5 + 32 K - 273.15 K Metric system handout Name: ___ Period: __ Mr. Gracias Temperature scales Page 1 of 6 This page is designed to help students practice written problems, and is meant to be printed out. Hit the print command and show all work in the spaces provided. Use the 5-step method, and be sure to round you answers correctly and include units where appropriate. K = C + 273 C = (F - 32) x 5/9 C = K - 273 F = (C x 9/5) + 32 Use the above formulas above to convert the following: 1) 250 Kelvin to Celsius 2) 339 Kelvin to Celsius 3) 17 Celsius to Kelvin 4) 55 Celsius to Kelvin 5) 89.5 Fahrenheit to Celsius 6) 383 Kelvin to Fahrenheit Mr. Gracias Temperature scales Page 2 of 6 TEMPERATURE SCALE PROBLEMS Name________ This page is designed to help students practice written problems, and is meant to be printed out. Hit the print command and show all work in the spaces provided. Use the 5-step method, and be sure to round you answers correctly and include units where appropriate. K = C + 273 C = (F - 32) x 5/9 C = K - 273 F = (C x 9/5) + 32 Use the above formulas above to convert the following: 1) Convert -200 Celsius to Kelvin 2) Convert 355 Kelvin to Celsius 3) Convert 230 Celsius to Fahrenheit 4) Convert 60 Fahrenheit to Kelvin 5) Convert 100 Fahrenheit to Celsius 6) Convert 150 Celsius to Fahrenheit Mr. Gracias Temperature scales Page 3 of 6 Mr. Gracias Temperature scales Page 4 of 6
14264	https://www.goodreads.com/book/show/187041.Problem_Solving_Strategies	Jump to ratings and reviews Rate this book Problem-Solving Strategies Arthur Engel 176 ratings 15 reviews Rate this book A unique collection of competition problems from over twenty major national and international mathematical competitions for high school students. Written for trainers and participants of contests of all levels up to the highest level, this will appeal to high school teachers conducting a mathematics club who need a range of simple to complex problems and to those instructors wishing to pose a "problem of the week", thus bringing a creative atmosphere into the classrooms. Equally, this is a must-have for individuals interested in solving difficult and challenging problems. Each chapter starts with typical examples illustrating the central concepts and is followed by a number of carefully selected problems and their solutions. Most of the solutions are complete, but some merely point to the road leading to the final solution. In addition to being a valuable resource of mathematical problems and solution strategies, this is the most complete training book on the market. GenresMathematicsTextbooksPuzzlesTechnologyScienceNonfictionPersonal Development 413 pages, Paperback First published December 12, 1997 35 people are currently reading 945 people want to read About the author Arthur Engel 49 books9 followers Ratings & Reviews What do you think? Rate this book Friends & Following Create a free account to discover what your friends think of this book! Community Reviews 4.47 176 ratings 15 reviews 5 stars 108 (61%) 4 stars 47 (26%) 3 stars 18 (10%) 2 stars 1 (<1%) 1 star 2 (1%) Displaying 1 - 15 of 15 reviews Murilo Andrade 43 reviews 20 followers September 7, 2015 Another gem every problem solver or math olympic should have on the shelf. It encapsulates problem solving techniques in a handful of concepts, providing several examples and a whole bunch of problems (around 1300 I guess), most of them with solutions or advanced hints. The book is divided into chapters consisting of a "strategy" or a subject. The range of problems is really wide, and sometimes are quite difficult. In my opinion there is a missing chapter on symmetry, for its applicability on problem solving. "An experienced problem solver can often infer the road to the solution from the result." - grab all the information the problem gives you. The chapters are: Chapter 1 - The invariance principle Here Engel describes invariants and monovariants. The idea is to construct a function ( often integer, positive) that does not change or is bounded and monotonic (in the latter case the algorithm must terminate) If you have (or can introduce a transformation), look for an invariant. Distance from the origin is frequently used on these problems parity is also useful If there is repetition, look for what does not change. Chapter 2 - Coloring proofs Weird chapter on solving board problems. Typical colorings are: chess-like, column (or row) based, etc. A few parity problems on this chapter. Chapter 3 - The extremal principle Very wide-applicable strategy. Many examples on Number Theory, Geometry, Combinatorics, etc are given. Pick an object that maximizes of minimizes some function. It frequently has nice properties, allows to simplify or even to arrive in a contradiction. Every finite nonempty set of nonnegative numbers has a min and a max Every nonempty set of positive integers has a min ( well ordering principle) Chapter 4 - The box principle ( pigeonhole ) If n+1 pearls are put into n boxes, then at least one box has more than one pearl. The difficulty on this chapter is to find the box and the pearls for each problem. Chapter 5 - Enumerative Combinatorics Count by bijection Recursion Divide and conquer : split a problem into smaller parts, solve the small problem and combine solutions. Sum rule, product rule, product-sum rule, sieving Construct of a graph which accepts the objects to be counted. Count in two ways If you cannot find the number of good objects, find the number of bad objects. Chapter 6 - Number Theory Basic introduction to NT. Lots of problems. Chapter 7 - Inequalities Basic Inequalities ( CS , x2> 0, AM-GM-HM, rearrangement, chebyshev, etc). A few hints for solving such problems: -does the expression remind you AM-GM-HM? Cauchy-Schwarz? -Can you apply rearrangement? -An inequality homogeneous on its variables can be normalized. -Is there any symmetry on the variables? In that case you can make additional hypothesis (e.g. a>= b >=c >=...) -Trigonometrical substitution? -Convexity and concavity ( Jensen) -Can you transform it into a simpler form? Chapter 8 - The induction Principle Small Chapter on induction. Chapter 9 - Sequences Nice chapter on sequences. Includes linear recurrence, josephus problem, etc. Chapter 10 - Polynomials Vieta's theorem Fundamental theorem of algebra Roots of unity reciprocal polynomials Symmetric polynomials Chapter 11 - Functional Equations Cauchy's functional equation and others. Small chapter Chapter 12 - Geometry Very long chapter, including: - geometry and complex numbers - transformation geometry - classical euclidean geometry Chapter 13 - Games Describes nim, Bachet, Wythoff, etc Chapter 14 - Further Strategies Graph theory Infinite descent Working backwards Conjugate numbers equations, functions and iterations integer funcitons (floor, ceiling, etc) math problem-solving Fri 4 reviews Read August 24, 2019 Number_theory problem-solving Jessada Karnjana 581 reviews 8 followers April 16, 2022 เป็นเล่มที่ใช้งานคุ้มมาก Arthur Engel บอกว่าตั้งใจใช้เป็นหนังสือสำหรับเทรนเข้มทีมนักเรียน IMO ของเยอรมัน Manik Dhama 5 reviews June 15, 2023 Summer vacation material Very fun! Saatvik Jha 1 review Read August 3, 2023 Very nice book Sabyasachi Mukherjee 3 reviews 54 followers April 13, 2014 I really liked this book when I read it in high school.It has numerous problems ranging from hard to routine and can be used effectively for training for mathematical contests and for problem solving in general. paniz 1 review March 14, 2007 I didn't read it yet! wannaread Raymond 2 reviews 2 followers June 8, 2008 Nice for training for IMO. Hessam 4 reviews Read August 22, 2008 Yeah! best book!! Maurizio Codogno Author 60 books 144 followers November 15, 2010 La mia recensione: finished math problems Ainun Najib 6 reviews 22 followers May 28, 2012 A must read for mathletes :-) Akhil 1 review December 5, 2012 Classic book for anyone interested in elementary mathematics and problem solving Anuwaya 1 review March 1, 2013 it's very pure and nice book thanks engel for this marvelous book Suraj Kumar 11 reviews 3 followers April 24, 2014 Too tough for me. :( Soumyadip Banik 1 review Want to read November 21, 2014 This entire review has been hidden because of spoilers. Displaying 1 - 15 of 15 reviews Join the discussion a quote a discussion Can't find what you're looking for? Get help and learn more about the design. Help center
14265	https://www.lecturio.com/concepts/hemostasis/	Achieve Mastery of Medical Concepts Study for medical school and boards with Lecturio Hemostasis Hemostasis refers to the innate, stepwise body processes that occur following vessel injury, resulting in clot formation and cessation of bleeding. Hemostasis occurs in 2 phases, namely, primary and secondary. Primary hemostasis involves platelet adhesion Adhesion The process whereby platelets adhere to something other than platelets, e.g., collagen; basement membrane; microfibrils; or other 'foreign' surfaces. Coagulation Studies, activation, and aggregation Aggregation The attachment of platelets to one another. This clumping together can be induced by a number of agents (e.g., thrombin; collagen) and is part of the mechanism leading to the formation of a thrombus. Coagulation Studies to the damaged vascular endothelium Endothelium A layer of epithelium that lines the heart, blood vessels (vascular endothelium), lymph vessels (lymphatic endothelium), and the serous cavities of the body. Arteries: Histology, forming a plug that stops the bleeding temporarily. Secondary hemostasis involves the activation of the coagulation cascade resulting in the formation of a more stable plug. Finally, as the vasculature is repaired, the clot is broken down in the fibrinolytic phase Fibrinolytic phase Coagulation Studies. Last updated: May 17, 2024 Contents Share this concept: Definition and Phases Definition Hemostasis refers to the innate, stepwise body processes that occur following vessel injury, resulting in clot formation. Phases of the hemostatic process Vessel wall injury and constriction: Site of injury Constriction caused by endothelin release Exposed collagen fibers Platelet aggregation: Platelets adhere to exposed collagen fibers. Chemicals are released by platelets to induce vasoconstriction and to attract more platelets. More platelets gather. Platelets cluster to repair the vessel wall. Platelet plug formation and coagulation: Tissue factor released Clotting factors released Blood clot formation: Red and white blood cells are trapped in mesh. Coagulation inhibitors and other chemicals are released. Vasoconstriction and Formation of the Platelet Plug Injured vessels vasoconstrict after endothelial injury. Additionally, exposure of blood to the subendothelial Subendothelial Membranoproliferative Glomerulonephritis components triggers formation of the platelet plug. Vasoconstriction Vasoconstriction The physiological narrowing of blood vessels by contraction of the vascular smooth muscle. Vascular Resistance, Flow, and Mean Arterial Pressure Endothelial injury results in a transient vasoconstriction Vasoconstriction The physiological narrowing of blood vessels by contraction of the vascular smooth muscle. Vascular Resistance, Flow, and Mean Arterial Pressure via: Steps in formation of the platelet plug Following an endothelial cell injury Cell injury The cell undergoes a variety of changes in response to injury, which may or may not lead to cell death. Injurious stimuli trigger the process of cellular adaptation, whereby cells respond to withstand the harmful changes in their environment. Overwhelmed adaptive mechanisms lead to cell injury. Mild stimuli produce reversible injury. If the stimulus is severe or persistent, injury becomes irreversible. Cell Injury and Death, the following processes occur with the platelets Platelets Platelets are small cell fragments involved in hemostasis. Thrombopoiesis takes place primarily in the bone marrow through a series of cell differentiation and is influenced by several cytokines. Platelets are formed after fragmentation of the megakaryocyte cytoplasm. Platelets: Histology to form a temporary platelet plug (also known as primary hemostasis): Formation of the temporary hemostatic plug: The disrupted endothelial surface exposes von Willebrand factor (vWF) to the passing blood. Platelets bind to vWF via their GpIb receptors and are activated. Platelet activation triggers the secretion of ADP, which stimulates the expression of the GpIIb/IIIa receptors on the platelets. The GpIIb/IIIa receptors bind to fibrinogen and a platelet on each end, causing platelets to aggregate. As more platelets bind to each other, a platelet plug is formed. As the coagulation cascade is activated, thrombin converts the weaker fibrinogen into the stronger fibrin, creating a much more stable clot. Platelet adhesion Adhesion The process whereby platelets adhere to something other than platelets, e.g., collagen; basement membrane; microfibrils; or other ‘foreign’ surfaces. Coagulation Studies Exposure of the blood to subendothelial Subendothelial Membranoproliferative Glomerulonephritis components at the site of injury causes platelets Platelets Platelets are small cell fragments involved in hemostasis. Thrombopoiesis takes place primarily in the bone marrow through a series of cell differentiation and is influenced by several cytokines. Platelets are formed after fragmentation of the megakaryocyte cytoplasm. Platelets: Histology to adhere to the injury site. Platelet activation Activated platelets Platelets Platelets are small cell fragments involved in hemostasis. Thrombopoiesis takes place primarily in the bone marrow through a series of cell differentiation and is influenced by several cytokines. Platelets are formed after fragmentation of the megakaryocyte cytoplasm. Platelets: Histology enhance further platelet adhesion Adhesion The process whereby platelets adhere to something other than platelets, e.g., collagen; basement membrane; microfibrils; or other ‘foreign’ surfaces. Coagulation Studies and aggregation Aggregation The attachment of platelets to one another. This clumping together can be induced by a number of agents (e.g., thrombin; collagen) and is part of the mechanism leading to the formation of a thrombus. Coagulation Studies, and stimulate secretion Secretion Coagulation Studies. Platelet aggregation Aggregation The attachment of platelets to one another. This clumping together can be induced by a number of agents (e.g., thrombin; collagen) and is part of the mechanism leading to the formation of a thrombus. Coagulation Studies Platelet secretion Secretion Coagulation Studies Platelets Platelets Platelets are small cell fragments involved in hemostasis. Thrombopoiesis takes place primarily in the bone marrow through a series of cell differentiation and is influenced by several cytokines. Platelets are formed after fragmentation of the megakaryocyte cytoplasm. Platelets: Histology contain 2 types of granules. These granules release various substances when platelets Platelets Platelets are small cell fragments involved in hemostasis. Thrombopoiesis takes place primarily in the bone marrow through a series of cell differentiation and is influenced by several cytokines. Platelets are formed after fragmentation of the megakaryocyte cytoplasm. Platelets: Histology are activated. Related videos Coagulation Cascade Overview The coagulation cascade is a series of reactions that ultimately generates a strong, cross-linked fibrin Fibrin A protein derived from fibrinogen in the presence of thrombin, which forms part of the blood clot. Rapidly Progressive Glomerulonephritis clot. This process is also known as secondary hemostasis. Overview of the coagulation cascade a: activated form PF3: platelet factor 3 (phospholipids) Coagulation factors Coagulation factors are trypsin-like serine Serine A non-essential amino acid occurring in natural form as the l-isomer. It is synthesized from glycine or threonine. It is involved in the biosynthesis of purines; pyrimidines; and other amino acids. Synthesis of Nonessential Amino Acids proteases Proteases Proteins and Peptides and are denoted with roman numerals. Vitamin K cycle: Vitamin K epoxide (1) is inactive and converted to its active, reduced form, vitamin K hydroquinone (2), by vitamin K epoxide reductase (VKOR; 3). Vitamin K hydroquinone is a cofactor in the carboxylation of specific glutamate residues within the vitamin K-dependent proteins (factors II, VII, IX, X, protein C and S), a process which is necessary to activate them. The carboxylation reaction is catalyzed by gamma-glutamyl carboxylase (4). Vitamin K hydroquinone is oxidized to the epoxide form when it acts as a cofactor, but is then recycled back to the hydroquinone form by VKOR. Warfarin inhibits VKOR (5) so that vitamin K cannot be recycled from its oxidized form to the reduced form. Thus, vitamin K-dependent proteins cannot be activated. Extrinsic pathway: The tissue factor pathway The extrinsic pathway is the primary physiological mechanism by which clotting is initiated. Intrinsic pathway: The contact pathway The intrinsic pathway is mainly responsible for the amplification of factor X activation. Factor X is activated by the initial thrombin generated by the extrinsic/common pathway, but also can be activated directly by endothelial injury. The extrinsic and intrinsic coagulation systems Common pathway The final common pathway a: activated form PF3: platelet factor 3 (phospholipids) Related videos Inhibition of Clotting and the Fibrinolytic Phase Inhibition of clotting The body produces several substances that inhibit platelet binding, aggregation Aggregation The attachment of platelets to one another. This clumping together can be induced by a number of agents (e.g., thrombin; collagen) and is part of the mechanism leading to the formation of a thrombus. Coagulation Studies, and secretion Secretion Coagulation Studies, as well as function as natural anticoagulants Anticoagulants Anticoagulants are drugs that retard or interrupt the coagulation cascade. The primary classes of available anticoagulants include heparins, vitamin K-dependent antagonists (e.g., warfarin), direct thrombin inhibitors, and factor Xa inhibitors. Anticoagulants. These mechanisms limit Limit A value (e.g., pressure or time) that should not be exceeded and which is specified by the operator to protect the lung Invasive Mechanical Ventilation clotting to specific focal sites and keep the blood fluid. Fibrinolytic phase Fibrinolytic phase Coagulation Studies The fibrinolytic system functions to remove the clot after the vasculature is repaired, and the process is accomplished primarily by plasmin. Related videos Laboratory Evaluation of Hemostasis Normal hemostasis laboratory evaluation Related videos Clinical Relevance Disorders of primary hemostasis (formation of the platelet plug) Disorders of secondary hemostasis (the coagulation cascade) Hemophilia Hemophilia The hemophilias are a group of inherited, or sometimes acquired, disorders of secondary hemostasis due to deficiency of specific clotting factors. Hemophilia A is a deficiency of factor VIII, hemophilia B a deficiency of factor IX, and hemophilia C a deficiency of factor XI. Patients present with bleeding events that may be spontaneous or associated with minor or major trauma. Hemophilia: a rare blood-clotting disorder in which the body lacks blood-clotting factors (factor VIII in hemophilia A Hemophilia A The classic hemophilia resulting from a deficiency of factor VIII. It is an inherited disorder of blood coagulation characterized by a permanent tendency to hemorrhage. Hemophilia; factor IX in hemophilia B Hemophilia B A deficiency of blood coagulation factor IX inherited as an X-linked disorder. (also known as Christmas disease, after the first patient studied in detail, not the holiday.) historical and clinical features resemble those in classic hemophilia, but patients present with fewer symptoms. Severity of bleeding is usually similar in members of a single family. Many patients are asymptomatic until the hemostatic system is stressed by surgery or trauma. Treatment is similar to that for hemophilia A. Hemophilia). Affected individuals present with abnormal bleeding that can occur spontaneously or after minor trauma. These individuals can bleed into joint spaces and develop life-threatening internal bleeding. Mixed disorders affecting both platelets Platelets Platelets are small cell fragments involved in hemostasis. Thrombopoiesis takes place primarily in the bone marrow through a series of cell differentiation and is influenced by several cytokines. Platelets are formed after fragmentation of the megakaryocyte cytoplasm. Platelets: Histology and coagulation factors Disorders of fibrinolysis Related videos References Study with Lecturio for Medical School Nursing School Lecturio Support Institutions Resources Follow Us USMLE™ is a joint program of the Federation of State Medical Boards (FSMB®) and National Board of Medical Examiners (NBME®). MCAT is a registered trademark of the Association of American Medical Colleges (AAMC). 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14266	https://pubmed.ncbi.nlm.nih.gov/31247585/	Molecular regulation of spermatogonial stem cell renewal and differentiation - 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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Molecular regulation of spermatogonial stem cell renewal and differentiation Juho-Antti Mäkelä1,Robin M Hobbs23 Affiliations Expand Affiliations 1 Research Centre for Integrative Physiology and Pharmacology, Institute of Biomedicine, University of Turku, Turku, Finland. 2 Australian Regenerative Medicine Institute, Monash University, Melbourne, Victoria, Australia. 3 Development and Stem Cells Program, Monash Biomedicine Discovery Institute and Department of Anatomy and Developmental Biology, Monash University, Melbourne, Victoria, Australia. PMID: 31247585 DOI: 10.1530/REP-18-0476 Free article Item in Clipboard Review Molecular regulation of spermatogonial stem cell renewal and differentiation Juho-Antti Mäkelä et al. Reproduction.2019 Nov. Free article Show details Display options Display options Format Reproduction Actions Search in PubMed Search in NLM Catalog Add to Search . 2019 Nov;158(5):R169-R187. doi: 10.1530/REP-18-0476. Authors Juho-Antti Mäkelä1,Robin M Hobbs23 Affiliations 1 Research Centre for Integrative Physiology and Pharmacology, Institute of Biomedicine, University of Turku, Turku, Finland. 2 Australian Regenerative Medicine Institute, Monash University, Melbourne, Victoria, Australia. 3 Development and Stem Cells Program, Monash Biomedicine Discovery Institute and Department of Anatomy and Developmental Biology, Monash University, Melbourne, Victoria, Australia. PMID: 31247585 DOI: 10.1530/REP-18-0476 Item in Clipboard Full text links Cite Display options Display options Format Abstract The intricate molecular and cellular interactions between spermatogonial stem cells (SSCs) and their cognate niche form the basis for life-long sperm production. To maintain long-term fertility and sustain sufficiently high levels of spermatogenesis, a delicate balance needs to prevail between the different niche factors that control cell fate decisions of SSCs by promoting self-renewal, differentiation priming or spermatogenic commitment of undifferentiated spermatogonia (Aundiff). Previously the SSC niche was thought to be formed primarily by Sertoli cells. However, recent research has indicated that many distinct cell types within the testis contribute to the SSC niche including most somatic cell populations and differentiating germ cells. Moreover, postnatal testis development involves maturation of somatic supporting cell populations and onset of cyclic function of the seminiferous epithelium. The stochastic and flexible behavior of Aundiff further complicates the definition of the SSC niche. Unlike in invertebrate species, providing a simple anatomical description of the SSC niche in the mouse is therefore challenging. Rather, the niche needs to be understood as a dynamic system that is able to serve the long-term reproductive function and maintenance of fertility both under steady-state and during development plus regeneration. Recent data from us and others have also shown that Aundiff reversibly transition between differentiation-primed and self-renewing states based on availability of niche-derived cues. This review focuses on defining the current understanding of the SSC niche and the elements involved in its regulation. PubMed Disclaimer Similar articles Mechanisms regulating mammalian spermatogenesis and fertility recovery following germ cell depletion.La HM, Hobbs RM.La HM, et al.Cell Mol Life Sci. 2019 Oct;76(20):4071-4102. doi: 10.1007/s00018-019-03201-6. Epub 2019 Jun 28.Cell Mol Life Sci. 2019.PMID: 31254043 Free PMC article.Review. Mast Cells as a Component of Spermatogonial Stem Cells' Microenvironment.Sadek A, Khramtsova Y, Yushkov B.Sadek A, et al.Int J Mol Sci. 2024 Dec 7;25(23):13177. doi: 10.3390/ijms252313177.Int J Mol Sci. 2024.PMID: 39684887 Free PMC article.Review. The Biology and Regulation of Spermatogonial Stem Cells in the Niche.Zheng T, Fok EKL.Zheng T, et al.Adv Exp Med Biol. 2025;1469:333-354. doi: 10.1007/978-3-031-82990-1_14.Adv Exp Med Biol. 2025.PMID: 40301263 Review. A multistate stem cell dynamics maintains homeostasis in mouse spermatogenesis.Nakagawa T, Jörg DJ, Watanabe H, Mizuno S, Han S, Ikeda T, Omatsu Y, Nishimura K, Fujita M, Takahashi S, Kondoh G, Simons BD, Yoshida S, Nagasawa T.Nakagawa T, et al.Cell Rep. 2021 Oct 19;37(3):109875. doi: 10.1016/j.celrep.2021.109875.Cell Rep. 2021.PMID: 34686326 WTAP Function in Sertoli Cells Is Essential for Sustaining the Spermatogonial Stem Cell Niche.Jia GX, Lin Z, Yan RG, Wang GW, Zhang XN, Li C, Tong MH, Yang QE.Jia GX, et al.Stem Cell Reports. 2020 Oct 13;15(4):968-982. doi: 10.1016/j.stemcr.2020.09.001.Stem Cell Reports. 2020.PMID: 33053361 Free PMC article. See all similar articles Cited by PRC1 directs PRC2-H3K27me3 deposition to shield adult spermatogonial stem cells from differentiation.Hu M, Yeh YH, Maezawa S, Nakagawa T, Yoshida S, Namekawa SH.Hu M, et al.Nucleic Acids Res. 2024 Mar 21;52(5):2306-2322. doi: 10.1093/nar/gkad1203.Nucleic Acids Res. 2024.PMID: 38142439 Free PMC article. The Intricate Functional Networks of Pre-mRNA Alternative Splicing in Mammalian Spermatogenesis.Jiang N, Li Y, Yin L, Yuan S, Wang F.Jiang N, et al.Int J Mol Sci. 2024 Nov 10;25(22):12074. doi: 10.3390/ijms252212074.Int J Mol Sci. 2024.PMID: 39596142 Free PMC article.Review. Altered Sertoli Cell Function Contributes to Spermatogenic Arrest in Dogs with Chronic Asymptomatic Orchitis.Rehder P, Packeiser EM, Körber H, Goericke-Pesch S.Rehder P, et al.Int J Mol Sci. 2025 Jan 27;26(3):1108. doi: 10.3390/ijms26031108.Int J Mol Sci. 2025.PMID: 39940876 Free PMC article. BRCA1 preserves genome integrity during the formation of undifferentiated spermatogonia.Li P, Song L, Ma L, Han C, Li L, Lu LY, Liu Y.Li P, et al.EMBO Rep. 2025 Aug;26(15):3747-3772. doi: 10.1038/s44319-025-00487-5. Epub 2025 May 28.EMBO Rep. 2025.PMID: 40437289 Free PMC article. Long-term effects of early postnatal stress on Sertoli cells.Thumfart KM, Lazzeri S, Manuella F, Mansuy IM.Thumfart KM, et al.Front Genet. 2022 Oct 24;13:1024805. doi: 10.3389/fgene.2022.1024805. eCollection 2022.Front Genet. 2022.PMID: 36353105 Free PMC article. See all "Cited by" articles Publication types Review Actions Search in PubMed Search in MeSH Add to Search MeSH terms Adult Germline Stem Cells / physiology Actions Search in PubMed Search in MeSH Add to Search Animals Actions Search in PubMed Search in MeSH Add to Search Cell Differentiation / genetics Actions Search in PubMed Search in MeSH Add to Search Cell Proliferation / genetics Actions Search in PubMed Search in MeSH Add to Search Cell Self Renewal / genetics Actions Search in PubMed Search in MeSH Add to Search Gene Expression Regulation Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Male Actions Search in PubMed Search in MeSH Add to Search Mice Actions Search in PubMed Search in MeSH Add to Search Spermatogenesis / genetics Actions Search in PubMed Search in MeSH Add to Search Spermatogonia / cytology Actions Search in PubMed Search in MeSH Add to Search Spermatogonia / physiology Actions Search in PubMed Search in MeSH Add to Search Stem Cell Niche / genetics Actions Search in PubMed Search in MeSH Add to Search Related information MedGen LinkOut - more resources Full Text Sources Research Portal of the University of Turku Sheridan PubFactory Full text links[x] Sheridan PubFactory Research Portal of the University of Turku [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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14267	https://www.youtube.com/watch?v=_18KHODPutM	PreCalculus - Section 2.3 - Modeling with Linear Functions mistapotta 3280 subscribers 8 likes Description 764 views Posted: 24 Sep 2018 PreCalculus In this video, we discuss a process to develop mathematical models to help solve practical real-world problems, and then use a couple of linear relationship situations to implement examples of this model. We're using the OpenStax Precalculus book available for free at This section: Assignment: p 170 6,8,9-18, 31-38, 46-56 even Transcript: Introduction howdy howdy this is mr. Potter we've been talking about linear functions for the past couple of episodes so I want to talk about how we can actually do modeling with linear functions and so I actually want to kind of talk about in general how we develop a mathematical model because I'll be honest because we're dealing with linear equations our models are going to be very simple but this technique that we're going to be talking about about how we actually develop a mathematical model can help you whether we're dealing with linear models or quadratic models or any other type of function so when we're developing a mathematical model one of the first things that we have to do is we have to identify our variable so we're going to have things that change in our problem then we're gonna have to identify what those variables are and we usually want to use a descriptive variable name because you know x and y while they're very good for graphing they're not very descriptive which will often see as you'll see t used for time or you'll see em used for money but you'll see something that's very descriptive and so the first thing that we want to do when we're dealing with a problem is try to identify what actually changes in the problem and once we've done that have a variable that's descriptive to kind of describe that action the second thing that we're supposed to look for is we need to look for properties look for properties so for example one of the things I might look for is initial condition because my initial condition is going to tell me about one aspect of my problem usually what happens when time equals zero or when the number of objects that I'm dealing with equals a particular value so this is probably going to be either my y-intercept or my point if I'm using a linear model because I've got the point-slope form or I've got the slope-intercept form I might also look at rates of change and of course rates of change are going to be very important when we're dealing with linear because that's going to be our slope so we're going to look for initial conditions rates of change or other important info so keep in mind that we know from geometry that two points determines a line and we've extrapolated in our algebra two classes that for a parabola we need three points and for a cubic we would need four points and for an integral polynomial we're always going to need one more point than the degree of the polynomial we're talking about so we're going to need at least that many bits of information in other words for a line I'm going to need two pieces of information one that's going to give me some initial condition and one that's going to lead me towards the rate of change the third thing that we're going to have to do is to identify what we're trying to find because I don't know if I'm going to necessarily have to come up with the model itself the equation or the table or the graph or if I'm gonna have to come up with an answer to what happens at a particular time or a particular number of objects or a particular amount of money or am I going to have to come up with some type of interpretation like what does the slope mean or what does the y-intercept mean or what does the model in general mean so I need to make sure that I identify what it is that I'm trying to find because I need to make sure that I answer the question ultimately every one of these problems that we have are going to deal with some question we need to make sure that we answer it adequately the next thing we have to do is we have to come up with a solution pathway in other words some way to come up with our model some way to come up with the equation or the table or the graph whatever it is we're trying to find I need some way to come up with that so I know how to find the slope if I'm given two points I know how to find the equation of a line if I have a slope in a point or slope and a y-intercept so I need to come up with some type of solution strategy some pathway that's gonna get me from the raw data that I have to a model that I can use and then of course I'm going to actually need to come up with a model so we're gonna have to make sure that we write a model and it's often going to be an equation equations are really the easiest thing to work with we've been working with equations since algebra 1 we're going to have to solve or evaluate using our formula so in other words I'm gonna have some formula we've come up with and I'm going to have to solve that equation or I'm going to have to evaluate at a particular time money number of objects and I have to consider how reasonable is my answer so I have to consider reasonability it doesn't make sense if I'm talking about you know I want to make 37 cars well is that what I should expect a car company to manufacture in a month is 37 a reasonable amount is it too low is it too high now if I'm talking about Tesla if I'm talking about some company that hand manufacturers these cars maybe 37 cars in a month is a reasonable amount but if I'm talking about a Ford or if I'm talking about a GM or if I'm talking about another major car company 37 may not be a reasonable amount I need to consider is the answer reasonable in the context of the problem and then finally I need to make sure that I convey my answer my solution in a complete sentence because if I cannot express myself in my language using words then my solution is nothing but a number and if it's nothing but a number then I'm not going to be able to prove to whoever's looking at my work what's going on keep in mind that oftentimes as a mathematician as a stem person you're gonna have to find a way to convey your information to someone who may not understand the mathematics who may not understand the process that you did to come up with your model you need to find a way to convey that information in a way that is simple and succinct something that uses verbage so I want to Example do a couple of examples and as I mentioned before these are gonna be fairly simple examples with the mathematics that's involved but I want to go through this process of how do I find all of these pieces how do I actually develop my mathematical model and how do I sufficiently answer the question in a way that's clear and concise so our problem here the water in a bathtub decreases from a height of 15 inches to a height of 8 inches over two minutes how much longer will it take to fully drain the tub so first thing I need to do is to identify Changez and so I'm letting that two things are changing here I'm noticing that the height is changing so height the height of the water which I'll call H and it's over two minutes so I know that I have time T and so what I really got here is height is going to depend on time and as an aside you're often going to have time be an independent variable just like X is our normal independent variable when you're graphing time is usually something that we don't have very much control over it's something that we can measure but it's not necessarily something that we can control and so we're very rarely going to have time as a function of something else we're often going to have something as a function of time and in this case we're dealing with height as a function of time so my first step is to identify what actually changes my second step is to say okay well what type of information do we have well I have initial conditions my initial condition is that my height equals 15 inches at the start and that really is where T equals zero and we also know that our height is eight inches when T equals two so from this I've got an initial condition which is most likely going to be my B my y-intercept and I have enough pieces of data to find my slope remember slope is the change in Y over the change in X or the change in my dependent variable or my independent variable 15 is the height at time 0 8 is the height at x 2 and so I have a negative slope so 15 minus 8 is 7 0 minus 2 is negative 2 and so it's really negative 3.5 and again I have to ask myself does this make sense in the context of the problem and my answer is yes the water is going down my rate of change should be negative if I was dealing with this in a verbal standpoint I would actually say that this is negative 3.5 inches per minute because that's really what's happening here notice my units for my numerator are inches 15 inches 8 inches and my units in the denominator are time minutes so once I have this I need to figure out ok what is it I'm trying to identify how much longer will it take to fully drain the tub what does it mean to fully drain the tub I need to find the time when the height equals zero inches that really is what I'm trying to find here so again I need to make sure that I read the question fully and I need to make sure that I understand what is it that the question is asking here so once I've done that now I'm ready to come up with some type of model okay so I have to find the rate of change which is my slope and for me I know that's negative 3.5 inches per minute and I'm going to have to find the the initial condition but this is really my y-intercept which for me is 15 inches so that is enough information for me to actually come up with my model my model here says that H of T where a height depends on time is going to be negative 3.5 times t plus 15 now you may see this written as 15 minus 3.5 T a lot of people don't want to put the leading negative here some people like to see it written as an expression where the minus is inside the expression but both of these are certainly fine and now what I've got is a function where the height depends on time now keep in mind what is it that I'm actually trying to find here well I'm trying to find where H of T equals zero so I want 15 minus 3.5 T to be equal to zero so really I've got 15 equals 3.5 T when I move this term to the other side and so my time is going to be 15 divided by 3.5 and this is time so this is minutes now by the way this is actually four point two eight six minutes I'm following the AP standard where they expect you to write your answer to three decimal places so four point two eight six minutes is about how long it would take for my height to reach zero given that my height is starting at fifteen inches and given that my height is decreasing at a rate of three point five inches per minute so now continuing on this the things I need to ask myself now because right now I've just found a solution my solution that T equals four point two eight six this is a solution this is not an answer and that's really important to understand when we're doing these mathematical problems it's often very easy to come up with a solution it's not easy to come up with an answer so now at this point I still have a couple more steps to actually get my answer the first thing I have to ask myself is is this reasonable well I know they get from fifteen inches to eight inches that drops seven inches in two minutes so it seems reasonable that it would drop fourteen inches in four minutes and if I want to fully dream the tub that I want to drop 15 inches in slightly more than four minutes so my answer of 4.28 six my solution that I've got up here seems reasonable it seems like it should take slightly more than four minutes to drain fifteen inches of height so now I'm ready to actually answer my question what is the answer on my question well the question is how much longer will it take to fully drain the tub my answer is it takes 4.28 six minutes to fully drain that tub that's my answer notice I have a solution T equals four point two eight six but without this minutes and without the idea that this actually takes time to fully drain the tub which is what it is we were asked to solve for point two eight six is just a number and it's really important for us to actually come up with an answer and not just a solution now some things that Considerations we really should consider in this problem keep in mind that we have talked about domain and range so for this particular problem what is my domain well I know that my domain is time my time starts at T equals zero and it really doesn't make sense to talk about what happens past four point two eight six because beyond four point two eight six my tub is fully drained there is no more water to get out the water is not going to go any lower than a height of zero or at least it shouldn't in the context of this problem so my domain is from zero to four point two eight six minutes and if I talk about my range well when I'm dealing with my range I know that the highest that the water is going to be in my tub is 15 inches and outside of this problem I don't care about what happens more than fifteen inches of water in the tub because that's really going back in time assuming that I've been draining the whole time fifteen was my initial condition and I went downhill from here and actually went down to a height of zero now I actually reached all of these numbers I reached zero for time and zero for height I reached 15 for height I reached the four point two eight six minutes so I'm including all of these points but these are important considerations what I'm talking about does the answer make sense is my answer reasonable so I want to do one more example here so Example Problem let's go on to our next example it says a phone company charges $75 for a data plan with 24 gigabytes of data and charges $5 for each extra one gigabyte used over the limit a mother notices her son over used his data and the total usage was 41 gigabytes how much of the mother expects the bill to be now keep in mind it's very nice to be able to plan ahead to kind of to figure out what should I expect to pay rather than waiting for the bill to come and reacting to it so you know we're being proactive here so what's happening here what are the things that change well I've got cost and data those really are the two things that are dealing with here and my cost is a function of data my cost is in dollars my data is in gigabytes and that's going to be important when I'm writing my sentence but I need to make sure that I'm using descriptive variable C and D work so much better than x and y X&Y we're going on a graph we're very poorly in other scenarios here C has a very clear meaning in my problem I'm trying to find out something about the cost cost is my dependent variable and it depends on data data's measured in gigabytes data is my independent variable so now what pieces of data can actually get out of here well I know that it's 24 gigabytes that costs $75 so in other words this is some kind of initial condition and the price goes up from here so the price should increase $5.00 per gigabyte now this is a rate of change so this is probably going to be my slope or my slopes can be related to this five my slope certainly going to be calculated using this five so what is it that we're trying to find here so we want C when D equals 41 gigabytes that's really what we're trying to find in this problem I need a model that will tell me what the cost is when we've reached 41 data bytes of data usage so what is our plan well of course we need to find the slope and we have a point and this is important because when I'm dealing with linear expressions I can either have y equals MX plus B I can have slope intercept form or I can have Y minus y1 equals M times X minus x1 I can have this point slope form to find out what my model is going to be and in this situation I don't have an initial condition I don't have a condition where time equals zero or money equals zero or in this case where the usage is zero what I have instead is some point here I know that at some point at some initial value twenty four gigabytes cost $75 and of course the cost depends on the data so this is really twenty four comma seventy-five this has really taken the place of my ax and seventy-five is taking the place of my Y so once I have this information now I'm ready to actually come up with a mathematical model and that's what we're going to do here I know that the cost minus the seventy five equals my slope which is five times data minus 24 now notice what I've got here is I've got the same point slope that we're dealing with just keep in mind that D is my independent variable and C is my dependent variable so I'm using those in place of x and y so if I do this that means that I've got five times D minus 120 and so if I add 75 to both sides I end up with my cost being five times D minus 45 and so now I've got a function cost as a function of data is equal to five times the amount of data used in gigabytes minus forty five dollars and one thing I really need to note is that this only works when D is at least twenty four gigabytes because if it's less than twenty four gigabytes that phone company is still going to charge me 75 bucks if all I use is two gigs of data they're still going to charge me the 75 bucks that's one of the ways that they make money so I have a model and I was interested in what happens when my data is 41 so I need to know what C of 41 is and for me that's going to be five times 41 minus 45 well five times 41 is 205 and when I subtract 45 I end up with a hundred and sixty keep in mind that 160 is a solution but what is 160 actually mean keep in mind this is a cost for using 41 gigabytes of data that's really what I'm dealing with here so now I need to ask myself does this make sense well I'm not saying that all phone companies are money-grubbing and eager to take as much money as possible but does it make sense that I would be charged that much more for 16 gigabytes of data well I don't have the luxury of really buying two of these data plans cuz two of these data plans should cost 150 dollars and give me 48 gigabytes but what's happening is that basically the first 24 gigabytes are supposed to be cheaper and then the fact that I'm over using my data plan means that I'm being penalized but really a hundred and fifty dollars and 160 dollars that's kind of reasonable it certainly seems that okay yeah I'm using more data than I should I'm I'm not quite using double but I'm pretty close to using double and so being really close to the price for double plus some type of penalty this answer does seem reasonable so finally I need to make sure that I actually answer the question the question is how much should the mother expect to build B she should expect a bill of 160 dollars for 41 gigabytes of data usage okay so again I'm answering the question I'm not just giving a solution but I'm actually coming up with an answer so finally in the context of this problem what are some of the things that we need to take away from this well in this particular problem what would my domain be well my domain is my independent variable my data my data really has to start at 24 because I'm being charged 75 dollars no matter how much data I use but really there's no limit no technically no limit as to how much data I could use some people may use 10 gigabytes a day in which case three hundred or three hundred ten gigabytes of data would be perfectly reasonable for them if you're live streaming your life at full HD you know who knows how much data you're gonna end up using this formula I have really does have a minimum though 24 gigabytes is the minimum anything less than that and I'm still being charged $75 $75 is the lowest value that I'll ever be charged but if I have no limit to the amount of data that I can use then technically I have no limit to the amount of money I can spend and I know that going through this process seems like overkill when we're just dealing with a linear relationship but really this the time to practice and get used to this skill set because this technique this algorithm of this process of going through and determining what my variables are determining what data the problem has and determining okay what do I need to do what's my plan for coming up with a model and then getting a model and using a model and then seeing if my answer is reasonable and then actually answering the question so just coming up with a solution this process is going to work whether we're dealing with linear or whether we're dealing with quadratic relationships or whether we're dealing with cubic relationships or square root racial ships or any of the relationships of our parent functions or other transformed functions that we've talked about in this course once again this is mr. Potter thank you for watching have a great day
14268	https://www.youtube.com/watch?v=OpPTTcRqP4w	Group Homomorphism Proof (Example with e^x) The Math Sorcerer 1220000 subscribers 49 likes Description 2799 views Posted: 21 Feb 2023 Consider the map f which takes x and sends it to e^x. This is a map from the group of reals under addition to the group of positive reals under multiplication. We prove this a group homomorphism. Useful Math Supplies My Recording Gear (these are my affiliate links) Math, Physics, and Computer Science Books Epic Math Book List Pre-algebra, Algebra, and Geometry College Algebra, Precalculus, and Trigonometry Probability and Statistics Discrete Mathematics Proof Writing Calculus Differential Equations Books Partial Differential Equations Books Linear Algebra Abstract Algebra Books Real Analysis/Advanced Calculus Complex Analysis Number Theory Graph Theory Topology Graduate Level Books Computer Science Physics These are my affiliate links. As an Amazon Associate I earn from qualifying purchases. If you enjoyed this video please consider liking, sharing, and subscribing. Udemy Courses Via My Website: Free Homework Help : My FaceBook Page: There are several ways that you can help support my channel:) Consider becoming a member of the channel: My GoFundMe Page: My Patreon Page: Donate via PayPal: Udemy Courses(Please Use These Links If You Sign Up!) Abstract Algebra Course Advanced Calculus Course Calculus 1 Course Calculus 2 Course Calculus 3 Course Calculus 1 Lectures with Assignments and a Final Exam Calculus Integration Insanity Differential Equations Course Differential Equations Lectures Course (Includes Assignments + Final Exam) College Algebra Course How to Write Proofs with Sets Course How to Write Proofs with Functions Course Trigonometry 1 Course Trigonometry 2 Course Statistics with StatCrunch Course Math Graduate Programs, Applying, Advice, Motivation Daily Devotionals for Motivation with The Math Sorcerer Thank you:) 9 comments Transcript: hello in this video we're going to do a very simple proof we're going to prove that a certain function is a group homomorphism so recall a function f from a group g into a group H is a group homomorphism if it has the property that for all x y in your group G when you look at F of the product of X and Y so f of x y That's equal to another product which is f of x times F of Y so the multiplication is preserved when you have multiplication in G right this multiplication is happening in G it corresponds to multiplication in h in a very cool way right it's f of x y is equal to f of x times F of Y and so in this particular example we're going to prove that f of x equals e to the x is a group homomorphism so what's all this this is just the notation for the function so the domain is the set of reals that's R and the plus indicates that this is a group under addition so the real numbers do form a group under addition the codomain is the set of positive reals that's what the little plus means here on top of the r and this is a group under multiplication so we're looking at a map or a function if from the real numbers under addition which is a group into the positive reals under multiplication which is a group and their function is defined by f of x equals e to the X and we're going to prove that this is a group homomorphism this is a really simple proof but it's kind of fun so proof and if you've never seen this maybe you'll learn something even if you learn one thing that's success so to start our proof we basically just to do the proof we basically just have to use this definition here so let's start by saying for all x y and R right because that's our domain that's our G is R the reason I have the parentheses in the plus here is because we're being extra descriptive we're specifying the operation and the reason we do that is um it's helpful and also a lot of times you can form a group using multiple operations on the same set so it's beneficial to write it that way so now we're going to look at f of x y except it's not f of x y and the reason is this is additive okay the the operation is addition here so it's actually F of X Plus y okay if you want the additive version of this it would be f of x plus y equal to f of x plus F of Y this is assuming that the operation in G is addition and the operation H is also addition right so here the operation is addition and r so this is equal to well let's see f of x is e to the X so this is e to the X Plus y and now we can use properties of exponents this is e to the x e to the Y very very powerful stuff there simple and basic but very powerful and then we know something about e to the X that's equal to f of x and we know something about e to the Y it's equal to F of Y and this makes sense right here the operation is multiplication right so it's multiplication so here it's addition here it's multiplication so you have to adjust the notation accordingly so we have that for all x y and r f of x plus Y is equal to f of x times F of Y so this shows that f is a groupomorphism so f is a group homomorphism homomorphism now let me talk about the notation a little bit more because um sometimes people get confused and it's important to understand it so in the first definition I gave you here we're assuming that the operation is multiplication and g and an H this is the standard way to do it most of the time this is how you write it this is how you see it in textbooks here I gave you the case where the operation in G and H are both addition in our particular example the operation in G right this is our G uh is addition and the operation in h is multiplication that's why it's a little bit different over here so this is an R Plus under multiplication and over here this is in r under addition so the operation changes right here it's addition on the set of reels and here with multiplication on the set of positive reels so yeah hopefully this video has taught you some mathematics until next time good luck
14269	https://www.uptodate.com/contents/clinical-presentation-and-diagnosis-of-von-willebrand-disease	Your Privacy To give you the best possible experience we use cookies and similar technologies. We use data collected through these technologies for various purposes, including to enhance website functionality, remember your preferences, and show the most relevant content. You can select your preferences by clicking the link. For more information, please review our Privacy & Cookie Notice Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device. Because we respect your right to privacy, you can choose not to allow certain types of cookies on our website. Click on the different category headings to find out more and manage your cookie preferences. However, blocking some types of cookies may impact your experience on the site and the services we are able to offer. Privacy & Cookie Notice Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function. They are usually set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, this may have an effect on the proper functioning of (parts of) the site. Performance Cookies These cookies support analytic services that measure and improve the performance of our site. They help us know which pages are the most and least popular and see how visitors move around the site.
14270	https://www.webpathology.com/images/gynecologic/uterus/endometrial-hyperplasia/38899	Atypical Endometrial Hyperplasia vs Adenocarcinoma Home Subspecialty AboutFeedbackSubmit Images Atypical Endometrial Hyperplasia vs Adenocarcinoma Home Gynecologic Uterus Endometrial Hyperplasia Atypical Endometrial Hyperplasia vs Adenocarcinoma Previous Image 31 of 34 Next Image Description Atypical Endometrial Hyperplasia/Endometrial Intraepithelial Neoplasia (AEH/EIN) vs Adenocarcinoma: AEH/EIN (left panel) should be differentiated from endometrial endometrioid carcinoma (right panel) and endocervical carcinoma. The morphologic features of AEH/EIN overlap those of well-differentiated adenocarcinoma and a distinction between the two may not always be possible on a biopsy, especially with scant or fragmented material. Features favoring carcinoma over AEH/EIN include: solid sheets of neoplastic cells; confluent, back-to-back glands not surrounded by stroma (most useful feature) or scant, threadlike stroma; extensive villoglandular, papillary, cribriform, or microacinar architecture; complex, maze-like glandular growth; desmoplastic stromal response adjacent to glands; marked cytologic atypia; necrosis and foamy macrophages. The cytologic and/or architectural pattern should be greater than 2 mm in size according to some experts. In challenging cases with feature intermediate between atypical hyperplasia and grade I endometrioid adenocarcinoma, a diagnosis of "atypical hyperplasia, cannot rule out well-differentiated adenocarcinoma" is recommended. Some pathologists use "at least AEH/EIN" or "AEH/EIN bordering on well-differentiated endometrioid carcinoma." Previous Image 31 of 34 Next We recommend Molecular classification and fertility-sparing outcomes in endometrial cancer and atypical endometrial hyperplasiaJiayi Wang, Guozhong Jiang, Shuping Yan, et al., Biomol Biomed, 2025 The impact of Ki-67 index, squamous differentiation, and several clinicopathologic parameters on the recurrence of low and intermediate-risk endometrial cancerBirol Ocak, Biomol Biomed, 2021 Morphometric Characterization of Staging Categories in Urothelial Cancer of the Urinary BladderGuy Hidas, Current Urology, 2009 Endometrial Cancer. You searched for endometrial carcinoma - BJBMS ViewpointsBJBMS Viewpoints Endometrial cancer epidemiology and prevention in Federation of Bosnia and Herzegovina, B&HIvan Vasilj, Biomol Biomed, 2004 Systemic analysis of the expression and prognostic significance of USP31 in endometrial cancerYuzhen Huang, Biomol Biomed, 2022 C-Erb-b2 Oncogene Expression in Intraductal Proliferative Lesions of the BreastFiliz Eren, Biomol Biomed, 2012 Epididymal Leiomyoadenomatoid Tumor: A Case Report and Review of LiteratureArnault Cazorla, Current Urology, 2014 Comprehensive genomic and transcriptomic analyses reveal prognostic stratification for esophageal squamous cell carcinomaJian Gao, Signal Transduction and Targeted Therapy, 2025 Powered by Targeting settings Do not sell my personal information Terms of UsePrivacy PolicySite Map © 2003-2025 WebPathology, LLC. All rights reserved We use cookies to improve your experience. By continuing to use our site, you agree to our use of cookies.
14271	https://www.geeksforgeeks.org/maths/monotonic-sequence/	Monotonic Sequence Monotonic sequence is one of the simplest terms used in mathematics to refer to a number sequence that moves from a smaller value to a bigger value or vice versa; that is, it only increases or decreases. Different fields of study where this type of sequence is important include calculus, probability and computer science. Mastering monotonically increasing and decreasing sequences is particularly important for studying the convergence and behavior of mathematical functions and series. In this article, we will learn in detail about monotonic sequence, theorem, types and examples. Table of Content What is a Monotonic Sequence? In some sequence theory, they define a monotonic sequence to be the sequence of numbers where the term is bigger or equal to the previous term or is lesser or equal to the previous term. Therefore if one aims at identifying if a sequence is monotonic what it means is whether the sequence is strictly increasing or decreasing. Formally, a sequence {an} is monotonic if either a{n+1} ≥ an for all n ≥ 1 (increasing) or a{n+1} ≤ an for all n ≥ 1 (decreasing). Specifically, monotonic sequences have the characteristic that the direction of their changes at any point is positively oriented, which implies that sequences of this type are either constantly on the rise or are progressively declining. Types of Monotonic Sequence Monotonic sequences are categorized into two main types: increasing monotonic sequences and decreasing monotonic sequences. Increasing Monotonic Sequence An increasing sequence is a sequence in mathematics in which the next term is greater than the previous term. Formally, a sequence {an} is increasing monotonic if: For example, the sequence {1, 2, 3, 4, 5, ...} is an increasing monotonic sequence because each term is greater than the previous term Decreasing Monotonic Sequence Therefore, the concept of the decreasing monotonic sequence can be defined as that each element of the sequence should not be greater than the previous element of the sequence. Formally, a sequence {an} is decreasing monotonic if: For example, the sequence {10, 8, 6, 4, 2, ...} is a decreasing monotonic sequence because each term is less than the previous term Monotonic Sequence Example and Graph Look at the sequence of numbers: 1, 2, 4, 8, 16, 0. . . This sequence is increasingly monotonic as is given by the fact that each element of the sequence is two times of previous element of the sequence. This preset sequence can be presented graphically on a coordinate plane to represent the terms. What this means is that a set of points will be established such that when the curve defined by these points is created, the entire curve will lie wholly in the first quadrant of the X-Y axis and will not extend downward. The graph of representation of the monotonic sequence is straight line or non-linear depending upon the nature of monotonicity or the kind of relation between the terms of the sequence. For instance, the sequence {1, 3, 5,/, 7, 9, . .. } is also increasingly monotonic although its graph is in just a straight line unlike the sequence {1, 2, 4, 8, 16, . .. } whose graph is non-linear. Monotonic Sequence Theorem The monotonic sequence theorem states that if a sequence is monotonic and bounded, then it converges. Formally, if {an} is a monotonic sequence and there exists M ∈ ℝ such that an ≤ M for all n ≥ 1 (or an ≥ M for all n ≥ 1), then {an} converges. Proof: The proof for decreasing monotonic sequences is similar, using the infimum instead of the supremum. Bounded and Monotonic sequence A sequence {an} is bounded if there exists M ∈ ℝ such that \|an\| ≤ M for all n ≥ 1. In other words, a bounded sequence is a sequence where the values of the terms, are all contained in a given interval. Furthermore, if a sequence is both monotonic and also a bounded sequence, then it is a convergent sequence by the monotonic sequence theorem. For instance, let us take the example of the increasing sequence, { 1, 1/2, 1/4, 1/8, . . . } This is a decreasing sequence, so it appears to be monotonic, and since it is also bounded above by 1, it converges to 0. The fact that the sequence is bounded implies that the terms of the sequence cannot diverge to infinity, while it's being monotonic implies that the sequence is either strictly increasing or strictly decreasing, thus it has to converge. Comparing Monotonic Sequences Monotonic sequences can be compared with other kinds of sequences, like arithmetic sequences, geometrical sequences, and Fibonacci sequences. All in all, there are certain similarities between these types of sequences, but each has distinct features and characteristics of its own. With Arithmetic Sequence While comparing the monotonic sequences with arithmetic sequences, the only distinguishable factor is that while forming the arithmetic sequences, a fixed difference is added to the previous term to obtain the succeeding term and, on the other hand, there is no fixed difference in monotonic sequences. For example, {1, 3, 5, 7, … } is an arithmetic progression with a constant difference of 2 while {1, 2, 4,8, …} is an increasing Monotonic sequence which has no constant difference. With Geometric Sequence Monotonic sequences can have a constant ratio between consecutive terms, similar to geometric sequences. However, monotonic sequences are not required to have this property. Additionally, geometric sequences can oscillate between increasing and decreasing, while monotonic sequences must either increase or decrease steadily. With Fibonacci sequence The Fibonacci sequence is defined by the recurrence relation an = a{n-1} + a{n-2}, with a1 = 0 and a2 = 1. Monotonic sequences do not like the Fibonacci sequence have specific rules for determining the nth term in the sequence. However, both sequences can share the same characteristics which are convergence or divergence, etc. based on the characteristics that the sequences may have. Conclusion Monotonic sequences are the sequences of numbers, either increasing or decreasing; these sequences are used everywhere in mathematics. They are used as a means of studying the behavior of sequences and series and are key to studying the convergence and characteristics of mathematical functions. The Monotonic sequence theorem is ‘if the sequence is monotonic and bounded; then; it is convergent’. There is therefore a need to distinguish one type of sequence from the other, as well as understand the characteristics of each sequence & how each of them differs from the others to most effectively solve problems in the areas of calculus, probability theory, and computer science. Also, Check Examples on Monotonic Sequences Example 1: Determine if the sequence {an} defined by an = 1 - 1/n is increasing, decreasing, or neither. Solution: To check if the sequence is increasing or decreasing, we need to compare consecutive terms. a{n+1} - an = (1 - 1/(n+1)) - (1 - 1/n) = 1/n - 1/(n+1) = (n+1 - n)/(n(n+1)) = 1/(n(n+1)) Since 1/(n(n+1)) > 0 for all n ≥ 1, we have a{n+1} - an > 0. This means each term is greater than the previous term, so the sequence is increasingly monotonic. Example 2: Let an = 2(-n). Prove that {an} is a decreasing monotonic sequence. Solution: To show that {an} is decreasing monotonic, we need to prove that a{n+1} ≤ an for all n ≥ 1. a{n+1} = 2(-(n+1)) = 2(-n)/2 = an/2 Since an/2 ≤ an for all n ≥ 1, we have a{n+1} ≤ an. Therefore, {an} is a decreasing monotonic sequence Practice Questions on Monotonic Sequence Q1. Determine whether the sequence {an} defined by an = n2 is increasing, decreasing, or neither. Q2. Prove that the sequence {bn} defined by bn = 3n - 2 is an increasing monotonic sequence. Q3. Consider the sequence {cn} defined by cn = (-1)n. Determine if {cn} is a bounded sequence and explain why or why not. T Explore Maths Basic Arithmetic What are Numbers? Arithmetic Operations Fractions - Definition, Types and Examples What are Decimals? 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14272	https://dictionary.cambridge.org/us/dictionary/english-polish/poverty	Translation of poverty – English-Polish dictionary poverty Your browser doesn't support HTML5 audio Your browser doesn't support HTML5 audio (Translation of poverty from the Cambridge English–Polish Dictionary © Cambridge University Press) Translation of poverty \| PASSWORD English-Polish Dictionary (Translation of poverty from the PASSWORD English-Polish Dictionary © 2022 K Dictionaries Ltd) Examples of poverty Translations of poverty Get a quick, free translation! Browse More translations of poverty in Polish Word of the Day Victoria sponge Your browser doesn't support HTML5 audio Your browser doesn't support HTML5 audio a soft cake made with eggs, sugar, flour, and a type of fat such as butter. It is made in two layers with jam or cream, or both, between them Blog Calm and collected (The language of staying calm in a crisis) New Words vibe coding © Cambridge University Press & Assessment 2025 © Cambridge University Press & Assessment 2025 Learn more with +Plus Learn more with +Plus To add poverty to a word list please sign up or log in. Add poverty to one of your lists below, or create a new one. {{message}} {{message}} Something went wrong. {{message}} {{message}} Something went wrong. {{message}} {{message}} There was a problem sending your report. {{message}} {{message}} There was a problem sending your report.
14273	https://www.khanacademy.org/math/statistics-probability/describing-relationships-quantitative-data/regression-library/a/introduction-to-residuals	Introduction to residuals (article) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. 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Skip to lesson content Statistics and probability Course: Statistics and probability>Unit 5 Lesson 4: Least-squares regression equations Introduction to residuals and least squares regression Introduction to residuals Calculating residual example Calculating and interpreting residuals Calculating the equation of a regression line Calculating the equation of the least-squares line Interpreting slope of regression line Interpreting y-intercept in regression model Interpreting a trend line Interpreting slope and y-intercept for linear models Math> Statistics and probability> Exploring bivariate numerical data> Least-squares regression equations © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Introduction to residuals Google Classroom Microsoft Teams Build a basic understanding of what a residual is. We run into a problem in stats when we're trying to fit a line to data points in a scatter plot. The problem is this: It's hard to say for sure which line fits the data best. For example, imagine three scientists, Andrea‍, Jeremy‍, and Brooke‍, are working with the same data set. If each scientist draws a different line of fit, how do they decide which line is best? 1‍2‍3‍4‍5‍6‍7‍8‍9‍10‍1‍2‍3‍4‍5‍6‍7‍8‍9‍10‍y‍x‍ If only we had some way to measure how well each line fit each data point... Residuals to the rescue! A residual is a measure of how well a line fits an individual data point. Consider this simple data set with a line of fit drawn through it 1‍2‍3‍4‍5‍6‍7‍8‍9‍10‍1‍2‍3‍4‍5‍6‍7‍8‍9‍10‍y‍x‍ and notice how point (2,8)‍ is 4‍ units above the line: 1‍2‍3‍4‍5‍6‍7‍8‍9‍10‍1‍2‍3‍4‍5‍6‍7‍8‍9‍10‍y‍x‍4‍ This vertical distance is known as a residual. For data points above the line, the residual is positive, and for data points below the line, the residual is negative. For example, the residual for the point (4,3)‍ is −2‍: 1‍2‍3‍4‍5‍6‍7‍8‍9‍10‍1‍2‍3‍4‍5‍6‍7‍8‍9‍10‍y‍x‍−2‍4‍ The closer a data point's residual is to 0‍, the better the fit. In this case, the line fits the point (4,3)‍ better than it fits the point (2,8)‍. Try to find the remaining residuals yourself What is the residual of the point (6,7)‍ in the graph above? Check What is the residual of the point (8,8)‍ in the graph above? Check What is the residual of the point (1,2)‍ in the graph above? Check Show answers 1‍2‍3‍4‍5‍6‍7‍8‍9‍10‍1‍2‍3‍4‍5‍6‍7‍8‍9‍10‍y‍x‍−2‍−1.5‍4‍1‍1‍ Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted imamulhaq 10 years ago Posted 10 years ago. Direct link to imamulhaq's post “How do you do this On a c...” more How do you do this On a calculator Answer Button navigates to signup page •Comment Button navigates to signup page (10 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer matepe 8 months ago Posted 8 months ago. Direct link to matepe's post “You can't, it's impossibl...” more You can't, it's impossible 1 comment Comment on matepe's post “You can't, it's impossibl...” (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Joona Rauhamäki 10 years ago Posted 10 years ago. Direct link to Joona Rauhamäki's post “This article does not exp...” more This article does not explain what to do with the residuals after calculating them. Are you supposed to sum them? When are you supposed to use them? Answer Button navigates to signup page •1 comment Comment on Joona Rauhamäki's post “This article does not exp...” (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Jo Long 10 years ago Posted 10 years ago. Direct link to Jo Long's post “Minimise them - the small...” more Minimise them - the smaller they are, the better the line fits. Also, graph them (on the y axis, with the original x values). They should be random. If there's a pattern, the line is not a good fit to the "cloud of dots". 1 comment Comment on Jo Long's post “Minimise them - the small...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more alyssah83 2 years ago Posted 2 years ago. Direct link to alyssah83's post “how can a residual be one...” more how can a residual be one sided? For example in the graphs, would being one sided mean the data points are not scattered? Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Parsa Abangah 2 years ago Posted 2 years ago. Direct link to Parsa Abangah's post “In statistics, resids (sh...” more In statistics, resids (short for residuals) are the differences between the predicted values and the actual values of the response variable. One-sided residuals can occur when a model is fitted to data with some specific characteristics. A one-sided residual plot is a plot of residual values against the fitted values of the model only for one side of the graph. For example, a one-sided residual plot can be observed when we have a regression model in which our residuals are constrained to be non-negative. In this case, we may have a one-sided residual plot resulting from the fact that only one side of the graph will have positive residuals, while the other side will have residuals of zero. In terms of scatterplots, being one-sided does not necessarily mean that the data points are not scattered. The scatter in the data points will still be visible in the one-sided residual plot. 1 comment Comment on Parsa Abangah's post “In statistics, resids (sh...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more yikian2021 2 years ago Posted 2 years ago. Direct link to yikian2021's post “So how do u decide which ...” more So how do u decide which line is best? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer daniella 2 years ago Posted 2 years ago. Direct link to daniella's post “Deciding which line is be...” more Deciding which line is best typically involves evaluating the residuals for each line. The line that minimizes the sum of squares of residuals is often considered the best fit to the data. This method, known as least squares regression, aims to minimize the overall discrepancy between the observed data points and the fitted line. Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Jamune 6 years ago Posted 6 years ago. Direct link to Jamune's post “In the article, it says t...” more In the article, it says that the closer the the data point's residual is to zero, it fits the line best. There's (4,3) and (2,8). The residuals are 4, and -2. It says 4 is closer ( aka (4,3) ) but isn't -2 closer to zero than 4? How is this possible? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Avi Mahajan 6 years ago Posted 6 years ago. Direct link to Avi Mahajan's post “The point (4,3) is two un...” more The point (4,3) is two units below the line. It has a residual of -2. However, the point (2,8) is four units above the line. It has a residual of 4. Since -2 is closer to zero than 4, the point (4,3) fits the line better than the point (2,8). I think you misunderstood that the residual of four is closer to the line. The article really meant that the point (4,3) is closer to the line. Hope this helped! Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Iustus82437 9 years ago Posted 9 years ago. Direct link to Iustus82437's post “in residuals how do you d...” more in residuals how do you determine which one is best? do you mean it or do you do something else this article did not tell me how to. Answer Button navigates to signup page •1 comment Comment on Iustus82437's post “in residuals how do you d...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Roman Gudkov 9 years ago Posted 9 years ago. Direct link to Roman Gudkov's post “Looks like you just sum t...” more Looks like you just sum them. Using example of this article, residual = - 1.5 + 4 - 2 + 1 + 1 = 2.5. That's the way I understood it. This way it would make intuitive sense: negative residual for a point means that line is 'underestimating' data set, positive - 'overestimating'. Summing shows you total over/under estimation 2 comments Comment on Roman Gudkov's post “Looks like you just sum t...” (0 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more raydot 8 years ago Posted 8 years ago. Direct link to raydot's post “When you are trying to fi...” more When you are trying to figure these out from an equation, how do you know which variable belongs on the x axis and which belongs on the y axis? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer makvik 8 years ago Posted 8 years ago. Direct link to makvik's post “There is no set way of as...” more There is no set way of assigning variables to axis', so you must determine from the information give, or decide for yourself. There are some common practices, such as putting Price and Age on the Y Axis, and Quantity on the X axis. But you could flip them. If you have an equation and two variables, try setting one of them to zero and see what you get. You might be able to deduce which variable is on which axis from the answer. 1 comment Comment on makvik's post “There is no set way of as...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Singh, Aarush 7 years ago Posted 7 years ago. Direct link to Singh, Aarush's post “What would happened if th...” more What would happened if the residual is close but not quite an exact number? would we just have to estimate? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer daniella 2 years ago Posted 2 years ago. Direct link to daniella's post “If a residual is close bu...” more If a residual is close but not an exact number, you would typically round it to the nearest whole number or decimal place, depending on the context of the problem and the level of precision required. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Richard 9 years ago Posted 9 years ago. Direct link to Richard's post “Why compute residuals usi...” more Why compute residuals using the vertical distance of the data points to the proposed line? My intuition for determining the line of best fit would be to minimize the sum of the squares of the perpendicular distances to the proposed line. Now obviously that is more complex, but with machine methods, why isn't it the logical goal? Best wishes to all. Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Goober5161 2 years ago Posted 2 years ago. Direct link to Goober5161's post “so would this be like an ...” more so would this be like an educated guess? or would this have to be exact? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer daniella 2 years ago Posted 2 years ago. Direct link to daniella's post “There is a formula for ca...” more There is a formula for calculating the sum of squares of residuals, which involves squaring each residual and then summing them up. The formula is commonly used in regression analysis to assess the goodness of fit of the regression line. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Up next: video Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. 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14274	https://www.stemlytutoring.com/blog/mastering-ap-physics-1-kinematic-equations	Skip to Content Book a Consultation Book a Consultation AP Physics 1: 4 Kinematic Equations to Know Inside and Out Written By Stemly Tutoring In AP Physics 1, the four kinematic equations are fundamental tools for analyzing motion under constant acceleration. With these formulas, you can connect displacement, velocity, acceleration, and time to solve a wide range of problems accurately. Here’s what you’ll get in this post: Clear breakdown of each equation and when to use it A handy comparison table for quick reference Pro tips to avoid common mistakes How targeted AP Physics 1 tutoring can help you master the material The Four Kinematic Equations—Explained Δx = v₀ t + ½ a t² Calculates displacement (Δx) when you know initial velocity, acceleration, and time. v = v₀ + a t Solves for final velocity when initial velocity, acceleration, and time are known. v² = v₀² + 2 a Δx Connects velocities and displacement—great when time isn’t given. Δx = ½ (v + v₀) t Calculates displacement using average velocity and time; especially handy when acceleration is constant. Quick Reference Table \| Equation \| Formula \| When to Use \| --- \| Displacement & Time \| Δx = v₀ t + ½ a t² \| Use when initial velocity, acceleration, and time are known and you want displacement. \| \| Final Velocity \| v = v₀ + a t \| When you know initial velocity, acceleration, and time; need final velocity. \| \| No Time Involved \| v² = v₀² + 2 a Δx \| Best when time is not given, but you know velocities and displacement. \| \| Average Velocity \| Δx = ½ (v + v₀) t \| Use if acceleration is constant and you know average velocity and time. \| Pro Tips for Students Choose your equation based on what variables are given. If time isn't given, avoid equations involving 't'. Double-check units (e.g. meters vs kilometers, seconds vs hours). Sketch the scenario to visualize initial/final positions and velocities. Plug and chug: write down known values before substituting to avoid mix-ups. How AP Physics 1 Tutoring Can Help Learning the equations is one thing—knowing how and when to apply them is another. That’s where a personalized tutor comes in: Identify where you're stuck: equation selection, algebra, unit conversion, or concept gaps Walk through tailored practice problems Help build problem-solving strategies and confidence Stemly’s AP Physics 1 tutoring offers one-on-one sessions that match your learning style, helping you internalize the “why” and “how,” not just the formula. FAQ Q: How many kinematic equations are there in AP Physics 1? A: Four key equations relate displacement, velocity, acceleration, and time under constant acceleration. Q: Which equation doesn’t use time? A: The third equation — v² = v₀² + 2aΔx — is used when time isn’t given. Q: When should you use Δx = ½ (v + v₀) t? A: When acceleration is constant and you know average velocity and time. City Selector AP Physics 1 Stemly Tutoring Stemly Tutoring is an online math and science tutoring company helping middle school, high school, and college students build confidence and excel in subjects like Algebra 2, Chemistry, Biology, and Physics. As a team, we are passionate about making STEM subjects approachable and empowering students to succeed. Learn more about one-on-one tutoring at Stemly Tutoring. Previous Previous AP Physics Made Easy: 7 Tips Your Tutor Won't Tell You Next Next Preparing for College: What to Expect from Your Freshman Year
14275	https://mathcentral.quora.com/Prove-that-each-odd-prime-number-p-can-be-written-as-the-difference-of-two-perfect-squares	Prove that each odd prime number p can be written as the difference of two perfect squares.? - Math Central - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Math Central Mathematics is the "Language of the universe" Follow · 125.5K 125.5K Prove that each odd prime number p can be written as the difference of two perfect squares.? Answer Request Follow 3 Answers Sort Recommended Eleftherios Argyropoulos B.S. in Mathematics&Physics, Northeastern University (Graduated 2002) ·4y We can prove this with at least two different ways. First proof: Observe that for every two consecutive natural numbers n n, n+1 n+1, we have: (n+1)2−n 2=2 n+1…(1)(n+1)2−n 2=2 n+1…(1) By (1)(1), we see that for every single n n, we take a unique odd number and hence we take all odd numbers, while each one of them is expressed as a difference of two squares. Finally, since the set of odd primes is a genuine subset of odd numbers, the first proof follows. Second proof: The second proof comes directly from the fact that the only integers, which cannot be expressed as difference of two squares, are those which are congruent to 2 m 2 m Continue Reading We can prove this with at least two different ways. First proof: Observe that for every two consecutive natural numbers n n, n+1 n+1, we have: (n+1)2−n 2=2 n+1…(1)(n+1)2−n 2=2 n+1…(1) By (1)(1), we see that for every single n n, we take a unique odd number and hence we take all odd numbers, while each one of them is expressed as a difference of two squares. Finally, since the set of odd primes is a genuine subset of odd numbers, the first proof follows. Second proof: The second proof comes directly from the fact that the only integers, which cannot be expressed as difference of two squares, are those which are congruent to 2 m o d 4 2 m o d 4. An odd prime cannot be congruent to 2 m o d 4 2 m o d 4, hence it is always expressible as difference of two squares. Conclusion: The only prime, which cannot be expressed as difference of two squares, is 2 2. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes 9 3 Lai Johnny M. Phil in Mathematics Major, The Chinese University of Hong Kong (Graduated 1985) ·4y For any odd integer p p, there exists an integer n n such that p=2 n+1.p=2 n+1. Since (n+1)2−n 2=2 n+1=p,(n+1)2−n 2=2 n+1=p, therefore p p can be written as the difference of two perfect squares. Note: p is not necessarily a prime. 9 1 Mohammad Afzaal Butt B.Sc in Mathematics&Physics, Islamia College Gujranwala (Graduated 1977) ·4y Let p be an odd prime, a and b are two natural numbers such that Let p be an odd prime, a and b are two natural numbers such that p=a 2−b 2 p=a 2−b 2 ⟹p=(a−b)(a+b)⟹p=(a−b)(a+b) But p is a prime number . The only factors of p are 1 and p But p is a prime number . The only factors of p are 1 and p Hence a−b=1 a+b=p Hence a−b=1 a+b=p ⟹(a−b)+(a+b)=p+1⟹2 a=p+1⟹a=p+1 2⟹(a−b)+(a+b)=p+1⟹2 a=p+1⟹a=p+1 2 similarly(a+b)−(a−b)=p−1⟹2 b=p−1⟹b=p−1 2 similarly(a+b)−(a−b)=p−1⟹2 b=p−1⟹b=p−1 2 ∴p=(p+1 2)2−(p−1 2)2∴p=(p+1 2)2−(p−1 2)2 It can be observed that writing a prime number as a difference of the squares of two It can be observed that writing a prime number as a difference of the squares of two \text{integers is uni\text{integers is uni Continue Reading Let p be an odd prime, a and b are two natural numbers such that Let p be an odd prime, a and b are two natural numbers such that p=a 2−b 2 p=a 2−b 2 ⟹p=(a−b)(a+b)⟹p=(a−b)(a+b) But p is a prime number . The only factors of p are 1 and p But p is a prime number . The only factors of p are 1 and p Hence a−b=1 a+b=p Hence a−b=1 a+b=p ⟹(a−b)+(a+b)=p+1⟹2 a=p+1⟹a=p+1 2⟹(a−b)+(a+b)=p+1⟹2 a=p+1⟹a=p+1 2 similarly(a+b)−(a−b)=p−1⟹2 b=p−1⟹b=p−1 2 similarly(a+b)−(a−b)=p−1⟹2 b=p−1⟹b=p−1 2 ∴p=(p+1 2)2−(p−1 2)2∴p=(p+1 2)2−(p−1 2)2 It can be observed that writing a prime number as a difference of the squares of two It can be observed that writing a prime number as a difference of the squares of two integers is unique.integers is unique. Related questions What is the general integral Z(x,y) of the PDE x(y+Z) Zx + Z (Z - y) Zy = y(y-Z)? Can you find the general integral of the PDE (2xy-1) zₓ + (z-2x^2) zᵧ = 2 (x-yz)? How do I solve the partial differential equation (D²-2DD'-15D'²) z=12xy? How do you graph the equation y=-4x^2? What are the intercepts and then use them to graph the equation 2y=-18+9x? How do we prove that ∫1 0 ln(1+x 2)x√x 2−1 d x=−i(a r c s i n h 2(1))?∫0 1 ln⁡(1+x 2)x x 2−1 d x=−i(a r c s i n h 2⁡(1))? How do we evaluate the integral ∫π 0 x cos x 1+sin 2 x d x?∫0 π x cos⁡x 1+sin 2⁡x d x? How do we evaluate the integral ∫2 π 0 x 2 sin x 8+sin 2 x d x?∫0 2 π x 2 sin⁡x 8+sin 2⁡x d x? How do we evaluate the integral ∫∞0 x 4(x 4−x 2+1)4 d x?∫0∞x 4(x 4−x 2+1)4 d x? A can company charges a $3 flat rate in addition to $1.50 per mile the domain of this function is The range of this function is? Agatha put $775 into a simple interest bearing account at a rate of 2.4% for 8 years. Her friend Dominic invested the same amount into an account that is compounded quarterly at the same rate for 6 years? Jeff has $28.75. He purchased three cookies that cost $1.50 each, five newspapers that each cost $0.50, five flowers for $1.25 each, and used the remainder of the cash on a pair of sunglasses. How much were the sunglasses? When a factory operates from 6 AM to 6 PM, its total fuel consumption varies according to the formula f(t) = 0.91 - 0.110.4 + 12, where t is the time in hours after 6 AM and f() is the number of barre? Cooper designs a photo album that he is going to give his parents as a gift. He created 15 pages of the album in June. He designed 1/10 of the album in August. He finishes the remaining 3/5 just in time for his parent's anniversary in November? The following scatterplot relates the life expectancy of animals to their heart rate. Ignoring humans (which are labeled "Man" in the scatterplot), which two conclusions can be made from the scatterplot? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
14276	https://www.khanacademy.org/math/cc-fifth-grade-math/imp-place-value-and-decimals/imp-decimals-in-written-form/v/decimal-place-value-2	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
14277	https://www.schott.com/shop/medias/tie-29-refractive-index-and-dispersion-eng.pdf?context=bWFzdGVyfHJvb3R8MTg4OTQ2NHxhcHBsaWNhdGlvbi9wZGZ8aDQ3L2hjYS84ODE3NDA5NTg5Mjc4LnBkZnxkMTc5MjMyYjI5YmRlZTAzZmFmYzIxZmVlNTMyMmE5NzBiNWI5OGJhODM3YWRjYmM5NmY1ZTczMjY1ZWM3NDdk	Published Time: Thu, 01 Jan 1970 00:00:01 GMT Introduction The most important property of optical glass is the refractive index and its dispersion behavior. This technical information gives an overview of the following topics: • Dispersion – Principal Dispersion – Secondary Spectrum – Sellmeier Dispersion Equation • Temperature Dependence of Refractive Index • Influence of the Fine Annealing Process on the Refractive Index and Abbe number • Tolerances • Refractive Index Measurement Refractive Index If light enters a non-absorbing homogeneous material reflection and refraction occurs at the boundary surface. The refractive index n is given by the ratio of the velocity of light in vacuum c to that of the medium v n = cv (1) The refractive index data given in the data sheets are measured relative to the refractive index measured in air. The refractive index of air is very close to 1. Refractive Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Wavelenght Dependence of Refractive Index: Dispersion . . . . . . . . . . . . . . . . . . . . . . 2 Temperature Dependence of Refractive Index . . . . . 6 Influence of the Fine Annealing Process on the Refractive Index and Abbe number . . . . . . . . . . . . . . . 7 Tolerances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Refractive Index Measurement . . . . . . . . . . . . . . . . . . . 11 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Refractive Index and Dispersion ➜ Practically speaking the refractive index is a measure for the strength of deflection occurring at the boundary surface due to the refraction of the light beam. The equation describing the amount of deflection is called Snell’s law: n 1 · sin( a1 ) = n 2 · sin( a2) (2) The refractive index is a function of the wavelength. The most common characteristic quantity for characterization of an optical glass is the refractive index n in the middle range of the visible spectrum. This principal refractive index is usually denoted as nd – the refractive index at the wavelength 587.56 nm or in many cases as ne at the wavelength 546.07 nm. Technical Information Advanced Optics TIE-29 Version February 2016 12 Wavelength Dependence of Refractive Index: Dispersion The dispersion is a measure of the change of the refractive index with wavelength. Dispersion can be explained by apply-ing the electromagnetic theory to the molecular structure of matter. If an electromagnetic wave impinges on an atom or a molecule the bound charges vibrate at the frequency of the incident wave. The bound charges have resonance frequency at a certain wavelength. A plot of the refractive index as a function of the wavelength for fused silica can be seen in figure 1. It can be seen that in the main spectral transmission region the refractive index increases towards shorter wavelength. Additionally the dotted line shows the absorption constant as a function of the wavelength. 2.1 Principal Dispersion The difference (n F – nC ) is called the principal dispersion. nF and nC are the refractive indices at the 486.13 nm and 656.27 nm wavelength. The most common characterization of the dispersion of optical glasses is the Abbe number. The Abbe number is defined as nd = (n d – 1)/(n F – nC ) (3) Sometimes the Abbe number is defined according to the e line as ne = (n e – 1)/(n F' – nC ' ) (4) Back to index ➜➜ Refractive Index and Dispersion Fig. 1: Measured optical constants of fused silica (SiO 2 glass) . Traditionally optical glasses in the range of nd > 50 are called crown glasses, the other ones as flint glasses. Glasses having a low refractive index in general also have a low dispersion behavior e. g. a high Abbe number. Glasses having a high refractive index have a high dispersion behavior and a low Abbe number. Wavelength λ/μm Refractive Index n (λ)Absorption Constant k (λ)3.5 3.0 2.5 2.0 1.5 1.0 0.5 010 110 010 –1 10 –2 10 –3 10 –4 10 –1 10 010 1nk Technical Information Advanced Optics TIE-29 3 2.2 Secondary Spectrum The characterization of optical glass through refractive index and Abbe number alone is insufficient for high quality optical systems. A more accurate description of the glass properties is achievable with the aid of the relative partial dispersions. The relative partial dispersion Px,y for the wavelengths x and y is defined by the equation: P x,y = (n x – ny )/(n F – n C ) (5) As Abbe demonstrated, the following linear relationship will approximately apply to the majority of glasses, the so-called ”normal glasses” Px,y ≈ a xy + b xy · nd (6) axy and bxy are specific constants for the given relative partial dispersion. In order to correct the secondary spectrum (i. e. color correc - tion for more than two wavelengths) glasses are required which do not conform to this rule. Therefore glass types having deviating partial dispersion from Abbe’s empirical rule are especially interesting. As a measure of the deviation of the partial dispersion from Abbe’s rule the ordinate difference ΔP is introduced. Instead of relation (6) the following generally valid equation is used: Px,y = a xy + b xy · nd + ΔP x,y (7) The term ΔPx,y therefore quantitatively describes a dispersion behavior that deviates from that of the ”normal glasses”. The deviations ΔPx,y from the ”normal lines” are listed for the following five relative partial dispersions for each glass type in the data sheets. PC,t = (nC – n t)/(nF – nC ) P C,s = (nC – n s)/(nF – n C) P F,e = (nF – ne )/(nF – n C) (8) P g,F = (ng – n F)/(nF – n C) Pi,g = (ni – n g)/(nF – n C)The position of the normal lines is determined based on value pairs of the glass types K7 and F2. The explicit formulas for the deviations ΔPx,y of the above-mentioned five relative partial dis-persions are: Δ PC,t = (nC – n t)/(nF – nC ) – (0.5450 + 0.004743 · nd) Δ PC,s = (nC – ns )/(nF – nC ) – (0.4029 + 0.002331 · nd) ΔPF,e = (nF – n e)/(nF – n C) – (0.4884 – 0.000526 · nd) (9) ΔPg,F = (ng – n F)/(nF – n C) – (0.6438 – 0.001682 · nd) ΔPi,g = (ni – ng )/(nF – nC ) – (1.7241 – 0.008382 · nd)Figure 2 shows the Pg,F versus the Abbe number nd diagram. The relative partial dispersions listed in the catalog were cal culated from refractive indices to 6 decimal places. The dispersion formula (10) can be used to interpolate additional unlisted refractive indices and relative partial dispersions (see chapter 2.3). Back to index ➜➜ Refractive Index and Dispersion Technical Information Advanced Optics TIE-29 4 0.65 90 100 80 70 60 50 40 30 20 90 100 80 70 60 50 40 30 20 10 10 0.63 0.62 0.61 0.60 0.59 0.58 0.57 0.56 0.55 0.54 0.53 0.52 0.51 0.65 0.63 0.64 0.64 0.62 0.61 0.60 0.59 0.58 0.57 0.56 0.55 0.54 0.53 0.52 0.51 Pg,F = ng – n Fn F – n CPg,F = ng – n Fn F – n C νd νd February 2016 P-SK57Q1 F2 N-KZFS8 N-KZFS5 LAFN7 SF56A SF6 N-LAF21 N-LAF35 N-LASF44 N-LASF31A P-LASF47 N-LAF34 N-LAK10 N-LAF2 N-LASF41 N-LAF33 N-KZFS4 N-KZFS2 N-KZFS11 F5 N-LASF40 N-LAK21 N-LAK14 N-LAK9 N-LAK34 N-LAK8 K10 N-LAK22 N-SK2 N-SK4 K7 N-BAK4 N-BALF4 N-KF9 N-SSK5 N-SSK8 N-FK58 N-BALF5 N-SSK2 N-LAK12 N-BAK1 P-SK57 N-LAK7 N-SK16 N-SK14 N-SK5 N-SK11 N-ZK7 N-BAK2 N-K5 N-BK10 N-FK5 N-PSK3 N-BK7 N-PSK53A LF5 N-BASF64 N-BASF2 N-SF5 N-SF1 SF10 N-SF8 N-SF15 P-SF8 N-SF10 N-SF4 SF57 N-SF6 N-SF11 N-SF57 N-SF14 N-LASF9 N-SF2 SF2 SF5 N-LAF7 N-LASF45 N-F2 LLF1 N-BAF51 N-BAF4 N-BAF52 N-BAF10 N-FK51A N-PK51 N-PK52A N-SF66 SF4 P-SF68 P-SK58A N-LASF43 P-LASF50 P-LASF51 P-LAK35 P-SK60 P-SF69 P-LAF37 N-LAK33B P-BK7 SF11 SF1 LASF35 N-LASF46B N-LASF46A Pg,F -Diagram Description of Symbols N-glasses P-glasses Classical crown and flint glasses Glasses suitable for precision molding HT – High transmittance glasses HTultra – Ultra high transmittance glasses Available in step 0.5 Back to index ➜➜ Refractive Index and Dispersion Fig. 2: Pg,F as a function of the Abbe number for SCHOTT’s optical glass assortment. Additionally the normal line is given. Technical Information Advanced Optics TIE-29 5 2.3 Sellmeier Dispersion Equation The Sellmeier Equation is especially suitable for the progres-sion of refractive index in the wavelength range from the UV through the visible to the IR area (to 2.3 μm). It is derived from the classical dispersion theory and allows the description of the progression of refractive index over the total transmis-sion region with one set of data and to calculate accurate intermediate values. n 2( λ) – 1 = B1 · λ2 B 2 · λ2 B 3 · λ2 (10) ( λ2 – C 1) ( λ2 – C 2) ( λ2 – C3 )The determination of the coefficients was performed for all glass types on the basis of precision measurements by fitting the dispersion equation to the measurement values. The coefficients are listed in the data sheets. The dispersion equation is only valid within the spectral region in which refractive indices are listed in the data sheet of each glass. Interpolation is possible within these limits. The wave-lengths used in the equation have to be inserted in μm with the same number of digits as listed in Table 1. For prac tical purposes Equation (10) applies to refractive indices in air at room temperature. The achievable precision of this calculation is generally better than 1 · 10 – 5 in the visible spectral range. The coefficients of the dispersion equation can be reported for individual glass parts upon request. This requires a precision measurement for the entire spectral region, provided the glass shows sufficient transmission. Back to index ➜➜ Refractive Index and Dispersion Wavelength [nm] Designation Spectral Line Used Element 2325.42 infrared mercury line Hg 1970.09 infrared mercury line Hg 1529.582 infrared mercury line Hg 1060.0 neodymium glass laser Nd 1013.98 t infrared mercury line Hg 852.11 s infrared cesium line Cs 706.5188 r red helium line He 656.2725 C red hydrogen line H 643.8469 C' red cadmium line Cd 632.8 helium-neon-gas-laser He-Ne 589.2938 D center of double sodium line Na 587.5618 d yellow helium line He 546.074 e green mercury line Hg 486.1327 F blue hydrogen line H 479.9914 F' blue cadmium line Cd 435.8343 g blue mercury line Hg 404.6561 h violet mercury line Hg 365.0146 i ultraviolet mercury line Hg 334.1478 ultraviolet mercury line Hg 312.5663 ultraviolet mercury line Hg 296.7278 ultraviolet mercury line Hg 280.4 ultraviolet mercury line Hg 248.3 ultraviolet mercury line Hg Tab. 1: Wavelengths for a section of frequently used spectral lines. Technical Information Advanced Optics TIE-29 6 Temperature [ºC] N-BK7 SF57 N-PK51 N-LAF2 F2 -100 -80 -60 -40 -20 020 40 60 80 100 120 140 15 10 50-5 -10 -15 temperature [ºC] ∆n rel [10 -4 ]N-BK7 SF57 N-PK51 N-LAF2 F2 -100 -80 -60 -40 -20 020 40 60 80 100 120 140 15 10 50-5 -10 -15 ∆n rel [10 -4 ] Temperature Dependence of Refractive Index The refractive index of glass is not only dependent on wavelength, but also on temperature. The relationship of refractive index change to temperature change is called the temperature coefficient of refractive index. This can be a positive or a negative value. The data sheets contain infor - mation on the temperature coefficients of refractive index for several temperature ranges and wavelengths. The temperature coefficients of the relative refractive indices Δnrel / ΔT apply for an air pressure of 0.10133 · 10 6 Pa. The coefficients of the absolute refractive indices dn abs /dT apply for vacuum. The temperature coefficients of the absolute refractive indices can be calculated for other temperatures and wavelengths values with the aid of Equation (11). Definitions: T0 Reference temperature (20 °C) T Temperature (in °C) ΔT Temperature difference versus T 0λ Wavelength of the electromagnetic wave in a vacuum (in μm) D 0, D 1, D 2, E0, E 1 and λTK : constants depending on glass type The constants of this formula given in the glass data sheet in general are valid for a temperature range from –100 °C to +140 °C and a wavelength range from 0.3650 μm to 1.014 μm .The temperature coefficients in the data sheets are guideline values. Upon request, measurements can be performed on individual melts in the temperature range from –100 °C to +140 °C and in the wavelength range from 0.365 μm to 1.014 μm with a precision better than ± 5 · 10 – 7/K. The constants of the disper-sion formula are also calculated from the measurement data and listed on the test certificate. dn abs ( λ,T )=n 2 ( λ,T 0) – 1 · (D 0 + 2 · D 1 · ΔT + 3 · D 2dT 2 · n ( λ,T 0) · ΔT 2 + E0 + 2 · E 1 · ΔT) (11) λ2 – λ22K The temperature coefficients of the relative refractive indices Δnrel / ΔT and the values for Δnabs can be calculated with the help of the equations listed in Technical Information TIE 19. Figure 3 shows the absolute temperature coefficient of refractive index for different glasses, temperatures and wavelengths. Back to index ➜➜ Refractive Index and Dispersion Fig. 3 : Change of the relative refractive index (catalog value at 20 °C) with temperature of some different glass types as shown in TIE-19. Technical Information Advanced Optics TIE-29 7 Influence of the Fine Annealing Process on the Refractive Index and Abbe number The optical data for a glass type are chiefly determined by the chemical composition and thermal treatment of the melt. The annealing rate in the transformation range of the glass can be used to influence the refractive index within certain limits (depending on the glass type and the allowable stress birefrin-gence). Basically slower annealing rates yield higher refractive indices. In practice, the following formula has proven itself. n d (h x) = n d (h 0) + m nd · log(h x/h 0) (12) h 0 Original annealing rate h x New annealing rate m nd Annealing coefficient for the refractive index depending on the glass type An analogous formula applies to the Abbe number. nd (h x) = nd (h 0) + m nd · log(h x/h 0) (13) mnd Annealing coefficient for the Abbe number depending on the glass type The annealing coefficient mvd can be calculated with the following equation: mnd = (m nd – nd [h 0] · mnF – nC ) / ((n F – nC ) + 2 · mnF – nC · log (h x /h 0 )) (14) The coefficient m nF – nC has to be determined experimentally. The refractive index and Abbe number dependence on annealing rate is graphically shown in Figure 4. Back to index ➜➜ Refractive Index and Dispersion Fig. 4 : Dependence of refractive index nd (a: upper) and Abbe number nd (b: lower) on the annealing rate for several glass types. Reference annealing rate is 2 °C/h . Figure 4b shows that individual glass types vary greatly in their dependence of the Abbe number on the annealing rate. In general also the Abbe number increases with decreasing annealing rate. High index lead free glass types such as N-SF5 show anomalous behavior. Anomalous behavior means that the Abbe number decreases with decreasing annealing rate. ∆n d [10 –5 ] Annealing rate [ºC/h] 300 200 100 0–100 0.1 1 10 N-BK7 N-KZFS4 N-SF5 SF2 ∆v d [%] 00– 0 –0 ∆n d [10 –5 ] Annealing rate [ºC/h] 300 200 100 0–100 0.1 1 10 N-BK7 N-KZFS4 N-SF5 SF2 ∆v d [%] Annealing rate [ºC/h] 0.3 0.0 – 0.3 –0.6 0.1 1 10 N-BK7 N-SF5 SF2 N-KZFS4 Technical Information Advanced Optics TIE-29 8 ∆n d[10 –5 ] ∆v d [%] –100 –80 –60 –40 –20 020 40 60 80 100 0.6 0.4 0.2 0.0 –0.2 –0.4 –0.6 glass for pressing glass for fine annealing 10 8 6410.6 0.8 0.1 0.2 0.4 0.6 0.8 146802°C/h 2°C/h Step 3 Step 2 Step 1 Step 0.5 Values for a annealing coefficients of some optical glasses are shown in Table 2. We will provide the values for the annealing coefficients of our glasses upon request. mnd mnF–nc mnd SCHOTT N-BK7 ®– 0.00087 – 0.000005 – 0.0682 SF2 – 0.00056 0.000013 – 0.0523 N-SF5 – 0.00242 – 0.000182 0.1654 Back to index ➜➜ Refractive Index and Dispersion for a given glass part can be adjusted by a fine annealing step along this characteristic line. Glass for cold processing has to be fine annealed to reduce internal stresses. During this fine annealing the annealing rate is in general lower than 2 °C/h. The initial refractive index has to be adjusted during melting in such a way that the desired tolerances can be reached during fine annealing. The initial refractive index of SCHOTT N-BK7 ® for example is in general lower than the target value. Glass for hot processing i.e reheat pressing is subjected to much more rapid annealing. The heat treatment processes used by the customer in general use annealing rates much higher than 2 °C/h. Therefore for SCHOTT N-BK7 ® pressings for example the initial refractive index needs to be higher than the target value. We deliver an annealing schedule for each batch of glass for hot processing purpose. This annealing schedule contains the initial refractive index at 2 °C/h and the limit annealing rates to stay within the tolerances. The annealing rate can be used to adjust the refractive index and Abbe number to the desired tolerance range. In practice the annealing rate influences the refractive index and the Abbe number simultaneously. Figure 5 shows a diagram of the Abbe number versus the refractive index for SCHOTT N-BK7 ® . The rectangular boxes indicate the toler-ance limits (steps) for the refractive index and the Abbe num-ber. For example the largest box indicates the tolerance bor-ders for step 3 in refractive index and step 3 in Abbe number. The smallest box indicates step 0.5 in refractive index and Abbe number. The center of the frames is defined by the nominal catalog value. After melting the optical glass is cooled down at a high anneal - ing rate. To control the refractive index during the melting process samples are taken directly from the melt after each casting. These samples are cooled down very fast together with a reference sample of the same glass. The reference sample has a known refractive index at an annealing rate of 2 °C/h. By measuring the change in refractive index of the reference sample the refractive index of the sample can be measured with moderate accuracy in the range of ± 10 – 4.The annealing rate dependence of the Abbe number and refractive index of each glass is represented by a line in the diagram having a slope that is characteristical for the glass type. For a given melt the position of the line in the diagram is given by the initial refractive index/Abbe number measure-ment for a cooling rate of 2 °C/h as a fix-point together with the glass typical slope. The refractive index and Abbe number Tab. 2: Annealing coefficients for several selected glass types. Fig. 5: The influence of the annealing rate on the refractive index and Abbe number of SCHOTT N-BK7 ®for different initial refractive indices. Technical Information Advanced Optics TIE-29 9 ∆n from catalogue value [10 – 5 ]35 30 25 20 15 10 50Batch No. 0510 15 20 25 Position of delivery lot = value in test report = (n max + n min )/2 (mid-level value of delivered batches) Variation in delivery lot: e.g. SN < ± 10 · 10 –5 around position mid-level value refractive index homogeneity within a single piece: e.g. H2 < 2 · 10 –5 Back to index ➜➜ Refractive Index and Dispersion Tolerances The refractive indices, which are listed to 5 decimal places in the data sheets, represent values for a melt with nominal n d- nd position for the glass type in question. The refractive index data are exact to five decimal places (for λ > 2 μm: ± 2 · 10 – 5). The accuracy of the data is less in wavelength regions with limited transmission. All data apply to room temperature and normal air pressure (0.10133 · 10 6 Pa). Defining tolerances for the refractive index of a glass the customer has to distinguish between the refractive index tolerance, the tolerance of refractive index variation within a lot and the refractive index homogeneity (Figure 6). All deliveries of fine annealed block glass and fabricated glass are made in lots of single batches. The batch may be a single block or some few strip sections. The refractive index and Abbe number tolerance is the maxi-mum allowed deviation of a single part within the delivery lot from nominal values given in the data sheets of the catalog. The refractive index of the delivery lot given in the standard test certificates is given by the following formulae: nlot = (n max + n min )/2 (15) n max is the maximum and nmin the minimum refractive index within the lot. Fig. 6: Refractive index variation from within a production sequence. Tab. 3: Refractive Index Tolerances. only for selected glass types The refractive index variation from part to part within a lot is always smaller than ± 5 · 10 – 5 . The refractive index homo - geneity within a single part is better than 4 · 10 – 5 ISO 12123 (equivalent to ± 2 · 10 – 5 according ISO 10110) in general. [4; TIE-26] A short summary of the refractive index tolerance, variation and homogeneity grades can be found in table 3. More information is given in the optical glass catalogue . Tolerance Grade Refractive Index [· 10 – 5] Abbe Number Absolute Step 3 ± 50 ± 0.5 %Step 2 ± 30 ± 0.3 %Step 1 ± 20 ± 0.2 %Step 0.5 ± 10 ± 0.1 % Variation SN ± 10 –S0 ± 5 –S1 ± 2 – Homogeneity ISO 12123 (ISO 10110) H1 4 (± 2) –H2 1 (± 0.5) –H3 0.4 (± 0.2) –H4 0.2 (± 0.1) –H5 0.1 (± 0.05) – Technical Information Advanced Optics TIE-29 10 Back to index ➜➜ Refractive Index and Dispersion 5.1 Tolerance Step 0.5 SCHOTT is the only manufacturer to offer optical glasses in the new tolerance step 0.5. For step 0.5 the maximum allowed deviation from the nominal values listed in the datasheet is only ± 0.0001 for the refractive index n d and ± 0.1 % for the Abbe number nd. This new tolerance grade can be offered only for selected glass types. Among these glass types are the following glasses: N-FK51A, N-PK51, N-FK5, SCHOTT N-BK7 ®,N-BK7HT, N-PSK53A, N-SK2, N-KZFS2, N-KZFS4, N-KZFS4HT, N-KZFS5, N-KZFS8, N-KZFS11, N-LAK9, N-LASF44, SF2, SF57HTUltra and N-SF5. The list comprises glass types, which are especially important for best color correction such as FK-, PK- and KZFS-types and which are often required in narrowest possible tolerances. Even though SCHOTT commits itself to be capable to deliver tolerance step 0.5 for the range of the listed 5.2 Tolerance of the refractive index of optical glasses in the near IR The question often arises if the tolerances for the refractive index at standard wavelength d can be transferred to longer wavelengths in the spectrum. This is approximately sure for wavelengths within the visible spectral range, whereas for wavelengths in the near IR range up to 1.7 μm one has to look a little bit more closely on the variation of the dispersion for different melts of different glass types. A first evaluation was published in , comparing extrapolations from v-block meas-urements in the visible to the IR range with precision spec-trometer measurements (URIS). Glas types evaluated were SCHOTT N-BK7 ®, N-KZFS4, N-PK51, N-PK52A, N-SF6, SF57, N-LAK22, N-LASF31A, F2 and N-SF57. The phenomenology of dispersion curves and the comparison of predicted with measured refractive index curves in the near infrared wavelength range lead to the following observations, glass types, it is possible that other glass types are available also in the narrowest tolerances. This will be checked on request. The fine annealing process has to be done with furnaces with spe-cial precision temperature control and high temperature field homogeneity. Therefore step 0.5 glasses are available as fine annealed cut blanks and most of the glass types are also avail-able as reheat pressings. All Step 0.5 tolerance grades are verified using the v-block with enhanced accuracy measurement procedure. For glasses with low dispersion (high Abbe number) like N-FK51A and N-PK51 the v-block contains an additional reference samples measured to an accuracy of ± 0.4 · 10 – 5 with the prism spec-trometer (compare table 4). • Melts with refractive index values very close to each other in the visible light range may have significant deviations in the near infrared wave length range. Slope and curvature can differ among melts and also with respect to the catalog curve. • Extrapolations from test certificate data obtained in the visible light range improved by using catalog relative partial dispersion data can lead to deviations from real measured values up to ± 5 · 10 – 5 or in some cases even more above 1.7 μm wavelength. • Using Abbe number nd, defined for the directly neighboring IR range with the spectral lines C, t and s deviations of extrapolated dispersion curves reduce roughly by a factor of two above 1.7 μm. • For better extrapolation it is recommended to perform a v-block measurement with enhanced accuracy because of its extended wavelength range. Technical Information Advanced Optics TIE-29 11 Back to index ➜➜ Refractive Index and Dispersion Refractive Index Measurement For refractive index measurement two different measurement setups are used: the v-block refractometer and the spectral goniometer. Figure 7 shows the principle of the v-block measurement. The samples are shaped in a nearly square shape. One sample is about 20 x 20 x 5 mm small. The sample will be placed in a v shaped block prism. The refractive index of this prism is known very precisely. The refraction of an in -coming light beam depends on the refractive index difference between the sample and the v-block-prism. The advantage of this method is that up to 10 samples can be glued together into one v-block stack. Therefore many samples can be meas-ured in a very short time. The relative measurement accuracy is very high therefore differences in refractive index within one v-block stack can be measured very accurately. Standard measurement temperature is 22 °C. The spectral goniometric method is based on the measure-ment of the angle of minimum refraction in a prism shaped sample. This is the most accurate absolute refractive index measurement method. In our laboratory we use an automated spectral goniometer with high accuracy and the ability to measure in the infrared and UV region (Figure 8). With the automated spectral goniometer, the Ultraviolett to infrared Refractive Index measurement System (URIS), the refractive index of optical glasses can be measured to an accuracy of ± 0.4 · 10 – 5 . The measurement accuracy for the dispersion ( nF-n C) is ± 2 · 10 – 6. These measurement accuracies can be achieved independent of the glass type and over the complete wavelength range from 185 nm to 2325 nm. The measurement is based on the minimum angle of refraction principle. The samples are prism shaped with dimensions of about 35 x 35 x 25 mm 3. The standard measurement tem - perature is 22 °C. The temperature can be varied between 18 to 28 °C on request. The standard measurement atmosphere is air. On special request also nitrogen is possible. Fig. 7: V-block refractormeter principle. Fig. 8: Automated spectral goniometer. Immersion oil n air n air θ n V – block nsample Incident light ray at wavelength λ Optical axis Sample with higher refractive index Sample with lower refractive index Sample V-Block-Prism with precisely known index of refraction Technical Information Advanced Optics TIE-29 12 Advanced Optics SCHOTT AG Hattenbergstrasse 10 55122 Mainz Germany Phone +49 (0)6131/66-1812 Fax +49 (0)3641/2888-9047 info.optics@schott.com www.schott.com/advanced_optics Literature The properties of optical glass H. Bach & N. Neuroth (Editors), Springer Verlag 1998 Temperature Coefficient of the Refractive Index SCHOTT Technical Information TIE-19 Optical glass with tightest refractive index and disper-sion tolerances for high-end optical designs, Ralf Jedamzik, Steffen Reichel, Peter Hartmann, SPIE Proceeding 8982-51, (2014) Homogeneity of Optical Glass SCHOTT Technical Information TIE-26 SCHOTT Optical Glass Pocket Catalogue Optical glass: dispersion in the near infrared Peter Hartmann, SPIE Proceedings Vol. 8167, (2011) V-Block refractometer for monitoring the production of optical glasses, U. Petzold, R. Jedamzik, P. Hartmann, and S. Reichel, Proc. SPIE 9628, (2015) Table 4 shows a summary of the refractive index measure-ments available at SCHOTT. The temperature coefficient of refractive index is measured using the automated spectral goniometer and a temperature controlled climate chamber with a temperature range from –100 °C up to +140 °C. The temperature coefficient can be measured with an accuracy of ± 5 · 10 –7 /K. With the introduction of step 0.5 tolerances also for low dis-persion glasses the question arises how accurate is the Abbe number measurement with the v-Block refractometer. Figure 9 shows the results of reproducibility measurement for the glass types N-FK58 and SF57. The plot shows the distribution of the n d and nd of SF57 (blue tilted crosses) and N-FK58 (green standing crosses) determined by V-block refractometer. The filled red signs correspond to the center of the V-block refrac-tometer measurements (SF57: circle and N-FK58: square). The average of several SCHOTT’s spectral goniometer URIS measurements (SF57: black diamants and N-FK58 orange circles) of the same piece of glass defines the origin of the plot. To give an impression of the deviation size the tolerance range of SCHOTT’s best commercially available quality step 0.5 (red dashed line) is added. The standard deviations of the V-block measurements of N-FK58 (SF57 respectively) are 0.2 · 10 – 5 (1.1 · 10 – 5) for the refractive index and 0.017 % (0.008 %) for the Abbe number. Systematic deviations can be corrected by implementing an URIS sample to the v-block stack. In the given example the systematic deviations are – 1.8 · 10 – 5 (0.2 · 10 – 5) for the refrac-tive index and – 0.011% (– 0.01 %) for the Abbe number of N-FK58 (SF57 respectively) . Back to index ➜ Refractive Index and Dispersion Measurement Measurement accuracy Wavelengths Method Refractive index Dispersion V-block standard ± 3 · 10 – 5 ± 2 · 10 – 5 g, F', F, e, d, C', C v-block refractometer V-block enhanced ± 2 · 10 – 5 ± 1 · 10 – 5 i, h, g, F', F, e, d, C', C, r, t Precision spectrometer ± 0.4 · 10 – 5 ± 0.2 · 10 – 7 185 nm – 2325 nm URIS auto-matic spectral goniometer Tab. 4: Absolute refractive index measurement accuracies. Fig.9: Reproducibility of V-block refractometer. Red box indicates the toler-ance step. In order to give an impression of the deviation size we added the tolerance range of SCHOTT’s best commercially available quality step 0.5 (red dashed line). Technical Information Advanced Optics TIE-29 Version April 2016 \| SCHOTT Advanced Optics reserves the right to make specification changes in this product flyer without notice.
14278	https://www.chegg.com/homework-help/questions-and-answers/exercise-215-let-denote-number-subsets-1--n-containing-two-consecutive-integers-thus-examp-q85990119	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Exercise 2.15 (a) Let In denote the number of subsets of {1,...,n} containing no two consecutive integers. Thus, for example, 91 = 2 (include the empty ( set!) and 92 = 3. Find a recurrence relation for In, and deduce that In = Fn+1. (b) A k-element subset of {1,...,n} can be considered as a binary se- quence of length n containing k ls and n- k 0s (see P.S. Please see the following C… Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
14279	https://arxiv.org/pdf/1503.01166	5th Fermi Symposium : Nagoya, Japan : 20-24 Oct, 2014 Primordial Black Holes Jane H. MacGibbon Dept of Physics, University of North Florida, Jacksonville, FL, 32224 USA Tilan N. Ukwatta Space and Remote Sensing (ISR-2) and Physics Division (P-23), Los Alamos National Laboratory, Los Alamos, NM, 87544, USA J.T. Linnemann, S.S. Marinelli, D. Stump, K. Tollefson Dept of Physics and Astronomy, Michigan State University, East Lansing, MI, 48824 USA Primordial Black Holes (PBHs) are of interest in many cosmological contexts. PBHs lighter than about 10 12 kg are predicted to be directly detectable by their Hawking radiation. This radiation should produce both a diffuse extragalactic gamma-ray background from the cosmologically-averaged distribution of PBHs and gamma-ray burst signals from individual light black holes. The Fermi, Milagro, Veritas, HESS and HAWC observatories, in combination with new burst recognition methodologies, offer the greatest sensitivity for the detection of such black holes or placing limits on their existence. INTRODUCTION A black hole (BH) is an object of classical gravity whose mass 𝑀𝑀 𝐵𝐵𝐵𝐵 is contained within its Schwarzschild volume which has radius 𝑟𝑟 𝐵𝐵𝐵𝐵 = 2𝐺𝐺𝑀𝑀 𝐵𝐵𝐵𝐵 𝑐𝑐 2 (1) . Here 𝐺𝐺 is the universal gravitational constant, 𝑐𝑐 is the speed of light and we have assumed that the BH has negligible rotation and/or electric charge. (Extension in General Relativity to include rotation and/or electric charge is straightforward.) Because Eq (1) implies that the average density inside a black hole goes as 𝜌𝜌 𝐵𝐵𝐵𝐵 ∝𝑀𝑀 𝐵𝐵𝐵𝐵 /𝑟𝑟 𝑠𝑠 3 ∝ 𝑀𝑀 𝐵𝐵𝐵𝐵 −2 , large mass black holes may be more easily produced than small mass black holes, at least in the present universe. In fact a 10 8 𝑀𝑀 ⊙ black hole has the density of water. Today there is strong evidence for the existence of stellar mass black holes (formed as supernova remnants) and 10 6 𝑀𝑀 ⊙-10 10 𝑀𝑀 ⊙ supermassive black holes in most galactic centers. There is also mounting evidence for black holes with masses intermediate between stellar mass black holes and supermassive black holes. ‘Primordial Black Hole’ (PBH) refers to a black hole of any size formed in the early universe (where by ‘early universe’ we mean before the formation of the first stars). Possible PBH formation mechanisms include the collapse of overdense regions arising from primordial density inhomogeneities (such as occur in many Inflation models, in particular those with a blue, peaked or ‘running index’ spectrum), an epoch of low pressure (soft equation of state), or cosmological phase transitions; and mechanisms involving topological defects, such as cosmic strings oscillating into their Schwarzschild volume or the collapse of domain walls. (For a recent review of PBH formation mechanisms and limits see and references therein.) In almost all scenarios, the PBH mass at the time of formation is roughly equal to, or smaller than, the cosmic horizon (or Hubble) mass 𝑀𝑀 𝐵𝐵 ≈ 10 15 (𝑡𝑡 /10 −23 s)g . Thus the range of possible PBH initial masses is enormous – from the Planck mass for PBHs forming around the Planck time, to 10 5 𝑀𝑀 ⊙ for PBHs forming around 1 s, or larger if forming later. Within a particular formation scenario, usually the PBHs are produced over a narrow initial mass range. An exception is scale-invariant cosmological primordial density perturbations which could produce PBHs over an extensive initial mass range with an initial mass spectrum of the form 𝑑𝑑𝑑𝑑 /𝑑𝑑𝑀𝑀 𝑖𝑖 ∝ 𝑀𝑀 𝑖𝑖 −𝛼𝛼 where 𝛼𝛼 = 5/2 for formation in the radiation era. Although scale-invariant density perturbations are not as well motivated in present cosmological models as they were a couple of decades ago, gamma-ray limits on the present cosmologically-averaged number density of PBHs were earlier derived assuming an 𝑀𝑀 𝑖𝑖 −5 /2 initial mass function. The formation constraints on PBHs inform us about cosmology. The PBHs themselves may also produce effects on cosmological scales. PBHs surviving today should behave as cold dark matter (CDM). (In fact, present limits allow 10 17 − 10 26 g PBHs to contribute all of Ω𝐶𝐶𝐶𝐶 𝑀𝑀 .) Like other CDM, PBHs should cluster in galactic haloes. They may also enhance the clustering of other dark matter, for example in WIMP and Ultra Compact Massive Halo scenarios. If a stable state such as a Planck mass relic remains after low mass PBHs have expired, the relics themselves are CDM candidates. PBHs may have played a role in the development of cosmological entropy, baryogenesis, the reionization of Universe in earlier epochs and producing observable annihilation lines. Very large PBHs may influence large scale structure development, seed SMBHs, or generate observable cosmic x-rays in their accretion disks. The number of PBHs formed with initial masses of 10 9 − 10 43 g have been constrained primarily by primordial nucleosynthesis, cosmic microwave background (CMB) anisotropies, MACHO searches and, in the case of 𝑀𝑀 𝐵𝐵𝐵𝐵 ≲ 10 17 g BHs, the search for Hawking radiation. Hawking radiation constraints derived from the 100 MeV extragalactic gamma-ray background and Galactic gamma-ray, e +, e - and anti-proton backgrounds place an upper limit on the background distribution of 𝑀𝑀 𝐵𝐵𝐵𝐵 ≈ 5 × 10 14 g PBHs of roughly Ω𝑃𝑃 𝐵𝐵 𝐵𝐵 ≲ 10 −9 . Direct searches for the final eConf C141020.1 5th Fermi Symposium : Nagoya, Japan : 20-24 Oct, 2014 gamma-ray burst of Hawking radiation from an expiring PBH allow us to directly constrain the local number density of 𝑀𝑀 𝐵𝐵𝐵𝐵 ≈ 5 × 10 14 g PBHs and much lighter BHs. BLACK HOLE BURSTS 2.1. Black Hole Thermodynamics The work by Hawking and Beckenstein in the 1970’s on extending the Laws of Classical Thermodynamics to include black holes (i.e. Classical Gravitation) resulted in the recognition of the Hawking (Gravitational) temperature 𝑇𝑇 𝐵𝐵𝐵𝐵 𝑘𝑘𝑇𝑇 𝐵𝐵𝐵𝐵 = ћ𝑐𝑐 38𝜋𝜋 𝐺𝐺 𝑀𝑀 𝐵𝐵𝐵𝐵 = 1.06 � 𝑀𝑀 𝐵𝐵𝐵𝐵 10 13 g �−1 GeV (2) where k and ћ are the Boltzmann and reduced Planck constants, respectively . An 𝑀𝑀 ⊙ black hole has a temperature of 10 -7 K; a 10 25 g black hole has the same temperature as the present CMB; and a 10 11 g black hole has a temperature of ~ 100 GeV. Hawking also derived the thermal flux radiating from a black hole of temperature 𝑇𝑇 𝐵𝐵𝐵𝐵 to be 𝑑𝑑 2𝑁𝑁 𝑠𝑠 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = Γ𝑠𝑠 2𝜋𝜋ћ �exp � 𝑑𝑑 𝑘𝑘𝑇𝑇 𝐵𝐵𝐵𝐵 � − (−1) 2𝑠𝑠 � −1 (3) per particle degree of freedom where Q is the energy of the Hawking-radiated particle, s is the particle spin and Γ𝑠𝑠 is the absorption probability . In the geometric optics (short-wavelength) limit, Γ𝑠𝑠 ≈ 27 𝐺𝐺 2 𝑀𝑀 𝐵𝐵𝐵𝐵 2 𝑄𝑄 2/ ћ 2 𝑐𝑐 6. Strictly Eqs (2) and (3) apply for a non-rotating, non-electrically charged black hole. Extension to a black hole with angular momentum and/or electric field is straightforward but because a small black hole emits its angular momentum and electric charge quickly compared to cosmological timescales we will assume PBHs surviving today have negligible angular momentum and electric field. In the standard (MacGibbon-Webber) emission picture, a black hole should directly Hawking-radiate those particles which appear non-composite compared to the wavelength of the radiated energy (or equivalently the black hole size) at a given 𝑇𝑇 𝐵𝐵𝐵𝐵 . In order of increasing 𝑇𝑇 𝐵𝐵𝐵𝐵 , a s 𝑇𝑇 𝐵𝐵𝐵𝐵 surpasses successive particle rest mass thresholds, the black hole initially directly emits photons (and gravitons), then neutrinos, electrons, muons and eventually direct pions. Once 𝑇𝑇 𝐵𝐵𝐵𝐵 ≿ 𝛬𝛬 𝑑𝑑𝐶𝐶𝐶𝐶 ≈ 200 − 300 MeV , the QCD confinement scale, the black hole should directly Hawking-radiate, not pions which are now composite at such temperatures, but quarks and gluons. Analogous to QCD jet behaviour in accelerators, the quarks and gluons will subsequently shower and hadronize into the astrophysically stable species 𝛾𝛾 , 𝜈𝜈 , 𝑝𝑝 , 𝑝𝑝 ̅ , 𝑒𝑒 − and 𝑒𝑒 + as they stream away from the black hole .Because of the large number of degrees of freedom for the fundamental QCD particles, the instantaneous emission spectra from 𝑇𝑇 𝐵𝐵𝐵𝐵 > 𝛬𝛬 𝑑𝑑𝐶𝐶𝐶𝐶 black holes are dominated by the component produced by the decay of the Hawking-radiated QCD particles. The instantaneous photon flux from a 𝑇𝑇 𝐵𝐵𝐵𝐵 > 𝛬𝛬 𝑑𝑑𝐶𝐶𝐶𝐶 black hole is dominated by this secondary QCD photon component while the directly Hawking-radiated photons contribute, at a given 𝑇𝑇 𝐵𝐵𝐵𝐵 , significantly only at the highest energies . For 𝑇𝑇 𝐵𝐵𝐵𝐵 = 0.3 − 100 GeV black holes, the total instantaneous fluxes of the final-state stable particles are 𝑁𝑁 ̇𝑝𝑝𝑝𝑝 ̅ ≈ 2.1(±0.4) × 10 23 �𝑇𝑇 𝐵𝐵𝐵𝐵 GeV �1.6±0.1 s−1 𝑁𝑁 ̇ 𝑒𝑒 ± ≈ 2.0(±0.6) × 10 24 �𝑇𝑇 𝐵𝐵𝐵𝐵 GeV �1.6±0.1 s−1 𝑁𝑁 ̇𝛾𝛾 ≈ 2.2(±0.7) × 10 24 �𝑇𝑇 𝐵𝐵𝐵𝐵 GeV �1.6±0.1 s−1 𝑁𝑁 ̇𝜈𝜈𝜈𝜈 � ≈ 5.6(±1.7) × 10 24 �𝑇𝑇 𝐵𝐵𝐵𝐵 GeV �1.6±0.1 s−1 And the average energies of the fluxes scale as roughly 𝑇𝑇 𝐵𝐵𝐵𝐵 0.5, not as 𝑇𝑇 𝐵𝐵𝐵𝐵 (as for the directly Hawking-radiated components) . Thus, even very high temperature black holes will produce significant fluxes of final state particles which have energies around 100 MeV – 1 TeV. As the black hole Hawking-radiates, its mass is carried off by the mass-energy of the emitted particles. The black holes mass loss rate is thus 𝑀𝑀̇ 𝐵𝐵𝐵𝐵 ≈ − 5.34 × 10 25 𝑓𝑓 (𝑀𝑀 𝐵𝐵𝐵𝐵 )( 𝑀𝑀 𝐵𝐵𝐵𝐵 /g) −2 g s −1 (4) where the weight 𝑓𝑓 (𝑀𝑀 𝐵𝐵𝐵𝐵 ) ac counts for the total number of directly emitted states and is normalized to unity for 𝑀𝑀 𝐵𝐵𝐵𝐵 ≫ 10 17 g black holes which emit only photons and the three neutrino species. The relativistic contributions to 𝑓𝑓 (𝑀𝑀 𝐵𝐵𝐵𝐵 ) per particle degree of freedom are 𝑓𝑓 𝑠𝑠=0 =0.267 , 𝑓𝑓 𝑠𝑠=1 /2 = 0.147 (uncharged), 𝑓𝑓 𝑠𝑠=1 /2 = 0.142 (charge 𝑒𝑒 ±), 𝑓𝑓 𝑠𝑠=1 = 0.060 , 𝑓𝑓 𝑠𝑠=3 /2 = 0.020 , and 𝑓𝑓 𝑠𝑠=2 = 0.007 . For a 𝑇𝑇 𝐵𝐵𝐵𝐵 ≈ 50 GeV black hole emitting all experimentally-confirmed Standard Model degrees of freedom including the 125 GeV Higgs boson, 𝑓𝑓 (𝑀𝑀 𝐵𝐵𝐵𝐵 ) ≈ 15 . Integrating Eq (4), the remaining evaporation lifetime of an 𝑀𝑀 𝑖𝑖 black hole is then 𝜏𝜏 𝑒𝑒𝑒𝑒𝑒𝑒 𝑝𝑝 ≈ 6.24 × 10 −27 𝑓𝑓 (𝑀𝑀 𝑖𝑖 )−1 (𝑀𝑀 𝑖𝑖 /g) 3s. (5) The mass of a PBH whose evaporation lifetime equals the age of the universe is 𝑀𝑀 ∗ ≈ 5.00(±0.04) × 10 14 g . Comparison of the observed diffuse extragalactic gamma-ray background around 100 MeV with the gamma-ray background that would be produced by a cosmological distribution of 𝑀𝑀 ∗ ≈ 5 × 10 14 g PBHs places the strictest limit on an cosmologically-averaged distribution of 𝑀𝑀 ∗ PBHs. The limit, updated in 2010 using the Fermi LAT data, is Ω𝑃𝑃 𝐵𝐵 𝐵𝐵 (𝑀𝑀 ∗) ≲ 5 × 10 −10 . (This 𝑀𝑀 ∗ limit is stricter and more robust eConf C121028 5th Fermi Symposium : Nagoya, Japan : 20-24 Oct, 2014 Figure 1: The instantaneous gamma-ray flux 𝑑𝑑 2 𝑁𝑁 /𝑑𝑑 𝑡𝑡 𝑑𝑑𝑑𝑑 detectable by Fermi-LAT from 𝑇𝑇 𝐵𝐵𝐵𝐵 = 0.1 − 50 GeV black holes . For 𝑇𝑇 𝐵𝐵𝐵𝐵 in this range, the flux should remain approximately constant over the lifetime of the Fermi Observatory. than the limits on the cosmological distribution of PBHs of any other mass derived by this or any other method.) Because PBHs should behave as CDM, however, they should not be uniformly distributed throughout the universe but should cluster in galactic halos (and possibly also on smaller scales). Assuming PBH clustering in the Galactic halo, the local number density of PBHs should be enhanced by a factor of 𝜂𝜂 𝑙𝑙𝑙𝑙 𝑐𝑐𝑒𝑒 𝑙𝑙 ~2 × 10 5( Ωℎ𝑒𝑒𝑙𝑙𝑙𝑙 /0.1) −1 where Ωℎ𝑒𝑒𝑙𝑙𝑙𝑙 is the cosmological density parameter associated with galactic halos . Clustering in the Galaxy leads to the possibility that PBHs are contributing to the Galactic halo gamma-ray background (as investigated by Wright using EGRET observations ), matter-antimatter interactions and microlensing events. Comparisons of the spectra from a Galactic distribution of PBHs with the observed Galactic antiproton and positron backgrounds around 100 MeV lead to limits on a Galactic distribution of 𝑀𝑀 ∗ ≈ 5 × 10 14 g PBHs which are similar or somewhat weaker than the extragalactic gamma-ray limit. These antiproton- and positron-derived limits, however, depend on the modeling of the propagation and leakage times of charged particles in the Galaxy and on the Galactic distribution of PBHs, and so are not as robust as the extragalactic 100 MeV gamma-ray limit on the cosmologically-averaged distribution of PBHs. We note that the extragalactic and Galactic limits are derived using the black hole emission spectra integrated over both a distribution of PBHs and Galactic or cosmological timescales. 2.2 Signatures of Black Hole Bursts Independently we can derive limits by directly searching for the present emission from an individual black hole. Equally importantly, we can predict the light curve that would be produced in a detector by an individual black hole and devise methodologies to distinguish the BH burst signal from other known gamma-ray source types. Burst searches are the direct method for detecting black hole Hawking radiation and do not depend on assumptions concerning the formation mechanism of the black hole. In fact, burst searches are equally searches for any local small black holes created in the present universe, as well as primordially-produced PBHs. Although there are no currently-fashionable theories predicting the production of such small black holes in the present Galaxy, we should not bias ourselves observationally against their possible existence, given the widespread acceptance of the existence in the Galaxy and beyond of stellar mass and higher mass black holes. We should investigate the black hole burst signature template so that we can recognize BH/PBH bursts if they are seen in a detector. Let us now predict the black hole burst signature. Re-writing Eq (5), a black hole with temperature 𝑇𝑇 𝐵𝐵𝐵𝐵 has a remaining evaporation lifetime of eConf C141020.1 5th Fermi Symposium : Nagoya, Japan : 20-24 Oct, 2014 𝜏𝜏 𝑒𝑒𝑒𝑒𝑒𝑒 𝑝𝑝 ≈ 5.0 × 10 11 �𝑓𝑓 (𝑇𝑇 𝐵𝐵𝐵𝐵 ) 15 � −1 � 𝑇𝑇 𝐵𝐵𝐵𝐵 GeV �−3 s. (6) A 𝑇𝑇 𝐵𝐵𝐵𝐵 ≈ 1 GeV black hole has a remaining lifetime of ~ 16,000 yr; a 𝑇𝑇 𝐵𝐵𝐵𝐵 ≈ 10 GeV black hole has a remaining lifetime of ~ 20 yr; a 𝑇𝑇 𝐵𝐵𝐵𝐵 ≈ 25 GeV black hole has a remaining lifetime of ~ 1 yr; a 𝑇𝑇 𝐵𝐵𝐵𝐵 ≈ 300 GeV black hole has a remaining lifetime of ~ 1 hr; a 𝑇𝑇 𝐵𝐵𝐵𝐵 ≈ 2 TeV black hole has a remaining lifetime of ~ 100 s; and a 𝑇𝑇 𝐵𝐵𝐵𝐵 ≈ 20 TeV black hole has a remaining lifetime of ~100 ms. As can be seen from Eqs (2) and (4), the 𝑀𝑀 𝐵𝐵𝐵𝐵 ≪ 𝑀𝑀 ∗ black hole’s mass quickly decreases as it radiates and its temperature increases at an accelerating pace. Recall that the photons produced from the decays of the directly Hawking-radiated QCD particles dominate the net instantaneous photon flux from a 𝑇𝑇 𝐵𝐵𝐵𝐵 > 𝛬𝛬 𝑑𝑑𝐶𝐶𝐶𝐶 black hole and have an average energy that scales as roughly 𝑇𝑇 𝐵𝐵𝐵𝐵 0.5,not as 𝑇𝑇 𝐵𝐵𝐵𝐵 . Thus substantial numbers of 100 MeV – 10 TeV photons will be produced even during the final explosive stage of the black hole’s evaporative lifetime. With respect to detecting gamma-ray black hole bursts with the Fermi Observatory, there are 3 cases of BH signals that we need to consider: Case (i) The gamma-ray spectrum from a 3 MeV < 𝑇𝑇 𝐵𝐵𝐵𝐵 < 12 GeV black hole will appear to be almost constant as a function of time over the lifetime of the Fermi Observatory. (Recall that the remaining evaporation lifetime of a 𝑇𝑇 𝐵𝐵𝐵𝐵 = 10 GeV black hole is ~ 20 yrs.) Case (ii) The gamma-ray spectrum from a 12 GeV < 𝑇𝑇 𝐵𝐵𝐵𝐵 < 50 GeV black hole will evolve significantly as a function of time over the lifetime of the Fermi Observatory but almost all its gamma-ray flux arriving over that time will lie within the LAT detector’s energy range, 20 MeV - 300 GeV . (Recall that the remaining evaporation lifetime of a 𝑇𝑇 𝐵𝐵𝐵𝐵 = 50 GeV black hole is ~ 50 days.) Case (iii) The gamma-ray spectrum from a 𝑇𝑇 𝐵𝐵𝐵𝐵 >50 GeV black hole will be a quickly evolving burst with part of its flux arriving in the LAT energy range and significant flux at energies above the LAT range. In the final stages of burst evolution, the incoming flux will not be resolvable as a function of time and the time-integrated flux will be deposited in one time interval in the detector. (Recall that the remaining evaporation lifetime of a 𝑇𝑇 𝐵𝐵𝐵𝐵 = 170 TeV black hole is ~ 100 μs.) In Figure 1, we show the instantaneous gamma-ray flux 𝑑𝑑 2 𝑁𝑁 /𝑑𝑑 𝑡𝑡 𝑑𝑑𝑑𝑑 which would be seen by the LAT from 𝑇𝑇 𝐵𝐵𝐵𝐵 = 0.1 − 50 GeV black holes , relevant to Cases (i) and (ii). For black holes with these temperatures the flux is dominated by the photons resulting from the Hawking-radiated QCD particles. The gamma-ray flux from a 𝑇𝑇 𝐵𝐵𝐵𝐵 = 20 MeV black hole, which is below the threshold to emit a QCD component and whose photons are all directly Hawking-radiated, is shown in Figure 2 . Figure 2: The instantaneous gamma-ray flux from a 𝑇𝑇 𝐵𝐵𝐵𝐵 = 20 MeV black hole, which is below the threshold to emit a QCD component . Figure 3: Preliminary calculation for the PBH burst light curve 𝑑𝑑 𝑁𝑁 /𝑑𝑑 𝑡𝑡 arriving in the detector with energy above a given threshold, here 𝑑𝑑 𝛾𝛾 = 100 GeV . Figure 4: The gamma-ray spectrum 𝑑𝑑 𝑁𝑁 /𝑑𝑑𝑑𝑑 time-integrated over various remaining black hole evaporation lifetimes . eConf C121028 5th Fermi Symposium : Nagoya, Japan : 20-24 Oct, 2014 For Case (iii), we show in Figure 3 our preliminary calculation for the PBH burst light curve, i.e. the number of photons arriving per unit time with energy above a given threshold. (In Figure 3, the energy threshold is taken to be 𝑑𝑑 𝛾𝛾 = 100 GeV ). In Figure 4, we plot the gamma-ray spectrum time-integrated over various BH remaining evaporation lifetimes . In Table 1, we list a number of distinguishing characteristics to discern a black hole burst from other known GRB source types. In particular, the BH burst will show a soft-to-hard (that is, low average energy to high average energy) time evolution and will be non-repeating. If it is bursting in free space, it should not be accompanied by an afterglow, but generation of an afterglow may be possible if the black hole is bursting in an ambient high density plasma or ambient high magnetic field. Table 1: Differences between black hole burst signals and GRBs of known source types. Gamma -Ray Bursts (known GRB types) BH Bursts Detected at cosmological distances Local, unlikely to be detected from beyond Galaxy Most GRBs show hard-to-soft evolution Hard-to-soft evolution expected Hadrons not expected from GRBs Accompanied by hadronic bursts which may be detectable if local Gravitational wave signal expected No accompanying gravitational wave signal Time duration ranges from fractions of second to hours Time duration of burst most likely 1-100 seconds Fast Rise Exponential Decay (FRED) light curve Exponential Rise Fast Fall (ERFF) light curve X-ray, optical, radio afterglows expected No multi-wavelength photon afterglows unless in exotic ambient environment TeV emission unknown TeV spectra predicted Multi-peak time profile Single-peak time profile May be repeating No burst repetition If no black hole bursts are observed by a detector, the null detection implies an upper limit on the local number density of small black holes. An amalgamation of recent limits and limits which would be set by null detection with HAWC are shown in Figure 5. As a general statement, the strongest limits have been set by searching for bursts of about 1 – 100 s duration because the detector signal weakens for bursts of shorter duration and the background dampens signal recognition at longer duration. The advantages of the Fermi Observatory, are that it is not background-limited, it has good angular and time resolution, a wide field of view and a low energy threshold, and it is anticipated to have a very long operational lifetime. Preliminary limits derived from a search of Fermi LAT data to date for pairs of photons with an arrival interval shorter than the time expected for a Poisson-distributed photon background give an upper limit of 2 × 10 3pc −3 yr −1 on BH bursts of 10 5 s duration (corresponding to 𝑇𝑇 𝐵𝐵𝐵𝐵 ≿ 200 GeV and 𝑀𝑀 𝐵𝐵𝐵𝐵 ≲ 6 × 10 10 g) . The 𝑀𝑀̇ 𝐵𝐵𝐵𝐵 ∝ 𝑀𝑀 𝐵𝐵𝐵𝐵 −2 dependence of Eq (4) means that, for any population of black holes that have masses today around some 𝑀𝑀 𝐵𝐵𝐵𝐵 ≪ 𝑀𝑀 ∗ (i.e. that have remaining lifetimes much less than the age of the universe), the number of black holes per mass interval around 𝑀𝑀 𝐵𝐵𝐵𝐵 today is 𝑑𝑑𝑑𝑑 𝑑𝑑𝑀𝑀 𝐵𝐵𝐵𝐵 ∝ 𝑀𝑀 𝐵𝐵𝐵𝐵 2 (7) independent of the BH formation time, formation mechanism or spatial distribution . For black holes recently created with mass 𝑀𝑀 𝑖𝑖 ≪ 𝑀𝑀 ∗, the distribution (7) applies around the evolved mass 𝑀𝑀 𝐵𝐵𝐵𝐵 even if the initial mass distribution had initially been almost a delta function at 𝑀𝑀 𝑖𝑖 (because in reality there is always some smearing of such a delta function). In the case of PBHs with initial masses of 𝑀𝑀 𝑖𝑖 ~𝑀𝑀 ∗ created in the early universe, the distribution (7) applies today up to 𝑀𝑀 𝐵𝐵𝐵𝐵 ~𝑀𝑀 ∗ but the mass distribution with which the PBHs were initially created would still apply above 𝑀𝑀 ∗ today because 𝑀𝑀 𝑖𝑖 > 𝑀𝑀 ∗ PBHs have lost little mass over the history of the universe. Therefore, using Eq (7), we can extrapolate the burst search limits to derive a limit on the number of 𝑀𝑀 𝑖𝑖 ~𝑀𝑀 ∗ PBHs created in the early universe. All of the BH burst search limits to date when extrapolated up to 𝑀𝑀 𝑖𝑖 ~𝑀𝑀 ∗ correspond to limits on the cosmologically-averaged number density of 𝑀𝑀 ∗ PBHs which are weaker than the limit derived from the 100 MeV extragalactic gamma-ray background. For reasonable values of the enhancement due to CDM clustering in the Galaxy, the 100 MeV extragalactic limit on the cosmologically-averaged number density of 𝑀𝑀 ∗ PBHs corresponds to a local BH burst limit of ~10 pc −3 yr −1 . It should be noted, however, that the BH burst search limits are robust limits on the number density of small black holes close to Earth, regardless of their formation epoch or formation mechanism. Such black holes, if they are observed, are not necessarily the evolved state of 𝑀𝑀 𝑖𝑖 ~𝑀𝑀 ∗ PBHs formed in the early universe. Also, the assumptions concerning the clustering or spatial distribution of local BHs/PBHs used in the analysis may be incorrect, making detection in a given scenario more or less likely. 2.3 Further Comments on the Black Hole Burst Spectra In the above analysis, the black hole is assumed to Hawking-radiate only the experimentally-confirmed fundamental particle species of the Standard Model of particle physics. If further fundamental modes beyond the Standard Model exist, the extra modes may enhance eConf C141020.1 5th Fermi Symposium : Nagoya, Japan : 20-24 Oct, 2014 Figure 5: Limits on the local number density of black hole bursts in pc −3 yr −1 set by null-detection in previous burst searches, together with projected limits which would be set by null-detection at the HAWC Observatory . both the instantaneous flux from the black hole and the rate at which 𝑀𝑀 𝐵𝐵𝐵𝐵 decreases and 𝑇𝑇 𝐵𝐵𝐵𝐵 increase. This will shorten the black hole’s remaining evaporation lifetime and the duration of the final burst. If new fundamental modes appear only at temperatures well above 100 TeV, the overall effect on the predicted observable spectra is, most likely, negligible. A significant number of new fundamental modes at lower energies are postulated, though, in some extensions to the Standard Model but it is expected that the weighting factor in Eqs – remains of order 𝑓𝑓 ( 𝑀𝑀 𝐵𝐵𝐵𝐵 ) ≲ 100 . For example in Supersymmetry models, each 𝑠𝑠 = 1/2 fundamental particle has an 𝑠𝑠 = 0 superpartner and each 𝑠𝑠 = 1 fundamental particle has an 𝑠𝑠 = 1/2 superpartner, giving 𝑓𝑓 ( 𝑀𝑀 𝐵𝐵𝐵𝐵 ) ≲ 45 . The accompanying enhancement to the instantaneous flux and 𝑑𝑑 𝑁𝑁 /𝑑𝑑𝑑𝑑 spectra at a particular energy would depend on the actual decay processes of the new modes. A number of PBH burst scenarios invoking significant self-interaction of the Hawking-radiated particles in the vicinity of the black hole after emission have been proposed. If such self-interaction did occur after emission, it would not change the remaining evaporation lifetime but would decrease the average energy of the photons arriving at the detector, i.e. decrease the expected observable spectra at high energies and increase the spectra at low energies . Such photosphere models have recently been re-analyzed in detail and it has been strongly argued that the conditions for photosphere or quark-gluon plasma development are not met in the vicinity of the evaporating black hole . Specifically, the time interval between successive Hawking emissions, the damping of Hawking emission and the limited amount of energy per emission near a species’ rest mass threshold, and the Lorentz-transformed distance over which a scattered particle becomes ‘on-shell’ are such as to prevent the Hawking-radiated particles undergoing a significant number of QED or QCD interactions in the neighbourhood of the black hole. Hagedorn models which invoke an exponential increase in fundamental hadronic states around a limiting temperature of 𝑇𝑇 𝐵𝐵𝐵𝐵 ~𝑚𝑚 𝜋𝜋 , and which would produce a more detectable BH burst signal peaking at lower photon energy, are inconsistent both with accelerator experiments at these and higher energies (which confirm the quark model interpretation) and with the gravitational definition of 𝑇𝑇 𝐵𝐵𝐵𝐵 whose evolution is determined by the BH mass-energy loss rate. Hagedorn-type behaviour which may occur at extremely high energies in string theories would have negligible effect on the BH burst signal. Although photospheres produced by intrinsic self-interaction of the radiated particles in the vicinity of a stand-alone BH appear to be ruled out, it may be possible to produce a non-intrinsic photosphere or distortion of the burst signal if the BH is embedded in an ambient high density plasma or strong magnetic field. Such scenarios have not yet been modeled in detail. The standard emission model BH gamma-ray spectra also do not yet incorporate the recently-recognized inner bremsstrahlung (single-vertex bremsstrahlung) component which is expected to dominate the directly Hawking-radiated photon component below about 50 MeV . eConf C141020.1 5th Fermi Symposium : Nagoya, Japan : 20-24 Oct, 2014 SUMMARY There is strong motivation for investigating the possibility of detecting black hole burst signals. Detection of an evaporating black hole burst would be definitive experimental proof of the amalgamation of classical gravity with classical and quantum thermodynamics, pioneered by Hawking and Bekenstein. Equally importantly, the final stages of the evaporation process would open a direct observational window into particle physics at energies higher than can ever be achieved with terrestrial accelerators. For example, the black hole evaporation rate will be significantly increased if the supersymmetry modes exist. Details of the final stage of the BH burst may give insight into a quantum aspect of gravitation. Deviations of the BH burst signature from the predicted standard emission model spectra could also be used a probe of ambient extreme astrophysical environments. Detection or non-detection of PBHs give important constraints on the conditions in the early universe, in particular the amplitude and spectral index of initial density perturbations on smaller scales than are probed by the cosmic microwave background measurements. Thus, even if there is null-detection of BH bursts, there is strong motivation for improving the search limits and the implied upper limits on the number density of PBHs. Updated detailed modeling of the BH burst signal that could be observed by the Fermi Observatory and exploration of new search methodologies is currently ongoing. Acknowledgments JHM wishes to thank the organizers of the 5th Fermi Symposium for their hospitality. References A. Einstein, “Die Feldgleichungen der Gravitation”, Sitzungsberichte der Preussischen Akademie der Wissenschaften zu Berlin, 844 (1915); A. Einstein “Die Grundlage der allgemeinen Relativitäts-theorie”, Annalen der Physik 49, 769 (1916) K. Schwarzschild, “Über das Gravitationsfeld eines Massenpunktes nach der Einsteinschen Theorie”, Sitzungsberichte der Deutschen Akademie der Wissenschaften zu Berlin, Klasse fur Mathematik, Physik, und Technik, 189 (1916); K. Schwarzschild, “Über das Gravitationsfeld einer Kugel aus inkompressibler Flussigkeit nach der Einsteinschen Theorie”, Sitzungsberichte der Deutschen Akademie der Wissenschaften zu Berlin, Klasse fur Mathematik, Physik, und Technik, 189 (1916) B.J. Carr et al., “New cosmological constraints on primordial black holes”, Phys.Rev.D 81, 104019 (May 2010) S.W. Hawking, “Black hole explosions?”, Nature 248, 30 (Mar 1974); S.W. Hawking, “Particle creation by black holes”, Commun.math.Phys. 43, 199 (Apr 1975) D.N. Page, “Particle emission rates from a black hole: Massless particles from an uncharged ,nonrotating hole”, Phys.Rev.D 13, 198 (Jan 1976); D.N. Page, “Particle emission rates from a black hole. II - Massless particles from a rotating hole”, Phys.Rev.D 14, 3260 (Dec 1976); D.N. Page, “Particle emission rates from a black hole. III – Charged leptons from a nonrotating hole”, Phys.Rev.D 16, 2402 (Oct 1977) J.H. MacGibbon and B.R. Webber, “Quark- and gluon-jet emission from primordial black holes: The instantaneous spectra” Phys.Rev.D 41, 3052 (May 1990) J.H. MacGibbon, “Quark- and gluon-jet emission from primordial black holes. II. The emission over the black-hole lifetime” Phys.Rev.D 44, 376 (July 1991) J.H. MacGibbon, B.J. Carr and D.N. Page, “Do evaporating black holes form photospheres?” Phys.Rev.D 78, 064043 (Sept 2008) J.H. MacGibbon and B.J. Carr, “Cosmic ray from primordial black holes” Ap.J. 371, 447 (Apr 1991) E.L. Wright, “On the density of primordial black holes in the Galactic halo” Ap.J. 459, 487 (Mar 1996) A.A. Abdo et al., “Milagro limits and HAWC sensitivity for the rate-density of evaporating primordial black holes”, Astropart.Phys. 64, 4 (Feb 2015) T.N. Ukwatta et al., “Sensitivity of the Fermi detectors to gamma-ray bursts from evaporating primordial black holes (PBHs)” Proc.12th Marcel Grossmann Meeting, Paris, Jul 2009 D. Malyshev et al., “Expected Fermi -LAT limits on primordial black hole evaporation”, 5th Fermi Symposium, Nagoya, Oct 2014 A.F. Heckler, “Formation of a Hawking-radiation photosphere around microscopic black holes”, Phys.Rev.D 55, 480 (Jan 1997); A.F. Heckler, “Calculation of the emergent spectrum and observation of primordial black holes”, Phys.Rev.Lett. 78, 3430 (May 1997); R. Hagedorn, “Statistical thermodynamics of strong interactions at high-energies”, Nuovo Cim.Suppl. 3, 147 (1965) D.N. Page, B.J. Carr and J.H. MacGibbon, “Bremsstrahlung effects around evaporating black holes” Phys.Rev.D 78, 064044 (Sept 2008) eConf C121028
14280	https://litfl.com/acute-dystonic-reaction/	Skip to content Stiff and Twisted Chris Nickson aka Toxicology Conundrum 030 A 21year old male with a history of schizophrenia and obsessive compulsive disorder has presented to the emergency department complaining of a stiff neck. He states that his neck is locked to the left and that he hasn’t been able to move it for the past hour. His usual medications are citalopram 20mg and haloperidol 1mg BD. On examination you note that he is alert and orientated, appears anxious and diaphoretic, and is tachycardic (115/min). His patient’s upper body and neck are rigid, with his neck locked in flexion and rotated to the left. His voice is normal and there is no airway compromise. On further questioning, the patient reports that 3 hours previously he took an additional 2mg of haloperidol orally to try to control the derogatory auditory hallucinations he was experiencing. Questions Q1. What is the likely diagnosis? Answer and interpretation Acute dystonic reaction to haloperidol, manifesting as torticollis Acute dystonic reactions are an extrapyramidal side effect of antipsychotic and certain other medications. 90% occur within 5 days of starting a new antipsychotic medication. as many as 1 in 3 patients experience at least a mild dystonic reaction in the first few days after starting an antipsychotic medication. occurs in 0.5-1% of people given metoclopramide or prochlorperazine Dystonia refers to sustained muscle contractions, frequently causing twisting, repetitive movements or abnormal postures. They may affect any part of the body. Patients experiencing acute dystonic reactions are often frightened and fearful, and may be in considerable pain. Acute dystonic reactions are rarely life threatening (due to airway or respiratory compromise). Q2. What is the underlying pathophysiology of this condition? Answer and interpretation Acute dystonic reactions result from an imbalance of dopaminergic and cholinergic neurotransmission. The dominant mechanism n acute dystonia is thought to be nigrostriatal dopamine D2 receptor blockade, which leads to an excess of striatal cholinergic output. High potency D2 receptor antagonists, such as the butyrophenone haloperidol, are most likely to produce acute dystonic reactions. Higher dosages are often linked to acute dystonic reactions, but the relationship is unpredictable and reactions are generally idiosyncratic. Q3. What is the natural history of this condition? Answer and interpretation Acute dystonic reactions usually occur within a few hours of taking a causative medication, but onset may be delayed a few days. Untreated the condition gradually resolves over a few days. It is rarely life-threatening (e.g. laryngeal dystonia). Q4. What are the risk factors for developing this condition? Answer and interpretation Suggested risk factors for acute dystonic reactions include: male gender young age (children are particularly susceptible) a previous episode of acute dystonia higher potency D2 receptor antagonists used in high doses family history of dystonia recent cocaine use Q5. What medications can cause this condition? Answer and interpretation Antipyschotics are the most important cause of acute dystonic reactions — all currently available antipsychotics (e.g. phenothiazines, butyrophenones and newer atypical agents) have the potential to cause acute dystonic reactions. Acute dystonic reactions can also be caused by drugs other than antipsychotics. They include: Antiemetics — e.g. metaclopramide, proclorperazine Antidepressants and serotonin receptor agonists — e.g. SSRIs, buspirone, sumitriptan Antibiotics — e.g. erythromycin Antimalarials — e.g. chloroquine Anticonvulsants — e.g. carbamazepine, vigabatrin H2 receptor antagonists — e.g. ranitadine, cimetidine Recreational drugs — e.g. cocaine Q6. What are the different ways that this condition may present? Answer and interpretation Acute dystonic reactions can present in a number of different ways. The diagnosis is not always obvious, but should be considered in patients who have been exposed to medications associated with the acute dystonic reactions. Layryngeal dystonia — a rare but potentially life-threatening variant characterised by throat pain, dyspnea, stridor and dysphonia. Oculogyric crisis — rotatory eye movements or deviated gaze Blepharospasm and other facial spasms — spasm of the eyelids (unable to open eyes) or other facial muscles Buccolingual crisis — protruding or pulling sensation of the tongue Torticollis, antecollis or retrocollis — twisting of the neck, or the head forced forwards or backwards Torticopelvic crisis — abdominal rigidity and pain Scoliosis or lordosis — lateral flexion of the spine or extension. Opisthotonic crisis — spasm of the entire body characterised by back arching, flexion of the upper limbs and extension of the lower limbs. Other characteristics of acute dystonic reactions include: Mental status is generally unaffected, anxiety & agitation are common Vital signs are often normal but tachycardia, tachypnoea and diaphoresis may be present. Q7. What is the differential diagnosis? Answer and interpretation Many conditions may resemble the different types of acute dystonic reaction. They include: Neurological: Status epilepticus stroke Stiff Man Syndrome other movement disorders Toxicological: strychnine serotonin toxicity anticholinergic syndrome other drug-induced movement disorders Infectious: Meningitis Tetanus Oropharyngeal infections Metabolic: Hypocalcaemia Hypomagnesaemia Metabolic or respiratory alkalosis Psychiatric: Conversion disorder Hyperventilation due to anxiety (carpopedal spasm) Q8. Describe the management of this condition. Answer and interpretation Resuscitation: attend to ABCs. on rare occasions acute dystonic reactions may be life-threatening: airway compromise e.g. laryngeal dysphonia respiratory compromise e.g. chest wall rigidity. administer oxygen, obtain IV access and assist ventilation as required. Specific treatments: Benztropine: first line treatment of acute dystonic reactions Response is often dramatic and generally occurs in 5-20mins if symptoms persist after 15-30mins a second dose can be given. if symptoms persist and are not improving after second dose consider the possibility of an alternative diagnosis. Adult: 1-2mg by slow IV injection Child: 0.02mg/kg to maximum of 1mg Benzodiazepines: Second line treatment help relieve muscle spasm and anxiety best used for acute dystonic reactions that are slow to resolve following benztropine administration — early use may lead to diagnostic confusion Midazolam dose 1-2mg IV/IM Diazepam dose 5-10mg IV/PO Antihistamines (H1 receptor antagonists) with anticholinergic activity: Can be used if benztropine not available. e.g. promethazine 25-50mg IV/IM or diphenhydramine 50mg IV/IM or 1 mg/kg in children) Q9. What is the appropriate emergency department disposition of patients with this condition? Answer and interpretation Patients can be discharge home when symptoms have resolved. Consider admitting patients that experienced airway or respiratory compromise to an observation ward for 24-48 hours. Discharge patients with at least 2-3 days supply of Benztropine 1-2mg PO BD. The half-lives of the agents that cause acute dystonic reactions generally exceed that benztropine. Acute dystonic reactions can recur, or mild symptoms may persist, for up to 3 days. Advise the patient to return if they have a recurrence and to avoid taking the offending medication in the future. Patients requiring ongoing antipyschotic treatment may require long-term anticholinergic treatment (e.g. benztropine) to prevent symptoms, or an alternative antipsychotic agent (e.g. a newer atypical agent) may be tried. References Campbell, D. (2001). The management of acute dystonic reactions. Australian Prescriber. 24(1), 19-20. Fines, R. Brady, W. & DeBehnke. (1997). Cocaine-Associated Dystonic Reaction. American Journal of Emergency Medicine. 15(5), 513-516. PMID: 9270394 Khan, N. & Razzak, J. (2006). Abdominal pain with rigidity secondary to the anti-emetic drug metaclopramide. The Journal of Emergency Medicine. 30(4), 411-413. PMID: 16740451 Nochimson, G. (2009). Toxicity, Medication-Induced Dystonic Reactions. Yis, U. et.al. (2005). Metaclopramide induced dystonia in children: two case reports. European Journal of Emergency Medicine. 12, 117-119. PMID: 15891443 CLINICAL CASES Toxicology Conundrum More Tox Cases Chris Nickson Chris is an Intensivist and ECMO specialist at The Alfred ICU, where he is Deputy Director (Education). He is a Clinical Adjunct Associate Professor at Monash University, the Lead for the Clinician Educator Incubator programme, and a CICM First Part Examiner. He is an internationally recognised Clinician Educator with a passion for helping clinicians learn and for improving the clinical performance of individuals and collectives. He was one of the founders of the FOAM movement (Free Open-Access Medical education)has been recognised for his contributions to education with awards from ANZICS, ANZAHPE, and ACEM. His one great achievement is being the father of three amazing children. On Bluesky, he is @precordialthump.bsky.social and on the site that Elon has screwed up, he is @precordialthump. \| INTENSIVE \| RAGE \| Resuscitology \| SMACC Leave a ReplyCancel reply This site uses Akismet to reduce spam. Learn how your comment data is processed.
14281	https://www.quora.com/How-do-you-sketch-the-curve-y-x-2-x-2-and-the-method	How to sketch the curve y=x^2(x+2) and the method - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Curve Sketching Plot Development Graph Functions Derivatives and Different... Polynomials Calculus 1 Differentiable Functions Graphing Equations 5 How do you sketch the curve y=x^2(x+2) and the method? All related (34) Sort Recommended Leila Schneps mathematician, expat, happy person · Author has 140 answers and 2.7M answer views ·6y This answer works for every polynomial. This is just one particular example. All polynomials essentially look like wavy curves (like a W for even degree or like an N for odd degree, but with a number of direction changes equal to the degree minus one). Since your polynomial expands to x^3+2x^2, it’s of degree three, so your curve will just look like a wavy N. To sketch your N in the proper position on the axes, you need only two pieces of information: Step 1) Find the roots (places where the function value is zero, which means that the curve crosses the x-axis). For your function, since you gave Continue Reading This answer works for every polynomial. This is just one particular example. All polynomials essentially look like wavy curves (like a W for even degree or like an N for odd degree, but with a number of direction changes equal to the degree minus one). Since your polynomial expands to x^3+2x^2, it’s of degree three, so your curve will just look like a wavy N. To sketch your N in the proper position on the axes, you need only two pieces of information: Step 1) Find the roots (places where the function value is zero, which means that the curve crosses the x-axis). For your function, since you gave it in its factored form, we see that f(0)=0 and f(-2)=0 and there are no other roots. So your curve touches the x-axis at the points (0,0) and (-2,0). Step 2) Find the maximum and minimum points where the curve changes direction. To do this you first need to find the roots of the derivative f’(x)=3x^2+4x. This factors to x(3x+4), so the roots are at x=0 and x=-4/3. Those are the x-coordinates of the maximum and the minimum, so the points where those are located are (0,f(0))=(0,0) and (-4/3,f(-4/3)) = (-4/3, 32/27). Ready to sketch. Plot the points where the curve touches the x-axis, the maximum and the minimum. (In your case these are three different points as the curve touches the x-axis at its minimum, (0,0).) Now sketch the N running through those points. Upvote · 9 1 9 1 Promoted by Grammarly Grammarly Great Writing, Simplified ·Aug 18 Which are the best AI tools for students? There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Do Continue Reading There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Docs – Your all-in-one writing surface Think of docs as your smart notebook meets your favorite editor. It’s a writing surface where you can brainstorm, draft, organize your thoughts, and edit—all in one place. It comes with a panel of smart tools to help you refine your work at every step of the writing process and even includes AI Chat to help you get started or unstuck. Expert Review – Your built-in subject expert Need to make sure your ideas land with credibility? Expert Review gives you tailored, discipline-aware feedback grounded in your field—whether you're writing about a specific topic, looking for historical context, or looking for some extra back-up on a point. It’s like having the leading expert on the topic read your paper before you submit it. AI Grader – Your predictive professor preview Curious what your instructor might think? Now, you can get a better idea before you hit send. AI Grader simulates feedback based on your rubric and course context, so you can get a realistic sense of how your paper measures up. It helps you catch weak points and revise with confidence before the official grade rolls in. Citation Finder – Your research sidekick Not sure if you’ve backed up your claims properly? Citation Finder scans your paper and identifies where you need sources—then suggests credible ones to help you tighten your argument. Think fact-checker and librarian rolled into one, working alongside your draft. Reader Reactions – Your clarity compass Writing well is one thing. Writing that resonates with the person reading it is another. Reader Reactions helps you predict how your audience (whether that’s your professor, a TA, recruiter, or classmate) will respond to your writing. With this tool, easily identify what’s clear, what might confuse your reader, and what’s most likely to be remembered. All five tools work together inside Grammarly’s document editor to help you grow your skills and get your writing across the finish line—whether you’re just starting out or fine-tuning your final draft. The best part? It’s built for school, and it’s ready when you are. Try these features and more for free at Grammarly.com and get started today! Upvote · 999 200 99 34 9 3 Alexey Godin Ph.D. in Mathematics&Economics, Moscow State University (Graduated 1998) · Author has 2.7K answers and 3.9M answer views ·6y The method: The function is defined everywhere It is neither odd nor even It has no asymptotes As f(x)=x 3(1+2 x)f(x)=x 3(1+2 x) we conclude that it behaves like x 3 x 3 for sufficiently large x. The zeroes of the function are - 2 and 0.The function changes its sign at - 2 and does not at 0 f′(x)=3 x 2+2 x=3 x(x+2 3)f′(x)=3 x 2+2 x=3 x(x+2 3) As f’(x) changes its sign in both points we conclude that it has local minimum at x=0 with f(0)=0 and local maximum at x=−2 3,f(−2 3)=16 27 x=−2 3,f(−2 3)=16 27 f′′(x)=6 x+2 f″(x)=6 x+2.Thus x=−1 3 x=−1 3 is and inflection point. Before it the functuon is concave down, after it the function is concave Continue Reading The method: The function is defined everywhere It is neither odd nor even It has no asymptotes As f(x)=x 3(1+2 x)f(x)=x 3(1+2 x) we conclude that it behaves like x 3 x 3 for sufficiently large x. The zeroes of the function are - 2 and 0.The function changes its sign at - 2 and does not at 0 f′(x)=3 x 2+2 x=3 x(x+2 3)f′(x)=3 x 2+2 x=3 x(x+2 3) As f’(x) changes its sign in both points we conclude that it has local minimum at x=0 with f(0)=0 and local maximum at x=−2 3,f(−2 3)=16 27 x=−2 3,f(−2 3)=16 27 f′′(x)=6 x+2 f″(x)=6 x+2.Thus x=−1 3 x=−1 3 is and inflection point. Before it the functuon is concave down, after it the function is concave up. This is fairly enough to draw a good sketch. Upvote · 9 1 Goh Kim Tee Former Tutor at Private (non-agency) (2008–2017) · Author has 4.2K answers and 2.4M answer views ·6y (i)For polynomial ,graph cuts/ touches the x-axis for non-repeated/repeated zeroes ② FOR cubic function,shape of graph is an inclined strectched-out S SO , let the snake cut through x-axis at x=-2 and comes down to touch the origin,then curve up to infinity . Upvote · Related questions More answers below How do you sketch the curve for y=x^2-6x? How can you sketch the curve of y=x^3-2x^2-x+2? How do you sketch the function f(x) =(x+1) /√ (x^2 -9)? If X^yy^x=256, then what is y^2-x^2(y>x)? What are the characteristics and sketch of the curve of y^2 (x^2-9) =x^4? Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views ·4y Related How do you sketch the graph of the curve y=4+3sin2x? I will actually SKETCH these diagrams without a graphing program… Continue Reading I will actually SKETCH these diagrams without a graphing program… Upvote · 99 10 Sponsored by Avnet Silica We're at the Pulse of the Market. Tap into emerging tech and market insights to stay informed, reduce risk, and make better decisions. Learn More Assistant Bot · 1y To sketch the curve y=x 2(x+2)y=x 2(x+2), follow these steps: Identify the Function The given function is: y=x 2(x+2)=x 3+2 x 2 y=x 2(x+2)=x 3+2 x 2 Determine Key Features a. Find the Roots Set y=0 y=0: x 2(x+2)=0 x 2(x+2)=0 This gives: x 2=0 x 2=0 → x=0 x=0 (double root) x+2=0 x+2=0 → x=−2 x=−2 (single root) The roots are x=0 x=0 and x=−2 x=−2. b. Find the Intercepts Y-intercept: Set x=0 x=0: y=0 2(0+2)=0 y=0 2(0+2)=0 So, the y-intercept is (0,0)(0,0). X-intercepts: Already found as (0,0)(0,0) and (−2,0)(−2,0). c. Determine the Behavior at Infinity As x→∞x→∞, y→∞y→∞ (since the leading term x 3 x 3 dominates). As x→−∞x→−∞, y→−∞y→−∞. d. Fin Continue Reading To sketch the curve y=x 2(x+2)y=x 2(x+2), follow these steps: Identify the Function The given function is: y=x 2(x+2)=x 3+2 x 2 y=x 2(x+2)=x 3+2 x 2 Determine Key Features a. Find the Roots Set y=0 y=0: x 2(x+2)=0 x 2(x+2)=0 This gives: x 2=0 x 2=0 → x=0 x=0 (double root) x+2=0 x+2=0 → x=−2 x=−2 (single root) The roots are x=0 x=0 and x=−2 x=−2. b. Find the Intercepts Y-intercept: Set x=0 x=0: y=0 2(0+2)=0 y=0 2(0+2)=0 So, the y-intercept is (0,0)(0,0). X-intercepts: Already found as (0,0)(0,0) and (−2,0)(−2,0). c. Determine the Behavior at Infinity As x→∞x→∞, y→∞y→∞ (since the leading term x 3 x 3 dominates). As x→−∞x→−∞, y→−∞y→−∞. d. Find the Derivative To find the critical points, calculate the first derivative: y′=3 x 2+4 x y′=3 x 2+4 x Set y′=0 y′=0: 3 x 2+4 x=0 3 x 2+4 x=0 Factoring gives: x(3 x+4)=0 x(3 x+4)=0 Thus, x=0 x=0 or x=−4 3 x=−4 3. e. Determine the Nature of Critical Points Evaluate the second derivative: y′′=6 x+4 y″=6 x+4 At x=0 x=0: y′′(0)=4>0(local minimum)y″(0)=4>0(local minimum) At x=−4 3 x=−4 3: y′′(−4 3)=6(−4 3)+4=−8<0(local maximum)y″(−4 3)=6(−4 3)+4=−8<0(local maximum) Calculate the Local Maximum and Minimum Local Maximum at x=−4 3 x=−4 3: y(−4 3)=(−4 3)2(−4 3+2)=16 9⋅2 3=32 27≈1.19 y(−4 3)=(−4 3)2(−4 3+2)=16 9⋅2 3=32 27≈1.19 Local Minimum at x=0 x=0: y(0)=0 y(0)=0 Sketch the Curve Plot the x-intercepts at (0,0)(0,0) and (−2,0)(−2,0). Plot the local maximum at (−4 3,32 27)(−4 3,32 27). Indicate that the curve approaches −∞−∞ as x→−∞x→−∞ and ∞∞ as x→∞x→∞. Use the information about the critical points and the behavior of the function to draw a smooth curve connecting these points. Final Sketch The curve will have a local maximum at (−4 3,32 27)(−4 3,32 27), cross the x-axis at (−2,0)(−2,0) and (0,0)(0,0), and will rise to infinity as x x moves away from −2−2 and 0 0 towards positive infinity. Summary By following these steps, including finding roots, intercepts, critical points, and evaluating local maxima and minima, you can effectively sketch the curve of the function y=x 2(x+2)y=x 2(x+2). Upvote · Gary Ward MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views ·1y Related How can I sketch the graph of this quadratic function: Y=x^2+3? How can I sketch the graph of this quadratic function: Y=x^2+3? x = -b/(2a) = -0/2 = 0 The axis of symmetry is the y-axis and the vertex is (0, 3). The coefficient of x² is plus one, so the parabola opens upwards. Plug in a few values for x; 1, 2, and 3. (1)² + 3 = 4 so (1, 4) and (-1, 4) are points on the parabola because it is symmetric about the y-axis, AoS. (2)² + 3 = 7 so (2, 7) and (-2, 7) (3)² + 3 = 12 so (3, 12) and (-3, 12) are points on the parabola. Plot the points and draw a line through them. Continue Reading How can I sketch the graph of this quadratic function: Y=x^2+3? x = -b/(2a) = -0/2 = 0 The axis of symmetry is the y-axis and the vertex is (0, 3). The coefficient of x² is plus one, so the parabola opens upwards. Plug in a few values for x; 1, 2, and 3. (1)² + 3 = 4 so (1, 4) and (-1, 4) are points on the parabola because it is symmetric about the y-axis, AoS. (2)² + 3 = 7 so (2, 7) and (-2, 7) (3)² + 3 = 12 so (3, 12) and (-3, 12) are points on the parabola. Plot the points and draw a line through them. Upvote · 99 12 Related questions More answers below How do you sketch a graph of the function y=x^3+3x^2, showing the positions of turning points if any? What is x 2−x 2 x 2−x 2? What is the curve for y^2 (a^2-x^2) =a^3x? How do you sketch the curve x^2+3Y^2-3=0? What is the curve y^2(a+x) =x^2 (a-x)? Peter Butcher Former Now Retired · Author has 1.4K answers and 4.6M answer views ·4y Related How would you sketch the curve y= (2x+1) (x–2) ²? Then draw the line y=x+2 on your graph and show that it intersects with the curve at the point x =1. (More in the answers, not enough space in the question) Question How would you sketch the curve y= (2x+1) (x–2) ²? Then draw the line y=x+2 on your graph and show that it intersects with the curve at the point x =1. Initial Thoughts I am happy to admit it. I submit a lot of responses to Quora questions and often insert a graph to illustrate analysis and conclusions. Most of my graphs have been plotted by one of the many Android Applications designed specifically for that purpose. Having said that, I have frequently plotted a curve the “old fashioned way" as I'll do to answer your question. If your question relates more to how to use a specific App, then I Continue Reading Question How would you sketch the curve y= (2x+1) (x–2) ²? Then draw the line y=x+2 on your graph and show that it intersects with the curve at the point x =1. Initial Thoughts I am happy to admit it. I submit a lot of responses to Quora questions and often insert a graph to illustrate analysis and conclusions. Most of my graphs have been plotted by one of the many Android Applications designed specifically for that purpose. Having said that, I have frequently plotted a curve the “old fashioned way" as I'll do to answer your question. If your question relates more to how to use a specific App, then I suggest that you reformat your question and post it accordingly. Analysis Consider y= (2x + 1)(x — 2)² We can tell immediately that this curve is: A Cubic Equation and will therefore have 3 Roots. We can also see some Roots simply by looking at the equation. For example when x = 2, then ( x — 2 ) (part of the Equation) will equal 0. Further when x = ( — 1/2 ), y is also 0. Already we have established 2 points on the graph! Step 1. Let's establish a Table of Values When: x = 0, y = (2x+1) (x–2) ² = (1)(-2)² = 4 x = 1, y = (2x+1) (x–2) ² = (2)(-1)² = 2 x = 2, y = (2x + 1) (x-2)² = (5)(0)² = 0 x = 3, y = (2x+1) (x–2) ² = (7)(1)². = 7 x = 4, y = (2x+ 1) (x-2 )² = (9)(2)² = 36 x = 5, y = (2x+ 1) (x- 2)² = (11)(3)² =99 x = -1, y= (2x+1) (x- 2)² = (-1)(-3)² = -9 x = -2, y= (2x+1) (x- 2)² = (-3)(-4)² = -48 x = -3, y= (2x+1 )(x- 2)² = (-5)(-5)²=-125 x = -4, y= (2x+1) (x-2 )² = (-7)(-6) = -252 x = -5, y= (2x+1) (x-2)² = (-9)(-7)²= -441 An obvious question Why did I choose “x" values in the range ( -5 </= x </= 5? No particular reason other than habit. If I found the curve varied a lot near +5 or -5, I would extend the Table accordingly. Step 2. Plot the Graph. Get some graph paper, with up to 10 graduations horizontally. Depending on how large we wish y values to be, perhaps up to 20 graduations top to bottom. If we choose to plot the Graph for values of greater than 5 or less than -5, need larger Graph paper or a different vertical scale. I've shown a (more interesting) section of the Graph I drew by hand, which illustrates some key points. Roots at x = -0.5 and x = 2 ( x y ) = (1, 3 ), (1.5, 1), (2, 0), (-0.5, 0) are all valid points on the curve. Step 3. Plot y = (x + 2) To plot this graph create a Table as above and plot y = ( x + 2 ) on the same axes as for the first graph. Consider this combined Graph. Note the following points: The 2 curves intersect at ( x, y ) = ( 1, 3 ), (-0.3508, 1.6492), ( 2.8508, 4.856). Some Closing Thoughts I hope what I've put together is something which will enable readers to confidently draw graphs of various curves. This question was good because it allowed me to go to first principles and do something I've always enjoyed. I hope it helps. PB PS: You didn't ask so I've given no explanation on how to identify and plot the coordinates of the maximum and minimum points on the curve. I've marked them on the curve, ( 2, 0 ), (0.3333′, 4.6296 ) If you wish to follow up on their derivation, please post another question on that topic. Thanks PB Upvote · Promoted by HP HP Tech Takes Tech Enthusiast \| Insights, Tips & Guidance ·Updated Sep 18 Which is a good affordable wireless laser printer that prints both sides of paper automatically? Finding an affordable wireless laser printer that supports automatic duplex printing can be a smart investment if you want to save paper and streamline your workflow. Duplex printing allows the printer to print on both sides of the page without manual intervention, which is especially useful for producing professional documents, reports, or booklets. Wireless connectivity adds convenience by enabling printing from multiple devices, including smartphones, tablets, and laptops. HP offers several models that combine these features with reliable performance and cost-efficiency, making them ideal f Continue Reading Finding an affordable wireless laser printer that supports automatic duplex printing can be a smart investment if you want to save paper and streamline your workflow. Duplex printing allows the printer to print on both sides of the page without manual intervention, which is especially useful for producing professional documents, reports, or booklets. Wireless connectivity adds convenience by enabling printing from multiple devices, including smartphones, tablets, and laptops. HP offers several models that combine these features with reliable performance and cost-efficiency, making them ideal for home offices or small businesses. One strong recommendation is the HP LaserJet Pro MFP 3102fdw. This model is priced around £229 to £329 and includes automatic duplex printing, wireless connectivity, and multifunction capabilities such as scanning and copying. It’s designed for users who need consistent, high-speed printing with minimal maintenance. The compact design makes it suitable for smaller workspaces, while the efficient toner system helps keep running costs low. Its compatibility with the HP Smart app allows for easy mobile printing, adding flexibility to your workflow. For a more budget-conscious alternative, the HP LaserJet M234sdw is another excellent option. Priced from £136 to £210, it offers automatic duplex printing and wireless functionality, along with fast print speeds of up to 29 pages per minute. This model is ideal for users who primarily need high-volume monochrome printing without the added features of scanning or copying. It supports mobile printing through platforms like Apple AirPrint and the HP Smart app, making it easy to print from various devices. It uses toner really efficiently, so you won’t be spending loads on refills, thus great for everyday printing. LaserJet Printers - Black & White or Color Document Printers In conclusion, if you need a multifunction printer with duplex printing and wireless capabilities, the HP LaserJet Pro MFP 3102fdw offers excellent value and performance. If your focus is on fast, reliable black and white printing with duplex support at a lower price point, the HP LaserJet M234sdw is a dependable and cost-effective solution. Both models can give you strong results without exceeding your budget, and the choice depends on whether you require additional features beyond printing. For more duplex printer model recommendations, check out this blog post: Top Multifunction Printers for Small Businesses Features And Recommendations \| HP® Tech Takes By Lizzie - HP Tech Expert Upvote · 99 15 9 5 Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views ·Updated 3y Related How do you sketch a curve such as y=(x-1) ²(x+3) ³ (2x–3) without using derivatives? Some basic general ideas… Firstly if x = 0 then y = – 81 so the scale on this sketch will not be very accurate! Just for interest, here is what my graphing program does… .These short video I made should be quite useful… TYPES OF CUBIC CURVES 2D TYPES OF QUARTIC CURVES 2D QUINTIC CURVE with 5 separate zeros transforming into y = x^5. 2D . Continue Reading Some basic general ideas… Firstly if x = 0 then y = – 81 so the scale on this sketch will not be very accurate! Just for interest, here is what my graphing program does… .These short video I made should be quite useful… TYPES OF CUBIC CURVES 2D TYPES OF QUARTIC CURVES 2D QUINTIC CURVE with 5 separate zeros transforming into y = x^5. 2D . Upvote · 99 15 9 7 9 8 Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views ·2y Related How can I sketch these curves on the same axes y=-(x-1/3) ^3&y=-2 (x-1/3) ^3? Continue Reading Upvote · 9 2 9 5 9 1 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 623 Mohammad Afzaal Butt B.Sc in Mathematics&Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views ·3y Related How do you sketch the curves y = x2 +3 and y = 7−3x and determine the area enclosed by them? y=x 2+3 y=7−3 x y=x 2+3 y=7−3 x ⟹x 2+3=7−3 x⟹x 2+3=7−3 x ⟹x 2+3 x−4=0⟹x 2+3 x−4=0 ⟹(x+4)(x−1)=0⟹(x+4)(x−1)=0 ⟹x=−4,1⟹x=−4,1 Area enclosed=∫1−4(7−3 x−x 2−3)d x Area enclosed=∫−4 1(7−3 x−x 2−3)d x =∫1−4(4−3 x−x 2)d x=∫−4 1(4−3 x−x 2)d x =4 x−3 x 2 2−x 3 3∣∣∣1−4=4 x−3 x 2 2−x 3 3\|−4 1 =4−3 2−1 3+16+24−64 3=4−3 2−1 3+16+24−64 3 =44−139 6=44−139 6 =125 6 sq units=125 6 sq units Continue Reading y=x 2+3 y=7−3 x y=x 2+3 y=7−3 x ⟹x 2+3=7−3 x⟹x 2+3=7−3 x ⟹x 2+3 x−4=0⟹x 2+3 x−4=0 ⟹(x+4)(x−1)=0⟹(x+4)(x−1)=0 ⟹x=−4,1⟹x=−4,1 Area enclosed=∫1−4(7−3 x−x 2−3)d x Area enclosed=∫−4 1(7−3 x−x 2−3)d x =∫1−4(4−3 x−x 2)d x=∫−4 1(4−3 x−x 2)d x =4 x−3 x 2 2−x 3 3∣∣∣1−4=4 x−3 x 2 2−x 3 3\|−4 1 =4−3 2−1 3+16+24−64 3=4−3 2−1 3+16+24−64 3 =44−139 6=44−139 6 =125 6 sq units=125 6 sq units Upvote · 9 2 9 3 9 1 Francesco Amato Studied at University of Bari (Graduated 1999) · Author has 4.5K answers and 1M answer views ·1y Related What is the sketch of the curve Y=(X+1) ^2(x+1)? Domain: x>−1 x>−1 y(x)=(x+1)2(x+1)>0 y(x)=(x+1)2(x+1)>0 y′(x)=2 y(x)[1+ln(x+1)]y′(x)=2 y(x)[1+ln⁡(x+1)] y′(x)=0→ln(x+1)=−1→x+1=e−1→x=−1+1 e=−0.632 y′(x)=0→ln⁡(x+1)=−1→x+1=e−1→x=−1+1 e=−0.632 The function decreases from x=1 x=1 to x=−0.632 x=−0.632 where a minimum y(−0.632)=0,479 y(−0.632)=0,479 is reached to increase monotonically toward infinity for x→∞x→∞. Continue Reading Domain: x>−1 x>−1 y(x)=(x+1)2(x+1)>0 y(x)=(x+1)2(x+1)>0 y′(x)=2 y(x)[1+ln(x+1)]y′(x)=2 y(x)[1+ln⁡(x+1)] y′(x)=0→ln(x+1)=−1→x+1=e−1→x=−1+1 e=−0.632 y′(x)=0→ln⁡(x+1)=−1→x+1=e−1→x=−1+1 e=−0.632 The function decreases from x=1 x=1 to x=−0.632 x=−0.632 where a minimum y(−0.632)=0,479 y(−0.632)=0,479 is reached to increase monotonically toward infinity for x→∞x→∞. Upvote · 9 1 Francesco Amato Studied at University of Bari (Graduated 1999) · Author has 4.5K answers and 1M answer views ·Aug 26 Related How do you sketch the curve y=12/(x-3) (x+4)? The function f(x)=12(x−3)(x+4)f(x)=12(x−3)(x+4) is characterized by two vertical asymptotes x=−4 x=−4 and x=3 x=3 where it behaves according to the limits lim x→−4±12(x−3)(x+4)=+(−)⋅(±)=∓∞lim x→−4±12(x−3)(x+4)=+(−)⋅(±)=∓∞ lim x→3±12(x−3)(x+4)=+(±)⋅(+)=±∞lim x→3±12(x−3)(x+4)=+(±)⋅(+)=±∞ At infinity it tends asymptotically to zero from above since on both sides lim x→±∞12(x−3)(x+4)=++∞=0+lim x→±∞12(x−3)(x+4)=++∞=0+ The value f(x=0)=−1 f(x=0)=−1 is not a local maximum since the first derivative f′(x)=−12(x+4)+(x−3)2(x+4)2=0.f′(x)=−12(x+4)+(x−3)2(x+4)2=0. at x=−1/2 x=−1/2 where f(x=−1/2)=−48 49>−1 f(x=−1/2)=−48 49>−1 The above data is Continue Reading The function f(x)=12(x−3)(x+4)f(x)=12(x−3)(x+4) is characterized by two vertical asymptotes x=−4 x=−4 and x=3 x=3 where it behaves according to the limits lim x→−4±12(x−3)(x+4)=+(−)⋅(±)=∓∞lim x→−4±12(x−3)(x+4)=+(−)⋅(±)=∓∞ lim x→3±12(x−3)(x+4)=+(±)⋅(+)=±∞lim x→3±12(x−3)(x+4)=+(±)⋅(+)=±∞ At infinity it tends asymptotically to zero from above since on both sides lim x→±∞12(x−3)(x+4)=++∞=0+lim x→±∞12(x−3)(x+4)=++∞=0+ The value f(x=0)=−1 f(x=0)=−1 is not a local maximum since the first derivative f′(x)=−12(x+4)+(x−3)2(x+4)2=0.f′(x)=−12(x+4)+(x−3)2(x+4)2=0. at x=−1/2 x=−1/2 where f(x=−1/2)=−48 49>−1 f(x=−1/2)=−48 49>−1 The above data is enough for sketching the graph without finding the second derivative signs in the different domain intervals between the asymptotes. Upvote · 9 6 Gary Ward MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views ·4y Related How do you sketch the curve y=(x-1) (X+1) (2-x)? How do you sketch the curve y=(x-1) (X+1) (2-x)? Multiply the coefficients of x and get -1. This tells you it is an inverted version of y = x^3. Set (x-1) (x+1) (2-x) = 0 which means x - 1 = 0; x + 1 = 0 and 2 - x = 0, so the graph crosses the x-axis at x = ±1 and x = 2. Multiply it out to get y = -x^3 +2x² + x - 2 ; y’ = -3x² + 4x + 1 = 0 0 = x² - (4/3)x - 1/3 =( x² - (4/3)x + 4/9) - 4/9 - 3/9 = (x - 2/3)² - 7/9 = 0 x = (2±√7)/3; x ≈ 1.55 and x ≈ -0.22 Feed into the main equation to get x ≈ 1.42 and x ≈ -2.08 Now you have 5 points (-1, 0), (-0.22, -2.08), (0, -2), (1.55, 1.42) and (2, 0) consisting Continue Reading How do you sketch the curve y=(x-1) (X+1) (2-x)? Multiply the coefficients of x and get -1. This tells you it is an inverted version of y = x^3. Set (x-1) (x+1) (2-x) = 0 which means x - 1 = 0; x + 1 = 0 and 2 - x = 0, so the graph crosses the x-axis at x = ±1 and x = 2. Multiply it out to get y = -x^3 +2x² + x - 2 ; y’ = -3x² + 4x + 1 = 0 0 = x² - (4/3)x - 1/3 =( x² - (4/3)x + 4/9) - 4/9 - 3/9 = (x - 2/3)² - 7/9 = 0 x = (2±√7)/3; x ≈ 1.55 and x ≈ -0.22 Feed into the main equation to get x ≈ 1.42 and x ≈ -2.08 Now you have 5 points (-1, 0), (-0.22, -2.08), (0, -2), (1.55, 1.42) and (2, 0) consisting of 3 x-intercepts, one y-intercept, a relative minimum and a relative maximum. If you need additional points. (-2 - 1)(-2 + 1)(2 +2) = 12, so (-2, 12) and (3 - 1)(3 + 1)(2 - 3) = -8, so (3, -8). That should take you off the little graphs on math papers. Upvote · 99 10 9 4 Related questions How do you sketch the curve for y=x^2-6x? How can you sketch the curve of y=x^3-2x^2-x+2? How do you sketch the function f(x) =(x+1) /√ (x^2 -9)? If X^yy^x=256, then what is y^2-x^2(y>x)? What are the characteristics and sketch of the curve of y^2 (x^2-9) =x^4? How do you sketch a graph of the function y=x^3+3x^2, showing the positions of turning points if any? What is x 2−x 2 x 2−x 2? What is the curve for y^2 (a^2-x^2) =a^3x? How do you sketch the curve x^2+3Y^2-3=0? What is the curve y^2(a+x) =x^2 (a-x)? How do you sketch a curve such as y=(x-1) ²(x+3) ³ (2x–3) without using derivatives? How can I sketch the graph of x^2+12y-2x-11=0? How do you sketch the graph of f(x,y) =x+y^2? How do I sketch y^2 = (x-1) /x^2? How do you prove that √(x 1−x 3)2+(y 1−y 3)2≤√(x 1−x 2)2+(y 1−y 2)2+√(x 2−x 3)2+(y 2−y 3)2(x 1−x 3)2+(y 1−y 3)2≤(x 1−x 2)2+(y 1−y 2)2+(x 2−x 3)2+(y 2−y 3)2? Related questions How do you sketch the curve for y=x^2-6x? How can you sketch the curve of y=x^3-2x^2-x+2? How do you sketch the function f(x) =(x+1) /√ (x^2 -9)? If X^yy^x=256, then what is y^2-x^2(y>x)? What are the characteristics and sketch of the curve of y^2 (x^2-9) =x^4? How do you sketch a graph of the function y=x^3+3x^2, showing the positions of turning points if any? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025 Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. 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14282	https://www.youtube.com/watch?v=62QL5Z1qjC0	PROVE LINES Are Parallel Using Their Normal Vectors & Find Value Of k For Lines To Be Coincident Homework Help 1420 subscribers 6 likes Description 183 views Posted: 26 May 2024 In this video, I share with you steps for using cartesian equation of line to solve the problem. Two lines have equations 2x - 3y + 6 = 0 and 4x - 6y + k = 0. Explain, with the use of normal vectors, why these lines are parallel? For what value of k will these lines be coincident? @24x7MathTutor #equationofline #equationofaline #equationofstraightline #vectoralgebra #vectors #vector #maths #math #mathematics #anilkumarmath #vectoralgebra #calculusandvectors #vectortipsandtricks #algebra #algebraicequations #algebraicexpression #khanacademy #khanacademyeducation #normalvector #cartesianequation #parallellines Transcript: Hello everyone, welcome to homework help. If you find the videos useful, please subscribe to the channel. So in this video, we'll solve this problem using the concepts of cartisian equation of a line. So as for the problem, two lines have equations 2x - 3 y + 6 = 0 and 4x - 6 y + k = 0. In part A, explain with the use of normal vectors why these lines are parallel. So let's start with part A. So we are given the two equations in cartisian form. The general cartisian equation of a line is the normal vector of this line is from the given equations of the two lines we can find their normal vector. The normal vector for the first line is 2, -3. The normal vector of the second line is 4 - 6. Now these normal vectors are scalar multiples of each other. If we divide their x and y components, we get the same value. That means these two normal vectors are parallel. And if for two lines the normal vectors are parallel that means the lines are also parallel to each other. So this should be the final answer for part A. Now we can analyze part B. For what value of K will these lines be coincident? For cartesian equation, the lines are coincident if their equations are multiples of each other. That means if we multiply the equation of one line by a number, we should get the equation of the other line. Or if we write the ratios of their x coefficients, y coefficients and number terms, we should get the same value. So the x coefficients in the two equations are 2 and 4. And the y coefficients are -3 and -6. And the number terms are 6 and k. We can reduce the first two ratios as 1 /2 and write an equation using these ratios to find the value of K. We can cross multiply the denominators. So 1 time k. So I get k= 12. So this is the value of k for which the lines will be coincident. So this should be the final answer for part B. So these are the steps you have to keep in mind to solve a problem like this one. So I hope you'll find this video useful. Please share it, like it, and subscribe to Homework Help. Thank you very much and have a nice day. Bye.
14283	https://flexbooks.ck12.org/cbook/ck-12-basic-geometry-concepts/section/3.9/related/plix/slopes-of-perpendicular-lines-5286a6695aa4137e691c50e9/?referrer=search	Slopes of Perpendicular Lines \| CK-12 Foundation AI Teacher Tools – Save Hours on Planning & Prep. Try it out! Skip to content What are you looking for? Search Math Grade 6 Grade 7 Grade 8 Algebra 1 Geometry Algebra 2 PreCalculus Science Earth Science Life Science Physical Science Biology Chemistry Physics Social Studies Economics Geography Government Philosophy Sociology Subject Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? Science Grade K to 5 Earth Science Life Science Physical Science Biology Chemistry Physics Advanced Biology FlexLets Math FlexLets Science FlexLets English Writing Spelling Social Studies Economics Geography Government History World History Philosophy Sociology More Astronomy Engineering Health Photography Technology College College Algebra College Precalculus Linear Algebra College Human Biology The Universe Adult Education Basic Education High School Diploma High School Equivalency Career Technical Ed English as 2nd Language Country Bhutan Brasil Chile Georgia India Translations Spanish Korean Deutsch Chinese Greek Polski Explore EXPLORE Flexi A FREE Digital Tutor for Every Student FlexBooks 2.0 Customizable, digital textbooks in a new, interactive platform FlexBooks Customizable, digital textbooks Schools FlexBooks from schools and districts near you Study Guides Quick review with key information for each concept Adaptive Practice Building knowledge at each student’s skill level Simulations Interactive Physics & Chemistry Simulations PLIX Play. Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up Back To Perpendicular Lines in the Coordinate PlaneBack 3.9 Slopes of Perpendicular Lines Written by:CK-12 Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Back to Slopes of Perpendicular Lines Ask me anything! CK-12 Foundation is a non-profit organization that provides free educational materials and resources. FLEXIAPPS ABOUT Our missionMeet the teamPartnersPressCareersSecurityBlogCK-12 usage mapTestimonials SUPPORT Certified Educator ProgramCK-12 trainersWebinarsCK-12 resourcesHelpContact us BYCK-12 Common Core MathK-12 FlexBooksCollege FlexBooksTools and apps CONNECT TikTokInstagramYouTubeTwitterMediumFacebookLinkedIn v2.11.10.20250923073248-4b84c670be © CK-12 Foundation 2025 \| FlexBook Platform®, FlexBook®, FlexLet® and FlexCard™ are registered trademarks of CK-12 Foundation. Terms of usePrivacyAttribution guide Curriculum Materials License Student Sign Up Are you a teacher? Sign up here Sign in with Google Having issues? Click here Sign in with Microsoft Sign in with Apple or Sign up using email By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Sign In Adaptive Practice I’m Ready to Practice! Get 10 correct to reach your goal Estimated time to complete: 8 min Start Practice Save this section to your Library in order to add a Practice or Quiz to it. Title (Edit Title)43/ 100 Save Go Back This lesson has been added to your library. Got It Searching in: CK-12 Looks like this FlexBook 2.0 has changed since you visited it last time. We found the following sections in the book that match the one you are looking for: Go to the Table of Contents Ok No Results Found Your search did not match anything in . Got It Are you sure you want to restart this practice? Restarting will reset your practice score and skill level.
14284	https://www.youtube.com/watch?v=F_5sV8s9ZEA	Electronics Tutorial #1 - Electricity - Voltage, Current, Power, AC and DC mjlorton 266000 subscribers 13165 likes Description 839194 views Posted: 1 Jan 2013 Visit my website for more Tips, Videos, DIY projects and more: --------------------- Click "Show more" ------------------------------- Basic / beginners Electronics Tutorial / course / lesson - Voltage, Current, Power, AC and DC Please support my content creation by using my Amazon Store: --------------------- Click "Show more" ------------------------------- My website and forum:- Donations and contributions:- My techie channel MJLorton - Solar Power and Electronic Measurement Equipment - My Techie Amazon Store: My other channel VBlogMag - For almost any topic under the sun! - My VBlogMag Amazon Store: Nikola Tesla - Thomas Edison - In this tutorial I cover the following: Some history about electricity / central power stations / electrification. Science of electron flow in a conductor / wire I use fluid dynamics in a pipe to explain voltage (pressure), current in Amps (flow) and consumption in Amp hours (rate of flow). We look at AC (alternating current) and DC (direct current) on the UNI-T UT81B scopemeter We look at how voltage / pressure is required to charge a 12 volt battery What is electricity? / How does electricity work? Topics for future videos in this series: Ohms law, resistance, power, energy Electronic components - diodes, transistors, FETS, capacitors, digital logic gates, integrated circuits, 555 timer. Series and parallel circuits Op amps and feedback 882 comments Transcript: hello everyone and welcome to this tutorial we're going to start getting to the basics of electronics now for this first video what we're going to do gonna and have try and have a basic understanding of electricity and how electricity works and the basic components of electricity now i'm going to start off with a little bit of history a little bit of science and also just give you a few basic analogies to understand the different types of electricity IE AC and DC so i'm going to talk through these just to start getting planted a few seeds in your head before we get down to some of the theory we're going to do a very basic theory on some paperwork and with some practical examples as well to show share your few bits and pieces now I don't want you to get too caught up i'll see on the history is therefore interest and the science side of it again it's just good to have a little bit of background but basically will get down to the fundamentals when we start looking at the bits of paper basically electricity was discovered a long time ago now it was discovered apparent key many many years bc in fact in the form people started seeing electricity in the form of static electricity or in fact even they started realizing with things like soup catfish and what happy when they touched them they've got a shock through their bodies so that's when the notion of electricity apparently started coming about static electricity obviously when certain objects with charge that and reacted that kind of led to describe you on electricity and static electricity and what have you and the first real work on electricity and calculations and theorizing came around in the sixteen and seventeen hundreds and kind of the more practical applications of electricity started happening in the 18 ribs so as an example let's say the practical like electrification of houses because of the invent the invention of the incandescent light bulb came around the eighteen eighties in fact in 1881 apparently this was news to me cycle it was America first but believe it or not in sorry in the UK believe or not that's where I was born and there was the first central power station which provided electricity to a little village a little village had a vote because at that point in time they were running on gas power the gas power wasn't economical so they decided to implement this new thing called electricity and they did implement that first power station but apparently it wasn't very economical so actually reverted back to guess but the main the kind of main power station to come online and start feeding streetlights and houses happened in the states in New York and if that was in eighteen eighty to say year later and that was the essence pull street station and they started off with using DC so direct current and they only later switch to AC now so let's just talk a little bit about AC and DC now basically once again once we get down to the piece of paper i'll start explaining this but DC direct current is electricity just flying in one direction constantly while AC is basically electricity in a conductor game backwards and forwards pushing backwards and forwards almost like a piston on a steam engine backwards and forwards and because of that reason there's far less effort involved pushing something back and forwards then constantly trying to push in one direction another analogy to give you an understanding of JC vs DC if you picture someone sewing a log you've got a saw and you're pushing it back and forth you push it you've got to pause you pull it back you've got a pause you push it you go to pause so there is this pause and that is a creation of a sinus hallway where electricity goes after certain value comes down to zero . and then goes negative and then comes back again i'm going to demonstrate this on a little scope as well we'll look at the mains electricity whereas if you can picture that same person are having a circular saw with a handle that they're having to crank around at constant pace constant pace your arms going to get a lot more tired doing that then had that sword that's going back and forth where you've got a pause but if you picture the saw blade on the log whether you cranking a handle round on that circular saw or you've got that saw blade coming back and forth it's still doing having the same result on the wood isn't doing the same that'll work and catching through the wood whether using and AC or DC to constantly cut on there so I just want to plant their season you're at the moment now to get down to the science and and again I certainly don't get bogged down in 22 minutes scientific stuff or even formulas at this point in time that's going to come in due course when we start working get down to working with components and putting circuits together that in essence electricity is a flow of electrons through a conductor now we're going to be using fluid dynamics of fluid mechanics to explain this because it is a very fluid mechanics and dynamics of I've got a very good correlation to the way electricity works but in a conductor and metallic conduct or something that can conduct electricity you've got atoms and electrons in there and basically it is the flow of electrons from negative to positive that provide at a current potential or electricity now so that is the converse of what we generally work in where conventional electricity we think safe flows from positive to negative so don't you get confused i just want you to understand the science behind it in fact that electrons flow from negative to positive that's the sign side of it but for all intents and purposes what people use day today and what we'll be using an explanation is conventional electricity flow which is from positive to negative in terms of direct current obviously as you'll see with alternating current it's a change it's actually going back and forth and cycling in two directions ok so let's get stuck down have a look at the piece of paper and I've got there ok so let me explain this diagram that i have here then and we'll start going through the individual components of it Bubba to have an understanding what it all means so basically this in essence is a big water pipe this big round circle rectangle over here this over here is a pump a water pump over here we have a valve which can be closed all open to allow water through over here is a paddle or basically in essence and appliance in this case it's a paddle driving a shaft into a mill well let's pretend that basically we milling or grinding flour or wheat into flour then we've got over here out gauges are pressure gauge is telling us the pressure of the water in the pipe as you might literally in a in a pipe system and then over here there's a paddle which basically is measuring the flow of water in the pipe and displaying it it's spinning around and it's displaying in essence like if you like on a rev counter on a meter over here right so let's first we'll start without our pump now over here you'll notice we if you like this to an electrical circuit the pump over here is in essence our power source naive note see here I've got power source and inverted commas because ultimately again if we talk about the science of electrons do you can't really create or destroy energy so it's not a source per se but it is the piece that actually generates the flow of energy in our circuit so I recorded the power source or the pump in essence it could be a battery or could be your mains outlet in your house but this is what in essence creates the pressure and the flow to get water flying around this pipe system and in essence that pressure and flow ultimately is the electricity flying around this circuit or this pipe system now on this pipe system as I said we've got these and pressure gauges over here and these present pressure gauges are measuring the pressure of the water they don't have to be inside the pipe itself you only need a small little tap into your pipe to actually pick pick up the pressure now the pressure of the system equates to your voltage so when we referred to a voltage and a circuit that in essence is the pressure which is there to push the water or in essence to push the electricity through your circuit as you can understand and this is something we're going to cover off probably in the next session where if you've got obstacles inside your pi e like a valve or and act hair and the plants or what have you they are going to create a restriction to the flow of the water around your circuit and that restriction in essence big is the resistance so the you need enough pressure to overcome that resistance they and their those restrictions to allow you to have water flowing or electricity flowing through your circuit now let's come down to this device over here so if we want to measure the flow of electricity through our system or the water flow throughout our pipe system we literally have to then we can't just have a little simple tap into the pipe to read the pressure as you can in a water hose or what have you you actually have to place something in a little pedal or a to buy a watch heavy inside the flow of the water to start measuring it and it's likewise when we have a multimeter and you want to measure current you need to place your multimeter in circuit to do this and it's exactly the same with measuring water you have this pedal in there and that spins around and that then on a dial gives us the flow now flow of electricity equates to the current which is aunt and so there's something quite important to understand let's pretend that I go and close this valve over here so that I have no flow of water do you understand that this this pressure gauge over here is still going to show you a pressure the water is pushing against it's going to be pushing up into that gage you still going to have pressure that pressure member is voltage so you still have a voltage potential pushing hard against over here that water pressure when you start opening this valve that starts aligned water flow then when you've got flow you've got current and current is amps so flow through a circuit is current which is amps so there's something important also to note about flow when you start talking about flow rate rate implies time and as soon as you start it adding time to the equation flow rate then you talking about consumption or amp hours and this is important because again what we're going to do I'm going to have a practical demonstration now also in some of the normal videos i'm going to be posting soon on batteries and battery testing all these components relate to each other and you will have a good understanding of what I'm doing for that battery testing if you're following what's going on here so the flow rate is basically your flow over time is an amp hour and that is what equates to consumption what you're using if you had a little bakit catching your water at the end of this and that bucket is filling up that feeling of the bakit would equate to your consumption how much electricity you're using or how much energy you using also so quiet it's quite quite a nice analogy here with using the pipe this closed pipe system with a pump and a valve and pressure gauges and something is measuring the flow it's like an electricity circuit it needs to be complete it needs to be intact that intactness is something we would relate to something called continuity and it's off see continuity is something that you can use a multimeter to measure now I i'm talking about a multi me to go and see my my video series on how to use a multimeter for beginners I then equate a lot of this and how you measure these bits and pieces with a multimeter so if you are new to this than those videos will help you as well and I'll annotate and with one of those here right so that going back to that continuity you need the pipe intact to lie or water to flow through the pipe system likewise you need your foot to allow electricity to flow you need an intact system of conductors IE your electrical wires if you have a break in your electrical wire over here just like--why's you broke the water pipe and the water the water would in essence start spilling out and then your plant or you're more over here would stop working because you wouldn't have water flow now in reality of you cut a while you're not going to see you like electrons flying out the edge hope hopefully but the analogies the same if you break the pipe and the waters basically flowing and not getting back up through the system then you're not going to gain to have the water flow to power your mole likewise with the wiring system if you have a breaking your wiring system the electrons are going to stop flowing your plants is going to stop working so let's now pretend we have we go back to having a complete working system let's go to the ultimate reason we have electricity and that's ultimately to power a device and to do work so here as I said we've got this pedal which is driving a shaft which is turning and milling stone here which is milling our flower let's pretend so in essence that equates to an appliance that you might have in your house now basically if you've got taking the energy out of the system to apply it that is the work being done by or water or the work being done by your electricity and that equates to power so there is a closed direct relationship between your pressure your flow and the work being done and that is a direct relationship between voltage current and power and we'll start looking at this in particular when we start looking at resistors and Ohm's law those relationships and how they interact but but from here all I want you to understand basically is that the end result and what you're looking for in terms of electricity and the flow of electricity is the work being done your plants being connected to your wiring system which in essence is using power which is a combination of your voltage and your current likewise although innocence in this system a pump let's pretend our pump is pumping our water around in a single direction that would equate to direct current a single direction and of your electricity flow constant push through the pipe system now equally could have you could still get your work done as I was explaining with a saw cutting the piece of wood if this pump had a piston in it that was pushing backwards and forwards instead of pumping in one direction as you can imagine your pressure gauges would be dancing around going back and forth and that's exactly in essence what happens with alternating current your voltage is in fluctuating and up and down positive and negative but even though you've got that flow of electricity going back and forth as you can see your pedal is still going to be flipped backwards and forwards it's still going to grind your flower at the end of the day so in electronics and that's why we are going to focus after this mainly on direct current because that's what we use for the most part in electronics but understanding electricity you can see that direct-current can power things and alternating current which is used in your house can power things even though it's going back and forth in essence going on and off because it it still does work at the end of the day now let's quickly actually discuss why alternating current is used as opposed to direct current for your house as I said there's a lot more takes a lot more energy to for to use direct carrot on a big system once you getting stocking to the high voltage higher voltages and what have you your losses across a piece of wire become far greater with direct current because the amount of energy required to push that constantly in one direction as opposed to back and forth so that's the one reason it takes a lot more energy and your losses are higher you need far thicker wire to make it easy for the electricity to go through for direct current then you do alternating current the other the other reason is the reason AC is used is that and you can generate far higher voltages with alternating current and that's far easier to transmit higher voltage over long distances and then transform that down to a lower voltage and get the current back once you get to a neighborhood or your house so you can transfer voltage at high voltage at lope at low current so you don't need thick wise you've got less loss and then you transform that back to a your normal voltage when you get to your house ok so let's actually put away from the paper quickly and have a quick chat about the mains electricity that comes into your house and the one thing obviously is it you realize a see besides being more efficient transmitted over a longer distance it's also safer because you've got that pause while it's going back and forth because it actually goes through 0 and back again it's safer than DC because DC is constantly honor if you if you touch your collection electrocuted by electricity and a see you've got potential it's gonna kick you and kick your way where is DC or grab onto because it's just constantly on any time to clinch up and hold where is AC tends to throw your way so DC will potentially impart far more energy into you as well then AC will if you have comparable voltage and current now the one interesting thing I discovered and it's something to hire obviously that around the world different countries use different voltages and frequencies now the frequency is obviously as i'll demonstrate is you'll see that that sinusoidal wave as the energy electricity is going on and off and that happens at a certain rate now here in South after we use 220 volts at 50 hertz so 50 times a second the electricity is going through a cycle on and off in the States it's 110 and and at a rate of its 110 120 and it's at a rate of 60 hertz so 60 times a second and one of the big reasons for having the use of 110 or 120 in the states is that it's off see far say that you've got low you've got a low voltage and the combination of the voltage and the current because the current supply is still roughly the same to the normal household outlet it's roughly 15 or 16 amps same and inside at the same in the state another one interesting thing to note is there are limitations with that safety factor in the states and that is you get particularly when it comes to tools which use need more power so the combination of your volts on your amps give you power and you've got less of that in the states from a standard 110 outlet and I when we lived in the states for two and a half years I came to realize as a snack I couldn't quite understand something I told her then clicks about not having enough power I i really enjoy drinking cup of tea I'm an Englishman even her live inside that guy enjoy my kapiti and for that reason when we got to the states and we had to buy our appliances I we went looking for a kittle now the states I couldn't understand in people's houses why everyone had skittles which they put on this stove on a hop on the top of a plate to heat up to heat the water just didn't make sense to me a cattle makes far more sense just ease it for laughs switch on it switches of automatically whereas a stove doesn't switch off automatically but then I realized generally speaking kettles require a lot of energy a lot of power to work often a cattle will be rated to run at 2,300 or 3000 watts and in the states if you look at your how much power you can get out of your 110 outlet often it's not enough to drive normal kettles which would find let's stay in south africa so for that reason I realized that's why people have to do their heat the water for their coffee or they are not that people have coffee makers but 14 you'd have something on a stone anyway I did eventually found a kettle but I hunted high and low in the state to find a simple plug-in kettle anyway let's get back to the energies and examples i'm going to show you what AC looks like and we're going to do a demonstration with DC to explain and pressure or voltage between two sources and this becomes important when again when we get to the battery testing because to charge a battery you've got a battery with a certain voltage or a certain holding pressure in it as a vessel with pressure to charge your charger needs to overcome that pressure it's like if you got a vessel with the pressure you got a pipe going into it and you want to push water into that whistle you have to have a greater pressure pushing to get inside there also is just going to push back likewise charging a battery or charger has to be at a greater pressure or greater voltage to charge your battery equated to pumping filling up a balloon with a when you feel that put blow air into that balloon you have to have a greater air pressure in your cheeks to get the end of the blue likewise pumping up aight i'll buy a car or a bicycle you're going to have a certain air pressure in that tire and you have to have a greater pressure that you're pushing in that pump to get a into that tire let's go have a look at some practical examples right so let's have a look at our mains electricity here in South Africa I've got this unity UT 81b if you are interested in it are there is a little review on it have you not seen one before man or annotate that here and basically it's a fairly useful device its scope meter its a mix of a multimeter and an oscilloscope so we can get to see wave and electricity signals basically now what I've done here I've these two probes as you can see i come from I've got a multi strip specially set up with a connecting block there where I take off the electricity or be very careful about working with mains electricity I don't I certainly don't and advise you to go get your multimeter and just willy-nilly sticking the probes into outlet mains electricity and bite you it can blow things so please be careful right uh-huh I'd rather hope that you're watching this example then trying to do this yourself because if you do it wrong and touch things that you shouldn't you can hurt and enjoy yourself that's right as you can see I've got this set to measure voltage I've set it to measure ac voltage and we've gotta figure of around 230 volts and now i'm going to change the mode of this so we go into it's a solution it's kind of scope mode or push the auto button so that it's sizes things up for us and as you can see there we've got this sinusoidal wave happening so let me just change the period that we sing so what we can see here is a sinusoidal way which is going positive we got the script in essence is cross hairs here and this is the zero line so this is the voltage going up positive and then coming back down negative a positive down- and that's that back-and-forth motion in essence which is being being measured and as you can see here we've got the voltage and we've got the frequency here in South Africa is at 50 Hertz and on this meter what you can do so that's the frequency and if you count the we've got the set to 10 milliseconds per division and if you count the divisions we've got a repeating cycle so their cycle starts that's half of the cycle and then that's the completed cycle over here it's two divisions that's 20 milliseconds that's the . or the time it takes to complete a cycle 20 minutes seconds and also you get 50 of those in a second so that is what ac voltage looks like now I'm not going to concentrate too much more on AC voltage because I've seen electronics we want to use DC but i'll show you what DC looks like on a scope right so I've got my variable DC power supply set up in the background its preset at 13 volts I've got a range i can vary that I've got a preset to a certain range of voltage because we're going to do it man demonstration with a battery after that I've got the unity skype meter linked up I've got to set two bolts at the moment it's set on AC summer change it so it's back to DC and then i'm going to now switch on the output of and my power supply and as you can see we're measuring showing as a reading of 13.1 volts and obviously as I change that the setting on the variable power supply you can see the voltage all the pressure changing now let's go over to the scope so as you can see there's that line a straight line appearing on the skirt now try and get that so conceived in the reflection and as I vary the voltage you can see the voltage level being noted there you can see that straight line just going its height is changing relative to the zero line but it is dead straight it is constantly on constantly pushing it is that is direct current it's not going on and off as we saw with with AC and that's what a DC and a DC signal looks like on an oscilloscope it is just straight on the whole time ok so what have we got here now we're moving over to our battery this is a 12 volt lead acid battery i've got it connected up a the positive opcion to the positive and the negative on to the negative that's the polarity now I've also got my variable power supply SAT here and I've got this what's up meet another what's up meter car keys actually showing you the voltage or the pressure the voltage potential that is sitting inside the battery it's sitting at about twelve point nine nine eight volts at the moment and as you can see here is the meter that also shows you with is a flow of electricity measured in amps and that is at zero at the moment so what I'm going to do I'm going to turn my power supply to roughly match the voltage or the pressure that's sitting in the battery and then I forget exactly right I'm gonna switch it on now and I've kind of got it spot-on right because if you look here the dial over here the digits over here show that basically there is no flow of electricity and likewise there's nothing going into the battery because the pressures are the same right so watch what happens as i start to step up the voltage and the pressure as you can see we start to get a flow of electricity which is being measured in amps in this case it's milliamps so it's only small you can see there's the power supply showing and the me to the what's up meat over here is showing that there's a flow so because now my voltage or my pressure at this point is higher than the battery I can actually start pushing back energy or electricity into the battery as obviously step up that pressure increase the pressure increase the voltage are likewise in the flow start increasing and I get more flow into the battery as i said just think about the blowing up of the balloon or the pumping up at the tire then if i drop the voltage actually let's just go take it up more so we can actually get a real representation of now got a kind of the highest safest pressure that I don't want to push back into this battery and again an analogy here is that if you blowing up that balloon you think of the walls of the balloon are are not that thick if you had to have a very high air pressure pumping into that blue and potentially going to rip and shred these walls of the balloon and burst it and likewise it's the same with electricity if you do things too quickly but i have to higher voltage or what have you can damage appliances so here has a safe voltage or pressure 14.4 volts I've set it up and we're pushing and we've got a flow of electricity at 276 milliamp going into the battery and slowly but surely that all bold up because we've got a in essence a closed I closed vessel here in essence with the battery so it's actually pushing more energy more pressure into it it's pressure is going to build up IE its voltage is going to build up and that will give us an indication or the level of energy storage and charge inside the battery now i'm going to do the the other I'm going to drop the pressure or the voltage on my power supply and just note what happens I'll see we've got a slightly higher we pushed up the pressure in the battery install got a slightly higher pressure on the on the power supply at the moment but look what happens as i drop below what the battery is I've got a negative reason reading on my pass by saying effect now what the battery is doing because the batteries got a higher pressure or higher voltage than my power supply it is trying to push electricity back into the power supply now I liquid basically you can't really push power back into the past players not design for that but in essence as you can see is doing it to a certain extent and that all i want to show is there is pressure differentials or voltage differentials determine the flow way how electricity is going to flow that's that will be important certainly once we start getting to looking at electric and electronic circuits with that different power sources and what have you I can it's just look at that in terms of this diagram again in in terms of different voltage potentials of different pressures in the system if I had to pretend that this valve over here was also a pump but it was pumping in the opposite direction to this pump you can understand that whichever pump was pumping had the highest pumping pressure they would be the one that would be winning winning the kind of direction in which the water would be flowing likewise if you had a second pump air flowing pushing in the same direction you then get another scenario where you going to get more higher pressure through the system and once we get down to looking at circuits with parallel and c series power sources and resistors or what have you will come back to that analogy to start explaining exactly how that all hangs together but just you know the important thing to walk away with from this session is under is correlating is that fluid dynamics and pipes are very good way to understand how electricity flows and wires and to understand that pressure equates to voltage that your flow is your current or your amps and it obviously those close to relate to the actual energy and parting to your appliance or whatever you trying to power ok so i hope you gained some value out of that first in the series thanks very much for watching again what I asked you to do is certainly if they're you picked up any mistakes or errors in that do post comments down below and I'll either annotate or correct them with the fallout for the video what have you also have a link to this video in my forum so that we can build up a series on the form words are easier to have open discussion it's far easier at the input from other people who are certainly that's wise than i am and can have had some good value to this series and there if you would like to support this work that I'm doing then certainly what you can do on my website and my you can there you can either make a donation you can be active in the forum I have an amazon store that you can purchase things from or literally just rating the videos and watching them supports what i do so I do appreciate your participation and let me just watching the videos we will continue this series are plant literally as long as there is support for them I continue and will continue the next thing we look at is obviously and resistance we started with the components like resistors Ohm's law and diodes transistors logic gates and even things like the triple five timer and then start putting circuits together and what have you so say if you are gaining any value out of this then certainly do subscribe and follow the series and I certainly hope you learned something from it thanks very much for watching and I'll catch you soon for the next one in the series
14285	https://www.reddit.com/r/calculus/comments/1fhu6fe/help_understanding_epsilonn_definition/	Help Understanding Epsilon-N Definition : r/calculus Skip to main contentHelp Understanding Epsilon-N Definition : r/calculus Open menu Open navigationGo to Reddit Home r/calculus A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to calculus r/calculus r/calculus Welcome to r/calculus - a space for learning calculus and related disciplines. Remember to read the rules before posting and flair your posts appropriately. 168K Members Online •1 yr. ago [deleted] Help Understanding Epsilon-N Definition Real Analysis I'm trying to wrap my head around the epsilon-N definition for the limit of a sequence. I'm trying to break down the components in simpler terms so that the concept sticks. So I know that for the formal definition: L is a limit of the sequence a_n if for all epsilon > 0, there exists a real number N such that n >N, then the distance between \|a_n - L\| < epsilon. Epsilon, if I'm understanding this right, is an arbitrary number that is the distance away from L. If we're looking at it from a graph, it's (L-e, L+e) or L-e < L < L+e. On a number line, it's the number of units to the left and right of L, with L being in the centre. I know that epsilon has to be greater than 0 because distance isn't negative and if epsilon did equal 0, it would be at the limit. If the limit exists, we should be able to find an x-value that has a corresponding y-value that is within epsilon. It doesn't matter if we change the value of epsilon, we can always find an x and y value within that range (L-e, L+e). If we're looking at it from a number line, epsilon is the boundary and we should be able to find as many points on the number line that gets closer and closer to L. I just don't know how N plays a factor in the definition. What is N? Since the definition says, "such that n > N," does it mean the range of x-values that correspond with all the y-values in epsilon? If N is the range of values that n can take on, wouldn't there come a point where n = N? Isn't n bound by the maximum in the range of N? Thank you and apologies for rambling. I've tried to read texts and watch Youtube videos, but it's just not sticking. Read more Share Related Answers Section Related Answers Applications of calculus in real life Visualizing multivariable calculus concepts Best methods to learn integration techniques Common mistakes in differential equations Calculus problems involving optimization New to Reddit? Create your account and connect with a world of communities. 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 September 16, 2024 Reddit reReddit: Top posts of September 2024 Reddit reReddit: Top posts of 2024 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
14286	https://www.geeksforgeeks.org/dsa/find-landaus-function-for-a-given-number-n/	Find Landau's function for a given number N Last Updated : 15 Jul, 2025 Suggest changes 2 Likes Given an integer N, the task is to find the Landau's Function of the number N. In number theory, The Landau's function finds the largest LCM among all partitions of the given number N. For Example: If N = 4, then possible partitions are: 1. {1, 1, 1, 1}, LCM = 1 2. {1, 1, 2}, LCM = 2 3. {2, 2}, LCM = 2 4. {1, 3}, LCM = 3 Among the above partitions, the partitions whose LCM is maximum is {1, 3} as LCM = 3. Examples: Input: N = 4 Output: 4 Explanation: Partitions of 4 are [1, 1, 1, 1], [1, 1, 2], [2, 2], [1, 3], among which maximum LCM is of the last partition 4 whose LCM is also 4. Input: N = 7 Output: 12 Explanation: For N = 7 the maximum LCM is 12. A slower (but easier to understand) algorithm Approach: The idea is to use Recursion to generate all possible partitions for the given number N and find the maximum value of LCM among all the partitions. Consider every integer from 1 to N such that the sum N can be reduced by this number at each recursive call and if at any recursive call N reduces to zero then find the LCM of the value stored in the vector. Below are the steps for recursion: Get the number N whose sum has to be broken into two or more positive integers. Recursively iterate from value 1 to N as index i: Base Case: If the value called recursively is 0, then find the LCM of the value stored in the current vector as this is the one of the way to broke N into two or more positive integers. if (n == 0) findLCM(arr); Recursive Call: If the base case is not met, then Recursively iterate from [i, N - i]. Push the current element j into vector(say arr) and recursively iterate for the next index and after the this recursion ends then pop the element j inserted previously: for j in range[i, N]: arr.push_back(j); recursive_function(arr, j + 1, N - j); arr.pop_back(j); After all the recursive call, print the maximum of all the LCM calculated. Below is the implementation of the above approach: C++ ```` // C++ program for the above approach include using namespace std; // To store Landau's function of the number int Landau = INT_MIN; // Function to return gcd of 2 numbers int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Function to return LCM of two numbers int lcm(int a, int b) { return (a b) / gcd(a, b); } // Function to find max lcm value // among all representations of n void findLCM(vector& arr) { int nth_lcm = arr; for (int i = 1; i < arr.size(); i++) nth_lcm = lcm(nth_lcm, arr[i]); // Calculate Landau's value Landau = max(Landau, nth_lcm); } // Recursive function to find different // ways in which n can be written as // sum of atleast one positive integers void findWays(vector& arr, int i, int n) { // Check if sum becomes n, // consider this representation if (n == 0) findLCM(arr); // Start from previous element // in the representation till n for (int j = i; j <= n; j++) { // Include current element // from representation arr.push_back(j); // Call function again // with reduced sum findWays(arr, j, n - j); // Backtrack - remove current // element from representation arr.pop_back(); } } // Function to find the Landau's function void Landau_function(int n) { vector arr; // Using recurrence find different // ways in which n can be written // as a sum of atleast one +ve integers findWays(arr, 1, n); // Print the result cout << Landau; } // Driver Code int main() { // Given N int N = 4; // Function Call Landau_function(N); return 0; } ```` // C++ program for the above approach // C++ program for the above approach ``` include #include ``` using namespace std; using namespace std // To store Landau's function of the number // To store Landau's function of the number int Landau = INT_MIN; int Landau = INT_MIN // Function to return gcd of 2 numbers // Function to return gcd of 2 numbers int gcd(int a, int b) int gcd int a int b { if (a == 0) if a == 0 return b; return b return gcd(b % a, a); return gcd b% a a } // Function to return LCM of two numbers // Function to return LCM of two numbers int lcm(int a, int b) int lcm int a int b { return (a b) / gcd(a, b); return a b/ gcd a b } // Function to find max lcm value // Function to find max lcm value // among all representations of n // among all representations of n void findLCM(vector<int>& arr) void findLCM vector< int>& arr { int nth_lcm = arr; int nth_lcm = arr 0 for (int i = 1; i < arr.size(); i++) for int i = 1 i< arr size i ++ nth_lcm = lcm(nth_lcm, arr[i]); nth_lcm = lcm nth_lcm arr i // Calculate Landau's value // Calculate Landau's value Landau = max(Landau, nth_lcm); Landau = max Landau nth_lcm } // Recursive function to find different // Recursive function to find different // ways in which n can be written as // ways in which n can be written as // sum of atleast one positive integers // sum of atleast one positive integers void findWays(vector<int>& arr, int i, int n) void findWays vector< int>& arr int i int n { // Check if sum becomes n, // Check if sum becomes n, // consider this representation // consider this representation if (n == 0) if n == 0 findLCM(arr); findLCM arr // Start from previous element // Start from previous element // in the representation till n // in the representation till n for (int j = i; j <= n; j++) {for int j = i j<= n j ++ // Include current element // Include current element // from representation // from representation arr.push_back(j); arr push_back j // Call function again // Call function again // with reduced sum // with reduced sum findWays(arr, j, n - j); findWays arr j n - j // Backtrack - remove current // Backtrack - remove current // element from representation // element from representation arr.pop_back(); arr pop_back } } // Function to find the Landau's function // Function to find the Landau's function void Landau_function(int n) void Landau_function int n { vector<int> arr; vector< int> arr // Using recurrence find different // Using recurrence find different // ways in which n can be written // ways in which n can be written // as a sum of atleast one +ve integers // as a sum of atleast one +ve integers findWays(arr, 1, n); findWays arr 1 n // Print the result // Print the result cout << Landau; cout<< Landau } // Driver Code // Driver Code int main() int main { // Given N // Given N int N = 4; int N = 4 // Function Call // Function Call Landau_function(N); Landau_function N return 0; return 0 } Java ```` // Java program for the above approach import java.util.; class GFG{ // To store Landau's function of the number static int Landau = Integer.MIN_VALUE; // Function to return gcd of 2 numbers static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Function to return LCM of two numbers static int lcm(int a, int b) { return (a b) / gcd(a, b); } // Function to find max lcm value // among all representations of n static void findLCM(Vector arr) { int nth_lcm = arr.get(0); for(int i = 1; i < arr.size(); i++) nth_lcm = lcm(nth_lcm, arr.get(i)); // Calculate Landau's value Landau = Math.max(Landau, nth_lcm); } // Recursive function to find different // ways in which n can be written as // sum of atleast one positive integers static void findWays(Vector arr, int i, int n) { // Check if sum becomes n, // consider this representation if (n == 0) findLCM(arr); // Start from previous element // in the representation till n for(int j = i; j <= n; j++) { // Include current element // from representation arr.add(j); // Call function again // with reduced sum findWays(arr, j, n - j); // Backtrack - remove current // element from representation arr.remove(arr.size() - 1); } } // Function to find the Landau's function static void Landau_function(int n) { Vector arr = new Vector<>(); // Using recurrence find different // ways in which n can be written // as a sum of atleast one +ve integers findWays(arr, 1, n); // Print the result System.out.print(Landau); } // Driver Code public static void main(String[] args) { // Given N int N = 4; // Function call Landau_function(N); } } // This code is contributed by amal kumar choubey ```` Python3 ```` Python3 program for the above approach import sys To store Landau's function of the number Landau = -sys.maxsize - 1 Function to return gcd of 2 numbers def gcd(a, b): if (a == 0): return b return gcd(b % a, a) Function to return LCM of two numbers def lcm(a, b): return (a b) // gcd(a, b) Function to find max lcm value among all representations of n def findLCM(arr): global Landau nth_lcm = arr for i in range(1, len(arr)): nth_lcm = lcm(nth_lcm, arr[i]) # Calculate Landau's value Landau = max(Landau, nth_lcm) Recursive function to find different ways in which n can be written as sum of atleast one positive integers def findWays(arr, i, n): # Check if sum becomes n, # consider this representation if (n == 0): findLCM(arr) # Start from previous element # in the representation till n for j in range(i, n + 1): # Include current element # from representation arr.append(j) # Call function again # with reduced sum findWays(arr, j, n - j) # Backtrack - remove current # element from representation arr.pop() Function to find the Landau's function def Landau_function(n): arr = [] # Using recurrence find different # ways in which n can be written # as a sum of atleast one +ve integers findWays(arr, 1, n) # Print the result print(Landau) Driver Code Given N N = 4 Function call Landau_function(N) This code is contributed by chitranayal ```` C# ```` // C# program for the above approach using System; using System.Collections.Generic; class GFG{ // To store Landau's function of the number static int Landau = int.MinValue; // Function to return gcd of 2 numbers static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Function to return LCM of two numbers static int lcm(int a, int b) { return (a b) / gcd(a, b); } // Function to find max lcm value // among all representations of n static void findLCM(List arr) { int nth_lcm = arr; for(int i = 1; i < arr.Count; i++) nth_lcm = lcm(nth_lcm, arr[i]); // Calculate Landau's value Landau = Math.Max(Landau, nth_lcm); } // Recursive function to find different // ways in which n can be written as // sum of atleast one positive integers static void findWays(List arr, int i, int n) { // Check if sum becomes n, // consider this representation if (n == 0) findLCM(arr); // Start from previous element // in the representation till n for(int j = i; j <= n; j++) { // Include current element // from representation arr.Add(j); // Call function again // with reduced sum findWays(arr, j, n - j); // Backtrack - remove current // element from representation arr.RemoveAt(arr.Count - 1); } } // Function to find the Landau's function static void Landau_function(int n) { List arr = new List(); // Using recurrence find different // ways in which n can be written // as a sum of atleast one +ve integers findWays(arr, 1, n); // Print the result Console.Write(Landau); } // Driver Code public static void Main(String[] args) { // Given N int N = 4; // Function call Landau_function(N); } } // This code is contributed by amal kumar choubey ```` JavaScript ```` // Javascript program for the above approach // To store Landau's function of the number var Landau = -1000000000; // Function to return gcd of 2 numbers function gcd(a, b) { if (a == 0) return b; return gcd(b % a, a); } // Function to return LCM of two numbers function lcm(a, b) { return (a b) / gcd(a, b); } // Function to find max lcm value // among all representations of n function findLCM(arr) { var nth_lcm = arr; for (var i = 1; i < arr.length; i++) nth_lcm = lcm(nth_lcm, arr[i]); // Calculate Landau's value Landau = Math.max(Landau, nth_lcm); } // Recursive function to find different // ways in which n can be written as // sum of atleast one positive integers function findWays(arr, i, n) { // Check if sum becomes n, // consider this representation if (n == 0) findLCM(arr); // Start from previous element // in the representation till n for (var j = i; j <= n; j++) { // Include current element // from representation arr.push(j); // Call function again // with reduced sum findWays(arr, j, n - j); // Backtrack - remove current // element from representation arr.pop(); } } // Function to find the Landau's function function Landau_function(n) { arr = []; // Using recurrence find different // ways in which n can be written // as a sum of atleast one +ve integers findWays(arr, 1, n); // Print the result document.write( Landau); } // Driver Code // Given N var N = 4; // Function Call Landau_function(N); // This code is contributed by rrrtnx. ```` Output: 4 Time Complexity: O(2N) Auxiliary Space: O(N2) A faster algorithm Lemma. There is an optimal answer where every number in partition is either 1 or of form p^a, where p is prime. Proof. Suppose that in an optimal partition we have some n which has more than one prime factors. Then we can write n=mk, where gcd(m, k) = 1 and m>1 and k>1. Here we should note than mk ≥ m+k, so (mk - m - k) ≥ 0. So let's remove n from partition and replace it with m and k and (mk - m - k) 1's (example : [..., 12, ...] -> [..., 3 4 1 1 1 1 1, ...]). The sum of the array hasn't changed, so it is still a partition of n, and the lcm, obviously, hasn't changed. So it is still an optimal partition. We do that replacement several times until there is no number with >1 prime factors. QED Lets define a function g(n, p), where n is prime, as the optimal answer where all numbers only consist of primes which are less than p. (Example: g(4, 3) = 4. The optimal answer is [2^2]. Note that the answer [2, 3] doesn't count since 3≥ p). Let's also say that g(0, p) = 1 and g(n, 2) = 1 so it's a bit easier for us. Then we can write a recursive formula for g(n, p): g(n, p) is maximum of: g(n, prev. prime of p) - we add nothing and restrict all next primes to be less than prev. prime of p g(n - p, prev. prime of p)p - we add p and restrict all next primes to be less than prev. prime of p g(n - p^2, prev. prime of p)p^2 - we add p^2 and restrict all next primes to be less than prev. prime of p g(n - p^3, prev. prime of p)p^3 - we add p^3 and restrict all next primes to be less than prev. prime of p etc... So g(n, min prime which is >n) is our answer. We can get the answer by using dynamic programming. C++ ```` include include using namespace std; int main() { int n = 4; vector<int> primes; vector<int> max_prime(2 n + 1, 0); for (int i = 2; i <= 2 n; ++i) { if (max_prime[i] == 0) { max_prime[i] = i; primes.push_back(i); } for (int j = 0; j < primes.size() && primes[j] <= max_prime[i] && 1ll primes[j] i <= 2 n; ++j) { max_prime[primes[j] i] = max_prime[i]; } } vector<vector<long long> > g( n + 1, vector<long long>(primes.size() + 1, 1ll)); for (int p = 1; p <= primes.size(); ++p) { for (int i = 0; i <= n; ++i) { g[i][p] = g[i][p - 1]; long long power = primes[p - 1]; while (power <= i) { long long new_option = g[i - power][p - 1] power; g[i][p] = max(g[i][p], new_option); power = primes[p - 1]; } } } cout << g[n][primes.size()] << endl; return 0; } ```` Java ```` import java.util.; import java.math.; public class Main { public static void main(String[] args) { int n = 4; List primes = new ArrayList<>(); int[] maxPrime = new int[2 n + 1]; Arrays.fill(maxPrime, 0); for (int i = 2; i <= 2 n; ++i) { if (maxPrime[i] == 0) { maxPrime[i] = i; primes.add(i); } for (int j = 0; j < primes.size() && primes.get(j) <= maxPrime[i] && (long) primes.get(j) i <= 2 n; ++j) { maxPrime[primes.get(j) i] = maxPrime[i]; } } long[][] g = new long[n + 1][primes.size() + 1]; for (int i = 0; i <= n; ++i) { Arrays.fill(g[i], 1); } for (int p = 1; p <= primes.size(); ++p) { for (int i = 0; i <= n; ++i) { g[i][p] = g[i][p - 1]; int power = primes.get(p - 1); while (power <= i) { long newOption = g[i - power][p - 1] power; g[i][p] = Math.max(g[i][p], newOption); power = primes.get(p - 1); } } } System.out.println(g[n][primes.size()]); } } ```` Python3 ```` import math n = 4 primes = [] max_prime = (2 n + 1) for i in range(2, 2 n + 1): if max_prime[i] == 0: max_prime[i] = i; primes.append(i); j = 0 while j < len(primes) and primes[j] <= max_prime[i] and primes[j] i <= 2 n: max_prime[primes[j] i] = max_prime[i] j += 1 g = [[1 for i in range(len(primes) + 1)] for j in range(n + 1)] for p in range(1, len(primes) + 1): for i in range(0, n + 1): g[i][p] = g[i][p - 1] power = primes[p - 1] while power <= i: new_option = g[i - power][p - 1] power; g[i][p] = max(g[i][p], new_option); power = primes[p - 1]; print(g[n][len(primes)]) ```` C# ```` // Importing required libraries using System; using System.Collections.Generic; // Defining the main class class MainClass { static void Main() { // Initializing the value of n int n = 4; // Declaring and initializing the lists List<int> primes = new List<int>(); List<int> max_prime = new List<int>(); for (int i = 0; i <= 2 n; ++i) { max_prime.Add(0); } // Finding the prime numbers for (int i = 2; i <= 2 n; ++i) { if (max_prime[i] == 0) { max_prime[i] = i; primes.Add(i); } for (int j = 0; j < primes.Count && primes[j] <= max_prime[i] && 1L primes[j] i <= 2 n; ++j) { max_prime[primes[j] i] = max_prime[i]; } } // Initializing the 2D list with all values as 1 List<List<long> > g = new List<List<long> >(); for (int i = 0; i <= n; ++i) { List<long> row = new List<long>(); for (int j = 0; j <= primes.Count; ++j) { row.Add(1L); } g.Add(row); } // Finding the maximum product of primes such that // the sum of the primes is n for (int p = 1; p <= primes.Count; ++p) { for (int i = 0; i <= n; ++i) { g[i][p] = g[i][p - 1]; long power = primes[p - 1]; while (power <= i) { long new_option = g[i - (int)power][p - 1] power; g[i][p] = Math.Max(g[i][p], new_option); power = primes[p - 1]; } } } // Printing the maximum product of primes such that // the sum of the primes is n Console.WriteLine(g[n][primes.Count]); } } ```` JavaScript ```` // Import required libraries // None needed in JavaScript // Define the main function function main() { const n = 4; const primes = []; const maxPrime = new Array(2 n + 1).fill(0); // Calculate the maximum prime factor for each number between 2 and 2n for (let i = 2; i <= 2 n; ++i) { if (maxPrime[i] === 0) { maxPrime[i] = i; primes.push(i); } for (let j = 0; j < primes.length && primes[j] <= maxPrime[i] && primes[j] i <= 2 n; ++j) { maxPrime[primes[j] i] = maxPrime[i]; } } // Create a 2D array g to store the maximum product for each i and j const g = new Array(n + 1).fill().map(() => new Array(primes.length + 1).fill(1)); // Fill in the g array for (let p = 1; p <= primes.length; ++p) { for (let i = 0; i <= n; ++i) { g[i][p] = g[i][p - 1]; let power = primes[p - 1]; while (power <= i) { const newOption = g[i - power][p - 1] power; g[i][p] = Math.max(g[i][p], newOption); power = primes[p - 1]; } } } // Print the maximum product for n console.log(g[n][primes.length]); } // Call the main function to run the program main(); ```` Time complexity: O(N^2 / logN) Proof. We use the fact that there are approx n/logn primes less than n and that log(sqrt(n)) = 1/2 log(n). For primes that are <= sqrt(N), we check no more than log2(N) powers. So the time complexity for these primes is O(N (sqrtN/0.5log(N)) logN) = O(N sqrtN) For primes that are > sqrt(N), we check exactly one power. So the time complexity for these primes is O(N (N/logN - sqrtN / 0.5logN) 1) = O(N N / logN) = O(N^2 / logN). So the time complexity is O(NsqrtN + N^2/logN) = O(N^2/logN). QED Space complexity: O(N^2 / logN). It should be noted that for large enough n you can't use standard integer types to get the answer. Instead, you should use long arithmetic. For multiplying you could use, for example, FFT. S spp____ Improve Article Tags : Mathematical Recursion DSA LCM Practice Tags : Mathematical Recursion Similar Reads Basics & Prerequisites Logic Building Problems Logic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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Here€™s the comparison of Linked List vs Arrays Linked List: 2 min readStack Data Structure A Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first 2 min readQueue Data Structure A Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems 2 min readTree Data Structure Tree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most 4 min readGraph Data Structure Graph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of 3 min readTrie Data Structure The Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this 15+ min read Algorithms Searching Algorithms Searching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input 2 min read[Sorting Algorithms A Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ 3 min read]( to Recursion The process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution 14 min readGreedy Algorithms Greedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get 3 min readGraph Algorithms Graph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net 3 min readDynamic Programming or DP Dynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of 3 min readBitwise Algorithms Bitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit 4 min read Advanced Segment Tree Segment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree 3 min readBinary Indexed Tree or Fenwick Tree Binary Indexed Trees are used for problems where we have following types of multiple operations on a fixed sized.Prefix Operation (Sum, Product, XOR, OR, etc). Note that range operations can also be solved using prefix. For example, range sum from index L to R is prefix sum till R (included minus pr 15 min readSquare Root (Sqrt) Decomposition Algorithm Square Root Decomposition Technique is one of the most common query optimization techniques used by competitive programmers. This technique helps us to reduce Time Complexity by a factor of sqrt(N) The key concept of this technique is to decompose a given array into small chunks specifically of size 15+ min readBinary Lifting Binary Lifting is a Dynamic Programming approach for trees where we precompute some ancestors of every node. It is used to answer a large number of queries where in each query we need to find an arbitrary ancestor of any node in a tree in logarithmic time.The idea behind Binary Lifting:1. Preprocess 15+ min readGeometry Geometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br 2 min read Interview Preparation Interview Corner This article serves as your one-stop guide to interview preparation, designed to help you succeed across different experience levels and company expectations. Here is what you should expect in a Tech Interview, please remember the following points:Tech Interview Preparation does not have any fixed s 3 min readGfG160 Are you preparing for technical interviews and would like to be well-structured to improve your problem-solving skills? Well, we have good news for you! GeeksforGeeks proudly presents GfG160, a 160-day coding challenge starting on 15th November 2024. 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14287	https://www.onlinemathlearning.com/integer-problems.html	Integer Word Problems Related Pages Consecutive Integer Word Problems Consecutive Integers 1 Consecutive Integers 2 More Algebra Word Problems In these lessons, we will look at Integer Word Problems that have more than two unknowns. In another set of lessons, we have some examples of Integer Word Problems that involve two unknowns. Share this page to Google Classroom Integer word problems with three unknowns involve setting up and solving a system of equations based on relationships between three integers. These problems often require finding three unknown integers that satisfy given conditions. Here are some examples of integer word problems with three unknowns, along with step-by-step solutions: The following diagram gives the steps to solve an integer word problem with three variables using Algebra. Integer Word Problems Consecutive Integers Problem Integer Word Problems with 2 Unknowns Integer Word Problems more than 2 Unknowns Integer Problems With More Than Two Unknowns Integer Problems with three unknowns are not necessarily more difficult than integer word problems with two unknowns. You just have to be careful when relating the different unknowns. Example: Jane and her friends were selling cookies. They sold 4 more boxes the second week than they did the first. On the third week, they doubled the sale of their second week. Altogether, they sold a total of 352 boxes. How many boxes did they sell in the third week? Solution: Step 1: Sentence: They sold 4 more boxes the second week than they did the first. On the third week, they doubled the sale of their second week. Assign variables: \| \| \| \| --- \| Let \| x = \| boxes sold in the first week \| \| \| x + 4 = \| boxes sold in the second week \| \| \| 2(x + 4) = \| boxes sold in the third week \| Sentence: She sold a total of 350 boxesx + x + 4 + 2(x + 4) = 352Remove the brackets and combine like terms x + x + 4 + 2x + 8 = 352 4x + 12 = 352 Isolate variable x 4x = 340 Step 2: The question asks for boxes sold in the third week. Plug x = 85 into 2(x + 4) = 178 Answer: In the third week, they sold 178 boxes. Example: The sum of three numbers is 12. The first is five times the second and the sum of the first and third is 9. Find the numbers. Advanced Consecutive Integer Problems Example: The largest of five consecutive even integers is 2 less than twice the smallest. Which of the following is the largest integer? Show Video Lesson Check out other Algebra Word Problems Age Word Problems, Average Word Problems, Coin Word Problems, Consecutive Integer Word Problems, Digit Word Problems, Distance Word Problems, Fraction Word Problems, Geometry Word Problems, Integer Word Problems, Interest Word Problems, Lever Word Problems, Mixture Word Problems, Money Word Problems, Motion & Distance Word Problems, Number Sequence Word Problems, Proportion Word Problems, Quadratic Equation Word Problems, Ratio Word Problems, Symbol Word Problems, Variation Word Problems, Work Word Problems. Try out our new and fun Fraction Concoction Game. Add and subtract fractions to make exciting fraction concoctions following a recipe. There are four levels of difficulty: Easy, medium, hard and insane. Practice the basics of fraction addition and subtraction or challenge yourself with the insane level. We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.
14288	http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/nuclidet.html	Radioactive Nuclides by Half-Life Radioactive Nuclides by Half-Life ================================= Nuclide Half-Life (years)Found in Nature? 50 V 6.0x10 15 yes 144 Nd 2.4x10 15 yes 174 Hf 2.0x10 15 yes 192 Pt 1x10 15 yes 115 In 6.0x10 14 yes 152 Gd 1.08x10 14 yes 123 Te 1.2x10 13 yes 190 Pt 6.9x10 11 yes 138 La 1.12x10 11 yes 147 Sm 1.06x10 11 yes 87 Rb 4.88x10 10 yes 187 Re 4.3x10 10 yes 176 Lu 3.5x10 10 yes 232 Th 1.40x10 10 yes 238 U 4.47x10 9 yes 40 K 1.25x10 9 yes 235 U 7.04x8 10 yes 244 Pu 8.2x10 7 yes 146 Sm 7.0x10 7 no 205 Pb 3.0x10 7 no 236 U 2.39x10 7 yes-P 129 I 1.7x10 7 yes-P 247 Cm 1.60x10 7 no 182 Hf 9.0x10 6 no 107 Pd 7x10 6 no 53 Mn 3.7x10 6 yes-P 135 Cs 3.0x10 6 no 97 Tc 2.6x10 6 no 237 Np 2.14x10 6 yes-P 150 Gd 2.1x10 6 no 10 Be 1.6x10 6 yes-P 93 Zr 1.5.0x10 6 no 98 Tc 1.5x10 6 no 153 Dy 1x10 6 no This table of radioactive nuclides sorted by their half-lives is taken from Miller, who uses it to point out that only those radionuclides with half-lives greater than 80 million years are found in nature. The exceptions are those marked with "Yes-P" indicating that they are being produced in nature, but would otherwise be missing. The point of Miller's use was to make the case that the Earth is more than 80 million years old since all isotopes with shorter halflives are no longer found in nature. It is also evident that the Earth is not infinitely old because all the radioactive series would have decay. There are various ways of modeling the age of the Earth, and the values center around 4.5 billion years. Beryllium-10 is found because it is continually produced by cosmic ray bombardment of the upper atmosphere. So very small amounts of this isotope are found in rainfall and sediment. Neptunium-237 is produced by cosmic ray bombardment of the moon. There are other short half-life isotopes found on the Earth from the natural radioactive series. Note: Information about isotopes may be found in tables linked to the Periodic table. Take the link to "Nuclear data" at the bottom of the display for any element. Modeling the age of the EarthIndex Nuclear Tables Reference Miller Ch 3 HyperPhysics NuclearR NaveGo Back
14289	https://www.themathdoctors.org/how-to-think-about-the-chain-rule/	Loading [MathJax]/jax/output/HTML-CSS/fonts/TeX/Main/Regular/GeometricShapes.js Skip to content How to Think About the Chain Rule Having recently helped some students (in person) with the rules of differentiation, I’m reminded to do so here, starting with the chain rule. It is easy to make this topic look harder than it really is; the two main ways to state the rule are often confusing, and different approaches fit different problems. We’ll try to untangle it. Confused by u’s We’ll start with a question from 1997: ``` Calculus Chain Rule I can't understand the chain rule. Every time I ask someone to explain it they use y's and u's, etc... could you give me the chain rule in easy terms, like how to do it, not just give me a formula like y=(U)^2? Thanks. Stu ``` A full statement of the chain rule tends to need lots of letters and tangled expressions. The short form Stu probably saw looks like \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} dydx=dydududx This says that if y is a function of u, and u is a function of x, then the derivative of the composite function y with respect to x is the product of the derivative of y with respect to u, and the derivative of u with respect to x. What this says is simple, and almost intuitive. For example, suppose your altitude, h, is a function of the distance, s, along a road, and your distance along the road is a function of time, t. The road has a certain slope at any time, which is the rate of change of altitude with respect to distance, \frac{dh}{ds}dhds meters up per meter forward; and your car has a certain speed, which is the rate of change of distance with respect to time, \frac{ds}{dt}dsdt meters per second. How fast are you going up? Every second you are moving \frac{ds}{dt}dsdt meters forward, and therefore \frac{dh}{ds}\cdot\frac{ds}{dt}dhds⋅dsdt meters up. That’s the derivative of the composite function h(t)h(t). But often we don’t have a “u” in the problem, or any other variable – just one big expression like \cos(\tan(5x-3))cos(tan(5x−3)). Then what? Doctor Scott answered: ``` Hi Stu! Good question. I was just skimming an excellent Calculus book written by Paul Foerster where this very question was addressed. His suggestion was that you should think of the chain rule as a process rather than a rule with a lot of du/dx and dy/dx's. So, here goes.... ``` It’s like using muscle memory rather than written instructions. Understanding composite functions ``` Remember that the chain rule is used to find the derivative of compositions of functions - that is, functions that have functions inside of them. For example, the function sin(x^2) can be thought of as a composition of two other functions, sin x and x^2, with the x^2 being INSIDE the sin function. Similarly, the function (x^2 - 5x + 8)^(1/2) is also a composition of two other functions, (x^2 - 5x + 8) and x^(1/2), with the first function being INSIDE the second. One more example? The function cos(tan(5x-3)) is the composition of three functions, 5x - 3 inside of tan x, inside of cos x. So the chain rule gets applied when there is some function INSIDE of another function. ``` We’ll be working all three of these examples. We traditionally represent composite functions as boxes connected by “plumbing” (or, if you prefer, “links in a chain”, the reason for the term “chain rule”): In particular, for our first example, we might think of it this way: This sort of diagram has to be read backward; the first function named in the expression is the last one in line. I like to think of it this way instead, which reads more naturally: Each function/fish “eats” the one in front of it, producing “meat” that is eaten by the one behind. Then, we have fish inside fish: We can see this in the expression: \require{AMSmath}y=\boxed{\sin\left(\,\boxed{x^2\strut}\,\right)} y=sin(x2) And we’ll apply the chain rule by following the “food chain” from the outside in. Understanding the chain rule The stuff that people have been telling you probably goes something like this: If y = sin(x^2), then we can write this function as the composition of y = sin u and u = x^2. (Again, notice that the x^2 is INSIDE of the sin function.) Then, dy/dx = dy/du du/dx. So, we have dy/dx = cos u 2x; but u = x^2, so we have dy/dx = 2x cos(x^2). Here, u is just a temporary name we’re giving to the result of the inside function, like this: This approach works fine when we are given named variables (as we’ll see later), but it gets in the way for problems like ours, where the function is written as one big expression. But we don’t need names; we can just do it: ``` How about another way? Let's think of the chain rule as a process. The derivative of a composite function is the DERIVATIVE OF THE OUTSIDE FUNCTION TIMES the DERIVATIVE OF THE INSIDE FUNCTION. In practice, here's how it works. Consider y = sin(x^2). The outside function is a sine function; its derivative is cosine, so we have (so far) cos(x^2). Now, INSIDE the sine function is x^2. Its derivative is 2x, so now we have 2x cos(x^2). Notice that there is no other function "inside" the x^2, so we are done. ``` The key idea is that we have to keep the same thing inside the derivative that was inside the function itself. I like to think of it like this, putting a box (or at least imagining it) around the inside function and thinking of it as a single entity (as if it were a variable): y=\boxed{\sin\left(\,\boxed{x^2\strut}\,\right)} y=sin(x2) y’=\cos\left(\,\boxed{x^2\strut}\,\right)\cdot\boxed{x^2\strut}\,’=\cos\left(\,\boxed{x^2\strut}\,\right)\cdot2x y′=cos(x2)⋅x2′=cos(x2)⋅2x To differentiate “sine of something”, we multiply “cosine of something” by the derivative of “something”. ``` Let's look at a couple more examples: y = (x^2 - 5x + 8)^(1/2). The OUTSIDE FUNCTION is basically a power rule problem, so we have 0.5(x^2 - 5x + 8)^(-1/2) using the power rule. The INSIDE FUNCTION is x^2 - 5x + 8; its derivative is 2x - 5, so we have y' = (2x - 5)(.5)(x^2 - 5x + 8)^(-1/2). ``` Here we have y=\boxed{\left(\,\boxed{x^2-5x+8\strut}\,\right)^{1/2}} y=(x2−5x+8)1/2 y’=\frac{1}{2}\left(\,\boxed{x^2-5x+8\strut}\,\right)^{-1/2}\cdot\boxed{x^2-5x+8\strut}\,’\=\frac{1}{2}\left(\,\boxed{x^2-5x+8\strut}\,\right)^{-1/2}\cdot(2x-5) y′=12(x2−5x+8)−1/2⋅x2−5x+8′=12(x2−5x+8)−1/2⋅(2x−5) This could also have been written as y=\boxed{\sqrt{\,\boxed{x^2-5x+8\strut}\,}\,} y=x2−5x+8−−−−−−−−−−−√ y’=\frac{1}{2\sqrt{\,\boxed{x^2-5x+8\strut}\,}\,}\cdot\,\boxed{x^2-5x+8\strut}\,’\=\frac{2x-5}{2\sqrt{\,\boxed{x^2-5x+8\strut}\,}\,} y′=12x2−5x+8−−−−−−−−−−−√⋅x2−5x+8′=2x−52x2−5x+8−−−−−−−−−−−√ I generally rewrite radicals as fractional powers, rather than memorize separate formulas for radicals. ``` y = cos(tan(5x-3)). The outermost function is a cosine, so its derivative is negative sine: -sin(tan(5x-3)). Inside the cosine is a tan function; its derivative is sec^2, so we now have sec^2 (5x-3) (-sin(tan(5x-3)) Finally, inside of the tan function is 5x-3; its derivative is 5. So, FINALLY, we have 5 sec^2 (5x-3) (-sin(tan(5x-3)) Or, simplifying, we get y' = -5 sec^2 (5x-3) sin(tan(5x-3)) ``` This has three layers (outside, middle, and inside): y=\boxed{\cos\left(\,\boxed{\tan\left(\,\boxed{5x-3\strut}\,\right)}\,\right)}\\y’=-\sin\left(\,\boxed{\tan\left(\,\boxed{5x-3\strut}\,\right)}\,\right)\cdot\boxed{\tan\left(\,\boxed{5x-3\strut}\,\right)}\,’\ =-\sin\left(\,\boxed{\tan\left(\,\boxed{5x-3\strut}\,\right)}\,\right)\cdot\sec^2\left(\,\boxed{5x-3\strut}\,\right)\cdot\boxed{5x-3\strut}\,’\=-\sin\left(\,\boxed{\tan\left(\,\boxed{5x-3\strut}\,\right)}\,\right)\cdot\sec^2\left(\,\boxed{5x-3\strut}\,\right)\cdot5 y=cos(tan(5x−3))y′=−sin(tan(5x−3))⋅tan(5x−3)′=−sin(tan(5x−3))⋅sec2(5x−3)⋅5x−3′=−sin(tan(5x−3))⋅sec2(5x−3)⋅5 So, it helps a lot to think of the chain rule as: The derivative of the outside TIMES the derivative of what's inside! Confused by function notation Consider this question from 1999, using another notation, which is technically more precise, but even more confusing to read: ``` Chain Rule Notation I'm trying to figure out these questions: Formula : (f◦g)'(x)= g'(x) · f'[g(x)] 1) f(x) = 2x+6 g(x) = 3x-4 (f◦g)'(x) = 3 · 2 = 6 I know that g'(x)= 3 but how about f'[g(x)]? How does 2 come about? I don't understand how it's done. 2) g(x) = 2x^2 + 5 h(x) = x^4 (g◦h)'(x) = 4x^3 · 4x^4 = 16x^7 It's the same here. I know how to differentiate h(x) but I got stuck on g'[h(x)]. How does 4x^4 come by? Please help me, Thanks. ``` The notation in the question means exactly what we’ve been doing. I prefer to write it in this order: (f\circ g)'(x)=f'(g(x))\cdot g'(x). (f∘g)′(x)=f′(g(x))⋅g′(x). This means that the derivative of a composite function h(x)=(f\circ g)(x)=f(g(x))h(x)=(f∘g)(x)=f(g(x)) is the derivative of the outside function, f, applied to the inside function, g, times the derivative of the inside function, g. The way I like to think about it, using the idea we saw above, is \require{AMSsymbols}\boxed{f\left(\square\right)}\,’=f\,’\left(\square\right)\cdot\square\,’. In the examples here, we are given the two functions separately (but with the same variable x, rather than an intermediate variable u). In the first question, functions were written as a single composite expression; in that form, respectively, these would be 2(3x-4)+6 and 2(x^4)^2+5. In this form, the inside functions could be marked like this: 2\left(\,\boxed{3x-4\strut}\,\right)+6 and 2\left(\,\boxed{x^4\strut}\,\right)^2+5 Doctor Mitteldorf answered, recommending the u formulation we avoided above: ``` Dear Eric, The chain rule can be taught in such a way that it's quite transparent, or it can be made utterly mysterious with bad notation. It looks as if you've been a victim of the latter. The chain rule is about taking the derivative of a function of a function. Instead of f being a function of x, we have f is a function of g, and g is a function of x. In this notation, the chain rule can be written: df/dx = df/dg · dg/dx It seems almost obvious when you write it that way. Just "cancel out" the dg's in the numerator and denominator. ``` In defense of the function notation form, that makes it explicit that the derivative of f is applied to g(x), not to x; and it emphasizes that the derivative is a new function f ‘, not a new variable. And this form is the most suitable for these problems, in which functions are named. Doctor Mitteldorf here used g not only as a function name, but also as a variable representing its output. He did this, presumably, to avoid bringing in another variable (the u that was confusing above) to represent g(x). And many authors prefer to avoid the d notation precisely because it looks so “obvious”, as if you are just canceling in a fraction. (See What Derivative Notations Mean.) The latter notation is very useful as a reminder of this rule, but mathematicians are uncomfortable talking as if that is all there is to it. (See What Do dx and dy Mean?) ``` In your example (1), f(x) = 2x+6 g(x) = 3x-4 The teacher gave you a notation that's deliberately confusing. You have to remember that the x in these equations is a dummy variable. The top equation just says f is a function that takes its argument, multiplies it by 2, then adds 6. The x is there just as a placeholder. You can replace it with a or b or theta or phi and the equation says exactly the same thing. ``` This is important: The x‘s in the two function definitions will represent different numbers. So giving them different names helps to differentiate them (no pun intended). ``` But in this case, you want to replace it with g: f(g) = 2g+6 Is it obvious why I want to replace the x by a g? It's because f◦g means "f composed with g," or "the function f taken of the function g." ``` Distinguishing the variables called x is essential. ``` Coming back now to problem 1, let's do it two ways. First, we'll actually find fg and differentiate it. Second, we'll use the chain rule. Then we'll be in a position to check that the two answers are the same. First, f(g) = 2g+6 g(x) = 3x-4 Substituting the second equation into the first, you have f(x) = 2(3x-4)+6 = 6x-2 It's obvious, then, that f'(x) = 6. ``` This process of expanding the composite function can be time-consuming; the chain rule is usually a time-saver. The point here is that it is not a necessity! It gives the same result as direct differentiation. ``` Second, we'll use the chain rule: df/dx = df/dg · dg/dx = 2 · 3 = 6 Hence, we get the same answer both ways. ``` This was a particularly simple example, where both derivatives are constant. Powers of trig functions Certain functions can make this harder. Here is a question from 1998: ``` Trigonometry and the Chain Rule I have three questions that have me stumped. I need to differentiate the following: y = 2 csc^3(sqrt(x)) y = x/2 - (sin(2x))/4 y = (1 - cos(x))/sin(x) ``` Doctor Santu answered, solving his own examples so Amanda could learn by doing her own homework: ``` Amanda: These all have to do with the Chain Rule. Here's the basic idea. Suppose: y = sin(x^3 + tan(x)). How do you find the derivative? Think of x^3 + tan x as a big BLOB. So we really need to find the derivative of: y = sin(BLOB) Well, the rule says that the derivative of sin(BLOB) is simply cos(BLOB) multiplied by the derivative of the BLOB itself. A word on notation: I'm going to write y' for the derivative of y (instead of dy/dx). ``` The BLOB idea is the same as my “something” or my boxes. I’ve been known to use the same word. ``` Now, in this case: y' = cos(x^3 + tan(x)) (3x^2 + sec^2(x)) because BLOB is x^3 + tan(x), and the derivative of the BLOB is 3x^2 + sec^2(x). ``` Move the exponent Now we come to something important: When we write a power of a trig function by putting an exponent on the function name (which, as I explained here, is a notation left over from before general function notation was introduced, as is permission to omit parentheses), we hide the fact that the power is the outside function, and the trig the inside: ``` Let's try another example: y = sin^3(x^3 + tan x) This is really: y = [ sin(x^3 + tan x) ]^3 Using our previous terminology, the derivative of (blob)^3 is simply: 3(blob)^2 (the derivative of blob itself). In this case: y' = 3[sin(x^3 + tan(x))]^2 (derivative of sin(x^3 + tan(x))) = 3[sin(x^3 + tan (x))]^2 cos(x^3 + tan(x)) (derivative of x^3 + tan(x)) = 3[sin(x^3 + tan(x))]^2 cos(x^3 + tan(x)) (3x^2 + sec^2(x)) ``` By writing the exponent on the outside, we make it easier to see that the sine is the inside function. In my formulation with boxes, this is y=\boxed{\,\left(\,\boxed{\sin\left(\,\boxed{x^3+\tan(x)}\,\right)}\,\right)^3}\\y’=3\left(\,\boxed{\sin\left(\,\boxed{x^3+\tan(x)}\,\right)}\,\right)^2\cdot\boxed{\sin\left(\,\boxed{x^3+\tan(x)}\,\right)}\,’\ =3\left(\,\boxed{\sin\left(\,\boxed{x^3+\tan(x)}\,\right)}\,\right)^2\cdot\cos\left(\,\boxed{x^3+\tan(x)}\,\right)\cdot\boxed{x^3+\tan(x)}\,’\=3\left(\,\boxed{\sin\left(\,\boxed{x^3+\tan(x)}\,\right)}\,\right)^2\cdot\cos\left(\,\boxed{x^3+\tan(x)}\,\right)\cdot\left(3x^2+\sec^2(x)\right) Although the exponent is after the parentheses, we can see it clearly as on the outside, which wasn’t obvious originally. Peeling the onion ``` In the chain rule, the basic idea is to peel the onion from the outside. You want to take the derivative of a function within a function within a function. You take the derivative of the outermost function relative to the stuff that's inside it, then multiply that by the derivative of the inside expression, relative to the expression inside the expression, and so on, all the way down to the tiniest little x all the way inside. (And some people even stick a "1" on at the end, because the derivative of an x is just 1. I think that's overdoing it a bit.) The Chain Rule needs quite a lot of imagination to see these formulas as expressions within expressions, and ideally you should have a friend sit by you and point out how to "peel the onion" layer by layer. ``` I’ve skipped a couple examples; he closed with an example five layers deep: ``` One final example: y = sin(tan(sin^2(x^7 + 3x))) y' = ...? You must first take the derivative of sin (expression), relative to the expression that's inside. You multiply that by the derivative of the tan (inside expression). You multiply that by the derivative of [sin (x^7 + 3x)]^2, because sin^2 (x^7+3x) means [sin(x^7+3x)]^2. That, in turn, will contain the derivative of sin(x^7 + 3x), which in turn will contain the derivative of (x^7 + 3x), which is 7x^6 + 3. It's important to put the proper expression inside the various partial expressions. So: y' = cos(tan ...) sec^2(sin ...) 2[sin ...] cos(x^7 + 3x) (7x^6 + 3) You have to know what I have left out, and you must know how to put it in. I suggest you complete the derivative of that derivative just above, inserting all the expressions that would take the place of the ...s, then try the problems you're interested in. (All of us at Dr. Math had to practice these too.) ``` The function can be seen as y=\boxed{\sin\left(\,\boxed{\tan\left(\,\boxed{\left(\,\boxed{\sin\left(\,\boxed{x^7+3x\strut}\right)}\,\right)^2}\,\right)}\,\right)} y’=\cos\left(\,\boxed{\tan\left(\,\boxed{\left(\,\boxed{\sin\left(\,\boxed{x^7+3x\strut}\right)}\,\right)^2}\,\right)}\,\right)\\cdot\sec^2\left(\,\boxed{\left(\,\boxed{\sin\left(\,\boxed{x^7+3x\strut}\right)}\,\right)^2}\,\right)\\cdot2\,\boxed{\sin\left(\,\boxed{x^7+3x\strut}\,\right)}\\cdot\cos\left(\,\boxed{x^7+3x\strut}\,\right)\\cdot\left(7x^6+3\strut\right) Exponential functions We’ll close with this, from 2004: ``` Chain Rule Applied to Exponential Functions At a time t hours after it was administered, the concentration of a drug in the body is f(t) = 27 e^(-0.14t) ng/ml. What is the concentration 4 hours after it was administered? At what rate is the concentration changing at that time? I got lost finding the derivative of the problem to find the rate of change. Part 1. f(4) = 27 e^(-0.14(4)) = 27e^(-0.56) = 15.42 ng/ml Part 2. Chain rule = f'(g(x)) x g'(x) = 27'(e^-0.56) x (e^-0.56) I get lost after that. I don't know if I am going in the right direction and I think the derivative of 27 = 0, so the whole first half of the problem would equal 0. ``` This is not really a hard application of the chain rule, but the notation is a little awkward. (A major error is in replacing t with 4 before differentiating, so there is no variable left!) Doctor Mike answered: ``` Hi Brendan, The derivative of a constant times a function is just that constant, times the derivative of the function. So, the derivative of 27 e^(-.14t) is 27 times the derivative of e^(-0.14t) . So, let's just concentrate on the derivative of e^(-0.14t) and you can put it all together later. OK? ``` We can think of the 27 as representing an outer function a(x)=27x, whose derivative is 27; but it’s easier just to let it pass through the process. ``` People often have problems with the Chain Rule applied to exponentials, because of not being clear of what is the "outside" function and what is the "inside" function. That's why I like to use the notation exp(x) in place of the notation e^x when we do problems like this. Also, let's give a function name "h" to what is in your original exponent. That is, define it like h(t) = -0.14t . Then, e^(-0.14t) can be written as exp( h(t) ) which clearly shows that the exponential is the outside function, and what we have called "h" is the inside function. ``` So we have the function f(t)=\exp(h(t)), where \exp(x)=e^x and h(t)=-0.14t. ``` What do we do with this now? To use the Chain Rule you have to know how to differentiate both functions that are involved. The exponential function is its own derivative. exp'(t) = exp(t). You should have seen this already. For the other one, h'(t) = -0.14 . That you should have seen a long time ago. Right? So, to use the Chain Rule on exp( h(t) ) you get exp'( h(t) ) h'(t) which is exp( h(t) ) (-0.14) . In this last expression, exp( h(t) ) is the derivative of the outside function evaluated at the inside function, and (-0.14) is the derivative of the inside function. ``` We don’t have to do this renaming, as long as we see the exponent as the inside function: f(t)=e^{\boxed{-0.14t}} If you find the “exp” function helpful, use it: f(t)=\exp\left(\,{\boxed{-0.14t}}\,\right) ``` The general notation for differentiating f(t) = g( h(t) ) with the C.R. is simply f'(t) = g'( h(t) ) g'(t) . If you spend some time to get your function expressed in that way, then the rest will be easier. ``` Carrying out the process, we have \boxed{27\boxed{e^{\boxed{-0.14t}}}}’\=27\boxed{e^{\boxed{-0.14t}}}’\=27e^{\boxed{-0.14t}}\cdot\boxed{-0.14t}’\=27e^{\boxed{-0.14t}}\cdot(-0.14) 3 thoughts on “How to Think About the Chain Rule” Pingback: How to Think About the Product and Quotient Rules – The Math Doctors Pingback: Implicit Differentiation: What to Do When It’s “Wrong” – The Math Doctors Pingback: Proving the Chain Rule: Details Matter – The Math Doctors Leave a Comment Cancel Reply
14290	https://chemcollective.org/activities/info/77	Determining Stoichiometric Coefficients Info Virtual Lab About Teachers Help Search Search Resources by Topic Stoichiometry Thermochemistry Kinetics Equilibrium Acid-Base Chemistry Solubility Oxidation/Reduction and Electrochemistry Analytical Chemistry/Lab Techniques Physical Chemistry Properties of Solutions Resources by Type Virtual Labs Autograded Problems Tutorials Scenario-Based Activities Online Courses Molecular Level Visualizations Simulations Concept Tests You are here: Home> Stoichiometry / Virtual Labs> Determining Stoichiometric Coefficients Determining Stoichiometric Coefficients Info Download AssignmentDownload Offline VersionGo to Activity Description In this activity, students use the virtual lab to determine how 4 unknown substances react with each other including their stoichiometric coefficients. \| Resource Type \| Virtual Lab The Virtual Lab is an online simulation of a chemistry lab. It is designed to help students link chemical computations with authentic laboratory chemistry. The lab allows students to select from hundreds of standard reagents (aqueous) and manipulate them in a manner resembling a real lab.More information and offline downloads. Please scroll below to find our collection of pre-written problems, they have been organized by concept and ranked by difficulty. \| \| Topic(s) \| Stoichiometry \| \| Subtopic(s) \| Reaction Stoichiometry and Limiting Reagents \| \| Education Level \| High School / Introductory Chemistry First-Year Undergraduate / General \| \| Difficulty \| Challenging \| \| Contributor(s) \| Mr. Donovan Lange, Dr. Dave Yaron, Carnegie Mellon University \| The ChemCollective site and its contents are licensed under a Creative Commons Attribution 3.0 NonCommercial-NoDerivs License.Contact Us
14291	https://runestone.academy/ns/books/published/pythonds/BasicDS/ConvertingDecimalNumberstoBinaryNumbers.html	4.8. Converting Decimal Numbers to Binary Numbers¶ In your study of computer science, you have probably been exposed in one way or another to the idea of a binary number. Binary representation is important in computer science since all values stored within a computer exist as a string of binary digits, a string of 0s and 1s. Without the ability to convert back and forth between common representations and binary numbers, we would need to interact with computers in very awkward ways. Integer values are common data items. They are used in computer programs and computation all the time. We learn about them in math class and of course represent them using the decimal number system, or base 10. The decimal number and its corresponding binary equivalent are interpreted respectively as But how can we easily convert integer values into binary numbers? The answer is an algorithm called “Divide by 2” that uses a stack to keep track of the digits for the binary result. The Divide by 2 algorithm assumes that we start with an integer greater than 0. A simple iteration then continually divides the decimal number by 2 and keeps track of the remainder. The first division by 2 gives information as to whether the value is even or odd. An even value will have a remainder of 0. It will have the digit 0 in the ones place. An odd value will have a remainder of 1 and will have the digit 1 in the ones place. We think about building our binary number as a sequence of digits; the first remainder we compute will actually be the last digit in the sequence. As shown in Figure 5, we again see the reversal property that signals that a stack is likely to be the appropriate data structure for solving the problem. The Python code in ActiveCode 1 implements the Divide by 2 algorithm. The function divideBy2 takes an argument that is a decimal number and repeatedly divides it by 2. Line 7 uses the built-in modulo operator, %, to extract the remainder and line 8 then pushes it on the stack. After the division process reaches 0, a binary string is constructed in lines 11-13. Line 11 creates an empty string. The binary digits are popped from the stack one at a time and appended to the right-hand end of the string. The binary string is then returned. 18 1 from pythonds.basic import Stack from pythonds basic import Stack 2 3 def divideBy2(decNumber): def divideBy2 decNumber 4 remstack = Stack() remstack = Stack 5 6 while decNumber > 0: while decNumber> 0 7 rem = decNumber % 2 rem = decNumber% 2 8 remstack.push(rem) remstack push rem 9 decNumber = decNumber // 2 decNumber = decNumber// 2 10 11 binString = "" binString = "" 12 while not remstack.isEmpty(): while not remstack isEmpty 13 binString = binString + str(remstack.pop()) binString = binString + str remstack pop 14 15 return binString return binString 16 17 print(divideBy2(42)) print divideBy2 42 18 Activity: 4.8.1 Converting from Decimal to Binary (divby2) The algorithm for binary conversion can easily be extended to perform the conversion for any base. In computer science it is common to use a number of different encodings. The most common of these are binary, octal (base 8), and hexadecimal (base 16). The decimal number and its corresponding octal and hexadecimal equivalents and are interpreted as and The function divideBy2 can be modified to accept not only a decimal value but also a base for the intended conversion. The “Divide by 2” idea is simply replaced with a more general “Divide by base.” A new function called baseConverter, shown in ActiveCode 2, takes a decimal number and any base between 2 and 16 as parameters. The remainders are still pushed onto the stack until the value being converted becomes 0. The same left-to-right string construction technique can be used with one slight change. Base 2 through base 10 numbers need a maximum of 10 digits, so the typical digit characters 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 work fine. The problem comes when we go beyond base 10. We can no longer simply use the remainders, as they are themselves represented as two-digit decimal numbers. Instead we need to create a set of digits that can be used to represent those remainders beyond 9. 21 1 from pythonds.basic import Stack from pythonds basic import Stack 2 3 def baseConverter(decNumber,base): def baseConverter decNumber base 4 digits = "0123456789ABCDEF" digits = "0123456789ABCDEF" 5 6 remstack = Stack() remstack = Stack 7 8 while decNumber > 0: while decNumber> 0 9 rem = decNumber % base rem = decNumber% base 10 remstack.push(rem) remstack push rem 11 decNumber = decNumber // base decNumber = decNumber// base 12 13 newString = "" newString = "" 14 while not remstack.isEmpty(): while not remstack isEmpty 15 newString = newString + digits[remstack.pop()] newString = newString + digits remstack pop 16 17 return newString return newString 18 19 print(baseConverter(25,2)) print baseConverter 25 2 20 print(baseConverter(25,16)) print baseConverter 25 16 21 Activity: 4.8.2 Converting from Decimal to any Base (baseconvert) A solution to this problem is to extend the digit set to include some alphabet characters. For example, hexadecimal uses the ten decimal digits along with the first six alphabet characters for the 16 digits. To implement this, a digit string is created (line 4 in Listing 6) that stores the digits in their corresponding positions. 0 is at position 0, 1 is at position 1, A is at position 10, B is at position 11, and so on. When a remainder is removed from the stack, it can be used to index into the digit string and the correct resulting digit can be appended to the answer. For example, if the remainder 13 is removed from the stack, the digit D is appended to the resulting string. Self Check Q-3: What is value of 25 expressed as an octal number? Activity: 4.8.3 Fill in the Blank (baseconvert1) Q-4: What is value of 256 expressed as a hexidecimal number? Activity: 4.8.4 Fill in the Blank (baseconvert2) Q-5: What is value of 26 expressed in base 26? Activity: 4.8.5 Fill in the Blank (baseconvert3) Activity: 4.8.6 YouTube (video_Stack2) You have attempted 1 of 7 activities on this page
14292	https://proofwiki.org/wiki/Cosine_of_Sum/Proof_1	Cosine of Sum/Proof 1 - ProofWiki Cosine of Sum/Proof 1 From ProofWiki <Cosine of Sum Jump to navigationJump to search Theorem cos(a+b)=cos a cos b−sin a sin b cos⁡(a+b)=cos⁡a cos⁡b−sin⁡a sin⁡b Proof cos(a+b)+i sin(a+b)cos⁡(a+b)+i sin⁡(a+b)==e i(a+b)e i(a+b)Euler's Formula ==e i a e i b e i a e i bExponential of Sum ==(cos a+i sin a)(cos b+i sin b)(cos⁡a+i sin⁡a)(cos⁡b+i sin⁡b)Euler's Formula ==(cos a cos b−sin a sin b)+i(sin a cos b+cos a sin b)(cos⁡a cos⁡b−sin⁡a sin⁡b)+i(sin⁡a cos⁡b+cos⁡a sin⁡b)Complex Numbers form Field The result follows by equating the real parts. ■◼ Retrieved from " Categories: Proven Results Cosine of Sum Navigation menu Personal tools Log in Request account Namespaces Page Discussion [x] English Views Read View source View history [x] More Search Navigation Main Page Community discussion Community portal Recent changes Random proof Help FAQ P r∞f W i k i P r∞f W i k i L A T E X L A T E X commands ProofWiki.org Proof Index Definition Index Symbol Index Axiom Index Mathematicians Books Sandbox All Categories Glossary Jokes To Do Proofread Articles Wanted Proofs More Wanted Proofs Help Needed Research Required Stub Articles Tidy Articles Improvements Invited Refactoring Missing Links Maintenance Tools What links here Related changes Special pages Printable version Permanent link Page information This page was last modified on 21 March 2019, at 15:18 and is 640 bytes Content is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Privacy policy About ProofWiki Disclaimers
14293	https://r-knott.surrey.ac.uk/PascalsTriangle/pascalsTriangle.html	An introduction to Pascal's Triangle of numbers and applications in algebra, coin tossing, with some basic theorems illustrated and proved. The calculators and other effects on this page require JavaScript but you appear to have switched JavaScript off (it is disabled) in this browser. Please go to the Preferences or Properties menu item for this browser and enable it and then Reload this page. Contents of this page The icon means there is a You Do The Maths... section of questions to start your own investigations. The calculator icon indicates that there is a live interactive calculator in that section. 1 Choosing a Selection 1.1 Notation 1.2 Pascal's Triangle 1.3 A recursive definition 1.4 Pascal's Triangle (Binomial) Calculator 2 Binomial expansion using Pascal's Triangle 2.1 The rows as coefficients 2.2 The columns as coefficients 3 Divisibility in Pascal's Triangle 3.1 You do the maths... 3.2 The Art of Pascal's Triangle Remainders 3.3 Remainder Patterns Calculator 4 How often does a number appear in Pascal's Triangle? 4.1 Pascal's Triangle (Binomial) Calculator 4.2 You do the maths... 5 The Fibonacci Numbers in Pascal's Triangle 5.1 Why do the Diagonals sum to Fibonacci numbers? 5.2 Another arrangement of Pascal's Triangle 5.3 You do the maths... 6 Galton's Quincunx 6.1 A Simulator 6.2 You do the maths... 6.3 Block Walking 6.4 Paths and Coins Calculator 7 Patterns in the Binomial Coefficients 7.1 Symmetry on rows 7.1.1 Row sums 7.1.2 Alternating sums along a row 7.2 Formulas for Binomial Coefficients 7.2.1 Recursive 7.2.2 Sums and differences of neighbours 7.2.3 Divisibility 7.3 The Hexagon or Star of David Products 7.3.1 The gcd of the triangles is the same 7.3.2 Hexagon Calculator 7.3.3 More recursions 7.4 Sum of squares of a row 7.4.1 Proof by "Committees" 7.4.2 Proof by "block walking" 7.5 The Hockey Stick Theorems 7.5.1 Hockey Sticks Demonstrator 7.6 Central Binomials 7.6.1 Central Binomials Calculator 8 Computing Binomials efficiently 8.1 Exact integer limits 8.2 Updating Binomials 8.2.1 The 8 Neighbours recursions 8.2.2 You do the maths... 8.3 Legendre's Prime Power Factors in Binomial(n,r) 8.3.1 Patterns in the Prime Factors of Binomial(2n,n) 8.3.2 Factorial Factors Calculator 8.3.3 You do the maths... 9 References 1 Choosing a Selection Blaise Pascal (1623 - 1662) was a French mathematician and philosopher who studied probabilites and took an interest in a triangle of numbers which, though named after him, had been known about for many centuries already. The Triangle contains the number of ways of choosing some objects out of a larger set of different things. These choices make a subset of the set of possible choices where a collection means a subset and the order of the items in the subset is of no importance. Sets are written as a list of objects inside curly brackets { and }. The empty set of no objects is also a set and is written as {}. We can standardise the choices by numbering the elements of the set of possible elements to choose and then just finding a subset of these indices to indicate which elements are in the collection (subset). For instance, if we use numbers 1 to n for the objects to choose from them the subsets are also the index numbers of the items in the set 2 objects: 1 and 2 : we can choose 1 of them in two ways {1} and {2} all of them {1,2} or none of them {} \| # of items seleted \| 0 \| 1 \| 2 \| --- --- \| \| # of ways \| 1 \| 2 \| 1 \| with a total of 4 ways of selecting a collection. 3 objects: 1, 2, 3 : \| # of items seleted \| 0 \| 1 \| 2 \| 3 \| --- --- \| selections \| {} \| {1} {2} {3} \| {1,2} {1,3} {2,3} \| {1,2,3} \| \| # of ways \| 1 \| 3 \| 3 \| 1 \| with a total of 8 = 2 ways of selecting a subset In general there are 2n ways of choosing a subset of n items. This is because each item in the master set may be in the selection or not in a selection, so each has 2 possibilities. With n objects to choose from then there will be have 2n subsets of choices in total. 1.1 Notation n! means "n factorial". It the number of ways to arrange n things, so the order of the items in the arrangement is important. n! is the product of all the numbers from 1 to n. For example: If we have 6 books to arrange on a shelf, then we can choose any of the 6 as the first on the shelf, followed by any of the remaining 5 for the second making 6×5=30 ways to fill the first two places on the shelf. The third place can take any of the remaining 4 books so there are 6×5×4 = 120 ways to fill the first 3 places. and the fourth choice is one of 3 books, the fifth is one of the remaining 2 leaving one left for the last place. Thus there are 6×5×4×3×2×1 = 720 orderings or arrangements of the 6 books on the shelf. This is denoted 6! . \| \| \| \| \| --- --- \| \| ( \| n \| ) \| = Binomial(n,r) = nCr = "n choose r" for 0≤r≤n \| \| r \| \| \| \| --- \| \| = \| n! \| \| (n-r)! r! \| \| \| \| --- \| \| = \| n(n−1)...(n−r+1) \| \| r(r−1)...3×2×1 \| 0! = 1 The reason for this formula for Binomial(n,r) (or n choose r meaning the number of collections (sets) of r things from n ) is that there are n×(n-1)×...×(r+1) (there are r numbers in this product) ways to pick the r objects where the order matters (so that AB is counted as well as BA) so if we are interested just in which are items chosen or not (subsets) and not in the order in which they are chosen, then every collection of r items will appear in r! different orders or arrangements. So for subsets - which is what Pascal's Triangle shows - we need to divide by r! to account for the the number of arrangements of each collection of r things chosen from n. Binomial(n,c) has other notations in maths books , for example: nCc, nCc or Cn,c All are pronounced "(from) n choose c". 1.2 Pascal's Triangle \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| n \| \| r= 0 \| 1 \| 2 \| 3 \| 4 \| ... \| \| 1 \| \| 0 \| 1 \| \| 1 1 \| 1 \| 1 \| 1 \| \| 1 2 1 \| 2 \| 1 \| 2 \| 1 \| \| 1 3 3 1 \| 3 \| 1 \| 3 \| 3 \| 1 \| \| 1 4 6 4 1 \| 4 \| 1 \| 4 \| 6 \| 4 \| 1 \| \| ... \| ... \| ... \| Each entry in the triangle on the left is the sum of the two numbers above it. If we re-align the table to look the one on the right then each number is the sum of the one above it and the one to the left of that one where a blank space can be taken as "0". Note that each row starts and ends with "1". 1.3 A recursive definition Each element in Pascal's Triangle above, which is left-justified here, is the sum of the element above and the element to the left of that one where blank entries mean "0". The right-hand elements in column 0 are always 1: The recursive definition \| nCc = \| ( \| n \| ) \| = \| ( \| n − 1 \| ) \| + \| ( \| n − 1 \| ) \| , c ≥ c > 0 \| \| x \| c \| x − 1 \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| ( \| n \| ) \| = 1; \| ( \| n \| ) \| = 0 otherwise \| \| 0 \| c \| Pascal's Triangle has lots of uses including Calculating probabilities. If you throw n coins randomly onto a table then the chance of getting H heads among them is the entry in row N, col H divided by 2n: for instance, for 3 coins, n=3 so we use row 3: 3 heads: H=3 is found in 1 way (HHH) 2 heads: H=2 can be got in 3 ways (HHT, HTH and THH) 1 head: H=1 is also found in 3 possible ways (HTT, THT, TTH) 0 heads: H=0 (i.e. all Tails) is also possible in just 1 way: TTT 1.4 Pascal's Triangle (Binomial) Calculator This calculator can compute elements of Pascal's triangle individually or by rows or show the subsets. Pascal's Triangle C A L C U L A T O R \| \| \| --- \| \| \| n= r= \| R E S U L T S \| \| \| --- \| \| \| \| Hockey Sticks Central binomials n! factors 2 Binomial expansion using Pascal's Triangle 2.1 The rows as coefficients Apart from counting subsets and additions using 1s and 2s, another application of Pascal's Triangle is to find the coefficients when we expand (1+x)n. We will find it convenient to use the triangle in its left-leaning form: \| n \| \| c= 0 \| 1 \| 2 \| 3 \| 4 \| ... \| --- --- --- --- \| \| \| 0 \| 1 \| \| 1 \| 1 \| 1 \| \| 2 \| 1 \| 2 \| 1 \| \| 3 \| 1 \| 3 \| 3 \| 1 \| \| 4 \| 1 \| 4 \| 6 \| 4 \| 1 \| \| ... \| ... \| (1+x)0 = 1 (1+x)1 = 1 + 1x (1+x)2 = (1+x)(1+x) = 1 + 2x + 1 x2 (1+x)3 = (1+x)(1+x)2 = 1 + 3x + 3x2 + 1x3 The coefficients of the powers of x, from x0 to xn The numbers on row n of Pascal's Triangle are the coefficients (multiples) of the powers of x The coefficient of xc is in column c of row n. Since (1+x)n = (1+x)(1+x)n-1, it is easy to show that the coefficient of xc in (1+x)n is made up of the coefficient of xr in the (1+x)c-1 row PLUS x times the coefficient of xc-1 from the same n-1 row also. This gives us the recursion formula that we found above: binomial(n,c) = binomial(n-1,c) + binomial(n-1,c-1) Since (1+x)0 = 1 then we have the same starting conditions too, and therefore the same triangle of numbers for the coefficients and for Pascal's triangle! We can generalise this to expansions of (a+b)n too where a+b has two variables, forming a binomial expression: (a+b)0 = 1 (a+b)1 = 1 a + 1 b (a+b)2 = 1 a2 + 2 ab + 1 b2 (a+b)3 = 1 a3 + 3 a2b + 3 ab2 + 1 b3 ... The terms of (a + b)n are made by choosing one of the two variables, a or b, form each bracket and multiplying them. So each term has a power of a and a power of b where the power must sum to n. Each has a multiple before it, called its coefficient whhich is a number on row n of Pascal's Triangle. The coefficient of acbn-c in the expansion of (a+b)n is binomial(n,c) The terms are Binomial(n,c) acbn-c 2.2 The columns as coefficients \| n \| \| c= 0 \| 1 \| 2 \| 3 \| 4 \| ... \| --- --- --- --- \| \| \| 0 \| 1 \| \| 1 \| 1 \| 1 \| \| 2 \| 1 \| 2 \| 1 \| \| 3 \| 1 \| 3 \| 3 \| 1 \| \| 4 \| 1 \| 4 \| 6 \| 4 \| 1 \| \| ... \| ... \| The columns also are important as coefficients of expansions of \| \| \| 1 \| \| (1−x)c + 1 \| Unlike the rows, the columns go on for ever and the powers of 1/(1+x)2 have a coefficient for each and every (non-negative) power of x. Since the powers go on for ever, we must ensure -1<x<1 or else when adding the terms up we will not get a finite number (the expansion will not converge). 1/(1−x) = 1 + 1 x + 1 x2 + 1 x3 + ... which is column c=0 of Pascal's Triangle 1/(1−x)2 = 1 + 2 x + 3 x2 + 4 x3 + ... which is column c=1 of Pascal's Triangle 1/(1−x)3 = 1 + 3 x + 6 x2 + 10 x3 + ... which is column c=2 An application of this is to let x=1/10 and look at decimal expansions of some fractions: 1/(1-0.1) = 10/(10-1) = 10/9 = 1 + 1/10 + 1/100 + 1/1000 + ... = 1.111... 1/(1-0.1)2 = 100/(92 = 100/81 = 1 + 2/10 + 3/100 + 4/1000 + ... = 1.23456... x=0.01: 1/(1-0.01) = 100/(100-1) = 100/99 = 1 + 1/100 + 1/10000 + ... = 1.010101... 1/(1-0.01)2 = 10000/(99×99) = 1 + 2/100 + 3/1000 + 4/10000 +... = 1.02030405... There is much more about decimal expansion of fractions on the Fractions and Decimals page. Note: We need to be careful about what values we use for x since sometimes the expansions do not converge to a number but keep increasing the more terms we add on. 3 Divisibility in Pascal's Triangle What do you notice about the prime-number rows of Pascal's Triangle? row 2: 1 2 1 row 3: 1 3 3 1 row 5: 1 5 10 10 5 1 row 7: 1 7 21 35 35 21 7 1 Apart from the starting and ending 1's, all the entries are divisible by the prime row number. This is not true for rows n where n is a non-prime number. 4: 1 4 6 4 1 6: 1 6 15 20 15 6 1 8: 1 8 28 56 70 56 28 8 1 9: 1 9 36 84 126 126 84 36 9 1 ... The more general result applicable to all rows is left as an investigation in the You Do The Maths... section below. The remainder after dividing n by d in mathematics is written n modulo d or n mod d. You can see the patterns and experiment for yourself in the Calculator below. 3.1 You do the maths... It looks like every element that is bigger than 1 on row n has a prime factor in common with the row number. Can you prove this or find a counter-example? Answer Here is a table of the gcds of the row number and the elements of that row: \| n: \| \| 1 \| 1 \| 1 \| \| 2 \| 1 \| 2 \| 1 \| \| 3 \| 1 \| 3 \| 3 \| 1 \| \| 4 \| 1 \| 4 \| 2 \| 4 \| 1 \| \| 5 \| 1 \| 5 \| 5 \| 5 \| 5 \| 1 \| \| 6 \| 1 \| 6 \| 3 \| 2 \| 3 \| 6 \| 1 \| \| 7 \| 1 \| 7 \| 7 \| 7 \| 7 \| 7 \| 7 \| 1 \| \| 8 \| 1 \| 8 \| 4 \| 8 \| 2 \| 8 \| 4 \| 8 \| 1 \| \| 9 \| 1 \| 9 \| 9 \| 3 \| 9 \| 9 \| 3 \| 9 \| 9 \| 1 \| \| 10 \| 1 \| 10 \| 5 \| 10 \| 10 \| 2 \| 10 \| 10 \| 5 \| 10 \| 1 \| \| 11 \| 1 \| 11 \| 11 \| 11 \| 11 \| 11 \| 11 \| 11 \| 11 \| 11 \| 11 \| 1 \| \| 12 \| 1 \| 12 \| 6 \| 4 \| 3 \| 12 \| 12 \| 12 \| 3 \| 4 \| 6 \| 12 \| 1 \| \| 13 \| 1 \| 13 \| 13 \| 13 \| 13 \| 13 \| 13 \| 13 \| 13 \| 13 \| 13 \| 13 \| 13 \| 1 \| \| 14 \| 1 \| 14 \| 7 \| 14 \| 7 \| 14 \| 7 \| 2 \| 7 \| 14 \| 7 \| 14 \| 7 \| 14 \| 1 \| \| 15 \| 1 \| 15 \| 15 \| 5 \| 15 \| 3 \| 5 \| 15 \| 15 \| 5 \| 3 \| 15 \| 5 \| 15 \| 15 \| 1 \| \| 16 \| 1 \| 16 \| 8 \| 16 \| 4 \| 16 \| 8 \| 16 \| 2 \| 16 \| 8 \| 16 \| 4 \| 16 \| 8 \| 16 \| 1 \| \| 17 \| 1 \| 17 \| 17 \| 17 \| 17 \| 17 \| 17 \| 17 \| 17 \| 17 \| 17 \| 17 \| 17 \| 17 \| 17 \| 17 \| 17 \| 1 \| \| 18 \| 1 \| 18 \| 9 \| 6 \| 18 \| 18 \| 6 \| 18 \| 18 \| 2 \| 18 \| 18 \| 6 \| 18 \| 18 \| 6 \| 9 \| 18 \| 1 \| \| 19 \| 1 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 19 \| 1 \| \| 20 \| 1 \| 20 \| 10 \| 20 \| 5 \| 4 \| 20 \| 20 \| 10 \| 20 \| 4 \| 20 \| 10 \| 20 \| 20 \| 4 \| 5 \| 20 \| 10 \| 20 \| 1 \| \| r: \| 0 \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| 13 \| 14 \| 15 \| 16 \| 17 \| 18 \| 19 \| 20 \| So yes, it is true that Every Binomial bigger than 1 has a factor in common with its row number: GCD(n,Binomial(n,r))>1 if Binomial(n,r)>1 3.2 The Art of Pascal's Triangle Remainders There are some fantastic patterns in Pascal's Triangle which we look at later on this page but one which is simple involves the remainders when we divide every element by a given number. For instance, dividing by 7 will leave us with 0,1,2,3,4,5 or 6 as the possible remainders for every element and dividing by 2 will give remainders 0 and 1 or the even and the odd numbers. If a remainder is 0 when dividing by d then the number is a multiple of d. We talk above looking just at remainders when dividing by d as "modulo d" or mod d for short. Here are rows 0 to 8 of Pascal's Triangle with the even number paler: \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| 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 \| \| \| \| \| \| 1 \| \| 7 \| \| 21 \| \| 35 \| \| 35 \| \| 21 \| \| 7 \| \| 1 \| \| \| \| 1 \| \| 8 \| \| 28 \| \| 56 \| \| 70 \| \| 56 \| \| 28 \| \| 8 \| \| 1 \| \| and here it is when we just coloured dots for the remainders and center each row, \| \| \| \| --- \| mod \| 0 \| 1 \| \| \| \| \| ▪ ▪▪ ▪▪▪ ▪▪▪▪ ▪▪▪▪▪ ▪▪▪▪▪▪ ▪▪▪▪▪▪▪ ▪▪▪▪▪▪▪▪ ▪▪▪▪▪▪▪▪▪ ▪▪▪▪▪▪▪▪▪▪ ▪▪▪▪▪▪▪▪▪▪▪ ▪▪▪▪▪▪▪▪▪▪▪▪ ▪▪▪▪▪▪▪▪▪▪▪▪▪ ▪▪▪▪▪▪▪▪▪▪▪▪▪▪ ▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪ ▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪ ▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪ ▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪ You can see the fractal nature of the pattern - that is, each part in repeated in the larger pattern. The patterns are beautiful and either simple or complex depending on the mod chosen. Here is a Calculator for you to investigate for yourself: 3.3 Remainder Patterns Calculator Remainders C A L C U L A T O R \| \| \| --- \| \| \| row: .. mod= \| \| bigger?: \| \| R E S U L T S \| \| \| --- \| \| \| \| Elements Mod Positions Quincunx Coins Hexagons Hockey Sticks Central binomials n! factors 4 How often does a number appear in Pascal's Triangle? Using a few facts: each row is a palindrome - that is, it is the same when reversed, called reflection symmetry or row symmetry. row n begins 1,n and ends n,1 the numbers increase on every row until the centre and then decrease (by the reflection symmetry rule) we can search Pascal's Triangle and find: 2 is the only number to appear just once. All the elements that appear more than 2 times will be in the inner part of Pascal's Triangle, between columns 2 and n-2 on row n. If we list these number in order we have: 6,10,15,20,21,28,35,36,45,55, 56,66,70,78,84,91, ... A006987 ... ... whereas the complement of that sequence is the numbers in Pascal's triangle that appear just twice: 3, 4, 5, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 22, 23, 24, 25, 26, 27, ... A137905 The first numbers to appear 3 times are 6, 70, 252, 924, 3432, 12870, ... which appears to be all the Central Binomial coefficients Binomial(2n,n) that are greater than 2: A000984 Occurring 4 times are 10, 15, 28, 35, 36,45, 55, 56, ... . A098564 No number has been found yet that appears just 5 times 6 times: 120, 210, 1540, 7140, 11628, 24310, ... A098565 These include the values of Binomial( Fib(2k)×Fib(2k+1), Fib(2k+1)2 ) see A090162 where Fib(n) is the n-th Fibonacci number These numbers get large very rapidly. 3003 has 4 digits and the next have 29, 205, 1412 and then 9688 digits! No number has been found that appears just 7 times 3003 appears 8 times but no other numbers are known. No number has been found that appears more than 8 times Interesting facts about 3003: 3003 is also the sum of the two previous binomials on the same row: with sum Are there any more? The next two are: with 29 digits each with 205 digits each 4.1 Pascal's Triangle (Binomial) Calculator This calculator can compute elements of Pascal's triangle individually or by rows or show the subsets. Pascal's Triangle C A L C U L A T O R \| \| \| --- \| \| \| n= r= \| \| of numbers \| .. \| \| R E S U L T S \| \| \| --- \| \| \| \| Elements Mod Positions Quincunx Coins Hexagons Hockey Sticks Central binomials n! factors 4.2 You do the maths... Use the Calculator above to find as many numbers as you can that occur just 4 times in Pascal's Triangle as you can. Can you spot any patterns? Note The list of these numbers is A098564 but no general formula (or even a formula for a subset) is known. 2. Use the Calculator above to find as many numbers as you can that occur 5 or more times in Pacal's Triangle as you can. Can you spot any patterns? Note The list of these numbers is 1, 120, 210, 1540, 3003, 7140, 11628, 24310, 61218182743304701891431482520 A003015 and it is known that there are infinitely many of them that occur just 6 times. There are no more numbers less than the last in the list above! 3. What is a formula for the central numbers, that is those that appear in the centre of an even-numbered row? Answer These are the Central Binomial Coefficients Binomial(2n,n): \| \| \| --- \| \| Binomial(2n,n) = \| (2n)! \| \| n!2 \| 1, 2, 6, 20, 70, 252, 924, 3432, ... A000984 4. No number is known that appears just 5 times. Since 5 is odd, it must appear once as a central number. Can you find one? (The Calculator above will compute numbers in Pascal's Triangle with a very large number of digits). 5. Which numbers in Pascal's Triangle are doubled on the next row, same column? Answer If we find a row with two identical central numbers then they are doubled on the next row by the The Recursive Definition. These are the Central Binomial Numbers: Binomial(2n,n): See the Question above about central numbers Binomial(5,2)=Binomial(5,3)=10, Binomial(6,3)=20 Binomial(7,3)=Binomial(7,4)=35, Binomial(8,4)=70 ... See A001700 6. Which numbers in Pascal's Triangle are followed by their double on the same row? Answer The general answer is Binomial(3n-1,n-1) and Binomial{3n-1,n): Binomial(8,2)= 28, Binomial(8,3)=56 Binomial(11,3)=165, Binomial(11,4)=330 Binomial(14,4)=1001, Binomial(14,5)=2002 ... See A025174 5 The Fibonacci Numbers in Pascal's Triangle Can we find the Fibonacci Numbers in Pascal's Triangle? Yes! The answer lies in the diagonals in the triangle: \| \| \| \| --- \| Sum \| \| 1 \| \| 1 ↗ \| 1 \| 1 \| \| 1 ↗ \| 1 \| 2 \| 1 \| \| 2 ↗ \| 1 \| 3 \| 3 \| 1 \| \| 3 ↗ \| 1 \| 4 \| 6 \| 4 \| 1 \| \| 5 ↗ \| 1 \| 5 \| 10 \| 10 \| 5 \| 1 \| \| 8 ↗ \| 1 \| 6 \| 15 \| 20 \| 15 \| 6 \| 1 \| \| 13↗ \| 1 \| 7 \| 21 \| 35 \| 35 \| 21 \| 7 \| 1 \| \| 21↗ \| ... \| 5.1 Why do the Diagonals sum to Fibonacci numbers? It is easy to see that the diagonal sums really are the Fibonacci numbers if we remember that each number in Pascal's triangle is the sum of two numbers in the row above it (blank spaces count as zero), so that 6 here is the sum of the two 3's on the row above. The numbers in any diagonal row are therefore formed from adding numbers in the previous two diagonal rows as we see here where all the blank spaces are zeroes and where we have introduced an extra column of zeros which we will use later: \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| 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 \| \| The green diagonal sums to 5; the blue diagonal sums to 8; the red diagonal sums to 13 Each red number is the sum of a blue and a green number on the row above. \| Notice that the GREEN numbers are on one diagonal and the BLUE ones on the next. The sum of all the green numbers is 5 and all the blue numbers add up to 8. Because all the numbers in Pascal's Triangle are made the same way - by adding the two numbers above and to the left on the row above, then we can see that each red number is just the sum of a green number and a blue number and we use up all the blue and green numbers to make all the red ones. The sum of all the red numbers is therefore the same as the sum of all the blues and all the greens: 5+8=13! The general principle that we have just illustrated is: The sum of the numbers on one diagonal is the sum of the numbers on the previous two diagonals. If we let D(i) stand for the sum of the numbers on the Diagonal that starts with one of the extra zeros at the beginning of row i, then D(0)=0 and D(1)=1 are the two initial diagonals shown in the table above. The green diagonal sum is D(5)=5 (since its extra initial zero is in row 5) and the blue diagonal sum is D(6) which is 8. Our red diagonal is D(7) = 13 = D(6)+D(5). We also have shown that this is always true: one diagonals sum id the sum of the previous two diagonal sums, or, in terms of our D series of numbers: D(i) = D(i-1) + D(i-2) But... D(0) = 1 D(1) = 1 D(i) = D(i-1) + D(i-2) is exactly the definition of the Fibonacci numbers! So D(i) is just F(i) and the sums of the diagonals in Pascal's Triangle are the Fibonacci numbers! 5.2 Another arrangement of Pascal's Triangle By drawing Pascal's Triangle with all the rows moved over by 1 place, we have a clearer arrangement which shows the Fibonacci numbers as sums of columns: \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| \| 1 \| . \| . \| . \| . \| . \| . \| . \| . \| . \| \| . \| 1 \| 1 \| . \| . \| . \| . \| . \| . \| . \| \| . \| . \| 1 \| 2 \| 1 \| . \| . \| . \| . \| . \| \| . \| . \| . \| 1 \| 3 \| 3 \| 1 \| . \| . \| . \| \| . \| . \| . \| . \| 1 \| 4 \| 6 \| 4 \| 1 \| . \| \| . \| . \| . \| . \| . \| 1 \| 5 \| 10 \| 10 \| 5 \| \| . \| . \| . \| . \| . \| . \| 1 \| 6 \| 15 \| 20 \| \| . \| . \| . \| . \| . \| . \| . \| 1 \| 7 \| 21 \| \| . \| . \| . \| . \| . \| . \| . \| . \| 1 \| 8 \| \| . \| . \| . \| . \| . \| . \| . \| . \| . \| 1 \| \| 1 \| 1 \| 2 \| 3 \| 5 \| 8 \| 13 \| 21 \| 34 \| 55 \| This table can be explained by referring to one of the (Easier) Fibonacci Puzzles - the one about Fibonacci for a Change. It asks how many ways you can pay n pence (in the UK) using only 1 pence and 2 pence coins. The order of the coins matters, so that 1p+2p will pay for a 3p item and 2p+1p is counted as a different answer. [We now have a new two pound coin that is increasing in circulation too!] Here are the answers for paying up to 5p using only 1p and 2p coins: \| 1p \| 2p \| 3p \| 4p \| 5p \| --- --- \| 1 \| 2 1+1 \| 1+2 2+1 1+1+1 \| 2+2 1+1+2 1+2+1 2+1+1 1+1+1+1 \| 1+2+2 2+1+2 2+2+1 1+1+1+2 1+1+2+1 1+2+1+1 2+1+1+1 1+1+1+1+1 \| \| 1 way \| 2 ways \| 3 ways \| 5 ways \| 8 ways \| Let's look at this another way - arranging our answers according to the number of 1p and 2p coins we use. Columns will represent all the ways of paying the amount at the head of the column, as before, but now the rows represent the number of coins in the solutions: \| cost: \| 1p \| 2p \| 3p \| 4p \| 5p \| --- --- --- \| \| 1 coin: \| 1 \| 2 \| \| \| \| \| 2 coins: \| \| 1+1 \| 1+2 2+1 \| 2+2 \| \| \| 3 coins: \| \| \| 1+1+1 \| 1+1+2 1+2+1 2+1+1 \| 1+2+2 2+1+2 2+2+1 \| \| 4 coins: \| \| \| \| 1+1+1+1 \| 2+1+1+1 1+1+1+2 1+1+2+1 1+2+1+1 \| \| 5 coins: \| \| \| \| \| 1+1+1+1+1 \| If you count the number of solutions in each box, it will be exactly the form of Pascal's triangle that we showed above! Fib(n) = 5.3 You do the maths... If you tossed a coin 10 times, how many possible sequences of Heads and Tails could there be in total (use Pascal's Triangle extending it to the row numbered 10)? In how many of these are there 5 heads (and so 5 tails)? What is the probability of tossing 10 coins and getting exactly 5 heads therefore - it is not 0·5! Draw up a table for each even number of coins from 2 to 10 and show the probability of getting exactly half heads and half tails for each case. What is happening to the probability as the number of coins gets larger? [HINT: Use the Calculator below.] Write out the first few powers of 11. Do they remind you of Pascal's triangle? Why? Why does the Pascal's triangle pattern break down after the first few powers? (Hint: consider (a+b)m where a=10 and b=1). 6 Galton's Quincunx This is a device with nails arranged in a regular hexagon pattern. Its name derives from the Latin word quincunx for the X-like shape of the spots on the 5-face of a dice: \| \| \| --- \| \| \| Hopper for balls balls fall onto nails with an equal chance of bouncing to left or right each time balls collect in bins \| When balls are poured onto the network of nails at the top, they fall through, bouncing either to the right or to the left and so hit another nail on the row below. Eventually they fall off the bottom row of nails and are collected in the bins. If we arrange pins on a board so that a ball, falling onto any pin has an equal chance of falling off to the left or to the right then we can simulate the selections of n objects into n bins under the final row of pins. Here is a simulator which lets you choose how many balls are dropped onto the top pin and how many pins are on the bottom row, counting the number of balls which land in each bin below the pins if they randomly bounce off to the left or right. In the Calculator here, the histogram of the number of balls in each box empty boxes are omitted at either side. 6.1 A Simulator Quincunx S I M U L A T O R \| \| \| --- \| \| Drop balls falling into bins Show pins? \| \| R E S U L T S Elements Mod Positions Quincunx Coins Hexagons Hockey Sticks Central binomials n! factors There is a very nice Maths is Fun simulation of a quincunx showing each ball bouncing as it falls. 6.2 You do the maths... Make a real Galton Quincunx. If you have a lot of nails and a lot of little balls (good sources for these are small steel ball-bearings from a bicycle shop or ping-pong balls for a large version or even dried peas or other cheap round seeds from the supermarket) then they end up forming a shape in the containers that is very much like the Bell curve of the previous exploration. You will need to space the nails so they are as far apart as about one and a half times the width of the balls you are using. This makes a good practical demonstration for a Science Fair or Open Day at your school or college. 6.3 Block Walking The Quincinx of the last section gives us another interpretation of the Binomial coefficients of Pascal's Triangle, called Block Walking invented by George Polya. \| x \| \| \| \| \| \| \| --- --- --- \| y \| 00 \| 1 \| 2 \| 3 \| 4 \| 5 \| \| 1 \| ABC \| \| \| \| \| \| 2 \| \| \| \| \| \| \| 3 \| \| \| \| \| \| \| 4 \| \| \| \| \| \| Imagine that the Quincunx is turned on its side so that the balls fall down on the sloping diagonal to the right. The board is then a map of part of a city where the roads are laid out in a grid, such as in New York. The pins become the (square) blocks and the roads are the edges around them that the falling balls travel along. We count the number of shortest paths from top left corner (n=0,r=0) to the intersection n rows down and r columns across travelling only to the right or down. There is clearly just 1 way to reach each intersection on the top edge and on the left edge. There is 1 way to reach point A (n=0,r=1): going to the next intersection right from the origin. There is 1 way to reach point B (n=1,r=0): going 1 block down from the origin. So there are 2 ways to reach point C (n=1,r=1). In general, the number of ways to reach a given intersection is the number of ways to get to the intersection above it (then go down) PLUS the number of ways to get to the intersection to its left (and then go right), which is just the Pascal's Triangle Addition (Recursion) formula. Here is a table of the number of different shortest paths from the Origin to each intersection: \| \| \| \| \| --- --- \| \| 0→ \| 1→ \| 1→ \| 1→ \| \| ↓ 1→ \| ↓ 2→ \| ↓ 3→ \| ↓ 4→ \| \| ↓ 1→ \| ↓ 3→ \| ↓ 6→ \| ↓ 10→ \| \| ↓ 1→ \| ↓ 4→ \| ↓ 10→ \| \| ↓ \| ↓ \| ↓ \| Each intersection is identified by the number of Downs and the number of Rights to reach it. Given the number of Ds and Rs then any order of Ds and Rs will get to the same location. The paths are found by computing the subsets of the numbers 1 to #D+#R ( the number of Ds plus the number of Rs taken to reach the location). So, for example, #D=4 and #R=2 intersection is reached in Binomial(6,4) ways to choose the Ds which is, of course, the same as the number Binomial(6,2) which chooses the Rs. 6.4 Paths and Coins Calculator Paths and Coins C A L C U L A T O R \| \| \| --- \| \| \| for n= and r= \| R E S U L T S \| \| \| --- \| \| \| \| Elements Mod Positions Quincunx Coins Hexagons Hockey Sticks Central binomials n! factors 7 Patterns in the Binomial Coefficients 7.1 Symmetry on rows If we think of the Pascal's triangle element Binomial(n,r) as the number of ways to get r heads when tossing n coins. For a fair coin this is the same number as tossing n coins and getting r tails. But getting r tails is the same as getting n−r heads and that is Binomial(n-r,r). For instance, tossing 6 coins there are Binomial(6.4) =15 ways to get 4 heads. Use the Binomial Calculator above to produce the 15 subsets of {1,2,3,4,5,6} which we can interpret as the number of the coins that are Heads, the missing numbers being Tails. \| subset \| coins 123456 \| --- \| \| {1,2,3,4} \| HHHHTT \| \| {1,2,3,5} \| HHHTHT \| \| {1,2,3,6} \| HHHTTH \| \| ... \| ... \| The number of ways of getting 4 heads is the same as the number of ways of getting 6-4=2 Tails so that Binomial(6,4) and Binomial(6,6-4) = Binomial(6,2) must be the same number. So any row can be reversed and still look the same: Binomial(n,r) = Binomial(n,n-r) nCr = nCn−r 7.1.1 Row sums If we interpret the Binomial coefficients as probabilities of getting a number (r) or Heads when throwing (n) coins, then there are 2n possibilities because each coin in the toss has two possibilities (Head or Tail). So the sum of all the entries in row n in Pascal's Triangle is 2n. 7.1.2 Alternating sums along a row The sum of the alternate elements along a row is the same as the sum of the elements not used: 7.2 Formulas for Binomial Coefficients 7.2.1 Recursive The formulas below take a smaller Binomial coefficient and update it to get the next one along a row or column of diagonal: From one row to the next in the same column: , n>0, 0≤r≤n From one column to the next on the same row: , n>0, 0<r<n From one row and column to the next row and column: , 0<n, 0<r 7.2.2 Sums and differences of neighbours 7.2.3 Divisibility , p prime Examples Examples is a multiple of : \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| n \| 7 \| 7 \| 7 \| 7 \| 7 \| 7 \| \| 8 \| 8 \| 8 \| 8 \| \| r \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 1 \| 2 \| 3 \| 4 \| \| \| \| 7 \| 21 \| 35 \| 35 \| 21 \| 7 \| 8 \| 28 \| 56 \| 70 \| \| GCD(n,r) \| 1 \| 1 \| 1 \| 1 \| 1 \| 1 \| 1 \| 2 \| 1 \| 4 \| \| n/GCD(n,r) \| 7 \| 7 \| 7 \| 7 \| 7 \| 7 \| 8 \| 4 \| 8 \| 2 \| \| / \| 1 \| 3 \| 5 \| 5 \| 3 \| 1 \| 1 \| 7 \| 7 \| 35 \| Examples Row 2: LCM(1,2,1) = 2 = LCM(1,2,3)/3=6/3 Row 3: LCM(1,3,3,1) = 3 = LCM(1,2,3,4)/4=12/4 Row 6: LCM(1,6,15,20,15,6,1) = 60 = LCM(1,2,3,4,5,6,7)/7=420/7=60 7.3 The Hexagon or Star of David Products \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| 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 \| \| \| \| \| \| 1 \| \| 7 \| \| 21 \| \| 35 \| \| 35 \| \| 21 \| \| 7 \| \| 1 \| \| \| \| 1 \| \| 8 \| \| 28 \| \| 56 \| \| 70 \| \| 56 \| \| 28 \| \| 8 \| \| 1 \| \| Let's look at the hexagon of 6 numbers surrounding an element in Pascal's triangle, the two above, the two on either side the same row and the two numbers below. For instance, the 4 on row 4, column 1 has these 6 neighbours: 1 3 1 4 6 5 10 We have grouped them into two triangles (to make a Star of David pattern), alternating round the hexagon, one red and the other green. If we multiply all those of the same colour, what happens? We get the same product for both triangles: in this case 1×3×10 = 1×6×5 = 30. This is not a coincidence: This works for any element of Pascal's triangle! n-1Cr nCr-1 n+1Cr+1 = n-1Cr-1 nCr+1 n+1Cr For the 1's at the ends of the rows, this product is 0 if we count a gap as having the value of 0. The product of all the 6 neighbours in the hexagon will always be a square number. A proof Using The recursive definition above, we want to show n-1Cr nCr-1 n+1Cr+1 = n-1Cr-1 nCr+1 n+1Cr \| \| \| \| \| --- --- \| \| LHS = \| n-1Cr \| nCr-1 \| n+1Cr+1 \| \| = \| (n-1)! \| n! \| (n+1)! \| \| r!(n-r-1)! \| (r-1)!(n-r+1)! \| (r+1)!(n-r)! \| \| Rearrange the factorials in the denominator... \| \| \| \| \| = \| (n-1)! \| n! \| (n+1)! \| \| (r-1)!(n-r)! \| (r+1)!(n-r-1)! \| r!(n-r+1)! \| \| RHS = \| n-1Cr-1 \| nCr+1 \| n+1Cr \| Since the product of the 3 numbers in both triangles is the same, the product of all 6 numbers will be a square number 7.3.1 The gcd of the triangles is the same Another property of these two triangles is The gcds of all the numbers in each triangle is the same Since there is a 1 in each of the triangles for the example above, let's try the hexagon around 126 on row 9, column 4: 56 70 84 126 126 210 252 gcd(70, 84, 252) = 14 = gcd(56, 126, 210) 7.3.2 Hexagon Calculator Hexagonal C A L C U L A T O R around row= col= R E S U L T S \| \| \| --- \| \| \| \| Elements Mod Positions Quincunx Coins Hexagons Hockey Sticks Central binomials n! factors The Hexagon Rule above is just a special case of the equation A K Gupta published in 1974 (see References)with r = s = 1. Here the Binomials rows and columns are shown in a table indicating if they appear on the Left Hand Side or the Right Hand Side of this equality: \| \| \| \| \| \| --- --- \| \| \| Column: \| \| \| \| \| \| m-r \| m \| m+s \| \| Row: \| n-s \| RHS \| LHS \| \| \| n \| LHS \| \| RHS \| \| n+r \| \| RHS \| LHS \| \| 7.3.3 More recursions which is easily proved by expanding each Binomial into its factorial form. 7.4 Sum of squares of a row If we square every element on a row n, the total is another element of Pascal's triangle, in the centre of row 2n, in column n: For example, row n=4 is 1 4 6 4 1 and 1² + 4² + 6² + 4²+ 1² = 70 = Binomial(8,4) 7.4.1 Proof by "Committees" One way to interpret binomial coefficients is as when choosing members of a committee. So Binomial(n,r) is the number of ways of selecting a committee of r people from n available to serve on it. As an example of proving results by such reasoning, let's take the Sum of Squares result in the last section. The RHS is Binomial(2n,n) so we are choosing a committee of n people from 2n available. If we think of those available as being from two schools, for instance, then we have n members from each school available. If we want to form a committee of n people to run a joint event, then we have Binomial(2n,n) ways to do this. We can also think of it as having from 0 to n members from the first school on the committee, the other n-k being chosen from the second school. This isBut, since the number of ways of choosing n-k from school 2 is exactly the same as the number of ways of not choosing k of the available n or, the Row Symmetry property that we found above, this is the same as as so 7.4.2 Proof by "block walking" We know that we can think of Binomial(n,r) as the number of different paths in a city map of a grid of roads to get from one intersection to another n blocks down and r blocks right by the shortest routes (always travelling towards the destination). So the Sum of Squares result is about the number of routes from an origin intersection to one 2n blocks right and n blocks down. \| x \| \| \| \| \| \| \| --- --- --- \| y \| 00 \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| \| 1 \| \| \| \| \| \| \| \| 2 \| \| \| \| \| \| \| \| 3 \| \| \| \| \| \| \| Clearly, we will have to cross the green vertical line of streets. We can go via the top intersection: there are Binomial(3,0) ways to get to the topmost intersection on this line, then a choice of Binomial(3,3) ways to get from there to our destination at the bottom right. ways or we go via the (3,1) intersection on the mid-line with Binomial(n,1) ways to get to it then Binomial(n,2) ways from it to our goal. Similarly there are Binomial(3,2) × Binomial(3,1) ways to go via the (3,2) intersection and finally Binomial(3,3)×Binomial(3,0) ways via the bottommost midline intersection. This gives us all the ways to get to our goal: Binomial(2n,n). By the same argument as above, applying the Row Symmetry rule, we change the second Binomials in each product to be identical to the first ones, and the result follows. 7.5 The Hockey Stick Theorems Every number in the triangle is the sum of all of those numbers in the diagonal leading up and left starting from the number above it and also the sum of all those numbers above it in the previous column Select which you direction you want for the hockey stick handle Click on a number in the triangle to see the hockey stick. 7.5.1 Hockey Sticks Demonstrator Hockey Stick handle: :Left diagonal OR :right diagonal \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- \| \| n \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| 0 \| \| \| \| \| \| \| \| 1 \| \| \| \| \| \| \| \| \| \| 1 \| \| \| \| \| \| \| 1 \| \| 1 \| \| \| \| \| \| \| \| \| 2 \| \| \| \| \| \| 1 \| \| 2 \| \| 1 \| \| \| \| \| \| \| \| 3 \| \| \| \| \| 1 \| \| 3 \| \| 3 \| \| 1 \| \| \| \| \| \| \| 4 \| \| \| \| 1 \| \| 4 \| \| 6 \| \| 4 \| \| 1 \| \| \| \| \| \| 5 \| \| \| 1 \| \| 5 \| \| 10 \| \| 10 \| \| 5 \| \| 1 \| \| \| \| \| 6 \| \| 1 \| \| 6 \| \| 15 \| \| 20 \| \| 15 \| \| 6 \| \| 1 \| \| \| \| 7 \| 1 \| \| 7 \| \| 21 \| \| 35 \| \| 35 \| \| 21 \| \| 7 \| \| 1 \| \| \| \| r= \| ↗ 0 \| \| ↗ 1 \| \| ↗ 2 \| \| ↗ 3 \| \| ↗ 4 \| \| ↗ 5 \| \| ↗ 6 \| \| ↗ 7 \| \| Elements Mod Positions Quincunx Coins Hexagons Hockey Sticks Central binomials n! factors The numbers used make a hockey stick shape in two orientations. For the handle going upwards to the left, we have for row n=7 and column r=3: 7C3 = 20 + 10 + 4 + 1 = 35: \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- \| ( \| 7 \| ) \| = \| ( \| 6 \| ) \| + \| ( \| 5 \| ) \| + \| ( \| 4 \| ) \| + \| ( \| 3 \| ) \| \| 3 \| 3 \| 2 \| 1 \| 0 \| The Hockey Stick Theorem 1: For the handle going up to the right, we have, for example, n=7,r=3: 7C3 = 15 + 10 + 6 + 3 + 1 + 0 + 0 = 35 \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| ( \| 7 \| ) \| = \| ( \| 6 \| ) \| + \| ( \| 5 \| ) \| + \| ( \| 4 \| ) \| + \| ( \| 3 \| ) \| + \| ( \| 2 \| ) \| + \| ( \| 1 \| ) \| + \| ( \| 0 \| ) \| \| 3 \| 2 \| 2 \| 2 \| 2 \| 2 \| 2 \| 2 \| The Hockey Stick Theorem 2: These are the same Theorem by row-symmetry Since every row is symmetrical in that it is the same when reversed, then by flipping the whole Triangle about a vertical axis and then choosing the mirror-image element on the same row, we see that each of the two Theorems is just a reflection of the other. 7.6 Central Binomials The binomial numbers in the centre of the even rows (those rows of Pascal's Triangle that begin: 1, 2n, ...) are: \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| 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 \| \| \| \| \| \| 1 \| \| 7 \| \| 21 \| \| 35 \| \| 35 \| \| 21 \| \| 7 \| \| 1 \| \| \| \| 1 \| \| 8 \| \| 28 \| \| 56 \| \| 70 \| \| 56 \| \| 28 \| \| 8 \| \| 1 \| \| 1, 2, 6, 20, 70, 252, 924, ... A000984 They have some interesting applications: The number of ways of getting exactly half Heads and half Tails when tossing an even number 2n of coins (or 1 single coin 2n times) is Binomial(2n,n). Dividing this by 22n (the total number of possibilities when tossing 2n coins) gives the probability of getting the same number of Heads as Tails For 4 tosses of a coin we have just 6 ways with half Heads and half Tails: HHTT HTHT HTTH TTHH THTH THHT out of the 24=16 possible arrangements of Heads and Tails. \| \| \| \| \| \| \| \| --- --- --- \| #coins \| 2 \| 4 \| 6 \| 8 \| 10 \| 12 \| \| #equal \| 2 \| 6 \| 20 \| 70 \| 252 \| 924 \| \| #possible \| 4 \| 16 \| 64 \| 256 \| 1024 \| 4096 \| \| Probability \| 0.5 \| 0.375 \| 0.3125 \| 0.2734 \| 0.2461 \| 0.2256 \| \| \| \| \| \| --- --- \| \| #coins \| 100 \| 1000 \| 1000000 \| \| #equal \| 1.001×1029 \| 2.70×10299 \| 7.89×10301026 \| \| #possible \| 1.27×1030 \| 1.07×10301 \| 9.90×10301029 \| \| Probability \| 0.0795 \| 0.02522 \| 0.0007979 \| Block sequences Suppose a city has a grid system of roads similar to that in Manhattan in New York, where avenues run north-south and streets run East-West and each is numbered. The number of shortest paths on a grid to go from one point (intersection of two roads) to another point n blocks East and n blocks North is Binomial(2n,n): Or think of a square lattice and paths from (0,0) to (n,n) with steps going East: 1 unit (in the positive x direction) (1,0) or one unit North (in the positive y direction) (0,1). Since the path is n blocks right and n blocks north we need 2n steps. We can choose which n of them are East then the remaining n will be North: Binomial(2n,n) possibilities. Here is a diagram for n=2 and there are 6 paths always going right or up each time. A convenient way to represent each path is to list the number of each E-W street on the path. Or, viewing the path as a bar-chart, what is the height of each bar from left to right. In a tournament of games where the winner is the "best of 3"="first to 2 wins", "best of 5"="first to 3 wins" and, in general, "best of (2n-1)" = "first to n wins", any odd number (2n-1) of games, how many patterns of game-wins are there before a winner is found or a draw declared? For an even number of games there may be a draw but not if the number of games is odd. For example, best of 3 games = first to 2 wins : we could have the following 6 possible match results before a winner is found: \| A wins: \| B wins: \| --- \| \| AA \| BB \| \| ABA \| BAB \| \| BAA \| ABB \| A Game Tree for 3 games (best of 3 games or first to 2 wins) : If we stop as soon as there is a clear winner (best of 3), we can show the games in a Game Tree where horizontal and vertical lines mark the individual game's winner: Here we have 2 wins out of 3 games for A and 2 wins out of 3 games for B; 1 way to win for each tam involving just 2 games. B wins, A wins . A game tree for "best of 5"or "first to 3 wins" : Here A wins in 1+3+6=10 cases, B wins in the same number of cases. Tracing the paths form the start in the lower-left corner to each of the numbers, with A wins vertically and B wins horizontally we have the following results: There is one win for A in 3 games AAA and also for B with BBB; There are 3 wins with the other team winning 1 and the winning team winning 3: BAAA, ABAA, AABA, for A and ABBB, BABB, BABB with B winning; There are 6 wins for each team where the other team wins 2: ABABA, ABBAA, AABBA, BABAA, BBAAA, BAABA where A is victor and BABAB, BAABB, BBAAB, ABABB, AABBB, ABBAB where B wins the tournament. Why are there Binomial(2n,n) games in these tournaments? If we use a block diagram with counts of the number of routes from the origin to that point, we get our usual Pascal's triangle. Here a left track means a win for team A and a right track a win for team B. We add the numbers above a location because those from the left on the row above are before B won the latest game and those from the right element are from before A won. If we want the best of (2n+1) which is the same as "the first to win n games" then we stop as soon as n+r is (2n+1), the winner being A for routes ending rn and a draw if r=n. For instance with n=1 for "best of 3" the games tree looks like this: \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| 1 \| \| \| \| \| \| \| \| \| \| 1 \| \| 1 \| \| \| \| \| \| \| \| 1 \| \| 2 \| \| 1 \| \| \| \| \| \| 1 \| \| 3 \| \| 3 \| \| 1 \| \| \| \| 1 \| \| 4 \| \| 6 \| \| 4 \| \| 1 \| \| ↙ A wins ↙↘ B wins ↘ So there are 1+3 game paths ending with a win for A on the left side and the same on the right hand side with wins for B. The 6 on the bottom row is the number of ways to get a draw. For "best of 5" (first to 3) we have: \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| 1 \| \| \| \| \| \| \| \| \| \| \| \| 1 \| \| 1 \| \| \| \| \| \| \| \| \| \| 1 \| \| 2 \| \| 1 \| \| \| \| \| \| \| \| 1 \| \| 3 \| \| 3 \| \| 1 \| \| \| \| \| \| 1 \| \| 4 \| \| 6 \| \| 4 \| \| 1 \| \| \| \| 1 \| \| 5 \| \| 10 \| \| 10 \| \| 5 \| \| 1 \| \| ↙ A wins ↙↘ B wins ↘ and, in general: 7.6.1 Central Binomials Calculator Central Binomials C A L C U L A T O R \| \| \| --- \| \| Binomial(2n,n) all block paths (0,0) to (n,n) games in first to win n games grid diagram for first to win n games \| for n= \| \| \| \| R E S U L T S \| \| \| --- \| \| \| \| Elements Mod Positions Quincunx Coins Hexagons Hockey Sticks Central binomials n! factors 8 Computing Binomials efficiently By efficiently we mean using only integers and exact division to avoid the approximations of real numbers and dividing. The definition of Binomial(n,r) is and this formula as it stands involves three factorials: , and as well as dividing the first by the product of the other two. In practice, we see that in all the bottom terms cancel with the last numbers in the numerator's factorial to give which involves no divisions. However for Binomial(n,r) we still need to divide by 8.1 Exact integer limits 253−1 ≈9.0×1015 is the maximum exact whole number in many programming languages (for example, JavaScript). Numbers larger than that are then automatically converted into real numbers (floating point numbers) and are not exact. For example: 18! ≈ 6.4×1015 but 19! ≈ 1.2×1017 and so Binomial(56,28) ≈ 7.6×1015 will be ok but Binomial(57,28) ≈ 1.5×1016 will exceed the exact-integer limit The 57-th row of Pascal's Triangle is the first row to have a value exceeding the 253−1 integer limit. This causes programming problems if we want accurate computations for large Binomial values. So we can ask the question: Can we compute Binomials efficiently, not involving floating-point (real) numbers resulting from division? Can we compute Binomials using only divisions that are exact? Here are two ideas: Is it possible to compute the terms by multiplying whilst ensuring that each divide will be exact? Can we avoid divisions altogether by computing directly the power of each of the primes in the prime factorization of Binomial(n,r)? We examine these questions now: 8.2 Updating Binomials Since there are the same number of numbers to multiply in the numerator of Binomial(n,r) as there are to divide by in the denominator, we can try Will this ensure that we never get a fractional value after a division? Yes! This is because if starting at the left of the expression above, having computed Binomial(n,i) we can update it to Binomial(n,i+1) by multiplying by (n-i+1) and dividing by (i+1). We start with i=1 and go up top i=r: Both of the above are integers and so we see that the following will always involve exact division: As i goes from 1 to r, starting from B=Binomial(n,1)=n, we multiply and divide to update B to Binomial(n,2) then Binomial(n,3) until we reach B = Binomial(n,r) and we can compute all these values as exact whole numbers. Here is a computer program to do this efficiently; first we deal with the values of n and r that give 0 or 1 in Pascal's Triangle, then we check if the Row Symmetry rule might reduce the number of times we multiply-and-divide: To compute Binomial(n,r) as a function whose result is the value to return Key \| . \| indicates that the program can exit the function \| \| ASSERT \| shows what limits there are on the values of n and r at that point in the code. \| \| ← \| copies the value on its right into the variable on its left. \| We must ensure that Binomial(0,0) = Binomial(0,r) = Binomial(n,0) = 1 for all r>0 and all n>0 Binomial(n,r) = 0 if r<0 or n<0 or r>n. \| \| \| --- \| \| Input n and r \| ASSERT r and n are integers \| \| If(n<0) result is 0. \| ASSERT n ≥ 0, r is an integer \| \| If(r<0) result is 0. \| ASSERT r ≥ 0 and n ≥ 0 \| \| If(r>n) result is 0. \| ASSERT 0 ≤ r ≤ n \| \| If(r=0) result is 1. \| ASSERT 0 < r ≤ n or n = 0 \| \| If(n=0) result is 1. \| ASSERT 0 < r ≤ n \| \| If(n=r) result is 1. \| ASSERT 0 < r < n \| \| If(n-r<r) r=n-r \| ASSERT 0 < r ≤ n/2 \| \| B ← 1; \| \| \| For i ← 0 up to r-1 in increments of +1: B ← B (n - i) / (i + 1); \| \| \| result is B. \| \| \| 8.2.1 The 8 Neighbours recursions Here are the formulas to update Binomial(n,r) to find each of its 8 neighbours: \| \| \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| 8.2.2 You do the maths... Prove each of the above formulas for the 8 neighbours. Hint: One way is to use the factorial formula for each and compare it to 8.3 Legendre's Prime Power Factors in Binomial(n,r) Our second idea is to work solely with the prime factorization of n! using an amazing result of A Legendre. For each prime in the factorization of Binomial(n,r) (each prime ≤ n) we write n in base p and let s be the sum of the base p digits of n, then the power of prime p that occurs in the prime factorization of n! is given by . . He gave a slightly different but equivalent definition: Legendre's Formula Eventually the prime power will become too big and resulting terms in the sum will be 0. This uses a beautifully simple way to calculate the power of each prime that factorizes n!. If p, a prime, is a factor of n! then it occurs as a factor exactly times, where is the sum of the digits of n in base p or, equivalently, the power of p is the sum of the quotients when computing n in base p For example, let's look at 12! and find the number of times 2 is a factor: To do this, write 12 in the base p, the prime whose power we want to find: \| \| \| --- \| \| 2 \| 12 \| \| 2 \| 6 \| remainder 0 \| \| 2 \| 3 \| remainder 0 \| \| 2 \| 1 \| remainder 1 \| \| \| 0 \| remainder 1 \| so 12 in base 2 (binary) is 1100 (reading the remainders upward). You can see that the sum of the quotients 6 + 3 + 1 + 0 = 10 PLUS the sum of the base 2 digits 0 + 0 + 1 + 1 = 2 is n = 12 itself. Legendre showed that this is always the case for any number in base 2. The maximum power of 2 that is a factor of 12! is 12 - (sum of binary digits) divided by one less than the prime 2 (so we divide by 1) and the sum of the digits 1 1 0 0 is 2 so the power of 2 involved in 12! is (12 - 2)/1 = 10. So the prime factorisation of 12! begins 210 For other primes the general formula is Legendre's Theorem The sum of base-p digits of n PLUS (p-1) times the sum of the quotients is always n Let's look at how many times the prime 3 is a factor of 12! Express 12 in base 3: \| \| \| \| --- \| 3 \| 12 \| \| \| 3 \| 4 \| remainder 0 \| \| 3 \| 1 \| remainder 1 \| \| \| 0 \| remainder 1 \| so 12 in base 3 (termary) is 110 Note that n = 12 = (1+1+0) + (3-1) × (4+1+0) The sum of the base 3 digits of 12 is 1 1 0 = 2 and so the power of 3 we need is (12-2)/(3-1) = 5. It is also the sum of the quotients: 4 + 1 + 0 = 5 So now we have 12! = 210 ×35. Repeating with p=5, p=7 and finally p=11 (all the primes up to 12) we get the full factorization of 12! as 12 ! = 210 35 52 71 111 which we can easily check by expanding 12! as \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| 12 ! = \| 12 \| ×11 \| ×10 \| ×9 \| ×8 \| ×7 \| ×6 \| ×5 \| ×4 \| ×3 \| ×2 \| ×1 \| \| \| 2×2×3 \| 11 \| 2×5 \| 3×3 \| 2×2×2 \| 7 \| 2×3 \| 5 \| 2×2 \| 3 \| 2 \| You can see the above working for any factorial in this Calculator which uses exact integers of many digits and is practically instantaneous even for 1000! 8.3.1 Patterns in the Prime Factors of Binomial(2n,n) All primes from n+1 up to 2n are factors of Binomial(2n,n) : This was known to Chebychev in 1850 ccording to Pomerance 2015 No prime from 2n/3 to n are factors of Binomial(2n,n) : Other patterns are observable in these graphs Graph of n against the prime factors of Binomial(2n,n) with y=n and y=2n/3 lines: The gap ends on y=n+1. 8.3.2 Factorial Factors Calculator This Calculator will illustrate Legendre's method of factorizing n! using the base p representations of n for each prime p: Factorial's Prime Factors C A L C U L A T O R \| \| \| --- \| \| of of prime in \| ! \| \| \| ! --- ! \| \| of leave lower input empty for ALL of the row \| \| \| \| \| --- \| ( \| \| ) \| \| \| \| R E S U L T S \| \| \| --- \| \| \| \| Elements Mod Positions Quincunx Coins Hexagons Hockey Sticks Central binomials n! factors 8.3.3 You do the maths... How many zeroes are there at the end of 100!? [Hint: There are more 2's in the factorisation of 100! than 5's] Answer The number of zeroes is the number of times 10 is a factor of 100! 10 is 2×5 so in the prime factorisation of 100! we count the pairs of a 2 with a 5. There are more 2s as factors than there are 5s so we just count the number of times 5 is a factor Using the calculator above -- or the formula counting the quotients when computing 100 in base 5: \| \| \| \| --- \| 5 \| 100 \| \| \| 5 \| 20 \| remainder 0 \| \| 5 \| 4 \| remainder 0 \| \| \| 0 \| remainder 4 \| Sum of quotients is 20 + 4 + 0 = 24 There are 24 zeroes at the end of 100! 2. Find a simple argument that shows Every product of n consecutive whole numbers is divisible exactly by n! Answer Suppose the n consecutive integers are x+1 up to x+n. Then is a whole number (because it is a binomial coefficient in Pascal's Triangle) so n! (the denominator) must divide exactly into the numerator which is the product of the n numbers (x+1) ... (x+n) 3. For which values of is an integer? Answer By testing to see if n+1 is a factor of n! using the Calculator, we find the first few values are 5, 7, 8, 9, 11, 13, 14, 15 Consulting the OEIS we find A118742 is Numbers n for which the expression n!/(n+1) is an integer. and is also the list of numbers after 3 for which the (n+1) is not a prime. For a proof see the link. 9 References Representing Numbers using Binomials The Hidden Hexagon Squares V E Hoggatt, W Hansell Fib Q 9 (1971) pages 120, 13. Generalised Hidden Hexagon Squares A K Gupta,Fib Q 12 (1974), pages 45-46. Concrete Mathematics (2nd edition, 1994) by Graham, Knuth and Patashnik, Addison-Wesley has a lot on binomials and formulas that is quite accessible even though much of the rest of the book is at university level. History of Pascal's Triangle by Mrs Wheeler is a good summary of earlier records of the Triangle before Pascal studied it Divisors of the Middle Binomial Coefficient C Pomerance, Amer. Math. Monthly 122 (2015) pp 636-644 PDF Computing Binomial Coefficients P Goetgheluck, Amer Math Monthly vol 94 (1987), pp 360-365 has a simple program to compute the power of each prime in binomial(n,r) Théorie des nombres A Legendre, (ed 2 1808, ed 3 1830) gives the formula for the highest power of prime p that is a factor of n! History of the Theory of Numbers, Vol 1:Divisibility and Primality L E Dickson (1919, Dover 2005 paperback) chapter 9 has a compact history of the Legendre theorem How Often Does an Integer Occur as a Binomial Coefficient? D Singmaster, American Mathematical Monthly 78 (1971) pages 385-386 © 2025 Created: 1 January 2025 updated: 28 June 2025 Dr Ron Knott Home page
14294	https://mathteachercoach.com/wp-content/uploads/2020/06/5-2-Lesson-Plan-Multiplying-Fractions-and-Whole-Number.pdf	UNIT 5 - LESSON PLANS Copyright © MathTeacherCoach.com Class Math 5 Topic Multiplication of Fractions and Whole Number Lesson 2 Of 8 Objective Students will:  Use fraction of a set to interpret fractions as division.  Use tape diagrams to multiply a whole number by a fraction.  Relate a fraction of a set to interpret fraction multiplication through repeated addition.  Solve problems involving multiplication of factions and whole number using visual models or equations. “I Can” Statement I can multiply fractions and whole numbers using tape diagrams or equations. I can solve problems involving multiplication of fractions and whole numbers. Common Core Standards CCSS.MATH.CONTENT.5.NF.B.4 Apply and extend previous understandings of m1ultiplication to multiply a fraction or whole number by a fraction. CCSS.MATH.CONTENT.5.NF.B.4.A Interpret the product (a/b) × q as a parts of a partition of q into b equal parts; equivalently, as the result of a sequence of operations a × q ÷ b. For example, use a visual fraction model to show (2/3) × 4 = 8/3, and create a story context for this equation. Do the same with (2/3) × (4/5) = 8/15. (In general, (a/b) × (c/d) = (ac)/(bd). CCSS.MATH.CONTENT.5.NF.B.6 Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem. Procedures 1. Start and lead student discussion related to the bell work. 2. Distribute the Guided Notes 3. Present lesson or play a video lesson. 4. Use an Online Activity if time permitted. 5. Distribute Lesson Assignment. Assessment Bell Work 5-2 Assignment 5-2 Exit Quiz 5-2 Bell Work See Bell Work 5-2 UNIT 5 - LESSON PLANS Copyright © MathTeacherCoach.com Additional Resources See Online Activities
14295	https://www.youtube.com/watch?v=9-pYzZLIjqU	Lesson: Symmetric Functions and Definite Integrals Divide and Conquer Math 3260 subscribers 33 likes Description 1472 views Posted: 21 Apr 2020 In this video, we look at the proof surrounding symmetric functions and definite integrals. 00:00 Theorem on symmetric integrals and integration 1:06 Test for symmetry and proof of theorem 11:04 Examples of what this theorem means More free math help at DivideAndConquerMath.com. Check out the playlist of integration topics here: 6 comments Transcript: Theorem on symmetric integrals and integration hey guys and welcome to this video on computing definite integrals of symmetric functions so in this video I'm actually gonna show you some examples but I'm also gonna show you the proof which i think is super fun and cute granted my definition of fun might be different from other people's fun but you know it's good to expand your brain sometimes it's clever it's clever so let's just get right into the theorem so all right you've got some function f of X that it's continuous on it's got to be an interval a closed interval that looks like this so going from something symmetric is what we call it so from the negative to the positive of the same number so here's the meet of that theorem so if you have an even function then you can compute the integral by taking two times that same integral but from zero to a instead of from negative a to a and then also if you know that it's odd then the integral will just equal zero so one thing I want to remind you of is just the test for symmetry so a lot of times people forget how do you even compute or figure out whether something symmetric so this is really the key that we want Test for symmetry and proof of theorem to remember and this will be very helpful as we go through the proof actually okay so let's let's talk about why this is true and we're gonna prove it from a mathematical standpoint so let's start with the case where f of X is even so I want to compute this integral well one property of definite integrals is that I can always break this up if I'd like so I'm gonna break this up going from negative a to zero and then I'm gonna do it again going from A oops from zero to a sorry okay so now what I'm going to do is I'm actually gonna flip my limits of integration here so watch this for a moment so I'm gonna take this piece here and I'm going to like I said just flip the limits of integration it will make sense why in a moment so that part's going to become that and then this part I didn't change okay so now I'm going to make a really odd substitution on this part here so you'll see why I'm doing this in a moment so if you're like white why would you even do that it's to actually help proof kind of what we're trying to get to so I'm actually gonna set my U equal to negative x so I know that that's not technically in here so we're gonna have to make some adjustments and that's okay so if my U is equal to negative x then therefore negative u will equal X okay so remember this part then also from this I have to compute my D U which will be negative DX and if I just wanted to solve for DX and this would say that negative D u equals DX okay so I just kind of threw a bunch of things at you so now what I want to do is I just want to kind of line up where where we're at so where I see this X here I'm gonna replace it with this piece here so this X will become negative U and then where I have this DX I'm going to replace it with this part here so with negative D U and so this is all going to mess altum Utley with the sign over here and so that's that's one part but we're still not done right so this is a definite integral so I also want to mess with my limits of integration so I've just a couple more pieces here to do so now if x equals zero that would mean that u equals zero so nothing's gonna change for that first limit of integration and then if x equals negative a you will equal negative negative a so you will just equal a so now the last piece of this is I basically need to go through now and change these limits of integration with these parts in here so hopefully you can kind of see all the different things I'm gonna do so now this becomes negative so first let's flip our limits of integration so zero stays at zero negative a will become positive a now I said for X here for f of X I need to replace this with negative u so this will be f of negative u and then DX I'm gonna replace with negative D u okay and then the other part still didn't change okay and so now I would actually encourage you maybe just to take a second just to sort this out algebraically just to convince yourself of this so looking at this these negatives here will cancel out so this becomes zero to a F of negative u du u plus 0 to a of f of X DX and now we can finally leverage the fact that we know that f of X is even so if f is an even function what would this mean that this equals this make this means then so if you just needed me to remind you it means that you get back to the function itself so now going back to this let's see I need to clear a little bit of space so let me move everything up okay so now I've cleared some space so I can rewrite this then as just F of U right that's looks a little funny doesn't it I'm not trying to offend anybody so okay so I've got this F of U du u so I got this part here and so now I can actually go back and go back and rewrite all of this really so this if my if I this is my F of U well this actually then if I were to place it back with my ex it would still be the same thing so because of just the substitutions that we've made this and this I've now basically manipulated it so these two things are equal to one another so this will become 2 x from 2 x from 0 to a of f of X DX and again if you're saying wait a sec what about this you you totally just can't waved your hand at this well if I just really quickly want to say so I've got my F of U so let's just replace u back with F sorry let's just replace u it back with X so using coming back here this was my original substitution right so I could rewrite F of U as F of negative X which we still know will be f of X because this is even and then this is just telling me what to integrate with respect to so if I'm going to flip this back to f of X then I would just I want to flip this back to DX because that's now what we're integrating with respect to so that's kind of gonna be the idea here so then that proves the first part of this theorem okay so now let's prove the other part of the theorem so this first one was this now let's go to this so let's go when the function is odd why does this equal zero so now I'm going to assume that my f of X is odd so I'll use this at some point well we're gonna use the exact same set up from before so I'm gonna go from negative a to a I've got f of X DX and so similar to before so I want to actually break this up and what I would encourage you to do here is so we're gonna actually just approach in a very very similar similar manner so just like in the last proof we're gonna flip our limits of integration here and so now let's goes from zero to negative a and you're gonna make the same substitutions from before and you're gonna make very similar like the same changes to your limits of integration so I think this is actually a really great place to pause and take a moment to really quiz yourself to see how well you understood that this really kind of is a good way to expand your thinking and challenge yourself a little bit this is how you grow as a math student so you've got all the work from the last part so you know kind of the substitutions that you need to make and you knew that at some point like in the last proof that we did that f of X being even came into play so see if you can kind of now wrap your head around the logic with this one and then when you think you're ready hit hit play okay so like I said so same set of tricks from before so my you is still this negative X um therefore my negative u will equal X therefore my d u is negative DX therefore my negative D u is DX all the same stuff and then just like before so if X is negative a then you will equal positive a and if x equals zero then you will equal zero so all that stuff is all the same so when I set this up so I'll change my limits of integration so now it goes from zero to a this will become F of negative u and then all of this times negative D u the other part is not changing at all okay so now I have to kind of sort out all the signs here again so notice that I've got this negative and this negative here so I can rewrite now this entire piece as just F of negative u du u Plus this other part that we didn't change and now here's where it matters that f of X is odd so when we clear a little space so recalling what it means to be odd so an odd function has this property if you plug in something negative it brings out the negative of that function so what this means then is I can rewrite this part here so maybe I'll just underline it just this part here I can now leverage the fact that this is odd so I know that this will equal negative the the negative of this so that's what this means and then using very similar logic to before now I can replace this with really just my my f of X so this will now be this so this is the original thing I really had right and now this becomes really obvious here's one integral here's the negative of the integral obviously now they're they're just additive opposites so the whole thing will equal zero so we've proved exactly what we want to prove so so this is actually a really cool thing then that you can do with symmetry so when you notice that things are symmetric you can just like boom go ahead and use these properties right away so now let me show you just a couple of examples of how you could bust this out I guess so Examples of what this theorem means when you're looking at something like this so probably in the back of your mind one big hint is that you want to look at your limits of kind of these limits of integration and if these are additive opposites so negative four and four or you know they're exactly opposite numbers that's probably a sign that you want to just test the symmetry and so then what you want to do is you just want to kind of evaluate this F of negative x to see what you get so in this case if I take negative x to the third plus negative X what will happen with the signs this becomes negative x to the third minus X so that is indeed negative f of X so because I can see that this is odd I don't even have to do any other work just equal zero done boom so that's the power of this theorem and that's why we kind of like it so so when you can actually figure this out like then you could just invoke this and be done okay so what about this problem here so now if I do the same thing so I'm gonna plug in negative x so this becomes negative x to the fourth minus so I'm gonna go ahead and take that to the fourth and to the second so this will become just X to the fourth minus x squared so I get back to f of X okay so now that tells me then that I could actually simplify this integral to this and you might think well why would I even care to do that so the the reason why that's nice is I personally when it when it comes to evaluating definite integrals gosh I cannot write over here like my tablets giving me all these problems back so when I am evaluating definite integrals I would rather plug in zero usually than anything else because zero usually is just very simple to work with right so that's kind of what we like it it can save you a little bit of work so I need to take ups and I need to write this all as two times all of this so this will be two times okay so now we can work this out so this is 1/5 X to the fifth minus 1/3 X to the third so now I need to evaluate that from 0 to 5 so this will be 2 times let's see 1/5 times 5 to the fifth minus 1/3 times 5 to the third but then this second part is all just going to be zero so all of that drops out so that's kind of why we like this and you can see that you're gonna get just a fraction here or if you want to just turn it into a decimal so I've got one thousand one hundred and sixty six point six seven so that would be what that would round to or you can get the fractional answer of that but so lady said that just saves you a little bit of work in this case so that that's why we like that so that that's it so that's just something that you want to watch out for now when you're integrating and working with definite integrals you can save yourself some work if you ever notice that so thanks for watching guys I'll see you next time
14296	https://www.youtube.com/watch?v=81aET7xfLoA	Subtraction of Decimals (With Regrouping) MatholiaChannel 60000 subscribers 477 likes Description 79136 views Posted: 2 Aug 2014 Matholia educational maths video on subtraction of decimals (with regrouping). matholia #singaporemath #subtraction #decimals Learning Targets: I can subtract decimals with regrouping. I can explain how I computed with decimals using concrete models or drawings. New videos added daily! Subscribe here: For more maths videos for kids and loads of interactive content, visit matholia.com. This video lesson is taken from Matholia.com – a world-class online learning resource for K-6 mathematics. Our international curriculum covers a comprehensive range of topics and concepts using the Singapore's teaching approach. Concepts are presented through Concrete, Pictorial, Abstract (CPA) approach. Resources available include dynamic practice, instructional videos, ebooks, interactive tools and progress tracking and recording – all adopting the Singapore pedagogy. Transcript: [Music] [Applause] [Music] subtract find 2.2 - 0.5 start by subtracting the 10th we cannot subtract 5/10 from 2/10 so we regroup 1 one into 10/10 we now have 1210 1210 - 510 = 710 write the decimal point now subtract the ones 1 1 - 0 1's = 1 1 so 2.2 - 0.5 = 1.7 subtract find 3.47 - 1.59 start by subtracting the hundreds we cannot subtract 9 hundreds from from 7 hundreds so we regroup 1110th into 10 hunds we now have 17 hunds 1700s - 900s = 800s subtract the tth we cannot subtract 5/10 from 3/10 so we regroup 1 1 into 1010 we now have 13/10 13/10 - 5/10 = 8/10 write the decimal point subtract the ones 2 1's - 1 1 = 1 1 so 3.47 - 1.59 = 1.88 subtract find 6051 - 4.87 we cannot subtract 7 hundreds from 100th so we regroup 1110th into 10 hundredths we now have 11 hundreds 11 hundreds - 700s = 400s subtract the TS we cannot subtract 8/10 from 4/10 so we regroup 110 into 9 1's and 1010 we now have 14/10 1410 - 810 equal 6/10 write the decimal point subtract the ones 9 1's - 4 1's = 5 1's subtract the 10 51 - 01 = 510 so 6051 - 4.87 = 5564
14297	http://learningradiology.com/archives2012/COW%20501-Rickets/ricketscorrect.html	LearningRadiology- rickets, osteomalacia, renal, bone, radiology, knee, wrist, fray, cupping, metaphysis Discover more cupping vitamin Cupping Vitamin Cupping therapy Learning Radiology: Recognizing the Basics Rickets Osteomalacia during enchondral bone growth 4-18 months Histology Zone of preparatory calcification does not form resulting in build-up of maturing cartilage cells Also occurs in shafts so that osteoid production elevates periosteum Clinical findings Irritability Bone pain Tenderness Craniotabes Rachitic rosary Bowed legs Delayed dentition Swelling of wrists and ankles Location Metaphyses of long bones subjected to stress are particularly involved Wrists Ankles Knees Imaging findings Cupping and fraying of metaphysis Poorly mineralized epiphyseal centers with delayed appearance Irregular widened epiphyseal plates (increased osteoid) Increase in distance between end of shaft and epiphyseal center Cortical spurs projecting at right angles to metaphysis Coarse trabeculation (not the ground-glass pattern found in scurvy) Periosteal reaction may be present Deformities common Bowing of long bones Molding of epiphysis Fractures Frontal bossing Causes Of Rickets Abnormality In Vitamin D Metabolism Associated with hyperparathyroidism Vitamin D deficiency Dietary lack of vitamin D Famine osteomalacia Lack of sunshine exposure Malabsorption of vitamin D Pancreatitis and biliary tract disease Steatorrhea, celiac disease, postgastrectomy Inflammatory bowel disease Defective conversion of vitamin D to 25-OH-cholecalciferol in liver Liver disease Anticonvulsant drug therapy (= induction of hepatic enzymes that accelerate degradation of biologically active vitamin D metabolites) Defective conversion of 25-OH-D3 to 1,25-OH-D3 in kidney Chronic renal failure = renal osteodystrophy Vitamin D-dependent rickets = autosomal recessive enzyme defect of 1-OHase Abnormality In Phosphate Metabolism Not associated with hyperparathyroidism secondary to normal serum calcium Phosphate deficiency Intestinal malabsorption of phosphates Ingestion of aluminum salts [Al(OH)2] forming insoluble complexes with phosphate Low phosphate feeding in prematurely born infants Severe malabsorption state Parenteral hyperalimentation Disorders of renal tubular reabsorption of phosphate Renal tubular acidosis (renal loss of alkali) deToni-Debré-Fanconi syndrome = hypophosphatemia, glucosuria, aminoaciduria Vitamin D-resistant rickets Cystinosis Tyrosinosis Lowe syndrome Hypophosphatemia with nonendocrine tumors Oncogenic rickets - elaboration of humeral substance which inhibits tubular reabsorption of phosphates Sclerosing hemangioma Hemangiopericytoma Ossifying mesenchymal tumor Nonossifying fibroma Hypophosphatasia Calcium Deficiency Dietary rickets = milk-free diet (extremely rare) Malabsorption Consumption of substances forming chelates with calcium Classification Of Rickets Primary vitamin D-deficiency rickets Gastrointestinal malabsorption Partial gastrectomy Small intestinal disease: gluten-sensitive enteropathy / regional enteritis Hepatobiliary disease: chronic biliary obstruction / biliary cirrhosis Pancreatic disease: chronic pancreatitis Primary hypophosphatemia; vitamin D-deficiency rickets Renal disease Chronic renal failure Renal tubular disorders: renal tubular acidosis Multiple renal defects Hypophosphatasia and pseudohypophosphatasia Fibrogenesis imperfecta osseum Axial osteomalacia Miscellaneous Hypoparathyroidism, hyperparathyroidism, thyrotoxicosis, osteoporosis, Paget disease, fluoride ingestion, ureterosigmoidostomy, neurofibromatosis, osteopetrosis, macroglobulinemia, malignancy Rickets of the knees demonstrates bowing of the femurs, metaphyseal cupping and fraying, coarsening of the trabecular pattern, increase in distance between end of shaft and epiphyseal center, poorly ossified epiphyseal centers. Rickets. There is cupping and fraying of all of the metaphyses (white arrows) in this skeletally-immature child. For more information, click on the link if you see this icon
14298	https://math.stackexchange.com/questions/2006159/leading-digits-of-large-power-of-2	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 leading digits of large power of 2 Ask Question Asked Modified 8 years, 10 months ago Viewed 5k times 10 $\begingroup$ How do I get the leading digits for a large power of 2 (say $2^{123456789}$, although I want a general method)? I was thinking about repeatedly dividing it by ten, but I don't know how to do that efficiently without calculating the number itself, and I was also thinking $\log(a) < 123456789\log(2) < b$ would imply that it would be the first three digits of $a$, but I don't know how I would find $a$ and $b$ in the first place. elementary-number-theory Share asked Nov 9, 2016 at 5:15 b_pcakesb_pcakes 1,57711 gold badge1515 silver badges3131 bronze badges $\endgroup$ 1 $\begingroup$ I sincerely doubt that there is any efficient way to do it (more efficient than calculating the decimal representation of the number). $\endgroup$ barak manos – barak manos 2016-11-09 05:21:39 +00:00 Commented Nov 9, 2016 at 5:21 Add a comment \| 2 Answers 2 Reset to default 21 $\begingroup$ Figured it out! We first need to find the number of digits of $2^n$, which can be done using logarithms: we need to solve (approximately) $2^n = 10^d$. In the end, the formula for the number of digits $d$ is $d = 1+\left\lfloor n\,\log_{10}2\right\rfloor$. Then, once we have $d$, then we see that if we can solve for $10^{d-k} \cdot t < 2^n < 10^{d-k} \cdot (t + 1)$ (where $k$ is the number of leading digits we want), then $t$ is the number we want. In this specific example, we find that the first three digits are $\lfloor 10 ^ {n \log2 - d + 3}\rfloor = 454$, where $n = 123456789$ and $d$ is calculated as above. Share edited Nov 9, 2016 at 5:48 answered Nov 9, 2016 at 5:34 b_pcakesb_pcakes 1,57711 gold badge1515 silver badges3131 bronze badges $\endgroup$ 15 $\begingroup$ 1. I hope you're aware of the fact that $d$ is not integer. 2. What is $a$? 3. How exactly does that get you the leading digits (and how many of them does it get you)??? $\endgroup$ barak manos – barak manos 2016-11-09 05:37:22 +00:00 Commented Nov 9, 2016 at 5:37 1 $\begingroup$ @barakmanos $\log(10^{d - k}\cdot t) < \log(2^n) < \log(10^{d-k}\cdot (t+1))$, which is equivalent to $d-k + \log t < n\log(2) $\endgroup$ b_pcakes – b_pcakes 2016-11-09 06:32:41 +00:00 Commented Nov 9, 2016 at 6:32 1 $\begingroup$ @barakmanos Okay, then according to my formula we have $n = 80$, $k = 5$, $d = 1 + \lfloor n \log 2\rfloor = 25$, and the answer is $\lfloor 10 ^ {n \log2 - d + k}\rfloor = 12089$. If you still have doubts I really suggest we continue this in chat. $\endgroup$ b_pcakes – b_pcakes 2016-11-09 06:43:10 +00:00 Commented Nov 9, 2016 at 6:43 3 $\begingroup$ OK, I've verified this with a Python script, very nice!!! I think that you just need to write it down explicitly as $\lfloor{10^{n\log2-\lfloor{n\log2}\rfloor+k-1}}\rfloor$, then accept your own answer. +1 from me, of course... $\endgroup$ barak manos – barak manos 2016-11-09 06:49:13 +00:00 Commented Nov 9, 2016 at 6:49 1 $\begingroup$ Note that, if this is done with floating-point numbers, you lose as many bits of precision as $n$ has. Regular IEEE-754 64-bit floats (doubles) have a 52-bit mantissa, so the usable bits of mantissa you'll get is approximately $52-log_2(n)$ (e.g. for $2^{100000}$ you'll get roughly 35 bits of precision) $\endgroup$ dzaima – dzaima 2021-10-30 17:53:47 +00:00 Commented Oct 30, 2021 at 17:53 \| Show 10 more comments 3 $\begingroup$ $123456789\log_{10} 2=37164196.65735903...$, so $2^{123456789}=10^ {.65735903}\times10^{37164196}$. Looking at $log_{10}d$ for single digit integers $d$, it turns out that $log_{10}4 < .65735903... < \log_{10}5$, so the first digit is $4$. To find more digits, narrow things down further or compute $10^.65735903...$ to as many digits as you need. Share answered Nov 9, 2016 at 5:32 Steve KassSteve Kass 15.4k11 gold badge2424 silver badges3434 bronze badges $\endgroup$ 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 elementary-number-theory See similar questions with these tags. 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14299	https://www.commoncoresheets.com/order-of-operations-worksheets	Home Favorites My Account Menu Common Core Sheets login Reading > comprehension - insects comprehension - arachnids New comprehension-reptiles New comprehension - mammals New comprehension - birds New Teacher Panel > My Account Distance Learning Assignments My Classes Grade Book New Grade Scale My Saved Sheets Site Options Spelling > Spelling Worksheets Maker Premade Spelling Worksheets Math > Daily Reviews Creator Create-A-Test Create-A-Flash Card Sort By Grade Addition Subtraction Multiplication Division Algebra Angles Area & Perimeter Balancing Equations Bar Graphs Box Plots Capacity Cheat Sheets Converting Forms Counting Decimals Division Drills Fact Families Factors Fractions Grids Lines Line Graphs Line Plots Mean, Median, Mode & Range Measurement Money Multiplication Multistep Problems Negative Numbers Order of Operations Patterns & Function Machines Percent Pictographs Pie Graphs Probability Properties Ratios Rounding Shapes Statistics Subtraction Tally Tape Diagrams Temperature Time Trigonometry Value & Place Value Variables Weight Venn Diagrams Volume Drills Search Premade Sheets social studies > Geography Primary & Secondary Sources Timelines Create-A-Sheet science > Tools Volume Create-A-Sheet Language Arts > Improving a Paragraph Create-A-Sheet Teacher Panel > My Account Distance Learning Assignments My Classes Grade Book Grade Scale My Saved Sheets Site Options Languages > spanish german russian italian vietnamese french english Help Support the Site > Patreon Paypal Other Stuff > Fun Sheets Coloring Sheets How to Use Worksheets Contact & Comment Order of Operations Worksheets Order of operations worksheets are a crucial resource for students learning about PEDMAS (Parentheses, Exponents, Division, Multiplication, Addition, Subtraction). These worksheets provide practice with solving mathematical expressions using the correct order of operations. Our order of operations worksheets are the best on the internet and are available for free to help students succeed in math. With a variety of exercises and levels of difficulty, our order of operations worksheets are perfect for students of all ages. So why wait? Get the best order of operations worksheets on the internet for free and start mastering PEDMAS today! Jump to a Heading > Filter By Grade > Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade No Filter Browse Sheets By Problem Type numericadditionsubtractionmultiplicationdivisionhorizontal List View (old) × Solving Order of Operations Link One Atta Time Flash Cards Distance Learning Create New Sheet Customize Problems Select a Worksheet > Version 1Version 2Version 3Version 4Version 5Version 6Version 7Version 8Version 9Version 10Grab 'em All Solving using Order of Operations 6ee2c × Description: "This worksheet is designed to enhance understanding of the order of operations in math. It features 10 intricate arithmetic problems, centering on logical methods of calculation for improvement of algebraic skills. Customizable for unique needs, this resource can transform into flash cards or be leveraged for distance learning, allowing versatile mathematical comprehension." × Student Goals: One Atta Time Flash Cards Distance Learning Create New Sheet Customize Problems Select a Worksheet > Version 1Version 2Version 3Version 4Version 5Version 6Version 7Version 8Version 9Version 10Grab 'em All Finding First Expression 6ee2c × Description: "This worksheet is designed to reinforce children's basic math skills, focusing on the concept of finding the first expression through 20 unique problems. It's highly flexible and can be customized, transformed into interactive flashcards, or incorporated into distance learning programs, creating a multifaceted approach to mathematics instruction. A must-have resource for effective learning." × Student Goals: Develop Mathematical SkillsUpon completion of the worksheet, students are expected to have honed their calculation skills. They will learn to apply basic math operations like addition, subtraction, multiplication, and division correctly and swiftly. It will also help develop their understanding of the order of operations in math problems and enable them to solve complex mathematical problems independently.Enhance Problem Solving AbilitiesThe worksheet is designed to challenge students to think critically and logically, thereby enhancing their problem-solving abilities. Through solving the 20 problems in the worksheet, students will be able to approach mathematical puzzles systematically. They will learn to break down complex problems into simple, manageable steps, enhancing their confidence in tackling mathematical challenges.Promote Analytical ThinkingThe problems in the worksheet are geared towards promoting students' analytical thinking skills. They will learn to analyze each math expression carefully, which will strengthen their understanding of math concepts as a whole. It allows them to visualize arithmetic as not just rote calculations, but as set of logical steps. Through the repeated practice, students will begin to recognize patterns in math expressions, which can strengthen their analytical skills.Improve Precision and AccuracyThis worksheet helps students to improve their precision and accuracy during mathematical computations. They become progressively adept at handling numbers and operations and can avoid common mathematical errors. Each correct answer provides positive reinforcement, increasing their competence in the subject.Increase Patience and PerseveranceCompleting the worksheet is expected to build students' patience and perseverance. As the problems become increasingly complex, students are needed to spend more time and thought in solving the problems. By working through these challenges, they will learn the importance of persistence in problem-solving activities. Even if they don't get the right answer the first time, they're encouraged to try again, instilling in them an understanding and acceptance of trial and error in learning.Boosts Confidence in MathematicsBy successfully completing and understanding the worksheet, students gain confidence in their mathematical abilities. This increased confidence can motivate students to participate more actively in class and take on more complex problems. Moreover, the confidence gained can also contribute to the positive attitude towards mathematics and academic achievement in general. One Atta Time Flash Cards Distance Learning Create New Sheet Customize Problems Select a Worksheet > Version 1Version 2Version 3Version 4Version 5Version 6Version 7Version 8Version 9Version 10Grab 'em All Using Order of Operations (No Negatives) × Description: "This worksheet is designed to enhance children's understanding of the order of operations in mathematics. Featuring 10 customizable problems encompassing concepts like the BODMAS rule, students learn to deal with complex calculations involving addition, subtraction, division, and multiplication. The content can be converted into flashcards for effective study or leveraged for distance learning, adapting to versatile teaching methods and learning styles." × Student Goals: Improved Mathematical SkillsUpon successful completion of the worksheet, students should demonstrate considerable mastery of the order of operations in mathematics. This fundamental mathematical concept will enable them to solve increasingly complex arithmetic equations accurately.Enhanced Logical ReasoningBy navigating through the tasks in the worksheet, students are expected to enhance their logical reasoning capabilities. The process of identifying the correct sequence of operations to apply in each problem facilitates the development of logical thought processes and critical thinking.Improved Problem SolvingAfter working on the worksheet, students should be able to better evaluate and solve problems. This inherent skill extends beyond math and into everyday life where solutions often require sequential steps or operations.Increased Confidence in MathematicsSuccessfully completing the problems set in the worksheet should give students increased confidence in tackling math problems, especially those involving order of operations. Confidence is key to nurturing a positive attitude towards mathematics, reducing math anxiety and fostering a growth mindset.Preparing for Advanced Math ConceptsMastering the order of operations is fundamental to understanding more advanced mathematical concepts. Upon finishing this worksheet, students should be well prepared to tackle higher-level mathematical equations and problems that require an in-depth understanding of the principles of operations and their hierarchical execution.Enhanced Mathematical FluencyArriving at correct solutions to the presented problems should signify that students are becoming more fluent in their math skills. Mathematical fluency entails precision, accuracy, and speed in solving math problems, and this worksheet is designed to foster such fluency.Comprehension of Mathematical SyntaxStudents will demonstrate that they have a grasp of the syntax of arithmetic. Understanding this syntax, as signified by the correct use and interpretation of parentheses, brackets and operation signs, is central to successfully encoding and decoding mathematical phrases and sentences. Math worksheets for kids. Created by educators, teachers and peer reviewed. Terms of Use FAQS Contact © 2012-2023, Common Core Sheets

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